Datasets:

HuggingFaceTB

/

finemath

like 160

Follow

Hugging Face TB Research 751

https://electrical-engineering-portal.com/download-center/books-and-guides/relays/centralized-protection

text/html

crawl-data/CC-MAIN-2023-50/segments/1700679100489.16/warc/CC-MAIN-20231203062445-20231203092445-00885.warc.gz

CYBER WEEK OFFER 💥 Save 20% on PRO Membership Plan and Video Courses with the coupon CYM23 and learn from experienced engineers. # Centralized substation protection and control system for HV/MV substation Home / Download Center / Electrical Engineering Books and Technical Guides / Relay control and protection guides / Centralized substation protection and control system for HV/MV substation ## The substation in the study The substation is one of seven substations in the surrounding grid. The grid is separated from the rest of the regional grid and its small size and autonomy has made it attractive for testing of the centralized protection and control system architecture. The grid has a partially radial structure and is fed at two points, one for normal operation and the other as backup. It has three voltage levels: 70 kV, 20 kV and 10 kV. A schematic of the grid is presented in figure below. The simple geometry of the grid means that, should the implementation of the new system into a single substation be deemed successful, the new system architecture could easily be tested on a grid where all of the substations were implemented with the new system. The types of consumers in the grid are for the most part residential, with some larger plants and factories. Thus, the grid is made up of several, small consumers rather than a few, heavy loads. This has the consequence that the load profile varies considerably during the course of a day, as the grid voltage spikes during mornings and evenings and drops during mid days and nights. A grid with a wider distribution of consumers would have smaller variations, as the factories and offices where the residents go to work each day would compensate for the voltage loss during the middle of the day. This consumer profile can potentially pose challenges for the voltage regulation of the substations in the grid as they must compensate for the variance in grid voltage several times a day. ### Single line diagram Figure below shows a single line diagram of the substation. It is a compact way of representing a complex coupling scheme by way of symbols for common power system components such as transformers and power lines. It also has the advantage of being capable of representing a three phase circuit as having only one phase. The main task of the substation is to transform incoming voltage from the transmission system at 70 kV to 10 kV for further urban distribution. As can be seen in the diagram, the incoming electricity is passed through a system of measuring and protection devices before encountering the transformers. The 20 MVA transformers transform voltage from 70 kV to 10 kV before passing it on to a busbar with 18 outgoing lines which distributes it to the grid. 14 of these lead to neighbourhood consumers, two distribute the voltage to secondary substations and the last two are a back up connections to a primary substation. Two capacitor banks with reactive loads at 2 and 3 Mvar are connected to the busbar to compensate for the reactive effect of the system. The transformers neutral is connected to reactive and resistive loads for compensating earth fault currents, as well as removing the risk for arcing earth faults. Title: Centralized substation protection and control system for HV/MV substation – ENERGY NETWORKS ASSOCIATION Format: PDF Size: 6.0 MB Pages: 81 Download: Right here | Video Courses | Membership | Download Updates

{"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}

CC-MAIN-2023-50

latest

en

http://www.jiskha.com/display.cgi?id=1299497981

text/html

crawl-data/CC-MAIN-2017-26/segments/1498128319902.52/warc/CC-MAIN-20170622201826-20170622221826-00585.warc.gz

posted by on . 1.) which of the following represents dy/dx when y=e^-2x Sec(3x)? A.3e^-2x sec(3x) tan (3x)-2e^-2x Sec(3x)<<<< my choice B.)3e^-2x sec(3x)tan (3x)-2xe^-2x sec(3x) C.)3e^-2x sec(3x)tan (x)-2e^-2x Sec(3x) D.)3e^-2x sec(3x)tan (x)-2xe^-2x sec(3x) 2.)calculate dy/dx if y=Ln(2x^3+3x) A.)1/2x^3+3x B.)1/6x^2+3 C.)2x^3+3x/6x^2+3 D.)6x^2+3/2x^3+3x <<<< my choice 3.)Given the curve that is described by the equation r=3 cos theta, find the angle that tangent lines makes with the radius vector when theta=120. A.) 30deg B.) 45deg C.) 60deg D.) 90deg << my choice 4.)A line rotates in a horizontal plane according to the equation theta=2^t-6t,where theta is the angular position of the rotating line, in radians ,and tis the time,in seconds. Determine the angular acceleration when t=2sec.

{"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}

CC-MAIN-2017-26

latest

en

https://pythonmana.com/2021/11/20211125094544874W.html

text/html

crawl-data/CC-MAIN-2022-27/segments/1656103516990.28/warc/CC-MAIN-20220628111602-20220628141602-00679.warc.gz

# RC4 Python implementation I'm cooking 2021-11-25 09:45:59 rc4 rc python implementation RC4 The algorithm is a stream cipher （ Byte oriented operations ）, Key length is variable （ Generate the key stream according to the plaintext length ） RC4 Encryption process ： 1, Initialize the status array S： length 256 byte , Every S[i] by 1 Bytes ,S[i] The value of is 0-255 2, Initialize temporary array T：T The length of is 256 byte , Every T[i] by 1 Bytes , First, generate a key array Key, Key The length of is 1~256 byte ( Random ), Every Key[i] by 1 Bytes , And each Key[i] The value of is randomly generated ; And then the array Key Each of the Key[i] Take turns and loop into the array T, such as Key[]=1,2,3,4,5 , Then put it into the array T：T[] =1,2,3,4,5,1,2,3,4,5…1,2,3,4,5,1,2 Will array T fill , This array T Finished initializing , And random ; 3, Initial permutation array S ： After initialization T, And a variable j=0, Scramble the array S, send S It's random ： Through for each S[i], adopt j = (j + S[i] + T[I]) mod 256 , In exchange for S[i] and S[j] Value ; My understanding of this step ： Because the initialized array T It's random , So it will T Element is added to get S[j] And replace S[i] and S[j] , Make the array S Pseudorandom ; 4, Generate key stream KeyStream： First, get the plaintext length , Decide to generate a key stream of the same length as plaintext , Through to each S[i],j=0,j=(j+S[i]) mod 256 take S[i] And S[j] substitution , When i>255, go back to S[0], Repeat the replacement operation again : j,t = 0,0 for i in range(txtlen): // txtlen Is the length of the plaintext , Through the loop , Generate equal length key stream i = i mod 256 // adopt mod 256 , Ensure that subsequent pairs of arrays S Repetitive operation j = (j+S[i]) mod 256 tmp = S[i] S[i] = S[j] // substitution S[i] And S[j] S[j] = tmp t = (S[i] + S[j]) mod 256 KeyStream.append(S[t]) //S[t] by A bit in the key stream My understanding of this step ： Suppose a box is called S, There is a row of 256 One says from 0-255 A number ball , And this 256 The ball was disrupted earlier , Now you need to start from S The first one in the box （i）, And through the formula j = (j+S[i]) mod 256 What you get is j individual , Write down the number on the ball and pass the formula t = (S[i] + S[j]) mod 256 Get number t, Then you go and find out the second t A little ball , The number on it S[t] Is a bit in the key stream , And you put the first （i） The ball and the second j Exchange a small ball and put it back S box , Then repeat the steps just now to touch the second （i） And the j And so on , Until you repeat this step txtlen Time ; notes ： When you touch the first 256 individual （ The last one ）, But you haven't finished yet txtlen Time , So the next time you touch the ball is to go back to the first start , This is the corresponding pair array S Repetitive operation ; Attach reference code , If there is any wrong , I would appreciate your comments. ^-^： ``````#encoding = utf-8 import random # Initialization vector S def init_S(): global S for i in range(256): S.append(i) # Initialization vector T def init_T(): global Key,T keylen = random.randint(1,256) for i in range(keylen): index = random.randint(0,61) # Get a random temporary key Key Key.append(WordList[index]) for i in range(256): tmp = Key[i % keylen] # Initialization vector T T.append(tmp) # The initial displacement S[i] def swap_S(): global S,T j = 0 for i in range(256): j = (j+S[i]+ord(T[i])) % 256 tmp = S[i] S[i] = S[j] S[j] = tmp # Key stream generation def Get_KeyStream(): global S,text,KeyStream txtlen = len(text) j,t = 0,0 for i in range(txtlen): i = i % 256 j = (j+S[i]) % 256 tmp = S[i] S[i] = S[j] S[j] = tmp t = (S[i] + S[j]) % 256 KeyStream.append(S[t]) # encryption & Decrypt def Get_code(): global PlainText,CryptoText,KeyStream,text for i in range(len(text)): CryptoText += chr(ord(text[i]) ^ KeyStream[i]) # encryption for i in range(len(text)): PlainText += chr(ord(CryptoText[i]) ^ KeyStream[i]) # Decrypt print('''[+]================= Began to encrypt ==================''') print('''[+]================= Start decrypting ==================''') if __name__ == '__main__' : T,S,Key = [],[],[] PlainText,CryptoText,KeyStream = '','',[] WordList = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789" # Used to generate temporary keys Key print(''' ——————— ———————— —————— | —— \ / _____| / | | | |__) | | / / —— | | | __ / | | / |_| | | ^__^ is easy ~, that right ? | | \ \ | |_____ |______| | |__| \__\ \________| |_| ██████╗ ██████╗██╗ ██╗ ███╗ ███╗ ██╗███████╗ ███████╗ █████╗ ███████╗██╗ ██╗ ██╔══██╗██╔════╝██║ ██║ ██╔██╗ ██╔██╗ ██║██╔════╝ ██╔════╝██╔══██╗██╔════╝╚██╗ ██╔╝ ██████╔╝██║ ███████║ ╚═╝╚═╝ ╚═╝╚═╝ ██║███████╗ █████╗ ███████║███████╗ ╚████╔╝ ██╔══██╗██║ ╚════██║ ██║╚════██║ ██╔══╝ ██╔══██║╚════██║ ╚██╔╝ ██║ ██║╚██████╗ ██║ ███████╗ ██║███████║ ███████╗██║ ██║███████║ ██║ ╚═╝ ╚═╝ ╚═════╝ ╚═╝ ╚══════╝ ╚═╝╚══════╝ ╚══════╝╚═╝ ╚═╝╚══════╝ ╚═╝ ''') text = input("[+]please input you Plaintext here : ") init_S() init_T() swap_S() Get_KeyStream() Get_code() `````` Screenshot of the code running correctly ： https://pythonmana.com/2021/11/20211109005248956u.html

{"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}

CC-MAIN-2022-27

latest

en

https://web2.0calc.com/questions/few-geo-questions-please-help

text/html

crawl-data/CC-MAIN-2020-40/segments/1600400221980.49/warc/CC-MAIN-20200925021647-20200925051647-00730.warc.gz

+0 0 127 4 1. A quarter-circle with radius $$5$$ is drawn. A circle is drawn inside the sector, which is tangent to the sides of the sector, as shown. Find the radius of the inscribed circle. 2. The sides of triangle $$XYZ$$ are $$XY = XZ = 25$$ and $$YZ=40.$$ A semicircle is inscribed in triangle $$XYZ$$ so that its diameter lies on $$\overline{YZ}$$, and is tangent to the other two sides. Find the area of the semicircle. Thank you very much for your help! May 25, 2020 #1 0 Anyone?? May 25, 2020 #2 +21953 0 #2)  Drop a perpendicular from X to YZ. Call this point O. O will be the midpoint of YZ. Since YZ = 40, OZ = 20. Use the Pythagorean Theorem on right trianle(XOZ) --->   XO2 + OZ2  =  XZ2   --->   XO2 + 202  =  252   --->   XO  =  15 Draw a perpendicular from O to XZ. Call the point of intersection P. Since this is perpendicular to XZ, this is the point of tangency to the semicircle. Consider triangle(OPZ); this right triangle is similar to triangle((XOZ) by AA. Therefore:  OP / OZ  =  XO / XZ. OP / 20  =  15 / 25   --->   OP  =  20 · 15 / 25   --->   OP  =  12. OP is the radius of the semicircle; now use the area formula: area semicircle  =  ½ · pi · radius2   to find the area of the semicircle. May 25, 2020 #3 +21953 0 #1)  Draw a line segment from O to the point of tangency of te quarter-circle (call that point T). The center of the circle will be on OT. Call the cdnter of the circle C. Drop perpendiculars from C to AO (call that point X) and from C to OB (call tht point Y). The distances from C to T, from C to X, and from C to Y are all equal and they are radii of the smaller circle. Let the radius of the smaller circle be x. All the distances:  XO, YO, CX, CY, and OT are equal to x. Since OXCY is a square with side = x, the distance from O to C  =  x·sqrt(2). The distance from  OT  =  OC + CT OT  =  x·sqrt(2) + x  =  x(sqrt(2) + 1) But, OT is a radius of the quarter-circle  =  5. 5 =  x(sqrt(2) + 1) x  =  5 / (sqrt(2) + 1) May 25, 2020 #4 0 Thank you. #1 was wrong though. Any ideas on where you went wrong? It's fine if you don't I might be able to figure it out. May 25, 2020

{"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}

CC-MAIN-2020-40

latest

en

http://perplexus.info/show.php?pid=10346&cid=56671

text/html

crawl-data/CC-MAIN-2021-21/segments/1620243989914.60/warc/CC-MAIN-20210516201947-20210516231947-00066.warc.gz

All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars perplexus dot info More bouncing (Posted on 2016-03-09) Place an isosceles right triangle with its right angle at the origin and legs along the positive x- and y-axes. Imagine this triangle is three mirrors and a laser is fired from the origin with some positive slope. The laser will bounce around and one of 3 possibilities will occur: 1. The laser will eventually come back to the origin. 2. The laser will eventually hit one of the other corners. 3. The laser will never hit any of the corners. For a given slope, how can you tell which will happen? No Solution Yet Submitted by Jer No Rating Comments: ( Back to comment list | You must be logged in to post comments.) proposed solution | Comment 1 of 3 As in Light Beam Reflection, a tesselation of the plane will allow treating the light beam as one straight ray. The triangle is reflected about each side to form extra triangles, and this process continues with each successive triangle so created. Consider the original triangle to have legs of unit length, with a hypotenuse of sqrt(2).  Each 2x2 block of the tesselation will contain 8 triangles; or you can consider it divided into 4 squares, overlaid with a diagonally oriented square with sides of length sqrt(2). If the slope is irrational, the beam will never hit a corner. If the slope is rational, reduce it to the lowest possible terms, such as 1/2, 1/3, 7/5, 8/7, etc. It looks, from a drawing on graph paper, that, if the numerator or denominator is a multiple of 2, then the beam eventually hits a different corner from the origin. If neither the numerator nor denominator is a multiple of 2, then it returns to the origin. Edited on March 10, 2016, 7:34 am Posted by Charlie on 2016-03-09 14:28:48 Search: Search body: Forums (0)

{"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}

CC-MAIN-2021-21

latest

en

https://tutorbin.com/questions-and-answers/a-what-is-the-mathematical-equation-for-the-ideal-gas-law-identify-eac

text/html

crawl-data/CC-MAIN-2022-27/segments/1656103640328.37/warc/CC-MAIN-20220629150145-20220629180145-00198.warc.gz

Question General Chemistry a. What is the mathematical equation for the ideal gas law? Identify each variable and giveits units. (2 points) b. Match the special cases of each gas law with its description. A law may be used morethan once. In the equations, k is a constant. (3 points) B. Charles's lawC. Avogadro's lawD. Dalton's law P_{1} V_{1}=P_{2} V_{2} V / T=k P V=k P_{\text {total }}=P_{1}+P_{2}+P_{3}+ V_{1} / T_{1}=V_{2} / T_{2} V=k n c. Conditions at STP (2 points) i. What is the temperature and pressure at STP? (1 point) ii. What is the volume of 1 mole of any kind of gas at STP? (1 point) Verified ### Question 45009 General Chemistry 15.9 g of CaCO3 decomposed and 7.31 g of CaO are obtained what is the percent yield of CaO obtained \mathrm{CaCO3}(\mathrm{s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO} 2(\mathrm{~g}) ### Question 45008 General Chemistry (20). After completing a titration, you determined 0.52 mol of sodium hydroxide were required to reach the endpoint of sulfuric acid sample of unknown concentration. Your acid sample was 45 mL. What is the concentration of your acid? (Hint: is the reaction balanced?) \mathrm{H} 2 \mathrm{SO} 4(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{Na2SO4}(\mathrm{aq})+\mathrm{H} 2 \mathrm{O}(1) ### Question 45007 General Chemistry (18). Find the mass of a metal object (in grams) that was heated to 105°C and placed in 27 g of water at 13°C The final temperature of the water is 40°C. The C water = 4.184 J/g°C and C metal = 0.184 J/g°C. ### Question 45006 General Chemistry (17). Determine the molarity of a solution formed by dissolving 48.9 g LiBr in enough water to yield a total solution volume of 750.0 mL. ### Question 45005 General Chemistry (12). A metal cube has an edge length of 2.44 cm and a mass of 16.28 g. Calculate the density of the metal. ### Question 44215 General Chemistry A lab technician had a sample of radioactive Americium. He knew it was one of the isotopeslisted in the table below but did not know which one. The sample originally contained 0.008g of the isotope, but 120 minutes later it contained 0.004 g. Use this information to answerthe following questions. (3 points) A. How many half-lives have passed? B. What is the half-life of the sample? C. Which isotope did he have? ( ### Question 44214 General Chemistry A. Name three kinds of radioactive decay and provide the symbol for each that is used to represent the product given off. (3 points) \text { i. } a \text { - decay (1 point) } \text { iii. Y - decay (1 point) } \text { ii. } \beta \text { - decay (1 point) } { }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow \frac{3}{2} \mathrm{He}+{ }_{0}^{1} \mathrm{n} ### Question 44213 General Chemistry A. What two competing fundamental forces are at work inside the nucleus of an atom? (1point) B. How can the ratio of number of protons to neutrons in the nucleus be used to predict thestability of a nucleus? (2 points) C. How much energy would be released if 0.00023 kg of plutoniumthrough radioactive decay? (2 points)were converted to energy ### Question 44212 General Chemistry Use the emission spectra shown to answer the following questions (note the direction of thewavelength scale): A. A high school student acquired an emission spectrum of an unknown sample. He knewthe unknown sample contained one of the elements, whose spectra are shown above. Theemission spectrum of his sample showed a strong emission at 610 nm but not at 480 nm.Which element did his unknown sample contain? (2 points) B. Element A above has a strong emission around 670 nm. Does this emission line representa lower energy or higher energy transition than the emission line at 428 nm? Explain youranswer. (2 points) C. Why does each element have its own unique emission spectrum? (2 points) ### Question 44211 General Chemistry A. Write a set of four possible quantum numbers for the circled electron. (There could bemore than one correct answer.) (4 points) n =| =m, =II B. The electron in a hydrogen atom can undergo a transition from n = 6 to n = 1, emitting aphoton with energy 2.11 x 10-18 J. (2 points) i. What is the frequency of this transition? (1 point) ii. How does this transition show that the energy of a photon is quantized? (1 point) C. Why is it impossible for an electron to have the quantum numbers n = 3, | = 0, m, = 1, m, = +\frac{1}{2} ?(2 \text { points })

{"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}

CC-MAIN-2022-27

latest

en

https://www.jiskha.com/display.cgi?id=1261761579

text/html

crawl-data/CC-MAIN-2017-30/segments/1500549428300.23/warc/CC-MAIN-20170727142514-20170727162514-00232.warc.gz

# Algebra posted by . define the following new variables X to A =W X to B =X Y to A =Y Y to B =Z w+x=32 y+z=8 w+y=22 x+z=18 You helped me solve for y Ijust added w+y=32 220w+300(32-w)+400(22-w)+180(8-y)=9280 , w+y=22 The answer I got was y=32 I solved for x and got x=42 I don't know if this is right. I am trying to solve for w and z but keep coming up with fractions. I am using the substitution method to calculate. Thank you Can you help me understand. • Algebra - It is not clear to me what this parts means: "define the following new variables X to A =W X to B =X Y to A =Y Y to B =Z " However, if we look at the system of 4x4 equations: w+x=32 y+z=8 w+y=22 x+z=18 Whichever method you use, you will find that one of the equations is a linear combination of the other three, and there is an infinite number of solutions, depending on the value of the "free variable". If we take the "free variable" as z, then we can express the solution in terms of z, as follows: w=t+14, x=18-t, y=8-t, z=t ## Similar Questions 1. ### Algebra Word Problems Barges from ports X and Y went to cities A and B. X sent 32 barges and Y sent 8. City A received 22 barges and B received 18 Shipping costs \$220 from X to A, \$300 from X to B, \$400 from Y to A, and \$180 from Y to B \$9280 was spent … 2. ### Algebra w+x=32 y+z=8 w+y=22 x+z=18 220w+300(32-w)+400(22-w)+180(8-y)=9280 , w+y=22 I was able to solve for w and y but keep coming up with fractions for x and z 3. ### Algebra Can someone check my work for the following? 4. ### math Barges from ports X and Y went to cities A and B. X sent 32 barges and Y sent 8. City A needs 22 barges and B needs 18. Shipping costs \$220 from X to A, \$300 from X to B, \$400 from Y to A, and \$180 from Y to B. \$9280 was spent. How … 5. ### accounting 1. Please refer to the worksheet below. Trial Balance Adjustments Adjusted Trial Balance Income Statement Balance Sheet Account Title Dr. Cr. Dr. Cr. Dr. Cr. Dr. Cr. Dr. Cr. Cash \$1,700 \$1,700 \$1,700 Accounts receivable 8,500 \$2,000 … 6. ### Algebra 2 Solve for B: P.S. How do you solve for variables? 7. ### math An auto manufacturer sends cars from two plants, I and II, to dealerships A and B located in a Midwestern city. Plant I has a total of 28 cars to send, and plant II has 8. Dealer A needs 20 cars, and dealer B needs 16. Transportation … 8. ### Electrophoresis You need to prepare 400 ml of a 4M glucose solution. It's chemical formula is C6H12O6 and its formula weight (FW) 180.16. Must show all work and calculation and then list what you will mix together to prepare the correct amount of … 9. ### statistics Consider the following situation, and answer each of the following multiple choice questions. In your write-up, just list the sub-question letter (A-J) and your answer – no need to restate the question or to justify your answer. … 10. ### algebra The volume V of a gas varies inversely as the pressure P on it. If the volume is 250 cm3 under a pressure of 35 kg/cm3, solve for the pressure applied to have a volume of 180 cm3. Be sure to define your variables, find your constant … More Similar Questions Post a New Question

{"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}

CC-MAIN-2017-30

longest

en

https://stockinvest.us/digest/price-to-earnings-ratio/

text/html

crawl-data/CC-MAIN-2024-30/segments/1720763857355.79/warc/CC-MAIN-20240725084035-20240725114035-00523.warc.gz

News Digest / Analysis & Ideas / What is the P/E ratio, and how can you use it to make better decisions? # What is the P/E ratio, and how can you use it to make better decisions? StockInvest.us 05:27am, Tuesday, Apr 26, 2022 The P/E (price-to-earnings) ratio is one of finance's essential valuation metrics. This blog post will explain what the P/E ratio is, how it's calculated, and why it's important. We will also give you a few rules on using the P/E ratio to make better investment decisions. Let's get started. ## How does the P/E ratio work? The price-to-earnings ratio is the ratio of a company's share (stock) price to the company's earnings per share. Usually is used to find out whether companies are overvalued or undervalued. The P/E ratio is determined by dividing the stock price of a company by its earnings per share (EPS). For example, if a company has a stock price of \$100 and its EPS is \$10, then its P/E ratio would be 10 (\$100/\$10). In a simplified way, you can say that: "P/E is the amount of years it will take to pay back the cost of buying the company based on the current profit." In the example above, this will be ten years. The P/E ratio gives you a better understanding of how much investors are willing to pay for each dollar of earnings. A higher P/E ratio means that investors are paying more for each dollar of earnings. For example, let's say two companies have the same EPS. Company A has a P/E ratio of 10, while Company B has a P/E ratio of 20. This means that investors are willing to pay twice as much for each dollar of earnings for Company B. The reason why similar companies may have different P/E is usually found in the expectation of future growth and your willingness to pay a higher premium now to be a part of the future income. An example of a company that has had a high P/E for a long time is Tesla Inc. which has grown rapidly, and future income is expected to be strong due to the growth. ### Why is the P/E ratio important? The P/E ratio is vital because it can give you an idea of whether a stock is undervalued or overvalued. A low P/E ratio (below 10) may mean an undervalued stock. A high P/E ratio (above 20) may mean overvalued stock. But keep in mind that P/E ratios vary from industry to industry. That's why it's important to compare companies within the same industry when using the P/E ratio as a valuation metric. It's also important to remember that the P/E ratio is just one metric and should not be used in isolation. Because sometimes, there is another reason behind the general rule. For example, a low P/E ratio could be because the company's business model itself is weak and does not have a promising future. It would be best to always look at a complete company analysis before making an investment decision. Are you ready to put all this information into practice? Subscribe to our daily newsletter if you want more alerts, signals, and posts like this. ### How can you use the P/E ratio to make better investment decisions? Here are a few tips on how you can use the P/E ratio to make better investment decisions: • Compare P/E ratios within the same industry: as we mentioned earlier, it's vital to analyze P/E ratios within the same industry because P/E ratios vary from industry to industry; • Look at other valuation metrics: the P/E ratio is just one metric, and you shouldn't rely on it exclusively. Make sure to look at other valuation metrics such as price-to-book value, price-to-sales ratio, and enterprise value-to-EBITDA ratio; • Consider the company's growth prospects: a company with high P/E ratios may be justified if it has strong growth prospects. Conversely, a company with low P/E ratios may not be as attractive if its growth prospects are weak; • Be aware that companies can manipulate P/E ratios through accounting techniques such as share buybacks. That's why make sure to do your own research before making any investment decisions. That's it for this introduction to the P/E ratio. In future articles, we will explore other aspects of this valuable metric and look at how you can use it to make better investment decisions. But for now, remember these key points: The P/E ratio is a measure of how much investors are willing to pay for each dollar of earnings a company generates. It can be used to compare different companies or industries. And finally, it's essential to watch the trend in the P/E ratio over time – especially when making long-term investments. We hope you found this blog post helpful. Thank you for reading. #### About The Author StockInvest.us StockInvest.us is a stock market research tool that provides daily stock signals and technical analysis for over 25 000 tickers on 38 exchanges. The company was founded in 2016 in Vilnius, Lithuania. Start Your Journey With: 0% Commission Stock Trading Free Insurance of up to 1 Million Regulated By FCA, ASIC & CySEC Ability to Copy Experienced Traders Your capital is at risk #### Read Next in Analysis & Ideas 09:59am, Wednesday, Jul 24, 2024 10:08am, Thursday, Jul 18, 2024 08:32am, Wednesday, Jul 17, 2024 09:18am, Monday, Jul 15, 2024

{"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}

CC-MAIN-2024-30

latest

en

http://mathhelpforum.com/calculus/114536-max-min-constraint.html

text/html

crawl-data/CC-MAIN-2016-50/segments/1480698541517.94/warc/CC-MAIN-20161202170901-00063-ip-10-31-129-80.ec2.internal.warc.gz

1. ## Max/Min with constraint I need help on the following problem: Find the absolute maximum and minimum of the function $f(x,y) = x^2 + y^2$ subject to the constraint $x^4 + y^4 = 625$. I tried using Lagrange's multiplier to this problem but here is where I'm stuck. $\frac{\partial f}{\partial x}= \lambda\frac{\partial g}{\partial x}$ $2x = 4x^3 \lambda$ $\frac{\partial f}{\partial y} = \lambda\frac{\partial g}{\partial y}$ $2y = 4y^3\lambda$ I just can't seem to be able to solve for $x,y, \lambda$,which I need for the critical points 2. Where is your third equation? $2x\;=\;4x^{3}\lambda$ $2y\;=\;4y^{3}\lambda$ $x^{4}\;+\;y^{4}\;=\;625$ That should do it.

{"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": 0, "codecogs_latex": 10, "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}

CC-MAIN-2016-50

longest

en

https://gamedev.stackexchange.com/questions/23748/testing-whether-two-cubes-are-touching-in-space/23749

text/html

crawl-data/CC-MAIN-2019-51/segments/1575540481281.1/warc/CC-MAIN-20191205164243-20191205192243-00067.warc.gz

# Testing whether two cubes are touching in space Does anyone have any clean ideas on testing whether two cubes in 3D space touch? By touch I mean, touch at corners or on a face or on an edge. Say that the cubes are axis aligned and there is no nesting of the cubes. So either the cubes touch or they don't. There is no overlap of the cubes at all (except of course possibly an edge or corner point). Everything I have thought of ends up needing a ton of cases. Also numerical stability is a factor. I don't want to just test == on all the edges/points. I guess I could utilize some sort of epsilon instead of testing whether something == 0. The main thing however is minimizing the number of cases to test. ## migrated from stackoverflow.comFeb 12 '12 at 8:03 This question came from our site for professional and enthusiast programmers. • Are all the cubes the same size? – brainjam Feb 10 '12 at 21:04 • Are the boxes axis aligned, or free orientation? – Jari Komppa Feb 13 '12 at 10:30 If your cubes overlap or they just touch, they have to overlap or touch in all three axis. In one axis it looks like this (for two intervals a and b): So what you have to test is: if ((min_a <= min_b && min_b <= max_a) || (min_b <= min_a && min_a <= max_b)) First part (before ||) is for case when min_a < min_b, second for case when min_b < min_a. For all three axis it looks like this: if ( ((min_x1 <= min_x2 && min_x2 <= max_x1) || (min_x2 <= min_x1 && min_x1 <= max_x2)) && ((min_y1 <= min_y2 && min_y2 <= max_y1) || (min_y2 <= min_y1 && min_y1 <= max_y2)) && ((min_z1 <= min_z2 && min_z2 <= max_z1) || (min_z2 <= min_z1 && min_z1 <= max_z2)) ) Where for example min_x and max_x are minimal and maximal coordinates of cube in x axis. x1 is for first cube, x2 for second. As I sad, this tests overlaping, but because there is "<=" and not only "<", it test also "touching". //EDIT: If you want to have more stable solution (there can be some floating point value errors), you should use epsilon. So it will look like: min_x1 - epsilon <= min_x2 && min_x2 <= max_x1 + epsilon • This only works for AABB collisions, naturally. – Jari Komppa Feb 13 '12 at 10:31 • I know. But author wrote: "Say that the cubes are axis aligned..." – zacharmarz Feb 13 '12 at 10:49 An alternate way to check for AABB collisions is to rule out all cases where you can't collide. if (min_x1 > max_x2) no collision if (max_x1 < min_x2) no collision if (min_y1 > max_y2) no collision if (max_y1 < min_y2) no collision if (min_z1 > max_z2) no collision if (max_z1 < min_z2) no collision otherwise, collision. You certainly want to minimise the number of tests being performed. You can do the test using only three comparisons. Let (x1,y1,z1) and (x2,y2,z2) be the centres of the cubes. Let a1 and a2 be their respective extents (ie. half their edge length). The cubes touch by your definition if and only if the distances between the projections of their centres is a1+a2 in each of the three directions of the coordinate system. /* Returns true if two axis-aligned cubes touch or overlap */ return fabs(x1-x2)-(a1+a2) <= EPSILON && fabs(y1-y2)-(a1+a2) <= EPSILON && fabs(z1-z2)-(a1+a2) <= EPSILON; Where EPSILON is a small number suitable for your typical object sizes and distances. Note that the above meethod will also return true if the cubes overlap, but since this does not happen, it is not a problem. If the platform you target has fast min/max selection, you can do it using only one test: /* Returns true if two axis-aligned cubes touch or overlap */ return max3(fabs(x1-x2),fabs(y1-y2),fabs(z1-z2))-(a1+a2) <= EPSILON; • Another example: pastie.org/3409055, where the extents in each direction are not identical (ie, there exists a1.x, a1.y, a1.z etc). – deceleratedcaviar Feb 18 '12 at 17:39 Various spatial tree algorithms are designed to prevent the extra checks you mention. You could easily add to said structures via a single distance (sphere) check from the cube's center to cull bad results. Or you could just sphere bound check each cube against the others (O(n^2)) to see if you need further tests. Note that test can be done without a sqrt if you compare via squared radii. The spatial trees (oct tree, quad tree, BST) take that O(N^2) term way, way down. Of course AA boxes means you only need compare 6 values for collisions /left,right, right/left, top/bottom, bottom/top, front/back, back/front. 6 'less than or equal to' compares isn't that bad! Most likely you will need a spatial data representation that allows for fast dismissal of uninteresting objects. The Octtree is one of the most used structures and fits perfectly with your requirements, plus they are easy to implement. A BSP Tree might be an alternative, depending how your data is set up. A fast and dirty solution would be to a hash-map to store all points and a reference to all cubes sharing that point. The query would be very fast but at the cost of (a lot of) memory. Although for a small number of cubes that might be applicable. if your cubes are axis aligned then you may construct tree interval Trees one for each axis. An intervall tree is a data structure that let you query for an intervall overllapping or a point hit of a set of intervall. If you can ensure that no cube will ever have an edge less than ε then you can easely see if a cube touches other cubes by querying your trees. To understand how this works lets scale the problem down to "1D cubes" i.e. segments It the testing segments is AB then take the points A-ε,A+ε,B-ε,B+ε check the intervall tree for the hit. For every segment that matches there are a few of possible relation with your testing segment: Testing Segment A- A+ B- B+ Case - - - - impossible : no match Case X - - - touch: the segments are disjoint but very near Case - X - - overlap Case X X - - overlap: segments touches for more that 2 epsilon Case - - X - overlap Case X - X - impossible: no holes in segments Case - X X - overlap: the segments are nearly the same Case X X X - overlap: the segments are nearly the same Case - - - X touch: disjointed segments but very near Case X - - X impossible: no holes in segments Case - X - X impossible: no holes in segments Case X X - X impossible: no holes in segments Case - - X X overlap: segments touches for more that 2 epsilon Case X - X X impossible: no holes in segments Case - X X X overlap: the segments are nearly the same Case X X X X overlap: the segments are nearly the same As you can see a segment touches our testing one if A- matches but A+ does not OR B+ matches but B- does not; in any other case (excluding the impossible ones) there is an overlap. In interval trees other data can be stored alongside the start and the end of the intervall. This possibility can be used to give names to the segments so you can wrap the query function so given a couple of points it gives you a list of (segment_name,"touch") or (segment_name,"overlap"). Now we need a way to use 1D results to get 2D results, a way that possibly scale up to 3D and over. We can say that a rectangle touches another if there is a touch in an axis and an overlap in the other (edge touch) or a touch in both axes (corner touch). To do this we need to store our intervals into two separate trees taking care of naming intervals correctly. Let say that we are looking if the rectangle (Ax,Ay)(Bx,By) touches some of the stored rectangle, then we have to query the X-tree for Ax,Bx and the Y-tree for Ay,By. Lets suppose that the result is [("alpha","touch"),("beta","touch")] by X-tree and [("alpha","overlap")] for Y-tree whe can say that our rectangle touches "alpha" ("edge" touches) but don't touches "beta" nor other rectangles. The test can be easely managed using a dictionary that keep the rectangle name as key and stores the X,Y values for example: {"alpha":[0,1],"beta":[0,-1]} means that "alpha" touches(0) in X and overlap(1) in Y while "beta" touches(0) in X but is disjoint in Y(-1). This reasoning can be extended further adding other dimensions: there is a touch if at least one dimension touches and there are no disjoint dimensions: 3D (0:touch, 1:overlap) X Y Z 0 0 0 corner touch 0 0 1 edge touch (over one of the four Z edge) 0 1 0 edge touch (over one of the four Y edge) 0 1 1 surface touch (over one of the two X surfaces) 1 0 0 edge touch (over one of the four X edge) 1 0 1 surface touch (over one of the two Y surfaces) 1 1 0 surface touch (over one of the two Z surfaces) 1 1 1 eerrr no touch... the cubes are nested (remeber, at least one dimension touch) [Appendix] To speed up the interval query match you can test only A- and B+; when you find a match with A- then you get the upper end of the segment that matches and measure the distance with A-: if the distance is greather then 2ε this means that the overlap is too large to be considered a touch so you don't have to continue. The same has to be done with B+ but this time you have to check the value of B+ with the lower end of the matching segments, again to see if the overlap you find with the extended segment is too large to be considered a simple touch. A balanced interval tree takes ~log(n) iterations to check if a point hits against n intervals, this means that if you keep (by using red black trees for example) your trees balanced you can check against a huge set in reasonable time

{"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}

CC-MAIN-2019-51

latest

en

http://ecoursesonline.iasri.res.in/mod/page/view.php?id=20802

text/html

crawl-data/CC-MAIN-2023-06/segments/1674764494852.95/warc/CC-MAIN-20230127001911-20230127031911-00240.warc.gz

## Relation between Respiratory Quotient and Energy output Human Nutrition 3(3+0) Lesson 6 : Energy – Determination of energy requirements and energy sources Relation between Respiratory Quotient and Energy output The term respiratory quotient refers to the ratio between the volume of Co2 given out and the volume of O2 consumed by the human subjects. When only carbohydrate is oxidized, the R.Q is 1.0 and when only fat is oxidized, the R.Q is 0.7 and when only protein is oxidized the R.Q. is 0.82. After an average meal containing 10 per cent protein, 20 per cent fat and 70 per cent carbohydrate, the R.Q. is about 0.82. The R.Q. is lowered to about 0.7 in diabetes mellitus as the body oxidizes large amounts of fat. Data regarding the quantities of carbohydrates and fats oxidized per liter of O2 with the corresponding non protein R.Q. and calories produced are given in table Non-protein respiratory quotient, quantities of carbohydrates and fats oxidized and heat produced per liter of O2 consumed (Data per liter of O2 consumed) Non-protein(RQ) Carbohydrate (g) Fat (g) Calories (kcal) 1.000.95 0.90 0.85 0.80 0.75 0.70 1.2321.010 0.793 0.580 0.375 0.173 0 00.091 0.180 0.267 0.350 0.433 0.502 5.0474.995 4.924 4.862 4.801 4.739 4.686 Calculation of amounts of proteins, fats and carbohydrates oxidized in the body From the quantity of nitrogen excreted in urine and the quantity of CO2 produced and O2 consumed in 24 hours it is possible to calculate the amounts of proteins, fats and carbohydrates oxidized in the body. The following example will illustrate the principles involved in such a calculation: Example: An adult excretes in 24 hours 8 g N (equivalent to 50 g proteins) in urine and uses 400 liters of O2 and gives out 320 liters of CO2. It is known that for 1 g of urinary N, 6.25g protein would oxidized and this will require 5.92 liters of O2 producing 38.0 liters of CO2. After deducting these from the total O2 consumed and CO2 produced, the O2 utilized and CO2 produced in the oxidation of fats and carbohydrates will be 400.0- 47.4 = 352.6 liters of O2 and 320.0-38.0 = 282.0 liters of CO2 It will be seen from table that at a R.Q of 0.80, 1 liter of O2 consumed would have oxidized 0.375 g carbohydrate and 0.350 g fat, 352.6 liters used would have oxidized 352.6x0.375g of Carbohydrate and 352.6x0.350g of fat. Therefore the quantities of nutrients oxidized and energy produced in 24 hours are as follows: Energy (Kcal) Protein = 50gx4.1 205.0 Carbohydrates = 132.1gx4.1 543.3 Fat = 123.4gx9.3 1147.6 Total energy (Kcal) 1895.9

{"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}

CC-MAIN-2023-06

latest

en

https://math.stackexchange.com/questions/365348/homework-prove-that-a-given-set-is-a-group

text/html

crawl-data/CC-MAIN-2020-05/segments/1579251690379.95/warc/CC-MAIN-20200126195918-20200126225918-00048.warc.gz

# Homework - Prove that a given set is a group [duplicate] I have a homework question and I don't know how to approach this exercise. The exercise is the following: Let's suppose $G$ is a set with binary function * defined for its members, which is: • closings; • associative; • there's $e\in G$, so that $a\star e=a$ where $a\in G$; • for each $a\in G$, there's a $b\in G$ so that $a\star b=e$. Prove that $G$ is a group. I have no Idea how approach this exercise. pay attention that 3,4 are Noncommutative. • What is the definition of a group? – user41442 Apr 18 '13 at 10:19 • @user41442 There's no definition – Billie Apr 18 '13 at 10:22 • Every mathematical structure has a definition, so you better find that definition. Try wikipedia for example. Then you can use that definition to see why your set G is a group. – Applied mathematician Apr 18 '13 at 10:26 • @user1798362: this could well be the definition of a group, but assuming there is something to prove, the definition you got in class should be that 3,4 are two sided, i.e there exists $e\in G$ such that $a*e=e*a=a$ for all $a\in G$ and for any $a \in G$ there exists $b\in G$ such that $a*b=b*a=e$ – Dennis Gulko Apr 18 '13 at 10:42 • This seems to be the same question as Right identity and Right inverse implies a group. – Martin Sleziak Oct 4 '13 at 14:05 First, let $a\in G$. By (4), there exists $b\in G$ such that $a*b=e$. Moreover, again by (4), there exists $c\in G$ such that $b*c=e$. Then: $$a=a*e=a*(b*c)=(a*b)*c=e*c$$ And thus $$b*a=b*(e*c)=(b*e)*c=b*c=e$$ Which proves the second direction of (4). This proof is independent of the finiteness of $G$. Now you should be able to complete the proof (you still need to show that $e*a=a$) This result is true if $G$ is a finite set, indeed: Fix $a\in G$ and define $\varphi_a:G\to G$, $x\mapsto x*a$ then if $\varphi_a(x)=x*a=\varphi_a(y)=y*a$ then multiply on RHS by $b$ and we find $x=y$ hence $\varphi_a$ is injective and by finite cardinality of $G$, $\varphi_a$ is bijective. Now there's $b'\in G$ such that $\varphi_a(b')=b'*a=e$ and multiply on the RHS by $b$ we find $b'=b$ so $b*a=a*b=e$ Finaly there's $e'$ such that $\varphi_a(e')=e'*a=a$ then we multiply on the RHS by $b$ and we have $e'=e$ so $e*a=a*e=a$. By assumption, $e$ is a right-neutral element for $*$ and, given $a\in G$, the element $a'\in G$ such that $a*a'=e$ is a right-inverse for $a$. All what you have to prove is that $e$ is also a left-neutral element and that $a'$ is also a left inverse of $a$. From $a*a'=e$ you can deduce: $$e*(a*a')=e*e=e$$ and by associativity $$(e*a)*a'=e=a*a'$$ Let $a''$ be the right-inverse of $a'$. Then $$[(e*a)*a']*a''=(a*a')*a''$$ Thus $$(e*a)*(a'*a'')=a*(a'*a'')$$ $$(e*a)*e=a*e$$ $$e*a=a$$ Again from $a*a'=e$ you deduce $$a'*(a*a')=a'*e$$ $$(a'*a)*a'=a'$$ $$[(a'*a)*a']*a''=a'*a''$$ $$(a'*a)*(a'*a'')=e$$ $$(a'*a)*e=e$$ $$a'*a=e$$

{"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}

CC-MAIN-2020-05

latest

en

https://www.physicsforums.com/threads/magnets-and-inertia.9600/

text/html

crawl-data/CC-MAIN-2017-51/segments/1512948581033.57/warc/CC-MAIN-20171216010725-20171216032725-00052.warc.gz

# Magnets and inertia 1. Nov 25, 2003 ### litewk Hi, Is it possible for two objects of nearly the exact same size, mass, and magnetic field strength to form something analogous to a curved space orbit in a relatively flat space by balancing inertia with magnetism? In other words, can two mutually magnetic objects orbit around a common center of magnetism for any length of time at some given distance scale? If so, can all four fundamental forces of physics form six relationships via their orbital time period ratios. It seems like macro orbits have no meaning in the quantum world but this thought occurred to me a while back and I am non-degreed so I'm unable to solve this one easily. Thanks for your time. 2. May 11, 2011 ### chrisbaird Any force that is central and attractive, like gravity, will lead to conservation of angular momentum, meaning that stable orbits are possible. The electrostatic force between two opposite charges is central and attractive, and the magnetostatic force between two ideal opposite magnetic poles is central and attractive, so you might think that electric charges or magnets could form stable orbits around each other. The problem is that orbits require motion, which mean we are no longer in statics. When motion is involved, the magnetic and electric field become coupled, and you enter the realm of electrodynamics. The electromagnetic (Lorentz) force is not central and therefore two electromagnetic objects cannot form a stable orbit with each other (classically speaking). This means classically that electrons should not form stable orbits around nuclei. This is the problem that lead to the need for a quantum description of atoms. 3. May 11, 2011 ### DrFurious I disagree with chrisbaird. Not all central force orbits are stable. In fact, (Bertrand's theorem) only TWO types of orbits are stable: inverse-square law (gravity, electrostatics) and a radial harmonic oscillator type. The attraction between two magnets is certainly NOT 1/r^2. I can't even remember what it is off the top of my head, but I remember trying to figure it out at some point and it was complicated... something like 1/r^3, or something similar. 4. May 11, 2011 ### pallidin If there is a rotational impulse provided, yes(this prevents/slows the collapse). If not, no. 5. May 11, 2011 ### Bill_K The pair of objects taken together will have a time-varying dipole moment. Therefore they will radiate electromagnetic waves, lose both energy and angular momentum and ultimately collapse. 6. May 12, 2011 ### chrisbaird DrFurious, Thank you for clarifying. I did not mean to imply that a central force is a sufficient condition, just a necessary condition. Hence the word 'possible'. Also, the attraction between two idealized point-like magnetic poles is an inverse square law, or similarly the force between two infinitesimal strips of current is an inverse square law. (See Jackson Eq. 5.10 for example.) Of course, idealizations like this are non-physical and when we go to actual macroscopic objects, the fields and forces get more complicated. - Chris 7. May 12, 2011 ### DrFurious Your reference is spot on. And the fields do get pretty complicated :) Word to the wise though: if you have to do an experiment, don't rely on the magnetic result from (e.g. Jackson). Cause you will go crazy. Other non-dipole forces behave a lot easier. Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

{"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}

CC-MAIN-2017-51

longest

en

http://at.metamath.org/ileuni/fndmdifcom.html

text/html

crawl-data/CC-MAIN-2020-50/segments/1606141718314.68/warc/CC-MAIN-20201203031111-20201203061111-00057.warc.gz

Intuitionistic Logic Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  ILE Home  >  Th. List  >  fndmdifcom Structured version   GIF version Theorem fndmdifcom 5216 Description: The difference set between two functions is commutative. (Contributed by Stefan O'Rear, 17-Jan-2015.) Assertion Ref Expression fndmdifcom ((𝐹 Fn A 𝐺 Fn A) → dom (𝐹𝐺) = dom (𝐺𝐹)) Proof of Theorem fndmdifcom Dummy variable x is distinct from all other variables. StepHypRef Expression 1 necom 2283 . . . 4 ((𝐹x) ≠ (𝐺x) ↔ (𝐺x) ≠ (𝐹x)) 21a1i 9 . . 3 (x A → ((𝐹x) ≠ (𝐺x) ↔ (𝐺x) ≠ (𝐹x))) 32rabbiia 2541 . 2 {x A ∣ (𝐹x) ≠ (𝐺x)} = {x A ∣ (𝐺x) ≠ (𝐹x)} 4 fndmdif 5215 . 2 ((𝐹 Fn A 𝐺 Fn A) → dom (𝐹𝐺) = {x A ∣ (𝐹x) ≠ (𝐺x)}) 5 fndmdif 5215 . . 3 ((𝐺 Fn A 𝐹 Fn A) → dom (𝐺𝐹) = {x A ∣ (𝐺x) ≠ (𝐹x)}) 65ancoms 255 . 2 ((𝐹 Fn A 𝐺 Fn A) → dom (𝐺𝐹) = {x A ∣ (𝐺x) ≠ (𝐹x)}) 73, 4, 63eqtr4a 2095 1 ((𝐹 Fn A 𝐺 Fn A) → dom (𝐹𝐺) = dom (𝐺𝐹)) Colors of variables: wff set class Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98   = wceq 1242   ∈ wcel 1390   ≠ wne 2201  {crab 2304   ∖ cdif 2908  dom cdm 4288   Fn wfn 4840  ‘cfv 4845 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bndl 1396  ax-4 1397  ax-14 1402  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019  ax-sep 3866  ax-pow 3918  ax-pr 3935 This theorem depends on definitions:  df-bi 110  df-3an 886  df-tru 1245  df-nf 1347  df-sb 1643  df-eu 1900  df-mo 1901  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-ne 2203  df-ral 2305  df-rex 2306  df-rab 2309  df-v 2553  df-sbc 2759  df-dif 2914  df-un 2916  df-in 2918  df-ss 2925  df-pw 3353  df-sn 3373  df-pr 3374  df-op 3376  df-uni 3572  df-br 3756  df-opab 3810  df-id 4021  df-xp 4294  df-rel 4295  df-cnv 4296  df-co 4297  df-dm 4298  df-iota 4810  df-fun 4847  df-fn 4848  df-fv 4853 This theorem is referenced by: (None) Copyright terms: Public domain W3C validator

{"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}

CC-MAIN-2020-50

longest

en

https://philoid.com/question/109179-answer-each-of-the-following-questions-in-one-word-or-one-sentence-or-as-per-exact-requirement-of-the-question-write-the-value-o

text/html

crawl-data/CC-MAIN-2024-38/segments/1725700651895.12/warc/CC-MAIN-20240918100941-20240918130941-00301.warc.gz

##### Answer each of the following questions in one word or one sentence or as per exact requirement of the question:Write the value of . Given tan-1 {tan (15π/4)} = tan-1 {tan (4π - π/4)} We know that tan (2π – θ) = -tan θ = tan-1 (-tan π/4) = tan-1 (-1) = -π/4 tan-1 {tan (15π/4)} = -π/4 1

{"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}

CC-MAIN-2024-38

latest

en

https://doc.qt.io/qt-6/qrotationreading.html

text/html

crawl-data/CC-MAIN-2023-23/segments/1685224646937.1/warc/CC-MAIN-20230531150014-20230531180014-00221.warc.gz

## Properties • x : const qreal • y : const qreal • z : const qreal ## Public Functions void setFromEuler(qreal x, qreal y, qreal z) qreal x() const qreal y() const qreal z() const ## Detailed Description The rotation reading contains 3 angles, measured in degrees that define the orientation of the device in three-dimensional space. These angles are similar to yaw, pitch and roll but are defined using only right hand rotation with axes as defined by the right hand cartesian coordinate system. The three angles are applied to the device in the following order. • Right-handed rotation z (-180, 180]. Starting from the y-axis and incrementing in the counter-clockwise direction. • Right-handed rotation x [-90, 90]. Starting from the new (once-rotated) y-axis and incrementing towards the z-axis. • Right-handed rotation y (-180, 180]. Starting from the new (twice-rotated) z-axis and incrementing towards the x-axis. Here is a visualization showing the order in which angles are applied. The 0 point for the z angle is defined as a fixed, external entity and is device-specific. While magnetic North is typically used as this reference point it may not be. Do not attempt to compare values for the z angle between devices or even on the same device if it has moved a significant distance. If the device cannot detect a fixed, external entity the z angle will always be 0 and the QRotationSensor::hasZ property will be set to false. The 0 point for the x and y angles are defined as when the x and y axes of the device are oriented towards the horizon. Here is an example of how the x value will change with device movement. Here is an example of how the y value will change with device movement. Note that when x is 90 or -90, values for z and y achieve rotation around the same axis (due to the order of operations). In this case the y rotation will be 0. ## Property Documentation ### `[read-only] `x : const qreal This property holds the rotation around the x axis. Measured as degrees. Access functions: qreal x() const ### `[read-only] `y : const qreal This property holds the rotation around the y axis. Measured as degrees. Access functions: qreal y() const ### `[read-only] `z : const qreal This property holds the rotation around the z axis. Measured as degrees. Access functions: qreal z() const ## Member Function Documentation Sets the rotation from three euler angles. This is to be called from the backend. The angles are measured in degrees. The order of the rotations matters, as first the z rotation is applied, then the x rotation and finally the y rotation.

{"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}

CC-MAIN-2023-23

latest

en

https://www.enotes.com/homework-help/finding-rate-depth-decrease-an-inverted-cone-272656

text/html

crawl-data/CC-MAIN-2017-04/segments/1484560280410.21/warc/CC-MAIN-20170116095120-00230-ip-10-171-10-70.ec2.internal.warc.gz

finding rate of depth decrease of an inverted coneA funnel has a circular top of diameter 20cm and a height of 30cm. When the depth of the liquid in the funnel is 12cm, the liquid is dropping at a... finding rate of depth decrease of an inverted cone A funnel has a circular top of diameter 20cm and a height of 30cm. When the depth of the liquid in the funnel is 12cm, the liquid is dropping at a rate of 0.2cm²/s. At what rate is the depth of the liquid in the funnel decreasing at this moment this was found in a chapter using the "chain rule" aka "composite function rule" or "function of a function rule" which states: if y = f(F(x)), and u = F(x) so that y=f(u), then  dy/dx = dy/du*du/dx (differentiation) Asked on by dark0flame giorgiana1976 | College Teacher | (Level 3) Valedictorian Posted on We'll start by recalling the formula that gives the volume of the cone: V = `pi` `r^(3)` h/3 In this case, w'ell consider V as being the volume of the liquid remained in the funnel, h is the depth of remaining liquid and r is the radius of the circle which represents the base of the new cone formed by the remaining liquid. Therefore, if h = 12, dV/dt = -0.2 Since in the formula that gives the volume of the cone, we have two variables, we'll express the radius in terms of the height. For this reason, we'll use similar triangles: r/h = (20/2)/30 = 1/3 r = h/3 We'll re-write the formula that gives the volume of cone, in terms of h. V = `pi` `h^(3)` /27 We'll differentiate with respect to t: dV/dt = (3`pi``h^(2)` /27)(dh/dt) dV/dt = (`pi` `h^(2)` /9)(dh/dt) But dV/dt = -0.2 -0.2 = (` `144`pi` /9)(dh/dt) -0.2 = (16`pi`) (dh/dt) (dh/dt) = -0.2/16`pi` (dh/dt) `~~` -0.003 cm/s Therefore, the depth of the liquid in the funnel is decreasing at a rate of 0.003 cm/s. We’ve answered 319,189 questions. We can answer yours, too.

{"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}

CC-MAIN-2017-04

longest

en

https://www.thestudentroom.co.uk/showthread.php?t=5329922

text/html

crawl-data/CC-MAIN-2021-31/segments/1627046153709.26/warc/CC-MAIN-20210728092200-20210728122200-00662.warc.gz

# If you thought you were good at maths... Watch Announcements #1 A school level competition for speed maths: The speed of these kids brains is beyond what I thought was possible. For GCSE/A Level students I recommend having a go at some of these problems but don't worry if they take you minutes instead of seconds Some highlights: 30:36 - This wasn't the fastest by a long way but I have no idea how he managed to work this out so quickly. They must practice techniques for similar questions? It took me a while to get this one - can anyone see a quick way of doing it? And then there's Luke who is in a different league: 43:20 - No idea how he processed the information so quickly 51:28 - Must be a speed reader. You may get a bit suspicious like I was just after seeing this. The whole final from the link above is worth watching. 11 3 years ago #2 Spoiler: Show With the (a-x)^2+(b-y)^2+(c-z)^2 problem, what you have to notice is that exactly one of a-x, b-y and c-z is 1 or -1 and the others are zero. WLOG we let a-x=1 (and then multiply answer by six afterwards -- by three for (b-y)^2 or (c-z)^2 being one and by two for the possibility that a-x=-1). We then have the following possibilities for a and x: 1, 0 2, 1 3, 2 4, 3 ... and the following possibilities for b and y: 0, 0 1, 1 2, 2 3, 3 4, 4 ... and the following possibilities for c and z: 0, 0 1, 1 2, 2 3, 3 4, 4 So that's 4*5*5 possibilities which is 100. As we said, we were going to multiply by six at the end, so we have 600 possibilities. I think that's the quickest way; how did you do it? 1 3 years ago #3 (Original post by Notnek) A school level competition for speed maths: The speed of these kids brains is beyond what I thought was possible. For GCSE/A Level students I recommend having a go at some of these problems but don't worry if they take you minutes instead of seconds Some highlights: 30:36 - This wasn't the fastest by a long way but I have no idea how he managed to work this out so quickly. They must practice techniques for similar questions? It took me a while to get this one - can anyone see a quick way of doing it? And then there's Luke who is in a different league: 43:20 - No idea how he processed the information so quickly 51:28 - Must be a speed reader. You may get a bit suspicious like I was just after seeing this. The whole final from the link above is worth watching. what a fascinating film ! i think the last question he must have done it recently.... or just a lucky guess 0 3 years ago #4 (Original post by the bear) what a fascinating film ! i think the last question he must have done it recently.... or just a lucky guess Almost certainly the former, if he skipped to "what is the digital root of " and used the well known trick that the digit sum of a number divisible by 9 is also divisible by 9 repeatedly.. It's a lot more believable. Of course, the speed he did it in is still impressive. However, I like to think the kids we have here on TSR doing the various olympiads are more impressive as this competition seems to emphasise speed and familiarity more than the creative process needed to solve an entirely unknown problem - to me it sends out wrong signals for what maths is all about. It's not about speed. You could spend weeks, months, years or even lifetimes on a good problem. 11 3 years ago #5 (Original post by Notnek) A school level competition for speed maths: The speed of these kids brains is beyond what I thought was possible. For GCSE/A Level students I recommend having a go at some of these problems but don't worry if they take you minutes instead of seconds Some highlights: 30:36 - This wasn't the fastest by a long way but I have no idea how he managed to work this out so quickly. They must practice techniques for similar questions? It took me a while to get this one - can anyone see a quick way of doing it? And then there's Luke who is in a different league: 43:20 - No idea how he processed the information so quickly 51:28 - Must be a speed reader. You may get a bit suspicious like I was just after seeing this. The whole final from the link above is worth watching. Luke was able to get full marks on AIME (somewhere between the level of our SMC and BMO) when he was in 5th grade (our Year 6). On AoPS he's universally regarded as a legend. 1 3 years ago #6 (Original post by I hate maths) Almost certainly the former, if he skipped to "what is the digital root of " and used the well known trick that the digit sum of a number divisible by 9 is also divisible by 9 repeatedly.. It's a lot more believable. Of course, the speed he did it in is still impressive. However, I like to think the kids we have here on TSR doing the various olympiads are more impressive as this competition seems to emphasise speed and familiarity more than the creative process needed to solve an entirely unknown problem - to me it sends out wrong signals for what maths is all about. It's not about speed. You could spend weeks, months, years or even lifetimes on a good problem. This, speed is not what ’being good at maths’ is about. 4 3 years ago #7 Did anyone actually bothered to watch the whole vid 6 3 years ago #8 It's quite impressive, in fact, very impressive. I don't think 43:20 was very fast, because that one was more of an obvious one. But I don't even get how they read the questions that quickly, probably loads of practise. 0 #9 (Original post by Thomazo) This, speed is not what ’being good at maths’ is about. (Original post by I hate maths) Almost certainly the former, if he skipped to "what is the digital root of " and used the well known trick that the digit sum of a number divisible by 9 is also divisible by 9 repeatedly.. It's a lot more believable. Of course, the speed he did it in is still impressive. However, I like to think the kids we have here on TSR doing the various olympiads are more impressive as this competition seems to emphasise speed and familiarity more than the creative process needed to solve an entirely unknown problem - to me it sends out wrong signals for what maths is all about. It's not about speed. You could spend weeks, months, years or even lifetimes on a good problem. I was waiting for someone to say this and I agree with it. I apologise for the clickbait title I rate Olympiad above this of course but I don’t see much of a problem with this competition. We can have speed maths as well as regular maths although this competition does feel very American! I wouldn’t be surprised if a lot of the competitors are current or future Olympiad students and I’m sure that the techniques they pick up will be useful for them so it’s not a complete waste of time. Being able to think fast is very useful in any timed maths competition. 0 #10 (Original post by I hate maths) Almost certainly the former, if he skipped to "what is the digital root of " and used the well known trick that the digit sum of a number divisible by 9 is also divisible by 9 repeatedly.. It's a lot more believable I don’t think he’s seen the question before but I do think that he knows what a digital root is so skipped straight to the actual question and answered it using the trick. He must be able to speed read and process information extremely quickly. 0 3 years ago #11 (Original post by Notnek) A school level competition for speed maths: The speed of these kids brains is beyond what I thought was possible. For GCSE/A Level students I recommend having a go at some of these problems but don't worry if they take you minutes instead of seconds Some highlights: 30:36 - This wasn't the fastest by a long way but I have no idea how he managed to work this out so quickly. They must practice techniques for similar questions? It took me a while to get this one - can anyone see a quick way of doing it? And then there's Luke who is in a different league: 43:20 - No idea how he processed the information so quickly 51:28 - Must be a speed reader. You may get a bit suspicious like I was just after seeing this. The whole final from the link above is worth watching. This is impressive. As someone who has always been very good at mental Maths and problem solving(can do 2^30 in my head, 1073741824),I'm Autistic) I still was no where near as fast as these kids! This is very inspirational and amazing! It's just a shame that there was talking going on as well as Maths, I'd have liked to see more of these guys! Especially Luke. 1 3 years ago #12 (Original post by .Iqra.) Did anyone actually bothered to watch the whole vid I did. 0 3 years ago #13 This may be impressive, but maths should not be a FLOPs test or a test of how quickly you can read the questions. All you have to do here is merely convince yourself of the correct answer. If it were up to me, the competition would require you to PROVE that it was the correct answer, but that would make for far less interesting TV. 0 3 years ago #14 (Original post by Notnek) 30:36 - This wasn't the fastest by a long way but I have no idea how he managed to work this out so quickly. They must practice techniques for similar questions? It took me a while to get this one - can anyone see a quick way of doing it? So, for what it's worth, here's a somewhat relevant couple of paragraphs from Feynman's autobiography: (Original post by Richard_Feynman) We had a thing at high school called the algebra team, which consisted of five kids, and we would travel to different schools as a team and have competitions. We would sit in one row of seats and the other team would sit in another row. A teacher, who was running the contest, would take out an envelope, and on the envelope it says "fortyfive seconds." She opens it up, writes the problem on the blackboard, and says, "Go!" ** so you really have more than forty*five seconds because while she's writing you can think. Now the game was this: You have a piece of paper, and on it you can write anything, you can do anything. The only thing that counted was the answer. If the answer was "six books," you'd have to write "6," and put a big circle around it. If what was in the circle was right, you won; if it wasn't, you lost. One thing was for sure: It was practically impossible to do the problem in any conventional, straightforward way, like putting "A is the number of red books, B is the number of blue books," grind, grind, grind, until you get "six books." That would take you fifty seconds, because the people who set up the timings on these problems had made them all a trifle short. So you had to think, "Is there a way to see it?" Sometimes you could see it in a flash, and sometimes you'd have to invent another way to do it and then do the algebra as fast as you could. It was wonderful practice, and I got better and better, and I eventually got to be the head of the team. So I learned to do algebra very quickly, and it came in handy in college. When we had a problem in calculus, I was very quick to see where it was going and to do the algebra **fast. On the other hand, if you look at the Putnam Exam Feynman famously took in 1939, you can see how much things have moved on: It's not really even STEP difficulty by today's standards. I thought it was interesting so have created a thread for discussion: https://www.thestudentroom.co.uk/sho....php?t=5330792 As far as this competition goes: it's frustrating the way they display the question - it seems the reveal for us is slower than the contestants see (since Luke answers before the question is even fully displayed), and also that it disappears as soon as someone tries to answer. I found it made it *much* harder to "try to compete" so to speak. And clearly the contestants have decided it's better be quick and wrong than slow and correct. Certainly with practice you can get a lot better at "intuitive guessing", and when they come off you look like a genius. That said, I'm sure everyone on there has perfectly legitimate mathematical skills as well. And that kind of speed practice (with a bit more emphasis on correctness) will certainly help in things like the MAT, STEP and BMO/IMO as well. 0 3 years ago #15 (Original post by Notnek) A school level competition for speed maths: The speed of these kids brains is beyond what I thought was possible. For GCSE/A Level students I recommend having a go at some of these problems but don't worry if they take you minutes instead of seconds Some highlights: 30:36 - This wasn't the fastest by a long way but I have no idea how he managed to work this out so quickly. They must practice techniques for similar questions? It took me a while to get this one - can anyone see a quick way of doing it? And then there's Luke who is in a different league: 43:20 - No idea how he processed the information so quickly 51:28 - Must be a speed reader. You may get a bit suspicious like I was just after seeing this. The whole final from the link above is worth watching. I managed to only do a few that they couldn't do xD. It's very impressive how fast they can do it like some of them answer the question before I have even read the damn question lol! It obviously takes a lot of practise to do that sort of problem solving efficiently. 0 3 years ago #16 They seemed to take a while for question 2... i thought that was the easiest I got it as soon as i read it! (almost none of the others though ) 0 3 years ago #17 Luke seems like another human calculator. 0 3 years ago #18 (Original post by DFranklin) So, for what it's worth, here's a somewhat relevant couple of paragraphs from Feynman's autobiography: On the other hand, if you look at the Putnam Exam Feynman famously took in 1939, you can see how much things have moved on: It's not really even STEP difficulty by today's standards. I thought it was interesting so have created a thread for discussion: https://www.thestudentroom.co.uk/sho....php?t=5330792 As far as this competition goes: it's frustrating the way they display the question - it seems the reveal for us is slower than the contestants see (since Luke answers before the question is even fully displayed), and also that it disappears as soon as someone tries to answer. I found it made it *much* harder to "try to compete" so to speak. And clearly the contestants have decided it's better be quick and wrong than slow and correct. Certainly with practice you can get a lot better at "intuitive guessing", and when they come off you look like a genius. That said, I'm sure everyone on there has perfectly legitimate mathematical skills as well. And that kind of speed practice (with a bit more emphasis on correctness) will certainly help in things like the MAT, STEP and BMO/IMO as well. I'm not sure how much this sort of speed practice is correlated with being good at BMO/ IMO. At the Trinity camp this year, ironically during the team relay competition (mainly focused on speed), the team with 3/8 of the IMO squad came last... 0 3 years ago #19 I got question 2 before them. waay before them. and they were taking so long over it that I thought I must be wrong and it's some kind of trick question lol it was easy. 0 #20 (Original post by laurawatt) They seemed to take a while for question 2... i thought that was the easiest I got it as soon as i read it! Yes it was strange that they took so long for that one. I think they both thought it couldn't be that easy and as time went on, the other person not answering made them even more paranoid 0 X new posts Back to top Latest My Feed ### Oops, nobody has postedin the last few hours. Why not re-start the conversation? see more ### See more of what you like onThe Student Room You can personalise what you see on TSR. Tell us a little about yourself to get started. ### Poll Join the discussion #### Poll: What factors affect your mental health most right now? Post-lockdown edition 5.62% Uncertainty around my education (69) 11.75% Uncertainty around my future career prospects (66) 11.24% Lack of purpose or motivation (75) 12.78% Lack of support system (eg. teachers, counsellors, delays in care) (33) 5.62% Impact lockdown had on physical health (31) 5.28% Social worries (incl. loneliness/making friends) (63) 10.73% Financial worries (38) 6.47% Concern about myself or my loves ones getting/having been ill (24) 4.09% Exposure to negative news/social media (34) 5.79% Difficulty accessing real life entertainment (15) 2.56% Lack of confidence in making big life decisions (59) 10.05% Worry about missed opportunities during the pandemic (47) 8.01%

{"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}

CC-MAIN-2021-31

latest

en

https://physics.stackexchange.com/questions/63101/infinite-reflection-of-light-and-the-conservation-of-energy-momentum/301942

text/html

crawl-data/CC-MAIN-2019-43/segments/1570986672723.50/warc/CC-MAIN-20191017045957-20191017073457-00477.warc.gz

# Infinite reflection of light and the conservation of energy / momentum First off, I confess I'm no physicist, but I have been asking people with a more extensive knowledge this one question, without a definitive answer so far. Basically, I'm playing around with the idea of photons having mass as this part of the wiki page shows, it's not an entirely new concept... Especially the latest (mangled) sentence is of interest to me. I have been reading up about light, the duality of it, photons and it being the only known gauge boson (massless thingamajig) and what have you. As a result, I'm even more confused, so I thought I'd ask my question here. Setup: Suppose we took two perfect mirrors, stretching out into infinity, perfectly parallel, facing each other. If I were to shoot a photon onto either one of these mirrors at a 45deg. angle, what would happen. Hypotheses: As far as I can know (or guess, or can imagine) either one of three things can happen: • The photon just bounces back and forth into infinity at that leisurely pace of $c$. • Calling on the particle part of light's duality: Every action causes an equal and opposite reaction. Upon colliding with the surface of the mirrors, energy is needed for the photon to change direction. As everything strides towards entropy, I'd assume there is some heat being released (photon's energy onto the mirror)? If that's the case, at some point the photon's electromagnetic "charge", ie energy-reserves should run out. What do I end up with? Slightly warmer mirrors and a massless, empty shell of a photon at the end? What is a photon that no longer has any energy anyway? Is that the famous dark-matter... or am I going all too scifi-crazy now? Because somewhere I did read that light, being massless, obviously has no rest-matter either, nor does it have an electric charge of its own. That causes me to think of a photon as some sort of carrier, an empty satchel and because it's not exactly huge, it can but contain a finite amount of energy (I think). • Last thing I can think of: because of my photon's bouncing, and my being at a terrible loss trying to grasp the formula's and theories about light's physical properties I've gotten the (perhaps silly) idea that the constant changing of the direction of propagation could affect the wavelength, essentially generating something more like gamma-rays. Again, I don't know what this entails for my mirror-setup, but when news breaks of an impending nuclear disaster, I don't think a mirror completely deflects gamma rays. In other words, I don't even think it unlikely if somebody told me that photon would just bugger off. I hope someone can make sense of the bizarre meanders of a non-physicist's mind, but I would like to know the answer to a question I came up with about 10 years ago. So far I've gotten the answers: • Oh, I'd have to check on that one. • Of course, they talk about the duality of light, but light is, essentially pure energy. they've developed this dual-character as a working model. Much like everything "'t is but a theory" (I particularly disliked this answer for some reason) • Do you know how they spot a black hole? (I replied: No) Because there is light, but none around it. All light is drawn to the black hole. (this was followed by an awkward silence, and a smug nod. Which met with a confused and monkey like gaze from my part) Any more confusing ideas are always welcome. Edit/recap: Thanks to all of you for the info. In response to the comments, the kernel of the question is this: If I were able to follow the afore mentioned proton in this setup, what changes, if any, will I see along the line? Heat being generated? The photon "disintegrating" or dissipating, nothing (just endlessly bouncing back and forth...? Reading the wiki on Total Internal Reflection, I noticed that this occurs with soundwaves, too. I immediately thought of that horrid screeching feedback noise you can get if you hold a mic to a speaker. I guess I sort of translated that phenomenon into the photon changing wavelengths. Funny, but true: I remember as a child asking my father if you were able to create an infinite broadcast of sorts using two transmitters and two receivers playing a sound back and forth to them. In some way or another, I've always wondered about stuff like this as it turns out... Mirror mass: I suppose the mirrors would have to have infinite mass for them to stretch out into infinity. Though after some more checking, that complicates things considering $E = pc$. I've added that to my many light-related bookmarks, and I'll get back to you on that. • +1 because I think the infinite reflection question is a neat concept. Your question is all over the place though and I think you should edit it down to just the kernel of what you want to know. – Brandon Enright May 2 '13 at 23:52 • Your infinite parallel mirrors setup is very similar to fiber optics and total internal reflection probably has a lot to say about the answer: en.wikipedia.org/wiki/Total_internal_reflection – Brandon Enright May 2 '13 at 23:53 • – dmckee May 3 '13 at 0:03 • Do the mirrors have infinite mass? – joshphysics May 3 '13 at 0:05 • @dmckee: Thanks for the link, I now know I have about a thousand more wiki pages to read/decipher ;) – Elias Van Ootegem May 3 '13 at 10:21 First, of course there's no perfect mirror. But let's assume there was one. Next, the question is: Is the bouncing off the mirrors elastic or inelastic. If the photon is absorbed and re-emitted with the same frequency, then the bouncing is elastic and no energy is lost by the photon. It would then go on forever and ever. But what if it does lose energy with each bounce? Well, your two mirrors form a cavity and if we appeal to the wave-aspect of light, only waves with wavelengths that "fit" into the cavity are allowed, so there'd be a minimum allowed wavelength, $\lambda_0$ with $\lambda_0 = L/2$ where $L$ is the distance between your mirrors and since energy and wavelength of a photon are intimately related, this means that the photon in your cavity has a minimum energy below which it cannot fall. If you add the concept of heat / temperature / entropy to the mix, what you will get is that the walls (mirrors) are in thermal equilibrium with the photons in your cavity: Some of the energy is then stored in the walls and some in the photons. In fact, considering the situation of taking a cavity at some temperature and looking at the nature of the light that comes out of it (if you poke a tiny hole in it) is one of the phenomena that led to the discovery of quantum physics. Some misconceptions: A photon has no "electromagnetic charge", it is a massless, chargeless particle. Now what if its energy "runs out"? Then it just ceases to exist. There is no photon without energy. • qualify "ceases to exist" by maybe :it is so low in the ifra red spctrum that it is absorbed in a vibrational qm transition of a molecule. from momentum conservation it will be losing energy at each bounce, lowering wavelength, ulsess as Josh is asking the mirrors have infinite mass? – anna v May 3 '13 at 4:15 • This might sound stupid, but I know a photon is massless and has no charge, but at the same time it's said to be "an elementary particle, the quantum of light and all other forms of electromagnetic radiation, and the force carrier for the electromagnetic force". So it has no charge, but carries energy... is energy even... what? who? how? *_- – Elias Van Ootegem May 3 '13 at 10:25 • Also: if a photon ceases to exist if the energy "runs out", shouldn't you be left with an empty gauge boson particle? – Elias Van Ootegem May 3 '13 at 10:27 • No, a photon cannot "run out" of energy independent of its other properties. Technically, if a photon "loses" energy, what really happens is that a photon of some initial energy $E_1$ is absorbed/destroyed and a new photon of some new energy $E_2$ is emitted. "Running out of energy" then just means that a new photon is never emitted. That can happen if a photon is absorbed by a crystal and the energy then re-emitted as lattice vibrations instead of a new photon. – Lagerbaer May 3 '13 at 15:10 • @Lagerbaer: I'm sorry to be this thick, I got hung up on that gauge boson being matter, and matter should be conserved at all time... I have found out, now, that matter is not perfectly conserved, though I've also learned that -even though they're massless- photons still add mass. Anyway, I've got enough material out of this to study this matter (no pun intended) for a couple of days/weeks... meanwhile: Thanks for the info! – Elias Van Ootegem May 3 '13 at 16:30 Pose the same question with slightly different parameters. Instead of parallel mirrors traveling to infinity, construct a sphere where the inside is reflective (read:mirror). Then pull a vacuum on the sphere so there is nothing to slow the photon. Then inject the photon at any angle and it will reflect forever at the speed of light and doing so allow the sphere to become incandescent (without the heat of course). • Why didn't I think of a sphere... would've made things so much easier. So a perfect sphere, with a perfect reflective surface on the inside would have the photon bouncing around for ever. There is no loss of energy caused by the photon, effectively, colliding with the sphere. If we had a device that was able to measure the light inside the sphere, without breaking the necessarily closed system, we'd see a constant radiant flux, correct? – Elias Van Ootegem Dec 30 '16 at 17:38

{"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}

CC-MAIN-2019-43

latest

en

http://krsna.lamost.org/engine/archive/blog/2005/3

text/html

crawl-data/CC-MAIN-2020-10/segments/1581875146485.15/warc/CC-MAIN-20200226181001-20200226211001-00475.warc.gz

Date ## 科学的边缘 I can smell wrong papers from a distance I have no time to read the paper. The paper is nonsense. The idea is utterly wrong. Sir, remember, a man like you will never understand the mystery of Nature. (A frog in the well will never see the whole sky. ---from an old Chinese aphorism) ## 半群的用处 <script type="text/javascript" src="http://krsna.lamost.org/ASCIIMathML.js"> </script> \Omega-代数第一同构定理 ## 郑重推荐WEB数学符号书写解决法 http://www1.chapman.edu/~jipsen/mathml/asciimath.html <script type="text/javascript" src="http://krsna.lamost.org/ASCIIMathML.js"> </script> alephoint\varphi\forall\nabla\subseteq E=mc^2e^(ipi)=-1 $AA x in CC (sin^2x+cos^2x=1)$ sum_(i=1)^ni^3=((n(n+1))/2)^2

{"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}

CC-MAIN-2020-10

longest

en

http://altrushare.com/tag/basic-electrical-wiring-manual/

text/html

crawl-data/CC-MAIN-2018-17/segments/1524125947803.66/warc/CC-MAIN-20180425115743-20180425135743-00471.warc.gz

# Basic Electrical Wiring Manual ## Basic Electricity TutorialBasic Electricity Tutorial By Lydie Honorine. Diagram. Publised at Sunday, November 19th 2017, 20:50:47 PM. A block diagram shows a higher level (or organizational layout) of functional units in a circuit (or a device, machine, or collection of these). It is meant to show data flow or organization between separate units of function. A block diagram gives you an overview of the interconnected nature of circuit assemblies or components. ## Reflection And Refraction Electric EquationsReflection And Refraction Electric Equations By Sasha Sara. Diagram. Published at Monday, December 25th 2017, 19:47:12 PM. Circuits often require inputs that come directly from users (as opposed to inputs that come from other devices). User-input devices can take many forms, among them keyboards (as on a PC), buttons (as on a calculator or telephone), rotary dials, switches and levers, etc. Digilent boards include several input devices, typically including push buttons and slide-switches. Since digital circuits operate with two voltage levels (LHV or Vdd, and LLV or GND), input devices like buttons and switches should be able to produce both of these voltages based on some user action. ## H Bridge Circuit SchematicH Bridge Circuit Schematic By Lydie Honorine. Diagram. Published at Monday, December 25th 2017, 19:45:06 PM. In contrast to digital circuits, analog circuits use signals whose voltage levels are not constrained to two distinct levels, but instead can assume any value between Vdd and GND. Many input devices, particularly those using electronic sensors (e.g., microphones, cameras, thermometers, pressure sensors, motion and proximity detectors, etc.) produce analog voltages at their outputs. In modern electronic devices, it is likely that such signals will be converted to digital signals before they are used within the device. For example, a digital voice-memo recording device uses an analog microphone circuit to convert sound pressure waves into voltage waves on an internal circuit node. A special circuit called an analog-to-digital converter, or ADC, converts that analog voltage to a binary number that can be represented as a bus in a digital circuit. An ADC functions by taking samples of the input analog signal, measuring the magnitude of the input voltage signal (usually with reference to GND), and assigning a binary number to the measured magnitude. Once an analog signal has been converted to a binary number, a bus can carry that digital information around a circuit. In a similar manner, digital signals can be reconstituted into analog signals using a digital-to-analog converter. Thus, a binary number that represents a sample of an audio waveform can be converted to an analog signal that can, for example, drive a speaker. ### Filter Inductor Regulated VoltageFilter Inductor Regulated Voltage By Cyrielle Marjolaine. Circuit. Published at Monday, December 25th 2017, 19:17:03 PM. Electronic signals are represented either by voltage or current. The timedependent characteristics of voltage or current signals can take a number of forms including DC, sinusoidal (also known as AC), square wave, linear ramps, and pulsewidth modulated signals. Sinusoidal signals are perhaps the most important signal forms since once the circuit response to sinusoidal signals are known, the result can be generalized to predict how the circuit will respond to a much greater variety of signals using the mathematical tools of Fourier and Laplace transforms. #### 7806 Voltage Regulator Circuit Diagram7806 Voltage Regulator Circuit Diagram By Bertille Solange. Power. Published at Monday, December 25th 2017, 18:57:16 PM. A variation of the Standard regulator is the quasi-LDO, which uses an NPN and PNP transistor as the pass device. The dropout voltage for a quasi-LDO delivering rated current is usually specified at about 1.5V(max). The actual dropout voltage is temperature and load current dependent, but could never be expected to go lower than about 0.9V (25°C) at even the lightest load. The dropout voltage for the quasi-LDO is higher than the LDO, but lower than the Standard regulator. ##### Images Of Current Flow CircuitImages Of Current Flow Circuit By Jessica Mireille. Diagram. Published at Monday, December 25th 2017, 18:39:55 PM. Now’s the fun stuff. Completing an electrical engineering degree and then getting a job in the field means you will see a lot a lot a lot of these schematics. It’s important to understand exactly what is going on with these. While they can (and will) get very complex, these are just a few of the common graphics to get your footing on. ###### Binary Math Problems PicturesBinary Math Problems Pictures By Sasha Sara. Diagram. Published at Monday, December 25th 2017, 17:26:43 PM. A schematic shows connections in a circuit in a way that is clear and standardized. It is a way of communicating to other engineers exactly what components are involved in a circuit as well as how they are connected. A good schematic will show component names and values, and provide labels for sections or components to help communicate the intended purpose. Note how connections on wires (or "nets") are shown using dots and non-connections are shown without a dot. ## Voltmeter Parts And CircuitsVoltmeter Parts And Circuits By Lydie Honorine. Circuit. Published at Monday, December 25th 2017, 17:00:36 PM. Electric circuits use electric power to perform some function, like energize a heating or lighting element, turn a motor, or create an electromagnetic filed. Electronic circuits differ from electric circuits in that they use devices that can themselves be controlled by other electric signals. Restated, electronic circuits are built from devices that use electricity to control electricity. Most electronic circuits use signals that are within 5 to 10 volts of ground; most circuits built within the past several years use signals that are within 3 to 5 volts from ground. Some electronic circuits represent information encoded as continuous voltage levels that can wander between the high and low voltage supply rails – these are called analog circuits. As an example, a sound pressure level transducer (i.e. a microphone) might drive a signal between 0V and 3.3V in direct proportion to the detected sound pressure level. In this case, the voltage signal output from the microphone is said to be an analog ifthe sound pressure wave itself. Other circuits use only two distinct voltage levels to represent information. Most often, these two voltage levels use the same voltages supplied by the power rails. In these circuits, called digital circuits, all information must be represented as binary numbers, with a signal at 0V (or ground) representing one kind of information, and a signal at 3.3V (or whatever the upper voltage supply rail provides) representing the other kind of information. In this series of modules, we will confine our discussions to digital circuits. Recent post Category Monthly Archives Static Page Altrushare - Wiring Diagram Gallery Copyright © 2003 - 2018 Domain Media. All sponsored products, company names, brand names, trademarks and logos arethe property of their respective owners.

{"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}

CC-MAIN-2018-17

longest

en

http://coolconversion.com/weight/7.5-ton+%5Bshort%2C+US%5D-to-grain

text/html

crawl-data/CC-MAIN-2018-22/segments/1526794864461.53/warc/CC-MAIN-20180521161639-20180521181639-00625.warc.gz

Please get in touch with us if you: 1. Have any suggestions 2. Have any questions 3. Have found an error/bug 4. Anything else ... # 7.5 Tons [short, US] to Grains (sh tn to gr) Conversion ## How many grains in 7.5 tons [short, US]? There are 105000000 grains in 7.5 tons [short, US] To convert any value in tons [short, US] to grains, just multiply the value in tons [short, US] by the conversion factor 14000000. So, 7.5 tons [short, US] times 14000000 is equal to 105000000 grains. See details below and use our calculator to convert any value in tons [short, US] to grains. To use this tons [short, US] to grains, converter simply type the sh tn value in the box at left (input). The conversion result in gr will immediately appear in the box at right. If you are looking for a BMI Calculator, please click here. ### Tons [short, US] to grains Converter Enter values here:   Results here: = Detailed result here To calculate a ton [short, US] value to the corresponding value in grain, just multiply the quantity in tons [short, US] by 14000000 (the conversion factor). Here is the tons [short, US] to grains conversion formula: Value in grains = value in tons [short, US] * 14000000 Supose you want to convert 7.5 tons [short, US] into grains. In this case you will have: Value in grains = 7.5 * 14000000 = 105000000 (grains) Using this converter you can get answers to questions like: 1. How many tons [short, US] are in 7.5 grains? 2. 7.5 tons [short, US] are equal to how many grains? 3. how much are 7.5 ton [short, US] in grains? 4. How to convert tons [short, US] to grains? 5. What is the conversion factor to convert from tons [short, US] to grains? 6. How to transform tons [short, US] in grains? 7. What is the tons [short, US] to grains conversion formula? Among others. ## Values Near 7.5 ton [short, US] in grain ton [short, US]grain 6.793800000 6.895200000 6.996600000 798000000 7.199400000 7.2100800000 7.3102200000 7.4103600000 7.5105000000 7.6106400000 7.7107800000 7.8109200000 7.9110600000 8112000000 8.1113400000 8.2114800000 8.3116200000 Note: some values may be rounded. ## Sample Weight / Mass Conversions ### Disclaimer While every effort is made to ensure the accuracy of the information provided on this website, we offer no warranties in relation to these informations.

{"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}

CC-MAIN-2018-22

latest

en

https://isabelle.in.tum.de/repos/isabelle/diff/300bd596d696/src/Provers/Arith/combine_numerals.ML?revcount=60

text/html

crawl-data/CC-MAIN-2022-05/segments/1642320304872.21/warc/CC-MAIN-20220125190255-20220125220255-00011.warc.gz

src/Provers/Arith/combine_numerals.ML changeset 9191 300bd596d696 parent 8816 31b4fb3d8d60 child 9571 c869d1886a32 ```--- a/src/Provers/Arith/combine_numerals.ML Thu Jun 29 12:19:27 2000 +0200 +++ b/src/Provers/Arith/combine_numerals.ML Thu Jun 29 16:50:52 2000 +0200 @@ -64,13 +64,18 @@ (*the simplification procedure*) fun proc sg _ t = - let val (u,m,n,terms) = find_repeated (Termtab.empty, [], Data.dest_sum t) + let (*first freeze any Vars in the term to prevent flex-flex problems*) + val rand_s = gensym"_" + fun mk_inst (var as Var((a,i),T)) = + (var, Free((a ^ rand_s ^ string_of_int i), T)) + val t' = subst_atomic (map mk_inst (term_vars t)) t + val (u,m,n,terms) = find_repeated (Termtab.empty, [], Data.dest_sum t') val reshape = (*Move i*u to the front and put j*u into standard form i + #m + j + k == #m + i + (j + k) *) if m=0 orelse n=0 then (*trivial, so do nothing*) raise TERM("combine_numerals", []) else Data.prove_conv [Data.norm_tac] sg - (t, + (t', Data.mk_sum ([Data.mk_coeff(m,u), Data.mk_coeff(n,u)] @ terms)) in @@ -78,7 +83,7 @@ (Data.prove_conv [Data.trans_tac reshape, rtac Data.left_distrib 1, Data.numeral_simp_tac] sg - (t, Data.mk_sum (Data.mk_coeff(m+n,u) :: terms))) + (t', Data.mk_sum (Data.mk_coeff(m+n,u) :: terms))) end handle TERM _ => None | TYPE _ => None; (*Typically (if thy doesn't include Numeral)```

{"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}

CC-MAIN-2022-05

latest

en

http://lists.contesting.com/_towertalk/2004-08/msg00382.html?contestingsid=5741s5mj6kca3ne39ljg9nt7f2

text/html

crawl-data/CC-MAIN-2016-44/segments/1476988722951.82/warc/CC-MAIN-20161020183842-00091-ip-10-171-6-4.ec2.internal.warc.gz

Towertalk [Top] [All Lists] ## RE: [BULK] - RE: [TowerTalk] ground strap width: 3" vs 6"... To: [email protected], "TowerTalk" RE: [BULK] - RE: [TowerTalk] ground strap width: 3" vs 6"... Jim Lux Wed, 18 Aug 2004 11:45:42 -0700 At 01:19 PM 8/18/2004 -0500, Keith Dutson wrote: ```What frequency? I have always looked at lightning as a low/no frequency flow of electrons, i.e. like a DC current.``` Keith NM5G ```-----Original Message----- From: Steve Katz [mailto:[email protected]] Sent: Wednesday, August 18, 2004 12:37 PM To: '[email protected]'; TowerTalk Subject: RE: [BULK] - RE: [TowerTalk] ground strap width: 3" vs 6"...``` ```Isn't most impedance in a length of copper inductance? The resistive component should be very, very small.....-WB2WIK/6 ```Lightning impulses are fairly fast pulses (rise time of a few microseconds, fall times of several tens of microseconds). Lightning impulses induced from nearby hits will be a bit different, of course. Since what cooks your gear is the high voltage between input and ground, you'd want to be concerned about the L*di/dt thing, so minimizing L is a "good thing". If your gear were all floated, using the exact same ground as the lightning impulse is travelling through, then everything would go up and down together, and you'd be safe. HOWEVER... there are always stray paths (capacitive, for instance) and for microsecond rise time pulses, the impedance of that stray path might be a bit low for comfort (i.e. the current through the path is high enough to produce voltage drops that will zap a junction). There's also the whole induced voltage problem. A wire carrying a current that changes quickly (high di/dt) can induce a voltage in an nearby conductor, thereby causing damage. For your basic lightning impulse, you're looking at 10kA in a microsecond, or a di/dt of 1E10 A/second. That will make a pretty healthy magnetic field transient to couple to the victim loop. In any case, far from a DC current. _______________________________________________ See: http://www.mscomputer.com for "Self Supporting Towers", "Wireless Weather Stations", and lot's more. Call Toll Free, 1-800-333-9041 with any questions and ask for Sherman, W2FLA. ```_______________________________________________ TowerTalk mailing list [email protected] http://lists.contesting.com/mailman/listinfo/towertalk``` Current Thread RE: [BULK] - RE: [TowerTalk] ground strap width: 3" vs 6"..., Steve Katz RE: [BULK] - RE: [TowerTalk] ground strap width: 3" vs 6"..., Keith Dutson RE: [BULK] - RE: [TowerTalk] ground strap width: 3" vs 6"..., Jim Lux <=

{"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}

CC-MAIN-2016-44

latest

en

https://apache.googlesource.com/impala/+/79aae231443a305ce8503dbc7b4335e8ae3f3946/docs/topics/impala_operators.xml

text/html

crawl-data/CC-MAIN-2020-45/segments/1603107880656.25/warc/CC-MAIN-20201023043931-20201023073931-00532.warc.gz

blob: 066645788dd11562e7247ec1e318062710d87b40 [file] [log] [blame] SQL Operators operators SQL operators are a class of comparison functions that are widely used within the WHERE clauses of SELECT statements. Arithmetic Operators arithmetic operators The arithmetic operators use expressions with a left-hand argument, the operator, and then (in most cases) a right-hand argument. left_hand_arg binary_operator right_hand_arg unary_operator single_arg • + and -: Can be used either as unary or binary operators. • With unary notation, such as +5, -2.5, or -col_name, they multiply their single numeric argument by +1 or -1. Therefore, unary + returns its argument unchanged, while unary - flips the sign of its argument. Although you can double up these operators in expressions such as ++5 (always positive) or -+2 or +-2 (both always negative), you cannot double the unary minus operator because -- is interpreted as the start of a comment. (You can use a double unary minus operator if you separate the - characters, for example with a space or parentheses.) • With binary notation, such as 2+2, 5-2.5, or col1 + col2, they add or subtract respectively the right-hand argument to (or from) the left-hand argument. Both arguments must be of numeric types. • * and /: Multiplication and division respectively. Both arguments must be of numeric types. When multiplying, the shorter argument is promoted if necessary (such as SMALLINT to INT or BIGINT, or FLOAT to DOUBLE), and then the result is promoted again to the next larger type. Thus, multiplying a TINYINT and an INT produces a BIGINT result. Multiplying a FLOAT and a FLOAT produces a DOUBLE result. Multiplying a FLOAT and a DOUBLE or a DOUBLE and a DOUBLE produces a DECIMAL(38,17), because DECIMAL values can represent much larger and more precise values than DOUBLE. When dividing, Impala always treats the arguments and result as DOUBLE values to avoid losing precision. If you need to insert the results of a division operation into a FLOAT column, use the CAST() function to convert the result to the correct type. • DIV: Integer division. Arguments are not promoted to a floating-point type, and any fractional result is discarded. For example, 13 DIV 7 returns 1, 14 DIV 7 returns 2, and 15 DIV 7 returns 2. This operator is the same as the QUOTIENT() function. • %: Modulo operator. Returns the remainder of the left-hand argument divided by the right-hand argument. Both arguments must be of one of the integer types. • &, |, ~, and ^: Bitwise operators that return the logical AND, logical OR, NOT, or logical XOR (exclusive OR) of their argument values. Both arguments must be of one of the integer types. If the arguments are of different type, the argument with the smaller type is implicitly extended to match the argument with the longer type. You can chain a sequence of arithmetic expressions, optionally grouping them with parentheses. The arithmetic operators generally do not have equivalent calling conventions using functional notation. For example, prior to , there is no MOD() function equivalent to the % modulo operator. Conversely, there are some arithmetic functions that do not have a corresponding operator. For example, for exponentiation you use the POW() function, but there is no ** exponentiation operator. See for the arithmetic functions you can use. The following example shows how to do an arithmetic operation using a numeric field of a STRUCT type that is an item within an ARRAY column. Once the scalar numeric value R_NATIONKEY is extracted, it can be used in an arithmetic expression, such as multiplying by 10: -- The SMALLINT is a field within an array of structs. describe region; +-------------+-------------------------+---------+ | name | type | comment | +-------------+-------------------------+---------+ | r_regionkey | smallint | | | r_name | string | | | r_comment | string | | | r_nations | array<struct< | | | | n_nationkey:smallint, | | | | n_name:string, | | | | n_comment:string | | | | >> | | +-------------+-------------------------+---------+ -- When we refer to the scalar value using dot notation, -- we can use arithmetic and comparison operators on it -- like any other number. select r_name, nation.item.n_name, nation.item.n_nationkey * 10 from region, region.r_nations as nation where nation.item.n_nationkey < 5; +-------------+-------------+------------------------------+ | r_name | item.n_name | nation.item.n_nationkey * 10 | +-------------+-------------+------------------------------+ | AMERICA | CANADA | 30 | | AMERICA | BRAZIL | 20 | | AMERICA | ARGENTINA | 10 | | MIDDLE EAST | EGYPT | 40 | | AFRICA | ALGERIA | 0 | +-------------+-------------+------------------------------+ BETWEEN Operator BETWEEN operator In a WHERE clause, compares an expression to both a lower and upper bound. The comparison is successful is the expression is greater than or equal to the lower bound, and less than or equal to the upper bound. If the bound values are switched, so the lower bound is greater than the upper bound, does not match any values. expression BETWEEN lower_bound AND upper_bound Data types: Typically used with numeric data types. Works with any data type, although not very practical for BOOLEAN values. (BETWEEN false AND true will match all BOOLEAN values.) Use CAST() if necessary to ensure the lower and upper bound values are compatible types. Call string or date/time functions if necessary to extract or transform the relevant portion to compare, especially if the value can be transformed into a number. Be careful when using short string operands. A longer string that starts with the upper bound value will not be included, because it is considered greater than the upper bound. For example, BETWEEN 'A' and 'M' would not match the string value 'Midway'. Use functions such as upper(), lower(), substr(), trim(), and so on if necessary to ensure the comparison works as expected. -- Retrieve data for January through June, inclusive. select c1 from t1 where month between 1 and 6; -- Retrieve data for names beginning with 'A' through 'M' inclusive. -- Only test the first letter to ensure all the values starting with 'M' are matched. -- Do a case-insensitive comparison to match names with various capitalization conventions. select last_name from customers where upper(substr(last_name,1,1)) between 'A' and 'M'; -- Retrieve data for only the first week of each month. select count(distinct visitor_id)) from web_traffic where dayofmonth(when_viewed) between 1 and 7; The following example shows how to do a BETWEEN comparison using a numeric field of a STRUCT type that is an item within an ARRAY column. Once the scalar numeric value R_NATIONKEY is extracted, it can be used in a comparison operator: -- The SMALLINT is a field within an array of structs. describe region; +-------------+-------------------------+---------+ | name | type | comment | +-------------+-------------------------+---------+ | r_regionkey | smallint | | | r_name | string | | | r_comment | string | | | r_nations | array<struct< | | | | n_nationkey:smallint, | | | | n_name:string, | | | | n_comment:string | | | | >> | | +-------------+-------------------------+---------+ -- When we refer to the scalar value using dot notation, -- we can use arithmetic and comparison operators on it -- like any other number. select r_name, nation.item.n_name, nation.item.n_nationkey from region, region.r_nations as nation where nation.item.n_nationkey between 3 and 5 +-------------+-------------+------------------+ | r_name | item.n_name | item.n_nationkey | +-------------+-------------+------------------+ | AMERICA | CANADA | 3 | | MIDDLE EAST | EGYPT | 4 | | AFRICA | ETHIOPIA | 5 | +-------------+-------------+------------------+ Comparison Operators comparison operators Impala supports the familiar comparison operators for checking equality and sort order for the column data types: left_hand_expression comparison_operator right_hand_expression • =, !=, <>: apply to all types. • <, <=, >, >=: apply to all types; for BOOLEAN, TRUE is considered greater than FALSE. Alternatives: The IN and BETWEEN operators provide shorthand notation for expressing combinations of equality, less than, and greater than comparisons with a single operator. Because comparing any value to NULL produces NULL rather than TRUE or FALSE, use the IS NULL and IS NOT NULL operators to check if a value is NULL or not. The following example shows how to do an arithmetic operation using a numeric field of a STRUCT type that is an item within an ARRAY column. Once the scalar numeric value R_NATIONKEY is extracted, it can be used with a comparison operator such as <: -- The SMALLINT is a field within an array of structs. describe region; +-------------+-------------------------+---------+ | name | type | comment | +-------------+-------------------------+---------+ | r_regionkey | smallint | | | r_name | string | | | r_comment | string | | | r_nations | array<struct< | | | | n_nationkey:smallint, | | | | n_name:string, | | | | n_comment:string | | | | >> | | +-------------+-------------------------+---------+ -- When we refer to the scalar value using dot notation, -- we can use arithmetic and comparison operators on it -- like any other number. select r_name, nation.item.n_name, nation.item.n_nationkey from region, region.r_nations as nation where nation.item.n_nationkey < 5 +-------------+-------------+------------------+ | r_name | item.n_name | item.n_nationkey | +-------------+-------------+------------------+ | AMERICA | CANADA | 3 | | AMERICA | BRAZIL | 2 | | AMERICA | ARGENTINA | 1 | | MIDDLE EAST | EGYPT | 4 | | AFRICA | ALGERIA | 0 | +-------------+-------------+------------------+ EXCEPT Operator EXCEPT operator EXISTS Operator EXISTS operator NOT EXISTS operator The EXISTS operator tests whether a subquery returns any results. You typically use it to find values from one table that have corresponding values in another table. The converse, NOT EXISTS, helps to find all the values from one table that do not have any corresponding values in another table. EXISTS (subquery) NOT EXISTS (subquery) The subquery can refer to a different table than the outer query block, or the same table. For example, you might use EXISTS or NOT EXISTS to check the existence of parent/child relationships between two columns of the same table. You can also use operators and function calls within the subquery to test for other kinds of relationships other than strict equality. For example, you might use a call to COUNT() in the subquery to check whether the number of matching values is higher or lower than some limit. You might call a UDF in the subquery to check whether values in one table matches a hashed representation of those same values in a different table. If the subquery returns any value at all (even NULL), EXISTS returns TRUE and NOT EXISTS returns false. The following example shows how even when the subquery returns only NULL values, EXISTS still returns TRUE and thus matches all the rows from the table in the outer query block. [localhost:21000] > create table all_nulls (x int); [localhost:21000] > insert into all_nulls values (null), (null), (null); [localhost:21000] > select y from t2 where exists (select x from all_nulls); +---+ | y | +---+ | 2 | | 4 | | 6 | +---+ However, if the table in the subquery is empty and so the subquery returns an empty result set, EXISTS returns FALSE: [localhost:21000] > create table empty (x int); [localhost:21000] > select y from t2 where exists (select x from empty); [localhost:21000] > Prior to , the NOT EXISTS operator required a correlated subquery. In and higher, NOT EXISTS works with uncorrelated queries also. The following examples refer to these simple tables containing small sets of integers or strings: [localhost:21000] > create table t1 (x int); [localhost:21000] > insert into t1 values (1), (2), (3), (4), (5), (6); [localhost:21000] > create table t2 (y int); [localhost:21000] > insert into t2 values (2), (4), (6); [localhost:21000] > create table t3 (z int); [localhost:21000] > insert into t3 values (1), (3), (5); [localhost:21000] > create table month_names (m string); [localhost:21000] > insert into month_names values > ('January'), ('February'), ('March'), > ('April'), ('May'), ('June'), ('July'), > ('August'), ('September'), ('October'), > ('November'), ('December'); The following example shows a correlated subquery that finds all the values in one table that exist in another table. For each value X from T1, the query checks if the Y column of T2 contains an identical value, and the EXISTS operator returns TRUE or FALSE as appropriate in each case. localhost:21000] > select x from t1 where exists (select y from t2 where t1.x = y); +---+ | x | +---+ | 2 | | 4 | | 6 | +---+ An uncorrelated query is less interesting in this case. Because the subquery always returns TRUE, all rows from T1 are returned. If the table contents where changed so that the subquery did not match any rows, none of the rows from T1 would be returned. [localhost:21000] > select x from t1 where exists (select y from t2 where y > 5); +---+ | x | +---+ | 1 | | 2 | | 3 | | 4 | | 5 | | 6 | +---+ The following example shows how an uncorrelated subquery can test for the existence of some condition within a table. By using LIMIT 1 or an aggregate function, the query returns a single result or no result based on whether the subquery matches any rows. Here, we know that T1 and T2 contain some even numbers, but T3 does not. [localhost:21000] > select "contains an even number" from t1 where exists (select x from t1 where x % 2 = 0) limit 1; +---------------------------+ | 'contains an even number' | +---------------------------+ | contains an even number | +---------------------------+ [localhost:21000] > select "contains an even number" as assertion from t1 where exists (select x from t1 where x % 2 = 0) limit 1; +-------------------------+ | assertion | +-------------------------+ | contains an even number | +-------------------------+ [localhost:21000] > select "contains an even number" as assertion from t2 where exists (select x from t2 where y % 2 = 0) limit 1; ERROR: AnalysisException: couldn't resolve column reference: 'x' [localhost:21000] > select "contains an even number" as assertion from t2 where exists (select y from t2 where y % 2 = 0) limit 1; +-------------------------+ | assertion | +-------------------------+ | contains an even number | +-------------------------+ [localhost:21000] > select "contains an even number" as assertion from t3 where exists (select z from t3 where z % 2 = 0) limit 1; [localhost:21000] > The following example finds numbers in one table that are 1 greater than numbers from another table. The EXISTS notation is simpler than an equivalent CROSS JOIN between the tables. (The example then also illustrates how the same test could be performed using an IN operator.) [localhost:21000] > select x from t1 where exists (select y from t2 where x = y + 1); +---+ | x | +---+ | 3 | | 5 | +---+ [localhost:21000] > select x from t1 where x in (select y + 1 from t2); +---+ | x | +---+ | 3 | | 5 | +---+ The following example finds values from one table that do not exist in another table. [localhost:21000] > select x from t1 where not exists (select y from t2 where x = y); +---+ | x | +---+ | 1 | | 3 | | 5 | +---+ The following example uses the NOT EXISTS operator to find all the leaf nodes in tree-structured data. This simplified tree of life has multiple levels (class, order, family, and so on), with each item pointing upward through a PARENT pointer. The example runs an outer query and a subquery on the same table, returning only those items whose ID value is not referenced by the PARENT of any other item. [localhost:21000] > create table tree (id int, parent int, name string); [localhost:21000] > insert overwrite tree values > (0, null, "animals"), > (1, 0, "placentals"), > (2, 0, "marsupials"), > (3, 1, "bats"), > (4, 1, "cats"), > (5, 2, "kangaroos"), > (6, 4, "lions"), > (7, 4, "tigers"), > (8, 5, "red kangaroo"), > (9, 2, "wallabies"); [localhost:21000] > select name as "leaf node" from tree one > where not exists (select parent from tree two where one.id = two.parent); +--------------+ | leaf node | +--------------+ | bats | | lions | | tigers | | red kangaroo | | wallabies | +--------------+ ILIKE Operator ILIKE operator A case-insensitive comparison operator for STRING data, with basic wildcard capability using _ to match a single character and % to match multiple characters. The argument expression must match the entire string value. Typically, it is more efficient to put any % wildcard match at the end of the string. This operator, available in and higher, is the equivalent of the LIKE operator, but with case-insensitive comparisons. string_expression ILIKE wildcard_expression string_expression NOT ILIKE wildcard_expression In the following examples, strings that are the same except for differences in uppercase and lowercase match successfully with ILIKE, but do not match with LIKE: select 'fooBar' ilike 'FOOBAR'; +-------------------------+ | 'foobar' ilike 'foobar' | +-------------------------+ | true | +-------------------------+ select 'fooBar' like 'FOOBAR'; +------------------------+ | 'foobar' like 'foobar' | +------------------------+ | false | +------------------------+ select 'FOOBAR' ilike 'f%'; +---------------------+ | 'foobar' ilike 'f%' | +---------------------+ | true | +---------------------+ select 'FOOBAR' like 'f%'; +--------------------+ | 'foobar' like 'f%' | +--------------------+ | false | +--------------------+ select 'ABCXYZ' not ilike 'ab_xyz'; +-----------------------------+ | not 'abcxyz' ilike 'ab_xyz' | +-----------------------------+ | false | +-----------------------------+ select 'ABCXYZ' not like 'ab_xyz'; +----------------------------+ | not 'abcxyz' like 'ab_xyz' | +----------------------------+ | true | +----------------------------+ For case-sensitive comparisons, see . For a more general kind of search operator using regular expressions, see or its case-insensitive counterpart . IN Operator IN operator NOT IN operator The IN operator compares an argument value to a set of values, and returns TRUE if the argument matches any value in the set. The NOT IN operator reverses the comparison, and checks if the argument value is not part of a set of values. expression IN (expression [, expression]) expression IN (subquery) expression NOT IN (expression [, expression]) expression NOT IN (subquery) The left-hand expression and the set of comparison values must be of compatible types. The left-hand expression must consist only of a single value, not a tuple. Although the left-hand expression is typically a column name, it could also be some other value. For example, the WHERE clauses WHERE id IN (5) and WHERE 5 IN (id) produce the same results. The set of values to check against can be specified as constants, function calls, column names, or other expressions in the query text. The maximum number of expressions in the IN list is 9999. (The maximum number of elements of a single expression is 10,000 items, and the IN operator itself counts as one.) In Impala 2.0 and higher, the set of values can also be generated by a subquery. IN can evaluate an unlimited number of results using a subquery. Any expression using the IN operator could be rewritten as a series of equality tests connected with OR, but the IN syntax is often clearer, more concise, and easier for Impala to optimize. For example, with partitioned tables, queries frequently use IN clauses to filter data by comparing the partition key columns to specific values. If there really is a matching non-null value, IN returns TRUE: [localhost:21000] > select 1 in (1,null,2,3); +----------------------+ | 1 in (1, null, 2, 3) | +----------------------+ | true | +----------------------+ [localhost:21000] > select 1 not in (1,null,2,3); +--------------------------+ | 1 not in (1, null, 2, 3) | +--------------------------+ | false | +--------------------------+ If the searched value is not found in the comparison values, and the comparison values include NULL, the result is NULL: [localhost:21000] > select 5 in (1,null,2,3); +----------------------+ | 5 in (1, null, 2, 3) | +----------------------+ | NULL | +----------------------+ [localhost:21000] > select 5 not in (1,null,2,3); +--------------------------+ | 5 not in (1, null, 2, 3) | +--------------------------+ | NULL | +--------------------------+ [localhost:21000] > select 1 in (null); +-------------+ | 1 in (null) | +-------------+ | NULL | +-------------+ [localhost:21000] > select 1 not in (null); +-----------------+ | 1 not in (null) | +-----------------+ | NULL | +-----------------+ If the left-hand argument is NULL, IN always returns NULL. This rule applies even if the comparison values include NULL. [localhost:21000] > select null in (1,2,3); +-------------------+ | null in (1, 2, 3) | +-------------------+ | NULL | +-------------------+ [localhost:21000] > select null not in (1,2,3); +-----------------------+ | null not in (1, 2, 3) | +-----------------------+ | NULL | +-----------------------+ [localhost:21000] > select null in (null); +----------------+ | null in (null) | +----------------+ | NULL | +----------------+ [localhost:21000] > select null not in (null); +--------------------+ | null not in (null) | +--------------------+ | NULL | +--------------------+ The following example shows how to do an arithmetic operation using a numeric field of a STRUCT type that is an item within an ARRAY column. Once the scalar numeric value R_NATIONKEY is extracted, it can be used in an arithmetic expression, such as multiplying by 10: -- The SMALLINT is a field within an array of structs. describe region; +-------------+-------------------------+---------+ | name | type | comment | +-------------+-------------------------+---------+ | r_regionkey | smallint | | | r_name | string | | | r_comment | string | | | r_nations | array<struct< | | | | n_nationkey:smallint, | | | | n_name:string, | | | | n_comment:string | | | | >> | | +-------------+-------------------------+---------+ -- When we refer to the scalar value using dot notation, -- we can use arithmetic and comparison operators on it -- like any other number. select r_name, nation.item.n_name, nation.item.n_nationkey from region, region.r_nations as nation where nation.item.n_nationkey in (1,3,5) +---------+-------------+------------------+ | r_name | item.n_name | item.n_nationkey | +---------+-------------+------------------+ | AMERICA | CANADA | 3 | | AMERICA | ARGENTINA | 1 | | AFRICA | ETHIOPIA | 5 | +---------+-------------+------------------+ -- Using IN is concise and self-documenting. SELECT * FROM t1 WHERE c1 IN (1,2,10); -- Equivalent to series of = comparisons ORed together. SELECT * FROM t1 WHERE c1 = 1 OR c1 = 2 OR c1 = 10; SELECT c1 AS "starts with vowel" FROM t2 WHERE upper(substr(c1,1,1)) IN ('A','E','I','O','U'); SELECT COUNT(DISTINCT(visitor_id)) FROM web_traffic WHERE month IN ('January','June','July'); INTERSECT Operator INTERSECT operator IREGEXP Operator IREGEXP operator Tests whether a value matches a regular expression, using case-insensitive string comparisons. Uses the POSIX regular expression syntax where ^ and \$ match the beginning and end of the string, . represents any single character, * represents a sequence of zero or more items, + represents a sequence of one or more items, ? produces a non-greedy match, and so on. This operator, available in and higher, is the equivalent of the REGEXP operator, but with case-insensitive comparisons. string_expression IREGEXP regular_expression The | symbol is the alternation operator, typically used within () to match different sequences. The () groups do not allow backreferences. To retrieve the part of a value matched within a () section, use the regexp_extract() built-in function. (Currently, there is not any case-insensitive equivalent for the regexp_extract() function.) The following examples demonstrate the syntax for the IREGEXP operator. select 'abcABCaabbcc' iregexp '^[a-c]+\$'; +---------------------------------+ | 'abcabcaabbcc' iregexp '[a-c]+' | +---------------------------------+ | true | +---------------------------------+ IS DISTINCT FROM Operator The IS DISTINCT FROM operator, and its converse the IS NOT DISTINCT FROM operator, test whether or not values are identical. IS NOT DISTINCT FROM is similar to the = operator, and IS DISTINCT FROM is similar to the != operator, except that NULL values are treated as identical. Therefore, IS NOT DISTINCT FROM returns true rather than NULL, and IS DISTINCT FROM returns false rather than NULL, when comparing two NULL values. If one of the values being compared is NULL and the other is not, IS DISTINCT FROM returns true and IS NOT DISTINCT FROM returns false, again instead of returning NULL in both cases. expression1 IS DISTINCT FROM expression2 expression1 IS NOT DISTINCT FROM expression2 expression1 <=> expression2 The operator <=> is an alias for IS NOT DISTINCT FROM. It is typically used as a NULL-safe equality operator in join queries. That is, A <=> B is true if A equals B or if both A and B are NULL. This operator provides concise notation for comparing two values and always producing a true or false result, without treating NULL as a special case. Otherwise, to unambiguously distinguish between two values requires a compound expression involving IS [NOT] NULL tests of both operands in addition to the = or != operator. The <=> operator, used like an equality operator in a join query, is more efficient than the equivalent clause: IF (A IS NULL OR B IS NULL, A IS NULL AND B IS NULL, A = B). The <=> operator can use a hash join, while the IF expression cannot. The following examples show how IS DISTINCT FROM gives output similar to the != operator, and IS NOT DISTINCT FROM gives output similar to the = operator. The exception is when the expression involves a NULL value on one side or both sides, where != and = return NULL but the IS [NOT] DISTINCT FROM operators still return true or false. select 1 is distinct from 0, 1 != 0; +----------------------+--------+ | 1 is distinct from 0 | 1 != 0 | +----------------------+--------+ | true | true | +----------------------+--------+ select 1 is distinct from 1, 1 != 1; +----------------------+--------+ | 1 is distinct from 1 | 1 != 1 | +----------------------+--------+ | false | false | +----------------------+--------+ select 1 is distinct from null, 1 != null; +-------------------------+-----------+ | 1 is distinct from null | 1 != null | +-------------------------+-----------+ | true | NULL | +-------------------------+-----------+ select null is distinct from null, null != null; +----------------------------+--------------+ | null is distinct from null | null != null | +----------------------------+--------------+ | false | NULL | +----------------------------+--------------+ select 1 is not distinct from 0, 1 = 0; +--------------------------+-------+ | 1 is not distinct from 0 | 1 = 0 | +--------------------------+-------+ | false | false | +--------------------------+-------+ select 1 is not distinct from 1, 1 = 1; +--------------------------+-------+ | 1 is not distinct from 1 | 1 = 1 | +--------------------------+-------+ | true | true | +--------------------------+-------+ select 1 is not distinct from null, 1 = null; +-----------------------------+----------+ | 1 is not distinct from null | 1 = null | +-----------------------------+----------+ | false | NULL | +-----------------------------+----------+ select null is not distinct from null, null = null; +--------------------------------+-------------+ | null is not distinct from null | null = null | +--------------------------------+-------------+ | true | NULL | +--------------------------------+-------------+ The following example shows how IS DISTINCT FROM considers CHAR values to be the same (not distinct from each other) if they only differ in the number of trailing spaces. Therefore, sometimes the result of an IS [NOT] DISTINCT FROM operator differs depending on whether the values are STRING/VARCHAR or CHAR. select 'x' is distinct from 'x ' as string_with_trailing_spaces, cast('x' as char(5)) is distinct from cast('x ' as char(5)) as char_with_trailing_spaces; +-----------------------------+---------------------------+ | string_with_trailing_spaces | char_with_trailing_spaces | +-----------------------------+---------------------------+ | true | false | +-----------------------------+---------------------------+ IS NULL Operator IS NULL operator IS NOT NULL operator IS UNKNOWN operator IS NOT UNKNOWN operator The IS NULL operator, and its converse the IS NOT NULL operator, test whether a specified value is NULL. Because using NULL with any of the other comparison operators such as = or != also returns NULL rather than TRUE or FALSE, you use a special-purpose comparison operator to check for this special condition. In and higher, you can use the operators IS UNKNOWN and IS NOT UNKNOWN as synonyms for IS NULL and IS NOT NULL, respectively. expression IS NULL expression IS NOT NULL expression IS UNKNOWN expression IS NOT UNKNOWN In many cases, NULL values indicate some incorrect or incomplete processing during data ingestion or conversion. You might check whether any values in a column are NULL, and if so take some followup action to fill them in. With sparse data, often represented in wide tables, it is common for most values to be NULL with only an occasional non-NULL value. In those cases, you can use the IS NOT NULL operator to identify the rows containing any data at all for a particular column, regardless of the actual value. With a well-designed database schema, effective use of NULL values and IS NULL and IS NOT NULL operators can save having to design custom logic around special values such as 0, -1, 'N/A', empty string, and so on. NULL lets you distinguish between a value that is known to be 0, false, or empty, and a truly unknown value. The IS [NOT] UNKNOWN operator, as with the IS [NOT] NULL operator, is not applicable to complex type columns (STRUCT, ARRAY, or MAP). Using a complex type column with this operator causes a query error. -- If this value is non-zero, something is wrong. select count(*) from employees where employee_id is null; -- With data from disparate sources, some fields might be blank. -- Not necessarily an error condition. select count(*) from census where household_income is null; -- Sometimes we expect fields to be null, and followup action -- is needed when they are not. select count(*) from web_traffic where weird_http_code is not null; IS TRUE Operator IS TRUE operator IS FALSE operator IS NOT TRUE operator IS NOT FALSE operator This variation of the IS operator tests for truth or falsity, with right-hand arguments [NOT] TRUE, [NOT] FALSE, and [NOT] UNKNOWN. expression IS TRUE expression IS NOT TRUE expression IS FALSE expression IS NOT FALSE This IS TRUE and IS FALSE forms are similar to doing equality comparisons with the Boolean values TRUE and FALSE, except that IS TRUE and IS FALSE always return either TRUE or FALSE, even if the left-hand side expression returns NULL These operators let you simplify Boolean comparisons that must also check for NULL, for example X != 10 AND X IS NOT NULL is equivalent to (X != 10) IS TRUE. The IS [NOT] TRUE and IS [NOT] FALSE operators are not applicable to complex type columns (STRUCT, ARRAY, or MAP). Using a complex type column with these operators causes a query error. select assertion, b, b is true, b is false, b is unknown from boolean_test; +-------------+-------+-----------+------------+-----------+ | assertion | b | istrue(b) | isfalse(b) | b is null | +-------------+-------+-----------+------------+-----------+ | 2 + 2 = 4 | true | true | false | false | | 2 + 2 = 5 | false | false | true | false | | 1 = null | NULL | false | false | true | | null = null | NULL | false | false | true | +-------------+-------+-----------+------------+-----------+ LIKE Operator LIKE operator A comparison operator for STRING data, with basic wildcard capability using the underscore (_) to match a single character and the percent sign (%) to match multiple characters. The argument expression must match the entire string value. Typically, it is more efficient to put any % wildcard match at the end of the string. string_expression LIKE wildcard_expression string_expression NOT LIKE wildcard_expression select distinct c_last_name from customer where c_last_name like 'Mc%' or c_last_name like 'Mac%'; select count(c_last_name) from customer where c_last_name like 'M%'; select c_email_address from customer where c_email_address like '%.edu'; -- We can find 4-letter names beginning with 'M' by calling functions... select distinct c_last_name from customer where length(c_last_name) = 4 and substr(c_last_name,1,1) = 'M'; -- ...or in a more readable way by matching M followed by exactly 3 characters. select distinct c_last_name from customer where c_last_name like 'M___'; For case-insensitive comparisons, see . For a more general kind of search operator using regular expressions, see or its case-insensitive counterpart . Logical Operators logical operators Logical operators return a BOOLEAN value, based on a binary or unary logical operation between arguments that are also Booleans. Typically, the argument expressions use comparison operators. boolean_expression binary_logical_operator boolean_expression unary_logical_operator boolean_expression The Impala logical operators are: • AND: A binary operator that returns true if its left-hand and right-hand arguments both evaluate to true, NULL if either argument is NULL, and false otherwise. • OR: A binary operator that returns true if either of its left-hand and right-hand arguments evaluate to true, NULL if one argument is NULL and the other is either NULL or false, and false otherwise. • NOT: A unary operator that flips the state of a Boolean expression from true to false, or false to true. If the argument expression is NULL, the result remains NULL. (When NOT is used this way as a unary logical operator, it works differently than the IS NOT NULL comparison operator, which returns true when applied to a NULL.) The following example shows how to do an arithmetic operation using a numeric field of a STRUCT type that is an item within an ARRAY column. Once the scalar numeric value R_NATIONKEY is extracted, it can be used in an arithmetic expression, such as multiplying by 10: -- The SMALLINT is a field within an array of structs. describe region; +-------------+-------------------------+---------+ | name | type | comment | +-------------+-------------------------+---------+ | r_regionkey | smallint | | | r_name | string | | | r_comment | string | | | r_nations | array<struct< | | | | n_nationkey:smallint, | | | | n_name:string, | | | | n_comment:string | | | | >> | | +-------------+-------------------------+---------+ -- When we refer to the scalar value using dot notation, -- we can use arithmetic and comparison operators on it -- like any other number. select r_name, nation.item.n_name, nation.item.n_nationkey from region, region.r_nations as nation where nation.item.n_nationkey between 3 and 5 or nation.item.n_nationkey < 15; +-------------+----------------+------------------+ | r_name | item.n_name | item.n_nationkey | +-------------+----------------+------------------+ | EUROPE | UNITED KINGDOM | 23 | | EUROPE | RUSSIA | 22 | | EUROPE | ROMANIA | 19 | | ASIA | VIETNAM | 21 | | ASIA | CHINA | 18 | | AMERICA | UNITED STATES | 24 | | AMERICA | PERU | 17 | | AMERICA | CANADA | 3 | | MIDDLE EAST | SAUDI ARABIA | 20 | | MIDDLE EAST | EGYPT | 4 | | AFRICA | MOZAMBIQUE | 16 | | AFRICA | ETHIOPIA | 5 | +-------------+----------------+------------------+ These examples demonstrate the AND operator: [localhost:21000] > select true and true; +---------------+ | true and true | +---------------+ | true | +---------------+ [localhost:21000] > select true and false; +----------------+ | true and false | +----------------+ | false | +----------------+ [localhost:21000] > select false and false; +-----------------+ | false and false | +-----------------+ | false | +-----------------+ [localhost:21000] > select true and null; +---------------+ | true and null | +---------------+ | NULL | +---------------+ [localhost:21000] > select (10 > 2) and (6 != 9); +-----------------------+ | (10 > 2) and (6 != 9) | +-----------------------+ | true | +-----------------------+ These examples demonstrate the OR operator: [localhost:21000] > select true or true; +--------------+ | true or true | +--------------+ | true | +--------------+ [localhost:21000] > select true or false; +---------------+ | true or false | +---------------+ | true | +---------------+ [localhost:21000] > select false or false; +----------------+ | false or false | +----------------+ | false | +----------------+ [localhost:21000] > select true or null; +--------------+ | true or null | +--------------+ | true | +--------------+ [localhost:21000] > select null or true; +--------------+ | null or true | +--------------+ | true | +--------------+ [localhost:21000] > select false or null; +---------------+ | false or null | +---------------+ | NULL | +---------------+ [localhost:21000] > select (1 = 1) or ('hello' = 'world'); +--------------------------------+ | (1 = 1) or ('hello' = 'world') | +--------------------------------+ | true | +--------------------------------+ [localhost:21000] > select (2 + 2 != 4) or (-1 > 0); +--------------------------+ | (2 + 2 != 4) or (-1 > 0) | +--------------------------+ | false | +--------------------------+ These examples demonstrate the NOT operator: [localhost:21000] > select not true; +----------+ | not true | +----------+ | false | +----------+ [localhost:21000] > select not false; +-----------+ | not false | +-----------+ | true | +-----------+ [localhost:21000] > select not null; +----------+ | not null | +----------+ | NULL | +----------+ [localhost:21000] > select not (1=1); +-------------+ | not (1 = 1) | +-------------+ | false | +-------------+ REGEXP Operator REGEXP operator Tests whether a value matches a regular expression. Uses the POSIX regular expression syntax where ^ and \$ match the beginning and end of the string, . represents any single character, * represents a sequence of zero or more items, + represents a sequence of one or more items, ? produces a non-greedy match, and so on. string_expression REGEXP regular_expression The RLIKE operator is a synonym for REGEXP. The | symbol is the alternation operator, typically used within () to match different sequences. The () groups do not allow backreferences. To retrieve the part of a value matched within a () section, use the regexp_extract() built-in function. The following examples demonstrate the identical syntax for the REGEXP and RLIKE operators. For regular expression matching with case-insensitive comparisons, see . RLIKE Operator RLIKE operator Synonym for the REGEXP operator. See for details. The following examples demonstrate the identical syntax for the REGEXP and RLIKE operators.

{"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}

CC-MAIN-2020-45

latest

en

https://memim.com/discrete-fourier-transform.html

text/html

crawl-data/CC-MAIN-2024-22/segments/1715971058822.89/warc/CC-MAIN-20240525100447-20240525130447-00243.warc.gz

 Discrete Fourier transform # Discrete Fourier transform The discrete Fourier transform or DFT is a transformation from the region of the Fourier analysis. It forms a time-discrete, finite signal which is periodically continued on a discrete periodic frequency spectrum, which is also referred to as the image area. The DFT has for signal analysis of great importance in the digital signal processing. Here you can find optimized variants in the form of the fast Fourier transform ( english Fast Fourier Transform, FFT) and its inverse application. DFT is used in the signal processing for many tasks, such as • To determine the in a sampled signal mainly occurring frequencies, • For the determination of the individual amplitudes of these frequencies • For the implementation of digital filters with large filter lengths Of the inverse DFT may be reconstructed in the time domain of the short IDFT frequency components again the signal. By coupling of the DFT and IDFT, a signal can be manipulated in the frequency domain as applied in the equalizer. The discrete Fourier transformation is related by the Fourier transform of time-discrete signals (English Discrete -Time Fourier Transform, DTFT ) to distinguish the form of time-discrete signals, a continuous frequency spectrum. • 3.1 Simple aperture • 3.2 Picture with periodic structures • 5.1 discretization of the Fourier transform • 5.2 Discretization of a Fourier series • 6.1 Spectrum of sampled functions • 6.2 aliasing • 6.3 DFT of a time-limited function • 6.4 leakage effect ( leakage effect) • 6.5 Sliding DFT as a band filter bank • 6.6 uncertainty relation of the sliding DFT ## Definition The discrete Fourier transform processing, a sequence of numbers which are formed, for example, as the measurement values ​​from a period of a periodic signal. The entries of the sequence are as values ​​of a polynomial Presented with complex coefficients. The arguments N points are chosen on the unit circle of the complex plane, which are uniformly distributed, that is, the N- th roots of unity If now the polynomial A (z) associated with a uniformly the unit circle rotating function, this results in a continuous-time periodic function Assumes the at times just the function values ​​. The powers of z (t) have the form of And hence the period T / k and the frequency k / T and the angular frequency. Thus the sequence of the measured values ​​has been shown and interpolated by the superposition of a constant level for k = 0, a basic vibration when k = 1, and harmonics at k > 1. These above mentioned interpolation is not the only one that can be constructed in this way. Each of the functions Has this interpolation property. ### DFT and IDFT of a complex vector The discrete Fourier transform of a complex vector, the coefficients This is called the even Fourier coefficients or Fourier components. The inverse DFT ( IDFT ) of the coefficients has ### Special case: DFT of a real vector Certain symmetry laws apply as for the Fourier transform for the DFT. Thus, a real signal in the period to a Hermitian signal () in the frequency domain: This means that in the frequency domain are present only independent complex coefficients. This fact can be utilized in the implementation of the DFT, it is known that the input signal is purely real. Are then used for the representation of the result do not (as with the full DFT ), but only complex numbers necessary. The other complex numbers can be reconstructed by elementary calculation (see formula above). The hermitian symmetry refers to the middle element of the signal. Conversely accordingly: Meets the requirement for all and so the inverse DFT is a real vector. ### Generalization: Mathematical definition of the DFT In mathematics, the discrete Fourier transform is considered in a very general context. It is among other things in the computer algebra in a variety of efficient algorithms for exact arithmetic application, for example in rapid multiplication of integers with Schönhage - road algorithm. Is a commutative unitary ring in which the number (which is the sum of the times ) is one unit. Furthermore, there is a primitive root of unity. At a " vector" is then the discrete Fourier transform by Explained. Under the given conditions in order to exist, the discrete inverse Fourier transform with the coefficients In the all-important special case of the root of unity is used for the DFT usually. This results in the formula in the first section. ### Multidimensional DFT The DFT can be easily extended to multi-dimensional signals. You will then be applied once on all coordinate directions. In the important special case of two dimensions ( image processing ) is approximately valid: For and The inverse transformation is accordingly: For and ## Shifting and scaling in the time and frequency In the calculation of DFT and IDFT formulas, the summation ( index variable above) rather than just running over a shifted range when the vector is periodically continued to all integer indices, because it is. So we can move the summation limits arbitrarily, as long as a segment of length N is swept into the integers. Let us now turn back to the complex case. In practical applications, one would like to combine the indices with an equidistant sequence of time points, Which also has the length N. It is also desirable to assign the frequencies calculated coefficients which are centered about 0 K near N / 2 One of the selected points in time " measured " function returns the observation vector with the coefficients, the DFT is taken in the Fourier analysis. Then And ## Examples The Fourier transform transforms a function f (t ) to g ( ν ) * t from a time chart in the reciprocal space frequency ν: = 1 / t. This also applies to local features that are defined on a (1D ), two ( 2D) or more spatial directions. These are converted by the Fourier transform, one by one in each direction in space frequencies. Diffraction phenomena in the optical or X-ray analysis can be interpreted directly as the intensity distribution of a Fourier transform. The phase relationship is what photography is normally lost. Only in the case of holography, the phase relationship is recorded by superimposition with a reference beam. ### Simple aperture The pictures on the right illustrate two-dimensional Fourier transformation ( 2D FFT) on geometric patterns, calculated for squares of discrete size of a × a pixel. The picture above left shows a gap of size e × f pixels, next to the intensity distribution of the diffraction pattern. The spatial variable r is converted into reciprocal complex values ​​r *. At the selected one pixel sizes is transferred to the reciprocal value of 1 / a reciprocal pixels. The width of the gap by e pixels appear in the reciprocal space as the value of the variable r * = a / e, the height r * = a / f, with harmonic frequencies of higher order. The calculated diffraction patterns indicate the intensity distributions of the complex variable r * again. That they carry only half of the image information can be identified by their rotational symmetry. The periodic peaks correspond to the spatial frequencies of higher order of a square wave. Similar examples can be found under the headings of Fourier analysis, Fourier transform, or Airy disk. In the second field a regular hexagon is diffracted. Again, the size of the figure appears as a period in the diffraction image to the right. 6- fold symmetry is clearly visible. A shift of the output image - as opposed to a rotation - would have an impact only in the phase relationship that is not visible in the chosen representation as intensity distribution. The lower part of image on the right shows the calculated diffraction pattern of a triangle. The 6- fold symmetry is only feigned, which can be seen in the lack of modulation of the diffraction star. The second image series compares the diffraction of two circular openings. A large circle creates a small diffraction pattern, and vice versa. In a telescope, the light diffraction limits the resolution of the lens opening. The larger the diameter, the smaller the diffraction image of a star, the better closely spaced star be differentiated. The image below is an example of a diffraction at a circular structure with no sharp boundary. With a sinusoidal intensity decrease at the wheel occur no higher order diffractions (see also zone plate ). ### Image with periodic structures The image on the left shows a SAR image of the Indian Ocean with water waves of different wavelengths. The internal waves top right have a wavelength of about 500 m. The surface waves generated by wind can not be seen in the reduced representation. The calculated diffraction pattern of the two dark reflections give (see short arrow ), both the direction and the mean wavelength of the regular long-period water waves. The wavelengths of the surface wave varies strongly, so they do not provide sharp reflexes. There are two excellent directions for the wave propagation, which are only dimly visible in the direct image. The wavelengths are approximately 150 m (long arrow ) and 160 m (slightly shorter arrow). ## Mathematical basis Occurring in the discrete Fourier transformation complex numbers Are the N-th roots of unity, that is, they are solutions of the equation. Be the "smallest ", ie primitive root in the first quadrant. This satisfies the following identity geometric sums of roots of unity: Because: for. This is the " deep reason " why the inverse DFT works. We define the vectors, n = 0, ..., N-1, they form an orthonormal basis for the inner product It is Each vector may be represented in the orthonormal basis: The coefficients are called ( in general with arbitrary orthonormal ) Fourier coefficients, ie the DFT maps a vector x to within an additive constant of the vector X = DFT ( x) of the Fourier coefficients. Y = DFT (y ) to a further vector, the Parseval equation for the Fourier coefficient is considered: ## Interpretations of the DFT ### Discretization of the Fourier transform The Fourier transform allows functions with real argument ( and various constraints such as: integrability, waste at infinity ) to think composed of vibrations: An important finding of the Fourier theory is that the amplitude can be determined similar to If we choose a radius R such that outside the interval [-R, R ] is only an insignificant part of f, f is also continuous and a number N chosen so large that T: = R / N is small enough, to f singular sense, i.e., the function values ​​f ( kT) to scan, the Fourier integral can be replaced useful in the transformation formula is represented by a sum of: This corresponds, up to a constant factor, the calculation formula of the DFT. The vector x = (f ( NT ), ..., f ( NT) ) has 2N 1 elements. We already know that it is sufficient that the frequency coefficients for the frequencies of 2N 1, n = -N, ..., -1,0,1, ..., to N, determined in order to reconstruct the function values ​​in the vector x. With the necessary adjustment of the constants in the IDFT we obtain Diskretisierungsabstand the frequency domain is proportional to 1 / R, that is, by assumption, is also small, so that the calculation of the discretization of the inverse Fourier transform corresponds. In the transition from the Fourier transform to the DFT so the following changes are noted: • The signal is discrete, equidistant time points before (T: distance between two consecutive time points), 0 is one of these time points. • The signal has a finite length (2N 1: the number of values ​​), which are interpreted as values ​​within a large interval [- NT, NT ]. • The integrals in the calculation of the Fourier coefficients in the DFT to totals. • The spectrum is calculated only for a finite number of (circular ) frequencies ( ω = (2 π ) · n / ( (2N 1) · T), n = -N, ..., -1,0, 1, 2, ..., N) and is periodic in frequency, the period (2 π ) / T by assumption (T small) is very large. ### Discretization of Fourier series Any periodic function with a real argument (and again restrictions as: integrability, no poles ) and period L can have as a function series with sinusoids that are fractions of L as period are shown ( so-called Fourier series ): Let's break the series expansion for large limits below 1-M and NM from above, we obtain with T: = L / N , That is, we obtain a form of the inverse DFT. Thus, the DFT coefficients can be approximated to In the limiting case of an infinitely large N, the known coefficients of integrals of Fourier series result: ## Properties ### Spectrum of sampled functions The discrete Fourier transform has a periodic spectrum which is repeated at the sampling frequency and is symmetrical with respect to the sampling frequency. The following applies: If the sampled signal frequency components above half the sampling frequency, the spectra of the original signal with the mirrored at the sampling frequency signal components overlap, and it comes to aliasing. ### Aliasing In general, the time-discrete signal is formed by discretization of a continuous signal. The products resulting from the DFT spectra are only identical to the spectra of the underlying continuous signal, if in sampling, the sampling was not injured. For signals in the baseband will be that the sampling frequency must be more than twice as large as the maximum occurring frequency ( Nyquist frequency). In violation of the sampling occurs a distortion of the original signal on (aliasing in the time domain ). A possibility of anti-aliasing, the band limiting of the signal at the input of the system, to avoid this effect. ### DFT of a time-limited function For periodic functions results ( in analogy to the continuous Fourier Transform) is a line spectrum with a frequency spacing of 1/Periodenlänge. A timed discrete function g ( kT) can be derived from a periodic discrete function f ( kT) by precisely cutting out a period over a time window w (t). Since the Fourier transform corresponds to a multiplication of functions in the time domain convolution of the Fourier transform in the frequency domain, the DFT of the time-limited function G ( ω ) is given by the convolution of the DFT of the periodic function F ( ω ) of the Fourier transform the time window W ( ω ). The result is a line spectrum, which is smeared by the Fourier transform of the time window. In Fig.3 shown on the right by broken lines, the influence of the time window on the DFT of the periodic function (thick lines). Due to the time limit frequency components are added between the analyzed frequency lines. By the transition of a periodic function of a time-limited function does not have to be changed, the calculation procedure for the determination of the spectrum. It will continue to be calculated discrete frequency lines, as if a periodic function stands behind it. As an effect of the time window each calculated frequency line now is representative of an entire frequency range, namely the frequency range which is added by the Fourier transform of the time window. This behavior is also called leakage effect. ### Leakage effect ( leakage effect) Due to the temporal limitation of the signal, it may happen that the input signal is cut off. A clipped input signal can only be transformed to the correct DFT if it is periodically resumable. If the signal is not periodically be continued, it contains frequencies that do not belong to the one calculated by the DFT discrete frequencies. The DFT " approaches " these frequencies by the adjacent frequencies, while the energy is distributed to these frequencies. This is called leakage effect (English leakage effect). The time limit is a multiplication with a square wave function equal to and corresponds to a convolution with a sinc function in the frequency domain. This is a different approach to the leakage effect to explain. This is also true in the case of other window functions (such as Hamming, Hann, Gaussian ). Thus, the spectrum of the window function ( or the width of the spectrum) is decisive for the leakage. The amplitude accuracy is the other criterion of a window function. ### Sliding DFT as a band filter bank A DFT of a time-limited function can also be regarded as a band filter bank. • The center frequencies of these band filter corresponding to the frequency lines of the function that arises when the considered period repeated periodically (multiples of 1/Fensterbreite ). • The width and slope of the bandpass filter is determined by the Fourier transform of the time window. (see Fig.3 ) By selecting an appropriate time window function can change the characteristics of the bandpass filter. • With a square -shaped window with points of discontinuity at the window boundaries frequencies are attenuated outside of the transmission range of the bandpass filter with 1/frequency; are obtained slopes between 6 dB / octave (see Fig.2) • If the window function continuous, frequencies are attenuated outside the transmission range of the bandpass filter with 1/Frequenz2; we achieved higher slope of 12 dB / octave • Is the 1st derivative of the window function continuous, frequencies are attenuated outside the transmission range of the bandpass filter with 1/Frequenz3; the slope is 18 dB / octave • Etc. By determining the Fourier transform of each of successive time intervals to obtain the sliding Fourier transform. With the analysis of a new time interval is then obtained new samples for the time course of spectral lines (i.e., the timing of the signals at the outputs of the " band filter "). ### Uncertainty relation of the sliding DFT Time and frequency resolution of the sliding DFT can not be chosen independently. • If you want to analyze signals with high frequency resolution, you have to make the time window very large, you get a low time resolution. • Do you need a high time resolution, you have to make the width of the time window is very short, but then one can determine only a few frequency lines. • The following applies: frequency resolution ≈ 1/Zeitfensterbreite ( a frequency resolution of 1 kHz is desired, the time window must be at least 1 ms ). ## FFT For block length N, which can be represented as a power of 2, the calculation using the algorithm of the fast Fourier transform ( FFT) can be done. In general: Can be factorized the block length, N = KM, so there is a decomposition of the DFT of length N into a product of DFTs of lengths K and M as well as two simple matrices. ## Goertzel algorithm For any block length N, and for determining a single or a few spectral components, the Goertzel algorithm can be used. The advantage is a very efficient implementation in a computer system, since the calculation of each spectral component includes only a complex multiplication and two complex additions. ## Applications • Calculating the Fourier transform of a signal. • Signal analysis. • Vibration analysis and modal analysis. • Processing of signals. • Calculation of correlations. • Calculation of Polynomprodukten in O (n * log ( n ) ) In the calculation of surface acoustic wave filters ( = SAW filters = SAW filter = surface acoustic wave - filter) - Fourier transform of the transfer function required ( represents the impulse response ), the inverse is. This task is handled by computers. 291683 de

{"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}

CC-MAIN-2024-22

latest

en

http://adam.chlipala.net/cpdt/repo/rev/fe6cfbae86b9

text/html

crawl-data/CC-MAIN-2018-39/segments/1537267160142.86/warc/CC-MAIN-20180924031344-20180924051744-00306.warc.gz

### changeset 109:fe6cfbae86b9 Recursive type definitions author Adam Chlipala Mon, 13 Oct 2008 14:04:39 -0400 7abf5535baab 4627b9caac8b src/DataStruct.v 1 files changed, 98 insertions(+), 0 deletions(-) [+] line wrap: on line diff --- a/src/DataStruct.v Mon Oct 13 13:20:57 2008 -0400 +++ b/src/DataStruct.v Mon Oct 13 14:04:39 2008 -0400 @@ -320,3 +320,101 @@ (** We are starting to develop the tools behind dependent typing's amazing advantage over alternative approaches in several important areas. Here, we have implemented complete syntax, typing rules, and evaluation semantics for simply-typed lambda calculus without even needing to define a syntactic substitution operation. We did it all without a single line of proof, and our implementation is manifestly executable. In a later chapter, we will meet other, more common approaches to language formalization. Such approaches often state and prove explicit theorems about type safety of languages. In the above example, we got type safety, termination, and other meta-theorems for free, by reduction to CIC, which we know has those properties. *) +(** * Recursive Type Definitions *) + +(** There is another style of datatype definition that leads to much simpler definitions of the [get] and [hget] definitions above. Because Coq supports "type-level computation," we can redo our inductive definitions as %\textit{%#<i>#recursive#</i>#%}% definitions. *) + +Section filist. + Variable A : Set. + + Fixpoint filist (n : nat) : Set := + match n with + | O => unit + | S n' => A * filist n' + end%type. + + (** We say that a list of length 0 has no contents, and a list of length [S n'] is a pair of a data value and a list of length [n']. *) + + Fixpoint findex (n : nat) : Set := + match n with + | O => Empty_set + | S n' => option (findex n') + end. + + (** We express that there are no index values when [n = O], by defining such indices as type [Empty_set]; and we express that, at [n = S n'], there is a choice between picking the first element of the list (represented as [None]) or choosing a later element (represented by [Some idx], where [idx] is an index into the list tail). *) + + Fixpoint fget (n : nat) : filist n -> findex n -> A := + match n return filist n -> findex n -> A with + | O => fun _ idx => match idx with end + | S n' => fun ls idx => + match idx with + | None => fst ls + | Some idx' => fget n' (snd ls) idx' + end + end. + + (** Our new [get] implementation needs only one dependent [match], which just copies the stated return type of the function. Our choices of data structure implementations lead to just the right typing behavior for this new definition to work out. *) +End filist. + +(** Heterogeneous lists are a little trickier to define with recursion, but we then reap similar benefits in simplicity of use. *) + +Section fhlist. + Variable A : Type. + Variable B : A -> Type. + + Fixpoint fhlist (ls : list A) : Type := + match ls with + | nil => unit + | x :: ls' => B x * fhlist ls' + end%type. + + (** The definition of [fhlist] follows the definition of [filist], with the added wrinkle of dependently-typed data elements. *) + + Variable elm : A. + + Fixpoint fmember (ls : list A) : Type := + match ls with + | nil => Empty_set + | x :: ls' => (x = elm) + fmember ls' + end%type. + + (** The definition of [fmember] follows the definition of [findex]. Empty lists have no members, and member types for nonempty lists are built by adding one new option to the type of members of the list tail. While for [index] we needed no new information associated with the option that we add, here we need to know that the head of the list equals the element we are searching for. We express that with a sum type whose left branch is the appropriate equality proposition. Since we define [fmember] to live in [Type], we can insert [Prop] types as needed, because [Prop] is a subtype of [Type]. + + We know all of the tricks needed to write a first attempt at a [get] function for [fhlist]s. + + [[ + + Fixpoint fhget (ls : list A) : fhlist ls -> fmember ls -> B elm := + match ls return fhlist ls -> fmember ls -> B elm with + | nil => fun _ idx => match idx with end + | _ :: ls' => fun mls idx => + match idx with + | inl _ => fst mls + | inr idx' => fhget ls' (snd mls) idx' + end + end. + + Only one problem remains. The expression [fst mls] is not known to have the proper type. To demonstrate that it does, we need to use the proof available in the [inl] case of the inner [match]. *) + + Fixpoint fhget (ls : list A) : fhlist ls -> fmember ls -> B elm := + match ls return fhlist ls -> fmember ls -> B elm with + | nil => fun _ idx => match idx with end + | _ :: ls' => fun mls idx => + match idx with + | inl pf => match pf with + | refl_equal => fst mls + end + | inr idx' => fhget ls' (snd mls) idx' + end + end. + + (** By pattern-matching on the equality proof [pf], we make that equality known to the type-checker. Exactly why this works can be seen by studying the definition of equality. *) + + Print eq. + (** [[ + +Inductive eq (A : Type) (x : A) : A -> Prop := refl_equal : x = x +]] + +In a proposition [x = y], we see that [x] is a parameter and [y] is a regular argument. The type of the constructor [refl_equal] shows that [y] can only ever be instantiated to [x]. Thus, within a pattern-match with [refl_equal], occurrences of [y] can be replaced with occurrences of [x] for typing purposes. All examples of similar dependent pattern matching that we have seen before require explicit annotations, but Coq implements a special case of annotation inference for matches on equality proofs. *) +End fhlist.

{"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}

CC-MAIN-2018-39

latest

en

https://m.dexlabanalytics.com/blog/category/data-science-certification/page/2

text/html

crawl-data/CC-MAIN-2021-17/segments/1618038061562.11/warc/CC-MAIN-20210411055903-20210411085903-00594.warc.gz

Data Science Certification Archives - Page 2 of 17 - DexLab Analytics | Big Data Hadoop SAS R Analytics Predictive Modeling & Excel VBA ## Linear Regression Part II: Predictive Data Analysis Using Linear Regression In our previous blog we studied about the basic concepts of Linear Regression and its assumptions and let’s practically try to understand how it works. Given below is a dataset for which we will try to generate a linear function i.e. y=b0+b1Xi Where, y= Dependent variable Xi= Independent variable b0 = Intercept (coefficient) b1 = Slope (coefficient) To find out beta (b0& b1) coefficients we use the following formula:- Let’s start the calculation stepwise. 1. First let’s find the mean of x and y and then find out the difference between the mean values and the Xi and Yie. (x-x ̅ ) and (y-y ̅ ). 2. Now calculate the value of (x-x ̅ )2 and (y-y ̅ )2. The variation is squared to remove the negative signs otherwise the summation of the column will be 0. 3. Next we need to see how income and consumption simultaneously variate i.e. (x-x ̅ )* (y-y ̅ ) Now all there is left is to use the above calculated values in the formula:- As we have the value of beta coefficients we will be able to find the y ̂(dependent variable) value. We need to now find the difference between the predicted y ̂ and observed y which is also called the error term or the error. To remove the negative sign lets square the residual. What is R2 and adjusted R2 ? R2 also known as goodness of fit is the ratio of the difference between observed y and predicted  and the observed y and the mean value of y. Hopefully, now you have understood how to solve a Linear Regression problem and would apply what you have learned in this blog. You can also follow the video tutorial attached down the blog. You can expect more such informative posts if you keep on following the DexLab Analytics blog. DexLab Analytics provides data Science certification courses in gurgaon. . ## ANOVA Part-II: What is Two-way ANOVA? In my previous blog, I have already introduced you to a statistical term called ANOVA and I have also explained you what one-way ANOVA is? Now in this particular blog I will explain the meaning of two-way ANOVA. The below image shows few tests to check the relationship/variation among variables or samples. When it comes to research analysis the first thing that we should do is to understand the sample which we have and then try to disintegrate the dataset to form and understand the relationship between two or more variables to derive some kind of conclusion. Once the relation has been established, our job is to test that relationship between variables so that we have a solid evidence for or against them. In case we have to check for variation among different samples, for example if the quality of seed is affecting the productivity we have to test if it is happening by chance or because of some reason. Under these kind of situations one-way ANOVA comes in handy (analysis on the basis of a single factor). #### Two-way ANOVA Two-way ANOVA is used when we are testing the variations among samples on the basis two factors. For example testing variation on the basis of seed quality and fertilizer. Hopefully you have understood what Two-way ANOVA is. If you need more information, check out the video tutorial attached down the blog. Keep on following the DexLab Analytics blog, to find more information about Data Science, Artificial Intelligence. DexLab Analytics offers data Science certification courses in gurgaon. . ## An Introduction to Sampling and its Types Sampling is a technique in which a predefined number of observation is taken from a large population for the purpose of statistical analysis and research. There are two types of sampling techniques:- #### Random Sampling Random sampling is a sampling technique in which each observation has an equal probability of being chosen. This kind of sample should be an unbiased representation of the population. #### Types of random sampling 1. Simple Random Sampling:- Simple random sampling is a technique in which any observation can be chosen and each observation has an equal probability of being selected. 2. Stratified Random Sampling:- In this sampling technique we create sub-group of the population with similar attributes and characteristics and then out of those sub-groups we then include each category in our sample with the probability of choosing each observation from the sub-group being equal. 3. Systematic Sampling:- This is a sampling technique where the first observation is selected randomly and then every kth element is chosen randomly to be included in our sample. k= 2, here the first observation is selected randomly and after that every second element is included in the sample. 4. Cluster Sampling:- This is a sampling technique in which the data is grouped into small sub-groups called clusters with random categories and then from those clusters random observation is selected which then is included in the sample. Two clusters are created from which then random observation will be chosen to form the sample. Non-Random Sampling :- It is a sampling technique in which an element of biasedness is introduced which means that an observation is selected for the sample on the basis of not probability but choice. Types of non-random sampling:- 1. Convenience Sampling:- When a sample observation is drawn from the population based on how comfortable it is for you to take the observation it is called convenience sampling. For example when you have a survey sheet that is to be filled by students from all the departments of your college but you only ask your friends to fill the survey sheet. 2. Judgment Sampling:– When the sample observation drawn from the population is based on your professional judgment or past experience, it is called judgment sampling. 3. Quota Sampling:– When you draw a sample observation from the population that is based on some specific attribute, it is called quota sampling. For example, taking sample of people over and above 50 years. 4. Snow Ball Sampling:– When survey subjects are selected based on referral from other survey respondents, it is called snow ball sampling. #### Sampling and Non-sampling errors Sampling errors:- It occurs when the sample is not representative of the entire population.  For example a sample of 10 people with or without COVID-19 cannot tell whether or not the entire population of a country is COVID positive. Non-sampling error:-This kind of error occurs during data collection. For example, during data collection if you falsely specified a name, it will be considered a non-sampling error. So, with that this discussion on Sampling wraps up, hopefully, at the end of this you have learned what Sampling is, what are its variations and how do they all work. If you need further clarification, then check out our video tutorial on Sampling attached down the blog. DexLab Analytics provides the best data science course in gurgaon, keep following the blog section to stay updated. . ## Hypothesis Testing: An Introduction You must be familiar with the phrase hypothesis testing, but, might not have a very clear notion regarding what hypothesis testing is all about. So, basically the term refers to testing a new theory against an old theory. But, you need to delve deeper to gain in-depth knowledge. Hypothesis are tentative explanations of a principal operating in nature. Hypothesis testing is a statistical method which helps you prove or disapprove a pre-existing theory. Hypothesis testing can be done to check whether the average salary of all the employees has increased or not based on the previous year’s data, testing can be done to check if the percentage of passengers increased or not in the business class due to introduction of a new service and testing can also be done to check the differences in the productivity varied land. #### There are two key concepts in testing of hypothesis:- Null Hypothesis:- It means the old theory is correct, nothing new is happening, the system is in control, old standard is correct etc. This is the theory you want to check if is true or not. For example if a ice-cream factory owner says that their ice-cream contains 90% milk, this can be written as:- Alternative Hypothesis:- It means new theory is correct, something is happening, system is out of control, there are new standards etc. This is the theory you check against the null hypothesis. For example you say that ice-cream does not contain 90% milk, this can be written as:- #### Two-tailed, right tailed and left tailed test Two-tailed test:- When the test can take any value greater or less than 90% in the alternative it is called two-tailed test ( H190%) i.e. you do not care if the alternative is more or less all you want to know is if it is equal to 90% or no. Right tailed test:-When your test can take any value greater than 90% (H1>90%) in the alternative hypothesis it is called right tailed test. Left tailed test:-When your test can take any value less than 90% (H1<90%) in the alternative hypothesis it is called left tailed test. #### Type I error and Type II error ->When we reject the null hypothesis when it is true we are committing type I error. It is also called significance level. ->When we accept the null hypothesis when it is false we are committing type II error. #### Steps involved in hypothesis testing 1. Build a hypothesis. 2. Collect data 3. Select significance level i.e. probability of committing type I error 4. Select testing method i.e. testing of mean, proportion or variance 5. Based on the significance level find the critical value which is nothing but the value which divides the acceptance region from the rejection region 6. Based on the hypothesis build a two-tailed or one-tailed (right or left) test graph 7. Apply the statistical formula 8. Check if the statistical test falls in the acceptance region or the rejection region and then accept or reject the null hypothesis Example:- Suppose the average annual salary of the employees in a company in 2018 was 74,914. Now you want to check whether the average salary  of the employees has increased or not in 2019. So, a sample of 112 people were taken and it was found out  that the average annual salary of the employees in 2019 is 78,795. σ=14.530. We will apply hypothesis testing of mean when  known with 5% of significance level. The test result shows that 2.75 falls beyond the critical value of 1.9 we reject the null hypothesis which basically means that the average salary has increased significantly in 2019 compared to 2018. So, now that we have reached at the end of the discussion, you must have grasped the fundamentals of hypothesis testing. Check out the video attached below for more information. You can find informative posts on Data Science course, on Dexlab Analytics blog. . ## ANOVA PART I: The Introductory Guide to ANOVA  In this blog, we are going to be discussing a statistical technique, ANOVA, which is used for comparison. The basic principal of ANOVA is to test for differences among the mean of different samples. It examines the amount of variation within each of these samples and the amount of variation between the samples. ANOVA is important in the context of all those situations where we want to compare more than two samples as in comparing the yield of crop from several variety of seeds etc. The essence of ANOVA is that the total amount of variation in a set of data is broken in two types:- 1. The amount that can be attributed to chance. 2. The amount which can be attributed to specified cause. #### One-way ANOVA Under the one-way ANOVA we compare the samples based on a single factor. For example productivity of different variety of seeds. Stepwise process involved in calculation of one-way ANOVA is as follows:- 1. Calculate the mean of each sample X ̅ 2. Calculate the super mean 3. Calculate the sum of squares between (SSB) samples 1. Divide the result by the degree of freedom between the samples to obtain mean square between (MSW) samples. 2. Now calculate variation within the samples i.e. sum of square within (SSW) 1. Calculate mean square within (MSW) 2. Calculate the F-ratio 3. Last but not the least calculate the total variation in the given samples i.e. sum of square for total variance. Lets now solve a one-way ANOVA problem. A,B and C are three different variety of seeds and now we need to check if there is any variation in their productivity or not. We will be using one-way ANOVA as there is a single factor comparison involved i.e. variety of seeds. The f-ratio is 1.53 which lies within the critical value of 4.26 (calculated from the f-distribution table). Conclusion:- Since the f-ratio lies within the acceptance region we can say that there is no difference in the productivity of the seeds and the little bit of variation that we see is caused by chance. Two-way ANOVA will be discussed in my next blog so do comeback for the update. Hopefully, you have found this blog informative, for more clarification watch the video attached down the blog. You can find more such posts on Data Science course topics, just keep on following the DexLab Analytics blog. . ## Introducing Automation: Learn to Automate Data Preparation with Python Libraries In this blog we are discussing automation, a function for automating data preparation using a mix of Python libraries. So let’s start. #### Problem statement A data containing the following observation is given to you in which the first row contains column headers and all the other rows contains the data. Some of the rows are faulty, a row is faulty if it contains at least one cell with a NULL value. You are supposed to delete all the faulty rows containing NULL value written in it. In the table given below, the second row is faulty, it contains a NULL value in salary column. The first row is never faulty as it contains the column headers. In the data provided to you every cell in a column may contain a single word and each word may contain digits between 0 & 9 or lowercase and upper case English letters. For example: In the above example after removing the faulty row the table looks like this: The order of rows cannot be changed but the number of rows and columns may differ in different test case. The data after preparation must be saved in a CSV format. Every two successive cells in each row are separated by a single comma ‘,’symbol and every two successive rows are separated by a new-line ‘\n’ symbol. For example, the first table from the task statement to be saved in a CSV format is a single string ‘S. No., Name, Salary\n1,Niharika,50000\n2,Vivek,NULL\n3,Niraj,55000’ . The only assumption in this task is that each row may contain same number of cells. Write a python function that converts the above string into the given format. #### Write a function: def Solution(s) Given a string S of length N, returns the table without the Faulty rows in a CSV format. Given S=‘S. No., Name, Salary\n1,Niharika,50000\n2,Vivek,NULL\n3,Niraj,55000’ The table with data from string S looks as follows: After removing the rows containing the NULL values the table should look like this: You can try a number of strings to cross-validate the function you have created. Let’s begin. • First we will store the string in a variable s • Now we will start by declaring the function name and importing all the necessary libraries. • Creating a pattern to separate the string from ‘\n’ . • Creating a loop to create multiple lists within a list. In the above code the list is converted to an array and then used to create a dataframe and stored as csv file in the default working directory. • Now we need to split the string to create multiple columns. The above code creates a dataframe with multiple columns. Now after dropping the rows with NaN values data looks like To reset the index we can now use .reset_index() method. • Now the problem with the above dataframe created is that the NULL values are in string format, so first we need to convert them into NaN values and then only we will be able to drop them. For that we will be using the following code. Now we will be able to drop the NaN values easily by using .dropna() method. In the above code we first dropped the NaN values  then we used the first row of the data set to create column names and then dropped the original row. We also made the first column as index. Hence we have managed to create a function that can give us the above data. Once created this function can be used to convert a string into dataframe with similar pattern. Hopefully, you found the discussion informative enough. For further clarification watch the video attached below the blog. To access more informative blogs on Data science using python training related topics, keep on following the Dexlab Analytics blog. Here’s a video introduction to Automation. You can check it down below to develop a considerable understanding of the same: . ## How The Industries Are Being Impacted By Data Science? The world has finally woken up and smelled the power of data science and now we are living in a world that is being driven by data. There is no denying the fact that new technologies are coming to the fore that are born out of data-driven insight and numerous sectors are also turning towards data science techniques and tools to increase their operational efficiency. This in turn is also pushing a demand for skilled people in various sectors who are armed with Data Science course or, Retail Analytics Courses to be able to sift through mountains of data to clean it, sort it and analyze it for uncovering valuable information. Decisions that were earlier taken often on the basis of erroneous data or, assumption can now be more accurate thanks to application of data science. Now  let’s take a look at which sectors are benefitting the most from data science #### Healthcare The healthcare industry has adopted the data science techniques and the benefits could already be perceived. Keeping track of healthcare records is easier not just that but digging through the pile of patient data and its analysis actually helps in giving hint regarding health issues that might crop up in near future. Preventive care is now possible and also monitoring patient health is easier than ever before. The development in the field  can also predict which medication would be suitable for a particular patient. Data analytics and data science application is also enabling the professionals in this sector to offer better diagnostic results. #### Retail This is one industry that is reaping huge benefits from the application of data science. Now sorting through the customer data, survey data it is easier to gauge the customers’ mindset. Predictive analysis is helping the experts in this field to predict the personal preference of the consumers and they are able to come up with personalized recommendations that is bound to help them retain customers.  Not just that they can also find the problem areas in their current marketing strategy to make changes accordingly. #### Transport Transport is another sector that is using data science techniques to its advantage and  in turn it is increasing its service quality. Both the public and private transportation services providers are keeping track of customer journey and getting the details necessary to develop personalized information, they are also helping people be prepared for unexpected issues and most importantly they are helping people reach their destinations without any glitch. #### Finance If so many industries are reaping benefits, Finance is definitely to follow suit. Dealing with  valuable data regarding banking transactions, credit history is essential. Based on the data insight it is possible to offer customers personalized financial advice. Also the credit risk issue could be minimized thanks to the insight derived from a particular customer’s credit history. It would allow the financial institute make an informed decision. However, credit risk analytics training would be required for personnel working in this field. #### Telecom The field of telecom is surely a busy sector that has to deal with tons of valuable data. With the application of data science now they are able to find a smart solution to process the data they gather from various call records, messages, social media platforms in order to design and deliver services that are in accordance with customers’ individualistic needs. Harnessing the power of data science is definitely going to impact all the industries in future. The data science domain is expanding and soon there would be more miracles to observe. Data Science training can help upskill the employees reduce the skill gap that is bugging most sectors. . ## Probability PART-II: A Guide To Probability Theorems This is the second part of the probability series, in the first segment we discussed the basic concepts of probability. In this second part we will delve deeper into the topic and discuss the theorems of probability. Let’s find out what these theorems are. • If A and B are two events and they are not necessarily mutually exclusive then the probability of occurrence of at least one of the two events A and B i.e. P(AUB) is given by Removing the intersections will give the probability of A or B  or both. Example:- From a deck of cards 1 card is drawn, what is the probability the card is king or heart or both? Total cards 52 P(KingUHeart)= P(King)+P(Heart) ─ P(King∩Heart) • If A and B are two mutually exclusive events then the probability that either A or B will occur is the sum of individual probabilities of the events A and B. P(A)+P(B), here the combined probability of the two will either give P(A) or P(B) • If A and B are two non mutually exclusive events then the probability of occurrence of event A is given by Where B’ is 1-P(B), that means probability of  A is calculated as P(A)=1-P(B) #### Multiplication Law The law of multiplication is used to find the joint probability or the intersection i.e. the probability of two events occurring together at the same point of time. In the above graph we see that when the bill is paid at the same time tip is also paid and the interaction of the two can be seen in the graph. #### Joint probability table A joint probability table displays the intersection (joint) probabilities along with the marginal probabilities of a given problem where the marginal probability is computed by dividing some subtotal by the whole. Example:- Given the following joint probability table find out the probability that the employee is female or a professional worker. Watch this video down below that further explains the theorems. At the end of this blog, you must have grasped the basics of the theorems discussed here. Keep on tracking the Dexlab Analytics blog where you will find more discussions on topics related to Data Science training. . ## What Role Does A Data Scientist Play In A Business Organization? The job of a data scientist is one that is challenging, exciting and crucial to an organization’s success.  So, it’s no surprise that there is a rush to enroll in a Data Science course, to be eligible for the job. But, while you are at it, you also need to have the awareness regarding the job responsibilities usually bestowed upon the data scientists in a business organization and you would be surprised to learn that the responsibilities of a data scientist differs from that of a data analyst or, a data engineer. So, what is the role and responsibility of a data scientist?  Let’s take a look. The common idea regarding a data scientist role is that they analyze huge volumes of data in order to find patterns and extract information that would help the organizations to move ahead by developing strategies accordingly. This surface level idea cannot sum up the way a data scientist navigates through the data field. The responsibilities could be broken down into segments and that would help you get the bigger picture. #### Data management The data scientist, post assuming the role, needs to be aware of the goal of the organization in order to proceed. He needs to stay aware of the top trends in the industry to guide his organization, and collect data and also decide which methods are to be used for the purpose. The most crucial part of the job is the developing the knowledge of the problems the business is trying solve and the data available that have relevance and could be used to achieve the goal. He has to collaborate with other departments such as analytics to get the job of extracting information from data. #### Data analysis Another vital responsibility of the data scientist is to assume the analytical role and build models and implement those models to solve issues that are best fit for the purpose. The data scientist has to resort to data mining, text mining techniques. Doing text mining with python course can really put you in an advantageous position when you actually get to handle complex dataset. #### Developing strategies The data scientists need to devote themselves to tasks like data cleaning, applying models, and wade through unstructured datasets to derive actionable insight in order to gauge the customer behavior, market trends. These insights help a business organization to decide its future course of action and also measure a product performance. A Data analyst training institute is the right place to pick up the skills required for performing such nuanced tasks. #### Collaborating Another vital task that a data scientist performs is collaborating with others such as stakeholders and data engineers, data analysts communicating with them in order to share their findings or, discussing certain issues. However, in order to communicate effectively the data scientists need to master the art of data visualization which they could learn while pursuing big data courses in delhi along with deep learning for computer vision course.  The key issue here is to make the presentation simple yet effective enough so that people from any background can understand it. The above mentioned responsibilities of a data scientist just scratch the surface because, a data scientist’s job role cannot be limited by or, defined by a couple of tasks. The data scientist needs to be in synch with the implementation process to understand and analyze further how the data driven insight is shaping strategies and to which effect. Most importantly, they need to evaluate the current data infrastructure of the company and advise regarding future improvement. A data scientist needs to have a keen knowledge of Machine Learning Using Python, to be able to perform the complex tasks their job demands. . +91 931 572 5902

{"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}

CC-MAIN-2021-17

latest

en

http://forums.wolfram.com/mathgroup/archive/1999/Feb/msg00308.html

text/html

crawl-data/CC-MAIN-2019-51/segments/1575541319511.97/warc/CC-MAIN-20191216093448-20191216121448-00092.warc.gz

Re: Pure Functions in rules • To: mathgroup at smc.vnet.net • Subject: [mg15983] Re: Pure Functions in rules • From: "Seth Chandler" <SChandler at uh.edu> • Date: Fri, 19 Feb 1999 03:27:03 -0500 • Organization: University of Houston • References: <7ag34l\[email protected]> • Sender: owner-wri-mathgroup at wolfram.com ```If you try {1,2,3}/.(m_List:>(2*#&/.m)) you get the behavior I believe you desire. Using Rule rather than RuleDelayed causes a problem because, until there is a specific value of m, Mathematica can't Map the way you want. Seth J. Chandler Associate Professor of Law University of Houston Law Center Will Self wrote in message <7ag34l\$aie at smc.vnet.net>... >It appears that I cannot depend on using a pure function >in a pattern-matching rule. > >Here I am trying to convince reluctant students that they're >better off learning to use Mathematica than doing things >by hand, and we run across something like this, and in a >much more complicated situation where the trouble was >hard to isolate. > >I am quite frankly incensed by the behavior shown in >In/Out 80, below. Look at these examples: > >In[73]:= {1,2,3}/.(m_List->7) >Out[73]= 7 > >In[74]:= {1,2,3}/.(m_List->(2*m)) >Out[74]= {2,4,6} > >In[75]:= 2*#& /@ {1,2,3} >Out[75]= {2,4,6} > >In[77]:= f[m_List]:=2*#& /@ m > >In[78]:= f[{1,2,3}] >Out[78]= {2,4,6} > >In[79]:= {1,2,3}/.m_List->f[m] >Out[79]= {2,4,6} > >Now try this: > >In[80]:= {1,2,3}/.(m_List->(2*#& /@ m)) >Out[80]= {1,2,3} > >Does anyone (say, at WRI for example) care to comment on >this? > >Will Self > ``` • Prev by Date: Re: escape-sequences in strings • Next by Date: Re: MyPrependTo • Previous by thread: Re: Pure Functions in rules • Next by thread: Re: Pure Functions in rules

{"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}

CC-MAIN-2019-51

longest

en

https://mathematica.stackexchange.com/questions/227893/the-numerics-of-fresnels/227894

text/html

crawl-data/CC-MAIN-2020-40/segments/1600400198887.3/warc/CC-MAIN-20200921014923-20200921044923-00033.warc.gz

# The numerics of FresnelS[] Bug introduced in 12.1 or earlier and persisting through 12.1.1 or later [CASE:4615361] Note: A worse problem existed in 12.0 for inputs greater than 8 and of precisions less than 43.66; 12.1 fixed the problem for precisions less than around 32, but the problem for precisions between ~32 and ~43 remains. The values of N[FresnelS[8 + 1*^-28], 32] and FresnelS[N[8 + 1*^-28, 32]] are surprisingly far apart: N[FresnelS[8 + 1*^-28], 32] - FresnelS[N[8 + 1*^-28, 32]] (* -0.0005 *) I suppose N[FresnelS[8 + 1*^-28], 32] is the more accurate value, but how can I be sure? What is it's true error? Block[{$MaxExtraPrecision = 500}, ListLinePlot[ Table[N[FresnelS[x], 32] - FresnelS[N[x, 32]] // RealExponent, {x, Subdivide[0, 15, 15*30]}], PlotRange -> {-36.5, 0.3}, DataRange -> {0, 15}] ] Is this a bug? Or some inevitable numerical difficulty? How to accurately evaluate FresnelS[x]? Update After @J.M.'s and @Carl's answers, I looked at Trace to see if there were any clue why the accuracy would jump around a precision of 43/44. I discovered the code for the Fresnel family of functions is exposed and can be inspected with GeneralUtilitiesPrintDefinitions, which is unusual for System functions. I'm not an expert on the Fresnel functions, so sorting it out will take much longer than it's worth to me. I'll happily leave that to the WRI developers. There seems to be a less egregious problem with the machine-precision computation of FresnelS[x] as x increases toward 4, which can be seen in the ramp of the first plot here. The cut-off prec = 43.66 for x = 8 + 1*^-28 in FresnelS[N[x, prec]] actually depends on x and can be found in the code for FresnelF: (InternalPrecAccur[x] * 2 * Log[10.]) / Pi <= N[x] ^ 2 • To repeat the observation I made in the chat room for people who don't follow it: the "bad" evaluation also happens if you try expressing the Fresnel integral in terms of the auxiliary functions, which are (in theory) supposed to be more stable: With[{x = N[8 + 1*^-28, 32]}, 1/2 - FresnelF[x] Cos[Pi x^2/2] - FresnelG[x] Sin[Pi x^2/2]]. – J. M.'s discontentment Aug 3 at 15:05 • Use higher precision. look at N[{#, N[FresnelS[8 + 1*^-28], #] - FresnelS[N[8 + 1*^-28, #]]} & /@ Range[30, 50, 2]] // Grid – Bob Hanlon Aug 3 at 15:17 • Wow, WRI support replied within two hours. They agree it's not behaving correctly. – Michael E2 Aug 3 at 21:12 • Nice. In versions 5.2 and 8.0.4 of Mathematica, N[FresnelS[8 + 1*^-28], 32] - FresnelS[N[8 + 1*^-28, 32]] is simply 0.*^-31. The progress cannot be stopped, however. – innaiz Aug 4 at 8:39 ## 3 Answers I think this is worth reporting to Support. For instance, using formula 7.5.8 from the DLMF: With[{x = N[8 + 1*^-28, 32]}, With[{ζ = Sqrt[π] (1 - I) x/2}, Im[(1 + I)/2 Erf[ζ]]]] 0.46021421439301448386198863207105 and the result is comparable to evaluating N[FresnelS[8 + 1*^-28], 32]. In theory, one is supposed to use the auxiliary functions $$f(z)$$ and $$g(z)$$ for computing Fresnel integrals of moderate or large arguments. However, With[{x = N[8 + 1*^-28, 32]}, {FresnelG[x], FresnelF[x]}] {0.00019781962280286444301613974000765, 0.0392} With[{x = N[8 + 1*^-28, 32]}, {gg, ff} = {FresnelG[x], FresnelF[x]}; 1/2 - ff Cos[π x^2/2] - gg Sin[π x^2/2]] 0.4608 Contrast this with (cf. formula 7.5.10): With[{x = N[8 + 1*^-28, 32]}, With[{ζ = Sqrt[π] (1 - I) x/2}, ReIm[(1 + I)/2 Exp[ζ^2] Erfc[ζ]]]] {0.000197819622802864443016139740, 0.039785785606985516138011367928} which works much better: With[{x = N[8 + 1*^-28, 32]}, With[{ζ = Sqrt[π] (1 - I) x/2}, {gg, ff} = ReIm[(1 + I)/2 Exp[ζ^2] Erfc[ζ]]]; 1/2 - ff Cos[Pi x^2/2] - gg Sin[Pi x^2/2]] 0.460214214393014483861988632071 • Actually, I just discovered the code for FresnelS[] is available via GeneralUtilitiesPrintDefinitions@FresnelS (or Trace). It does just what you show. You can also inspect FresnelF. (I think there's a problem with FresnelF and maybe G.) – Michael E2 Aug 3 at 17:38 • It does look to be so when I inspected it myself, and now the fault is not overly surprising. Hopefully it gets resolved soon. – J. M.'s discontentment Aug 4 at 1:20 • I wonder whether special functions are even something of a priority to WRI, given that the broken Mathieu functions remain broken for years (even after reporting). – Ruslan Aug 4 at 9:43 The incorrect result is: FresnelS[N[8+1*^-28, 32]] //InputForm 0.46075248359440792463.970167826243401 Note that the precision of the output is 3.97, indicating that the 4th digit may not be accurate, which is exactly what you observe. If you increase the precision: FresnelS[N[8+1*^-28, 43]] //InputForm 0.46075248359440792463.970167826243401 You still only get approximately 4 digits of precision. A further increase to 44 does produce a better answer: FresnelS[N[8+1*^-28, 44]] //InputForm 0.460214214393014483861988632071052433913260535977611360507240.31143196057397 I think having the same result for 32 and 43 digits of precision looks suspicious, and the fact that approximately an additional 16 digits of precision (beyond the 28) are needed to get a correct result suggests that there may be some incorrect machine number approximation being used under the hood. I would also suggest reporting this issue to support. • +1 & thanks. I will point out that the uncertainty 10^-Accuracy[FresnelS[N[8 + 1*^-28, 32]]] is a bit less than the error, so I think the precision 3.97 is higher than it should be. The constant result for precision between 32 and 43 is suspicious. I'll report it. – Michael E2 Aug 3 at 17:17 ListPlot[Table[10^-Accuracy[FresnelS[N[8 + 1*^-28, n]]], {n, 0, 99}]] This alters the cognition of what is really going on hopefully. This plot causes some more sensation: ListPlot[Table[10^-Accuracy[FresnelS[N[8 + 1*^-28, n]]], {n, 0, 10}]] How can this happen? It is the very same function on a subinterval. ListPlot[Table[10^-Accuracy[FresnelS[N[8 + 1*^-28, n]]], {n, 11, 99}]] More as expected the Accuracy falls monotonically but slow. ListPlot[Table[{n, 10^-Accuracy[FresnelS[N[8 + 1*^-28, n]]]}, {n, 31, 45, 1}]] ListPlot[Table[{n, 10^-Accuracy[FresnelS[N[8 + 1*^-28, n]]]}, {n, 32, 45, 1}]] 10^-Accuracy[FresnelS[N[8 + 1*^-28, 32]]] 0.0000493514 ListPlot[Table[{n, 10^-Accuracy[FresnelS[N[8 + 1*^-28, n]]]}, {n, 44, 48, 1}]] This is rapid decreasing and not some magnitudes smaller but 10^-Accuracy[FresnelS[N[8 + 1*^-28, 32]]]/ 10^-Accuracy[FresnelS[N[8 + 1*^-28, 44]]] 2.19671*10^36 36 magnitudes smaller. ListPlot[Table[{n, Accuracy[FresnelS[N[8 + 1*^-28, n]]]}, {n, 1, 31, 1}]] FresnelS[] is on the summary of new features of version 12. So this can not be a bug. It is intended. The annotation is "(updated) — substantially expanded examples for 250 special functions". On the page linked from there, Guide: Mathematical Functions they mention "The Wolfram Language has the most extensive collection of mathematical functions ever assembled. Often relying on original results and algorithms developed at Wolfram Research over the past two decades, each function supports a full range of symbolic operations, as well as efficient numerical evaluation to arbitrary precision, for all complex values of parameters." So this behavior gives a look into the efficient numerical evaluation to arbitrary precision of the Wolfram Language. MathematicalFunctionData["FresnelS", "RelatedFunctionRepresentations"] {Function[{\[FormalZ]}, Inactivate[ FresnelS[\[FormalZ]] == 1/4 (1 + I) (Erf[1/2 (1 + I) Sqrt[\[Pi]] \[FormalZ]] - I Erf[1/2 (1 - I) Sqrt[\[Pi]] \[FormalZ]])]], Function[{\[FormalZ]}, Inactivate[ FresnelS[\[FormalZ]] == 1/4 (I - 1) (Erfi[1/2 (1 - I) Sqrt[\[Pi]] \[FormalZ]] + I Erfi[1/2 (1 + I) Sqrt[\[Pi]] \[FormalZ]])]], Function[{\[FormalZ]}, Inactivate[ FresnelS[\[FormalZ]] == 1/2 + 1/4 (1 + I) (-Erfc[1/2 (1 + I) Sqrt[\[Pi]] \[FormalZ]] + I Erfc[1/2 (1 - I) Sqrt[\[Pi]] \[FormalZ]])]], Function[{\[FormalZ]}, Inactivate[ FresnelS[\[FormalZ]] == ((1 - I) (DawsonF[1/2 (I - 1) Sqrt[\[Pi]] \[FormalZ]] - I E^(I \[Pi] \[FormalZ]^2) DawsonF[ 1/2 (1 + I) Sqrt[\[Pi]] \[FormalZ]]))/((2 Sqrt[\[Pi]]) E^( 1/2 I \[Pi] \[FormalZ]^2))]], Function[{\[FormalZ]}, Inactivate[ FresnelS[\[FormalZ]] == FresnelF[\[FormalZ]] (-Cos[(\[Pi] \[FormalZ]^2)/2]) - FresnelG[\[FormalZ]] Sin[(\[Pi] \[FormalZ]^2)/2] + 1/2]]} So the behavior is brand new and at the kernel of the innovations. The page MathematicalFunctionData states all definitions are now direct Digital Library of Mathematics Functions standard. MathematicalFunctionData["FresnelS", "ArgumentPattern"] Inactive[FresnelS][_] MathematicalFunctionData["FresnelS", "WolframFunctionsSiteLink"] Hyperlink["http://functions.wolfram.com/GammaBetaErf/FresnelS/", \ "http://functions.wolfram.com/GammaBetaErf/FresnelS/"] This shows another hierarchical depends used internally to calculate the values the accuracy is derived from. The overwhelming power of mathematical knowledge representation leads to the Mathematica input: MathematicalFunctionData["FresnelS", "PropertyAssociation"] ... For numerical evaluation the definitions in GeneralUtilitiesPrintDefinitions@FresnelS ... are much more relevant. GeneralUtilitiesPrintDefinitions@FresnelS ... HoldPattern[FresnelS][ Pattern[TrigExpIntegralDumpx, Blank[Real]]] := Module[{z = Pi * x ^ 2, ax}, If[Less[z, 8.], Times[(z * x) / 6, HypergeometricPFQ[{3 / 4}, {3 / 2, 7 / 4}, -(z / 4) ^ 2 ] ], DivideBy[z, 2]; ax = Abs @ x; Times[Sign @ x, (1 / 2) - ((FresnelG[ax] * Sin[z]) + FresnelF[ax] * Cos[z]) ] ] ]; ... This has a conditional at z=8.. At this point on the Reals representation to calculate the numerical values is changed. That holds for the jump in Accuracy. fslist = {FresnelS[N[8 - 1*^-43 + 8 1*^-44, 100]] // InputForm, FresnelS[N[8 - 1*^-43 + 9 1*^-44, 100]] // InputForm, FresnelS[N[8 - 1*^-43 + 1*^-44, 100]] // InputForm, FresnelS[N[8 - 1*^-43 + 11 1*^-44, 100]] // InputForm, FresnelS[N[8 - 1*^-43 + 12 1*^-44, 100]] // InputForm} // Dataset There is a jump in accuracy and in the value at 8. To get this Dataset it is necessary to use the 8 and the notation 1*^-43 otherwise the represetion is not that exact or change the corresponding values in the notebooks settings. (FresnelS[N[8 - 1*^-43 + 1*^-44, 100]] - FresnelS[N[8 - 1*^-43 + 11 1*^-44, 100]]) // InputForm 054.488171823879036 A very accurate zero. But not machine presicion or arbitrary precision. There is evidence that the considerations of the designer were reduce the time for calculations first then presicion at 8 the situation changes and the higher precision approximation for lower values is replaced by the representation with higher precision. Since all other have less precision this suffices to be market leader again disregading Fortran implementation for arbitrary presicion. Since these values can not be exact to arbitrary presision but just to a very high precision this choice was there and is acceptable. I get a bigger problem with this ListPlot[Table[{n, Accuracy[FresnelS[N[8 + 1*^-150, n]]]}, {n, 1, 46, 1}]] ListPlot[Table[{n, 10^-Accuracy[FresnelS[N[8 + 1*^-150, n]]]}, {n, 1, 46, 1}]] ListPlot[Table[{n, 10^-Accuracy[FresnelS[N[8 + 1*^-150, n]]]}, {n, 1, 46, 1}], PlotRange -> Full] So the accuracy get smaller the higher the value for Accurcay is selected. It has a clear maximum at the Accuracy n=4 at 5 10^41 and crosses zero between n=45 and n=46. Expected is an Accuracy that is negative and gets more negative. This is a behaviour forced by the function representation to the Gaussian error function with pure complex arguments. Per definitionem this is so. So to work with FresnelS it is advisable to set the Accuracy parameter bigger than 46 to calculate with sensible values. So the plot under consideration looks different: Block[{$MaxExtraPrecision = 500}, ListLinePlot[ Table[N[FresnelS[x], 50] - FresnelS[N[x, 50]] // RealExponent, {x, Subdivide[0, 15, 15*30]}], PlotRange -> {-60, 0.3}, DataRange -> {0, 15}]] This shows there is a strategy necessary to calculate the values the function reaches -50 almost regularly. It is never worse than -3.25. Block[{$MaxExtraPrecision = 500}, ListLinePlot[ Table[N[FresnelS[x], 50] - FresnelS[N[x, 50]] // RealExponent, {x, Subdivide[0, 15, 15*30]}], PlotRange -> {-56, -3.25}, DataRange -> {0, 15}]] This about -3 is the difference that almost reached with the -0.0005 in the question. The behavior is even more brave for Block[{$MaxExtraPrecision = 500}, ListLinePlot[ Table[N[FresnelS[x], 100] - FresnelS[N[x, 100]] // RealExponent, {x, Subdivide[0, 15, 15*30]}], PlotRange -> {-108, -3.5}, DataRange -> {0, 15}]] So the advice can go further to set the Accuracy parameter as high as possible to avoid the divergence with upper limit and set the algorithm in operation as seldom as possible. The x=8 remains for all Accuracy parameters this way but the extent of the first value at which the curve comes close to about -3.5 can be extended. The relation seems to be something like square root between 50 and 100. The minimium is always taken for very small x and reaches if the correction algorithm is set into operation to calculate the values. Block[{\$MaxExtraPrecision = 500}, ListLinePlot[ Table[N[FresnelS[x], 500] - FresnelS[N[x, 500]] // RealExponent, {x, Subdivide[0, 15, 15*30]}](*,PlotRange\[Rule]{-108,-3.5}*), DataRange -> {0, 15}]] I wish at this point I am able to look deeper into the details, but higher value in N is not better is required for high demands. This even gets the values at the peak very low.

{"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": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2020-40

latest

en

https://opiatalk.com/where-is-countif-sumif-and-averageif/

text/html

crawl-data/CC-MAIN-2024-38/segments/1725700651614.9/warc/CC-MAIN-20240915020916-20240915050916-00327.warc.gz

Opiatalk.com Best tips and tricks from media worldwide # Where is Countif Sumif and Averageif? ## Where is Countif Sumif and Averageif? SUMIF – Add values if a condition is met, such as adding up all purchases from one category. COUNTIF – Count up the number of items that meet a condition, such as counting the number of times a name appears in a list. AVERAGEIF – Conditionally average values; for instance, you could average your grades for only exams. How do you find the Sumif average? The SUMIF() function returns the total sold per region; COUNTIF() returns the total transactions per region. The simple expression, F3/G3, used in column H returns the average sold per region. In other words, there were 3 transactions made in the Frankfort region for a total of 244 units sold. ### How do you use the Averageif function in Excel? Type an = sign, AVERAGEIF, opening parenthesis (in this example, we are going to evaluate and average the same range of cells, C2 through C5, in the Sales column), comma, then we type the criteria against which the range is evaluated, enclosed in quotes (you put it in quotes so Excel interprets the operator and value … What’s the difference between Sumif and Countif? COUNTIFS applies criteria to cells across multiple ranges and counts the number of times all criteria are met. SUMIFS adds the cells in a range that meet multiple criteria. ## What would be a correct formula for sum in Excel? The SUM function adds values. You can add individual values, cell references or ranges or a mix of all three. For example: =SUM(A2:A10) Adds the values in cells A2:10. What is the formula for average in Excel? Syntax: AVERAGE(number1, [number2].)Example: =AVERAGE(A2:A6)Description: Returns the average (arithmetic mean) of the arguments. See More… ### What is the average formula for Microsoft Excel? Excel Weighted Average. The simplest Excel Average Formula for a set of values consists of the = sign, followed a the sum of the values, all divided by the number of values in the group. A simple Excel average formula, that calculates the average of the three values 5, 10 and 15, is shown in cell A1 of the above spreadsheet on the right. What is SumIf formula? 1) Open Excel sheet and from Row 1, create three columns named Customer, Product Price and Payment Status. 2) In cell B10, enter the following formula: =SUMIF (C3:C7, TRUE, B3:B7) 3) Press Enter. 4) In cell B11, enter the following formula: =SUMIF (C3:C7, FALSE, B3:B7) 5) Press ENTER. 6) In cell B12, enter the following formula: =SUMIF (B3:B7,”>100″) 7) Press ENTER. ## What is the different between sum and SumIf? Difference between SUM, SUMIF, SUMIFS, DSUM SUM. Sum function will get you a sum from multiple individual cells, a single range, or even multiple ranges in 1 shot. SUMIF. Sumif function gets you a sum based on 1 condition (only). SUMIFS. Sumifs (notice the ‘s’) gets you a sum based on multiple conditions. DSUM.

{"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}

CC-MAIN-2024-38

latest

en

http://server.chessvariants.com/index/listcomments.php?itemid=OctahedralChess&limit=100&order=DESC&first=3137&last=20023

text/html

crawl-data/CC-MAIN-2022-49/segments/1669446711162.52/warc/CC-MAIN-20221207121241-20221207151241-00098.warc.gz

The site has moved to a new server, and there are now some issues to fix. Please report anything needing fixing with a comment to the homepage. [ List Latest Comments Only For Pages | Games | Rated Pages | Rated Games | Subjects of Discussion ] Later Earlier Octahedral Chess. 3d-board in octahedral form. (x9, Cells: 340) George Duke wrote on 2008-07-02 UTCGood ★★★★ From 1996 this is like my idea at Chessboard Math for 1x1 over 3x3 over 5x5 over 7x7, all centered totalling 84 squares. Ward's Octahedral carries on the other way, and its 10x10 become too many. Octahedral would be pretty good with 2x2 over 4x4 over 6x6 over 8x8, totalling 120. Opposed to write-ups, we accept the Pyramid board-space that flashed across the mind as hybrid of Thompson's Tetrahedral and this Octahedral probably noticed then. xxx John Lawson wrote on 2003-06-15 UTC ```Charles, Are you aware of the Yahoo group for 3d Chess? http://groups.yahoo.com/group/3-d-chess/ There are links there to other 3-d chess sites as well.``` Charles Gilman wrote on 2003-06-14 UTC Further to my comments on the 2:1:1 leaper, the 2:2:1 one also has some interesting, and quite different, characteristics. Most obviously its leap length is an integer (2²+2²+1²=3²). The smallest integer leaper with two nonzero coordinates is the 4:3 Antelope (4²+3²=5²). Secondly a move 2 forward, 2 left, and 1 up followed by 2 forward, 2 down, and 1 right adds up to 4 forward, 1 left, and 1 down - and the leap of the 4:1:1 leaper is root 18. This means that the 2:2:1 has moves at right angles to each other, usual among leapers with two nonzero coordinates but rare among those with three. Finally it has no colourbinding. Indeed if you divide the square of any leap length by four remainders indicate: 1 no colourbinding, 2 diagonal colourbinding, and 3 triagonal colourbinding. Those dividing exactly are non-coprime and therefore even more bound. Charles Gilman wrote on 2003-05-17 UTCGood ★★★★ It is great to see constructive comments taken on board so quickly. I recently discovered an extraordinary feature common to the leaper you call a Camel (2:1:1) and the one more commonly called a Camel (3:1:0). Both can lose the move in 3d, that is, return to a square in an odd number of moves (5 minimum). It is notable because the 3:1:0 leape rcannot lose the move on a 2d board! In general root-odd leapers cannot even lose the move in 3d, and root-even ones can lose it in 3d if their leap does not pass through the centre of a cell of the other Bishop colour, but not in 2d. Thus the Ferz and Alfil can lose the move in 3d but the Dabbaba cannot. Charles Gilman wrote on 2003-04-19 UTC ```1 You rightly point out that it takes 4 of your Elephants (called Unicorns in e.g. Raumschach) to cover the board, yet you have only 2 aside, and they cannot capture each other. Is this an oversight or a deliberate attempt to emulate the original Chaturanga Elephants which cannot capture each other? 2 Camel usually means a 3:1:0 leaper, although that piece might also be useful. I have seen the 2:1:1 called the Sexton (a pun on the leap length of root 6, although the word actually derives from sacristan). 3 Finally, combinations of 2 as well as 3 elemental pieces are valid in 3d Chess. Indeed you could even have a hybrid of long-range radial and short-range oblique. I could list some stanadrd and suggested names if you are interested.```

{"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}

CC-MAIN-2022-49

latest

en

https://mathematica.stackexchange.com/questions/17885/finding-resolution-of-an-optical-system-using-rayleigh-criterion

text/html

crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00851.warc.gz

# Finding resolution of an optical system using Rayleigh criterion I want to use the Rayleigh criterion to find the resolution of an optical system. I have two functions f1 = Sinc[x]^2 and f2 = Sinc[x-T]^2, T being the translation. I want to change the value of T until the first minimum of f2 coincides with the maximum of f1 (Rayleigh criterion). That being achieved, the value x and T should be returned. Pseudocode: • for x=100 translate T until the first minimum of f2 coincides with the peak of f1; print {x, T} • repeat the same procedure for x=105 ... up to x=200. So, what I would like to do is basically start the program and generate a table with the values x and T. • Welcome to Mathematica.SE! I think you would find it useful to review the documentation and basic tutorials. Mathematica uses square brackets for functions so your f1 and f2 definitions have the wrong syntax. Also, it is not good practice to use uppercase single letters to name expressions as they can clash with built-in functions. Try using lower case f instead. Commented Jan 16, 2013 at 11:59 • Just find the distance between the max and min ... Commented Jan 16, 2013 at 12:07 • The Rayleigh criterion is for the first zero (not the first minimum) of f2 to coincide with the maximum of f1. Commented Jan 16, 2013 at 13:01 • Oops, I just noticed that you were using $sinc(x)^2$, in which case the first minimum and first zero are the same thing. Commented Jan 23, 2013 at 17:12 You may do it the smart way just looking at the distance between max and min, but anyway this is the procedure for doing what you asked for Maximize[Sinc[x - t /. (NMinimize[Sinc[t]^2, t][[2]])]^2, x] {1., {x -> -3.14159}} Edit Plot[{Sinc[x]^2, Sinc[x - t /. (NMinimize[Sinc[t]^2, t][[2]])]^2}, {x, -10, 10}, Evaluated -> True] • Hi, I have two functions Sinc[x1] and Sinc[x2-T], x100=1, x2 =200, T a real number; Pseudocode: for x=100 translate T till the first min of f2 coincide with peak of f1 print print {x1, T} repeat the same procedure for x=105 That would be all! Thanks in advance everyone Commented Jan 16, 2013 at 14:49

{"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}

CC-MAIN-2024-30

latest

en

https://www.math.ias.edu/seminars/abstract?event=125655

text/html

crawl-data/CC-MAIN-2017-51/segments/1512948608836.84/warc/CC-MAIN-20171218044514-20171218070514-00416.warc.gz

# The many forms of rigidity for symplectic embeddings Princeton/IAS Symplectic Geometry Seminar Topic: The many forms of rigidity for symplectic embeddings Speaker: Felix Schlenk Affiliation: University of Neuchâtel Date: Thursday, March 30 Time/Room: 9:30am - 10:30am/S-101 Video Link: https://video.ias.edu/puias/2017/0330-FelixSchlenk We look at the following chain of symplectic embedding problems in dimension four. $E(1, a) \to Z_4(A),\ E(1, a) \to C_4(A),\ E(1, a) \to P(A, ba) (b \in {\mathbb N}_{\geq 2}),\ E(1, a) \to T_4(A).$ Here $E(1, a)$ is a symplectic ellipsoid, $Z_4(A)$ is the symplectic cylinder $D_2(A) \times R_2$, $C_4(A) = D_2(A) \times D_2(A)$ is the cube and $P(A, bA) = D_2(A) \times D_2(bA)$ the polydisc, and $T_4(A) = T_2(A) \times T_2(A)$, where $T_2(A)$ is the 2-torus of area $A$. In each problem we ask for the smallest $A$ for which $E(1, a)$ symplectically embeds. The answer is very different in each case: total rigidity, total flexibility with a hidden rigidity, and a two-fold subtle transition between them. The talk is based on works by Cristofaro-Gardiner, Frenkel, Latschev, McDuff, Muller, and myself.

{"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": 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}

CC-MAIN-2017-51

longest

en

http://www.foxbusiness.com/markets/2016/03/05/how-to-calculate-interest-rate-using-present-and-future-value.html

text/html

crawl-data/CC-MAIN-2017-43/segments/1508187823255.12/warc/CC-MAIN-20171019065335-20171019085335-00120.warc.gz

# How to Calculate Interest Rate Using Present and Future Value By Calculating the interest rate using the present value formula can at first seem impossible. However, with a little math and some common sense, anyone can quickly calculate an investment's interest rate with just its price, its face value, and its duration. The theoretical formula is kind of intense First, let's break down the formula for the present value of an investment based on future cash flows. From this fundamental formula, we'll rearrange the terms to give us a formula to use when we want to calculate the interest rate. In this equation, the present value of the investment is its price today and the future value is its face value. The number of period terms should be calculated to match the interest rate's period, generally annually. Six months would, therefore, be 0.5 periods. Brushing off some algebra, we can rearrange this formula to solve for the interest rate term. That process results in this formula. With this formula, all we need to do is plug the known terms into the equation and we'll have the interest rate. That said, this isn't the easiest formula in the world, and a scientific or financial calculator will be required to reach the final result. Thankfully, the process is much easier than this equation for many finance purposes Fortunately, we don't need to do any of this fancy math to calculate the interest rate in many cases in the financial world. For short term investments, like Treasury bills, this formula is accurate and far simpler than our first equation. In the case of a T-bill, we know our purchase price, or present value, its face value or future value, and how long until it matures. For short-term Treasuries, this duration could be 30 to 182 days depending on the specific note. Let's assume that our bond cost us \$4,900 to purchase, has a face value of \$5,000, and will mature in 182 days. The first step is to subtract the present value from the future value to determine the actual cash return we'll receive over this period. In this case, that works out to \$100. Next, divide that difference by the face value of the Treasury bill. \$100 divided by \$6,000 is 0.0167. The final step is to multiply that result by the 360 divided by the days to maturity, 182 in this case. 360 divided by 182, multiplied by 0.0167 from above gives us 0.033, or a 3.3% annual interest rate on this Treasury bill. When in doubt, use the Internet to find and use an interest rate calculator If you're ever in doubt as to any interest rate calculation, then don't worry. There is an easier way. Thanks to modern computing and the Internet, there are countless interest rate and present value calculators available online. A quick search will turn up dozens of options, providing you the convenience of plug and play math without any of the algebra above. To make it even easier for you, the Motley Fool maintains this list of common calculators you may need for all aspects of your financial life. It's one stop shopping! This article is part of The Motley Fool's Knowledge Center, which was created based on the collected wisdom of a fantastic community of investors. We'd love to hear your questions, thoughts, and opinions on the Knowledge Center in general or this page in particular. Your input will help us help the world invest, better! Email us at [email protected]. Thanks -- and Fool on! The article How to Calculate Interest Rate Using Present and Future Value originally appeared on Fool.com. Try any of our Foolish newsletter services free for 30 days. We Fools may not all hold the same opinions, but we all believe that considering a diverse range of insights makes us better investors. The Motley Fool has a disclosure policy.

{"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}

CC-MAIN-2017-43

latest

en

https://www.holidayeducationist.com/cool-math-solver-games-for-kids-download-for-free/

text/html

crawl-data/CC-MAIN-2023-40/segments/1695233510501.83/warc/CC-MAIN-20230929090526-20230929120526-00366.warc.gz

# Math Solver Games for Kindergarten Kids-Cool Math Game Cool Math Solver Games for Kids gives students chances to investigate principal number ideas, for example, the checking succession, coordinated correspondence, and calculation systems. Connecting with numerical games can likewise urge students to investigate number mixes, place worth, designs, and other significant numerical ideas. Further, they manage the cost of chances for students to extend their numerical understanding and thinking. Instructors ought to give rehashed chances to students to mess around, at that point let the numerical thoughts develop as students notice new examples, connections, and systems. Games are a significant device for learning in primary school science classrooms: ## Cool Math Solver Games! 1. Calculate the Following 2. How do we make Numbers ### 1) Calculate the Following Generates a very simple question and asks the user to drag the correct answer in the Answer Box. Also, it generates a math question that is primarily built for early learners (grade 1). The question is so simple that the kid doesn’t have to calculate above number 10. However, the operation is limited to addition and subtraction only. ### 2) How do we make Numbers On the other hand, is flexible to be used from 1st grade to any age group. Even more, the app generates a math equation where the user has to guess the missing number. Still, depending on the difficulty level, the operation can involve addition, subtraction, multiplication, and division of higher numbers. Also, a question might have two blank pieces to fill. The app starts with a simple math question and automatically generates difficult questions over progression. ### Why play Cool Math Solver Games for Kids? NOTE: 1. Cool math solver games use left to right sequence for calculation and don’t use DMAS rule for a few reasons. However, we are working to provide it as an additional difficulty level in future versions. A reset button is provided so the kids can use the app at initial levels. We look forward to your input to make the app even better. Please consider supporting us by rating Cool Math Solver Games for Kids and writing a small review. It also helps us to produce new and better apps Cool Math Solver Games for Kids Available on IOS and Android. #### More Apps 1 - 2 Years 2 - 3 Years 3 - 6 Years (Kindergartener)

{"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}

CC-MAIN-2023-40

latest

en

https://www.johndcook.com/blog/2010/01/05/how-the-central-limit-theorem-began/

text/html

crawl-data/CC-MAIN-2024-30/segments/1720763514917.3/warc/CC-MAIN-20240719170235-20240719200235-00189.warc.gz

# How the central limit theorem began The Central Limit Theorem says that if you average enough independent copies of a random variable, the result has a nearly normal (Gaussian) distribution. Of course that’s a very rough statement of the theorem. What are the precise requirements of the theorem? That question took two centuries to resolve. You can see the final answer here. The first version of the Central Limit Theorem appeared in 1733, but necessary and sufficient conditions weren’t known until 1935. I won’t recap the entire history here. I just want to comment briefly on how the Central Limit Theorem began and how different the historical order of events was from the typical order of presentation. A typical probability course might proceed as follows. 1. Define the normal distribution. 2. State and prove a special case of the Central Limit Theorem. 3. Present the normal approximation to the binomial as a corollary. This is the opposite of the historical order of events. Abraham de Moivre discovered he could approximate binomial distribution probabilities using the integral of exp(-x2) and proved an early version of the Central Limit Theorem in 1733. At the time, there was no name given to his integral. Only later did anyone think of exp(-x2) as the density of a probability distribution. De Moivre certainly didn’t use the term “Gaussian” since Gauss was born 44 years after de Moivre’s initial discovery. De Moivre also didn’t call his result the “Central Limit Theorem.” George PĆ³lya gave the theorem that name in 1920 as it was approaching its final form. For more details, see The Life and Times of the Central Limit Theorem.

{"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}

CC-MAIN-2024-30

latest

en

https://lists.nongnu.org/archive/html/help-gsl/2006-07/msg00006.html

text/html

crawl-data/CC-MAIN-2021-25/segments/1623488525399.79/warc/CC-MAIN-20210622220817-20210623010817-00370.warc.gz

help-gsl [Top][All Lists] ## [Help-gsl] Matrix memory deallocation problem From: Wei Cheng Subject: [Help-gsl] Matrix memory deallocation problem Date: Thu, 6 Jul 2006 15:05:23 +0100 ```I have a matrix-matrix multiplication function based on BLAS matrix-matrix mutiplication function,just for the ease of my own use as follow: //****************************************************************************** gsl_matrix* matrix_matrix_mul(CBLAS_TRANSPOSE_t TransA,CBLAS_TRANSPOSE_t TransB,const gsl_matrix* X1,const gsl_matrix* X2) //matrix matrix multiplication, return the result { int N1=(*X1).size1; int N2=(*X2).size1; int D1=(*X1).size2; int D2=(*X2).size2; gsl_matrix* Y;// to store the product in Y //determine the dimension of product matrix Y if(TransB==CblasTrans) //X2 is transposed Y=gsl_matrix_alloc (N1,N2); else if(TransA==CblasTrans) //X1 is transposed Y=gsl_matrix_alloc (D1,D2); else //No transpose Y=gsl_matrix_alloc (N1,D2); //calling BLAS general matrix multiplication function , product is stored in Y gsl_blas_dgemm (TransA, TransB, 1.0, X1, X2, 0.0, Y); return Y; } //***************************************************************************** The function returns the matrix product of X1 and X2. I notice that whenever we create matrix we have to free the memory with gsl_matrix_free. However in the above function I can't free Y before I return it and i am not able to free Y after return statement. Memory leak problem occurs when I call the above function 5000000 times in a loop like this: //************************************************************************************ void testing() { gsl_matrix* A=rand(50,100);//create random 50 by 100 matrix gsl_matrix* B=rand(100,50);//create random 100 by 50 matrix for(int i=0;i<5000000;i++) matrix_matrix_mul(CblasNoTrans ,CblasNoTrans ,A ,B );//calculate the project of matrix A and B} //************************************************************************************* The problem I am having is that during the loop, my RAM usage shoot up and eventually run out. I am certain the problem is with the the deallocation of Y in mutiplication function. Can anybody help me with this matrix deallocation problem. This is just a test, in my application the matrices are thousands by thousands. Wei Thanks ``` reply via email to

{"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}

CC-MAIN-2021-25

latest

en

http://www.mathematicsgre.com/viewtopic.php?f=1&t=653

text/html

crawl-data/CC-MAIN-2019-04/segments/1547583660020.5/warc/CC-MAIN-20190118090507-20190118112507-00132.warc.gz

## Problems from various topics ... Forum for the GRE subject test in mathematics. pavan4850 Posts: 1 Joined: Thu Oct 13, 2011 12:16 am ### Problems from various topics ... 1. Profit and loss problem.. Two mercants sold a good of worth Rs 1000 from which one merchant A got profit from selling prices and merchant B got profit form cost price and profit for both is 25% what is difference amount.? 2.5lts solution bottles in which 4 lts of solution is filled in which 15 % of salt is mixed,1.5 lits solution is fell out and it is filled with water how much of salt content is present..? 3.Square of 21 cms and bull was tied at corner with rope of 7 cms how much area it can graze... 4.A and B can do a work 15 and 20 days respectively form which A as done half of the work and then b joind then it done 1/20th work how much many days in which a complted work which left by B. Napot Posts: 9 Joined: Sat Apr 09, 2011 10:03 pm ### Re: Problems from various topics ... I'm sorry. Those questions were not coherently stated. I'm not even sure they're even GRE Math Subject level questions. If your English is this bad, you might want to focus your attention on the TOEFL.

{"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}

CC-MAIN-2019-04

latest

en

way2h.blogspot.com

text/html

crawl-data/CC-MAIN-2018-30/segments/1531676589251.7/warc/CC-MAIN-20180716095945-20180716115945-00164.warc.gz

// # Cryptography 102: History Substitution Cipher One of the most readily known Ciphers is called the substitution cipher, in fact you probably played with one as a kid. However what you didn't know is they are all very badly broken. A Substitution Cipher is essentially where you create a table where you would say A = C, B = W, C = N all the way to Z = A. This table becomes your key for the cipher, we discussed keys in my earlier lesson. For instance is we had the word "ABBA" and we wanted to encrypt that using our cipher it would come out as "CWWC". To decrypt it we would just use the key in reverse. Caesar Cipher Back in the days of Rome when Caesar wanted to send secret messages to people he would use something he developed called the Caesar Cipher. This cipher essentially involves a shift in the letters. For example your Caesar Cipher may be a shift of 3, A = D, B = E, C = F... Y = B, Z = C. In short this shifts all of the letters values over by 3. Again we we had a message of "Hello" it would become, "Khoor". We do this by again shifting the letters by 3. They key in this case would be the shift of 3. Breaking The Substitution Cipher The first question to ask is what is the size of the Key Space? In other words how many keys could their possibly be assuming there are 26 letters. The key is simply a permutation of all of the letters, if you are a math student you should know that to calculate this you use something called a Factorial, denoted by 26!. 26! is equal to about roughly 2^88, meaning our key is an 88 bit key. A key of this size is very adequate, in fact we will be going over very secure ciphers in a bit with key spaces no bigger than that. However because of the way the Substitution Cipher works, it is very easy to break. To break The Substitution Cipher we use letter frequencies, this is essentially using the probability of a letter occurring in the English language to break the Cipher. The most common letter in English text is the letter "e" at 12.7%, followed by "t" at 9.1% and then "a" at 8.1%. So for example, if you gave me a Cipher Text to decrypt that was encrypted using a substitution cipher and I know the original plain text was English, I can crack it with only the cipher text and probability. I would count the frequency of letters in the cipher text, lets say "t" is the most common letter. I can say that the decrypted version of "t" is "e" with high probability. We can then continue to do this with more letters as well as letter pairs. Frequent letters pairs like "he", "an", "in", "th" and so forth can help us even more to create the decryption table. Next think you know your message is completely decrypted using only the cipher text, we call this kind of attack a Cipher Text only Attack. This is the lowest level of attack and if your cipher is susceptible to it, well it sucks Vigener Cipher The Vigener Cipher combines two messages to create a new decrypted one. k = C R Y P T O C R Y P T O C R Y P T m = W H A T A N I C E D A Y T O D A Y __________________________________ C = Z Z Z J U C L U D T U N W G C Q S Here is an example of a Vigener Cipher. Pretend every letter has a numerical value, a=1, b=2, c=3 and so forth. In the vigener cipher we add up these values and create a new message with it. C+W=Z, R+H=Z, Y+A=Z, and so forth. So the key in this case is the word "Crypto", replicated as many times as needed to fit the message. Again breaking this is very easy, but we must first assume we know the length of the key, in this case 6. We would break up the cipher text into groups of 6. ZZZJUC|LUDTUN|WGCQS Than we look at the first letter of every group. We know that the most common letter is "e" so lets suppose we went and counted all of the first letters and it turned out to be "h". We know that H is most likely the encrypted version of "e" so theoretically if we subtract "E" from "H" we should get the first letter in the key. H-E= C which is the first letter in the key. You would continue to do this for the second letter, and then third letter until the entire set is done using letter and letter pair frequencies. Lets say for instance that you did not know the length of the key, you would simply start and assume the key is the length of 1, solve for that.When it doesn't work you would then go to a length of 2, and so forth until it is decrypted. Rotor Machines Another famous example is the Rotor Machine. In a rotor machine the key changes after a letter is pressed. The key actually rotates. If we had a key that said Z = E, we then hit a number, that key would cycle over and A = E now. You can decrypt these using similar Cipher Text Attacks.

{"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}

CC-MAIN-2018-30

latest

en

https://brainly.ph/question/295010

text/html

crawl-data/CC-MAIN-2017-04/segments/1484560282202.61/warc/CC-MAIN-20170116095122-00112-ip-10-171-10-70.ec2.internal.warc.gz

# What is the common difference of the arithmetic sequence whose first and 11th terms are 5 and 15 respectively 1 by Rachel111 • Brainly User 2016-02-27T14:45:21+08:00 ### This Is a Certified Answer Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest. Arithmetic Sequence Rule: = a₁ + (n-1) (d) a₁ = 5 a₁₁ = 15 n = 11 d = ? 15 = 5 + (11-1) (d) 15 = 5 + 10 (d) 15 = 15 (d) 15/15 = 15d/15 d = 1 Common difference (d) = 1 Check: 15 = 5 + (11-1) (1) 15 = 5 + 10 (1) 15 = 5 + 10 15 = 15   (True)

{"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}

CC-MAIN-2017-04

latest

en

http://www.jiskha.com/display.cgi?id=1334759127

text/html

crawl-data/CC-MAIN-2017-22/segments/1495463613796.37/warc/CC-MAIN-20170530051312-20170530071312-00310.warc.gz

physics posted by on . To illustrate the effect of ice on the aluminum cooling plate, consider the data shown in the drawing below. The length of the ice block is L1 = 0.00476 m and its temperature is T1 = -11.9 °C. The aluminum block is L2 = 0.00131 m long and has a temperature of T2 = -23.8 °C. Ignore any limitations due to significant figures. Calculate the heat per second per square meter that is conducted through the ice-aluminum combination. Do not enter unit. I'm not able to submit the URL for the picture so here's my best attempt at drawing it |--------|--| T1 | ice | | <---aluminum |_________|___| T2 ---L1---|-L2- • physics - , the picture got messed up. theres a block of ice on the left and the strip of aluminum on the right. the base of the ice is L1 and the base of the aluminum is L2. T1 is the ice and T2 is the aluminum. Please help me with this problem!!!! • physics - , Isn't this a simple heat flow problem. Boundry coditions: At the ice/Al interface, temp 0deg. At the right side of the aluminum, Temp is T2. Heat flow through the ice, and then in aluminum will be equal. You cannot assume anything about the temperature of the ice/aluminum interface. You need the thermal conductivity of ice, and aluminum, look them up. Then start with the heat flow through the ice (deltaTemp T1-Tinterface), setting it equal to heat flow through the aluminum, (deltatemp Tinerface-T2). Solve for the unknown Temp at interface, then you can solve for heat flow.

{"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}

CC-MAIN-2017-22

latest

en

http://www.chegg.com/homework-help/questions-and-answers/using-kirchhoff-s-rules-given-1-705-v-2-607-v-3-788-v-calculate-following-figure-p1823-fin-q197414

text/html

crawl-data/CC-MAIN-2015-40/segments/1443736679281.15/warc/CC-MAIN-20151001215759-00120-ip-10-137-6-227.ec2.internal.warc.gz

Using Kirchhoff's rules (and given 1 =70.5 V, 2 =60.7 V, 3 =78.8 V) calculate the following. Figure P18.23 (a) Find the current in each resistor shown inFigure P18.23. IR1 = .535 Your answer is within 10% of the correctvalue. This may be due to roundoff error, or you could have amistake in your calculation. Carry out all intermediate results toat least four-digit accuracy to minimize roundoff error.mA IR2 = Your answer differs from the correct answerby 10% to 100%. mA IR3 = Your answer differs from the correct answerby 10% to 100%. mA (b) Find the potential difference between points c andf. cf = V Whichpoint is at the higher potential?

{"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}

CC-MAIN-2015-40

latest

en

http://www.wyzant.com/resources/answers/2541/if_abe_comes_to_the_party_then_bill_comes_to_the_party_and_if_bill_comes_to_the_party_then_carol_comes_to_the_party_a_b_c

text/html

crawl-data/CC-MAIN-2014-41/segments/1410657134514.93/warc/CC-MAIN-20140914011214-00184-ip-10-234-18-248.ec2.internal.warc.gz

Search 74,112 tutors 0 0 # If Abe comes to the party, then Bill comes to the party; and if Bill comes to the party, then Carol comes to the party. (A, B, C) Symbolic Logic , translate single statements, whole arguments, and to use the comparative method to determine validity or invalidity. I have to symbolize these statements by using the Letters provided in the parentheses. I am not completely sure of the notation required by your teacher, but here is some explanation of what is meant by this question. Use A to mean 'Abe comes to the party'. Use B to mean 'Bill comes to the party'. Use C to mean 'Carol comes to the party'. Now we can translate the statements above as follows: If A then B, and if B then C. From these two statements we can conclude 'if A then C'. You might have a notation similar to this: (A>B)U(B>C)>(A>C) What these arguments mean is that although we don't know whether Abe is coming to the party, we do know that if he does come, we can expect to see Bill and Carol as well.  However, if Abe doesn't come, Bill and Carol, or just Carol, might be there without Abe.

{"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}

CC-MAIN-2014-41

latest

en

http://www.nelson.com/nelson/school/elementary/mathK8/math7/quizzes/math7quizzes/gr7_ch2_les4-edited.htm

text/html

crawl-data/CC-MAIN-2018-05/segments/1516084887832.51/warc/CC-MAIN-20180119065719-20180119085719-00754.warc.gz

Name:    Try It Out -- Chapter 2, Lesson 4: Communicating about Ratio and Rate Problems Matching Last season Kayla played 594 games and won 440 of those games Next season she is scheduled to play 540 games. She wants to know how many games she needs to win next season, to keep the same rate of games won. Order the steps she would take to solve the problem. A. She wrote a proportion with a missing term for the number of games she needs to win next season. B. The answer is 400. This is correct because 400 x 1.1 is 440 games. C. She knows that she won 440 out of 594 games last season, and she will play 540 games next season. D. The scale factor is 1.1 because 594 ¸ 1.1 is 540. 1. What is her first step? 2. What is her second step? 3. What is her third step? 4. What is her final step?

{"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}

CC-MAIN-2018-05

latest

en

https://www.jiskha.com/display.cgi?id=1372624033

text/html

crawl-data/CC-MAIN-2018-05/segments/1516084890187.52/warc/CC-MAIN-20180121040927-20180121060927-00748.warc.gz

# math posted by . x-5y=-21 9y-8x=44 substitution method • math - x-5y=-21 9y-8x=44 x = 5y - 21 9y - 8(5y - 21) = 44 9y - 40y + 168 = 44 -31y = 44 - 168 -31y = -124 y = 4 Take it from there. • math - x = 5y -21 9y-8(5y-21) = 44 9y -40y +168 = 44 -31y + 168 = 44 -31y + 168 -168 = 44 -168 -31y = -124 y = 4 x = 5(4)-21 = -1 (-1, 4) ## Similar Questions 1. ### math Systems of equations can be solved by graphing or by using substitution or elimination. What are the pros and cons of each method? 2. ### Math 1. Explain the advantages and disadvantages of both the substitution method and the elimination method for solving a system. Help please. The only reasons I can think of is because in substitution it might be more difficult cause there's … 3. ### Algebra Solving system of linear equations by using the apropiate method name the method and the x,y coordinates 1)X=4y+8 2x-8y=-3 2)3x-7y=6 elimination method 2x+7y=4 (2,0) 3)5x-2y=12 elimination method 3x-2y=-2 (7,23/2) 4)9x-8y=42 elimination … 4. ### MATH SYSTEMS OF EQUATIONS CAN BE SOLVED BY GRAPHING OR BY USING SUBSTITUTION OR ELIMINATION. WHAT ARE THE PROS AND CONS OF EACH METHOD? 5. ### math substitution method use the substitution method to solve 7x+5y=-26 x=28-6y 6. ### substitution method Solve by substitution method please help me solve this. I do not know how to do this at all. The question is 3x+8y=33 and x=31-6y I know that I have to plug x into the first x like 3(31-6y)+ 8y = 33 but i do not know what to do from … 7. ### algebra use the Substitution Method. Please give it a try. y = 3x + 4 5x + 3y = 12 Why is the substitution method preferred in this case?

{"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}

CC-MAIN-2018-05

latest

en

https://mph-to-kmh.com/196-inches-to-feet/

text/html

crawl-data/CC-MAIN-2024-10/segments/1707947475833.51/warc/CC-MAIN-20240302152131-20240302182131-00253.warc.gz

# Convert 196 inches to feet: Learn How to Easily Convert Inches to Feet ## 196 inches to feet ### Converting inches to feet Calculating the conversion from inches to feet is a fundamental task in many fields, whether you are a builder, an architect, or simply measuring length for projects at home. Understanding how to convert inches to feet is essential for accuracy and precision. In particular, converting 196 inches to feet can be useful when working with objects or spaces that are measured in inches but need to be presented in feet. The conversion factor for inches to feet is 12. To convert 196 inches to feet, divide the length in inches by the conversion factor. In this case, dividing 196 by 12 gives us 16.33 feet. However, since we are dealing with measurements, it is important to round the final result to the appropriate decimal place. Therefore, 196 inches is equivalent to approximately 16 feet and 4 inches. ### Applications of converting inches to feet Being able to convert inches to feet has practical applications in various fields. For example, in construction, it allows workers to accurately measure and cut materials to the required lengths. Similarly, in interior design, converting inches to feet helps determine the appropriate dimensions for furniture and decor placement in a room. Additionally, when working with architectural drawings or blueprints, understanding the conversion allows for precise scaling and measurements. Converting inches to feet is also relevant in everyday situations. For instance, when purchasing new flooring or carpeting, the dimensions are often given in feet. By knowing how to convert inches to feet, you can determine whether a certain amount of flooring will be sufficient for a given space. Additionally, if you enjoy DIY projects such as building shelves or fitting objects into tight spaces, converting inches to feet will aid in measuring and planning. You may also be interested in:  Converting Height: Discover the Easy Way to Convert 194 cm to Feet ### Tips for converting inches to feet accurately To ensure accurate conversions, here are some tips and tricks to keep in mind when converting inches to feet: 1. Use a calculator: Precision is key when dealing with measurements, so using a calculator to avoid errors is recommended. 2. Round to the appropriate decimal place: Depending on the context, you may need to round the final result to the nearest whole number or a specific decimal place. 3. Double-check your calculations: Mistakes happen, especially when working with numbers. Always double-check your calculations to avoid any inaccuracies. 4. Understand the context: Knowing when and how to convert inches to feet is crucial. Take into account the specific requirements and conventions of the project or industry you are working in. By following these guidelines, you can confidently convert inches to feet and ensure your measurements are accurate and precise.

{"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}

CC-MAIN-2024-10

longest

en

https://www.911metallurgist.com/weighting-core-cuttings-diamond-drilling/

text/html

crawl-data/CC-MAIN-2023-14/segments/1679296943749.68/warc/CC-MAIN-20230322020215-20230322050215-00523.warc.gz

# Weighting Core and Cuttings in Diamond Drilling A drill hole is bored in iron ore exploration principally to test variations in rock composition with depth and is usually directed as nearly normal to the bedding of a horizon to be tested as possible. This practice has a tendency to minimize variation in composition laterally which in any event is not likely to be great. It is obvious that the opportunity for change in analysis of a particular rock is not statistically as great radially in a diamond drill hole where the distance in which such a change may occur is from 0.719 in. (EX bit) to 1.469 in. (NX bit) as there would be longitudinally even in a run as short as 5 ft. Variations in composition of bedded or layered rocks are usually greater normal to the bedding than parallel thereto. Even in massive rocks, like porphyries, variations are functions of distance. Hence, in either case, variations along the hole are of greater effect than across it. The Longyear formula involves a weighting of the core and sludge in proportion to the theoretical volume of each, that is, where the volume of the core recovered is one-third of the total cubic volume of the cylindrical hole made by the drill, the analysis of the core is given one-third of the weight in the final analysis. This approach to the question has certain disadvantages, particularly in the high core recovery brackets. Even when 100 pct of the core is recovered, the volume of the core is only from 35.5 to 51.2 pct of the volume of the hole, depending on the bit size, and is weighted accordingly. This runs counter to our basic principle that when all the core is recovered it should receive all the weight in the final analysis. Now we are ready to analyze the volumetric or Longyear method of weighting: Let a = the figure for percentage of the total volume of the hole occupied by core when core recovery is 100 pct. This figure varies between 35.5 and 51.2 and is a function of the relation between the second power of the outside and inside diameters of the particular bit in question. It can be seen that in this direct proportion if core recovery is slight, the core still has a strong influence in the final analysis so that when anomolous beds only are cored, it might prove to give undue weight to the core. Since the only defect of the Longyear  or volumetric formula in the high brackets of recovery is due to the factor  “a,” an empirical solution might be to substitute “i” for “a” so that as recovery increases the core analysis may be multiplied by a factor greater than from 35.5 to 51.2 pct. The formula would then become : At = S(100 – i²/100) + i²/100 C/100

{"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}

CC-MAIN-2023-14

latest

en

https://preparingtobecome.com/elementary-math/

text/html

crawl-data/CC-MAIN-2024-18/segments/1712297295329.99/warc/CC-MAIN-20240425130216-20240425160216-00312.warc.gz

# Elementary Math /10 0 1 / 10 37 + 51 = 2 / 10 69 + 99 = 3 / 10 53 + 39 = 4 / 10 24 + 48 = 5 / 10 48 + 77 = 6 / 10 66 + 97 = 7 / 10 78 + 73 = 8 / 10 55 + 93 = 9 / 10 47 + 44 = 10 / 10 39 + 32 = The average score is 0% 0% /10

{"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}

CC-MAIN-2024-18

latest

en

https://finance.yahoo.com/news/does-mtr-corporation-limiteds-hkg-221136972.html

text/html

crawl-data/CC-MAIN-2019-35/segments/1566027331485.43/warc/CC-MAIN-20190826085356-20190826111356-00390.warc.gz

U.S. Markets open in 2 hrs 38 mins # Does MTR Corporation Limited's (HKG:66) P/E Ratio Signal A Buying Opportunity? This article is written for those who want to get better at using price to earnings ratios (P/E ratios). We'll look at MTR Corporation Limited's (HKG:66) P/E ratio and reflect on what it tells us about the company's share price. Looking at earnings over the last twelve months, MTR has a P/E ratio of 18.31. That corresponds to an earnings yield of approximately 5.5%. ### How Do You Calculate MTR's P/E Ratio? The formula for P/E is: Price to Earnings Ratio = Share Price ÷ Earnings per Share (EPS) Or for MTR: P/E of 18.31 = HK\$48.4 ÷ HK\$2.64 (Based on the year to December 2018.) ### Is A High Price-to-Earnings Ratio Good? A higher P/E ratio means that buyers have to pay a higher price for each HK\$1 the company has earned over the last year. All else being equal, it's better to pay a low price -- but as Warren Buffett said, 'It's far better to buy a wonderful company at a fair price than a fair company at a wonderful price.' ### How Does MTR's P/E Ratio Compare To Its Peers? The P/E ratio indicates whether the market has higher or lower expectations of a company. If you look at the image below, you can see MTR has a lower P/E than the average (20.6) in the transportation industry classification. This suggests that market participants think MTR will underperform other companies in its industry. Since the market seems unimpressed with MTR, it's quite possible it could surprise on the upside. If you consider the stock interesting, further research is recommended. For example, I often monitor director buying and selling. ### How Growth Rates Impact P/E Ratios Earnings growth rates have a big influence on P/E ratios. When earnings grow, the 'E' increases, over time. That means even if the current P/E is high, it will reduce over time if the share price stays flat. Then, a lower P/E should attract more buyers, pushing the share price up. MTR shrunk earnings per share by 6.6% last year. But over the longer term (5 years) earnings per share have increased by 3.3%. ### A Limitation: P/E Ratios Ignore Debt and Cash In The Bank Don't forget that the P/E ratio considers market capitalization. So it won't reflect the advantage of cash, or disadvantage of debt. The exact same company would hypothetically deserve a higher P/E ratio if it had a strong balance sheet, than if it had a weak one with lots of debt, because a cashed up company can spend on growth. Such expenditure might be good or bad, in the long term, but the point here is that the balance sheet is not reflected by this ratio. ### Is Debt Impacting MTR's P/E? MTR has net debt worth 10% of its market capitalization. This could bring some additional risk, and reduce the number of investment options for management; worth remembering if you compare its P/E to businesses without debt. ### The Bottom Line On MTR's P/E Ratio MTR's P/E is 18.3 which is above average (10.4) in its market. With modest debt but no EPS growth in the last year, it's fair to say the P/E implies some optimism about future earnings, from the market. When the market is wrong about a stock, it gives savvy investors an opportunity. As value investor Benjamin Graham famously said, 'In the short run, the market is a voting machine but in the long run, it is a weighing machine.' So this free visualization of the analyst consensus on future earnings could help you make the right decision about whether to buy, sell, or hold. But note: MTR may not be the best stock to buy. So take a peek at this free list of interesting companies with strong recent earnings growth (and a P/E ratio below 20). We aim to bring you long-term focused research analysis driven by fundamental data. Note that our analysis may not factor in the latest price-sensitive company announcements or qualitative material. If you spot an error that warrants correction, please contact the editor at [email protected]. This article by Simply Wall St is general in nature. It does not constitute a recommendation to buy or sell any stock, and does not take account of your objectives, or your financial situation. Simply Wall St has no position in the stocks mentioned. Thank you for reading.

{"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}

CC-MAIN-2019-35

latest

en

https://discourse.julialang.org/t/julia-matrix-norm-is-different-matlab/72500

text/html

crawl-data/CC-MAIN-2022-27/segments/1656104597905.85/warc/CC-MAIN-20220705174927-20220705204927-00180.warc.gz

# JULIA matrix norm is different MATLAB Hi, when I calculate the norm of vector a3, I get different values ​​in julia vs MatLab If I copied your matrix correctly: julia> a3 = [0 sqrt(2)/2 sqrt(2) sqrt(2)/2 0 -sqrt(2)/2 -sqrt(2) -sqrt(2)/2; -sqrt(2)/2 -sqrt(2) 0 sqrt(2)/2 sqrt(2) -sqrt(2)/2 0 -sqrt(2)/2] 2×8 Matrix{Float64}: 0.0 0.707107 1.41421 0.707107 0.0 -0.707107 -1.41421 -0.707107 -0.707107 -1.41421 0.0 0.707107 1.41421 -0.707107 0.0 -0.707107 julia> norm(a3) 3.464101615137755 the result looks correct to me: the sum of the squares of the elements is 12, it’s square root is 3.464…, which is the 2-norm. Does Matlab give the 2-norm as well with the norm function? 1 Like https://www.mathworks.com/help/matlab/ref/norm.html#bvhji30-1 wait , no n = norm( X ) returns the 2-norm or maximum singular value of matrix X , which is approximately max(svd(X)) . 1 Like Matlab’s norm gives you the operator norm for matrices, whereas Julia’s norm will always give you the standard L2 norm. The equivalent in Julia would be opnorm: julia> using LinearAlgebra julia> a3 = [0 sqrt(2)/2 sqrt(2) sqrt(2)/2 0 -sqrt(2)/2 -sqrt(2) -sqrt(2)/2; -sqrt(2)/2 -sqrt(2) 0 sqrt(2)/2 sqrt(2) -sqrt(2)/2 0 -sqrt(2)/2] 2×8 Matrix{Float64}: 0.0 0.707107 … -1.41421 -0.707107 -0.707107 -1.41421 0.0 -0.707107 julia> opnorm(a3) 2.5495097567963927 2 Likes Perhaps this might be worth mentioning in Noteworthy Differences from other Languages · The Julia Language, since it is a bit of a gotcha, if anyone cares to make a PR. 3 Likes Actually, the previous posts have it backwards, as far as the names of the norms are concerned. For matrices, it’s opnorm that gives you the induced matrix 2-norm, \|A\|_2 = \sup_{x\neq0} \|Ax\|_2/\|x\|_2 = max singular value, whereas norm gives the root-sum-squares Frobenius norm, \|A\|_F = \sqrt{\sum_{i,j} A^2_{ij}}. E.g. julia> A = [1 1; 2 0] 2×2 Matrix{Int64}: 1 1 2 0 julia> opnorm(A) 2.2882456112707374 julia> svdvals(A) 2-element Vector{Float64}: 2.2882456112707374 0.8740320488976421 julia> norm(A) 2.449489742783178 julia> sqrt(1^2 + 1^2 + 2^2 + 0^2) 2.449489742783178 That’s what documentation says as well help?> opnorm search: opnorm opnorm(A::AbstractMatrix, p::Real=2) Compute the operator norm (or matrix norm) induced by the vector p-norm, where valid values of p are 1, 2, or Inf. (Note that for sparse matrices, p=2 is currently not implemented.) Use norm to compute the Frobenius norm. and help?> norm search: norm normpath normalize normalize! opnorm issubnormal UniformScaling ColumnNorm set_zero_subnormals norm(A, p::Real=2) For any iterable container A (including arrays of any dimension) of numbers (or any element type for which norm is defined), compute the p-norm (defaulting to p=2) as if A were a vector of the corresponding length. Note the “as if A were a vector of the corresponding length.” I.e. unpack A into a vector and compute the 2-norm of that vector. That is totally different from the induced matrix 2-norm. 4 Likes Matlab’s norm applied to a matrix gives the induced matrix 2-norm, equal to the matrices’ largest singular value. Julia’s norm applied to a matrix gives the Frobenius norm, equal to the root sum of squares of the matrix elements. Presumably Julia uses the Frobenius norm because it’s way cheaper to compute root sum of squares than an SVD. And also, the title of the OP is wrong. This is a different of matrix norms, not vector norms. 3 Likes

{"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}

CC-MAIN-2022-27

latest

en

https://math.stackexchange.com/questions/2102133/double-counting-the-number-of-proper-divisors

text/html

crawl-data/CC-MAIN-2019-51/segments/1575540551267.14/warc/CC-MAIN-20191213071155-20191213095155-00382.warc.gz

# Double counting the number of proper divisors Suppose $n$ is a composite natural number. Then $n$ has unique prime factorization. To count the number of proper divisors, simply take the product of the exponents +1 in the prime factorization. $$n = \prod_{i = 1}^{n} p_i^{a_i}$$ $$\mbox{proper divisors} = \prod_{i = 1}^{n}(a_i+1)$$ Is 1 counted multiple times by doing this? For instance, I can choose from $a_0+1$ factors contributed from $p_0$, namely $1, p_0, p_0^2, \dots , p_0^{a_0}$ Don't I count 1 multiple times? • Your notation is suspect because $n$ is used both as the left hand side and as the limit of the index. Rarely does $n$ have $n$ distinct prime factors, if that is what you want to convey. – hardmath Jan 17 '17 at 21:04 • With this product you count all divisors, not only the proper ones. – ajotatxe Jan 17 '17 at 21:07 You don't count 1 multiple times: The divisor 1 is only obtained if you choose the exponent $0$ for every prime factor at the same time and the only way of doing that is by choosing every exponent to be $0$; there is only one way of doing it. Hence you do not double count $1$, nor any other divisor. In order to count one, the exponent on each prime factor has to be zero. So you are only counting it once. As an example of how this works, consider the factors of $72 = 2^3\cdot 3^2$. The factors are: $$\begin {array}{c} 1 & 2 & 4 & 8 \\ 3 & 6 & 12 & 24 \\ 9 & 18 & 36 & 72 \end{array}$$ As you can see in this $(3{+}1)\times (2{+}1)$ array, $1$ only occurs once as the multiplication process produces another prime power factors otherwise.

{"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}

CC-MAIN-2019-51

latest

en

https://boredofstudies.org/search/328608/

text/html

crawl-data/CC-MAIN-2019-43/segments/1570986661296.12/warc/CC-MAIN-20191016014439-20191016041939-00496.warc.gz

# Search results 1. ### PDF files of new textbook. Hey so what textbooks would you recommend with the new Stage 6 math new syllabus (2U/3U/4U) Also, does anyone have any pdf files of those? Alright guys I'm nearly at the end of my degree and for my econ degree, I need 2 more business elective courses. So ironically, I tend to do better in conceptual courses like math than courses that involve quite a bit of memorisation. So what would be recommended for a guy who's good at maths... 3. ### Dropping from a major to a minor So it works like this, I'm currently studying Adv Math/ Economics at UNSW, I decided to combine my math degree with eco since it would give me better job prospects (well that's what I think) and it seemed interesting to me. However as I'm as going through halfway of eco, I'm starting to hate... 4. ### What makes someone a BAD tutor? So I stumbled an article about good and bad tutors, but I'd like so hear some personal experiences from my fellow peers here. So have you had a terrible tutor? if so then how? I'm a tutor myself, and let's just say that greed hasn't overcome my morals. 5. ### Positive Matrix proof $a) Prove that a Square Matrix$ \ G \ $is a positive definite if and only if$ \ G+G^T \ $is a positive definite$ $b) Prove that an$ \ n \ times \ n \ $symmetric matrix$ \ G \ $is a positive definite if and only if$ \ G^{-1} \ $is positive definite$ 6. ### log base 1 Yeah sorry if this was asked before, but this just crossed my mind. y = log_1(x) \Rightarrow x = 1^y = 1 So what I'm getting is that this particular log is only defined at x = 1? But if we use the change base formula y = ln(x)/ln(1) = \frac{ln(x)}{0} We can see that it's not... 7. ### Moment Generating Function $For any$ \ \ 0<p<1 \ \ $and$ \ \ r \ \ $a positive integer, the probability function$ \\ \\ f(x) = \binom{r+x+1}{x}p^r(1-p)^x \ \ \ x = 0,1,2 ... \\ \\ $defines a random variable X$ \\ \\ $Show that the moment generating function$ \ \ m_X(u) = \mathbb{E}(e^{ux}) = \\... 8. ### Gamma distribution function Yeah my mind's probably not working or I'm missing something important but how do you evaluate this \frac{1}{\beta^2} \int_0^{\infty}e^{-\frac{y^2}{2\beta^2}} dy 12. ### Partial derivative question Ok I maybe I'm derping and stuff but i can't seem to get this one $Suppose that function f is a differentiable function of a single variable and F(x,y) is defined by$ \ \ F(x,y) = \ f(x^2-y) \\ $a) Show that$ \frac{\partial F}{\partial x} + 2x\frac{\partial F}{\partial y} = 0 \\ \\ b) \ \... 13. ### Going home at 6pm Hey guys, so I have like 3 days where i finish my last class at 6pm. The thing is, there's heaps of people crowding around the bus stop to central at that time, and it took me around 40minute to get on my bus. So just wondering, is an alternate way of getting back to central faster? 14. ### Driving Manual cars Hey so I just got my Ls and like my mum owns a manual car (a 15 year old toyota starlett lol), while my dad owns a toyota corrola which is automatic Now the thing is my dad is mostly out at work so most of the time i could only drive my mum's manual car. I only just started learning to... Sorry if there's already a thread like this (Ｉcan't find one like this haha) any ways post your highscore (must be legit i.e not hacks/cheats) and your thoughts about the game ahah EDIT: My best is 62 16. ### Class clashes I don't know if there's already a thread like this, and if there is I'm sorry lols So I applied in 4 subjects under B science Maths 1141 Matsh 1081 Chem 1031 Phys 1111 Unfortunately my enrollment could not be processed because there's a clash in the lectures Maths 1081 and Chem 1031... 17. ### B science with another degree? So I'm planning to study Bachelor of science (maths major, still deciding whether UNSW or Usyd, could you suggest which one is better) But I'm told that there are limited job opportunities with that degree itself, so therefore I should combine with something. Now I'm not really a fan of... 18. ### ATAR Estimate Hey Guys just need a rough estimate now 2011 School Rank: around 200s 2012 School Rank : 600 (it was the worst year ever with only 1 band 6, but I'm confident this year we have a much stronger cohort) Maths Ext 1 1/12 (beating the 2nd guy by at least 15%) Maths Ext 2 2/5 (Joined classes... 19. ### Standard English dragging down my ATAR So yeah I'm doing some of the higher scaling subjects like 4U maths and chemistry and I'm doing great with those, but I can't simply deal with english. I'm doing English standard and my rank is maybe just in the top 15 after doing pretty bad in my half yearlies. I currently attend a school that...

{"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}

CC-MAIN-2019-43

latest

en

https://rightbankwarsaw.com/players/which-is-bigger-a-football-or-rugby-pitch.html

text/html

crawl-data/CC-MAIN-2021-31/segments/1627046154277.15/warc/CC-MAIN-20210801221329-20210802011329-00519.warc.gz

# Which is bigger a football or rugby pitch? Contents Rugby pitches (often called fields) are bigger than American football fields. … A rugby pitch is 112-122 meters (122.5-133.4 yards) long and 68m (74.3 yards wide). Attached to the ends of this pitch are a pair of in-goal areas, the equivalent of an end zone, which can be from 5-22 meters (5.4-24.1 yards) deep. ## What’s bigger a rugby or football pitch? A football (soccer) pitch is 90-120m along the touchline (side line) and 45-90m wide with a total playing area of 4,050-10,800sq m. A rugby pitch is 106-144m long (including the in-goal area) by 68-70m wide with a playing area of 7,208-10,080sq m. ## How big is a rugby pitch? The maximum playing area is 120m long and 60m wide (full field minus 10m width). The field of play is a maximum of 100m long. Each in‑goal is a maximum of 10m long. There is a 5m run‑off from playing area to a roped boundary. ## Which is tougher football or rugby? Football is for sure way more violent. Both games require different things from their athletes. A rugby player would probably need to bulk up to play football, while the football player would need to trim his weight down to play rugby. … Rugby definitely comes out as the tougher sport to play. ## How big is the football pitch? Standard pitch measurements. Not all pitches are the same size, though the preferred size for many professional teams’ stadiums is 105 by 68 metres (115 yd × 74 yd) with an area of 7,140 square metres (76,900 sq ft; 1.76 acres; 0.714 ha). ## How many football fields make a mile? How many football field [U.S.] in 1 miles? The answer is 17.6. ## How long is a football field in yards? When the “football field” is used as unit of measurement, it is usually understood to mean 100 yards (91.44 m), although technically the full length of the official field, including the end zones, is 120 yards (109.7 m). New Zealand have been the most consistently ranked #1 team since the introduction of IRB World Rankings, having held the #1 ranking for more than 85 percent of the time during this period. South Africa, England, Wales and Ireland make up the remainder. ## What is the 22 in rugby? BBC SPORT | Rugby Union | Laws & Equipment | The 22-metre drop-out. This is one of the methods used to restart play when the ball has gone over a team’s dead ball line. For example, if the attacking team kicks the ball beyond the dead ball line, a member of the defending team can touch it down for a 22-metre drop-out. ## What is not allowed in rugby? Players can only tackle by wrapping their arms around their opponents to bring them to the ground and players are not allowed to tackle opponents above the shoulder,or to use their legs to tackle or trip them. IT IS INTERESTING:  Who has the most Golden Gloves in football? ## What sport has most deaths? Here are the 5 most deadly sports in the world. 1. Base Jumping. Deaths per 100,000 population: 43.17. Odds of dying: 1 in 2,317. … 2. Swimming. Deaths per 100,000 population: 1.77. … 3. Cycling. Deaths per 100,000 population: 1.08. … 4. Running. Deaths per 100,000 population: 1.03. … 5. Skydiving. Deaths per 100,000 population: 0.99. ## What is the toughest sport in the world? Degree of Difficulty: Sport Rankings SPORT END RANK Boxing 8.63 1 Ice Hockey 7.25 2 Football 5.38 3 ## Do football players hit harder than rugby players? Both camps are right. Football players do indeed hit quite a lot harder than rugby players, who have to be more careful not to injure themselves or the other player, but the net result of the collisions are probably about the same for each sport. ## Who has the biggest football pitch? Answer has 6 votes. Manchester City (Maine Road) has the largest pitch in the league at 107 metres by 73 metres (approximately). HalifaxTown (The Shay) and Swansea City (Vetch Field) have the smallest pitches in the league at 100 metres by 64 metres. ## How big is an 11 aside football pitch? 11-a-side football pitch size 11-a-side pitch measurements can vary considerably, but according to FA regulations, should fall between 100-130 yards (90-120m) long and 50-100 yards (45-90m) wide. ## What is the biggest pitch in the Premier League? At 116 x 77 yards, the Etihad remains the largest pitch English football’s top-flight has seen since the Premier League began in 1992. Old Trafford runs it extremely close at 116 x 76 yards, as does the new Wembley (used for Tottenham Hotspur during the 2017-18 season and most of 2018-19). IT IS INTERESTING:  Your question: How do you motivate a soccer player?

{"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}

CC-MAIN-2021-31

latest

en

https://brainly.in/question/104577

text/html

crawl-data/CC-MAIN-2017-04/segments/1484560279657.18/warc/CC-MAIN-20170116095119-00368-ip-10-171-10-70.ec2.internal.warc.gz

# The digit in the unit’s place of 7171 + (177)! is 1 by akashgowdap16 2015-05-02T22:26:18+05:30 ### This Is a Certified Answer Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest. Any number  which is  n !    for  n >= 5  will have  10 as one of its factors... 5!  = 5 * 4 * 3 * 2 * 1 = 4 * 3 * 5*2 =  4* 3 * 10 so 177 !    will have  0 at the units place. 7171 + 177 !    will have   1 at the units place. ======================= 10!  has  two zeros at the end. 15!  has to have 3 zeros at the end. 20!  has to have 4 zeros  at the end. 25!  has to have 6 zeros at the end, as  25 = 5 * 5,  there are two 5s. 30 !  has 7 zeros. 35! 40!  has  9 zeros. 50!  has  12 zeros,  as  50 = 5 * 5 * 10

{"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}

CC-MAIN-2017-04

latest

en

https://wearethearchetype.com/qa/what-3-coins-make-80-cents.html

text/html

crawl-data/CC-MAIN-2020-50/segments/1606141188800.15/warc/CC-MAIN-20201126142720-20201126172720-00673.warc.gz

# What 3 Coins Make 80 Cents? ## How much is 99 cents? 99 cents (99/100 of one cent) to the base unit price for most merchandise. As a result, currently, the most common price point for items in our stores is 99.99 cents. In almost all instances, this price will round up to one dollar at the register and that is the amount a customer will be charged.. ## How many cents are in 3 dimes? 30 centsCounting MoneyAB3 dimes =30 cents1 dime =10 cents1 quarter =25 centshalf dollar =50 cents36 more rows ## What coins make 99 cents? The answer is 10 coins, 3 Quarters, 1 dime, 2 nickel, and 4 pennies. With this combination you can produce any number between 1-99 cents. An alternative answer would be 6 coins, 3 Quarters, 2 dimes and 1 nickel. ## Is a penny 1 cent? The penny is the United States’ one-cent coin. Every penny you’ve ever spent probably had Abraham Lincoln on it. He’s been on the front (obverse) of the penny since 1909! ## What are 5 cents called? nickelThe nickel is the United States’ five-cent coin. We know the five-cent coin as a nickel, but it wasn’t always so. ## Which set of coins is 3 100 of a dollar? The United States three cent piece was a unit of currency equaling ​3⁄100 of a United States dollar. ## What are 2 cent coins worth? Two Cent Coin Values Are RisingDATEGOODUNCIRCULATED1866 2-Cent\$19\$801867 2-Cent\$20\$801868 2-Cent\$20\$1101869 2-Cent\$25\$1257 more rows ## How much money should you keep in your wallet? However \$50 is not a reasonable amount to have with you in case of emergency, let alone \$10. On the other hand, \$500 is quite a lot to lose if your wallet gets stolen or lost. That’s how experts came to the conclusion that you should always have \$200 in your wallet. ## Do you keep coins in your wallet? Coins take up a lot of space and add a lot of weight to your wallet. Take them out and put them in a jar to save up for a rainy day. If you actually use change and insist on carrying it, then use a wallet that has a small zippered pouch for coins. ## What 3 coins make 65 cents? Half-dollar, dime, and a nickel. ## What is a 3 cent coin worth? Three Cent Nickel Values Are Always ChangingDATEGOODUNCIRCULATED1869 3-Cent Nickel\$15\$651870 3-Cent Nickel\$20\$651871 3-Cent Nickel\$20\$651872 3-Cent Nickel\$20\$6520 more rows ## How do you make 95 cents? Expert Answers info Use a fifty cent piece, one quarter, three nickels, and five pennies. That gives you 10 coins, and a total of ninety five cents. ## How do you make 25 cents? There are 13 ways to create 25 cents from quarters, dimes, nickels, and pennies. nickels, and pennies. It will take eight hours for the faucet to leak a gallon. ## How many different ways can you make 21 cents with coins? nine waysnine ways : 2 dimes , 0 nickels , 1 penny . 1 dime , 2 nickels , 1 penny . 1 dime , 1 nickel , six pennies . ## How can you make change for \$1 using exactly 50 coins? Therefore one way of choosing to make a 1-dollar change is 45-Penny,1-Quarter,2-Dime,2-Nickle coins. we need to choose 10 more coins to make the total 50 coins. Therefore we need to get the rest amount 1-0.4=0.6 dollars from the remaining 6 coins. ## What 21 coins make a dollar? 2 QUARTERS, 3 DIMES,1 NICKEL,15 PENNIES ANOTHER WAY……. 7 DIMES, 4 NICKELS,10 P. ## How many dollars is 5 cents? The nickel is a US coin worth five cents. Twenty nickels make a dollar. ## How many different combinations of pennies nickels dimes quarters and half dollars can a piggy bank contain if it has 21 coins in it? Neither of the top two answers are correct. The correct answer is C( 24 choose 20)= 10626 different combinations. ## How many \$20 bills does it take to make \$500? 25 x \$20 bills = \$500. ## What year is a penny worth 1 million dollars? 1943A 1943 Lincoln penny that soared in value because it was made from the “wrong” material reportedly has sold for \$1 million. ## How do I make 49 cents? Here you will find the answer to the question “What is least amount of coins to make 49 cents?” To make 49 cents, you use one quarter, two dimes, and four pennies. Total amount is seven coins.

{"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}

CC-MAIN-2020-50

latest

en

https://math.stackexchange.com/questions/3512721/a-conjecture-in-number-theory-with-twin-primes

text/html

crawl-data/CC-MAIN-2023-50/segments/1700679099892.46/warc/CC-MAIN-20231128151412-20231128181412-00612.warc.gz

# A conjecture in number theory with twin primes It's a conjecture found with the help of Wolfram Alpha : Let $$p_i$$ be the first primes and $$n> 5$$ with $$n$$ an odd natural number numbers we are interested by the quantity : $$A=(1+p_1\times p_2(1+p_3\times p_4(1+p_5\times p_6(\cdots(1+p_n\times p_{n+1})^{\frac{1}{2}})\cdots)$$ Or $$A=(1+2\times 3(1+5\times 7(1+11\times 13(\cdots(1+p_n\times p_{n+1})^{\frac{1}{2}})\cdots)$$ Where $$p_n$$ and $$p_{n+1}$$ are twin primes numbers ## Example $$(1+2\times3(1+5\times7(1+11\times13(1+17\times19(1+23\times29(1+31\times37(1+41\times43)^{\frac{1}{2}}))))))=311677481085187=7×43×433×2391393439$$ ## Conjecture 1 The last digit of $$A$$ is seven . ## Conjecture 2 If $$A$$ is not a prime number then $$A$$ is divisible by $$7$$. I try to work with divisibility rule for small numbers but it becomes insane with big numbers . I'm a very beginners in number theory so if you could use elementary tools it will be cool . Thanks a lot for your time and patience . • How far have you tested this? Your formula appears to assume that $n$ is odd...are you assuming that as a requirement? If not, what does your formula mean when $n$ is even? For instance, $p_{18}=71$ is the least of a twin prime pair. What is $A$ in this case? – lulu Jan 17, 2020 at 18:20 • Typo: meant to write $p_{20}=71$. Same question, though. – lulu Jan 17, 2020 at 18:27 • Ok let me try it . Jan 17, 2020 at 18:29 • Well, my point was the parity. Since you multiply your primes in pairs, it looks like you need $n$ to be odd. I don't understand what your formula means if $n$ is even. – lulu Jan 17, 2020 at 18:30 • I asked you before, how far have you checked this? – lulu Jan 17, 2020 at 18:37 We know that $$p_3 \times p_4 \times (\text{stuff})$$ is a multiple of $$5$$, so it ends in a $$0$$ or a $$5$$. We then get that $$1 + p_3 \times p_4 \times (\text{stuff})$$ ends in a $$1$$ or a $$6$$, and so $$p_1 \times p_2 (1 + p_3 \times p_4 \times (\text{stuff})) = 6 \times (\text{something ending in } 1 \text{ or } 6)$$ which means that it ends in a $$6$$, or a $$6$$. Add $$1$$ to that and you get something that ends in a $$7$$. Here we have that $$p_3 \times p_4 \times (\text{stuff})$$ is a multiple of $$7$$, so $$1 + p_3 \times p_4 \times (\text{stuff})$$ is $$1$$ more than a multiple of $$7$$. It follows that $$p_1 \times p_2 (1 + p_3 \times p_4 \times (\text{stuff})) = 6 \times (\text{something } \equiv 1 \pmod 7)$$ is $$6$$ more than a multiple of $$7$$. Add $$1$$ to that and you get a multiple of $$7$$.

{"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": 34, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}

CC-MAIN-2023-50

latest

en

https://www.numbersaplenty.com/145160662

text/html

crawl-data/CC-MAIN-2023-40/segments/1695233506329.15/warc/CC-MAIN-20230922034112-20230922064112-00197.warc.gz

Search a number 145160662 = 2312341301 BaseRepresentation bin10001010011011… …11100111010110 3101010010220222211 420221233213112 5244130120122 622223144034 73411562633 oct1051574726 9333126884 10145160662 1174a37399 124074501a 13240c62aa 14153c918a 15cb25877 hex8a6f9d6 145160662 has 8 divisors (see below), whose sum is σ = 224764992. Its totient is φ = 70239000. The previous prime is 145160629. The next prime is 145160669. The reversal of 145160662 is 266061541. 145160662 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times. It is a sphenic number, since it is the product of 3 distinct primes. It is a Harshad number since it is a multiple of its sum of digits (31). It is a congruent number. It is not an unprimeable number, because it can be changed into a prime (145160669) by changing a digit. It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1170589 + ... + 1170712. It is an arithmetic number, because the mean of its divisors is an integer number (28095624). Almost surely, 2145160662 is an apocalyptic number. 145160662 is a deficient number, since it is larger than the sum of its proper divisors (79604330). 145160662 is a wasteful number, since it uses less digits than its factorization. 145160662 is an evil number, because the sum of its binary digits is even. The sum of its prime factors is 2341334. The product of its (nonzero) digits is 8640, while the sum is 31. The square root of 145160662 is about 12048.2638583325. The cubic root of 145160662 is about 525.5527505564. The spelling of 145160662 in words is "one hundred forty-five million, one hundred sixty thousand, six hundred sixty-two".

{"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}

CC-MAIN-2023-40

latest

en

https://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_1&oldid=96299

text/html

crawl-data/CC-MAIN-2020-50/segments/1606141708017.73/warc/CC-MAIN-20201202113815-20201202143815-00318.warc.gz

# 1999 AMC 8 Problems/Problem 1 ## Problem $(6?3) + 4 - (2 - 1) = 5.$ To make this statement true, the question mark between the 6 and the 3 should be replaced by $\text{(A)} \div \qquad \text{(B)}\ \times \qquad \text{(C)} + \qquad \text{(D)}\ - \qquad \text{(E)}\ \text{None of these}$ ## Solution Simplifying the given expression, we get: $(6?3)=2.$ At this point, it becomes clear that it should be $\div (A)$.

{"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": 4, "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}

CC-MAIN-2020-50

latest

en

http://hoven.in/Home/Aptitude/Simple-Interest/maths-aptitude-quiz-questions-and-mock-test-on-simple-interest-004.html

text/html

crawl-data/CC-MAIN-2017-22/segments/1495463607862.71/warc/CC-MAIN-20170524211702-20170524231702-00299.warc.gz

# Simple Interest Quiz Set 004 ### Question 1 A sum of Rs. 270 is divided into two parts such that simple interest on these parts at 10% p.a. after 1 and 2 years, respectively, is same. What is the amount of the smaller part? A Rs. 90. B Rs. 190. C Rs. 140. D Rs. 290. Soln. Ans: a We should use the shortcut technique here. If r1, t1, r2, t2 be the rates and times for two parts with same interest amount, then the two parts must be in the ratio \$1/{r_1 t_1} : 1/{r_2 t_2}\$. In our case r1 = r2 = 10, which cancels, so the ratio is \$1/t_1 : 1/t_2\$. Thus, the two parts are in the ratio \$t_2 : t_1\$. The parts are: 270 × \$2/{1 + 2}\$, and 270 × \$1/{1 + 2}\$, which are 180 and 90. The smaller is Rs. 90. ### Question 2 A certain amount is split into two parts. The first part is invested at 10% p.a. and the second at 3% p.a. What is the total amount if the total simple interest at the end of 4 years is Rs. 220, and if the amount invested at 10% is Rs. 700? A Rs. 200. B Rs. 300. C Rs. 250. D Rs. 400. Soln. Ans: a Let the amount x be invested at r1% and remaining (P - x) at r2%, and let the sum of interests be I. Then, I = \${x × r_1 × 4}/100\$ + \${(P - x) × r_2 × 4}/100\$, which simplifies to 100I = 4 × (\$x × (r_1 - r_2) + P × r_2\$). Putting r1 = 10, r2 = 3, x = 700, I = 220, we get P = Rs. 200. ### Question 3 A sum of Rs. 57970 is divided into three parts such that simple interest on these parts at 10% p.a. after 13, 17 and 4 years, respectively, is same. What is the amount of the smallest part? A Rs. 8840. B Rs. 8940. C Rs. 8740. D Rs. 9040. Soln. Ans: a We should use the shortcut technique here. If r1, t1, r2, t2 and r3, t3 be the rates and times for three parts with same interest amount, then the three parts must be in the ratio \$1/{r_1 t_1} : 1/{r_2 t_2} : 1/{r_3 t_3}\$. In our case r1 = r2 = r3 = 10, which cancels, so the ratio is \$1/t_1 : 1/t_2 : 1/t_3\$. The product of denominators is 13 × 17 × 4 = 884. Thus, the three parts are in the ratio \$68 : 52 : 221\$. The parts are: 57970 × \$221/{68 + 52 + 221}\$, 57970 × \$52/{68 + 52 + 221}\$ and 57970 × \$68/{68 + 52 + 221}\$, which are 11560, 8840 and 37570. The smaller is Rs. 8840. ### Question 4 What is the interest on Rs. 12500 @6% for 73 days starting from Jan 1, 2601? A Rs. 150. B Rs. 250. C Rs. 200. D Rs. 350. Soln. Ans: a The given year is not a leap year. So it has 365 days. Since 73 X 5 = 365, t = 1/5 year. We have r = 6%, t = 1/5, P = 12500, so I = \${12500 × 6 × 1}/{5 × 100}\$ = Rs. 150. ### Question 5 The interest on a certain principal sum is 4/9 times the sum. What is R, the rate of interest, if the time is R years? A \$6{2/3}\$%. B 7%. C \$7{1/3}\$%. D \$6{1/3}\$%. Soln. Ans: a I = P × (4/9), so we can write P × (4/9) = P × (R/100) × R. Cancelling P and solving for R, we get, R = \$√{100 × 4/9}\$ = \$6{2/3}\$%.

{"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}

CC-MAIN-2017-22

latest

en

http://deeplearning.stanford.edu/wiki/index.php?title=Fine-tuning_Stacked_AEs&diff=2303&oldid=729

text/html

crawl-data/CC-MAIN-2020-50/segments/1606141187753.32/warc/CC-MAIN-20201126084625-20201126114625-00487.warc.gz

# Fine-tuning Stacked AEs Revision as of 00:25, 12 May 2011 (view source)Jngiam (Talk | contribs) (→Recap of the Backpropagation Algorithm)← Older edit Latest revision as of 04:04, 8 April 2013 (view source)Kandeng (Talk | contribs) Line 5: Line 5: Fortunately, we already have all the tools necessary to implement fine tuning for stacked autoencoders! In order to compute the gradients for all the layers of the stacked autoencoder in each iteration, we use the [[Backpropagation Algorithm]], as discussed in the sparse autoencoder section. As the backpropagation algorithm can be extended to apply for an arbitrary number of layers, we can actually use this algorithm on a stacked autoencoder of arbitrary depth. Fortunately, we already have all the tools necessary to implement fine tuning for stacked autoencoders! In order to compute the gradients for all the layers of the stacked autoencoder in each iteration, we use the [[Backpropagation Algorithm]], as discussed in the sparse autoencoder section. As the backpropagation algorithm can be extended to apply for an arbitrary number of layers, we can actually use this algorithm on a stacked autoencoder of arbitrary depth. - === Finetuning for Classification Error === + === Finetuning with Backpropagation === For your convenience, the summary of the backpropagation algorithm using element wise notation is below: For your convenience, the summary of the backpropagation algorithm using element wise notation is below: Line 34: Line 34: {{Quote| {{Quote| - Note: While one could consider the softmax classifier as an additional layer, the derivation above does not. Specifically, we consider the "last layer" of the network to be the features that goes into the softmax classifier. Therefore, the derivatives (in Step 2) are computed using $\delta^{(n_l)} = - (\nabla_{a^{n_l}}J) \bullet f'(z^{(n_l)})$. + Note: While one could consider the softmax classifier as an additional layer, the derivation above does not. Specifically, we consider the "last layer" of the network to be the features that goes into the softmax classifier. Therefore, the derivatives (in Step 2) are computed using $\delta^{(n_l)} = - (\nabla_{a^{n_l}}J) \bullet f'(z^{(n_l)})$, where  $\nabla J = \theta^T(I-P)$. }} }} + + + {{CNN}} + + + {{Languages|微调多层自编码算法|中文}}

{"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}

CC-MAIN-2020-50

latest

en

http://www.mathisfunforum.com/viewtopic.php?pid=313516

text/html

crawl-data/CC-MAIN-2018-51/segments/1544376824912.16/warc/CC-MAIN-20181213145807-20181213171307-00546.warc.gz

Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ ° You are not logged in. ## #1 2014-05-15 01:27:27 mittalanirudh.17 Member Registered: 2014-05-15 Posts: 4 ### Need Urgent Help!! Can anyone explain me how to solve the following problem Factorise -  x^12 - y^12 Would be really obliged Offline ## #2 2014-05-15 01:29:00 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: Need Urgent Help!! Hi; Is that a minus in front of the x^12? In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #3 2014-05-15 01:33:55 ShivamS Member Registered: 2011-02-07 Posts: 3,648 ### Re: Need Urgent Help!! If you mean to factor -x^12-y^12 , it is If you mean to factor x^12-y^12, it is Last edited by ShivamS (2014-05-15 01:35:22) Offline ## #4 2014-05-15 01:36:06 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: Need Urgent Help!! That is what I mean. The first one maybe can be done by pen and paper but the second one needs a CAS. In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #5 2014-05-15 01:37:58 ShivamS Member Registered: 2011-02-07 Posts: 3,648 ### Re: Need Urgent Help!! I don't want to do even the first one without a CAS. Offline ## #6 2014-05-15 01:40:26 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: Need Urgent Help!! I agree. That is no picnic. In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #7 2014-05-15 01:43:12 Agnishom Real Member From: Riemann Sphere Registered: 2011-01-29 Posts: 24,858 Website ### Re: Need Urgent Help!! Here is how: x^12 - y^12 = (x^6)^2 - (y^6)^2 = (x^6 + y^6)(x^6 - y^6) = ((x^2)^3 + (y^2)^3)((x^3)^2 - (y^3)^2) = ((x^2)^3 + (y^2)^3)(x^3 + y^3) (x^3 - y^3) = (x^2 + y^2) (x^4 - x^2 y^2 + y^4)(x + y) (x^2 - x y + y^2)(x - y) (x^2 + x y + y^2) We are using a few  identities here: Last edited by Agnishom (2014-05-15 01:56:50) 'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.' 'God exists because Mathematics is consistent, and the devil exists because we cannot prove it' I'm not crazy, my mother had me tested. Offline ## #8 2014-05-15 01:44:18 ShivamS Member Registered: 2011-02-07 Posts: 3,648 ### Re: Need Urgent Help!! You can use sum of cubes formula for the first one. Agnishom wrote: @bobbym, it is not. Here is how: x^12 - y^12 = (x^6)^2 - (y^6)^2 = (x^6 + y^6)(x^6 - y^6) = ((x^2)^3 + (y^2)^3)((x^3)^2 - (y^3)^2) = ((x^2)^3 + (y^2)^3)(x^3 + y^3) (x^3 - y^3) = (x^2 + y^2) (x^4 - x^2 y^2 + y^4)(x + y) (x^2 - x y + y^2)(x - y) (x^2 + x y + y^2) But a CAS did that for you. Of course it is possible to do by hand but will take a long time. Last edited by ShivamS (2014-05-15 01:45:03) Offline ## #9 2014-05-15 01:45:23 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: Need Urgent Help!! That is algebraically correct! In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #10 2014-05-15 01:46:42 Agnishom Real Member From: Riemann Sphere Registered: 2011-01-29 Posts: 24,858 Website ### Re: Need Urgent Help!! Please do not confuse the newbie. I agree that it is tedious but this kind of questions are asked in Middle School and you are considered a dummy if you do not answer them. 'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.' 'God exists because Mathematics is consistent, and the devil exists because we cannot prove it' I'm not crazy, my mother had me tested. Offline ## #11 2014-05-15 01:48:13 ShivamS Member Registered: 2011-02-07 Posts: 3,648 ### Re: Need Urgent Help!! Agnishom wrote: Please do not confuse the newbie. I agree that it is tedious but this kind of questions are asked in Middle School and you are considered a dummy if you do not answer them. Please tell me which Middle School. Offline ## #12 2014-05-15 01:50:05 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: Need Urgent Help!! in Middle School and you are considered a dummy if you do not answer them In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #13 2014-05-15 01:52:53 Agnishom Real Member From: Riemann Sphere Registered: 2011-01-29 Posts: 24,858 Website ### Re: Need Urgent Help!! In the seventh and eighth grade, they can ask you to do this by hand. But a CAS did that for you. Of course it is possible to do by hand but will take a long time. I really did it by hand so that the OP gets how to do it. 'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.' 'God exists because Mathematics is consistent, and the devil exists because we cannot prove it' I'm not crazy, my mother had me tested. Offline ## #14 2014-05-15 01:54:26 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: Need Urgent Help!! Okay, that is a very nice solution then. I did not see that. In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #15 2014-05-15 01:55:02 ShivamS Member Registered: 2011-02-07 Posts: 3,648 ### Re: Need Urgent Help!! Well, not here in the US or Canada. But even in India, there is little thinking involved in doing this question by hand. It is just a rote computation problem, which is a waste of one's time. I don't know how you did it in one minute Last edited by ShivamS (2014-05-15 01:55:51) Offline ## #16 2014-05-15 01:59:18 Agnishom Real Member From: Riemann Sphere Registered: 2011-01-29 Posts: 24,858 Website ### Re: Need Urgent Help!! Thank you, bobbym. Well, in the begining, in order to see if you can actually recognise those patterns or not, these problems are given to you. 'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.' 'God exists because Mathematics is consistent, and the devil exists because we cannot prove it' I'm not crazy, my mother had me tested. Offline ## #17 2014-05-15 02:01:38 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: Need Urgent Help!! Interesting to see what different places put emphasis on. In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #18 2014-05-15 02:04:49 Agnishom Real Member From: Riemann Sphere Registered: 2011-01-29 Posts: 24,858 Website ### Re: Need Urgent Help!! But being able to recognise where a^3 - b^3 and a^2 - b^2 occurs is important. 'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.' 'God exists because Mathematics is consistent, and the devil exists because we cannot prove it' I'm not crazy, my mother had me tested. Offline ## #19 2014-05-15 02:07:22 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: Need Urgent Help!! It gets used in problems. In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #20 2014-05-15 02:08:42 Agnishom Real Member From: Riemann Sphere Registered: 2011-01-29 Posts: 24,858 Website ### Re: Need Urgent Help!! That is why it is important. Is it not? 'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.' 'God exists because Mathematics is consistent, and the devil exists because we cannot prove it' I'm not crazy, my mother had me tested. Offline ## #21 2014-05-15 02:10:45 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: Need Urgent Help!! It is good that you are so proficient with this type of problem. But as you know I would have turned to the heavy machinery after about a minute... In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #22 2014-05-15 03:53:33 mittalanirudh.17 Member Registered: 2014-05-15 Posts: 4 ### Re: Need Urgent Help!! Thank you very much, btw all of you are Indians Offline ## #23 2014-05-15 05:24:23 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: Need Urgent Help!! 2 out of 3 ain't bad. In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #24 2014-05-15 05:47:46 Agnishom Real Member From: Riemann Sphere Registered: 2011-01-29 Posts: 24,858 Website ### Re: Need Urgent Help!! I think 3 out of 4 'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.' 'God exists because Mathematics is consistent, and the devil exists because we cannot prove it' I'm not crazy, my mother had me tested. Offline ## #25 2014-05-15 05:49:28 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: Need Urgent Help!! Counting him, yes. In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline

{"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}

CC-MAIN-2018-51

latest

en

http://www.deltalab-smt.com/teaching-mechanical-engineering/mechanics-and-technology/mechanical-engineering/hydraulic-gate-opener-ex700

text/html

crawl-data/CC-MAIN-2017-51/segments/1512948512208.1/warc/CC-MAIN-20171211052406-20171211072406-00223.warc.gz

Sections You are here: Home / HYDRAULIC GATE OPENER EX700 HYDRAULIC GATE OPENER EX700 This is a commonly used gate opener, adapted to investigate studies in mechanical and energetic engineering. The system permits different but complementary studies: experimentation, modelling and computer aided design with special CAD software in 2D (AUTOCAD®) and 3D (SolidWorks®). PRINCIPAL TEACHING OBJECTIVES The equipment covers the following areas of study : • Energetic • Fluid mechanics • Functional analysis of an industrial product • Technical communication tools • Relationship between product - manufacturing process - material • Measurement of physical values and calculation of performance These calculations are entered into a spreadsheet file (model supplied) to allow interpretation of the test data and results. Using the measurements and experimental data, calculations will among other give : • The theoretical capacity of the pump (using CAD software) • The determination of the following parameters, based on the measurements obtained with the equipment: - Translation speed of jack rod (time, distance) - Theoretical flow rate of the pump (capacity, rotation speed) - Oil flow rate (piston diameter, rod diameter, rod speed) - Released mechanical power (mass lifted, rod speed) - Resultant pressure forces on the piston (piston diameter, rod diameter, pressure) - Jack efficiency and global efficiency ( if functioning in load) - Flow measurement comparison: pump - jack and jack - pump. • Analysis of the functions provided by the different valves and safety pressure valve used in the hydraulic circuit • Highlighting of the change in efficiency and friction as a function of the applied load and temperature increase • Analysis of the design methods associated to prevent leaks and to facilitate assembly… TECHNICAL SPECIFICATIONS The electro-hydraulic gate opener is mounted on a plastic PVC base together with the related instrumentation. It includes : - An electrical motor - An oil gear pump - A lift device for the mass ( the masses can be hanged in a driven position or a braking position) - A complete instrumentation allowing the measurement of: • Motor speed using an induction sensor and signal conditioner / indicator display • Power consumed using an optional Clamp-type ammeter • Oil pressure in the two supply circuits of the cylinder using 2 pressure gauges • Time taken for the shaft to reach a predetermined position using an optional stopwatch Certain essential specifications for the equipment are provided, including : • Pump dimensions, measured on a tri-axis measurement machine and profile projector, supplied on CAD files • Performance specifications of the hydraulic jack • Intensity of the lifted mass (system can also be used empty) EQUIPMENT PACKAGE Standard equipment EX700: Table top test bench (the table is not supplied) The masses are not included as standard Essential requirements : • Voltage : 220V, 50 Hz, other voltages available on demand • Clamp-type ammeter, • Stopwatch, • Set of masses : 4 x 1 5 kg • 100 MHz PC with Windows*, Excel*, AutoCAD** and SolidWorks*** * Windows, Excel are a registered trademark of Microsoft. ** AutoCAD is a registered trademark of Autodesk Inc. *** SolidWorks is a registered trademark of SolidWorks Corporation. Dimensions and weight - 1200 x 300 x 200 mm - Weight : 20 kg

{"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}

CC-MAIN-2017-51

latest

en

https://hub.jmonkeyengine.org/t/vehicle-spinning-around/12286

text/html

crawl-data/CC-MAIN-2023-23/segments/1685224649193.79/warc/CC-MAIN-20230603101032-20230603131032-00741.warc.gz

# Vehicle spinning around Hey, I just startet playing around with jbullet-jme and it looks really cool, but i got a Problem with the PhysicsVehicleNode. So this is how i try to use it: ```physicsCar=new PhysicsVehicleNode(carBody,CollisionShape.ShapeTypes.BOX);         physicsCar.setFrictionSlip(0.8f);         physicsCar.setMaxSuspensionTravelCm(20);         physicsCar.setMass(250);                  float stiffness=190;//200=f1 car         float compValue=0.2f; //(lower than damp!)         float dampValue=0.5f;         physicsCar.setSuspensionCompression(compValue*2.0f*FastMath.sqrt(stiffness));         physicsCar.setSuspensionDamping(dampValue*2.0f*FastMath.sqrt(stiffness));         physicsCar.setSuspensionStiffness(stiffness);                                    Vector3f wheelDirection=new Vector3f(0,-1,0);         Vector3f wheelAxle=new Vector3f(-1,0,0);                  // Front left         physicsCar.addWheel(wheel1, new Vector3f(0.7f,0.4f,1.2f), wheelDirection, wheelAxle, 0.13f, 0.34f, true);         physicsCar.setRollInfluence(0, 1);         // Front right         physicsCar.addWheel(wheel2, new Vector3f(-0.7f,0.4f,1.2f), wheelDirection, wheelAxle, 0.13f, 0.34f, true);         physicsCar.setRollInfluence(1, 1);         physicsCar.addWheel(wheel3, new Vector3f(0.7f,0.4f,-1.2f), wheelDirection, wheelAxle, 0.13f, 0.34f, false);         physicsCar.setRollInfluence(2, 1);         physicsCar.addWheel(wheel4, new Vector3f(-0.7f,0.4f,-1.2f), wheelDirection, wheelAxle, 0.13f, 0.34f, false);         physicsCar.setRollInfluence(3, 1);                  physicsCar.setLocalScale(3.5f);       physicsCar.setLocalTranslation(new Vector3f(-20,2,0));       rootNode.attachChild(physicsCar);         physicsCar.updateRenderState();                     pSpace.add(physicsCar);``` The Node carBody, as the name already says, only contains the Car-Body and the wheel-Nodes are all clones from one wheel. First it starts falling until it hits the ground. This works now, after it took me some time to get the (hopefully) right values for the suspension-stuff. But as soon as i try to accelerate the vehicle just starts spinning around like crazy and falls through the ground. Do you have any Idea what i'm doing wrong? Dennis normen said: Simple solution is dont change the mass after creation, set it in the constructor already. Setting the mass after creation is bound to produce problems atm. I'm not changing the mass after creation it's done in the constructor. BoxCollisionShape boxcs = new BoxCollisionShape((Node) vehicle); vehiclePhysics = new PhysicsVehicleNode(vehicle, boxcs, 600); What is the acceleration value? Maybe it is set too high? Also, when using a big box (few vertices) as floor the vehicle might get problems, try if it works better on a terrain or something with more vertexes. Getting bullet vehicles to work smoothly can be a bit hard I have to admit. Hi, i just tried to change the acceleration-value but i didn't change. And i already use some kind of Terrain. It's a model i made with Cheetah3d and the ground has 1024 Polygons, so i think that should be enough. And with the Vehicle from the Test everything works fine on the Terrain. I think the problem lies somewhere else. I made small movie, to show you whats happening: http://bazinga.ld-network.de/vehicleProblem.mov There you can see what happens as soon as i try to accelerate. It still looks as if the acceleration was way too high… Try reducing the weight of the car and/or make the suspensionRestLength longer. Strange things are also more likely to happen when the stiffness is set too high. I played around with the values, but it still didn't change. Currently i think that it's gotta be something with my model, because i tried putting the Vehicle into your Test. The Tires worked, but as soon as i add the Car-model instead of the Box it starts going crazy. And if i replace the carmodel in my example with a box it works just fine. So i guess there is something wrong with the model. Are there any requirements at the Car-Model? If somebody wants to check out the Model, I got it from here: PSIONIC GAMES & RETRO VANDAL ART The only thing i changed is, that i removed the tires. Ah, might be that when you create the PhysicsNode, the BoundingVolume of the model is not computed right. Try creating the BoxCollisionShape before and use that to create the PhysicsNode. And dont use setMass() on the PhysicsNode, set the mass in the constructor, there is some bullet problem with setting the mass after constructing the RigidBody. Thanks for your help, but it still doesnt work… This is how I create the Vehicle currently: ```BoxCollisionShape boxCollisionShape=new BoxCollisionShape((BoundingBox)body.getWorldBound());         physicsCar=new PhysicsVehicleNode(body,boxCollisionShape,250);``` The BoundingBox looks also right around the Car, but it still doesn't really work. Any other Idea? Maybe if you got some time you can look over it. I integrated it into your TestSimplePhysicsCar example. Dennis As said, the error is probably within the bounds of the car, so dont use it, create your BoxCollisionShape like this: ``` BoxCollisionShape boxCollisionShape=new BoxCollisionShape(new Vector3f(x,y,z)); ``` Hi, i'm sorry to bother you again, but it still won't work. I did like you said and created my own BoxCollisionShape with the x/y/z-Values: `BoxCollisionShape boxCollisionShape=new BoxCollisionShape(new Vector3f(0.23f,.14f,.5f));` In the javadoc it says, that this should be the halfExtent, so i figured it should be half of the size. Is this right? Anyways, i tried both ways, but with both basically the same thing happened. The Vehicle was first falling down and as soon as it hits the ground it spins around a little (not as bad as before) and then stops in the air. It looks as if the Vehicle is flying above the ground. So i tried changing the values so that it would hit the ground, but as soon as it hits the ground, is goes crazy again. But the x/y/z-Values should be right. The old ModelBox has the same extends and in Cheetah3d i also got the same values, only doubled, so i guess they should be right. Yes, thats right, the halfextents are half the box's size. Dont know right now whats the problem there, maybe I can have a look at your project when I come home from work. One last thing you could try is lowering the weight of the car and see if that solves the problem. Ok, I looked at the thing and there are several problems. First one is when you add the wheels: `physicsCar.addWheel(wheel3, new Vector3f(0.7f,0.4f,-1.2f), wheelDirection, wheelAxle, 0.13f, 0.34f, false);` The vector has a positive y-value which means that the wheel is attached *above* the suspension, not below, set that to a negative value and the car stops jumping around. Second, you should not scale your PhysicsNode after you have created the CollisionShape for it, scale the attached nodes before and then create the CollisionShape from it. Although jbullet-jme will try to adapt the CollisionShape size currently when setLocalScale is used, this feature will be removed because the CollisionShape might be shared with other PhysicsNodes. Hope this helps, Normen Hi, it finally works Thank you! You're doing a great job around here! I think i got about 10 years older while working on it, i just couldn't figure out what i did wrong. As you said i moved the Wheels down and scaled the Nodes up first and then created the CollisionShape and now it works. Dennis Hello I'm trying to do simple car simulation with JMonkey and JBullet-Jme. I'm very new to JMonkey and having many problems, but hopefully everyday I'm getting experienced but there is still a lot of things to learn. I've read almost every posts related to the PhysicsVehicle and followed Normen's suggestion to apply physics to my car. But still getting some strange results. Everytime I change some values (like mass, comp or damp etc …) I get different results like spinning, sinking, sometimes it just doesn't move. I'm really stuck with this. Here is my code. BoxCollisionShape boxcs = new BoxCollisionShape((Node) vehicle); vehiclePhysics = new PhysicsVehicleNode(vehicle, boxcs, 600); vehiclePhysics.setFriction(0.8f); vehiclePhysics.setMaxSuspensionTravelCm(20); // 0.0 - Offroad buggy, 50.0 - Sports car, 200.0 - F1 Car float stiffness = 80.0f;// 200=f1 car float compValue = 0.2f; // (lower than damp!) float dampValue = 0.5f; vehiclePhysics.setSuspensionCompression(compValue * 2.0f * FastMath.sqrt(stiffness)); vehiclePhysics.setSuspensionDamping(dampValue * 2.0f * FastMath.sqrt(stiffness)); vehiclePhysics.setSuspensionStiffness(stiffness); // Should be rotated before adding wheels vehicle.rotateUpTo(new Vector3f(0, 0, -1)); // Getting Tires Node a = (Node) vehicle; Spatial lfw = a.getChild("VPRedT01"); Spatial rfw = a.getChild("VPRedT03"); Spatial lbw = a.getChild("VPRedT00"); Spatial rbw = a.getChild("VPRedT02"); // Scaling tires lfw.setLocalScale(0.1f); rfw.setLocalScale(0.1f); lbw.setLocalScale(0.1f); rbw.setLocalScale(0.1f); Vector3f wheelDirection = new Vector3f(0, -1, 0); Vector3f wheelAxle = new Vector3f(-1, 0, 0); BoundingBox wheelBoundry = (BoundingBox) lfw.getWorldBound(); float wheelRadius = wheelBoundry.yExtent / 2; // It's around 0.3 float suspRestLength = 0.15f; // Left Forward Wheel = lfw vehiclePhysics.setLocalTranslation(new Vector3f(20, 30, 20)); rootNode.attachChild(vehiclePhysics); vehiclePhysics.updateRenderState();

{"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}

CC-MAIN-2023-23

latest

en

http://www.cram.com/flashcards/cyberphysics-waves-revision-297709

text/html

crawl-data/CC-MAIN-2018-13/segments/1521257646176.6/warc/CC-MAIN-20180318204522-20180318224522-00243.warc.gz

• Shuffle Toggle On Toggle Off • Alphabetize Toggle On Toggle Off • Front First Toggle On Toggle Off • Both Sides Toggle On Toggle Off • Read Toggle On Toggle Off Reading... Front ### How to study your flashcards. Right/Left arrow keys: Navigate between flashcards.right arrow keyleft arrow key Up/Down arrow keys: Flip the card between the front and back.down keyup key H key: Show hint (3rd side).h key A key: Read text to speech.a key Play button Play button Progress 1/13 Click to flip ### 13 Cards in this Set • Front • Back What does a wave do? A wave transports energy from one place to another What are mechanical waves? Mechanical waves use particles to transfer energy (neighbouring particles bump into each other and set their neighbours moving). Eg. sound waves use air particles so it cannot travel through a vacuum (a vacuum is empty space with no particles in it at all!). Sound travels faster through solids than liquids than gases because the particles are more closely packed. What are electromagnetic waves? Electromagnetic waves don't use particle vibration to transfer their energy (in fact particles interrupt their progress through a material) and can therefore travel through a vacuum. e.g. light What is the unit of frequency? Frequency is measured in hertz (Hz) What is the wave equation? wavespeed = frequency x wavelength Define a longitudinal wave. The vibrations in a longitudinal wave are parallel to the direction in which the energy is travelling. Eg. sound. Define a transverse wave. The vibrations in a transverse wave are perpendicular (at right angles) to the direction in which the energy is travelling. Eg. light. Define wavelength. Wavelength (lamda a Greek letter 'l') is the shortest distance between two particles that are oscillating in phase. (Distance between two crests on a displacement/distance graph). It is measured in metres (m) What is the period of a wave? Period (T) is the time taken for one complete oscillation of a particle in the wave. (Distance between two crests on a displacement/time graph). It is measured in seconds (s). What is amplitude? Amplitude (A) is the maximum displacement from the mean position. (To the top of a crest or bottom of a trough from the middle line in either graph). What is frequency? Frequency (f) is the number of oscillations per second. It cannot be read directly off a graph. You need to find (T) from the displacement/time graph and then find its reciprocal (f =1/T) What are analogue signals? Analogue signals contain all the information as a continuously varying wave What are digital signals? Digital signals are a series of pulses - either high or low - on or off - sometimes expressed as binary code 1s and 0s.

{"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}

CC-MAIN-2018-13

latest

en

http://mathhelpforum.com/pre-calculus/64946-more-inverse.html

text/html

crawl-data/CC-MAIN-2018-26/segments/1529267860776.63/warc/CC-MAIN-20180618183714-20180618203714-00427.warc.gz

# Thread: More Inverse... 1. ## More Inverse... I'm sorry I have to keep pestering everyone about inverse. I ran into a problem on my homework that is frustrating me. I still kind of don't get it.... If anyone could put it put it step by step it would be greatly appreciated. Find the Inverse: y= 2x^2 + 6x - 4 2. To do this you need to isolate x then switch y for x and vice versa. So let's isolate x: $\displaystyle y=2x^2+6x-4$ $\displaystyle y=2(x^2+3x)-4$ $\displaystyle y=2(x^2+3x+\frac{9}{4}-\frac{9}{4})-4$ (complete the square: 3 divided by 2 squared) $\displaystyle y=2(x+\frac{3}{2})^2-\frac{9}{2}-4$ $\displaystyle y=2(x+\frac{3}{2})^2-\frac{17}{2}$ $\displaystyle \frac{1}{2}(y+\frac{17}{2})=(x+\frac{3}{2})^2$ $\displaystyle x=\sqrt{\frac{1}{2}(y+\frac{17}{2})}-\frac{3}{2}$ So then the inverse of y is $\displaystyle y^{-1}(x)=\sqrt{\frac{1}{2}(x+\frac{17}{2})}-\frac{3}{2}$

{"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}

CC-MAIN-2018-26

latest

en

http://mathhelpforum.com/calculus/75400-derivative-help-print.html

text/html

crawl-data/CC-MAIN-2018-26/segments/1529267860041.64/warc/CC-MAIN-20180618031628-20180618051628-00313.warc.gz

# Derivative HELP • Feb 23rd 2009, 04:22 PM Derivative HELP I have absolutely no idea how to solve these: SHOW THAT 1. \$\displaystyle y=1/x\$ (x doesNOT equal 0) satisfies the equation \$\displaystyle x^3y^(2derivative) + x^2y^(1st derivative)- xy=0\$ 2.y=xe^-x satisfies xy^(1st derivative)= (1-x)y 3. y=xe^-x^2/2 satisfies xy^(1st derivative)= (1-x^2)y • Feb 23rd 2009, 04:38 PM ccross You should calculate the derivatives and then plug them into the given equations to see if they are satisfied. For y=1/x, you have y'=-1/x^2 y''=2/x^3. Just substitute these into the equation. It works.

{"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}

CC-MAIN-2018-26

latest

en

http://talkstats.com/threads/how-to-find-the-probability-of-a-sampling-error-made-in-estimating-the-population-mean.71254/

text/html

crawl-data/CC-MAIN-2019-04/segments/1547583700734.43/warc/CC-MAIN-20190120062400-20190120084400-00057.warc.gz

# How to find the probability of a sampling error made in estimating the population mean? #### annehenderson ##### New Member Alright, I have been working on my statistics assignment since Monday and I only have two questions left, dealing with the probability of a sampling error. Unfortunately, I cannot find anything in my notes and only a brief example in my textbook that does not explain the concepts behind the answer. I'm going to post the question below for clarification but I was hoping for a bit of an explanation so that I can figure out my second question on my own! mean = 6.74 million standard deviations = 15.37 million variable is known to have a right-skewed distribution What is the probability that the sampling error made in estimating the population mean loan amount by the mean loan amount of a simple random sample of 200 loans will be at most 1 million? I figured it has to do something with z-scores but that as far as I got as I do not understand how to do this problem at all! Thank you for the help! #### hlsmith ##### Not a robit Yeah these are tricking given wording. I would if you had a null hypothesis that the sample mean equals the population means if the respective pvalue for having a sample mean the large or larger would translate to what they are calling a sampling error. Yes, traditionally when you are dealing also with the population values you use z-scores. #### annehenderson ##### New Member thank you for your reply! so how would i find the z-score since i am not given an x? the only formula i know for finding z-scores is subtracting x by the mean and dividing that by the standard deviation. my professor hasn't gone over anything about p-values or null hypothesis either so i'm not sure how to proceed with that approach

{"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}

CC-MAIN-2019-04

longest

en

http://www.net-texts.com/Course/MTH6465/Rose%20DeLuccio-%20Grade%208%20Unit%202

text/html

crawl-data/CC-MAIN-2019-35/segments/1566027330786.8/warc/CC-MAIN-20190825173827-20190825195827-00221.warc.gz

Mathematics MTH6465 ccs Rose DeLuccio- Grade 8 Unit 2 This course contains CCSS aligned materials that correlate with the ADNY Grade 8 Math Unit 2. 7 Units ### CCSS.MATH.CONTENT.8.G.A.1 This course contains materials aligned to the 8th Grade Math Common Core Standard CCSS.MATH.CONTENT.8.G.A.1. Verify experimentally the properties of rotations, reflections, and translations: items (5) WEB RESOURCE Assessment: Properties of rigid transformations WEB RESOURCE Introduction to congruence and similarity through ... WEB RESOURCE Video: Identify what changes and what stays the sa... WEB RESOURCE Video: Identify what changes and what stays the sa... WEB RESOURCE Video: Identify what changes and what stays the sa... ### CCSS.MATH.CONTENT.8.G.A.1... This course contains materials aligned to the 8th Grade Math Common Core Standard CCSS.MATH.CONTENT.8.G.A.1.A Lines are taken to lines, and line segments to line segments of the same length. items (5) DOCUMENT Basics Geometry VIDEO Performing rotations (old) VIDEO Performing reflections: line (old) WEB RESOURCE Introduction to congruence and similarity through ... WEB RESOURCE Reflections 2 ### CCSS.MATH.CONTENT.8.G.A.1... This course contains materials aligned to the 8th Grade Math Common Core Standard CCSS.MATH.CONTENT.8.G.A.1.B Angles are taken to angles of the same measure. items (0) ### CCSS.MATH.CONTENT.8.G.A.2 This course contains materials aligned to the 8th Grade Math Common Core Standard CCSS.MATH.CONTENT.8.G.A.2. Understand that a two-dimensional figure is congruent to another if the second can be obtained from the first by a sequence of rotations, reflections, and translations; given two congruent figures, describe a sequence that exhibits the congruence between them. items (5) VIDEO CA Geometry: More on congruent and similar triangl... VIDEO Intro to triangle similarity DOCUMENT Basics Geometry DOCUMENT Geometric Transformations WEB RESOURCE "16. Exploring Angle Relationships Through Transfo... ### CCSS.MATH.CONTENT.8.G.A.5 This course contains materials aligned to the 8th Grade Math Common Core Standard CCSS.MATH.CONTENT.8.G.A.5. Use informal arguments to establish facts about the angle sum and exterior angle of triangles, about the angles created when parallel lines are cut by a transversal, and the angle-angle criterion for similarity of triangles. For example, arrange three copies of the same triangle so that the sum of the three angles appears to form a line, and give an argument in terms of transversals why this is so. items (10) DOCUMENT Area of Polygons DOCUMENT Parallel and Perpendicular Lines WEB RESOURCE "9. Building Triangles with Cubes" WEB RESOURCE Equation practice with vertical angles - interacti... VIDEO Measures of angles formed by a transversal VIDEO Triangle exterior angle example VIDEO Triangle angle challenge problem VIDEO Triangle angle challenge problem 2 WEB RESOURCE Assessment: Finding angle measures between interse... WEB RESOURCE Assessment: Angles 2 ### CCSS.MATH.CONTENT.8.G.B.6 This course contains materials aligned to the 8th Grade Math Common Core Standard CCSS.MATH.CONTENT.8.G.B.6. Explain a proof of the Pythagorean Theorem and its converse. items (2) WEB RESOURCE "1. Introduction to Pythagorean Theorem" WEB RESOURCE "4. Solving the Pythagorean Theorem Algebraically" ### CCSS.MATH.CONTENT.8.G.B.7 This course contains materials aligned to the 8th Grade Math Common Core Standard CCSS.MATH.CONTENT.8.G.B.7. Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions. items (8) WEB RESOURCE Pythagorean Theorem Interactive Assessment DOCUMENT Applying the Pythagorean Theorem DOCUMENT SE Right Triangle Trigonometry - TI VIDEO Pythagorean theorem word problem: fishing boat VIDEO Pythagorean theorem example VIDEO Pythagorean theorem word problem: carpet WEB RESOURCE Assessment: Use Pythagorean theorem to find right ... WEB RESOURCE Assessment: Special right triangles

{"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}

CC-MAIN-2019-35

longest

en

https://qna.talkjarvis.com/2477/how-many-comparisons-will-be-made-to-sort-the-array-arr-1-5-3-8-2-using-counting-sort

text/html

crawl-data/CC-MAIN-2024-10/segments/1707947473824.45/warc/CC-MAIN-20240222193722-20240222223722-00427.warc.gz

# How many comparisons will be made to sort the array arr={1,5,3,8,2} using counting sort? +1 vote How many comparisons will be made to sort the array arr={1,5,3,8,2} using counting sort? (a) 5 (b) 7 (c) 9 (d) 0 This question was posed to me in an interview for internship. This question is from Sorting in chapter Sorting of Data Structures & Algorithms II +1 vote by (962k points) selected by The best I can explain: As counting sort is an example of non comparison sort so it is able to sort an array without making any comparison. +1 vote +1 vote +1 vote +1 vote

{"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}

CC-MAIN-2024-10

latest

en

https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-5-review-page-404/50

text/html

crawl-data/CC-MAIN-2019-04/segments/1547583657907.79/warc/CC-MAIN-20190116215800-20190117001800-00188.warc.gz

## Algebra: A Combined Approach (4th Edition) $868,000 = 8.68 \times 10^5$ $868,000$ Step 1: Move the decimal point until the number is between 1 and 10. Step 2: The decimal point is moved 5 places and the original number is 10 or greater, so the count is positive 5. Step 3: The scientific notation is: $868,000 = 8.68 \times 10^5$

{"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}

CC-MAIN-2019-04

latest

en

https://amp.doubtnut.com/question-answer/if-x3-a-n-d-y-1-find-the-values-of-each-of-the-using-identity-5-x-5x25-x2-25-25-x2-1409096

text/html

crawl-data/CC-MAIN-2019-51/segments/1575540527620.19/warc/CC-MAIN-20191210123634-20191210151634-00125.warc.gz

IIT-JEE Apne doubts clear karein ab Whatsapp (8400400400) par bhi. Try it now. Click Question to Get Free Answers This browser does not support the video element. PLAY in APP PLAY HERE 2:16 Question From class 9 Chapter ALGEBRAIC IDENTITIES If find the values of each of the using identity: Find the values of in each of the following : (i) (ii) 2:31 If find the values of each of the using identity: 2:08 If find the values of each of the using identity: 2:15 If , then find the values of x and y 5:07 If find the values of each of the using identity: 2:04 13. 14. 2:47 In the identity then sum of digits of number is 2:49 If find the values of each of the using identity: 2:36 - 1:32 At what values of parameter are there values of such that the numbers: form an ? 6:29 Find the products: 1:09 The equation of the ellipse whose vertices are and the distance between the foci is 8 unit is (A) (B) (C) (D) 4:21 If , the numbers form an A.P. then must lie in the interval 6:57 If , the numbers form an then must lie in the interval 3:20 2:39 Latest Blog Post CBSE Announces Key Changes in Paper Pattern for Class 10 and 12 Board Exams CBSE to introduce major changes in exam pattern of Class 10 and Class 12 board classes by 2023. Get here complete updates on the CBSE 2020 exam. NTA NEET 2020 Exam Schedule Released, Registration Last Date 31st Dec 2019 NEET 2020 exam schedule has been released by NTA. Check registration date, application procedure, NEET application fee, and important dates here! JEE Main to be Multilingual, Priority to States with Majority Applications JEE main paper will be in multilingual and preference will be given to states with maximum number of applications. Know the details and benefits of this update. Top 5 Best Physics Books for JEE Advanced 2020 Preparations Know top 5 best Physics books for JEE advanced preparation in 2020. Check basic comparison among best JEE physics books and crack JEE advanced 2020. Why Pre Boards are Important Before Board Exam CBSE 2020 Know why pre board exam is important before CBSE board exam 2020. Pre board exam help students to boost confidence and score better in board exams. Top 5 Best Maths Books for JEE Advanced 2020 Preparation Know top 5 best math books for JEE advanced preparation in 2020. Also check basic comparison among those math books and crack JEE advanced exam 2020. Microconcepts Latest from Forum What are some popular medical examinations? Shubham 3   Replies 06-11-2019 Which books are good for the JEE Advanced? quora writer 0 8   Replies 05-09-2019 Which one is the best way to memorize your subject for exams? prachi verma 4   Replies 10-09-2019 What are the signs that you are wasting your life ?

{"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}

CC-MAIN-2019-51

latest

en

http://qs321.pair.com/~monkads/?node_id=782900

text/html

crawl-data/CC-MAIN-2023-50/segments/1700679101282.74/warc/CC-MAIN-20231210060949-20231210090949-00359.warc.gz

There's more than one way to do things PerlMonks Re^2: factorial through recursion by lidden (Curate) on Jul 24, 2009 at 09:36 UTC ( #782900=note: print w/replies, xml ) Need Help?? in reply to Re: factorial through recursion Finally, it's very weird that your factorial function takes two inputs. That's inconsistent with the mathematical function. Yes, but I note that his function looks similar to what a helper function would look like in a language with tail recursion optimization. It should still return a value instead of printing it though. Replies are listed 'Best First'. Re^3: factorial through recursion by ikegami (Patriarch) on Jul 24, 2009 at 16:10 UTC I have no problem with a helper function having more arguments. ```sub fact { my (\$n) = @_; local *_fact = sub { my (\$n, \$prod) = @_; return \$prod if \$n == 0; return _fact(\$n-1, \$n*\$prod); }; return _fact(\$n, 1); } fact(\$n); In this case, you could even fake it as follows: ```sub fact { my (\$n, \$prod) = @_; \$prod ||= 1; return \$prod if \$n == 0; return _fact(\$n-1, \$n*\$prod); } fact(\$n); But since Perl doesn't do tail recursion optimisation, and since sub calls are relatively slow, you're better off with: ```sub fact { my (\$n) = @_; my \$prod = 1; \$prod *= \$n-- while \$n > 0; return \$prod; } fact(\$n); Create A New User Domain Nodelet? Node Status? node history Node Type: note [id://782900] help Chatterbox? and the web crawler heard nothing... How do I use this?Last hourOther CB clients Other Users? Others avoiding work at the Monastery: (4) As of 2023-12-10 08:27 GMT Sections? Information? Find Nodes? Leftovers? Voting Booth? What's your preferred 'use VERSION' for new CPAN modules in 2023? Results (39 votes). Check out past polls. Notices?

{"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}

CC-MAIN-2023-50

latest

en

https://gitlab.mpi-sws.org/iris/iron/-/blame/8aacd0985b3c7f33dcc4d4bbe9b8489da9f507da/theories/heap_lang/adequacy.v

text/html

crawl-data/CC-MAIN-2020-34/segments/1596439735916.91/warc/CC-MAIN-20200805065524-20200805095524-00283.warc.gz

Robbert Krebbers committed Oct 31, 2018 1 2 3 4 5 6 ``````(** This file proves the basic and correct usage of resources adequacy statements for [heap_lang]. It does do so by appealing to the generic results in [iron_logic/adequacy]. Note, we only state adequacy for the lifted logic, because in the Coq formalization we state all specifications in terms of the lifted logic. *) `````` Robbert Krebbers committed Jun 11, 2019 7 ``````From iris.algebra Require Import big_op gmap ufrac excl ufrac_auth. `````` Robbert Krebbers committed Oct 31, 2018 8 9 10 11 12 13 14 ``````From iron.iron_logic Require Export weakestpre adequacy. From iron.heap_lang Require Import heap. From iris.proofmode Require Import tactics. Set Default Proof Using "Type". Class heapPreG Σ := HeapPreG { heap_preG_iris :> ironInvPreG Σ; `````` Robbert Krebbers committed Feb 03, 2019 15 16 `````` heap_preG_inG :> inG Σ (ufrac_authR heapUR); heap_preG_fork_post_inG :> inG Σ (authR (gmapUR positive (exclR (optionC ufracC)))); `````` Robbert Krebbers committed Oct 31, 2018 17 18 19 20 ``````}. Definition heapΣ : gFunctors := #[ ironInvΣ; `````` Robbert Krebbers committed Feb 03, 2019 21 22 `````` GFunctor (ufrac_authR heapUR); GFunctor (authR (gmapUR positive (exclR (optionC ufracC))))]. `````` Robbert Krebbers committed Oct 31, 2018 23 24 25 26 27 28 29 30 31 32 33 34 35 ``````Instance subG_heapPreG {Σ} : subG heapΣ Σ → heapPreG Σ. Proof. solve_inG. Qed. (** A generic helper *) Theorem heap_adequacy Σ `{heapPreG Σ} s e σ φ : (∀ `{heapG Σ}, ([∗ map] l ↦ v ∈ σ, l ↦ v) ⊢ WP e @ s; ⊤ {{ v, |={⊤,∅}=> ⎡∀ h m, heap_ctx h [] m -∗ ⌜φ v h⌝⎤ }}%I) → adequate s e σ φ. Proof. (* TODO: refactor this proof so we don't have the two cases *) destruct (decide (σ = ∅)) as [->|Heq]. - intros Hwp; eapply (iron_wp_adequacy _ _ _ _ _ _ (1 / 2) ε); iIntros (? κs) "". `````` Robbert Krebbers committed Jun 11, 2019 36 `````` iMod (own_alloc (●U{1} ∅ ⋅ (◯U{1/2} ∅ ⋅ ◯U{1/2} ε))) as (γ) "[Hσ [Hp Hp']]". `````` Robbert Krebbers committed Feb 03, 2019 37 38 `````` { rewrite -ufrac_auth_frag_op Qp_div_2 right_id. by apply ufrac_auth_valid. } `````` Hai Dang committed May 24, 2019 39 `````` iMod (own_alloc (● ∅)) as (γf) "Hf"; first by apply auth_auth_valid. `````` Robbert Krebbers committed Oct 31, 2018 40 41 42 43 44 45 46 47 48 49 50 51 52 `````` iModIntro. pose (HeapG _ _ _ γ _ _ γf). iExists heap_perm, heap_ctx, (λ _, heap_fork_post), _, _, True%I. iFrame "Hp". iSplitL "Hσ Hf". { iExists ∅. rewrite /= to_heap_empty fmap_empty big_opM_empty. by iFrame. } iPoseProof (Hwp _) as "Hwp"; clear Hwp. iApply (iron_wp_wand with "[Hwp]"). { iApply "Hwp". by rewrite fracPred_at_emp. } iIntros (v π1) "Hupd". iExists π1, ε; repeat (iSplit || iModIntro)=> //. iIntros (π2 π3 h2 Qs Hπ) "Hctx _ _". iMod ("Hupd" with "Hp'") as (π4 π5 _) "[_ H]". iModIntro. iApply ("H" with "Hctx"). - intros Hwp; eapply (iron_wp_adequacy _ _ _ _ _ _ (1 / 2 / 2) (Some (1 / 2)%Qp)). iIntros (? κs) "". `````` Robbert Krebbers committed Jun 11, 2019 53 54 `````` iMod (own_alloc (●U{1} (to_heap σ) ⋅ (◯U{1/2} (to_heap σ) ⋅ ◯U{1/2} ε))) as (γ) "[Hσ [Hσ' [Hp Hp']]]". `````` Robbert Krebbers committed Feb 03, 2019 55 56 `````` { rewrite -ufrac_auth_frag_op Qp_div_2 right_id. apply ufrac_auth_valid; by apply to_heap_valid. } `````` Hai Dang committed May 24, 2019 57 `````` iMod (own_alloc (● ∅)) as (γf) "Hf"; first by apply auth_auth_valid. `````` Robbert Krebbers committed Oct 31, 2018 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 `````` iModIntro. pose (HeapG _ _ _ γ _ _ γf). iExists heap_perm, heap_ctx, (λ _, heap_fork_post), _, _, True%I. iFrame "Hp". iSplitL "Hσ Hf". { iExists ∅. rewrite /= fmap_empty big_opM_empty. by iFrame. } iPoseProof (Hwp _) as "Hwp"; clear Hwp. iDestruct (big_mapsto_alt_2 with "Hσ'") as "Hσ'"; first done. iDestruct ("Hwp" with "Hσ'") as "Hwp'". iApply (iron_wp_wand with "Hwp'"). iIntros (v π1) "Hupd". iExists π1, ε; repeat (iSplit || iModIntro)=> //. iIntros (π2 π3 h2 Qs Hπ) "Hctx _ _". iMod ("Hupd" with "Hp'") as (π4 π5 _) "[_ H]". iModIntro. iApply ("H" with "Hctx"). Qed. (** The basic adequacy statement: the program is safe & when the main thread terminates, the postcondition hold. *) Theorem heap_basic_adequacy Σ `{heapPreG Σ} s e σ φ : (∀ `{heapG Σ}, {{{ [∗ map] l ↦ v ∈ σ, l ↦ v }}} e @ s; ⊤ {{{ v, RET v; ⌜φ v⌝ }}}%I) → adequate s e σ (λ v _, φ v). Proof. intros Hwp. apply (heap_adequacy Σ). iIntros (?) "Hall". iApply (Hwp with "Hall"). iIntros "!>" (v) "Hφ". iMod (fupd_intro_mask' ⊤ ∅) as "H"; first done. iModIntro. iDestruct "Hφ" as %Hφ. by iIntros "!>" (σ' Qs) "_". Qed. (** Adequacy for correct usage of resources: when all threads terminate, and the post-condition exactly characterizes the heap, we know that the heap is in fact of the given shape. *) Theorem heap_all_adequacy Σ `{heapPreG Σ} s e σ1 σ2 σ2' v vs φ : (∀ `{heapG Σ}, ([∗ map] l ↦ v ∈ σ1, l ↦ v) ⊢ WP e @ s; ⊤ {{ v, ([∗ map] l ↦ v ∈ σ2, l ↦ v) ∧ ⌜φ v⌝ }}%I ) → rtc erased_step ([e], σ1) (of_val <\$> v :: vs, σ2') → σ2' = σ2. Proof. (* TODO: refactor this proof so we don't have the two cases *) destruct (decide (σ1 = ∅)) as [->|Heq]. - intros Hwp Hsteps. eapply (iron_wp_all_adequacy _ heap_lang _ _ ∅ σ2' _ _ (λ _, σ2' = σ2) _ ε); [|done]; iIntros (? κs) "". `````` Hai Dang committed May 24, 2019 100 `````` iMod (own_alloc (● ∅)) as (γf) "Hf"; first by apply auth_auth_valid. `````` Robbert Krebbers committed Jun 11, 2019 101 `````` iMod (own_alloc (●U{1} (to_heap ∅) ⋅ (◯U{1/2} (to_heap ∅) ⋅ ◯U{1/2} ε))) `````` Robbert Krebbers committed Oct 31, 2018 102 `````` as (γ) "[Hσ [Hσ' Hp]]". `````` Robbert Krebbers committed Feb 03, 2019 103 104 `````` { rewrite -ufrac_auth_frag_op Qp_div_2 right_id. apply ufrac_auth_valid; by apply to_heap_valid. } `````` Robbert Krebbers committed Oct 31, 2018 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 `````` iModIntro. pose (HeapG _ _ _ γ _ _ γf). iExists heap_perm, heap_ctx, (λ _, heap_fork_post), _, _. iFrame "Hp"; iExists ([∗ map] l ↦ v ∈ σ2, l ↦ v)%I. iSplitL "Hσ Hf". { iExists ∅. rewrite /= to_heap_empty fmap_empty big_opM_empty. by iFrame. } iPoseProof (Hwp _) as "Hwp"; clear Hwp. iDestruct ("Hwp" with "[]") as "Hwp'". { rewrite big_opM_empty fracPred_at_emp; auto. } iApply (iron_wp_wand with "Hwp'"); iIntros (v' π) "[Hσ2 _] /=". iExists π, ε; iSplit; first (by rewrite left_id_L right_id_L). iFrame "Hσ2". iModIntro; iSplit; first done. iIntros (π1 π2 Hπ) "Hσ2' HQs Hp' Hσ2". rewrite to_heap_empty. iMod (heap_thread_adequacy _ _ _ (1/2 + π1)%Qp with "Hσ2' [HQs] [Hσ' Hp'] Hσ2") as "\$". { by rewrite -frac_op' cmra_op_opM_assoc_L -Hπ frac_op' Qp_div_2. } { by iApply big_sepL_replicate. } { by iSplitL "Hσ'". } iMod (fupd_intro_mask' ⊤ ∅) as "_"; auto. - intros Hwp Hsteps. eapply (iron_wp_all_adequacy _ heap_lang _ _ σ1 σ2' _ _ (λ _, σ2' = σ2) _ (Some (1 / 2)%Qp)); [|done]; iIntros (? κs) "". `````` Hai Dang committed May 24, 2019 125 `````` iMod (own_alloc (● ∅)) as (γf) "Hf"; first by apply auth_auth_valid. `````` Robbert Krebbers committed Jun 11, 2019 126 `````` iMod (own_alloc (●U{1} (to_heap σ1) ⋅ (◯U{1/2} (to_heap σ1) ⋅ ◯U{1/2} ε))) `````` Robbert Krebbers committed Oct 31, 2018 127 `````` as (γ) "[Hσ [Hσ' Hp]]". `````` Robbert Krebbers committed Feb 03, 2019 128 129 `````` { rewrite -ufrac_auth_frag_op Qp_div_2 right_id. apply ufrac_auth_valid; by apply to_heap_valid. } `````` Robbert Krebbers committed Oct 31, 2018 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 `````` iModIntro. pose (HeapG _ _ _ γ _ _ γf). iExists heap_perm, heap_ctx, (λ _, heap_fork_post), _, _. iFrame "Hp"; iExists ([∗ map] l ↦ v ∈ σ2, l ↦ v)%I. iSplitL "Hσ Hf". { iExists ∅. rewrite /= fmap_empty big_opM_empty. by iFrame. } iPoseProof (Hwp _) as "Hwp"; clear Hwp. iDestruct ("Hwp" with "[Hσ']") as "Hwp'"; first by iApply big_mapsto_alt_2. iApply (iron_wp_wand with "Hwp'"). iIntros (v' π) "[Hσ2 _]". iExists π, ε; iSplit; first (erewrite left_id_L, right_id_L; auto; apply _). iFrame "Hσ2". iModIntro; iSplit; first done. iIntros (π1 π2 Hπ) "Hσ2' HQs Hp' Hσ2". iMod (heap_thread_adequacy with "Hσ2' [HQs] Hp' Hσ2") as "\$". { by rewrite -Hπ /= frac_op' Qp_div_2. } { by iApply big_sepL_replicate. } iMod (fupd_intro_mask' ⊤ ∅) as "_"; auto. Qed.``````

{"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}

CC-MAIN-2020-34

latest

en

https://www.physicsforums.com/threads/differential-equation.337354/

text/html

crawl-data/CC-MAIN-2017-43/segments/1508187825575.93/warc/CC-MAIN-20171023035656-20171023055656-00048.warc.gz

# Differential equation 1. Sep 14, 2009 ### intenzxboi Give the general solution of: dy/dx = 6x y^2 -6x + 3 y^2 -3. No clue how to start. Guessing you would use Bernoulli's equation but i don't see how it is possible to put in the y'(x) + p(x)y(x) = q(x) y(x)^n form. 2. Sep 14, 2009 ### Gregg Bernoulli Equation $$\frac{dy}{dx} = f(x)y+g(x)y^k$$ $$y^{1-k}=y_1 + y_2$$ where $$\phi(x) = (1-k)\int f(x) dx$$ $$y_1 = Ce^\phi$$ $$y_2 = (1-k)e^\phi \int e^{-\phi} g(x) dx$$ 3. Sep 14, 2009 ### Gregg Actually don't bother, you can rearrange to get this: $$\frac{dy}{dx} = (6x+3)(y^2-1)$$ you can take it from there, no? 4. Sep 14, 2009 ### intenzxboi Yes i can thank you 5. Sep 14, 2009 ### intenzxboi actually im still having trouble do i still use the bernoulli method? so it is a simple separation problem? 6. Sep 14, 2009 ### Gregg Sorry I didn't realise you replied. $$\frac{dy}{dx} = (6x+3)(y^2-1)$$ $$\int \frac{dy}{y^2-1} = \int (6x+3) dx$$ $$\frac{1}{2}log\left(\frac{1-y}{1+y}\right) = 3x^2+3x + C$$ $$\sqrt{\frac{1-y}{1+y}} = Ae^{3x^2+3x}$$ $$y=\frac{1-Ae^{6x^2+6x}}{1+Ae^{6x^2+6x}}$$ Something like this I think? Last edited: Sep 14, 2009

{"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}

CC-MAIN-2017-43

longest

en

https://visionpune.com/other/660-probability-and-combinatorics.html

text/html

crawl-data/CC-MAIN-2018-09/segments/1518891814538.66/warc/CC-MAIN-20180223075134-20180223095134-00046.warc.gz

Popular Tags # Probability and Combinatorics BI 1077 New According to New Revised Syllabus w.e.f. 2008 PUNE, Maharashtra (INDIA) Text Book of Probability and Combinatorics MCA Sem-II Author : Deshmukh Book ID : 1077 Rs.150 Contents 1. Permutations and Combinations 1.1 Introduction 1.2 Principles of counting 1.3 Permutations 1.4 Combinations 2. Integer Solutions of Linear Equations 2.1 Introduction 2.2 Integer solutions of linear equations 2.3 Binomial identities 3. Principle of Inclusion and Exclusion 3.1 Introduction 3.2 Principle of inclusion and exclusion 3.3 Derangements 3.4 Generating functions 4. Recurrence Relations 4.1 Introduction 4.2 The process of obtaining a recurrence relation 4.3 The homogeneous equation 4.4 The non-homogeneous equation 4.5 Pigeonhole principle 5. Probability and Probability Distributions 5.1 Introduction 5.2 Deterministic and non deterministic models 5.3 Sample space 5.4 Events 5.5 Probability 5.6 Random variables 5.7 Continuous random variables 6. Special Probability Distributions 6.1 Introduction 6.2 Discrete distributions 6.3 Continuous distributions 7. Expectation 7.1 Introduction 7.2 Expectation of a discrete random variable 7.3 Expectation of continuous random variable 8. Moment and Cumulant Generating Functions 8.1 Moment Generating Functions (M.G.F.) 8.2 Cumulant Generating Function (C.G.F.) 8.3 Characteristic functions 8.4 M(T) and K(T) for some common distributions Solved Question Paper No customer comments for the moment. • ## Principles and Practices of... Rs.144 Rs.160 ## Principles and Practices of... Rs.144 Rs.160 Sale • ## Discrete Mathematics Rs.126 Rs.140 ## Discrete Mathematics Rs.126 Rs.140 Sale • ## Operating System Concepts Rs.144 Rs.160 ## Operating System Concepts Rs.144 Rs.160 Sale • ## Database Management System Rs.144 Rs.160 ## Database Management System Rs.144 Rs.160 Sale • ## Data Structure Using C Rs.140 Rs.155 ## Data Structure Using C Rs.140 Rs.155 Sale Rs.170 Rs.170 • ## Web Technologies Rs.144 Rs.160 ## Web Technologies Rs.144 Rs.160 Sale • ## Data Communication & Computer... Rs.180 Rs.200 ## Data Communication & Computer... Rs.180 Rs.200 Sale • ## Object Oriented Programming... Rs.171 Rs.190 ## Object Oriented Programming... Rs.171 Rs.190 Sale • ## Java Programming Rs.162 Rs.180 ## Java Programming Rs.162 Rs.180 Sale • ## Software Testing and Quality... Rs.126 Rs.140 ## Software Testing and Quality... Rs.126 Rs.140 Sale • ## Object Oriented Analysis and Design Rs.171 Rs.190 ## Object Oriented Analysis and Design Rs.171 Rs.190 Sale • ## Optimization Technique Rs.153 Rs.170 ## Optimization Technique Rs.153 Rs.170 Sale • ## Emerging Trends In Information... Rs.171 Rs.190 ## Emerging Trends In Information... Rs.171 Rs.190 Sale Rs.171 Rs.190 Rs.171 Rs.190 Sale • ## Human Computer Interface Rs.162 Rs.180 ## Human Computer Interface Rs.162 Rs.180 Sale • ## Management Support System & IS... Rs.108 Rs.120 ## Management Support System & IS... Rs.108 Rs.120 Sale • ## C Programming Rs.171 Rs.190 ## C Programming Rs.171 Rs.190 Sale Rs.205 Rs.205 Rs.135 ## Principles & Practices Of... Rs.135 ### 20 other products in the same category: • ##### Principles... Rs.144 -10% Rs.160 • ##### Discrete... Rs.126 -10% Rs.140 • ##### Operating... Rs.144 -10% Rs.160 • ##### Database... Rs.144 -10% Rs.160 • ##### Data... Rs.140 -10% Rs.155 Rs.170 • ##### Web... Rs.144 -10% Rs.160 • ##### Data... Rs.180 -10% Rs.200 • ##### Object... Rs.171 -10% Rs.190 • ##### Java... Rs.162 -10% Rs.180 • ##### Software... Rs.126 -10% Rs.140 • ##### Object... Rs.171 -10% Rs.190 • ##### Optimizatio... Rs.153 -10% Rs.170 • ##### Emerging... Rs.171 -10% Rs.190 Rs.171 -10% Rs.190 • ##### Human... Rs.162 -10% Rs.180 • ##### Management... Rs.108 -10% Rs.120 • ##### C Programming Rs.171 -10% Rs.190 Rs.205 Rs.135

{"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}

CC-MAIN-2018-09

longest

en

http://dropdoc.ru/doc/8950/zanyatiya-v-dome-predprinimatelya--ul.-mayakovskogo--d.46-%3Bpdf

text/html

crawl-data/CC-MAIN-2020-40/segments/1600402118004.92/warc/CC-MAIN-20200930044533-20200930074533-00070.warc.gz

Забыли? # Занятия в Доме предпринимателя (ул. Маяковского, д.46);pdf код для вставкиСкачать ```CUBO A Mathematial Journal Vol.15, No 03, (4550). Otober 2013 On entralizers of standard operator algebras with involution Maja Fo²ner, Benjamin Maren Nej ’irovnik Faulty of Logistis, University of Maribor, Mariborska esta 7 3000 Celje Slovenia, Faulty of Natural Sienes and Mathematis, University of Maribor, Koro²ka esta 160 2000 Maribor Slovenia. maja.fosnerfl.uni-mb.si, benjamin.marenfl.uni-mb.si nej.sirovnikuni-mb.si ABSTRACT The purpose of this paper is to prove the following result. Let X be a omplex Hilbert spae, let L(X) be the algebra of all bounded linear operators on X and let A(X) ⊂ L(X) be a standard operator algebra, whih is losed under the adjoint operation. Let T : A(X) → L(X) be a linear mapping satisfying the relation 2T (AA∗ A) = T (A)A∗ A + AA∗ T (A) for all A ∈ A(X). In this ase T is of the form T (A) = λA for all A ∈ A(X), where λ is some xed omplex number. RESUMEN El propósito de este artíulo es probar el siguiente resultado. Sea X un espaio de Hilbert omplejo, sea L(X) el álgebra de todos los operadores lineales aotados sobre X y sea A(X) ⊂ L(X) la álgebra de operadores lásia, la ual es errada bajo la operaión adjunto. Sea T : A(X) → L(X) una apliaión lineal satisfaiendo la relaión 2T (AA∗ A) = T (A)A∗ A + AA∗ T (A) para todo A ∈ A(X). En este aso, T es de la forma T (A) = λA para todo A ∈ A(X), donde λ es un número omplejo jo. Keywords and Phrases: ring, ring with involution, prime ring, semiprime ring, Banah spae, Hilbert spae, standard operator algebra, H∗ -algebra, left (right) entralizer, two-sided entralizer. 2010 AMS Mathematis Subjet Classiation: 16N60, 46B99, 39B42. 46 Maja Fo²ner, Benjamin Maren & Nej ’irovnik CUBO 15, 3 (2013) This researh has been motivated by the work of Vukman, Kosi-Ulbl [5℄ and Zalar [13℄. Throughout, R will represent an assoiative ring with enter Z(R). Given an integer n ≥ 2, a ring R is said to be n-torsion free if for x ∈ R, nx = 0 implies x = 0. An additive mapping x 7→ x∗ on a ring R is alled involution if (xy)∗ = y∗ x∗ and x∗∗ = x hold for all pairs x, y ∈ R. A ring equipped with an involution is alled a ring with involution or ∗ -ring. Reall that a ring R is prime if for a, b ∈ R, aRb = (0) implies that either a = 0 or b = 0, and is semiprime in ase aRa = (0) implies a = 0. We denote by Qr and C the Martindale right ring of quotients and the extended entroid of a semiprime ring R, respetively. For the explanation of Qr and C we refer the reader to [2℄. An additive mapping T : R → R is alled a left entralizer in ase T (xy) = T (x)y holds for all pairs x, y ∈ R. In ase R has the identity element, T : R → R is a left entralizer i T is of the form T (x) = ax for all x ∈ R, where a is some xed element of R. For a semiprime ring R all left entralizers are of the form T (x) = qx for all x ∈ R, where q ∈ Qr is some xed element (see Chapter 2 in [2℄). An additive mapping T : R → R is alled a left Jordan entralizer in ase T (x2 ) = T (x)x holds for all x ∈ R. The denition of right entralizer and right Jordan entralizer should be self-explanatory. We all T : R → R a two-sided entralizer in ase T is both a left and a right entralizer. In ase T : R → R is a two-sided entralizer, where R is a semiprime ring with extended entroid C, then T is of the form T (x) = λx for all x ∈ R, where λ ∈ C is some xed element (see Theorem 2.3.2 in [2℄). Zalar [13℄ has proved that any left (right) Jordan entralizer on a semiprime ring is a left (right) entralizer. Let us reall that a semisimple H∗ -algebra is a omplex semisimple Banah∗-algebra whose norm is a Hilbert spae norm suh that (x, yz∗ ) = (xz, y) = (z, x∗ y) is fullled for all x, y, z ∈ A. For basi fats onerning H∗ -algebras we refer to [1℄. Vukman [10℄ has proved that in ase there exists an additive mapping T : R → R, where R is a 2-torsion free semiprime ring satisfying the relation 2T (x2 ) = T (x)x + xT (x) for all x ∈ R, then T is a two-sided entralizer. Kosi-Ulbl and Vukman [9℄ have proved the following result. Let A be a semisimple H∗ −algebra and let T : A → A be an additive mapping suh that 2T (xn+1 ) = T (x)xn + xn T (x) holds for all x ∈ R and some xed integer n ≥ 1. In this ase T is a two-sided entralizer. Reently, Benkovi£, Eremita and Vukman [3℄ have onsidered the relation we have just mentioned above in prime rings with suitable harateristi restritions. Kosi-Ulbl and Vukman [9℄ have proved that in ase there exists an additive mapping T : R → R, where R is a 2-torsion free semiprime ∗ -ring, satisfying the relation T (xx∗ ) = T (x)x∗ (T (xx∗ ) = xT (x∗ )) for all x ∈ R, then T is a left (right) entralizer. For results onerning entralizers on rings and algebras we refer to [413℄, where further referenes an be found. Let X be a real or omplex Banah spae and let L(X) and F (X) denote the algebra of all bounded linear operators on X and the ideal of all nite rank operators in L(X), respetively. An algebra A(X) ⊂ L(X) is said to be standard in ase F (X) ⊂ A(X). Let us point out that any standard operator algebra is prime, whih is a onsequene of a Hahn-Banah theorem. In ase X is a real or omplex Hilbert spae, we denote by A∗ the adjoint operator of A ∈ L(X). We denote CUBO 15, 3 (2013) On entralizers of standard operator algebras with involution 47 by X∗ the dual spae of a real or omplex Banah spae X. Vukman and Kosi-Ulbl [5℄ have proved the following result. Theorem 0.1. Let R be a 2-torsion free semiprime ring and let T Suppose that : R → R be an additive mapping. 2T (xyx) = T (x)yx + xyT (x) (1) holds for all x, y ∈ R. In this ase T is a two-sided entralizer. In ase we have a ∗ -ring, we obtain, after putting y = x∗ in the relation (1), the relation 2T (xx∗ x) = T (x)x∗ x + xx∗ T (x). It is our aim in this paper to prove the following result, whih is related to the above relation. Theorem 0.2. Let X be a omplex Hilbert spae and let A(X) be a standard operator algebra, whih is losed under the adjoint operation. Suppose T : A(X) → L(X) is a linear mapping satisfying the relation 2T (AA∗ A) = T (A)A∗ A + AA∗ T (A) (2) for all A ∈ A(X). In this ase T is of the form T (A) = λA, where λ is a xed omplex number. Proof. Let us rst onsider the restrition of T on F (X). Let A be from F (X) (in this ase we have A∗ ∈ F (X)). Let P ∈ F (X) be a self-adjoint projetion with the property AP = PA = A (we also have A∗ P = PA∗ = A∗ ). Putting P for A in (2) we obtain 2T (P) = T (P)P + PT (P). Left multipliation by P in the above relation gives PT (P) = PT (P)P. Similarly, right multipliation by P in the above relation leads to T (P)P = PT (P)P. Therefore T (P) = T (P)P = PT (P) = PT (P)P. (3) Putting A + P for A in the relation (2) we obtain 2T (A2 ) + 2T (AA∗ + A∗ A) + 4T (A) + 2T (A∗ ) = = T (A)(A + A∗ ) + T (A)P + T (P)A∗ A + T (P)(A + A∗ )+ + (A + A∗ )T (A) + PT (A) + AA∗ T (P) + (A + A∗ )T (P). Putting −A for A in the above relation and omparing the relation so obtained with the above relation gives 2T (A2 ) + 2T (AA∗ + A∗ A) = = T (A)(A + A∗ ) + T (P)A∗ A + (A + A∗ )T (A) + AA∗ T (P) (4) 48 Maja Fo²ner, Benjamin Maren & Nej ’irovnik CUBO 15, 3 (2013) and 4T (A) + 2T (A∗ ) = = T (A)P + PT (A) + T (P)(A + A∗ ) + (A + A∗ )T (P). (5) So far we have not used the assumption of the theorem that X is a omplex Hilbert spae. Putting iA for A in the relations (4) and (5) and omparing the relations so obtained with the above relations, respetively, we obtain 2T (A2 ) = T (A)A + AT (A), (6) 4T (A) = T (A)P + PT (A) + T (P)A + AT (P). (7) Putting A∗ for A in the relation (5) gives 4T (A∗ ) + 2T (A) = = T (A∗ )P + PT (A∗ ) + T (P)(A + A∗ ) + (A + A∗ )T (P). Putting iA for A in the above relation and omparing the relation so obtained with the above 2T (A) = T (P)A + AT (P). Comparing the above relation and (7), we obtain 2T (A) = T (A)P + PT (A). (8) Right (left) multipliation by P in the above relation gives T (A)P = PT (A)P and PT (A) = PT (A)P, respetively. Hene, PT (A) = T (A)P, whih redues the relation (8) to T (A) = T (A)P. From the above relation one an onlude that T maps F (X) into itself. We therefore have a linear mapping T : F (X) → F (X) satisfying the relation (6) for all A ∈ F (X). Sine F (X) is prime, one an onlude, aording to Theorem 1 in [10℄ that T is a two-sided entralizer on F (X). We intend to prove that there exists an operator C ∈ L(X), suh that T (A) = CA (9) for all A ∈ F (X). For any xed x ∈ X and f ∈ X∗ we denote by x ⊗ f an operator from F (X) dened by (x ⊗ f)y = f(y)x, y ∈ X. For any A ∈ L(X) we have A(x ⊗ f) = (Ax) ⊗ f. Now let us hoose suh f and y that f(y) = 1 and dene Cx = T (x ⊗ f)y. Obviously, C is linear and applying the fat that T is a left entralizer on F (X), we obtain (CA)x = C(Ax) = T ((Ax) ⊗ f)y = T (A(x ⊗ f))y = T (A)(x ⊗ f)y = T (A)x for any x ∈ X. We therefore have T (A) = CA for any A ∈ F (X). As T is a right entralizer on F (X), we obtain C(AB) = T (AB) = AT (B) = ACB. We therefore have [A, C]B = 0 for any CUBO 15, 3 (2013) On entralizers of standard operator algebras with involution 49 A, B ∈ F (X), whene it follows that [A, C] = 0 for any A ∈ F (X). Using losed graph theorem one an easily prove that C is ontinuous. Sine C ommutes with all operators from F (X), we an onlude that Cx = λx holds for any x ∈ X and some xed omplex number λ, whih gives together with the relation (9) that T is of the form T (A) = λA (10) for any A ∈ F (X) and some xed omplex number λ. It remains to prove that the relation (10) holds on A(X) as well. Let us introdue T1 : A(X) → L(X) by T1 (A) = λA and onsider T0 = T −T1 . The mapping T0 is, obviously, additive and satises the relation (2). Besides, T0 vanishes on F (X). It is our aim to show that T0 vanishes on A(X) as well. Let A ∈ A(X), let P ∈ F (X) be a onedimensional self-adjoint projetion and S = A + PAP − (AP + PA). Suh S an also be written in the form S = (I − P)A(I − P), where I denotes the identity operator on X. Sine S − A ∈ F (X), we have T0 (S) = T0 (A). It is easy to see that SP = PS = 0. By the relation (2) we have T0 (S)S∗ S + SS∗ T0 (S) = = 2T0 (SS∗ S) = = 2T0 ((S + P)(S + P)∗ (S + P)) = = T0 (S + P)(S + P)∗ (S + P) + (S + P)(S + P)∗ T0 (S + P) = T0 (S)S∗ S + T0 (S)P + SS∗ T0 (S) + PT0 (S). We therefore have T0 (S)P + PT0 (S) = 0. Considering T0 (S) = T0 (A) in the above relation, we obtain T0 (A)P + PT0 (A) = 0. (11) Multipliation from both sides by P in the above relation leads to PT0 (A)P = 0. Right multipliation by P in the relation (11) and onsidering the above relation gives T0 (A)P = 0. Sine P is an arbitrary one-dimensional self-adjoint projetion, it follows from the above relation that T0 (A) = 0 for all A ∈ A(X), whih ompletes the proof of the theorem. We onlude the paper with the following onjeture. Conjeture 0.3. Let R be a semiprime ∗ -ring with suitable torsion restritions and let T be an additive mapping satisfying the relation 2T (xx∗ x) = T (x)x∗ x + xx∗ T (x) for all x ∈ R. In this ase T is a two-sided entralizer. Reeived: April 2013. Aepted: September 2013. :R→R 50 Maja Fo²ner, Benjamin Maren & Nej ’irovnik CUBO 15, 3 (2013) Referenes [1℄ W. Ambrose: Struture theorems for a speial lass of Banah algebras, Trans. Amer. Math. So. 57 (1945), 364-386. [2℄ K. I. Beidar, W. S. Martindale 3rd, A. V. Mikhalev: Rings with generalized identities, Marel Dekker, In., New York, (1996). [3℄ D. Benkovi£, D. Eremita, J. Vukman: A haraterization of the entroid of a prime ring, Studia Si. Math. Hungar. 45 (3) (2008), 379-394. [4℄ I. Kosi-Ulbl, J. Vukman: An equation related to entralizers in semiprime rings, Glas. Mat. 38 (58) (2003), 253-261. [5℄ I. Kosi-Ulbl, J. Vukman: On entralizers of semiprime rings, Aequationes Math. 66 (2003), 277-283. [6℄ I. Kosi-Ulbl, J. Vukman: On ertain equations satised by entralizers in rings, Internat. Math. J. 5 (2004), 437-456. [7℄ I. Kosi-Ulbl, J. Vukman: Centralizers on rings and algebras, Bull. Austral. Math. So. 71 (2005), 225-234. [8℄ I. Kosi-Ulbl, J. Vukman: A remark on a paper of L. Molnár, Publ. Math. Debreen. 67 (2005), 419-421. [9℄ I. Kosi-Ulbl, J. Vukman: On entralizers of standard operator algebras and semisimple H∗ algebras, Ata Math. Hungar. 110 (3) (2006), 217-223. [10℄ J. Vukman: An identity related to entralizers in semiprime rings, Comment. Math. Univ. Carol. 40 (1999), 447-456. [11℄ J. Vukman: Centralizers of semiprime rings, Comment. Math. Univ. Carol. 42 (2001), 237245. [12℄ J. Vukman: Identities related to derivations and entralizers on standard operator algebras, Taiwan. J. Math. Vol. 11 (2007), 255-265. [13℄ B. Zalar: On entralizers of semiprime rings, Comment. Math. Univ. Carol. 32 (1991), 609614. ``` 1/--страниц Пожаловаться на содержимое документа

{"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}

CC-MAIN-2020-40

latest

en

https://brainmass.com/business/arbitrage-pricing-theory/interest-arbitrage-151077

text/html

crawl-data/CC-MAIN-2021-25/segments/1623488259200.84/warc/CC-MAIN-20210620235118-20210621025118-00120.warc.gz

Explore BrainMass Interest Arbitrage Not what you're looking for? Search our solutions OR ask your own Custom question. This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here! Assume the following information: Spot rate of Mexican peso =\$.100 180 day forward rate of Mexican peso = \$.098 180 day Mexican interest rate = 6% 180 day U.S. interest rate=5% Given this information, is covered interest arbitrage worthwhile for Mexican investors who have pesos to invest? Explain the answer. Solution Preview To answer, begin with an assumed amount of pesos and determine the yield to Mexican investors who attempt ... Solution Summary This solution calculates the yield to Mexican investors who attempt covered interest arbitrage and explains if it is worthwhile for the investors. \$2.49

{"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}

CC-MAIN-2021-25

longest

en

https://en.eclass.cc/courses/coursera_linearprogramming_linear_and_integer_programming

text/html

crawl-data/CC-MAIN-2019-30/segments/1563195524879.8/warc/CC-MAIN-20190716201412-20190716223412-00406.warc.gz

# Linear and Integer Programming This course will cover the very basic ideas in optimization. Topics include the basic theory and algorithms behind linear and integer linear programming along with some of the important applications. We will also explore the theory of convex polyhedra using linear programming. Linear Programming (LP) is arguably one of the most important optimization problems in applied mathematics and engineering. The Simplex algorithm to solve linear programs is widely regarded as one among the "top ten" algorithms of the 20th century. Linear Programs arise in almost all fields of engineering including operations research, statistics, machine learning, control system design, scheduling, formal verification and computer vision. It forms the basis for numerous approaches to solving hard combinatorial optimization problems through randomization and approximation. The primary goals of this course will be to: 1. Understand the basic theory behind LP, algorithms to solve LPs, and the basics of (mixed) integer programs (ILP). 2. Understand important and emerging applications of LP and ILPs to economic problems (optimal resource allocation, scheduling problems), machine learning (SVM), and combinatorial optimization problems. At the end of the course, the successful student will be able to cast various problems that may arise in her research as optimization problems, understand the cases where the optimization problem will be linear, choose appropriate solution methods and interpret results appropriately. This is generally considered a useful ability in many research areas. ### Syllabus Introductory Material • Introduction to Linear Programming. Week #1: • The Diet Problem. • Linear Programming Formulations. • Tutorials on using GLPK (AMPL), Matlab, CVX and Microsft Excel. • The Simplex Algorithm (basics). Week #2: • Handling unbounded problems • Degeneracy • Geometry of Simplex • Initializing Simplex. • Cycling and the Use of Bland's rule. Week #3: • Duality: dual variables and dual linear program. • Strong duality theorem. • Complementary Slackness. • KKT conditions for Linear Programs. • Understanding the dual problem: shadow costs. • Extra: The revised simplex method. Week #4: • Advanced LP formulations: norm optimization. • Least squares, and quadratic programming. • Applications #1: Signal reconstruction and De-noising. • Applications #2: Regression. Week #5: • Integer Linear Programming. • Integer vs. Real-valued variables. • NP-completeness: basic introduction. • Reductions from Combinatorial Problems (SAT, TSP and Vertex Cover). • Approximation Algorithms: Introduction. Week #6: • Branch and Bound Method • Cutting Plane Method Week #7: • Applications: solving puzzles (Sudoku), reasoning about systems and other applications. • Classification and Machine Learning ### Recommended Background Mathematical Maturity (undergraduate algorithms or CS theory) and basic programming ability. A background in engineering or applied sciences could be useful, as well. The prerequisites for this class include: • Basic College Level mathematics: calculus and some linear algebra: matrices, matrix operations and solving linear equations. • Programming skills:  However,   students without a programming background  can pass the course by solving all the weekly assignments. Textbook: Linear Programming by Robert J. Vanderbei (available online  from author's webpage). Alternative: Linear Programming by Vasek Chvatal, W.H. Freeman published, 1983. This book still remains a very clear and concise presentation. Other texts: We may borrow material that is covered in some of the texts below. As you go down this list, the texts become less relevant for this class but remain very important for the broader field of optimization that we seek to introduce through this class. 1. Combinatorial optimization: Algorithms & Complexity, by Christos Papadimitrou and Ken Steglitz. 2. Linear Programming series of two books by G.B. Danzig and M.K. Thapa. 3. Schrijver's book on the Theory of Integer Linear Programming and Integer, Integer and Combinatoral Optimization book by Wolsey and Nemhauser are also good references. 4. Convex Optimization by Boyd and Vandenbreghe is a good reference for the more general area of convex optimization. 5. Numerical Optimization by Nocedal and Wright is a great reference for solving non-linear optimization problems. ### Course Format Each class will consist of many short lecture videos between 10-15 minutes in length with nearly 1.5-2 hours of  lectures each week. We will have weekly assignments involving 5-10 problems each week. These assignments will either be multiple choice or ask you to compute something and enter the answer. We will have four programming assignments (one every two weeks) that will incrementally help you build an ILP solver. Programming assignments can be done in virtually any general purpose language. The course will be structured with two interactive tracks: algorithms/theory for linear optimization problems and applications. The applications will include ideas on modeling real-life optimization problems as linear programs and the appropriate use of concepts like duality in finding and interpreting solutions. ### FAQ • Will I get a statement of accomplishment after completing this class? Yes. • What textbooks will I need for this class? No textbook is officially required. However, there are numerous great textbooks on this topic. We are hoping you will be able to find one. You can always cross check whether your textbook has adequate coverage for the topics to be taught in this class. We recommend a great textbook by Prof. Vanderbei (see http://www.princeton.edu/~rvdb/LPbook/). It is available as a free download. The classic book by Chvatal is an excellent textbook but sadly out-of-print. • What are the pre-requisites? The prerequisites for this class include: • Basic College Level mathematics: calculus and some knowledge of linear algebra. • Some programming skills:  However,   students without a programming background  can pass the course by solving the weekly assignments. • What will the level of the class be? Normally, this will be a junior/senior level undergraduate class or even a beginning graduate class, depending on the major and the university. A highly motivated high school student with advanced mathematical skills can follow along. • What skills will I learn? The class is about Linear and Integer Programming. You will definitely learn: what optimization problems are, what are linear optimization problems, what are the applications, what are the algorithms used for solving LPs, and how do they work? In addition, we found through our previous offering in 2013 that the course helps many students think mathematically and improve the programming skills. We were proud of the many students from last year who took on some of the challenging programming assignments and reported a big confidence boost in their skills. • Can I get university course credits for this class? Unfortunately, not at this time. University of Colorado, Boulder may be considering ways of recognizing Coursera classes, but there is no consensus at this point in time. However, Prof. Sankaranarayanan teaches an on-campus class (CSCI 5654) at the University of Colorado, Boulder, and it will be available simultaneously on-line for credit through CAETE (see http://cuengineeringonline.colorado.edu/ ). You can enroll for our class CSCI 5654 through CAETE from anywhere in the world. • Are we going to do rigorous mathematical proofs? How much programming do you expect to do? We will cover proofs for interested students. But we will never have proofs in the assignments. Assignments will test the conceptual ideas in the class including algorithms, and many assignments may ask you to use an existing LP solver (open source or commercial) to solve a LP/ILP and interpret its answer. We do not consider this part as programming: any computer user should be able to manage this. We will have programming assignments: in fact, students will get to build their own LP and later ILP solver in stages. But the programming is not going to be for everyone. The last assignment (ILP) will be quite intense and definitely challenging to students. • Is there a particular programming language that you expect for the assignments? No. We will allow students to program in most general purpose languages. However, by the very nature of the assignments, some languages such as python, C/C++, Java, Haskell, Scheme, Lisp, MATLAB/Octave, ... will be more suitable than others. You can judge for yourselves as soon as the assignments are posted on week 1. • What will the coursework involve? Coursework will involve watching videos, solving weekly assignments and tackling four programming assignments. See below on how the grading will work. • How do I pass? How does distinction work? To pass the class: you will need to get at least 35% of the total grade.  The pass cutoff is  designed so that students can pass either by solving some/all of the programming assignments OR by solving some/all of the weekly assignments. To score a distinction: you will need to get at least 85% of the total grade. To score a distinction, students must do well in ALL aspects of the course: programming as well as weekly assignments. Dates: • 20 October 2014, 7 weeks • 2 September 2013, 9 weeks Course properties: • Free: • Paid: • Certificate: • MOOC: • Video: • Audio: • Email-course: • Language: English ### Reviews No reviews yet. Want to be the first? Register to leave a review Included in selections: Optimization Optimization is for optimistic people! More on this topic: Combinatorial Optimization Combinatorial Optimization provides a thorough treatment of linear programming... Linear and Discrete Optimization The course is an introduction to linear and discrete optimization - an important... Integer Programming and Combinatorial Optimization The course is a comprehensive introduction to the theory, algorithms and applications... Discrete Optimization Tired of solving Sudokus by hand? This class teaches you how to solve complex... Intro Algebra Review. A Quick Refresher A brief review of introductory algebra topics including integer operations,... More from 'Coursera': First Year Teaching (Secondary Grades) - Success from the Start Success with your students starts on Day 1. Learn from NTC's 25 years developing... Understanding 9/11: Why Did al Qai’da Attack America? This course will explore the forces that led to the 9/11 attacks and the policies... Aboriginal Worldviews and Education This course will explore indigenous ways of knowing and how this knowledge can... Analytic Combinatorics Analytic Combinatorics teaches a calculus that enables precise quantitative... Accountable Talk®: Conversation that Works Designed for teachers and learners in every setting - in school and out, in...

{"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}

CC-MAIN-2019-30

longest

en

https://samacheerkalviguru.com/samacheer-kalvi-9th-maths-chapter-1-ex-1-3/

text/html

crawl-data/CC-MAIN-2024-30/segments/1720763860413.86/warc/CC-MAIN-20240725145050-20240725175050-00801.warc.gz

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations. ## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.3 9th Maths Exercise 1.3 Samacheer Kalvi Question 1. Using the given venn diagram, write the elements of (i) A (ii) B (iii) A ∪ B (iv) A ∩ B (v) A – B (vi) B – A (vii) A’ (viii) B’ (ix) U Solution: (i) A = {2, 4, 7, 8, 10} (ii) B = {3, 4, 6, 7, 9, 11} (iii) A ∪ B = {2, 3, 4, 6, 7, 8, 9, 10, 11} (iv) A ∩ B = {4, 7} (v) A – B = {2, 8, 10} (vi) B – A = {3, 6, 9, 11} (vii) A’ = {1, 3, 6, 9, 11, 12} (viii) B’ = {1, 2, 8, 10, 12} (ix) U = {1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12}. 9th Maths Set Language Exercise 1.3 Question 2. Find A ∪ B, A ∩ B, A – B and B – A for the following sets. (i) A = {2, 6, 10, 14} and B = {2, 5, 14, 16} (ii) A = {a, b, c, e, u} and B = {a, e, i, o, u] (iii) A = {x : x ∈ N, x ≤ 10} and B = {x : x ∈ W, x < 6} (iv) A = Set of all letters in the word “mathematics” and B = Set of all letters in the word “geometry” Solution: (i) A = {2, 6, 10, 14} and B = {2, 5, 14, 16} A ∪ B = {2, 6, 10, 14} ∪ {2, 5, 14, 16} = {2, 5, 6, 10, 14, 16} A ∩ B = {2, 6, 10, 14} ∩ {2, 5, 14, 16} = {2, 14} A – B = {2, 6, 10, 14} – {2, 5, 14, 16} = {6, 10} B – A = {2, 5, 14, 16} – {2, 6, 10, 14}  = {5, 16} (ii) A = {a, b, c, e, u} and B = {a, e, i, o, u} A ∪ B = {a, b, c, e, u) ∪ {a, e, i, o, u) = {a, b, c, e, i, o, u} A ∩ B = {a, b, c, e, u} ∩ {a, e, i, o, u} {a, e, u} A – B = {a, b, c, e, u) – {a, e, i, o, u) = {b, c} B – A = {a, e, i, o, u} – {a, b, c, e, u} =  {i, o} (iii) x ∈ {1, 2, 3, ……..} ; x ∈ {0, 1, 2, 3, 4, 5, ……..} A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} B = {0, 1, 2, 3, 4, 5} A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ∪ {0, 1, 2, 3, 4, 5} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} A ∩ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ∩ {0, 1, 2, 3, 4, 5} = {1, 2, 3, 4, 5} A – B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {0, 1, 2, 3, 4, 5} = {6, 7, 8, 9, 10} B – A = {0, 1, 2, 3, 4, 5} – {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} = {0} (iv) A= {m, a, t, h, e, i, c, s), B = {g, e, o, m, t, r, y) A ∪ B = {m ,a, t, h, e, i, c, s} ∪ {g, e, o, m, t, r, y} = {m, a, t, h, e, i, c, s, g, o, r, y) A ∩ B = {m, a, t, h, e, i, c, s} ∩ {g, e, o, m, t,r,y} = {m, t, e} A – B = {m ,a, t, h, e, i, c, s} ∪ {g, e, o, m, t, r, y} = {a, h, i, c, s) B – A = {m, a, t, h, e, i, c, 5} ∩ {g, e, o, m, t,r,y} = {g, o, r, y} 9th Std Maths Exercise 1.3 Question 3. If U = {a, b, c, d, e,f g ,h}, A = {b, d, f, h} and B = {a, d, e, h}, find the following sets. (i) A’ (ii) B’ (iii) A’ ∪ B’ (iv) A’ ∩ B’ (v) (A ∪ B)’ (vi) (A ∩ B)’ (vii) (A’)’ (viii) (B’)’ Solution: (i) A’ = U – A = {a, b, c, d, e, f, g, y} – {b, d, f, h} = {a, c, e, g} (ii) B’ = U – B = {a, b, c, d, e, f, g, y) – {a, d, e, h] = {b, c, f, g} (iii) A’ ∪ B’= {a, c, e, g} ∪ {b, c, f, g} = {a, b, c, e, f g} (iv) A’ ∩ B’= {a, c, e, g} ∩ {b, c, f, g} = {c, g} (v) (A ∪ B)’ = U – (A ∪ B) = {a, b, c, d, e, f, g, y) – {a, b, d, e, f, h} = {c, g} (vi) (A ∩ B)’ = U – (A ∩B) = {a, b, c, d, e, f, g, y} – {d, h} = {a, b, c, e, f, g} (vii) (A’)’ = U – A’ = {a, b, c, d, e, f, g, h} – {a, c, e, g} = {b, d, f, h) (viii) (B’)’ = U – B’ = {a, b, c, d, e, f, g, h} – {b, c, f, g} = {a, d, e, h} 9th Standard Maths Exercise 1.3 Question 4. Let U = {0, 1, 2 , 3, 4, 5, 6, 7}, A = {1, 3, 5, 7} and B = {0, 2, 3, 5, 7}, find the following sets. (i) A’ (ii) B’ (iii) A ‘ ∪ B’ (iv) A’ ∩ B’ (v) (A ∪ B)’ (vi) (A ∩ B)’ (vii) (A’)’ (viii) (B’)’ Solution: (i) A’ = U – A = {0, 1 ,2, y, 4, 5, 6, 7} – {1, 3, 5, 7} = {0, 2, 4, 6} (ii) B’ = U – B = {0, 1, 2, 3, 4, 5, 6 ,7} – {0, 2, 3, 5, 7} = {1, 4, 6} (iii) A’ ∪ B’ = {0, 2, 4, 6} ∪ {1, 4, 6} = {0, 1, 2, 4, 6} (iv) A’ ∩ B’ = {0, 2, 4, 6} ∩ {1, 4, 6} = {4, 6} (v) (A ∪ B)’ = U – (A ∪ B) = {0, 1, 2, 3, 4, 5, 6, 7} – {0, 1, 2, 3, 5, 7} = {4, 6} (vi) (A ∩ B)’ = U – (A ∩ B)= {0, 1, 2, 3, 4, 5, 6, 7} – {3,5,7} = {0, 1, 2, 4, 6} (vii) (A’)’ = U – A’ = {0, 1, 2, 3, 4, 5, 6, 7} – {0, 2, 4, 6} = {1, 3, 5, 7} (viii) (B’)’ = U – B’ = {0, 1, 2, 3, 4, 5, 6, 7} – {1, 4, 6} = {0, 2, 3, 5, 7}. 9th Maths Exercise 1.3 In Tamil Question 5. Find the symmetric difference between the following sets. (i) P = {2, 3, 5, 7, 11} and Q = {1, 3, 5, 11} (ii) R = {l, m, n, o, p} and S = {j, l, n, q) (iii) X = {5, 6, 7} and Y = {5, 7, 9, 10} Solution: (i) P = {2, 3, 5, 7, 11} Q= {1, 3, 5, 11} P – Q = {2, 3, 5, 7, 11} – {1, 3, 5, 11} = {2, 7} Q – P = {1, 3, 5, 11} – {2, 3, 5, 7, 11} = {1} P ∆ Q = (P – Q) ∪ (Q – P) = {2, 7} ∪ {1} = {1, 2, 7} (ii) R = {l, m, n, o, p} S = {j, l, n, q} R – S = {l, m, n, o, p) – {j, l, n, q} = {m, o, p) s – R = {j, l, n, q) – {l, m, n, o, p}= {j, q} R ∆ S = (R – S) ∪ (S – R) = {m, o, p) ∪ {j, q} = {j, m, o, p, q) (iii) X = {5, 6, 7} Y = {5, 7, 9, 10} X – Y = {5, 6, 7} – {5, 7, 9, 10} – {6} Y – X = {5, 6, 9, 10} – {5, 6, 7} = {9, 10} X ∆ Y = (X – Y) ∪ (Y – X) = {6} ∪ {9, 10} = {6, 9, 10}. 9th Maths 1.3 Question 6. Using the set symbols, write down the expressions for the shaded region in the following (i) (ii) (iii) Solution: (i) X – Y (ii) (X ∪ Y)’ (iii) (X – Y) ∪ (X – Y) 9th Maths Exercise 1.3 Question 7. Let A and B be two overlapping sets and the universal set U. Draw appropriate Venn diagram for each of the following, (i) A ∪ B (ii) A ∩ B (iii) (A ∩ B)’ (iv) (B – A)’ (v) A’ ∪ B’ (vi) A’ ∩ B’ (vii)What do you observe from the diagram (iii) and (v)? Solution: (i) A ∪ B (ii) A ∩ B (iii) (A ∩ B)’ (iv) (B – A)’ (v) A’ ∪ B’ (vi) A’ ∩ B’ (vii) From the diagram (iii) and (v) we observe that (A ∩ B)’ = A’ ∪ B’.

{"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}

CC-MAIN-2024-30

latest

en

https://www.geeksforgeeks.org/gate-gate-cs-2018-question-30/

text/html

crawl-data/CC-MAIN-2023-50/segments/1700679099942.90/warc/CC-MAIN-20231128183116-20231128213116-00189.warc.gz

# GATE | GATE CS 2018 | Question 30 Consider a system with 3 processes that share 4 instances of the same resource type. Each process can request a maximum of K instances. Resource instances can be requested and released only one at a time. The largest value of K that will always avoid deadlock is _______ . Note – This was Numerical Type question. (A) 1 (B) 2 (C) 3 (D) 4 Explanation: Given, Number of processes (P) = 3 Number of resources (R) = 4 ```R ≥ P(N − 1) + 1 ``` Where R is total number of resources, P is the number of processes, and N is the max need for each resource. ```4 ≥ 3(N − 1) + 1 3 ≥ 3(N − 1) 1 ≥ (N − 1) N ≤ 2 ``` Therefore, the largest value of K that will always avoid deadlock is 2. Option (B) is correct. Quiz of this Question

{"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}

CC-MAIN-2023-50

latest

en

https://docs.xilinx.com/r/2022.1-English/ug1483-model-composer-sys-gen-user-guide/Matrix-Multiply

text/html

crawl-data/CC-MAIN-2024-10/segments/1707947474784.33/warc/CC-MAIN-20240229035411-20240229065411-00085.warc.gz

# Matrix Multiply - 2022.1 English ## Vitis Model Composer User Guide (UG1483) Document ID UG1483 Release Date 2022-05-26 Version 2022.1 English Compute matrix product of two input signals. The first operand is the top input on the block. ## Library Math Functions / Matrices and Linear Algebra ## Description The Matrix Multiply block has two input ports and one output port. The output signal is the matrix product of the input signals where the first operand corresponds to the top input. ## Data Type Support The data type of the input signals can be any floating-point, fixed-point, integer, or Boolean. The input signals can be real or complex. The input signals can be scalar, vector, or matrix, but they do need to be such that mathematically, the matrix product is defined. The table below shows valid combinations. Combinations that do not match any row in the table result in an error. Table 1. Data Type Combinations Dimensions of First Operand Dimensions of Second Operand Dimensions of Matrix Product Conditions K x L L x M K x M K >= 1, L >= 1, M >= 1 K x L L K K >= 1, L > 1 K x 1 1 K x 1 K >= 1 K 1 K K >= 1 K 1 x M K x M K >= 1, M >= 1 The output data type is determined according to the following rules, in the order listed. T1 is a variable representing the type of the first operand; T2 is a variable representing the type of the second operand. These rules were chosen for maximum alignment with Vitis HLS, which may not correspond to the output data type computed via the internal rule of the Simulink® Matrix Product block. Table 2. Output Data Type Data Type of First Operand Data Type of Second Operand Data Type of Matrix Product T1: floating-point T2 The widest floating-point type between T1 and T2 if T2 is a floating-point type; otherwise T1 T1 T2: floating-point The widest floating-point type between T1 and T2 if T1 is a floating-point type; otherwise T2 fixed-point fixed-point The smallest fixed-point type capable of representing the product without loss of precision fixed-point integer The smallest fixed-point type capable of representing the product without loss of precision integer fixed-point The smallest fixed-point type capable of representing the product without loss of precision T1: integer T2: integer Let W1 be the bit width of T1 and W2 be the bit width of T2. The product is the integer type with bit width max (W1,W2) and it is signed if either T1 or T2 are signed. boolean T2 T2 T1 boolean T1 ## Parameters The Matrix Multiply block has no parameters to set.

{"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}

CC-MAIN-2024-10

latest

en

https://www.learncs.online/solve/java/caught-speeding/[email protected]?returnTo=functions

text/html

crawl-data/CC-MAIN-2024-30/segments/1720763518579.47/warc/CC-MAIN-20240725053529-20240725083529-00353.warc.gz

KotlinJava ## Solve: Caught Speeding Imported By: Geoffrey Challen / Version: 2021.6.0 You are driving a little too fast, and a police officer stops you. Write code to compute the result, encoded as an `int` value: 0=no ticket, 1=small ticket, 2=big ticket. If speed is 60 or less, the result is 0. If speed is between 61 and 80 inclusive, the result is 1. If speed is 81 or more, the result is 2. Unless it is your birthday! On that day, your speed can be 5 higher in all cases. int caughtSpeeding(int speed, boolean isBirthday) { return 0; // You may need to remove this starter code } ### Related Lessons Stuck? You may find these lessons helpful:

{"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}

CC-MAIN-2024-30

latest

en

https://officialsavanarae.com/dermatology/frequent-question-how-many-moles-are-in-a-ml.html

text/html

crawl-data/CC-MAIN-2022-05/segments/1642320306346.64/warc/CC-MAIN-20220128212503-20220129002503-00294.warc.gz

Frequent question: How many moles are in a mL? Contents What unit is mol mL? In chemistry, the most commonly used unit for molarity is the number of moles per liter, having the unit symbol mol/L or mol⋅dm3 in SI unit. A solution with a concentration of 1 mol/L is said to be 1 molar, commonly designated as 1 M. Is mol L the same as mL? mol/L↔umol/mL 1 mol/L = 1000 umol/mL. How many moles are in 1 mL of water? of moles of water in 1 ml = 1/18 mol. How do you find moles with molarity and mL? To calculate the number of moles in a solution given the molarity, we multiply the molarity by total volume of the solution in liters. Is mM MOL mL? How many Mole/Milliliter are in a Millimolar? The answer is one Millimolar is equal to 0.000001 Mole/Milliliter. How do you convert Millimolar to mg mL? So, to convert mg/mL to mM, I divided the concentration in mg/mL (20 mg/mL) by the molecular weight of the sample (232,278 g/mole) and multiplied by 1,000. How many moles are in a Nanomole? Nanomole is a unit of measurement for amount of substance. Nanomole is a decimal fraction of amount of substance unit mole. One nanomole is equal to 1e-9 moles. Does mL mean microliter? One megaliter (ML) equals one million liters. One microliter (µL, lowercase mu) equals one millionth of a liter. What is a 1 molar solution? Molarity is another standard expression of solution concentration. … A 1 molar (M) solution will contain 1.0 GMW of a substance dissolved in water to make 1 liter of final solution. Hence, a 1M solution of NaCl contains 58.44 g. Example: HCl is frequently used in enzyme histochemistry. How many liters are in a mole? As long as the gas is ideal, 1 mole = 22.4L.

{"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}

CC-MAIN-2022-05

latest

en

https://gmatclub.com/forum/gmat-score-710-q50-v37-planning-to-retake-help-needed-132734.html?fl=similar

text/html

crawl-data/CC-MAIN-2017-09/segments/1487501171053.19/warc/CC-MAIN-20170219104611-00040-ip-10-171-10-108.ec2.internal.warc.gz

GMAT Score: 710 (Q50/V37) - Planning to retake - help needed : General GMAT Questions and Strategies Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack It is currently 22 Feb 2017, 15:24 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # GMAT Score: 710 (Q50/V37) - Planning to retake - help needed Author Message Intern Joined: 08 Feb 2012 Posts: 5 Location: India GMAT 1: 710 Q50 V37 GMAT 2: 730 Q50 V39 GPA: 2.96 WE: Consulting (Consulting) Followers: 0 Kudos [?]: 5 [0], given: 0 GMAT Score: 710 (Q50/V37) - Planning to retake - help needed [#permalink] ### Show Tags 17 May 2012, 06:48 Hi Guys, I really need some help now! I gave my GMAT on 3rd May and scored a 710 (Q50/V35). however, I am not very happy with it and want to retake the exam in another 30 or 45 days. I feel pretty confident in the quant section, but very vulnerable in verbal. Kindly someone guide me with a study plan for next 30 days to improve my verbal score to over 40. In the mean time, I have started studying Manhattan GMAT SC book. I have created detailed notes for the entire book and am hoping to digest it completely in another week's time. What else do I need to do? What more do I do for SC? What about CR and RC? Would my joining a coaching center be helpful? The reason I am not too keen on that is because (1) I do not have too much time; (2) I dont think going through complete basics as they are taught in classes would be very helpful. Also pls suggest me on a good test series to join. I might be repeating the questions already discussed somewhere on GMATClub but if someone can provide me a consolidated view, it would be great! Manhattan GMAT Discount Codes Economist GMAT Tutor Discount Codes Magoosh Discount Codes GMAT Forum Moderator Status: Accepting donations for the mohater MBA debt repayment fund Joined: 05 Feb 2008 Posts: 1884 Location: United States Concentration: Operations, Finance Schools: Ross '14 (M) GMAT 1: 610 Q0 V0 GMAT 2: 710 Q48 V38 GPA: 3.54 WE: Accounting (Manufacturing) Followers: 59 Kudos [?]: 803 [0], given: 234 Re: GMAT Score: 710 (Q50/V37) - Planning to retake - help needed [#permalink] ### Show Tags 20 May 2012, 05:20 Re: GMAT Score: 710 (Q50/V37) - Planning to retake - help needed   [#permalink] 20 May 2012, 05:20 Similar topics Replies Last post Similar Topics: Low GMAT Score- Planning for retake: Need advice 4 06 Oct 2016, 06:31 GMAT : Scored a 720 (Q 50, V 37, IR 8) against a target score of 750+ 12 11 Sep 2016, 01:21 GMAT : Scored a 720 (Q 50, V 37, IR 8) against a target score of 750+ 0 11 Sep 2016, 01:16 2 Risk in retake after 720 Q50 V37 IR 4 2 28 Aug 2016, 19:53 I recently gave my Gmat and scored 720 (Q50 V37 IR6 AWA6) 2 02 Oct 2013, 01:03 Display posts from previous: Sort by # GMAT Score: 710 (Q50/V37) - Planning to retake - help needed Moderators: WaterFlowsUp, HiLine Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.

{"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}

CC-MAIN-2017-09

longest

en

http://swift60k.club/excel-formula-to-calculate-percentage-increase/excel-formula-to-calculate-percentage-increase-how-to-calculate-percentages-in-excel-example-formula-to-calculate-percentage-increase-in-excel-how-do-you-calculate-percentage-increase-between-two-numb/

text/html

crawl-data/CC-MAIN-2019-13/segments/1552912202924.93/warc/CC-MAIN-20190323181713-20190323203713-00536.warc.gz

# Excel Formula To Calculate Percentage Increase How To Calculate Percentages In Excel Example Formula To Calculate Percentage Increase In Excel How Do You Calculate Percentage Increase Between Two Numb excel formula to calculate percentage increase how to calculate percentages in excel example formula to calculate percentage increase in excel how do you calculate percentage increase between two numb. how to calculate percentage difference between two numbers in excel 2010 figure formula increase decrease times for image,excel calculate percentage increase between two numbers magic trick change formula chart how to determine annual,excel formula percentage increase between two numbers formulas for row to determine change how calculate difference in 2010,excel formula to determine percentage change for calculating difference between two numbers what is the in percent how calculate increase,how to calculate percentage growth between two numbers in excel increase formulas for percentages formula decrease times,excel how to determine percentage increase formula for calculating difference between two numbers calculate values percent in 2,excel formula to calculate percentage increase decrease between times annual how two numbers applying a percent or values in dummies,excel formula percentage increase between two numbers for calculating difference how to calculate in 2010 percent drag your across,how to calculate percentage difference between two numbers in excel 2010 do you increase determine formulas for spreadsheet,excel formula calculate percentage increase in how to a percent between two numbers for calculating difference.

{"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}

CC-MAIN-2019-13

latest

en

https://zbmath.org/1215.11014

text/html

crawl-data/CC-MAIN-2023-23/segments/1685224647409.17/warc/CC-MAIN-20230531182033-20230531212033-00515.warc.gz

## Integrals of products of Bernoulli polynomials.(English)Zbl 1215.11014 Summary: Extending known results on integrals of products of two or three Bernoulli polynomials with limits of integration 0 and 1, we obtain identities for such integrals with limits of integration 0 and $$x$$, for a variable $$x$$. Proposition 1. For all $$k,m\geq 0$$ we have $\int_0^x B_k(t)B_m(t)\,dt=\frac{k!m!}{(k+m+1)!} \sum_{j=0}^k (-1)^j\binom{k+m+1}{k-j} \left(B_{k-j}(x)B_{m+j+1}(x)-B_{k-j}B_{m+j+1}\right).$ As applications we obtain certain quadratic and cubic identities for Bernoulli polynomials. ### MSC: 11B68 Bernoulli and Euler numbers and polynomials Full Text: ### References: [1] Agoh, T.; Dilcher, K., Convolution identities and lacunary recurrences for Bernoulli numbers, J. number theory, 124, 105-122, (2007) · Zbl 1137.11014 [2] Agoh, T.; Dilcher, K., Reciprocity relations for Bernoulli numbers, Amer. math. monthly, 115, 237-244, (2008) · Zbl 1204.11048 [3] Carlitz, L., Note on the integral of the product of several Bernoulli polynomials, J. London math. soc., 34, 361-363, (1959) · Zbl 0086.05801 [4] Dilcher, K., Sums of products of Bernoulli numbers, J. number theory, 60, 23-41, (1996) · Zbl 0863.11011 [5] Espinosa, O.; Moll, V.H., The evaluation of Tornheim double sums. I, J. number theory, 116, 200-229, (2006) · Zbl 1168.11033 [6] Hansen, E.R., A table of series and products, (1975), Prentice-Hall, Inc. Englewood Cliffs, NJ · Zbl 0302.65039 [7] Mikolás, M., Integral formulae of arithmetical characteristics relating to the zeta-function of Hurwitz, Publ. math. debrecen, 5, 44-53, (1957) · Zbl 0081.27403 [8] Mordell, L.J., Integral formulae of arithmetical character, J. London math. soc., 33, 371-375, (1958) · Zbl 0081.27404 [9] Nielsen, N., Traité élémentaire des nombres de Bernoulli, (1923), Gauthier-Villars Paris · JFM 50.0170.04 [10] Nörlund, N.E., Vorlesungen über differenzenrechnung, (1924), Springer-Verlag Berlin · JFM 50.0315.02 [11] Wilson, J.C., On franel-kluyver integrals of order three, Acta arith., 66, 71-87, (1994) · Zbl 0807.11013 This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.

{"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}

CC-MAIN-2023-23

longest

en

http://xine.org/~jwest/documentation/alg/node159.html

text/html

crawl-data/CC-MAIN-2019-13/segments/1552912203021.14/warc/CC-MAIN-20190323201804-20190323223804-00322.warc.gz

Search Next: 0.8.4 Knapsack Problem Up: 0.8.3 Determining Whether a Previous: 0.8.3 Determining Whether a ### 0.8.3.1 Source Code ```#include <iostream.h> #include <math.h> #include <stdlib.h> // // Input parameters: // npol : number of vertices // xp[npol], yp[npol] : x,y coord of vertices // x,y : coord of point to test // // Return Value: // 0 : test point is outside polygon // 1 : test point is inside polygon // // Notes: // if test point is on the border, 0 or 1 is returned. // If there exists an adiacent polygon, the point is // _in_ only in one of the two. // int pnpoly(int npol, float *xp, float *yp, float x, float y) { int i, j, c = 0; for (i = 0, j = npol-1; i < npol; j = i++) { if ( ( ((yp[i]<=y) && (y<yp[j])) || ((yp[j]<=y) && (y<yp[i])) ) && (x < (xp[j] - xp[i]) * (y - yp[i]) / (yp[j] - yp[i]) + xp[i])) c = !c; } return c; } ``` Scott Gasch 1999-07-09

{"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}

CC-MAIN-2019-13

latest

en

https://www.gamedev.net/forums/topic/611070-leveling-increases-grinding-for-next-level/

text/html

crawl-data/CC-MAIN-2019-43/segments/1570987829507.97/warc/CC-MAIN-20191023071040-20191023094540-00220.warc.gz

# Leveling increases grinding for next level This topic is 2948 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic. ## Recommended Posts Lets suppose player A is lv 2304, in a normal game with linear formulas he would be able to kill enemies up to level 4000 ( after than he begins to do 0 dmg ). The problem is the following : 1) Player has to grind for 3 months non stop to get to level 4000 to beat that enemy. After that the game creates enemies higher level so that he doesn't feel "bored". So even if he goes to level 4000, his next enemy will be 8000 = 3 years grinding. This leads to a point where the player no longer levels up fast enough to reach his next "kill enemy objective" thus quits. ps. The next enemy it generates is strong enough to force the player to grind in order to be able to beat him. (You remember disgaia grinding ?) 2) The actual problem is the following : "the time to double your power" increases exponentially(doubles each time), which means that the player gets less satisfaction as time passes. You can notice this in all games, for example in wow how much time does it take to double your hp : lv1-2 : 5 min, lv2-6: 40 min, lv6-12: 6 hours, lv12-20: 10 hours. lv 20-30: 24 hours. lv 30-40: 2 days. lv 40 -50: 5 days. lv 50-60 : 10 days. This means that leveling in this system has "diminishing returns" which prevents the player for embracing the leveling up system. Solutions : 1) Exponential xp gain. Player levels up faster the higher level they are. He may get 1% of his total levels per enemy kill, so that after 100 enemies he doubles in power. This system is flawed as numbers have limited precision thus after 64 (double ups) all numbers will overflow and the game would crash. Also numbers are hard to read i deal 12312323123 damage out of your 976421367899 hp pool. 2) Level has hidden exponential stats, for example the enemies deals 10% extra dmg per level difference. Lv 2304 Player A will now die from lv 2334 enemy ( he deals 300% more dmg and receives only 30% dmg). Pros: 1) it works. It forces grinding for a specified amount of time to battle next enemy. Cons: 1) The status screen doesn't display big numbers to make the player "happy". Thus reduces his will to grind. 2) Hidden stats will confuse the player and make them ask why they died to an enemy with 10 hp higher than them. 3) Wow style : Xp requirements increase exponentially every time you level up (200% more time), however when you level up you get exponential stat bonuses. For instance In wow lv 1 gives 10 hp and lv 80 gives 20000 hp. Cons : 1) User has to grind too much to get a feedback of his overall stat increase. This will lead to a point where he has to grind for 3 years to reach next level. Any alternative ideas for making a "level grind" system ? ##### Share on other sites A level cap? The numbers just... hurt my head. Each level gained provides the same increase in power. Each level gained is slightly longer to accomplish than the previous. Leveling to the cap shouldn't take forever, but it should certainly take more than two days played. Don't rely on "levels" to determine the full capabilities of a character. ##### Share on other sites I think this is an unavoidable problem with leveling systems of any sort (and the same could probably be said about pretty much any stat progression system). Balancing it against time invested and power gained is going to feel wrong to someone, somewhere. I'm a fan of linear progression though. X time invested = Y power gained, no matter how much Y you already have. This helps ease the pain of exponential time needed to gain a level, but has its own issues. Of the games that use this system in which I'm familiar, the game developers introduced monster leveling to keep enemies at a similar power to the player, which I detest. Why on Earth is that wolf now level 45 when yesterday they were all level 25? I'm not really sure why this is necessary. I think the XP system of Final Fantasy VIII was close to what I'd aim for: it always took 1000 XP to level up, no matter what level you were. Killing enemies close to your level would give between 50 and 150 XP, depending on exactly what the enemy was. Killing weaker enemies gave less, but it would never take more than a couple hundred enemies to level up unless you were deliberately finding very weak things to kill (which was hard, as most enemies leveled with you). The problem with the game was that enemies leveled with the player, which as I said, I hate. Stat progression in FFVIII was very strange, and I wouldn't recommend using it in any other game. Leveling up didn't do much actually (which is why some players played through the game without doing so, it made it easier because enemies were weaker), and most of your stats came from binding magic to them. You could have max stats at level 1 if you tried, and enemies would be crushed under your attacks in an instant. It was tremendously abusable. Ok, rambling aside, how about something like this? If you have a straight level = power system, it takes a set amount of XP to level up. Killing things close to your level gives X XP, so grinding takes about the same amount of time no matter what level you are. Gaining a level gives the same benefit at level 30 as it does at 2, adding the same amount of stats. Enemies are balanced for your expected level when you encounter them. If you have an alternate state progression system (i.e. levels of strength, HP, whatever), then you do the same thing. You would encourage players to diversify their progression by keeping enemies from leveling with them, since eventually they will get diminishing returns on improving a stat. If you can kill all the enemies in an area in 1 hit, there's no point in improving damage any more. ##### Share on other sites your solution # 1 is basically like... saying each level takes 100 xp, and each kill nets you 1 xp. Why won't you do that instead, and make weaker enemies return no hp? that removes a lot of the complexity. Feels to me like you've had too much WoW in you ;) ##### Share on other sites Think about it, the reason most rpg players quit a game after finishing its story, is because they "maxed" their character and there isn't anything else to do. If you could continue leveling forever, with no caps, they will never get bored. However this will happen if the stats don't start to converge to a certain number. If it takes 6 months to double my overall power then chances is that the player will give up the level grind. So what we need is a controlled environment, where : 1) punishment : you cant kill target enemy if you arent level X. 2) motivation : Achieving level Y allows you to kill enemy X, however there is now enemy X2 to force to continue leveling. The motivation part must happen in a reasonable amount of time that is always constant, e.g 1 day to reach this amount of power. @Telgin Yes its good to level at a guaranteed pace, but since stats per level are linear and you get the same amount per level, it means that the time needed to double your overall power will increase as time passes. For example lets say i played for 8 months and i am level 5400, in 2 days i will go to level 5520. Total power increase 0.022% in 2 days. Now lets say i am level 54000, in 2 days i will go to level 54120. Total power increase 0.0022% Thus it cannot achieve the motivation aspect since the numbers "converge". @Orymus Basically what i meant with 1 is the following : My lv 2304 player will gain 40-50 levels per enemy kill so that he reaches double potential at the set amount of expected time. Ideas : 1) Maybe player chooses enemy level at stage start ? ##### Share on other sites Think about it, the reason most rpg players quit a game after finishing its story, is because they "maxed" their character and there isn't anything else to do. If you could continue leveling forever, with no caps, they will never get bored. However this will happen if the stats don't start to converge to a certain number. If it takes 6 months to double my overall power then chances is that the player will give up the level grind. So what we need is a controlled environment, where : 1) punishment : you cant kill target enemy if you arent level X. 2) motivation : Achieving level Y allows you to kill enemy X, however there is now enemy X2 to force to continue leveling. The motivation part must happen in a reasonable amount of time that is always constant, e.g 1 day to reach this amount of power. @Telgin Yes its good to level at a guaranteed pace, but since stats per level are linear and you get the same amount per level, it means that the time needed to double your overall power will increase as time passes. For example lets say i played for 8 months and i am level 5400, in 2 days i will go to level 5520. Total power increase 0.022% in 2 days. Now lets say i am level 54000, in 2 days i will go to level 54120. Total power increase 0.0022% Thus it cannot achieve the motivation aspect since the numbers "converge". @Orymus Basically what i meant with 1 is the following : My lv 2304 player will gain 40-50 levels per enemy kill so that he reaches double potential at the set amount of expected time. Ideas : 1) Maybe player chooses enemy level at stage start ? I saw what I put in bold and your thought process is extremely flawed. Leveling forever =/= unlimited fun. What game should EVER have 54,000 levels? Not even arcade games go that far. ##### Share on other sites Well a few things here. For finite RPG's (A game that's meant to end, ala Final Fantasy sans 11), there's motivation beyond grinding to the next level to play. Typically there's some sort of story, or perhaps some sort of engaging mechanic that keeps the player interested in the play itself. For MMORPG's, particularly EQ derived MMO's, the character progression is the prime motivation, or at least an inseparable aspect of motivation for play. If you want to competitively do arenas, well guess what you need to grind. Raids? Grinding. Battlegrounds? Hope you like grinding. Yeah, you can screw around in PvP on WoW for a while without high level gear and what not, but to truly be competitive there's a grinding aspect involved. I think the issue here stems from gameplay itself in a lot of RPG's being fairly shallow or repetitive. Players do not particularly enjoy the act of playing the game, thus grinding is introduced to give a concrete goal that players can latch onto. MMO's in particular are often guilty of this as a result of their technical limitations, not always lazy design. While we'd all love the MMOFPS with 20k people on the same server providing a truly massive experience and skill based gameplay, that's not easy to execute. Anyway, if you want grinding to be less of a pain, then you can do a couple things: 1) Rest Experience: I love this mechanic. It gives people who can only play for a little bit of time every day some extra incentive and reward without directly punishing people who can afford to play constantly. 2) Forgiving Death: This may come as a shock, but players hate dying all their own. Even if you don't put in any penalty for death, players will still avoid it due to the psychological aversion humans have to failure. Making heavy death penalties, especially those that affect your experience, are a big no no in my mind. It's just an underhanded way to force even more grinding. 3) Offloading progression into other character aspects: Ok, so maybe it does take you a while to level up. How about giving your players something else in the meantime? Look at the elements that makeup your character and see if there's a way you can bust some of those out of the leveling system so they can be dolled out an independent schedule. Loot and equipment is a good example of this. You could have bonus talents/feats that occur off-level, or skills that you purchase with some sort of currency, etc. 4) Flattening the Experience Curve: I don't like the FFVIII system, or really anything from FFVIII, so I wouldn't go with a totally flat leveling curve, but one that doesn't have an insane increase can be more accessible to players. Really, the only number that matters in any leveling system is TTNL (time to next level), and if you make sure this doesn't wildly grow out of control you'll make the grind less intense for the player. Of course, my philosophy on grinding in general is to make grinding a tertiary incentive to playing the game. The player's primary motivation should not be to grind, it should be to play the game. I never had to grind in Call of Duty for unlocks because the act of playing the game was enjoyable in and of itself. Leveling up and getting a new piece of equipment was just gravy on an already fun experience. ##### Share on other sites (I may have got the wrong end of the stick so sorry in advance if I did) I would never say that "most" players quit because they hit the level cap. Instead they quite because there is nothing else for them to do. Now depending on the genre those two things go hand in hand but my point is that never having a player hit the level cap doesn't mean they won't quite. If there's nothing for them to do but keep killing the same enemies over and over then i can't see a reason for the vast majority to keep playing. I don't really see the problem you put forward. Surely the reason for such long "grinds" would be to keep the player playing said game for as long as possible, changing this formula may let the player kill said monster earlier but then what's the point of having such a system in the first place. You can simply put rewards closer together. If you think about what the levelling system represents, the "WoW" system as you called it, makes perfect sense to me. It doesn't take long for me to become proficient at the process of stabbing someone with a knife but it takes a lot longer for me to master that process. I would also like to point out that the stats a person gains each level are not as important to the player as say the new things they can do/wear. If your intent is to go for the classic levelling system, i.e. kill mob = exp, enough exp and gain a level which means you become more powerful, I don't think you can really improve on the basic mechanics. Instead its up to you to balance time/reward as best as possible and thus make it worth the players while to keep playing the game. Choosing an enemy's level at the start of a stage is essentially changing the difficulty of the game (which most games currently do anyway). If it's not for the challenge but for maximum exp gain with the least hassle then you may as well choose the level for the player since they will most likely go for the most optimal. (I could go on about theories as to which method is best depending on the situation but i think that would end up being dangerously off topic.) ##### Share on other sites Well I found that oversimplification made sense. Rather than level, have the amount of experience points directly matter in your equations. Damage = (insert math formula including experience as a variable). etc. Basically, your leveling is turned into millions of levels instead, and each matters. ##### Share on other sites Well I found that oversimplification made sense. Rather than level, have the amount of experience points directly matter in your equations. Damage = (insert math formula including experience as a variable). etc. Basically, your leveling is turned into millions of levels instead, and each matters. Not a bad approach, but very fine incremental gains can feel unrewarding to the player. It's the whole frog in a boiling pot thing, it's often more gratifying to players to receive +20 hp after killing 20 monsters than it is to receive 1 HP per monster they kill. You're spending the same amount of time to get the same reward, but its presentation matters. Not to say an incremental power progression can't work, it totally can, but you have to make sure the player actually feels like they're getting a reward for their effort, which is easier to accomplish with big leves. • ### Game Developer Survey We are looking for qualified game developers to participate in a 10-minute online survey. Qualified participants will be offered a \$15 incentive for your time and insights. Click here to start! • 15 • 12 • 9 • 11 • 15

{"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}

CC-MAIN-2019-43

latest

en

http://www.math.utah.edu/software/lapack/lapack-s/sorm2l.html

text/html

crawl-data/CC-MAIN-2018-34/segments/1534221212040.27/warc/CC-MAIN-20180817104238-20180817124238-00377.warc.gz

Previous: sorgtr Up: ../lapack-s.html Next: sorm2r # sorm2l ``` NAME SORM2L - overwrite the general real m by n matrix C with Q * C if SIDE = 'L' and TRANS = 'N', or Q'* C if SIDE = 'L' and TRANS = 'T', or C * Q if SIDE = 'R' and TRANS = 'N', or C * Q' if SIDE = 'R' and TRANS = 'T', SYNOPSIS SUBROUTINE SORM2L( SIDE, TRANS, M, N, K, A, LDA, TAU, C, LDC, WORK, INFO ) CHARACTER SIDE, TRANS INTEGER INFO, K, LDA, LDC, M, N REAL A( LDA, * ), C( LDC, * ), TAU( * ), WORK( * ) PURPOSE SORM2L overwrites the general real m by n matrix C with where Q is a real orthogonal matrix defined as the product of k elementary reflectors Q = H(k) . . . H(2) H(1) as returned by SGEQLF. Q is of order m if SIDE = 'L' and of order n if SIDE = 'R'. ARGUMENTS SIDE (input) CHARACTER*1 = 'L': apply Q or Q' from the Left = 'R': apply Q or Q' from the Right TRANS (input) CHARACTER*1 = 'N': apply Q (No transpose) = 'T': apply Q' (Transpose) M (input) INTEGER The number of rows of the matrix C. M >= 0. N (input) INTEGER The number of columns of the matrix C. N >= 0. K (input) INTEGER The number of elementary reflectors whose product defines the matrix Q. If SIDE = 'L', M >= K >= 0; if SIDE = 'R', N >= K >= 0. A (input) REAL array, dimension (LDA,K) The i-th column must contain the vector which defines the elementary reflector H(i), for i = 1,2,...,k, as returned by SGEQLF in the last k columns of its array argument A. A is modified by the routine but restored on exit. LDA (input) INTEGER The leading dimension of the array A. If SIDE = 'L', LDA >= max(1,M); if SIDE = 'R', LDA >= max(1,N). TAU (input) REAL array, dimension (K) TAU(i) must contain the scalar factor of the elemen- tary reflector H(i), as returned by SGEQLF. C (input/output) REAL array, dimension (LDC,N) On entry, the m by n matrix C. On exit, C is overwritten by Q*C or Q'*C or C*Q' or C*Q. LDC (input) INTEGER The leading dimension of the array C. LDC >= max(1,M). WORK (workspace) REAL array, dimension (N) if SIDE = 'L', (M) if SIDE = 'R' INFO (output) INTEGER = 0: successful exit < 0: if INFO = -i, the i-th argument had an illegal value ```

{"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}

CC-MAIN-2018-34

latest

en

https://cacm.acm.org/magazines/2014/9/177933-optimality-in-robot-motion/fulltext

text/html

crawl-data/CC-MAIN-2022-33/segments/1659882572215.27/warc/CC-MAIN-20220815235954-20220816025954-00374.warc.gz

# Communications of the ACM Review articles ## Optimality in Robot Motion: Optimal Versus Optimized Motion The first book dedicated to robot motion was published in 1982 with the subtitle "Planning and Control."5 The distinction between motion planning and motion control has mainly historical roots. Sometimes motion planning refers to geometric path planning, sometimes it refers to open loop control; sometimes motion control refers to open loop control, sometimes it refers to close loop control and stabilization; sometimes planning is considered as an offline process whereas control is real time. From a historical perspective, robot motion planning arose from the ambition to provide robots with motion autonomy: the domain was born in the computer science and artificial intelligence communities.22 Motion planning is about deciding on the existence of a motion to reach a given goal and computing one if this one exists. Robot motion control arose from manufacturing and the control of manipulators30 with rapid effective applications in the automotive industry. Motion control aims at transforming a task defined in the robot workspace into a set of control functions defined in the robot motor space: a typical instance of the problem is to find a way for the end-effector of a welding robot to follow a predefined welding line. ### Key Insights What kind of optimality is about in robot motion? Many facets of the question are treated independently in different communities ranging from control and computer science, to numerical analysis and differential geometry, with a large and diverse corpus of methods including, for example, the maximum principle, the applications of Hamilton-Jacobi-Bellman equation, quadratic programming, neural networks, simulated annealing, genetic algorithms, or Bayesian inference. The ultimate goal of these methods is to compute a so-called optimal solution whatever the problem is. The objective of this article is not to overview this entire corpus that follows its own routes independently from robotics, but rather to emphasize the distinction between "optimal motion" and "optimized motion." Most of the time, robot algorithms aiming at computing an optimal motion provide in fact an optimized motion that is not optimal at all, but is the output of a given optimization method. Computing an optimal motion is mostly a challenging issue as it can be illustrated by more than 20 years of research on wheeled mobile robots (as we discuss later). Note that the notion of optimality in robot motion as it is addressed in this article is far from covering all the dimensions of robot motion.7 It does not account for low-level dynamical control, nor for sensory-motor control, nor for high level cognitive approaches to motion generation (for example, as developed in the context of robot soccer or in task planning). ### What Is Optimal in Robot Motion Planning and Control? Motion planning explores the computational foundations of robot motion by facing the question of the existence of admissible motions for robots moving in an environment populated with obstacles: how to transform the continuous problem into a combinatorial one? This research topic22,26 evolved in three main stages. In the early 1980s, Lozano-Perez first transformed the problem of moving bodies in the physical space into a problem of moving a point in some so-called configuration space.28 In doing so, he initiated a well-defined mathematical problem: planning a robot motion is equivalent to searching for connected components in a configuration space. Schwartz and Sharir then showed the problem is decidable as soon as we can prove that the connected components are semi-algebraic sets.35 Even if a series of papers from computational geometry explored various instances of the problem, the general "piano mover" problem remains intractable.14 Finally by relaxing the completeness exigence for the benefit of probabilistic completeness, Barraquand and Latombe introduced a new algorithmic paradigm3 in the early 1990s that gave rise to the popular probabilistic roadmap20 and rapid random trees31 algorithms. Motion planning solves a point-to-point problem in the configuration space. Whereas the problem is a difficult computational challenge that is well understood, optimal motion planning is a much more difficult challenge. In addition to finding a solution to the planning problem (that is, a path that accounts for collision-avoidance and kinematic constraints if any), optimal motion planning refers to finding a solution that optimizes some criterion. These can be the length, the time or the energy (which are equivalent criteria under some assumption), or more sophisticated ones, as the number of maneuvers to park a car. In such a context many issues are concerned with optimization: • For a given system, what are the motions optimizing some criteria? Do such motions exist? The existence of optimal motion may depend either on the presence of obstacles or on the criterion to be optimized. • When optimal motions exist, are they computable? If so, how complex is their computation? How to relax exactness constraints to compute approximated solutions? We will address the combinatorial structure of the configuration space induced by the presence of obstacles and by the metric to be optimized. Time criterion is also discussed, as are practical approaches to optimize time along a predefined path. • Apart from finding a feasible solution to a given problem, motion planning also wants to optimize this solution once it has been found. The question is particularly critical for the motions provided by probabilistic algorithms that introduce random detours. The challenge here is to optimize no more in the configuration space of the system, but in the motion space. In this article, optimal motion planning is understood with the underlying hypothesis that the entire robot environment is known and the optimization criterion is given: the quest is to find a global optimum without considering any practical issue such as model uncertainties or local sensory feedback. ### Optimal Motion Existence Before trying to compute an optimal motion, the first question to ask is about its existence. To give some intuition about the importance of this issue, consider a mobile robot moving among obstacles. For some rightful security reason, the robot cannot touch the obstacles. In mathematical language, the robot must move in an open domain of the configuration space. Yet, an optimal motion to go from one place to another one located behind some obstacle will necessarily touch the obstacle. So this optimal motion is not a valid one. It appears as an ideal motion that cannot be reached. The best we can do is to get a collision-free motion whose length approaches the length of this ideal shortest (but non-admissible) motion. In other words, there is no optimal solution to the corresponding motion planning problem. The question here is of topological nature: combinatorial data structures (for example, visibility graphs) may allow us to compute solutions that are optimal in the closure of the free space, and that are not solutions at all in the open free space. Even without obstacle, the existence of an optimal motion is far from being guaranteed. In deterministic continuous-time optimal control problems we usually search for a time-dependent control function that optimizes some integral functional over some time interval. Addressing the issue of existence requires us to resort to geometric control theory;18 for instance, Fillipov's theorem proves the existence of minimum-time trajectories,a whereas Prontryagin Maximum Principle (PMP) or Boltyanskii's conditions give respectively necessary and sufficient conditions for a trajectory to be optimal. However it is usually difficult to extract useful information from these tools. If PMP may help to characterize optimal trajectories locally, it generally fails to give their global structure. Later, we show how subtle the question may be in various instances of wheeled mobile robots. The class of optimal control problems for which the existence of an optimal solution is guaranteed, is limited. The minimum time problems for controllable linear systems with bounded controls belong to this class: optimal solutions exist and optimal controls are of bang-bang type. However, the so-called Fuller problem may arise: it makes the optimal solution not practical at all as it is of bang-bang type with infinitely many switches. Other examples include the famous linear-quadratic-Gaussian problem (the cost is quadratic and the dynamics is linear in both control and state variables), and systems with a bounded input and with a dynamics that is affine in the control variables. In the former a closed loop optimal solution can be computed by solving algebraic Riccati equations, whereas in the latter the existence of an optimal control trajectory is guaranteed under some appropriate assumptions. Even without an obstacle, the existence of an optimal motion is far from being guaranteed. In more general cases, we can only hope to approximate as closely as desired the optimal value via a sequence of control trajectories. There is indeed no optimal solution in the too restricted space of considered control functions. This has already been realized since the 1960s. The limit of such a sequence can be given a precise meaning as soon as we enlarge the space of functions under consideration. For instance, in the class of problems in which the control is affine and the integral functional is the L1-norm, the optimal control is a finite series of impulses and not a function of time (for example, see Neustadt29). In some problems such as the control of satellites, such a solution makes sense as it can approximately be implemented by gas jets. However, in general, it cannot be implemented because of the physical limitations of the actuators. Changing the mathematical formulation of the problem (for example, considering a larger space of control candidates) may allow the existence of an optimal solution. In the former case of satellite control, the initial formulation is coherent as an "ideal" impulse solution can be practically approximated by gas jets. However, in other cases the initial problem formulation may be incorrect as an ideal impulse solution is not implementable. Indeed, if we "feel" a smooth optimal solution should exist in the initial function space considered and if in fact it does not exist, then either the dynamics and/or the constraints do not reflect appropriately the physical limitations of the system or the cost functional is not appropriate to guarantee the existence of an optimal solution in that function space. To the best of our knowledge, this issue is rarely discussed in textbooks or courses in optimal control. ### Optimal Path Planning Considering a motion is a continuous function of time in the configuration (or working) space, the image of a motion is a path in that space. The "piano mover" problem refers to the path-planning problem, that is, the geometric instance of robot motion planning (see Figure 1). The constraint of obstacle avoidance is taken into account. In that context, optimality deals with the length of the path without considering time and control. The issue is to find the shortest path between two points. Depending on the metric that equips the configuration space, a shortest path may be unique (for example, for the Euclidean metric) or not unique (for example, for the Manhattan metric). All configuration space metrics are equivalent from a topological point of view (that is, if there is a sequence of Euclidean paths linking two points, then there is also a sequence of Manhattan paths linking these two points). However, different metrics induce different combinatorial properties in the configuration space. For instance, for a same obstacle arrangement, two points may be linked by a Manhattan collision-free path, while they cannot by a collision-free straight-line segment: both points are mutually visible in a Manhattan metric, while they are not in the Euclidean one. So, according to a given metric, there may or may not exist a finite number of points that "watch" the entire space.25 These combinatorial issues are particularly critical to devise sampling-based motion planning algorithms. Now, consider the usual case of a configuration space equipped with a Euclidean metric. Exploring visibility graph data structures easily solves the problem of finding a bound on the length of the shortest path among polygonal obstacles. This is nice, but it is no longer true if we consider three-dimensional spaces populated with polyhedral obstacles. Indeed, finding the shortest path in that case becomes a NP-hard problem.14 So, in general, there is no hope to get an algorithm that computes an optimal path in presence of obstacles, even if the problem of computing an optimal path in the absence of obstacle is solved and even if we allow the piano-robot to touch the obstacles. As a consequence of such poor results, optimal path planning is usually addressed by means of numerical techniques. Among the most popular ones are the discrete search algorithms operating on bitmap representations of work or configuration spaces.3 The outputs we only obtain are approximately optimal paths, that is, paths that are "not so far" from a hypothetical (or ideal) estimated optimal path. Another type of methods consists in modeling the obstacles by repulsive potential. In doing so, the goal is expressed by an attractive potential, and the system tends to reach it by following a gradient descent.21 The solution is only locally optimal. Moreover, the method may get stuck in a local minimum without finding a solution, or that a solution actually exists or not. So it is not complete. Some extensions may be considered. For instance, exploring harmonic potential fields8 or devising clever navigation functions34 allow providing globally optimal solutions; unfortunately, these methods require an explicit representation of obstacles in the configuration space, which is information that is generally not available. At this stage, we can see how the presence of obstacles makes optimal path planning a difficult problem. ### Optimal Motion Planning In addition to obstacle avoidance, constraints on robot controls or robot dynamics add another level of difficulties. The goal here is to compute a minimal-time motion that goes from a starting state (configuration and velocity) to a target state while avoiding obstacles and respecting constraints on velocities and acceleration. This is the so-called kinodynamic motion planning problem.12 The seminal algorithm is based on discretizations of both the state space and the workspace. It gave rise to many variants including nonuniform discretization, randomized techniques, and extensions of A* algorithms (see LaValle26). Today, they are the best algorithms to compute approximately optimal motions. Less popular in the robot motion planning community are numerical approaches to optimal robot control.11 Numerical methods to solve optimal control problems fall into three main classes. Dynamic programming implements the Bellman optimality principle saying that any sub-motion of an optimal motion is optimal. This leads to a partial differential equation (the so-called Hamilton-Jacobi-Bellman equation in continuous time) whose solutions may sometimes be computed numerically. However, dynamic programming suffers from the well-known curse of the dimensionality bottleneck. Direct methods constitute a second class. They discretize in time both control and state trajectories so that the initial optimal control problem becomes a standard static non-linear programming (optimization) problem of potentially large size, for which a large variety of methods can be applied. However, local optimality is generally the best one can hope for. Moreover, potential chattering effects may appear hidden in the obtained optimal solution when there is no optimal solution in the initial function space. Finally, in the third category are indirect methods based on optimality conditions provided by the PMP and for which, ultimately, the resulting two-point boundary value problem to solve (for example, by shooting techniques) may be extremely difficult. In addition, the presence of singular arcs requires specialized treatments. So direct methods are usually simpler than indirect ones even though the resulting problems to solve may be very large. Indeed, their structural inherent sparsity can be taken into account efficiently. At this stage, we can conclude that exact solutions for optimal motion planning remain out of reach. Only numerical approximate solutions are conceivable. ### Optimal Motion Planning Along a Path A pragmatic way to bypass (not overcome) the intrinsic complexity of the kinodynamic and numerical approaches is to introduce a decoupled approach that solves the problem in two stages: first, an (optimal) path planning generates a collision-free-path; then a time-optimal trajectory along the path is computed while taking into account robot dynamics and control constraints. The resulting trajectory is of course not time-optimal in a global sense; it is just the best trajectory for the predefined path. From a computational point of view, the problem is much simpler than the original global one because the search space (named phase plane) is reduced to two dimensions: the curvilinear abscissa along the path and its time-derivative. Many methods have been developed since the introduction of dynamic programming approaches by Shin and McKay36 in configuration space and simultaneously by Bobrow et al.4 in the Cartesian space. Many variants have been considered including the improvement by Pfeiffer and Johanni31 that combines forward and backward integrations, and the recent work by Verscheure et al.39 who transform the problem into a convex optimization one. ### Optimization in Motion Space Extensions of path-tracking methods may be considered as soon as we allow the deformations of the supporting paths. Here, we assume some motion planner provides a first path (or trajectory). Depending on the motion planner, the path may be far from being optimal. For instance, probabilistic motion planners introduce many useless detours. This is the price to pay for their effectiveness. So, the initial path must be reshaped, that is, optimized with respect to certain criteria. Geometric paths require to be shortened according to a given metric. The simplest technique consists in picking pairs of points on the path and linking them by a shortest path: if the shortest path is collision-free, it replaces the corresponding portion of the initial path. Doing so iteratively, the path becomes shorter and shorter. The iterative process stops as soon as it does not significantly improve the quality of the path. The technique gives good results in practice. Beside this simple technique, several variational methods operating in the trajectory space have been introduced. Among the very first ones, Barraquand and Ferbach2 propose to replace a constrained problem by a convergent series of less constrained subproblems increasingly penalizing motions that do not satisfy the constraints. Each sub-problem is then solved using a standard motion planner. This principle has been successfully extended recently to humanoid robot motion planning.9 Another method introduced by Quinlan and Khatib consists in modeling the motion as a mass-spring system.32 The motion then appears as an elastic band that is reshaped according to the application of an energy function optimizer. The method applies for nonholonomic systems as soon as the nonholonomic metric is known16 as well as for real-time obstacle avoidance in dynamic environments.6 Recently, successful improvements have been introduced by following the same basic principle of optimizing an initial guess in motion space. Zucker et al. take advantage of a simple functional expressing a combination of smoothness and clearance to obstacles to apply gradient descent in the trajectory space.40 A key point of the method is to model a trajectory as a geometric object, invariant to parametrization. In the same framework, Kalakrishman et al. propose to replace the gradient descent with a derivative-free stochastic optimization technique allowing us to consider non-smooth costs.19 ### What We Know and What We Do Not Know About Optimal Motion for Wheeled Mobile Robots Mobile robots constitute a unique class of systems for which the question of optimal motion is best understood. Since the seminal work by Dubins in the 1950s,13 optimal motion planning and control for mobile robots has attracted a lot of interest. We briefly review how some challenging optimal control problems have been solved and which problems still remain open. Let us consider four control models of mobile robots based on the model of a car (Figure 2). Two of them are simplified models of a car: the so-called Dubins (Figure 3) and Reeds-Shepp (Figure 4) cars respectively. The Dubins car moves only forward. The Reeds-Shepp car can move forward and backward. Both of them have a constant velocity of unitary absolute value. Such models account for a lower bound on the turning radius, that is, the typical constraint of a car. Such a constraint does not exist for a two-wheel differentially driven mobile robot. This robot may turn on the spot while a car cannot. Let us consider two simple control schemes of a two-driving wheel mobile robot:b in the first one (Hilare-1), the controls are the linear velocities of the wheels; in the second one (Hilare-2), the controls are the accelerations (that is, the second system is a dynamic extension of the first). Time-optimal trajectories. The car-like robots of figures 3 and 4 represent two examples of non-linear systems for which we know exactly the structure of the optimal trajectories. Note that in both examples the norm of the linear velocity is assumed to be constant. In those cases, time-optimal trajectories are supported by the corresponding shortest paths. Dubins solved the problem for the car moving only forward.13 More than 30 years later, Reeds and Shepp33 solved the problem for the car moving both forward and backward. The problem has been completely revisited with the modern perspective of geometric techniques in optimal control theory:37,38 the application of PMP shows that optimal trajectories are made of arcs of a circle of minimum turning radius (bang-bang controls) and of straight-line segments (singular trajectories). The complete structure is then derived from geometric arguments that characterize the switching functions. The Dubins and Reeds-Shepp cars are among the few examples of nonlinear systems for which optimal control is fully understood. The same methodology applies for velocity-based controlled differential drive vehicles (Hilare-1 in Figure 6). In that case, optimal trajectories are bang-bang, that is, made of pure rotations and straight-line segments. The switching functions are also fully characterized.1 This is not the case for the dynamic extension of the system, that is, for acceleration-based controlled differential drive vehicles (Hilare-2 as shown in Figure 7). Only partial results are known: optimal trajectories are bang-bang (that is, no singular trajectory appears) and are made of arcs of clothoid and involutes of a circle.15 However, the switching functions are unknown. The synthesis of optimal control for the Hilare-2 system the remain an open problem. While the existence of optimal trajectories is proven for the four systems shown here, a last result is worth mentioning. If we consider the Reeds-Shepp car optimal control problem in presence of an obstacle, even if we allow the car touching the obstacles, it has been proven that a shortest path may not exist.10 Motion planning. These results are very useful for motion planning in the presence of obstacles. In figures 3 and 4 we display the reachable domain for both the Dubins and Reeds-Shepp cars. While the reachable set of the Reeds-Shepp car is a neighborhood of the origin, it is not the case for the Dubins car. Stated differently, the Reeds-Shepp car is small-time controllable, while the Dubins car is only controllable. The consequence in terms of motion planning is important. In the case of the Reeds-Shepp car, any collision-freenot necessarily feasiblepath can be approximated by a sequence of collision-free feasible paths. Optimal paths allow building the approximation, giving rise to an efficient motion-planning algorithm.24 Not only does such an algorithm not apply for the Dubins car, we still do not know whether the motion-planning problem for Dubins car is decidable or not. In Laumond et al.,24 we prove the number of maneuvers to park a car varies as the inverse of the square of the clearance. This result is a direct consequence of the shape of the reachable sets. So, the combinatorial complexity of (nonholonomic) motion planning problems is strongly related to optimal control and the shape of the reachable sets in the underlying (sub-Riemannian) geometry.17 ### Conclusion When optimal solutions cannot be obtained for theoretical reasons (for example, nonexistence) or for practical ones (for example, untractability), we have seen how the problem can be reformulated either by considering a discrete representation of space and/or time, or by slightly changing the optimization criterion, or by resorting to numerical optimization algorithms. In all these cases, the resulting solutions are only approximated solutions of the original problem. In conclusion, it appears the existence of optimal robot motions is rarely guaranteed. When it is, finding a solution has never been proven to be a decidable problem as is the motion-planning problem. So, "optimal motion" is most often an expression that should be understood as "optimized motion," that is, the output of an optimization numerical algorithm. However, motion optimization techniques follow progress in numerical optimization with effective practical results on real robotic platforms, if not with new theoretical results. The distinction between optimal and optimized motions as it is addressed in this article does not cover all facets of optimality in robot motion. In a companion article,23 we consider the issue of motion optimal as an action selection principle and we discuss its links with machine learning and recent approaches to inverse optimal control. ### Acknowledgments The article benefits from comments by Quang Cuong Pham, from a careful reading by Joel Chavas, and above all, from the quality of the reviews. The work has been partly supported by ERC Grant 340050 Actanthrope, by a grant of the Gaspar Monge Program for Optimization and Operations Research of the Fédération Mathématique Jacques Hadamard (FMJH) and by the grant ANR 13-CORD-002-01 Entracte. ### References 1. Balkcom, D. and Mason, M. Time optimal trajectories for differential drive vehicles. Int. J. Robotics Research 21, 3 (2002), 199217. 2. Barraquand, J. and Ferbach, P. A method of progressive constraints for manipulation planning. IEEE Trans. on Robotics and Automation 13, 4 (1997) 473485. 3. Barraquand, J. and Lacombe, J.-C. Robot motion planning: A distributed representation approach. Intern. J. of Robotics Research 10, 6 (1991), 628649. 4. Bobrow, J., Dubowsky, S. and Gibson, J. Time-optimal control of robotic manipulators along specified paths. Int. J. of Robotics Research 4, 3 (1985), 317 5. Brady, M., Hollerbach, J., Johnson, T., Lozano-Pérez, T. and Masson, M.T. Robot Motion: Planning and Control. MIT Press, 1983. 6. Brock, O. and Khatib, O. Elastic strips: A framework for motion generation in human environments. Int. J. of Robotics Research 21, 12 (2002), 10311052. 7. Choset, H., Lynch, K.M., Hutchinson, S., Kantor, A., Burgard, W., Kavraki, L.E. and Thrun, S. Principles of Robot Motion: Theory, Algorithms, and Implementations. MIT Press, Cambridge, MA, June 2005. 8. Connolly, C. and Grupen, R. Applications of harmonic functions to robotics. J. of Robotic Systems 10, 7 (1992), 931946. 9. Dalibard, S., Nakhaei, A., Lamiraux, F. and Laumond, J.-P. Whole-body task planning for a humanoid robot: A way to integrate collision avoidance. In Proceedings of IEEE-RAS Int. Conference on Humanoid Robots, 2009. 10. Desaulniers, G. On shortest paths for a car-like robot maneuvering around obstacles. Robotics and Autonomous Systems 17 (1996), 139148. 11. Diehl, M. and Mombaur, K., eds. Fast Motions in Biomechanics and Robotics. Lecture Notes in Control and Information Sciences, vol 340. Springer, 2006. 12. Donald, B., Xavier, P., Canny, J. and Reif, J. Kinodynamic motion planning. JACM 40, 5 (1993), 10481066. 13. Dubins, L. On curves of minimal length with a constraint on average curvature and with prescribed initial and terminal positions and tangents. Amer. J. of Mathematics 79 (1957), 497516. 14. Hopcroft, J. Schwartz, J. and Sharir, M. Planning, Geometry, and Complexity of Robot Motion. Ablex, 1987. 15. Jacobs, P., Laumond, J.-P. and Rege, A. Nonholonomic motion planning for HILARE-like mobile robots. Intelligent Robotics. M. Vidyasagar and M. Trivedi, eds. McGraw Hill, 1991. 16. Jaouni, H., Khatib, M. and Laumond, J.-P. Elastic bands for nonholonomic car-like robots: algorithms and combinatorial issues. In Robotics: The Algorithmic Perspective. P. Agarwal, L. Kavraki, and M. Mason, eds. A.K. Peters, 1998. 17. Jean, F. Complexity of nonholonomic motion planning. Int. J. of Control 74, 8 (2001), 776782. 18. Jurdjevic, V. Geometric Control Theory. Cambridge University Press, 1996. 19. Kalakrishnan, M., Chitta, S., Theodorou, E., Pastor, P. and Schaal, S. Stomp: Stochastic trajectory optimization for motion planning. In Proceedings of the IEEE Int. Conf. on Robotics and Automation (2011). 20. Kavraki, L., Svestka, P., Latombe, J.-C. and Overmars, M. Probabilistic roadmaps for path planning in high-dimensional configuration spaces. IEEE Trans. on Robotics and Automation 12, 4 (1996), 566580. 21. Khatib, O. Real-time obstacle avoidance for manipulators and mobile robots. The Int. J. of Robotics Research 5, 1 (1986), 9098, 1986. 22. Latombe, J.-C. Robot Motion Planning. Kluwer Academic Press, 1991. 23. Laumond, J., Mansard, N. and Lasserre, J. Robot motion optimization as action selection principle. Commun. ACM, (to appear). 24. Laumond, J.-P., Jacobs, P., Taix, M. and Murray, R. A motion planner for nonholonomic mobile robots. IEEE Trans. on Robotics and Automation, 10, 5 (1994), 577593. 25. Laumond, J.-P. and Siméon, T. Notes on visibility roadmaps and path planning. New Directions in Algorithmic and Computational Robotics. B. Donald, K. Lynch, and D. Rus, eds. A.K. Peters, 2001. 26. LaValle, S. Planning Algorithms. Cambridge University Press, 2006. 27. LaValle, S. and Kuffner, J. Rapidly-exploring random trees: Progress and prospects. Algorithmic and Computational Robotics: New Directions. B. Donald, K. Lynch and D. Rus, eds. A.K. Peters, 2001, 293308. 28. Lozano-Pérez, T. Spatial planning: A configuration space approach. IEEE Trans. on Computer 32, 2 (1983), 108120. 29. Neustadt, L. Optimization, a moment problem and non linear programming. SIAM J. Control (1964), 3353. 30. Paul, R. Robot Manipulators: Mathematics, Programming, and Control. MIT Press, Cambridge, MA, 1st edition, 1982. 31. Pfeiffer, F. and Johanni, R. A concept for manipulator trajectory planning. IEEE Journal of Robotics and Automation 3, 2 (1987). 32. Quinlan, S. and Khatib, O. Elastic bands: Connecting path planning and control. In Proceedings of the IEEE Int. Conf. on Robotics and Automation (1993). 33. Reeds, J. and Shepp, L. Optimal paths for a car that goes both forwards and backwards. Pacific Journal of Mathematics 145, 2 (1990), 367393. 34. Rimon, E. and Koditschek. Exact robot navigation using artificial potential fields. IEEE Trans. on Robotics and Automation 8, 5 (1992), 501518. 35. Schwartz, J. and Sharir, M. On the piano movers problem II: General techniques for computing topological properties of real algebraic manifolds. Advances of Applied Mathematics 4 (1983), 298351. 36. Shin, K.G. and McKay, N.D. Minimum-time control of robotic manipulators with geometric path constraints. IEEE Trans. on Automatic Control 30, 6 (1985), 531541. 37. Souères, P. and Laumond, J.-P. Shortest paths synthesis for a car-like robot. IEEE Trans. on Automatic Control 41, 5 (1996), 672688. 38. Sussmann, H. and Tang, G. Shortest paths for the Reeds-Shepp car: A worked out example of the use of geometric techniques in nonlinear optimal control. Rutgers Center for Systems and Control Technical Report 91-10, 1991. 39. Verscheure, D., Demeulenaere, B., Swevers, J., Schutter, J.D. and Diehlm M. Time-optimal path tracking for robots: A convex optimization approach. IEEE Trans. on Automatic Control 54, 10 (2009), 23182327. 40. Zucker, M., Ratliff, N., Dragan, A., Pivtoraiko, M., Klingensmith, M., Dellin, C., Bagnell, J. and Srinivasa, S. Chomp: Covariant Hamiltonian optimization for motion planning. Int. J. of Robotics Research 32, 9-10 (2013), 11641193. ### Authors Jean-Paul Laumond ([email protected]) is a CNRS director of research at LAAS, Toulouse, France. Nicolas Mansard ([email protected]) is a CNRS researcher at LAAS, Toulouse, France. Jean-Bernard Lasserre ([email protected]) is CNRS director of research at LAAS, Toulouse, France. ### Footnotes a. Here and in the following, we treat trajectory and motion as synonyms. b. The distance between the wheels is supposed to be 2. ### Figures Figure 1. A modern view of the "piano mover" problem: two characters have to move a shared piano while avoiding surrounding obstacles. Figure 2. A car (logo of the European Project Esprit 3 PRO-Motion in the 1990s) together with the unicycle model equations. Figure 3. Dubins car. Figure 4. Reeds-Shepp car. Figure 5. The Hilare robot at LAAS-CNRS in the 1990s. Figure 6. Hilare-1: A different drive mobile robot. First model: The controls are the velocities of the wheels. The optimal controls are bang-bang. Optimal trajectories are made of pure rotations and of straight-line segments. Figure 7. Hilare-2: A different drive mobile robot. Second model: The controls are the acceleration of the wheels. The optimal controls are bang-bang. Optimal trajectories are made of arcs of clothoids and of arcs of involute of a circle.

{"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}

CC-MAIN-2022-33

latest

en

http://docplayer.net/13320877-Withdrawing-from-your-ira-a-guide-to-the-basic-distribution-rules.html

text/html

crawl-data/CC-MAIN-2019-04/segments/1547583671342.16/warc/CC-MAIN-20190119160425-20190119182425-00520.warc.gz

# WITHDRAWING FROM YOUR IRA: A GUIDE TO THE BASIC DISTRIBUTION RULES Save this PDF as: Size: px Start display at page: ## Transcription 2 TABLE 1. SUBSTANTIALLY EQUAL PERIODIC PAYMENTS: A COMPARISON OF CALCULATION METHODS For a \$500,000 account, 7% interest rate, owner is age 50 and first-year withdrawal is in Minimum Annuity Distribution Factor Amortization First Year Withdrawal Amount \$12,755 \$38,118 \$37,654 the balance of all her IRAs combined is \$100,000, then 6% (\$6,000 \$100,000) of her distribution that year would be tax exempt. If she takes \$10,000, then only \$9,400 is taxable. In the second year, the numerator in that same equation would be \$5,400 (\$6, ), and the denominator is the market value of her combined IRAs as of the beginning of the next year she takes a distribution. Roth IRA: The other type of IRA is the Roth IRA, which allows for tax-free growth over the life of the account. There is no tax deduction in the year of contribution, but the contributions and all earnings in the account are distributed tax free upon withdrawal. Another attractive feature of a Roth is that it escapes the required minimum distribution rules at age 70½ that are discussed below. With that basic background information regarding the types of IRAs, you can better understand the various distribution rules and tax consequences of the timing of distributions from IRAs. EARLY DISTRIBUTIONS Since IRAs are vehicles to encourage saving for retirement, penaltyfree regular distributions cannot start until after the account owner turns age 59½. Generally speaking, a withdrawal prior to age 59½ will be subject to the ordinary income tax and a 10% early distribution penalty. Certain exceptions that allow you to avoid the 10% penalty will be discussed later. All taxable distributions from traditional IRAs will be subject to ordinary income tax i.e., no capital gains rates are applicable on distribution. You might be able to use one of the following exceptions to avoid the 10% penalty: Distributions due to death; Distributions due to disability; Distributions in the form of a series of substantially equal periodic payments (discussed below); Distributions for qualified deductible medical expenses; Distributions for qualified higher education expenses; and Distributions for qualified firsttime homebuyer distributions, up to a lifetime maximum of \$10,000. AVOIDING THE PENALTY A substantially equal periodic payment plan is the distribution option that allows an account owner who is younger than 59½ to receive distributions from an IRA without the 10% early distribution penalty. The substantially equal periodic payment amount must be withdrawn each year for the greater of five years or until the account owner attains age 59½. For example, someone who begins a substantially equal periodic payment at age 50 would be required to take out the equal amounts each year for 9½ years. Someone who begins a at age 58 would be required to take out that amount for five full years, or until they attained age 63. It is also important to note that a distribution only avoids the 10% penalty; the distributions will still be subject to ordinary income tax. The substantially equal periodic payment is a financial calculation best determined by a financial adviser. There are three methods to calculate an allowable withdrawal amount minimum distribution, annuity and amortization. (The mechanics of these calculations appear in our October 2000 AAII Journal article, Early Plan Distributions: How to Avoid the 10% Penalty. ) The minimum distribution method takes into consideration the account owner s age, account balance, and life expectancy. The annuity and amortization methods also factor in a reasonable interest rate. Typically, a larger withdrawal amount results with the amortization and annuity methods. The dollar amount yielded by the formula is the amount the account owner may withdraw from the account without paying the 10% early distribution penalty. Table 1 shows the withdrawal amounts for the first year under the three substantially equal periodic payment methods. The amounts assume a \$500,000 account, 7% interest rate, that the account owner is age 50 and that the first withdrawal year is Once an account owner starts a, the amount withdrawn cannot be changed until the greater of five years or when the owner reaches age 59½. The dollar amount that comes out of the IRA must be the same each year (although under the minimum distribution method the amount varies based on the changing year-end account balance and owner s life expectancy each year). Let s say you start a substantially equal periodic payment at age 50, and the penalty-free dollar amount to be withdrawn is \$50,000 (based on either the amortization or annuity methods). In this example, you would be required to withdraw \$50,000 each year until age 59½. AAII Journal/August ### Wealth Strategies. www.rfawealth.com. Saving For Retirement: Tax Deductible vs Roth Contributions. www.rfawealth.com www.rfawealth.com Wealth Strategies Saving For Retirement: Tax Deductible vs Roth Contributions Part 2 of 12 Your Guide to Saving for Retirement WEALTH STRATEGIES Page 1 Saving For Retirement: Tax Deductible Leaving your employer? Options for your retirement plan Contents Evaluating your options 1 The benefits of tax-deferred investing 4 Flexibility offered by an IRA rollover 6 How to get started 9 Evaluating ### 8 Years Later, Same Old Questions: Evaluating Roth IRAs 8 Years Later, Same Old Questions: Evaluating Roth IRAs By Mark H. Gaudet, CPA, CFP The Taxpayer Relief Act of 1997 introduced a new type of IRA for taxpayers and investors the Roth IRA. Eight years have ### Considering IRA Rollovers. Making the right distribution decision now can make a big difference down the road. Considering IRA Rollovers Making the right distribution decision now can make a big difference down the road. Considering IRA Rollovers Are you changing jobs? careers? Retiring? If you are planning to ### THE POWER OF AN IRA. THE POWER OF AN IRA. Put the Power of an IRA to Work For You Whether retirement seems like a lifetime away or is just around the corner, getting started on your retirement savings plan is one of the smartest ### Retirement Savings Options For Plan Participants. know your options when leaving wju s retirement plan. Revised 3/24/14 Retirement Savings Options For Plan Participants know your options when leaving wju s retirement plan You ve been planning for your future and saving for retirement. Now that you re leaving Jessup, you ### chart retirement plans 8 Retirement plans available to self-employed individuals include: retirement plans Contributing to retirement plans can provide you with financial security as well as reducing and/or deferring your taxes. However, there are complex rules that govern the type of plans ### Retirement Plan Distributions Choices & Opportunities Retirement Plan Distributions Choices & Opportunities Leaving Your Job: Things to Think About» What you want to do next Work full time? Part time? Retire? How much will your lifestyle cost?» Continuing ### EXPLORING YOUR IRA OPTIONS. Whichever you choose traditional or Roth investing in an IRA is a good step toward saving for retirement. EXPLORING YOUR IRA OPTIONS Whichever you choose traditional or Roth investing in an IRA is a good step toward saving for retirement. 2 EXPLORING YOUR IRA OPTIONS Planning for retirement can be a challenging ### Your Guide to. Individual Retirement Accounts Your Guide to Individual Retirement Accounts Your Guide to Individual Retirement Accounts (IRAs) TRADITIONAL AND ROTH IRAs ROLLOVERS A CONSUMER S GUIDE TO IRA ELIGIBILITY AND BENEFIT CHART Only 10% of ### ira individual retirement accounts Traditional IRA ira individual retirement accounts Traditional IRA Grow dollars for tomorrow, save on taxes today. A traditional IRA may provide you significant immediate tax savings, and due to the deferral of all taxes ### COLLIERS INTERNATIONAL USA, LLC And Affiliated Employers 401(K) Plan DISTRIBUTION ELECTION 1. EMPLOYEE INFORMATION (Please print) COLLIERS INTERNATIONAL USA, LLC And Affiliated Employers 401(K) Plan DISTRIBUTION ELECTION Name: Address: Social Security No.: Birth Date: City: State: Zip: Termination Traditional and Roth IRAs Invest for retirement with tax-advantaged accounts Your Retirement It is your ultimate reward for a lifetime of hard work and dedication. It is a time when you should have the ### MFS Retirement Strategies Stretch IRA and distribution options READY, SET, RETIRE. Taking income distributions during retirement MFS Retirement Strategies Stretch IRA and distribution options READY, SET, RETIRE Taking income distributions during retirement ASSESS YOUR NEEDS INCOME WHEN YOU NEED IT Choosing the right income distribution Traditional and s Invest for retirement with tax-advantaged accounts Your Retirement It is your ultimate reward for a lifetime of hard work and dedication. It is a time when you should have the financial ### The IRA Rollover. Making Sense Out of Your Retirement Plan Distribution The IRA Rollover Making Sense Out of Your Retirement Plan Distribution Expecting a Distribution? You have been a participant in your employer s retirement plan for a number of years, and you have earned ### NOT FDIC INSURED NO BANK GUARANTEE MAY LOSE VALUE. Tax-advantaged IRAs. Invest in your retirement savings while reducing taxes NOT FDIC INSURED NO BANK GUARANTEE MAY LOSE VALUE Taxadvantaged IRAs Invest in your retirement savings while reducing taxes Find the answers inside Why invest for retirement? p. 1 Discover three good reasons ### IRA opportunities at UBS IRA opportunities at UBS IRAs are highly popular and effective retirement savings vehicles that give your investment earnings the benefit of tax-favored treatment and provide an ideal supplement to employer-sponsored ### Rollover IRAs. Maximize the potential of your retirement savings Rollover IRAs Maximize the potential of your retirement savings Consolidate Your Retirement Savings With a Rollover IRA If you have changed jobs, left the workforce or plan to retire, you likely have ### WHICH TYPE OF IRA MAKES THE MOST SENSE FOR YOU? WHICH TYPE OF IRA MAKES THE MOST SENSE FOR YOU? In 1974, when IRAs were first created, they were rather simple and straightforward. Now, 35 years later, it s challenging to know the best way to save more Page 1 of 6 Frequently asked questions Distributions and rollovers from retirement accounts Choosing what to do with your retirement savings is an important decision. Tax implications are just one of several ### SHENANDOAH LIFE ANNUITIES ANNUITIES AGENT GUIDE SHENANDOAH LIFE ANNUITIES FOR AGENT USE ONLY This piece is not intended to create public interest in an insurance product, an insurer, or an agent. SHENANDOAH LIFE ANNUITIES For Your Retirement Plan Distribution How Your Decisions Today Affect Your Future Retirement Plans About Stifel Nicolaus Stifel Nicolaus is a full-service Investment firm with a distinguished history of providing ### LIQUIDATING RETIREMENT ASSETS LIQUIDATING RETIREMENT ASSETS IN A TAX-EFFICIENT MANNER By William A. Raabe and Richard B. Toolson When you enter retirement, you retire from work, not from decision-making. Among the more important decisions ### Chapter 14. Agenda. Individual Annuities. Annuities and Individual Retirement Accounts Chapter 14 Annuities and Individual Retirement Accounts Agenda 2 Individual Annuities Types of Annuities Taxation of Individual Annuities Individual Retirement Accounts Individual Annuities 3 An annuity ### Tax-smart ways to save and invest. TIAA-CREF Financial Essentials Tax-smart ways to save and invest TIAA-CREF Financial Essentials Today s agenda: 1. Finding funds for saving 2. Tax law provisions promoting saving 3. TIAA-CREF savings opportunities 4. TIAA-CREF can help ### Part VII Individual Retirement Accounts Part VII are a retirement planning tool that virtually everyone should consider. The new IRA options also have made selecting an IRA a bit more complicated. IRA Basics The Traditional IRA is an Individual ### Franklin Templeton IRA Investor s Guide Franklin Templeton IRA Traditional IRA Roth IRA Whether you are just starting to save or entering retirement, an IRA can be an important part of a sound financial strategy to meet your ### Distributions and Rollovers from Page 1 of 6 Frequently Asked Questions about Distributions and Rollovers from Retirement Accounts Choosing what to do with your retirement savings is an important decision. Tax implications are just one ### Chapter 19 Retirement Products: Annuities and Individual Retirement Accounts Chapter 19 Retirement Products: Annuities and Individual Retirement Accounts Overview Thus far we have examined life insurance in great detail. Life insurance companies also market a product that addresses ### Retirement Plan Contribution Limits and Withdrawal Requirements Manning & Napier Advisors, LLC Retirement Plan Contribution Limits and Withdrawal Requirements April 2011 Approved CAG-CM PUB030-R (10/11) Introduction The following report provides general information ### Withdrawal Request Form Without Spousal Consent Section I: Plan Information Plan Name: Plan Sponsor Name: Section II: Participant Information **If you have a P.O. Box, U.S. tax laws require a street address to be indicated, or ### An IRA can put you in control of your retirement, whether you IRAs: Powering Your Retirement One of the most effective ways to build and manage funds to help you meet your financial goals is through an Individual Retirement Account (IRA). An IRA can put you in control ### PERSONAL FINANCE. individual retirement accounts (IRAs) PERSONAL FINANCE individual retirement accounts (IRAs) 1 our mission To lead and inspire actions that improve financial readiness for the military and local community. table of contents The Basics Of IRAs... october 2012 Understanding Traditional and Roth IRAs summary An Individual Retirement Account (IRA) is a powerful savings vehicle that can help you meet your financial goals. As shown in the chart on page ### AN ANALYSIS OF ROTH CONVERSIONS 1 AN ANALYSIS OF ROTH CONVERSIONS In 1997, Congress introduced the Roth IRA, giving investors a new product for retirement savings. The Roth IRA is essentially a mirror image of the Traditional IRA, but ### Caution: Special rules apply to certain distributions to reservists and national guardsmen called to active duty after September 11, 2001. Thorsen Clark Tracey Wealth Management 301 East Pine Street Suite 1100 Orlando, FL 32801 407-246-8888 407-897-4427 [email protected] tctwealthmanagement.com Roth IRAs 2 Your pension benefit options Traditional pension plans generally provide the option of a lump-sum payment or a fixed monthly payment for life through an annuity. The fixed monthly payment amount is usually ### Roth IRAs and Conversions Roth IRAs and Conversions When the law that created Roth IRAs was originally enacted there were two eligibility limits placed on them. The first affected an individual s ability to contribute to the Roth- ### IRAs, Roth IRAs and the Conversion Decision for Americans Living Abroad IRAs, Roth IRAs and the Conversion Decision for Americans Living Abroad David Kuenzi, CFP, Thun Financial Advisors Updated, October 2014 Introduction Expat IRAs and Roth IRAs Even under the most conventional ### Making the Most of IRA Opportunities Making the Most of IRA Opportunities Why Is Saving for Retirement So Important? Increasing life expectancies mean more time spent in retirement. Aging population puts added strain on Social Security and ### Is the Roth IRA Conversion For Me? Is the Roth IRA Conversion For Me? By Dennis A. Suckstorf, CFP, ChFC, AFC. Financial Advantage, Inc. (www.financialadvantageinc.com) Prior to 2010, an individual could only convert a traditional IRA (qualified ### Comparison Chart: Pre-Tax Deferrals, Designated Roth Deferrals, In-Plan Roth Rollovers, & Roth IRAs Comparison Chart: Pre-Tax Deferrals, Designated Roth Deferrals, In-Plan Roth Rollovers, & s Caution: Please note that some of the items included on this grid are the result of Lincoln National Corporation ### IRA Opportunities. Traditional IRA vs. Roth IRA: Which is right for you? What kind of retirement funding vehicle is right for you? IRA Opportunities. Traditional IRA vs. Roth IRA: Which is right for you? What kind of retirement funding vehicle is right for you? Now more than ever, an Individual Retirement Account (IRA) may help provide ### Basics of IRAs ING FINANCIAL SOLUTIONS. Your future. Made easier. SM Basics of IRAs t FDIC/NCUA Insured t A Deposit Of A Bank t Bank Guaranteed May Lose Value t Insured By Any Federal Government Agency ING FINANCIAL SOLUTIONS Your future. Made easier. SM Traditional IRA ### Converting or Rolling Over Traditional IRAs to Roth IRAs Davis & Graves CPA LLP Jerry Davis, CPA/PFS 700 N Main Gresham, OR 97009 503-665-0173 [email protected] www.jjdcpa.com Converting or Rolling Over Traditional IRAs to Roth IRAs Page 1 of 9, see disclaimer ### 16. Individual Retirement Accounts 16. Individual Retirement Accounts Introduction Through enactment of the Employee Retirement Income Security Act of 1974 (ERISA), Congress established individual retirement accounts (IRAs) to provide workers ### To Roth or Not Revised September 2013 Introduction To Roth or Not Revised September 2013 Tax law allows all taxpayers (without income limitation) to convert all or part of their traditional IRAs to Roth IRAs. Even though conversion to Roth ### Roth IRA. Answers. To Your. Questions Roth IRA Answers To Your Questions A wealth of information to plan your future The Roth IRA offers unique and exciting savings opportunities. Not only can it help with retirement needs, but also a first-time ### Roth Conversion Frequently Asked Questions Roth Conversion Frequently Asked Questions Brian Dobbis QPA, QKA, QPFC Retirement Analyst, Private Wealth Group 888-522-2388 A Roth individual retirement account (IRA) is a tax-deferred and potentially ### Retirement Plans Guide Facts at a glance Retirement Plans Guide Facts at a glance Contents 1 What s Your Plan? 2 Small Business/Employer Retirement Plans 4 IRAs 5 Retirement Plan Distributions 7 Rollovers and Transfers 8 Federal Tax Rates and ### Six Best and Worst IRA Rollover Decisions Six Best and Worst IRA Rollover Decisions Provided to you by: Yvette Scanlon President, Scanlon Financial Services, LLC Six Best and Worst IRA Rollover Decisions Written by Financial Educators Provided ### 1040 Review Guide: MARKETS ADVANCED. Using Your Clients 1040 to Identify Planning Opportunities 1040 Review Guide: Using Your Clients 1040 to Identify Planning Opportunities ADVANCED MARKETS Producers Guide to a 1040 Review At Transamerica, we re committed to providing you and your clients with the ### Hardship distributions. A hardship distribution is not eligible for rollover. SPECIAL TAX NOTICE REGARDING PLAN PAYMENTS 1 (Alternative to IRS Safe Harbor Notice - For Participant) This notice explains how you can continue to defer federal income tax on your retirement plan savings UNDERSTANDING YOUR RETIREMENT CHOICIES If you re thinking about retiring or changing jobs: You may be receiving the largest check you ll ever see, and You have a major decision to make. Leave retirement ### IRAs. Investing in Your Future. Retirement Plans IRAs Investing in Your Future Retirement Plans About Stifel Nicolaus Stifel Nicolaus is a full-service Investment firm with a distinguished history of providing securities brokerage, investment banking, ### Six Best and Worst IRA Rollover Decisions Six Best and Worst IRA Rollover Decisions Provided to you by: Dayton Tatman Independent Agent / Broker Six Best and Worst IRA Rollover Decisions Written by Financial Educators Provided to you by Dayton ### The Federal Thrift Savings Plan Vertex Wealth Management, LLC Michael J. Aluotto, CRPC President Private Wealth Manager 1325 Franklin Ave., Ste. 335 Garden City, NY 11530 516-294-8200 [email protected] www.vertexwm.com The Federal ### BMO Funds State Street Bank and Trust Company Universal Individual Retirement Account Disclosure Statement. Part One: Description of Traditional IRAs BMO Funds State Street Bank and Trust Company Universal Individual Retirement Account Disclosure Statement Part One: Description of Traditional IRAs Part One of the Disclosure Statement describes the rules ### SPECIAL TAX NOTICE REGARDING PLAN PAYMENTS SPECIAL TAX NOTICE REGARDING PLAN PAYMENTS This notice explains how you can continue to defer federal income tax on your retirement savings and contains important information you will need before you decide ### MOAA Caregivers Guide: Planning For Retirement MOAA Caregivers Guide: Planning For Retirement To obtain financial security during your retirement years, start planning for it now. The most powerful factor is time. Starting to invest now for retirement ### OPTION SELECT ROTH IRA DISCLOSURE STATEMENT OPTION SELECT ROTH IRA DISCLOSURE STATEMENT Group Policy Form No. 01-1115-98 Group Certificate Form No. 01-1115C-98 and state variations thereof. Not for use in Florida. Regarding Roth Individual Retirement premiere select Traditional IRA/Roth IRA Invest in your retirement today. Saving for your retirement. 01 Important Section in head any lorem market. ipsum dolore sit amet If you re planning for your future, ### SPECIAL TAX NOTICE REGARDING PLAN PAYMENTS SPECIAL TAX NOTICE REGARDING PLAN PAYMENTS This notice explains how you can continue to defer federal income tax options for your distribution from the Plan and contains important information you will ### A. TYPES OF PLAN DISTRIBUTIONS SPECIAL TAX NOTICE REGARDING PLAN PAYMENTS 1 (Alternative to IRS Safe Harbor Notice - For Participant) This notice explains how you can continue to defer federal income tax on your retirement plan savings ### IMPORTANT INFORMATION ABOUT ROTH CONTRIBUTIONS IMPORTANT INFORMATION ABOUT ROTH CONTRIBUTIONS What is a Roth 401(k)? Is it a new type of plan? No, it is not a new type of plan. Designated Roth contributions are a type of contribution that new or existing ### ROTH 401(k) FEATURE QUESTION & ANSWER (Q&A) ROTH 401(k) FEATURE QUESTION & ANSWER (Q&A) Purpose of Q&A: Beginning January 1, 2006, employers that sponsor 401(k) retirement plans may offer a new plan design feature after-tax Roth deferrals. The purpose ### Beginning in 2010, the Tax Increase Prevention and ROTH IRA CONVERSION ROTH IRA CONVERSION Assessing Suitability of the Strategy for Individuals and their Heirs Executive Summary A Roth IRA conversion may benefit individuals during their retirement years by potentially reducing ### Roth IRAs The Roth IRA Roth IRAs The Roth IRA 2010 and 2011 Questions & Answers What is a Roth Individual Retirement Account (Roth IRA)? A Roth IRA is a type of tax-preferred savings and investment account authorized by Internal ### H.T. BAILEY INSURANCE GROUP 401(k) PLAN Case # 943-80987. ELECTION OF PAYMENT METHOD (Please Print Clearly) H.T. BAILEY INSURANCE GROUP 401(k) PLAN Case # 943-80987 ELECTION OF PAYMENT METHOD (Please Print Clearly) PARTICIPANT NAME: DATE OF BIRTH: SOCIAL SECURITY NUMBER: DATE OF HIRE: DATE OF TERMINATION: DO ### TOWN OF NATICK OBRA 457 DEFERRED COMPENSATION GOVERNMENTAL PLAN DISTRIBUTION FORM TOWN OF NATICK OBRA 457 DEFERRED COMPENSATION GOVERNMENTAL PLAN DISTRIBUTION FORM PARTICIPANT/ ALTERNATE PAYEE INFORMATION DISTRIBUTION REASON PAYMENT METHOD SPOUSE S CONSENT TO DISTRIBUTION (not applicable ### P. O. Box 2069 Woburn, MA 01801-1721 (781) 938-6559 NOTICE TO PARTICIPANTS SEPARATED FROM SERVICE P. O. Box 2069 Woburn, MA 01801-1721 (781) 938-6559 NOTICE TO PARTICIPANTS SEPARATED FROM SERVICE Under the terms of the SBERA 401 (k) Plan, you may now elect to withdraw your total account balance. Your ### IRAs Traditional Individual Retirement Accounts. 2014 and 2015. Questions & Answers IRAs Traditional Individual Retirement Accounts 2014 and 2015 Questions & Answers What is the purpose of this brochure? It summarizes the primary laws governing traditional IRAs for 2014 and 2015. What ### IRAs Traditional Individual Retirement Accounts. 2013 and 2014. Questions & Answers IRAs Traditional Individual Retirement Accounts 2013 and 2014 Questions & Answers What is the purpose of this brochure? It summarizes the primary laws governing traditional IRAs for 2013 and 2014. What ### ! There are currently two types of IRAs. An IRA can be established and funded at any time from January 1 of the current year and up to and including the date an individual s income tax return is due (generally, April 1 of the following year), Table of Contents Frequently Asked Questions... 2 Individual Retirement Arrangements (IRAs)... 2 IRA Basics... 2 Why should I name beneficiaries for my IRA?... 2 Traditional IRAs... 9 Roth IRAs... 13 SEP ### Sample. Table of Contents. Introduction... 1. What is the difference between a regular 401(k) deferral (pre-tax) and a Roth 401(k) deferral?... Table of Contents Introduction... 1 What is the difference between a regular 401(k) deferral (pre-tax) and a Roth 401(k) deferral?... 2 Who is eligible to make a Roth 401(k) deferral?... 3 Roth IRAs have ### EXPLANATION OF DIRECT ROLLOVER OPTION EXPLANATION OF DIRECT ROLLOVER OPTION You have the right to consider the decision whether or not to elect a direct rollover for at least 30 days after this Explanation is provided to you. This notice explains ### Annuities. Introduction 2. What is an Annuity?... 2. How do they work?... 3. Types of Annuities... 4. Fixed vs. Variable annuities... An Insider s Guide to Annuities Whatever your picture of retirement, the best way to get there and enjoy it once you ve arrived is with a focused, thoughtful plan. Introduction 2 What is an Annuity?... ### Table of contents. 2 Federal income tax rates. 12 Required minimum distributions. 4 Child credits. 13 Roths. 5 Taxes: estates, gifts, Social Security 2015 Tax Guide Table of contents 2 Federal income tax rates 4 Child credits 5 Taxes: estates, gifts, Social Security 6 Rules on retirement plans 8 Saver s credit 12 Required minimum distributions 13 Roths ### Retirement Plans: Evaluating the New Roth IRA Conversion Opportunity Retirement Plans: Evaluating the New Roth IRA Conversion Opportunity By Christine Fahlund Starting next year, investors will have the option of converting all or part of their money in a traditional IRA THE RETIREMENT GROUP AT MERRILL LYNCH Q A Questions and Answers about the Roth 401(k) How the Roth 401(k) Works Q. What is the Roth 401(k) contribution option? A. The Roth 401(k) contribution option allows ### Six Best & Worst IRA Rollover Decisions Six Best & Worst RA Rollover Decisions Provided to you by - Jay Kemmerer Berkshire 6t~ Advisors, nc. :i~~~ REGSTERE)f~.h!EST~lE.!\JAD\rSOR Six Best & Worst RA Rollover Decisions Written by Financial Educators THE VARIABLE ANNUITY LIFE INSURANCE COMPANY SEPARATE ACCOUNT A UNITS OF INTEREST UNDER GROUP AND INDIVIDUAL FIXED AND VARIABLE DEFERRED ANNUITY CONTRACTS INDEPENDENCE PLUS CONTRACT SERIES STATEMENT OF ### Tax Strategies From a Financial Planning Perspective Tax Strategies From a Financial Planning Perspective Spectrum Advisors Don Goerner offers securities through Purshe Kaplan Sterling Investments Member FINRA/ SIPC Headquartered at 18 Corporate Woods Blvd., ### Guide to Individual Retirement Accounts. Make a secure retirement yours Guide to Individual Retirement Accounts Make a secure retirement yours Retirement means something different to everyone. Some dream of stopping employment completely and some want to continue working. ### Raymond James Financial Services September 06, 2011 Raymond James Financial Services Bob Alft Wealth Advisor 220 E. Nine Mile Road Pensacola, FL 32534 850-479-7190 [email protected] www.raymondjames.com/bobalft Understanding IRAs Understanding IRAs Rollover IRAs Consider the advantages of consolidating your retirement savings Consider the Advantages of Consolidating Your Retirement Savings If you have changed jobs, left the workforce or plan to ### What is an annuity? The basics Part 1 of 8 What is an annuity? The basics Part 1 of 8 You may be considering an annuity, and, if that s the case, it is important that know what an annuity is and isn t. Basically, an annuity is a contract with an ### Nonqualified annuities can be classified in a number of ways: The term annuity refers to any situation where principal and interest are paid out in a series of regular payments. A nonqualified annuity, generally, is an annuity purchased by an individual from a life ### SPECIAL TAX NOTICE REGARDING PLAN PAYMENTS (Alternative to IRS Safe Harbor Notice - For Participant) SPECIAL TAX NOTICE REGARDING PLAN PAYMENTS (Alternative to IRS Safe Harbor Notice - For Participant) This notice explains how you can continue to defer federal income tax on your retirement plan savings ### NOTICE OF HARDSHIP WITHDRAWAL NOTICE OF HARDSHIP WITHDRAWAL The current Plan provides that certain amounts may be withdrawn if you have a financial hardship. This hardship distribution is not in addition to your other benefits and ### Participant Name (First) (Middle Initial) (Last) Social Security Number I.D. Number. Participant Address (Street) City State ZIP Code + 4 Mailing Address: Des Moines, IA 50392-0001 Principal Life Insurance Company Early Withdrawal of Benefits Without Guaranteed Accounts No Spousal Consent Needed CTD00603 Complete this form to withdraw part ### IRAs. AKD Consultants Adam Dworkin CPA 188 Whiting Street Suite 10 Hingham, MA 02043 781-556-5554 [email protected] AKD Consultants Adam Dworkin CPA 188 Whiting Street Suite 10 Hingham, MA 02043 781-556-5554 [email protected] IRAs October 01, 2013 Page 1 of 8, see disclaimer on final page Both traditional and ### Supplement to IRA Custodial Agreements Supplement to IRA Custodial Agreements Effective December 31, 2014, the update below will be made to the American Century Custodial agreements for the following retirement accounts: Traditional IRAs, Roth

{"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}

CC-MAIN-2019-04

latest

en

http://physics.stackexchange.com/questions/48586/is-the-heat-required-to-alter-the-higgs-field-an-absolute-heat/48603

text/html

crawl-data/CC-MAIN-2013-20/segments/1368707439012/warc/CC-MAIN-20130516123039-00066-ip-10-60-113-184.ec2.internal.warc.gz

# Is the heat required to alter the Higgs field an 'absolute heat'? I have read and heard that manipulating the Higgs field would require heating up a local geometry to ridiculous temperature. I am trying to understand if there are stars or places in the universe which have this level of heat. Would this be the upper limit on heat, an 'absolute heat'? - ## 1 Answer We need to be clear about what temperature means in this context. For gases the temperature is linked to the particle velocities by the Maxwell-Boltzmann distribution, so we can convert a temperature to an average particle velocity and hence average particle energy, and vice versa. The electroweak transition happened when the average particle energy was in the range 100GeV to 1TeV. We could take this energy and use a Maxwell-Boltzmann distribution to convert it back to a temperature, and if you do this the temperature comes out at about 10$^{16}$K. However it's debatable how much sense it makes to apply the concept of temperature to the universe at these high energies, so don't take this number too seriously. The temperature at the core of a star is in the range 10$^7$ to 10$^8$K (depending on the size of the star) so it's unlikely there is anywhere in the universe that gets anywhere near the electroweak transition temperature. Individual cosmic rays can have energies far above 1TeV, but you can't convert this to a temperature because by definition only an assemblage of particles can have a temperature. - "it's debatable how much sense it makes to apply the concept of temperature to the universe at these high energies" Why is that? I've always heard that quasi-equilibrium holds throughout the expansion (except for the odd phase transition), so temperature definitely applies. What is the argument against it? – Michael Brown Feb 8 at 0:25 Lots of things have temperature that aren't gases. It can be defined as $\partial U/\partial S$ or as energy per degree of freedom, the two being very close approximations of one another except at low temperatures. Without knowing much about particle physics, I would imagine that energy per degree of freedom is easy to calculate and relatively unproblematic as a definition. – Nathaniel Mar 9 at 14:49

{"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}

CC-MAIN-2013-20

latest

en

https://hardware.slashdot.org/story/12/07/16/1617202/record-setting-500-trillion-watt-laser-shot-achieved/informative-comments

text/html

crawl-data/CC-MAIN-2018-22/segments/1526794867417.71/warc/CC-MAIN-20180526112331-20180526132331-00271.warc.gz

typodupeerror ## Record Setting 500 Trillion-Watt Laser Shot Achieved252 cylonlover writes "Researchers at the Lawrence Livermore National Laboratory's National Ignition Facility (NIF) have achieved a laser shot which boggles the mind: 192 beams delivered an excess of 500 trillion-watts (TW) of peak power and 1.85 megajoules (MJ) of ultraviolet laser light to a target of just two millimeters in diameter. To put those numbers into perspective, 500 TW is more than one thousand times the power that the entire United States uses at any instant in time." This discussion has been archived. No new comments can be posted. ## Record Setting 500 Trillion-Watt Laser Shot Achieved • #### Oww, it burns! (Score:4, Informative) by Anonymous Coward on Monday July 16, 2012 @01:41PM (#40664551) 500 TW is more than one thousand times the *average* power that the entire United States uses at any instant in time. • #### Re:To put that in perspective (Score:5, Informative) on Monday July 16, 2012 @01:45PM (#40664597) Journal I heard a radio program (NPR I think) talking about this. The entire energy was about the same as rubbing your hands together for a few seconds. Can anyone verify? It was early on a Monday morning, so it could ahve been the haze of the weekend... • #### Putting the hyperbole in perspective... (Score:3, Informative) by Anonymous Coward on Monday July 16, 2012 @01:49PM (#40664649) Off the cuff, 500 TW divided by 1.58 MJ implies the beam lasted only a few nanoseconds. So, "To put those numbers into perspective", 500 TW is more than one thousand times the power that the entire United States uses for a few nanoseconds." • #### Re:To put that in perspective (Score:5, Informative) on Monday July 16, 2012 @01:51PM (#40664675) It's a bit more energy than that, but it's not a remarkable amount of energy. 1.85MJ is enough to turn just under 1L of water from 100C liquid phase to 100C vapour phase. ie - it's enough to boil 1L of water, if the water is already at the boiling point. Latent heat of vapourization for H2O is about 2200 kJ/kg. • #### Re:Putting the hyperbole in perspective... (Score:5, Informative) on Monday July 16, 2012 @02:19PM (#40665049) Journal Off the cuff, 500 TW divided by 1.58 MJ implies the beam lasted only a few nanoseconds. So, "To put those numbers into perspective", 500 TW is more than one thousand times the power that the entire United States uses for a few nanoseconds." Sigh... You are conflating power with energy. Don't feel bad: the press gets it wrong more than half the time. Energy is a bulk quantity: a total amount. Power is a rate: how energy over how much time. Because this is /., I'll use a car analogy: energy is analogous to how large the gas tank is (gallons, liters, etc.), power is how quickly that gas gets consumed (g/sec, mL/sec, L/100km, mpg). The average power consumption of the U.S. is a few hundred gigawatts...period. There is no gigawatts per second, or any other monstrous measure that pretends to be power, because the "per second" is already built into the Watt unit. Correcting your statement: 1.85 MJ is more than one thousand times the energy that the entire United States uses in a few nanoseconds The original statement comparing 500 TW to the (average) power consumption of the U.S. was correct. • #### Re:Fusion Ignition (Score:5, Informative) on Monday July 16, 2012 @02:38PM (#40665251) Lasers are not normally used in Tokamak reactors. In those systems, the idea is to use magnetic fields to hold a plasma tight enough (and long enough) for fusion to initiate. The energy input (i.e. "heating") is done ohmically, that is, by radio waves that induce electric currents in the gas. The NIF pursues a different approach, called "inertial confinement fusion." The idea in these systems is to supply a whole load of energy in a very short time, so the hydrogen nuclei don't have time to move apart before the fusion reaction takes place. That is, their inertia is what confines them long enough for the reaction to go. In order to do this, you need a giant load of energy delivered into a very small volume in a very short time. That's why they quote the number as terawatts. The interesting part of this announcement is not just the TW energy rate, but the nanosecond-scale pulse width. This is actually pretty cool news... • #### Re:One Thousand Times (Score:4, Informative) <[email protected]> on Monday July 16, 2012 @07:13PM (#40667591) Homepage Nope. The original definition (by some French guy, according to Wikipedia) was that 1 calorie heats one gram of water by 1 *C. I remember learning that the original definition was for one kilogram, that's why it was called kilocalorie, and was first measured by Joule. Wikipedia contradicts my history knowledge, but not my numbers. #### Related LinksTop of the: day, week, month. In English, every word can be verbed. Would that it were so in our programming languages. Working...

{"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}

CC-MAIN-2018-22

latest

en

https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-6th-edition/chapter-2-section-2-5-compound-inequalities-exercise-set-page-94/21

text/html

crawl-data/CC-MAIN-2019-04/segments/1547584331733.89/warc/CC-MAIN-20190123105843-20190123131843-00617.warc.gz

## Intermediate Algebra (6th Edition) The solution is $(-\infty,-3]\cap(-\infty,0]$ $4x+2\le-10$ and $2x\le0$ Solve both inequalities separately: $4x+2\le-10$ Take $2$ to the right side: $4x\le-10-2$ $4x\le-12$ Take $4$ to divide the right side: $x\le-\dfrac{12}{4}$ $x\le-3$ The solution is $(-\infty,-3]$ $2x\le0$ Take $2$ to divide the right side: $x\le\dfrac{0}{2}$ $x\le0$ The solution is $(-\infty,0]$ Since the compound inequality has the word "and", the solution is the intersection of the two solution sets found. The solution is $(-\infty,-3]\cap(-\infty,0]$

{"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}

CC-MAIN-2019-04

latest

en

http://www.jiskha.com/display.cgi?id=1299533757

text/html

crawl-data/CC-MAIN-2017-17/segments/1492917122159.33/warc/CC-MAIN-20170423031202-00519-ip-10-145-167-34.ec2.internal.warc.gz

# algebra 2 posted by on . how to find the median of x,2x+1,x/2-13,45 and x+22 if the mean of this set of numbers is 83? • algebra 2 - , first solve: (x + 2x+1 + (x/2)-13 + 45 + x+22)/5 = 83 everything times 10 2x + 4x + 2 + x - 26 + 90 + 2x + 44 = 830 9x = 720 x = 80 so #'s are: 80 ,161, 27, 45 , and 102 Can you find their median?

{"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}

CC-MAIN-2017-17

latest

en

http://encyclopedia.kids.net.au/page/co/Complex_numbers

text/html

crawl-data/CC-MAIN-2019-43/segments/1570987836295.98/warc/CC-MAIN-20191023201520-20191023225020-00354.warc.gz

## Encyclopedia > Complex numbers Article Content # Complex number Redirected from Complex numbers The complex numbers are an extension of the real numbers: the real number line is enlarged to get the complex number plane. The complex numbers contain a number i, the imaginary unit, with i2= -1. Every complex number can be represented in the form x+iy, where x and y are real numbers called the real part and the imaginary part of the complex number respectively. The sum and product of two complex numbers are: (a+ib) + (c+id) = (a+c) + i(b+d) (a+ib) · (c+id) = ac-bd + i (bc+ad) Complex numbers were first introduced in connection with explicit formulas for the roots of cubic[?] polynomials. See the section History below. Table of contents Formally we may define complex numbers as ordered pairs of real numbers (a, b) together with the operations: • (a, b) + (c, d) = (a + c, b + d) • (a, b) · (c, d) = (ac - bd, bc + ad) So defined, the complex numbers form a field, the complex number field, denoted by C (or $\mathbb{C}$ in blackboard bold). We identify the real number a with the complex number (a, 0), and in this way the field of real numbers R becomes a subfield of C. The imaginary unit i is the complex number (0,1). A complex number can also be viewed as a point or a position vector on the two dimensional Cartesian coordinate system. This representation is sometimes called an Argand diagram. In the figure, we have z = x + iy = r (cos φ + i sin φ). The latter expression is sometimes shorthanded as r cis φ, where r is called the absolute value of z and φ is called the complex argument of z. By simple trigonometric identities, we see that r1 cis φ1 · r2 cis φ2 = r1r2 cis (φ12); r1 cis φ1 / r2 cis φ2 = r1 / r2 cis (φ12); Now the addition of two complex numbers is just the vector addition of two vectors, and the multiplication with a fixed complex number can be seen as a simultaneous rotation and stretching. Multiplication with i corresponds to a counter clockwise rotation by 90 degrees. The geometric content of the equation i2 = -1 is that a sequence of two 90 degree rotation results in a 180 degree rotation. Even the fact (-1) · (-1) = +1 from arithmetic can be understood geometrically as the combination of two 180 degree turns. Euler's formula states that ei φ = cisφ. The exponential form gives us a better insight then the shorthand rcisφ, which is almost never used in serious mathematical articles. Recall that the absolute value (or modulus or magnitude) of a complex number z = r e is defined as |z| = r. Algebraically, if z = a + ib, then |z| = (a2 + b2 ). One can check readily that the absolute value has three important properties: |z + w| ≤ |z| + |w| |z w| = |z| |w| |z / w| = |z| / |w| for all complex numbers z and w. By defining the distance function d(z, w) = |z - w| we turn the complex numbers into a metric space and we can therefore talk about limits and continuity. The addition, subtraction, multiplication and division of complex numbers are then continuous operations. Unless anything else is said, this is always the metric being used on the complex numbers. The complex conjugate of the complex number z = a + ib is defined to be a - ib, written as $\bar{z}$ or z*. As seen in the figure, $\bar{z}$ is the "reflection" of z about the real axis. The following can be checked: $\overline{z+w} = \bar{z} + \bar{w}$ $\overline{zw} = \bar{z}\bar{w}$ $\overline{(z/w)} = \bar{z}/\bar{w}$ $\bar{\bar{z}}=z$ $\bar{z}=z$ if and only if z is real $|z|=|\bar{z}|$ $|z|^2 = z\bar{z}$ $z^{-1} = \bar{z}|z|^{-2}$ if z is non-zero The latter formula is the method of choice to compute the inverse of a complex number if it is given in rectangular coordinates. The complex argument of z=re is φ. Note that the complex argument is unique up to modulo 2π. C is a two-dimensional real vector space. Unlike the reals, complex numbers cannot be ordered in any way that is compatible with its arithmetic operations: C cannot be turned into an ordered field. A root of the polynomial p is a complex number z such that p(z) = 0. A most striking result is that all polynomials of degree n with real or complex coefficients have exactly n complex roots (counting multiple roots according to their multiplicity). This is known as the Fundamental Theorem of Algebra, and shows that the complex numbers are an algebraically closed field. Indeed, the complex number field is the algebraic closure of the real number field. It can be identified as the quotient ring of the polynomial ring R[X] by the ideal generated by the polynomial X2 + 1: C = R[X] / (X2 + 1). This is indeed a field because X2 + 1 is irreducible. The image of X in this quotient ring becomes the imaginary unit i. The study of functions of a complex variable is known as complex analysis and has enormous practical use in applied mathematics as well as in other branches of mathematics. Often, the most natural proofs for statements in real analysis or even number theory employ techniques from complex analysis (see prime number theorem for an example). Unlike real functions which are commonly represented as two dimensional graphs, complex functions have four dimensional graphs and may usefully be illustrated by color coding a three dimensional graph to suggest four dimensions, or by animating the complex function's dynamic transformation of the complex plane. While usually not useful, alternative representations of complex field can give some insight into their nature. One particularly elegant representation interprets every complex number as 2x2 matrix with real entries which stretches and rotates the points of the plane. Every such matrix has the form $\begin{pmatrix} a&&-b\\ b&&a \end{pmatrix}$ with real numbers a and b. The sum and product of two such matrices is again of this form. Every non-zero such matrix is invertible, and its inverse is again of this form. Therefore, the matrices of this form are a field. In fact, this is exactly the field of complex numbers. Every such matrix can be written as $\begin{pmatrix} a&&-b\\ b&&a \end{pmatrix} # a \begin{pmatrix} 1&&0\\ 0&&1 \end{pmatrix} + b \begin{pmatrix} 0&&-1\\ 1&&0 \end{pmatrix}$ which suggests that we should identify the real number 1 with the matrix $\begin{pmatrix} 1&&0\\ 0&&1 \end{pmatrix}$ and the imaginary unit i with $\begin{pmatrix} 0&&-1\\ 1&&0 \end{pmatrix}$ a counter-clockwise rotation by 90 degrees. Note that the square of this latter matrix is indeed equal to -1. The absolute value of a complex number expressed as a matrix is equal to the square root of the determinant of that matrix. If the matrix is viewed as a transformation of a plane, then the transformation rotates points through an angle equal to the argument of the complex number and scales by a factor equal to the complex number's absolute value. The conjugate of the complex number z corresponds to the transformation which rotates through the same angle as z but in the opposite direction, and scales in the same manner as z; this can be described by the transpose of the matrix corresponding to z. Applications Complex numbers are used in signal analysis[?] and other fields as a convenient description for periodically varying signals. The absolute value |z| is interpreted as the amplitude and the argument arg(z) as the phase of a sine wave[?] of given frequency. If Fourier analysis is employed to write a given real-valued signal as a sum of periodic functions, these periodic functions are often written as the real part of complex valued functions of the form f(t) z eiωt where ω represents the angular frequency and the complex number z encodes the phase and amplitude as explained above. In electrical engineering, this is done for varying voltages and currents. The treatment of resistors, capacitors and inductors can then be unified by introducing imaginary frequency-dependent resistances for the latter two and combining all three in a single complex number called the impedance. (Electrical engineers and some physicists use the letter j for the imaginary unit since i is typically reserved for varying currents.) The residue theorem of complex analysis is often used in applied fields to compute certain improper integrals. The complex number field is also of utmost importance in quantum mechanics since the underlying theory is built on (infinite dimensional) Hilbert spaces over C. In Special and general relativity, some formulas for the metric on spacetime become simpler if one takes the time variable to be imaginary. In differential equations, it is common to first find all complex roots r of the characteristic equation of a linear differential equation[?] and then attempt to solve the system in terms of base functions of the form f(t) = ert. The earliest fleeting reference to square roots of negative numbers occurred in the work of the Greek mathematician and inventor Heron of Alexandria in the 1st century AD, when he considered the volume of an impossible frustum of a pyramid. They became more prominent when in the 16th century closed formulas for the roots of third and fourth degree polynomials were discovered by Italian mathematicians (see Tartaglia, Cardano). It was soon realized that these formulas, even if one was only interested in real solutions, sometimes required the manipulation of square roots of negative numbers. This was doubly unsettling since not even negative numbers were considered to be on firm ground at the time. The term "imaginary" for these quantities was coined by René Descartes in the 17th century and was meant to be derogatory. The existence of complex numbers was not completely accepted until the above mentioned geometrical interpretation had been described by Caspar Wessel in 1799; it was rediscovered several years later and popularized by Carl Friedrich Gauss. The formally correct definition using pairs of real numbers was given in the 19th century. • An Imaginary Tale, by Paul J. Nahin; Princeton University Press; ISBN 0691027951 (hardcover, 1998). A gentle introduction to the history of complex numbers and the beginnings of complex analysis. All Wikipedia text is available under the terms of the GNU Free Documentation License Search Encyclopedia Search over one million articles, find something about almost anything! Featured Article Lona ... - Wikipedia <

{"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}

CC-MAIN-2019-43

longest

en

https://www.glassdoor.com/Interview/test-systems-engineer-interview-questions-SRCH_KO0,21.htm

text/html

crawl-data/CC-MAIN-2021-21/segments/1620243991378.52/warc/CC-MAIN-20210515192444-20210515222444-00128.warc.gz

Test systems engineer Interview Questions | Glassdoor # Test systems engineer Interview Questions 147 test systems engineer interview questions shared by candidates ## Top Interview Questions Sort: RelevancePopular Date ### Senior Systems Test Engineer at Qualcomm was asked... Jun 7, 2010 There is a body of water that starts with 1 square unit, and doubles in size every day (2 units after 2 days, 4 units after 4 days). It takes 100 days to fill up. How many days would it take to fill if you started with 2 square units?6 Answers100 - 1 = 99 daysThis question is phrased incorrectly. I think you meant "4 units after 3 days". Which makes your answer wrong as well. This is not helpful at all.if you start with 1sq unit - lets say you end up with 'x' amount to fill up - takes 100 days if you start with 2 sq unit - you will have to fill up '2x' amount thus it will take - 200 days.Show More ResponsesStarting with 2 square units at time t=0 is like 1 square unit at t = 1. [this logic is the key to answering the question]. Now let's do the first few cases. t = 0: size = 1+1 = 2 t = 1: size = 2(1+1) = 2+2 = 4 = f(2) in the 1 unit case. Pretty easy to see it only requires 1 time period less from here. The OP was right.1 day less that is 99 days.It is still 100 days. Starting from 2^0, and going all the way to 2^100, the reservoir has a capacity of 2^101 - 1 units. Now in the second case, starting from 2^1, in 99 days there will be 2^101-2 units of water in the reservoir, which is one unit short of the capacity of the reservoir. So we need the 100th day to fill it up! ### Modem Systems Test Engineer at Qualcomm was asked... Dec 13, 2015 All about resume, why SCFDMA in uplink, PHFICH channel frequencies, what happens in MRC, LTE call procedure.2 AnswersNeed to know wireless communications, digital communications concepts in and out.The key in these questions is to cover the fundamentals, and be ready for the back-and-forth with the interviewer. Might be worth doing a mock interview with one of the Qualcomm or ex-Qualcomm Modem Systems Test Engineer experts on Prepfully? They give real-world practice and guidance, which is pretty helpful. prepfully.com/practice-interviews ### Wireless Systems Test Engineer at Qualcomm was asked... Mar 23, 2011 When you get higher throughput: while going at 5mph and while going at 50 mph? And Why?2 Answershigh mobility communication is more difficult.You will get higher a throughput while going at 5mph. Higher speeds indicates that the RF channel conditions such as delay and Doppler spread are changing at fast rate and its a fast fading channel. It is difficult to achieve equalization in such situation. Hence the modulation scheme can be shot down to, say QPSK(2 bits/symbol) from maybe 16QAM(4 bits/symbol) to provide a more robust transmission. This brings down the user throughput. ### Systems Integration In Test Engineer at ADVA was asked... Jul 19, 2019 How would you go about writing test cases to thoroughly test a piece of optical equipment?1 AnswerTesting over the working range to ensure proper functionality. Then, testing outside of the range to ensure any failures that could happen in the field were not catastrophic. ### Systems Test Engineer, Wireless at ABB was asked... Sep 16, 2016 onisite : 1st round Basic python scripting. 2 coding question,, Test case, SQL,(RDBMS), UI testing (Which idint know selenium). 2nd round: All basic network protocols, and some senario based questions. interviewer was good.1 Answercoding was not up to mark, Networking part was good ### Systems Test Lead Engineer at Modis was asked... Oct 1, 2015 Detail your past experience1 Answergave information of job details ### Systems Test Engineer at Qualcomm was asked... Dec 7, 2012 To draw complete block diagram of OFDM2 AnswersHi, can I ask you some detailed question about the interview? Can you send me your email, mine is [email protected], thank you so much! One or more comments have been removed. Please see our Community Guidelines or Terms of Service for more information. ### Systems Integration and Test Engineer at Space Telescope Science Institute was asked... Aug 27, 2015 What I can bring to the table.1 AnswerExperience. Years of spacecraft knowledge. Excellent multi tasking skills, great precision, detail oriented. ### Senior Systems Test Engineer at Illumina was asked... Oct 19, 2020 What was something that challenged me and how did I cope with it?1 AnswerDescribed my thesis and the duties I took to done problems. ### Associate Systems Integration/Test Engineer at Omnitracs was asked... Sep 29, 2015 If I give you a pen, describe how you would test it? 1 AnswerTo test the pen, I need to understand the functionality of the product. Then I would design different scenarios to test these functions. 110 of 147 Interview Questions

{"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}

CC-MAIN-2021-21

latest

en

https://www.assignmentexpert.com/homework-answers/physics/mechanics-relativity/question-128859

text/html

crawl-data/CC-MAIN-2023-06/segments/1674764499934.48/warc/CC-MAIN-20230201112816-20230201142816-00081.warc.gz

107 139 Assignments Done 100% Successfully Done In January 2023 # Answer to Question #128859 in Mechanics | Relativity for Ariyo Emmanuel Question #128859 1. A particle is travelling along a straight path described as = 0.75x . If its position along the x axis is x= (8t)m, where t is in seconds, determine its speed when t=2s. 1 2020-08-26T11:23:22-0400 Particle A particle is travelling along a straight path described as "y= 0.75x" . And, "x=8t \\implies y=0.75\\times 8t=6t" So particle path is given by "s=(8t) \\ \\hat{i} +(6t)\\ \\hat{j}" Velocity of Particle "(v)=\\frac{ds}{dt}=\\frac{d}{dt} (8t) \\hat{i} +\\frac{d}{dt} (6t) \\hat{j}=8\\hat{i}+ 6\\hat{j}\\ m\/s" Or Magnitude will be "|v| =\\sqrt{8^2+6^2}=10 \\ m\/s" Need a fast expert's response? Submit order and get a quick answer at the best price for any assignment or question with DETAILED EXPLANATIONS!

{"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}

CC-MAIN-2023-06

latest

en

http://2012.igem.org/wiki/index.php?title=Team:SUSTC-Shenzhen-B/future_plan&diff=prev&oldid=145781

text/html

crawl-data/CC-MAIN-2020-24/segments/1590348504341.78/warc/CC-MAIN-20200605205507-20200605235507-00018.warc.gz

# Team:SUSTC-Shenzhen-B/future plan (Difference between revisions) Revision as of 10:22, 23 September 2012 (view source)Water (Talk | contribs)← Older edit Revision as of 10:23, 23 September 2012 (view source)Water (Talk | contribs) Newer edit → Line 24: Line 24: Sometimes the user wants a terminator with specific efficiency. If the current efficiency of users terminator is not the one he want, our program can do mutation on terminator sequences and evolve a terminator that meets the requirement of efficiency. Sometimes the user wants a terminator with specific efficiency. If the current efficiency of users terminator is not the one he want, our program can do mutation on terminator sequences and evolve a terminator that meets the requirement of efficiency. + - http://2012.igem.org/wiki/images/0/0a/Future_plan.1.JPG ## Revision as of 10:23, 23 September 2012 SUSTC iGEM Theme - Free CSS Template SUSTC iGEM Theme - Free CSS Template ## Future Version of Our Software ### Introduction Now our pc software and online tool can predict the efficiency of rho-independent terminator. There are two algorithms, algorithm 1 and algorithm 2, which are based on the model created by d'Aubenton Carafa and Elena A Lesnik respectively. But when users input the sequence of their terminator, they have to choose one of the algorithms without any idea of which one is better. What’s more, if the efficiency of their terminator is not suitable for them, they have to modify the terminator themselves. Thus, we want to solve these problems in the future. ### Future Plan 1. Change the coefficient of the scoring formular and find the best algorithm D Score = nT * 18.16 + deltaG / LH * 96.59 – 116.87 (nT is the score of t tail, deltaG is the energy change of the stemloop formation and LH is the length of stemloop sequence.) Here the coefficient of each terms can be adjusted to help make the best simulation between score and efficiency. Similarly, the calculation of E Score can also be improved. What’s more, we want to find out which algorithm can better predict the efficiency based on our experiment data.Then users don’t need to hesitate to choose algorithm. 2. Modify the terminator sequence to match the user selected efficiency Sometimes the user wants a terminator with specific efficiency. If the current efficiency of users terminator is not the one he want, our program can do mutation on terminator sequences and evolve a terminator that meets the requirement of efficiency.

{"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}

CC-MAIN-2020-24

latest

en