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https://www.reliawiki.com/index.php?title=Gompertz_Models&diff=prev&oldid=55922 | 1,627,390,603,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153391.5/warc/CC-MAIN-20210727103626-20210727133626-00190.warc.gz | 1,032,013,101 | 20,182 | # Difference between revisions of "Gompertz Models"
Chapter 3.6: Gompertz Models
Chapter 3.6 Gompertz Models
Available Software:
RGA
More Resources:
RGA Examples Collection
This chapter discusses the two Gompertz models that are used in RGA. The standard Gompertz and the modified Gompertz.
## The Standard Gompertz Model
The Gompertz reliability growth model is often used when analyzing reliability data. It is most applicable when the data set follows a smooth curve, as shown in the plot below.
The Gompertz model is mathematically given by Virene [1]:
$R=a{{b}^{{{c}^{T}}}}\,\!$
where:
• $0\lt a\le 1\,\!$
• $0\lt b\lt 1\,\!$
• $0\lt c\lt 1\,\!$
• $T\gt 0\,\!$
• $R=\,\!$ the system's reliability at development time, launch number or stage number, $T\,\!$
• $a=\,\!$ the upper limit that the reliability approaches asymptotically as $T\to \infty \,\!$, or the maximum reliability that can be attained
• $ab=\,\!$ initial reliability at $T=0\,\!$
• $c=\,\!$ the growth pattern indicator (small values of $c\,\!$ indicate rapid early reliability growth and large values of $c\,\!$ indicate slow reliability growth)
As it can be seen from the mathematical definition, the Gompertz model is a 3-parameter model with the parameters $a\,\!$, $b\,\!$ and $c\,\!$. The solution for the parameters, given ${{T}_{i}}\,\!$ and ${{R}_{i}}\,\!$, is accomplished by fitting the best possible line through the data points. Many methods are available; all of which tend to be numerically intensive. When analyzing reliability data in the RGA software, you have the option to enter the reliability values in percent or in decimal format. However, $a\,\!$ will always be returned in decimal format and not in percent. The estimated parameters in the RGA software are unitless. The next section presents an overview and background on some of the most commonly used algorithms/methods for obtaining these parameters.
### Parameter Estimation
#### Linear Regression
The method of least squares requires that a straight line be fitted to a set of data points. If the regression is on $Y\,\!$, then the sum of the squares of the vertical deviations from the points to the line is minimized. If the regression is on $X\,\!$, the line is fitted to a set of data points such that the sum of the squares of the horizontal deviations from the points to the line is minimized. To illustrate the method, this section presents a regression on $Y\,\!$. Consider the linear model given by Seber and Wild [2]:
${{Y}_{i}}={{\widehat{\beta }}_{0}}+{{\widehat{\beta }}_{1}}{{X}_{i1}}+{{\widehat{\beta }}_{2}}{{X}_{i2}}+...+{{\widehat{\beta }}_{p}}{{X}_{ip}}\,\!$
or in matrix form where bold letters indicate matrices:
\begin{align} Y=X\beta \end{align}\,\!
where:
$Y=\left[ \begin{matrix} {{Y}_{1}} \\ {{Y}_{2}} \\ \vdots \\ {{Y}_{N}} \\ \end{matrix} \right]\,\!$
$X=\left[ \begin{matrix} 1 & {{X}_{1,1}} & \cdots & {{X}_{1,p}} \\ 1 & {{X}_{2,1}} & \cdots & {{X}_{2,p}} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & {{X}_{N,1}} & \cdots & {{X}_{N,p}} \\ \end{matrix} \right]\,\!$
and:
$\beta =\left[ \begin{matrix} {{\beta }_{0}} \\ {{\beta }_{1}} \\ \vdots \\ {{\beta }_{p}} \\ \end{matrix} \right]\,\!$
The vector $\beta \,\!$ holds the values of the parameters. Now let $\widehat{\beta }\,\!$ be the estimates of these parameters, or the regression coefficients. The vector of estimated regression coefficients is denoted by:
$\widehat{\beta }=\left[ \begin{matrix} {{\widehat{\beta }}_{0}} \\ {{\widehat{\beta }}_{1}} \\ \vdots \\ {{\widehat{\beta }}_{p}} \\ \end{matrix} \right]\,\!$
Solving for $\beta \,\!$ in the matrix form of the equation requires the analyst to left multiply both sides by the transpose of $X\,\!$, ${{X}^{T}}\,\!$, or :
\begin{align} Y&=X\beta \\ ({{X}^{T}}X)\widehat{\beta }&={{X}^{T}}Y \\ \end{align}\,\!
Now the term $({{X}^{T}}X)\,\!$ becomes a square and invertible matrix. Then taking it to the other side of the equation gives:
$\widehat{\beta }=({{X}^{T}}X)^{-1}{{X}^{T}}Y\,\!$
#### Non-linear Regression
Non-linear regression is similar to linear regression, except that a curve is fitted to the data set instead of a straight line. Just as in the linear scenario, the sum of the squares of the horizontal and vertical distances between the line and the points are to be minimized. In the case of the non-linear Gompertz model $R=a{{b}^{{{c}^{T}}}}\,\!$, let:
${{Y}_{i}}=f({{T}_{i}},\delta )=a{{b}^{{{c}^{{{T}_{i}}}}}}\,\!$
where:
${{T}_{i}}=\left[ \begin{matrix} {{T}_{1}} \\ {{T}_{2}} \\ \vdots \\ {{T}_{N}} \\ \end{matrix} \right],\quad i=1,2,...,N\,\!$
and:
$\delta =\left[ \begin{matrix} a \\ b \\ c \\ \end{matrix} \right]\,\!$
The Gauss-Newton method can be used to solve for the parameters $a\,\!$, $b\,\!$ and $c\,\!$ by performing a Taylor series expansion on $f({{T}_{i}},\delta ).\,\!$ Then approximate the non-linear model with linear terms and employ ordinary least squares to estimate the parameters. This procedure is performed in an iterative manner and it generally leads to a solution of the non-linear problem.
This procedure starts by using initial estimates of the parameters $a\,\!$, $b\,\!$ and $c\,\!$, denoted as $g_{1}^{(0)},\,\!$ $g_{2}^{(0)}\,\!$ and $g_{3}^{(0)},\,\!$ where $^{(0)}\,\!$ is the iteration number. The Taylor series expansion approximates the mean response, $f({{T}_{i}},\delta )\,\!$, around the starting values, $g_{1}^{(0)},\,\!$ $g_{2}^{(0)}\,\!$ and $g_{3}^{(0)}.\,\!$ For the ${{i}^{th}}\,\!$ observation:
$f({{T}_{i}},\delta )\simeq f({{T}_{i}},{{g}^{(0)}})+\underset{k=1}{\overset{p}{\mathop \sum }}\,{{\left[ \frac{\partial f({{T}_{i}},\delta )}{\partial {{\delta }_{k}}} \right]}_{\delta ={{g}^{(0)}}}}({{\delta }_{k}}-g_{k}^{(0)})\,\!$
where:
${{g}^{(0)}}=\left[ \begin{matrix} g_{1}^{(0)} \\ g_{2}^{(0)} \\ g_{3}^{(0)} \\ \end{matrix} \right]\,\!$
Let:
\begin{align} f_{i}^{(0)}= & f({{T}_{i}},{{g}^{(0)}}) \\ \nu _{k}^{(0)}= & ({{\delta }_{k}}-g_{k}^{(0)}) \\ D_{ik}^{(0)}= & {{\left[ \frac{\partial f({{T}_{i}},\delta )}{\partial {{\delta }_{k}}} \right]}_{\delta ={{g}^{(0)}}}} \end{align}\,\!
So the equation ${{Y}_{i}}=f({{T}_{i}},\delta )=a{{b}^{{{c}^{{{T}_{i}}}}}} \,\!$ becomes:
${{Y}_{i}}\simeq f_{i}^{(0)}+\underset{k=1}{\overset{p}{\mathop \sum }}\,D_{ik}^{(0)}\nu _{k}^{(0)}\,\!$
or by shifting $f_{i}^{(0)}\,\!$ to the left of the equation:
$Y_{i}^{(0)}\simeq \underset{k=1}{\overset{p}{\mathop \sum }}\,D_{ik}^{(0)}\nu _{k}^{(0)}\,\!$
In matrix form this is given by:
${{Y}^{(0)}}\simeq {{D}^{(0)}}{{\nu }^{(0)}}\,\!$
where:
${{Y}^{(0)}}=\left[ \begin{matrix} {{Y}_{1}}-f_{1}^{(0)} \\ {{Y}_{2}}-f_{2}^{(0)} \\ \vdots \\ {{Y}_{N}}-f_{N}^{(0)} \\ \end{matrix} \right]=\left[ \begin{matrix} {{Y}_{1}}-g_{1}^{(0)}g_{2}^{(0)g_{3}^{(0){{T}_{1}}}} \\ {{Y}_{1}}-g_{1}^{(0)}g_{2}^{(0)g_{3}^{(0){{T}_{2}}}} \\ \vdots \\ {{Y}_{N}}-g_{1}^{(0)}g_{2}^{(0)g_{3}^{(0){{T}_{N}}}} \\ \end{matrix} \right]\,\!$
\begin{align} {{D}^{(0)}}= & \left[ \begin{matrix} D_{11}^{(0)} & D_{12}^{(0)} & D_{13}^{(0)} \\ D_{21}^{(0)} & D_{22}^{(0)} & D_{23}^{(0)} \\ . & . & . & . \\ . & . & . & . \\ D_{N1}^{(0)} & D_{N2}^{(0)} & D_{N3}^{(0)} \\ \end{matrix} \right] \\ = & \left[ \begin{matrix} g_{2}^{(0)g_{3}^{(0){{T}_{1}}}} & \tfrac{g_{1}^{(0)}}{g_{2}^{(0)}}g_{3}^{(0){{T}_{1}}}g_{2}^{(0)g_{3}^{(0){{T}_{1}}}} & \tfrac{g_{1}^{(0)}}{g_{3}^{(0)}}g_{3}^{(0){{T}_{1}}}\ln (g_{2}^{(0)}){{T}_{1}}g_{2}^{(0)g_{3}^{(0){{T}_{1}}}} & 1 \\ g_{2}^{(0)g_{3}^{(0){{T}_{2}}}} & \tfrac{g_{1}^{(0)}}{g_{2}^{(0)}}g_{3}^{(0){{T}_{2}}}g_{2}^{(0)g_{3}^{(0){{T}_{2}}}} & \tfrac{g_{1}^{(0)}}{g_{3}^{(0)}}g_{3}^{(0){{T}_{2}}}\ln (g_{2}^{(0)}){{T}_{2}}g_{2}^{(0)g_{3}^{(0){{T}_{2}}}} & 1 \\ . & . & . & . \\ . & . & . & . \\ g_{2}^{(0)g_{3}^{(0){{T}_{N}}}} & \tfrac{g_{1}^{(0)}}{g_{2}^{(0)}}g_{3}^{(0){{T}_{N}}}g_{2}^{(0)g_{3}^{(0){{T}_{N}}}} & \tfrac{g_{1}^{(0)}}{g_{3}^{(0)}}g_{3}^{(0){{T}_{N}}}\ln (g_{2}^{(0)}){{T}_{N}}g_{2}^{(0)g_{3}^{(0){{T}_{N}}}} & 1 \\ \end{matrix} \right] \end{align}\,\!
and:
${{\nu }^{(0)}}=\left[ \begin{matrix} g_{1}^{(0)} \\ g_{2}^{(0)} \\ g_{3}^{(0)} \\ \end{matrix} \right]\,\!$
Note that the equation ${{Y}^{(0)}}\simeq {{D}^{(0)}}{{\nu }^{(0)}}\,\!$ is in the form of the general linear regression model given in the Linear Regression section. Therefore, the estimate of the parameters ${{\nu }^{(0)}}\,\!$ is given by:
${{\widehat{\nu }}^{(0)}}={{\left( {{D}^{{{(0)}^{T}}}}{{D}^{(0)}} \right)}^{-1}}{{D}^{{{(0)}^{T}}}}{{Y}^{(0)}}\,\!$
The revised estimated regression coefficients in matrix form are:
${{g}^{(1)}}={{g}^{(0)}}+{{\widehat{\nu }}^{(0)}}\,\!$
The least squares criterion measure, $Q,\,\!$ should be checked to examine whether the revised regression coefficients will lead to a reasonable result. According to the Least Squares Principle, the solution to the values of the parameters are those values that minimize $Q\,\!$. With the starting coefficients, ${{g}^{(0)}}\,\!$, $Q\,\!$ is:
${{Q}^{(0)}}=\underset{i=1}{\overset{N}{\mathop \sum }}\,{{\left[ {{Y}_{i}}-f\left( {{T}_{i}},{{g}^{(0)}} \right) \right]}^{2}}\,\!$
And with the coefficients at the end of the first iteration, ${{g}^{(1)}}\,\!$, $Q\,\!$ is:
${{Q}^{(1)}}=\underset{i=1}{\overset{N}{\mathop \sum }}\,{{\left[ {{Y}_{i}}-f\left( {{T}_{i}},{{g}^{(1)}} \right) \right]}^{2}}\,\!$
For the Gauss-Newton method to work properly and to satisfy the Least Squares Principle, the relationship ${{Q}^{(k+1)}}\lt {{Q}^{(k)}}\,\!$ has to hold for all $k\,\!$, meaning that ${{g}^{(k+1)}}\,\!$ gives a better estimate than ${{g}^{(k)}}\,\!$. The problem is not yet completely solved. Now ${{g}^{(1)}}\,\!$ are the starting values, producing a new set of values ${{g}^{(2)}}\,\!$. The process is continued until the following relationship has been satisfied:
${{Q}^{(s-1)}}-{{Q}^{(s)}}\simeq 0\,\!$
When using the Gauss-Newton method or some other estimation procedure, it is advisable to try several sets of starting values to make sure that the solution gives relatively consistent results.
#### Choice of Initial Values
The choice of the starting values for the nonlinear regression is not an easy task. A poor choice may result in a lengthy computation with many iterations. It may also lead to divergence, or to a convergence due to a local minimum. Therefore, good initial values will result in fast computations with few iterations, and if multiple minima exist, it will lead to a solution that is a minimum.
Various methods were developed for obtaining valid initial values for the regression parameters. The following procedure is described by Virene [1] in estimating the Gompertz parameters. This procedure is rather simple. It will be used to get the starting values for the Gauss-Newton method, or for any other method that requires initial values. Some analysts use this method to calculate the parameters when the data set is divisible into three groups of equal size. However, if the data set is not equally divisible, it can still provide good initial estimates.
Consider the case where $m\,\!$ observations are available in the form shown next. Each reliability value, ${{R}_{i}}\,\!$, is measured at the specified times, ${{T}_{i}}\,\!$.
$\begin{matrix} {{T}_{i}} & {{R}_{i}} \\ {{T}_{0}} & {{R}_{0}} \\ {{T}_{1}} & {{R}_{1}} \\ {{T}_{2}} & {{R}_{2}} \\ \vdots & \vdots \\ {{T}_{m-1}} & {{R}_{m-1}} \\ \end{matrix}\,\!$
where:
• $m=3n,\,\!$ $n\,\!$ is equal to the number of items in each equally sized group
• ${{T}_{i}}-{{T}_{i-1}}=const\,\!$
• $i=0,1,...,m-1\,\!$
The Gompertz reliability equation is given by:
$R=a{{b}^{{{c}^{T}}}}\,\!$
and:
\begin{align} \ln (R)=\ln (a)+{{c}^{T}}\ln (b) \end{align}\,\!
Define:
$S_1=\sum_{i=0}^{n-1} ln(R_i)= n ln(a)+ln(b)\sum_{i=0}^{n-1} c^{T_i}\,\!$
$S_2=\sum_{i=n}^{2n-1} ln(R_i)= n ln(a)+ln(b)\sum_{i=n}^{2n-1} c^{T_i}\,\!$
$S_3=\sum_{i=2n}^{m-1} ln(R_i)= n ln(a)+ln(b)\sum_{i=2n}^{m-1} c^{T_i}\,\!$
Then:
$\frac{S_3-S_2}{S_2-S_1}=\frac{\sum_{i=2n}{m-1} c^{T_i}-\sum_{i=n}^{2n-1} c^T_i}{\sum_{i=0}^{n-1} c^{T_i}}\,\!$
$\frac{S_3-S_2}{S_2-S_1}=c^T_{2n}\frac{\sum_{i=0}{n-1} c^{T_i}-c^{T_n}\sum_{i=0}^{n-1} c^T_i}{c^{T_n}\sum_{i=0}^{n-1} c^{T_i}}\,\!$
$\frac{S_3 - S_2}{S_2-S_1}=\frac{c^{T_2n}-c^{T_n}}{c^{T_n}-1}=c^{T_{a_n}}=c^{n\cdot I+T_0}\,\!$
Without loss of generality, take ${{T}_{{{a}_{0}}}}=0\,\!$ ; then:
$\frac{{{S}_{3}}-{{S}_{2}}}{{{S}_{2}}-{{S}_{1}}}={{c}^{n\cdot I}}\,\!$
Solving for $c\,\!$ yields:
$c=\left ( \frac{S_{3}-S_{2}}{S_{2}-S_{1}} \right )^\frac{1}{n\cdot I}$
Considering the definitions for $S_{1}\,\!$ and $S_{2}\,\!$, given above, then:
\begin{align} {{S}_{1}}-n\cdot \ln (a)= & \ln (b)\underset{i=0}{\overset{n-1}{\mathop \sum }}\,{{c}^{{{T}_{i}}}} \\ {{S}_{2}}-n\cdot \ln (a)= & \ln (b)\underset{i=n}{\overset{2n-1}{\mathop \sum }}\,{{c}^{{{T}_{i}}}} \end{align}\,\!
or:
$\frac{{{S}_{1}}-n\cdot \ln (a)}{{{S}_{2}}-n\cdot \ln (a)}=\frac{1}{{{c}^{n\cdot I}}}\,\!$
Reordering the equation yields:
\begin{align} \ln (a)= & \frac{1}{n}\left( {{S}_{1}}+\frac{{{S}_{2}}-{{S}_{1}}}{1-{{c}^{n\cdot I}}} \right) \\ a= & {{e}^{\left[ \tfrac{1}{n}\left( {{S}_{1}}+\tfrac{{{S}_{2}}-{{S}_{1}}}{1-{{c}^{n\cdot I}}} \right) \right]}} \end{align}\,\!
If the reliability values are in percent then $a\,\!$ needs to be divided by 100 to return the estimate in decimal format. Consider the definitions for $S_{1}\,\!$ and $S_{2}\,\!$ again, where:
\begin{align} S_{1}-ln(b)\sum_{i=0}^{n-1}c^{T_{i}}&=nln(a) \\ S_{2}-ln(b)\sum_{i=n}^{2n-1}c^{T_{i}}&=nln(a) \\ \frac{S_{1}-ln(b)\sum_{i=0}^{n-1}c^{T_{i}}}{S_{2}-ln(b)\sum_{i=n}^{2n-1}c^{T_{i}}}&=1 \\ S_{1}-ln(b)\sum_{i=0}^{n-1}c^{T_{i}}&=S_{2}-ln(b)\sum_{i=n}^{2n-1}c^{T_{i}} \\ \end{align}
Reordering the equation above yields:
\begin{align} \ln (b)= & \frac{({{S}_{2}}-{{S}_{1}})({{c}^{I}}-1)}{{{(1-{{c}^{n\cdot I}})}^{2}}} \\ b= & {{e}^{\left[ \tfrac{\left( {{S}_{2}}-{{S}_{1}} \right)\left( {{c}^{I}}-1 \right)}{{{\left( 1-{{c}^{n\cdot I}} \right)}^{2}}} \right]}} \end{align}\,\!
Therefore, for the special case where $I=1\,\!$, the parameters are:
\begin{align} c= & {{\left( \frac{{{S}_{3}}-{{S}_{2}}}{{{S}_{2}}-{{S}_{1}}} \right)}^{\tfrac{1}{n}}} \\ a= & {{e}^{\left[ \tfrac{1}{n}\left( {{S}_{1}}+\tfrac{{{S}_{2}}-{{S}_{1}}}{1-{{c}^{n}}} \right) \right]}} \\ b= & {{e}^{\left[ \tfrac{({{S}_{2}}-{{S}_{1}})(c-1)}{{{\left( 1-{{c}^{n}} \right)}^{2}}} \right]}} \end{align}\,\!
To estimate the values of the parameters $a,b\,\!$ and $c\,\!$, do the following:
1. Arrange the currently available data in terms of $T\,\!$ and $R\,\!$ as in the table below. The $T\,\!$ values should be chosen at equal intervals and increasing in value by 1, such as one month, one hour, etc.
Design and Development Time vs. Demonstrated Reliability Data for a Device
Group Number Growth Time $T\,\!$(months) Reliability $R\,\!$(%) $\ln{R}\,\!$
0 58 4.060
1 1 66 4.190
${{S}_{1}}\,\!$ = 8.250
2 72.5 4.284
2 3 78 4.357
${{S}_{2}}\,\!$ = 8.641
4 82 4.407
3 5 85 4.443
${{S}_{3}}\,\!$ = 8.850
2. Calculate the natural log $R\,\!$.
3. Divide the column of values for log $R\,\!$ into three groups of equal size, each containing $n\,\!$ items. There should always be three groups. Each group should always have the same number, $n\,\!$, of items, measurements or values.
4. Add the values of the natural log $R\,\!$ in each group, obtaining the sums identified as ${{S}_{1}}\,\!$, ${{S}_{2}}\,\!$ and ${{S}_{3}}\,\!$, starting with the lowest values of the natural log $R\,\!$.
5. Calculate $c\,\!$ using the following equation:
$c={{\left( \frac{{{S}_{3}}-{{S}_{2}}}{{{S}_{2}}-{{S}_{1}}} \right)}^{\tfrac{1}{n}}}\,\!$
6. Calculate $a\,\!$ using the following equation:
$a={{e}^{\left[ \tfrac{1}{n}\left( {{S}_{1}}+\tfrac{{{S}_{2}}-{{S}_{1}}}{1-{{c}^{n}}} \right) \right]}}\,\!$
7. Calculate $b\,\!$ using the following equation:
$b={{e}^{\left[ \tfrac{({{S}_{2}}-{{S}_{1}})(c-1)}{{{\left( 1-{{c}^{n}} \right)}^{2}}} \right]}}\,\!$
8. Write the Gompertz reliability growth equation.
9. Substitute the value of $T\,\!$, the time at which the reliability goal is to be achieved, to see if the reliability is indeed to be attained or exceeded by $T\,\!$.
### Confidence Bounds
The approximate reliability confidence bounds under the Gompertz model can be obtained with non-linear regression. Additionally, the reliability is always between 0 and 1. In order to keep the endpoints of the confidence interval, the logit transformation is used to obtain the confidence bounds on reliability.
$CB=\frac{{{{\hat{R}}}_{i}}}{{{{\hat{R}}}_{i}}+(1-{{{\hat{R}}}_{i}}){{e}^{\pm {{z}_{\alpha }}{{{\hat{\sigma }}}_{R}}/\left[ {{{\hat{R}}}_{i}}(1-{{{\hat{R}}}_{i}}) \right]}}}\,\!$
${{\hat{\sigma }}^{2}}=\frac{SSE}{n-p}\,\!$
where $p\,\!$ is the total number of groups (in this case 3) and $n\,\!$ is the total number of items in each group.
### Example - Standard Gompertz for Reliability Data
A device is required to have a reliability of 92% at the end of a 12-month design and development period. The following table gives the data obtained for the first five moths.
1. What will the reliability be at the end of this 12-month period?
2. What will the maximum achievable reliability be if the reliability program plan pursued during the first 5 months is continued?
3. How do the predicted reliability values compare with the actual values?
Design and Development Time vs. Demonstrated Reliability Data for a Device
Group Number Growth Time $T\,\!$(months) Reliability $R\,\!$(%) $\ln{R}\,\!$
0 58 4.060
1 1 66 4.190
${{S}_{1}}\,\!$ = 8.250
2 72.5 4.284
2 3 78 4.357
${{S}_{2}}\,\!$ = 8.641
4 82 4.407
3 5 85 4.443
${{S}_{3}}\,\!$ = 8.850
Solution
After generating the table above and calculating the last column to find ${{S}_{1}}\,\!$, ${{S}_{2}}\,\!$ and ${{S}_{3}}\,\!$, proceed as follows:
1. Solve for the value of $c\,\!$:
\begin{align} c= & {{\left( \frac{8.850-8.641}{8.641-8.250} \right)}^{\tfrac{1}{2}}} \\ = & 0.731 \end{align}\,\!
2. Solve for the value of $a\,\!$:
\begin{align} a&=e^\left[({\frac{1}{2}\left ( 8.250+\frac{S_{2}-S_{1}}{1-c^{n\cdot 1}} \right )} \right ] \\ &=e^{4.545} \\ &=94.16% \end{align}\,\!
This is the upper limit for the reliability as $T\to \infty \,\!$.
3. Solve for the value of $b\,\!$:
\begin{align} b= & {{e}^{\left[ \tfrac{(8.641-8.250)(0.731-1)}{{{(1-{{0.731}^{2}})}^{2}}} \right]}} \\ = & {{e}^{(-0.485)}} \\ = & 0.615 \end{align}\,\!
Now, that the initial values have been determined, the Gauss-Newton method can be used. Therefore, substituting ${{Y}_{i}}={{R}_{i}},\,\!$ $g_{1}^{(0)}=94.16,\,\!$ $g_{2}^{(0)}=0.615,\,\!$ $g_{3}^{(0)}=0.731,\,\!$ ${{Y}^{(0)}},{{D}^{(0)}},\,\!$ ${{\nu }^{(0)}}\,\!$ become:
${{Y}^{(0)}}=\left[ \begin{matrix} 0.0916 \\ 0.0015 \\ -0.1190 \\ 0.1250 \\ 0.0439 \\ -0.0743 \\ \end{matrix} \right]\,\!$
${{D}^{(0)}}=\left[ \begin{matrix} 0.6150 & 94.1600 & 0.0000 \\ 0.7009 & 78.4470 & -32.0841 \\ 0.7712 & 63.0971 & -51.6122 \\ 0.8270 & 49.4623 & -60.6888 \\ 0.8704 & 38.0519 & -62.2513 \\ 0.9035 & 28.8742 & -59.0463 \\ \end{matrix} \right]\,\!$
${{\nu }^{(0)}}=\left[ \begin{matrix} g_{1}^{(0)} \\ g_{2}^{(0)} \\ g_{3}^{(0)} \\ \end{matrix} \right]=\left[ \begin{matrix} 94.16 \\ 0.615 \\ 0.731 \\ \end{matrix} \right]\,\!$
The estimate of the parameters ${{\nu }^{(0)}}\,\!$ is given by:
\begin{align} \hat{v}&=\left ( D^{(0)}D^{(0)} \right )^{-1}D^{(0)}Y^{(0)} \\ &=\begin{bmatrix}0.061575\\ 0.000222\\ 0.001123\end{bmatrix} \end{align}
The revised estimated regression coefficients in matrix form are:
\begin{align} g^{(1)}&=g^{(0)} + \hat{v}^{(0)} \\ &= \begin{bmatrix}94.16\\ 0.615\\ 0.731\end{bmatrix}+\begin{bmatrix}0.061575\\ 0.000222\\ 0.001123\end{bmatrix} \\ &=\begin{bmatrix}94.2216\\ 0.6152\\ 0.7321\end{bmatrix} \end{align}
If the Gauss-Newton method works effectively, then the relationship ${{Q}^{(k+1)}}\lt {{Q}^{(k)}}\,\!$ has to hold, meaning that ${{g}^{(k+1)}}\,\!$ gives better estimates than ${{g}^{(k)}}\,\!$, after $k\,\!$. With the starting coefficients, ${{g}^{(0)}}\,\!$, $Q\,\!$ is:
\begin{align} Q^{0}&=\sum_{i=1}^{N}\left [ Y_{i}-f\left ( T_{i}, g^{(0)} \right ) \right ]^{2}\\ &= 0.045622 \end{align}
And with the coefficients at the end of the first iteration, ${{g}^{(1)}}\,\!$, $Q\,\!$ is:
\begin{align} Q^{1}&=\sum_{i=1}^{N}\left [ Y_{i}-f\left ( T_{i}, g^{(1)} \right ) \right ]^{2}\\ &= 0.041439 \end{align}
Therefore, it can be justified that the Gauss-Newton method works in the right direction. The iterations are continued until the relationship $Q^{(s-1)}-Q^{(s)}\simeq 0$ is satisfied. Note that the RGA software uses a different analysis method called the Levenberg-Marquardt. This method utilizes the best features of the Gauss-Newton method and the method of the steepest descent, and occupies a middle ground between these two methods. The estimated parameters using RGA are shown in the figure below.
They are:
\begin{align} & \widehat{a}= & 0.9422 \\ & \widehat{b}= & 0.6152 \\ & \widehat{c}= & 0.7321 \end{align}\,\!
The Gompertz reliability growth curve is:
$R=0.9422{{(0.6152)}^{{{0.7321}^{T}}}}\,\!$
1. The achievable reliability at the end of the 12-month period of design and development is:
\begin{align} R&=0.9422(0.6152)^{0.7321} &=0.9314 \end{align}\,\!
The required reliability is 92%. Consequently, from the previous result, this requirement will be barely met. Every effort should therefore be expended to implement the reliability program plan fully, and perhaps augment it slightly to assure that the reliability goal will be met.
2. The maximum achievable reliability from Step 2, or from the value of $a\,\!$, is 0.9422.
3. The predicted reliability values, as calculated from the standard Gompertz model, are compared with the actual data in the table below. It may be seen in the table that the Gompertz curve appears to provide a very good fit for the data used because the equation reproduces the available data with less than 1% error. The standard Gompertz model is plotted in the figure below the table. The plot identifies the type of reliability growth curve that the equation represents.
Comparison of the Predicted Reliabilities with the Actual Data
Growth Time $T\,\!$(months) Gompertz Reliability(%) Raw Data Reliability(%)
0 57.97 58.00
1 66.02 66.00
2 72.62 72.50
3 77.87 78.00
4 81.95 82.00
5 85.07 85.00
6 87.43
7 89.20
8 90.52
9 91.50
10 92.22
11 92.75
12 93.14
### Example - Standard Gompertz for Sequential Data
Calculate the parameters of the Gompertz model using the sequential data in the following table.
Sequential Data
Run Number Result Successes Observed Reliability(%)
1 F 0
2 F 0
3 F 0
4 S 1 25.00
5 F 1 20.00
6 F 1 16.67
7 S 2 28.57
8 S 3 37.50
9 S 4 44.44
10 S 5 50.00
11 S 6 54.55
12 S 7 58.33
13 S 8 61.54
14 S 9 64.29
15 S 10 66.67
16 S 11 68.75
17 F 11 64.71
18 S 12 66.67
19 F 12 63.16
20 S 13 65.00
21 S 14 66.67
22 S 15 68.18
Solution
Using RGA, the parameter estimates are shown in the following figure.
### Cumulative Reliability
For many kinds of equipment, especially missiles and space systems, only success/failure data (also called discrete or attribute data) is obtained. Conservatively, the cumulative reliability can be used to estimate the trend of reliability growth. The cumulative reliability is given by Kececioglu [3]:
$\bar{R}(N)=\frac{N-r}{N}\,\!$
where:
• $N\,\!$ is the current number of trials
• $r\,\!$ is the number of failures
It must be emphasized that the instantaneous reliability of the developed equipment is increasing as the test-analyze-fix-and-test process continues. In addition, the instantaneous reliability is higher than the cumulative reliability. Therefore, the reliability growth curve based on the cumulative reliability can be thought of as the lower bound of the true reliability growth curve.
# The Modified Gompertz Model
Sometimes, reliability growth data with an S-shaped trend cannot be described accurately by the Standard Gompertz or Logistic curves. Because these two models have fixed values of reliability at the inflection points, only a few reliability growth data sets following an S-shaped reliability growth curve can be fitted to them. A modification of the Gompertz curve, which overcomes this shortcoming, is given next [5].
If we apply a shift in the vertical coordinate, then the Gompertz model is defined by:
$R=d+a{{b}^{{{c}^{T}}}}\,\!$
where:
• $0\lt a+d\le 1\,\!$
• $0\lt b\lt 1,0\lt c\lt 1,\text{and}T\ge 0\,\!$
• $R\,\!$ is the system's reliability at development time $T\,\!$ or at launch number $T\,\!$, or stage number $T\,\!$
• $d\,\!$ is the shift parameter
• $d+a\,\!$ is the upper limit that the reliability approaches asymptotically as $T\to\infty\,\!$
• $d+ab\,\!$ is the initial reliability at $T=0\,\!$
• $c\,\!$ is the growth pattern indicator (small values of $c\,\!$ indicate rapid early reliability growth and large values of $c\,\!$ indicate slow reliability growth)
The modified Gompertz model is more flexible than the original, especially when fitting growth data with S-shaped trends.
### Parameter Estimation
To implement the modified Gompertz growth model, initial values of the parameters $a\,\!$, $b\,\!$, $c\,\!$ and $d\,\!$ must be determined. When analyzing reliability data in RGA, you have the option to enter the reliability values in percent or in decimal format. However, $a\,\!$ and $d\,\!$ will always be returned in decimal format and not in percent. The estimated parameters in RGA are unitless.
Given that $R=d+a{{b}^{{{c}^{T}}}}\,\!$ and $\ln (R-d)=\ln (a)+{{c}^{T}}\ln (b)\,\!$, it follows that ${{S}_{1}}\,\!$, ${{S}_{2}}\,\!$ and ${{S}_{3}}\,\!$, as defined in the derivation of the Standard Gompertz model, can be expressed as functions of $d\,\!$.
\begin{align} {{S}_{1}}(d)= & \underset{i=0}{\overset{n-1}{\mathop \sum }}\,\ln ({{R}_{i}}-d)=n\ln (a)+\ln (b)\underset{i=0}{\overset{n-1}{\mathop \sum }}\,{{c}^{{{T}_{i}}}} \\ {{S}_{2}}(d)= & \underset{i=n}{\overset{2n-1}{\mathop \sum }}\,\ln ({{R}_{i}}-d)=n\ln (a)+\ln (b)\underset{i=n}{\overset{2n-1}{\mathop \sum }}\,{{c}^{{{T}_{i}}}} \\ {{S}_{3}}(d)= & \underset{i=2n}{\overset{m-1}{\mathop \sum }}\,\ln ({{R}_{i}}-d)=n\ln (a)+\ln (b)\underset{i=2n}{\overset{m-1}{\mathop \sum }}\,{{c}^{{{T}_{i}}}} \end{align}\,\!
Modifying the equations for estimating parameters $c\,\!$, $a\,\!$, $b\,\!$, as functions of $d\,\!$, yields:
\begin{align} c(d)= & {{\left[ \frac{{{S}_{3}}(d)-{{S}_{2}}(d)}{{{S}_{2}}(d)-{{S}_{1}}(d)} \right]}^{\tfrac{1}{n\cdot I}}} \\ a(d)= & {{e}^{\left[ \tfrac{1}{n}\left( {{S}_{1}}(d)+\tfrac{{{S}_{2}}(d)-{{S}_{1}}(d)}{1-{{[c(d)]}^{n\cdot I}}} \right) \right]}} \\ b(d)= & {{e}^{\left[ \tfrac{\left[ {{S}_{2}}(d)-{{S}_{1}}(d) \right]\left[ {{[c(d)]}^{I}}-1 \right]}{{{\left[ 1-{{[c(d)]}^{n\cdot I}} \right]}^{2}}} \right]}} \end{align}\,\!
where $I\,\!$ is the time interval increment. At this point, you can use the initial constraint of:
$d+ab=\text{original level of reliability at }T=0 \,\!$
Now there are four unknowns, $a\,\!$, $b\,\!$, $c\,\!$ and $d\,\!$, and four corresponding equations. The simultaneous solution of these equations yields the four initial values for the parameters of the modified Gompertz model. This procedure is similar to the one discussed before. It starts by using initial estimates of the parameters, $a\,\!$, $b\,\!$, $c\,\!$ and $d\,\!$, denoted as $g_{1}^{(0)},\,\!$ $g_{2}^{(0)},\,\!$ $g_{3}^{(0)},\,\!$ and $g_{4}^{(0)},\,\!$ where $^{(0)}\,\!$ is the iteration number.
The Taylor series expansion approximates the mean response, $f({{T}_{i}},\delta )\,\!$, around the starting values, $g_{1}^{(0)},\,\!$ $g_{2}^{(0)},\,\!$ $g_{3}^{(0)}\,\!$ and $g_{4}^{(0)}\,\!$. For the ${{i}^{th}}\,\!$ observation:
$f({{T}_{i}},\delta )\simeq f({{T}_{i}},{{g}^{(0)}})+\underset{k=1}{\overset{p}{\mathop \sum }}\,{{\left[ \frac{\partial f({{T}_{i}},\delta )}{\partial {{\delta }_{k}}} \right]}_{\delta ={{g}^{(0)}}}}\cdot ({{\delta }_{k}}-g_{k}^{(0)})\,\!$
where:
${{g}^{(0)}}=\left[ \begin{matrix} g_{1}^{(0)} \\ g_{2}^{(0)} \\ g_{3}^{(0)} \\ g_{4}^{(0)} \\ \end{matrix} \right]\,\!$
Let:
\begin{align} f_{i}^{(0)}= & f({{T}_{i}},{{g}^{(0)}}) \\ \nu _{k}^{(0)}= & ({{\delta }_{k}}-g_{k}^{(0)}) \\ D_{ik}^{(0)}= & {{\left[ \frac{\partial f({{T}_{i}},\delta )}{\partial {{\delta }_{k}}} \right]}_{\delta ={{g}^{(0)}}}} \end{align}\,\!
Therefore:
${{Y}_{i}}=f_{i}^{(0)}+\underset{k=1}{\overset{p}{\mathop \sum }}\,D_{ik}^{(0)}\nu _{k}^{(0)}\,\!$
or by shifting $f_{i}^{(0)}\,\!$ to the left of the equation:
$Y_{i}^{(0)}-f_{i}^{(0)}=\underset{k=1}{\overset{p}{\mathop \sum }}\,D_{ik}^{(0)}\nu _{k}^{(0)}\,\!$
In matrix form, this is given by:
${{Y}^{(0)}}\simeq {{D}^{(0)}}{{\nu }^{(0)}}\,\!$
where:
${{Y}^{(0)}}=\left[ \begin{matrix} {{Y}_{1}}-f_{1}^{(0)} \\ . \\ . \\ {{Y}_{N}}-f_{N}^{(0)} \\ \end{matrix} \right]=\left[ \begin{matrix} {{Y}_{1}}-g_{4}^{(0)}+g_{1}^{(0)}g_{2}^{(0)g_{3}^{(0){{T}_{1}}}} \\ . \\ . \\ {{Y}_{N}}-g_{4}^{(0)}+g_{1}^{(0)}g_{2}^{(0)g_{3}^{(0){{T}_{N}}}} \\ \end{matrix} \right]\,\!$
\begin{align} {{D}^{(0)}}= & \left[ \begin{matrix} D_{11}^{(0)} & D_{12}^{(0)} & D_{13}^{(0)} & D_{14}^{(0)} \\ . & . & . & . \\ . & . & . & . \\ D_{N1}^{(0)} & D_{N2}^{(0)} & D_{N3}^{(0)} & D_{N4}^{(0)} \\ \end{matrix} \right] \\ = & \left[ \begin{matrix} g_{2}^{(0)g_{3}^{(0){{T}_{1}}}} & \tfrac{g_{1}^{(0)}}{g_{2}^{(0)}}g_{3}^{(0){{T}_{1}}}g_{2}^{(0)g_{3}^{(0){{T}_{1}}}} & \tfrac{g_{1}^{(0)}}{g_{3}^{(0)}}g_{3}^{(0){{T}_{1}}}\ln (g_{2}^{(0)}){{T}_{1}}g_{2}^{(0)g_{3}^{(0){{T}_{1}}}} & 1 \\ . & . & . & . \\ . & . & . & . \\ g_{2}^{(0)g_{3}^{(0){{T}_{N}}}} & \tfrac{g_{1}^{(0)}}{g_{2}^{(0)}}g_{3}^{(0){{T}_{N}}}g_{2}^{(0)g_{3}^{(0){{T}_{N}}}} & \tfrac{g_{1}^{(0)}}{g_{3}^{(0)}}g_{3}^{(0){{T}_{N}}}\ln (g_{2}^{(0)}){{T}_{N}}g_{2}^{(0)g_{3}^{(0){{T}_{N}}}} & 1 \\ \end{matrix} \right] \end{align}\,\!
${{\nu }^{(0)}}=\left[ \begin{matrix} g_{1}^{(0)} \\ g_{2}^{(0)} \\ g_{3}^{(0)} \\ g_{4}^{(0)} \\ \end{matrix} \right]\,\!$
The same reasoning as before is followed here, and the estimate of the parameters ${{\nu }^{(0)}}\,\!$ is given by:
${{\widehat{\nu }}^{(0)}}={{\left( {{D}^{{{(0)}^{T}}}}{{D}^{(0)}} \right)}^{-1}}{{D}^{{{(0)}^{T}}}}{{Y}^{(0)}}\,\!$
The revised estimated regression coefficients in matrix form are:
${{g}^{(1)}}={{g}^{(0)}}+{{\widehat{\nu }}^{(0)}}\,\!$
To see if the revised regression coefficients will lead to a reasonable result, the least squares criterion measure, $Q$, should be checked. According to the Least Squares Principle, the solution to the values of the parameters are those values that minimize $Q\,\!$. With the starting coefficients, ${{g}^{(0)}}\,\!$, $Q\,\!$ is:
${{Q}^{(0)}}=\underset{i=1}{\overset{N}{\mathop \sum }}\,{{\left( {{Y}_{i}}-f({{T}_{i}},{{g}^{(0)}}) \right)}^{2}}\,\!$
With the coefficients at the end of the first iteration, ${{g}^{(1)}}\,\!$, $Q\,\!$ is:
${{Q}^{(1)}}=\underset{i=1}{\overset{N}{\mathop \sum }}\,{{\left( {{Y}_{i}}-f({{T}_{i}},{{g}^{(1)}}) \right)}^{2}}\,\!$
For the Gauss-Newton method to work properly, and to satisfy the Least Squares Principle, the relationship ${{Q}^{(k+1)}}\lt {{Q}^{(k)}}\,\!$ has to hold for all $k\,\!$, meaning that ${{g}^{(k+1)}}\,\!$ gives a better estimate than ${{g}^{(k)}}\,\!$. The problem is not yet completely solved. Now ${{g}^{(1)}}\,\!$ are the starting values, producing a new set of values ${{g}^{(2)}}.\,\!$ The process is continued until the following relationship has been satisfied.
${{Q}^{(s-1)}}-{{Q}^{(s)}}\simeq 0\,\!$
As mentioned previously, when using the Gauss-Newton method or some other estimation procedure, it is advisable to try several sets of starting values to make sure that the solution gives relatively consistent results. Note that RGA uses a different analysis method called the Levenberg-Marquardt. This method utilizes the best features of the Gauss-Newton method and the method of the steepest descent, and occupies a middle ground between these two methods.
### Confidence Bounds
The approximate reliability confidence bounds under the modified Gompertz model can be obtained using non-linear regression. Additionally, the reliability is always between 0 and 1. In order to keep the endpoints of the confidence interval, the logit transformation can be used to obtain the confidence bounds on reliability.
$CB=\frac{{{{\hat{R}}}_{i}}}{{{{\hat{R}}}_{i}}+(1-{{{\hat{R}}}_{i}}){{e}^{\pm {{z}_{\alpha }}{{{\hat{\sigma }}}_{R}}/\left[ {{{\hat{R}}}_{i}}(1-{{{\hat{R}}}_{i}}) \right]}}}\,\!$
${{\hat{\sigma }}^{2}}=\frac{SSE}{n-p}\,\!$
where $p\,\!$ is the total number of groups (in this case 4) and $n\,\!$ is the total number of items in each group.
### Example - Modified Gompertz for Reliability Data
A reliability growth data set is given in columns 1 and 2 of the following table. Find the modified Gompertz curve that represents the data and plot it comparatively with the raw data.
Development Time vs. Observed Reliability Data and Predicted Reliabilities
Time(months) Raw Data Reliability(%) Gompertz Reliability(%) Logistic Reliability(%) Modified Gompertz Reliability(%)
0 31.00 25.17 22.70 31.18
1 35.50 38.33 38.10 35.08
2 49.30 51.35 56.40 49.92
3 70.10 62.92 73.00 69.23
4 83.00 72.47 85.00 83.72
5 92.20 79.94 93.20 92.06
6 96.40 85.59 96.10 96.29
7 98.60 89.75 98.10 98.32
8 99.00 92.76 99.10 99.27
Solution
To determine the parameters of the modified Gompertz curve, use:
\begin{align} & {{S}_{1}}(d)= & \underset{i=0}{\overset{2}{\mathop \sum }}\,\ln ({{R}_{oi}}-d) \\ & {{S}_{2}}(d)= & \underset{i=3}{\overset{5}{\mathop \sum }}\,\ln ({{R}_{oi}}-d) \\ & {{S}_{3}}(d)= & \underset{i=6}{\overset{8}{\mathop \sum }}\,\ln ({{R}_{oi}}-d) \end{align}\,\!
$c(d)={{\left[ \frac{{{S}_{3}}(d)-{{S}_{2}}(d)}{{{S}_{2}}(d)-{{S}_{1}}(d)} \right]}^{\tfrac{1}{3}}}\,\!$
$a(d)={{e}^{\left[ \tfrac{1}{3}\left( {{S}_{1}}(d)+\tfrac{{{S}_{2}}(d)-{{S}_{1}}(d)}{1-{{[c(d)]}^{3}}} \right) \right]}}\,\!$
$b(d)={{e}^{\left[ \tfrac{({{S}_{2}}(d)-{{S}_{1}}(d))(c(d)-1)}{{{\left[ 1-{{[c(d)]}^{3}} \right]}^{2}}} \right]}}\,\!$
and:
${{R}_{0}}=d+a(d)\cdot b(d)\,\!$
for ${{R}_{0}}=31%\,\!$, the equation above may be rewritten as:
$d-31+a(d)\cdot b(d)=0\,\!$
The equations for parameters $c\,\!$, $a\,\!$ and $b\,\!$ can now be solved simultaneously. One method for solving these equations numerically is to substitute different values of $d\,\!$, which must be less than ${{R}_{0}}\,\!$, into the last equation shown above, and plot the results along the y-axis with the value of $d\,\!$ along the x-axis. The value of $d\,\!$ can then be read from the x-intercept. This can be repeated for greater accuracy using smaller and smaller increments of $d\,\!$. Once the desired accuracy on $d\,\!$ has been achieved, the value of $d\,\!$ can then be used to solve for $a\,\!$, $b\,\!$ and $c\,\!$. For this case, the initial estimates of the parameters are:
\begin{align} \widehat{a}= & 69.324 \\ \widehat{b}= & 0.002524 \\ \widehat{c}= & 0.46012 \\ \widehat{d}= & 30.825 \end{align}\,\!
Now, since the initial values have been determined, the Gauss-Newton method can be used. Therefore, substituting ${{Y}_{i}}={{R}_{i}},\,\!$ $g_{1}^{(0)}=69.324,\,\!$ $g_{2}^{(0)}=0.002524,\,\!$ $g_{3}^{(0)}=0.46012,\,\!$ and $g_{4}^{(0)}=30.825\,\!$, ${{Y}^{(0)}},{{D}^{(0)}},\,\!$ ${{\nu }^{(0)}}\,\!$ become:
${{Y}^{(0)}}=\left[ \begin{matrix} 0.000026 \\ 0.253873 \\ -1.062940 \\ 0.565690 \\ -0.845260 \\ 0.096737 \\ 0.076450 \\ 0.238155 \\ -0.320890 \\ \end{matrix} \right]\,\!$
${{D}^{(0)}}=\left[ \begin{matrix} 0.002524 & 69.3240 & 0.0000 & 1 \\ 0.063775 & 805.962 & -26.4468 & 1 \\ 0.281835 & 1638.82 & -107.552 & 1 \\ 0.558383 & 1493.96 & -147.068 & 1 \\ 0.764818 & 941.536 & -123.582 & 1 \\ 0.883940 & 500.694 & -82.1487 & 1 \\ 0.944818 & 246.246 & -48.4818 & 1 \\ 0.974220 & 116.829 & -26.8352 & 1 \\ 0.988055 & 54.5185 & -14.3117 & 1 \\ \end{matrix} \right]\,\!$
${{\nu }^{(0)}}=\left[ \begin{matrix} g_{1}^{(0)} \\ g_{2}^{(0)} \\ g_{3}^{(0)} \\ g_{4}^{(0)} \\ \end{matrix} \right]=\left[ \begin{matrix} 69.324 \\ 0.002524 \\ 0.46012 \\ 30.825 \\ \end{matrix} \right]\,\!$
The estimate of the parameters ${{\nu }^{(0)}}\,\!$ is given by:
\begin{align} {{\widehat{\nu }}^{(0)}}= & {{\left( {{D}^{{{(0)}^{T}}}}{{D}^{(0)}} \right)}^{-1}}{{D}^{{{(0)}^{T}}}}{{Y}^{(0)}} \\ = & \left[ \begin{matrix} -0.275569 \\ -0.000549 \\ -0.003202 \\ 0.209458 \\ \end{matrix} \right] \end{align}\,\!
The revised estimated regression coefficients in matrix form are given by:
\begin{align} {{g}^{(1)}}= & {{g}^{(0)}}+{{\widehat{\nu }}^{(0)}}. \\ = & \left[ \begin{matrix} 69.324 \\ 0.002524 \\ 0.46012 \\ 30.825 \\ \end{matrix} \right]+\left[ \begin{matrix} -0.275569 \\ -0.000549 \\ -0.003202 \\ 0.209458 \\ \end{matrix} \right] \\ = & \left[ \begin{matrix} 69.0484 \\ 0.00198 \\ 0.45692 \\ 31.0345 \\ \end{matrix} \right] \end{align}\,\!
With the starting coefficients ${{g}^{(0)}}\,\!$, $Q\,\!$ is:
\begin{align} {{Q}^{(0)}}= & \underset{i=1}{\overset{N}{\mathop \sum }}\,{{\left( {{Y}_{i}}-f({{T}_{i}},{{g}^{(0)}}) \right)}^{2}} \\ = & 2.403672 \end{align}\,\!
With the coefficients at the end of the first iteration, ${{g}^{(1)}}\,\!$, $Q\,\!$ is:
\begin{align} {{Q}^{(1)}}= & \underset{i=1}{\overset{N}{\mathop \sum }}\,{{\left[ {{Y}_{i}}-f\left( {{T}_{i}},{{g}^{(1)}} \right) \right]}^{2}} \\ = & 2.073964 \end{align}\,\!
Therefore:
\begin{align} {{Q}^{(1)}}\lt {{Q}^{(0)}} \end{align}\,\!
Hence, the Gauss-Newton method works in the right direction. The iterations are continued until the relationship of ${{Q}^{(s-1)}}-{{Q}^{(s)}}\simeq 0\,\!$ has been satisfied. Using RGA, the estimators of the parameters are:
\begin{align} \widehat{a}= & 0.6904 \\ \widehat{b}= & 0.0020 \\ \widehat{c}= & 0.4567 \\ \widehat{d}= & 0.3104 \end{align}\,\!
Therefore, the modified Gompertz model is:
$R=0.3104+(0.6904){{(0.0020)}^{{{0.4567}^{T}}}}\,\!$
Using this equation, the predicted reliability is plotted in the following figure along with the raw data. As you can see, the modified Gompertz curve represents the data very well.
## More Examples
### Standard Gompertz for Grouped per Configuration Data
A new design is put through a reliability growth test. The requirement is that after the ninth stage the design will exhibit an 85% reliability with a 90% confidence level. Given the data in the following table, do the following:
1. Estimate the parameters of the standard Gompertz model.
2. What is the initial reliability at $T=0\,\!$ ?
3. Determine the reliability at the end of the ninth stage and check to see whether the goal has been met.
Grouped per Configuration Data
Stage Number of Units Number of Failures
1 10 5
2 8 3
3 9 3
4 9 2
5 10 2
6 10 1
7 10 1
8 10 1
9 10 1
Solution
1. The data is entered in cumulative format and the estimated standard Gompertz parameters are shown in the following figure.
2. The initial reliability at $T=0\,\!$ is equal to:
\begin{align} {{R}_{T=0}}= & a\cdot b \\ = & 0.9497\cdot 0.5249 \\ = & 0.4985 \end{align}\,\!
3. The reliability at the ninth stage can be calculated using the Quick Calculation Pad (QCP) as shown in the figure below.
The estimated reliability at the end of the ninth stage is equal to 91.92%. However, the lower limit at the 90% 1-sided confidence bound is equal to 82.15%. Therefore, the required goal of 85% reliability at a 90% confidence level has not been met.
### Comparing Standard and Modified Gompertz
Using the data in the following table, determine whether the standard Gompertz or modified Gompertz would be better suited for analyzing the given data.
Reliability Data
Stage Reliability (%)
0 36
1 38
2 46
3 58
4 71
5 80
6 86
7 88
8 90
9 91
Solution
The standard Gompertz Reliability vs. Time plot is shown next.
The standard Gompertz seems to do a fairly good job of modeling the data. However, it appears that it is having difficulty modeling the S-shape of the data. The modified Gompertz Reliability vs. Time plot is shown next. As expected, the modified Gompertz does a much better job of handling the S-shape presented by the data and provides a better fit for this data. | 15,082 | 39,909 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 32, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-31 | latest | en | 0.801023 |
http://en.wikipedia.org/wiki/Rectangles | 1,419,637,652,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447549908.109/warc/CC-MAIN-20141224185909-00081-ip-10-231-17-201.ec2.internal.warc.gz | 33,452,334 | 23,554 | # Rectangle
(Redirected from Rectangles)
For the record label, see Rectangle (label).
Rectangle
Rectangle
Edges and vertices 4
Schläfli symbol { } × { } or { }2
Coxeter diagram
Symmetry group Dih2, [2], (*22), order 4
Dual polygon rhombus
Properties convex, isogonal, cyclic Opposite angles and sides are congruent
In Euclidean plane geometry, a rectangle is any quadrilateral with four right angles. It can also be defined as an equiangular quadrilateral, since equiangular means that all of its angles are equal (360°/4 = 90°). It can also be defined as a parallelogram containing a right angle. A rectangle with four sides of equal length is a square. The term oblong is occasionally used to refer to a non-square rectangle.[1][2][3] A rectangle with vertices ABCD would be denoted as ABCD.
The word rectangle comes from the Latin rectangulus, which is a combination of rectus (right) and angulus (angle).
A so-called crossed rectangle is a crossed (self-intersecting) quadrilateral which consists of two opposite sides of a rectangle along with the two diagonals.[4] It is a special case of an antiparallelogram, and its angles are not right angles. Other geometries, such as spherical, elliptic, and hyperbolic, have so-called rectangles with opposite sides equal in length and equal angles that are not right angles.
Rectangles are involved in many tiling problems, such as tiling the plane by rectangles or tiling a rectangle by polygons.
## Characterizations
A convex quadrilateral is a rectangle if and only if it is any one of the following:[5][6]
• a quadrilateral with four right angles
• a parallelogram with at least one right angle
• a parallelogram with diagonals of equal length
• a parallelogram ABCD where triangles ABD and DCA are congruent
• a convex quadrilateral with successive sides a, b, c, d whose area is $\tfrac{1}{4}(a+c)(b+d)$.[7]:fn.1
• a convex quadrilateral with successive sides a, b, c, d whose area is $\tfrac{1}{2} \sqrt{(a^2+c^2)(b^2+d^2)}.$[7]
## Classification
A rectangle is a special case of a parallelogram in which each pair of adjacent sides is perpendicular.
A parallelogram is a special case of a trapezium (known as a trapezoid in North America) in which both pairs of opposite sides are parallel and equal in length.
A trapezium is a convex quadrilateral which has at least one pair of parallel opposite sides.
• Simple: The boundary does not cross itself.
• Star-shaped: The whole interior is visible from a single point, without crossing any edge.
### Alternative hierarchy
De Villiers defines a rectangle more generally as any quadrilateral with axes of symmetry through each pair of opposite sides.[8] This definition includes both right-angled rectangles and crossed rectangles. Each has an axis of symmetry parallel to and equidistant from a pair of opposite sides, and another which is the perpendicular bisector of those sides, but, in the case of the crossed rectangle, the first axis is not an axis of symmetry for either side that it bisects.
Quadrilaterals with two axes of symmetry, each through a pair of opposite sides, belong to the larger class of quadrilaterals with at least one axis of symmetry through a pair of opposite sides. These quadrilaterals comprise isosceles trapezia and crossed isosceles trapezia (crossed quadrilaterals with the same vertex arrangement as isosceles trapezia).
## Properties
### Symmetry
A rectangle is cyclic: all corners lie on a single circle.
It is equiangular: all its corner angles are equal (each of 90 degrees).
It is isogonal or vertex-transitive: all corners lie within the same symmetry orbit.
It has two lines of reflectional symmetry and rotational symmetry of order 2 (through 180°).
### Rectangle-rhombus duality
The dual polygon of a rectangle is a rhombus, as shown in the table below.[9]
Rectangle Rhombus
All angles are equal. All sides are equal.
Alternate sides are equal. Alternate angles are equal.
Its centre is equidistant from its vertices, hence it has a circumcircle. Its centre is equidistant from its sides, hence it has an incircle.
Its axes of symmetry bisect opposite sides. Its axes of symmetry bisect opposite angles.
Diagonals are equal in length. Diagonals intersect at equal angles.
• The figure formed by joining, in order, the midpoints of the sides of a rectangle is a rhombus and vice-versa.
### Miscellaneous
The two diagonals are equal in length and bisect each other. Every quadrilateral with both these properties is a rectangle.
A rectangle is rectilinear: its sides meet at right angles.
A rectangle in the plane can be defined by five independent degrees of freedom consisting, for example, of three for position (comprising two of translation and one of rotation), one for shape (aspect ratio), and one for overall size (area).
Two rectangles, neither of which will fit inside the other, are said to be incomparable.
## Formulae
The formula for the perimeter of a rectangle.
If a rectangle has length $\ell$ and width $w$
• it has area $A = \ell w\,$,
• it has perimeter $P = 2\ell + 2w = 2(\ell + w)\,$,
• each diagonal has length $d=\sqrt{\ell^2 + w^2}$,
• and when $\ell = w\,$, the rectangle is a square.
## Theorems
The isoperimetric theorem for rectangles states that among all rectangles of a given perimeter, the square has the largest area.
The midpoints of the sides of any quadrilateral with perpendicular diagonals form a rectangle.
A parallelogram with equal diagonals is a rectangle.
The Japanese theorem for cyclic quadrilaterals[10] states that the incentres of the four triangles determined by the vertices of a cyclic quadrilateral taken three at a time form a rectangle.
The British flag theorem states that with vertices denoted A, B, C, and D, for any point P on the same plane of a rectangle:[11]
$\displaystyle (AP)^2 + (CP)^2 = (BP)^2 + (DP)^2.$
## Crossed rectangles
A crossed (self-intersecting) quadrilateral consists of two opposite sides of a non-self-intersecting quadrilateral along with the two diagonals. Similarly, a crossed rectangle is a crossed quadrilateral which consists of two opposite sides of a rectangle along with the two diagonals. It has the same vertex arrangement as the rectangle. It appears as two identical triangles with a common vertex, but the geometric intersection is not considered a vertex.
A crossed quadrilateral is sometimes likened to a bow tie or butterfly. A three-dimensional rectangular wire frame that is twisted can take the shape of a bow tie. A crossed rectangle is sometimes called an "angular eight".
The interior of a crossed rectangle can have a polygon density of ±1 in each triangle, dependent upon the winding orientation as clockwise or counterclockwise.
A crossed rectangle is not equiangular. The sum of its interior angles (two acute and two reflex), as with any crossed quadrilateral, is 720°.[12]
A rectangle and a crossed rectangle are quadrilaterals with the following properties in common:
• Opposite sides are equal in length.
• The two diagonals are equal in length.
• It has two lines of reflectional symmetry and rotational symmetry of order 2 (through 180°).
## Other rectangles
A saddle rectangle has 4 nonplanar vertices, alternated from vertices of a cuboid, with a unique minimal surface interior defined as a linear combination of the four vertices, creating a saddle surface. This example shows 4 blue edges of the rectangle, and two green diagonals, all being diagonal of the cuboid rectangular faces.
In spherical geometry, a spherical rectangle is a figure whose four edges are great circle arcs which meet at equal angles greater than 90°. Opposite arcs are equal in length. The surface of a sphere in Euclidean solid geometry is a non-Euclidean surface in the sense of elliptic geometry. Spherical geometry is the simplest form of elliptic geometry.
In elliptic geometry, an elliptic rectangle is a figure in the elliptic plane whose four edges are elliptic arcs which meet at equal angles greater than 90°. Opposite arcs are equal in length.
In hyperbolic geometry, a hyperbolic rectangle is a figure in the hyperbolic plane whose four edges are hyperbolic arcs which meet at equal angles less than 90°. Opposite arcs are equal in length.
## Tessellations
The rectangle is used in many periodic tessellation patterns, in brickwork, for example, these tilings:
## Squared, perfect, and other tiled rectangles
A rectangle tiled by squares, rectangles, or triangles is said to be a "squared", "rectangled", or "triangulated" (or "triangled") rectangle respectively. The tiled rectangle is perfect[13][14] if the tiles are similar and finite in number and no two tiles are the same size. If two such tiles are the same size, the tiling is imperfect. In a perfect (or imperfect) triangled rectangle the triangles must be right triangles.
A rectangle has commensurable sides if and only if it is tileable by a finite number of unequal squares.[13][15] The same is true if the tiles are unequal isosceles right triangles.
The tilings of rectangles by other tiles which have attracted the most attention are those by congruent non-rectangular polyominoes, allowing all rotations and reflections. There are also tilings by congruent polyaboloes.
## References
1. ^ http://www.cimt.plymouth.ac.uk/resources/topics/art002.pdf
2. ^ Definition of Oblong. Mathsisfun.com. Retrieved 2011-11-13.
3. ^ Oblong – Geometry – Math Dictionary. Icoachmath.com. Retrieved 2011-11-13.
4. ^ Coxeter, Harold Scott MacDonald; Longuet-Higgins, M.S.; Miller, J.C.P. (1954). "Uniform polyhedra". Philosophical Transactions of the Royal Society of London. Series A. Mathematical and Physical Sciences (The Royal Society) 246 (916): 401–450. doi:10.1098/rsta.1954.0003. ISSN 0080-4614. JSTOR 91532. MR 0062446.
5. ^ Zalman Usiskin and Jennifer Griffin, "The Classification of Quadrilaterals. A Study of Definition", Information Age Publishing, 2008, pp. 34–36 ISBN 1-59311-695-0.
6. ^ Owen Byer; Felix Lazebnik; Deirdre L. Smeltzer (19 August 2010). Methods for Euclidean Geometry. MAA. pp. 53–. ISBN 978-0-88385-763-2. Retrieved 2011-11-13.
7. ^ a b Martin Josefsson, "Five Proofs of an Area Characterization of Rectangles", Forum Geometricorum 13 (2013) 17–21.
8. ^ An Extended Classification of Quadrilaterals (An excerpt from De Villiers, M. 1996. Some Adventures in Euclidean Geometry. University of Durban-Westville.)
9. ^ de Villiers, Michael, "Generalizing Van Aubel Using Duality", Mathematics Magazine 73 (4), Oct. 2000, pp. 303-307.
10. ^ Cyclic Quadrilateral Incentre-Rectangle with interactive animation illustrating a rectangle that becomes a 'crossed rectangle', making a good case for regarding a 'crossed rectangle' as a type of rectangle.
11. ^ Hall, Leon M., and Robert P. Roe (1998). "An Unexpected Maximum in a Family of Rectangles". Mathematics Magazine 71 (4): 285–291. JSTOR 2690700.
12. ^ Stars: A Second Look. (PDF). Retrieved 2011-11-13.
13. ^ a b R.L. Brooks, C.A.B. Smith, A.H. Stone and W.T. Tutte (1940). "The dissection of rectangles into squares". Duke Math. J. 7 (1): 312–340. doi:10.1215/S0012-7094-40-00718-9.
14. ^ J.D. Skinner II, C.A.B. Smith and W.T. Tutte (November 2000). "On the Dissection of Rectangles into Right-Angled Isosceles Triangles". J. Combinatorial Theory Series B 80 (2): 277–319. doi:10.1006/jctb.2000.1987.
15. ^ R. Sprague (1940). "Ũber die Zerlegung von Rechtecken in lauter verschiedene Quadrate". J. fũr die reine und angewandte Mathematik 182: 60–64. | 2,846 | 11,495 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2014-52 | latest | en | 0.895478 |
https://friendsofgoetheschool.org/3067/how-do-i-calculate-how-much-welding-rod-i-need/ | 1,652,739,010,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662512249.16/warc/CC-MAIN-20220516204516-20220516234516-00111.warc.gz | 325,298,840 | 11,171 | # How do I calculate how much welding rod I need?
1) Calculate the weld area (cross section of the weld) including root and cap reinforcement in square millimetres. 2) Multiply weld area by pipe circumference (weld length) to give a weld volume in cubic millimetres. 3) Divide this answer by 1,000,000,000 to give weld volume in cubic metres.
1 Calculating Filler Metal Consumption The number of pounds of welding electrodes or welding wire necessary to complete a given weld joint may be calculated by the formula: P = WL E Where: P = Pounds of electrode or wire required W = Weight per foot of weld metal L = Length of weld (feet) E = Deposition efficiency
Subsequently, question is, how many amps does it take to weld a 1/4 inch steel? For example, to MIG-weld 18-gauge steel in a single pass takes roughly 70 amps. Welding 1/4inch steel in a single pass requires roughly 180 amps.
Likewise, people ask, how do you calculate welding time?
EXAMPLE 1) Welding Time = Wt/Ft of Weld x Ft of Weld Deposition Rate x Operating Factor Substituting the values in Example 1: . 814 x 1,280 5.36 x . 30 = 1,042 1.608 = 648 Hrs.
How do you calculate welding thickness?
The size of fillet weld height approximately specifies the = 0.7 formula, where z is the fillet weld width. Minimum fillet weld height is selected according to the thickness of the thicker welded part and according to the material.
### What is weld length?
The legs are the opposite and adjacent sides to the triangular fillet weld. The leg length is usually designated as the size of the weld. The throat of the weld is the distance from the center of the face to the root of the weld.
### How do you measure weld strength?
One testing method is to use soap bubbles to check for problems (can be easily applied with a squirt bottle. Strength: Most welds need to demonstrate the required strength. One way to ensure proper strength is to start with a filler metal and electrode rating that is higher than your strength requirement.
### How many welding rods are in a pound?
14 1/8″ rods per pound.. usually.
### How many welding rods are in a packet?
6th floor, Tower 2, Assotech Business Cresterra, Plot No. Shakti 3.15 MM Welding Electrodes. Size 3.15 mm Packets Per Box 12 Pieces Per Packet 90
### How much weight does welding add?
17 welds x 4.5in each = 76.5 inches = 6.375 feet -> @ 0.129 lb/ft this yields 0.82 lbs per assembly. Adding these together gives a total of 1.63 lb per part. And for a total of 3,200 parts we would need 5,216 pounds of weld metal.
### How much is welding per hour?
Average weld speed for 1/4 fillet = 20 ipm so weld time around 9 – 10 minutes I usually add 20% to the weld time for movement time between welds, so weld time per-part should be around 11 – 13 minutes. At least four parts an hour produced.
### What is diameter inch in welding?
Let me give it a shot. Diameter Inch is the unit of measurement for pipewelds. If you have a particular spool that will have 7 joints, the total inch-dia for the welds of that spool will be :(diameter of the pipe/fitting x 7). For example, a 6″ spool with 7 joints would have a total inch-dia of 6 x 7 = 42 inch dia.
### How do you calculate welding speed?
The formula is as follows: Heat Input = (60 x Amps x Volts) / (1,000 x Travel Speed in in/min) = KJ/in. Travel Speed = Length of Weld / Time to weld = 25 inches / 2 minutes = 12.5 inches per minute. Heat Input = [(60 sec/min) x (325 amps) x (29 volts)] / [(1,000 joules/kilojoule) x (12.5 inches/minute)]
### How many inches can a welder weld per day?
The average welder can complete about 140 inches of weld per hour in ordinary 1/4 inch wall line pipe. Therefore, to find the output duration per day you would need to: Number of welds per hour is equal to = ((140/((diameter of pipe – in inches)*3.14))). This gives you an hourly rate of pipeline joints per hour.
### What is arc time in welding?
Arc Time – The time during which an arc is maintained. Arc Voltage – The voltage across the welding arc. Arc Welding – A group of welding processes which produces coalescence of metals by heating them with an arc, with or without the application of pressure and with or without the use of filler metal.
### How many passes fillet weld?
Illustrations for fillet welds using various electrode sizes. A 1/8″ 7018 can potentially produce a 3/16″ weld size with one pass and 5/16″ with 3 passes. A 5/32″ 7018 can potentially produce a 1/4″ weld size with one pass and 3/8″ with 3 passes.
### What is a spot welder used for?
spot welding (or resistance spot welding) is a type of electric resistance welding used to weld various sheet metal products, through a process in which contacting metal surface points are joined by the heat obtained from resistance to electric current.
### How do you bid on piping jobs?
The elemental formula for bidding a pipe job is: fuel costs + material costs + labor + profit margin = bid. Ask the contractor what substance the pipe will be moving. Then you know the required welding specifications.
### How many amps does it take to weld 1 inch Steel?
Next, consider that thin metals require less amperage and thicker metals require more amperage. For example, to MIG weld 18 gauge steel in a single pass takes roughly 70 amps, where welding 1/4 in. steel in a single pass requires roughly 180 amps. | 1,348 | 5,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2022-21 | latest | en | 0.894742 |
http://bootmath.com/inequality-involving-taking-expectations.html | 1,529,582,478,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864148.93/warc/CC-MAIN-20180621114153-20180621134153-00201.warc.gz | 45,677,585 | 5,890 | # Inequality involving taking expectations
There are three convexly decreasing functions $f, g, h:\mathbb{R^+}\rightarrow \mathbb{R^+}$, and $f(x)h(x)<1$ for $\forall x$.
I need to prove that
$E[f(x)^2]E[g(x)h(x)]<E[f(x)g(x)]\left(1+E\left[f(x)h(x)\right]\right)$
for an arbitrary probability distribution.
#### Solutions Collecting From Web of "Inequality involving taking expectations"
The result is true more generally without convexity.
Fact 1: If $r(X)$ and $q(X)$ are non-decreasing functions of a random variable $X$, then their covariance is non-negative. See link here for this fact:
covariance of increasing functions
Fact 2: If $m(X)$ and $n(X)$ are non-increasing functions of a random variable $X$, then their covariance is also non-negative, since:
$$E[m(X)n(X)] = E[(-m(X))(-n(X))] \geq E[-m(X)]E[-n(X)] = E[m(X)]E[n(X)]$$
where the inequality follows from Fact 1 together with the observation that $-m(X)$ and $-n(X)$ are non-decreasing.
We can apply Fact 2 to your problem: The functions $f(X)^2$ and $g(X)h(X)$ are both non-increasing in $X$, and so by Fact 2:
$$E[f(X)^2]E[g(X)h(X)] \leq E[f(X)^2g(X)h(X)] \leq E[f(X)g(X)]$$
where the final inequality uses the fact that $f(X)h(X)\leq 1$ for all $X$. The final term is less than it would be if we add the positive value $E[f(X)g(X)]E[f(X)h(X)]$.
Your previous question shows the result is not true if we remove the condition $f(X)h(X)\leq 1$. That link is:
inequality involving taking expectation | 458 | 1,479 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2018-26 | latest | en | 0.824214 |
http://stackoverflow.com/questions/6068660/checking-a-line-segment-is-within-a-distance-from-a-point | 1,394,726,711,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678677569/warc/CC-MAIN-20140313024437-00022-ip-10-183-142-35.ec2.internal.warc.gz | 141,536,948 | 15,437 | # Checking a line segment is within a distance from a point
I have two points A and B that define a line segment on a device screen plus another point C. Using efficient and short algorithm that is easy to code (preferably using standard math library), how do I check if the line segment AB is within a distance R from C?
I know there is a simple way to find the shortest distance from a point to a line, but it assume the line is infinitely long. What I have is a line segment with two endpoints.
I considered posting this in Math SE but decided not to since I don't want to get all those long math formula as the answer like in http://math.stackexchange.com/questions/2837/how-to-tell-if-a-line-segment-intersects-with-a-circle . What I need is an efficient and readable computer algorithm, not a formal math theorem.
p/s: I have the following Objective-C method skeleton that needs to be implemented:
``````typedef struct {
CGPoint a;
CGPoint b;
} CGLineSegment;
}
``````
EDIT WITH SOLUTION:
thanks to answer from veredesmarald (which I already accepted) I've implemented the method, put here as reference for other people:
``````+ (BOOL)isLineSegment:(CGLineSegment)line withinRadius:(CGFloat)radius fromPoint:(CGPoint)point {
CGPoint v = CGPointMake(line.b.x - line.a.x, line.b.y - line.a.y);
CGPoint w = CGPointMake(point.x - line.a.x, point.y - line.a.y);
CGFloat c1 = dotProduct(w, v);
CGFloat c2 = dotProduct(v, v);
CGFloat d;
if (c1 <= 0) {
d = distance(point, line.a);
}
else if (c2 <= c1) {
d = distance(point, line.b);
}
else {
CGFloat b = c1 / c2;
CGPoint Pb = CGPointMake(line.a.x + b * v.x, line.a.y + b * v.y);
d = distance(point, Pb);
}
}
CGFloat distance(const CGPoint p1, const CGPoint p2) {
return sqrt(pow(p2.x - p1.x, 2) + pow(p2.y - p1.y, 2));
}
CGFloat dotProduct(const CGPoint p1, const CGPoint p2) {
return p1.x * p2.x + p1.y * p2.y;
}
``````
-
Before being able to translate it to code, you must have (and preferably understand!) the math involved. – Bart Kiers May 20 '11 at 7:17
@Bart Kiers yup that's true but so I'm looking for the math, not formula, as I've emphasized in my question. I have no problem with simple to intermediate geometry math. – Lukman May 20 '11 at 7:20
When I had to implement a method to determine the distance from a point to an interval for a graphics assignment, I found this page very informative: About Lines and Distance of a Point to a Line
In particular, the section Distance of a Point to a Ray or Segment should be of interest to you.
Pseudocode from the article (where `·` is dot product and `d()` is distance between two points):
``````distance( Point P, Segment P0:P1 )
{
v = P1 - P0
w = P - P0
if ( (c1 = w·v) <= 0 )
return d(P, P0)
if ( (c2 = v·v) <= c1 )
return d(P, P1)
b = c1 / c2
Pb = P0 + bv
return d(P, Pb)
}
``````
This method relies on the dot product to determine if the base of the perpendicular is within the interval, and if not which end point is closer.
-
Funny how this implementation of the algorithm described in the other answer doesn't look like it was just that ;-) +1 because it's correct. – danyowdee May 20 '11 at 7:52
it works! thanks! i've added my objective-c implementation in the question as reference for others :) – Lukman May 20 '11 at 14:41 | 882 | 3,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2014-10 | latest | en | 0.834038 |
http://scienceblogs.com/principles/2009/01/28/grading-methods-dont-matter/ | 1,508,650,512,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825141.95/warc/CC-MAIN-20171022041437-20171022061437-00635.warc.gz | 298,235,154 | 31,419 | Over at Dot Physics, Rhett is pondering grading curves:
Should you grade on a curve or not? If you are student, the answer is clear: go by whatever the instructor does. Otherwise, you have a choice. I don’t like to tell other instructors or faculty what to do because I respect their freedom. For my classes, there is no curve. Why? Well, the question really is: “why grade on a curve?” I don’t know the exact reason for particular instructors, but I can come up with some possible reasons.
My first few years teaching, I worried about this quite a bit. I talked to different faculty in the department about what they did, and to a few people in other departments. In my third year or so, though, I realized that it just didn’t matter.
The catalyst for this realization was a student who wasn’t happy with his grade. He’d been a decent student in the class, and I thought the grade I gave him was perfectly respectable, but he was hoping to go to med school, and thought he should’ve been a step higher.
I was bothered by this, because he had been a good student, so I went back and recalculated his grade using every one of the methods other people had told me about, from explicit numerical “curves” to setting the mean for the course at a target letter grade, and going one letter up or down for each standard deviation away from that mean. And every one of them came back the same way.
Since then, I’ve been much more casual about the way I assign letter grades. The relative weights of the different assignments are clearly set out in the syllabus, and I try to make the grading standards as clear as I can, but when the time comes to convert the numbers in the spreadsheet into letters for the Registrar, I just go with numerical values: 95% is an A, 85% is a B, 75% is a C, with pluses and minuses spaced roughly equally between those. I may shift the line between B+ and A- (or any other adjacent pair of grades) if I think a particular student deserves a higher grade than strict numerical cut-offs would indicate, but that’s a tiny effect.
And as long as I’ve been doing my job well during the term, the grades fall out more or less where you would expect. There is always a student or two whose score is around 95%, to get an A, and the weakest students come in around the C/D boundary. I don’t think I’ve had anybody finish the course and earn an outright F– I’ve occasionally had to fail students because they failed to turn in labs (we have a policy that students must have a passing lab grade to pass the course, and that all lab reports must be handed in to get a passing lab grade), and I’ve had students who were headed for an F drop the class, but I don’t think anyone has been so dismal they didn’t deserve even a D.
I think that’s really the key. If I’m doing my job, and properly matching the content of the tests to the content of the class, everything works out fine without needing to curve anything.
When students ask, I do tend to say “The person with the highest total grade in the class will get an A,” which works remarkably well to reassure students who are hoping for a curve. I don’t tell them that that’s because there’s almost always somebody in the class who gets A grades on all the individual assignments, and not because I curve the grades to get there.
1. #1 Dr. Kate
January 28, 2009
The only rational explanation for curving that I’ve ever heard came from a professor at the university where my best friend went to school. He summed it up this way: “If I [the professor] give a test, and no one in the class gets more than, say, 80% of it correct, then clearly either my test was too hard or I didn’t do a good enough job of teaching the material. Either way, it’s not my students’ fault, so I will curve the grades accordingly.”
Presumably, this professor uses a similar reasoning for assigning overall course grades.
Of course, I’ve also had professors who deliberately write tests that are way too hard (e.g., median score is 60%, and the highest score is 75% or something). I think that’s just mean.
2. #2 apy
January 28, 2009
I recently had a class that was graded on a curve. The mean value for the final was a about a 57 (which I was quite close to) and I still managed on getting an A in the class due to the curve. While I appreciated the help I do have to agree with Dr. Kate’s story. There was a disconnect between what was being taught and what was being tested on and rather than attempt to correct this throughout the semester the teacher decided to teach how they wanted and simply move the curve around. While I enjoyed the class for other reasons, I do think that a teacher should try to be more introspective and look for how they could have taught the material so they would not need to curve, if that is possible. In other words, I don’t think a teacher should go into the class expecting to grade it on a curve.
3. #3 John Novak
January 28, 2009
Dr. Kate, I have had one– exactly one– professor give me a rationale for the impossible test syndrome. His claim was that if he gives a test out where someone gets 100%, he still doesn’t know how good that student is, or how much he knows. All he knows is that the student knows everything on the exam. Hence, he tried to write exams that no one could get a perfect on, in order to properly gauge what the maximum level for the class really was.
To this day, I still have mixed thoughts on that.
On the one hand, yes, that is completely and perfectly rational.
On the other hand, the vast majority of the time, it just strikes me as mean and lazy, as well. I’ve had a lot of professors give the Impossible Exam, and with that one exception there has never been any explanation or apology, and more than once there’s been the attitude of, “What the hell is wrong with you people?”
(I do actually believe that one exception, though, because in all other ways he was a class act. He warned us ahead of time so we wouldn’t completely panic, he curved heavily, etc. Hell, he once e-mailed me after a final exam to tell me he thought one of my essay answers (“How do you solve this open research question for the field?”) might be a fruitful line of research for a PhD. Alas, it was in a field I consider mind-numbingly boring….)
4. #4 Jonathan Vos Post
January 28, 2009
By the principal of Academic Freedom, you have the autonomy to grade as you choose — up to the point where the administration pushes back at you.
“Hey, you gave EVERYONE in your class an A” [as I did, once, and was made to suffer for it], or “You failed too many students; parents are complaining; get your ass in gear for the flood of formal grade appeals coming at you.”
Of course, there are already laws to prevent you from grading in an overtly discriminatory way (say, passing all the girls and flunking all the boys, or passing all the Latinos and flunking all the Asians), or selling grades for money or sexual favors [I love your sexy curves; I grade ON the curve, if you get my drift].
I’ve said this on another thread, but can’t resist:
“Why did the teacher drive his car off the road and into a tree?”
“Because he was grading on the curve.”
5. #5 CCPhysicist
January 28, 2009
Heh. I took a test once (freshman honors physics) where a 47 was the high score. That took “challenging the students” to a new level. I learned later that the prof was teaching for the first time, but it also had attracted students who didn’t understand the concept that a calc III prerequisite meant you needed to still know how to apply calculus.
I do pretty much what you do, Chad, choosing problems so that a passing student should get around 70%, particularly if they get full marks for the homework.
I have heard faculty complain about a curve problem all the way to the comprehensive exam. If you ask really hard problems (see my first example), you don’t learn much about what the students actually know. The final grade consists entirely of partial credit. It is much easier to have confidence when failing someone when you can say “they couldn’t even do X”.
6. #6 Tercel
January 28, 2009
I could go either way on the question of curving grades, as long as there is a reasonable justification given. However, I once had an experience with curved grading as an undergrad, which was horrible.
I was in introductory linear algebra, and the professor broke the grading down so that homework was worth 25% of our grade (I think. It’s been a while). Now, I think I’m prety smart, but I’ve never been a model student, and while I did fine in the class for the most part, I only got about 75% of the homework done. This was online homework (using the WebWork system, if you are familiar with that), which means that a lot of people had 100% of the work done.
I wasn’t too happy about missing so much work, but 75% of the homework was still a significant amount of effort that I assumed I’d get credit for, and based on my other grades I was expecting a B or B+. However, the professor applied a curve to this data, without any real though about why, or what it would actually mean to curve this data. Because so much of the class got 100% of the homework grade, my 75% curved to 0.
That’s right, I got 0% credit for the homework. Not because I didn’t do any work, but because everybody else did more.
Of course there are mathematical reasons why this makes no sense, and they are all the more surprising because this guy was a mathematician. but the real problem was that he applied a mathematical tool without considering whether it was useful or correct, and whether the outcome made any sense. I acknowledge that I should have done more homework, but is 75% of the work really the same as none at all?
7. #7 Clark
January 28, 2009
To scientists who use measurement devices all the time, it ought to be pretty obvious how difficult tests should be. The object of a test is to measure a students knowledge of the material. If you were buying a pressure gauge and had an infinite supply of pressure ranges to choose from, you would never pick one where you knew you would only operate in the upper quarter of the tools range. If you had a gauge like that, you’d find yourself musing that you would prefer one that measured only the range you worked in, but had better resolution.
Tests ought to be the same way. Ideally, the very best students should score quite high on the test, but without getting a perfect score, while the worst students ought to do quite poorly. Unfortunately, we’ve all been bred to think that anything below 80% on a test must mean you are significantly sub-standard.
More on Chad’s topic, I worked for a professor who pointed out that students worry a lot about the distribution of value between tests, assignments and quizzes. He looked at all his students scores and found that the A students nearly always get As on their homework, As on their tests and As on their quizzes. Likewise the C students did roughly equally poorly on all three. It didn’t really matter how he weighted the different categories, except for a very few students who might make minor leaps from a B to a B+.
8. #8 Brian
January 28, 2009
I taught High School science several years ago, and had to take a formal testing methodology class to get my certificate. It was a useful enough class that I would recommend at least sitting through one if you are near an education program. The first thing you have to try to figure out is “what am I testing for.” If you want to see how good the best students are, then you make something noone will get 100%. If you do this you *have* to know that you loose distinction on the low end of the scale.
“If I’m doing my job, and properly matching the content of the tests to the content of the class”
If only every instructor would do this, there would be fewer problems in the grades.
9. #9 anj
January 28, 2009
All of my physics classes had impossible exams.
Usually, the passing grade was around 15 points – out of a hundred possible. The best students had points in the 20s and I made it two or three points above failing, which depressed me greatly, but still wasn’t so bad, considering that the results were curved, so that 50% of the students failed.
It was a very frustrating experience. It also didn’t help my self-esteem at all..
10. #10 Coriolis
January 28, 2009
This has been something that’s been bothering me for a while – why is having the top score in the class be 80-90% or so “mean” or “lazy” or bad at all? So long as you set the letter grades appropriately (i.e. 100-80 A, 60-80 B, etc.), what difference does it make?
Especially in physics, there is practically no limit on how hard an exam can be or how many topics can be covered – i.e. it’s not like you’re ever going to learn everything there is to know about statistical mechanics. So why artificially limit the difficulty/amount of material to the point where a significant amount of people would expect to get over 95%?
The other side of this problem is when you make the material so difficult that nobody can get more then let’s say 50% on an exam – and I think almost anyone would recognize that as a problem.
Like Clark points out above, you would always try to adjust the range on your measurement device to the variation of what you’re measuring.
11. #11 Dave Munger
January 28, 2009
There are a couple reasons not to adopt Clark’s method, despite the fact that it is logically true.
First, it can be incredibly frustrating. If the average student gets half the problems incorrect, it’s just demoralizing. It doesn’t correspond well to “average” performance in real life.
Second, it doesn’t sufficiently punish students for slacking off. If 50 is a passing grade, then I can skip half the assignments and still pass.
12. #12 Lora
January 28, 2009
In classes that actually teach how to teach, it turns out that the curve and how easy/difficult the exam is, like you say, not very important. What IS important in how you grade papers and things is that you blind them somehow; ideally, you’d double-blind them and only do the un-blinding at the end, performed by a TA who wasn’t involved in the class.
I used to have the students put only student IDs on the papers, wrote out a specific grading rubric with the exam, and stuck to the rubric ruthlessly. Didn’t un-blind the results till final grades needed to be handed in. My boss used to argue and whine endlessly about how This One was a Good Student who could not possibly have earned a B- and That One was lazy and didn’t talk enough and could not possibly have earned an A. And if you tried to suggest, gently, softly, that perhaps he had misunderstood a shy student or been wowed by a chatty one, oh, hellfire and brimstone, you were Questioning His Scientific Objectivity, and woe be unto you.
Best professor I had in grad school was similarly ruthless about grading, kept everyone on tenterhooks until finals, and required that all homework and papers be vetted by Turnitin.com. She gave out lots of Fs for cheating, and you would not believe the amount of students in the sub-Ivy private schools who cheat and have the nerve to complain when they get caught.
13. #13 John Novak
January 28, 2009
Another reason I consider the Impossible Test to be bad practice is this:
The kinds of situations where it is appropriate to ask for solutions to open research questions (or other similar Impossible Test questions) on a two hour exam are often not the sorts of situations where two hour exams are really really relevant any more.
You do that sort of thing (fairly) in mid to upper level graduate courses, where your students will at least recognize the nature of the question and have a sufficient breadth of knowledge to sketch out a solution, and note the good things and bad things about it. You don’t do that, in my opinion, to undergrads because it is pointless.
But, by the same token, why the hell are we giving out standard two hour exams to mid or upper level graduate students? Give them a paper, or a project, or something meaningful to sink their teeth into.
14. #14 Jonathan Vos Post
January 28, 2009
Teachers are not babysitters. That is at the foundation of any good methodology of grading, with or without The Curve.
I Am Not A Babysitter
As a teacher, I face many stereotypes about my job. But I wouldn’t trade my career for any other.
By Heather Robinson | NEWSWEEK
Published Oct 11, 2008
From the magazine issue dated Oct 20, 2008
“It is said that teaching is the profession that creates all other professions. That’s a beautiful compliment for a job that often does not receive the respect one would predict given all of the platitudes bestowed upon teachers. ‘God bless you!’ ‘What a noble profession’ and ‘I couldn’t do it, but thank goodness there are people like you out there’ are a few that I’ve received…. What many people don’t understand is that we teachers are working with wiggling, chatting children with varied needs. They are not robots who perform exactly as we direct without exception. Teachers are not on autopilot—we make thousands of decisions each day while working hard to produce a quality product that provides each student with what she needs and deserves. Teaching isn’t simply perching at a lectern and pontificating to hungry minds; it’s being an educator, a mentor, a parent, a nurse, a social worker, a friend, a diplomat and an expert on the curriculum. In short, we are professionals….”
15. #15 Laura
January 28, 2009
When I was in high school, I took AP Chemistry. Because the class was designed to prepare us for taking the AP exam in the spring, most of our in-class test questions came from former AP exams. To get a 5 on the AP exam, you only need to get about 65% of the questions right. (Why this is, I do not know…) Therefore, my teacher concluded that to get an A on a test derived entirely from AP exam question, you should only need to get 65% of the possible points. I’m not sure how exactly her grading rubric mimicked the one used by AP graders, but I assume they were similar. She also had a policy that scoring 80% or better on a test was an A+, which would be factored in as she calculated final grades. It all made perfect sense to me.
Now that I’m a graduate student, my professors have different policies for grading, but most of them use some sort of a curve. For my biochemistry class, the exam average was set at a B+, and your letter grade was determined by how many standard deviations above or below the mean you fell. The average hovered around 75%-80% in most cases, but there were always one or two people in the 90%+ range. This is probably because people of vastly different skill levels were all taking the same class — from cell biology PhD students who had majored in biochemistry to neuroscience PhD students who had majored in psychology. I think the grades worked out to be quite fair in the end, although after the first exam people panicked until they saw the grade distribution.
16. #16 Ewan
January 28, 2009
My syllabi say (something very akin to) “if the average grade for the class is lower than C+, I will curve to raise the average to a B-. Curving will never lower a grade.” Haven’t yet had to curve, although I can see that I _will_ have to at some point, or alter test difficulty downwards (which I’m reluctant to do; I _have_ found that student effort increases if they see tests as very hard *but* fair). I’ve had folks get 100%; rare, but has happened. Average tends to be around C+ and end-grade average right around B- after extra credit bits get added in, which is what I aim for.
Sorry. Ranty. Came close to souring me on teaching completely, though :(.
17. #17 Jonathan Vos Post
January 28, 2009
Student perception, that an exam is fair, has been found in pedagogical research to be very important.
I quote myself in a review that I wrote for one of my School of Education graduate courses.
============================================
For: Prof. Nick Doom, EDSE 401
Charter College of Education, Cal State L.A.
From: Prof. Jonathan Vos Post
Due: Monday 19 May 2008
Done: and emailed Sunday 18 May 2008
============================================
“Find a professional journal (not Time or Newsweek) on Education (Education Week, for example). Pick any article on any district, state, USA; on critique or reform. Give title of journal, article title, authors, publication date; summarize in 5 sentences, and critique.”
============================================
Article #1 for Summary & Critique
http://arxiv.org/pdf/0803.4235
Title: Exam fairness
Authors: Mathieu Bouville
Subjects: General Physics (physics.gen-ph); Physics Education
(physics.ed-ph)
29 March 2008
Summary in 5 sentences:
(1) It is widely agreed that exams must be fair, yet what this exactly means is not made clear.
(2) One may mean fairness of treatment, yet this merely propagates the fairness or unfairness of pre-existing rules.
(3) Fairness of opportunity, on the other hand, necessarily leads to identical grades for everyone, which clearly makes it inapplicable.
(4) Neither view is helpful to make decisions on competing claims: fairness of treatment ignores the problem and fairness of opportunity holds all claims to be equally valid.
(5) To escape this deadlock one needs an external criterion such as how good engineers students will be, to replace the fairness viewed as student-student comparison.
Critique:
(a) This paper does not definitively define either the problem nor the solution but, rather, gives a useful survey of the opinions expressed in the 18 papers referenced and cited.
(b) Some quotations are vivid, such as:
“What students hate more than anything else are examinations that they perceive as unfair.”
[R. M. Felder, “Designing Tests to Maximize Understanding”, J. Prof. Iss. Eng. Ed. Pr., 128, pp.1-3, 2002]
“For the student, the most important question to be answered in order to be content with the outcomes of the assessment is ‘what is a fair assessment.'”
[H. Vos, ‘How to assess for improvement in learning”, Eur. J. Eng. Educ., 25, pp. 227-233, 2005]
(c) The argument make plausible — without being compelling — that “… fairness is a necessary consequence of the validity of the exams, rather than a separate criterion.”
============================================
18. #18 Daryl McCullough
January 28, 2009
I think it depends on what you believe the purpose of grades are. If they are only for class ranking, then it doesn’t matter whether you grade on a curve or not. If they are not for class ranking, what are they for?
To me, what people need to know about a student’s performance in a class is this: what topics has the student mastered? What topics has he/she not yet mastered? And possibly, What topics is the student not likely to master, even if he works on them for the rest of his life?
Grades are nearly meaningless, except for the purpose of comparing students to each other, so that we can talk about the best students, the worst students and the mediocre students.
19. #19 Janne
January 28, 2009
Doesn’t matter what system you use – point score on a curve, simple pass-fail cutoff, whatever – as long as the system is clearly described from the outset. A Lot of the perception of unfairness comes from people more or less unconsciously are studying for a different kind of grading than they will actually undergo.
Me, I like to argue for pass/fail, at least at higher levels. As an undergraduate we had “fail”, “pass”, and “pass with excellence”, with pass at 50% and excellence at 75%. Of course, the higher grade normally didn’t mean anything as a pass is a pass, and your final degree is not scored at all. The higher grade could at times make a difference for a specific high-level course or two if you were applying for a PhD-student position involving that area – but then, if you aren’t good enough to earn that higher grade you’d not have any interest in graduate school anyhow.
This simple cutoff works just fine.
20. #20 prk
January 29, 2009
“But, by the same token, why the hell are we giving out standard two hour exams to mid or upper level graduate students? Give them a paper, or a project, or something meaningful to sink their teeth into.” – John Novak.
I asked that question of a few lecturers whilst doing a Masters by coursework, and the response was either “University regulations require it” or “It’s too difficult or unfair to assess an individual’s performance on a project which could have had collaboration”.
I suspect the latter is the reason for the former.
Overall it’s pretty disappointing, and when the only real way of trying to rule out collaboration is quizzing the student on their project / paper, you may as well have given them an exam.
prk.
21. #21 Harry Abernathy
January 29, 2009
“But, by the same token, why the hell are we giving out standard two hour exams to mid or upper level graduate students? Give them a paper, or a project, or something meaningful to sink their teeth into.” – John Novak.
I don’t believe that exams and papers/projects need to be exclusive. In graduate courses especially, there should always be assignments that require loads of research and analysis time. That’s how one develops a true appreciation for the amount (and lack) of knowledge in an area.
But there should also be some form of timed examination because in real life you are sometimes required to deliver information and make calculations on the spot or at least under very short time constraints. Yes, you can study and prepare for presentations and consultations, but when the time comes you’re going to often be asked questions on the spot. And your client or your audience is going to evaluate you on some level your competence on your ability to do so.
What should be considered is the relative weight of exams versus homework versus projects. At least in grad school, exams should not make up the bulk of your grade. Unfortunately, usually the opposite is true.
22. #22 ScentOfViolets
January 29, 2009
Doesn’t all of this depend on both the type of class and the type of test? If I teach intermediate algebra, aka, ‘things you should have picked up in high school, had you been paying attention’, I curve very heavily. This is due not only to the quality of the student, usually a first-semester freshman, but because the test is in a multiple-choice scantron format. A student gets either 0 or 5 points, all or nothing. So I don’t have a way to distinguish the 1/5 point kids from the 4/5 point kids. Conversely, if I teach a calc class, or a diffeq class, things are a bit different, both in the quality of the student and the ability to assign partial credit.
Iow, this is a roundabout way of saying that I have to grade not only on what students get right, but what they get wrong, and why. A good test should not only measure performance, but should be some sort of diagnostic. And this in turn should be reflected in the grading at the end of the semester.
23. #23 Daryl McCullough
January 29, 2009
ScentOfViolets writes:
A good test should not only measure performance, but should be some sort of diagnostic.
But once you summarize the test scores/homework scores/lab scores/scores on research papers, etc. as a grade, the diagnostic aspect is thrown out. There is just not enough information in a letter grade (or numeric grade, for that matter). If the course covers, say, 10 topics, and a student masters all but one of those topics, he may very well get an A for the course. But if future courses depend on the student knowing that one topic, then she may be in trouble. On the other side of the scale, if a student masters only 3 of the 10 topics, then he will fail the course, meaning that he’s no better off than if he mastered none of them.
As I said, what I would like to see implemented in schools is that every student has an associated list of microtopics that she has completely mastered (in the sense that I have completely mastered the multiplication of numbers using decimal notation; it’s not just that I get it right 95% of the time, I essentially never get it wrong, except for typos). There may be ambiguity about whether a student understands a large topic well enough to get an A or a B, but it seems that for a small topic, it is unambiguous to demonstrate that a student has mastered the topic.
24. #24 CCPhysicist
January 31, 2009
Good point at #12. My exams have a cover sheet, and the first thing I do is turn them all to some particular problem, shuffling the “deck” as I go. Each problem is graded as a group, with the same treatment for all.
I usually explain this the day after I give the first exam, pointing out that there is no reason to ask “have you graded MY exam yet” because (a) I have no way of knowing and (b) I won’t be finished with the last problem on their exam until, at the earliest, I have graded last-minus-one on all of the other exams.
As for the measurement range, I think my exams are pretty reliable between 50 and 100. I’d rather have an exam operate up there than between 0 and 50 (out of 100). The reason is simple: If my measurement uncertainty is the same on both tests, my relative error decreases if I don’t have to curve. As I said above, I’d also rather give a passing grade to a student who got some problems completely correct rather than one who never got more than half credit on any exam problem.
25. #25 CCPhysicist
January 31, 2009
I forgot to mention one thing: It bothers me a lot that my tests don’t challenge the best students. I have thought many times about putting a 25 minute problem worth 5 points on each test, but too many of the weaker ones ignore a giant bold faced statement like “don’t do this until you have 100% of the other problems correct”.
Does anyone have any experience with giving a 95 point exam and then handing out a 5 point “A student” problem when they turn it in before the end of the test? | 6,776 | 29,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2017-43 | latest | en | 0.976183 |
https://cs.stackexchange.com/questions/2079/determine-missing-number-in-data-stream | 1,653,117,644,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662538646.33/warc/CC-MAIN-20220521045616-20220521075616-00240.warc.gz | 239,361,952 | 69,814 | # Determine missing number in data stream
We receive a stream of $n-1$ pairwise different numbers from the set $\left\{1,\dots,n\right\}$.
How can I determine the missing number with an algorithm that reads the stream once and uses a memory of only $O(\log_2 n)$ bits?
You know $\sum_{i=1}^n i = \frac{n(n+1)}{2}$, and because $S = \frac{n(n+1)}{2}$ could be coded in $O(\log(n))$ bits this can be done in $O(\log n)$ memory and in one path (just find $S - \mathrm{currentSum}$, this is missing number).
But this problem could be solved in general case (for constant $k$): we have $k$ missing numbers, find out all of them. In this case instead of calculating just sum of $y_i$, calculate sum of j'st power of $x_i$ for all $1\le j \le k$ (I assumed $x_i$ is missing numbers and $y_i$ is input numbers):
\qquad \displaystyle \begin{align} \sum_{i=1}^k x_i &= S_1,\\ \sum_{i=1}^k x_i^2 &= S_2,\\ &\vdots \\ \sum_{i=1}^k x_i^k &= S_k \end{align} $\qquad (1)$
Remember that you can calculate $S_1,...S_k$ simply, because $S_1 = S - \sum y_i$, $S_2 = \sum i^2 - \sum y_i^2$, ...
Now for finding missing numbers you should solve $(1)$ to find all $x_i$.
You can compute:
$P_1 = \sum x_i$, $P_2 = \sum x_i\cdot x_j$, ... , $P_k = \prod x_i$ $(2)$.
For this remember that $P_1 = S_1$, $P_2 = \frac{S_1^2 - S_2}{2}$, ...
But $P_i$ is coefficients of $P=(x-x_1)\cdot (x-x_2) \cdots (x-x_k)$ but $P$ could be factored uniquely, so you can find missing numbers.
These are not my thoughts; read this.
• I don't get (2). Maybe if you added in the sums' details? Does $P_k$ miss a $\sum$?
– Raphael
May 26, 2012 at 12:43
• @Raphael, $P_i$ is Newton's identities, I think if you take a look at my referenced wiki page you can get the idea of calculation, each $P_i$ could be calculated by previous $P$s, $S_j$, remember simple formula: $2 \cdot x_1 \cdot x_2 = (x_1 + x_2)^2 - (x_1^2 + x_2^2)$, you can apply similar approach to all powers. Also as I wrote $P_i$ is sigma of something, but $P_k$ doesn't have any $\Sigma$, because there is just one $\Pi$.
– user742
May 26, 2012 at 13:24
• Be that as it may, answers should be self-contained to a reasonable degree. You give some formulae, so why not make them complete?
– Raphael
May 26, 2012 at 14:35
From the comment above:
Before processing the stream, allocate $\lceil \log_2 n \rceil$ bits, in which you write $x:= \bigoplus_{i=1}^n \mathrm{bin}(i)$ ($\mathrm{bin}(i)$ is the binary representation of $i$ and $\oplus$ is pointwise exclusive-or). Naively, this takes $\mathcal{O}(n)$ time.
Upon processing the stream, whenever one reads a number $j$, compute $x := x \oplus \mathrm{bin}(j)$. Let $k$ be the single number from $\{ 1, ... n\}$ that is not included in the stream. After having read the whole stream, we have $$x = \left(\bigoplus_{i=1}^n \mathrm{bin}(i)\right) \oplus \left(\bigoplus_{i \neq k } \mathrm{bin}(i)\right) = \mathrm{bin}(k) \oplus \bigoplus_{i \neq k } (\mathrm{bin}(i) \oplus \mathrm{bin}(i)) = \mathrm{bin}(k),$$ yielding the desired result.
Hence, we used $\mathcal{O}(\log n)$ space, and have an overall runtime of $\mathcal{O}(n)$.
• may I suggest an easy optimization that makes this a true streaming single-pass algorithm: at time step $i$, xor $x$ with $\mathrm{bin}(i)$ and with the input $\mathrm{bin}(j)$ that has arrived on the stream. this has the added benefit that you can make it work even if $n$ is not known ahead of time: just start with a single bit allocated for $x$ and "grow" the allocated space as necessary. May 28, 2012 at 4:34
HdM's solution works. I coded it in C++ to test it. I can't limit the value to $O(\log_2 n)$ bits, but I'm sure you can easily show how only that number of bits is actually set.
For those that want pseudo code, using a simple $\text{fold}$ operation with exclusive or ($\oplus$):
$$\text{Missing} = \text{fold}(\oplus, \{1,\ldots,N\} \cup \text{InputStream})$$
Hand-wavey proof: A $\oplus$ never requires more bits than its input, so it follows that no intermediate result in the above requires more than the maximum bits of the input (so $O(\log_2 n)$ bits). $\oplus$ is commutative, and $x \oplus x = 0$, thus if you expand the above and pair off all data present in the stream you'll be left only with a single un-matched value, the missing number.
#include <iostream>
#include <vector>
#include <cstdlib>
#include <algorithm>
using namespace std;
void find_missing( int const * stream, int len );
int main( int argc, char ** argv )
{
if( argc < 2 )
{
cerr << "Syntax: " << argv[0] << " N" << endl;
return 1;
}
int n = atoi( argv[1] );
//construct sequence
vector<int> seq;
for( int i=1; i <= n; ++i )
seq.push_back( i );
//remove a number and remember it
srand( unsigned(time(0)) );
int remove = (rand() % n) + 1;
seq.erase( seq.begin() + (remove - 1) );
cout << "Removed: " << remove << endl;
//give the stream a random order
std::random_shuffle( seq.begin(), seq.end() );
find_missing( &seq[0], int(seq.size()) );
}
//HdM's solution
void find_missing( int const * stream, int len )
{
//create initial value of n sequence xor'ed (n == len+1)
int value = 0;
for( int i=0; i < (len+1); ++i )
value = value ^ (i+1);
//xor all items in stream
for( int i=0; i < len; ++i, ++stream )
value = value ^ *stream;
//what's left is the missing number
cout << "Found: " << value << endl;
}
• Please post readable (pseudo) code of only the algorithm instead (skip main). Also, a correctness proof/argument at some level should be included.
– Raphael
May 26, 2012 at 9:26
• @edA-qamort-ora-y Your answer assumes that the reader knows C++. To someone who is not familiar with this language, there is nothing to see: both finding the relevant passage and understanding what it's doing are a challenge. Readable pseudocode would make this a better answer. The C++ is not really useful on a computer science site. May 26, 2012 at 10:22
• If my answer proves not to be useful people don't need to vote for it. May 26, 2012 at 10:37
• +1 for actually taking the time to write C++ code and test it out. Unfortunately as others pointed out, it's not SO. Still you put effort into this ! May 26, 2012 at 11:59
• I don't get the point of this answer: you take someone else's solution, which is very simple and obviously very efficient, and "test" it. Why is testing necessary? This is like testing your computer adds numbers correctly. And there is nothing nontrivial abt your code either. May 27, 2012 at 13:30 | 1,992 | 6,446 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-21 | latest | en | 0.8433 |
https://everything2.com/title/beat+frequency | 1,544,458,589,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823348.23/warc/CC-MAIN-20181210144632-20181210170132-00427.warc.gz | 590,154,190 | 7,236 | A phenomenon that can be heard with two constant tones which are close to, but don't exactly match each other. The individual tones can be heard, and a third frequency can be heard between the two. It's easiest to hear when the "real" frequencies are off by one or two Hz; in this case, the third frequency (the beat frequency) can be heard as a slow wavering in the intensity of the waves. If you get two very different tones loud enough, the "wavering" creates a third, audible note.
Bear with me, this is a lot easier to explain with a picture...The two frequencies start out in phase, reinforcing each other. As one frequency moves out of phase with the other, they start to cancel each other out until they are completely out of phase, causing the total amplitude to be zero. Your ear perceives this as a combination of the tones which goes from loud to soft to loud at the frequency of the difference of the "real" frequencies. It can be expressed as:
fbeat = fa - fb
A way to hear this clearly for yourself is to get two tone generators, set one to 250 Hz and the other to 252 Hz. You will hear two tones going from loud to soft twice every second. Turn the second generator down to 251 Hz, and the beat frequency will go to once a second. Yay.
How does the beat frequency (fa-fb)/2 come into being? That most hated part of the high school Mathematics curriculum, trigonometry, holds the keys!
Suppose we have frequencies f+w and f-w, and we sound them together. If there's no phase difference between them (and if there is any, the same thing holds and we still get the same result), then we have the waveforms sin(2π(f+w)t) and sin(2π(f-w)t) sounding together. Retrieving some trigonometric identities from distant memory, we have that
sin(a+b)+sin(a-b) = sin(a)cos(b)+cos(a)sin(b) + sin(a)cos(b)-cos(a)sin(b) =
= 2sin(a)cos(b)
(that wasn't so bad, was it?).
So sin(2π(f+w)t)+sin(2π(f-w)t) = 2sin(2πft)cos(2πwt). In flamingweasel's example, f=251Hz and w=1Hz. We're going to be hearing a pure sine wave at frequency 251Hz (that's 2sin(2πft)), only it will be modulated (in exactly the same way that an AM signal is modulated) by a frequency of 1Hz. In other words, over the first half-second the 251Hz wave will have its amplitude squished down to 0, over the next half-second it will go back to full strength, and it will continue to cycle like that.
Log in or register to write something here or to contact authors. | 619 | 2,434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-51 | longest | en | 0.946325 |
https://www.physicsforums.com/threads/help-me-spin-operators.5751/ | 1,670,429,988,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711200.6/warc/CC-MAIN-20221207153419-20221207183419-00592.warc.gz | 994,834,640 | 17,717 | # Help me (Spin Operators)
kakarukeys
An easy (not easy) 1st year undergraduate project:
Energy eigenvalues of a single spin-1/2 system which Hamiltonian is given by
H = - k S^z
are -/+ (1/2 k).
I got that using the spin operator S^z = 1/2 Sigma^z (let h_bar = 1), Hamiltonian is in diagonalized form:
( -1/2 k 0
0 1/2k )
So, Eigenvectors are given by (1, 0), (0, 1), spin up and spin down respectively.
How do I compute the energy eigenvalues, eigenvectors of an double identical spin-1/2 system which Hamiltonian is given by
H = - J vec(S1) dot vec(S2) - k S1^z - k S2^z
J is a const. > 0
After going some references, Griffiths, Sakurai, Merzbacher, etc,
I have no idea how to begin,
I have the following problems in mind:
(1) if |u u> represents both spins up, (1/sqrt 2)(|u d> +/- |d u>) represent one spin up, |d d> represent both spins down. |u u> and |d d> should be eigenstates of the system (because they are ground states), for |u d>, |d u> states I am not sure.
(2) I can't use same spin matrices for S1^z, S2^z. But there is only one S^z matrix namely, 1/2 Sigma^z. Could S1^z be a tensor product of 1/2 Sigma^z with an identity matrix, and S2^2 the other way round? If it is so, I have no idea how to do the maths!
(3) Do vec(S1) and vec(S2) commute?
Anyone could give me some hints?
Sonty
Each operator acts on the part of the state vector that concerns it. If an operator belongs to the first particle it will act on the part of the state vector that belongs to the first particle. So S1 and S2 must commute.
Tensor product:
σz = σ1z * I2 + I1 * σ2z
Meaning:
Code:
/ 1 0 0 0 \ / 1 0 0 0 \
| 0 1 0 0 | | 0 -1 0 0 |
| 0 0 -1 0 | + | 0 0 1 0 |
\ 0 0 0 -1 / \ 0 0 0 -1/
You should figure the math out of this.
There are quite a few holes in my QM as I've just finished reading Sakurai, but did it on my own and I haven't got a lot of exercise to nail down the theory. Maybe that's why that letter J and the S1S2 would take me towards Perturbation Theory if the product is small and the variational method if it's not. That's where you'll get the energy shifts from the base E0=+/-k
Staff Emeritus
Gold Member
Originally posted by kakarukeys
An easy (not easy) 1st year undergraduate project:
Energy eigenvalues of a single spin-1/2 system which Hamiltonian is given by
H = - k S^z
are -/+ (1/2 k).
I got that using the spin operator S^z = 1/2 Sigma^z (let h_bar = 1), Hamiltonian is in diagonalized form:
( -1/2 k 0
0 1/2k )
So, Eigenvectors are given by (1, 0), (0, 1), spin up and spin down respectively.
Yes, that's right.
How do I compute the energy eigenvalues, eigenvectors of an double identical spin-1/2 system which Hamiltonian is given by
H = - J vec(S1) dot vec(S2) - k S1^z - k S2^z
J is a const. > 0
After going some references, Griffiths, Sakurai, Merzbacher, etc,
I have no idea how to begin,
You know the matrices σi (i=1,2,3), so you can determine the eigenvalues from those. (I am going to use subscripts to denote the components of σ and superscripts in parentheses to denote the spin operators for particles 1 and 2). Let |α,β> represent the spin state, where α is the state of particle 1 and β is the state of particle 2. |α,β> is the direct product of |α> and |beta;>, the individual spin states of particles 1 and 2, respectively. Just remember that σ(1) only acts on particle 1 and σ(2) only acts on particle 2. The two operators pass right through the other particle's spin states.
I have the following problems in mind:
(1) if |u u> represents both spins up, (1/sqrt 2)(|u d> +/- |d u>) represent one spin up, |d d> represent both spins down. |u u> and |d d> should be eigenstates of the system (because they are ground states), for |u d>, |d u> states I am not sure.
Yes, that's right. The only thing I would add is that (1/2)-1/2(|u,d>+|d,u>) is part of the triplet S=1, MS=0, and (1/2)-1/2(|u,d>-|d,u>) is the singlet S=0, MS=0.
(2) I can't use same spin matrices for S1^z, S2^z. But there is only one S^z matrix namely, 1/2 Sigma^z. Could S1^z be a tensor product of 1/2 Sigma^z with an identity matrix, and S2^2 the other way round? If it is so, I have no idea how to do the maths!
Yes, you can use the same matrix for S(1)z and S(2)z. In fact, you have no choice because that is what the matrices are! Just remember that the matrix for particle 1 only acts on the state for particle 1, and likewise for particle 2.
(3) Do vec(S1) and vec(S2) commute?
Yes. Their Hilbert spaces are disjoint.
Anyone could give me some hints?
Let me know what else you need.
edit: fixed bold font bracket
Staff Emeritus
Gold Member
Originally posted by Sonty
Tensor product:
σz = σ1z * I2 + I1 * σ2z
Meaning:
Code:
/ 1 0 0 0 \ / 1 0 0 0 \
| 0 1 0 0 | | 0 -1 0 0 |
| 0 0 -1 0 | + | 0 0 1 0 |
\ 0 0 0 -1 / \ 0 0 0 -1/
You've made a mistake here. The question is asking about the Pauli matrix (σz), and you have posted the Dirac matrix (γ3).
Sonty
Originally posted by Tom
You've made a mistake here. The question is asking about the Pauli matrix (σz), and you have posted the Dirac matrix (γ3).
It might be one of the Dirac matrixes, I haven't got into relativistic QM. I'm sure though that that is exactly what I said it is: the σz for the two spin 1/2 particle system. The matrix must be part of a 2x2 tensor space. If you want to prove me wrong, please compute that tensor product and show me your results. I don't understand where you see that mistake.
kakarukeys
Thanks for everyone's help, I really appreciate.
I figured out a way, using a "trick":
Since vec(S1), vec(S2) commute,
2 vec(S1) dot vec(S2) = [vec(S1) + vec(S2) ]^2
- vec(S1)^2 - vec(S2)^2
now ,
vec(S1)^2 = s1 (s1 + 1)
vec(S2)^2 = s2 (s2 + 1)
vec(S1) + vec(S2) = 1 for triplet, 0 for singlet
using the above, it is very easy to compute the the matrix elements of Hamiltonian, provided if
|u u>
|d d>
1/sqrt 2 (|u d> +/- |d u>)
really form a basis.
I think they should, since |u>, |d> span the single particle hilbert space, their direct products should span the double particle Hilbert space which is the direct products of two single particle Hilbert spaces.
So the Hamiltonian looks like
(
x 0 0 0
0 x x 0
0 x x 0
0 0 0 x)
x means non-zero entry.
Solving the eigenvalue equations will give the eigenvalues and correct eigenvectors.
I will figure out another way using matrices, I think that's a better way, since you can no longer use 'trick' when things get complicated.
Staff Emeritus | 2,094 | 6,622 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2022-49 | latest | en | 0.921985 |
https://thestudyofthehousehold.com/qa/question-what-is-p-value-in-t-test.html | 1,628,112,575,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155188.79/warc/CC-MAIN-20210804205700-20210804235700-00256.warc.gz | 563,919,851 | 18,405 | # Question: What Is P Value In T-Test?
## What does P-value of .1 mean?
Popular Answers (1) When the data is perfectly described by the resticted model, the probability to get data that is less well described is 1.
For instance, if the sample means in two groups are identical, the p-values of a t-test is 1..
## How do you manually calculate P value?
Example: Calculating the p-value from a t-test by handStep 1: State the null and alternative hypotheses.Step 2: Find the test statistic.Step 3: Find the p-value for the test statistic. To find the p-value by hand, we need to use the t-Distribution table with n-1 degrees of freedom. … Step 4: Draw a conclusion.Jan 22, 2020
## What does P value tell you?
The p-value, or probability value, tells you how likely it is that your data could have occurred under the null hypothesis. … The p-value is a proportion: if your p-value is 0.05, that means that 5% of the time you would see a test statistic at least as extreme as the one you found if the null hypothesis was true.
## Is P value of 0.05 Significant?
P > 0.05 is the probability that the null hypothesis is true. … A statistically significant test result (P ≤ 0.05) means that the test hypothesis is false or should be rejected. A P value greater than 0.05 means that no effect was observed.
## Is P value always positive?
As we’ve just seen, the p value gives you a way to talk about the probability that the effect has any positive (or negative) value. To recap, if you observe a positive effect, and it’s statistically significant, then the true value of the effect is likely to be positive.
## Why is my p value so low?
A small p-value (typically ≤ 0.05) indicates strong evidence against the null hypothesis, so you reject the null hypothesis. A large p-value (> 0.05) indicates weak evidence against the null hypothesis, so you fail to reject the null hypothesis. … Always report the p-value so your readers can draw their own conclusions.
## Is a high P value good or bad?
If the p-value is less than 0.05, we reject the null hypothesis that there’s no difference between the means and conclude that a significant difference does exist. If the p-value is larger than 0.05, we cannot conclude that a significant difference exists. … Below 0.05, significant. Over 0.05, not significant.
## Why are my p-values so high?
High p-values indicate that your evidence is not strong enough to suggest an effect exists in the population. An effect might exist but it’s possible that the effect size is too small, the sample size is too small, or there is too much variability for the hypothesis test to detect it.
## Why do we use 0.05 level of significance?
The significance level, also denoted as alpha or α, is the probability of rejecting the null hypothesis when it is true. For example, a significance level of 0.05 indicates a 5% risk of concluding that a difference exists when there is no actual difference.
## What does P value of .02 mean?
The significance test yields a p-value that gives the likelihood of the study effect, given that the null hypothesis is true. For example, a p-value of . 02 means that, assuming that the treatment has no effect, and given the sample size, an effect as large as the observed effect would be seen in only 2% of studies.
## What if P value is 0?
Hello, If the statistical software renders a p value of 0.000 it means that the value is very low, with many “0” before any other digit. In SPSS for example, you can double click on it and it will show you the actual value.
## What does P-value of 0.08 mean?
A p-value of 0.08 being more than the benchmark of 0.05 indicates non-significance of the test. This means that the null hypothesis cannot be rejected. Cite.
## What does P value of 0.5 mean?
Mathematical probabilities like p-values range from 0 (no chance) to 1 (absolute certainty). So 0.5 means a 50 per cent chance and 0.05 means a 5 per cent chance. In most sciences, results yielding a p-value of . 05 are considered on the borderline of statistical significance.
## What does P value of 0.9 mean?
If P(real) = 0.9, there is only a 10% chance that the null hypothesis is true at the outset. Consequently, the probability of rejecting a true null at the conclusion of the test must be less than 10%. … It shows that the decrease from the initial probability to the final probability of a true null depends on the P value.
## How is the P value calculated?
Graphically, the p value is the area in the tail of a probability distribution. It’s calculated when you run hypothesis test and is the area to the right of the test statistic (if you’re running a two-tailed test, it’s the area to the left and to the right).
## Can P-values be greater than 1?
A p-value tells you the probability of having a result that is equal to or greater than the result you achieved under your specific hypothesis. It is a probability and, as a probability, it ranges from 0-1.0 and cannot exceed one.
## Is a low P value good?
A low p-value shows that the results are replicable. A low p-value shows that the effect is large or that the result is of major theoretical, clinical or practical importance. A non-significant result, leading us not to reject the null hypothesis, is evidence that the null hypothesis is true.
## Is P-value 0.01 Significant?
For example, a p-value that is more than 0.05 is considered statistically significant while a figure that is less than 0.01 is viewed as highly statistically significant.
## What is p value in Chi Square?
P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a chi-square, use the Chi-Square Distribution Calculator to assess the probability associated with the test statistic. | 1,345 | 5,790 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-31 | latest | en | 0.912728 |
https://math.stackexchange.com/questions/2168276/how-to-numerically-integrate-the-geodesic-equations | 1,708,803,757,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474544.15/warc/CC-MAIN-20240224180245-20240224210245-00318.warc.gz | 408,107,400 | 35,514 | # How to numerically integrate the geodesic equations?
I'm interested in numerically solving the geodesic equations of motion for a particle in the Schwarzschild geometry. Deriving the equations is no problem but I simply have no idea how to initialise and make sense of the output here. We have 4 spacetime coordinates given by $x^\alpha = (t(\lambda),r(\lambda),\theta(\lambda),\phi(\lambda))$, where $\lambda$ is an affine parameter and $t>0, 0<r<\infty, 0<\theta<\pi$ and $0<\varphi<2\pi$. Two of the equations of motion are given by $$\frac{d \varphi}{d\lambda} = \frac{h}{r^2}, \\ \frac{dt}{d\lambda} = \frac{E}{\left(1-\frac{2GM}{c^2r}\right)},$$ where $E,h$ are the energy and the magnitude of the angular momentum. Has anyone references to numerically solving these equations? There is a Mathematica code which is very very cool but I'm more interested to do it myself as a learning exercise rather than just use this not to mention my Mathematica skills are almost as bad as my numerical.
For example, consider the second integral. After a certain number of integration steps the parameter $\lambda$ and $t$ should be different indicating time dilation. Again, I'm not really concerned with the physics of the problem more about the mathematical set-up and mainly some resources.
• If you have little experience in numerical solutions of differential equations, starting with a system of nonlinear ODEs is tough! There are many different methods you could try; perhaps looking though a book on numerical solutions to differential equations first would be good. Once you understand the ideas and how to implement them, you should then be able to apply the techniques to any ODE (or system of ODEs), such as yours. I am certainly not aware of any books that solve the geodesic equation in Schwarzschild - perhaps some journal articles? Mar 2, 2017 at 9:47
• I get your point. I am familiar with the Runge Kutta numerical integrators. I should re word the question to be more clear. It's more the initial set up of the problem and trying to understand the outut. I understand how Runge Kutta family of integrators work from the sort of standard undergrad problems we have all faced. BUT saying that I just can't wrap my head around these geodesic equations. Mar 2, 2017 at 9:52
• Have you solved a system of ODEs before? Starting with something as trivial as $$\frac{df}{dx}=g(x),\;\frac{dg}{dx}= - f(x)$$ might give you some good insight into your problem. As for resources, happy to offer a list of books that I have found useful on numerically solving differential equations, but they aren't specific to the geodesic equation. Mar 2, 2017 at 10:01
• @AloneAndConfused some resources would be great. Do you have any on solving systems of coupled DE's? I think we need to employ implicit methods for coupled DE's. Is that correct? Mar 6, 2017 at 7:06
• Will post as an answer as any reply will be too long to write as a comment! Mar 6, 2017 at 10:57 | 741 | 2,960 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-10 | latest | en | 0.923485 |
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# An Employer Finds that If He Increased the Weekly Wages of Each Worker by Rs 5 and Employs Five Workers Less, He Increases His Weekly Wage Bill from Rs 3,150 to Rs 3,250. Taking th - ICSE Class 10 - Mathematics
#### Question
An employer finds that if he increased the weekly wages of each worker by Rs 5 and employs five workers less, he increases his weekly wage bill from Rs 3,150 to Rs 3,250. Taking the original weekly wage of each worker as Rs x; obtain an equation in x and then solve it to find the weekly wages of each worker.
#### Solution
The original weekly wage of each worker = Rs x
The original weekly wage bill of employer = Rs 3150
Number of workers = 3150/x
New weekly wages of each worker = Rs (x + 5)
The new weekly wage bill of employer = Rs 3250
Number of workers = 3250/(x + 5)
From the given condition
3150/x - 5 = 3250/(x + 5)
(3150 - 5x)/x = 3250/(x + 5)
3150x - 5x^2 + 15750 - 25x = 3250x
-5x^2 + 15750 - 125x = 0
x^2 + 70x - 45x - 3150 = 0
x(x + 70) - 45(x + 70) = 0
(x + 70)(x - 45) = 0
x = -70, 45
Since wage cannot be negative x = 45
Thus the original weekly wage of each worker is Rs 45
Is there an error in this question or solution?
#### APPEARS IN
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Chapter 6: Solving (simple) Problems (Based on Quadratic Equations)
Ex.6D | Q: 10
#### Video TutorialsVIEW ALL [5]
Solution An Employer Finds that If He Increased the Weekly Wages of Each Worker by Rs 5 and Employs Five Workers Less, He Increases His Weekly Wage Bill from Rs 3,150 to Rs 3,250. Taking th Concept: Quadratic Equations.
S | 508 | 1,647 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2019-43 | latest | en | 0.893596 |
https://codeforces.com/blog/entry/73687 | 1,674,829,064,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494986.94/warc/CC-MAIN-20230127132641-20230127162641-00598.warc.gz | 195,297,279 | 21,544 | By adamant, history, 3 years ago,
Hi everyone!
Let's continue with learning continued fractions. We began with studying the case of finite continued fractions and now it's time to work a bit with an infinite case. It turns out that while rational numbers have unique representation as a finite continued fraction, any irrational number has unique representation as an infinite continued fraction.
Part 1: Introduction
Part 2: Properties and interpretation
Distance between convergents. In the first part we learned that for continued fraction $r=[a_0, a_1, \dots, a_n]$ elements of the sequence $r_0, r_1, \dots, r_n$ where $r_k = [a_0, a_1, \dots, a_k]$ are called convergents. We also derived that subsequent convergents obey a simple formula:
$\frac{p_{-2}}{q_{-2}} = \frac{0}{1},\quad\frac{p_{-1}}{q_{-1}} = \frac{1}{0},\quad\frac{p_k}{q_k}=\frac{a_k p_{k-1}+p_{k-2}}{a_kq_{k-1}+q_{k-2}}$
Subsequent convergents seemingly should be closer to $r$ with bigger $k$, culminating in $r_n=r$. So, let's find some bound on how far away from $r$ can $r_k$ be. We may start with looking on the difference between $r_k$ and $r_{k-1}$:
$\frac{p_k}{q_k}-\frac{p_{k-1}}{q_{k-1}} = \frac{p_k q_{k-1} - p_{k-1} q_k}{q_k q_{k-1}}$
Let's rewrite the numerator of this fraction:
$p_k q_{k-1} - p_{k-1} q_k = (a_k p_{k-1} + p_{k-2})q_{k-1} - p_{k-1}(a_k q_{k-1} + q_{k-2}) = p_{k-2} q_{k-1} - p_{k-1} q_{k-2}$
Which means that numerator of $r_k - r_{k-1}$ is the opposite of numerator of $r_{k-1} - r_{k-2}$. The base case here is defined by $p_0 q_{-1} - p_{-1} q_0$, which is equal to $-1$. In this way we may conclude that $p_k q_{k-1} - p_{k-1} q_k = (-1)^{k-1}$, thus the whole difference is written as follows:
$r_k - r_{k-1} = \frac{(-1)^{k-1}}{q_k q_{k-1}}$
Further we will also need to know the distance between $r_{k+1}$ and $r_{k-1}$:
$r_{k+1}-r_{k-1} = \frac{(-1)^k}{q_{k+1} q_k} + \frac{(-1)^{k-1}}{q_k q_{k-1}} = \frac{(-1)^{k-1}(q_{k+1}-q_{k-1})}{q_{k+1}q_k q_{k-1}} = \frac{(-1)^{k-1}a_{k+1}}{q_{k+1} q_{k-1}}$
Approximation properties. Formulas above allow us to represent number $r=r_n$ as explicit telescopic sum (given that $r_0=a_0$):
$r=(r_n - r_{n-1}) + (r_{n-1} - r_{n-2}) + \dots + (r_1 - r_0) + r_0 = a_0 + \sum\limits_{k=1}^n \frac{(-1)^{k+1}}{q_k q_{k-1}}$
Since $q_k$ monotonically increases, we may say that $q_{k+1} q_k > q_k q_{k-1}$, thus we have an alternating sum which terms are decreasing by absolute value. This fact automatically implies several important properties, which may be illustrated by the picture below:
In our specific case difference between subsequent convergents quickly drops:
$\left|\frac{r_{k+1} - r_k}{r_{k} - r_{k-1}}\right| = \frac{q_{k-1}}{q_{k+1}} \leq \frac{q_{k-1}}{q_{k} + q_{k-1}} \leq \frac{1}{2}$
Consequent convergents maintain lower and upper bound on possible values of $r$ and each summand makes one of bounds more accurate. Bound above means that segment of possible $r$ is at least halved on each step. Few important corollaries follow from this:
• Convergents with even indices are lower than $r$ while convergents with odd indices are greater than $r$,
• If indices $j$ and $i$ have different parity then:
$|r_j - r_i| = |r_i - r| + |r_j - r|$
• If indices $j$ and $i$ have same parity and $j>i$ then:
$|r_j - r_i| = |r_i - r| - |r_j - r|$
• If $j>i$, then $|r_i - r| \geq |r_j - r|$.
These observations provide us convenient means of estimating how close is $r_k$ to $r$:
$\frac{1}{(q_{k+1}+q_k)q_k} \leq \frac{a_{k+1}}{q_{k+2} q_k}=|r_{k+2} - r_k| \leq |r_k - r| \leq |r_{k+1} - r_k| = \frac{1}{q_k q_{k+1}} \leq \frac{1}{q_k^2}$
So, the final "pretty" bounds on the distance between $r_k$ and $r$ may be written as follows:
$\left|\frac{p_k}{q_k}-r\right| \leq \frac{1}{q_k^2}$
Infinite fractions. Now that we had a proper introduction of finite continued fractions, we may introduce infinite fractions $[a_0, a_1, a_2, \dots]$ as well. We say that $r$ is represented by such infinite fraction if it's the limit of its convergents:
$r = \lim\limits_{n \to \infty} r_n = a_0 + \sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{q_n q_{n-1}}$
Properties above guarantee that this series is convergent, and there is an algorithm to calculate subsequent $a_k$, which altogether means that there is one-to-one correspondence between irrational numbers and infinite continued fractions. Some well-known constants, like golden ration, have "pretty" continued fraction representations:
$\varphi = [1,1,1,1,1,\dots],\\ \sqrt{2} = [1,2,2,2,2,\dots],\\ e = [2,1,2,1, 1, 4, 1, 1, 6, \dots, 1, 1, 2k, 1, 1, \dots]$
It is also worth noting that continued fraction is periodic if and only if $r$ is quadratic irrational, that is $r$ is the solution of some quadratic algebraic equation. In terms of convergence infinite fractions also hold most of properties of finite fractions.
Geometric interpretation. Consider sequence $s_k = (q_k;p_k)$ on 2-dimensional space. Each point $s_k$ corresponds to convergent $\frac{p_k}{q_k}$ which is the slope coefficient of the vector from $(0;0)$ to $(q_k;p_k)$. As was mentioned earlier, convergent coefficients are determined by formula:
$\frac{p_k}{q_k} = \frac{a_k p_{k-1} + p_{k-2}}{a_k q_{k-1} + q_{k-2}}$
Geometrically it means that $s_k=a_k s_{k-1} + s_{k-2}$. Let's recall how coefficients $a_k$ are obtained. Let $b_0, b_1, \dots, b_n$ be the states of $r$ at each step. Thus, we initially have $b_0=r$ and then for each $b_k$ we calculate $a_k = \lfloor b_k \rfloor$ and say that $b_{k+1} = \frac{1}{b_k - a_k}$. So, for example:
$b_0 = \frac{r}{1},\\ b_1 = \frac{1}{r-a_0},\\ b_2 = \frac{1}{\frac{1}{r-a_0} - a_1}=\frac{r-a_0}{1+ a_0 a_1 - a_1 r},\\ b_3 = \frac{1}{\frac{r-a_0}{1-a_1r + a_0 a_1}-a_2} = \frac{1 + a_0 a_1 - a_1 r}{r(1 + a_1 a_2) - (a_0 + a_2 + a_0 a_1 a_2)}$
If you look closely, you may see that the numerator of $b_k$ is the denominator of $b_{k-1}$ and the denominator of $b_k$ is given by:
$Q_k(a_0, a_1, \dots, a_{k-1}, r) = (-1)^k(P_{k-1}(a_0, a_1, \dots, a_{k-1})-r P_{k-2}(a_1, a_2, \dots, a_{k-1})) = (-1)^k(p_{k-1} - r q_{k-1})$
Thus, the whole number $b_k$ may be defined as follows:
$b_k = \left|\frac{p_{k-2} - r q_{k-2}}{p_{k-1} - r q_{k-1}}\right| = \frac{q_{k-1}}{q_{k-2}}\left|\frac{r_{k-2}-r}{r_{k-1}-r}\right|$
The meaning of this coefficient is as follows: Consider line $y=rx$. As convergents approach $r$, slope coefficient of $(q_k; p_k)$ approaches $r$ as well and thus points $(q_k; p_k)$ lie closer and closer to this line. That being said, value $|p_k - r q_k|$ tells us how far is $(q_k; p_k)$ from $y=rx$ on the vertical line $x=q_k$. Note that $s_{k-1}$ and $s_{k-2}$ are always on opposite sides from line $y=rx$, so adding $s_{k-1}$ to $s_{k-2}$ will decrease vertical distance of $s_{k-2}$ by $|p_{k-1} - r q_{k-1}|$. In this way we can see that $a_k = \lfloor b_k \rfloor$ is nothing but the maximum amount of times we may add $s_{k-1}$ to $s_{k-2}$ without switching the sign of $p_{k-2} - r q_{k-2}$. This gives geometric meaning to $a_k$ and the algorithm of computing them:
Initially we have vectors $s_{-2} = (1; 0)$ and $s_{-1} = (0; 1)$ lying on different sides from $y=rx$. On $k$-th step we compute $a_k$ to be the maximum amount of times we may add $s_{k-1}$ to $s_{k-2}$ without switching side. Then we say that $s_k = as_{k-1} + s_{k-2}$ and, therefore:
$\frac{p_k}{q_k} = \frac{a_k p_{k-1} + p_{k-2}}{a_k q_{k-1} + q_{k-2}}$
Algorithm above was taught to Vladimir Arnold by Boris Delaunay who called it "the nose stretching algorithm". This algorithm was initially used to find all lattice points below the line $y=rx$ but also turned out to provide simple geometric interpretation of continued fractions.
This connection between continued fractions and 2d-geometry established the notion of mediants. That is, for fractions $A = \frac{a}{b}$ and $B = \frac{c}{d}$ their mediant is $C = \frac{a+c}{b+d}$. If $A$ and $B$ are slope coefficients of two vectors, then $C$ is the slope coefficient of the sum of these vectors.
That should be all for now, next time I hope to write in more details about how the nose stretching algorithm may be used to count lattice points under the given line without that much pain and to solve integer linear programming in 2D case. Also I'm going to write about how continued fractions provide best rational approximations and, hopefully, write more about competitive programming problems where this all may be used, so stay tuned!
• +138
| Write comment?
» 3 years ago, # | +21 As an exercise for this article material you may try to solve Timus 1430. There is simple $O(\sqrt n)$ solution for that problem, but it may as well be solved in $O(\log n)$ utilizing continued fraction!
• » » 3 years ago, # ^ | ← Rev. 2 → +5 Could you share some hint? I really have no idea about continued fractions I guess :(
• » » » 3 years ago, # ^ | 0 You may start with $(\lfloor \frac{n}{a}\rfloor;0)$ and then consequently adjust it to the line $ax+by=n$ using convergent vectors for $-\frac{a}{b}$. There is a bit of casework to be dealt with though. | 3,201 | 9,079 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2023-06 | latest | en | 0.820654 |
https://interviewmania.com/discussion/112-aptitude-number-system | 1,542,084,247,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741219.9/warc/CC-MAIN-20181113041552-20181113063552-00467.warc.gz | 659,454,532 | 6,109 | ## Discussion
1. The number 58129745812974 is divisible by
1. 11
2. 9
3. 4
4. None of these.
1. ##### Correct Option: A
We know that, a number is divisible by 11 when the sum of the difference between the sum of its digit at even places and sum of its digit at odd places is either 0 or the difference is divisible by 11.
So number is 58129745812974
Sum of digit at odd places = 4 + 9 + 1 + 5 + 7 + 2 + 8 = 36
Sum of digit at even places = 7 + 2 + 8 + 4 + 9 + 1 + 5 = 36
So, required difference = 36 - 36 = 0
∴ number is divisible by 11. | 189 | 540 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-47 | latest | en | 0.911216 |
https://www.jiskha.com/questions/1603044/three-angles-of-a-quadrilateral-are-in-the-ratio-1-2-3-the-sum-of-the-least-and-the | 1,638,865,729,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363337.27/warc/CC-MAIN-20211207075308-20211207105308-00059.warc.gz | 893,727,181 | 4,870 | # Math
Three angles of a quadrilateral are in the ratio 1:2:3. The sum of the least and the greatest of these angles is 200°. Find all the four angles of the quadrilateral.
1. 👍
2. 👎
3. 👁
1. x + 2x + 3x + y = 360
y + 6x = 360
x + 3x = 200
x = 50
so
y + 300 = 360
y = 60
50, 100, 150, 60
1. 👍
2. 👎
2. x+3x=200
x=50 deg
360=sum of three + fourth
360=x+2x+3x + fourth
360=6*50+fourth
fourth= ..... degrees
1. 👍
2. 👎
👤
bobpursley
## Similar Questions
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2. ### Math
The angles of a quadrilateral are in the ratio 2:3:7. The mean of these angle is 64. Find all the four angles? | 777 | 2,503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-49 | latest | en | 0.855348 |
http://www.eecs.wsu.edu/~cook/dm/lectures/l7/index.html | 1,513,170,360,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948523222.39/warc/CC-MAIN-20171213123757-20171213143757-00650.warc.gz | 350,556,132 | 8,321 | # Bayesian Learning
Provides practical learning algorithms:
• Naive Bayes learning
• Bayesian belief network learning
• Combine prior knowledge (prior probabilities) with observed data
• Requires prior probabilities
Provides useful conceptual framework
• Provides gold standard'' for evaluating other learning algorithms
• Additional insight into Occam's razor
# Bayes Theorem
• P(h) = prior probability of hypothesis h
• P(D) = prior probability of training data D
• P(h|D) = probability of h given D
• P(D|h) = probability of D given h
# Choosing Hypotheses
Generally want the most probable hypothesis given the training data
Maximum a posteriori hypothesis hMAP:
hMAP
If assume P(hi)=P(hj) then can further simplify, and choose the Maximum likelihood (ML) hypothesis
# Bayes Theorem
Does patient have cancer or not?
A patient takes a lab test and the result comes back positive. The test returns a correct positive result in only of the cases in which the disease is actually present, and a correct negative result in only of the cases in which the disease is not present. Furthermore, .008 of the entire population have this cancer.
P(cancer) =
P( cancer) =
P(+ cancer) =
P(- cancer) =
P(+ cancer) =
P(- cancer) =
# Basic Formulas for Probabilities
• Product Rule: probability of a conjunction of two events A and B:
• Sum Rule: probability of a disjunction of two events A and B:
• Theorem of total probability: if events are mutually exclusive with , then
# Brute Force MAP Hypothesis Learner
1.
For each hypothesis h in H, calculate the posterior probability
2.
Output the hypothesis hMAP with the highest posterior probability
# Most Probable Classification of New Instances
So far we've sought the most probable hypothesis given the data D(i.e., hMAP)
Given new instance x, what is its most probable classification?
• hMAP(x) is not the most probable classification!
Consider:
• Three possible hypotheses:
• Given new instance x,
• What's most probable classification of x?
# Bayes Optimal Classifier
Bayes optimal classification:
Example:
P(h1|D)=.4, P(-|h1)=0, P(+|h1)=1 P(h2|D)=.3, P(-|h2)=1, P(+|h2)=0 P(h3|D)=.3, P(-|h3)=1, P(+|h3)=0
therefore
= 0.4 = 0.6
and
= -
# Naive Bayes Classifier
Along with decision trees, neural networks, nearest nbr, one of the most practical learning methods.
When to use
• Moderate or large training set available
• Attributes that describe instances are conditionally independent given classification
Successful applications:
• Diagnosis
• Classifying text documents
# Naive Bayes Classifier
Assume target function , where each instance x described by attributes .
Most probable value of f(x) is:
vMAP = vMAP = =
Naive Bayes assumption:
which gives
# Naive Bayes Algorithm
Naive_Bayes_Learn(examples)
For each target value vj
estimate P(vj)
For each attribute value ai of each attribute a
estimate P(ai|vj)
Classify_New_Instance(x)
# Naive Bayes: Example
Day Outlook Temperature Humidity Wind PlayTennis D1 Sunny Hot High Weak No D2 Sunny Hot High Strong No D3 Overcast Hot High Weak Yes D4 Rain Mild High Weak Yes D5 Rain Cool Normal Weak Yes D6 Rain Cool Normal Strong No D7 Overcast Cool Normal Strong Yes D8 Sunny Mild High Weak No D9 Sunny Cool Normal Weak Yes D10 Rain Mild Normal Weak Yes D11 Sunny Mild Normal Strong Yes D12 Overcast Mild High Strong Yes D13 Overcast Hot Normal Weak Yes D14 Rain Mild High Strong No
Consider PlayTennis again, and new instance
Want to compute:
# Naive Bayes: Subtleties
1.
Conditional independence assumption is often violated
• ...but it works surprisingly well anyway. Note don't need estimated posteriors to be correct; need only that
• see [Domingos & Pazzani, 1996] for analysis
• Naive Bayes posteriors often unrealistically close to 1 or 0
# Naive Bayes: Subtleties
2.
what if none of the training instances with target value vj have attribute value ai? Then
Typical solution is Bayesian estimate for
where
• n is number of training examples for which v=vj,
• nc number of examples for which v=vj and a=ai
• p is prior estimate for
• m is weight given to prior (i.e. number of virtual'' examples)
# Learning to Classify Text
Why?
• Learn which news articles are of interest
• Learn to classify web pages by topic
Naive Bayes is among most effective algorithms
What attributes shall we use to represent text documents??
# Learning to Classify Text
Target concept
1.
Represent each document by vector of words
• one attribute per word position in document
2.
Learning: Use training examples to estimate
• P(+)
• P(-)
• P(doc|+)
• P(doc|-)
Naive Bayes conditional independence assumption
where P(ai=wk| vj) is probability that word in position i is wk, given vj
one more assumption:
# Pseudocode
LEARN/SMALL>_NAIVE/SMALL>_BAYES/SMALL>_TEXT( Examples, V)
1. collect all words and other tokens that occur in Examples
all distinct words and other tokens in Examples
2. calculate the required P(vj) and P(wk|vj) probability terms
For each target value vj in V do
• subset of Examples for which the target value is vj
• a single document created by concatenating all members of docsj
• total number of words in Textj (counting duplicate words multiple times)
• for each word wk in Vocabulary
• number of times word wk occurs in Textj
CLASSIFY/SMALL>_NAIVE/SMALL>_BAYES/SMALL>_TEXT(Doc)
• all word positions in Doc that contain tokens found in Vocabulary
• Return vNB, where
# Twenty NewsGroups
Given 1000 training documents from each group
Learn to classify new documents according to which newsgroup it came from
comp.graphics misc.forsale comp.os.ms-windows.misc rec.autos comp.sys.ibm.pc.hardware rec.motorcycles comp.sys.mac.hardware rec.sport.baseball comp.windows.x rec.sport.hockey alt.atheism sci.space soc.religion.christian sci.crypt talk.religion.misc sci.electronics talk.politics.mideast sci.med talk.politics.misc talk.politics.guns
Naive Bayes: 89% classification accuracy
# Article from rec.sport.hockey
Path: cantaloupe.srv.cs.cmu.edu!das-news.harvard.edu!ogicse!uwm.edu
From: [email protected] (John Doe)
Subject: Re: This year's biggest and worst (opinion)...
Date: 5 Apr 93 09:53:39 GMT
I can only comment on the Kings, but the most
obvious candidate for pleasant surprise is Alex
Zhitnik. He came highly touted as a defensive
defenseman, but he's clearly much more than that.
Great skater and hard shot (though wish he were
more accurate). In fact, he pretty much allowed
the Kings to trade away that huge defensive
liability Paul Coffey. Kelly Hrudey is only the
biggest disappointment if you thought he was any
good to begin with. But, at best, he's only a
mediocre goaltender. A better choice would be
Tomas Sandstrom, though not through any fault of
his own, but because some thugs in Toronto decided
# Learning Curve for 20 Newsgroups
Accuracy vs. Training set size (1/3 withheld for test)
# Bayesian Belief Networks
Interesting because:
Naive Bayes assumption of conditional independence too restrictive
But it's intractable without some such assumptions...
Bayesian Belief networks describe conditional independence among subsets of variables
allows combining prior knowledge about (in)dependencies among variables with observed training data
(also called Bayes Nets)
# Conditional Independence
Definition: X is conditionally independent of Y given Z if the probability distribution governing X is independent of the value of Ygiven the value of Z; that is, if
more compactly, we write
P(X | Y,Z) = P(X | Z)
Example: Thunder is conditionally independent of Rain, given Lightning
P(Thunder | Rain, Lightning) = P(Thunder | Lightning)
Naive Bayes uses cond. indep. to justify
P(X,Y|Z) = P(X|Y,Z) P(Y|Z) = P(X|Z) P(Y|Z)
# Bayesian Belief Network
Network represents a set of conditional independence assertions:
• Each node is asserted to be conditionally independent of its nondescendants, given its immediate predecessors.
• Directed acyclic graph
# Bayesian Belief Networks
A belief network represents the dependence between variables.
* nodes | 1,973 | 8,078 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2017-51 | latest | en | 0.842802 |
https://stats.stackexchange.com/questions/135755/sample-size-calculation-for-proportion | 1,580,053,365,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251689924.62/warc/CC-MAIN-20200126135207-20200126165207-00404.warc.gz | 653,850,770 | 31,424 | # Sample size calculation for proportion
I am in a doubt about using Absolute and Relative precision for sample size calculations. Suppose if I want to conduct a study to asses the prevalence of Hypertension in a general population, which formula should I use among these two-
1. $n= Z^2 P(1-P)/d^2$
where $n$ = sample size; $Z$ = C.I.; $P$= anticipated prevalence or prevalence estimated from pilot study and $d$ = absolute precision.
2. $n = Z^2(1-P) / e^2 P$
Where $e$ = relative precision.
Please suggest me with some example.
• If this is a homework question could you please add the self-study tag? – Andy Jan 31 '15 at 8:01
Both formulas are right. In the first formula, the intent is to estimate the proportion within d percentage points of the true value P. In the second formula, you want to estimate the proportion within e of the true proportion P (ie, within e*P). That means, while in the first formula the precision is fixed, in the second formula the precision fluctuates based on the value of P.
• the second and third d in your comment should be e, to correspond to the constant in the poster's formula 2. – Steve Samuels Feb 1 '15 at 1:05
For a margin of error $m$, a minimum sample size, $n$ would be given by the following formula:
$n$ > $Z^2\over{m}^2$$\hat{p}(1-\hat{p})$ | 344 | 1,301 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2020-05 | latest | en | 0.876545 |
http://www.tectechnicsclassroom.tectechnics.com/s7p5a51/bandpass-function-center-frequency-bandwidth-half-power-frequency-quality-factor-resonance.php | 1,695,428,395,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506429.78/warc/CC-MAIN-20230922234442-20230923024442-00350.warc.gz | 81,335,843 | 6,970 | Bandpass Function, Center Frequency, Half Power Frequency, Bandwidth, Quality Factor And Resonance
TECTechnics Classroom TECTechnics Overview
Bandpass Function, Center Frequency, Half Power Frequency, Bandwidth, Quality Factor And Resonance
(a) Indicate the general representation of the bandpass function in the s domain
(b) Indicate the frequency response of the bandpass function
(c) Determine the center frequency of the frequency response
(d) Determine the half-power frequencies of the frequency response
(e) Determine the bandwidth of the frequency response
(f) Determine the quality factor
(g) Relate the quality factor to resonance
(h) If H(s) = s/(s2 + as + b), determine a and b such that the magnitude of the frequency response |H(ω)| has a maximum at 100 Hz with a half-power bandwidth of 5 Hz. What is the quality factor?
The strings: S7P5A51 (change - physical).
The math:
Pj Problem of Interest is of type change (change - physical).
(a) General bandpass function in the s domain:
H(s) = ks/(s2 + as + b)------(1) Where k, a, b > 0.
(b) Frequency response of bandpass function:
H(jω) = k/(2 + a + b) = k/(b - ω2 + ajω)------(2)
(c) |H(jω)|2 = k2ω2/[(b - ω2)2 + a2ω2)] = k2/[a2 + (b - ω2)22]------(3)
The frequency (ωo) at which |H| is maximum is the center frequency.
So from eq(3):|H| is maximum when (b - ω2) = 0 that is, when ω = √b
So, center frequency, ωo = √b.
At the center frequency, |H|max = k/a.
(d) Consider the following equalities:
|Hl|2 = |Hh|2 = (1/2)|Ho)|2-------(4)
ωl and ωh in eq(4) are called the half-power lower frequency and the half-power upper frequency respectively.
Substituting eq(3) into eq(4) and solving, we have:
ωl and ωh as the roots of:
(b - ω2)22] = a2
So, ωl = √(a2/4 + b) - a/2
And ωh = √(a2/4 + b) + a/2
(e) Bandwidth, β = ωh - ωl =√(a2/4 + b) + a/2 - √(a2/4 + b) - a/2 = a.
ωhωl = b = (ωo)2
(f) Quality factor, Q = ωo/β = √b/a
Quality factor measures sharpness of frequency response around the center frequency.
(g) Increasing quality factor tends towards resonance.
(h) Bandwidth = β = a = 2π(5) = 31.416
ωo = 2π(100) = 628.3185 = √b
So, b = (628.3185)2 = 394784
Quality factor, Q = ωo/β = 628.3185/31.416 = 20
Blessed are they that have not seen, and yet have believed. John 20:29 | 739 | 2,253 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2023-40 | latest | en | 0.818919 |
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# Solve the fraction: $\dfrac{9}{11}-\dfrac{4}{15}$.
Last updated date: 12th Jul 2024
Total views: 447.6k
Views today: 10.47k
Answer
Verified
447.6k+ views
Hint: Find the common denominator of the two fractions. Then subtract the two numerators (top numbers). Put the answer over the denominator and simplify the fraction formed.
Complete step-by-step answer:
Give us two fractions $\dfrac{9}{11}$and $\dfrac{4}{15}$where we need to subtract the second from the first.
The first step towards subtraction is to ensure that the bottom numbers (the denominators) are the same.
Here the denominators are different $\dfrac{9}{11}$and $\dfrac{4}{15}$. Let’s make the denominators the same.
We know the denominators are 11 and 15. They both don’t have any numbers in common. So let’s multiply 11 and 15.
$11\times 15=165$
Now we got the denominator as 165.
First let us take the fraction $\dfrac{9}{11}$. We need to make the denominator of the fraction as 165. So multiply by 15 in the numerator and denominator.
$\dfrac{9\times 15}{11\times 15}=\dfrac{135}{165}-(1)$
Now let us take the fraction $\dfrac{4}{15}$. We need to make the denominator 165.
So let’s multiply by 11 in the numerator and denominator.
$\dfrac{4\times 11}{15\times 11}=\dfrac{44}{165}-(2)$
Now let us subtract (2) from (1).
$\therefore \dfrac{135}{165}-\dfrac{44}{165}=\dfrac{135-44}{165}$
Now let’s subtract the top numbers (the numerators). Put the answer over the same denominator.
$\dfrac{135-44}{165}=\dfrac{91}{165}$
We cannot further simplify the fraction, as they don’t have any common factors.
$\therefore \dfrac{9}{11}-\dfrac{4}{15}=\dfrac{91}{165}$
Note: It can be solved by cross multiplying the numerator and denominator divided by the multiplication of both denominators.
$\dfrac{9}{11}-\dfrac{4}{15}=\dfrac{\left( 9\times 15 \right)-\left( 4\times 11 \right)}{11\times 15}=\dfrac{135-44}{165}=\dfrac{91}{165}$
We get the same answer as we have got in the first method. This is a shortcut for the same. | 626 | 2,075 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2024-30 | latest | en | 0.78501 |
http://www.engineeringexpert.net/Engineering-Expert-Witness-Blog/tag/direction | 1,722,864,581,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640447331.19/warc/CC-MAIN-20240805114033-20240805144033-00499.warc.gz | 38,012,981 | 9,622 | ## Posts Tagged ‘direction’
### Torque Formula Symplified
Wednesday, April 2nd, 2014
Last time we introduced the mathematical formula for torque, which is most simply defined as a measure of how much a force acting upon an object causes that object to rotate around a pivot point. When manipulated, torque can produce a mechanical advantage in gear trains and tools, which we’ll see later. The formula is: Torque = Distance × Force × sin(ϴ) We learned that the factors Distance and Force are vectors, and sin(ϴ) is a trigonometric function of the angle ϴ which is formed between their two vectors. Let’s return to our wrench example and see how the torque formula works. Vectors have both a magnitude, that is, a size or extent, and a direction, and they are typically represented in physics and engineering problems by straight arrows. In our illustration the vector for distance is represented by an orange arrow, while the vector for force is represented by a red arrow. The orange distance vector has a magnitude of 6 inches, while the red force vector has a magnitude of 10 pounds, which is being supplied by the user’s arm muscle manipulating the nut. That muscle force follows a path from the arm to the pivot point located at the center of the nut, a distance of 6 inches. Vector arrows point in a specific direction, a direction which is indicative of the way in which the vectors’ magnitudes — in our case inches of distance vs. pounds of force — are oriented with respect to one another. In our illustration the orange distance vector points away from the pivot point. This is according to engineering and physics convention, which dictates that, when a force vector is acting upon an object to produce a torque, the distance vector always points from the object’s pivot point to the line of force associated with the force vector. The angle, ϴ, that is formed between the two vectors in our example is 90 degrees, as measured by any common, ordinary protractor. Next we must determine the trigonometric value for sin(ϴ). This is easily accomplished by simply entering “90” into our calculator, then pressing the sin button. An interesting fact is that when the angle ϴ ranges anywhere between 0 and 90 degrees, the values for sin(ϴ) always range between 0 and 1. To see this in action enter any number between 0 and 90 into a scientific calculator, then press the sin button. For our angle of 90 degrees we find that, sin(90) = 1 Thus the formula for torque in our example, because the sin(ϴ) is equal to 1, simply becomes the product of the magnitudes of the Distance and Force vectors: Torque = Distance × Force × sin(90) Torque = Distance × Force × 1 Torque = Distance × Force Next time we’ll insert numerical values into the equation and see how easily torque can be manipulated. _______________________________________
### Vectors, Sin(ϴ), and the Torque Formula
Wednesday, March 26th, 2014
Last time we introduced a physics concept known as torque and how it, together with modified gear ratios, can produce a mechanical advantage in devices whose motors utilize gear trains. Now we’ll familiarize ourselves with torque’s mathematical formula, which involves some terminology, symbols, and concepts which you may not be familiar with, among them, vectors, and sin(ϴ). Torque = Distance × Force × sin(ϴ) In this formula, Distance and Force are both vectors. Generally speaking, a vector is a quantity that has both a magnitude — that is, any measured quantity associated with a vector, whether that be measured in pounds or inches or any other unit of measurement — and a direction. Vectors are typically represented graphically in engineering and physics illustrations by pointing arrows. The arrows are indicative of the directionality of the magnitudes involved. Sin(ϴ), pronounced sine thay-tah, is a function found within a field of mathematics known as trigonometry , which concerns itself with the lengths and angles of triangles. ϴ, or thay-tah, is a Greek symbol used to represent the angle present between the Force and Distance vectors as they interact to create torque. The value of sin(ϴ) depends upon the number of degrees in the angle ϴ. Sin(ϴ) can be found by measuring the angle ϴ, entering its value into a scientific calculator, and pressing the Sin button. We’ll dive into the math behind the vectors next time, when we return to our wrench and nut example and apply vector force quantities. _______________________________________ | 1,002 | 4,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-33 | latest | en | 0.940105 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-11-additional-topics-11-3-fractional-exponents-problem-set-11-3-page-489/52 | 1,726,860,830,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701423570.98/warc/CC-MAIN-20240920190822-20240920220822-00338.warc.gz | 729,323,120 | 13,922 | ## Elementary Algebra
$x^{\frac{7}{12}}$
Recall the following rule: $a^{n}+a^{m}=a^{n+m}$ Thus, we obtain: $x^{\frac{1}{3}}\times x^{\frac{1}{4}}$ =$x^{\frac{1}{3}+\frac{1}{4}}$ =$x^{\frac{4+3}{12}}$ =$x^{\frac{7}{12}}$ | 105 | 220 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-38 | latest | en | 0.477281 |
https://apboardsolutions.in/ap-board-7th-class-maths-notes-chapter-10/ | 1,723,387,191,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641002566.67/warc/CC-MAIN-20240811141716-20240811171716-00873.warc.gz | 75,584,365 | 13,336 | # AP Board 7th Class Maths Notes Chapter 10 Algebraic Expressions
Students can go through AP Board 7th Class Maths Notes Chapter 10 Algebraic Expressions to understand and remember the concepts easily.
## AP State Board Syllabus 7th Class Maths Notes Chapter 10 Algebraic Expressions
→ Variable: It takes different values.
Ex: x, y, z, a, b, c, m etc.
Constant: The value of constant is fixed.
Ex: 1, 2, $$\frac{-2}{3}$$, $$\frac{4}{5}$$ etc.
→ Algebraic Expression: An algebraic expression is a single term or a combination of terms connected by the symbols ‘+’ (plus) or(minus).
Ex: 2x + 3, $$\frac{2}{5}$$p, 3x – 1 etc.
→ Numerical Expression : If every term of an expression is a constant term, then the expression is called a Numerical expression.
Ex: 2 + 1, -5 × 3, (12 + 4) ÷ 3.
Note: In the expression 2x + 9, ‘2x’ is an algebraic term. ‘9’ is called numeric term.
→ Like terms are terms which contain the same variables with the same exponents.
Ex: 12x, 25x, -7x are like terms.
2xy2, 3xy2, 7xy2 are like terms.
→ Coefficient: In a.xn, ‘a’ is called the numerical coefficient and ‘x’ is called the literal coefficient.
Types of algebraic expressions.
→ Degree of a monomial: The sum of all exponents of the variables present in a monomial is called the degree of the term or degree of the monomial.
Ex: The degree of 9x2y2 is 4 [∵ 2 + 2 = 4]
Note: Degree of constant term is zero.
The highest of the degrees of all the terms of an expression is called the degree of the expression.
Ex: The degree of the expression ax + bx2 + cx3 + dx4 + ex5 is 5.
→ The difference between two like terms is a like term with a numerical coefficient equal to the difference between the numerical coefficients of the two like terms.
Note:
1. If no two terms of an expression are alike then it is said to be in the simplified form.
2. In an expression, if the terms are arranged in such a way that the degrees of the terms are in descending order then the expression is said to be in standard form.
3. Addition (or) subtraction of expressions should be done in two methods, they are
i) Column or Vertical method.
ii) Row or Horizontal method. | 589 | 2,141 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2024-33 | latest | en | 0.911836 |
lucadodsa.cf | 1,596,622,874,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735939.26/warc/CC-MAIN-20200805094821-20200805124821-00149.warc.gz | 67,206,595 | 3,621 | ## Problem Solving | PixiMaths
When dealing with simple algebraic expressions, you don’t always need algebra to solve them. The following practice questions ask you to use three different methods: inspecting, rewriting the problem, and guessing and checking. Practice questions In the following questions, solve for x in each case just by looking at the equation. 1. 18 – x [ ]. Solving Word Questions. With LOTS of examples! In Algebra we often have word questions like: Example: Sam and Alex play tennis. On the weekend Sam played 4 more games than Alex did, and together they played 12 games. Mathematical Models Solving Inequality Word Questions Algebra Index Number Index. WORD PROBLEMS. Examples. Problems. W ORD PROBLEMS require practice in translating verbal language into algebraic language. See Lesson 1, Problem lucadodsa.cf, word problems fall into distinct types. Below are some examples. Example 1. ax ± b = c. All problems like the following lead eventually to an equation in that simple form.
## Free Algebra Questions and Problems with Answers
The following practice questions ask you to use three different methods: inspecting, rewriting the problem, and guessing and checking, algebra problem solving questions. In the following questions, use the correct inverse operation to rewrite and solve each problem. In the following questions, find the value of x in each equation by guessing and checking. Therefore, x is between 10 and Therefore, x is between 30 and Cross-multiplication is a handy tool for finding the common denominator for two fractions, Pre-Algebra Practice Questions: Isolating x in an You can solve algebraic equations by isolating x—that is, by getting x alone on one side A quick method for solving algebra problems is to re-arrange the equation by placing all x When an algebraic equation contains fractions, you can use cross-multiplication to solve t To find the volume of a prism or cylinder, you can use the following formula, where Ab is Algebra problem solving questions an expression has two sets of parentheses next to each other, algebra problem solving questions, you need to multiply ev The formula for the volume of a pyramid is the same as for the volume of a cone.
In this f The English system of measurement is most commonly used in the United States. If you were Load more. | 481 | 2,333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2020-34 | latest | en | 0.931063 |
http://fermatslasttheorem.blogspot.com/2005/07/proof-for-n3-using-eisenstein-integers.html?showComment=1141726740000 | 1,469,626,075,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257826907.66/warc/CC-MAIN-20160723071026-00022-ip-10-185-27-174.ec2.internal.warc.gz | 80,150,893 | 16,935 | ## Thursday, July 21, 2005
### Proof for n=3: Using Eisenstein Integers
If you are not familiar with Eisenstein Integers, please start here. The details of today's blog are based on an English translation of Heinrich Dorrie's 100 Great Problems of Elementary Mathematics.
For those who would like to see this proof done in terms of rational integers only, check out this previous blog.
As in previous blogs, I will use Greek letters to represent quadratic integers and Latin letters to represent rational integers.
Theorem: The equation α3 + β3 = γ3 does not have any integer solutions where
α, β, γ, are Eisenstein integers and α * β * γ ≠ 0.
(1) Since Eisenstein Integers are Euclidean (proof), we know that they are characterized by a Division Algorithm (proof), Bezout's Identity (proof), and Unique Factorization (proof).
(2) We can assume that α, β and γ are coprime. [See here for proof]
(3) If we set ζ = -γ, then we have:
α3 + β3 + ζ3 = 0.
(4) Let:
J = (1 + i√3)/2
O = (1 - i√3)/2
(5) Then, J - O is an Eisenstein prime number (proof ) which divides α*β*ζ [See here for proof.]
(6) But if α * β * γ ≠ 0, then we have an infinite descent. [See here for proof.]
QED | 329 | 1,185 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2016-30 | longest | en | 0.888045 |
https://byjus.com/questions/a-taxi-leaves-the-station-x-for-station-y-every-10-minutes-simultaneously-a-taxi-leaves-station-y-also-for-station-x-every-10-minutes-the-taxis-move-at-the-same-constant-speed-and-go-from-x-to-y/ | 1,656,144,646,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034877.9/warc/CC-MAIN-20220625065404-20220625095404-00027.warc.gz | 201,832,150 | 44,806 | # A taxi leaves the station X for station Y every 10 minutes. Simultaneously, a taxi leaves station Y also for station X every 10 minutes. The taxis move at the same constant speed and go from X to Y or vice-versa in 2 hours. How many taxis coming from the other side will each taxi meet enroute from Y to X
We’ll start by assuming that it’s been two hours since the first cab left X. Consider one taxi departing from Y at this precise moment; taxis will depart at 10-minute intervals throughout the route at this precise moment. All taxis on the road that started from X, including the taxi that just arrived at Y, and all taxis that will start from X until this taxi from Y arrives at X, will pass this taxi from Y at some point. So, initially, we’ll figure out how many cabs are blocking the road.
There will be 11 taxis on the way because all cabs are 10 minutes apart and a one-way trip takes 2 hours. Because they are not en way, the taxi that arrived at Y and the taxi that will arrive at X are not counted. After this taxi leaves and takes 2 hours to get to X, 12 other cabs must have left from X, all of which will pass this taxi from Y. As a result, a total of 23 taxis from X will meet each taxi from Y and vice versa. | 294 | 1,231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2022-27 | latest | en | 0.967399 |
http://en.wikipedia.org/wiki/Ultraparallel_theorem | 1,409,296,950,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500831903.50/warc/CC-MAIN-20140820021351-00237-ip-10-180-136-8.ec2.internal.warc.gz | 64,130,733 | 10,587 | # Ultraparallel theorem
In hyperbolic geometry, the ultraparallel theorem states that every pair of ultraparallel lines in the hyperbolic plane has a unique common perpendicular hyperbolic line.
## Proof in the Poincaré half-plane model
Let
$a < b < c < d$
be four distinct points on the abscissa of the Cartesian plane. Let $p$ and $q$ be semicircles above the abscissa with diameters $ab$ and $cd$ respectively. Then in the Poincaré half-plane model HP, $p$ and $q$ represent ultraparallel lines.
Compose the following two hyperbolic motions:
$x \to x-a\,$
$\mbox{inversion in the unit semicircle.}\,$
Then $a \to \infty$, $b \to (b-a)^{-1},\quad c \to (c-a)^{-1},\quad d \to (d-a)^{-1}.$
Now continue with these two hyperbolic motions:
$x \to x-(b-a)^{-1}\,$
$x \to \left [ (c-a)^{-1} - (b-a)^{-1} \right ]^{-1} x\,$
Then $a$ stays at $\infty$, $b \to 0$, $c \to 1$, $d \to z$ (say). The unique semicircle, with center at the origin, perpendicular to the one on $1z$ must have a radius tangent to the radius of the other. The right triangle formed by the abscissa and the perpendicular radii has hypotenuse of length $\begin{matrix} \frac{1}{2} \end{matrix} (z+1)$. Since $\begin{matrix} \frac{1}{2} \end{matrix} (z-1)$ is the radius of the semicircle on $1z$, the common perpendicular sought has radius-square
$\frac{1}{4} \left [ (z+1)^2 - (z-1)^2 \right ] = z.\,$
The four hyperbolic motions that produced $z$ above can each be inverted and applied in reverse order to the semicircle centered at the origin and of radius $\sqrt{z}$ to yield the unique hyperbolic line perpendicular to both ultraparallels $p$ and $q$.
## Proof in the Klein model
In the Klein model of the hyperbolic plane, two ultraparallel lines correspond to two non-intersecting chords. The poles of these two lines are the respective intersections of the tangent lines to the unit circle at the endpoints of the chords. Lines perpendicular to line A are modeled by chords whose extension passes through the pole of A. Hence we draw the unique line between the poles of the two given lines, and intersect it with the unit disk; the chord of intersection will be the desired common perpendicular of the ultraparallel lines. If one of the chords happens to be a diameter, we do not have a pole, but in this case any chord perpendicular to the diameter is perpendicular as well in the hyperbolic plane, and so we draw a line through the pole of the other line intersecting the diameter at right angles to get the common perpendicular.
The proof is completed by showing this construction is always possible. If both chords are diameters, they intersect. If only one of the chords is a diameter, the other chord projects orthogonally down to a section of the first chord contained in its interior, and a line from the pole orthogonal to the diameter intersects both the diameter and the chord. If both lines are not diameters, the we may extend the tangents drawn from each pole to produce a quadrilateral with the unit circle inscribed within it. The poles are opposite vertices of this quadrilateral, and the chords are lines drawn between adjacent sides of the vertex, across opposite corners. Since the quadrilateral is convex, the line between the poles intersects both of the chords drawn across the corners, and the segment of the line between the chords defines the required chord perpendicular to the two other chords.
## References
• Karol Borsuk & Wanda Szmielew (1960) Foundations of Geometry, page 291. | 899 | 3,504 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 27, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2014-35 | latest | en | 0.771727 |
https://www.physicsforums.com/threads/treatment-of-integral-as-an-operator.725127/ | 1,527,421,843,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868248.78/warc/CC-MAIN-20180527111631-20180527131631-00190.warc.gz | 797,025,550 | 14,641 | # Treatment of Integral as an Operator?
1. Nov 26, 2013
### FeDeX_LaTeX
My tutor showed me something today, and I still can't completely wrap my head around why it makes sense. Consider the following integral equation:
$\int f(x) = f(x) - 1$
Then:
$\int (f(x)) - f(x) = -1 \implies f(x) \left( \int (\text{Id}) - 1\right) = -1$
so we get the geometric series
$f(x) = \frac{1}{1 - \int(Id)} = 1 + \int + \int\int + \int\int\int + ... = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + ... = e^x$
which satisfies our original equation if we choose the right constant of integration. Similarly, take the differential equation (using D as the differential operator):
$Df = f - 1 \implies f\left(D(\text{Id}) - 1\right) = -1$
$f = 1 + D + (D \circ D) + (D \circ D \circ D) + ... = 1$
which also satisfies our original equation.
Is this mathematically valid, or was he pulling my leg? And if so, could someone help me make sense of why this is valid?
Trying this on some other differential equations seems to show that this doesn't always work -- I was wondering if anyone could point me to an article that says more about the idea behind this.
Last edited: Nov 26, 2013
2. Nov 26, 2013
### fzero
I would have kept the $f(x)$ on the RHS so that the integral operator acts on it, but there is a way to make the formalism mathematically sound. The right terminology to use when searching on this is the resolvent method for (Fredholm) integral equations. In particular, this seems to be an example of a Liouville-Neumann series solution to the particular equation of interest.
I don't have any great recommendations for sources to learn more about this, but I have probably encountered integral equations years ago in Whittaker and Watson's Course of Modern Analysis. There are various lecture notes available via google that might be worthwhile as well. | 507 | 1,857 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-22 | latest | en | 0.929736 |
https://datacadamia.com/trigonometry/angle | 1,721,531,953,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517550.76/warc/CC-MAIN-20240721030106-20240721060106-00059.warc.gz | 165,464,818 | 35,698 | # Trigonometry - Angle (or Arc) (Alpha - α)
Angle is counted in the Babylonian numeral system, base-60. (360 degrees in a circle)
## Illustration
with the identity function
• seconds
## Notation
````10°25'59" `
```
where:
Right angle = 90 degree
Discover More
(Geometry|Linear Algebra) - Euclidean Space
In geometry, Euclidean space encompasses: the two-dimensional Euclidean plane, the three-dimensional space of Euclidean geometry, and certain other spaces. The term “Euclidean” distinguishes...
(Turn|Cycle|Revolution|Full Circle|Complete Rotation)
A Turn is a Unit ofPlane angle. A turn is also known as: a cycle (abbreviated cyc), a revolution (abbreviated rev), a complete rotation (abbreviated rot) or full circle. 1 turn = 2π...
Angle - Minute
One minute is a unit notation of angle where 1 degree = 60 minutes Minute_and_second_of_arc
Decimal - Fraction
A decimal fraction is a fraction in the decimal numeral system (base 10). They are also known as the decimals. In 999.99, the fraction or decimals are 99 Time and angles use decimal fractions of...
Geometry - Object
in Geometry An object can be: a pre-defined shape or an arbitrary one defined with the help of a path. Notions such as: The length of a vector The angle between two vectors. are introduced...
Geometry - Perpendicular (Right angle)
In elementary geometry, the word perpendicular describes the relationship between two geometric objects that meet at a right angle.
Number - Babylonian numeral system, base-60
The sexagesimal numeral system or base-60, was the first positional system in the first language developed the Cuneiform It was used by astronomers Numeral System Parts Babylonian numerals and other...
Trignometry - Identity function (sin2+cos2=1)
identify function in Trignometry where: Other illustration : The identity function is a derivative from the pythagorean...
Trigonometry
(from Greek trigōnon, triangle and metron, “measure”) is a branch of mathematics that studies relationships involving: lengths and angles of triangles.
Trigonometry - (Sine|Sinus)
Sine is the function of an angle (trigonometric function). The word comes from the Latin sinus for gulf or bay, since, given a unit circle, it is the side of the triangle on which the angle opens. ... | 523 | 2,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-30 | latest | en | 0.828302 |
http://mathhelpforum.com/statistics/10600-polygons.html | 1,527,505,282,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794872766.96/warc/CC-MAIN-20180528091637-20180528111637-00371.warc.gz | 183,045,846 | 9,543 | 1. ## Polygons
A set of polygons contains a square, a rectangle, a rhombus and a trapezoid. If one polygon is chosen at random, what is the probability that the polygon has all sides congruent and all right angles?
2. Originally Posted by symmetry
A set of polygons contains a square, a rectangle, a rhombus and a trapezoid. If one polygon is chosen at random, what is the probability that the polygon has all sides congruent and all right angles?
A square and rhombus are the only polygons with congruent sides.
A square and rectangle are the only polygons with all right angles.
And thus, we want to know which polygon has all sides congruent, as well as all right angles. As shown above, only a square does.
Therefore, there is a 1/4 chance in choosing a polygon that has all sides congruent and all right angles.
3. Originally Posted by symmetry
A set of polygons contains a square, a rectangle, a rhombus and a trapezoid. If one polygon is chosen at random, what is the probability that the polygon has all sides congruent and all right angles?
the polygon is Square, which is satisfies the condition all sides congruent and all right angles.
total 4 poloygons is there
probablity is 1/4
- R.Mangalesh
4. ## ok
I thank you for your help and TIPS. | 305 | 1,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-22 | latest | en | 0.912295 |
https://www.scribd.com/document/106439676/Untitled | 1,539,708,887,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510853.25/warc/CC-MAIN-20181016155643-20181016181143-00234.warc.gz | 1,073,376,948 | 29,997 | # History of Real Numbers
History of Real Numbers In mathematics, a real number is a value that represents a quantity along a continuous line. The real numbers include all the rational numbers, such as the integer −5 and the fraction 4/3, and all the irrational numbers such as √2 (1.41421356... the square root of two, an irrational algebraic number) and π (3.14159265..., a transcendental number). Real numbers can be thought of as points on an infinitely long line called the number line or real line, where the points corresponding to integers are equally spaced. Any real number can be determined by a possibly infinite decimal representation such as that of 8.632, where each consecutive digit is measured in units one tenth the size of the previous one. These descriptions of the real numbers are not sufficiently rigorous by the modern standards of pure mathematics. The discovery of a suitably rigorous definition of the real numbers — indeed, the realization that a better definition was needed — was one of the most important developments of 19th century mathematics. The currently standard axiomatic definition is that real numbers form the unique Archimedean complete totally ordered field (R,+,·,<), up to isomorphism, whereas popular constructive definitions of real numbers include declaring them as equivalence classes of Cauchy sequences of rational numbers.
Know More About :- History of whole Numbers
Math.Edurite.com
Page : 1/3
HISTORY Vulgar fractions had been used by the Egyptians around 1000 BC; the Vedic "Sulba Sutras" ("The rules of chords") in, ca. 600 BC, include what may be the first 'use' of irrational numbers. The concept of irrationality was implicitly accepted by early Indian mathematicians since Manava (c. 750–690 BC), who were aware that the square roots of certain numbers such as 2 and 61 could not be exactly determined.[5][verification needed] Around 500 BC, the Greek mathematicians led by Pythagoras realized the need for irrational numbers, in particular the irrationality of the square root of 2. The Middle Ages brought the acceptance of zero, negative, integral, and fractional numbers, first by Indian and Chinese mathematicians, and then by Arabic mathematicians, who were also the first to treat irrational numbers as algebraic objects,[6] which was made possible by the development of algebra. Arabic mathematicians merged the concepts of "number" and "magnitude" into a more general idea of real numbers.[7] The Egyptian mathematician Abū Kāmil Shujā ibn Aslam (c. 850–930) was the first to accept irrational numbers as solutions to quadratic equations or as coefficients in an equation, often in the form of square roots, cube roots and fourth roots. In the 16th century, Simon Stevin created the basis for modern decimal notation, and insisted that there is no difference between rational and irrational numbers in this regard. In the 17th century, Descartes introduced the term "real" to describe roots of a polynomial, distinguishing them from "imaginary" ones. In the 18th and 19th centuries there was much work on irrational and transcendental numbers. Johann Heinrich Lambert (1761) gave the first flawed proof that π cannot be rational;[citation needed] Adrien-Marie Legendre (1794) completed the proof, and showed that π is not the square root of a rational number. Paolo Ruffini (1799) and Niels Henrik Abel (1842) both constructed proofs of Abel–Ruffini theorem: that the general quintic or higher equations cannot be solved by a general formula involving only arithmetical operations and roots. | 770 | 3,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-43 | latest | en | 0.941332 |
https://socratic.org/questions/how-do-you-solve-2-x-7#240233 | 1,638,007,061,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358153.33/warc/CC-MAIN-20211127073536-20211127103536-00238.warc.gz | 603,949,162 | 5,893 | # How do you solve 2^x = 7?
Mar 16, 2016
$x \approx 2.81$
#### Explanation:
$1$. Since the left and right sides of the equation do not have the same base, start by taking the log of both sides.
${2}^{x} = 7$
$\log \left({2}^{x}\right) = \log \left(7\right)$
$2$. Use the log property, ${\log}_{\textcolor{p u r p \le}{b}} \left({\textcolor{red}{m}}^{\textcolor{b l u e}{n}}\right) = \textcolor{b l u e}{n} \cdot {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{m}\right)$, to simplify $\log \left({2}^{x}\right)$.
$x \log \left(2\right) = \log \left(7\right)$
$3$. Solve for $x$.
$x = \log \frac{7}{\log} \left(2\right)$
$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x \approx 2.81 \textcolor{w h i t e}{\frac{a}{a}} |}}}$ | 313 | 777 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2021-49 | latest | en | 0.514852 |
https://flippiefanus.wordpress.com/tag/recovery-rate/ | 1,674,818,591,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494976.72/warc/CC-MAIN-20230127101040-20230127131040-00395.warc.gz | 281,958,971 | 17,535 | ## COVID-19 numbers
In a time like this, there is much uncertainty. It may be the first time that the world is experiencing such a pandemic, but similar situations have been encountered before. People have lots of questions and they are looking for information. Unfortunately, there is a lot of misleading information available. It seems that most people don’t check such information. They just believe what they read. It is for this reason tragic when “official” websites provide misleading information.
One example is the incorrect representations of numerical quantities about the COVID-19 pandemic. Two such quantities that is for obvious reasons quite important for people is the recovery rate and the mortality rate. They give an estimate, based on the currently available statistics, of the chances for a person to recover or die from COVID-19, given that the person has contracted the decease.
The statistics, which is generally available (see for example Wikipedia), consist of three numbers provided for every country on a daily basis. These number are: the number of confirmed cases (CC), the number of deaths (D) and the number of recoveries (R). For example, on 14 June 2020, the quoted number for the whole world are:
CC = 7 763 921
D = 429 632
R = 3 682 950
From these numbers, one can now compute the mortality rate and the recovery rate. The mistake that one often finds is that these rates are computed by dividing D or R by CC. That gives a misleading result, because CC also contains the currently active cases that does not form part of D or R yet.
The correct way to compute the mortality rate is to divide D by the sum of D and R and multiply the result by 100 to express it as a percentage. In a similar way, the recovery rate is obtained to dividing R by the sum of D and R and then multiply it by 100. You will note that when you add up the mortality rate and the recovery rate you get 100%. That makes sense, because one would either die or recover. There is no other option.
Applying these calculations to the above statistics for the world, we find the mortality rate to be just over 10% and the recovery rate just under 90%. These rates differ from country to country. For instance, the current morality rate in the USA is about 15%, while for SA it is only 3.7%.
Why is it different for the different countries? This is an important question, because it affects people’s behavior. There are many possible reasons, including the age demographic of a country and the availability of medical facilities in the country. The mortality rates for different age groups are different: it increases for older people. If the number of active cases becomes too high, there may not be enough hospital beds and equipment to treat all those that need treatment. As a result, one can expect the morality rate to increase.
The government of a country needs to try and keep the number of active cases low enough so that those that need to treatment can get it. For that reason, they impose restrictions that are aimed to reduce the rate at which the virus is spreading. Restrictions may seem to be a violation of people’s freedom, but in this case it is necessary. However, a government can only do so much. If the people decide to ignore the regulations imposed by the government, because they want to exercise their freedom, then the virus would spread too fast, with the result that the number of active cases can increase above the level where the country would have enough medical facilities.
There are more numbers that are important, for instance the growth rate in the number of confirmed cases. That is a topic for another day. | 757 | 3,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-06 | latest | en | 0.950825 |
http://2048game.mobi/video/craps-principle.php | 1,582,510,941,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145869.83/warc/CC-MAIN-20200224010150-20200224040150-00397.warc.gz | 1,588,540 | 12,288 | 4692
# Craps principle
The Five Principle Bets Made at the Craps Table. The five main bets one makes at the craps table are these: Pass; Don't Pass; Come; Don't Come; Place; Let's explore each . In probability theory, the craps principle is a theorem about event probabilities under repeated iid trials. Let and denote two mutually exclusive events . If the game being played is craps, then this principle can greatly simplify the computation of the probability of winning in a certain the craps principle applies.
## Craps principle: Wikis
This page was last edited on 27 October , at The pair of 6 pips resembles a pair of boxcars on a freight train, in modern parlance, it refers to such a roll in any game involving 6 sided dice which are marked with pips. For example, rolling an honest die produces one of six possible results, one collection of possible results corresponds to getting an odd number. This can create a cohesive look to a website. Thanks for taking the time to explain C.
• The casino is the most common place in which suicides are committed
• The annual profit from the gaming industry in the US is 18 billion dollars.
## The Five Principle Bets Made at the Craps Table
In probability theory , the craps principle is a theorem about event probabilities under repeated iid trials. Invalid response "Math extension cannot connect to Restbase. By conditional probability ,. Then the craps principle gives the respective conditional probabilities of each player winning a certain repetition, given that someone wins i. In fact, the result is only affected by the relative marginal probabilities of winning Failed to parse MathML with SVG or PNG fallback recommended for modern browsers and accessibility tools: If the game is played repeatedly until someone wins, then the conditional probability above turns out to be the probability that the player wins the game.
If the game being played is craps , then this principle can greatly simplify the computation of the probability of winning in a certain scenario. Specifically, if the first roll is a 4, 5, 6, 8, 9, or 10, then the dice are repeatedly re-rolled until one of two events occurs:. For example, if the original roll was a 4, then the probability of winning is. This avoids having to sum the infinite series corresponding to all the possible outcomes:. The summation becomes an infinite geometric series:.
Retrieved from " https: Articles lacking sources from April All articles lacking sources Articles with invalid date parameter in template Probability and statistics Statistics Pages with math render errors. Namespaces Home Page Discussion. Views Read View source View history Help. This page was last edited Based on work by wikidoc anonymous user QmunkE. Privacy policy About wikidoc Disclaimers Mobile view.
SLEAZY CDC CAMPAIGN MISSTATES RISK OF AIDS Remember those TV ads featuring the Baptist minister's son, who said, "If I can get AIDS, anyone can". Одного ее взгляда достаточно, чтобы понять, что она точно знает, что нужно настоящему мужику. Yet public health officials, afraid they couldn't honestly generate support, deliberately frightened and deceived American taxpayers to get them to cough up the dough.
Мы расселись по местам. Jeff moved the knife vertically, straight up, slicing though her stomach.
### Details
There are four principles of design that we want you to get under your belt. Watch the video, read the stuff below, and suggestion: Creating contrast for elements means that discrete elements stand out. Contrast is all about making things stand out. A call to action that is red on a black and white website will stand out in a big way.
Users know where to look first, second, third, last. Repetition, for instance making a header and footer the same color, makes scanning a website easier. Bullet lists are easy to scan, right? Because the repetition of the little dots makes it so.
This can create a cohesive look to a website. Obviously, color and shapes are important with repetition. Columns within a page makes it easier to scan horizontally. Newspapers use this to great effect. Aligning a whole bunch of elements with one another makes them scan faster. Alignment makes things easier to read. Proximity means that things are associated with one another — or not.
The closer things are, the more they are associated. The farther things are away from one another, the less they are associated. This helps immensely in navigation. Tell you what, just watch the C. Data Visualisation — The Power of Infographics Morgan Edwards March 21, […] ICTs enable our students to look at things in new ways and create things that would otherwise be impossible, and infographics are just one example of this.
Infographics also require the use of programs such as Photoshop and Illustrator, along with a clear knowledge of how to apply design principles such as the CRAP principles.
## Craps chips sale
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## Video
### Free Roulette
The thrill of watching the spinning red and black Roulette wheel has long served to grip many avid gamblers around the g... | 1,197 | 5,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2020-10 | latest | en | 0.93477 |
https://alternavision.com/5oqrcf/8ab246-what-is-the-opposite-of-a-factor-in-math | 1,631,917,702,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055808.78/warc/CC-MAIN-20210917212307-20210918002307-00256.warc.gz | 150,009,468 | 11,283 | # what is the opposite of a factor in math
How many right angels can be in an obtuse triangle? is an unsaturated acid. If a number is not prime, then it is composite. It can be expressed by the formula y=a(1-b) x wherein y is the final amount, a is the original amount, b is the decay factor, and x … Factoring is the opposite of distributing. The centre of enlargement is (1,0) Opposite Numbers Calculator: Simply enter a postive or negative integer, and the calculator will determine the opposite number Solve your math problems using our free math solver with step-by-step solutions. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. Synonyms: agent, assignee, attorney… Antonyms: whole… Find the right word. This is a glossary of math definitions for common and important mathematics terms used in arithmetic, geometry, and statistics. Note that 6 is both a factor and a multiple of itself. A factor is a part or element that contributes to a result. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. What is 7,198 rounded to the nearest thousand? What is the perimeter of a heptagon with a side length of 14cm? So if 6 is divisible by 1, 2, 3 and 6, these positive integers are all factors of 6. A factor of n is a positive integer that n is divisible by with no remainder. How many right angels can be in an obtuse angle? An expression that represents repeated multiplication of the same factor is called a power. When distributing, you multiply a series of terms by a common factor. A math factor is a number you multiply with others to get to a desired number. expanding; when you factor, you are taking common factors and separating an equation. Such as −6 and +6 Or −150 and +150 When we add opposite numbers we get 0 Example: adding +6 and −6 gives 0 Numbers that are in opposite positions on the number line. According to the stuff explained on "Divisors and factors", 7 has no factor. 2 1. monica. Sal simplifies and states the domain of (x^2-36)/(6-x). Copyright © 2020 Multiply Media, LLC. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Find 134 opposite words and antonyms for factor based on 17 separate contexts from our thesaurus. A number and its additive inverse equal zero when added. What is 7,198 rounded to the nearest thousand? Need antonyms for divisor?Here's a list of opposite words from our thesaurus that you can use instead. Opposites : Opposite number is a number which having same value but different sign. Enter a polynomial, or even just a number, to see its factors. What is a Factor… How many right angels can be in an obtuse angle? Note: When the common factor is less than 1 the transformation is called a compression. If you’re walking to school and realize you forgot your math homework, turn around and walk in the opposite direction to go back and get it. Mickey L. Lv 4. Factors of composite numbers. What is the solutions to y plus 3 squared minus 81? If you only fly budget airlines and refuse to stay anywhere but youth hostels, then money is probably a big factor in your travel plans. How long will the footprints on the moon last? 1 0. What is the perimeter of a heptagon with a side length of 14cm? 1 decade ago. The exponent corresponds to the number of times the base is used as a factor. The number 5 is called the base, and the number 2 is called the exponent. When factoring, you seek to find what a series of terms have in common and then take it away, dividing the common factor out from each term. The main basic facts encountered in math are "whole-number" basic facts, in particular multiplication, addition, division and subtraction. The Princeton's WordNet (2.50 / 4 votes) Rate these antonyms: factor (noun) anything that contributes causally to a result "a number of factors determined the outcome" Synonyms: Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. With a negative scale factor the number part of it tells us if the shape is to be made bigger or smaller, while the negative sign tells us that the image and object are on opposite sides of the centre of enlargement. what shape that has 4 sides and 4 angels? For example, 3 and 6 are factors of 12 because 12 ÷ 3 = 4 exactly and 12 ÷ 6 = 2 exactly. The opposite number for n is written as −n. Factor definition, one of the elements contributing to a particular result or situation: Poverty is only one of the factors in crime. Consider \left(x+3\right)\left(x-3\right). Copyright © 2020 Multiply Media, LLC. What Does the Opposite of a Number Mean? Examples: if you mean to factor a polynomial (verb), then the opposite is FOIL if you mean a factor of an integer (noun), then the opposite is multiple 1 0 Mickey L Lv 4 1 decade ago Product 2 1 monica 1 decade ago multiple. Favorite Answer. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. Definition 1: Opposite number or additive inverse of any number (n) is a number which, if added to , results in 0, the identity element of addition. Definition 3: The opposite of a number is the number on the other side of the 0 number line, and the same distance from 0. So, prime number has no factors and it will have only two divisors. Here you can learn what is the opposite of a number as use our step-by-step calculator to find the opposite of any real number or fraction'. Another word for factor. They are 1 and the number itself. One operation that is frequently used to form new sets from old ones is called the union. Factor, in mathematics, a number or algebraic expression that divides another number or expression evenly—i.e., with no remainder. Think of each term as a numerator and then find […] Opposite means “other." Mean: The mean is the same as the average. What is the solutions to y plus 3 squared minus 81? Whatever numbers you use to multiply together to get to a number of your choice are factors. Who is the longest reigning WWE Champion of all time? if you mean a factor of an integer (noun), then the opposite is multiple. 1 0. smert96. The other factors of 12 are 1, 2, 4, and 12. Factoring (called "Factorising" in the UK) is the process of finding the factors: It is like "splitting" an expression into a multiplication of simpler expressions. Suppose you have a preimage of a polygon with a side C D that is 9 c m long and a scale factor of 1 3 . Find more ways to say factor, along with related words, antonyms and example phrases at Thesaurus.com, the world's most trusted free thesaurus. Why don't libraries smell like bookstores? Why don't libraries smell like bookstores? Simplify the rational expression and state the domain. factor synonyms, factor pronunciation, factor translation, English dictionary definition of factor. When factoring, you seek to find what a series of terms have in common and then take it away, dividing the common factor out from each term. when you expand, you are multiplying through (thereby doing the opposite of separating) so 5(x+3) is the factored version or 5x + 15 . When distributing, you multiply a series of terms by a common factor. If you roll a regular 6-sided cube 1200 times about how many times will you get 2? Free online factoring calculator that factors an algebraic expression. n. See inverse. I am studying math, and there's a term I don't know how to express, I just know that is the opposite of the operation of distribution of terms in an algebraic expression. multiple. Factoring is the opposite of distributing. Due to this property it is said as 'Additive Inverse'. When did organ music become associated with baseball? Also learn the facts to easily understand math glossary with fun math worksheet online at SplashLearn. Need antonyms for factor?Here's a list of opposite words from our thesaurus that you can use instead. 1 decade ago. When two opposite numbers are added it will result in 0. Solve your math problems using our free math solver with step-by-step solutions. This new side will be opposite the center of dilation from the preimage, as will all the other sides of the polygon since the scale factor was negative. In common usage, the word union signifies a bringing together, such as unions in organized labor or the State of the Union address that the U.S. President makes before a joint session of Congress. For example, 3 and 6 are factors of 12 because 12 ÷ 3 = 4 exactly and 12 ÷ 6 = 2 exactly. This diagram shows the blue shaded shape enlarged by a scale factor of -2. Factor, in mathematics, a number or algebraic expression that divides another number or expression evenly—i.e., with no remainder. Find 5 opposite words and antonyms for divisor based on 2 separate contexts from our thesaurus. In the near term, a more likely contributing factor would be that consumers would have redirected funds for holiday trips and travel to sending holiday gifts. and 5x + 15 is the expanded version of 5(x+3). Only \displaystyle{C}. Few more examples 5^3 =5X5X5= 125 2^4 = 2X2X2X2 =16 If any One last example: factor 2x 3 - 6x 2 - 80x.This one looks quite a bit different. A "basic fact" in math is defined as any mathematical number, fact or idea instantly recalled without resorting to strategies, according to NZCER.org. what shape that has 4 sides and 4 angels? Antonyms for factor ˈfæk tər This page is about all possible antonyms and opposite words for the term factor. You may have to use Map, or apply Factor again, to reach other levels. Who is the longest reigning WWE Champion of all time? Factor, in the design of experiments, a phenomenon presumed to affect an experiment Human factors , a profession that focuses on how people interact with products, tools, or procedures Sun protection factor , a unit describing reduction in transmitted ultraviolet light How long will the footprints on the moon last? If a number has more than two divisors, then it has factor or factors and it is composite. Factoring is also the opposite of Expanding: Factor [poly, GaussianIntegers->True] factors allowing Gaussian integer coefficients. Opposites : Opposite number is a number which having same value but different sign. What is the opposite of divisor? When two opposite numbers are added it will result in 0. Factor applies only to the top algebraic level in an expression. The math factor pairs of 18 are (1,18) , (2,9) , (3,6) . It means raised to the power (^) For eg:- 3^2 =3X3, here 3 is the base and 2 is the power of 3 (power is also called as exponent in mathematical term). American Heritage® Dictionary of the English In math, what is the opposite of the word "FACTOR"? When did organ music become associated with baseball? The 'opposite' status of a number that is not prime is a composite number.Except for 0 and 1, all numbers that are not prime (i.e. x 3?We don't know how to do that. What is the opposite of pity? You may be familiar with the word divisor - for the purpose of the GMAT, factors and divisors are quite similar, with a slight difference: a divisor can be positive or negative, while a factor is always positive. the opposite of pity is like when you say that is amazing for example: that's a pity that you failed your math test, the opposite is that is awesome that you past the math test. All Rights Reserved. Translation means moving an object without rotation, and can be described as “sliding”. All Rights Reserved. Which Institute is learned mobile repairing course in 40 days? Opposite (mathematics) synonyms, Opposite (mathematics) pronunciation, Opposite (mathematics) translation, English dictionary definition of Opposite (mathematics). Due to this property it is said as 'Additive Inverse'. Define factor. 3 1: 3 to the first power: 3: 4 2: How many right angels can be in an obtuse triangle? Example: 2 is a factor of 6, which means that 6 is a multiple of 2 - 6=2⋅3. make sense? If you roll a regular 6-sided cube 1200 times about how many times will you get 2? Product. One last important bit, which again demonstrates the fact that factors and multiple are opposite terms: If x is a factor of 6 then 6 is a multiple of x. In mathematics, exponential decay describes the process of reducing an amount by a consistent percentage rate over a period of time. ... Common Factors: ... Logarithm is the opposite of exponentiation. See more. depends on your definition of "FACTOR" if you mean to factor a polynomial (verb), then the opposite is FOIL. Definition 2: The opposite of a number is its additive inverse, which means that its sign is reversed. Which Institute is learned mobile repairing course in 40 days? What is the opposite of factor? Definition of Factor explained with real life illustrated examples. Factor: a person who acts or does business for another. Add up a series of numbers and divide the sum by the total number of values to find the mean. In describing transformations of graphs, some textbooks use the formal term “translate”, while others use an informal term like “shift”.Our first question comes from 1998:These examples represent the three main transformations: translation (shifting), reflection (flipping), and dilation (stretching). Signup for detailed step-by-step solutions. Let's see if we can start with the domain part of the question, if … Think of each term as a numerator and then find […] A factor of -2 addition, division and subtraction american Heritage® Dictionary of factors! Factor translation, English Dictionary definition of factor '' if you mean to a..., trigonometry, calculus and more step-by-step solutions the factors in crime are all factors of 12 12., 2, 3 and 6 are factors equal zero when added, then the opposite of....: this is a factor transformation is called the exponent opposite numbers are added it will result in 0 (... 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Reigning WWE Champion of all time to get to a number is a glossary of definitions. Are ( 1,18 ), ( 3,6 ), assignee, attorney…:! Examples 5^3 =5X5X5= 125 2^4 = 2X2X2X2 =16 if any a factor 6. In math are whole-number '' basic facts, in mathematics, a number of choice! Use instead definition of factor explained with real life illustrated examples if any factor... Whole-Number '' basic facts encountered in math are whole-number '' basic facts, in mathematics a. Of 5 ( x+3 ) of 18 are ( 1,18 what is the opposite of a factor in math, then the opposite is multiple )... Your definition of factor explained with real life illustrated examples True ] factors Gaussian. 17 separate contexts from our thesaurus that you can use instead are 1, 2, 3 and are... X+3\Right ) \left ( x-3\right ) get to a result factor translation, English Dictionary definition of factor explained real. 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Because 12 ÷ 3 what is the opposite of a factor in math 4 exactly and 12 ÷ 6 = exactly! | 5,259 | 22,242 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-39 | latest | en | 0.907833 |
https://quizizz.com/admin/quiz/5e25dd38a9521f001b77e9e0/algebra-2-unit-9-review | 1,642,863,809,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303864.86/warc/CC-MAIN-20220122134127-20220122164127-00650.warc.gz | 542,571,682 | 7,266 | Algebra 2 Unit 9 Review
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• Question 1
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Q.
Graph the function
f(x) = 6(1/2)x - 5
• Question 2
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Q. Describe the transformation, Domain & Range of the function:
g(x) = (1/6)x-3 + 5
Decay; 3 Right, 5 Up
D: ℜ
R: y > 5
Decay; 5 Left, 3 Up
D: ℜ
R:y > 3
Growth; 3 Left 5 Up
D: ℜ
R: y < 5
Decay; 5 Right, 3 Up
D: ℜ
R: y > 3
• Question 3
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Q. Describe the transformation, Domain & Range:
f(x) = (1/3) x+2
Growth; Up 2
D: ℜ
R: y > 2
Decay; 2 Left
D: ℜ
R: y > 0
Decay; Right 2
D: ℜ
R: y > 0
Decay; Up 3
D: ℜ
R: y > 2
• Question 4
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Q. Describe Domain & Range:
f(x)= - (3/4)x+1 - 7
D: ℜ
R: y > -7
D: ℜ
R: y > 7
D: ℜ
R: y < -7
D: ℜ
R: y < 7
• Question 5
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Q. Describe the transformation:
g(x) = 2x-4 + 1
Growth; Right 4
Up 1
Decay; Right 4
Up 1
Growth; Left 4
Up 1
Decay; Left 4
Up 1
• Question 6
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Q. The production of a new car model in the year 2000 was 1.35 million units. In the year 2010, it rose to 3.7 million. Write an exponential function to model the cars y produced after x years. | 553 | 1,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2022-05 | latest | en | 0.534489 |
https://math.stackexchange.com/questions/2332013/trouble-understanding-baby-rudin-2-38 | 1,660,078,250,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571086.77/warc/CC-MAIN-20220809185452-20220809215452-00505.warc.gz | 358,824,255 | 64,043 | # Trouble Understanding Baby Rudin 2.38
Theorem 2.38 states the following: If $\{I_n\}$ is a sequence of intervals in $R^1$, such that $I_n$ contains $I_{n+1}$ $(n=1,2,3,…),$ then the infinite intersection of the sets $\{I_n\}$ is not empty.
In the proof, Rudin then defines $\{I_n\} = [a_n, b_n]$, and set $E$ as the set of all $a_n$. Letting $x =$ sup $E$ and arbitrary $m, n \in \mathbf{N}$, the following is true $$a_n \leq a_{m+n} \leq b_{m+n} \leq b_{m}.$$
He then establishes the fact that $x \leq b_m$ for any $m \in \mathbf{N}$. This is the step that I fall short on. Why is this statement true?
• From the above $a_n\le b_m$, so $b_m$ is an upper bound of $E$, and so $\ge$ the least upper bound of $E$. Jun 22, 2017 at 6:18
• Oh that was easy Jun 22, 2017 at 6:20
Suppose there exists $n\in\mathbb N$ such that $x > b_n$. Letting $\varepsilon = x-b_n>0$, there then exists $m\in\mathbb N$ such that $x-a_m < \varepsilon/2$. But then $a_m-b_n = a_m-x+x-b_n>\varepsilon/2$, and so $a_m > b_n$, which is impossible. | 380 | 1,028 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2022-33 | latest | en | 0.780536 |
https://algebra-class-ecourse.com/question/describe-how-to-solve-problems-for-the-area-of-a-reduced-or-enlarged-figure-when-the-scale-facto-8875926-72/ | 1,637,971,794,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358074.14/warc/CC-MAIN-20211126224056-20211127014056-00604.warc.gz | 156,421,830 | 12,768 | Describe how to solve problems for the area of a reduced or enlarged figure when the scale factor and dimensions of the original figure are
Question
Describe how to solve problems for the area of a reduced or enlarged figure when the scale factor and dimensions of the original figure are given.
0
1. To solve problems for the area of a reduced or enlarged figure, first square the scale factor. Then use the given dimensions to find the area of the original figure. Multiply the square of the scale factor by the area of the original to get the reduced or enlarged figure.
2. We are going to define the following variables:
A: Area of the original figure.
k: scale factor.
We have that the area of the new figure (reduced or enlarged, depends on the value of k) is:
A ‘= k ^ 2 * A
Example: rectangle of dimensions (2) * (3) A factor of k = 2 is extended. Find the new area.
The original area is:
A = (2) * (3) = 6
The new area will be:
A ‘= (2 * 2) * (2 * 3) = 4 * 6 = 24
Or equivalently:
A ‘= k ^ 2 * A
A ‘= (2) ^ 2 * (6) = 4 * 6 = 24 | 287 | 1,042 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2021-49 | latest | en | 0.897419 |
https://www.coursehero.com/file/5920135/21-Cost-Curves-Solutions/ | 1,516,505,706,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889917.49/warc/CC-MAIN-20180121021136-20180121041136-00219.warc.gz | 985,110,398 | 24,033 | 21. Cost Curves - Solutions
# 21. Cost Curves - Solutions - Chapter 21 NAME Cost Curves...
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Unformatted text preview: Chapter 21 NAME Cost Curves Introduction. Here you continue to work on cost functions. Total cost can be divided into fixed cost, the part that doesn’t change as output changes, and variable cost. To get the average (total) cost, average fixed cost, and average variable cost, just divide the appropriate cost function by y , the level of output. The marginal cost function is the derivative of the total cost function with respect to output—or the rate of increase in cost as output increases, if you don’t know calculus. Remember that the marginal cost curve intersects both the average cost curve and the average variable cost curve at their minimum points. So to find the minimum point on the average cost curve, you simply set marginal cost equal to average cost and similarly for the minimum of average variable cost. Example: A firm has the total cost function C ( y ) = 100 + 10 y . Let us find the equations for its various cost curves. Total fixed costs are 100, so the equation of the average fixed cost curve is 100 /y . Total variable costs are 10 y , so average variable costs are 10 y/y = 10 for all y . Marginal cost is 10 for all y . Average total costs are (100 + 10 y ) /y = 10 + 10 /y . Notice that for this firm, average total cost decreases as y increases. Notice also that marginal cost is less than average total cost for all y . 21.1 (0) Mr. Otto Carr, owner of Otto’s Autos, sells cars. Otto buys autos for \$ c each and has no other costs. (a) What is his total cost if he sells 10 cars? 10 c . What if he sells 20 cars? 20 c . Write down the equation for Otto’s total costs assuming he sells y cars: T C ( y ) = cy . (b) What is Otto’s average cost function? AC ( y ) = c . For every additional auto Otto sells, by how much do his costs increase? c . Write down Otto’s marginal cost function: MC ( y ) = c ....
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21. Cost Curves - Solutions - Chapter 21 NAME Cost Curves...
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Ask a homework question - tutors are online | 590 | 2,462 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-05 | latest | en | 0.8905 |
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Solution: 39
Surface area is directly proportional to the combined length of all fish) (given in question)
First Area = 16 * 18
Therefore, 16* 18 = k * 24 (where k is the constant of proportionality)
Also, for the second case
36 * 13 = k * Combined Length
Therefore, Combined Length = (36 * 13) / k
= (36 * 13 * 24) / (16 * 18)
= 39.
Hence, the tank can support 39 inches of rainbow fish. | 128 | 415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2021-25 | longest | en | 0.833107 |
https://quant.stackexchange.com/questions/10323/summary-statistic-for-the-average-probability-of-default | 1,701,935,782,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100650.21/warc/CC-MAIN-20231207054219-20231207084219-00135.warc.gz | 535,468,687 | 42,008 | # Summary statistic for the average probability of default?
I have the following scenario:
Let $X_i$ denote the event where some institution $i$ 'defaults' (don't worry about the exact definition of a default here, it is not relevant to the question at hand). Now, I have 10 institutions in my sample and I have calculate the following probabilities:
$P(X_1 \cap X_2 \cap \cdots \cap X_9 | X_{10})$
$P(X_1 \cap X_2 \cap \cdots \cap X_{8} \cap X_{10} | X_{9})$
$\vdots$
$P(X_2 \cap \cdots \cap X_{9} \cap X_{10} | X_{1})$
In words, the above expressions represent the (joint) probability of default of the 'remaining' institutions given that a particular institution has defaulted.
To calculate these probabilities, I know the underlying probability distribution that describes the entire system, i.e., $p(x_1, x_2, \cdots. x_{10})$, so for example, calculating the probability $P(X_1 \cap X_2 \cap \cdots \cap X_9 | X_{10})$ just simply requires me to find $\frac{P(X_1 \cap \cdots \cap X_{10})}{P(X_{10})}$ where both the denominator and numerator can be calculated by integrating over certain regions of the probability density function $p(x_1, x_2, \cdots. x_{10})$.
My question is, I wish to find one summary value that describes the average probability of default of this system, e.g. say $P(X_1 \cap X_2 \cap \cdots \cap X_9 | X_{10})= 0.4$, $P(X_1 \cap X_2 \cap \cdots \cap X_{8} \cap X_{10} | X_{9})= 0.3$ etc, how can I "combine" this $0.4$, $0.3$, etc into one value that describes the "average" probability of default of this system? My initial method is just to take the arithmetic average of each conditional probability, but that is mathematically incorrect conditional probabilities aren't summable (except when they are conditioned on the same event). So are there any other measures/techniques I can use to somehow "combine" these single probabilities into "one" value?
Let me put this into context to make things more concrete.
Pretend each event is the event where a bank defaults. By default, I mean that the bank's assets drop below some pre-determined threshold. Then, say we have a sample of 10 banks. The entire "system" is the universe of these 10 banks. I want to find what is the "contribution" of each bank's default on the rest of the system, that is, given that one bank defaults, how does that affect the probability of default of the rest of the system (i.e., the remaining 9 banks). To do this, I have modeled the underlying (joint) asset distribution of this system of banks. Then, computing the probability $P(X_1 \cap X_2 \cap \cdots \cap X_9 | X_{10})$ represents the "contribution" of the default of the 10th bank on the rest of the system. Similarly, $P(X_1 \cap X_2 \cap \cdots \cap X_{8} \cap X_{10} | X_{9})$ represents the impact of the default of the 9th bank on the rest of the system. Now given I have computed the "contribution" of each bank's default, I want to find a single value that describes the average "contribution" of each bank's default on the rest of the system.
• What do you mean by the ""average" probability of default of this system"? What is wrong with the probability that all of them will default? Feb 21, 2014 at 19:53
• Sorry, my question was a bit vague, I have made an edit, putting it into context, hopefully that helps! Feb 21, 2014 at 19:58
• There is no right answer to this question since you haven't defined "average". Feb 21, 2014 at 20:03
• You need to explain more precisely what properties your "average" should have. Let us say you have two banks, each with a 10 percent chance of default, and they are either fully correlated or fully independent. What answers do you want? Feb 22, 2014 at 1:05
• To both above comments: Therein lies the problem, I'm not exactly sure what kind of properties I should have. Any suggestions? I know the correlation between each and every one of the banks. So if this "average" can incorporate those correlations, that would be good. Feb 22, 2014 at 6:54
I have an Idea perhaps it helps you a bit (even though it deviates somewhat from your original setup). Let's assume you know the "anaffected" default probabilities for each bank $P(X_1<=C_1), \dots, P(X_n<=C_n)$. (Here I assumed that bank $i$ defaults when it's value falls below a certain value $C_i$)
Now e.g. for bank $n$ you can calulate $P(X_1<=C_1|X_n<=C_n), \dots, P(X_{n-1}<=C_{n-1}|X_{n}<=C_{n})$. Thus the default probabilites conditioned on the default of bank $n$.
Now you could calculate $d_i(n)=P(X_1<=C_1|X_n<=C_n) - P(X_1<=C_1)$ for $i\in\{1,\dots , n-1\}$ to determine whether the default probability of $X_1$ increased or decreased after the default of $X_n$.
Now add some weights $(w_1, \dots , w_{n-1})$ bcause the default of one bank might be seen as having more impact than the default of another.
Finally by taking $\sum_{i=1}^{n-1} w_i d_i(n)$ you would get a number telling you how the default of $X_n$ "impacts" the system.
Here I still haven't used all the "correlation-information" into account (some of it is already contained in the probablities $P(X_1<=C_1|X_n<=C_n)$) This information (or at least some of it) could also be included into the weights $w_i$.
One way could be to calculate the average number of banks that would default if e.g. $X_i$ defaulted. To do that you will have to calculate the conditional expectation $\mathbb{E}[ \sum_{j=1,j\neq i}^{n} 1_{X_j<=C_j}|X_i<=C_i]$.
Setting
$w_i=\mathbb{E}[ \sum_{j=1,j\neq i}^{n} 1_{X_j<=C_j}|X_i<=C_i]$
will give you weights that incoporate more information on the correlation among the banks. | 1,577 | 5,571 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2023-50 | longest | en | 0.844675 |
https://mathproblems123.wordpress.com/2017/03/05/seemous-2017-problems/ | 1,531,944,094,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590329.25/warc/CC-MAIN-20180718193656-20180718213656-00074.warc.gz | 710,526,415 | 23,516 | Home > Uncategorized > SEEMOUS 2017 Problems
## SEEMOUS 2017 Problems
Problem 1. Let ${A \in \mathcal M_2(\Bbb{R})}$. Suppose
$\displaystyle A =\begin{pmatrix} a& b\\ c & d \end{pmatrix}$
satisfies
$\displaystyle a^2+b^2+c^2+d^2<1/5.$
Show that ${I+A}$ is invertible.
Problem 2. Let ${A,B \in \mathcal M_n(\Bbb{R})}$.
• a) Prove that there exists ${a>0}$ such that for every ${\varepsilon \in (-a,a),\varepsilon\neq 0}$ the matrix equation
$\displaystyle AX+\varepsilon X = B,\ X \in \mathcal M_n(\Bbb{R})$
has a unique solution ${X(\varepsilon) \in \mathcal M_n(\Bbb{R})}$.
• b) Prove that if ${B^2 = I_n}$ and ${A}$ is diagonalizable then
$\displaystyle \lim_{\varepsilon \rightarrow 0} \text{tr}(BX(\varepsilon)) = n-\text{rank}(A).$
Problem 3. Let ${f: \Bbb{R} \rightarrow \Bbb{R}}$ be a continuous function. Prove that
$\displaystyle \int_0^4 f(x(x-3)^2)dx = 2 \int_1^3 f(x(x-3)^2)dx$
Problem 4. a) Let ${n \geq 0}$ be an integer. Compute ${\displaystyle \int_0^1 (1-t)^n e^t dt}$.
b) Let ${k \geq 0}$ be a fixed integer and let ${(x_n)_{n \geq k}}$ be the sequence defined by
$\displaystyle x_n = \sum_{i=k}^n {i \choose k}\left(e-1-\frac{1}{1!}-\frac{1}{2!}-...-\frac{1}{i!}\right).$
Prove that the sequence converges and find its limit.
Hints: 1. This must be a joke 🙂 We know that if for a norm matrix we have ${\|A\|<1}$ then ${I-A}$ is invertible. The sum of squares is a norm, so ${\|A\|<1/5}$ implies that ${I+A}$ is invertible.
2. The equation is ${(A+\varepsilon I)X = B}$. We know that there are at most ${n}$ values ${\varepsilon_i}$ for ${\varepsilon}$ when ${A+\varepsilon I}$ is not invertible (related to the eigenvalues of ${A}$). We have two cases. If one of the eigenvalues is zero, then pick ${a}$ smaller than the distance to the closest eigenvalue to ${0}$. If ${0}$ is not an eigenvalue, then choose ${a}$ again smaller than the closest eigenvalue to ${0}$. For ${b}$ note that ${BX(\varepsilon) = B(A+\varepsilon I)^{-1}B}$ and since ${\text{tr}(XY)=\text{tr}(YX)}$ and ${B^2=I}$ we get that ${\text{tr}(BX(\varepsilon))= \text{tr}((A+\varepsilon I)^{-1})}$. Since ${A}$ is diagonalizable we conclude that
$\displaystyle \text{tr}(BX(\varepsilon)) = \sum_{i=1}^n \frac{1}{\lambda_i+\varepsilon}$
Thus, multiplying by ${\varepsilon}$ and letting ${\varepsilon}$ go to zero we obtain ${1}$ for zero eigenvalues and zero for non-zero ones. Thus, the limit counts the number of zero eigenvalues, which is precisely ${n-\text{rank}(A)}$ (since ${A}$ is diagonalizable).
3. Classic SEEMOUS stuff. Prove that this works when ${f(x) = x^n}$, then extend to polynomials, and then, by density, to continuous functions. Proving this for monomial might get involved, but initial computations show that we have the same terms when performing the integrations by parts on the two sides…
4. coming… | 973 | 2,836 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 47, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2018-30 | latest | en | 0.710771 |
http://www.algebra.com/algebra/homework/percentage/percentage.faq.question.201735.html | 1,369,331,837,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703662159/warc/CC-MAIN-20130516112742-00038-ip-10-60-113-184.ec2.internal.warc.gz | 311,448,366 | 4,541 | # SOLUTION: If \$8,000.00 is deposited in a bank that compounds interest anually at 5%, then what is the interest paid at the end of the second year?
Algebra -> Algebra -> Percentages: Solvers, Trainers, Word Problems and pie charts -> SOLUTION: If \$8,000.00 is deposited in a bank that compounds interest anually at 5%, then what is the interest paid at the end of the second year? Log On
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Algebra: Percentage and Pie Charts Solvers Lessons Answers archive Quiz In Depth
Click here to see ALL problems on percentage Question 201735: If \$8,000.00 is deposited in a bank that compounds interest anually at 5%, then what is the interest paid at the end of the second year?Answer by stanbon(57347) (Show Source): You can put this solution on YOUR website!If \$8,000.00 is deposited in a bank that compounds interest anually at 5%, then what is the interest paid at the end of the second year? -------- A(t) = P(1+(r/n))^(nt) A(2) = 8000(1+(0.05)/2))^(1*2) A(2) = 8000(1+0.025)^2 A(2) = \$8405.00 ---------------------- Amount of interest earned is \$405.00 ============================================== Cheers, Stan H. | 344 | 1,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2013-20 | latest | en | 0.893592 |
https://www.gradesaver.com/textbooks/math/other-math/thinking-mathematically-6th-edition/chapter-10-geometry-10-3-polygons-perimeter-and-tessellations-exercise-set-10-3-page-636/19 | 1,537,671,069,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158958.72/warc/CC-MAIN-20180923020407-20180923040807-00287.warc.gz | 759,065,459 | 12,972 | # Chapter 10 - Geometry - 10.3 Polygons, Perimeter, and Tessellations - Exercise Set 10.3 - Page 636: 19
perimeter = 18 yards
#### Work Step by Step
RECALL: (1) The perimeter of a polygon is equal to the sum of the lengths of all its sides. (2) An equilateral triangle has three side with equal lengths. This means that the two unknown side lengths of the triangle are both $6$ yards. Add the lengths of all the sides to find the perimeter: perimeter = $6+6+6= 18$ yards
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 165 | 634 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2018-39 | longest | en | 0.906677 |
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# plusmul-need an easy tutorial
0 can somebody post an easy tutorial/editorial for the snackdown elimination round 2k17 problem:plusmul I found the editorials tough to understand. asked 23 Jun '17, 22:59 183●8 accept rate: 14%
4 I havn't read the previous editorial, but i will try to explain my approach i think it's an easy approach. So let's start, we have n elements in an array. Now, let dp[i] be the required result as asked in the question for first i (just like for n) elements of the array. Now we will add the next element ( a[i+1] ), and will see how the result will be affacted, so we have to calculate dp[i + 1]. Before this it should be clear that there will be 2 ^ (i - 1) total combinations taking i elements all together because we will have two options ( * and +) for each (i - 1) places. That means dp[i] is the summation of all these 2 ^ (i - 1) combinations. Now let's calculate dp[i + 1], we now included the next element a[i + 1] and for this either we can add ( '+' ) or multiply ( '*' ) as per rules given in question. dp[i + 1] = dp[i] + a[i + 1] * (2 ^ (i - 1)) + dp[i] * a[i + 1] This equation is not entirely correct because the last term we have multiplied the element a[i + 1] to dp[i]. For example: lets the current elements of array be 2, 3, 5, 8, 9. All 2 ^ 3 = 8 combinations from dp[4] (taking first 4 elements 2, 3, 5, 8) are given as : 2 + 3 + 5 + 8 2 + 3 + 5 * 8 2 + 3 * 5 + 8 2 + 3 * 5 * 8 2 * 3 + 5 + 8 2 * 3 + 5 * 8 2 * 3 * 5 + 8 2 * 3 * 5 * 8 So basically, dp[i] is the sum of all these values. Now if we multiply a[i + 1] ie 9 to dp[i] then we will get summation of all these terms (2 + 3 + 5 + 8) * 9 (2 + 3 + 5 * 8) * 9 (2 + 3 * 5 + 8) * 9 (2 + 3 * 5 * 8) * 9 (2 * 3 + 5 + 8) * 9 (2 * 3 + 5 * 8) * 9 (2 * 3 * 5 + 8) * 9 (2 * 3 * 5 * 8) * 9 But we only need these 8 combinations 2 + 3 + 5 + 8 * 9 2 + 3 + 5 * 8 * 9 2 + 3 * 5 + 8 * 9 2 + 3 * 5 * 8 * 9 2 * 3 + 5 + 8 * 9 2 * 3 + 5 * 8 * 9 2 * 3 * 5 + 8 * 9 2 * 3 * 5 * 8 * 9 So subtracting this from the above one, we will get the extra terms like (2 + 3 + 5) * 8 (2 + 3) * 8 (2 + 3 * 5) * 8 (2) * 8 (2 * 3 + 5) * 8 (2 * 3) * 8 (2 * 3 * 5) * 8 which we have to subtract if u rearrange this u can get this: 2 2 + 3 2 * 3 2 + 3 + 5 2 + 3 * 5 2 * 3 + 5 2 * 3 * 5 all terms multiplied with 8 at end. and then u can say that the first term is dp[1]. the summation of 2nd and 3rd term is dp[2]. the summation of last 4 terms is dp[3]. hence the overall summation is dp[1] + dp[2] + dp[3] multiplied with 8 (that is 9 - 1) at end. So, we have to subtract the term sum_dp[3] * (9 - 1) where sum_dp[3] = dp[1] + dp[2] + dp[3]. In general we have to subtract the term sum_dp[i - 1] * (a[i + 1] - 1) that's it. where sum_dp[i] = dp[1] + dp[2] + dp[3] + ... + dp[i]. hence the one line dp equation will be now, dp[i + 1] = dp[i] + a[i + 1] * (2 ^ (i - 1)) + dp[i] * a[i + 1] - sum_dp[i - 1] * (a[i + 1] - 1) That's it. Here u can see my solution based on this approach : link. And sorry for bad English. :) answered 24 Jun '17, 00:09 116●3 accept rate: 16% how did u get extra terms? (24 Jun '17, 01:16) 1 I have updated it. Hope this helps. (24 Jun '17, 03:20) yup thanks a lot.. (24 Jun '17, 11:25)
4 This is what I did: I'll explain with an example, n=4, arr[]={1,2,3,4}, the expressions corresponding to this example are: 1 + 2 + 3 + 4 1 + 2 + 3 * 4 1 + 2 * 3 + 4 1 + 2 * 3 * 4 1 * 2 + 3 + 4 1 * 2 + 3 * 4 1 * 2 * 3 + 4 1 * 2 * 3 * 4 We need to find the sum of all these expressions, one thing that you need to observe is that all of these expressions are made up of several multiplication terms (group of numbers with only '*' in between them and '+' at the beginning and the end of the group) that are added up. Eg. 3 * 4 appears in the 2nd and the 6th expression among the ones I mentioned above. Now, if we are able to count how many times each term appears in the expression, we can multiply the value of each term with the number of times it appears, and then summing this up for all possible terms will give us the answer. Notice that every possible term is a continuous sequence of numbers in the input array. So, there will be N*(N+1)/2 terms. This can be done in O(N^3) or O(N^2), and both these ways are not effecient. But the idea itself can be used with a slight modification. First of all, lets see how to count the number of times some term appears in the expressions. I'll explain this with an example: Suppose that our input is {1,2,3,4,5,6}, and we need to count the number of times the term 2 * 3 appears. The expressions where 2 * 3 appears as a separate term will be of this form: 1 + 2 * 3 + 4 _ 5 _ 6 ( '_' means that the position can be filled with any symbol). So, the number of times this term appears will be the number of ways in which all the positions with '_' can be filled. In this case, its 2^2 because there are 2 positions and each position has 2 choices. The number of such positions for each term can be easily calculated. If the term is made up of 'k' numbers, then there will be (N-k-2) positions. This is because there will be N-1 positions for symbols in an expression with N numbers. Since our term has 'k' numbers in it, the k-1 symbols between them will be fixed to * and two more symbols, the one before and after the term will be fixed to +. So the remaining positions will be = N-1 - (k-1+2) = N-k-2. Now, as I mentioned, calculating this for every term will be really slow, so we need to do something better. Observe that the terms ending at position 'i' are obtained by multiplying the terms ending at position 'i-1' by the number at position 'i'. I will explain this with an example : N=5 Arr={1,2,3,4,5}, consider the terms ending at position 3 (1-based indexing) the terms will be: 3 3 * 2 3 * 2 * 1 Now, take a look at the terms ending at position 4: 4 4 * (3) 4 * (3 * 2) 4 * (3 * 2 * 1) Observe that, except the first term, all the others are obtained by multiplying the terms at the previous position with 4. Lets also take a look at the counts of the number of times each of these terms appears: Terms ending at position 3: 3 : 2^2 3 * 2 : 2^1 3 * 2 * 1 : 2^1 (Terms containing the first/last number will have N-K-1 symbols to fix because there is no symbol before/after those terms) Similarly, terms ending at position 4: 4 : 2^2 4 * (3) : 2^1 4 * (3 * 2) : 2^0 4 * (3 * 2 * 1) : 2^0 Observe that every term ending at position 4 (except the first) is obtained by multiplying 4 to some term ending at position 3, and its count is same as the count of that term divided by 2. This is always true because the only difference between the terms ending at positions 3 and 4 is that an additional symbol is fixed to '*'. Now, if we define ans[i] to be the sum of all the terms ending at position 'i', then ans[i] will be given by multiplying the previous answer with the number at this position, and then dividing by 2. You will have to consider the term containing the single number ending at this position separately because it does not depend on the terms ending at the previous position. So, ans[i] = (ans[i-1]*arr[i])/2 + arr[i]*(2^(N-3)) (Becomes 2^(N-2) if i=1 or i=N) The final answer will be sum of ans[i] for i from 1 to N. answered 24 Jun '17, 06:26 1.4k●11 accept rate: 28%
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University of Maryland at College Park Dept. of Aerospace Engineering ENAE 432: Aerospace Control Systems Problem Set #7 Issued: 26 Mar. 2011 Due By: 1 Apr 2011 Question 1: Use Matlab to obtain Bode diagrams for the system G ( s ) below when K = 2 , 8 and 20. Use these to sketch by hand the corresponding polar plots for each case (three different plots). Accurately label the intersections with the unit circle and real axis. Hint: right click on the Bode diagrams to have Matlab show you the crossover frequencies and margins. Alternately, you can use the margin command to get the Bode diagrams with both margins displayed. G ( s ) = K ( s/ 10 + 1) 3 Question 2: Repeat Question #1 for the system G ( s ) = 1 + τs s ( s + 1) 2 when τ = 10 and τ = 0 . 1. Label also the point where the diagram crosses the imaginary axis in each case (if applicable). Question 3: Repeat Question #2 for the system G ( s ) = 1 + τs 50 s 2 ( s + 1) 2 Use the same values for τ .
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Question 4: Use the nichols command in Matlab to draw the Nichols chart for the two systems in
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Unformatted text preview: Question #2. Show and label on this graph the gain and phase margins of the system. Question 5: Suppose the denominator of the transfer function for a physical system T ( s ) can be written as 1 + L ( s ), where L ( s ) is a different transfer function. Prove the following: if the polar diagram of L ( jω ) passes through the -1 point in the complex plane, then T ( s ) has a pole on the imaginary axis. Hint: the poles of T ( s ) here are the solutions of 1 + L ( s ) = 0. Question 6: Suppose that L ( s ) = 8 ( s/ 10 + 1) 3 Use the results from Question #5 and the Bode diagrams of L ( s ) to show that T ( s ) has a pole on the imaginary axis, and determine the specific location of this pole. Note: this can be easily done by explicit factorization, but to receive credit you must show how the results of Question #5 and the Bode diagrams lead to the desired information....
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https://mathematica.stackexchange.com/questions/125693/why-does-this-whole-number-have-a-recurring-value?noredirect=1 | 1,656,516,813,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103640328.37/warc/CC-MAIN-20220629150145-20220629180145-00792.warc.gz | 445,714,741 | 61,413 | # Why does this whole number have a recurring value? [duplicate]
I have a math question which that may seem simple to some but not to me if you where to do the calculation 12.30 * 12 it would equal 147.60 but why is that when I show the answer to more then two decimal place it has a 2 at the end 147.60000000000002 where did the 2 come from if 12.30 is a whole number ?
Again Im sorry if this sounds dumb but Im curious and thought this is the best place to ask.
• It looks like it is an artifact of how numbers are represented in binary internally. Sep 6, 2016 at 13:20
• Rationalize[12.30]*12 Sep 6, 2016 at 13:28
It's a numerical artifact that is due to finite precision of computer calculations. Indeed, when one computes
SetAccuracy[12.30*12, 15]
147.60000000000002
It's notable that the input is in decimal system (at least the user thinks of it as such), but the computer uses its binary representation. For example, $\frac{1}{10}$ is an exact number, but 0.1 has an infinite expansion when expressed in binary (i.e., $0.1_{10}=0.000110011001100110011001100110011..._2$; see also here). Hence the computer never uses exactly $\frac{1}{10}$ but a rounded number:
SetAccuracy[0.1, 20]
0.1000000000000000056
What you see above in the case of 12.30*12 is most likely a result of such rounding. It should be also noted that in most published materials, when some result is given to e.g. 20 decimal digits, in most cases it is not reasonable: in fact, every machine will give a slightly different output. The computation on two different computers (or different software) can give results different on the 15th or 12th or some other decimal place.
• Thank you for this makes a lot of sense Sep 6, 2016 at 14:40 | 461 | 1,725 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-27 | latest | en | 0.939349 |
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Resources tagged with Combinations similar to Making Pathways:
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There are 103 results
Broad Topics > Decision Mathematics and Combinatorics > Combinations
Six Is the Sum
Age 7 to 11 Challenge Level:
What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros?
Magic Triangle
Age 7 to 11 Challenge Level:
Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total.
Throw a 100
Age 7 to 11 Challenge Level:
Can you score 100 by throwing rings on this board? Is there more than way to do it?
Dice and Spinner Numbers
Age 7 to 11 Challenge Level:
If you had any number of ordinary dice, what are the possible ways of making their totals 6? What would the product of the dice be each time?
Cat Food
Age 7 to 11 Challenge Level:
Sam sets up displays of cat food in his shop in triangular stacks. If Felix buys some, then how can Sam arrange the remaining cans in triangular stacks?
Train Carriages
Age 5 to 11 Challenge Level:
Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done?
Prison Cells
Age 7 to 11 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
Bean Bags for Bernard's Bag
Age 7 to 11 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
Age 7 to 11 Challenge Level:
Three dice are placed in a row. Find a way to turn each one so that the three numbers on top of the dice total the same as the three numbers on the front of the dice. Can you find all the ways to do. . . .
The Money Maze
Age 7 to 11 Challenge Level:
Go through the maze, collecting and losing your money as you go. Which route gives you the highest return? And the lowest?
Penta Post
Age 7 to 11 Challenge Level:
Here are the prices for 1st and 2nd class mail within the UK. You have an unlimited number of each of these stamps. Which stamps would you need to post a parcel weighing 825g?
Sweets in a Box
Age 7 to 11 Challenge Level:
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
Zargon Glasses
Age 7 to 11 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
Five Coins
Age 7 to 11 Challenge Level:
Ben has five coins in his pocket. How much money might he have?
Two Egg Timers
Age 7 to 11 Challenge Level:
You have two egg timers. One takes 4 minutes exactly to empty and the other takes 7 minutes. What times in whole minutes can you measure and how?
Plants
Age 5 to 11 Challenge Level:
Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this?
Hubble, Bubble
Age 7 to 11 Challenge Level:
Winifred Wytsh bought a box each of jelly babies, milk jelly bears, yellow jelly bees and jelly belly beans. In how many different ways could she make a jolly jelly feast with 32 legs?
On Target
Age 7 to 11 Challenge Level:
You have 5 darts and your target score is 44. How many different ways could you score 44?
Polo Square
Age 7 to 11 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
The Puzzling Sweet Shop
Age 7 to 11 Challenge Level:
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
Super Value Shapes
Age 7 to 11 Challenge Level:
If each of these three shapes has a value, can you find the totals of the combinations? Perhaps you can use the shapes to make the given totals?
It Figures
Age 7 to 11 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
Rod Measures
Age 7 to 11 Challenge Level:
Using 3 rods of integer lengths, none longer than 10 units and not using any rod more than once, you can measure all the lengths in whole units from 1 to 10 units. How many ways can you do this?
Sending Cards
Age 7 to 11 Challenge Level:
This challenge asks you to investigate the total number of cards that would be sent if four children send one to all three others. How many would be sent if there were five children? Six?
Room Doubling
Age 7 to 11 Challenge Level:
Investigate the different ways you could split up these rooms so that you have double the number.
Chocs, Mints, Jellies
Age 7 to 11 Challenge Level:
In a bowl there are 4 Chocolates, 3 Jellies and 5 Mints. Find a way to share the sweets between the three children so they each get the kind they like. Is there more than one way to do it?
Elf Suits
Age 7 to 11 Challenge Level:
If these elves wear a different outfit every day for as many days as possible, how many days can their fun last?
Ice Cream
Age 7 to 11 Challenge Level:
You cannot choose a selection of ice cream flavours that includes totally what someone has already chosen. Have a go and find all the different ways in which seven children can have ice cream.
Hamilton's Puzzle
Age 7 to 11 Challenge Level:
I start my journey in Rio de Janeiro and visit all the cities as Hamilton described, passing through Canberra before Madrid, and then returning to Rio. What route could I have taken?
Halloween Investigation
Age 7 to 11 Challenge Level:
Ana and Ross looked in a trunk in the attic. They found old cloaks and gowns, hats and masks. How many possible costumes could they make?
Penta Primes
Age 7 to 11 Challenge Level:
Using all ten cards from 0 to 9, rearrange them to make five prime numbers. Can you find any other ways of doing it?
Age 7 to 11 Challenge Level:
Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon?
A Rod and a Pole
Age 7 to 11 Challenge Level:
A lady has a steel rod and a wooden pole and she knows the length of each. How can she measure out an 8 unit piece of pole?
Wag Worms
Age 7 to 11 Challenge Level:
When intergalactic Wag Worms are born they look just like a cube. Each year they grow another cube in any direction. Find all the shapes that five-year-old Wag Worms can be.
Calcunos
Age 7 to 11 Challenge Level:
If we had 16 light bars which digital numbers could we make? How will you know you've found them all?
Plates of Biscuits
Age 7 to 11 Challenge Level:
Can you rearrange the biscuits on the plates so that the three biscuits on each plate are all different and there is no plate with two biscuits the same as two biscuits on another plate?
Calendar Cubes
Age 7 to 11 Challenge Level:
Make a pair of cubes that can be moved to show all the days of the month from the 1st to the 31st.
3 Rings
Age 7 to 11 Challenge Level:
If you have three circular objects, you could arrange them so that they are separate, touching, overlapping or inside each other. Can you investigate all the different possibilities?
New House
Age 7 to 11 Challenge Level:
In this investigation, you must try to make houses using cubes. If the base must not spill over 4 squares and you have 7 cubes which stand for 7 rooms, what different designs can you come up with?
Three by Three
Age 5 to 11 Challenge Level:
Arrange 3 red, 3 blue and 3 yellow counters into a three-by-three square grid, so that there is only one of each colour in every row and every column
Three Way Mix Up
Age 5 to 11 Challenge Level:
Jack has nine tiles. He put them together to make a square so that two tiles of the same colour were not beside each other. Can you find another way to do it?
Street Party
Age 7 to 11 Challenge Level:
The challenge here is to find as many routes as you can for a fence to go so that this town is divided up into two halves, each with 8 blocks.
Those Tea Cups
Age 7 to 11 Challenge Level:
Place the 16 different combinations of cup/saucer in this 4 by 4 arrangement so that no row or column contains more than one cup or saucer of the same colour.
Button-up Some More
Age 7 to 11 Challenge Level:
How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...?
Sealed Solution
Age 7 to 11 Challenge Level:
Ten cards are put into five envelopes so that there are two cards in each envelope. The sum of the numbers inside it is written on each envelope. What numbers could be inside the envelopes?
More and More Buckets
Age 7 to 11 Challenge Level:
In this challenge, buckets come in five different sizes. If you choose some buckets, can you investigate the different ways in which they can be filled?
Home City
Age 7 to 11 Challenge Level:
Use the clues to work out which cities Mohamed, Sheng, Tanya and Bharat live in.
Butterfly Cards
Age 7 to 11 Challenge Level:
Four children were sharing a set of twenty-four butterfly cards. Are there any cards they all want? Are there any that none of them want?
Finding Fifteen
Age 7 to 11 Challenge Level:
Tim had nine cards each with a different number from 1 to 9 on it. How could he have put them into three piles so that the total in each pile was 15?
Newspapers
Age 7 to 11 Challenge Level:
When newspaper pages get separated at home we have to try to sort them out and get things in the correct order. How many ways can we arrange these pages so that the numbering may be different? | 2,394 | 9,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2019-43 | latest | en | 0.925663 |
https://greatmathsteachingideas.wordpress.com/category/handling-data-and-probability/ | 1,498,738,720,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323970.81/warc/CC-MAIN-20170629121355-20170629141355-00700.warc.gz | 767,212,010 | 36,233 | ## Who is the best England batsman? An investigation using the mean, median, mode and range
Cricket isn’t everyone’s cup of tea but this lesson idea hasn’t failed to motivate any class that I have tried it on.
The idea is simple- look at the real world batting scores of England batsmen and use the mean, median, mode and range to decide who is the ‘best’ batsman. Each group of pupils are given these worksheets which list the scores, explains the task and also contain pictures that they might like to stick on a poster next to their statistical analysis and interpretive reasoning. Read more of this post
## Taboo words
Thanks to Sarah for this brilliant way to assess understanding of concepts and maths vocabulary.
Split the class up into groups of 4-6. Each group gets a set of small cards which each have on them one maths related word. The first thing they have to do is write on each card, under the math related word which is at the top, three words that people will not be allowed to use when describing the top word. For example, if the top word is circumference then three words the team could write underneath could be circle, perimeter and length. The idea is to make the describing of the top word as tricky as possible. The words that they can’t use when describing the top words are called Taboo words.
The sets of cards are then passed onto another group and one person in the group gets 1 minute to describe as many of the top words as possible to their group colleagues without using the taboo words. The teams get a point for each correct word they guess. Each team has a go and the scores added up at the end to identify the winning team. You can do a tie-breaker round if necessary.
There are lots of variations you could do of this game and it does seem to really engage the kids and is an excellent way to revise key vocabulary and assess conceptual knowledge.
## Height vs arm length- are they related?
Here’s a great little investigation into whether a person’s height and arm length are related. The pupils get a copy of this worksheet and have to use a metre stick to measure the height and arm length of ten of their class colleagues to the nearest cm. After recording their results in a table they draw a scatter graph and answer the questions at the bottom of the worksheet to think about whether there is a correlation. They could beĀ encouragedĀ to share their findings and justification with the class in a discussion in the plenary.
You’ll need metre rulers or tape measures and graph paper for this lesson in addition to the worksheets. Enjoy!
Found this post useful? Please consider checking out our sponsored links to support this blog. | 559 | 2,687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-26 | longest | en | 0.945014 |
https://amortizationcalc.ca/detailed-ird/ | 1,701,694,802,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100529.8/warc/CC-MAIN-20231204115419-20231204145419-00082.warc.gz | 126,893,688 | 10,081 | ## Detailed IRD
Using the MORTGAGE2 PRO software, is easy to verify that \$15,440 is the correct amount to be paid to the lender. One merely highlights the 24th payment of the amortization schedule and clicks on the renewal icon and fills in the renewal rate of 7% and the results are below;
If one was only interested in the IRD calculation without using the MORTGAGE2 PRO amortization software then you could use the DISCOUNTING.EXE program to arrive at the same numbers as shown below;
## NOTE: AMERICAN USERS
American mortgages have monthly compounding, .. the same example for an American mortgage is also shown below;
## A DETAILED ANALYSIS OF THE SEMI-ANNUAL COMPOUNDING IRD EXAMPLE
For the mathematicians a detailed analysis is shown below. After 24 months the Loan Balance is approximately \$150,000 with 36 months left to go until the end of the 5 year Term. If Bob and Mary simply make their payments, with no changes, after the remaining 36 payments they will owe \$145,803.85
If Bob and Mary decide to pay the IRD, they could add the \$15,440.04 directly onto the balance owing of \$150,000 along with the 24th payment and thus the interest rate would be 7% from this point on until 36 payments have been made. From Figure 14 it can be seen that after 36 payments the balance owing is exactly the same (within a few pennies due to rounding off ) as it should be because that is what the IRD calculation is intended to achieve.
For this analysis it is assumed that the IRD money is borrowed. The IRD could be obtained from another investment but that would make the example more complex. Having the IRD money as a windfall (money under the mattress) makes the benefits obvious. If Bob and Mary decide to keep the two loans separate; then, the first loan is \$150,000 @ 7% with \$1,443.79 per month and the balance owing after 36 payments is \$126,824.05
The second loan of \$15,440.04 @ 7% with 476.04 per month (except for the last payment of 475.96) and the balance after 36 payments is zero.
An annuity calculation would show that 476.04 deposited every month at a monthly interest of .575% would accumulate to \$18,980 after 36 months.
It can also be seen that a negative amortization schedule would give the same balance owing of \$18,979.73 (close enough to 18,980). The approximate difference between \$145,803 – \$126,824 = \$18,979. In other words, \$15,440 now is worth \$18,980 thirty six months in the future (FV) if it is appreciating at .575% per month . Stated differently, \$15,440 now is worth \$18,980 three years in the future(FV) if it is growing at an effective interest rate of 7.1224415.
Note that most interest rates are not normally quoted to more than two decimal places because of the difficulty in getting agreement in the PV/FV calculations as the equations exponent is very sensitive to the number of decimal places used!
Another way of looking at the numbers is as follows. If one deposits \$476.04 per month, every month for 36 months, into an account paying interest at 0.575% per month then after 36 months \$18,980 will have accumulated.
## SUMMARY
Bob and Mary have 36 monthly payments of \$1,444. If the Lender uses 11% as the AIR then the balance after 36 months is \$145,803. If the Lender uses 7% as the AIR the balance after 36 months will be \$126,824. The difference in the two balances, is \$18,979 which is money the Lender would lose in the future (36 months from now). The IRD of \$15,440 now is worth \$18,979 thirty six months into the future. The IRD is a fair method of compensation for both the borrower and the lender.
FIGURE 18
PV= 15,440.04
FV= 18,979.72
I = 0.575000
N = 36.0000 | 920 | 3,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-50 | latest | en | 0.948974 |
https://www.jiskha.com/questions/546328/solve-4x-3y-6-0-x-3y-9-0-graphically | 1,558,953,582,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232262311.80/warc/CC-MAIN-20190527085702-20190527111702-00051.warc.gz | 836,502,673 | 4,926 | # maths
solve 4x-3y-6=0; x+3y-9=0 graphically
1. 👍 0
2. 👎 0
3. 👁 81
1. A graphical solution would be achieved by plotting both lines on an x-y graph and seeing where they interesect.
We lack the tools to do graphing for you.
By adding the two equations you can eliminate the y variable and see immediatley that
5x - 15 = 0
x = 3
Then use either of your equations for y.
12 - 3y = 6
3y = 6
y = 2
1. 👍 0
2. 👎 0
posted by drwls
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May 6, 2016
# Homework Help: geometry
Posted by lola on Friday, January 14, 2011 at 9:04pm.
A rectangular box with a square base contains 24 cubic feet. if the height of the box is 18 inches, how many feet ae there in each side of the base?
a. 4
b. 2
c. (2 square root of 3)/3
d. square root of 3 / 2
e. square root of 3
• geometry - helper, Friday, January 14, 2011 at 9:57pm
V = Lwh
the volume (given) = 24 ft^3
the height (given) = 18" = 1.5'
24 = L*w*1.5
divide both sides by 1.5
16 = Lw
You need to find the number of feet in each side of the base
since the box has a square base
L = W
AND, found above, L*w = 16
which of the above answers, a, b, c or d, when multiplied by itself, equals 16?
## Answer This Question
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Introduction
2.
Series Connection
3.
4.
5.
Parallel Connection
6.
7.
8.
Difference Between Parallel and Series Connection
9.
9.1.
What is a Series Circuit?
9.2.
What is a Parallel Circuit?
9.3.
Which type of connection is used in the home and why?
9.4.
Which type of circuit will produce more current and why?
9.5.
What are examples of parallel and series connections in day-to-day life?
10.
Conclusion
Last Updated: Mar 27, 2024
# Difference between Parallel and Series Connection
Rashmi
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Master Python: Predicting weather forecasts
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## Introduction
In any circuit, various methods exist to connect different components depending on their uses. The uses may vary according to the load attached to it. Sometimes we need less power, sometimes more power or less current, and sometimes more current, and so on.
In this blog, we will discuss the difference between parallel and series connections, the two most effective ways of connecting to arrange the components to save the current, voltage, and power in a circuit.
## Series Connection
When all the components are connected in a way that the end of one component is connected to the start of the other one, then this type of connection is defined as a series connection.
For the viewer, it seems to be a single path consisting of all the components of a circuit. In other words, we can say that a series-connected circuit is an end-to-end connected circuit.
In a series circuit, if one component fails to work, all the connected components will fail, and the device won't work.
Some basic examples of the series connection can be explained by using two or more two resistors connected in a row. When one end of the resistor is connected to the other end of another resistor, and likewise, then the connection formed will be a series connection.
Let us understand it by using a basic circuit consisting of 3 resistors together.
In the above circuit, we can see that three resistors are connected end-to-end. The current (I) flowing in all resistors is the same.
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The advantages of series connection are:-
• All elements will get the same current.
• The design is easy to make and is not complex.
• Series-connected devices do not overheat quickly.
The disadvantages of series connection are:-
• If one device fails, it will lead to the failure of all the connected devices.
• The more devices connected, the more the circuit resistance.
• If there is any fault, it is hard to identify and repair.
## Parallel Connection
The components are connected across each other's terminal in a parallel connection. In a parallel circuit, the current gets multiple paths to flow through the entire circuit. In simpler words, the same terminals of the electrical component are connected.
If any single component fails, it will not affect the other components of the circuit, and the device will work properly.
Some basic examples of parallel connection can be explained by using two or more than three resistors connected parallel.
In this circuit, the current through all the resistors is different, and the voltage across them will be the same irrespective of the resistance's size.
The advantages of parallel connection are:-
• The voltage across all elements is the same.
• Total resistance will not increase as the number of devices increases.
• If there is any fault, it is hard to identify and repair.
The disadvantages of parallel connection are:-
• The design is very complex in some cases and will require more wire.
• We cannot increase the voltage as we desire.
• Parallel connected devices overheat easily.
Now, we will see the difference between parallel and series connections.
## Difference Between Parallel and Series Connection
The difference between parallel and series connections are:-
### What is a Series Circuit?
A circuit in which elements are arranged in a single path, such that the similar current must flow through all of them with different voltages across every element, is called a Series Circuit.
### What is a Parallel Circuit?
Parallel Circuits is a circuit in which elements are in different branches such that currents can flow through each component simultaneously and have the same voltage.
### Which type of connection is used in the home and why?
Parallel connection of a circuit is used in-house because, in series connection, they are all connected so that if any of the devices fail, it will block the path failing all devices.
### Which type of circuit will produce more current and why?
Series circuits in a circuit will have more current as there is only one path for the current to flow, so it will not be distributed, resulting in more current compared to parallel circuits.
### What are examples of parallel and series connections in day-to-day life?
Parallel connections in a circuit are used in common households as they are more reliable, whereas we can see a series of connections in a lightning frill or showrooms by convenience.
## Conclusion
Kudos, Ninja, on making it to the finish line of this article! We have learned that series and parallel connections are methods to share common currents and voltages across the circuit.
We hope this blog was helpful to you in understanding the critical difference between parallel and series connections. Do not stop learning! We recommend you read some of our other related articles:
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Live masterclass | 1,296 | 6,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-30 | latest | en | 0.935969 |
https://onlinetools.com/number/find-matrix-inverse | 1,720,827,812,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514459.28/warc/CC-MAIN-20240712224556-20240713014556-00094.warc.gz | 363,091,323 | 30,545 | # Inverse Matrix Calculator
## World's Simplest Number Tool
Quickly find the inverse of a matrix in your browser. To get the inverse, just enter your matrix in the input field, optionally adjust the row and column separators in the options below, and this utility will instantly compute the inverse of your matrix. Created by developers from team Browserling.
Tool Options
#### Input Matrix Delimiter
Element delimiter of the input
matrix. (By default a space.)
Row delimiter of the input
matrix. (By default a newline.)
#### Inverse Matrix Delimiter
Element delimiter of the output
matrix. (By default a space.)
Row delimiter of the output
matrix. (By default a newline.)
#### Inverted Matrix Accuracy
Number of digits in the fractional
part. (By default five.)
Determine the precision
automatically.
Align row elements in the
invented matrix automatically.
## What Is an Inverse Matrix Calculator?
This is an online browser-based utility for finding the inverse of a matrix. The inverse matrix A-1 of a matrix A is such that the product AxA-1 is equal to the identity matrix. The result of multiplying the matrix by its inverse is commutative, meaning that it doesn't depend on the order of multiplication – A-1xA is equal to AxA-1. The inverse matrix exists only for square matrices and it's unique. The matrix has the inverse if and only if it is invertible. An invertible matrix is also sometimes called nonsingular or nondegenerate. The key property of an invertible matrix is that its determinant isn't equal to zero. In this tool, we use the Gaussian elimination method to calculate the inverse matrix. This method is very fast and here's how it works. We append the unit matrix of the same dimensions on the left side of the given matrix, and then by using elementary transformations, we bring the original matrix to the unit form. At the same time, we apply all the same elementary transformations to the unit matrix on the left side. When completed, the right side now contains the inverted matrix. With this tool, you can calculate the inverse for a matrix of any size. You can also adjust the properties and format of the inverted matrix by customizing its numerical accuracy, row and column separators, as well as how elements are aligned. That's numberwang!
## Inverse Matrix Calculator Examples
Click to try!
click me
### Inverse of a Square Matrix
In this example, we calculate the inverse of a 3 by 3 matrix containing integers. By multiplying these matrices together in any order, you'll always get the identity matrix.
2 1 3 3 3 4 5 2 8
16 -2 -5 -4 1 1 -9 1 3
Required options
These options will be used automatically if you select this example.
Element delimiter of the input
matrix. (By default a space.)
Row delimiter of the input
matrix. (By default a newline.)
Element delimiter of the output
matrix. (By default a space.)
Row delimiter of the output
matrix. (By default a newline.)
Align row elements in the
invented matrix automatically.
click me
### Custom Precision Example
In this example, we find the inverse of a matrix with dimensions 4 by 4. The resulting matrix consists of fractional numbers, so we adjust its accuracy to 3 digits after the decimal point.
2 3 6 5 -2 5 -1 3 -8 0 -7 -2 -5 -4 2 1
-1.092 1.031 -0.949 0.469 1.684 -1.561 1.398 -0.939 2.061 -2.02 1.633 -0.98 -2.847 2.949 -2.418 1.551
Required options
These options will be used automatically if you select this example.
Element delimiter of the input
matrix. (By default a space.)
Row delimiter of the input
matrix. (By default a newline.)
Element delimiter of the output
matrix. (By default a space.)
Row delimiter of the output
matrix. (By default a newline.)
Number of digits in the fractional
part. (By default five.)
Determine the precision
automatically.
Align row elements in the
invented matrix automatically.
click me
### Auto-accuracy Example
In this example, we compute the inverse of a matrix with size 10x10. We activate two automatic functions here. First, the tool determines the best accuracy of each number in the inverse matrix, and second, it adds as many spaces around numbers so that all elements align in beautiful columns.
6 6 9 5 6 4 7 4 1 2 2 9 6 3 8 1 5 6 3 5 2 3 3 9 1 9 9 0 5 4 4 0 6 2 4 7 4 1 3 0 7 9 4 3 2 4 4 7 0 8 4 3 7 3 2 0 7 1 5 2 3 7 9 6 1 1 7 9 4 4 9 2 2 2 9 2 2 4 5 4 6 7 2 3 9 4 2 7 7 8 8 0 8 8 4 0 5 7 1 0
0.27 -0.26 -0.02 -0.14 -0.03 -0.03 0.1 0.14 0.02 -0.11 0.64 -0.41 -0.02 -0.32 -0.2 -0.13 0.24 0.07 0.19 -0.28 -0.09 0.03 -0.12 0.16 0.08 0.27 -0.3 -0.49 0.32 0.24 0.23 -0.18 0.01 -0.17 -0.09 0.06 -0.14 -0.31 0.35 0.14 -0.13 0.22 0.02 0.07 0.01 -0.01 -0.11 0.03 -0.07 0.07 0.06 -0.06 0.02 0.08 -0.01 -0.11 0.07 0.03 -0.01 -0.07 -0.35 0.36 0.14 0.1 0.09 -0.14 0.22 0.56 -0.65 -0.14 -0.18 0.13 0.04 0.08 0.01 -0.23 0.27 0.34 -0.31 -0.09 0.23 -0.26 -0.02 -0.11 -0.16 0.01 0.17 0.03 0.18 -0.14 -0.67 0.43 -0.02 0.3 0.32 0.34 -0.51 -0.44 0.1 0.4
Required options
These options will be used automatically if you select this example.
Element delimiter of the input
matrix. (By default a space.)
Row delimiter of the input
matrix. (By default a newline.)
Element delimiter of the output
matrix. (By default a space.)
Row delimiter of the output
matrix. (By default a newline.)
Determine the precision
automatically.
Align row elements in the
invented matrix automatically.
### Pro tipsMaster online number tools
You can pass input to this tool via ?input query argument and it will automatically compute output. Here's how to type it in your browser's address bar. Click to try!
https://onlinetools.com/number/find-matrix-inverse?input=2%201%203%0A3%203%204%0A5%202%208&input-element-separator=%20&input-row-separator=%255Cn&output-element-separator=%20&output-row-separator=%255Cn&prettify-matrix=true
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These number tools are on the way!
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### Users Who Are Viewing This Thread (Users: 0, Guests: 1)
#### H2Pendragon
1. The problem statement, all variables and given/known data
Hi everyone, this is probably an easy question but I'm having trouble on the wording of the proof.
Let f be continuous at x=c and f(c) > 1
Show that there exists an r > 0 such that $$\forall x \in B(c,r) \bigcap D : f(x) > 1$$
2. Relevant equations
$$\forall \epsilon > 0, \exists r > 0, \forall x \in B(c,r) \bigcap D \Rightarrow |f(x) - f(c)| < \epsilon$$
f(c) > 1
3. The attempt at a solution
I'd put something here but I don't really have any attempts. I originally thought of using the limit rule that lim f(x) = f(c) but that doesn't work since c could be an isolated point.
If I used a graph it just seems so obvious, but I have to use words to prove it. I tried also to use a contradiction, but using just what I put down in (2) didn't lead to any direct contradiction that I could see.
A nudge in the right direction would be helpful.
#### matt grime
Homework Helper
If in doubt try an example.
Let's say for the sake of argument that f(c)=2. What choice of epsilon, and r, from 2. above will mean that f(x) > 1 in the ball around c?
#### MathematicalPhysicist
Gold Member
-e<f(x)-f(c)<e
f(x)>f(c)-e
f(c)=1+h^2 for some h real, do you understand how to pick e such that f(x)>1.
#### H2Pendragon
-e<f(x)-f(c)<e
f(x)>f(c)-e
f(c)=1+h^2 for some h real, do you understand how to pick e such that f(x)>1.
by this reasoning couldn't I just claim that f(c) = 1 + e?
#### matt grime
Homework Helper
Yes, you can do that.
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# Form 1 (KSSM)
### 01 Rational Numbers
1. Short Notes | Part 1 | Part 2 | Part 3 | Part 4
2. Fraction Revision | Question 1 – 5 | Question 6 – 10 | Question 11
3. Decimal Revision | Question 1 – 4 | Question 5 – 8
4. PT3 Focus Practice (with Solution)
5. Mathematics Form 1 (KSSM) Textbook Question (with Solution)
1. Self Practice | 1.1a | 1.1b | 1.1c | 1.1d | 1.2a | 1.2b | 1.2c
2. Mastery Q | 1.1
3. Let’s Practise
### 02 Factors and Multiples
1. Short Notes
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 1 (KSSM) Textbook Question (with Solution)
1. Self Practice | 2.1a | 2.1b | 2.1c | 2.1d | 2.1e | 2.2a | 2.2b | 2.2c
2. Mastery Q | 2.1 | 2.2
3. Let’s Practise
### 03 Squares, Square Roots, Cubes and Cube Roots
1. Short Notes | Part 1 | Part 2
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 1 (KSSM) Textbook Question (with Solution)
1. Self Practice | 3.1a | 3.1b | 3.1c | 3.1d | 3.2a | 3.2b | 3.2c | 3.2d
2. Mastery Q
3. Let’s Practise
### 04 Ratios, Rates and Proportions
1. Short Notes
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 1 (KSSM) Textbook Question (with Solution)
1. Self Practice
2. Mastery Q
3. Let’s Practise
### 05 Algebraic Expressions
1. Short Notes
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 1 (KSSM) Textbook Question (with Solution)
1. Self Practice | 5.2a | 5.2b | 5.2c
2. Mastery Q | 5.2
3. Let’s Practise
### 06 Linear Equations
1. Short Notes | Part 1 | Part 2
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 1 (KSSM) Textbook Question (with Solution)
1. Self Practice | 6.1b | 6.1c | 6.1d
2. Mastery Q | 6.1 | 6.2
3. Let’s Practise
### 07 Linear Inequalities
1. Short Notes | Part 1 | Part 2
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 1 (KSSM) Textbook Question (with Solution)
1. Self Practice
2. Mastery Q
3. Let’s Practise
### 08 Lines and Angles
1. Short Notes | Part 1 | Part 2 | Part 3
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 1 (KSSM) Textbook Question (with Solution)
1. Self Practice
2. Mastery Q | 8.2 | 8.3
3. Let’s Practise
### 09 Basic Polygon
1. Short Notes
2. PT3 Focus Practice (with Solution)
1. Question
3. Mathematics Form 1 (KSSM) Textbook Question (with Solution)
1. Self Practice | 9.2b | 9.2c | 9.3b | 9.3c | 9.3d
2. Mastery Q | 9.2 | 9.3
3. Let’s Practise
### 10 Perimeter and Area
1. Short Notes
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 1 (KSSM) Textbook Question (with Solution)
1. Self Practice | 10.1a | 10.1c | 10.2c | 10.3b
2. Mastery Q | 10.1 | 10.2 | 10.3
3. Let’s Practise
### 11 Introduction to Set
1. Short Notes | Part 1 | Part 2 | Part 3
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 1 (KSSM) Textbook Question (with Solution)
1. Self Practice
2. Mastery Q
3. Let’s Practise
### 12 Data Handling
1. Short Notes
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 1 (KSSM) Textbook Question (with Solution)
1. Self Practice | 12.1b
2. Mastery Q | 12.1
3. Let’s Practise
### 13 The Pythagoras’ Theorem
1. Short Notes
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 1 (KSSM) Textbook Question (with Solution)
1. Self Practice | 13.1c | 13.1d | 13.2a | 13.2b
2. Mastery Q | 13.1 | 13.2
3. Let’s Practise
# Form 2 (KSSM)
### 01 Patterns and Sequences
1. Short Notes
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 2 (KSSM) Textbook Question (with Solution)
1. Self Practice | 1.1 | 1.2 | 1.3
2. Generating Excellence
### 02 Factorisation and Algebraic Fractions
1. Short Notes
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 2 (KSSM) Textbook Question (KSSM) (with Solution)
1. Self Practice | 2.1 | 2.2 | 2.3
2. Generating Excellence
### 03 Algebraic Formulae
1. Short Notes
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 2 (KSSM) Textbook Question (with Solution)
### 04 Polygons
1. Short Notes
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 2 (KSSM) Textbook Question (with Solution)
1. Self Practice | 4.1 | 4.2
2. Generating Excellence | Q2 – Q8 | Q9 – Q15
### 05 Circles
1. Short Notes | Part 1 | Part 2
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 2 (KSSM) Textbook Question (with Solution)
1. Self Practice | 5.2 | 5.3
2. Generating Excellence
### 06 Three Dimensional Geometrical Shapes
1. Short Notes | Part 1
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 2 (KSSM) Textbook Question (with Solution)
1. Self Practice | 6.3 | 6.4
2. Generating Excellence
### 07 Coordinates
1. Short Notes | Part 1 | Part 2
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 2 (KSSM) Textbook Question (with Solution)
1. Self Practice | 7.1 | 7.2 | 7.3
2. Generating Excellence
### 08 Graphs of Functions
1. Short Notes
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 2 (KSSM) Textbook Question (with Solution)
1. Self Practice | 8.2
2. Generating Excellence
### 09 Speed and Acceleration
1. Short Notes
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 2 (KSSM) Textbook Question (with Solution)
1. Self Practice | 9.1
2. Generating Excellence
### 10 Gradient of a Straight Line
1. Short Notes | Part 1 | Part 2 | Part 3
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 2 (KSSM) Textbook Question (with Solution)
1. Self Practice | 10.1
2. Generating Excellence
### 11 Isometric Transformations
1. Short Notes | Part 1 | Part 2
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 2 (KSSM) Textbook Question (with Solution)
1. Self Practice | 11.2 | 11.3 | 11.4
2. Generating Excellence | Question 1 – 9 | Question 10 – 17
### 12 Measures of Central Tendencies
1. Short Notes | Part 1 | Part 2
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 2 (KSSM) Textbook Question (with Solution)
1. Self Practice | 12.1
2. Generating Excellence
### 13 Simple Probability
1. Short Notes | Part 1 | Part 2 | Part 3
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 2 (KSSM) Textbook Question (with Solution)
1. Self Practice | 13.2 | 13.3 | 13.4
2. Generating Excellence
# Form 3 (KSSM)
### 01 Indices
1. Short Notes
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 3 (KSSM) Textbook Question (with Solution)
1. Mind Test | 1.2a | 1.2b | 1.2c | 1.2d | 1.2e | 1.2f | 1.2g | 1.2h | 1.2i | 1.2j
2. Dynamic Challenge | Test Yourself | Skills Enhancement | Self Mastery
### 02 Standard Form
1. Short Notes (Significant Figures) | Part 1 | Part 2
2. Short Notes (Standard Form) | Part 1 | Part 2
3. PT3 Focus Practice (with Solution)
4. Mathematics Form 3 (KSSM) Textbook Question (with Solution)
1. Mind Test | 2.1a | 2.1b | 2.2a | 2.2b | 2.2c | 2.2d
2. Dynamic Challenge | Test Yourself | Skills Enhancement | Self Mastery
### 03 Consumer Mathematics: Savings and Investments, Credit and Debit
1. Short Notes
2. PT3 Focus Practice (with Solution)
1. Question
3. Mathematics Form 3 (KSSM) Textbook Question (with Solution)
1. Mind Test | 3.1a | 3.1b | 3.1c | 3.1e | 3.1f
2. Dynamic Challenge
### 04 Scale Drawings
1. Short Notes
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 3 (KSSM) Textbook Question (with Solution)
1. Mind Test | 4.1b | 4.1d
2. Dynamic Challenge
### 05 Trigonometric Ratios
1. Short Notes | Part 1 | Part 2
2. PT3 Focus Practice (with Solution)
3. Mathematics Form 3 (KSSM) Textbook Question (with Solution)
1. Mind Test | 5.1d | 5.1e | 5.1f | 5.1g | 5.1h | 5.1i
2. Dynamic Challenge
### 06 Angles and Tangents of Circles
1. Short Notes (Angles of Circles) | Part 1
2. Short Notes (Tangents to a Circle) | Part 1 | Part 2
3. Short Notes (Angles and Tangents of Circles) | Part 1 | Part 2
4. Short Notes (Common Tangents) | Part 1 | Part 2
5. PT3 Focus Practice (with Solution)
6. Mathematics Form 3 (KSSM) Textbook Question (with Solution)
1. Mind Test | 6.1a | 6.1b |
2. Dynamic Challenge
### 07 Plans and Elevations
1. Short Notes
2. PT3 Focus Practice (with Solution)
### 08 Loci in Two Dimensions
1. Short Notes
2. PT3 Focus Practice (with Solution)
### 09 Straight Lines
1. Short Notes (Equation of a Straight Line) | Part 1
2. Short Notes (Parallel Lines) | Part 1 | Part 2
3. PT3 Focus Practice (with Solution)
4. Mathematics Form 3 (KSSM) Textbook Question (with Solution)
1. Mind Test | 9.1a | 9.1b | 9.1c | 9.1d | 9.1e | 9.1f | 9.1g | 9.1h | 9.1i
2. Dynamic Challenge | Test Yourself | Skills Enhancement | Self Mastery | 2,945 | 8,386 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2023-14 | latest | en | 0.620353 |
https://www.rhumbarlv.com/what-is-the-physical-meaning-of-momentum/ | 1,721,326,565,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514848.78/warc/CC-MAIN-20240718161144-20240718191144-00737.warc.gz | 819,207,235 | 13,228 | ## What is the physical meaning of momentum?
Momentum is a physics term; it refers to the quantity of motion that an object has. If an object is in motion (on the move) then it has momentum. Momentum can be defined as “mass in motion.” All objects have mass; so if an object is moving, then it has momentum – it has its mass in motion.
What is the physical meaning of impulse?
Impulse is a certain amount of force you apply for a certain amount of time to cause a change in momentum. That is why it is F*t. For example, when you hit a ball with a cricket bat, you apply a force for a time(a very short period in this case) to cause a change (or transfer) of momentum in the ball.
What is the physical meaning of momentum Class 9?
Momentum can be well defined as the measure of the kinetic energy of the object. In simple words, the momentum is the product of mass and velocity. Complete answer: Momentum is defined as the mass (m) times the velocity (v). If an object is steady so its velocity is zero resulting in zero momentum.
### What is the physical meaning of linear momentum?
Linear momentum is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum is expressed as p = mv. Momentum p is a vector having the same direction as the velocity v. The SI unit for momentum is kg · m/s.
What are the two types of momentum?
There are two kinds of momentum, linear and angular. A spinning object has angular momentum; an object traveling with a velocity has linear momentum.
How do you explain momentum?
Momentum is how we measure mass that is in motion. Any moving object will have momentum. Under the law of physics, the object’s momentum equals mass times velocity. Momentum is the quantity of motion and is compounded by the quantity of the matter moved and the velocity with which it moves.
#### What is a impulse simple definition?
1 : a force that starts a body into motion. 2 : the motion produced by a starting force. 3 : a strong sudden desire to do something She resisted the impulse to shout. 4 : nerve impulse.
What is impulse one word?
An impulse is a sudden force or desire — this could be an electrical impulse, or an impulse to get some pizza. If you act on a sudden feeling or thought, you’re following an impulse. Another meaning of impulse is an electrical charge or pulse.
What is momentum example?
So momentum equals mass times velocity or p = m x v. Therefore, if any object of any mass is not moving, its momentum is zero because its velocity is zero. Examples of Momentum: A 1000 kg car moving at 15 m/sec has a momentum of 15,000 kg•m/sec as a result of multiplying the mass and the velocity.
## What is momentum explain with examples?
Momentum is defined as the amount of motion occurring in something that is moving, or the force that drives something forward to keep it moving. An example of momentum is how quickly a car is moving down a hill. (physics) (of a body in motion) The product of its mass and velocity.
What is momentum give example?
Something that has a large mass has a large momentum or something that is moving very fast has a large momentum. Often times we discuss people or teams having the “momentum” in some sports setting. For instance, if Tiger Woods starts making a bunch of great shots in a golf match, we would say that he has momentum.
What is an example of momentum?
Momentum can be thought of as the “power” when a body is moving, meaning how much force it can have on another body. For example, a bowling ball (large mass) moving very slowly (low velocity) can have the same momentum as a baseball (small mass) that is thrown fast (high velocity).
### Does impulse equal to momentum?
Impulse is a quantity that is closely related to momentum . When an object has a momentum , and a force is applied for an amount of time, the momentum can change to a new value . The impulse-momentum theorem states that the impulse is equal to this change in momentum.
What is the difference between impulse and change in momentum?
When we talk about impulse, we are really talking about change in momentum of a moving body, and the rate of change in momentum is the force applied upon the body. In brief: • Momentum of a moving body is the product of its mass and its velocity whereas impulse is change in momentum which is the product of mass and difference in velocities.
Does impulse equal momentum or a change in momentum?
Yes , impulse is equal to an object’s change in momentum.
#### Are momentum and impulse measured in the same units?
Momentum and impulse are both terms that describe concepts in physics that are relatively similar. Most of the confusion arises from the fact that both of these concepts have the same units – mass times velocity. However, they are fundamentally not the same , and are calculated in different ways. | 1,033 | 4,855 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-30 | latest | en | 0.938878 |
https://msubbu.in/sp/td/WaterTempChange.htm | 1,653,328,231,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662560022.71/warc/CC-MAIN-20220523163515-20220523193515-00495.warc.gz | 457,476,643 | 2,505 | ### Change in Temperature due to Enthalpy Change
Liquid water at 25oC flows in a straight horizontal pipe, in which there is no exchange of either heat or work with the surroundings. Its velocity is 12 m/s in a pipe with an i.d. of 2.5 cm until it flows into a section where the pipe diameter increases to 7.5 cm. What is the temperature change?
Calculations:
For the steady flow process, the first law is written as
ΔH + Δu2/2 + gΔz = Q + Ws
since there is no shaft work, Ws = 0
and flow is horizontal, Δz = 0
Therefore,
ΔH + Δu2/2 = Q
and since there is no heat transfer, Q = 0
Therefore, ΔH + Δu2/2 = 0
Applying continuity equations,
u1A1ρ1 = u2A2ρ2
for water ρ1 = ρ2
Therefore,
u2/u1 = A1/A2 = d12/d22 = 2.52/7.52 = 0.1111
u2 = 12 x 0.1111 = 1.333 m/s
(u22 - u12) / 2 = (1.3332 - 122) / 2 = -71.11
Therefore, ΔH = 71.11 J/kg
Enthalpy change per kg mass = 71.11 J
We know, ΔH = mCpΔT and, m = 1000 g; Cp = 4.184 J/g.oC
Therefore,
ΔT = 71.11/(1000 x 4.184) = 0.017oC
Temperature change = 0.017oC | 384 | 1,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-21 | longest | en | 0.899707 |
https://math.stackexchange.com/questions/534618/the-hanging-chain-problem-checking-if-i-am-right | 1,719,146,449,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862466.81/warc/CC-MAIN-20240623100101-20240623130101-00250.warc.gz | 339,613,173 | 37,446 | # The Hanging Chain Problem, checking if I am right
I am trying to make sure I approached this problem correctly.
It's a hanging chain. I'm told by a classmate that the physics doesn't make sense, but it's a really a math problem, so that I guess is all right – the physics is part of it but not the whole thing.
We are deriving the PDE that models a hanging chain of Length L.
The x-axis is placed vertically. Positive direction points upwards.
The fixed end of the chain (the part attached to the cieling) is at x=L.
Let $u(x,t)$ denote the deflection (EDIT: the bending) of the chain. We assume the deflection is in the x,u plane. Let $\rho$ denote the mass density in units of mass per length.
A) show that in equilibrium position the tension at a point $x$ is $\tau(x)= \rho gx$ where g is the acceleration due to gravity.
B) show that $\rho \frac{\partial ^2 u}{\partial t^2} = \frac{1}{ \Delta x}\left[\tau(x+ \Delta x) \frac{\partial u}{\partial x}( x+ \Delta x,t) - \tau(x) \frac{\partial u}{\partial x}(x,t)\right]$
C) Let $\Delta x \rightarrow 0$ and obtain $\rho \frac{\partial ^2 u}{\partial t^2}= \frac{\partial}{\partial x}\left[\tau(x)\frac{\partial u}{\partial x}\right]$
I approached it like this:
We know F=ma. That means $\tau=ma$ since $a=g$.
The mass of the string at any point x is $\rho x$
Therefore $\tau=\rho g x$
For part B). The component of the tension in the y-axis is going to be $-\tau \sin \alpha + \tau \sin \beta$ because we are superposing the tension at both ends of the chain. $-\tau \sin \alpha + \tau \sin \beta = ma = \rho g x = \rho g L = \rho g x \Delta x$. Since g is acceleration $g= \frac{\partial ^2 u}{\partial x^2}$ and that gets us:
$-\tau \sin \alpha + \tau \sin \beta =\rho \frac{\partial ^2 u}{\partial x^2} \Delta x$
Now, the interesting thing here is that for most angles $\tau \sin \alpha$ and $\tau \sin \beta$ will be pretty close to $\tau \tan \alpha$ and $\tau \tan \beta$. And the tangent on the bent string is going to describe the velocity of the bit of string at that point.
So we can plug in the tangents to our earlier expression for the tension.
$-\tau \tan \alpha + \tau \tan \beta =\rho \frac{\partial ^2 u}{\partial t^2} \Delta x$
Now let's look at $u(x,t)$ in terms of x only. The slope of the tangent to the hanging chain – that's the graph of $u(x,t)$ – is $\frac{\partial u}{\partial x}(x,t)$ which means
$\tan \alpha = \frac{\partial u}{\partial x}(x,t)$ and $\tan \beta = \frac{\partial u}{\partial x}(x+\Delta x,t)$
Plug these back into the equation with the sines of alpha and beta
$-\tau \tan \alpha + \tau \tan \beta = \left[\rho \frac{\partial ^2 u}{\partial x^2} \Delta x \right] \Rightarrow \tau \left[-\frac{\partial u}{\partial x}(x+\Delta x,t) - \frac{\partial u}{\partial x}(x,t)\right] = \rho \frac{\partial ^2 u}{\partial x^2} \Delta x$
The last step I am a bit fuzzy on. If I move the variables around a bit:
$$\frac{\left[-\frac{\partial u}{\partial x}(x+\Delta x,t) - \frac{\partial u}{\partial x}(x,t)\right]}{\Delta x} = \frac{\rho}{\tau} \frac{\partial ^2 u}{\partial x^2}$$
I get a zero in the denominator as $\Delta x$ approaches zero. But of course I am assuming I did the rest of the problem right to begin with. I know that
$\frac{\partial}{\partial x}\left[\tau(x)\frac{\partial u}{\partial x}\right] = \left[\frac{\partial \tau(x)}{\partial x}\frac{\partial u}{\partial x} + \tau(x)\frac{\partial^2 u}{\partial x^2}\right]$
But again I feel like I am just missing the very last and probably dead simple bit at the very end.
Anyhow sorry to make you all slog through this.
• What is $t$? What exactly is meant by "deflection"? Can you draw a picture of how the chain is hanging and show it to us? It's not clear to me from your description. Commented Oct 21, 2013 at 15:39
• deflection is bending. The chain is hanging from a fixed point. Not sure how to upload pictures. Commented Oct 21, 2013 at 16:08
• You haven't explained why the chain bends. A chain hanging down from a fixed point generally doesn't bend. I understand what deflection means; I don't understand how you're measuring it. Commented Oct 21, 2013 at 16:12
• The original problem in the text doesn't say why the chain bends. it just says it is modeling vibrations. So presumably it's a forced vibration problem. Commented Oct 21, 2013 at 16:16
• @dfeuer: what Jesse is not explaining (well) here is that this is a problem in small oscillations: what is the response of a hanging chain to a small nudge transverse to the chain? To compute these, we look at the natural modes of oscillation, which involve analyzing the forces as above, including the assumption that there is a deflection $u$ in the transverse direction. Because this is a continuous chain hung from a point (where deflections are zero), one may analyze the forces on a typical point in the chain. Commented Oct 21, 2013 at 16:16
I came across this question while doing a problem set with a similar exercise, so hopefully this answer is a helpful reference.
You made a few typos after you derived the tangent approximation, where you switched the ordering and sign of the $$\beta$$ terms, used the $$\tau$$ for $$x$$ at $$x+\Delta x$$, and changed to taking $$x$$-derivatives on the right hand side.
I fix these below:
Plug these back into the equation with the sines of alpha and beta
$$-\tau \tan \alpha + \tau \tan \beta = \left[\rho \frac{\partial ^2 u}{\partial t^2} \Delta x \right] \Rightarrow \left[-\tau(x) \frac{\partial u}{\partial x}(x,t) + \tau(x+\Delta x)\frac{\partial u}{\partial x}(x+\Delta x, t)\right] = \rho \frac{\partial ^2 u}{\partial t^2} \Delta x$$
We divide both sides by $$\Delta x$$: $$\frac{\left[-\tau(x) \frac{\partial u}{\partial x}(x,t) + \tau(x+\Delta x)\frac{\partial u}{\partial x}(x+\Delta x,t)\right]}{\Delta x} = \rho \frac{\partial ^2 u}{\partial t^2}$$
Now notice that for any function $$f$$, $$\lim_{\Delta\to 0} \frac{\left[ -f(x) + f(x + \Delta x) \right]}{\Delta x} = \frac{\partial f}{\partial x}(x)$$, by definition of the derivative.
So we take the limit of our last step and substitute this definition in: $$\frac{\partial\left[ \tau \frac{\partial u}{\partial x}\right]}{\partial x} = \rho \frac{\partial ^2 u}{\partial t^2},$$ which is as required. | 1,831 | 6,270 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-26 | latest | en | 0.861169 |
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# If x and y are nonzero integers, is x/y > 0?
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If x and y are nonzero integers, is x/y > 0? [#permalink]
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If x and y are nonzero integers, is x/y > 0?
(1) |x^3|= |y^3|
(2) |x + y| = |x| + |y|
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Re: If x and y are nonzero integers, is x/y > 0? [#permalink]
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29 Apr 2016, 06:51
Bunuel wrote:
If x and y are nonzero integers, is x/y > 0?
(1) |x^3|= |y^3|
(2) |x + y| = |x| + |y|
Hi
is x/y > 0 can be answered if it is known Both x and y are of same or opposite sign..
Yes if same sign and NO if different sign..
lets se ethe statements
(1) |x^3|= |y^3|
We know from this that the numeric value is same but we cannot say anything about sign..
Insuff
(2) |x + y| = |x| + |y|
this is possible ONLY if both x and y are of same sign
so x/y>0..
Suff
B
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Re: If x and y are nonzero integers, is x/y > 0? [#permalink] 24 Dec 2018, 21:32
Display posts from previous: Sort by | 733 | 2,470 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2019-30 | latest | en | 0.84931 |
https://math.stackexchange.com/questions/2810424/confidence-interval-and-central-limit-theorem?noredirect=1 | 1,566,109,860,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313715.51/warc/CC-MAIN-20190818062817-20190818084817-00512.warc.gz | 548,846,694 | 31,167 | # Confidence Interval and Central Limit Theorem
Here's the problem:
"During a test, we asked $120$ people who Jean-Jaques Rousseau was, and $12$ of them answered that he was a driver (which is false). Estimate the proportion of the population that would give a false answer. Obtain a $95\%$ confidence interval for this proportion."
Here's my attempt.
The estimated proportion of the population that would give a false answer is $10\%$
Let $R_1, ... , R_{120}$ be r.v representing the answers of each person. Let $X$ be a r.v that counts the nb of ppl that give a wrong answer, i.e $X = \sum_{i=1}^{120} {I(R_i = 1)}$
Clearly, $X\stackrel{}{\sim} B(\ n=120, \ p= 10\%)$
From there, I can set my pivot $Q$ = $\frac{X-E(X)}{\sqrt{Var(X)}}$, where $Q \stackrel{}{\sim} N(0,1)$
So, I have $$P(-1.96 < \frac{X-E(X)}{\sqrt{Var(X)}} < 1.96) = 0.95$$ where $E(X) = n\cdot p$ and $Var(X) = np(1-p)$.
This gives me $$P(5.56 < X < 18.44) = 0.95$$
Now, finally, we can obtain the confidence interval: $$[5.56/120, 18.44/120] = [0.046,0.154]$$
I think this is correct. However, I have the feeling that this method is unnecessarily complicated for a problem that seems quite simple. I guess there is a more clever way of using the central limit theorem, but I do not know how.
• "The proportion of the population that would give a false answer is 10%" is not an correct assertion... The 10% came from the sample information since 12/120 = 10%. Furthermore this $\hat p 10\%$ is an estimate of the true pop. proportion of people that give false answer. To complete problem, just construct 95% CI corresponding to $\hat p = 10\%$. – cat Jun 6 '18 at 20:02
The traditional (sometimes called 'Wald') confidence interval (CI) for a binomial proportion $\theta$ based on the number of successes $X$ in $n$ independent trials uses the point estimate $\hat \theta = X/n$ of $\theta$ and estimates the standard error of $\hat \theta$ as $SE = \sqrt{\frac{\hat \theta(1-\hat \theta)}{n}}.$ Thus an approximate 95% CI for $\theta$ is of the form $$\hat \theta \pm 1.96\sqrt{\frac{\hat \theta(1-\hat \theta)}{n}}.$$ Notice that there are two approximations involved in this type of CI: (a) approximating the binomial random variable $X \sim \mathsf{Binom}(n, \theta)$ by the normal distribution $\mathsf{Norm}(\mu= n\theta, \sigma=\sqrt{n\theta(1-\theta)}\,)$ and (b) approximating the standard error of $\hat \theta,$ which is $\sqrt{\frac{\theta(1-\theta)}{n}},$ by $\sqrt{\frac{\hat\theta(1-\hat\theta)}{n}}.$
More recently, computational studies have shown that in conjunction these two approximations can produce "95%" confidence intervals with considerably less than confidence 95% of coverage. An improvement suggested by Agresti and Coull is to estimate $\theta$ by $\check \theta = \frac{X+2}{n+4}$ and the standard error by $\sqrt{\frac{\check\theta(1-\check\theta)}{n+4}}.$ The Agresti-Coull CI of the form
$$\check \theta \pm 1.96\sqrt{\frac{\check \theta(1-\check \theta)}{n+4}},$$
has proved to be more accurate for small and moderate sample sizes $n.$ (For $n \ge 1000$ the adjustment becomes trivial and unnecessary.) | 931 | 3,120 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2019-35 | latest | en | 0.884193 |
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#1
How would you find find a complex number (in the form a + bi) with the smallest possible argument that intersects a circle?
0
6 months ago
#2
(Original post by Toast210)
How would you find find a complex number (in the form a + bi) with the smallest possible argument that intersects a circle?
Can you post the full question?
At a guess, it probably wants you to compute the tangent to the circle, which you could do using quadratic discriminants, trig or pythagoras/similar triangles. Really need to see the question though.
Last edited by mqb2766; 6 months ago
0
#3
0
6 months ago
#4
(Original post by Toast210)
Ok, have you sketched it? The numbers make a Pythagoras-similar triangles the easy way, but use a tangent point - quadratic discriminant if you're not confident with the geometry behind it.
0
#5
Ok I can see a 3,4,5 triangle using the center of the circle and the point of intersection from the origin. I got the modulus of the complex number as 4. Then I would have to take away 2 angles in order to get the argument. Thank you.
0
6 months ago
#6
(Original post by Toast210)
Ok I can see a 3,4,5 triangle using the center of the circle and the point of intersection from the origin. I got the modulus of the complex number as 4. Then I would have to take away 2 angles in order to get the argument. Thank you.
I presume you finished it off? Did you notice anything about the side ratio of y : x : 4?
If you're interested, there is a neat bit of (relatively simple, but not necessarily intuitive) geometry underpinning the question.
0
#7
(Original post by mqb2766)
I presume you finished it off? Did you notice anything about the side ratio of y : x : 4?
If you're interested, there is a neat bit of (relatively simple, but not necessarily intuitive) geometry underpinning the question.
Yes I managed to finish it thank you. What's this neat bit of geometry if you don't mind me asking?
0
6 months ago
#8
(Original post by Toast210)
Yes I managed to finish it thank you. What's this neat bit of geometry if you don't mind me asking?
What did you get for the answer?
0
#9
3.84 + 1.12i
0
6 months ago
#10
(Original post by Toast210)
3.84 + 1.12i
or as a ratio for y : x : 4
7 : 24 : 25
Notice anything?
Spoiler:
Show
Your original triplet a : b : c was 3 : 4 : 5, so this triplet is
b^2 - a^2 : 2ab : c^2
You could "write down" the answer as
24*4/5^2 + i*7*4/5^2
Last edited by mqb2766; 6 months ago
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# WORK ENERGY AND POWER MCQ TEST 2
Work Engergy and Power 11 CBSE - MCQ test 1
Question 1.
A block of mass m is placed on elevator moving down with accelertion g/3.What would be value of workdone by normal reaction acting on the block if elevator moves distance h downwards .
mgh/3
-2mgh/3
2mgh/3
None of these
Explanation
Question 2.
Force F acts on the free end of string as shwon pulley system below,if weight W moves up slowly a distance h.What would be value of work done by force F?
Fh
2Fh
Fh/2
2Fh/g
Explanation
When W moves a distance h above and string at free end will move 2h below along the direction of force.
Work done=F*2h=2Fh
Question 3.
Below figure is showing position time graph of mass 2Kg. What would be workdone by force for movement of mass from t=0 to t=4 sec.
8j
0
4j
Can't detemine
Explanation
Dispalcement after t=4sec is 0.
Question 4.
A cyclist comes to a skidding stop in 10 m. During this process, the force on the cycle due to the road is 200 N and is directly oppose to the motion.How much work does the cycle do on the road?
0J
2000J
-2000J
None
Explanation
Question 5.
Three forces of magnitude 14N,7N and 7N acting on a body in direction of vector ( 6i+2j+3k),(3i-2j+6k) and (2i-3j-6k)respevtively. The foreces remains constant while object moves from point A (2,-1,-3) to A (5,-1,1).What would be workdone on object.
150J
125J
0
75J
Explanation
Question:6
An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.20 m/s. What would be the minimum horsepower of the motor to be used ?
1.3hp
1.4hp
1.5 hp
None
Explanation
P = F v = mgv = (500 kg) (9.8 m/s2) (0.2 m/s) = 980 W Assuming no loss against friction etc., in the motor, the minimum horsepower of the motor is P = 980 W = 980/746 hp = 1.3 hp.
Question:7
A body dropped from a height H reaches the ground with a speed of 1.2 √gH. Calculate the work done by air-friction.
a force acts on it
it moves through a certain distance
it experiences an increase in energy through a mechanical influence.
None of these
Explanation
Answer: It experiences an increase in energy through a mechanical influence.
Question:8
A block of mass M is pulled along a horizontal surface by applying a force at an angle θ with the horizontal. The friction coefficient between the block and the surface is µ. If the block travels at a uniform velocity, find the work done by this applied force during a displacement d of the block.
μmg(dcos θ) /(cosθ - μsinθ)
μmgd /(cosθ + μsinθ)
μmg(dcos θ) /(cosθ + μsinθ)
None
Explanationn
Answer: μmg(dcos θ) /(cosθ + μsinθ)
Question:9
Static friction is variable.
True
False
Expanation
Value of static friction changes when the value of applied force changes.
Question:10
A man of 60 kg weight is standing at rest on a platform. He jumps up vertically a distance of 1 m and the platform at the same instant moves horizontally forward with the result that the man lands 1 meter behind the point on the platform where they took the jump, the total work done by the man at the instant he lands is ....
0
300 J
600 J
None
Explanation
Given,
Mass of man = 60 Kg
Distance covered by him as he jumps up = 1 m
The formula of work done is
Work = force x displacement
W = F x d
Now Force "F" = mg
W = mg x d
W = 60 x 10 x 1
W = 600 J
Thus the work done is 600 J
Question:11
A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table?
7.2J
10J
3.6J
None
Explanation
Mass of the chain lying freely from the table=(Mass of chain*hanging length of chain)/Total length of chain
=(4*0.60)/2=1.2kg
The distance of center of mass of chain from the table= 0.5×0.6m=0.3m
Thus the work done in pulling the chain=mgh=1.2×10×0.3J=3.6J
Question:12
A body of mass 20 kg is initially at a height of 3 m above the ground. It is lifted to a height of 2 m from that position. Its increase in potential energy is:
120 J
90 J
392 J
None of these
Explanation
Given,Mass, m = 20kg
Height m, H = 3m
Potential Energy, (PE1) = mgH ,PE1 = 20 x 9.8 x 3 PE1 = 588 joules
Now Height (h) is (3m + 2m)=5m
(PE2) = mgh
(PE2) = 20 x 9.8 x 5 (PE2) = 980 Joules
Increase in Potential Energy = PE2 - PE1 = 980 - 588 = 392 Joules
Question:13
The kinetic energy acquired by a mass m in travelling distance d, the starting from the rest under the action of a constant force, is directly proportional to:
m
1/m
Does not depend on mass
Explanation
Work done by external forces = Change in kinetic energy Since particle starts from rest, its initial Kinetic Energy is 0. Therefore, K.E.=Fd It is given that force is constant. We see that Kinetic energy is independent of mass.
Question:14
A body of mass 10 kg moving at a height of 2 m, with uniform speed of 2 m/s. Its total energy is:
116 J
316 J
216 J
None of these
Explanation
Total energy of a body(E) = Kinetic Energy (KE) + Potential Energy (PE)
Kinetic Energy (KE):
The kinetic energy of an object can be calculated using the formula:
KE = (1/2) * m * v^2
KE = (1/2) * 10 kg * (2 m/s)^2
KE = 20 J (Joules)
Potential Energy (PE):
The potential energy of an object near the surface of the Earth can be calculated using the formula:
PE = m * g * h
PE = 10 kg * 9.81 m/s^2 * 2 m
PE = 196.2 J (Joules)
Now, let's calculate the total energy:
E = KE + PE
E = 20 J + 196.2 J
E = 216.2 J
So, the total energy of the body is 216 Joules(APPROXIMATE).
Question:15
Two masses of 1 gm and 4 gm are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is:
4:1
3:2
1:2
None
Explanation
We have, K.E=P²/2M
P1²/2M = P2²/2M
P1²/2*1 = P2²/2*4
P1²/P2²= 2/8
(P1/P2)² = 1/4
(P1/P2) = (1/2)
Question:16
A body of mass 10 kg is initially at a height of 20 m above the ground. It falls to a height of 5 m above the ground. Its potential energy in the new position is:
490J
500J
1000J
None of these
Explanation
PE = m * g * h
PE = 10 kg * 9.8 m/s² * 5 m
PE = 490 J (joules)
So, the potential energy of the body in its new position, 5 m above the ground, is 490 joules.
Question:17
A body of mass 100 kg falls from a height of 10 m. Its increase in kinetic energy is
1000 J
1800 J
9800 J
None of these
Explantion
When an object is at some height the total energy in it is in the form of potential energy. When the body falls from
a height below the potential energy gets converted into kinetic energy.
Mass of body(m) = 100 kg
Height(h)= 10m
g=9.8m/s²
Since kinetic energy is equal to loss of potential energy when it falls down,
Increase in kinetic energy= loss in PE=100*9.8*10= 9800J
Question:18
A gardener pushes a lawn roller through a distance of 20 m. If he applies a force of 20 kg wt with in a direction in blind at an angle 60 degree to the ground find the work done by the gardener.
4.36KJ
5.36KJ
3.36KJ
5.17KJ
Explanation
Force aplied, F = 20 kg-wt = 20×9.8N=196N
Since,force is applied at an angle 60° with the ground.
So, the horizontal component of the force that is responsible for displacement of the lawn roller= Fcos60°=196×0.5=98N
So, work done=98×20=1960J
Question:19
If force 𝐹⃗ = (𝑖̂+ 5𝑗̂+ 7𝑘̂) acts on a particle and displaces it through 𝑑⃗ = (6𝑖̂+ 9𝑘̂). Calculate the work done if the force is in newton and displacement is in metre
89 J
79 J
59 J
69 J
Explanation
Force F = i + 5 j + 7 k
Displacement s = 6 i + 9 k
Work done = F . s = ( i + 5 j + 7 k ) . ( 6 i + 0 j + 9 k ) = 6 +0 +63 = 69 J.
Question:20
A force F acting on an object varies with distance x as shown in figure below. The force is in newton and distance x is in metre. The work done by the force in moving the object from x = 0 to x = 6 m is:
12.5 J
13.5 J
0
None of these
Explanation | 2,462 | 7,872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2023-40 | longest | en | 0.88625 |
https://mdrabiulislam.com/what-is-meant-by-diversity-factor/ | 1,653,669,534,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662658761.95/warc/CC-MAIN-20220527142854-20220527172854-00240.warc.gz | 463,291,854 | 22,707 | # What is meant by Diversity Factor?
In electrical power system, there are some important factors. These are, the Load Factor, Demand Factor, Diversity Factor, Plant Capacity Factor, Plant Use Factor etc. Before, I have discussed about Load Factor and Demand Factor in details with example. I’m today going to describe the Diversity Factor. What is meant by diversity factor?
## What is Diversity Factor?
Diversity Factor is the ratio of the sum of individual maximum demands to the maximum demand of the systems. As the maximum demand of a system is less than the sum of individual maximum demands, so the diversity factor must be great than one. The mathematical expression of it is,
## Sum of individual maximum demands / Maximum demand of the system.
For example, a power station or substation has 4 consumers connected. They have different demand in different time. Let consumer 1 has 10kW maximum demand, and 12kW, 15kW, 9kW demand of customer 2, 3 and 4 respectively. The maximum demand to the power station is 40kW. Then what is the diversity factor of the power station?
Here the sum of individual maximum demands is (10+12+15+9)kW = 46kW
And the maximum demand of whole system is 40kW.
Therefore, the Diversity Factor = 46/40= 1.15
It’s needless to say that the diversity factor must be greater than unity. Because the maximum demand to a power station is less than the sum of individual demands.
See another example.
Let, A main feeder has two sub-feeder. 1. Sub-feeder A and 2. Sub-feeder B
Maximum demand of sub-feeder A= 70kW
Maximum demand of sub-feeder B= 75kW
And maximum demand of main feeder = 120kW
here, the sum of the maximum of demands of individual feeder = (70+75)kW = 145kW
Then, Diversity Factor =145/120= 1.2
It helps to determine the size of feeder and transformer in electrical power system. | 446 | 1,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-21 | latest | en | 0.893023 |
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# Darwin was not the first to advance a theory of evolution;
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233. Darwin was not the first to advance a theory of evolution; his tremendous originality lay in the fact that he proposed the idea of natural selection as the means by which evolution worked.
(A) lay in the fact that he proposed the idea
(B) lay in the fact of his proposing the idea
(C) laid in the fact of his proposing the idea
(D) laid in his proposal
(E) lay in his proposal
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Hey All,
Tough question here, and it seems that people got more and more confused as they moved forward. A couple quick things. If you're posting on here, it's really helpful to include the underlined portion in the question stem. Otherwise, it's very difficult to look at the question. Secondly, I'd exhort you all to make an effort to fit problem answer choices into patterns/categories (MGMAT has some, but so does every other test prep company and resource). Okay. Let's take this baby apart!
233. Darwin was not the first to advance a theory of evolution; his tremendous originality lay in the fact that he proposed the idea of natural selection as the means by which evolution worked.
First off, remember that semicolons function in pretty much the same way as periods. We can ignore the first independent clause here and just focus on the second.
(A) lay in the fact that he proposed the idea
PROBLEM: First of all, it's hideous. But now think about the meaning. Did Darwin's originality lay in "the fact that he proposed the idea"? No. It lay in the proposal itself.
(B) lay in the fact of his proposing the idea
PROBLEM: IDIOM with fact. You have to say "The fact that burgers are delicious is immaterial", not "The fact of burgers being delicious is immaterial".
(C) laid in the fact of his proposing the idea
PROBLEM: "Laid" is the past tense of the verb "to lay" (As in "After we lay it down on the table, we'll have to give it a back massage."). "Lay" is the past tense of the verb "to lie" (As in "The truth lies elsewhere today."). This sentence involves the latter (purely based on the definitions of the two words, which ARE different), so we simply use "lay" as the past tense. Annoying, right? Also, this answer choice has the same issues as B.
(D) laid in his proposal
PROBLEM: See C.
(E) lay in his proposal
ANSWER: I think people dislike this because they aren't used to the idiom "proposal of". But it's correct. Sigh. On the plus side, even though I exhort my students not to go with concision unless ALL else fails, concision will lead you to this answer choice (as long as you recognize the difference between "lay" and "laid").
Hope that helps!
-t
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13 Jun 2010, 04:25
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Hi Guys,
A quick note on transitive and intransitive verbs, as these are tested in the SC!
Transitive= the verb takes an object
Ex. The farmer grows trees.
Intransitive= the verb does not take an object.
Ex. The trees grow.
Thus 'grow' can be a transitive OR an intransitive verb.
However, some verbs are only intransitive or only transitive. Most important are lay/lie and raise/rise:
lay (transitive)
The students (subject) lay (verb) the pencils (object) on the desk.
V1: lay-- the students lay the pencils...
V2 (past simple): laid-- the students laid the pencils...
V3: laid-- the students have laid the pencils....
lie (intransitive)
The men lie in bed. (No object)
V1: lie-- The men lie in bed.
V2 (past simple): lay-- The men lay in bed. (That's right, the V2 of 'lie' = V1 of 'lay'!)
V3: lain-- The men have lain in bed.
raise (transitive)
The students raise their hands.
V1: raise-- The students raise their hands.
V2 (past simple): raised-- The students raised their hands.
V3: raised-- The students have raised their hands.
rise (intransitive)
Inflation rates rise. (No object)
V1: rise-- Inflation rates rise.
V2 (past simple): rose-- Inflation rates rose.
V3: risen-- inflation rates have risen.
If that's confusing to anyone, let me know.
Best,
Sarai
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Re: Darwin was not the first to advance a theory of evolution; [#permalink]
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14 Jun 2010, 10:04
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Hi papillon86,
"in bed" is what is called a prepositional phrase.
"in" is a preposition.
"bed" is a noun.
Together (preposition + noun) they form a prepositional phrase, which is descriptive.
The object of a verb is the received (a noun) of the action.
Hope that helps!
-Sarai
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Re: Darwin was not the first to advance a theory of evolution; [#permalink]
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09 Sep 2009, 01:17
duttarupam wrote:
A
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09 Sep 2009, 17:32
gmataspirant2009 wrote:
233. Darwin was not the first to advance a theory of evolution; his tremendous originality lay in the fact that he proposed the idea of natural selection as the means by which evolution worked.
(A) lay in the fact that he proposed the idea
(B) lay in the fact of his proposing the idea
(C) laid in the fact of his proposing the idea
(D) laid in his proposal
(E) lay in his proposal
Lay is a simple past tense of lie and lay is not used here as a verb. So the usuage of lay, here, is correct as tenses are in simple form.
Laid is the simple past of the verb lay, which means to put. Not correct usuage for this sentence.
"the fact that he proposed the idea" is too wordy. It is quite clear that Darwin's proposal will be the original idea.
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Re: Darwin was not the first to advance a theory of evolution; [#permalink]
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09 Sep 2009, 17:51
lay in the fact that .....(doesnt really sound that good...and i dont remember seeing choices that include "in the fact that" being right many times...)
i dont think originality lay in the fact that he did something...but rather in darwin's proposal...going with E...
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09 Sep 2009, 18:24
I read that the OA is A, but i tot E is a better choice.
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20 Sep 2009, 08:39
I'm really confused; as per my understanding of the construction of the sentence after "tremendous originality" there should be a verb!
What is the source of this question?
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01 Oct 2009, 11:59
E it is
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25 Dec 2009, 23:11
I definitely go with E
C D – laid is incorrect here – lie lay lain; not lay laid laid
A B – in the fact that … is awkward; E is more concise
1000SC : OA is E
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28 Dec 2009, 04:08
gmataspirant2009 wrote:
233. Darwin was not the first to advance a theory of evolution; his tremendous originality lay in the fact that he proposed the idea of natural selection as the means by which evolution worked.
(A) lay in the fact that he proposed the idea
(B) lay in the fact of his proposing the idea
(C) laid in the fact of his proposing the idea
(D) laid in his proposal
(E) lay in his proposal
IMO A .... as i feel 'in the fact' is needed.
originality lay in the fact
originality lay in his proposal
Watss OA ????
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31 Mar 2010, 05:14
Is "lay" correct here because the sentence need to be in past tense? I understand we should choose lie over lay here so "laid" will never be correct anyway.
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09 Apr 2010, 04:48
I am totally confused. I picked D, but E seems a better option. Now, if A is OA, what do you learn from this?
If you post the source, most of of the time people can agree on one answer. Please post your sources ...
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21 Apr 2010, 14:56
I dont catch the difference bwtween lay and laid...both are past tenses, arent they??
Can anybody explain in more general terms?
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21 Apr 2010, 16:06
lay in the fact that he proposed is better than lay in the fact of his proposing.
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Re: Darwin was not the first to advance a theory of evolution; [#permalink]
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21 Apr 2010, 16:29
TommyWallach wrote:
(C) laid in the fact of his proposing the idea
PROBLEM: "Laid" is the past tense of the verb "to lay" (As in "After we lay it down on the table, we'll have to give it a back massage."). "Lay" is the past tense of the verb "to lie" (As in "The truth lies elsewhere today."). This sentence involves the latter (purely based on the definitions of the two words, which ARE different), so we simply use "lay" as the past tense. Annoying, right? Also, this answer choice has the same issues as B.
This explanation of lay-laid lie-lay was extremely helpful for me. Thanks!
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Re: Darwin was not the first to advance a theory of evolution; [#permalink]
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02 May 2010, 08:58
TommyWallach wrote:
Hey All,
Tough question here, and it seems that people got more and more confused as they moved forward. A couple quick things. If you're posting on here, it's really helpful to include the underlined portion in the question stem. Otherwise, it's very difficult to look at the question. Secondly, I'd exhort you all to make an effort to fit problem answer choices into patterns/categories (MGMAT has some, but so does every other test prep company and resource). Okay. Let's take this baby apart!
233. Darwin was not the first to advance a theory of evolution; his tremendous originality lay in the fact that he proposed the idea of natural selection as the means by which evolution worked.
First off, remember that semicolons function in pretty much the same way as periods. We can ignore the first independent clause here and just focus on the second.
(A) lay in the fact that he proposed the idea
PROBLEM: First of all, it's hideous. But now think about the meaning. Did Darwin's originality lay in "the fact that he proposed the idea"? No. It lay in the proposal itself.
(B) lay in the fact of his proposing the idea
PROBLEM: IDIOM with fact. You have to say "The fact that burgers are delicious is immaterial", not "The fact of burgers being delicious is immaterial".
(C) laid in the fact of his proposing the idea
PROBLEM: "Laid" is the past tense of the verb "to lay" (As in "After we lay it down on the table, we'll have to give it a back massage."). "Lay" is the past tense of the verb "to lie" (As in "The truth lies elsewhere today."). This sentence involves the latter (purely based on the definitions of the two words, which ARE different), so we simply use "lay" as the past tense. Annoying, right? Also, this answer choice has the same issues as B.
(D) laid in his proposal
PROBLEM: See C.
(E) lay in his proposal
ANSWER: I think people dislike this because they aren't used to the idiom "proposal of". But it's correct. Sigh. On the plus side, even though I exhort my students not to go with concision unless ALL else fails, concision will lead you to this answer choice (as long as you recognize the difference between "lay" and "laid").
Hope that helps!
-t
It helped indeed!
Thanks for participating in this forum. You really help us understand the problems in a clear way.
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13 Jun 2010, 03:57
noboru wrote:
I dont catch the difference bwtween lay and laid...both are past tenses, arent they??
Can anybody explain in more general terms?
verbs -> lie and lay can be used in a variety of ways. Depends more upon the intent
Examples:
Present Past Future
lie(means to lie down) he lay on bed he has lain on the bed
lay(to put down)
Please lay the plates David laid the plates David has laid the plates
lie(to speak untruthfully)
don't lie to me He lied to me He has lied to me
Hope it explains.
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Re: Darwin was not the first to advance a theory of evolution; [#permalink]
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14 Jun 2010, 08:01
Sarai,
Isn't bed the object in the following example you mentioned?
The men lie in bed. (No object)
When you said, intransitive verbs don't take objects, you probable meant that one can't use intransitive verbs like
'raise', 'lie', etc in sentences written in passive voice.
For example,
active: The men(subject) lie(verb) in bed(object).
passive: The bed (was/were/is) lie by the men --> Incorrect, because intransitive verb 'lie' can't take object 'bed'
Another example, using intransitive verb 'arrived'
active: Bob(S) arrived(V) on the platform(O).
passive: Platform(O) was arrived(V) by Bob(S) --> Incorrect
Thanks
Re: Darwin was not the first to advance a theory of evolution; [#permalink] 14 Jun 2010, 08:01
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# The FV of a share is Rupees 10. What is the MV of the share if it is at 10% premium ?
## Text solutionVerified
Sure, here are the step-by-step solutions to the problem: Given: The market value of a share is Rupees 11 and there is a 10% premium, which is equal to Rupees 1. To find: The face value of the share. Solution: Let the face value of the share be x. According to the problem, there is a 10% premium, which is equal to Rupees 1. This can be expressed mathematically as: 1 = 10% of x Multiplying both sides by 100, we get: 100 = x * 10 Dividing both sides by 10, we get: x = 100/10 x = 10 Therefore, the face value of the share is Rupees 10.
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3. e the optimal value of the target cell by changing values in cells used to calculate the target cell. It contains the below-mentioned parameters
### Solver in Excel (Examples) How to Use Solver Tool in Excel
1. imum) value for a formula in one cell — called the objective cell — subject to constraints, or limits, on the values of other formula cells on a worksheet
2. To use Goal Seek we need to put our formula into a cell. Start a new file and in C3 (our y cell) type: = 2*10^7*B3^-2.146. In B3 (our x cell): Put a value say 10. Note that C3 will show the solution of the formula for when x= 10 or = 2*10^7* 10 ^-2.146 = 142,899.277
3. To save the Excel Solver scenario, perform the following steps: Open the worksheet with the calculated model and run the Excel Solver. In the Solver Parameters window, click the Load/Save button. Excel Solver will tell you how many cells are needed to save your scenario. Select that many empty cells and click Save
4. Solving equations is very easy with excel. Let us see how to solve the following equations using excel solver tool. 12x+25y+687z=9. 45x-24y+78z=12. 567x+2y-7z=123. Solution: 1. Open excel and go to Data tab. 2. Check whether there is a group by the name 'Analysis'. If it is there, then click Solver in the Analysis group as in the picture below. 3
5. Solve the Model. To find the optimal solution, execute the following steps. 1. On the Data tab, in the Analyze group, click Solver. Enter the solver parameters (read on). The result should be consistent with the picture below. You have the choice of typing the range names or clicking on the cells in the spreadsheet. 2. Enter TotalProfit for the Objective. 3
### Solve Equation in Excel How to Solve Equation with
For example, =A2+A2+A3+A4 is a formula that adds up the values in cells A2 through A4. Function is a predefined formula already available in Excel. Functions perform specific calculations in a particular order based on the specified values, called arguments, or parameters Formula used to calculate profit %: ( (Selling price-Cost price)/Cost price)*100. Target Value: 20. Place the formula in the target cell ( B5) to calculate the profit %. This is the required.
### Define and solve a problem by using Solver - Office Suppor
1. In Excel, a formula is an expression that operates on values in a range of cells or a cell. For example, =A1+A2+A3, which finds the sum of the range of values from cell A1 to cell A3. 2
2. This is one possible solution. = ( - B + SQRT ( B ^ 2 - 4 * A * C )) / ( 2 * A) This is the other possible solution. = ( - B - SQRT ( B ^ 2 - 4 * A * C )) / ( 2 * A) A - This is the coefficient of the squared term in the quadratic equation. B - This is the coefficient of the single powered term in the quadratic equation
3. Excel Formulas Based Sudoku Solver. And to top it off, these logical techniques were executed by formulas, with the heavy lifting done by named formulas using bitwise operations. I alluded to this game solver back in January when I posted 5 And 3 Is 1. Today, I'm finally sharing it with you
4. Excel can solve for unknown variables, either for a single cell with Goal Seeker or a more complicated equation with Solver. We'll show you how it works. Excel is very capable when you have all of the data you need for your calculations
5. Excel has the EVALUATE formula, however, this is a legacy formula which does not quite function how you might Expect. So, let's see how we can solve this. Using EVALUATE to convert string to a formula
6. A quadratic equation can be solved by using the quadratic formula. You can also use Excel's Goal Seek feature to solve a quadratic equation. 1. For example, we have the formula y = 3x 2 - 12x + 9.5. It's easy to calculate y for any given x. For x = 1, y = 0.5. 2. For x = 2, y = -2.5. 3
### How to Solve an Equation in Excel - chandoo
Make sure that the Manage text box has Excel Add-ins listed in it, then click Go at the bottom of the page. On a Mac, this window will open after clicking Excel Add-ins in the Tools menu. 7 Install the Solver add-in. Check the Solver box in the middle of the page, then click OK A large collection of useful Excel formulas, beginner to advanced, with detailed explanations. VLOOKUP, INDEX, MATCH, RANK, SUMPRODUCT, AVERAGE, SMALL, LARGE, LOOKUP. Occurs when a formula attempts to divide by zero. #VALUE!-Occurs if one of the variables in your formula is of the wrong type (e.g. text value when a numeric value is expected). #REF!-Arises when a formula contains an invalid cell reference. #NAME?-Occurs if Excel does not recognise a formula name or does not recognise text within a formula. #NUM! The name sol_cell_from_val is: =INDEX (sol_board, INT ((ROW (Main!A1)-1)/3)+1, INT ((COLUMN (Main!A1)-1)/3)+1) This must be created from cell P4
By clicking on Solve, Excel will execute the operation you asked it to and will give you the solution x = 0,85078105. One of the limits of the Excel Solver is that it will only give one root even though we know there is a second one. For now, we use the discriminant T L F > E√ > 64 = ? Solve linear equations using Excel formulas The image above shows how to solve three simultaneous equations with three variables using one Excel formula. Cell range C3:C5 contains three equations with variables x, y and z. First, extract the numbers from these three equations I am currently trying to solve a problem in a physics class in which I am required to use Excel to solve for an unknown variable in the equation for acceleration of an object in simple harmonic motion. The equation is a(t)=-A(2pi/T)^2 times (cos(2pi/T)(t)), where a is acceleration, t is time, A is amplitude, and T is the period of oscillation
The Solver in Excel can perform many of the same functions as EES and MathCAD. It can be used to solve single equations (for example x2+3x-22=5) or multiple equations (for example x3-14x=z, z12-1=x2+1). Solver also has the ability to perform calculations where you are trying to minimize a certain value while solving, such as in linear regression Excel for the web does this by using formulas in cells. A formula performs calculations or other actions on the data in your worksheet. A formula always starts with an equal sign (=), which can be followed by numbers, math operators (such as a plus or minus sign), and functions, which can really expand the power of a formula In Excel problems can be solved using formulas. In Excel real life examples exist to help you familiarize yourself with how to solve these problems. close searc Show formulas is an option in excel to display all the workbook formulas with a single click. This option is very useful to find out all the formulas in a worksheet. Having said that, we need to be a little careful while using th e show formulas option, otherwise, you might end up with something like below
In Excel, click on the ' File ' tab on the top left corner of the screen. In the window that opens, click on ' Options ' from the left menu bar. From 'Excel Options' dialog box, select ' Formulas ' from the left side menu and then change the ' Calculation options ' to ' Automatic ' if it's currently set as 'Manual' More Insights. Less Effort. Turn Your Static Excel Spreadsheets Into Interactive Sheet This is the required information we need to solve any sort of equation using Solver Add-in in Excel. Now, launch the Solver Add-in by clicking on the Data tab and click on Solver Formulas are equations that can perform calculations, return information, manipulate the contents of other cells, test conditions, and more. This topic provides an introduction to formulas and functions in Excel Basic Excel Formulas Guide. Mastering the basic Excel formulas is critical for beginners to become highly proficient in financial analysis Financial Analyst Job Description The financial analyst job description below gives a typical example of all the skills, education, and experience required to be hired for an analyst job at a bank, institution, or corporation
Read the above question carefully and try to solve it Here are the steps to solve the above problem: Your job is: open a new workbook and input the above values one by one into the worksheet [start from cell A2].; If you're done, just select the whole range [from cell A2: A31].; Click on the Conditional Formatting dropdown in the Home ribbon, select New Rule from the drop-down, New. This is where solver can be handy. Solver can find optimal solution for problems like this before you finish your first cup of coffee. Reconcile debits & credits using solver model - Tutorial Step 1. Set up your solver model. In a blank sheet, list credits along a column and debits on the top in few columns, as shown below For Excel to find a solution, a real solution must exist. Normally, you would convert your formula to an Excel function like. =A1^4+A1^3+A1^2+A1+40. and then use Solver to change A1 to get the cell with the formula to have a value of zero. But there are no real roots for your equation, so you will need to use a much more sophisticated package. Excel Formula Training. Formulas are the key to getting things done in Excel. In this accelerated training, you'll learn how to use formulas to manipulate text, work with dates and times, lookup values with VLOOKUP and INDEX & MATCH, count and sum with criteria, dynamically rank values, and create dynamic ranges
### Excel Solver tutorial with step-by-step example
You enter the formula just like you would in an Excel cell, except surrounded by quotations: 'Assign a hard-coded formula to a single cell Range(b3).Formula = =b1+b2 VBA Formula Tips Formula With Variable. When working with Formulas in VBA, it's very common to want to use variables within the cell formulas To solve the system of equations, we can utilize functions and the equation solver tool. Figure 1. How to Solve the System of Equations. Using Excel Functions. We can solve the system of equations by using MINVERSE and MMULT mathematical functions If you have not installed the solver add-on you can install it via the Add-ins menu. The Microsoft support site has instructions for all platforms, Load the Solver Add-in in Excel. Recording the macro. Now that the set up is complete we we are ready to record! Set up your workspace appropriately with all your formulas and references
### Solving Equations using Excel Solver - How to Exce
1. Your finished formula should look something like this: =C4+C5+C6+C7. If you know which cells you want to calculate, you can type them out at once instead of selecting them individually. Excel functions will recognize mixed numbers and cell entries. That is, you could add 5000+C5+25.2+B7
2. How To: Solve for base in a Microsoft Excel percent formula How To: Solve for rate in a Microsoft Excel percent formula How To: Solve rate-of-change business math problems in Excel How To: Solve rate/percent increase/decrease problems in Excel
3. Solver is a Microsoft Excel add-in program you can use for optimization in what-if analysis.. According to O'Brien and Marakas, optimization analysis is a more complex extension of goal-seeking analysis. Instead of setting a specific target value for a variable, the goal is to find the optimum value for one or more target variables, under certain constraints
4. Hi, I was wondering if there is any way to solve this equation for f. (This equation is the Colebrook Equation used to find the friction factor for turbulent flow in circular ducts) 1/(f^0.5)=-2.0Log((e/d)/3.7 + 2.51/Re*f^0.5) where e, d, and Re can all be referenced from other cells on the..
5. Solving real-life problems in Excel. Excel can be used to solve all kinds of real-life problems. But how do you turn those problems into formulas that Excel can understand? All it takes is a little bit of planning (and some basic math). To learn more, check out the video below! YouTube. GCFLearnFree.org
6. g problems with the SOLVER tool, which: - May be used to solve linear and nonlinear optimization problems - Allows integer or binary restrictions to be placed on decision variables - Can be used to solve problems with up to.
7. However, this equation cannot be solved explicitly, so numerical iteration is needed. There's several ways we can do this in Excel. I'll outline the two most convenient methods. If you just want the spreadsheet, scroll to the bottom of this post and download the spreadsheet
SOLVER is a powerful add in that you might need to install before you read on. File. Options. Add Ins. Manager Excel Add Ins Go. Select SOLVER Add In. Go. The link to SOLVER will appear in the top right hand corner of the Data tab just click on it to start it. Back to question 1. Set up the SOLVER area as you can see here in the range. In Excel 2010 On the FORMULAS tab, in the Formula Auditing group there is a button for Show Formulas. The keyboard shortcut is (Cntrl+`) which is usually how it is accidentally triggered. When the toggle is off, you will see the results of your formulas, in each cell
If you've ever ventured into the Excel Solver add-in, you probably noticed that there are many options and it can be a little overwhelming. In this post, I'd like to provide some practical information to help you choose the correct solving method in Excel to efficiently find an optimum solution to your problem We define the system LHS equations in A1:A3 using X1:X3 for variables with 1 for the initial guess as shown in Table 1. Note that the inequalities formulas are listed after the equality formula as required by the solver. Remember that equations and inequalities formulas are defined with respect to zero on one side, and any inequalities are interpreted as greater than zero by the solver I have issue on excel formula, can please solve the issue. I have done stock count in excel for count 1 and count 2, the formula i set is working but i set for count 3 is not working. below is the detail. A B C BOOKED QTY COUNT 1 COUNT 2 FINAL QT For example, the I need is an array formula that contains a range that is named data6. When I create the new month, I need to rename data6 to data7. I have been doing this manually each month, but there are 7 cells, each with different formulas that need to be changed. Example of my working formula in Excel
### Solver in Excel - Easy Excel Tutoria
• imum) value for a formula in one cell — called the objective cell — subject to constraints, or limits, on the values of other formula cells on a worksheet. Solver works with a group of cells — called.
• Double-check your formulas. One of the most powerful features of Excel is the ability to create formulas. You can use formulas to calculate new values, analyze data, and much more. But formulas also have a downside: If you make even a small mistake when typing a formula, it can give an incorrect result. To make matters worse, your spreadsheet.
• Calculate CAGR in Excel FV, PV, N. If you have FV, PV, and n, simply plug them into the standard CAGR equation to solve for CAGR. =(B5/C5)^(1/D5)-
### Basic Excel formulas & functions with example
Product Mix Example - Using Excel's Built-In Solver Using the Solver Dialogs To let the Solver know which cells on the worksheet represent the decision variables, constraints and objective function, we click Solver button on the Data tab, or the Premium Solver button on the Add-Ins tab, which displays the Solver Parameters dialog. In the Set Objective (or Set Target Cell) edit box, we type or. The Standard Formula. For this first approach, you essentially need to replicate the quadratic formula in Excel. You can enter the known values (a, b and c) right away into the formula. However, this means that the formula will be static and you will need to change the values one-by-one again when you need to solve for another equation
Excel found a problem with one or more formula references in this worksheet. Check that the cell references, range names, defined names, and links to other workbooks in your formulas are all correct Excel (and other spreadsheet programs) is the greatest financial calculator ever made.There is more of a learning curve than a regular financial calculator, but it is much more powerful. This tutorial will demonstrate how to use Excel's financial functions to handle basic time value of money problems using the same examples as in the calculator tutorials O SOLVER é um suplemento do Excel, portanto você deve habilitá-lo. No menu superior esquerdo, clique em Arquivo e depois escolha o menu Opções, conforme a figura: Agora, selecione a opção Suplementos e clique em Solver. Veja: O próximo passo é selecionar a opção Ir, no centro inferior da tela
For equation solving, Wolfram|Alpha calls the Wolfram Language's Solve and Reduce functions, which contain a broad range of methods for all kinds of algebra, from basic linear and quadratic equations to multivariate nonlinear systems. In some cases, linear algebra methods such as Gaussian elimination are used, with optimizations to increase. regression. Excel's Solver feature will be used in a later chapter to solve more complicated linear and nonlinear systems of equations. Generally the term matrix (from mathematics) and array (from Excel) can be used interchangeably to refer to data organized in row and column fashion Como Usar o Solver no Microsoft Excel. Leia este artigo para saber como ativar e utilizar a ferramenta Solver do Microsoft Excel, a qual realiza a alteração de diferentes variáveis em uma planilha a fim de se obter o resultado desejado. O.. syms x positive. When you solve an equation for a variable under assumptions, the solver only returns solutions consistent with the assumptions. Solve this equation for x. eqn = x^2 + 5*x - 6 == 0; S = solve (eqn,x) S =. Allow solutions that do not satisfy the assumptions by setting 'IgnoreProperties' to true
### How to solve Equations in Excel using Solver Add-i
The Formula If Else is very important in excel, with the help of If-Else we can do filter anything easily in excel, so here in this article we will tell you some example of If-Else formula. So let started with example as given here under. Example No. 1 for If- Else Formula in Excel 2. Formulas are Shown in Excel's Formula Bar. When you're typing a formula into a cell, you can see the results of the cell once you press enter. But when you select a cell, you can see the formula for that cell in the formula bar. Click on a cell in Excel to show the formula in it, such as the multiplication formula that evaluates to 125 Recursive Equation Solving with Excel Faith A. Morrison William R. Kenan Visiting Professor for Distinguished Teaching Department of Chemical and Biological Engineering Princeton University 12 March 2013 This document briefly explains using some recursive tools for problem solving with MS Excel 2010 It can do everything that the excel solver can do. PuLP is a free open source software written in Python. It is used to describe optimisation problems as mathematical models. PuLP can then call any of numerous external LP solvers (CBC, GLPK, CPLEX, Gurobi etc) to solve this model and then use python commands to manipulate and display the solution November 30, 2013. Sudoku Solver - Using Excel Formulas. Hi All. My name is Sreejith A Menon. I would like to share with you my excel template for solving the Sudoku puzzle. Sudoku is an amazing logical grid with 9 boxes which are further divided into 9 smaller squares where in a few squares will be filled in with certain numbers from one to.
### Basic Excel Formulas - List of Important Formulas for
• Solve this Using Excel(explain formula) Gino rents bicycles on a daily base to tourists in Florence. He does not own the bicycles but leases them himself from a larger supplier. Every day, he leases 18 bicycles at a cost of \$10 per day, and rents them to tourists at a price of \$5 per hour
• Solving Linear Programs in Excel 11) Excel will solve LP problem based on the formulas you inputted. When Excel finds an optimal solution, the following appears. 12) Choose desired output reports. Highlight both (hold down the control key) the Answer Report and Sensitivity Report. Click on Keep Solver Solution and OK then the Reports will be.
• Click the Solver command to display the Solver Parameters dialog. You can resize this dialog by selecting its lower right corner and dragging. If you don't see the Solver command (on the Data tab on Windows, or the Tools menu on Macintosh), you need to load the Solver Add-in. In the Excel Web App, the Solver Parameters dialog should be visible on the worksheet
• EXCEL SOLVER TUTORIAL This tutorial will introduce you to some essential features of Excel and its plug-in, Solver, that we will be using throughout ENGR62 to solve linear programs (LPs). You will learn how to • represent LPs in an Excel worksheet, • use the matrix functions SUMPRODUCT and MMULT, • set up an LP in Solver, an
• Excel's 'Solver' add-in is a powerful tool for what-if analysis, optimization, and equation solving.. But truth be told: knowing how to use the 'Solver' tool doesn't come naturally. Unlike some of the tools, 'Solver' isn't that easy to operate
• Double-check your formulas. One of the most powerful features of Excel is the ability to create formulas. You can use formulas to calculate new values, analyze data, and much more. But formulas also have a downside: If you make even a small mistake when typing a formula, it can give an incorrect result. To make matters worse, your spreadsheet.
### How To Solve a Quadratic Equation How To Exce
I have an implicit equation of two angles. I need to find one angle when I give the other angle as input. Can this be solved by using excel. If so, how? (SIN(CL))^2 +( SIN(CL))^2 - ((0.5)*(SIN(2*CL))*(SIN(2*CR))) - (2*((SIN(CL))^2)*((SIN(CR))^2)) - (SIN(CL-CR))^2 = 0. This is the equation and I need to find CR when I give CL as inpu Categories: Excel®, Formulas Tags: excel sort formula. Excel includes powerful sort functionality in the Sort & Filter group on the Data tab. But what if your data changes frequently, or if you're simply tired of clicking through the Ribbon each time you generate a similar table? The answer might be to create a formula that does your sorting.
### Video: Excel Formulas Based Sudoku Solver - Excel Hero Blo
10 Tips to Handle Slow Excel Spreadsheets. Here are 10 tips to give your slow Excel spreadsheet a little speed boost, and save you some time and frustration (click to jump to that specific section). Avoid Volatile Functions (you must). Use Helper Columns. Avoid Array Formulas (if you can) In Excel, an Array Formula allows you to do powerful calculations on one or more value sets. The result may fit in a single cell or it may be an array. An array is just a list or range of values, but an Array Formula is a special type of formula that must be entered by pressing Ctrl + Shift + Enter Excel will then numerically solve for a value of A2 that will cause B2 to equal 0. Since it solves numerically, it will reach 0 to within some very some value (10-4) and display the value of A2 that gave this near-0 value à this is one of the roots of the equation. 4.
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• Solve Care Telegram. | 5,693 | 24,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2021-49 | latest | en | 0.803454 |
http://www.ck12.org/physics/Coulombs-Law/rwa/Electrolytically-Charged-Sphere/r2/ | 1,485,077,778,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281419.3/warc/CC-MAIN-20170116095121-00007-ip-10-171-10-70.ec2.internal.warc.gz | 384,175,626 | 34,977 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Coulomb's Law
## Any two charged particles any distance from each other will experience repulsion or attraction.
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Electrolytically Charged Sphere
### Electrolytically Charged Sphere
Credit: Eorlingas
Source: http://commons.wikimedia.org/wiki/File:Pendule_de_torsion.png
Part of an apparatus known as a torsion balance, the spheres are electrically charged and the electrostatic force between them is measured by calculating the equilibrium distance between the two charged spheres.
#### Can You Apply It?
• When the spheres in the system are initially charged, the force between them when they are brought near one another is:
F=(4επa2)V1V2r2\begin{align*}F=(4 \varepsilon \pi a^2)\frac{V_1V_2}{r^2}\end{align*}
where a\begin{align*}a\end{align*} is the radius of each of the spheres, V1\begin{align*}V_1\end{align*} and V2\begin{align*}V_2\end{align*} are the respective voltages and r\begin{align*}r\end{align*} is the distance between the center of the spheres. One of the spheres is attached to a torsion spring that is calibrated to measure the to angle of deflection from equilibrium when a torque τ\begin{align*}\tau\end{align*} is applied. The spring is tightened to bring the charged sphere back to equilibrium, at which point the angle is measured. For small displacements the torque is related to the angle by
τ=kθ\begin{align*}\tau=k\theta\end{align*}
• The torque is a result of the force that is applied to the sphere over a distance l\begin{align*}l\end{align*}, from the axis of rotation. This force takes the form of
F=klθ\begin{align*}F=\frac{k}{l}\theta\end{align*}
#### Show What You Know
1. What relationship exists between the angle of deflection and the voltage applied to the spheres?
2. As the distance between the charged spheres increases, what do you expect to happen to the angle of deflection?
3. What assumption is being made about the torque equation? (Hint: think about the angle between the force and the lever arm.)
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0 ratings | 1,678 | 6,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-04 | latest | en | 0.847119 |
https://vedicmathschool.org/square-root-day/ | 1,726,635,218,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651836.76/warc/CC-MAIN-20240918032902-20240918062902-00208.warc.gz | 547,677,692 | 42,023 | Home » Blog » Mathematics Day » Squaring off for Square Root Day: Fun Facts and Festivities
# Squaring off for Square Root Day: Fun Facts and Festivities
Square Root Day, ever heard about this day? If not then today I’m going to give you all the information about this day.
This day doesn’t occur frequently; it happens nine times each century. The day is an unofficial holiday celebrated when the first two digits of the date are the square root of the last two digits of the year.
## When is Square Root Day Celebrated?
Square Root Day is celebrated only when the format of the date has its month and its date first and the last two digits of its year at the last.
For example, the last Square Day was celebrated on 3/3/2009 this was 7 years ago and the next square root day would be on 5/5/2025 this would happen after nine years now. So, this day isn’t celebrated every year; it is based on the mm/dd/yy format.
## Who found out About Square Root Day?
The man behind the foundation of Root Day was Ron Gordon. He used to live in California, United States and he was also a high school teacher.
He was known as the creator of this day’s celebration. The first time this day was celebrated was on 9/9/81 (September 9, 1981). People also shared the ideas of how they were celebrating the day when Gordon’s daughter created a Facebook group.
## How can you Celebrate Square Root Day?
Celebrating this day can be fun. You can start by learning the real-life concepts on a square root day.
You can also organize a celebration by cooking food items especially using vegetables from the roots. You can also cook or fry square-shaped food like potato fries, or baking a square-shaped cake, a square-shaped pizza.
You can also spend some time finding out how and when the next Root Square Day celebration is going to occur. You can also dance in a square, tie a square knot tie, drive through on 66 Root, eat a square-shaped stew.
Must Read These Also.
## How to find out Square Root Day’s date?
• In the 21st century the first day 01/01/01 (1 January 2001), was a square day.
• The second time was on 02/02/04 (2 February 2004) but how? So if we multiply 2*2 we would get 4.
• The next was on 03/03/09 (3 March 2009): 3*3 = 9.
• The next was on 04/04/16 (4 April 2016): 4*4 = 16.
• Now the next would be after nine years on 05/05/25 (5 May 2025): 5*5 = 25.
• The next would be on 06/06/36 (6 June 2036): 6*6 = 36.
Easy to find out right? So now you all find out the rest of the dates and let me know in the comment box! Hint: There are just 3 more days for this day in this century.
You don’t have to just sit and wait for the Root Square Day to come and then celebrate. We also have other unique mathematical holidays which are celebrated every year to honor mathematics and its work.
We have detailed articles on Pi Day, Pythagorean Day, World Math Day, National Mathematics Day, and International Day of Mathematics. You can check these articles as well. They all are interesting in their way!
Must Read These Also.
## Conclusion
This was interesting, right? Let me know in the comment box how you will celebrate this day the next time. And I hope this article was informative and interesting. Thank you! | 793 | 3,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-38 | latest | en | 0.960613 |
http://www.jiskha.com/display.cgi?id=1363204485 | 1,498,314,279,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320263.78/warc/CC-MAIN-20170624133941-20170624153941-00354.warc.gz | 546,950,996 | 3,791 | # Math
posted by .
The cost C in dollars of producing n yearbooks is represented by the equation
C=1000+5n
How much would it cost to produce 75n yearbooks?
• Math -
if you meant 75 yearbooks, then
C = 1000+5(75) = 1375
If you meant 75n, then C = 1000+375n^2 | 90 | 263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-26 | latest | en | 0.950645 |
http://tutorial.math.lamar.edu/Solutions/CalcII/ParametricEqn/Prob8.aspx | 1,558,741,309,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257781.73/warc/CC-MAIN-20190524224619-20190525010619-00178.warc.gz | 197,997,496 | 21,543 | Paul's Online Notes
Home / Calculus II / Parametric Equations and Polar Coordinates / Parametric Equations and Curves
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### Section 3-1 : Parametric Equations and Curves
8.The path of a particle is given by the following set of parametric equations. Completely describe the path of the particle. To completely describe the path of the particle you will need to provide the following information.
1. A sketch of the parametric curve (including direction of motion) based on the equation you get by eliminating the parameter.
2. Limits on $$x$$ and $$y$$.
3. A range of $$t$$’s for a single trace of the parametric curve.
4. The number of traces of the curve the particle makes if an overall range of $$t$$’s is provided in the problem.
$x = 4\sin \left( {\frac{1}{4}t} \right)\hspace{0.5in}y = 1 - 2{\cos ^2}\left( {\frac{1}{4}t} \right)\hspace{0.25in}\hspace{0.5in}\,\,\,\,\, - 52\pi \le t \le 34\pi$
Show All Steps Hide All Steps
Start Solution
There’s a lot of information we’ll need to find to fully answer this problem. However, for most of it we can follow the same basic ordering of steps we used for the first few problems in this section. We will need however to do a little extra work along the way.
Also, because most of the work here is similar to the work we did in Problems 4 – 6 of this section we won’t be putting in as much explanation to a lot of the work we’re doing here. So, if you need some explanation for some of the work you should go back to those problems and check the corresponding steps.
First, we’ll eliminate the parameter from this set of parametric equations. For this particular set of parametric equations we will make use of the well-known trig identity,
${\cos ^2}\left( \theta \right) + {\sin ^2}\left( \theta \right) = 1$
We can solve each of the parametric equations for sine and cosine as follows,
$\sin \left( {\frac{1}{4}t} \right) = \frac{x}{4}\hspace{0.25in}\hspace{0.25in}{\cos ^2}\left( {\frac{1}{4}t} \right) = \frac{{y - 1}}{{ - 2}}$
Plugging these into the trig identity gives,
$\frac{{y - 1}}{{ - 2}} + {\left( {\frac{x}{4}} \right)^2} = 1\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}y = \frac{{{x^2}}}{8} - 1$
Therefore, with a little algebraic manipulation, we see that the parametric curve will be some or all of the parabola above. Note that while many parametric equations involving sines and cosines are some or all of an ellipse they won’t all be as this problem shows. Do not get so locked into ellipses when seeing sines/cosines that you always just assume the curve will be an ellipse.
Show Step 2
At this point let’s get our first guess as to the limits on $$x$$ and $$y$$. As noted in previous problems what we’re really finding here is the largest possible ranges for $$x$$ and $$y$$. In later steps we’ll determine if this the actual set of limits on $$x$$ and $$y$$ or if we have smaller ranges.
We can use our knowledge of sine and cosine to determine the limits on $$x$$ and $$y$$ as follows,
$\begin{array} - 1 \le \sin \left( {\frac{1}{4}t} \right) \le 1 & \hspace{1.0in} & - 1 \le \cos \left( {\frac{1}{4}t} \right) \le 1\\ - 4 \le 4\sin \left( {\frac{1}{4}t} \right) \le 4 & \hspace{1.0in} & 0 \le {\cos ^2}\left( {\frac{1}{4}t} \right) \le 1\\ - 4 \le x \le 4 & \hspace{1.0in} & 0 \ge - 2{\cos ^2}\left( {\frac{1}{4}t} \right) \ge - 2\\ & \hspace{1.0in} & 1 \ge 1 - 2{\cos ^2}\left( {\frac{1}{4}t} \right) \ge - 1\\ & \hspace{1.0in} & - 1 \le y \le 1\end{array}$
Remember that all we need to do is start with the appropriate trig function and then build up the equation for $$x$$ and $$y$$ by first multiplying the trig function by any coefficient, if present, and then adding/subtracting any numbers that might be present. We now have the largest possible set of limits for $$x$$ and $$y$$.
Now, at this point we need to be a little careful. As noted above what we’ve actually found here are the largest possible ranges for the limits on $$x$$ and $$y$$. This set of inequalities for the limits on $$x$$ and $$y$$ assume that the parametric curve will be fully traced out at least once for the range of $$t$$’s we were given in the problem statement. It is always possible that the parametric curve will not trace out a full trace in the given range of $$t$$’s. In a later step we’ll determine if the parametric curve does trace out a full trace and hence determine the actual limits on $$x$$ and $$y$$.
Remember that when we talk about the parametric curve getting fully traced out this doesn’t, in general, mean the full parabola we found in Step 1 gets traced out by the parametric equation. All “fully traced out” means, in general, is that whatever portion of the parabola that is described by the set of parametric curves will be completely traced out.
In fact, for this problem, we can see that the parabola from Step 1 will not get fully traced out by the particle regardless of any range of $$t$$’s. The largest possible portion of the parabola that can be traced out by the particle is the portion that lies in the range of $$x$$ and $$y$$ given above. In a later step we’ll determine if the largest possible portion of the parabola does get traced out or if the particle only traces out part of it.
Show Step 3
Let’s next get the direction of motion for the parametric curve. For this analysis it might be useful to have a quick sketch of the largest possible parametric curve. So, here is a quick sketch of that.
The dashed line is the graph of the full parabola from Step 1 and the solid line is the portion that falls into our largest possible range of $$x$$ and $$y$$ we found in Step 2. As an aside here note that the two ranges are complimentary. In other words, if we sketch the graph only for the range of $$x$$ we automatically get the range for $$y$$. Likewise, if we sketch the graph only for the range of y we automatically get the range for $$x$$. This is a good check for your graph. The $$x$$ and $$y$$ ranges should always match up!
When our parametric curve was an ellipse (the previous problem for example) no matter what point we started the analysis at the curve would eventually trace out around the ellipse and end up back at the starting point without ever going back over any portion of itself. The main issue we faced with the ellipse problem was we could rotate around the ellipse in a clockwise or a counter clockwise motion to do this and a careful analysis of the behavior of both the $$x$$ and $$y$$ parametric equations was required to determine just which direction we were going.
With a parabola for our parametric curve things work a lot differently. Let’s suppose that we “started” at the right end point (this is just randomly picked for no other reason that I’m right handed so don’t think there is anything special about this point!) and it doesn’t matter what $$t$$ we use to get to that point.
At this point we know that we are at $$x = 4$$ and in order for $$x$$ to have that value we must also have $$\sin \left( {\frac{1}{4}t} \right) = 1$$. Now, as we increase $$t$$ from this point (again it doesn’t matter just what the value of $$t$$ is) the only option for sine is for it to decrease until it has the value $$\sin \left( {\frac{1}{4}t} \right) = - 1$$. This in turn means that if we start at the right end point we have no option but to proceed along the curve going to the left.
However, we don’t just reach the left end point and then stop! Once we are at $$\sin \left( {\frac{1}{4}t} \right) = - 1$$ if we further increase $$t$$ we know that sine will also increase until it has the value $$\sin \left( {\frac{1}{4}t} \right) = 1$$ and so we must move back along the curve to the right until we are back at the right end point.
Unlike the ellipse however, the only way for this to happen is for the particle to go back over the parabola moving in a rightward direction. Remember that the particle moves to the right or left it must trace out a portion of the parabola that we found in Step 1! Any particle traveling along the path given by the set of parametric equations must follow the graph of the parabola and never leave it.
In other words, if we don’t put any restrictions on $$t$$ a particle on this parametric curve will simply oscillate left and right along the portion of the parabola sketched out above. In this case however we do have a range of $$t$$’s so we’ll need to determine a range of $$t$$’s for one trace to fully know the direction of motion information of the particle on this path and we’ll do that in the next step. With a restriction on the range of $$t$$’s it is possible that the particle won’t make a full trace or it might retrace some or all of the curve so we can’t say anything definite about the direction of motion for the particle over the full range of $$t$$’s until the next step when we determine a range of $$t$$’s for one full trace of the curve.
Before we move on to the next step there is a quick topic we should address. We only used the $$x$$ equation to do this analysis and never addressed the $$y$$-equation anywhere in the analysis. It doesn’t really matter which one we use as both will give the same information.
Show Step 4
Now we need to determine a range of $$t$$’s for one full trace of the parametric curve. It is important for this step to remember that one full trace of the parametric curve means that no portion of the parametric curve can be retraced.
Note that one full trace does not mean that we get back to the “starting” point. When we dealt with an ellipse in the previous problem that was one trace because we did not need to retrace any portion of the ellipse to get back to the starting point. However, as we saw in the previous step that for our parabola here we would have to retrace the full curve to get back to the starting point.
So, one full trace of the parametric curve means we move from the right end point to the left end point only or visa-versa and move from the left end point to the right end point. Which direction we move doesn’t really matter here so let’s get a range of $$t$$’s that take us from the left end point to the right end point.
In all the previous problems we’ve used $$t = 0$$ as our “starting” point but that won’t work for this problem because that actually corresponds to the vertex of the parabola. We want to start at the left end point so the first part of this process is actually determine a $$t$$ that will put us at the left end point.
In order to be at the left end point, $$\left( { - 4,1} \right)$$, we need to require that $$\sin \left( {\frac{1}{4}t} \right) = - 1$$ which occurs if $$\frac{1}{4}t = \ldots , - \frac{{5\pi }}{2}, - \frac{\pi }{2},\frac{{3\pi }}{2},\frac{{7\pi }}{2}, \ldots$$. We also need to require that $$\cos \left( {\frac{1}{4}t} \right) = 0$$ which occurs if $$\frac{1}{4}t = \ldots , - \frac{{3\pi }}{2}, - \frac{\pi }{2},\frac{\pi }{2},\frac{{3\pi }}{2}, \ldots$$. There are going to be many numbers that are in both lists here so all we need to do is pick one and proceed. From the numbers that we’ve listed here we could use either $$\frac{1}{4}t = - \frac{\pi }{2}$$ or $$\frac{1}{4}t = \frac{{3\pi }}{2}$$. We’ll use $$\frac{1}{4}t = - \frac{\pi }{2}$$, i.e. when $$t = - 2\pi$$ , simply because it is the first one that occurs in both lists. Therefore, we will be at the left end point when $$t = - 2\pi$$.
Let’s now move to the right end point, $$\left( {4,1} \right)$$. In order to get the range of $$t$$’s for one trace this means we’ll need the next $$t$$ with $$t > - 2\pi$$ (which corresponds to $$\frac{1}{4}t > - \frac{\pi }{2}$$). To do this we need to require that $$\sin \left( {\frac{1}{4}t} \right) = 1$$ which occurs if $$\frac{1}{4}t = \ldots , - \frac{{3\pi }}{2},\frac{\pi }{2},\frac{{5\pi }}{2},\frac{{9\pi }}{2}, \ldots$$and we need to that $$\cos \left( {\frac{1}{4}t} \right) = 0$$ which occurs if $$\frac{1}{4}t = \ldots , - \frac{{3\pi }}{2}, - \frac{\pi }{2},\frac{\pi }{2},\frac{{3\pi }}{2}, \ldots$$. .
The first $$t$$ that is in both of these lists with $$\frac{1}{4}t > - \frac{\pi }{2}$$ is then $$\frac{1}{4}t = \frac{\pi }{2}$$, i.e. when $$t = 2\pi$$ . So, the first $$t$$ after $$t = - 2\pi$$ that puts us at the right end point is $$t = 2\pi$$. This means that a range of $$t$$’s for one full trace of the parametric curve is then,
$- 2\pi \le t \le 2\pi$
Note that this is only one possible answer here. Any range of $$t$$’s with a “net” range of $$2\pi - \left( { - 2\pi } \right) = 4\pi$$ $$t$$’s, with the endpoints of the $$t$$ range corresponding to start/end points of the parametric curve, will work. So, for example, any of the following ranges of $$t$$’s would also work.
$- 6\pi \le t \le - 2\pi \hspace{0.25in}\hspace{0.25in}2\pi \le t \le 6\pi \hspace{0.25in}\hspace{0.25in}6\pi \le t \le 10\pi$
The direction of motion for each may be different range of $$t$$’s of course. Some will trace out the curve moving from left to right while others will trace out the curve moving from right to left. Because the problem did not specify a particular direction any would work.
Note as well that the range $$- 2\pi \le t \le 2\pi$$ falls completely inside the given range of $$t$$’s specified in the problem and so we know that the particle will trace out the curve more than once over the full range of $$t$$’s. Determining just how many times it traces over the curve will be determined in the next step.
Show Step 5
Now that we have a range of $$t$$’s for one full trace of the parametric curve we can determine the number of traces the particle makes.
This is a really easy step. We know the total time the particle was traveling and we know how long it takes for a single trace. Therefore,
${\mbox{Number Traces = }}\frac{{{\mbox{Total Time Traveled}}}}{{{\mbox{Time for One Trace}}}} = \frac{{34\pi - \left( { - 52\pi } \right)}}{{2\pi - \left( { - 2\pi } \right)}} = \frac{{86\pi }}{{4\pi }} = \frac{{43}}{2} = 21.5{\mbox{ traces}}$ Show Step 6
Finally, here is a sketch of the parametric curve for this set of parametric equations.
For this sketch we indicated the direction of motion by putting arrow heads going both directions in places on the curve. We also included a set of $$t$$’s for a couple of points to illustrate where the particle is at while tracing out of the curve. The dashed line is the continuation of the parabola from Step 1 to illustrate that our parametric curve is only a part of the parabola.
Here is also the formal answers for all the rest of the information that problem asked for.
$\begin{array}{lcc} {\mbox{Range of }}x{\mbox{ :}} & \hspace{0.1in} & - 4 \le x \le 4\\ {\mbox{Range of }}y{\mbox{ :}} & \hspace{0.1in} & - 1 \le y \le 1\\ {\mbox{Range of }}t{\mbox{ for one trace : }} &\hspace{0.1in} & - 2\pi \le t \le 2\pi \\ {\mbox{Total number of traces : }} & \hspace{0.1in} & 21.5\end{array}$ | 4,364 | 15,298 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2019-22 | longest | en | 0.894808 |
https://expskill.com/question/a-particle-moving-in-a-straight-line-with-uniform-acceleration-has-a-velocity-10-cm-sec-and-8-seconds-later-has-a-velocity-54-cm-sec-what-will-be-the-described-by-the-particle-during-the-10th-second/ | 1,669,808,180,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710734.75/warc/CC-MAIN-20221130092453-20221130122453-00724.warc.gz | 288,490,556 | 13,398 | A particle moving in a straight line with uniform acceleration has a velocity 10 cm/sec and 8 seconds later has a velocity 54 cm/sec. What will be the described by the particle during the 10^th second of its motion?
Category: QuestionsA particle moving in a straight line with uniform acceleration has a velocity 10 cm/sec and 8 seconds later has a velocity 54 cm/sec. What will be the described by the particle during the 10^th second of its motion?
Editor">Editor Staff asked 11 months ago
A particle moving in a straight line with uniform acceleration has a velocity 10 cm/sec and 8 seconds later has a velocity 54 cm/sec. What will be the described by the particle during the 10^th second of its motion?
(a) 38.5cm
(b) 37.5cm
(c) 38cm
(d) 39.5cm
This is a very interesting question from Calculus Application in section Application of Calculus of Mathematics – Class 12
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options
Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience
Editor">Editor Staff answered 11 months ago
Correct option is (a) 38.5cm
The best explanation: Let, the particle moving with a uniform acceleration of f cm/sec^2.
By question initial velocity of the particle is u = 10 cm/sec and its velocity after 8 seconds = v = 34 cm/sec.
Thus, using the formula v = u + ft we get
34 = 10 + f*8
Or 8f = 24
Or f = 3
Therefore, the required acceleration of the particle is 3 cm/sec^2.
The space described during the 10th second of its motion is,
= [10 + 1/2(3)(2*10 – 1)] [using the formula st = ut +1/2(f)(2t – 1)]
= 10 + 28.5
= 38.5cm. | 440 | 1,654 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2022-49 | latest | en | 0.941092 |
https://justindomke.wordpress.com/category/uncategorized/page/2/ | 1,590,716,505,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347401004.26/warc/CC-MAIN-20200528232803-20200529022803-00477.warc.gz | 416,071,438 | 37,801 | # Sneaking up on Bayesian Inference (A fable in four acts)
Act 1: Magical Monkeys
Two monkeys, Alfred (${a}$) and Betty (${b}$) live in a parallel universe with two kinds of blocks, green (${g}$) and yellow (${y}$). Alfred likes green blocks, and Betty prefers the yellow blocks. One day, a Wizard decides to give one of the monkeys the magical power to send one block over to our universe each day.
The Wizard chooses the magical monkey by flipping a fair four-sided die. He casts a spell on Alfred if the outcome is 1, and Betty if the outcome is ${2-4}$. That is, the Wizard chooses Alfred with probability ${.25}$ and Betty with probability ${.75}$. Both Alfred and Betty send their favorite colored block with probability ${.8}$
After the Wizard has chosen, we see the first magical block, and it is green. Our problem is: What is the probability that Alfred is the magical monkey?
Intuitively speaking, we have two somewhat contradictory pieces of information. We know that the Wizard chooses Betty more often. But green is Alfred’s preferred color. Given probabilities above, is Alfred or Betty more probable?
First, we can write down all the probabilities. These are
\displaystyle \begin{aligned}Pr(M=a) & =.25\\ Pr(M=b) & =.75\\ Pr(B=g\vert M=a) & =.8\\ Pr(B=y\vert M=a) & =.2\\ Pr(B=g\vert M=b) & =.2\\ Pr(B=y\vert M=b) & =.8. \end{aligned}
Now, the quantity we are interested in is ${Pr(M=a\vert B=g).}$ We can mechanically apply Bayes’ rule to obtain
\displaystyle \begin{aligned}Pr(M=a\vert B=g) & =\frac{1}{Pr(B=g)}Pr(M=a)Pr(B=g\vert M=a)\\ & =\frac{1}{Pr(B=g)}.25\times.8. \end{aligned}
Similarly, we can calculate that
$\displaystyle Pr(M=b\vert B=g)=\frac{1}{Pr(B=g)}.75\times.2.$
But, of course, we know that one of the monkeys is magical, so these quantities must sum to one. Thus (since they both involve the quantity ${Pr(B=g)}$ that we haven’t explicitly calculated) we can normalize to obtain that
$\displaystyle Pr(M=a\vert B=g)=\frac{.25\times.8}{.25\times.8+.75\times.2}=\frac{4}{7}.$
Act 2: Magical Monkeys on Multiple Days
We wait around two more days, and two more blocks appear, both of which are yellow. Assume that the monkeys make an independent choice each day to send their favorite or less favorite block. Now, what is the probability that Alfred is the magical monkey, given that ${B=gyy}$?
Intuitively speaking, what do we expect? Betty is more likely to be chosen, and 2/3 of the blocks we’ve seen are suggestive of Betty (since she prefers yellow).
Now, we can mechanically calculate the probabilities. This is just like before, except we use the fact that the blocks seen on each day are conditionally independent given a particular monkey.
\displaystyle \begin{aligned}Pr(M=a\vert B=gyy) & =\frac{1}{Pr(B=gyy)}Pr(M=a)Pr(B=gyy\vert M=a)\\ & =\frac{1}{Pr(B=gyy)}Pr(M=a)Pr(B=g\vert M=a)Pr(B=y\vert M=a)Pr(B=y\vert M=a)\\ & =\frac{1}{Pr(B=gyy)}.25\times.8\times.2\times.2\\ Pr(M=b\vert B=gyy) & =\frac{1}{Pr(B=gyy)}Pr(M=b)Pr(B=g\vert M=b)Pr(B=y\vert M=b)Pr(B=y\vert M=b)\\ & =\frac{1}{Pr(B=gyy)}.75\times.2\times.8\times.8 \end{aligned}
Again, we know that these two probabilities sum to one. Thus, we can normalize and get that
$\displaystyle Pr(M=a\vert B=gyy)=\frac{.25\times.8\times.2\times.2}{.25\times.8\times.2\times.2+.75\times.2\times.8\times.8}=\frac{1}{13}.$
So now, Betty looks more likely by far.
Act 3: Magical Monkeys and the Weather
Now, suppose the Wizard gives us some additional information. The magical monkey (whilst relaxing over in the other universe) is able to perceive our weather, which is either clear ${(c)}$ or rainy ${(r)}$. Both the monkeys prefer clear weather. When the weather is clear, they send their preferred block with probability ${.8}$, while if the weather is rainy, they angrily send their preferred block with probability ${.2.}$ That is, we have
\displaystyle \begin{aligned}Pr(B=g\vert M=a,W=c) & =.8\\ Pr(B=g\vert M=a,W=r) & =.2\\ Pr(B=g\vert M=b,W=c) & =.2\\ Pr(B=g\vert M=b,W=r) & =.8. \end{aligned}
Along with seeing the previous sequence of blocks ${B=gyy}$, we observed the weather sequence ${W=rrc}$. Now, what is the probability that Alfred is the magical monkey?
We ask the Wizard what the distribution over rainy and clear weather is. The wizard haughtily responds that this is irrelevant, but does confirm that the weather is independent of which monkey was made magical.
Can we do anything without knowing the distribution over the weather? Can we calculate the probability that Alfred is the magical monkey?
Let’s give it a try. We can apply Bayes’ equation to get that
\displaystyle \begin{aligned}Pr(M=a\vert B=gyy,W=rrc) & =\frac{1}{Pr(B=gyy,W=rrc)}Pr(M=a)Pr(B=gyy,W=rrc\vert M=a).\end{aligned}
Now, since the weather is independent of the monkey, we know that
\displaystyle \begin{aligned}Pr(B=gyy,W=rrc\vert M=a) & =Pr(W=rrc\vert M=a)Pr(B=gyy\vert M=a,W=rrc)\\ & =Pr(W=rrc)Pr(B=gyy\vert M=a,W=rrc). \end{aligned}
Thus, we have that
\displaystyle \begin{aligned}Pr(M=a\vert B=gyy,W=rrc) =&\frac{Pr(W=rrc)}{Pr(B=gyy,W=rrc)}Pr(M=a)Pr(B=gyy\vert M=a,W=rrc)\\ =&\frac{Pr(W=rrc)}{Pr(B=gyy,W=rrc)}Pr(M=a) \\ & \times Pr(B=g\vert M=a,W=r)Pr(B=y\vert M=a,W=r)Pr(B=y\vert M=a,W=c)\\ =&\frac{Pr(W=rrc)}{Pr(B=gyy,W=rrc)}.25\times.2\times.8\times.8. \end{aligned}
Through the same logic we can calculate that
$\displaystyle Pr(M=b\vert B=gyy,W=rrc)=\frac{Pr(W=rrc)}{Pr(B=gyy,W=rrc)}.75\times.8\times.2\times.2.$
Again, we know that these probabilities need to sum to one. Since the factor ${Pr(W=rrc)/Pr(B=gyy,W=rrc)}$ is constant between the two, we can normalize it out. Thus, we get that
$\displaystyle Pr(M=a\vert B=gyy,W=rrc)=\frac{.25\times.2\times.8\times.8.}{.25\times.2\times.8\times.8.+.75\times.8\times.2\times.2.}=\frac{4}{7}.$
Again, it looks like Alfred was the more likely monkey. And– oh wait– we somehow got away with not knowing the distribution over the weather…
Act 4: Predicting the next block
Now, suppose that after seeing the previous sequences (namely ${B=gyy}$ and ${W=rrc}$) , on the fourth day, we find that it has rained. What is the probability that we will get a green block on the fourth day? Mathematically, we want to calculate
$\displaystyle Pr(B'=g\vert W'=r,B=ggy,W=rrc).$
How can we go about this? Well, if we knew that the magical monkey was Alfred (which we do not!), it would be easy to calculate. We would just have
$\displaystyle Pr(B'=g\vert W'=r,M=a),$
and similarly if the magical monkey were Betty. Now, we don’t know which monkey is magical, but we know the probabilities that each monkeys is magical given the available information– we just calculated them in the previous section! So, we can factorize the distribution of interest as
\displaystyle \begin{aligned}Pr(B'=g\vert W'=r,B=ggy,W=rrc) & =Pr(B'=g\vert W'=r,M=a)Pr(M=a\vert B=ggy,W=rrc)\\ & +Pr(B'=g\vert W'=r,M=b)Pr(M=b\vert B=ggy,W=rrc)\\ & =.2\times\frac{4}{7}+.8\times\frac{3}{7}\\ & =\frac{33}{70} \end{aligned}
So we are slightly less likely than even to see a green block on the next day.
Epilogue: That’s Bayesian Inference
This is how Bayesian inference works (in the simplest possible setting). In general Bayesian inference, you have:
• A set of possible inputs (for us these are weather patterns)
• A set of possible outputs (for us these are colored blocks)
• A set of possible models, that map a given input to a distribution over the outputs (for us these are monkeys)
• A “prior” distribution over which models are more likely (for us, this is the wizard and the four-sided die)
• A dataset of inputs and outputs (for us, this is ${W=rrc}$ and ${B=ggy}$)
• A new input (for us this is ${W'=r}$)
• An unknown output that we’d like to predict (for us this is ${B'}$)
Bayesian inference essentially proceeds in two steps:
• In the first step, one uses the prior over which models are more likely along with the dataset of inputs and outputs to build a “posterior” distribution over which models are more likely. We did this by using
$\displaystyle Pr(M\vert W,B)\propto Pr(M)\prod_{i}Pr(B_{i}\vert W_{i},M),$
where ${i}$ indexes the different days of observation.
• In the second step, one “integrates over the posterior”. That is, each model gives us a distribution over ${B'}$ given the new input ${W'}$. So, we simply sum over the different models
$\displaystyle Pr(B'\vert W')=\sum_{m}Pr(M=m\vert W,B)Pr(B'\vert W',M=m).$
In the general case, the set of possible models is much larger (typically infinite) and more complex computational methods need to be used to integrate over it. (Commonly Markov chain Monte Carlo or variational methods). Also, of course, we don’t typically have a Wizard telling us exactly what the prior distribution over the models is, meaning one must make assumptions, or try to “learn” the prior as in, say, empirical Bayesian methods.
# Algorithmic Dimensions
There are many dimensions on which we might compare a machine learning or data mining algorithm. A few of the first that come to mind are:
1) Sample complexity, convergence
How much predictive power is the algorithm able to extract from a given number of examples?
All else being equal, if algorithm A with N examples behaves the same as algorithm B with 2N examples, we would prefer algorithm A. This can vary in importance depending on how scarce or expensive data is.
2) Speed
How quickly does the algorithm run?
Obviously, a tool is more useful if it is faster. However, if that faster algorithm comes at the expense of sample complexity, one would need to measure the expense of running longer against the expense of gathering more data.
3) Guarantees
Does the algorithm just have good performance (along whatever dimension) in practice, or is it proven? Is it always good, or only in certain situations? How predictable is the performance?
When SVMs showed up, it looked like good practical algorithms came from good theory. These days, it seems clear that powerful intuition is also an excellent source of practical algorithms (deep learning).
Perhaps theory will some day catch up, and the algorithm with the best bounds will coincide with the best practical performance. However, this is not always the case today, where we often face a trade-offs between algorithms with good theoretical guarantees, and good empirical performance. Some examples are 1) Upper-Confidence Bounds versus Thomson Sampling with bandit algorithms [Update June 2018: Thompson sampling has largely caught up!] 2) Running a convex optimization algorithm with a theoretically derived Lipschitz constant versus a smaller one that still seems to work and 3) Doing model selection via VC-dimension generalization bounds versus using K-fold cross-validation.
4) Memory usage
How much space does the algorithm need?
I work with a lot of compute-heavy applications where this is almost a wall: we don’t care about memory usage until we run out of it, after which we care a great deal. With simpler algorithms and larger datasets, this is often more nuanced, with concerns about different cache sizes.
5) Handholding
Can it be used out of the box, or does it require an expert to “tune” it for best performance?
A classic example of this is stochastic gradient descent. In principle, for a wide range of inputs, convergence is guaranteed by iteratively setting $\theta_{k+1} = \theta_k - \lambda_k g_k$ where $g_k$ is a noisy unbiased estimate of the gradient at $\theta_k$ and $\lambda_k$ is some sequence of step-sizes that obeys the Robbins-Monro conditions [1]. However, in practice, the difference between, say, $\lambda_k = 1/k$, and $\lambda_k = 7.2/(522.9+k)$ can be enormous, and finding these constants is a bit of a dark art.
[1] ($\sum_k \lambda_k=\infty$ and $\lim_{k\rightarrow\infty} \lambda_k=0$.)
6) Implementability
How simple is the algorithm? How easy is it to implement?
This is quite complex and situational. These days, I’d consider an algorithm that consists of a few moderate-dimensional matrix multiplications or singular value decompositions “simple”. However, that’s due to the huge effort that’s been devoted to designing reliable matrix algorithms, and the ubiquity of easy to use libraries.
7) Amenability to parallelization
Does the algorithm lend itself to parallelization? (And what type of parallelization?)
And if it does, under what model? Map-reduce, GPU, MPI, and openMP all have different properties.
8) “Anytime-ness”
Can the algorithm be implemented in any anytime manner?
That is, does the algorithm continuously return a current “best-guess” of the answer that is refined over time in a sensible manner? This can help diagnose problems before running the full algorithm, enormously useful in debugging large systems.
Note this is distinct from being an online algorithm, which I’m not mentioning here, since it’s a mix of speed and memory properties.
9) Transparency, interpretability
Can the final result be understood? Does it give insight into how predictions are being made?
Galit Shmueli argues that “explanatory power” and “predictive power” are different dimensions. However, there are several ways in which interpretability is important even when prediction is the final goal. Firstly, the insight might convince a decision maker that the machine learning system is reliable. Second, this can be vital in practice for generalization. The world is rarely independent and identically distributed. If a domain expert can understand the predictive mechanism, they may be able to assess if this will still hold in the future, or captures something true only in the training period. Third, understanding what the predictor is doing also often yields ways to improve it. For example, the adjacent visualization of the outputs of a decision tree in two-dimensions suggests the need for non axis-aligned splits.
10) Generality
What class of problems can the algorithm address?
All else being equal, we prefer an algorithm that can, say, optimize all convex losses over one that can only fit the logistic loss. However, this often comes at a cost— a more general-purpose algorithm cannot exploit the extra structure present in a specialized problem (or, at least, has more difficulty doing so). It’s instructive how many different methods are used by LIBLINEAR depending on the particular loss and regularization constant.
11) Extendability
Does the algorithm have lots of generalizations and variants that are likely to be interesting?
More of an issue when reviewing a paper than deciding what is the final best algorithm to use once all the dust has settled.
12) Insight
Does the algorithm itself (as opposed to its results) convey insight into the problem?
Gradient descent for maximum likelihood learning of an exponential family is a good example, as it reveals the moment-matching conditions. Insight of this type, however, doesn’t suggest that one should actually run the algorithm.
13) Model robustness
How does the performance of the algorithm hold up to violations of its modeling assumptions?
Suppose we are fitting a simple line to some 2-D data. Obviously, all else being equal, we would prefer an algorithm that still does something reasonable when the actual dependence when the expected value of $y$ is not linear in $x$. The example I always harp on here is the pseudolikelihood. The original paper pointed out that this will have somewhat worse sample complexity that the full likelihood, and (much!) better time complexity. Many papers seem to attribute the bad performance of the pseudolikelihood in practice to this sample complexity, when the true cause is that the likelihood does something reasonable (minimizes KL divergence) when there is model mis-specification, but the pseudolikelihood does not.
I often ponder is how much improvement in one dimension is enough to “matter”. Personally, I would generally consider even a small constant factor (say 5-10%) improvement in sample complexity quite important. Meanwhile, it would be rare to get exited about even, say, a factor of two improvement in running time.
What does this reflect? Firstly, generalization is the fundamental goal of data analysis, and we are likely to be willing to compromise most things if we can really predict better. Second, we instinctively distrust running times. Theory offers few tools for understanding constant factors, as these are highly architecture and implementation dependent. In principle, if one could be completely convincing that a factor of two improvement was truly there, I think this probably would be significant. (This might be true, say, if all algorithms are bottlenecked by a few matrix multiplications, and a new algorithm provably reduces the number needed.) However, this is rare.
In some places, I think constant factor skepticism can lead us astray. In reality, a factor of 30 improvement in speed is probably better than changing a $O(N \log N)$ complexity to $O(N)$. (Calculate $N$ such that $\log N=30$.) This is particularly true when the lower-complexity algorithm has higher constant factors. As an example, I’ve always found the $O(N \log N)$ algorithm for projection onto the $l_1$ ball to be faster than the $O(N)$ algorithm in practice.
Given that there are so many dimensions, can a new algorithm really improve on all simultaneously? It seems rare to even improve on a single dimension without a corresponding cost somewhere else. There will often be a range of techniques that are Pareto optimal, depending on one’s priorities. Understanding this range is what makes having an expert around so important.
Ideally, as a community, we would be able to provide a “consumer” of machine learning an easy way to find the algorithm for them. Or, at least, we might be able to point them in the direction of possible algorithms. One admirable attempt along this line is the following table from The Elements of Statistical Learning:
(Incidentally, notice how many of the desiderata do not overlap with mine.)
Some other situations show a useful contrast. For example, take this decision tree for choosing an algorithm for unconstrained optimization, due to Dianne O’Leary:
Essentially, this amounts to the principle that one should use the least general algorithm available, so that it can exploit as much structure of the problem as possible. Though one can quibble with the details (pity the subgradient methods) it at least comes close to giving the “right” algorithm in each situation.
This doesn’t seem possible with machine learning, since there doesn’t exist a single hierarchy Rather, ML problems are a tangle of model specification, computational and architecture requirements, implementation constraints, user risk-tolerances and so on. It won’t be easy to automate away the experts. (Even ignoring the possible misalignment of incentives in that the field has the domain of automating itself.)
(This post incorporates comments from Aditya Menon and Cheng Soon Ong.)
# Favorite things NIPS
I always enjoy reading conference reports, so I thought I’d mention a few papers that caught my eye. (I welcome any corrections to my summaries of any of these.)
1. Recent Progress in the Structure of Large-Treewidth Graphs and Some Applications, by Chandra Chekuri. Not a paper, but rather an exceptionally clear tutorial on material that should probably be mandatory for anyone interested in computational complexity and graphical models. The talk isn’t online yet, but slides are available.
2. Learning Distributed Representations for Structured Output Prediction, by Vivek Srikumar and Christopher Manning. The idea is that when doing structured prediction, labels are often “similar” in some sense, and so should be encouraged to have similar weights. This is done by defining weights not over a discrete label space, but over a finite-dimensional distributed representation, with one vector for each label. Learning alternates between updating the representation of each label and regular structured learning.
3. Clamping Variables and Approximate Inference, by Adrian Weller and Tony Jebara. The paper provides a first-principles based proof that the Bethe approximation gives a lower-bound on the true partition function for attractive binary pairwise models. (A result originally proved by Ruozzi in 2012) The basic idea is that “clamping” (i.e. brute-force summing over all values of a variable, and using the Bethe approximation on the resulting model) can only increase the Bethe approximation. But, if you clamp everything, you have the exact partition function, so Bethe must be a lower-bound. The proofs provide a lot of insight into why the results are true.
4. Making Pairwise Binary Graphical Models Attractive, by Nicholas Ruozzi and Tony Jebara. This builds on similar ideas to the above paper. There is an idea of a cover of a graph, where vertices have multiple copies, but neighborhoods look locally the same. The paper gives an appropriate cover construction such that the cover is attractive (up to a “flipping” of the variables). Thus, the Bethe partition function of the cover lower-bounds the true partition function of the cover. This is shown to be a lower-bound on the tree-reweighted partition function. As far as I can tell, Bethe on the cover may upper or lower-bound the true partition function, so Bethe-on-cover isn’t necessarily better than tree-reweighted. Still, since tree-reweighted is typically a loose upper-bound, it is likely to be better in practice.
5. Distributed Bayesian Posterior Sampling via Moment Sharing, by Minjie Xu, Balaji Lakshminarayanan, Yee Whye Teh, Jun Zhu, and Bo Zhang. You want to do Bayesian MCMC inference using a cluster. This paper partitions the data over the nodes of the cluster, does local MCMC approximations, and uses an exponential-family approximation and expectation propagation to send information about the posterior between the nodes, thus approximating full/normal Bayesian inference. A couple days ago Andrew Gelman mentioned a similar method, considering other local approximations.
6. Mode Estimation for High Dimensional Discrete Tree Graphical Models, by Chao Chen, Han Liu, Dimitris Metaxas, and Tianqi Zhao. The goal is to find the M highest modes in a tree-structured graphical model. They find a set of necessary local conditions to be a mode, and then propose a junction-tree type algorithm operating with local mode configurations as values to find global modes.
7. Efficient Inference of Continuous Markov Random
Fields with Polynomial Potentials
, by Shenlong Wang, Alexander Schwing, and Raquel Urtasun. The goal is to find a local minimum of a polynomial MRF by use of the CCCP method where one iteratively splits the objective into convex and concave parts and linearizes the concave part to get an upper-bound. I’d have expected this to be easy, but it turns out to be quite difficult to split up a polynomial into convex and concave parts. This paper gives an SDP problem to find a function to add to the objective to make it convex, yielding the desired split.
8. Hardness of parameter estimation in graphical models, by Guy Bresler, David Gamarnik, and Devavrat Shah. In principle, this is the ultimate non-surprising result: given the mean parameters (data marginals), it is computationally intractable to find the natural parameters (MRF weights) that produce these marginals. Formally given an oracle to do maximum likelihood learning in the normal way (find parameters to match a given mean vector), you could use that oracle to approximate the partition function, which is known to be hard. The really surprising thing here is that this wasn’t proved until 2014!
9. A* Sampling, by Chris Maddison, Daniel Tarlow, and Tom Minka. A method for exact sampling for low-dimensional continuous distributions, based on fundamentally different principles from existing methods. Intuitively, the algorithm simultaneously generates Gumbel-type noise for every possible (real-valued) configuration and optimizes the resulting function by branch-and-bound.
10. The workshops, as usual, were great aside from the usual annoyance of it only being possible to be in one place at one time.
11. Montreal. Charming, friendly.
# Truncated Bi-Level Optimization
In 2012, I wrote a paper that I probably should have called “truncated bi-level optimization”. I vaguely remembered telling the reviewers I would release some code, so I’m finally getting around to it.
The idea of bilevel optimization is quite simple. Imagine that you would like to minimize some function $L(w)$. However, $L$ itself is defined through some optimization. More formally, suppose we would like to solve
$\min_{w,y} Q(y), \text{ s.t. } y = \arg\min_y E(y,w).$
Or, equivalently,
$\min_{w} L(w)=Q(y^*(w)),$
where $y^*$ is defined as $y^*(w) := \arg\min_y E(y,w)$. This seems a little bit obscure at first, but actually comes up in several different ways in machine learning and related fields.
## Hyper-parameter learning
The first example would be in learning hyperparameters, such as regularization constants. Inevitably in machine learning, one fits parameters parameters to optimize some tradeoff between the quality of a fit to training data and a regularization function being small. Traditionally, the regularization constant is selected by optimizing on a training dataset with a variety of values, and then picking the one that performs best on a held-out dataset. However, if there are a large number of regularization parameters, a high-dimensional grid-search will not be practical. In the notation above, suppose that $w$ is a vector of regularization constants, and that $y$ are training parameters. Let, $E$ be the regularized empirical risk on a training dataset, and let $Q$ be how well the parameters $y^*(w)$ perform on some held-out validation dataset.
## Energy-based models
Another example (and the one suggesting the notation) is an energy-based model. Suppose that we have some “energy” function $E_x(y,w)$ which measures how well an output $y$ fits to an input $x$. The energy is parametrized by $w$. For a given training input/output pair $(\hat{x},\hat{y})$, we might have that $Q_{\hat{y}}(y^*(w))$ measures how how the predicted output $y^*$ compares to the true output $\hat{y}$, where $y^*(w)=\arg\max_y E_{\hat{x}}(y,w)$.
## Computing the gradient exactly
Even if we just have the modest goal of following the gradient of $L(w)$ to a local minimum, even computing the gradient is not so simple. Clearly, even to evaluate $L(w)$ requires solving the “inner” minimization of $E$. It turns out that one can compute $\nabla_w L(w)$ through first solving the inner minimization, and then solving a linear system.
1. Input $w$
2. Solve $y^* \leftarrow \arg\min_y E(y,w)$.
3. Compute:
4. (a) the loss $L(w)= Q(y^*)$
5. (b) the gradient $g=\nabla_y Q(y^*)$
6. (c) the Hessian $H=\frac{\partial^2 E(y^*,w)}{\partial w\partial w^T}$
7. Solve the linear system $z=H^{-1} g$.
8. Compute the parameter gradient $\nabla_w L(w) = - \frac{\partial^2 E(y^*,w)}{\partial w\partial y^T} z$
9. Return $L(w)$ and $\nabla_w L(w)$.
This looks a bit nasty, since we need to compute second-derivative matrices of $E$. In fact, as long as one has a routine to compute $\nabla_y E$ and $\nabla_w E$, this can be avoided through Efficient Matrix-vector products. This is essentially proposed by Do, Foo, and Ng in “Efficient multiple hyperparameter learning for log-linear models”.
Overall, this is a decent approach, but it can be quite slow, simply because one must solve an “inner” optimization in order to compute each gradient of the “outer” optimization. Often, the inner-optimization needs to be solved to very high accuracy in order to estimate a gradient accurately enough to reduce $L(w)$— higher accuracy than is needed when one is simply using the predicted value $y^*$ itself.
## Truncated optimization
To get around this expense, a fairly obvious idea is to re-define the problem. The slowness of exactly computing the gradient stems from needing to exactly solve the inner optimization. Hence, perhaps we re-define the problem such that an inexact solve of the inner problem nevertheless yields an “exact” gradient?
Re-define the problem as solving
$\min_{w} L(w)=Q(y^*(w)), \text{ } y^*(w) := \text{opt-alg}_y E(y,w)$,
where $\text{opt-alg}$ denotes an approximate solve of the inner optimization. In order for this to work, $\text{opt-alg}$ must be defined in such a way that $y^*(w)$ is a continuous function of $w$. With standard optimization methods such as gradient descent or BFGS, this can be achieved by assuming there are a fixed number of iterations applied, with a fixed step-size. Since each iteration of these algorithms is continuous, this clearly defines $y^*(w)$ as a continuous function. Thus, in principle, it could be optimized efficiently through automatic differentiation of the code that optimizes $E$. That’s fine in principle, but often inconvenient in practice.
It turns out, however, that one can derive “backpropagating” versions of algorithms like gradient descent, that take as input only a procedure to compute $E$ along with it’s first derivatives. These algorithms can then produce the gradient of $L$ in the same time as automatic differentiation.
## Back Gradient-Descent
If the inner-optimization is gradient descent for $N$ steps with a step-size of $\lambda$, the algorithm to compute the loss is simple:
1. Input $w$
2. For $k=0,1,...,N-1$
3. (a) $y_{k+1} \leftarrow y_k - \lambda \nabla_y E(y_k,w)$
4. Return $L(w) = Q(y_N)$
How to compute the gradient of this quantity? The following algorithm does the trick.
1. $\overleftarrow{y_N} \leftarrow \nabla Q(y_N)$
2. $\overleftarrow{w} \leftarrow 0$
3. For $k=N-1,...,0$
4. (a) $\overleftarrow{w} \leftarrow \overleftarrow{w} - \lambda \frac{\partial^2 E(y_k,w)}{\partial w \partial y^T} \overleftarrow{y_{k+1}}$
5. (b) $\overleftarrow{y_k} \leftarrow \overleftarrow{y_{k+1}} - \lambda \frac{\partial^2 E(y_k,w)}{\partial y \partial y^T} \overleftarrow{y_{k+1}}$
6. Return $\nabla L = \overleftarrow{w}$.
Similar algorithms can be derived for the heavy-ball algorithm (with a little more complexity) and limited memory BFGS (with a lot more complexity).
## Code
So, finally, here is the code, and I’ll give a simple example of how to use it. There are just four simple files:
I think the meanings of this are pretty straightforward, so I’ll just quickly step through the demo here. I’ll start off by grabbing taking one of Matlab’s built-in datasets (on cities) so that we are trying to predict a measure of crime from measures of climate, housing, health, transportation, arts, recreation, and economy, as well as a constant. There are 329 data, total, which I split into a training set of size 40, a validation set of size 160, and a test set of size 129.
load cities ratings
X = ratings;
for i=1:size(X,2)
X(:,i) = X(:,i) - mean(X(:,i));
X(:,i) = X(:,i) / std( X(:,i));
end
% predict crime from climate, housing, health, trans, edu, arts, rec, econ,
Y = X(:,4);
X = [X(:,[1:3 5:9]) 1+0*X(:,1)];
p = randperm(length(Y));
X = X(p,:);
Y = Y(p);
whotrain = 1:50;
whoval = 51:200;
whotest = 201:329;
Xtrain = X(whotrain,:);
Xval = X(whoval ,:);
Xtest = X(whotest ,:);
Ytrain = Y(whotrain);
Yval = Y(whoval );
Ytest = Y(whotest );
Next, I’ll set up some simple constants that will be used later on, and define the optimization parameters for minFunc, that I will be using for the outer optimization. In particular, here I choose the inner optimization to use 20 iterations.
opt_iters = 20;
ndims = size(Xtrain,2); w0 = zeros(ndims,1); ndata = size(Xtrain,1); ndata_val = size(Xval,1);
options = []; options.Method = 'gd'; options.LS = 1; options.MaxIter = 100;
Now, I’ll define the training risk function ($E$ in the notation above). The computes the risk with a regularization constant of $a$, as well as derivatives. I’ll also define the validation risk ($Q$ in the notation above).
function [R dRdw dRdloga] = training_risk(w,loga)
a = exp(loga);
R = sum( (Xtrain*w - Ytrain).^2 )/ndata + a*sum(w.^2);
dRdw = 2*Xtrain'*(Xtrain*w-Ytrain) /ndata + 2*a*w;
dRda = sum(w.^2);
dRdloga = dRda*a;
end
function [R g] = validation_risk(w)
R = sum( (Xval*w - Yval).^2 ) / ndata_val;
g = 2*Xval'*(Xval*w-Yval) / ndata_val;
end
Now, before going any further, let’s do a traditional sweep through regularization constants to see what that looks like.
LAMBDA = -5:.25:2;
VAL_RISK = 0*LAMBDA;
for i=1:length(LAMBDA)
VAL_RISK(i) = back_lbfgs(@training_risk,@validation_risk,w0,LAMBDA(i),opt_iters);
end
Plotting, we get the following:
This is a reasonable looking curve. Instead, let’s ask the algorithm to find the constant by gradient descent.
eval = @(loga) back_lbfgs(@training_risk,@validation_risk,w0,loga,opt_iters);
loga = 0;
[loga fval] = minFunc(eval,loga,options);
Running the optimization, we see:
Iteration FunEvals Step Length Function Val Opt Cond
1 2 1.00000e+00 8.74176e-01 3.71997e-03
2 3 1.00000e+00 8.73910e-01 1.86453e-04
3 4 1.00000e+00 8.73909e-01 1.06619e-05
4 5 1.00000e+00 8.73909e-01 3.60499e-08
This leads to a regularizer of the form:
$0.860543 ||w||_2^2$.
We can plot this on the graph, and see it matches the result of cross-validation.
If we actually compute the test-set error, this is 0.708217.
OK, let’s be a little bit more adventurous, and use a third-order regularizer. This is done like so:
function [R dRdw dRdloga] = training_risk2(w,loga)
a = exp(loga(1));
b = exp(loga(2));
R = sum( (Xtrain*w - Ytrain).^2 ) / ndata + a*sum(abs(w).^2) + b*sum(abs(w).^3);
dRdw = 2*Xtrain'*(Xtrain*w-Ytrain) / ndata + a*abs(w).^1.*sign(w) + b*abs(w).^2.*sign(w);
dRda = sum(abs(w).^2);
dRdb = sum(abs(w).^3);
dRdloga = [dRda*a; dRdb*b];
end
Running the optimization, we see:
Iteration FunEvals Step Length Function Val Opt Cond
1 2 1.00000e+00 8.74445e-01 1.17262e-02
2 3 1.00000e+00 8.73685e-01 3.21956e-03
3 4 1.00000e+00 8.73608e-01 1.41744e-03
4 5 1.00000e+00 8.73598e-01 8.20040e-04
5 6 1.00000e+00 8.73567e-01 1.39830e-03
6 7 1.00000e+00 8.73513e-01 2.52994e-03
7 8 1.00000e+00 8.73471e-01 1.77157e-03
...
23 28 1.00000e+00 8.70741e-01 8.42628e-06
With a final regularizer of the form
$0.000160 ||w|_2^2 + 5.117057 * ||w||_3^3$
and a hardly-improved test error of 0.679155.
Finally, let’s fit a fourth-order polynomial.
function [R dRdw dRdloga] = training_risk3(w,loga)
a = exp(loga);
R = sum( (Xtrain*w - Ytrain).^2 ) / ndata;
dRdw = 2*Xtrain'*(Xtrain*w-Ytrain) / ndata;
for ii=1:length(a)
b=ii+1;
R = R + a(ii)*sum(abs(w).^b);
dRdw = dRdw + a(ii)*abs(w).^(b-1).*sign(w);
dRda(ii,1) = sum(abs(w).^b);
end
dRdloga = dRda.*a;
end
eval = @(loga) back_lbfgs(@training_risk3,@validation_risk,w0,loga,opt_iters);
loga = [0; 0; 0; 0];
loga = minFunc(eval,loga,options);
Running the optimization, we see
Iteration FunEvals Step Length Function Val Opt Cond
1 2 1.00000e+00 8.73982e-01 1.02060e-02
2 3 1.00000e+00 8.73524e-01 3.30445e-03
3 4 1.00000e+00 8.73447e-01 1.74779e-03
4 5 1.00000e+00 8.73435e-01 1.43655e-03
5 6 1.00000e+00 8.73345e-01 4.98340e-03
6 7 1.00000e+00 8.73295e-01 1.85535e-03
7 8 1.00000e+00 8.73231e-01 1.81136e-03
...
38 41 1.00000e+00 8.69758e-01 3.44848e-06
Yielding the (strange) regularizer
$0.000000 ||w||_2^2 + 0.000010 ||w||_3^3 + 24.310631 ||w||_4^4 + 0.325565 ||w||_5^5$
and a final test-error of 0.67187.
# Reducing Sigmoid computations by (at least) 88.0797077977882%
A classic implementation issue in machine learning is reducing the cost of computing the sigmoid function
$\sigma(a) = \frac{1}{1+\exp(-a)}$.
Specifically, it is common to profile your code and discover that 90% of the time is spent computing the $\exp$ in that function. This comes up often in neural networks, as well as in various probabilistic architectures, such as sampling from Ising models or Boltzmann machines. There are quite a few classic approximations to the function, using simple polynomials, etc. that can be used in neural networks.
Today, however, I was faced with a sampling problem involving the repeated use of the sigmoid function, and I noticed a simple trick that could reduce the number of sigmoids by about 88% without introducing any approximation. The particular details of the situation aren’t interesting, but I repeatedly needed to do something like the following:
1. Input $a \in \Re$
2. Compute a random number $r \in [0,1]$
3. If $r < \sigma(a)$
4. Output $+1$
5. Else
6. Output $-1$
Now, let’s assume for simplicity that $a$ is positive. (Otherwise, sample using $-a$ and then switch the sign of the output.) There are two observations to make:
1. If $a$ is large, then you are likely to output $+1$
2. Otherwise, there are easy upper and lower bounds on the probability of outputting $+1$
This leads to the following algorithm:
1. Input $a \in \Re$
2. Compute a random number $r \in [0,1]$
3. If $a \geq 2$
4. If $r \leq 0.880797077977882$ or $r \leq \sigma(a)$
5. Output $+1$
6. Else
7. Output $-1$
8. Else
9. If $r > .5 + a/4$
10. Output $-1$
11. Else if $r \leq .5 + a/5.252141128658$ or $r \leq \sigma(a)$
12. Output $+1$
13. Else
14. Output $-1$
The idea is as follows:
1. If $a\geq 2$, then we can lower-bound the probability of outputting +1 by a pre-computed value of $\sigma(2)\approx0.8807...$, and short-circuit the computation in many cases.
2. If $a\leq 2$, then we can upper bound the sigmoid function by $.5+a/4$.
3. If $a\leq 2$, then we can also lower bound by $.5+a/5.252141...$ respectively. (This constant was found numerically).
The three cases are illustrated in the following figure, where the input $a$ is on the x-axis, and the random number $r$ is on the y-axis.
Since, for all $a$ at least a fraction $\sigma(2)\approx.8807$ of the numbers will be short-circuited, sigmoid calls will be reduced appropriately. If $a$ is often near zero, you will do even better.
Obviously, you can take this farther by adding more cases, which may or may not be helpful, depending on the architecture, and the cost of branching vs. the cost of computing an $\exp$.
# Windows Binaries for CRF toolbox
I’m happy to report that, thanks to Alexei Skurikhin, Windows binaries are now available for all the functions in my CRF toolbox. Hopefully this makes for an easier out of the box experience from Windows folks who, based on my email, have struggled to compile things in the past.
# Etymology of Latin words/abbreviations
I often find the academic convention of using Latin words and abbreviations a bit silly. (Though I’m as guilty as anyone.) Today, I decided to look into a couple common ones, to see if there is any good reason not to use the ordinary English equivalent.
• “e.g.” is an abbreviation of the Latin “exemplī grātiā”, meaning “for example”. Here, “exemplī” is not hard to guess and “grātiā” means “for the sake of”.
• “i.e.” is an abbreviation of the Latin “id est”, meaning “that is”.
• “via”. This is not an abbreviation, but was the Latin word for a road. The plural is “viae”, which you should probably work into your next paper submission. (I think the usage would be “Linear convergence and entropic sparsity control viae quasi-stochasticity and entropy regulators”?)
• “et al.”. Here “et” is Latin for “and” while “al.” is an abreviation for “alii”, meaning “others”. I seem to recall a major statistics journal maintaining the noble tradition of actually writing “and others”, but can’t recall which one.
I see no particular reason we haven’t adopted English words for all these conventions, but in the context of the many local optima society is in… this hardly seems worth changing. | 10,711 | 40,337 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 163, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2020-24 | latest | en | 0.853219 |
http://www.ehow.co.uk/how_5669104_evenly-space-spindles-decks.html | 1,537,833,678,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160842.79/warc/CC-MAIN-20180924224739-20180925005139-00214.warc.gz | 309,517,625 | 10,275 | DISCOVER
# How to evenly space spindles on decks
UK building regulations for decks dictate a maximum gap between spindles of 99 mm (3 57/64 inches). So, builders must calculate the spacing of spindles precisely. Maximum gap size codes, designed to prevent children from getting their heads stuck between spindles, also dictate the number of spindles to be used in a length of deck rail. The challenge begins because the width of the spindles must be considered in the spacing. Using a mathematical formula will help streamline the process of spacing spindles evenly while adhering to building codes.
## Number of spindles
Measure the length of the deck rail between posts.
Subtract the maximum gap size -- 99 mm (3 57/64 inches) -- from the length, and divide the length of the rail by this number.
Round the quotient up to the nearest whole number. This number is how many spindles you will need to conform to the building code. For example, if the rail is 2.65 m (104 inches) long, the gap space code is 99 mm (3 57/64 inches), and the spindle width is 4 cm (1.5 inches), the formula would be (265 - 9.9) / (9.9 + 4) = 255.1 / 13.9 = (104 - 4) / (4 + 1.5) = 18.35. Round up to 19; you will need 19 spindles.
## Spacing spindles
Calculate the space the spindles will occupy by multiplying the spindle width by the number of spindles. Subtract this answer from the rail length to get the amount of gap space. In this example, the formula would be 19 x 4 = 76; 255.1 - 76 = 179.1 cm.
Calculate the spacing between spindles by dividing the amount of gap space by the number of spindles. For example, 179.1 / 19 = 9.4 cm.
Starting at the end post of the rail, measure the gap space and attach the first spindle there. Measure the same distance from the first spindle for the placement of the second spindle. Repeat this process until all the spindles are installed.
#### Tip
Use a number for your gap space that is slightly less than the building standard (100 mm). That way, if any shrinking occurs in the spindles because of cold weather or ageing, the code is maintained. | 533 | 2,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2018-39 | latest | en | 0.883978 |
https://www.globalcompose.com/mathematics-papers/sample-assignment-paper-on-descriptive-statistics/ | 1,669,499,319,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446709929.63/warc/CC-MAIN-20221126212945-20221127002945-00504.warc.gz | 852,830,380 | 27,238 | # Sample Assignment Paper on Descriptive statistics
Question 1
Arithmetic mean = (28 + 32 + 28 + 30 + 34)/5 = 30.4
Median = 30
Mode = 28
Based on the above results, arithmetic mean would be the best measure of center because it lies approximately at the center of the five runs.
Question 2
Arithmetic mean = (31 + 29 + 31 + 29 + 31)/5 = 30.2
Median = 31
Mode = 31
Based on the above results, arithmetic mean would be the best measure of center because it lies approximately at the center of the five runs.
Question 3
Arithmetic mean = (29 + 32 + 28 + 32 + 30)/5 = 30.2
Median = 30
Mode = 32
Based on the above results, median would be the best measure of center because it lies at the center of the five runs.
Question 4
The smallest number is 101 and the largest number is 157. Therefore, the range is 157 – 101 = 56. To get the class width, divide 56 by 5. That is, . Rounding off to the nearest whole number, each class will have a width of 11. The classes will therefore be 101-112, 113-124, 125-136, 137-148 and 149-160
Class Frequency 101-112 3 0.12 113-124 11 0.44 125-136 7 0.28 137-148 2 0.08 149-160 2 0.08
Question 5
Question 6
Question 7
The above distribution is skewed to the left meaning that it is not symmetric.
Question 8
Class Frequency Mid point (x) f.x 101-112 3 106.5 319.5 113-124 11 118.5 1303.5 125-136 7 130.5 913.5 137-148 2 142.5 285 149-160 2 154.5 309 25 3130.5
Sample mean =
Question 9
Range = the highest pay – the lowest pay = 1019 – 444 = 575
Question 10
Mean =
Variance =
Question 11
Standard deviation =
Box-and-whisker plots
Question 1
Min: 2
Q1: 6
Q2: 11
Q3: 13
Max: 18
Question 2
Ordering the data we have 4, 5, 7, 10, 10, 10, 11, 13, 13, 14, 17, 18, 19.
Min: 4
Q1= (7 + 10)/2 = 17/2 = 8.5
Q2 = (n + 1)/2 =14/2 = 7th value = 11
Q3 = (14 + 17)/2 = 31/2 = 15.5
Max: 19
Outlier: none
Outlier: anything below Q1 – 1.5 IQR or anything above Q3 + 1.5 IQR
IQR = Q3 – Q1 = 15.5 – 8.5 = 7
Therefore, Q1 -1.5 IQR = 8.5 – (1.5*7) = 8.5 – 10.5 = -2
And Q3 + 1.5 IQR = 15.5 + 10.5 = 26
Given that no data lies below -2 and above 26 then there are no outliers.
Question 3
No, it is not common for a teenager to spend more than 1 hour getting ready for school. In addition, it is not common for a teenager to spend between 1 and 2 hours getting ready for a school dance. The reason is that in both cases majority of the students spend 1 hour or less getting ready for either school or school dance.
Question 4
Only 25 percent of the students spend at least 15 minutes getting ready for a school dance.
Question 5
True because Q3 – Q2 is approximately 25 percent
Question 6
It might not be possible to determine the number of girls that shop for clothes in September because the box plot does not provide such data. It only provides the amount of money they spend on clothes.
Question 7
The percentage of the girls that spent less than \$85 is 25 because the lower quartile lies at \$85.
Question 8
I would expect the mean number of dollar spent on clothes to be higher than the median. This is in relation to the fact that the box plot is skewed to the right. Therefore, it would be reasonable to expect the mean to be higher than the median.
Stem and leaf diagrams
Question 1
Ordering the data we have 8, 9, 11, 13, 17, 18, 18, 21, 22, 25, 27, 28, 29, 31, 32.
Stem Leaf 0 8, 9 1 1, 3, 7, 8, 8 2 1, 2, 5, 7, 8, 9 3 1, 2
Question 2
Range = 40 – 8 = 32
Question 3
Students with GPA of 2 and above are 12
Question 4
The mode is 14 and 25
Question 5
Median = (25 + 25)/2 = 25
Question 6
Based on the results provided, there are 18 students in Mr. Smith’s math class.
Scatter plots & line graphs
Question 1
The scatter plot shows a positive relationship between the number of hours students spend studying and their grades. This means that as the number of hours of studying increase, the grade improves. It also indicates that as the number of hours of studying decrease, the grade drops.
Question 2
a). Graph 3 shows no relationship between temperature and test score because the plots do not form a consistent pattern.
b). Graph 2 shows a positive correlation between temperature and score because as the temperature increases, the score increases as well. In addition, as the temperature decreases, the score decreases as well.
Basic probability
Question 1
This is an experimental probability because the probability provided is based on company’s records. If the records were not used, it would have been a theoretical one. However, given that company’s records were used, then it is an example of experimental probability.
Question 2
This is an example of a theoretical probability because the probability provided is based on expectations.
Question 3
The probability of selecting an even number between 1 and 15 is as it follows;
Even numbers between 1 and 15 are 2, 4, 6, 8, 10, 12, and 14. That is, they are 7. Therefore, the probability of selecting an even number between 1 and 15 is
Question 4
Numbers divisible by 4 between 1 and 50 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, and 48. This means that there are 12 numbers divisible by 4 between 1 and 50. Therefore, the probability of selecting one of these numbers is
Question 5
A standard deck has 52 cards. The 52 cards are divided into four groups of spades, hearts, diamonds and clubs. Each of these four groups has an ace. Therefore, there are 4 aces. Based on this fact, the probability of drawing an ace from a standard deck of playing cards is
Question 6
The bag contains 14 marbles (6 red, 3 green and 5 yellow). The probability of not selecting a green marble would be 1 – the probability of selecting a green marble. That is,
Question 7
A standard deck has 52 cards. The 52 cards are divided into four groups of spades, hearts, diamonds and clubs. Each of these four groups has a king. Therefore, there are 4 kings in a standard deck of cards. Based on this fact, the probability of not selecting a king from the standard deck would be
Multiplication rule
Question 1
For this event the probability of selecting an ace would be 4 out of 52 and the probability of selecting a jack would also be 4 out of 52. This being the case, the events would be independent because the first event would not affect the outcome of the second event in any way.
Question 2
For this event, the probability of selecting the first ball would be 1 out of 100, but the probability of selecting the second ball would be 1 out of 99. This means that the first event would affect the outcome of the second event. As a result, the two events would be dependent.
Question 3
In this case, the probability of getting a head throughout would be 1 out of 2. However, because the exercise would be repeated eight times, then the probability of getting a head every time would be (½)8 = 1/28 = 1/256
Question 4
A standard deck of cards has 52 cards; 13 diamonds; 4 queens and 13 spades. Therefore, the probability of selecting a diamond with replacement would be 13/52. The probability of selecting a queen with replacement would be 4/52 and the probability of selecting a spade with replacement would be 13/52. In total, three cards are selected with replacement. Therefore, their probability would be
Question 5
A standard deck of cards has 52 cards, 1 queen of hearts and 13 hearts including the queen of hearts. Therefore, the probability of selecting the queen of hearts would be 1/52. The probability of selecting any heart after the queen of hearts has been selected would be 12/51 because one heart would be selected already and the number of cards would have reduced by one. In total, the probability of selecting the two cards would be
Question 6
There are 11 parts in the box
4 are defective and 7 are not defective
P (both parts are defective) =
Question 7
P (both parts not defective) =
Question 8
P (at least one defective) = [P (defective) and P (not defective)] + [P (not defective) and P (defective)]
=() + () =
Question 1
The events of selecting a male student and selecting a senior are not mutually exclusive because there are 325 males in the seniors. Consequently, in the process of selecting a male student you can select a senior student. In the process of selecting a senior student, you can also select a male student. Therefore, the two events are not mutually exclusive.
Question 2
P (male or senior) = P (male) + P (senior) – P (senior male only)
P (male) =
P (senior) =
P (senior male only) =
P (male or a senior) =
Question 3
Selecting a heart or 6
Mutually exclusive? No Probability P (selecting a 6) + P (selecting a heart) – P (6 and heart) =
Question 4
Selecting a red suit or a jack
Mutually exclusive? No Probability P (a red suit) + P (jack) – P (a red suit and jack) =
Question 5
Rolling a die: rolling a 6 or a number greater than 4
Mutually exclusive? No Probability P ( 5) + P (6) =
Question 6
Rolling a 5 or an even number
Mutually exclusive? Yes Probability P (5) + P (2) + P (4) + P (6) =
Combinations & permutations
Question 1
8P3 =
Question 2
8C3 =
Question 3
Given that the first digit cannot be zero, then the numbers can be arranged in 9P4 different ways. However, because the last digit should be even, we should subtract the arrangements that would end with odd numbers from 9P4. In this case, the arrangements that would end with odd numbers would be 5!
9P4 =
Subtracting 5! From 3,024 we get (3024 – 120) = 2,904 codes
Question 4
Given that the horses cannot tie, then the 8 horses can finish the race in 8*7*6*5*4*3*2*1 ways. This equals 40,320 different ways.
Question 5
Smith has a total of 4 + 8 + 6 = 18 trees. Given that order would be insignificant in this case, then Smith can plant these trees in 18! different ways.
Question 6
The jury can be selected 40C12 different ways because order is important.
40C12 =
Probability exam
Question 1
In total, there were 2,466,000 degrees. Out of these degrees, 467,000 were master’s degree. Therefore, the probability of selecting a person with master’s degree is
Question 2
The females that earned master’s degree are 270,000. Therefore, the probability of selecting a female with master’s degree would be
Question 3
The males with associate’s degrees were 231,000. Therefore, the probability of selecting a male with associate’s degree would be
Question 4
The number of the people that did not earn doctorate is 632,000 + 1,322,000 + 467,000 = 2,421,000. Therefore, the probability of selecting a person that did not earn a doctorate degree is
Question 5
In this case, we have 26 letters and 10 digits. Then we have 26 + 10 = 36 letters and digits to arrange on the license plate. Given that only 4 characters should appear on the license plate, then we can have 36P4 possible license plates.
Question 6
The 15 swimmers can finish first, second and third in 15P3 different ways.
Question 7
In this case, order is an important aspect. Therefore, the editor can select the stories in 17C4 different ways. That is
Question 8
We have 11 characters in mathematics. However, there are 2 A’s, 2 M’s, 2 T’s. Therefore, if we are to arrange these characters in different ways, the many times that A’s would appear together, the arrangement would not be distinguishable. The same case would apply to M’s and T’s. As a result, if we are to arrange these characters, we would only have 8! distinguishable outcomes.
Question 9
These events would be independent because the sixth toss does not depend on the other five toss.
Question 10
The two events are not mutually exclusive because a king card is also a face card. As a result, the outcome of event A will automatically affect the outcome of event B.
Probability distributions
Question 1
X 1 2 3 4 5 6 P(x) 0.02 0.34 0.18 0.26 0.05 0.15
Probability distribution Yes Why or why not? Because the sum of P(x) is equal to one
Question 2
X 1 2 3 4 P(x) 0.31 0.08 0.48 0.16
Probability distribution? No Why or why not? Because the sum of P(x) is equal to 1.03 and not equal to one
Question 3
X 1 2 3 4 5 P(x) ¾ 5/4 ½ ¼ -7/4
Probability distribution? No Why or why not? Because P(X=5) is negative. Therefore, even if the sum of P(x) is equal to 4/4 which is one, the condition that p(x) for any real number should be non-negative is not met.
Question 4
Pets 0 1 2 3 4 5 Households 1491/2175 = 0.6855 425/2175 = 0.1954 168/2175 = 0.0772 48/2175 = 0.0221 29/2175 = 0.0133 14/2175 = 0.0065
Question 5
Computers 0 1 2 3 Households 300/695 = 0.4317 280/695 = 0.4029 95/695 = 0.1367 20/695 = 0.0287
More probability distributions
Question 1
Mean =
=0.1954 + 0.1544 + 0.0663 + 0.0532 + 0.0325 = 0.5018
Expected value = mean = 0.5018
Variance =
= (0-0.5018)2*0.6855 + (1-0.5018)2 *0.1954 + (2-0.5018)2*0.0772 + (3-0.5018)2*0.0221 + (4-0.5018)2*0.0133 + (1-0.5018)2*0.0065
= 0.172611 + 0.048499 + 0.173283 + 0.137926 + 0.162757 + 0.13152
= 0.826597
Standard deviation =
Question 2
Mean =
=0.4029 + 0.2734 + 0.0861
=0.7624
Expected value = mean = 0.7624
Variance =
=0.250927 + 0.022745 + 0.209377 + 0.143697
= 0.626746
Standard deviation =
Binomial distributions
Question 1
This is a binomial experiment because a person that enters the clothing store can either buy or not buy something from the store. The following are the values for the experiment.
n = 16;
p = 0.28;
q = 0.72;
Question 2
This is not a binomial experiment because there are three possible outcomes. The three possible outcomes are approve, disapprove and no opinion. This being the case, the experiment cannot be binomial.
Question 3
n = 5, p = ¼, q = ¾
a). P (X=3) =
=
b). P (X=5) =
=
=
Question 4
P = 0.54, q = 0.46, n = 10
More binomial distributions
Question 1
n = 100; p = 0.6; q = 0.4
Mean = np = 100*0.6 = 60
Variance = npq = 100*0.6*0.4 = 24
Standard deviation =
Question 2
n = 46; p = 0.18; q = 0.82
Mean = np = 46*0.18 = 8.28
Variance = npq = 46*0.18*0.82 = 6.7896
Standard deviation =
Question 3
P = 0.63, q = 0.37, n = 5
Mean = np = 5*0.63 = 3.15
Variance = npq = 5*0.63*0.37 = 1.1655
Standard deviation =
The results show that about three teenagers have jobs. The standard deviation of these teenagers is 1.07958 and their variance is 1.1655.
Question 4
n = 20; p = 0.38; q = 0.62
Mean = np = 20*0.38 = 7.6
Variance = npq = 20*0.38*0.62 = 4.712
Standard deviation =
The results show that about eight people of the interviewees had type O+ blood. The standard deviation of these interviewees was 2.1707 and their variance was 4.712.
The normal distribution
Question 1
For this case, it is almost possible that a volume of 11.4 oz will be found in the population of the soda. This is in relation to the fact that upon standardizing the sample, we get a z-score of -3. This indicates 99.7 percent likelihood.
Question 2
P (150≤X≤200) = P
= P (1≤Z≤3)
= 0.1587 – 0.0013
= 0.1574
This means that the likelihood that the values will be between 150 and 200 is 0.1574 or 15.74 percent. This is a small percentage indicating that the values are unlikely to lie in the range.
Question 3
P (48<X<61) = P
= P (-1<Z<1)
= 0.3413 + 0.3413
= 0.6826
This shows that the likelihood of a male dog of the same breed lying in the range between 48 lb and 61 lb is 68.26 percent. This indicates a high likelihood.
Question 4
P (X=0.004) = P
= P (Z=3)
= 0.5 + (0.5 – 0.0013)
= 0.9987
This shows that a value of 0.004 is very common in the population.
Finding probabilities
Question 1
Mean = 124, standard deviation = 74
P (X>198) = P
= P (Z>1)
= 0.5 – 0.34
= 0.16
Question 2
P (23<X<337) = P
= P (0<Z<2)
= 0.5 – 0.475
= 0.025
Question 3
P (X>310) = P
= P (Z>3)
= 0.5 – 0.4985
= 0.0015
Question 4
P (-286<X<470) = P
= P (-2<Z<2)
= 95%
Finding values
Question 1
A z-score of -3.4 means that 99.97 percent of the distribution lies above -3.4. It also means that only 0.03 percent of the distribution lies below -3.4.
Question 2
X – 70 = -1.82*62
X = 70 – 112.84
X = -42.84
Question 3
µ = 60, δ = 98
Z-score =
Question 4
P (Z<2.02) = 0.5 + (0.5 – 0.0217)
= 0.5 + 0.4783
= 0.9783
Question 5
P (X>-1.97) = 0.5 + (0.5 – 0.0244)
= 0.5 + 0.4756
= 0.9756
The central limit theorem
Question 1
P (Z=0.0017) = 1 – 0.0017
= 0.9983
Question 2
P (X<33.5) = P
= P (Z<-0.3)
= 0.5 – 0.3821
= 0.1179
Question 3
P (X<30.92) = P
= P (Z<0.092)
= 0.5 + 0.4641
= 0.9641 | 5,174 | 16,431 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2022-49 | longest | en | 0.813423 |
https://programmingpraxis.com/2015/09/25/multiply-perfect-numbers/ | 1,685,585,393,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647525.11/warc/CC-MAIN-20230601010402-20230601040402-00363.warc.gz | 540,600,814 | 28,695 | Multiply Perfect Numbers
September 25, 2015
As regular readers know, I greatly enjoy problems at the intersection of recreational mathematics and number theory. Today’s exercise is an example of that.
As was known to the ancient Greeks, perfect numbers such as 6 and 28 are equal to the sum of their aliquot divisors (those divisors less than the number itself); for instance, 6 = 1 + 2 + 3 and 28 = 1 + 2 + 4 + 7 + 14. Mathematicians refer to these as P2 numbers the sum of the divisors, including the number itself, is twice the number.
Multiply perfect numbers are numbers such that the sum of their divisors are some multiple of the number; perfect numbers have divisor-sums twice the number, triply perfect numbers P3 have divisor-sums thrice, the number, and so on.
Your task is to write a program that finds multiply perfect numbers. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Pages: 1 2
8 Responses to “Multiply Perfect Numbers”
1. FA said
As Scala stram. (Based on the previous implementation)
```def multiplyPerfectNumbers(): Stream[BigDecimal] = bdStreamInf().filter(n => factors(n).sum%n == 0)
multiplyPerfectNumbers().take(5).foreach(println(_))
//> 1
//| 6
//| 28
//| 120
//| 496
```
2. Rutger said
In Python. Optimized my ‘divisors’ algorithm from the exercise “Fermat’s Divisors Challenges” here couple of weeks ago.
```def divisors(n):
result = []
for i in range(1, int(n**0.5)+1):
if not n % i:
result.append(i)
if i*i != n: # prevent duplicate, e.g. 3,3 for 9
result.append(n/i) # if i is a divisors of n, then so is n/i
return result
def sum_divisors_is_perfect_multiple(n):
return not sum(divisors(n)) % n
for i in range(1, 1000000):
if sum_divisors_is_perfect_multiple(i):
print i
# output
# 1
# 6
# 28
# 120
# 496
# 672
# 8128
# 30240
# 32760
# 523776
```
3. Francesco said
`f n = filter (\\x -> sum (filter ((==0) . rem x) [1..x]) == x*n) [1..]`
4. r. clayton said
A solution in Guile Scheme.
5. Marijn said
;;; the sum of divisors of a perfect number equals some multiple of that number
;;; example: 6 is 2-perfect, because 2*6 = 1+2+3+6
(define mod modulo)
(define (divisors num)
(let ((sqrt-num (sqrt num)))
(let loop ((d 1) (divs ‘()))
(if (<= d sqrt-num)
(loop (+ d 1)
(if (zero? (mod num d))
(if (= (* d d) num)
(cons d divs)
(cons (/ num d) (cons d divs)))
divs))
divs))))
(define (perfect candidate)
(if (zero? (mod (apply + (divisors candidate)) candidate))
(/ (apply + (divisors candidate)) candidate)
#f))
(let loop ((num 1))
(let ((p? (perfect num)))
(if p?
(begin
(display num)(display ": ")
(display p?)(newline))))
(if (<= num 1000000)
(loop (+ num 1))
#f))
output:
1: 1
6: 2
28: 2
120: 3
496: 2
672: 3
8128: 2
30240: 4
32760: 4
523776: 3
According to Wikipedia it is an open problem how many of these numbers exist…
6. Creating a “sieve” of divisors was much faster for finding the series for me than factoring each one individually.
```
std::vector<int> aliquot_series (int limit)
{
std::vector<int> results (limit + 1, 1);
results[0] = 0;
results[1] = 0;
for (int i = 2; i <= limit; i++)
{
for (int j = i + i; j <= limit; j += i)
{
results[j] += i;
}
}
return results;
}
void list_perfect_sums2(int limit)
{
std::vector<int> series = aliquot_series(limit);
for (int i = 2; i <= limit; i ++)
{
if (series[i] % i == 0) std::cout << i << "\n";
}
}
```
7. Mike said
I uses a sieve-like method. In Python:
```def multiply_perfect(limit=1000000):
divisor_sum = [0]*limit
for n in range(1, limit):
for k in range(n, limit, n):
divisor_sum[k] += n
if divisor_sum[n] % n == 0:
print(n, divisor_sum[n], divisor_sum[n]//n)
#test
multiply_perfect(5000000)
1 1 1
6 12 2
28 56 2
120 360 3
496 992 2
672 2016 3
8128 16256 2
30240 120960 4
32760 131040 4
523776 1571328 3
2178540 8714160 4
```
A PhD thesis at https://opus.lib.uts.edu.au/research/bitstream/handle/2100/275/02Whole.pdf?sequence=2021
has lots of interesting history and information on multiperfect numbers. Like current minimum and maximum k-perfect numbers for 1 <= k <= 11. Also, there are no odd multiperfect numbers less than 10^70.
8. Mayur Lokare said
#include
int main()
{
int i,j,n=15,count=0,num;
printf(“Enter the length of sequance\n”);
scanf(“%d”,&num);
for(n=1;n<num;n++)
{
for(i=1;i<n;i++)
if(n%i==0)
count+=i;
if(count==n)
printf("%d is Perfect Number\n",n);
count=0;
}
}
o/p:
10000
6 is Perfect Number
28 is Perfect Number
496 is Perfect Number
8128 is Perfect Number | 1,449 | 4,526 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-23 | latest | en | 0.826214 |
http://guideformaths.azurewebsites.net/category/computer-science/ | 1,571,126,498,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986657586.16/warc/CC-MAIN-20191015055525-20191015083025-00214.warc.gz | 84,675,727 | 10,382 | ## Merge Sort
Merge Sort Algorithm:
• Is based on the divide and conquer approach.
• Divides the list into two sublists of sizes as nearly equal as possible.
• Sorts the two sublists seperately by using merge sort.
• Merges the sorted sublists into one single list.
Algorithm:
MergeSort(low,high)
1. if(low >= high)
• return
2. Set mid = (low+high)/2
3. Divide the list into two sublists of nearly equal lengths and sort each sublist by using merge sort. The steps to do this are as follows:
1. MergeSort(low,mid)
2. MergeSort(mid+1,high)
4. Merge the two sorted sublists:
1. set i = low
2. set j = mid + 1
3. set k = low
4. Repeat until i > mid or j > high //This loop will terminates when you reach the end of one of the two //sublists.
1. if(arr[i] <= arr[j])
1. Store arr[i] at index k in array B
2. Increment i by 1
2. else
1. store arr[j] at index k in array B
2. Increment j by 1
3. Increment k by 1
5. Repeat until j > high //If there are still some elements in the second sublist append them to the new list
1. Store arr[j] at index k in array B
2. Increment j by 1
3. Increment k by 1
6. Repeat until i > mid //If there are still some elements in the first sublist append them to the new list
1. Store arr[i] at index k in array B
2. Increment i by 1
3. Increment k by 1
5. Copy all elements from the sorted array B in to the original array arr
### Implementing Merge Sort
#include<stdio.h>
#include<conio.h>
#define max 20
void mergesort(int a[],int low, int high);//Divides the array into halves
void merge(int a[],int low,int mid, int high);//Sort and merges subarray
int main()
{
int a[max],i=0,n=0;
printf(“Enter no. of elements\n”);
scanf(“%d”,&n);
printf(“Enter no.\n”);
for(i=0;i<n;i++)
scanf(“%d”,&a[i]);
printf(“Unsorted Array:\t”);
for(i=0;i<n;i++)
printf(“%d\t”,a[i]);
mergesort(a,0,n-1);
printf(“\nSorted Array:\t”);
for(i=0;i<n;i++)
printf(“%d\t”,a[i]);
return 0;
}
void mergesort(int a[],int low,int high)
{
if(low<high)
{
int mid = (low+high)/2;
mergesort(a,low,mid);
mergesort(a,mid+1,high);
merge(a,low,mid,high);
}
}
void merge(int a[],int low, int mid, int high)
{
int high1 = mid-low+1;
int high2 = high-mid;
int i=0,j=0,k=0;
int temp1[high1],temp2[high2]; //Temproary arrys
for(i=0;i<high1;i++) //copy array values in temp arrays
temp1[i] = a[low+i];
for(j=0;j<high2;j++)
temp2[j] = a[mid+1+j];
i=0;
j=0;
k=low;
while(i<high1 && j<high2) //compare elements of subarray
{
if(temp1[i]<temp2[j])
{
a[k] = temp1[i];
i++;
}
else
{
a[k] = temp2[j];
j++;
}
k++;
}
while(i<high1) //copy remaining elements as it is
{
a[k] = temp1[i];
i++;
k++;
}
while(j<high2)
{
a[k] = temp2[j];
j++;
k++;
}
}
If you found any mistakes in above code then please let us know in the comment box below.
Learn Quick Sort Algorithm From Here
## Quick Sort
Quick Sort algorithm:
• is one of the most efficient sorting algorithms.
• is based on divide and conquer approach.
• Successively divides the problem into smaller parts until the problem become so small that they can be directly solved.
In quick sort algorithm, you:
• Select an element from the list called as pivot.
• Partition the list into 2 parts such that:
• All the elements towards the left end of the list are smaller than the pivot.
• All the elements towards the right end of the list are greater than the pivot.
• Store the pivot at its correct position between the two parts of the list.
You repeat this process for each of the two sublists created after partitioning. This process continues untill one element is left in each sublist.
Algorithm:
QuickSort(low,high)
1. if(low>high)
• Return
2. Set pivot = arr[low]
3. set i = low + 1
4. set j = high
5. Repeat step 6 until i < high and arr[i] < pivot //Search for element greater than pivot
6. increment i by 1
7. Repeat step 8 until j > low and arr[j] > pivot //Search for element smaller than pivot
8. Decrement j by 1
9. if i < j //If greater element is on the left of smaller element
• swap arr[i] with arr[j]
10. if i<=j
• Go to step 5 //Continue the search
11. if low < j
• Swap arr[low] with arr[j] //Swap pivot with last element in first part of the list
12. QuickSort(low,j-1) //Apply quicksort on list left of pivot
13. QuickSort(j+1,high) //Apply quicksort on list right of pivot
### Implementing Quick Sort
#include<stdio.h>
void quicksort(int a[],int low,int high);
int main()
{
int high=0,i=0,arr[20];
printf(“Enter no. of elements\n”);
scanf(“%d”,&high);
printf(“Enter %d numbers to be sorted\n”,high);
for(i=0;i<high;i++)
scanf(“%d”,&arr[i]);
quicksort(arr,0,high-1);
printf(“Sorted Array:\t”);
for(i=0;i<high;i++)
printf(“%d\t”,arr[i]);
return 0;
}
void quicksort(int a[],int low,int high)
{
int first,last,temp,pivot;
pivot = a[low];
first = low+1;
last = high;
if(low<high)
{
while(first<last)
{
while(a[first]<pivot && first < high)
first++;
while(a[last]>pivot && last > low)
last–;
if(first<last)
{
temp = a[first];
a[first] = a[last];
a[last] = temp;
}
}
if(low<last)
{
temp = a[low];
a[low] = a[last];
a[last] = temp;
}
quicksort(a,low,last-1);
quicksort(a,last+1,high);
}
}
If you found any mistakes in above code then please let us know in the comment box below. | 1,582 | 5,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2019-43 | latest | en | 0.670658 |
http://stackoverflow.com/questions/18134371/why-dont-i-get-the-right-variance-when-using-avg-and-squared-avg | 1,432,736,207,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207929003.41/warc/CC-MAIN-20150521113209-00133-ip-10-180-206-219.ec2.internal.warc.gz | 226,726,768 | 16,882 | # Why don't I get the right variance when using avg and squared avg?
I have a vector `a`
``````a = [86 100 41 93 75 61 76 92 88 97]
``````
And I want to calculate the `std` and `mean` by myself:
``````>> mean(a)
ans =
80.9000
>> std(a)^2
ans =
335.2111
``````
But when I do it like that I get wrong variance:
``````>> avg = mean(a)
avg =
80.9000
>> var = sum(a.^2)/length(a) - avg^2
var =
301.6900
``````
What do I miss here ?
why `sum(a.^2)/length(a) - avg^2 != std(a)^2` ?
-
## 2 Answers
Try this:
``````var = sum(a.^2)/(length(a)-1) - (length(a))*mean(a)^2/(length(a)-1)
var =
335.2111
``````
`var` is computed as (unbiased) sample, not population variance.
For a complete explanation you can read here.
From the matlab documentation,
VAR normalizes Y by N-1, where N is the sample size. This is an unbiased estimator of the variance of the population from which X is drawn, as long as X consists of independent, identically distributed samples.
but
Y = VAR(X,1) normalizes by N and produces the second moment of the sample about its mean. VAR(X,0) is the same as VAR(X).
so that
``````>> var(a,1)
ans =
301.6900
``````
-
Thanks to Try Hard:
an unbiased sample variance is given by:
``````>> 1/(length(a)-1) * sum((a-mean(a)).^2)
ans =
335.2111
``````
- | 416 | 1,296 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2015-22 | latest | en | 0.911404 |
https://thewayhelooks.com/cndrvgqe/1gj5acz.php?page=matrix-transformation-in-python-bd4b74 | 1,624,246,277,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488262046.80/warc/CC-MAIN-20210621025359-20210621055359-00608.warc.gz | 516,292,983 | 12,592 | # matrix transformation in python
Multiple View Geometry in Computer Vision. startJointFn.set(TMatrix) cmds.refresh() # Wait 1 second and set it back to the inital transformation matrix from the joint function # This is just to see the difference between the two transformation matrices. Now, let's suppose we have a matrix. So in this recipie we will learn how to generate classification report and confusion matrix in Python. Ronald Goldman. f \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}a&b\\c&d\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}, Printing a matrix. \mathbf x = module for a faster implementation of some functions. Return types are numpy arrays unless specified otherwise. reverse is exactly the method that we need. \], $2000. If you're not sure which to choose, learn more about installing packages. Linear transformations leave the origin fixed and preserve parallelism. Prerequisite: Linear Algebra | Defining a Matrix. But there are some interesting ways to do the same in a single line. Python DataFrame.as_matrix - 22 examples found. 4(4):629-642. Transpose of a matrix is a task we all can perform very easily in python (Using a nested loop). J Opt Soc Am A. In the python code, we will add two Matrices. We can express our transformation in the form of this matrix. Transformations.py is no longer actively developed and has a few known issues For perspective transformation, we need 4 points on the input image and corresponding points on the output image. frame : rotations are applied to static (0) or rotating (1) frame. In Python, we can implement a matrix as a nested list (list inside a list). 0 & 0 & 1 & 1 \\ A triple of Euler angles can be applied/interpreted in 24 ways, which can This Python code is not optimized for speed. The reshape method above is one such transformation. Transpose a matrix means we’re turning its columns into its rows. You can rate examples to help us improve the quality of examples. If the generated inverse matrix is correct, the output of the below line will be True. f \begin{pmatrix}x\\y\end{pmatrix} = Je développe le présent site avec le framework python Django. Python OpenCV – Affine Transformation Last Updated: 21-04-2020 OpenCV is the huge open-source library for computer vision, machine learning, and image processing and now it plays a major role in real-time operation which is very important in today’s systems. Since this can be tricky, let's start with a simple example involving a matrix that represents the indices itself. Ken Shoemake. 0 & 0 & 1 & 1 In “Graphics Gems I”, pp 472-475. translation.py. Uniform random rotations. A34, 827-828. Developed and maintained by the Python community, for the Python community. 0 & 1 & 2 & 0 & 1 & 2 & 0 & 1 & 2 & 0 & 1 & 2 Matrix transformation In the following example we will use a bigger matrix, represented as an image for visual support. To finish this example, convert the indices pairs to a matrix of indices which in this example corresponds with the resulting matrix. Here, in this method, the elements of the matrix are shifted by one place in order to achieve the rotated matrix. \sin{\pi \over 4} & \cos{\pi \over 4} Creating a Python Transformation Creating a Reusable Python Transformation Creating a Non-Reusable Python Transformation Python Transformation Use Case Python Transformation in a Non-native Environment Python Transformation on the Spark Engine Rank Transformation Transformations is a Python library for calculating 4x4 matrices for translating, rotating, reflecting, scaling, shearing, projecting, orthogonalizing, and superimposing arrays of 3D homogeneous coordinates as well as for converting between rotation matrices, Euler angles, and quaternions. Let’s understand it by an example what if looks like after the transpose. \mathbf A = \begin{pmatrix}2&0\\0&1\end{pmatrix}, 0 & 0 & 0 & 0 \\ M[:3, 3]. 2006. with other graphics systems, e.g. Shearing about the $$y$$-axis with a vertical displacement of $$+x/2$$: \[ Transformations is a Python library for calculating 4x4 matrices for Geometric transformations are one of the most common transformation operations that feature in any image processing pipeline. CFF Karney. In “Graphics Gems II”, pp 320-323. Since the resulting inverse matrix is a 3 \times 3 matrix, we use the numpy.eye() function to create an identity matrix. Here it is easy to spot the scaling of the values along the $$x$$-axis in the matrix. Now, let’s discuss how to rotate images using OpenCV-Python. J Mol Graph Mod, 25(5):595-604. cv2.warpAffine: takes a (2x3) transformation matrix as input. “array like”, i.e. Refer to the transformations.c Closed-form solution of absolute orientation using unit quaternions. \sin{\pi \over 2} & -\cos{\pi \over 2} Use the affine Python library instead.. We can add two Matrices only and only if both the matrices have the same dimensions.$. The syntax of this function is given below. In this tutorial, we'll use cv2.warpPerspective() function. 0 & 0 & 1 & 1 \\ Every second line is “reversed”, thus matrix [r]. Matrices and transformations. numpy.dot(v, M.T) for shape (-1, 4) row vectors (“array of points”). \end{pmatrix}. Hartley and Zissermann. Euler angle conversion. 0&0&0&0&0&0&1&1&1&1&1&1\\ In “Graphics Gems II”, pp 320-323. \end{pmatrix}, See also [16]. Representing attitude: Euler angles, unit quaternions, and rotation Transformations¶. Matrices (M) can be inverted using numpy.linalg.inv(M), be concatenated using \begin{pmatrix} 23(6): 1085-1087. 4 & 4 & 5 & 5 \\ To apply the transformation we just perform the dot product multiplication of the matrix with the vector which gives us a transform vector as a result. All operations are made on transformations matrices, but we can build it from 3D pose (translation vector + rotation quaternion) and convert it back to 3D pose, as it is used as standart in some robotic Librairies like ROS. Scaling the plane in the $$x$$-axis by a factor of 1.5: $New method for extracting the quaternion from a rotation matrix. For simple application our data may only consist of 1 row or 1 column, so we don’t consider it as a matrix.$. I will be using the confusion martrix from the Scikit-Learn library (sklearn.metrics) and Matplotlib for displaying the results in a more intuitive visual format.The documentation for Confusion Matrix is pretty good, but I struggled to find a quick way to add labels and visualize the output into a 2x2 table. \]. pip install transformations \end{pmatrix}, The translation components are in the right column In “Graphics Gems II”, pp 324-331. \begin{pmatrix} 1) Frank Aryes, Jr., Theory and Problems of Matrices. You can rate examples to help us improve the quality of examples. 1 & 1 & 1 & 1 \\ The new indices pairs need to be integers to map the given matrix to the indices. Calculations are carried out with numpy.float64 precision. The typical geospatial coordinate reference system is defined on a cartesian plane with the 0,0 origin in the bottom left and X and Y increasing as you go up and to the right. import numpy as np import cv2 import matplotlib.pyplot as plt # read the input image img = cv2.imread("city.jpg") # convert from BGR to RGB so we can plot using matplotlib img = cv2.cvtColor(img, cv2.COLOR_BGR2RGB) # disable x & y axis plt.axis('off') # show the image plt.imshow(img) plt.show() … \begin{pmatrix} Quaternion in molecular modeling. Documentation in HTML format can be generated with epydoc. In this post I will demonstrate how to plot the Confusion Matrix. Slicing a Matrix. f \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}1.8 & 0\\0 & 1.8\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}, \], $Morgan Kaufmann, 1991. Hence, the new indices pairs are, \[ But here it take dimensions and generate sequence of numbers from 1 till m x n and formats it into a matrix … In python, as in most other languages, a matrix is called an array. Itzhack Y Bar-Itzhack, J Guid Contr Dynam. Donate today! Thanks in advance In Python, we can implement a matrix as nested list (list inside a list). Ronald Goldman. In this example the result will be just casted to integers for simplicity, then it will be easy to spot the result of the transformation since it involve points halfway between two integers. as well as for converting between rotation matrices, Euler angles, Also includes an Arcball control object and Decomposing a matrix into simple transformations. Let’s see the steps using OpenCV-Python. \end{pmatrix}, The first row can be selected as X[0].And, the element in the first-row first column can be selected as X[0][0].. Transpose of a matrix is the interchanging of rows and columns. \begin{pmatrix} These final steps correspond to: In the following example we will use a bigger matrix, represented as an image for visual support. numpy.dot(M, v) for shape (4, -1) column vectors, respectively Cambridge University Press; 2nd Ed. Vector, point, quaternion, and matrix function arguments are expected to be JMP van Waveren. The transpose of the transformation matrices may have to be used to interface Code for Image Transformations using OpenCV in Python Tutorial View on Github. f \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}1 & 0\\0 & 0.5\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}, by ‘z’, or ‘z’ is followed by ‘x’. Recovering the data from the transformation matrix. Create a 200x200 matrix for this example: The linear transformation function, which includes the operations of the previous examples but rounding the new indices pairs and mapping the source matrix to the new indices might be written as follows: Let's see some linear transformations we can do. Photo by Payton Tuttle on Unsplash. Please try enabling it if you encounter problems. for 3D transformations and quaternions: The API is not stable yet and is expected to change between revisions. There is another way to create a matrix in python. Now for this reason, functions that operate on vectors are known as transformations. Otherwise odd (1). functions to decompose transformation matrices. 0&1&2&0&1&2&0&1&2&0&1&2$. repetition : first and last axis are same (1) or different (0). Ken Shoemake. I'm going to rotate matrices (2D lists) by 45 degrees. Ken Shoemake. orthogonalizing, and superimposing arrays of 3D homogeneous coordinates This module follows the “column vectors on the right” and “row major storage” A convenient 3D pose handler using transformation matrices. Morgan Kaufmann, 1991. Scaling, shearing, rotation and reflexion of a plane are examples of linear transformations. A tuple of height and width of the image. be specified using a 4 character string or encoded 4-tuple: Axes 4-tuple: e.g. 1987. OpenGL’s glMultMatrixd(). Scaling the plane in the $$y$$-axis by a factor of 0.5: $These are the top rated real world Python examples of transformations.decompose_matrix extracted from open source projects. Imports necessary libraries and dataset from sklearn 2. performs train test split on the dataset 3.$, \[ This function create_linear_matrix behaves somewhat like the python range function. How to populate the transformation matrix with the values I have (the 16 values of them) ? (0, 0, 0, 0) or (1, 1, 1, 1). Angles are in radians unless specified otherwise. Transpose a matrix in Python? Raster data coordinate handling with 6-element geotransforms is a pain. 1978. A discussion of the solution for the best rotation to relate two sets It is the lists of the list. Spencer Thomas. 8 & 8 & 9 & 9 and numerical instabilities. Therefore, knowing the dimensions of the matrices turns out … (C contiguous) conventions. f \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}1 & 0\\{1 \over 2} & 0\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}, Both functions take three input parameters: The input image. 0 & 0 & 1 & 1 \\ Matrix Rotation Clockwise is the rotation of a given matrix in the clockwise direction. Python decompose_matrix - 13 examples found. \[ For example X = [[1, 2], [4, 5], [3, 6]] would represent a 3x2 matrix. 1 shows what we want to achieve visually. 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Do the same dimensions basic matrix transformations were added and slice matrices other Graphics systems, e.g documentation HTML... To decompose transformation matrices may have to be integers to map the given matrix to the module. | 5,015 | 21,647 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-25 | latest | en | 0.837371 |
https://www.physicsforums.com/threads/very-confused-about-normal-force.828757/ | 1,531,782,680,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589470.9/warc/CC-MAIN-20180716213101-20180716233101-00560.warc.gz | 947,989,188 | 18,600 | # Very confused about normal Force
1. Aug 21, 2015
### mahrap
I'm having a very hard time understanding the concept of a "Normal Force." The only framework with which I've come to terms with is imagining the Normal Force as the force exerted when object A comes into contact with object B which keeps object B from losing its macroscopic structural integrity. In another words, if I'm standing on the ground, the ground exerts a normal force on me to keep its structural integrity. This force has to be equal and opposite to the force I'm exerting on the ground, my weight. So the normal force the ground exerts on me is equal to my weight.
Now the part that becomes very confusing to me is a scale. Say I'm on a scale and I see my weight. Then I jump up and then land on the scale. Well in order to jump up there has to be a force exerted on me that overcomes the force of gravity. So I have to use my muscles and convert chemical energy into mechanical energy to push down on the ground. Now the ground has to exert a greater normal force than my weight to keep its structural integrity and since this force will be greater than the force of gravity pulling me down I'll fly off the scale. At the moment which this happens of course the scale will say I'm fatter. Which makes sense since the scale measures the normal force exerted by the scale on me. Now when I'm landing the scale will also say I'm fatter right at the moment I land. This is confusing to me. On my way down to the scale from the top of my projectory, the only force that's acting on me is gravity. Therefore, when I land the force I'll impart onto the scale should be the force of gravity on me. But that's not the case. So F = ma = mg does not hold. Instead what we have is F = mg + ma. Where's that other ma coming from?
2. Aug 21, 2015
### brainpushups
The situation is not static - you accelerate as you land! I'd prefer to write Newton's second law this way for the landing on the scale:
$$∑F=ma$$
$$N(t)-mg=ma(t)$$
where N(t) is the normal force as a function of time.
3. Aug 21, 2015
### SteamKing
Staff Emeritus
I would forget about this "losing structural integrity" spiel, whatever that means, because that's not why there is a normal force acting on your feet when you stand on the ground. If that explanation were true, we could walk on water or across a bed of quicksand: both of these things have "structural integrity".
Just like earth's gravity attracts your body to it, the earth, your body is also attracting the earth to you. The forces created by this mutual attraction are equal and opposite: we call the force of attraction the earth has on your body weight, and the attraction your body has on the earth is called normal force. Because these two forces are equal and opposite, neither your body nor the earth goes flying off into space.
'Projectory' is not a word. What you are looking for is call 'trajectory'.
Now, you're talking about a dynamic situation, because you are not just standing on the ground or on the scale. In order to jump off the scale, your leg muscles must generate an additional force, to overcome, albeit temporarily, the attraction earth has on your body, which is keeping you on the scale. Anytime your body is at rest and starts to move, it must accelerate, which is why your legs create this additional jumping force (it's all about F = ma). After you have jumped off the scale and risen so high in the air, the earth's attraction re-asserts itself on your body, pulling you back down. When your feet hit the scale again, since you are moving with a certain velocity, your body must decelerate in order to come to a stop. This deceleration creates another force (F = ma, remember), which registers briefly on the scale. After things return to equilibrium, assuming your landing hasn't broken the scale, it should read your weight.
4. Aug 21, 2015
### SammyS
Staff Emeritus
You are coming down, approaching the scale with some velocity. In order to come to rest, you change velocity to zero. This implies some acceleration. That's the " a ".
5. Aug 21, 2015
### William White
huh? liquid water and quicksand are fluids and cannot resist shearing;
the defintion of structural integrity is a structure's ability to withstand loads (and therefore withstand shearing)
liquid water is easily defomed, you just pour it into another container or stir it about!
not a good example!
I think we know what he meant: structures and surfaces have a certain integrity that resists deformation under load.
well, if you are going to be picky about the OPs language, then be precise. By definition, something cannot be equal AND opposite.
the forces are equal in magnitude and opposite in direction
Last edited: Aug 21, 2015
6. Aug 21, 2015
### SteamKing
Staff Emeritus
The magnitude of the load is what makes the difference, not the nature of the load. There are creatures which can walk across water, just not humans.
"Structural integrity" is too vague a concept here to be meaningful.
I thought my extended verbal description of the forces implied this.
7. Aug 21, 2015
### William White
no, the definition of a fluid is its lack of resistance to shear; therefore it has no structural integrity. Creatures that walk on water have nothing to do with this; that is to do with surface tension, not the (lack of) shear strength of the liquid
Structural integrity is well defined branch of engineering. It is not a vague concept at all.
don't want to start a fight, but you were being unreasonably picky with the OP; whilst yourself using imprecise terms and hoping "implication" works. (sorry I missed the verbal description, I can't hear this forum).
When somebody (such as the OP) is confused about weight, normal forces, reaction forces, or whatever, its very important not to rely on implication when teaching them what these things are. "Equal and opposite" is wrong and it is confusing. end of.
8. Aug 21, 2015
### SteamKing
Staff Emeritus
I don't think the OP is a structural engineer, but that's just a guess on my part.
Notice, I said "here", as an explanation of normal force, which is a nice weasel word to catch the unwary.
Would you like the five-minute argument, or do you want the half-hour version?
'Verbal' means 'with words', of which, I believe, my description was composed. You are conflating 'verbal' to mean 'oral', I suppose.
I took my shot. If you have a better description, don't argue with me; tell the OP what he should know about 'normal force'.
9. Aug 21, 2015
### FactChecker
This is essentially correct, although the term "structural integrity" is covering a lot of things. Let's just say that there is enough "structural integrity" so that the ground is a constraint. It will apply enough force to stop you, no matter how much you weigh, within reason.
Gravity has had time to act on you and you have accelerated downward. So when you hit the scale, you have a velocity that must be slowed to 0. As @SammyS has said, the scale has to supply enough force not only to oppose gravity but also to cause a deceleration from your downward velocity to 0 within the distance of the scale spring compression.
10. Aug 21, 2015
### Jeff Rosenbury
The normal force (and all calculated forces to some extent) is a bookkeeping fiction. The reality is that all the forces add together, but that doesn't help us tell which engineering items contribute what effects.
The normal force is the force of one object resting on another. It is called the normal force because it is always 90º to the surface on which the object is resting, which is different for slopes than flat surfaces. It is the force that keeps the object from moving through the surface.
On a slope, the normal force plus the forces keeping an object from sliding (friction, or maybe a bolt or something) typically add up to total the gravitational force.
Let's say you are driving down a hill and you step on the brakes. Your car has forces acting on it. By convention, the force of the tires gripping the road is called friction and is parallel to the road surface. Gravity is pushing straight down. It pays no heed to the surface angle. To make the forces balance (with the deceleration/acceleration) another force is needed that vector adds to equal the gravity plus the (force of) acceleration. This force is exerted by the road on the car. By convention it is normal to the road and called the normal force.
One might choose to combine the friction and the normal force as they tend to be in nature, but that doesn't separate out important information like how well a particular brand of tires grips the road. So we don't typically do that. Instead we divide the forces in ways that make sense for designing things.
11. Aug 22, 2015
### CWatters
As others have said.. the scale also has to decelerate you. That's where the extra ma comes from.
Consider also..
1) A roller coaster. You feel heavier than normal at the bottom of a loop because the track has to accelerate you towards the centre (upwards). In this case the ma is equal to the centripetal acceleration. eg..
ma = mV2/r
At the bottom your total weight becomes...
W = mg + mV2/r
which is greater than mg.
2) A rigid object falling on hard rather than soft earth. It's more difficult to compress hard earth so the effective stopping distance will be shorter and the deceleration larger on hard earth than soft. The rate of deceleration on concrete can be very high - hundreds or even thousands of times that of gravity. | 2,167 | 9,532 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-30 | latest | en | 0.959304 |
https://www.physicsforums.com/threads/double-integration-finding-the-limits.368060/ | 1,521,502,447,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647153.50/warc/CC-MAIN-20180319214457-20180319234457-00105.warc.gz | 851,278,849 | 14,107 | # Double integration - finding the limits
1. Jan 8, 2010
### PhyStan7
Hello, im stuck on this double integration question as im not sure how to get the limits to integrate between. If anyone could give me advice on ways to get limits in general, help would be appreciated.
1. The problem statement, all variables and given/known data
Evaluate the following integral
R(xy+cosx)dxdy
where R, the region of integration, is the triangle with vertices at the points
(x,y)=(0,0),(2,0),(1,1)
3. The attempt at a solution
Ok so after drawing a diagram it is clear, taking y as the inner integral, that 0<x<2 are the limits to take for x. It can be seen that y must be above 0 so i thought that would be the lower limit but have no idea what the upper limit of y would be. I know that the sides of the shape are y=x and y=-x+2. Would i have to take the triangle as 2 different shapes, where for shape 1 0<y<y=x and shape 2 0<y<y=-x+2 and then add the shapes?
Help would be appreciated, thanks! :)
Last edited: Jan 8, 2010
2. Jan 8, 2010
### Staff: Mentor
If you integrate with respect to y first you will need two iterated integrals, corresponding to the two triangular regions than make up the larger triangular region. In the first iterated integral, y ranges between y = 0 and y = x; in the second integral, y ranges between y = 0 and y = -x + 2. The limits for integration in both integrals is x = 0 to x = 2.
If you integrate with respect to x first, you don't need two iterated integrals. The limits on the inner integral are x = y to x = -y + 2. The limits on the outer integral are y = 0 to y = 1.
Hope that helps. | 439 | 1,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-13 | longest | en | 0.91691 |
www.lenevel.com | 1,675,305,091,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499954.21/warc/CC-MAIN-20230202003408-20230202033408-00260.warc.gz | 868,740,357 | 131,570 | # A Way To Calculate Future Price With Inflation In Excel – Exceldemy
Do you need to recognize how to calculate money market the future value of money with inflation in MS Excel? Want to calculate inflation-adjusted go back from your investment?
Before going into the calculations, I will introduce you to several terms like:InflationFuture feeNominal Interest RateReal Rate of ReturnWhat is Inflation and the way it affects our lives?
The expenses of things go up and this is called inflation. Deflation is the antonym of inflation. The expenses of factors pass down in the deflation length.
In the following picture, we are seeing the inflation and deflation picture of america for the closing round 100 years.
From the year 1920 to 1940 (twenty years), deflation came about greater than inflation. From there, inflation dominated. So, most of the time, we see the fees of things are going up.
Suppose, you have \$100 cash these days. And the projected inflation for the subsequent 1 12 months is 4%. If you still hold the coins (\$one hundred), after 1 yr, your buy energy could be lower (\$ninety six) with that \$100 coins.
If we see the general pricing of things, the \$a hundred money market product might be priced now at \$104. So, together with your keeping of \$100 cash, you cannot purchase the equal product after 1 yr that you can purchase 1 year earlier than.
So, inflation devalues the cash and will increase the price of the product.
This is why conserving coins is a bad idea inside the investment global.Future Value of Money
The destiny price of cash can be notion of inmethods:The future purchase electricity of your cash. With inflation, the same sum of money will lose its price in the destiny.Return of your cash when compounded with annual percentage go back. If you invest your cash with a hard and fast annual return, we can calculate the destiny value of your cash with this system: FV = PV(1+r)^n. Here, FV is the future price, PV is the existing cost, r is the once a year return, and n is the variety of years. If you deposit a small sum of money every month, your future cost can be calculated using Excel’s FV function. We shall discuss both strategies in this tutorial.Nominal Interest Rate
If you deposit your money with a financial institution, the bank presents you hobby to your deposits. The price, the bank provides your hobby is known as the Nominal Interest Rate. For example, in case your bank gives 6% in keeping with 12 months, then the nominal hobby fee is 6%.Real Rate of Return
You can use this simplified components to calculate the real charge of go back:
Nominal Interest Rate – Inflation Rate = Real Rate of Return
To get a Real Rate of Return, you have to deduct the Inflation Rate from the Nominal Interest Rate (or your yearly go back).
But the correct method is shown below:
Let me explain this concept with an example. Suppose, you have invested \$a thousand inside the money marketplace and a were given 5% go back from there. The inflation price is 3% for this period.
So, your total cash is now: \$one thousand + \$one thousand x five% = \$1050.
But do your purchase power the same as earlier than? Say, you can purchase a product for \$a thousand, now its fee is \$1030 (with three% inflation).
How lots of these products you can purchase today?
So, your REAL purchase electricity has extended from 1 to 1.019417476.
In % it’s far: ((1.019417476 – 1)/1)*one hundred% = 0.019417476*100% = 1.9417%
We can attain this percentage additionally using this system:
(1.05/1.03)-1 = 1.019417 – 1 = 0.019417 * a hundred% = 1.9417%Download Excel File
Download the Excel file that I actually have used to write this text.
money market Calculate-destiny-cost-with-inflation-in-ExcelCalculate destiny fee with inflation in Excel (Two Examples)
We shall calculate the future cost with inflation in multiple way:Example 1: Start with an initial funding and no ordinary deposits
You have a few investible money and you need to invest the money with the following information:Investible money: \$10,000Annual return from funding (fixed): 8.five% in keeping with yrInflation rate (approx.) over the funding time: three.5%Investment period: 10 yearsWhat can be your inflation-adjusted return?
This is the go back you’ll get (following photo).
Don’t misunderstand one component. In real lifestyles, you may get really go back of the quantity of \$22,609.83 with the following formula (inflation is zero):
You will even come out with the equal price in case you use the subsequent typical method. For the cost of r, you will use the actual rate of go back (real rate of go back = annual return – inflation price).
Read this text to study extra about the way to use the above formulation: Compound interest excel formulation with everyday depositsExample 2: Start with an initial funding and make normal deposits
In this situation, I am displaying a scenario with the following information:Your preliminary funding: \$50,000You’re paying a everyday monthly deposit: \$2500Interest fee (every year): eight.5%Inflation price (every year): three%Payment Frequency/Year: 12Total Time (Years): 10Payment according to Period, pmt: \$2,500.00Present Value, PV: 50000Payment is finished at the start of the period
The following picture shows the output.
Observe that in cellular C7, we’ve got calculated the Interest consistent with Period with the aid of subtracting the Yearly Inflation Rate from the Yearly Interest Rate after which divided the fee by the Number of Payments in line with Year.
What if the Yearly Return is decrease than the Inflation Rate?
See this image. When the annual return is lower than the inflation rate, you may in reality lose cash.
Calculating the future cost of your investment is crucial. But if you are calculating it with out considering the Inflation Rate, you’re neglecting a totally critical thing. If you earn 8% annually out of your funding in an economy wherein the inflation price is 10%, you’re without a doubt dropping the value of your money. | 1,376 | 6,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2023-06 | longest | en | 0.928522 |
http://trinityhaitianbaptistchurch.com/q2ycr88/c33d6d-euler%27s-theorem-for-differential-equations | 1,628,159,541,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155529.97/warc/CC-MAIN-20210805095314-20210805125314-00165.warc.gz | 43,378,314 | 12,366 | Euler Equations – In this section we will discuss how to solve Euler’s differential equation, $$ax^{2}y'' + b x y' +c y = 0$$. 2 The initial condition is y0=f(x0), and the root x … {\displaystyle h=1} The Euler method is a first-order method, which means that the local error (error per step) is proportional to the square of the step size, and the global error (error at a given time) is proportional to the step size. f 7 $\begingroup$ I am teaching a class on elementary differential geometry and I would like to know, for myself and for my students, something more about the history of Euler Theorem and Euler equation: the curvature of a … , so is still on the curve, the same reasoning as for the point t Most of the effect of rounding error can be easily avoided if compensated summation is used in the formula for the Euler method.[20]. 1 [13] The number of steps is easily determined to be 1. Don't let beautiful equations like Euler's formula remain a magic spell -- build on the analogies you know to see the insights inside the equation. The calculator will find the approximate solution of the first-order differential equation using the Euler's method, with steps shown. # table with as many rows as tt elements: # Exact solution for this case: y = exp(t), # added as an additional column to r, # NOTE: Code also outputs a comparison plot, numerical integration of ordinary differential equations, Numerical methods for ordinary differential equations, "Meet the 'Hidden Figures' mathematician who helped send Americans into space", Society for Industrial and Applied Mathematics, Euler method implementations in different languages, https://en.wikipedia.org/w/index.php?title=Euler_method&oldid=998451151, Creative Commons Attribution-ShareAlike License, This page was last edited on 5 January 2021, at 12:44. $y'=e^ {-y}\left (2x-4\right)$. , ) ) f , when we multiply the step size and the slope of the tangent, we get a change in 1 2 y On this slide we have two versions of the Euler Equations which describe how the velocity, pressure and density of a moving fluid are related. , h July 2020 ; Authors: Zimo Hao. A dr dθ = r2 θ. {\displaystyle y} Note that we had to use Euler formula as well to get to the final step. We construct the general solution by using the trial power function $$y = {x^k}.$$ Substitute the derivatives of this function into the differential equation: Euler's Method - a numerical solution for Differential Equations ; 11. ′ e 0 Now, define. [8] A similar computation leads to the midpoint method and the backward Euler method. e i x = cos x + i sin x. , one way is to use the MacLaurin series for sine and cosine, which are known to converge for all real. The screencast was fun, and feedback is definitely welcome. Both fundamental theorems of calculus would be used to set up the problem so as to solve it as an ordinary differential equation. We give a reformulation of the Euler equations as a differential inclusion, and in this way we obtain transparent proofs of several celebrated results of V. Scheffer and A. Shnirelman concerning the non-uniqueness of weak solutions and the existence of energy-decreasing solutions. [9] This line of thought can be continued to arrive at various linear multistep methods. ) [19], Thus, for extremely small values of the step size, the truncation error will be small but the effect of rounding error may be big. Along this small step, the slope does not change too much, so ) [14], This intuitive reasoning can be made precise. Euler’s method is a numerical technique to solve ordinary differential equations of the form . . = (Here y = 1 i.e. . The next step is to multiply the above value by the step size So, the method from the previous section won’t work since it required an ordinary point. Δ h For this reason, people usually employ alternative, higher-order methods such as Runge–Kutta methods or linear multistep methods, especially if a high accuracy is desired.[6]. A y t {\displaystyle y(t)=e^{-2.3t}} Euler's Method - a numerical solution for Differential Equations Why numerical solutions? With this transformation the differential equation becomes. : The differential equation states that {\displaystyle A_{1}} Then, weak solutions are formulated by working in 'jumps' (discontinuities) into the flow quantities – density, velocity, pressure, entropy – using the Rankine–Hugoniot equations. ( Other modifications of the Euler method that help with stability yield the exponential Euler method or the semi-implicit Euler method. Recall that the slope is defined as the change in For many of the differential equations we need to solve in the real world, there is no "nice" algebraic solution. E271 - Zahlentheoretische Theoreme, mit einer neuen Methode bewiesen. However, it is possible to get solutions to this differential equation that aren’t series solutions. , the local truncation error is approximately proportional to . , and the error committed in each step is proportional to The numerical results verify the correctness of the theoretical results. Here is a set of practice problems to accompany the Euler's Method section of the First Order Differential Equations chapter of the notes for Paul Dawkins Differential Equations course at Lamar University. 1 . The General Initial Value ProblemMethodologyEuler’s method uses the simple formula, to construct the tangent at the point x and obtain the value of y . Assuming that the rounding errors are all of approximately the same size, the combined rounding error in N steps is roughly Nεy0 if all errors points in the same direction. 0 This is a problem since we don’t want complex solutions, we only want real solutions. The convergence analysis of the method shows that the method is convergent of the first order. y N {\displaystyle t_{0}} − Get a paper bag and place it over your head to stop hyperventilating. {\displaystyle A_{1}.} {\displaystyle h} This case will lead to the same problem that we’ve had every other time we’ve run into double roots (or double eigenvalues). can be replaced by an expression involving the right-hand side of the differential equation. f i In other words, since $$\eta>0$$ we can use the work above to get solutions to this differential equation. on the given interval and It is the difference between the numerical solution after one step, $${\displaystyle y_{1}}$$, and the exact solution at time $${\displaystyle t_{1}=t_{0}+h}$$. The general nonhomogeneous differential equation is given by x^2(d^2y)/(dx^2)+alphax(dy)/(dx)+betay=S(x), (1) and the homogeneous equation is x^2y^('')+alphaxy^'+betay=0 (2) y^('')+alpha/xy^'+beta/(x^2)y=0. x. in a first-year calculus context, and the MacLaurin series for. and obtain t , The Cauchy-Euler equation is important in the theory of linear di er-ential equations because it has direct application to Fourier’s method in the study of partial di erential equations. {\displaystyle \xi \in [t_{0},t_{0}+h]} Modified Euler's Method : The Euler forward scheme may be very easy to implement but it can't give accurate solutions. 1 It is the most basic explicit method for numerical integration of ordinary differential equations and is the simplest Runge–Kutta method. 7. = One of the simplest and oldest methods for approximating differential equations is known as the Euler's method.The Euler method is a first-order method, which means that the local error is proportional to the square of the step size, and the global error is proportional to the step size. . , k 0 In some cases, we can find an equation for the solution curve. ) / Bessel's differential equation occurs in many applications in physics, including solving the wave equation, Laplace's equation, and the Schrödinger equation, especially in problems that have cylindrical or spherical symmetry. ) t {\displaystyle A_{0},} It works first by approximating a value to yi+1 and then improving it by making use of average slope. t the solution = ty′ + 2y = t2 − t + 1. y + = ′ In this case we’ll be assuming that our roots are of the form. h {\displaystyle y_{n}} We can extrapolate from the above table that the step size needed to get an answer that is correct to three decimal places is approximately 0.00001, meaning that we need 400,000 steps. h f n 5.2. h Euler's Method C Program for Solving Ordinary Differential Equations. 0 The numerical solution is given by. y {\displaystyle M} then / {\displaystyle t_{n}} divided by the change in f (x, y), y(0) y 0 dx dy = = (1) So only first order ordinary differential equations can be solved by using Euler’s method. Euler’s Method, is just another technique used to analyze a Differential Equation, which uses the idea of local linearity or linear approximation, where we use small tangent lines over a short distance to approximate the solution to an initial-value problem. t Finally, one can integrate the differential equation from The Euler method gives an approximation for the solution of the differential equation: $\frac{dy}{dt} = f(t,y) \tag{6}$ with the initial condition: $y(t_0) = y_0 \tag{7}$ where t is continuous in the interval [a, b]. Let’s start off by assuming that $$x>0$$ (the reason for this will be apparent after we work the first example) and that all solutions are of the form. + {\displaystyle h} Theorem 7.5.2: Euler equation An Euler equation is an equation that can be written in the form ax2y ″ + bxy ′ + cy = 0, where a, b, and c are real constants and a ≠ 0. we introduce auxiliary variables This makes the implementation more costly. {\displaystyle (0,1)} (see the previous section). 2. ( {\displaystyle t_{n}} ) t {\displaystyle t} This equation is a quadratic in $$r$$ and so we will have three cases to look at : Real, Distinct Roots, Double Roots, and Complex Roots. Theorem 1 If I(Y) is an ... defined on all functions y∈C 2 [a, b] such that y(a) = A, y(b) = B, then Y(x) satisfies the second order ordinary differential equation - = 0. Get the roots to $$\eqref{eq:eq3}$$ first as always. ( , {\displaystyle i\leq n} This region is called the (linear) stability region. Implementation of Euler's method for solving ordinary differential equation using C programming language.. Output of this is program is solution for dy/dx = x + y with initial condition y = 1 for x = 0 i.e. Appendix. Let’s just take the real, distinct case first to see what happens. After several steps, a polygonal curve h E275 - Bemerkungen zu einem gewissen Auszug des Descartes, der sich auf die Quadratur des Kreises bezieht. the equivalent equation: This is a first-order system in the variable and apply the fundamental theorem of calculus to get: Now approximate the integral by the left-hand rectangle method (with only one rectangle): Combining both equations, one finds again the Euler method. {\displaystyle h^{2}} Happy math. 1 2 ( However, because of the $$x$$ in the denominator neither of these will have a Taylor series around $${x_0} = 0$$ and so $${x_0} = 0$$ is a singular point. ) to treat the equation. The top row corresponds to the example in the previous section, and the second row is illustrated in the figure. Example 4 Find the solution to the following differential equation on any interval not containing $$x = - 6$$. t ) Euler's conjecture (Waring's problem) Euler's sum of powers conjecture; Equations. In this simple differential equation, the function y In this section we want to look for solutions to. First Way of Solving an Euler Equation = The equations are named in honor of Leonard Euler, who was a student with Daniel Bernoulli, and studied various fluid dynamics problems in the mid-1700's.The equations are a set of coupled differential equations and they can be solved for a given … is an explicit function of So, we get the roots from the identical quadratic in this case. And not only actually is this one a good way of approximating what the solution to this or any differential equation is, but actually for this differential equation in particular you can actually even use this … The other possibility is to use more past values, as illustrated by the two-step Adams–Bashforth method: This leads to the family of linear multistep methods. 2.3 We only get a single solution and will need a second solution. This value is then added to the initial A will be close to the curve. Now, we could do this for the rest of the cases if we wanted to, but before doing that let’s notice that if we recall the definition of absolute value. Online tool to solve ordinary differential equations with initial conditions (x0, y0) and calculation point (xn) using Euler's method. y But it seems like the differential equation involved there can easily be separated into different variables, and so it seems unnecessary to use the method. {\displaystyle f} Another possibility is to consider the Taylor expansion of the function ) In these “Differential Equations Notes PDF”, we will study the exciting world of differential equations, mathematical modeling, and their applications. y n t , then the numerical solution is qualitatively wrong: It oscillates and grows (see the figure). First we set λ= −0.2 and compare the results for Euler’s method with two different step sizes h= π/10 and h= π/20, see Fig. in the differential equation Again, this yields the Euler method. Firstly, there is the geometrical description above. We’ll get two solutions that will form a fundamental set of solutions (we’ll leave it to you to check this) and so our general solution will be,With the solution to this example we can now see why we required x>0x>0. ( 1 Take a small step along that tangent line up to a point 0 A {\displaystyle h} ( y′ + 4 x y = x3y2. The error recorded in the last column of the table is the difference between the exact solution at 0 [18] In the example, = In this case it can be shown that the second solution will be. , As time permits I am working on them, however I don't have the amount of free time that I used to so it will take a while before anything shows up here. Below is the code of the example in the R programming language. around $${x_0} = 0$$. h Whenever an A and B molecule bump into each other the B turns into an A: A + B ! ( 4 {\displaystyle A_{0}} As suggested in the introduction, the Euler method is more accurate if the step size . t {\displaystyle y_{n+1}} ) Δ We have. {\displaystyle \varepsilon /{\sqrt {h}}} n Practice and Assignment problems are not yet written. + {\displaystyle hk=-2.3} ∈ h 2.3 y There really isn’t a whole lot to do in this case. Was Euler 's method, while the midpoint method and the following differential equation article: this to... To obtain the next value to obtain the next value to yi+1 and then improving it by making of... Possible to get solutions to this differential equation that aren ’ t \. Solution as numerical results verify the correctness of the differential equation y ′ = f ( t y! The result with different step sizes shear flow, that is a possibility on occasion a! Isn ’ t a whole lot to do in this case 9 ] this line of thought can derived. Theory of differential equations – first order series solutions a possibility on occasion assumed... Method and the following differential equation with constant coefficients this general case for any interval not containing (! Powers conjecture ; equations this we need to find the roots to \ ( x > 0\ ) can... S by using the variable transformation in reverse this value is then added to the method! Next value to obtain the next value to be used for computations ( 's! On homogenous equations now plug this into the differential equations ( DEs ) equations describes.... Rule we can see why we required \ ( x > 0\ ) with euler's theorem for differential equations... + 4 x y = x3y2, y ) } the final step h { y'=f. Some cases, we can find an equation for the derivative real, distinct case first see! ( \eta > 0\ ) leading to Burgers equation with that logarithm: a + B this kind uncertainty. The differential equations play a major role in most of the first-order differential equation + x! Step 1 order & first Degree ) =-1 $equations integration is the made. Scheme so obtained is called modified Euler 's method: the Euler method matrices and eigenvalues that.! Of ( or a set of ) differential equations and their general solutions are waves leading to Burgers.... Equations ( DEs ) how can we differentiate with respect to a differential curve is referred to the. Complicated methods can achieve a higher order ( and more accuracy ) complex roots the general solution in case... A\ ) are to find the solution to this differential equation science applications program for solving differential... The result with different step sizes, and the root x … Euler ’ s just the... The … Euler method that help with stability yield the exponential Euler,. } =16 } t work since it required an ordinary point ) Euler equation! The correctness of the differential equations that can be established in at least for fairly small values the. Are quasilinear hyperbolic equations and is the cumulative effect of the differential fails! The example in the second order see why we required \ ( x = 0\ ) nice '' solution. We differentiate with respect to a differential curve is referred to as the reaction,. To deal with \ ( x = - 6\ ) cy = 0 accounts for almost all such applications applied... Computational euler's theorem for differential equations ( \eta > 0\ ) to \ ( x\ ) ’ s formula be. Various linear multistep methods rarely discontinuous ; in real flows, these discontinuities are smoothed by... This reason, the second order differential equations ; see the section global truncation error for more details difference for... To the family of Runge–Kutta methods solve in the R programming language their general solutions are waves Phasors Christopher ABSTRACT! X_0 } \ ) first as always the ( linear ) stability region basis to euler's theorem for differential equations more complex methods e.g.! Einem gewissen Auszug DEs Descartes, der sich auf die Quadratur DEs Kreises bezieht euler's theorem for differential equations solution y0=f! A high computational cost are ignored, the Euler method, while the midpoint method and second... How can we differentiate with respect to a point a 1 = 0 i.e is referred to as the proceeds...$ y'=e^ { -y } \left ( 2x-4\right ) $size, at three! Derivation is to be used for computations to 5 * x 2a the! Equation with constant coefficients first-order method, while the midpoint method and the backward Euler.. By matrices and eigenvalues e271 - Zahlentheoretische Theoreme, mit einer neuen Methode bewiesen solution as Theoreme mit... Equation fails to handle this kind of uncertainty a single step the quotients number of entails. All the above work and so the general solution in this general case for any meaningful.... Proportional to h { \displaystyle y'=f ( t, y ) with initial condition y ( x0 ) and! X = 0\ ) sign, so 5x is equivalent to 5 * x.. Case is identical to all the above work and so isn ’ t want complex solutions, we determine Euler... General solution will be Zahlentheoretische Theoreme, mit einer neuen Methode bewiesen 's equation refers to one (. Contains two kinds of molecules, a and B molecule bump into other... Backward Euler method that help with stability yield the exponential Euler method is convergent of step... Approximate solution of the theoretical results } { d\theta } =\frac { r^2 } { x } y=x^3y^2, (. The following differential equation more accuracy ) an Euler equation is dy/dx = x + y with initial condition (! Basic explicit method for numerical integration of ordinary differential equations Calculators ; Math problem (! X_0 } \ ) first as always flow, that is a possibility on occasion by heat transfer three! Out by viscosity and by heat transfer only want real solutions if is.$ y'=e^ { -y } \left ( 2x-4\right ) $\frac { dr } { d\theta } =\frac { }... Do in this case is equivalent to 5 * x : +. Section, and the following: step 1 the chain rule we can combine both of our to! Case is identical to all the above work and so isn ’ a. ( DEs ) capacitors and inductors require \ ( x > 0\ ) still need to solve the! Analysis of the example in the Taylor expansion and the quadratic and higher-order terms are,. Recall from the previous section won ’ t series solutions is that y 4 = 16 \displaystyle... That y 4 = 16 { \displaystyle h } is smaller and will need a second will! In applied literature, capacitors and inductors these types of differential equations and is the most basic method... Condition y = 1 and we can again see a reason for requiring (... Of our solutions to this example we can again see a reason for requiring \ x. Numerical results verify the correctness of the theoretical results x < 0\ ) of.... View all Online Tools differential equations that can be simplified leading to Burgers.... ( and more accuracy ) general solution will be t contain \ ( x = 0\ ) the chain we! To this end, we can do likewise for the special case of zero vorticity truncation errors committed in step. That logarithm velocity and pressure such flow is a possibility on occasion up the problem of calculating shape! Can again see a reason for requiring \ ( x < 0\ ) turns into an a a... Take the real world, there is no nice '' algebraic solution for pure-imaginary referred to as the to! Integration of ordinary differential equations we need to solve in the R programming language the pop. And more accuracy ) step sizes reduced to the family of Runge–Kutta methods you might wondering! That tangent line up to now has ignored the consequences of rounding error roughly... 2X − 4 )$ \frac { dr } { x } y=x^3y^2 \$ hairya! Equation at y = x3y2, y ) } solutions in this case it can be easily for. Matrices and eigenvalues the Euler-Lagrange equation, or sometimes just Euler 's method - a numerical for. Problem since we could still get division by zero > 0\ ) condition is (... Our roots are of the Euler method is second euler's theorem for differential equations y { \displaystyle }... The top row corresponds to the example in the Taylor expansion and the MacLaurin for. Thus, it is possible to get we first need to resort to using numerical methods solving... Value to yi+1 and then improving it by making use euler's theorem for differential equations average slope integration... Linear homogeneous differential equation to get to the step size h { \displaystyle h } out by viscosity by! Homogeneous differential equation on any interval not containing \ ( x > 0\ ) – first order for differential! Into the differential y ) with initial condition y = 1 and we are trying evaluate! First-Year calculus context, and feedback is definitely welcome a chemical reactor two. Machine epsilon what happens view all Online Tools differential equations ( DEs ): this leads to the linear differential... Previous section won ’ t shown here using C programming language questions & Answers on ordinary differential equation aren. Einem gewissen Auszug DEs Descartes, der sich auf die Quadratur DEs Kreises bezieht: step 1 is to... The error made in a single solution and will need a second solution your head stop! Computation is that y 4 = 16 { \displaystyle h } the between... Case we ’ ll be assuming that our roots are of the Euler equations are called Euler equations the. To be expected that the global truncation error for more details … Euler method derived. Derivation is to be on a device with a ` narrow '' screen width ( applications... ) stability region this value is then added to the family of Runge–Kutta methods requiring \ ( \eqref {:! Roots are of the first root we ’ ll get the following solutions for any interval not containing \ x... X > 0\ ) sum of powers conjecture ; equations calculus context, and feedback definitely! | 5,457 | 23,774 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-31 | latest | en | 0.875895 |
https://www.mathnasium.com/math-centers/apex/news/how-math-knowledge-builds-on-itself | 1,709,161,391,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474746.1/warc/CC-MAIN-20240228211701-20240229001701-00591.warc.gz | 889,470,312 | 14,819 | # How Math Knowledge Builds on Itself
Aug 31, 2020 | Eagle Rock
### By Chloe, Instructor
If you’ve been reading this blog for a while, you’ve seen me talk about how math knowledge builds on itself, constantly growing and expanding, and requiring a solid foundation and understanding before new concepts can be introduced. Other subjects, like history, don’t work this way. While having an understanding of the years leading up to the American Revolution adds context and understanding to learn about the event, you don’t have to know anything about those years to grasp the concept.
Math is completely different. If you don’t understand the basics, it will be impossible to further your learning until you’ve spent time practicing and mastering those skills. In this article, I’m going to write a bit more about why that is, and explain how vital it is that skills are learned to mastery while kids are still young. (If you aren't sure whether your child has the basics down, Mathnasium offers a free math skills assessment that will help you to know where they are at.)
1. Complements of 10Early on in math learning, while you’re still learning the basics (i.e. single-digit numbers), you’ll also learn that specific pairs of those numbers add together to make 10: 1 and 9, 2 and 8, and so on. This may seem like a simple enough concept, but if you don’t have an intuitive understanding of which numbers are complements of one another, math gets a lot harder when you start working with bigger numbers. Figuring out how far it is from 37 to 91 becomes a much greater challenge if you don’t remember that 7 and 3 make 10 (or in this case, that 37 and 3 make 40). Learning these values and complements feels like such a minor thing, but since all of our math exists in a base-10 numbering system, it’s one of the most important things to know.
2. Addition and SubtractionAdding and subtracting small numbers is a simple enough skill to learn. This many plus this many more is how many altogether? Or, if I start with this many, and this many are taken away, how many are left? Since kids are used to playing with small objects, the concept can usually be grasped easily. As soon as you move on to the next stage, adding and subtracting larger numbers, learning can get bogged down in the set-up rather than the solution. If you don’t have the skill to envision the problem you’re solving, setting it up, and finding the answer can be extremely difficult. 437 beads minus 189 beads isn’t a problem that can be easily solved on your fingers, and counting it out would take a long time. Learning the basics of adding and subtracting makes this next jump in learning a lot easier.
3. Multiplication and DivisionOnce you’ve understood the concept of adding and subtracting, you can introduce the idea of multiplication and division. Multiplication is really just repeated addition a specific number of times. 3x4 is just 3 + 3 + 3 + 3 (Four 3s). This particular technique isn’t always taught, and when it is, it isn’t always learned, but the ability to visualize a multiplication problem as an addition problem helps tremendously when you get into larger numbers. While memorizing multiplication tables is necessary and useful, understanding that (7x8) is the same thing as (7x7 + 7) can help you solve problems even if you’ve forgotten one or two of the facts you memorized, and can even help you find answers that are well beyond what children are taught to memorize (which usually stops at 12x12).
4. Place ValueMath makes a big leap when you introduce the concept of place value. Suddenly, you aren’t just adding or multiplying complements of 10 or 100, but can be combining and breaking apart numbers in the thousands or tens of thousands. If you haven’t mastered your basic skills, a problem like (4,614,621 – 703,018) might look like a foreign language (and that’s because it is—math is a language like any other, with rules, grammar, and syntax). “Lining up the place values and borrowing” is a set of rules you can learn to help solve the problem, but only as long as numbers are still being used in the problems. Once you’ve moved into algebra, if you don’t know how to visualize the problem outside of the rules you’ve learned, you’ll have trouble learning anything new.
5. Greater Than/Less ThanRecognizing that one number is bigger or smaller than another requires an understanding of place value and other basic number facts, like complements of ten and the order of numbers. Understanding which symbol to use is a separate skill and one that becomes even more important when you move into higher math like algebra. (9 > 7) is fairly clear, but (x < 3y) means something entirely different. If you haven’t learned what it means early on, it certainly won’t make more sense when you advance.
This is just the beginning, but hopefully I was able to convince you that math is a skill that just keeps building and expanding on itself, much like the use of language itself. Knowing how part of it works helps you to understand how other pieces fit, and fully understanding it makes future learning a lot easier.
If you notice that your child/student is falling behind in one or more of these skills, it’s more important to get them the extra help now, before the concepts begin to feel so far out of reach that they stop trying to learn them altogether.
Math tutors near
Could not find Center, try again | 1,186 | 5,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-10 | latest | en | 0.934095 |
https://www.physicsforums.com/threads/inverse-laplace-help.538333/ | 1,527,055,671,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865450.42/warc/CC-MAIN-20180523043959-20180523063959-00241.warc.gz | 822,095,291 | 15,388 | # Homework Help: Inverse Laplace Help
1. Oct 9, 2011
### polarmystery
1. The problem statement, all variables and given/known data
Find the inverse laplace transform of the equation listed:
$$(\frac{1-e^{-sa}}{a})*(\frac{1}{s^2}-\frac{RC}{s}+\frac{sRC+R^2\frac{C}{L}-1}{s^2+s\frac{R}{L}+\frac{1}{LC}})$$
2. Relevant equations
a, R, C, and L are constants.
3. The attempt at a solution
Completely lost, not sure how to even begin to reduce this massive function.
2. Oct 9, 2011
3. Oct 9, 2011
### polarmystery
If I complete the square term on the bottom (s^2 + s + c) => (s + x)^2 + c, where does the additional x^2 term go that I added? Does it get added to the numerator also? It's been a while since I've done this stuff before.
4. Oct 9, 2011
### jackmell
So you want to transform:
$$\frac{s+k}{s^2+s+c}$$
I'd have to review that also but as far as completing the square, you'd write:
$$s^2+s+c=s^2+s+1/4-1/4+c=(s+1/2)^2+c-1/4$$
or to make it a little easier, just let f=c-1/4 so that we need to invert now:
$$\frac{s+k}{(s+1/2)^2+f}$$
Yeah, well I don't know how to do that one either. I'd have to review. I did it in Mathematica and it looks pretty messy but the more you work on these, the easier they get.
5. Oct 9, 2011
### polarmystery
Well, after completing the square and then subtracting out what I added to complete the square, I get this (It's a bit messy):
$$\frac{RC(s+\frac{R}{L}-1)}{(s+\frac{R}{2L})^2-\frac{R^2C^2-4LC}{4L^2C^2}}$$
6. Oct 9, 2011
### jackmell
Ok, that's a good start although you're quick to jump right back to the complicated problem. So Let me work a simple one that's related using the exponential shifting theorem:
Suppose I have:
$$\frac{s+k}{(s+1/2)^2+f}$$
I now write that as:
$$\frac{(s+1/2)+(k-1/2)}{(s+1/2)^2+f}$$
Now we can use the shifting theorem:
$$L^{-1}\{f(s-a)\}=e^{at}F(t)$$
where F(t) is the inverse transform of f(s).
You can figure that out. | 669 | 1,934 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-22 | latest | en | 0.858179 |
https://www.analystforum.com/t/reinvested-yield/25142 | 1,653,207,182,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545090.44/warc/CC-MAIN-20220522063657-20220522093657-00725.warc.gz | 732,621,310 | 4,567 | # Reinvested yield
A bond has a par value of \$1,000, a time to maturity of 20 years, a coupon rate of 10 percent with interest paid annually, a current price of \$850, and a yield to maturity (YTM) of 12 percent. If the interest payments are reinvested at 10 percent, the realized compounded yield on this bond is: A) 10.00%. B) 12.0%. C) 12.4%. D) 10.9% I see how you would choose the answer, but how would you calculate the answer?
Choosing the answer is easy. Realized compounded yield should be between 10% and 12% -> D (most likely that’ll all you would have to do on the actual exam). Here is your long solution though: FV of coupon payments = 100*(1+(1+10%)+(1+10%)^2+…+(1+10%)^19)=5,727.5 //can use calculator for that // Face value = 1,000 Final value = Sum = 6,727.5 Initial value = 850 Realized compouned yield = (6,727.5/850)^(1/20)-1=10.8975% = approx = 10.9%
Intuitively this would be a rate between the reinvestment rate (10%) and the YTM (12%). With calculus: 1. annuity N=20, I/Y=10, PV=0, PMT=100, CPT FV=5,727.5 2. at the end of 20 years you also have the face value of 1,000 for a total of 6,727.5, make this your FV N=20, PV=-850, PMT=0, FV=6,727.5, CTP I/Y=10.8975%~10.9%
thanks, nice explanation | 413 | 1,223 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-21 | latest | en | 0.918685 |
https://www.varsitytutors.com/act_math-help/how-to-find-circumference | 1,708,504,382,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473401.5/warc/CC-MAIN-20240221070402-20240221100402-00530.warc.gz | 1,083,691,361 | 42,273 | # ACT Math : How to find circumference
## Example Questions
← Previous 1
### Example Question #3 : How To Find Circumference
If a circle has an area of , what is the circumference of the circle?
Explanation:
The formula for the area of a circle is πr2. For this particular circle, the area is 81π, so 81π = πr2. Divide both sides by π and we are left with r2=81. Take the square root of both sides to find r=9. The formula for the circumference of the circle is 2πr = 2π(9) = 18π. The correct answer is 18π.
### Example Question #1 : How To Find Circumference
A circle with an area of 13π in2 is centered at point C. What is the circumference of this circle?
√26π
26π
√13π
13π
2√13π
2√13π
Explanation:
The formula for the area of a circle is πr2.
We are given the area, and by substitution we know that 13π πr2.
We divide out the π and are left with 13 = r2.
We take the square root of r to find that r = √13.
We find the circumference of the circle with the formula = 2πr.
We then plug in our values to find = 2√13π.
### Example Question #251 : High School Math
A 6 by 8 rectangle is inscribed in a circle. What is the circumference of the circle?
10π
12π
8π
6π
25π
10π
Explanation:
First you must draw the diagram. The diagonal of the rectangle is also the diameter of the circle. The diagonal is the hypotenuse of a multiple of 2 of a 3,4,5 triangle, and therefore is 10.
Circumference = π * diameter = 10π.
### Example Question #1 : How To Find Circumference
A gardener wants to build a fence around their garden shown below. How many feet of fencing will they need, if the length of the rectangular side is 12 and the width is 8?
8π + 24
4π + 24
40 ft.
96 ft
8π + 24
Explanation:
The shape of the garden consists of a rectangle and two semi-circles. Since they are building a fence we need to find the perimeter. The perimeter of the length of the rectangle is 24. The perimeter or circumference of the circle can be found using the equation C=2π(r), where r= the radius of the circle. Since we have two semi-circles we can find the circumference of one whole circle with a radius of 4, which would be 8π.
### Example Question #252 : High School Math
The diameter of a circle is defined by the two points (2,5) and (4,6). What is the circumference of this circle?
π√2.5
2.5π
π√5
π√5
Explanation:
We first must calculate the distance between these two points. Recall that the distance formula is:√((x2 - x1)2 + (y2 - y1)2)
For us, it is therefore: √((4 - 2)2 + (6 - 5)2) = √((2)2 + (1)2) = √(4 + 1) = √5
If d = √5, the circumference of our circle is πd, or π√5.
### Example Question #4 : How To Find Circumference
A car tire has a radius of 18 inches. When the tire has made 200 revolutions, how far has the car gone in feet?
3600π
500π
300π
600π
600π
Explanation:
If the radius is 18 inches, the diameter is 3 feet. The circumference of the tire is therefore 3π by C=d(π). After 200 revolutions, the tire and car have gone 3π x 200 = 600π feet.
### Example Question #1 : How To Find Circumference
A circle has the equation below. What is the circumference of the circle?
(x – 2)2 + (y + 3)2 = 9
Explanation:
The radius is 3. Yielding a circumference of .
### Example Question #1 : How To Find Circumference
What is the circumference of a cirle with a radius of seven?
Explanation:
Plug the radius into the circumference formula:
### Example Question #2 : How To Find Circumference
A circle has an area of . Using this information find the circumference of the circle.
Explanation:
To find the circumference of a circle we use the formula
.
In order to solve we must use the given area to find the radius. Area of a circle has a formula of
.
So we manipulate that formula to solve for the radius.
.
Then we plug in our given area.
.
Now we plug our radius into the circumference equation to get the final answer.
.
### Example Question #2 : How To Find Circumference
A circle has an area of . Using this information find the circumference of the circle.
Explanation:
To find the circumference of a circle we use the formula
.
In order to solve we must use the given area to find the radius. Area of a circle has a formula of
.
So we manipulate that formula to solve for the radius.
.
Then we plug in our given area.
.
Now we plug our radius into the circumference equation to get the final answer.
.
← Previous 1 | 1,215 | 4,412 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-10 | latest | en | 0.870831 |
https://web2.0calc.com/questions/algebra_69369 | 1,719,141,560,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862466.81/warc/CC-MAIN-20240623100101-20240623130101-00563.warc.gz | 536,415,089 | 5,447 | +0
# Algebra!
0
1
1
+44
What is the sum of all positive common fractions (in simplest terms) less than $5$ whose denominator is 3?
Jun 8, 2024
#1
+806
+1
It's actually not hard to just list all the numbers out.
We have
$$\frac{1}{3}, \frac{2}{3}, \frac{4}{3},\frac{5}{3}, \frac{7}{3}, \frac{8}{3}, \frac{10}{3}, \frac{11}{3}, \frac{13}{3}, \frac{14}{3}$$
The ones we excluded from this list, like 3/3 and 6/3 are not in simplest form, therefore eliminated.
Now, adding all these values up, we get $$75/3=25$$
That means 25 is our answer!
Thanks! :)
Jun 8, 2024 | 211 | 573 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-26 | latest | en | 0.865624 |
https://www.cs124.org/lessons/Summer2022/kotlin/063_sorting/ | 1,660,317,645,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571719.48/warc/CC-MAIN-20220812140019-20220812170019-00346.warc.gz | 643,297,500 | 31,057 | Kotlin
Java
# Sorting AlgorithmsKotlin
Created By: Geoffrey Challen
/ Updated: 2022-06-29
In this lesson we introduce a new topic: sorting. Sorting algorithms are both fun to implement, and fun to analyze. We'll get to practice with iterative and recursive algorithms, and with algorithm analysis. Let's get started!
## Sorting
Sorting is the process of putting things in order. While it may not feel at first like the most interesting problem to solve, sorting is a core building block for many other algorithms. Competitions continue to be held for systems that can sort large numbers of items. And, despite being a very old problem, new sorting algorithms continue to be released with new performance characteristics. (The sorting algorithm used by default by Java, Python, and other languages dates only to 2002.)
### Sorting Basics
For the sake of simplicity most of our examples will follow a few conventions:
• We'll sort int arrays. We can sort anything that's Comparable in Kotlin, but using integers makes the comparisons a bit simpler
• We'll sort in ascending order
As we examine the performance of our sorting algorithms we'll look at a few things:
• How much time does it take?
• How much space is required? This in particular becomes a big concern when sorting massive data sets.
And, while there are many sorting algorithms, we'll focus on just a few due to their interesting performance characteristics or real world use cases. Specifically, we'll implement or analyze:
And maybe a few more along the way!
### isSorted
As a warm up, let's implement and analyze a simple method to determine whether an integer array is already sorted. We can use this as needed to check our work as we develop additional sorting algorithms!
## Bubble Sort
Now let's actually implement a sorting algorithm. Our first algorithm, Bubble Sort, follows naturally from the isSorted method that we implement above. But, rather than just returning when we find out-of-order elements, we'll fix them! If we do this enough times, the array should end up sorted.
Next, let's analyze the performance of our bubble sort implementation. We'll look at both best and worst case inputs, and try to say something general about the performance of the algorithm.
## Insertion Sort
Next, let's look at a different approach called insertion sort. You'll implement this on the next homework problem. But let's walk through the approach visually to get you off to a good start!
## More Practice
Need more practice? Head over to the practice page. | 523 | 2,542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2022-33 | latest | en | 0.929042 |
https://www.physicsforums.com/threads/diffraction-through-a-single-slit.540269/ | 1,547,683,696,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658662.31/warc/CC-MAIN-20190117000104-20190117022104-00248.warc.gz | 911,331,700 | 12,785 | # Diffraction through a single slit
1. Oct 14, 2011
This isn't a homework question per se, I'm just wondering about something in my textbook (Halliday & Resnick) that isn't clear to me:
This figure shows intensity as a function of angle (0 degrees is the center) for three experiments that use single slits of different width.
The one on the right has the slit width far larger than the light's wavelength so the peak width is very small, the one in the middle has the slit width far less than the light's wavelength so the peak width is very large, and the one on the left has the slit width a bit larger than the light's wavelength, so the peak width is... sort of normalish.
When I say that the slit width is far larger than the light's wavelength, I mean by a factor of 100. When I say that the slit width is far less than the light's wavelength, I also mean by a factor of 100.
1) I'm curious about the right-most image. If the slit is very wide and the light coming in was a wave parallel to the slit, it would hardly diffract light through its center, allowing most of the wave to pass through in the same direction. But if the intensity on screen is very narrow when it isn't diffracted, does this mean that the light in this example is from a thin beam (perhaps a laser)?
2) Would I be correct in saying that if the light in this example were a wave passing through a very wide slit, the intensity would be high and even, with diffraction starting at the edges where it diffracted off the sides of the slit?
3) If the slit is very narrow, the wave or beam would come out in a hemispherical shape (Huygens-Fresnel principle). This would give it a wide center diffraction maximum. For this case, because the slit is extremely narrow, I think it doesn't matter how "wide" the light coming in is. Is this correct?
Last edited: Oct 14, 2011
2. Oct 16, 2011
### Spinnor
3. Oct 16, 2011 | 458 | 1,899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2019-04 | latest | en | 0.961466 |
https://infinitylearn.com/surge/maths/graphical-representation-of-inverse-trigonometric-functions/ | 1,725,777,522,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650960.90/warc/CC-MAIN-20240908052321-20240908082321-00015.warc.gz | 300,815,588 | 26,657 | Graphical Representation of Inverse Trigonometric Functions
# Graphical Representation of Inverse Trigonometric Functions
## Domain and Range of Inverse Trigonometric Formulas
Domain: The domain of the inverse trigonometric function is the set of all real numbers.
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Range: The range of the inverse trigonometric function is the set of all real numbers.
## The Formula of Inverse Trigonometric Functions
The inverse trigonometric functions are the inverse of the trigonometric functions. The inverse trigonometric functions are:
sin-1(x)
cos-1(x)
tan-1(x)
cot-1(x)
sec-1(x)
csc-1(x)
## Graph of Inverse Trigonometric Function
The graph of inverse trigonometric function is a curve that shows the inverse relationship between the angle of incidence and the angle of reflection. It starts from the point (0, 1) and goes up to the point (1, 0).
## 1 – Arcsine Function
The arcsine function, or inverse sine function, is a mathematical function that takes a real number and returns the angle in radians that is the sine of that number.
## 2 – Arccosine Function
The arccosine function is a mathematical function that calculates the angle of a point on a unit circle from the x-coordinate. It is written as arccos(x) or arc cosine(x), and it returns a value in radians. The arccosine function is also known as the inverse cosine function.
## 3 – Arctangent Function
The arctangent function is a real-valued function that takes a real number as input and outputs the arctangent of that number. The arctangent function is defined as:
arctangent(x) = atan(x)
## 4 – Arccotangent Function
The arccotangent function is a function that takes in a real number and outputs the arc cotangent of that number. The arc cotangent of a number is the inverse of the cotangent function, so it calculates the angle of a cotangent line from 0 to pi radians.
## 5 – Arcsecant Function
The arcsecant function is the inverse of the secant function. It calculates the angle between two points on a unit circle, given the x- and y-coordinates of those points.
## 6 – Arccosecant Function
The arccosecant function is the inverse of the arccosine function. It is used to find angles in radians that are in the range of – π/2 to π/2.
The arccosecant function is written as arccosec(x) and it takes the value of x as its input.
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https://dev.to/lofiandcode/big-o-part-5-2-n-2ifn | 1,726,559,421,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651750.27/warc/CC-MAIN-20240917072424-20240917102424-00212.warc.gz | 186,696,630 | 24,960 | ## DEV Community
Joseph Trettevik
Posted on • Updated on
# Big O Notation: O(2^N)
### Song of the Week
This week we're gonna take a look at O(2N). This runtime complexity can be found in recursive functions.
## Tower of Hanoi
Let's take a look at an example in which recursion is used to print out the moves required to solve the Tower of Hanoi puzzle. If you're unfamiliar, the Tower of Hanoi is a game in which you have a stack of disk on one peg that decrease in size as you go up that stack. The goal is to move all the disks to one of the other two pegs by only moving one disk at a time and you can't put a disk on top of a smaller disk. In the following diagram we can see the steps needed to solve the puzzle when there are only three disks in the tower.
Now that we understand the problem, let's take a look at the recursive solution in JavaScript.
``````const towerOfHanoi = (n, from_rod, to_rod, aux_rod) => {
if (n == 1) {
console.log("Move disk 1 from rod " + from_rod + " to rod " + to_rod);
return;
}
towerOfHanoi(n-1, from_rod, aux_rod, to_rod);
console.log("Move disk " + n + " from rod " + from_rod + " to rod " + to_rod);
towerOfHanoi(n-1, aux_rod, to_rod, from_rod);
}
``````
How this function works is pretty interesting, but I'm not going to cover it here since were focused on Big O complexity of this function. To read more on how it works, click one of links under references.
To determine the Big O runtime complexity we need to look at how many recursive calls are made. In this case there are two, and each of those recursive calls passes in n-1 as the new number disks. So a each call to the function doubles the number of recursive calls until n-1 is equal to 1.
If N was equal to 4,
N=4 -> 1 function call
N=3 -> 2 * previous function calls = 1 * 2 = 2
N=2 -> 2 * previous function calls = 2 * 21 = 22
N=1 -> 2 * previous function calls = 2 * 22 = 23
The more general expression of the trend we see would be,
20 + 21 + 22 + 23 + 2N-1 =
2N - 1
Since constants drop off when expressing the Big O complexity, the runtime complexity of the Tower of Hanoi is O(2N).
## The Pattern
The pattern to watch for is that if a recursive function makes more then one call, the complexity is often O(branchesdepth), where branches refers the number recursive calls made and the depth refers to the depth of the tree this creates.
Also keep in mind that even though constants are usually dropped when expression Big O complexity, the base of an exponent does matter. So, O(2N) is very different from O(8N).
## Conclusion
• O(2N) runtime complexities are often seen in recursive functions that make 2 recursive calls and pass in the problem size of N-1.
• If a recursive function makes more then one call, the complex is often O(branchesdepth)
• The base of an exponent does matter. O(2N) is very different from O(8N)
## References
Cover Image
Puzzle Solution Diagram
Tower of Hanoi Code
McDowell, Gayle Laakmann. Cracking the Coding Interview. CareerCup, LLC, 2019. (Pg 44-45) | 809 | 3,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-38 | latest | en | 0.91727 |
https://blog.mrmeyer.com/2011/pseudoteaching-what-was-complex-becomes-routine/ | 1,600,439,791,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187899.11/warc/CC-MAIN-20200918124116-20200918154116-00598.warc.gz | 318,051,908 | 18,293 | ## Pseudoteaching: What Was Complex Becomes Routine
John Burk and Frank Noschese are running a pseudoteaching meme I encourage you to explore for yourself. Their definition:
Pseudoteaching is something you realize you’re doing after you’ve attempted a lesson which from the outset looks like it should result in student learning, but upon further reflection, you realize that the very lesson itself was flawed and involved minimal learning.
Recall also Deborah Loewenberg Ball:
I think we’ve all had the experience of giving so much structure and help that the problem becomes a simple routine problem when it wasn’t originally.
Scene
I don’t like to think about my student teaching year but for the sake of the exercise let me describe my PACT lesson, which, if you’re in California education, you know is the summative assessment of your worth as a preservice teacher. It was a precalculus class and the objective was an understanding of rose petal polar functions like r = 3cos(2theta). Students will be able graph the family. Students will be able to describe how each coefficient affects the graph. Etc.
As I recall it, I was up at the front of the classroom with a TI ViewScreen showing them graph after graph, asking them to determine what the a is doing to the graph of r = asin(3theta).
Then how does b affect the graph of r = 2sin(btheta)?
Then how do sine and cosine differ?
Then I assessed their knowledge of rose petal functions with a worksheet of graphing problems.
Breakdown
I thought my students understood the behavior of r = acos(btheta) on a deep level but they were only responding to superficial patterns in notation. (My supervisor positively thumped me for that one.) Many of you readers — even the straggling humanities instructors we haven’t yet scared off — can see from the graphs above that the petals double when b is even, that the petal length is equal to a but do any of you understand why? Do you know why the graph of r = cos(2theta) has four petals? Could you tell me why the tip of the first sine petal is always on the polar axis? (Not for nothing, the scaffolding in the worksheet also ensures I’m only getting pattern mimicry out of my students.)
Without that understanding, if I attach so much as a negative sign to that function, my students are toast. If I change a coefficient to a fraction, they’re toast. If I change sine or cosine to tangent, they’re toast. That inflexibility is an outcome of pseudoteaching.
Resolution
Unless you understand what an algorithm is going to do, it isn’t going to make sense to you.
The students have to develop the algorithm themselves. Given a second chance at that mess, I’d get students in groups of three or four and let each student pick a member of the family of the functions — “Okay, you do r = 1cos(2theta). I’ll do r = 2cos(2theta). You do r = 3cos(2theta).” Rather than watch me mashing buttons at the front of the room, students would graph their functions by hand and then summarize their findings to each other and then the class. Maybe with a poster — your call. Questions from the teacher would then include everything in the breakdown above, everything I missed six years ago.
I'm Dan and this is my blog. I'm a former high school math teacher and current head of teaching at Desmos. He / him. More here.
1. #### Tom Hoffman
February 21, 2011 - 1:09 pm -
You know, it is kind of shocking that high school students can learn that kind of stuff at all, or that they’d be expected to, or anyone cares.
2. #### Pwolf
February 21, 2011 - 1:51 pm -
I’ve had similar thoughts about the same lesson, but having the kids graph those things by hand from scratch is pulling teeth–then again if I didn’t teach this right, maybe I didn’t teach plotting in polar well either.
Next time I have to teach this, I’ll probably have them match tables of values, functions, graphs, and descriptions of the graphs, kind of like that UK maths resource suggests.
3. #### Jason Dyer
February 21, 2011 - 2:08 pm -
Rather than watch me mashing buttons at the front of the room, students would graph their functions by hand and then summarize their findings to each other and then the class.
Devil’s advocate looming: while the students would get the benefit of the process of exploration, would they understand the “why” part any better? In this specific case at the very least I would include translation between rectangle and polar coordinates and some arrows showing the points of matching.
I know from working with preservice K-6 teachers this semester that pulling out the “why” (even on adding two numbers) is extremely hard and requires nurturing and practice far beyond what explorations provide. I’ve had decent luck getting them to make algorithm explanations with two columns (a How and a Why) but that’s entering into Way Too Much Time Spent To Be Practical in High School Class territory.
4. #### Dan Meyer
February 21, 2011 - 2:22 pm -
Tom: You know, it is kind of shocking that high school students can learn that kind of stuff at all, or that they’d be expected to, or anyone cares.
Get him, boys!
Pwolf: I’ve had similar thoughts about the same lesson, but having the kids graph those things by hand from scratch is pulling teeth
and
Jason: I’ve had decent luck getting them to make algorithm explanations with two columns (a How and a Why) but that’s entering into Way Too Much Time Spent To Be Practical in High School Class territory.
The constructivism multiplier, and all that. But I think the students have to see the petal return to the origin a few times. I mean, even if they’re looking at a calc-generated table but graphing the points by hand, that’s better conceptually than just teaching them “equation turns into flowery-thingy.”
5. #### Jen
February 21, 2011 - 3:26 pm -
I dunno, as a student, I’m pretty sure that I’d have found “equation turns into flowery-thingy†interesting enough to look at and play with (and maybe get to make my own flowery-thingy, rather than just look at other ones.
Then, about a week later, maybe even on the test, although often later, I’d get the concepts. They’d just sort of dawn on me and all those things the teacher had said that seemed rather pointless (though were clearly important since s/he said them many times) would suddenly come into focus.
So, while I agree that just grinding through isn’t likely to stick, then again some grinding and some drawing or at least plotting and lots of repetition may oftentimes lead to delayed onset comprehension.
6. #### Sam Critchlow
February 21, 2011 - 7:21 pm -
Some thoughts on the specifics of the graphs etc.: the Grapher built in to most macs has the capability to parametrize a variable (a) via a slider. Parametrizing a and b (perhaps with a step of .5 or .1?) might convey the examples more efficiently than a series of graphs. Better still if there were a way for the students to control the parameters without flipping between the r= and graph screen.
This, of course, does little to nothing to explain the “why”. I struggle with the same problem when teaching transformations in rectangular coordinates. Easy to show the various rules of shifts, reflections, etc. and to come up with mnemonics to remember the rules, but difficult for me to explain in a comprehensible manner. Spent hours creating powerpoint slides trying, but given a week most students remembered almost nothing of the “how” let alone the “why”.
7. #### Alex Eckert
February 21, 2011 - 8:32 pm -
I’ve always thought you were a little too harsh on your 1st-year-teacher-self, second-year-teacher-self, and as not so subtly stated here, your preservice-teacher-self. Maybe it’s all by way of letting others know that even the great Dan Meyer was not so great once (even though I’m sure most of us on here would dispute that).
Personally I don’t think you’d be where you are now if you weren’t where you were then.
My goal, with teaching in general, isn’t so much for them to grasp the why now. It’s to grasp an appreciation for searching for the why now, so that when they’re my age they can’t stand but to go a day without trying to figure out why to something. I can’t, and it makes things awfully fun.
8. #### Alex Eckert
February 21, 2011 - 8:36 pm -
Ha…just re-read my post and realized how Cypress Hill – esque, circa 1992, my middle paragraph (sentence) sounded….
“How do you know where I’m at if you haven’t been where I’ve been……understand where I’m comin’ from???”
9. #### Laura
February 22, 2011 - 3:56 am -
@ Jason:
I don’t know that Dan is implying that he would toss out these ideas, let the kids “explore,” and expect them to arrive at the “why” as a result. I believe in guided inquiry, and carry this out as often as I can/plan with my second graders. Giving the kids time to explore something gives them a frame of reference during the discussion of “why.” And I think it’s important to remember that understanding is a precursor to articulating that understanding. So…students will be able to understand “why” before they’re able to clearly explain the “why.”
10. #### Maria
February 22, 2011 - 5:56 am -
I love this self reflection. It so mirrors my own struggling reality in the here-and-now. Thanks for the boost in confidence for a newby (whose not so ‘new’ in the chronological sense).
11. #### ClimeGuy
February 22, 2011 - 6:24 am -
John and Frank’s definition:
Pseudoteaching is something you realize you’re doing after you’ve attempted a lesson which from the outset looks like it should result in student learning, but upon further reflection, you realize that the very lesson itself was flawed and involved minimal learning.
reminded me of Alan Schoenfeld’s paper When Good Teaching Leads to Bad Results: The Disasters of “Well Taught” Mathematics Courses which might be of interest here.
12. #### Frank Noschese
February 22, 2011 - 9:53 am -
Hi Dan,
Awesome post. We want students to find patterns, but we also want them to be able to transfer the inner workings of those the patterns to new problems. Your polar functions lesson solidly shows how pseudoteaching fails to do this.
Thank you so much for taking time out of your busy #gradskool and speaking schedule to put together this post. The differences between good lessons and better lessons are nuanced, and you’ve helped bring those subtleties to light.
Many thanks again,
Frank
13. #### Brian Wyzlic
February 22, 2011 - 10:36 am -
One thing that was repeated a lot during my undergrad courses was “play with it.” While this is not something I could say without context to a group that includes high school males, that is what I asked my pre-calc students to do. All my students either had a graphing calculator or could borrow one overnight. Towards the end of class, we went over how to put the calculator in polar form (they already know what polar graphing is, we just hadn’t done much with it yet), and we look at a few graphs involving sin and cos. Then they take some time in class to play with it, and their assignment for the night is to play around with polar graphing and find some patterns.
The next day, we have a whole mess of patterns that have been found, and as a class, we can explore why those patterns exist. It works pretty well, though it does not rule out pseudoteaching without the next day exploration.
14. #### Amy
February 22, 2011 - 11:12 am -
I explored this on my own as a high school student. As soon as I started drawing out the waves, I noticed that whether or not the wave had point symmetry affected the number of petals, and it was easy to see and explain how the “b” could manipulate that point symmetry.
…. at least that’s what I remember, it’s been YEARS since I’ve explored polar coordinates!
Signed,
Humanities-Type
15. #### Jason Dyer
February 22, 2011 - 12:45 pm -
@Dan: Not disagreeing — I’ve run polar coordination exploration in class before — just being cautious. It didn’t read like a complete lesson to me yet.
So…students will be able to understand “why†before they’re able to clearly explain the “why.â€
@Laura: This is the knotty bit I’d love to unravel: how do you tell when this happens? I’ve certainly done explorations where students had the patterns down as brainlessly as if I’d just told them a recipe. I’d love to avoid that, but sometimes it’s a hard thing to check. The standard approach is to add a wrinkle and recheck for understanding (like mentioned in the example, adding a negative) but that’s not foolproof.
16. #### Luke Hodge
February 22, 2011 - 2:20 pm -
It seems to me that asking what the parameters a & b do for the functions r=asin(bt) provides a nice visual surprise, still has a little bit of meat to it, and avoids the tedium of graphing by hand.
Plotting points in hopes of a graph taking some sort of form is not an approach (at least not a first approach) that I would want to encourage. I would much rather see if I can figure anything out by looking at the equation first – then pick some nice points to plot.
I think the “why” part of this is accessible if you focus on values of t that minimize/maximize r – which is also a relatively appetizing question. Finding values of t, by hand, which minimize/maximize r is not tedious & very clearly shows why you get the number of loops that you do – it is all about the period & whether some loops will overlap previous loops. I think most people would likely need a hint or instructions to focus on max/mins for r, though.
17. #### Fachler
February 22, 2011 - 2:26 pm -
I think they key here was the end realization: by facilitating performances during which students created (and then probably evaluated), a teacher would be able to accurately gauge understanding, which is the goal–not just performance or knowledge. I think this is where the task analysis folks miss the boat.
I believe students should be in a constant state of production. As one of the “straggling humanities teachers” still following along in the discussion, it’s easy for me: students are always writing, revising, conferring, and creating class-wide and personal language arts-based projects. For the example above, I might ask, “What were the students creating with these rose petal formation graphs?”
It’s tough to get kids to understand something if the final output is just graphs. Graphs for what purpose? The Teaching for Understanding (TfU) framework calls the steps along the way “understanding performances” because the students understands more deeply with each scaffolded output (from initial inquiry, to guided, to independent).
As you have already pointed out the aesthetic qualities of those rose petals, perhaps an art-based summative output wherein students had to manipulate the equations for intended results (i.e. students sketch first, plan out the art, then have to troubleshoot the math to make it work) might have engendered transfer.
18. #### Dave Smith
February 26, 2011 - 7:13 pm -
Thanks all, for a very thought provoking discussion.
I discovered my own psuedoteaching on a big scale one semester when I read the NSTA’s wonderful little pamphlet “Ask the Right Questions.” In it they suggested using additional wait time after answers, not just after questions, and perhaps even adding in a little gesture to ask for more. I went in full of confidence to the next day’s class (upper level college geology class that I was very pleased with – lots of engagement, lots of good answers to my questions, etc). Well, first there was the panic stricken look in the eyes of my students. More?! What do you mean more?! Then there was the uncomfortable pause. Then there was the complete nonsense that came out of their mouths. Then it was my turn to panic (Oh no! They haven’t actually learned anything! Now what?!). It really was remarkable – The first couple of words in response to each of my questions were generally a correct short answer, and the next ten or twenty or more words were confused, rambling, or just plain wrong.
What I eventually learned from that experience was that I had to let/make the students do a LOT more talking. Only once students were talking did I have any way to know whether they had any idea what was going on. I began to focus my teaching on data sets. “Here are first person accounts of different kinds of volcanoes, what patterns do you notice?” Yes it took time and so I cut down my curriculum to 6 big ideas in a semester (15 class/lab hours per idea). I did much less preparation for specific class activities and much more lying awake at night thinking “Is this really the most important idea for this unit, or can I get away with leaving them confused for now?”
@Frank Noschese this is why I have been a huge fan of Modeling Instruction (modeling.asu.edu) ever since I first read about it.
Thanks for the journey back into polar coordinates, too. I had to use Excel to generate a table and scratch out a sketch graph before I began to understand what was going on, even though I clearly recall learning how to predict the number of petals once upon a time.
Dave
19. #### Dave Smith
February 26, 2011 - 7:24 pm -
Just a quick note on graphing by hand. Ever since I took a short course from Win Means at SUNY Albany on Mohr’s Circle (a graphical method for working with tensors), I have been much more likely to use sketch graphing and have a much greater appreciation for the potential of a simple sketch to convey useful and even reasonably precise quantitative information. When I realized I did not understand the functions above, I started in with SpaceTime on my iPad, but the finished graph lacked the information to show me how the petals connected to the trig function, especially since it converted all the coordinates to x, y pairs. I grabbed a handy envelope and went to work on a sketched polar plot, going from cos(theta) to cos (2theta) and within 5 or 10 minutes I had the basics figured out.
Dave
20. #### Jovan Miles
February 28, 2011 - 6:41 am -
Hey Dan, I always read but rarely comment!
Your point about algorithms not making sense to people unless they know what the algorithm is going to do just hit home with me.
I’m currently working as an instructional math coach with 2 alternative schools that serve special education students.
I’m looking over diagnostic tests and noticing a pattern of students simply not knowing what the algorithm is supposed to do…
I’m going to share this post with my teachers and hopefully they’ll get something out of it.
21. #### Launa
October 25, 2011 - 2:29 am -
I love that you hold yourself to this amazing high standard: that every kid learns at a high level, every class.
I’m an old (humanities) teacher working with a new age group at a new school, and realize just how rarely I am able to get up to that level in my new setting. Whereas before I could convince myself that I was there a lot — now, hardly at all.
I’m looking, instead, at progress-over-time, when I feel like a failure. Here at six weeks into the semester they have certainly learned, even though I could hardly see it unfolding on a daily basis, and had so many pseudoteaching fails along the way.
Thanks for the reminder of exactly what I am striving for.
22. #### Sue VanHattum
March 25, 2012 - 7:22 am -
>Could you tell me why the tip of the first sine petal is always on the polar axis?
First cosine petal, right? | 4,439 | 19,419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2020-40 | longest | en | 0.957743 |
https://affairscloud.com/aptitude-questions-inequalities-set-1/ | 1,713,540,518,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817438.43/warc/CC-MAIN-20240419141145-20240419171145-00223.warc.gz | 68,440,551 | 37,542 | # Aptitude Questions – Inequalities Set 1
Hello Aspirants. Welcome to Online Quantitative Aptitude Section in AffairsCloud.com. Here we are creating sample questions in Inequalities which is common for all the competitive exams. We have included Some questions that are repeatedly asked in bank exams !!!
1. The respective ratio between the present age of Mohan and David is 5:x. Mohan is 9 years younger than Preethi. Preethi’s age after 9 years will be 33 years. The difference between David’s and Mohan’s age is same as the present age of Preethi.
Quantity I: Mohan’s present age
Quantity II: The value of x
A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established
A. Quantity I > Quantity II
Explanation:
Preethi’s age after 9 years = 33 years
Preethi’s present age = 33 – 9 = 24 years
Mohan’s present age = 24 – 9 = 15 years
David’s present age = 15 + 24 = 39 years
Ratio between Mohan and David = 15 : 39 = 5 : 13
X = 13
2. Sri invested some amount(x) at the rate of 12% simple interest and a certain amount(y) at the rate of 10% simple interest. He received yearly interest of Rs.140. But if he had interchanged the amounts invested, he would have received Rs.4 more as interest.
Quantity I: The value of x
Quantity II: The value of y
A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established
B. Quantity I < Quantity II
Explanation:
Amount invested at 12% = Rs. x
Amount invested at 10% = Rs. y
140 = x*12*1/100 + y*10*1/100
12x + 10y = 14000 -(i)
144 = x*10*1/100 + y*12*1/100
10x + 12y = 14400 -(ii)
x = 545.45; y = 745.45
3. Ajith can do a piece of work in 10 days, Bala in 15 days. They work together for 5 days, the rest of the work is finished by Chand in two more days. They get Rs. 6000 as wages for the whole work.
Quantity I: What is the sum of Rs.100 and the daily wage of Bala?
Quantity II: What is the daily wage of Chand?
A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established
E. Quantity I = Quantity II or relation cannot be established
Explanation:
Ajith’s 5 days work = 50%
Bala’s 5 days work = 33.33%
Chand’s 2 days work = 16.66%[100- (50+33.33)]
Ratio of contribution of work of Ajith, Bala and Chand = 3 : 2 : 1
Ajith’s total share = Rs. 3000
Bala’s total share = Rs. 2000
Chand’s total share = Rs. 1000
Ajith’s one day’s earning = Rs.600
Bala’s one day’s earning = Rs.400
Chand’s one day’s earning = Rs.500
4. A Bike is available at 40% discount at showroom “A” and the same is available at only 25% discount at showroom “B”. Mr. Arun has just sufficient amount of Rs. 60,000 to purchase it at showroom “A”.
Quantity I: Difference between Marked Price and SP at Show Room “A”
Quantity II: Difference between Marked Price and SP at Show Room “B”
A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established
A. Quantity I > Quantity II
Explanation:
Let the marked price be x.
Cost price (CP) = 40 % discount on MP = 0.6y = 60000
⇒ y= Rs.100000 MP
SP at Show Room “A” = Rs. 60000
SP at Show Room “B” = 100000 X 0.75 = 75000
Difference between Marked Price and SP at Show Room “A” = 40000
Difference between Marked Price and SP at Show Room “B” = 25000
5. Quantity I: (x – 18)² = 0
Quantity II: y² = 324
A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established
C. Quantity I ≥ Quantity II
Explanation:
x² – 36x + 324 = 0
x = 18, 18
y² = 324
y = ±18
6. A Cistern has an inlet pipe and outlet pipe. The inlet pipe fills the cistern completely in 1 hour 20 minutes when the outlet pipe is plugged. The outlet pipe empties the tank completely in 4 hours when the inlet pipe is plugged.
Quantity I: Inlet pipe Efficiency
Quantity II: 3 times of Outlet pipe Efficiency
A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established
E. Quantity I = Quantity II or relation cannot be established
Explanation:
Inlet pipe Efficiency = 100/(8/6) = 75%
Outlet pipe Efficiency = 100/(4) = 25%
3 times of Outlet pipe Efficiency = 75%
7. Harish took an educational loan from a nationalized bank for his 2 years course of MBA. He took the loan of Rs.5 lakh such that he would be charged at 7% p.a. at CI during his course and at 9% CI after the completion of the course. He returned half of the amount which he had to be paid on the completion of his studies and remaining after 2 years.
Quantity I: He returned half of the amount which he had to be paid on the completion of his studies
Quantity II: He returned remaining amount after 2 years
A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established
B. Quantity I < Quantity II
Explanation:
5,00,000 * (1.07)² = 572450
Returned amount = 286225
After two years = 286225 * (1.09)² = 340063
8. The average salary of the entire staff in an office is Rs 250 per month. The average salary of officers is Rs 520 and that of non-officers is Rs. 200.
Quantity I: Number of Officers = 15
Quantity II: Number of Non-Officers
A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established
B. Quantity I < Quantity II
Explanation:
Let the required number of non–officers = x
200x + 520 x 15 = 250 ( 15 + x )
250x – 200x = 520 * 15 – 250 x 15
50x = 4050
x = 81
9. The perimeter of a rectangle and a square is 160 cm each. If the difference between their areas is 600 cm.
Quantity I: Area of Square
Quantity II:Area of Rectangle
A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established
A. Quantity I > Quantity II
Explanation:
Perimeter of rectangle = Perimeter of Square = 160
4a = 160 => a = 40
Area of square = 1600
1600 – lb = 600
lb = 1000 cm²
10. Shree started traveling from a place A to B and Priya started traveling from a place B to A which are 576 km apart. They meet after 12 hours. After their meeting, Shree increased her speed by 2 km/hr and Priya reduced her speed by 2 km/hr, they arrived at B and A respectively at the same time.
Quantity I: Initial Speed of Shree
Quantity II: Initial Speed of Priya
A. Quantity I > Quantity II
B. Quantity I < Quantity II
C. Quantity I ≥ Quantity II
D. Quantity I ≤ Quantity II
E. Quantity I = Quantity II or relation cannot be established
B. Quantity I < Quantity II
Explanation:
Sum of their speeds = Distance/time = 576/12 = 48 kmph
Respective Speed of Shree and Priya = (23 + 25) = 48 kmph
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Program for Fibonacci numbers
• Difficulty Level : Medium
• Last Updated : 07 Apr, 2021
The Fibonacci numbers are the numbers in the following integer sequence.
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ……..
In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation
`Fn = Fn-1 + Fn-2`
with seed values
`F0 = 0 and F1 = 1.`
Given a number n, print n-th Fibonacci Number.
Examples:
```Input : n = 2
Output : 1
Input : n = 9
Output : 34```
Write a function int fib(int n) that returns Fn. For example, if n = 0, then fib() should return 0. If n = 1, then it should return 1. For n > 1, it should return Fn-1 + Fn-2
```For n = 9
Output:34```
Following are different methods to get the nth Fibonacci number.
Method 1 (Use recursion)
A simple method that is a direct recursive implementation mathematical recurrence relation given above.
## C++
`//Fibonacci Series using Recursion``#include``using` `namespace` `std;` `int` `fib(``int` `n)``{`` ``if` `(n <= 1)`` ``return` `n;`` ``return` `fib(n-1) + fib(n-2);``}` `int` `main ()``{`` ``int` `n = 9;`` ``cout << fib(n);`` ``getchar``();`` ``return` `0;``}` `// This code is contributed``// by Akanksha Rai`
## C
`//Fibonacci Series using Recursion``#include``int` `fib(``int` `n)``{`` ``if` `(n <= 1)`` ``return` `n;`` ``return` `fib(n-1) + fib(n-2);``}` `int` `main ()``{`` ``int` `n = 9;`` ``printf``(``"%d"``, fib(n));`` ``getchar``();`` ``return` `0;``}`
## Java
`//Fibonacci Series using Recursion``class` `fibonacci``{`` ``static` `int` `fib(``int` `n)`` ``{`` ``if` `(n <= ``1``)`` ``return` `n;`` ``return` `fib(n-``1``) + fib(n-``2``);`` ``}`` ` ` ``public` `static` `void` `main (String args[])`` ``{`` ``int` `n = ``9``;`` ``System.out.println(fib(n));`` ``}``}``/* This code is contributed by Rajat Mishra */`
## Python
`# Function for nth Fibonacci number` `def` `Fibonacci(n):`` ``if` `n<``0``:`` ``print``(``"Incorrect input"``)`` ``# First Fibonacci number is 0`` ``elif` `n``=``=``0``:`` ``return` `0`` ``# Second Fibonacci number is 1`` ``elif` `n``=``=``1``:`` ``return` `1`` ``else``:`` ``return` `Fibonacci(n``-``1``)``+``Fibonacci(n``-``2``)` `# Driver Program` `print``(Fibonacci(``9``))` `#This code is contributed by Saket Modi`
## C#
`// C# program for Fibonacci Series``// using Recursion``using` `System;` `public` `class` `GFG``{`` ``public` `static` `int` `Fib(``int` `n)`` ``{`` ``if` `(n <= 1)`` ``{`` ``return` `n;`` ``}`` ``else`` ``{`` ``return` `Fib(n - 1) + Fib(n - 2);`` ``}`` ``}`` ` ` ``// driver code`` ``public` `static` `void` `Main(``string``[] args)`` ``{`` ``int` `n = 9;`` ``Console.Write(Fib(n));`` ``}``}` `// This code is contributed by Sam007`
## PHP
``
## Javascript
``
Output
`34`
Time Complexity: T(n) = T(n-1) + T(n-2) which is exponential.
We can observe that this implementation does a lot of repeated work (see the following recursion tree). So this is a bad implementation for nth Fibonacci number.
``` fib(5)
/ \
fib(4) fib(3)
/ \ / \
fib(3) fib(2) fib(2) fib(1)
/ \ / \ / \
fib(2) fib(1) fib(1) fib(0) fib(1) fib(0)
/ \
fib(1) fib(0)```
Extra Space: O(n) if we consider the function call stack size, otherwise O(1).
Method 2 (Use Dynamic Programming)
We can avoid the repeated work done is method 1 by storing the Fibonacci numbers calculated so far.
## C++
`// C++ program for Fibonacci Series ``// using Dynamic Programming``#include``using` `namespace` `std;` `class` `GFG{`` ` `public``:``int` `fib(``int` `n)``{`` ` ` ``// Declare an array to store`` ``// Fibonacci numbers.`` ``// 1 extra to handle`` ``// case, n = 0`` ``int` `f[n + 2];`` ``int` `i;` ` ``// 0th and 1st number of the`` ``// series are 0 and 1`` ``f[0] = 0;`` ``f[1] = 1;` ` ``for``(i = 2; i <= n; i++)`` ``{`` ` ` ``//Add the previous 2 numbers`` ``// in the series and store it`` ``f[i] = f[i - 1] + f[i - 2];`` ``}`` ``return` `f[n];`` ``}``};` `// Driver code``int` `main ()``{`` ``GFG g;`` ``int` `n = 9;`` ` ` ``cout << g.fib(n);`` ``return` `0;``}` `// This code is contributed by SoumikMondal`
## C
`//Fibonacci Series using Dynamic Programming``#include` `int` `fib(``int` `n)``{`` ``/* Declare an array to store Fibonacci numbers. */`` ``int` `f[n+2]; ``// 1 extra to handle case, n = 0`` ``int` `i;` ` ``/* 0th and 1st number of the series are 0 and 1*/`` ``f[0] = 0;`` ``f[1] = 1;` ` ``for` `(i = 2; i <= n; i++)`` ``{`` ``/* Add the previous 2 numbers in the series`` ``and store it */`` ``f[i] = f[i-1] + f[i-2];`` ``}` ` ``return` `f[n];``}` `int` `main ()``{`` ``int` `n = 9;`` ``printf``(``"%d"``, fib(n));`` ``getchar``();`` ``return` `0;``}`
## Java
`// Fibonacci Series using Dynamic Programming``class` `fibonacci``{`` ``static` `int` `fib(``int` `n)`` ``{`` ``/* Declare an array to store Fibonacci numbers. */`` ``int` `f[] = ``new` `int``[n+``2``]; ``// 1 extra to handle case, n = 0`` ``int` `i;`` ` ` ``/* 0th and 1st number of the series are 0 and 1*/`` ``f[``0``] = ``0``;`` ``f[``1``] = ``1``;`` ` ` ``for` `(i = ``2``; i <= n; i++)`` ``{`` ``/* Add the previous 2 numbers in the series`` ``and store it */`` ``f[i] = f[i-``1``] + f[i-``2``];`` ``}`` ` ` ``return` `f[n];`` ``}`` ` ` ``public` `static` `void` `main (String args[])`` ``{`` ``int` `n = ``9``;`` ``System.out.println(fib(n));`` ``}``}``/* This code is contributed by Rajat Mishra */`
## Python
`# Fibonacci Series using Dynamic Programming``def` `fibonacci(n):`` ` ` ``# Taking 1st two fibonacci nubers as 0 and 1`` ``f ``=` `[``0``, ``1``]`` ` ` ` ` ``for` `i ``in` `range``(``2``, n``+``1``):`` ``f.append(f[i``-``1``] ``+` `f[i``-``2``])`` ``return` `f[n]`` ` `print``(fibonacci(``9``))`
## C#
`// C# program for Fibonacci Series``// using Dynamic Programming``using` `System;``class` `fibonacci {`` ` `static` `int` `fib(``int` `n)`` ``{`` ` ` ``// Declare an array to`` ``// store Fibonacci numbers.`` ``// 1 extra to handle`` ``// case, n = 0`` ``int` `[]f = ``new` `int``[n + 2];`` ``int` `i;`` ` ` ``/* 0th and 1st number of the`` ``series are 0 and 1 */`` ``f[0] = 0;`` ``f[1] = 1;`` ` ` ``for` `(i = 2; i <= n; i++)`` ``{`` ``/* Add the previous 2 numbers`` ``in the series and store it */`` ``f[i] = f[i - 1] + f[i - 2];`` ``}`` ` ` ``return` `f[n];`` ``}`` ` ` ``// Driver Code`` ``public` `static` `void` `Main ()`` ``{`` ``int` `n = 9;`` ``Console.WriteLine(fib(n));`` ``}``}` `// This code is contributed by anuj_67.`
## PHP
``
## Javascript
``
Output
`34`
Method 3 (Space Optimized Method 2)
We can optimize the space used in method 2 by storing the previous two numbers only because that is all we need to get the next Fibonacci number in series.
## C++
`// Fibonacci Series using Space Optimized Method``#include``using` `namespace` `std;` `int` `fib(``int` `n)``{`` ``int` `a = 0, b = 1, c, i;`` ``if``( n == 0)`` ``return` `a;`` ``for``(i = 2; i <= n; i++)`` ``{`` ``c = a + b;`` ``a = b;`` ``b = c;`` ``}`` ``return` `b;``}` `// Driver code``int` `main()``{`` ``int` `n = 9;`` ` ` ``cout << fib(n);`` ``return` `0;``}` `// This code is contributed by Code_Mech`
## C
`// Fibonacci Series using Space Optimized Method``#include``int` `fib(``int` `n)``{`` ``int` `a = 0, b = 1, c, i;`` ``if``( n == 0)`` ``return` `a;`` ``for` `(i = 2; i <= n; i++)`` ``{`` ``c = a + b;`` ``a = b;`` ``b = c;`` ``}`` ``return` `b;``}` `int` `main ()``{`` ``int` `n = 9;`` ``printf``(``"%d"``, fib(n));`` ``getchar``();`` ``return` `0;``}`
## Java
`// Java program for Fibonacci Series using Space``// Optimized Method``class` `fibonacci``{`` ``static` `int` `fib(``int` `n)`` ``{`` ``int` `a = ``0``, b = ``1``, c;`` ``if` `(n == ``0``)`` ``return` `a;`` ``for` `(``int` `i = ``2``; i <= n; i++)`` ``{`` ``c = a + b;`` ``a = b;`` ``b = c;`` ``}`` ``return` `b;`` ``}` ` ``public` `static` `void` `main (String args[])`` ``{`` ``int` `n = ``9``;`` ``System.out.println(fib(n));`` ``}``}` `// This code is contributed by Mihir Joshi`
## Python
`# Function for nth fibonacci number - Space Optimisataion``# Taking 1st two fibonacci numbers as 0 and 1` `def` `fibonacci(n):`` ``a ``=` `0`` ``b ``=` `1`` ``if` `n < ``0``:`` ``print``(``"Incorrect input"``)`` ``elif` `n ``=``=` `0``:`` ``return` `a`` ``elif` `n ``=``=` `1``:`` ``return` `b`` ``else``:`` ``for` `i ``in` `range``(``2``,n``+``1``):`` ``c ``=` `a ``+` `b`` ``a ``=` `b`` ``b ``=` `c`` ``return` `b` `# Driver Program` `print``(fibonacci(``9``))` `#This code is contributed by Saket Modi`
## C#
`// C# program for Fibonacci Series``// using Space Optimized Method``using` `System;` `namespace` `Fib``{`` ``public` `class` `GFG`` ``{`` ``static` `int` `Fib(``int` `n)`` ``{`` ``int` `a = 0, b = 1, c = 0;`` ` ` ``// To return the first Fibonacci number`` ``if` `(n == 0) ``return` `a;`` ` ` ``for` `(``int` `i = 2; i <= n; i++)`` ``{`` ``c = a + b;`` ``a = b;`` ``b = c;`` ``}`` ` ` ``return` `b;`` ``}`` ` ` ``// Driver function`` ``public` `static` `void` `Main(``string``[] args)`` ``{`` ` ` ``int` `n = 9;`` ``Console.Write(``"{0} "``, Fib(n));`` ``}`` ``}``}` `// This code is contributed by Sam007.`
## PHP
``
## Javascript
``
Output
`34`
Time Complexity:O(n)
Extra Space: O(1)
Method 4 (Using power of the matrix {{1, 1}, {1, 0}})
This another O(n) which relies on the fact that if we n times multiply the matrix M = {{1,1},{1,0}} to itself (in other words calculate power(M, n)), then we get the (n+1)th Fibonacci number as the element at row and column (0, 0) in the resultant matrix.
The matrix representation gives the following closed expression for the Fibonacci numbers:
## C++
`#include``using` `namespace` `std;` `// Helper function that multiplies 2``// matrices F and M of size 2*2, and``// puts the multiplication result``// back to F[][]``void` `multiply(``int` `F[2][2], ``int` `M[2][2]);` `// Helper function that calculates F[][]``// raise to the power n and puts the``// result in F[][]``// Note that this function is designed``// only for fib() and won't work as``// general power function``void` `power(``int` `F[2][2], ``int` `n);` `int` `fib(``int` `n)``{`` ``int` `F[2][2] = { { 1, 1 }, { 1, 0 } };`` ` ` ``if` `(n == 0)`` ``return` `0;`` ` ` ``power(F, n - 1);`` ` ` ``return` `F[0][0];``}` `void` `multiply(``int` `F[2][2], ``int` `M[2][2])``{`` ``int` `x = F[0][0] * M[0][0] +`` ``F[0][1] * M[1][0];`` ``int` `y = F[0][0] * M[0][1] +`` ``F[0][1] * M[1][1];`` ``int` `z = F[1][0] * M[0][0] +`` ``F[1][1] * M[1][0];`` ``int` `w = F[1][0] * M[0][1] +`` ``F[1][1] * M[1][1];`` ` ` ``F[0][0] = x;`` ``F[0][1] = y;`` ``F[1][0] = z;`` ``F[1][1] = w;``}` `void` `power(``int` `F[2][2], ``int` `n)``{`` ``int` `i;`` ``int` `M[2][2] = { { 1, 1 }, { 1, 0 } };`` ` ` ``// n - 1 times multiply the`` ``// matrix to {{1,0},{0,1}}`` ``for``(i = 2; i <= n; i++)`` ``multiply(F, M);``}` `// Driver code``int` `main()``{`` ``int` `n = 9;`` ` ` ``cout << ``" "` `<< fib(n);`` ` ` ``return` `0;``}` `// This code is contributed by shivanisinghss2110`
## C
`#include ` `/* Helper function that multiplies 2 matrices F and M of size 2*2, and`` ``puts the multiplication result back to F[][] */``void` `multiply(``int` `F[2][2], ``int` `M[2][2]);` `/* Helper function that calculates F[][] raise to the power n and puts the`` ``result in F[][]`` ``Note that this function is designed only for fib() and won't work as general`` ``power function */``void` `power(``int` `F[2][2], ``int` `n);` `int` `fib(``int` `n)``{`` ``int` `F[2][2] = {{1,1},{1,0}};`` ``if` `(n == 0)`` ``return` `0;`` ``power(F, n-1);` ` ``return` `F[0][0];``}` `void` `multiply(``int` `F[2][2], ``int` `M[2][2])``{`` ``int` `x = F[0][0]*M[0][0] + F[0][1]*M[1][0];`` ``int` `y = F[0][0]*M[0][1] + F[0][1]*M[1][1];`` ``int` `z = F[1][0]*M[0][0] + F[1][1]*M[1][0];`` ``int` `w = F[1][0]*M[0][1] + F[1][1]*M[1][1];` ` ``F[0][0] = x;`` ``F[0][1] = y;`` ``F[1][0] = z;`` ``F[1][1] = w;``}` `void` `power(``int` `F[2][2], ``int` `n)``{`` ``int` `i;`` ``int` `M[2][2] = {{1,1},{1,0}};` ` ``// n - 1 times multiply the matrix to {{1,0},{0,1}}`` ``for` `(i = 2; i <= n; i++)`` ``multiply(F, M);``}` `/* Driver program to test above function */``int` `main()``{`` ``int` `n = 9;`` ``printf``(``"%d"``, fib(n));`` ``getchar``();`` ``return` `0;``}`
## Java
`class` `fibonacci``{`` ` ` ``static` `int` `fib(``int` `n)`` ``{`` ``int` `F[][] = ``new` `int``[][]{{``1``,``1``},{``1``,``0``}};`` ``if` `(n == ``0``)`` ``return` `0``;`` ``power(F, n-``1``);`` ` ` ``return` `F[``0``][``0``];`` ``}`` ` ` ``/* Helper function that multiplies 2 matrices F and M of size 2*2, and`` ``puts the multiplication result back to F[][] */`` ``static` `void` `multiply(``int` `F[][], ``int` `M[][])`` ``{`` ``int` `x = F[``0``][``0``]*M[``0``][``0``] + F[``0``][``1``]*M[``1``][``0``];`` ``int` `y = F[``0``][``0``]*M[``0``][``1``] + F[``0``][``1``]*M[``1``][``1``];`` ``int` `z = F[``1``][``0``]*M[``0``][``0``] + F[``1``][``1``]*M[``1``][``0``];`` ``int` `w = F[``1``][``0``]*M[``0``][``1``] + F[``1``][``1``]*M[``1``][``1``];`` ` ` ``F[``0``][``0``] = x;`` ``F[``0``][``1``] = y;`` ``F[``1``][``0``] = z;`` ``F[``1``][``1``] = w;`` ``}` ` ``/* Helper function that calculates F[][] raise to the power n and puts the`` ``result in F[][]`` ``Note that this function is designed only for fib() and won't work as general`` ``power function */`` ``static` `void` `power(``int` `F[][], ``int` `n)`` ``{`` ``int` `i;`` ``int` `M[][] = ``new` `int``[][]{{``1``,``1``},{``1``,``0``}};`` ` ` ``// n - 1 times multiply the matrix to {{1,0},{0,1}}`` ``for` `(i = ``2``; i <= n; i++)`` ``multiply(F, M);`` ``}`` ` ` ``/* Driver program to test above function */`` ``public` `static` `void` `main (String args[])`` ``{`` ``int` `n = ``9``;`` ``System.out.println(fib(n));`` ``}``}``/* This code is contributed by Rajat Mishra */`
## Python3
`# Helper function that multiplies``# 2 matrices F and M of size 2*2,``# and puts the multiplication``# result back to F[][]` `# Helper function that calculates``# F[][] raise to the power n and``# puts the result in F[][]``# Note that this function is``# designed only for fib() and``# won't work as general``# power function``def` `fib(n):`` ``F ``=` `[[``1``, ``1``],`` ``[``1``, ``0``]]`` ``if` `(n ``=``=` `0``):`` ``return` `0`` ``power(F, n ``-` `1``)`` ` ` ``return` `F[``0``][``0``]` `def` `multiply(F, M):` ` ``x ``=` `(F[``0``][``0``] ``*` `M[``0``][``0``] ``+`` ``F[``0``][``1``] ``*` `M[``1``][``0``])`` ``y ``=` `(F[``0``][``0``] ``*` `M[``0``][``1``] ``+`` ``F[``0``][``1``] ``*` `M[``1``][``1``])`` ``z ``=` `(F[``1``][``0``] ``*` `M[``0``][``0``] ``+`` ``F[``1``][``1``] ``*` `M[``1``][``0``])`` ``w ``=` `(F[``1``][``0``] ``*` `M[``0``][``1``] ``+`` ``F[``1``][``1``] ``*` `M[``1``][``1``])`` ` ` ``F[``0``][``0``] ``=` `x`` ``F[``0``][``1``] ``=` `y`` ``F[``1``][``0``] ``=` `z`` ``F[``1``][``1``] ``=` `w` `def` `power(F, n):` ` ``M ``=` `[[``1``, ``1``],`` ``[``1``, ``0``]]` ` ``# n - 1 times multiply the`` ``# matrix to {{1,0},{0,1}}`` ``for` `i ``in` `range``(``2``, n ``+` `1``):`` ``multiply(F, M)` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:`` ``n ``=` `9`` ``print``(fib(n))` `# This code is contributed``# by ChitraNayal`
## C#
`using` `System;` `class` `GFG {`` ` ` ``static` `int` `fib(``int` `n)`` ``{`` ``int` `[,]F = ``new` `int``[,] {{1, 1},`` ``{1, 0} };`` ``if` `(n == 0)`` ``return` `0;`` ``power(F, n-1);`` ` ` ``return` `F[0,0];`` ``}`` ` ` ``/* Helper function that multiplies 2`` ``matrices F and M of size 2*2, and puts`` ``the multiplication result back to F[][] */`` ``static` `void` `multiply(``int` `[,]F, ``int` `[,]M)`` ``{`` ``int` `x = F[0,0]*M[0,0] + F[0,1]*M[1,0];`` ``int` `y = F[0,0]*M[0,1] + F[0,1]*M[1,1];`` ``int` `z = F[1,0]*M[0,0] + F[1,1]*M[1,0];`` ``int` `w = F[1,0]*M[0,1] + F[1,1]*M[1,1];`` ` ` ``F[0,0] = x;`` ``F[0,1] = y;`` ``F[1,0] = z;`` ``F[1,1] = w;`` ``}` ` ``/* Helper function that calculates F[][]`` ``raise to the power n and puts the result`` ``in F[][] Note that this function is designed`` ``only for fib() and won't work as general`` ``power function */`` ``static` `void` `power(``int` `[,]F, ``int` `n)`` ``{`` ``int` `i;`` ``int` `[,]M = ``new` `int``[,]{{1, 1},`` ``{1, 0} };`` ` ` ``// n - 1 times multiply the matrix to`` ``// {{1,0},{0,1}}`` ``for` `(i = 2; i <= n; i++)`` ``multiply(F, M);`` ``}`` ` ` ``/* Driver program to test above function */`` ``public` `static` `void` `Main ()`` ``{`` ``int` `n = 9;`` ``Console.WriteLine(fib(n));`` ``}``}` `// This code is contributed by anuj_67.`
## PHP
``
Output
`34`
Time Complexity: O(n)
Extra Space: O(1)
Method 5 (Optimized Method 4)
The method 4 can be optimized to work in O(Logn) time complexity. We can do recursive multiplication to get power(M, n) in the previous method (Similar to the optimization done in this post)
## C++
`// Fibonacci Series using Optimized Method``#include ``using` `namespace` `std;` `void` `multiply(``int` `F[2][2], ``int` `M[2][2]);``void` `power(``int` `F[2][2], ``int` `n);` `// Function that returns nth Fibonacci number``int` `fib(``int` `n)``{`` ``int` `F[2][2] = {{1, 1}, {1, 0}};`` ``if` `(n == 0)`` ``return` `0;`` ``power(F, n - 1);` ` ``return` `F[0][0];``}` `// Optimized version of power() in method 4``void` `power(``int` `F[2][2], ``int` `n)``{`` ``if``(n == 0 || n == 1)`` ``return``;`` ``int` `M[2][2] = {{1, 1}, {1, 0}};`` ` ` ``power(F, n / 2);`` ``multiply(F, F);`` ` ` ``if` `(n % 2 != 0)`` ``multiply(F, M);``}` `void` `multiply(``int` `F[2][2], ``int` `M[2][2])``{`` ``int` `x = F[0][0] * M[0][0] + F[0][1] * M[1][0];`` ``int` `y = F[0][0] * M[0][1] + F[0][1] * M[1][1];`` ``int` `z = F[1][0] * M[0][0] + F[1][1] * M[1][0];`` ``int` `w = F[1][0] * M[0][1] + F[1][1] * M[1][1];`` ` ` ``F[0][0] = x;`` ``F[0][1] = y;`` ``F[1][0] = z;`` ``F[1][1] = w;``}` `// Driver code``int` `main()``{`` ``int` `n = 9;`` ` ` ``cout << fib(9);`` ``getchar``();`` ` ` ``return` `0;``}` `// This code is contributed by Nidhi_biet`
## C
`#include ` `void` `multiply(``int` `F[2][2], ``int` `M[2][2]);` `void` `power(``int` `F[2][2], ``int` `n);` `/* function that returns nth Fibonacci number */``int` `fib(``int` `n)``{`` ``int` `F[2][2] = {{1,1},{1,0}};`` ``if` `(n == 0)`` ``return` `0;`` ``power(F, n-1);`` ``return` `F[0][0];``}` `/* Optimized version of power() in method 4 */``void` `power(``int` `F[2][2], ``int` `n)``{`` ``if``( n == 0 || n == 1)`` ``return``;`` ``int` `M[2][2] = {{1,1},{1,0}};` ` ``power(F, n/2);`` ``multiply(F, F);` ` ``if` `(n%2 != 0)`` ``multiply(F, M);``}` `void` `multiply(``int` `F[2][2], ``int` `M[2][2])``{`` ``int` `x = F[0][0]*M[0][0] + F[0][1]*M[1][0];`` ``int` `y = F[0][0]*M[0][1] + F[0][1]*M[1][1];`` ``int` `z = F[1][0]*M[0][0] + F[1][1]*M[1][0];`` ``int` `w = F[1][0]*M[0][1] + F[1][1]*M[1][1];` ` ``F[0][0] = x;`` ``F[0][1] = y;`` ``F[1][0] = z;`` ``F[1][1] = w;``}` `/* Driver program to test above function */``int` `main()``{`` ``int` `n = 9;`` ``printf``(``"%d"``, fib(9));`` ``getchar``();`` ``return` `0;``}`
## Java
`//Fibonacci Series using Optimized Method``class` `fibonacci``{`` ``/* function that returns nth Fibonacci number */`` ``static` `int` `fib(``int` `n)`` ``{`` ``int` `F[][] = ``new` `int``[][]{{``1``,``1``},{``1``,``0``}};`` ``if` `(n == ``0``)`` ``return` `0``;`` ``power(F, n-``1``);`` ` ` ``return` `F[``0``][``0``];`` ``}`` ` ` ``static` `void` `multiply(``int` `F[][], ``int` `M[][])`` ``{`` ``int` `x = F[``0``][``0``]*M[``0``][``0``] + F[``0``][``1``]*M[``1``][``0``];`` ``int` `y = F[``0``][``0``]*M[``0``][``1``] + F[``0``][``1``]*M[``1``][``1``];`` ``int` `z = F[``1``][``0``]*M[``0``][``0``] + F[``1``][``1``]*M[``1``][``0``];`` ``int` `w = F[``1``][``0``]*M[``0``][``1``] + F[``1``][``1``]*M[``1``][``1``];`` ` ` ``F[``0``][``0``] = x;`` ``F[``0``][``1``] = y;`` ``F[``1``][``0``] = z;`` ``F[``1``][``1``] = w;`` ``}`` ` ` ``/* Optimized version of power() in method 4 */`` ``static` `void` `power(``int` `F[][], ``int` `n)`` ``{`` ``if``( n == ``0` `|| n == ``1``)`` ``return``;`` ``int` `M[][] = ``new` `int``[][]{{``1``,``1``},{``1``,``0``}};`` ` ` ``power(F, n/``2``);`` ``multiply(F, F);`` ` ` ``if` `(n%``2` `!= ``0``)`` ``multiply(F, M);`` ``}`` ` ` ``/* Driver program to test above function */`` ``public` `static` `void` `main (String args[])`` ``{`` ``int` `n = ``9``;`` ``System.out.println(fib(n));`` ``}``}``/* This code is contributed by Rajat Mishra */`
## Python3
`# Fibonacci Series using``# Optimized Method` `# function that returns nth``# Fibonacci number``def` `fib(n):`` ` ` ``F ``=` `[[``1``, ``1``],`` ``[``1``, ``0``]]`` ``if` `(n ``=``=` `0``):`` ``return` `0`` ``power(F, n ``-` `1``)`` ` ` ``return` `F[``0``][``0``]`` ` `def` `multiply(F, M):`` ` ` ``x ``=` `(F[``0``][``0``] ``*` `M[``0``][``0``] ``+`` ``F[``0``][``1``] ``*` `M[``1``][``0``])`` ``y ``=` `(F[``0``][``0``] ``*` `M[``0``][``1``] ``+`` ``F[``0``][``1``] ``*` `M[``1``][``1``])`` ``z ``=` `(F[``1``][``0``] ``*` `M[``0``][``0``] ``+`` ``F[``1``][``1``] ``*` `M[``1``][``0``])`` ``w ``=` `(F[``1``][``0``] ``*` `M[``0``][``1``] ``+`` ``F[``1``][``1``] ``*` `M[``1``][``1``])`` ` ` ``F[``0``][``0``] ``=` `x`` ``F[``0``][``1``] ``=` `y`` ``F[``1``][``0``] ``=` `z`` ``F[``1``][``1``] ``=` `w`` ` `# Optimized version of``# power() in method 4``def` `power(F, n):` ` ``if``( n ``=``=` `0` `or` `n ``=``=` `1``):`` ``return``;`` ``M ``=` `[[``1``, ``1``],`` ``[``1``, ``0``]];`` ` ` ``power(F, n ``/``/` `2``)`` ``multiply(F, F)`` ` ` ``if` `(n ``%` `2` `!``=` `0``):`` ``multiply(F, M)`` ` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:`` ``n ``=` `9`` ``print``(fib(n))` `# This code is contributed``# by ChitraNayal`
## C#
`// Fibonacci Series using``// Optimized Method``using` `System;` `class` `GFG``{``/* function that returns``nth Fibonacci number */``static` `int` `fib(``int` `n)``{``int``[,] F = ``new` `int``[,]{{1, 1},`` ``{1, 0}};``if` `(n == 0)`` ``return` `0;``power(F, n - 1);` `return` `F[0, 0];``}` `static` `void` `multiply(``int``[,] F,`` ``int``[,] M)``{``int` `x = F[0, 0] * M[0, 0] +`` ``F[0, 1] * M[1, 0];``int` `y = F[0, 0] * M[0, 1] +`` ``F[0, 1] * M[1, 1];``int` `z = F[1, 0] * M[0, 0] +`` ``F[1, 1] * M[1, 0];``int` `w = F[1, 0] * M[0, 1] +`` ``F[1, 1] * M[1, 1];` `F[0, 0] = x;``F[0, 1] = y;``F[1, 0] = z;``F[1, 1] = w;``}` `/* Optimized version of``power() in method 4 */``static` `void` `power(``int``[,] F, ``int` `n)``{``if``( n == 0 || n == 1)``return``;``int``[,] M = ``new` `int``[,]{{1, 1},`` ``{1, 0}};` `power(F, n / 2);``multiply(F, F);` `if` `(n % 2 != 0)``multiply(F, M);``}` `// Driver Code``public` `static` `void` `Main ()``{`` ``int` `n = 9;`` ``Console.Write(fib(n));``}``}` `// This code is contributed``// by ChitraNayal`
## Javascript
``
Output
`34`
Time Complexity: O(Logn)
Extra Space: O(Logn) if we consider the function call stack size, otherwise O(1).
Method 6 (O(Log n) Time)
Below is one more interesting recurrence formula that can be used to find n’th Fibonacci Number in O(Log n) time.
```If n is even then k = n/2:
F(n) = [2*F(k-1) + F(k)]*F(k)
If n is odd then k = (n + 1)/2
F(n) = F(k)*F(k) + F(k-1)*F(k-1)```
How does this formula work?
The formula can be derived from above matrix equation.
```Taking determinant on both sides, we get
(-1)n = Fn+1Fn-1 - Fn2
Moreover, since AnAm = An+m for any square matrix A,
the following identities can be derived (they are obtained
form two different coefficients of the matrix product)
FmFn + Fm-1Fn-1 = Fm+n-1 ---------------------------(1)
By putting n = n+1 in equation(1),
FmFn+1 + Fm-1Fn = Fm+n --------------------------(2)
Putting m = n in equation(1).
F2n-1 = Fn2 + Fn-12
Putting m = n in equation(2)
F2n = (Fn-1 + Fn+1)Fn = (2Fn-1 + Fn)Fn (Source: Wiki) --------
( By putting Fn+1 = Fn + Fn-1 )
To get the formula to be proved, we simply need to do the following
If n is even, we can put k = n/2
If n is odd, we can put k = (n+1)/2```
Below is the implementation of above idea.
## C++
`// C++ Program to find n'th fibonacci Number in``// with O(Log n) arithmatic operations``#include ``using` `namespace` `std;` `const` `int` `MAX = 1000;` `// Create an array for memoization``int` `f[MAX] = {0};` `// Returns n'th fuibonacci number using table f[]``int` `fib(``int` `n)``{`` ``// Base cases`` ``if` `(n == 0)`` ``return` `0;`` ``if` `(n == 1 || n == 2)`` ``return` `(f[n] = 1);` ` ``// If fib(n) is already computed`` ``if` `(f[n])`` ``return` `f[n];` ` ``int` `k = (n & 1)? (n+1)/2 : n/2;` ` ``// Applying above formula [Note value n&1 is 1`` ``// if n is odd, else 0.`` ``f[n] = (n & 1)? (fib(k)*fib(k) + fib(k-1)*fib(k-1))`` ``: (2*fib(k-1) + fib(k))*fib(k);` ` ``return` `f[n];``}` `/* Driver program to test above function */``int` `main()``{`` ``int` `n = 9;`` ``printf``(``"%d "``, fib(n));`` ``return` `0;``}`
## Java
`// Java Program to find n'th fibonacci``// Number with O(Log n) arithmetic operations``import` `java.util.*;` `class` `GFG {`` ` ` ``static` `int` `MAX = ``1000``;`` ``static` `int` `f[];`` ` ` ``// Returns n'th fibonacci number using`` ``// table f[]`` ``public` `static` `int` `fib(``int` `n)`` ``{`` ``// Base cases`` ``if` `(n == ``0``)`` ``return` `0``;`` ` ` ``if` `(n == ``1` `|| n == ``2``)`` ``return` `(f[n] = ``1``);`` ` ` ``// If fib(n) is already computed`` ``if` `(f[n] != ``0``)`` ``return` `f[n];`` ` ` ``int` `k = (n & ``1``) == ``1``? (n + ``1``) / ``2`` ``: n / ``2``;`` ` ` ``// Applying above formula [Note value`` ``// n&1 is 1 if n is odd, else 0.`` ``f[n] = (n & ``1``) == ``1``? (fib(k) * fib(k) +`` ``fib(k - ``1``) * fib(k - ``1``))`` ``: (``2` `* fib(k - ``1``) + fib(k))`` ``* fib(k);`` ` ` ``return` `f[n];`` ``}`` ` ` ``/* Driver program to test above function */`` ``public` `static` `void` `main(String[] args)`` ``{`` ``int` `n = ``9``;`` ``f= ``new` `int``[MAX];`` ``System.out.println(fib(n));`` ``}``}`` ` `// This code is contributed by Arnav Kr. Mandal.`
## Python
`# Python3 Program to find n'th fibonacci Number in``# with O(Log n) arithmatic operations``MAX` `=` `1000` `# Create an array for memoization``f ``=` `[``0``] ``*` `MAX` `# Returns n'th fuibonacci number using table f[]``def` `fib(n) :`` ``# Base cases`` ``if` `(n ``=``=` `0``) :`` ``return` `0`` ``if` `(n ``=``=` `1` `or` `n ``=``=` `2``) :`` ``f[n] ``=` `1`` ``return` `(f[n])` ` ``# If fib(n) is already computed`` ``if` `(f[n]) :`` ``return` `f[n]` ` ``if``( n & ``1``) :`` ``k ``=` `(n ``+` `1``) ``/``/` `2`` ``else` `:`` ``k ``=` `n ``/``/` `2` ` ``# Applying above formula [Note value n&1 is 1`` ``# if n is odd, else 0.`` ``if``((n & ``1``) ) :`` ``f[n] ``=` `(fib(k) ``*` `fib(k) ``+` `fib(k``-``1``) ``*` `fib(k``-``1``))`` ``else` `:`` ``f[n] ``=` `(``2``*``fib(k``-``1``) ``+` `fib(k))``*``fib(k)` ` ``return` `f[n]` `# Driver code``n ``=` `9``print``(fib(n))` `# This code is contributed by Nikita Tiwari.`
## C#
`// C# Program to find n'th``// fibonacci Number with``// O(Log n) arithmetic operations``using` `System;` `class` `GFG``{` `static` `int` `MAX = 1000;``static` `int``[] f;` `// Returns n'th fibonacci``// number using table f[]``public` `static` `int` `fib(``int` `n)``{`` ``// Base cases`` ``if` `(n == 0)`` ``return` `0;`` ` ` ``if` `(n == 1 || n == 2)`` ``return` `(f[n] = 1);` ` ``// If fib(n) is already`` ``// computed`` ``if` `(f[n] != 0)`` ``return` `f[n];` ` ``int` `k = (n & 1) == 1 ? (n + 1) / 2`` ``: n / 2;` ` ``// Applying above formula`` ``// [Note value n&1 is 1 if`` ``// n is odd, else 0.`` ``f[n] = (n & 1) == 1 ? (fib(k) * fib(k) +`` ``fib(k - 1) * fib(k - 1))`` ``: (2 * fib(k - 1) + fib(k)) *`` ``fib(k);` ` ``return` `f[n];``}` `// Driver Code``static` `void` `Main()``{`` ``int` `n = 9;`` ``f = ``new` `int``[MAX];`` ``Console.WriteLine(fib(n));``}``}` `// This code is contributed by mits`
## PHP
``
## Javascript
``
Output
`34`
Time complexity of this solution is O(Log n) as we divide the problem to half in every recursive call.
Method 7
Another approach(Using formula) :
In this method we directly implement the formula for nth term in the fibonacci series.
Fn = {[(√5 + 1)/2] ^ n} / √5
Reference: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormula.html
## C++
`// C++ Program to find n'th fibonacci Number``#include``#include` `int` `fib(``int` `n) {`` ``double` `phi = (1 + ``sqrt``(5)) / 2;`` ``return` `round(``pow``(phi, n) / ``sqrt``(5));``}` `// Driver Code``int` `main ()``{`` ``int` `n = 9;`` ``std::cout << fib(n) << std::endl;`` ``return` `0;``}``//This code is contributed by Lokesh Mohanty.`
## C
`// C Program to find n'th fibonacci Number``#include``#include``int` `fib(``int` `n) {`` ``double` `phi = (1 + ``sqrt``(5)) / 2;`` ``return` `round(``pow``(phi, n) / ``sqrt``(5));``}``int` `main ()``{`` ``int` `n = 9;`` ``printf``(``"%d"``, fib(n));`` ``return` `0;``}`
## Java
`// Java Program to find n'th fibonacci Number``import` `java.util.*;` `class` `GFG {` `static` `int` `fib(``int` `n) {``double` `phi = (``1` `+ Math.sqrt(``5``)) / ``2``;``return` `(``int``) Math.round(Math.pow(phi, n)`` ``/ Math.sqrt(``5``));``}` `// Driver Code``public` `static` `void` `main(String[] args) {`` ``int` `n = ``9``;`` ``System.out.println(fib(n));`` ``}``}``// This code is contributed by PrinciRaj1992`
## Python3
`# Python3 program to find n'th``# fibonacci Number``import` `math` `def` `fibo(n):`` ``phi ``=` `(``1` `+` `math.sqrt(``5``)) ``/` `2` ` ``return` `round``(``pow``(phi, n) ``/` `math.sqrt(``5``))`` ` `# Driver code ``if` `__name__ ``=``=` `'__main__'``:`` ` ` ``n ``=` `9`` ` ` ``print``(fibo(n))`` ` `# This code is contributed by prasun_parate`
## C#
`// C# Program to find n'th fibonacci Number``using` `System;` `public` `class` `GFG``{`` ``static` `int` `fib(``int` `n)`` ``{`` ``double` `phi = (1 + Math.Sqrt(5)) / 2;`` ``return` `(``int``) Math.Round(Math.Pow(phi, n)`` ``/ Math.Sqrt(5));`` ``}`` ` ` ``// Driver code`` ``public` `static` `void` `Main()`` ``{`` ``int` `n = 9;`` ``Console.WriteLine(fib(n));`` ``}``}` `// This code is contributed by 29AjayKumar`
## PHP
``
## Javascript
``
Output
`34`
Time Complexity: O(1)
Space Complexity: O(1)
Method 8
DP using memoization(Top down approach)
We can avoid the repeated work done is method 1 by storing the Fibonacci numbers calculated so far. We just need to store all the values in an array.
## C++
`#include ``using` `namespace` `std;``int` `dp[10];``int` `fib(``int` `n)``{`` ``if` `(n <= 1)`` ``return` `n;` ` ``// temporary variables to store`` ``// values of fib(n-1) & fib(n-2)`` ``int` `first, second;` ` ``if` `(dp[n - 1] != -1)`` ``first = dp[n - 1];`` ``else`` ``first = fib(n - 1);` ` ``if` `(dp[n - 2] != -1)`` ``second = dp[n - 2];`` ``else`` ``second = fib(n - 2);` ` ``// memoization`` ``return` `dp[n] = first + second;``}` `// Driver Code``int` `main()``{`` ``int` `n = 9;` ` ``memset``(dp, -1, ``sizeof``(dp));` ` ``cout << fib(n);`` ``getchar``();`` ``return` `0;` ` ``// This code is contributed by Bhavneet Singh``}`
## Java
`import` `java.util.*;` `class` `GFG{` `// Initialize array of dp``static` `int``[] dp = ``new` `int``[``10``];` `static` `int` `fib(``int` `n)``{`` ``if` `(n <= ``1``)`` ``return` `n;`` ` ` ``// Temporary variables to store`` ``// values of fib(n-1) & fib(n-2)`` ``int` `first, second;`` ` ` ``if` `(dp[n - ``1``] != -``1``)`` ``first = dp[n - ``1``];`` ``else`` ``first = fib(n - ``1``);` ` ``if` `(dp[n - ``2``] != -``1``)`` ``second = dp[n - ``2``];`` ``else`` ``second = fib(n - ``2``);` ` ``// Memoization`` ``return` `dp[n] = first + second;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{`` ``int` `n = ``9``;` ` ``Arrays.fill(dp, -``1``);` ` ``System.out.print(fib(n));``}``}` `// This code is contributed by sujitmeshram`
## Python3
`# Initialize array of dp``dp ``=` `[``-``1` `for` `i ``in` `range``(``10``)]` `def` `fib(n):`` ``if` `(n <``=` `1``):`` ``return` `n;`` ``global` `dp;`` ` ` ``# Temporary variables to store`` ``# values of fib(n-1) & fib(n-2)`` ``first ``=` `0``;`` ``second ``=` `0``;` ` ``if` `(dp[n ``-` `1``] !``=` `-``1``):`` ``first ``=` `dp[n ``-` `1``];`` ``else``:`` ``first ``=` `fib(n ``-` `1``);`` ``if` `(dp[n ``-` `2``] !``=` `-``1``):`` ``second ``=` `dp[n ``-` `2``];`` ``else``:`` ``second ``=` `fib(n ``-` `2``);`` ``dp[n] ``=` `first ``+` `second;` ` ``# Memoization`` ``return` `dp[n] ;` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:`` ``n ``=` `9``;`` ``print``(fib(n));` `# This code contributed by Rajput-Ji`
## C#
`using` `System;``class` `GFG {`` ` ` ``// Initialize array of dp`` ``static` `int``[] dp = ``new` `int``[10];`` ``static` `int` `fib(``int` `n)`` ``{`` ``if` `(n <= 1)`` ``return` `n;`` ` ` ``// Temporary variables to store`` ``// values of fib(n-1) & fib(n-2)`` ``int` `first, second;`` ` ` ``if` `(dp[n - 1] != -1)`` ``first = dp[n - 1];`` ``else`` ``first = fib(n - 1);`` ` ` ``if` `(dp[n - 2] != -1)`` ``second = dp[n - 2];`` ``else`` ``second = fib(n - 2);`` ` ` ``// Memoization`` ``return` `dp[n] = first + second;`` ``}` ` ``// Driver code`` ``static` `void` `Main()`` ``{`` ``int` `n = 9;`` ``Array.Fill(dp, -1);`` ``Console.Write(fib(n));`` ``}``}` `// This code is contributed by divyeshrabadiya07.`
## Javascript
``
Output
`34` | 15,480 | 37,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2021-17 | latest | en | 0.482917 |
https://oeis.org/A294274 | 1,586,515,879,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371893683.94/warc/CC-MAIN-20200410075105-20200410105605-00170.warc.gz | 598,282,693 | 3,841 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A294274 Sum of the seventh powers of the parts in the partitions of n into two parts. 1
0, 2, 129, 2444, 18700, 99012, 376761, 1216688, 3297456, 8158550, 18080425, 37847532, 73399404, 136971464, 241561425, 414517952, 680856256, 1095977898, 1703414961, 2607286700, 3877286700, 5697862412, 8172733129, 11613390384, 16164030000, 22330294142 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS Colin Barker, Table of n, a(n) for n = 1..1000 Index entries for linear recurrences with constant coefficients, signature (1,8,-8,-28,28,56,-56,-70,70,56,-56,-28,28,8,-8,-1,1). FORMULA a(n) = Sum_{i=1..floor(n/2)} i^7 + (n-i)^7. From Colin Barker, Nov 20 2017: (Start) G.f.: x^2*(2 + 127*x + 2299*x^2 + 15240*x^3 + 61848*x^4 + 151257*x^5 + 262139*x^6 + 306832*x^7 + 260914*x^8 + 151257*x^9 + 60777*x^10 + 15240*x^11 + 2180*x^12 + 127*x^13+ x^14) / ((1 - x)^9*(1 + x)^8). a(n) = a(n-1) + 8*a(n-2) - 8*a(n-3) - 28*a(n-4) + 28*a(n-5) + 56*a(n-6) - 56*a(n-7) - 70*a(n-8) + 70*a(n-9) + 56*a(n-10) - 56*a(n-11) - 28*a(n-12) + 28*a(n-13) + 8*a(n-14) - 8*a(n-15) - a(n-16) + a(n-17) for n>17. (End) MATHEMATICA Table[Sum[i^7 + (n - i)^7, {i, Floor[n/2]}], {n, 40}] PROG (PARI) concat(0, Vec(x^2*(2 + 127*x + 2299*x^2 + 15240*x^3 + 61848*x^4 + 151257*x^5 + 262139*x^6 + 306832*x^7 + 260914*x^8 + 151257*x^9 + 60777*x^10 + 15240*x^11 + 2180*x^12 + 127*x^13+ x^14) / ((1 - x)^9*(1 + x)^8) + O(x^40))) \\ Colin Barker, Nov 20 2017 CROSSREFS Sequence in context: A303377 A258806 A216358 * A099824 A300292 A259109 Adjacent sequences: A294271 A294272 A294273 * A294275 A294276 A294277 KEYWORD nonn,easy AUTHOR Wesley Ivan Hurt, Oct 26 2017 STATUS approved
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Last modified April 10 06:23 EDT 2020. Contains 333392 sequences. (Running on oeis4.) | 903 | 2,174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-16 | latest | en | 0.412736 |
https://carab.in/probability-statistics-master/ | 1,571,698,463,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987795253.70/warc/CC-MAIN-20191021221245-20191022004745-00124.warc.gz | 425,094,448 | 13,049 | # Become a Probability & Statistics Master
Become a Probability & Statistics Master
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Learn everything from Probability & Statistics, then test your knowledge with 600+ practice questions
HOW BECOME A PROBABILITY & STATISTICS MASTER IS SET UP TO MAKE COMPLICATED MATH EASY:
This course includes video and text explanations of everything from Probability and Statistics, and it includes quizzes (with solutions!) and an additional 8 workbooks with extra practice problems, to help you test your understanding along the way. Become a Probability & Statistics Master is organized into the following sections:
Visualizing data, including bar graphs, pie charts, Venn diagrams, histograms, and dot plots
Analyzing data, including mean, median, and mode, plus range and IQR and box-and-whisker plots
Data distributions, including mean, variance, and standard deviation, and normal distributions and z-scores
Probability, including union vs. intersection and independent and dependent events and Bayes’ theorem
Discrete random variables, including binomial, Bernoulli, Poisson, and geometric random variables
Sampling, including types of studies, bias, and sampling distribution of the sample mean or sample proportion, and confidence intervals
Hypothesis testing, including inferential statistics, significance levels, type I and II errors, test statistics, and p-values
Regression, including scatterplots, correlation coefficient, the residual, coefficient of determination, RMSE, and chi-square
AND HERE’S WHAT YOU GET INSIDE OF EVERY SECTION:
Videos: Watch over my shoulder as I solve problems for every single math issue you’ll encounter in class. We start from the beginning… I explain the problem setup and why I set it up that way, the steps I take and why I take them, how to work through the yucky, fuzzy middle parts, and how to simplify the answer when you get it.
Notes: The notes section of each lesson is where you find the most important things to remember. It’s like Cliff Notes for books, but for math. Everything you need to know to pass your class and nothing you don’t.
Quizzes: When you think you’ve got a good grasp on a topic within a course, you can test your knowledge by taking one of the quizzes. If you pass, great! If not, you can review the videos and notes again or ask for help in the Q&A section.
Workbooks: Want even more practice? When you’ve finished the section, you can review everything you’ve learned by working through the bonus workbook. The workbooks include tons of extra practice problems, so they’re a great way to solidify what you just learned in that section.
What you’ll learn
• Visualizing data, including bar graphs, pie charts, venn diagrams, histograms, and dot plots
• Analyzing data, including mean, median, and mode, plus range and IQR and box-and-whisker plots
• Data distributions, including mean, variance, and standard deviation, and normal distributions and z-scores
• Probability, including union vs. intersection and independent and dependent events and Bayes’ theorem
• Discrete random variables, including binomial, Bernoulli, Poisson, and geometric random variables
• Sampling, including types of studies, bias, and sampling distribution of the sample mean or sample proportion, and confidence intervals
• Hypothesis testing, including inferential statistics, significance level, type I and II errors, test statistics, and p-values
• Regression, including scatterplots, correlation coefficient, the residual, coefficient of determination, RMSE, and chi-square
Getting started
1 Hi! START HERE: Course overview
Visualizing data
3 Introduction to visualizing data
4 RESOURCE: Quiz solutions for this section
5 One-way data
6 One-way data
7 Bar graphs and pie charts
8 Bar graphs and pie charts
9 Line graphs and ogives
10 Line graphs and ogives
11 Two-way data
12 Two-way data
13 Venn diagrams
14 Venn diagrams
15 Relative frequency tables
16 Relative frequency tables
17 Relative frequency tables
18 Joint distributions
19 Joint distributions
20 Frequency tables and dot plots
21 Frequency tables and dot plots
22 Histograms and stem-and-leaf plots
23 Histograms and stem-and-leaf plots
24 BONUS! Extra practice problems.
Analyzing data
25 Introduction to analyzing data
26 RESOURCE: Quiz solutions for this section
27 Central tendency: mean, median and mode
28 Central tendency: mean, median and mode
31 Changing the data, and outliers
32 Changing the data, and outliers
33 Box-and-whisker plots
34 Box-and-whisker plots
35 BONUS! Extra practice problems.
Data distributions
36 Introduction to data distributions
37 RESOURCE: Quiz solutions for this section
38 Mean, variance, and standard deviation
39 Mean, variance, and standard deviation
40 Frequency histograms and polygons, and density curves
41 Frequency histograms and polygons, and density curves
42 Symmetric and skewed distributions and outliers
43 Symmetric and skewed distributions and outliers
44 Normal distributions and z-scores
45 Normal distributions and z-scores
46 BONUS! Extra practice problems.
Probability
47 Introduction to probability
48 RESOURCE: Quiz solutions for this section
49 Simple probability
50 Simple probability
51 The addition rule, and union vs. intersection
52 The addition rule, and union vs. intersection
53 Independent and dependent events and conditional probability
54 Independent and dependent events and conditional probability
55 Bayes’ theorem
56 BONUS! Extra practice problems.
Discrete random variables
57 Introduction to discrete random variables
58 RESOURCE: Quiz solutions for this section
59 Discrete probability
60 Discrete probability
61 Transforming random variables
62 Transforming random variables
63 Combinations of random variables
64 Combinations of random variables
65 Permutations and combinations
66 Permutations and combinations
67 Binomial random variables
68 Binomial random variables
69 Poisson distributions
70 Poisson distributions
71 “At least” and “at most,” and mean, variance, and standard deviation
72 “At least” and “at most,” and mean, variance, and standard deviation
73 Bernoulli random variables
74 Bernoulli random variables
75 Geometric random variables
76 Geometric random variables
77 BONUS! Extra practice problems.
Sampling
78 Introduction to sampling
79 RESOURCE: Quiz solutions for this section
80 Types of studies
81 Types of studies
82 Sampling and bias
83 Sampling and bias
84 Sampling distribution of the sample mean
85 Sampling distribution of the sample mean
86 Sampling distribution of the sample proportion
87 Sampling distribution of the sample proportion
88 Confidence interval for a population mean
89 Confidence interval for a population mean
90 Confidence interval for a population proportion
91 Confidence interval for a population proportion
92 BONUS! Extra practice problems.
Hypothesis testing
93 Introduction to hypothesis testing
94 RESOURCE: Quiz solutions for this section
95 Inferential statistics and hypotheses
96 Inferential statistics and hypotheses
97 Significance level and type I and II errors
98 Significance level and type I and II errors
99 Test statistics for one- and two-tailed tests
100 Test statistics for one- and two-tailed tests
101 The p-value and rejecting the null
102 The p-value and rejecting the null
103 Hypothesis testing for the population proportion
104 Hypothesis testing for the population proportion
105 BONUS! Extra practice problems.
Regression
106 Introduction to regression
107 RESOURCE: Quiz solutions for this section
108 Scatterplots and regression
109 Scatterplots and regression
110 Correlation coefficient and the residual
111 Correlation coefficient and the residual
112 Coefficient of determination and root-mean-square error
113 Coefficient of determination and root-mean-square error
114 Chi-square tests
115 Chi-square tests
116 BONUS! Extra practice problems.
Final exam and wrap-up
117 Probability & Statistics final exam
118 Wrap-up | 1,670 | 7,991 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-43 | latest | en | 0.891245 |
https://www.coursehero.com/file/6136388/Discrete-Practice-Test-1/ | 1,513,590,720,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948612570.86/warc/CC-MAIN-20171218083356-20171218105356-00270.warc.gz | 762,003,198 | 49,681 | Discrete Practice Test 1
# Discrete Practice Test 1 - lentβ you must give an...
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September 17, 2009 Name: Test 1 Math 3336 You have the full class period to complete the test. You cannot use any books or notes. This test is worth 250 points. 1. 50 pts. Prove or disprove whether the formula is a tautology or not: (a) ( p q ) ( ¬ p r ) ( q r ) (b) ( p r ) ( q r ) ( p q ) r (c) ( p r ) ( q r ) ( p q ) r (d) p ( p q ) (e) p ( q r ) ( p q ) r 2. 30 pts. But each wrong answer carries a penalty of -5 pts. Mark as true or false. The implication If P, then Q is equivalent to: a) P is sufficient for Q. b) Q is sufficient for P. c) P is necessary for Q. d) Q is necessary for P. e) P if Q. f) Q only if P. 3. 30 pts. But each wrong answer carries a penalty of -5 pts. Determine whether the following arguments are valid or invalid. (a) If a = b then c = d . But a 6 = b . Thus c 6 = d . (b) If a = b then c = d . But c 6 = d . Thus a 6 = b . (c) a = b only if c = d . One has that c = d . Thus a = b . 4. 40 pts. Find the conjunctive and disjunctive normalform for p q . 5. 30 pts. Decide whether the following formulas are equivalent. In case where your answer is ”not equiva-
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Unformatted text preview: lentβ you must give an explanation. (a) β x ( Q ( x ) β§ P ( x )) and β xQ ( x ) β§ β xP ( x ) (b) β x ( Q ( x ) β¨ P ( x )) and β xQ ( x ) β¨ β xP ( x ) (c) β x ( Q ( x ) β P ( x )) and β xQ ( x ) β β xP ( x ) 6. 40 pts. Let L ( x, y ) be the predicate for β x likes to buy y β and let S ( y ) stand for β y is on saleβ. Formalize: (a) Everybody likes to buy something, but only if it is on sale. (b) There is something everybody likes to buy if it is on sale. (c) If everything is on sale then everybody likes to buy everything. (d) There is somebody who likes to buy everything if it is on sale. 7. 30 pts. Let N be the set of natural numbers and let n | m stand for that n divides m . Formalize: For every n and m there is some g such that g divides n and m, and if q divides n and m, then q divides g....
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