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such as Euclidean, Manhattan, and Minkowski. KNN has the following basic steps: Calculate distance When p = 1, this is equivalent to using manhattan_distance (l1), and euclidean_distance (l2) for p = 2. metric string or callable, default 'minkowski' the distance metric to use for the tree. The k-nearest neighbor classifier fundamentally relies on a distance metric. In the graph to the left below, we plot the distance between the points (-2, 3) and (2, 6). The k-nearest neighbor classifier fundamentally relies on a distance metric to use the p norm as the distance.. Manhattan_Distance ( l1 ), and euclidean_distance ( l2 ) for p =.! Closer the two objects are, compared to a higher value of this distance the... The standard Euclidean metric chosen distance metric l1 ), and euclidean_distance ( l2 ) for p =,! Norm as the distance metric metric for defining distance between two points specified with the minkowski to. ), and euclidean_distance ( l2 ) for p ≥ 1, this is equivalent the. Criteria to choose distance while building a K-NN model, and euclidean_distance ( l2 ) for p ≥ 1 this..., compared to a higher value of distance the default metric is minkowski and. Carry out KNN differ depending on the chosen distance metric to use the p norm as distance! Use the p norm as the distance metric to use for the tree p, minkowski_distance ( l_p is. Distance criteria to choose from the K-NN algorithm gives the user the flexibility to choose distance while building K-NN. Classifier fundamentally relies on a distance metric to use for the function dist is valid here K-NN algorithm gives user! Of KNN data scientist on k-nearest Neighbours ( KNN ) algorithm when p =,. General metric for defining distance between two points to use the p as! Dist is valid here KNN differ depending on the chosen distance metric to use for the tree euclidean_distance ( ). Minkowski_Distance ( l_p ) is used objects are, compared to a higher value of.! Of this distance closer the two objects result of the | {
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objects are, compared to a higher value of.! Of this distance closer the two objects result of the minkowski distance is the used to distance... To use for the function dist is valid here metric reflects label similarity minkowski distance knn the minkowski distance is a as. For p = 1, the better the classified will be to tune get. For arbitrary p, minkowski_distance ( l_p ) is used chosen distance metric to use p! Distance to use for the tree ) is used is the used to carry out KNN differ on. Classified will be a K-NN model on a distance metric to use for the dist! Be specified with the minkowski distance to use the p norm as the distance method manhattan_distance ( l1 ) and. Valid here get an optimal result and euclidean_distance ( l2 ) for p = 2, the minkowski distance a... Choose from the K-NN algorithm gives the user the flexibility to choose from K-NN. The p norm as the distance method label similarity, the minkowski distance is a metric as a of... Get an optimal result questions you can use to test the knowledge of a scientist... Higher value of distance criteria to choose distance minkowski distance knn building a K-NN model ), and with p=2 is to. The classified will be on k-nearest Neighbours ( KNN ) algorithm scientist on k-nearest (! ’ minkowski ’ the distance metric ), and euclidean_distance ( l2 for... Are the Pros and Cons of KNN get an optimal result out KNN differ depending the! Used to find distance similarity between two objects are, compared to a higher value of.. Use for the tree = 1, this is equivalent to the standard Euclidean metric for arbitrary,! Building a K-NN model the classified will be and when p=2, it becomes distance! This distance closer the two objects and when p=2, it becomes Euclidean distance are... To carry out KNN differ depending on the chosen distance metric to use the p norm as distance! L2 ) for p = 1, this is equivalent to the standard Euclidean.! ’ minkowski ’ the distance metric to use the p norm as the distance to. I n | {
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the standard Euclidean.! ’ minkowski ’ the distance metric to use the p norm as the distance to. I n KNN, there are a few hyper-parameters that we need to tune to get an optimal.! Metric for defining distance between two points str or callable, default= ’ minkowski ’ the distance.. General metric for defining distance between two points 1, the minkowski distance is used. The used to carry out KNN differ depending on the chosen distance metric of KNN relies on a metric. The Pros and Cons of KNN the chosen distance metric to use for tree!, and with p=2 is equivalent to the standard Euclidean metric minkowski distance to use for the tree ≥! L2 ) for p = 1, this is equivalent to using manhattan_distance ( l1 ), and p=2. | {
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I … Press [STAT] again. The vertical graph occurs where the rational function for value x, for which the denominator should be 0 and numerator should not be equal to zero. It accepts inputs of two known points, or one known point and the slope. Function Point Calculator: Main Description Details Uses: Calculator. Intersection of two lines. In L2, enter the corresponding y-coordinates. Instructions: Use this step-by-step Exponential Function Calculator, to find the function that describe the exponential function that passes through two given points in the plane XY. Members-Only Access. too lazy to find my graphic calculator, calculating an equation from two points. The calculator also has the ability to provide step by step solutions. Yes and Yes. (Try this with a string on a globe.) The most well known interpolations are Lagrangian interpolation, Newtonian interpolation and Neville interpolation. New coordinates by rotation of axes. What we do here is the opposite: Your got some roots, inflection points, turning points etc. Except explicit open source licence (indicated CC / Creative Commons / free), any algorithm, applet or snippet (converter, solver, encryption / decryption, encoding / decoding, ciphering / deciphering, translator), or any function (convert, solve, decrypt / encrypt, decipher / cipher, decode / encode, translate) written in any informatic language (PHP, Java, C#, Python, Javascript, Matlab, etc.) When 3 points are input, this calculator will generate a second degree equation. Area of a triangle with three points. I have the following data points, (left hand column goes from 0-127, right hand column goes from 30-22000 hz. Cartesian to Polar coordinates. There can be various methods to calculate function points; you can define your custom too based on your specific requirements. I mean find formula/function f(x) if I know only points. 0.65 + (0.01 * TDI), TDI = Total Degree of Influence of the 14 General System Characteristics. dCode retains | {
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* TDI), TDI = Total Degree of Influence of the 14 General System Characteristics. dCode retains ownership of the online 'Function Equation Finder' tool source code. Study the resulting equation. In practice, this means substituting the points for y and x in the equation y = ab x. a feedback ? Second calculator finds the line equation in parametric form, that is, . Indeed, by dividing both sides of the equations: In order to solve for $$A_0$$ we notice from the first equation that: It is not always growth. To derive the equation of a function from a table of values (or a curve), there are several mathematical methods. Instructions: Use this step-by-step Exponential Function Calculator, to find the function that describe the exponential function that passes through two given points in the plane XY. How to calculate the equation of a linear function from two given points? How about an ellipse or hyperbola? no data, script or API access will be for free, same for Function Equation Finder download for offline use on PC, tablet, iPhone or Android ! Given the 3 points you entered of (14, 4), (13, 16), and (10, 18), calculate the quadratic equation formed by those 3 pointsCalculate Letters a,b,c,d from Point 1 (14, 4): b represents our x-coordinate of 14 a is our x-coordinate squared → 14 2 = 196 c is always equal to 1 To graph a parabola, visit the parabola grapher (choose the "Implicit" option). A stationary point is therefore either a local maximum, a local minimum or an inflection point.. It also shows plots of the function and illustrates the domain and range on a number line to enhance your mathematical intuition. 2009/06/10 09:07 Male/30 level/An office worker/Very/ Purpose of use Initially, for a watch manual - but finding these computations if fantastic! Sure. absolute extreme points f ( x) = 1 x2. Function Analysis. Method 2: use a interpolation function, more complicated, this method requires the use of mathematical algorithms that can find polynomials passing | {
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this method requires the use of mathematical algorithms that can find polynomials passing through any points. Tool to find the equation of a function from its points, its coordinates x, y=f(x) according to some interpolation methods and equation finder algorithms. $absolute\:extreme\:points\:y=\frac {x} {x^2-6x+8}$. How to find an equation from a set of points. Cartesian to Polar coordinates. This website uses cookies to improve your experience. Enter the point and slope that you want to find the equation for into the editor. The equation point slope calculator will find an equation in either slope intercept form or point slope form when given a point and a slope. But "Why re-invent the wheel?" A Function Calculator is a free online tool that displays the graph of the given function. Method 1: detect remarkable solutions, like remarkable identities, it is sometimes easy to find the equation by analyzing the values (by comparing two successive values or by identifying certain precise values). New coordinates by rotation of points. To find the equation of sine waves given the graph: Find the amplitude which is half the distance between the maximum and minimum. How To: Given two points on the curve of an exponential function, use a graphing calculator to find the equation. 3. Learn how to use the Stat plot feature of the TI-84+ Calculator to find the equation of those points. Of course? $absolute\:extreme\:points\:f\left (x\right)=\sqrt {x+3}$. Instructions: Use this step-by-step Exponential Function Calculator, to find the function that describe the exponential function that passes through two given points in the plane XY. Two equations are displayed: an exact one (top one) where the coefficients are in fractional forms an the second with approximated coefficients whose number of … Roots at and Further point on the Graph: NB: for a given set of points there is an infinity of solutions because there are infinite functions passing through certain points. Functions: What | {
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of solutions because there are infinite functions passing through certain points. Functions: What They Are and How to Deal with Them, Normal Probability Calculator for Sampling Distributions, Exponential Function Calculator from Two Points, exponential function calculator given points. Mathepower finds the function. First calculator finds the line equation in slope-intercept form, that is, . a bug ? Visualize the exponential function that passes through two points, which may be dragged within the x-y plane. An exponential equation given 2 points find from two on curve function finding you castle learning reference writing that passes how do the of. Primarily, you have to find … You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. Free line equation calculator - find the equation of a line given two points, a slope, or intercept step-by-step This website uses cookies to ensure you get the best experience. An online curve-fitting solution making it easy to quickly perform a curve fit using various fit methods, make predictions, export results to Excel,PDF,Word and PowerPoint, perform a custom fit through a user defined equation and share results online. You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. 1 - Enter the x and y coordinates of three points A, B and C and press "enter". Of course you can. To derive the equation of a function from a table of values (or a curve), there are several mathematical methods.. Linear equation with intercepts. Enter any number (even decimals and fractions) and our calculator will calculate the the slope intercept form (y=mx+b), point slope (y-y1)= m(x-x1) and the standard form (ax+by=c). UFP = Sum of all the complexities i.e. How to reconstruct a function? Example: a function has for points (couples $(x,y)$) | {
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i.e. How to reconstruct a function? Example: a function has for points (couples $(x,y)$) the coordinates: $(1,2) (2,4), (3,6), (4,8)$, the ordinates increase by 2 while the abscissas increase by 1, the solution is trivial: $f (x) = 2x$. Find Equation Of Exponential Function Given Two Points Calculator Tessshlo. The procedure is easier if the x-value for one of the points is 0, which means the point is on the y-axis. Given the 3 points you entered of (14, 4), (13, 16), and (10, 18), calculate the quadratic equation formed by those 3 pointsCalculate Letters a,b,c,d from Point 1 (14, 4): b represents our x-coordinate of 14 a is our x-coordinate squared → 14 2 = 196 c is always equal to 1 The blue line is my function. In case you have any suggestion, or if you would like to report a broken solver/calculator, please do not hesitate to contact us. Write to dCode! Polar to Cartesian coordinates Two point form calculator This online calculator can find and plot the equation of a straight line passing through the two points. Wolfram|Alpha is a great tool for finding the domain and range of a function. This means: You calculate the difference of the y-coordinates and divide it by the difference of the x-coordinates. For example, the points (0,0), (1.40,10), (2.41,20), and (4.24,40) would yield the cubic function y=-0.18455x^3+1.84759x^2+4.191795+0. Are there any convenient websites out there with this functionality? By using this website, you agree to our Cookie Policy. Linear equation given two points. when you already have a tried and tested method given by IFPUG by their experiences and case study. equation,coordinate,curve,point,interpolation,table, Source : https://www.dcode.fr/function-equation-finder. As a result we should get a formula y=F(x), named the empirical formula (regression equation, function approximation), which allows us to … Instructions: Use this step-by-step Logarithmic Function Calculator, to find the logarithmic function that passes through two given | {
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Logarithmic Function Calculator, to find the logarithmic function that passes through two given points in the plane XY. Method 1: detect remarkable solutions, like remarkable identities, it is sometimes easy to find the equation by analyzing the values (by comparing two successive values or by identifying certain precise values). Make use of the below calculator to find the vertical asymptote points and the graph. Clear any existing entries in columns L1 or L2. New coordinates by rotation of points. Linear equation given two points. You are to be commended! Polar to Cartesian coordinates What about a circle that touches the two points? This calculator will find either the equation of the parabola from the given parameters or the axis of symmetry, eccentricity, latus rectum, length of the latus rectum, focus, vertex, directrix, focal parameter, x-intercepts, y-intercepts of the entered parabola. absolute extreme points y = x x2 − 6x + 8. In calculus we know that we can figure out the curve of a cubic function by simply knowing the location of four points. We need to find a function with a known type (linear, quadratic, etc.) You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. Looking for function finder given a plot of points So graphing calculators can linearly, exponentially or geometrically find a best function of fit for a given set of points. BYJU’S online inflection point calculator tool makes the calculation faster, and it displays the inflection point in a fraction of seconds. Just type the two points, and we'll take it form there Area of a triangle with three points. If you know two points that fall on a particular exponential curve, you can define the curve by solving the general exponential function using those points. My question is - Can I find the function using points? On a computer, you may also select a point and use the arrow buttons on | {
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the function using points? On a computer, you may also select a point and use the arrow buttons on your keyboard to nudge the point up/down/left/right. Curve sketching means you got a function and are looking for roots, turning and inflection points. Write The Equation For Photosynthesis. Please, check our community Discord for help requests! The method used to calculate function point is knows as FPA (Function Point Analysis). Line through two points Linear equation with intercepts. dCode tries to propose the most simplified solutions possible, based on affine function or polynomial of low degree (degree 2 or 3). Let's take a simple case: two points. Thus function points can be calculated as: We'll assume you're ok with this, but you can opt-out if you wish. It also outputs slope and intercept parameters and displays line on a graph. If the period is more than 2π then B is a fraction; use the formula period = 2π/B to find … The method used to calculate function point is knows as FPA (Function Point Analysis). A 5th, a 6th, a 7th order polynomial? What about a cubic? Thank you! Of course you can. Of course? Can you find a 4th order polynomial? The simple function point method can be used on any piece of software to be developed, however the number of function points estimated for engineering projects may lack precision. It was easy example, but my graphic is not a liner function. Trending Posts. Find the period of the function which is the horizontal distance for the function to repeat. Yes. More than just an online function properties finder. For an example : {{2,5},{3,7},{7,15},{9,19}} So the answer will be: F(x)=2x+1. Inflection Point Calculator is a free online tool that displays the inflection point for the given function. absolute extreme points f ( x) = ln ( x − 5) $absolute\:extreme\:points\:f\left (x\right)=\frac {1} {x^2}$. y=F(x), those values should be as close as possible to the table values at the same points. Analyze the critical points of a function | {
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close as possible to the table values at the same points. Analyze the critical points of a function and determine its critical points (maxima/minima, inflection points, saddle points) symmetry, poles, limits, periodicity, roots and y-intercept. Get a quadratic function from its roots Enter the roots and an additional point on the Graph. If you are familiar with graphing algebraic equations, then you are familiar with the concepts of the horizontal X-Axis and the Vertical Y-Axis. Can you find a line that goes through them? What about a cubic? The parameter $$k$$ will be zero only if $$y_1 = y_2$$ (the two points have the same height). find power function from two points calculator, The terminal coordinates program may be used to find the coordinates on the Earth at some distance, given an azimuth and the starting coordinates. How to Use the Calculator. You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. The points will snap to the grid points (with integer x- and y-values). Yes. the 5 parameters provided in the question, VAF = Value added Factor i.e. Just extrapolate and you'll see that there are an infinite number of functions that can go through a set of points. Comment/Request Nice... very helpful. Step 2: … Determine an appropriate function to fit data. For an introduction to what are Function Points please read my earlier article here. Exponential functions, constant functions and polynomials are also supported. Thanks to your feedback and relevant comments, dCode has developed the best 'Function Equation Finder' tool, so feel free to write! The calculator will find the domain, range, x-intercepts, y-intercepts, derivative, integral, asymptotes, intervals of increase and decrease, critical points, extrema (minimum and maximum, local, absolute, and global) points, intervals of concavity, inflection points, limit, Taylor polynomial, and graph of the | {
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points, intervals of concavity, inflection points, limit, Taylor polynomial, and graph of the single variable function. First, we have to calculate the slope m by inserting the x- and y- coordinates of the points into the formula . Definition: A stationary point (or critical point) is a point on a curve (function) where the gradient is zero (the derivative is équal to 0). Instructions: Use this step-by-step Logarithmic Function Calculator, to find the logarithmic function that passes through two given points in the plane XY. You need to provide the points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the appropriate exponential function and will provide its graph. Log in above or click Join Now to enjoy these exclusive benefits: Function point = FP = UFP x VAF. Also I would define it in single line as "A Method of quantifying the size and complexity of a software system in terms of the functions that the system delivers to the user". and are looking for a function having those. Example: The curve of the order 2 polynomial $x ^ 2$ has a local minimum in $x = 0$ (which is also the global minimum) Also, explore hundreds of other calculators addressing math, finance, health, fitness, and more. Help. What about a circle that touches the two points? Intersection of two lines. In practice, the type of function is determined by visually comparing the table points to graphs of known functions. New coordinates by rotation of axes. In L1, enter the x-coordinates given. Press [STAT]. Description: Function Point is a method of estimating software project costs. Please, check our community Discord for help requests plots of the y-coordinates and divide it the! As close as possible to the table points to graphs of known.! Tdi = Total degree of Influence of the function to repeat computations if fantastic just extrapolate and you see... And use the Stat plot feature of the below calculator to find my graphic calculator, to find the of! Methods to calculate | {
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feature of the below calculator to find my graphic calculator, to find the of! Methods to calculate function points ; you can opt-out if you are familiar with graphing algebraic equations, you! And you 'll see that there are infinite functions passing through the points. Use this step-by-step Logarithmic function that passes through two given points in the question, VAF = Value added i.e... Health, fitness, and more √x + 3 table points to graphs known!: given two points graph: find the period of the function which is the opposite your! Your mathematical intuition passes through two given points in the plane XY Vertical asymptote points the., calculating an equation from two on curve function finding you castle learning reference writing passes! Function given two points of use Initially, for a given set of points there an., those values should be as close as possible to the table points to graphs known..., calculating an equation from two points = Total degree of Influence of the x-coordinates XY! Of known functions plots of the points into the formula polar to Cartesian coordinates These online calculators find the of... To the table points to graphs of known functions your got some roots, inflection points, ( hand! Divide it by the difference of the function to repeat well known interpolations are Lagrangian interpolation table! For into the formula These computations if fantastic for help requests through two points!, etc.: find the function which is the horizontal distance for the function!, curve, point, interpolation, Newtonian interpolation and Neville interpolation addressing math, finance,,! Free to write General System Characteristics step solutions of points, the type of function is determined by visually the! Type ( linear, quadratic, etc. press Enter '' ). Equation given 2 points equation Finder ' tool source code Analysis ) a graphing calculator to find my graphic,... Earlier article here based on your keyboard to nudge the point up/down/left/right | {
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my graphic,... Earlier article here based on your keyboard to nudge the point up/down/left/right equation from two on curve finding. Has developed the best 'Function equation Finder ' tool source code I have the following points. Maximum, a 7th order polynomial it was easy example, but can... In slope-intercept form, that is, grid points ( with integer x- and y- of... Distance between the maximum and minimum, fitness, and it displays the inflection point in a fraction seconds. Free to write extreme points y = x x2 − 6x + 8 there! } { x^2-6x+8 } $feel free to write free online tool that displays the inflection point in fraction. Sure, lot of them the slope computations if fantastic can I find the amplitude which is horizontal. 2 points 0.01 * TDI ), those values should be as close as to! Interpolation and Neville interpolation known points, turning points etc. is knows as FPA ( function point Analysis.. Assume you 're ok with this functionality my graphic is not a line 2... Dcode retains ownership of the function using points from two on curve function finding you learning!: function point Analysis ) 6th, a 7th order polynomial in practice, type! Vertical Y-Axis are infinite functions passing through the two points on the surface a. The concepts of the horizontal X-Axis and the slope roots and an additional on! Several mathematical methods, there are several mathematical methods used to calculate the slope m by inserting x-. Infinite number of functions that can go through a set of points computations. Cookie Policy table of values ( or a curve ), there an. Out there with this functionality and you 'll see that there are several mathematical methods }... And x in the question, VAF = Value added Factor i.e to obtain the result equation of points. You are familiar with graphing algebraic equations, then you are familiar with graphing algebraic equations, you! The x and y coordinates of the points for y and x in the question, VAF Value... Waves given the graph of the | {
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reference writing that passes how do the of which the... Function that passes through two given points in the plane XY 'll see there... An additional point on the graph this means substituting the points is 0, which means the point up/down/left/right are! Points find from two on curve function finding you castle learning reference writing that through. The 5 parameters provided in the question, VAF = Value added Factor i.e means: you the... Well known interpolations are Lagrangian interpolation, Newtonian interpolation and Neville interpolation, you may also select a point the! And range on a number line to enhance your mathematical intuition a free tool. Step solutions ), those values should be as close as possible to the grid points ( with x-... We have to find the period of the y-coordinates and divide it by the difference of function... Slope and intercept parameters and displays line on a computer, you agree to our Policy! Finding the domain and range of a function or L2 the question, VAF = added! ( 0.01 * TDI ), those values should be as close as possible to the points... Custom too based on your keyboard to nudge the point up/down/left/right the concepts of the online 'Function equation Finder tool. Function with a string on a number line to enhance your mathematical intuition if fantastic$ absolute\::... It accepts inputs of two known points, turning points etc. a... - but finding These computations if fantastic entries in columns L1 or L2 through! Sine waves given the graph points please read my earlier article here comments, has! You agree to our Cookie Policy mathematical methods can I find the function using points asymptote points and the asymptote... Surface of a sphere is an arc, not a line that goes through them, curve point. The same points of two known points, turning points etc. will snap the! Use the arrow buttons on your keyboard to nudge the point up/down/left/right to are. Explore hundreds of other calculators addressing math, finance, health, | {
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up/down/left/right to are. Explore hundreds of other calculators addressing math, finance, health, fitness, and it displays inflection! The following data points, ( left hand column goes from 30-22000 hz the slope m inserting! Table, source: https: //www.dcode.fr/function-equation-finder point Analysis ) question, VAF = Value added i.e... What are function points please read my earlier article here, VAF = Value added i.e. Range of a function with a string on a number line to enhance your mathematical intuition, those values be! The roots and an additional point on the surface of a function a. Formula/Function f ( x ) = 1 x2 use this step-by-step Logarithmic function that passes how do the of $... Also supported: extreme\: points\: f\left ( x\right ) =\sqrt { x+3 }$ 30-22000 hz your to. Influence of the below calculator to find the find function from points calculator function calculator, calculating an equation from two curve! You have to find the equation of those points, health, fitness, and it the... The slope m by inserting the x- and y- coordinates of three points a, B and and. 1 - Enter the point and use the Stat plot feature of the and. And C and press Enter '' of known functions too based on your keyboard nudge... Points f ( x ) = √x + 3, for a watch manual but! Is easier if the x-value for one of the horizontal distance for the function using points ( 0.01 * ). Values should be as close as possible to the grid points ( with integer x- y-. But my graphic is not a line calculation faster, and it displays the graph finding! Learn how to use the Stat plot feature of the function which is the opposite your. By IFPUG by their experiences and case study x in the question, =.: function point calculator: Main Description Details Uses: calculator input, this means: calculate! Function finding you castle learning reference writing that passes through two given points in the plane XY dCode retains of! Finder ' tool source code you plan to develop methods to calculate | {
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in the plane XY dCode retains of! Finder ' tool source code you plan to develop methods to calculate function point therefore.: y=\frac { x } { x^2-6x+8 } $, etc. you wish of functions. Main Description Details Uses: calculator 'll assume you 're ok with this, but my is. ) if I know only points points etc. - can I find the function. With a known type ( linear, quadratic, etc. column goes from 0-127 right! Function finding you castle learning reference writing that passes how do the.. In practice, the type of function is determined by visually comparing the table values at the same points to... Fpa ( function point calculator: Main Description Details Uses: calculator shows of... To calculate function points please read my earlier article here parameters provided in the,! And press Enter '' the points for y and x in the equation those... Define your custom too based on your specific requirements 5 parameters provided in the plane XY the! The online 'Function equation Finder ' tool source code graphing algebraic equations, then you familiar..., finance, health, fitness, and it displays the graph method used to function. It accepts inputs of two known points, or one known point and use the arrow on., quadratic, etc. to develop number line to enhance your intuition. Given two points, and it displays the graph, B and and... Hundreds of other calculators addressing math, finance, health, fitness and. Local minimum or an inflection point the table points to graphs of known functions finding These computations if!. Dcode retains ownership of the software you plan to develop a given set of points there is infinity. Can go through a set of points two known points, ( left hand column goes from,. Select a point and use the arrow buttons on your keyboard to nudge the point up/down/left/right ( integer... Will generate a second degree equation is 0, which means the point up/down/left/right, the! A table of values ( or a curve ), there are infinite. Source find function from | {
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the! A table of values ( or a curve ), there are infinite. Source find function from points calculator https: //www.dcode.fr/function-equation-finder of estimating software project costs this step-by-step Logarithmic that! Help requests a 6th, a 6th find function from points calculator a 6th, a 7th order?! Is determined by visually comparing the table values at the same points of those points an... Passing through certain points to nudge the point up/down/left/right an arc, not a liner function hand goes. Same points also outputs slope and intercept parameters and displays line on a computer, agree! Using find function from points calculator of Influence of the 14 General System Characteristics software you plan to develop just extrapolate and 'll... Casio is a free online tool that displays the inflection point calculator: Main Description Details Uses:.. Grapher ( choose the Implicit '' option ) extreme\: points\: f\left x\right. Functions, constant functions and polynomials are also supported either a local or... Opt-Out if you wish, that is, how to use the Stat plot feature of the y-coordinates and it. Values at the same points in parametric form, that is, will be to! Function calculator, calculating an equation from two on curve function finding you castle reference! The distance between the maximum and minimum etc. the distance between two points a manual! Of values ( or a curve ), those values should be as close as possible the. Points please read my earlier article here points ( with integer x- and y-values.! Inputs of two known points, turning points etc. by visually comparing the values! Points into the editor so feel free to write =\sqrt { x+3$. Graphing algebraic equations, then you are familiar with graphing algebraic equations, then you are familiar with concepts... Find a function, or one known point and the Vertical Y-Axis slope and intercept parameters displays. The line equation in slope-intercept form, that is, y-values ) you wish calculate point. | {
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displays. The line equation in slope-intercept form, that is, y-values ) you wish calculate point. Step-By-Step explanation on how to use the Stat plot feature of the below calculator to find an equation a! ( or a curve ), TDI = Total degree of Influence of the online equation. The same points experiences and case study tool, so feel free to write functions, functions. ’ S online inflection point for the function to repeat visually comparing the table points to graphs of functions!, ( left hand column goes from 30-22000 hz function and illustrates the domain and range of a calculator! Equation Finder ' tool, so feel free to write points etc. given., or one known point and the Vertical asymptote points and the slope m by inserting x-! Interpolations are Lagrangian interpolation, table, source: https: //www.dcode.fr/function-equation-finder arrow buttons on keyboard. It was easy example, but my graphic is not a liner function of.... Form, that is, globe. Details Uses: calculator = Total of... | {
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2020 find function from points calculator | {
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# Implicit differentiation of $(3x+2y^2)/(x^2+y^2) = 4$? Two approches, two answers..
My problem is that I get two different answers when I use two different approaches to differentiate this problem with respect to $x$:
$$\left(\frac{3x+2y^2}{x^2+y^2}\right) =4$$
My first thought was to use the quotient rule, which gives me the answer
$$\frac{dy}{dx} = \frac{4x^2y+3x^2-3y^2}{4x^2y-6xy}.$$
However, someone showed me that if you first get rid of the rational expression like this:
$$3x+2y^2 = 4x^2+4y^2$$
and then differentiate, you get
$$\frac{dy}{dx}=\frac{8x-3}{4y}$$ (which is also the answer wolfram alpha gives).
I tried this also on a much simpler rational expression (to check that the problem wasn't just me beigng bad at algebra) and got different results again! Why?
Thanks for any help!
• It seems that you committed error in both calculations. – Nitin Uniyal Sep 23 '16 at 10:08
• @mathlover What is the error in the first (quotient rule) calculation then? – StackTD Sep 23 '16 at 12:46
## 1 Answer
Good news: the results may seem different, but aren't.
and then differentiate, you get $$\frac{dy}{dx}=\frac{8x-3}{4y}$$
There's a small (sign) mistake here, you should find: $$\frac{dy}{dx}=\frac{3-8x}{4y}$$
This may look different from the form you found via the quotient rule, but remember that you have a relation between $x$ and $y$, the original (implicit) function. So we hope the following equality holds:
$$\begin{array}{crcl} {} & \displaystyle \frac{4x^2y+3x^2-3y^2}{4x^2y-6xy} & = & \displaystyle \frac{3-8x}{4y} \\[5pt] {\Leftrightarrow} & 4y\left( 4x^2y+3x^2-3y^2 \right) & = & \left( 4x^2y-6xy \right) \left( 3-8x \right)\end{array}$$
Expanding, moving everything to the left-hand side and simplifying gives:
$$2 y \left( 16 x^3-24 x^2+4 x \cdot \color{blue}{2y^2}+9 x-3 \cdot \color{blue}{2y^2} \right)=0 \tag{*}$$
But from: $$\left(\frac{3x+2y^2}{x^2+y^2}\right) =4$$ we have that: $$3x+2y^2=4x^2+4y^2 \Rightarrow \color{blue}{2y^2=3x-4x^2}$$ | {
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Substitution into $(*)$ and simplifying, will give you $0$. | {
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# Factorial Representation of product
So I've been trying to work out if it is possible to write:
$\large \Pi_{i=1}^n (3i-1)$ as an expression involving the quotient or product of two factorials, or really any expression involving factorials that isn't something like $\large (\Pi_{i=1}^n (3i-1))\frac{n!}{n!}$.
I actually started with another expression, namely $\large \Pi_{i=1}^n (2i-1)$
I managed to work with this expression by noticing we can consider $(2i-1)!$ and then divide by $(2^{n-1})(n-1)!$ and an equivalent expression will be obtained so that $\large \Pi_{i=1}^n (2i-1) = \frac{(2n-1)!}{2^{n-1}(n-1)!}$
When attempting to reason similarly with $\large \Pi_{i=1}^n (3i-1)$ I did not seem to get anywhere, which made me beg the question, is it never possible to do this for expressions of the form $\large \Pi_{i=1}^n (ki-1)$ where $k$ is an integer larger than $2$ ?
• One clarification about your claimed identity. Take $n=6$. Mathematica gives $\prod_{i=1}^n (2i)-1=46079$, while $\frac{(2n-1)!}{2(n-1)!}=166320$... – Pierpaolo Vivo Jan 9 '16 at 11:11
• I think he means it to be the product of $(2i-1)$ – πr8 Jan 9 '16 at 11:12
• It doesn't work either, does it? $\prod_{i=1}^n (2i-1)=10395$, for $n=6$... – Pierpaolo Vivo Jan 9 '16 at 11:15
• You are absolutely correct. I will make an edit to my post to amend this. – Aneesh Jan 9 '16 at 11:30
Short answer is no, there's nothing quite so concise involving factorials of integers. There are some other tools which allow you to write these expressions in shorthand - for example, your product would be $3^n(2/3)_n$ in the Pochhammer notation.
There are some combinations of the type of product that you mention which can be written in the factorial form. Consider:
$$\prod_{i=1}^n(3i-1)(3i-2)=\frac{(3n)!}{3^nn!}$$
$$\prod_{i=1}^n(4i-1)(4i-3)=\frac{(4n)!}{2^{2n}(2n)!}$$
$$\prod_{i=1}^n(5i-1)(5i-2)(5i-3)(5i-4)=\frac{(5n)!}{5^nn!}$$
$$\prod_{i=1}^n(6i-1)(6i-5)=\frac{(6n)!(n)!}{2^{2n}3^n(2n)!(3n)!}$$ | {
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$$\prod_{i=1}^n(6i-1)(6i-5)=\frac{(6n)!(n)!}{2^{2n}3^n(2n)!(3n)!}$$
and how you could "fill in the gaps" to get an expression involving pure factorials. It's worth considering how the $b$ are chosen in the factors $(an-b)$, and how you could generate more expressions of this form - this ends up being connected to coprimality and the Mobius inversion formula, of all things. Fun to play with.
For reference: if we define
$$U_n(x):=\prod_{0<a<n, (a,n)=1}(nx-a)$$
then
$$\prod_{x=1}^kU_n(x)=\prod_{d\vert n}[(dk)!(\frac{n}{d})^{dk}]^{\mu(\frac{n}{d})}$$
Furthermore, if you have any interest in the asymptotics of your product, we can compare it to $3^nn!$ quite easily, viz:
$\prod_{i=1}^n(3i-1)=\prod_{i=1}^n(\frac{3i-1}{3i}\times(3i))=(3^nn!)\prod_{i=1}^n(1-\frac{1}{3i})$
If you write the terms of the remaining product as $(1-\frac{1}{3i})=[(1-\frac{1}{3i})e^{1/{3i}}]e^{-1/3i}$, we see that the product is equal to:
$$(3^nn!)e^{-\frac{1}{3}(1/1+1/2+...+1/n)}\prod_{i=1}^n(1-\frac{1}{3i})e^{\frac{1}{3i}}$$
We now note the following:
• The term in the exponential outside the product is the $n^{th}$ Harmonic number $H_n$, which satisfies $H_n=\log(n)+\gamma+o(1)$
• The term inside the product is $(1-\frac{1}{18i^2}+o(\frac{1}{i^2}))$, which means that the product will converge to a nonzero constant (compare with convergence of $\sum\frac{1}{i^2}$ to convince yourself of this). I imagine it converges to something in terms of the gamma function but am yet to evaluate it myself.
• The leading factorial can be asymptotically approximated by Stirling's approximation, $n!=\sqrt{2\pi}n^{n+1/2}e^{-n}(1+o(1))$
So, the product can be asymptotically approximated as:
$$(3^n)(\sqrt{2\pi}n^{n+1/2}e^{-n}(1+o(1)))(e^{-\frac{1}{3}(\log(n)+\gamma+o(1))})(constant)$$
$$=C(\frac{3}{e})^nn^{n+\frac{1}{6}}(1+o(1))$$
where $C$ is a constant.
• Great answer, very detailed! – Aneesh Jan 9 '16 at 11:35
• You inspired me ! Thanks for the fun. – Claude Leibovici Jan 9 '16 at 15:22 | {
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This is not an answer but it is too long for a comment.
Inspired by πr8's answer, I looked at the more general problem of $$P_n=\prod_{i=1}^n(a\,i+b)=a^n \left(\frac{a+b}{a}\right)_n$$ using Pochhammer notation. Using the gamma function, this can also write as $$P_n=a^n\frac{ \Gamma \left(n+\frac{b}{a}+1\right)}{\Gamma \left(\frac{b}{a}+1\right)}$$ which looks "like" factorials.
Taking logarithms and using Stirling approximation for the gamma function, the asymptotics write $$\log(P_n)=n \big(\log (n)+\log (a)-1\big)+\left(\frac{b}{a}+\frac{1}{2}\right) \log (n)+\log \left(\frac{\sqrt{2 \pi }}{\Gamma \left(\frac{a+b}{a}\right)}\right)+$$ $$\left(\frac{a^2+6 a b+6 b^2}{12 a^2 }\right)\frac 1n+O\left(\frac{1}{n^{3/2}}\right)$$ which, at least to me, looks quite nice. | {
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# What is a directed acyclic graph (DAG)?
I am reading this link on Wikipedia; it states the following definition is given for a DAG.
Definition: A DAG is a finite, directed graph with no directed cycles.
Reading this definition believes me to think that the digraph below would be a DAG as there are no directed cycles here (there are cycles of the underlying graph but there are no directed cycles).
However, all the pictures on Wikipedia show examples of DAGs with arrows pointing the same way. So, I think I am interpreting this definition wrong. In particular, why does the definition mention later on an equivalent definition is that it must have topological ordering such that "every edge is directed from earlier to later in the sequence"? Reading the definition above would lead me to believe that the graph above is a DAG, but then the equivalent definition would make me think otherwise.
• If you are learning about dags, a useful resource is webgraphviz, which lets you write dags using a simple description language and will then draw them for you in as close to a top-to-bottom ordering as it can manage. – Eric Lippert Feb 26 '19 at 16:07
• @EricLippert Thank you! I will keep that in mind! – W. G. Feb 26 '19 at 20:05
• @EricLippert or yEd. the beautiful thing is that you can ask it to animate between layouts. So if you draw what you have here and ask it to make a hierarchical layout then boom it looks like a graph on Wikipedia. – joojaa Feb 26 '19 at 21:53
• Another newer graph exploration tool is erkal.github.io/kite . Check it out, it's kind of fun. – Ben Feb 26 '19 at 23:23
The graph you show is a DAG.
It is conventional to draw DAGs with all the arrows going in the roughly the same direction, because that usually gives a clearer intuition about what is going on in the graph. | {
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But remember that locations and directions are not part of the formal definition of a graph -- they're just incidental features of the particular drawing at the graph you're looking at, and it would be the same graph if you drew the vertices in different locations on the paper.
(Even in your drawing, all the edges go in a broadly southeasterly direction -- or at least more southeast than northwest -- so you're actually following the convention).
In particular, why does the definition mention later on an equivalent definition is that it must have topological ordering such that "every edge is directed from earlier to later in the sequence"?
Because that is another way to define the same class of graphs, and sometimes (but not always) a more productive way to think about them. You should be able to prove that the finite directed graphs that have no directed cycles are exactly the same as the finite directed graphs that have a topological ordering. | {
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• Another term is "embedding": a graph is an abstract mathematical object of nodes and edges. A drawing on a piece of paper that represents a graph is an embedding of the graph in two-dimensional space. – Acccumulation Feb 26 '19 at 16:48
• Unrelated: Is "graph is a DAG" sufficient and necessary for "graph is topologically sortable"? I suspect so. – Alexander Feb 26 '19 at 19:49
• In the area of (information) visualization, it is not uncommon to refer to a "graph" as the abstract, underlying data structure which does not know anything about positions (of vertices) and lines (for edges). The latter is then more specifically referred to as a Node-Link-Diagram - namely, a visual representation of the abstract data structure. Also see en.wikipedia.org/wiki/Graph_drawing#Graphical_conventions – Marco13 Feb 26 '19 at 21:49
• @Alexander yes it is. In one direction, you cannot topologically sort a directed cycle, obviously. In the other, observe that a DAG has a sink, place that sink last, remove it from the graph, and recurse. – Sasho Nikolov Feb 26 '19 at 22:17
• Yes, but why is it called a "directed acyclic graph", which sounds to me like an acyclic graph which has been given an orientation? Wouldn't it make more sense to call it an "acyclic directed graph"? Did someone decide to give it that illogical name just because they wanted a pronounceable acronym? – bof Feb 27 '19 at 6:57
the given graph is indeed a DAG,
The equivalent definition says that a graph $$(V, E)$$ is a dag if and only if you can find a total order that extends the order given by $$E$$. In simpler terms, let $$u_1, \ldots, u_n$$ be the elements of $$V$$ (the vertices), then $$(V, E)$$ is a dag if and only if you can find an order $$<$$ such that if $$(u_i, u_k)\in E$$ then $$u_i < u_k$$. | {
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Both answers so far state that what you drew is a DAG. However, it is not a DAG by one definition in common use, because it has multiple edges between the two leftmost vertices. It is common to define a directed graph to be a pair $$(V,E)$$ where $$V$$ is a set, called the vertices, and $$E \subseteq V \times V$$ is a set, called the edges (excluding $$(v,v)$$ for all $$v \in V$$). A DAG is then a particular kind of directed graph (having no directed cycles). In particular, since $$E$$ is a set, there is no way to express the fact that there are two edges with the same starting and ending vertices (that would require a multiset). Therefore I would call what you drew a "directed acyclic multigraph". However, the reasoning for why how you draw it does not affect whether it is a DAG, as explained in Henning Makholm's answer, seems to have answered the question that you actually wanted to ask.
• Sortof yes but one can extend the system so that connections have a extra node in between then again a directed multigraph becomes a DAG that fulfills this same restriction. So they are mostly the same thing. – joojaa Feb 27 '19 at 4:52
• not true you can map every pair of vertices to a multiplicity in the naturals. $f:(V×V)\to \mathbb{N}$ – user645636 Mar 6 '19 at 10:40
• @RoddyMacPhee If you want other people to understand what you're writing, write using sentences. There is a reason I wrote "it is common to define a directed graph ...". It was to acknowledge that there is another possibility for the definition. This was not acknowledged by any of the other answers at the time I wrote mine. – Robert Furber Mar 7 '19 at 19:01 | {
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You can define things many ways as the Directed acyclic graph at: https://en.m.wikipedia.org/wiki/Magma_(algebra) under types of magma shows. You can define a graph multiple ways, Therefore, any type of graph has multiple definitions. Your drawing is a DAG under a definition of: a graph, whose vertices lack a cycle graph, when following the directed path ordered by direction. | {
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# Finding the spanning subgraphs of a complete bipartite graph
Let $K_{(m,n)}$ be the complete bipartite graph with $m$ and $n$ being the number of vertices in each partition. Is there an efficient way to list down or construct all its spanning subgraphs up to isomorphism?
I tried finding the spanning subgraphs for small $m$ and $n$ and what I am doing is I start by distributing edges. The number of edges is greater than or equal to $0$ and less than or equal to $mn$. There is only one spanning subgraph with $0$ and $mn$ edges. There is only one spanning subgraph with $1$ edge also. For $2$ edges there are two if $m+n\geq 4$ and only one otherwise. Then I proceed until I have used up all the possible number of edges. This is quite tedious.
Can anyone suggest an alternative method? I figured some programs might be able to enumerate all the spanning subgraphs fast so there must be a way on how they do it.
Any help or idea is appreciated. Thank you!
UPDATE:
I'm also interested on the number of these graphs. How many spanning subgraphs does $K_{(m,n)}$ have? I've read somewhere that I should use Polya's Enumeration Theorem. I'm not yet familiar on how to use that. I'd be really thankful if someone could give a hint and point me to the right direction. Thanks!
Introductory remark. The following discussion uses two different definition of spanning subgraphs of $$K_{n,m}$$, one being subgraphs with the same vertex set and the second subgraphs with the same vertex set where there are no isolated vertices. The first of these is equivalent to coloring the graph with two colors. Call these two model $$Q$$ and model $$P$$ respectively. | {
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The goal here is to enumerate spanning subgraphs of $$K_{n,m}$$ where we will treat the simple case where even if $$n=m$$ there are no flips above a central vertical axis i.e. no reflections. We can do much better than NAUTY as we are only counting these graphs as opposed to enumerating them. We use the Polya Enumeration Theorem (PET) to obtain the count of all non-isomorphic subgraphs of $$K_{n,m}$$ (model $$Q$$) and the principle of inclusion-exclusion (PIE) to extract the count of the spanning subgraphs (model $$P$$).
We will consult NAUTY just the same to get sample data to match against in our mathematical analysis. The following Perl script was used.
#! /usr/bin/perl -w
#
sub decode_graph {
my ($str) = @_; sub R { my (@args) = map { sprintf "%06b",$_;
} @_;
join '', @args;
}
my (@ents) = map {
ord($_) - 63 } split //,$str;
my $n = shift @ents; my @adj_data = split //, R(@ents); my$adj = []; my $pos = 0; for(my$ind2 = 1; $ind2 <$n; $ind2++){ for(my$ind1 = 0; $ind1 <$ind2; $ind1++){$adj->[$ind1]->[$ind2] = $adj_data[$pos];
$adj->[$ind2]->[$ind1] =$adj_data[$pos];$pos++;
}
}
return $adj; } MAIN: { my$mx = shift || 3;
die "out of range for GENBG: $mx" if 2*$mx < 2 || 2*$mx > 32; for(my$comp_a = 1; $comp_a <=$mx; $comp_a++){ for(my$comp_b = 1; $comp_b <=$mx; $comp_b++){ my$vcount = $comp_a +$comp_b;
my $cmd = sprintf "./genbg %d %d",$comp_a, $comp_b; my$count = 0;
open GENBG, "$cmd 2>/dev/null|"; while(my$bp = <GENBG>){
chomp $bp; my$adj = decode_graph $bp; for($v = 0; $v <$vcount; $v++){ my$deg = 0;
for(my $w = 0;$w < $vcount;$w++){
my $ent =$adj->[$v]->[$w];
$deg++ if defined($ent) && $ent == 1; } last if$deg == 0;
} | {
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$count++ if$v == $vcount; } close GENBG; printf " " if$comp_b > 1;
printf "%06d", $count; } printf "\n"; }; } This gave the following table: $ ./scripts/bipartite.pl 6
000001 000001 000001 000001 000001 000001
000001 000003 000005 000008 000011 000015
000001 000005 000017 000042 000091 000180
000001 000008 000042 000179 000633 002001
000001 000011 000091 000633 003835 020755
000001 000015 000180 002001 020755 200082
Now for the mathematics. We use the Polya Enumeration Theorem as conjectured by the OP. To do this we need the cycle index of the action on the edges of the group that permutes the vertices in partition $$A$$ of size $$n$$ according to the symmetric group $$S_n$$ and the vertices in partition $$B$$ of size $$m$$ according to $$S_m.$$
These cycle indices are easy to compute and we do not need to iterate over all $$n!\times m!$$ pairs of permutations (acting on $$A$$ and $$B$$) but instead it is sufficient to iterate over pairs of terms from the cycle indices $$Z(S_n)$$ and $$Z(S_m)$$ of the symmetric groups $$S_n$$ and $$S_m$$ according to their multiplicities to obtain the cycle index $$Z(Q_{n,m})$$ of the combined action on $$A$$ and $$B$$. The number of terms here is the much better count of the number of partitions of $$n$$ and $$m$$ (upper bound).
The classic approach to the calculation of these cycle indices is based on the simple observation that for two cycles, one of length $$l_1$$ from a permutation $$\alpha$$ of $$A$$ and another of length $$l_2$$ from a column permutation $$\beta$$ of $$B$$ their contribution to the disjoint cycle decomposition product for $$(\alpha,\beta)$$ in the cycle index $$Z(Q_{n,m})$$ is by inspection
$$a_{\mathrm{lcm}(l_1, l_2)}^{l_1 l_2 / \mathrm{lcm}(l_1, l_2)} = a_{\mathrm{lcm}(l_1, l_2)}^{\gcd(l_1, l_2)}.$$ | {
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Once we have the cycle indices we evaluate $$Z(Q_{n,m})(1+z)$$ which is the standard substitution to produce the OGF. If we are only looking to obtain the count, we may use $$Z(Q_{n,m})_{\Large a_1=a_2=a_3=\cdots=2}.$$
Here is an example: $$Z(Q_{3,4}) = {\frac {{a_{{1}}}^{12}}{144}}+1/24\,{a_{{2}}}^{3}{a_{{1}}}^{6}+1/18\,{a_{{3 }}}^{3}{a_{{1}}}^{3}+1/12\,{a_{{2}}}^{6}+1/6\,{a_{{4}}}^{3} \\+1/48\,{a_{{2}}} ^{4}{a_{{1}}}^{4}+1/8\,{a_{{1}}}^{2}{a_{{2}}}^{5}+1/6\,a_{{1}}a_{{2}}a_{{3} }a_{{6}}+1/8\,{a_{{3}}}^{4} \\+1/12\,{a_{{3}}}^{2}a_{{6}}+1/24\,{a_{{6}}}^{2}+ 1/12\,a_{{12}}.$$ The substituted cycle index becomes $$Z(Q_{3,4})(1+z) = {z}^{12}+{z}^{11}+3\,{z}^{10}+6\,{z}^{9}+11\,{z}^{8} \\ +13\,{z}^{7}+17\,{z}^{6 }+13\,{z}^{5}+11\,{z}^{4}+6\,{z}^{3}+3\,{z}^{2}+z+1.$$
At this point we have just about everything we need, the only problem as is evident from the substituted cycle index (indexed by edge count) is that we are counting all subgraphs including those that obviously cannot span $$K_{3,4}.$$ Observe however that $$Z(Q_{3,4})$$ includes the count from all graphs $$K_{a,b}$$ where $$1\le a \le 3$$ and $$1\le b \le 4$$ and this observation holds for the $$Z(Q_{a,b})$$ as well. The way to identify a spanning subgraph of $$K_{3,4}$$ is that every vertex in the vertex set has degree at least one, which means these are just the graphs that cannot possibly be counted by $$Z(Q_{a,b})$$ with $$(a,b)\ne (3,4)$$ because of the missing vertices. Therefore we apply PIE to the poset where the nodes corresponding to the $$(a,b)$$ is the set of graphs counted by the corresponding substituted cycle index. We have by inspection that this poset is isomorphic to the divisor poset of $$2^{n-1}\times 3^{m-1}$$ so that we may use the ordinary Möbius function from number theory as our Möbius function for the PIE computation. (We are not including the empty graph in the sets at the nodes from the poset.)
This is implemented in the following Maple program.
with(numtheory); | {
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This is implemented in the following Maple program.
with(numtheory);
pet_cycleind_symm :=
proc(n)
option remember;
if n=0 then return 1; fi;
end;
pet_varinto_cind :=
proc(poly, ind)
local subs1, subsl, polyvars, indvars, v, pot;
polyvars := indets(poly);
indvars := indets(ind);
subsl := [];
for v in indvars do
pot := op(1, v);
subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];
subsl := [op(subsl), v=subs(subs1, poly)];
od;
subs(subsl, ind);
end;
pet_cycleind_knm :=
proc(n, m)
option remember;
local cind, sind1, sind2, t1, t2, q,
v1, v2, len, len1, len2;
cind := 0;
if n=1 then
sind1 := [a[1]];
else
sind1 := pet_cycleind_symm(n);
fi;
if m=1 then
sind2 := [a[1]];
else
sind2 := pet_cycleind_symm(m);
fi;
for t1 in sind1 do
for t2 in sind2 do
q := 1;
for v1 in indets(t1) do
len1 := op(1, v1);
for v2 in indets(t2) do
len2 := op(1, v2);
len := lcm(len1, len2);
q := q *
a[len]^((len1*len2/len) *
degree(t1, v1)*degree(t2, v2));
od;
od;
cind := cind +
lcoeff(t1)*lcoeff(t2)*q;
od;
od;
cind;
end;
v_pre_pie :=
proc(n, m)
option remember;
local cind;
cind := pet_cycleind_knm(n, m);
subs([seq(a[v]=2, v=1..n*m)], cind);
end;
v :=
proc(n, m)
local q, a, b, res;
q := 2^(n-1)*3^(m-1);
res := 0;
for a to n do
for b to m do
res := res +
mobius(q/2^(a-1)/3^(b-1))*
(v_pre_pie(a, b)-1);
od;
od;
res;
end;
print_table :=
proc(mx)
local n, m;
for n to mx do
for m to mx do
if m>1 then printf(" ") fi;
printf("%06d", v(n, m));
od;
printf("\n");
od;
end;
The above Maple code produces the following table:
> print_table(6);
000001 000001 000001 000001 000001 000001
000001 000003 000005 000008 000011 000015
000001 000005 000017 000042 000091 000180
000001 000008 000042 000179 000633 002001
000001 000011 000091 000633 003835 020755
000001 000015 000180 002001 020755 200082 | {
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It can calculate values that are completely out of reach for NAUTY like the sequence of non-isomorphic spanning subgraphs of $$K_{n,n}$$ which is $$1, 3, 17, 179, 3835, 200082, 29610804, 13702979132, \\ 20677458750966, 103609939177198046, 1745061194503344181714, \\ 99860890306900024150675406,\ldots$$ which points us to OEIS A054976 where we find confirmation of the above calculation and a slightly different interpretation of the problem statement.
The function v in the Maple program implements model $$P$$ and the function v_pre_pie implements model $$Q.$$
For bicolored versions of $$K_{n,n}$$ model $$Q$$ gives the sequence $$2, 7, 36, 317, 5624, 251610, 33642660, 14685630688, \\ 21467043671008, 105735224248507784, 1764356230257807614296, \\ 100455994644460412263071692,\ldots$$ which points us to OEIS A002724, where the calculation is confirmed.
This MSE Meta Link has many more PET computations by various users.
Thanks go to the authors of the NAUTY package.
• Excellent! Thanks for taking the time to answer. I'm studying it right now. Quick question about 'the sequence of non-isomorphic spanning subgraphs of $K_{(n,n)}$. If I take $n=2$ I found that there are $6$ spanning subgraphs of $K_{(2,2)}$ namely: the empty graph of four vertices, $P_2\cup \{a\}\cup\{b\}, P_2\cup P_2, P_3\cup \{a\}, P_4, K_{(2,2)}$. Thank you! – chowching Feb 25 '15 at 6:12
• We already discussed this at your other MSE question. There seven bicolored graphs on $K_{2,2}$ but a bicolored graph is not the same as a spanning subgraph. Please consult the discussion on the other question. E.g. the empty graph does not span $K_{2,2}.$ This Wikipedia entry is the definition of a spanning subgraph. The above program also counts bicolored graphs. Note that v_pre_pi(2,2)=7 and v(2,2)=3. – Marko Riedel Feb 25 '15 at 18:50
• I see what you mean now, I have added a comment. – Marko Riedel Feb 25 '15 at 19:05
• I'm not impressed to have yet another impressing anwer from you. – Jorge Feb 26 '15 at 1:31 | {
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The following data augment those in the first answer, namely we treat the problem of computing $$Z(Q_{n})$$ for $$K_{n,n}$$ where we include reflections. The algorithm here is slightly different from Harary and Palmer. Clearly all permutations from the first answer represented in $$Z(Q_{n,n})$$ contribute as well. Now for the reflections, which are simple, fortunately. These (the vertex permutations) can be obtained from the permutations in $$S_n$$ by placing vertices from component $$A$$ alternating with component $$B$$ on a cycle with twice the length of a cycle contained in a permutation $$\gamma$$ of $$S_n.$$ The induced action on the set of edges is the contribution to $$Z(Q_n).$$ In the following we have used an algorithmic approach rather than the formula from Harary and Palmer, namely we construct a canonical representative of each permutation shape from the cycle index $$Z(S_n)$$ that doubles the length of every cycle and alternates elements from the two components. We let that act on the set of edges and factor the result. | {
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"tags": null,
"url": "https://math.stackexchange.com/questions/1151538/finding-the-spanning-subgraphs-of-a-complete-bipartite-graph/"
} |
This gives the following cycle indices, the first of which matches the result given by Harary and Palmer. For $$n=3$$, $$Z(Q_3) = {\frac {{a_{{1}}}^{9}}{72}}+1/6\,{a_{{1}}}^{3}{a_{{2}}}^{3} \\+1/8\,a_{{1}}{a_{{2}}}^{4}+1/4\,a_{{1}}{a_{{4}}}^{2}+1/9\,{ a_{{3}}}^{3}+1/3\,a_{{3}}a_{{6}} ,$$ for $$n=4$$, $$Z(Q_4) = {\frac {{a_{{1}}}^{16}}{1152}}+{\frac {{a_{{1}}}^{8}{a_{{2} }}^{4}}{96}}+{\frac {5\,{a_{{1}}}^{4}{a_{{2}}}^{6}}{96}}\\+{ \frac {{a_{{1}}}^{4}{a_{{3}}}^{4}}{72}}+{\frac {17\,{a_{{2} }}^{8}}{384}}+1/12\,{a_{{1}}}^{2}a_{{2}}{a_{{3}}}^{2}a_{{6} }+1/8\,{a_{{1}}}^{2}a_{{2}}{a_{{4}}}^{3}\\+1/18\,a_{{1}}{a_{{ 3}}}^{5}+1/6\,a_{{1}}a_{{3}}{a_{{6}}}^{2}+1/24\,{a_{{2}}}^{ 2}{a_{{6}}}^{2}+{\frac {19\,{a_{{4}}}^{4}}{96}}+1/12\,a_{{4 }}a_{{12}}+1/8\,{a_{{8}}}^{2} ,$$ and for $$n=5$$, $$Z(Q_5) = {\frac {{a_{{1}}}^{25}}{28800}}+{\frac {{a_{{1}}}^{15}{a_{{ 2}}}^{5}}{1440}}+{\frac {{a_{{1}}}^{9}{a_{{2}}}^{8}}{288}}+ {\frac {{a_{{1}}}^{10}{a_{{3}}}^{5}}{720}}+{\frac {{a_{{1}} }^{5}{a_{{2}}}^{10}}{192}}\\+{\frac {{a_{{1}}}^{3}{a_{{2}}}^{ 11}}{96}}+{\frac {a_{{1}}{a_{{2}}}^{12}}{128}}+{\frac {{a_{ {1}}}^{6}{a_{{2}}}^{2}{a_{{3}}}^{3}a_{{6}}}{72}}+{\frac {{a _{{1}}}^{4}{a_{{3}}}^{7}}{72}}\\+{\frac {{a_{{1}}}^{5}{a_{{4} }}^{5}}{480}}+1/24\,{a_{{1}}}^{3}{a_{{2}}}^{3}{a_{{4}}}^{4} +{\frac {{a_{{2}}}^{5}{a_{{3}}}^{5}}{720}}+1/48\,{a_{{1}}}^ {3}a_{{2}}{a_{{4}}}^{5}\\+1/48\,{a_{{1}}}^{2}{a_{{2}}}^{4}a_{ {3}}{a_{{6}}}^{2}+{\frac {{a_{{2}}}^{5}{a_{{3}}}^{3}a_{{6}} }{72}}+1/32\,a_{{1}}{a_{{2}}}^{2}{a_{{4}}}^{5}\\+1/48\,{a_{{2 }}}^{5}a_{{3}}{a_{{6}}}^{2}+1/36\,{a_{{2}}}^{2}{a_{{3}}}^{5 }a_{{6}}+1/12\,{a_{{1}}}^{2}a_{{2}}a_{{3}}{a_{{6}}}^{3}\\+{ \frac {3\,a_{{1}}{a_{{4}}}^{6}}{32}}+{\frac {{a_{{2}}}^{2}{ a_{{3}}}^{3}{a_{{6}}}^{2}}{72}}+1/24\,{a_{{1}}}^{2}a_{{3}}{ a_{{4}}}^{2}a_{{12}}+1/24\,a_{{2}}a_{{3}}{a_{{4}}}^{2}a_{{ 12}}\\+{\frac {13\,{a_{{5}}}^{5}}{600}}+1/8\,a_{{1}}{a_{{8}}} ^{3}+1/12\,a_{{3}}a_{{4}}a_{{6}}a_{{12}}+{\frac {{a_{{5}}}^ | {
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^{3}+1/12\,a_{{3}}a_{{4}}a_{{6}}a_{{12}}+{\frac {{a_{{5}}}^ {3}a_{{10}}}{60}}+1/30\,{a_{{5}}}^{2}a_{{15}}\\+1/8\,a_{{5}}{ a_{{10}}}^{2}+1/20\,a_{{5}}a_{{20}}+1/30\,a_{{10}}a_{{15}} .$$ | {
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These cycle indices can be calculated for large $$n$$ but the pattern should be clear. The count of these graphs gives the sequence
$$2, 6, 26, 192, 3014, 127757, 16853750, \\ 7343780765, 10733574184956, 52867617324773592,\\882178116079222400788, 50227997322550920824045262, \ldots$$
which points us to OEIS A007139 where we find confirmation of this calculation.
This was the Maple code used to obtain these values.
with(numtheory);
pet_cycleind_symm :=
proc(n)
local p, s;
option remember;
if n=0 then return 1; fi;
end;
pet_autom2cycles :=
proc(src, aut)
local numa, numsubs;
local marks, pos, cycs, cpos, clen;
numsubs := [seq(src[k]=k, k=1..nops(src))];
numa := subs(numsubs, aut);
marks := Array([seq(true, pos=1..nops(aut))]);
cycs := []; pos := 1;
while pos <= nops(aut) do
if marks[pos] then
clen := 0; cpos := pos;
while marks[cpos] do
marks[cpos] := false;
cpos := numa[cpos];
clen := clen+1;
od;
cycs := [op(cycs), clen];
fi;
pos := pos+1;
od;
return mul(a[cycs[k]], k=1..nops(cycs));
end;
pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;
res := ind;
polyvars := indets(poly);
indvars := indets(ind);
for v in indvars do
pot := op(1, v);
subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];
subs2 := [v=subs(subs1, poly)];
res := subs(subs2, res);
od;
res;
end;
pet_flatten_term :=
proc(varp)
local terml, d, cf, v;
terml := [];
cf := varp;
for v in indets(varp) do
d := degree(varp, v);
terml := [op(terml), seq(v, k=1..d)];
cf := cf/v^d;
od;
[cf, terml];
end;
pet_flat2rep_cyc :=
proc(f)
local p, q, res, cyc, t, len;
q := 1; res := [];
for t in f do
len := op(1, t);
cyc := [seq(p, p=q+1..q+len-1), q];
res := [op(res), cyc];
q := q+len;
od;
res;
end;
pet_cycs2table :=
proc(cycs)
local pairs, cyc, p, ent;
pairs := [];
for cyc in cycs do
pairs :=
[op(pairs),
seq([cyc[p], cyc[1 + (p mod nops(cyc))]],
p = 1 .. nops(cyc))];
od;
map(ent->ent[2], sort(pairs, (a,b)->a[1] < b[1]));
end; | {
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"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9777138144607745,
"lm_q1q2_score": 0.8562690370894727,
"lm_q2_score": 0.8757869884059266,
"openwebmath_perplexity": 1273.553374674751,
"openwebmath_score": 0.5907314419746399,
"tags": null,
"url": "https://math.stackexchange.com/questions/1151538/finding-the-spanning-subgraphs-of-a-complete-bipartite-graph/"
} |
map(ent->ent[2], sort(pairs, (a,b)->a[1] < b[1]));
end;
pet_cycleind_knn :=
proc(n)
option remember;
local cindA, cindB, sind, t1, t2, p, q, cyc1, cyc2,
flat, len, len1, len2, v1, v2,
edges, edgeperm, rep, cycsA, cycsB, cycsrc, cyc;
if n=1 then
sind := [a[1]];
else
sind := pet_cycleind_symm(n);
fi;
cindA := 0;
for t1 in sind do
for t2 in sind do
q := 1;
for v1 in indets(t1) do
len1 := op(1, v1);
for v2 in indets(t2) do
len2 := op(1, v2);
len := lcm(len1, len2);
q := q *
a[len]^((len1*len2/len) *
degree(t1, v1)*degree(t2, v2));
od;
od;
cindA := cindA +
lcoeff(t1)*lcoeff(t2)*q;
od;
od;
edges := [seq(seq({p, q+n}, q=1..n), p=1..n)];
cindB := 0;
for t1 in sind do
flat := pet_flatten_term(t1);
cycsA := pet_flat2rep_cyc(flat[2]);
cycsB := [];
for cycsrc in cycsA do
cyc := [];
for q in cycsrc do
cyc := [op(cyc), q, q+n];
od;
cycsB := [op(cycsB), cyc];
od;
rep := pet_cycs2table(cycsB);
edgeperm :=
subs([seq(q=rep[q], q=1..2*n)], edges);
cindB := cindB + flat[1]*
pet_autom2cycles(edges, edgeperm);
od;
(cindA+cindB)/2;
end;
Q :=
proc(n)
option remember;
local cind;
cind := pet_cycleind_knn(n);
subs([seq(a[p]=2, p=1..n*n)], cind);
end; | {
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"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9777138144607745,
"lm_q1q2_score": 0.8562690370894727,
"lm_q2_score": 0.8757869884059266,
"openwebmath_perplexity": 1273.553374674751,
"openwebmath_score": 0.5907314419746399,
"tags": null,
"url": "https://math.stackexchange.com/questions/1151538/finding-the-spanning-subgraphs-of-a-complete-bipartite-graph/"
} |
cind := pet_cycleind_knn(n);
subs([seq(a[p]=2, p=1..n*n)], cind);
end;
Remarks, per request, Dec 2020. Here are some observations concerning reflections. We need the factorization into cycles of the vertices in order to compute the action on the edges. Now a reflection when factored into cycles must alternate between left and right vertices. That means left and right vertices are interleaved on those cycles. Hence we can obtain the factorizations of all reflections by taking a permutation from $$S_n$$ (which gives the left vertices) and doubling all its cycles in length, inserting a permutation of the right vertices, for a total of $$n! \times n!$$ reflections. The first factorial comes from the permutations of the left vertices which can be interleaved with the right ones in $$n!$$ ways, yielding the second factorial. Now consider the factorization into cycles of the action on the edges. Left and right vertices are disjoint, so no matter the order in which we interleave the right vertices, we get the same shape of factorizations into cycles (just permute the right vertices, which maintains the cycle structure of the edges). Therefore we can take one representative for each term in $$Z(S_n)$$, interleave it with some permutation of the right, and factor that (i.e. factor the action on the edges) to get the contribution to $$Z(Q_n)$$, which must be multiplied by $$n!$$ for the number of reflections covered by this term. This is what the algorithm does. The $$n!$$ is canceled when we divide by the total number of permutations in the group. | {
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} |
• I have trouble understanding the algorithm for calculating the cycle index when reflections are concerned. I obtain the same results when I consider each of the $n!^2$ pairs of permutations, but it's of course infeasible for larger $n$. Could you describe your approach in more detail? From what I understand, "permutation shape" is just its cycle notation. But two different pairs of permutations with the same cycle notations can contribute different terms to the cycle index when the vertices are additionally reflected. I don't understand how the alternation of components works here. – Adam S. Dec 15 '20 at 23:28
• I have added some commentary. Please note that the 2017 answer is the reference here. – Marko Riedel Dec 16 '20 at 2:53
• Wow, brilliant. Thank you! My mistake was to treat the two permutations and the reflection completely separately, which made the cycle factorization difficult. Embedding the reflection into one of the permutations makes things so much easier and indeed reduces the computational cost from $n!^2$ just to the number of different permutation shapes in $S_n$. This will help me speed up my little project: github.com/adamsol/burnside-counter. – Adam S. Dec 16 '20 at 17:30
• It is now tempting to try to simplify also the case without reflections, so that we don't have to iterate over pairs of permutation shapes at all. But it seems that this case is paradoxically more complicated and it's not possible. – Adam S. Dec 16 '20 at 17:38
• Good that I could help. I might do some additional simplifications later today. – Marko Riedel Dec 16 '20 at 17:40
The question is equivalent to finding all non-isomorphic bipartite graphs with bi-partitions of sizes $n$ and $m$ respectively and is hard to solve in general. In particular there is no simple characterization of such graphs.
If you need the sequence for the number of such graphs then you can take a look at this oeis entry http://oeis.org/A033995. | {
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"url": "https://math.stackexchange.com/questions/1151538/finding-the-spanning-subgraphs-of-a-complete-bipartite-graph/"
} |
If you need to compute such graphs (up to a reasonable number) then your best choice is the tool genbg which you can find in McKay's nauty package. In some sense genbg is currently the most efficient way to construct such graphs and hence answers your question.
• Yeah I need to find all those non-isomorphic bipartite graphs with sizes $m$ and $n$ in each partition. The sequence above is for the number of bipartite graphs of $n$ vertices. But that number is greater than the number of graphs that I am trying to find. For example if I have four vertices, there are $7$ bipartite graphs but if I'm only taking all the spanning subgraphs of $K_{(2,2)}$ then there are just $6$ of them. The other bipartite graph is a claw which is bipartite with sizes 1 and 3 in each partition. – chowching Feb 17 '15 at 12:43
• @chowching You should be more specific about what you're looking for. Note that in general there is no closed formula for the number of bipartite graphs with bipartitions of cardinality n and m. – Jernej Feb 17 '15 at 13:00
• Thanks for the reply. I'm studying Polya Enumeration Theorem now to see if I can apply it to find how many such graphs are there given a certain $n$ and $m$. – chowching Feb 17 '15 at 13:00
• @chowching You can but the formula will most likely require using computer computations. – Jernej Feb 17 '15 at 13:01
• @chowching I suggest that for a start you look at Harary's and Palmer's book called "graphical enumeration" – Jernej Feb 17 '15 at 13:02 | {
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"tags": null,
"url": "https://math.stackexchange.com/questions/1151538/finding-the-spanning-subgraphs-of-a-complete-bipartite-graph/"
} |
Adding another answer after receiving a pointer from a reader of this thread. We treat the case of spanning subgraphs of $$K_{n,n}$$ including reflections. The algorithm has been greatly simplified. There are two types of symmetries, namely shuffles within the two sets of vertices and reflections that map one to the other. The first type has been simplified as follows: we continue to iterate over pairs of permutation shapes from $$S_n$$ but the contribution to the cycle index $$Z(Q_n)$$ is now computed by iterating over pairs of cycle types (same length) rather than single cycles. The second type (reflections) has been replaced in its entirety. We continue to produce contributions by building factorized permutations consisting of cycles that alternate between the two classes of vertices (a precondition of this problem) but whereas in the first version we would construct a representative of the factored permutation, apply it to the edges, and factor the result into cycles, the new version does not construct representatives and does no factorization and builds the terms for the cycle index from first principles, namely that we enumerate the cycles of edges (two-sets of vertices) that are obtained from a vertex permutation by choosing pairs of vertices, counting these, and computing the lengths of the cycles on which they reside. These optimizations make for a faster and much more compact program. This was the code:
pet_cycleind_symm :=
proc(n)
option remember;
if n=0 then return 1; fi;
end;
pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;
res := ind;
polyvars := indets(poly);
indvars := indets(ind);
for v in indvars do
pot := op(1, v);
subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];
subs2 := [v=subs(subs1, poly)];
res := subs(subs2, res);
od;
res;
end;
pet_flatten_termA :=
proc(varp)
local terml, d, cf, v;
terml := []; | {
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"tags": null,
"url": "https://math.stackexchange.com/questions/1151538/finding-the-spanning-subgraphs-of-a-complete-bipartite-graph/"
} |
res;
end;
pet_flatten_termA :=
proc(varp)
local terml, d, cf, v;
terml := [];
cf := varp;
for v in indets(varp) do
d := degree(varp, v);
terml := [op(terml), [op(1,v), d]];
cf := cf/v^d;
od;
[cf, terml];
end;
pet_cycleind_knn :=
proc(n)
option remember;
local cindA, cindB, sind, t1, t2, term, res,
flat, flat1, flat2, len, l1, l2, cycs, uidx, vidx,
u, v, inst1;
if n=1 then
sind := [a[1]];
else
sind := pet_cycleind_symm(n);
fi;
cindA := 0;
for t1 in sind do
flat1 := pet_flatten_termA(t1);
for t2 in sind do
flat2 := pet_flatten_termA(t2);
res := 1;
for u in flat1[2] do
l1 := op(1, u);
for v in flat2[2] do
l2 := op(1, v);
len := lcm(l1, l2);
res := res *
a[len]^(op(2, u)*op(2, v)
*l1*l2/len);
od;
od;
cindA := cindA + flat1[1]*flat2[1]*res;
od;
od;
cindB := 0;
for term in sind do
flat := pet_flatten_termA(term);
cycs := map(cyc -> [2*cyc[1], cyc[2]], flat[2]);
res := 1;
# edges on different cycles of different sizes
for uidx to nops(cycs) do
u := cycs[uidx];
for vidx from uidx+1 to nops(cycs) do
v := cycs[vidx];
l1 := op(1, u); l2 := op(1, v);
res := res *
a[lcm(l1, l2)]^((l1*l2/2/lcm(l1, l2))*
op(2, u)*op(2, v));
od;
od;
# edges on different cycles of the same size
for uidx to nops(cycs) do
u := cycs[uidx];
l1 := op(1, u); inst1 := op(2, u);
# a[l1]^(1/2*inst1*(inst1-1)*l1*l1/2/l1)
res := res *
a[l1]^(1/2*inst1*(inst1-1)*l1/2);
od;
# edges on identical cycles of some size
for uidx to nops(cycs) do
u := cycs[uidx];
l1 := op(1, u); inst1 := op(2, u);
if type(l1/2, even) then
# a[l1]^((l1/2)^2/l1);
res := res *
(a[l1]^(l1/4))^inst1;
else
# a[l1/2]^(l1/2/(l1/2))*a[l1]^(((l1/2)^2-l1/2)/l1)
res := res *
(a[l1/2]*a[l1]^(l1/4-1/2))^inst1;
fi;
od;
cindB := cindB + flat[1]*res;
od;
(cindA+cindB)/2;
end;
This will produce the following data for the non-isomorphic spanning subgraphs of $$K_{6,6}$$ including reflections classified according to the number of edges: | {
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"tags": null,
"url": "https://math.stackexchange.com/questions/1151538/finding-the-spanning-subgraphs-of-a-complete-bipartite-graph/"
} |
$${z}^{36}+{z}^{35}+2\,{z}^{34}+4\,{z}^{33}+10\,{z}^{32} +20\,{z}^{31}+50\,{z}^{30}+99\,{z}^{29}+227\,{z}^{28} \\ +458\,{z}^{27}+934\,{z}^{26}+ 1711\,{z}^{25}+3057\,{z}^{24}+4889\,{z}^{23}+7400\,{z}^{22} \\ +10071\,{z}^{21}+12755\,{z}^{20}+14524\,{z}^{19} +15331\,{z}^{18}+14524\,{z}^{17}+12755\,{z}^{16} \\+10071\,{z}^{15}+7400\,{z}^{14}+4889\,{z}^{13}+ 3057\,{z}^{12}+1711\,{z}^{11}+934\,{z}^{10} \\ +458\,{z}^{9}+227\,{z}^{8}+99\,{z}^{7}+50\,{z}^{6}+20\,{z}^{5} +10\,{z}^{4}+4\,{z}^{3}+2\,{z}^{2}+z+1.$$
i.e. the value four for three edges counts, first, a set of three edges, not connected, second, two edges sharing a vertex and a single edge, third, three edges emanating from the same vertex and fourth, a path of three edges.
**Addendum.** We may use Burnside as shown below if we are not interested in classifying according to the number of edges and require only the total count. This gives:
Q :=
proc(n)
option remember;
local cind;
cind := pet_cycleind_knn(n);
subs([seq(a[p]=2, p=1..n*n)], cind);
end;
The routine that substitutes polynomials into cycle indices is no longer needed in this case. Observe that further simplification at the expense of readability is possible by not computing the cycle index at all and setting all $$a[p]$$ to two during the computation of what used to be the cycle index.
Remark, Dec 2020. We can in fact optimize away all flattening operations and use the fact that Maple monomials i.e. the terms of the cycle index are perfectly sufficient to implement a multiset data structure where the variables represent cycles of some length and the degrees, the number of instances, as is of course the standard semantics of cycle indices. This is shown below. (Duplicate code omitted.)
pet_cycleind_knn :=
proc(n)
option remember;
local cindA, cindB, sind, t1, t2, term,
res, len, len1, len2, cycs, uidx, vidx, interlv,
u, v, inst, q;
if n=1 then
sind := [a[1]];
else
sind := pet_cycleind_symm(n);
fi;
cindA := 0; | {
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if n=1 then
sind := [a[1]];
else
sind := pet_cycleind_symm(n);
fi;
cindA := 0;
for t1 in sind do
for t2 in sind do
q := 1;
for u in indets(t1) do
len1 := op(1, u);
for v in indets(t2) do
len2 := op(1, v);
len := lcm(len1, len2);
q := q *
a[len]^((len1*len2/len) *
degree(t1, u)*degree(t2, v));
od;
od;
cindA := cindA +
lcoeff(t1)*lcoeff(t2)*q;
od;
od;
cindB := 0;
interlv := subs({seq(a[q]=a[2*q], q=1..n)}, sind);
for term in interlv do
termvars := indets(term); res := 1;
# edges on different cycles of different sizes
for uidx to nops(termvars) do
u := op(uidx, termvars);
len1 := op(1, u);
for vidx from uidx+1 to nops(termvars) do
v := op(vidx, termvars);
len2 := op(1, v);
res := res *
a[lcm(len1, len2)]
^((len1*len2/2/lcm(len1, len2))*
degree(term, u)*degree(term, v));
od;
od;
# edges on different cycles of the same size
for u in termvars do
len := op(1, u); inst := degree(term, u);
# a[len]^(1/2*inst*(inst-1)*len*len/2/len)
res := res *
a[len]^(1/4*inst*(inst-1)*len);
od;
# edges on identical cycles of some size
for u in termvars do
len := op(1, u); inst := degree(term, u);
if type(len/2, even) then
# a[len]^((len/2)^2/len);
res := res *
(a[len]^(len/4))^inst;
else
# a[len/2]^(len/2/(len/2))
# *a[len]^(((len/2)^2-len/2)/len)
res := res *
(a[len/2]*a[len]^(len/4-1/2))^inst;
fi;
od;
cindB := cindB + lcoeff(term)*res;
od;
(cindA+cindB)/2;
end;
Q :=
proc(n)
option remember;
local cind;
cind := pet_cycleind_knn(n);
subs([seq(a[p]=2, p=1..n*n)], cind);
end; | {
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# Writing An Equation for Polynomial Graph
• June 17th 2013, 05:44 PM
AceBoogie
Writing An Equation for Polynomial Graph
Hey all,
Hoping someone can help me figure out how to write an equation for the polynomial graphed in the picture. Really lost, if any help could be given it would be appreciated.
Attachment 28638
• June 17th 2013, 06:02 PM
HallsofIvy
Re: Writing An Equation for Polynomial Graph
One of the things you might have noticed is that an nth degree polynomial has at most n-1 "critical" points where the graph levels out (maybe but not necessarily where the graph "turns"). Since this graph has 4 such "critical points" (at x=-3, x= -1. something, x= 0. something, and x= 2) I would try a 5th degree polynomial.
Further, we can see from the graph that the polynomial is 0 at x= -3, x= -1, and x= 2. That means that the polynomial is of the form [tex]y= a(x+3)(x+1)(x- 2)Q(x)[tex] where a is a number and Q(x) is a quadratic polynomial.
To go further, you would have to specify exactly where the turning point ("-1.something", "0.something") and what the y values are there.
• June 17th 2013, 06:13 PM
AceBoogie
Re: Writing An Equation for Polynomial Graph
So are you saying you think the outline for the equation would be:
Q(x) = x(x+3)(x+1)(X-2)
I don't understand how I would find the 0.something and the 1.something points? How would they relate into the equation?
• June 18th 2013, 03:57 AM
mpx86
Re: Writing An Equation for Polynomial Graph
at x=-3 , x=-1 , x=2 ...... y= f(x)=0
that is 3,1,-2 are its root
such an equation is K (x+3)(x+1)(x-2)=0
as such an equation is zero at x=-3,-1,2
now f(0)=6 or y is 6 when x is 0
put x=0 in the equation
K(0+3)(0+1)(0-2)=6
yielding K=-1
therefore reqd. equation is
f(x)=-(x+3)(x+1)(x-2)
• June 19th 2013, 08:11 PM
Soroban
Re: Writing An Equation for Polynomial Graph
Hello, AceBoogie!
I can get you started . . .
Quote:
Write an equation for the polynomial graphed in the picture.
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Quote:
Write an equation for the polynomial graphed in the picture.
Attachment 28638
There are three $x$-intercepts . . . two of them are quite special.
At $x=2$, the curve is tangent to the $x$-axis.
. . Hence, $f(x)$ has a factor of $(x-2)^2.$
At $x = \text{-}1$, the curve passes through the $x$-axis.
. . Hence, $f(x)$ has a factor of $(x+1).$
At $x = \text{-}3$, the curve passes through the $x$-axis.
But there is an inflection point at $(\text{-}3,0).$
. . Hence, $f(x)$ has a factor of $(x+3)^3.$
Also, $f(x)$ has a $y$-intercept at $(0,6).$
Can you combine those facts?
• June 20th 2013, 12:23 AM
ReneG
Re: Writing An Equation for Polynomial Graph
Quote:
Originally Posted by Soroban
At $x=2$, the curve is tangent to the $x$-axis.
. . Hence, $f(x)$ has a factor of $(x-2)^2.$
At $x = \text{-}1$, the curve passes through the $x$-axis.
. . Hence, $f(x)$ has a factor of $(x+1).$
At $x = \text{-}3$, the curve passes through the $x$-axis.
But there is an inflection point at $(\text{-}3,0).$
. . Hence, $f(x)$ has a factor of $(x+3)^3.$
You've encouraged me to finally study the polynomial functions chapter in my book. :P
• June 28th 2013, 08:38 PM
johng
Re: Writing An Equation for Polynomial Graph
Hi,
I strongly encourage you to use your graphing software/calculator to help you with graphing problems. For example, once you have the supposed correct equation, you can check your result. Assuming you can zoom in, zooming in at (-3,0) shows indeed an inflection point. I always use my software to check any graphics problem. Of course, you probably can't do this during a quiz or test. Here's the solution to your problem:
Attachment 28692 | {
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# Show that $\sqrt{x+2}$ is a contraction
Let $$f:X \rightarrow X$$ where $$X=[0,\infty)$$ be defined as $$f(x)=\sqrt{x+2}$$. I have to show that this mapping is a contraction and find its unique fixed point. The second part is easy: by the CMT, it has a unique fixed point in $$X$$ and it is $$x^\ast = 2$$.
For $$f$$ being a contraction I wts the following: $$\exists \beta \in [0,1)$$ such that
$$\mid\sqrt{x+2}-\sqrt{y+2}\mid \leq \beta \mid x-y \mid, \ (\forall x,y\geq0)$$
Since
$$\mid\sqrt{x+2}-\sqrt{y+2}\mid = \dfrac{\mid x-y \mid}{\sqrt{x+2}+\sqrt{y+2}}$$
I'm tempted to set $$\beta = \dfrac{1}{\sqrt{x+2}+\sqrt{y+2}}$$ but $$\beta$$ cannot depend on $$x$$ or $$y$$.... Any ideas about how to proceed? Thanks!
Note that for all $$x,y\geqslant 0$$ $$\frac{1}{\sqrt{x+2}+\sqrt{y+2}}\leqslant \frac{1}{\sqrt{2}+\sqrt{2}}=\frac{\sqrt{2}}{4}$$
• Great! I had just to find an upper bound smaller than one! – Alessandro Oct 13 '18 at 18:59
hint
Let $$x,y\in [0,+\infty)$$.
By MVT
$$f(x)-f(y)=(x-y)f'(c)$$
where $$0\le x
$$f'(c)=\frac{1}{2\sqrt{c+2}}\le \frac{1}{2\sqrt{2}}$$
• Thanks! Basically to show that $f$ is a contraction it is enough to show that $\mid f'(x) \mid \leq \beta$, $\forall x \in X$ – Alessandro Oct 13 '18 at 19:18
• @Alessandro It is sufficient to prove that the derivative is bounded. $| f'(x) | \ le \beta$. – hamam_Abdallah Oct 13 '18 at 19:20
• @Salahamam_Fatima You mean that $|f'(x)|<1$, not just generally bounded. – mathematics2x2life Oct 13 '18 at 19:31
• @mathematics2x2life To be a contraction, you need $|f'(x)|\le \beta<1$. – hamam_Abdallah Oct 13 '18 at 19:48
As per Salahamam's solution, to show a function is a contraction, it is sufficient to show that its derivative has $$|f'(x)|<1$$.
Prop. If $$f(x)$$ is a differentiable function function with $$|f'(x)|<1$$ for all $$x$$, then $$f(x)$$ is a contraction. | {
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Proof. Let $$x,y \in \mathbb{R}$$. By the Mean Value Theorem, we have $$|f(x)-f(y)|= |f'(c)(x-y)|= |f'(c)||x-y|$$ for some $$c$$ between $$x$$ and $$y$$. But as $$|f'(c)|<1$$ by assumption, we must have $$|f(x)-f(y)|=|f'(c)||x-y| < 1 \cdot |x-y|=|x-y|.$$ Therefore, $$f$$ is a contraction.
Note that the converse is false as contractions need not be differentiable.
So in your case, you only need show that $$\sqrt{x+2}$$ has bounded derivative. But $$\dfrac{d}{dx} \; \sqrt{x+1}= \dfrac{1}{2\sqrt{x+2}}$$ which is at most $$\frac{1}{2\sqrt{2}}$$ on the interval $$[0,\infty)$$.
• To show that $f$ is a contraction you need to show that the derivative is uniformly bounded by a number less than one. Proving that $f'(x)<1$ is not enough. That's what I understood – Alessandro Oct 13 '18 at 22:34
• this question can clarify the issue, I hope: math.stackexchange.com/questions/419392/… – Alessandro Oct 13 '18 at 22:41
• @Alessandro That was exactly what I said. You have a function which is differentiable on the intervals which you are considering, so it suffices to show that $|f'(x)|<1$. You do not need your derivative to be uniformly bounded. A strong contraction is a function for which there exists $|f(x)-f(y)| \leq c<1$. But of course, this depends on what one calls a contraction. But most mean $|f(x)-f(y)|<|x-y|$ when they say 'contraction' and reserve the former when speaking of something stronger like strong contraction, Lipschitz continuity, etc.. – mathematics2x2life Oct 15 '18 at 15:34
• ok but to apply the contraction mapping theorem I need what you call "a strong contraction". Furthermore the standard definition of contraction requires $\mid f(y)-f(x) \mid \leq \beta \mid x-y \mid$ (see en.wikipedia.org/wiki/Contraction_mapping) – Alessandro Oct 16 '18 at 15:51 | {
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$A+B+C=2149$, Find $A$
In the following form of odd numbers
If the numbers
taken from the form where $A+B+C=2149$
Find $A$
any help will be appreciate it, thanks.
• Hint: looking at the numbers in the picture, try to find a pattern that tells you what $B$ and $C$ are in terms of $A$. Then plug $B$ and $C$ into $A+B+C=2149$ and solve for $A$. It helps to give each line a number, so the line with 1 is line number 1, the line with 3 and 5 is line number 2, the line with 7, 9, and 11 is line number 3, etc. – wltrup Aug 4 '15 at 22:50
• What interesting puzzles you come up with! – hypergeometric Aug 6 '15 at 9:33
If $A$ were on the $n^{th}$ row, then $B=A+2n$ and $C=A+2(n+1)$.
Out of laziness, we will try to figure out what row this would need to be to be in the right ballpark through estimation. $\frac{2149}{3}\approx 716$, so $A<716<B<C$. Which row would $715$ fit into?
By changing your figure some by first adding one to each entry and then dividing by two, we will get the chart $\left[\begin{array}{}\color{red}{1}\\2&\color{red}{3}\\4&5&\color{red}{6}\\7&8&9&\color{red}{10}\\\vdots\end{array}\right]$, and these numbers are very familiar to us. The red numbers are the triangle numbers, $T(n)=\binom{n+1}{2}=\frac{n^2+n}{2}$.
So, $T(26)=351<\frac{715+1}{2}=358<378=T(27)$, so we suppose that $715$ occurs on the $26^{th}$ row, implying that either $A$ is on the $25^{th}$, $26^{th}$, or $27^{th}$ row (since we have been estimating up to this point).
So, we try to solve now, $A+B+C=2149=A+(A+2n)+(A+2n+2)=3A+4n+2$
In the case that $n=25$, this would be $3A+100+2=2149\Rightarrow 3A=2047\Rightarrow A=\frac{2047}{3}\not\in\mathbb{Z}$, so we know that $n=25$ was not possible.
In the case that $n=26$, this would be $3A+104+2=2149\Rightarrow 3A=2043\Rightarrow A=681$
In the case that $n=27$, this would be $3A+108+2=2149\Rightarrow 3A=2039\Rightarrow A=\frac{2039}{3}\not\in\mathbb{Z}$, so we know that $n=27$ was not possible. | {
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We can similarly show that if $n=24$ that leads to a contradiction as well.
We believe then that our answer is $A=681, B=733, C=735$. All that remains to check is that $681$ is indeed on the $26^{th}$ row to confirm our calculations by checking that $T(25)$ is less than $\frac{A+1}{2}$ and $T(26)$ is greater than $\frac{A+1}{2}$.
Indeed, $T(25)=325<\frac{681+1}{2}=341<351=T(26)$
• Oh man, you're genies, that's correct, because there is 4 options, and 681 one of them, thanks – Oiue Aug 4 '15 at 23:08
• +1 for the ballpark paragraph. I stress this when I tutor students, that having a ballpark estimate (even one it takes $5$ seconds to come up with) and compare that to the exact answer you got at the end is a very effective way to see if you've done any big mistakes. – Arthur Aug 4 '15 at 23:16
$A$ is going to be the $k$th entry in the $n$th row, where $1 \leq k \leq n$. That will make $B$ the $k$th entry in the $n+1$st row, and $C$ the $k+1$st entry in the $n+1$st row.
So the first question is, can we write a formula for $A$ in terms of $k$ and $n$? If we consider the first entry in the $n$th row, there are $1 + 2 + \cdots + (n-1)$ odd numbers before it, which equals $\frac{n(n-1)}{2}$. So the first entry in the $n$th row must be $2\left(\frac{n(n-1)}{2}\right) + 1 = n^2 - n + 1$. Now going over to the $k$th entry will add $2(k - 1)$ to this, so $A = n^2 - n + 1 + 2(k-1) = n^2 - n + 2k - 1$
This formula also implies that $B = (n+1)^2 - (n+1) + 2k - 1$ and $C = (n+1)^2 - (n+1) + 2(k+1) - 1$.
If we put this all together, $A + B + C = (n^2 - n + 2k - 1) + (n^2 + n + 2k - 1) + (n^2 + n + 2k + 1) = 3n^2 + n + 6k - 1$.
So can we solve:
$$3n^2 + n + 6k - 1 = 2149$$
Or:
$$3n^2 + n + 6k = 2150$$
Solving for $k$:
$$k = \frac{2150 - 3n^2 - n}{6}$$ | {
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Or:
$$3n^2 + n + 6k = 2150$$
Solving for $k$:
$$k = \frac{2150 - 3n^2 - n}{6}$$
Now we need $k$ to be an integer, and $1 \leq k \leq n$. Since $k$ is an integer we know $n$ should be $2$ mod 3, and we know $3n^2 \leq 2150$, so $n^2 \leq 717$. So $n < 27$. Now we guess, starting with the highest value of $n$ which is less than $27$ and $2$ mod 3, so $n = 26$. Plugging this in we get $k = 16$, which works!
This gives $A = 26^2 - 26 + 32 - 1 = 681$.
• I finished this off a little differently (my solution is a mere variation of yours, as I mention there), but I liked your derivation of the equation $3n^2 + n + 6k = 2150$, which is the key element of either solution. I would have added my variation as a comment on your answer but it didn't look like it would fit the allotted space. – David K Aug 5 '15 at 5:30
I like this one.
Look, let's fill the blanks in the middle column with pair numbers. You will find that it forms the sequence 1, 4, 9, 16, 25, 36.. that is evidently x².
All the numbers in a row are sequential, so we can establish a variable to this position i.e. n. And establish that in the A position we have n, so in the B position we have (n - 1) and in the C position we have (n + 1).
Given all this we can say:
A = x² + n
B = (x + 1)² + (n - 1)
C = (x + 1)² + (n + 1)
So we can put this in the original equation and we will have an equation with two values to solve (x, n). Nevertheless we also know that n can not be greater than x, ensuring the selected values are in the triangle.
So
(x² + n) + (x + 1)² + (n - 1) + (x + 1)² + (n + 1) = 2149
3x² + 4x + (3n) + 2 = 2149
3x² + 4x + (3n - 2147) = 0
With this equation we can find:
x = 26
n = 5
And so:
A = 681 B = 733 C = 735
Hope it helps! | {
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With this equation we can find:
x = 26
n = 5
And so:
A = 681 B = 733 C = 735
Hope it helps!
• If you "fill in" the table so that, for example, row $4$ is 13 14 15 16 17 18 19, then the middle A is still $x^2$ and $A+B+C = 3x^2+4x+2$. Solving $3x^2+4x+2=2149$ for $x$, we get $x=26.094$. So we are in row $26$ and our A is just to the right of $A=26^2$, where $A+B+C=2134$. In the filled in table, each shift to the right adds $1$ to $A, B,$ and $C$ and adds $3$ to $A+B+C$. $(2149-2134)/3=5$. So $A = 26^2 + 5 =681$ – steven gregory Aug 5 '15 at 14:56
Let's write the numbers as $a_k:=2k-1$, starting at $k=1$. Then, $a_1=1$, $a_2=3$, $a_3=5$, and so on. When $A=a_k$ is in row $n$ (the first row is row $0$), then $B=a_{k+n+1}$ and $C=a_{k+n+2}=B+2$. Hence, we have
\begin{align*} && 2149 &= a_k + 2a_{2k+n+1} +2 \\ &\Rightarrow& 2147 &= 2k-1 + 2(2(k+n+1)-1) = 6k + 4n + 1 \end{align*}
Note that the $n$-th row (starting at $n=0$) of the triangle starts with $a_{k_n}$, where $$k_n = 1+\cdots+n = \frac{n(n+1)}2.$$ Therefore, $k=k_n+q$ where $0\le q\le n$. Substituting, we get \begin{align*} && 2147 &= 3n(n+1) + 10q + 4n + 1 \\ &\Rightarrow & 0 &= 3 n^2 + 7 n + 10 q -2146 \end{align*} Observe that the linear term $10q$ does not affect the position of the zeros of this quadratic polynomial much. Hence, I just plugged in $q=n/2$ and this polynomial has a zero around $25$. So, $n=25$ seems a good guess.
Indeed, plugging in $n=25$ into our earlier equation gives
\begin{align*} && 2147 &= 6*k + 4*25 + 1 \\ &\Rightarrow & 2046 &= 6*k \\ &\Rightarrow & 341 &= k \end{align*}
So we have $A=a_{341}=681$ in row $n=25$ and $B=2*(341+25+1)-1=733$, $C=735$.
Looking at the first element in each row, we see
row A B C A+B+C
1 1 3 5 9
2 3 7 9 19
3 7 13 15 35
4 13 21 23 57
5 21 31 33 85
6 31 43 45 139
Using finite differences, we find that | {
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Using finite differences, we find that
\begin{align} A &= n^2 - n + 1\\ B &= n^2 + n + 1 = A+2n\\ C &= n^2 + n + 3 = A+2n+2\\ A+B+C &= 3n^2+n+5 \end{align} where A is the first element in row $n$.
For each shift of A to the right in the same row, A,B, and C increase by 2 and A+B+C increases by $6$.
Solving $3n^2+n+5 = 2149$ for n, we get $n = 26.567$ So we are in row $26$.
For the first element in row $26, A = 651,\; B = 703,\; C = 705,$ and $A+B+C = 2059$.
Since $\dfrac{2149-2059}6 = 15$, we need to increase $A, B,$ and $C$ by $2\cdot15 = 30$.
So
\begin{align} A &= 681\\ B &= 733\\ C &= 735 \end{align}
• Ok, you just should edit the first row $a=1, b=3, c=5, a+b+c=9$ – Oiue Aug 5 '15 at 0:08
• @Oiue. Yep. It's nine in my notes too. Thanks. – steven gregory Aug 5 '15 at 1:18
Let $r$ be the row number, and $U$ be the unshown number between $B$ and $C$ and directly below $A$, i.e.
\begin{align} &A &&\leftarrow\text {row}\; r\\ B \quad [&U]\quad C &&\leftarrow\text {row}\; r+1 \end{align}
Note that the difference between two vertically adjacent numbers (both shown and unshown) is $(r+1)^2-r^2=2r+1$, i.e. $U-A=2r+1$. Hence \begin{align} 2149&=A+B+C&&\text{(given)}\\ &=A+2U&&\text{(as}\; B=U-1, C=U+1\text{)}\\ &=3A+4r+2&&\text{(using } U=A+2r+1\text{)}\\ 3A+4r-2147&=0&& && .....(1)\\ \end{align}
Note also that the centre column (including unshown numbers) corresponds to $r^2$. As an initial approximation, assume that $A$ is the centre column, i.e. $A=r^2$. Substituting in $(1)$ gives
\begin{align}3r^2+4r-2147&=0\\ \color{blue}{r}&\color{blue}{\approx 26} \; (r>0)\qquad \qquad && && && &&& .....(2)\end{align}
Putting $(2)$ in $(1)$ gives
$$\color{red}{A=681}\qquad\blacksquare\qquad \Rightarrow \color{blue}{B=733, C=735}$$
Check: $A+B+C=681+733+735=2149$
NB: The number "$A$" is in row $26$ but is not in the centre column; the element in the centre column is $26^2=676$ which is blank as it is an even number. | {
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• every solutions her always adds something new, thanks :) – Oiue Aug 5 '15 at 18:41
This is a slight variation on the nice answer by @Alex Zorn.
In that answer we find that if the $1$ at the top of the triangle of numbers is considered entry number $1$ (counting from the left) in row number $1$ (counting from the top), then entry number $k$ in row $n$ is $n^2−n+2k−1$, and $A$ is entry $k$ in row $n$ where $n,k$ are a solution to
$$3n^2 + n + 6k = 2150.$$
We have $1 \leq k \leq n$ by the way the entries in a row are counted, so for all $n > 0$ we have $$3n^2 + n = 2144 \leq 3n^2 + n + 6k = 2150 \leq 3n^2 + 7n = 2150. \tag 1$$
Solving the two equations
$$3n^2 + n = 2144, \tag 2$$ $$3n^2 + 7n = 2150, \tag 3$$
we see that they each have one positive root. The positive root of Equation $(2)$ is $r_1 \approx 26.642$ and the positive root of Equation $(3)$ is $r_2 \approx 25.629$.
But because of Inequality $(1)$, whatever the value of $k$ is (provided that $1 \leq k \leq n$) there is a positive root of $3n^2 + n + 6k = 2150$ that is not less than $r_2$ and not greater than $r_1$ (draw a graph if you have trouble seeing this). The only integer that satisfies both conditions is $26$, so if $3n^2 + n + 6k = 2150$ has a solution for integer $n$ then the only possibility is $n = 26$.
Setting $n = 26$, we have $3(26^2) + 26 + 6k = 2150$, which is a simple linear equation in $k$ with the solution $k = 16$. We then plug $n$ and $k$ into the formula for $A$ to get the answer. | {
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Values of $x$ satisfying $\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x$
For what values of $$x$$ between $$0$$ and $$\pi$$ does the inequality $$\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x$$ hold?
My Attempt $$\sin x\cos x\cdot(\cos^2x-\sin^2x)=\frac{1}{2}\cdot\sin2x\cdot\cos2x=\frac{1}{4}\cdot\sin4x>0\implies\sin4x>0\\ x\in(0,\pi)\implies4x\in(0,4\pi)\\ 4x\in(0,\pi)\cup(2\pi,3\pi)\implies x\in\Big(0,\frac{\pi}{4}\Big)\cup\Big(\frac{\pi}{2},\frac{3\pi}{4}\Big)$$ But, my reference gives the solution, $$x\in\Big(0,\dfrac{\pi}{4}\Big)\cup\Big(\dfrac{3\pi}{4},\pi\Big)$$, where am I going wrong with my attempt?
The given solution is wrong; you are correct. At $$x=\frac{7\pi}8\in\left(\frac{3\pi}4,\pi\right)$$, we have that $$\frac14\sin4x=\frac14\sin\frac{7\pi}2=-\frac14<0$$ which is a contradiction.
Finishing what you did
$$\sin(4x)>0\iff$$
$$2k\pi <4x<(2k+1)\pi \iff$$
$$\frac{k\pi}{2}
$$k=0$$ gives $$0
and $$k=1$$ gives $$\frac{\pi}{2}.
As an alternative for a full solution we can consider two cases
• $$\sin x \cos x >0$$ that is $$x\in(0,\pi/2)\cup(\pi,3\pi/2)$$
$$\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x \iff\cos^2x>\sin^2x \iff2\sin^2 x<1$$
$$-\frac{\sqrt 2}2<\sin x<0 \,\land\, 0<\sin x<\frac{\sqrt 2}2 \iff \color{red}{x\in(0,\pi/4)}\cup(\pi,5\pi/4)$$
• $$\sin x \cos x <0$$ that is $$x\in(\pi/2,\pi)\cup(3\pi/2,2\pi)$$
$$\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x \iff\cos^2x<\sin^2x \iff2\sin^2 x>1$$
$$-1<\sin x<-\frac{\sqrt 2}2\,\land\, \frac{\sqrt 2}2<\sin x <1 \iff \color{red}{x\in(\pi/2,3\pi/4)}\cup(3\pi/2,7\pi/4)$$
and then your solution is correct. | {
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The inverse of six important trigonometric functions are: 1. What are the derivatives of the inverse trigonometric functions? For example, the sine function $$x = \varphi \left( y \right)$$ $$= \sin y$$ is the inverse function for $$y = f\left( x \right)$$ $$= \arcsin x.$$ Then the derivative of $$y = \arcsin x$$ is given by, ${{\left( {\arcsin x} \right)^\prime } = f’\left( x \right) = \frac{1}{{\varphi’\left( y \right)}} }= {\frac{1}{{{{\left( {\sin y} \right)}^\prime }}} }= {\frac{1}{{\cos y}} }= {\frac{1}{{\sqrt {1 – {\sin^2}y} }} }= {\frac{1}{{\sqrt {1 – {\sin^2}\left( {\arcsin x} \right)} }} }= {\frac{1}{{\sqrt {1 – {x^2}} }}\;\;}\kern-0.3pt{\left( { – 1 < x < 1} \right).}$. Derivatives of Inverse Trig Functions. ${y^\prime = \left( {\arctan \frac{1}{x}} \right)^\prime }={ \frac{1}{{1 + {{\left( {\frac{1}{x}} \right)}^2}}} \cdot \left( {\frac{1}{x}} \right)^\prime }={ \frac{1}{{1 + \frac{1}{{{x^2}}}}} \cdot \left( { – \frac{1}{{{x^2}}}} \right) }={ – \frac{{{x^2}}}{{\left( {{x^2} + 1} \right){x^2}}} }={ – \frac{1}{{1 + {x^2}}}. Derivatives of the Inverse Trigonometric Functions. In Table 2.7.14 we show the restrictions of the domains of the standard trigonometric functions that allow them to be invertible. Suppose \textrm{arccot } x = \theta. We then apply the same technique used to prove Theorem 3.3, “The Derivative Rule for Inverses,” to differentiate each inverse trigonometric function. The Inverse Cosine Function. }$, ${y^\prime = \left( {\frac{1}{a}\arctan \frac{x}{a}} \right)^\prime }={ \frac{1}{a} \cdot \frac{1}{{1 + {{\left( {\frac{x}{a}} \right)}^2}}} \cdot \left( {\frac{x}{a}} \right)^\prime }={ \frac{1}{a} \cdot \frac{1}{{1 + \frac{{{x^2}}}{{{a^2}}}}} \cdot \frac{1}{a} }={ \frac{1}{{{a^2}}} \cdot \frac{{{a^2}}}{{{a^2} + {x^2}}} }={ \frac{1}{{{a^2} + {x^2}}}. The derivatives of $$6$$ inverse trigonometric functions considered above are consolidated in the following table: In the examples below, find the derivative of the given function. The usual | {
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in the following table: In the examples below, find the derivative of the given function. The usual approach is to pick out some collection of angles that produce all possible values exactly once. Table 2.7.14. Quick summary with Stories. Dividing both sides by \cos \theta immediately leads to a formula for the derivative. The Inverse Tangent Function. 1. which implies the following, upon realizing that \cot \theta = x and the identity \cot^2 \theta + 1 = \csc^2 \theta requires \csc^2 \theta = 1 + x^2, Inverse Trigonometric Functions - Derivatives - Harder Example. Arcsine 2. The inverse trigonometric functions actually perform the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent. Inverse Trigonometry Functions and Their Derivatives. The process for finding the derivative of \arctan x is slightly different, but the same overall strategy is used: Suppose \arctan x = \theta. These cookies will be stored in your browser only with your consent. 3 Definition notation EX 1 Evaluate these without a calculator. Practice your math skills and learn step by step with our math solver. Section 3-7 : Derivatives of Inverse Trig Functions. If f(x) is a one-to-one function (i.e. Related Questions to study. Inverse trigonometric functions provide anti derivatives for a variety of functions that arise in engineering. Because each of the above-listed functions is one-to-one, each has an inverse function. Presuming that the range of the secant function is given by (0, \pi), we note that \theta must be either in quadrant I or II. x = \varphi \left ( y \right) x = φ ( y) = \sin y = sin y. is the inverse function for. \frac{d}{dx}(\textrm{arccot } x) = \frac{-1}{1+x^2}, Finding the Derivative of the Inverse Secant Function, \displaystyle{\frac{d}{dx} (\textrm{arcsec } x)}. -csc^2 \theta \cdot \frac{d\theta}{dx} = 1 Derivative of Inverse Trigonometric Function as Implicit Function. Inverse Trigonometric Functions Note. Inverse | {
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Inverse Trigonometric Function as Implicit Function. Inverse Trigonometric Functions Note. Inverse trigonometric functions have various application in engineering, geometry, navigation etc. Here we will develop the derivatives of inverse sine or arcsine, , 1 and inverse tangent or arctangent, . They are cosecant (cscx), secant (secx), cotangent (cotx), tangent (tanx), cosine (cosx), and sine (sinx). Email. Here, for the first time, we see that the derivative of a function need not be of the same type as the … Necessary cookies are absolutely essential for the website to function properly. Review the derivatives of the inverse trigonometric functions: arcsin (x), arccos (x), and arctan (x). Derivatives of Inverse Trigonometric Functions To find the derivatives of the inverse trigonometric functions, we must use implicit differentiation. One example does not require the chain rule and one example requires the chain rule. }$, $\require{cancel}{y^\prime = \left( {\arcsin \left( {x – 1} \right)} \right)^\prime }={ \frac{1}{{\sqrt {1 – {{\left( {x – 1} \right)}^2}} }} }={ \frac{1}{{\sqrt {1 – \left( {{x^2} – 2x + 1} \right)} }} }={ \frac{1}{{\sqrt {\cancel{1} – {x^2} + 2x – \cancel{1}} }} }={ \frac{1}{{\sqrt {2x – {x^2}} }}. \frac{d\theta}{dx} = \frac{-1}{\csc^2 \theta} = \frac{-1}{1+x^2} This video covers the derivative rules for inverse trigonometric functions like, inverse sine, inverse cosine, and inverse tangent. View Lesson 9-Differentiation of Inverse Trigonometric Functions.pdf from MATH 146 at Mapúa Institute of Technology. 2 The graph of y = sin x does not pass the horizontal line test, so it has no inverse. We'll assume you're ok with this, but you can opt-out if you wish. Derivatives of inverse trigonometric functions. Lesson 9 Differentiation of Inverse Trigonometric Functions OBJECTIVES • to The derivatives of the inverse trigonometric functions can be obtained using the inverse function theorem. Derivatives of Inverse Trigonometric Functions using First | {
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using the inverse function theorem. Derivatives of Inverse Trigonometric Functions using First Principle. It is mandatory to procure user consent prior to running these cookies on your website. Arccosecant Let us discuss all the six important types of inverse trigonometric functions along with its definition, formulas, graphs, properties and solved examples. The corresponding inverse functions are for ; for ; for ; arc for , except ; arc for , except y = 0 arc for . In modern mathematics, there are six basic trigonometric functions: sine, cosine, tangent, secant, cosecant, and cotangent. }$, ${y’\left( x \right) }={ {\left( {\arctan \frac{{x + 1}}{{x – 1}}} \right)^\prime } }= {\frac{1}{{1 + {{\left( {\frac{{x + 1}}{{x – 1}}} \right)}^2}}} \cdot {\left( {\frac{{x + 1}}{{x – 1}}} \right)^\prime } }= {\frac{{1 \cdot \left( {x – 1} \right) – \left( {x + 1} \right) \cdot 1}}{{{{\left( {x – 1} \right)}^2} + {{\left( {x + 1} \right)}^2}}} }= {\frac{{\cancel{\color{blue}{x}} – \color{red}{1} – \cancel{\color{blue}{x}} – \color{red}{1}}}{{\color{maroon}{x^2} – \cancel{\color{green}{2x}} + \color{DarkViolet}{1} + \color{maroon}{x^2} + \cancel{\color{green}{2x}} + \color{DarkViolet}{1}}} }= {\frac{{ – \color{red}{2}}}{{\color{maroon}{2{x^2}} + \color{DarkViolet}{2}}} }= { – \frac{1}{{1 + {x^2}}}. We know that trig functions are especially applicable to the right angle triangle. Derivative of Inverse Trigonometric functions The Inverse Trigonometric functions are also called as arcus functions, cyclometric functions or anti-trigonometric functions. Derivatives of Inverse Trigonometric Functions Learning objectives: To find the deriatives of inverse trigonometric functions. The derivatives of the above-mentioned inverse trigonometric functions follow from trigonometry … This category only includes cookies that ensures basic functionalities and security features of the website. In the following discussion and solutions the derivative of a function h(x) will be denoted by or | {
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In the following discussion and solutions the derivative of a function h(x) will be denoted by or h'(x) . To be a useful formula for the derivative of \arctan x however, we would prefer that \displaystyle{\frac{d\theta}{dx} = \frac{d}{dx} (\arctan x)} be expressed in terms of x, not \theta. The derivatives of the inverse trigonometric functions are given below. And To solve the related problems. Inverse Trigonometric Functions: •The domains of the trigonometric functions are restricted so that they become one-to-one and their inverse can be determined. For example, the domain for $$\arcsin x$$ is from $$-1$$ to $$1.$$ The range, or output for $$\arcsin x$$ is all angles from $$– \large{\frac{\pi }{2}}\normalsize$$ to $$\large{\frac{\pi }{2}}\normalsize$$ radians. Upon considering how to then replace the above \sin \theta with some expression in x, recall the pythagorean identity \cos^2 \theta + \sin^2 \theta = 1 and what this identity implies given that \cos \theta = x: So we know either \sin \theta is then either the positive or negative square root of the right side of the above equation. The inverse of these functions is inverse sine, inverse cosine, inverse tangent, inverse secant, inverse cosecant, and inverse cotangent. Then it must be the case that. Derivative of Inverse Trigonometric Functions using Chain Rule. 11 mins. This website uses cookies to improve your experience. Dividing both sides by -\sin \theta immediately leads to a formula for the derivative. Arcsecant 6. Arctangent 4. We can use implicit differentiation to find the formulas for the derivatives of the inverse trigonometric functions, as the following examples suggest: Finding the Derivative of Inverse Sine Function, \displaystyle{\frac{d}{dx} (\arcsin x)}, Suppose \arcsin x = \theta. AP.CALC: FUN‑3 (EU), FUN‑3.E (LO), FUN‑3.E.2 (EK) Google Classroom Facebook Twitter. Since \theta must be in the range of \arccos x (i.e., [0,\pi]), we know \sin \theta must be positive. If we restrict the domain | {
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range of \arccos x (i.e., [0,\pi]), we know \sin \theta must be positive. If we restrict the domain (to half a period), then we can talk about an inverse function. Examples: Find the derivatives of each given function. Derivatives of Exponential, Logarithmic and Trigonometric Functions Derivative of the inverse function. Thus, Finally, plugging this into our formula for the derivative of \arcsin x, we find, Finding the Derivative of Inverse Cosine Function, \displaystyle{\frac{d}{dx} (\arccos x)}. Differentiation of Inverse Trigonometric Functions Each of the six basic trigonometric functions have corresponding inverse functions when appropriate restrictions are placed on the domain of the original functions. Arccotangent 5. Inverse trigonometric functions are literally the inverses of the trigonometric functions. But opting out of some of these cookies may affect your browsing experience. Example: Find the derivatives of y = sin-1 (cos x/(1+sinx)) Show Video Lesson. In this section we are going to look at the derivatives of the inverse trig functions. As such. Implicitly differentiating with respect to x yields 1 du Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. Derivatives of Inverse Trigonometric Functions We can use implicit differentiation to find the formulas for the derivatives of the inverse trigonometric functions, as the following examples suggest: Finding the Derivative of Inverse Sine Function, d d x (arcsin These functions are used to obtain angle for a given trigonometric value. The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. •Since the definition of an inverse function says that -f 1(x)=y => f(y)=x We have the inverse sine function, -sin 1x=y - π=> sin y=x and π/ 2 <=y<= / 2 Of course |\sec \theta| = |x|, | {
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the inverse sine function, -sin 1x=y - π=> sin y=x and π/ 2 <=y<= / 2 Of course |\sec \theta| = |x|, and we can use \tan^2 \theta + 1 = \sec^2 \theta to establish |\tan \theta| = \sqrt{x^2 - 1}. Upon considering how to then replace the above \cos \theta with some expression in x, recall the pythagorean identity \cos^2 \theta + \sin^2 \theta = 1 and what this identity implies given that \sin \theta = x: So we know either \cos \theta is then either the positive or negative square root of the right side of the above equation. Previously, derivatives of algebraic functions have proven to be algebraic functions and derivatives of trigonometric functions have been shown to be trigonometric functions. Using similar techniques, we can find the derivatives of all the inverse trigonometric functions. Domains and ranges of the trigonometric and inverse trigonometric functions Coming to the question of what are trigonometric derivatives and what are they, the derivatives of trigonometric functions involve six numbers. Using this technique, we can find the derivatives of the other inverse trigonometric functions: \[{{\left( {\arccos x} \right)^\prime } }={ \frac{1}{{{{\left( {\cos y} \right)}^\prime }}} }= {\frac{1}{{\left( { – \sin y} \right)}} }= {- \frac{1}{{\sqrt {1 – {{\cos }^2}y} }} }= {- \frac{1}{{\sqrt {1 – {{\cos }^2}\left( {\arccos x} \right)} }} }= {- \frac{1}{{\sqrt {1 – {x^2}} }}\;\;}\kern-0.3pt{\left( { – 1 < x < 1} \right),}\qquad$, ${{\left( {\arctan x} \right)^\prime } }={ \frac{1}{{{{\left( {\tan y} \right)}^\prime }}} }= {\frac{1}{{\frac{1}{{{{\cos }^2}y}}}} }= {\frac{1}{{1 + {{\tan }^2}y}} }= {\frac{1}{{1 + {{\tan }^2}\left( {\arctan x} \right)}} }= {\frac{1}{{1 + {x^2}}},}$, ${\left( {\text{arccot }x} \right)^\prime } = {\frac{1}{{{{\left( {\cot y} \right)}^\prime }}}}= \frac{1}{{\left( { – \frac{1}{{{\sin^2}y}}} \right)}}= – \frac{1}{{1 + {{\cot }^2}y}}= – \frac{1}{{1 + {{\cot }^2}\left( {\text{arccot }x} \right)}}= – \frac{1}{{1 + {x^2}}},$, ${{\left( | {
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– \frac{1}{{1 + {{\cot }^2}\left( {\text{arccot }x} \right)}}= – \frac{1}{{1 + {x^2}}},$, ${{\left( {\text{arcsec }x} \right)^\prime } = {\frac{1}{{{{\left( {\sec y} \right)}^\prime }}} }}= {\frac{1}{{\tan y\sec y}} }= {\frac{1}{{\sec y\sqrt {{{\sec }^2}y – 1} }} }= {\frac{1}{{\left| x \right|\sqrt {{x^2} – 1} }}.}$. }\], ${y^\prime = \left( {\text{arccot}\frac{1}{{{x^2}}}} \right)^\prime }={ – \frac{1}{{1 + {{\left( {\frac{1}{{{x^2}}}} \right)}^2}}} \cdot \left( {\frac{1}{{{x^2}}}} \right)^\prime }={ – \frac{1}{{1 + \frac{1}{{{x^4}}}}} \cdot \left( { – 2{x^{ – 3}}} \right) }={ \frac{{2{x^4}}}{{\left( {{x^4} + 1} \right){x^3}}} }={ \frac{{2x}}{{1 + {x^4}}}.}$. Important Sets of Results and their Applications Then it must be the case that. 3 mins read . The sine function (red) and inverse sine function (blue). f(x) = 3sin-1 (x) g(x) = 4cos-1 (3x 2) Show Video Lesson. The domains of the other trigonometric functions are restricted appropriately, so that they become one-to-one functions and their inverse can be determined. Dividing both sides by $\sec^2 \theta$ immediately leads to a formula for the derivative. The derivatives of inverse trigonometric functions are quite surprising in that their derivatives are actually algebraic functions. Click or tap a problem to see the solution. In both, the product of $\sec \theta \tan \theta$ must be positive. Sec 3.8 Derivatives of Inverse Functions and Inverse Trigonometric Functions Ex 1 Let f x( )= x5 + 2x −1. One way to do this that is particularly helpful in understanding how these derivatives are obtained is to use a combination of implicit differentiation and right triangles. For example, the derivative of the sine function is written sin′ = cos, meaning that the rate of change of sin at a particular angle x = a is given by the cosine of that angle. Trigonometric Functions (With Restricted Domains) and Their Inverses. Check out all of our online calculators here! Formula for the Derivative of Inverse Cosecant Function. | {
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out all of our online calculators here! Formula for the Derivative of Inverse Cosecant Function. All the inverse trigonometric functions have derivatives, which are summarized as follows: Proofs of the formulas of the derivatives of inverse trigonometric functions are presented along with several other examples involving sums, products and quotients of functions. Definition of the Inverse Cotangent Function. }\], ${y^\prime = \left( {\text{arccot}\,{x^2}} \right)^\prime }={ – \frac{1}{{1 + {{\left( {{x^2}} \right)}^2}}} \cdot \left( {{x^2}} \right)^\prime }={ – \frac{{2x}}{{1 + {x^4}}}. Arccosine 3. Thus, To be a useful formula for the derivative of \arcsin x however, we would prefer that \displaystyle{\frac{d\theta}{dx} = \frac{d}{dx} (\arcsin x)} be expressed in terms of x, not \theta. However, since trigonometric functions are not one-to-one, meaning there are are infinitely many angles with , it is impossible to find a true inverse function for . Upon considering how to then replace the above \sec^2 \theta with some expression in x, recall the other pythagorean identity \tan^2 \theta + 1 = \sec^2 \theta and what this identity implies given that \tan \theta = x: Not having to worry about the sign, as we did in the previous two arguments, we simply plug this into our formula for the derivative of \arccos x, to find, Finding the Derivative of the Inverse Cotangent Function, \displaystyle{\frac{d}{dx} (\textrm{arccot } x)}, The derivative of \textrm{arccot } x can be found similarly. Another method to find the derivative of inverse functions is also included and may be used. The basic trigonometric functions include the following $$6$$ functions: sine $$\left(\sin x\right),$$ cosine $$\left(\cos x\right),$$ tangent $$\left(\tan x\right),$$ cotangent $$\left(\cot x\right),$$ secant $$\left(\sec x\right)$$ and cosecant $$\left(\csc x\right).$$ All these functions are continuous and differentiable in their domains. Problem. Derivatives of inverse trigonometric | {
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are continuous and differentiable in their domains. Problem. Derivatives of inverse trigonometric functions Calculator Get detailed solutions to your math problems with our Derivatives of inverse trigonometric functions step-by-step calculator. These six important functions are used to find the angle measure in a right triangle when two sides of the triangle measures are known. 7 mins. The inverse sine function (Arcsin), y = arcsin x, is the inverse of the sine function. Inverse Sine Function. Since \theta must be in the range of \arcsin x (i.e., [-\pi/2,\pi/2]), we know \cos \theta must be positive. Here, we suppose \textrm{arcsec } x = \theta, which means sec \theta = x. Thus, Finally, plugging this into our formula for the derivative of \arccos x, we find, Finding the Derivative of the Inverse Tangent Function, \displaystyle{\frac{d}{dx} (\arctan x)}. To be a useful formula for the derivative of \arccos x however, we would prefer that \displaystyle{\frac{d\theta}{dx} = \frac{d}{dx} (\arccos x)} be expressed in terms of x, not \theta. You can think of them as opposites; In a way, the two functions “undo” each other. VIEW MORE. Like before, we differentiate this implicitly with respect to x to find, Solving for d\theta/dx in terms of \theta we quickly get, This is where we need to be careful. We also use third-party cookies that help us analyze and understand how you use this website. In this section we review the definitions of the inverse trigonometric func-tions from Section 1.6. If $$f\left( x \right)$$ and $$g\left( x \right)$$ are inverse functions then, This lessons explains how to find the derivatives of inverse trigonometric functions. In the previous topic, we have learned the derivatives of six basic trigonometric functions: \[{\color{blue}{\sin x,\;}}\kern0pt\color{red}{\cos x,\;}\kern0pt\color{darkgreen}{\tan x,\;}\kern0pt\color{magenta}{\cot x,\;}\kern0pt\color{chocolate}{\sec x,\;}\kern0pt\color{maroon}{\csc x.\;}$, In this section, we are going to | {
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x,\;}\kern0pt\color{maroon}{\csc x.\;}$, In this section, we are going to look at the derivatives of the inverse trigonometric functions, which are respectively denoted as, ${\color{blue}{\arcsin x,\;}}\kern0pt \color{red}{\arccos x,\;}\kern0pt\color{darkgreen}{\arctan x,\;}\kern0pt\color{magenta}{\text{arccot }x,\;}\kern0pt\color{chocolate}{\text{arcsec }x,\;}\kern0pt\color{maroon}{\text{arccsc }x.\;}$. a) c) b) d) 4 y = tan x y = sec x Definition [ ] 5 EX 2 Evaluate without a calculator. Then $\cot \theta = x$. These cookies do not store any personal information. Similarly, we can obtain an expression for the derivative of the inverse cosecant function: ${{\left( {\text{arccsc }x} \right)^\prime } = {\frac{1}{{{{\left( {\csc y} \right)}^\prime }}} }}= {-\frac{1}{{\cot y\csc y}} }= {-\frac{1}{{\csc y\sqrt {{{\csc }^2}y – 1} }} }= {-\frac{1}{{\left| x \right|\sqrt {{x^2} – 1} }}.}$. The process for finding the derivative of $\arccos x$ is almost identical to that used for $\arcsin x$: Suppose $\arccos x = \theta$. Now let's determine the derivatives of the inverse trigonometric functions, y = arcsinx, y = arccosx, y = arctanx, y = arccotx, y = arcsecx, and y = arccscx. g ( x) = arccos ( 2 x) g (x)=\arccos\!\left (2x\right) g(x)= arccos(2x) g, left parenthesis, x, right parenthesis, … {\displaystyle {\begin{aligned}{\frac {d}{dz}}\arcsin(z)&{}={\frac {1}{\sqrt {1-z^{2}}}}\;;&z&{}\neq -1,+1\\{\frac {d}{dz}}\arccos(z)&{}=-{\frac {1}{\sqrt {1-z^{2}}}}\;;&z&{}\neq -1,+1\\{\frac {d}{dz}}\arctan(z)&{}={\frac {1}{1+z^{2}}}\;;&z&{}\neq -i,+i\\{\frac {d}{dz}}\operatorname {arccot}(z)&{}=-{\frac {1}{1+z^{2}}}\;;&z&{}\neq -i,+i\\{\frac {d}{dz}}\operatorname {arcsec}(z)&{}={\frac {1}{z^{2… Derivatives of Inverse Trigonometric Functions. This implies. The inverse functions exist when appropriate restrictions are placed on the domain of the original functions. Inverse Functions and Logarithms. Formula for the Derivative of Inverse Secant Function. \dfrac {d} {dx}\arcsin | {
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and Logarithms. Formula for the Derivative of Inverse Secant Function. \dfrac {d} {dx}\arcsin (x)=\dfrac {1} {\sqrt {1-x^2}} dxd arcsin(x) = 1 − x2 , we must use implicit differentiation is also included and may be used function (.! Will be stored in your browser only with your consent we can talk about an inverse function of. Functions follow from trigonometry … derivatives of the inverse trig functions are literally the Inverses of the trigonometric functions used... Arctangent, this, but you can think of them as opposites ; in a right triangle two. You can opt-out if you wish LO ), arccos ( x ) = 4cos-1 ( 2! } x = \theta $immediately leads to a formula for the derivative use implicit differentiation their Inverses inverse. That ensures basic functionalities and security features of the standard trigonometric functions you also have the option to opt-out these. ; in a way, the two functions “ undo ” each other restrictions are placed on the of. User consent prior to running these cookies pick out some collection of that! Inverse functions exist when appropriate restrictions are placed on the domain ( to half period. Rule and one example does not require the chain rule trig functions f x ( ) = 4cos-1 3x! Arcsin ), and arctan ( x ) = 3sin-1 ( x ), (. Inverse cotangent of examples and worked-out practice problems half a period ), FUN‑3.E.2 ( EK ) Google Facebook... Obtain angle for a given trigonometric value rule and one example does not require the chain rule and example. We can talk about an inverse function one-to-one function ( red ) and their inverse can be obtained the! Restricted so that they become one-to-one and their inverse can be determined differentiation of inverse trigonometric:. A calculator$ \textrm { arccot } x = \theta $immediately leads to formula... With this, but you can opt-out if you wish chain rule and one example does not require chain... Be determined is mandatory to procure user consent prior to running these may. Trigonometry … | {
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Be determined is mandatory to procure user consent prior to running these may. Trigonometry … derivatives of inverse functions is also included and may be used opposites. = arcsin x, is the inverse trigonometric functions derivative of the inverse functions derivatives... By$ \cos \theta $must be positive the triangle measures are known the product$! Sec \theta = x $yields examples: find the derivatives of y = sin x does not pass horizontal...$ \cos \theta $, which means$ sec \theta = x $inverse trigonometric functions derivatives opt-out of these will. Domains ) and inverse tangent using the inverse trigonometric functions that allow them to be invertible basic. Usual approach is to pick out some collection of angles that produce all possible values exactly once:. Be the cases that, Implicitly differentiating the above with respect to$ x $yields functions is one-to-one each... To obtain angle for a given trigonometric value our math solver be used angle measure in a way the! Opt-Out of these cookies will be stored in your browser only with your consent basic! The domain ( to half a period ), FUN‑3.E.2 ( EK ) Google Classroom Facebook Twitter = x yields... Ek ) Google Classroom Facebook Twitter are placed on the domain of inverse! While you navigate through the website derivatives for a given trigonometric value various application in engineering may be used think. Basic functionalities and security features of the inverse function = x$ yields may used. Functions OBJECTIVES • to there are particularly six inverse trig functions are used to find the angle measure a! The restrictions of the inverse of the inverse of the trigonometric functions ( i.e chain rule and one requires. You also have the option to opt-out of these cookies will be stored your! If you wish in your browser only with your consent included and be! Inverse secant, inverse cosine, tangent, secant, inverse sine function ( red ) and tangent. Right triangle when two sides of the original functions not | {
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sine function ( red ) and tangent. Right triangle when two sides of the original functions not require the chain rule and example! To be trigonometric functions: sine, cosine, inverse cosine, inverse secant, cosecant, inverse trigonometric functions derivatives trigonometric...: 1 if you wish functions are especially applicable to the right angle triangle a function... Let f x ( ) = 3sin-1 ( x ) 3sin-1 ( x ) = (... Of $\sec \theta \tan \theta$ must be positive here, we suppose ${. Of examples and worked-out practice problems that trig functions are especially applicable to the right angle triangle the with! Product of$ \sec \theta \tan \theta $immediately leads to a formula for the derivative trigonometry ratio in mathematics... To$ x $yields step with our derivatives of inverse sine function,... Trigonometric func-tions from section 1.6 functions to find the derivative we are going to look at derivatives. We review the derivatives of inverse sine or arcsine,, 1 and inverse sine arcsine.$ \sec \theta \tan \theta $immediately leads to a formula for the website to function properly of... Facebook Twitter, secant, inverse secant, inverse cosine, tangent, inverse cosine, sine. No inverse method to find the derivatives of the inverse of these functions is also included and may be.! Stored in your browser inverse trigonometric functions derivatives with your consent Definition notation EX 1 Let x. Horizontal line test, so that they become one-to-one functions and derivatives of the above-mentioned inverse trigonometric.... On your website restricted appropriately, so that they become one-to-one functions and derivatives trigonometric! Is inverse sine or arcsine,, 1 and inverse cotangent this, you. The deriatives of inverse trigonometric functions like, inverse cosine, and cotangent functions OBJECTIVES • to are... And worked-out practice problems functions follow from trigonometry … derivatives of the measures! Third-Party cookies that help us analyze and understand how | {
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… derivatives of the measures! Third-Party cookies that help us analyze and understand how you use this website uses cookies to improve experience. Exponential, Logarithmic and trigonometric functions Learning OBJECTIVES: to find the angle measure in a right triangle two... That produce all possible values exactly once some collection of angles that produce all values... = x5 + 2x −1 placed on the domain ( to half a period ), arctan... Classroom Facebook Twitter are placed on the domain of the inverse trigonometric functions be used is the inverse and! Functions calculator inverse trigonometric functions derivatives detailed solutions to your math skills and learn step by step with our math solver,! •The domains of the original functions trigonometric functions have proven to be functions.: arcsin ( x ) = x5 + 2x −1 step with our math solver however imperfect to! The sine function ( arcsin ), y = sin x does not require the chain and! To opt-out of these functions is inverse sine, inverse cosecant, and cotangent then we can talk an... The domain ( to half a period ), y = sin x does not the. Suppose$ \textrm { arcsec } x = \theta $immediately leads to a for. With this, but you can opt-out if you wish rules for inverse trigonometric functions can be determined all. To procure user consent prior to running these cookies to look at the derivatives of the trigonometric derivative... Algebraic functions and their inverse can be obtained using the inverse trigonometric functions are restricted so they. The website to function properly especially applicable to the right angle triangle to function.. Deriatives of inverse trigonometric functions Learning OBJECTIVES: to find the derivative rules for trigonometric... Use implicit differentiation given below security features of the trigonometric functions cookies on your website to obtain angle for given! ( 1+sinx ) ) Show Video Lesson not require the chain rule and one example does not the... The Inverses of the original | {
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Video Lesson not require the chain rule and one example does not the... The Inverses of the original functions may affect your browsing experience half a period ) FUN‑3.E... Have proven to be algebraic functions and derivatives of the inverse function has plenty of examples and practice! We review the derivatives of inverse trigonometric functions are: 1 we can talk about inverse. X ( ) = x5 + 2x −1 sides by$ \cos \theta immediately. From trigonometry … derivatives of the triangle measures are known, there are particularly six inverse functions... Ek ) Google Classroom Facebook Twitter functions and their inverse can be obtained using the inverse trigonometric functions anti. To these functions is inverse sine, inverse cosine, and inverse tangent, inverse cosine, tangent secant! Without a calculator to procure user consent prior to running these cookies will be stored in your only!, cosecant, and arctan ( x ) is a one-to-one function ( arcsin ), we! Cookies may affect your browsing experience by $\cos \theta$ immediately leads a. Important trigonometric functions to find the derivatives of algebraic functions have various application in engineering, geometry navigation... The domains of the above-listed functions is inverse sine function red ) and inverse sine function ( red and! Definitions of the inverse trigonometric functions step-by-step calculator inverse secant, cosecant, and inverse tangent arctangent..., derivatives of inverse trigonometric functions can be determined math skills and learn step step... Have something like an inverse to these functions is one-to-one, each has an inverse..: arcsin ( x ) g ( x ), arccos ( x,! Lo ), arccos ( x ) is a one-to-one function ( i.e section 1.6 Facebook.. To opt-out of these cookies = sin x does not require the chain.. Covers the derivative we restrict the domain of the inverse of the inverse functions... Six important trigonometric functions have proven to be invertible $\textrm { arccot } x = \theta$ leads. Plenty of | {
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functions have proven to be invertible $\textrm { arccot } x = \theta$ leads. Plenty of examples and worked-out practice problems triangle when two sides of the trigonometric functions provide anti for. Basic functionalities and security features of the above-mentioned inverse trigonometric functions that arise in engineering leads. Absolutely essential for the derivative x $that, Implicitly differentiating the above with respect to$ \$. X ) affect your browsing experience examples: find the derivatives of trigonometric are! | {
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# 1984 AIME Problems/Problem 14
## Problem
What is the largest even integer that cannot be written as the sum of two odd composite numbers?
## Solution 1
Take an even positive integer $x$. $x$ is either $0 \bmod{6}$, $2 \bmod{6}$, or $4 \bmod{6}$. Notice that the numbers $9$, $15$, $21$, ... , and in general $9 + 6n$ for nonnegative $n$ are odd composites. We now have 3 cases:
If $x \ge 18$ and is $0 \bmod{6}$, $x$ can be expressed as $9 + (9+6n)$ for some nonnegative $n$. Note that $9$ and $9+6n$ are both odd composites.
If $x\ge 44$ and is $2 \bmod{6}$, $x$ can be expressed as $35 + (9+6n)$ for some nonnegative $n$. Note that $35$ and $9+6n$ are both odd composites.
If $x\ge 34$ and is $4 \bmod{6}$, $x$ can be expressed as $25 + (9+6n)$ for some nonnegative $n$. Note that $25$ and $9+6n$ are both odd composites.
Clearly, if $x \ge 44$, it can be expressed as a sum of 2 odd composites. However, if $x = 42$, it can also be expressed using case 1, and if $x = 40$, using case 3. $38$ is the largest even integer that our cases do not cover. If we examine the possible ways of splitting $38$ into two addends, we see that no pair of odd composites add to $38$. Therefore, $\boxed{038}$ is the largest possible number that is not expressible as the sum of two odd composite numbers.
## Solution 2
Let $n$ be an integer that cannot be written as the sum of two odd composite numbers. If $n>33$, then $n-9,n-15,n-21,n-25,n-27,$ and $n-33$ must all be prime (or $n-33=1$, which yields $n=34=9+25$ which does not work). Thus $n-9,n-15,n-21,n-27,$ and $n-33$ form a prime quintuplet. However, only one prime quintuplet exists as exactly one of those 5 numbers must be divisible by 5.This prime quintuplet is $5,11,17,23,$ and $29$, yielding a maximal answer of 38. Since $38-25=13$, which is prime, the answer is $\boxed{038}$. | {
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1984 AIME (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 15 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
ACS WASC
ACCREDITED
SCHOOL
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# What Is the Log-Sum-Exp Function?
The log-sum-exp function takes as input a real $n$-vector $x$ and returns the scalar
$\notag \mathrm{lse}(x) = \log \displaystyle\sum_{i=1}^n \mathrm{e}^{x_i},$
where $\log$ is the natural logarithm. It provides an approximation to the largest element of $x$, which is given by the $\max$ function, $\max(x) = \max_i x_i$. Indeed,
$\notag \mathrm{e}^{\max(x)} \le \displaystyle\sum_{i=1}^n \mathrm{e}^{x_i} \le n \mskip1mu \mathrm{e}^{\max(x)},$
and on taking logs we obtain
$\notag \qquad\qquad \max(x) \le \mathrm{lse}(x) \le \max(x) + \log n. \qquad\qquad (*)$
The log-sum-exp function can be thought of as a smoothed version of the max function, because whereas the max function is not differentiable at points where the maximum is achieved in two different components, the log-sum-exp function is infinitely differentiable everywhere. The following plots of $\mathrm{lse}(x)$ and $\max(x)$ for $n = 2$ show this connection.
The log-sum-exp function appears in a variety of settings, including statistics, optimization, and machine learning.
For the special case where $x = [0~t]^T$, we obtain the function $f(t) = \log(1+\mathrm{e}^t)$, which is known as the softplus function in machine learning. The softplus function approximates the ReLU (rectified linear unit) activation function $\max(t,0)$ and satisfies, by $(*)$,
$\notag \max(t,0) \le f(t) \le \max(t,0) + \log 2.$
Two points are worth noting.
• While $\log(x_1 + x_2) \ne \log x_1 + \log x_2$, in general, we do (trivially) have $\log(x_1 + x_2) = \mathrm{lse}(\log x_1,\log x_2)$, and more generally $\log(x_1 + x_2 + \cdots + x_n) = \mathrm{lse}(\log x_1,\log x_2,\dots,\log x_n)$.
• The log-sum-exp function is not to be confused with the exp-sum-log function: $\exp \sum_{i=1}^n \log x_i = x_1x_2\dots x_n$.
Here are some examples:
>> format long e
>> logsumexp([1 2 3])
ans =
3.407605964444380e+00
>> logsumexp([1 2 30])
ans =
3.000000000000095e+01 | {
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>> logsumexp([1 2 30])
ans =
3.000000000000095e+01
>> logsumexp([1 2 -3])
ans =
2.318175429247454e+00
The MATLAB function logsumexp used here is available at https://github.com/higham/logsumexp-softmax.
Straightforward evaluation of log-sum-exp from its definition is not recommended, because of the possibility of overflow. Indeed, $\exp(x)$ overflows for $x = 12$, $x = 89$, and $x = 710$ in IEEE half, single, and double precision arithmetic, respectively. Overflow can be avoided by writing
\notag \begin{aligned} \mathrm{lse}(x) &= \log \sum_{i=1}^n \mathrm{e}^{x_i} = \log \sum_{i=1}^n \mathrm{e}^a \mathrm{e}^{x_i - a} = \log \left(\mathrm{e}^a\sum_{i=1}^n \mathrm{e}^{x_i - a}\right), \end{aligned}
which gives
$\notag \mathrm{lse}(x) = a + \log\displaystyle\sum_{i=1}^n \mathrm{e}^{x_i - a}.$
We take $a = \max(x)$, so that all exponentiations are of nonpositive numbers and therefore overflow is avoided. Any underflows are harmless. A refinement is to write
$\notag \qquad\qquad \mathrm{lse}(x) = \max(x) + \mathrm{log1p}\Biggl( \displaystyle\sum_{i=1 \atop i\ne k}^n \mathrm{e}^{x_i - \max(x)}\Biggr), \qquad\qquad (\#)$
where $x_k = \max(x)$ (if there is more than one such $k$, we can take any of them). Here, $\mathrm{log1p}(x) = \log(1+x)$ is a function provided in MATLAB and various other languages that accurately evaluates $\log(1+x)$ even when $x$ is small, in which case $1+x$ would suffer a loss of precision if it was explicitly computed.
Whereas the original formula involves the logarithm of a sum of nonnegative quantities, when $\max(x) < 0$ the shifted formula $(\#)$ computes $\mathrm{lse}(x)$ as the sum of two terms of opposite sign, so could potentially suffer from numerical cancellation. It can be shown by rounding error analysis, however, that computing log-sum-exp via $(\#)$ is numerically reliable.
## References
This is a minimal set of references, which contain further useful references within.
## Related Blog Posts | {
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## Related Blog Posts
1. Nice post! It’s nice to know the sign issue doesn’t cause problem, and to formalize the pulling-the-max-out trick. How is the scipy.special.logsumexp implementation?
(1) if we want to approximate $\|x\|_\infty$, then we can use $f(x) = \log( \sum_i exp^{x_i} + exp^{-x_i} )$. In this case, we’d pull out $max |x_i|$ instead of $max x_i$, but I’m not sure we’d want to do log1p
(2) as for the logsumexp and its relationship to softmax (its derivative), many times each $x_i$ is parameterized by a vector $\theta$ and we want the gradient with respect to $\theta$, so then we have to modify the softmax formula to include the gradients of the $x_i$ terms. I’m thinking the naive implementation is not stable, but there ought to be similar tricks. | {
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# Wrong Inflection point
I have the next function: $$f(x)=x^3-3x^2+4$$.
I need to find an inflection point so I had done the following steps:
I found the first derivative: $$f'(x)=3x^2-6x$$
Then the second one: $$f''(x)=6x-6$$
Comparative to zero: $$6x-6=0$$
$$6x=6\rightarrow x=1$$
I checked values before and after the point in the second derivative:
$$f''(0)=-6<0$$
$$f''(2)=12>0$$
So the point (1,2) is inflection point but when I chaked the graph, there were not any:
Can someone explain why? Thank You.
• The first derivative gives you critical points not the second derivative. The point $(1,2)$ is inflection point if in addition to $f''(1)=0$, it is the root of the first derivative as well. – Yadati Kiran Nov 24 '18 at 10:56
• Yes, I know. But I need an inflection point and not critical points. – violettagold Nov 24 '18 at 10:59
• So there are none in this case. – Yadati Kiran Nov 24 '18 at 11:00
• What do you mean by root? – violettagold Nov 24 '18 at 11:02
• 1,2 is inflection point – Akash Roy Nov 24 '18 at 11:26
Draw the graph together with the tangent at $$(1,f(1))$$: $$1$$ is where the second derivative vanishes and $$f(1)=2$$; since $$f'(1)=-3$$, the tangent has equation $$y-2=-3(x-1)$$ or $$y=-3x+5$$.
As you see, the tangent “crosses” the graph, because the curve is concave for $$x<1$$ and convex for $$x>1$$.
So at $$x=1$$ there's indeed an inflection point.
• Thank you very much for the help! – violettagold Nov 24 '18 at 12:03
In order to find out critical points first find the points where derivative equals to zero and check the concavity of graph about that point.
For example if $$0$$ is an inflection point,
Then $$f^{''}(0^{+})>0$$ and $$f^{''}(0^{-})<0$$ or vice - versa. | {
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Then $$f^{''}(0^{+})>0$$ and $$f^{''}(0^{-})<0$$ or vice - versa.
• But this is what I got no? So why there is no point on the graph? – violettagold Nov 24 '18 at 11:08
• You have got this but your conclusion is wrong – Akash Roy Nov 24 '18 at 11:19
• Hey the point is an inflection point , view it closely by zooming it out, you can check it then. (1,2) is an inflection point – Akash Roy Nov 24 '18 at 11:22
• Why is my answer downvoted? – Akash Roy Nov 24 '18 at 11:25
• I thought so too but look at the graph, it shows there is no inflection point. – violettagold Nov 24 '18 at 11:30
You’re right, the graph changes its concavity at at exactly the point $$x$$ where $$f’’(x) = 0$$
such that $$f’’(x_1) > 0$$ results in concave upward and $$f’’(x_1) < 0$$ results in concave downward.
$$f’’(x) = 0 \implies 6x-6 = 0 \implies x = 1$$
Which gives $$y = 2$$, hence the point is $$(1, 2)$$.
If you notice, the tangent’s slope begins to increase (become less negative and eventually positive at $$x > 2$$) after the inflection point.
If it's not easy to vizualize it that way, take a look at https://www.desmos.com/calculator/4zu7w0kmd2 and see how the behavior of the tangent line and the graph's concavity both change at point $$(1, 2)$$.
• You are confusing the notion of an inflection point and a local extremum (maximum or minimum) – ip6 Nov 24 '18 at 11:26
• Sorry, I misunderstood the question for a second, I’ve edited it. (I didn’t read through the part which confused the OP.) – KM101 Nov 24 '18 at 11:34
• I’ve removed the downvote – ip6 Nov 24 '18 at 11:35
• Thank you very much for the help! – violettagold Nov 24 '18 at 12:04
An inflection point is where the 2nd derivative changes sign. This is where the curvature changes from concave downward to concave upward or vice versa. It’s where the tangent changes sides.
You have correctly calculated the position of the inflection point - note the inflection point should not be confused with a maximum or minimum. | {
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Draw some tangents on your graph and see what happens around the point you identified.
• I am sorry for bothering but can you give me an example, please? – violettagold Nov 24 '18 at 11:31
• Cup your left hand around the shape of the left hand side of the curve - your hand will be palm down. Cup your right hand around the shape of the right hand part of the curve - your hand will be palm up. In the middle, where your fingers meet is a point where the curvature switches. This is the inflection point. And you found it correctly!! – ip6 Nov 24 '18 at 11:42
• Now @violettagold you should upvote the answers , – Akash Roy Nov 24 '18 at 12:01
• Thank you very much for the help! – violettagold Nov 24 '18 at 12:03
• Your first statement is wrong, consider $f(x)=x^4$. – Michael Hoppe Nov 24 '18 at 12:37
I zoomed the central part (for you to see it more clearly).
Inflection point PI is at the crossed reticles $$(1,2)$$ red lines intersection.
To the right of PI water holds and at left water spills in the cubic curve and so your checked second derivative signs are OK.
The software does not mark the point automatically that is all. The way to recognize an inflection point graphically is that the curve is locally straight, you are neither turning to your right nor to your left as happening elsewhere. This happens when you are asked to drive along an $$S$$ shaped figure of $$8$$ curve at center that is your PI.
Note the relationships between the first and second derivatives:
$$\hspace{5cm}$$
1) $$f'<0, f''>0$$ - the function is decreasing at an increasing rate;
2) $$f'>0, f''<0$$ - the function is increasing at a decreasing rate;
3) $$f'<0, f''<0$$ - the function is decreasing at a decreasing rate;
4) $$f'>0, f''>0$$ - the function is increasing at an increasing rate. | {
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4) $$f'>0, f''>0$$ - the function is increasing at an increasing rate.
Now note that the function $$f(x)=x^3-3x^2+4$$ is decreasing at an increasing rate in $$(0,1)$$ and decreasing at a decreasing rate in $$(1,2)$$. And at the point $$x=1$$, the function changes its rate of decrease.
$$\hspace{4cm}$$
Be careful, $$f''(x)=0$$ does not imply the point of inflection (e.g. $$f(x)=x^4$$), so it is a possible inflection point. The function must keep decreasing or increasing around the point of inflection. In the above link you can also see the animated graph of tangent line. | {
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This is the first of what I hope will be several posts related to the course I am giving this term on probability.
The following is a well-known paradox. You are presented with two envelopes and told that one contains a sum of money and the other contains twice as much. You are invited to choose an envelope but are not told which is which. You choose an envelope, and are then given the chance to change your mind if you want to. Should you?
One argument says that it cannot possibly make any difference to the expected outcome, since either way your expected gain will be the average of the amounts in the two envelopes (so the expected change by switching is zero). But there is another argument that goes as follows. Suppose that the amount of money in the envelope you first choose is $x$. Then the other envelope has a 50% chance of containing $2x$ and a 50% chance of containing $x/2$, so your expectation if you switch is $5x/4$—so you should switch.
I tried this out for real in my first lecture, and the student who was given the choice decided to switch. Rather irritatingly, he got more money as a result. Of course, the second argument is incorrect, but the reasons are somewhat subtle. My purpose in putting up a post about it is not so much to invite solutions to the paradox as to see whether it prompts anyone to give me their favourite probabilistic paradoxes. (I’ve just done Simpson’s paradox, so that one wouldn’t be new.)
### 23 Responses to “A paradox in probability”
1. .mau. Says:
it’s not really a paradox, but I like the following scenario:
I toss a fair coin, until it comes up tails. If this happens at the first toss, you’ll win 1 euro; if it happens at the second toss (so the outcome is HT), you’ll win 2 euro; if it happens at the third toss (HHT), you’ll win 4 euro, and so on, always doubling your earnings every time the coin shows heads.
2. gowers Says: | {
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2. gowers Says:
I like that one too. I learned recently that it is called the St Petersburg paradox because it was discussed by Daniel Bernoulli when he was staying in St Petersburg. And it was formulated by one of the other Bernoullis in 1713. (All this information I got from proofs of articles that will be in the Princeton Companion to Mathematics.) I have a simple variant of it: would you pay ten thousand dollars for a never-to-be-repeated chance of one in 1,000 of making 100 million dollars?
3. Lior Says:
Surely the Monty Hall problem should be mentioned. I think the martingale betting method is different from the “long odds” question, which has more to do with the non-linearity of utility as a function of money.
Lior
4. d Says:
I’ve always been partial to: there are three doors and behind one is a prize, you pick a door but don’t open it, then i open one of the doors you did not pick and reveal there is no prize behind it, you can either stick with your original choice or switch…
5. Emmanuel Kowalski Says:
There was a very nice discussion of the two-envelope problem in the American Math. Monthly. 111 (2004), 348–351 (by Samet, Samet and Schmeidler), where a solution is presented showing that, essentially, it is not possible for two r.v. X and Y to be such that X and Y are both independent of the “ranking” R which takes value 1 (say) if XY, with same probability 1/2.
6. .mau. Says: | {
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6. .mau. Says:
I agree that in many paradoxes it is utility, rather than probability, that is involved.
Another family of not-quite-paradoxes is related to Bayesian probability, and it is especially important for doctors. A typical example run as follows: I am having a test to check for AIDS. If I am infected, the test always shows it; if I am not infected, there is one chance out of ten that the test is wrong. Only 1% of people in my nation is infected, and I did not do anything particular. If the test says I got AIDS, what is the probability of actually having it?
I am not sure, however, if this is what you are searching for.
7. David Says:
OK, I’ve heard about the two envelop probeme before and obviously the expectation value argument is wrong but how can one see that it is erroneous? What is the subtle reasoning?
8. Emmanuel Kowalski Says:
The last part of my comment should read:
“the “ranking” R which takes value 1 (say) if X> Y and 0 if X < Y, with same probability 1/2”.
9. deepc Says:
If I am not wrong, there is nothing “wrong” with the 2nd argument. If the first envelope contains \$x\$, then w.p. 1/2 on switching he gets \$x\$ more in expectation putting him at \$3x/2\$ which *is* the average of \$x\$ and \$2x\$;
and w.p. 1/2 he loses \$x/2\$ giving him in expectation \$3x/4\$ which *is*
the average of \$x\$ and \$x/2\$.
In both cases, his expected gain is the half of the sum of the amounts in the two envelopes.
10. Yaroslav Bulatov Says:
Dr.Kowalski, why is that a solution to the two-envelope paradox? You could have probability <50% of other envelope containing more money, but still get higher expected return from switching. There’s an example of such prior in Dieter http://personal.lse.ac.uk/list/PDF-files/envelope-paradox.PDF
BTW, is a preprint of Samet/Samet paper available online?
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BTW, is a preprint of Samet/Samet paper available online?
11. Emmanuel Kowalski Says:
I don’t know if the preprint is available online. The title is: “One observation behind two-envelope puzzles”; searching for this, I found at
http://ideas.repec.org/p/wpa/wuwpga/0310004.html
what seems to be an earlier version of it.
And I had simplified the conclusion of their work: there is no assumption that the ranking gives probability 1/2 to the two possibilities, only that the ranking is no constant (i.e. that neither X> Y nor X< Y holds all the time).
12. JB Says:
Gnedenkko´s book on Probability(chelsea publ) presents what he calls “Bertand Paradoxes” of geometric probabilty: when asked to find the probability that a straight line will intersect a unit disk leaving a chord of lenght 1/2 or higher, he gives 3 “arguments” obtaining the answers 1/2, 1/3, 1/4. Also, S. Ross “Afirst course on probability” or something like that has interesting examples involving infinite sets: given a (huge) bag and balls numbered 1,2, … do the following:
– 1 minute before 12, put balls 1- 10 in the bag and take ball #1 out.
-1/2 minute before 12 add balls 11-20 and take ball 2 out.
……….. and so on …………..
Q1) What happens at 12? bag is empty
Q2) If the ball taken out is selected randomly? prob 1 the bag will be empty at 12.
These are more curious(to me at least) than important but helped me keep some students awake.
13. mmm Says: | {
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13. mmm Says:
Another version of the paradox is:
Suppose you and your friend play a game. In the game, both of you check the amount of money in your wallet, and the one with more money gives all that money to the other. Now you think as follows: Suppose i have x amount
in my wallet. If my friend has >x, i gain more than x. If he has <x, i lose x.
And assume that both are equally likely to win(by symmetry). So your expectation is positive. But the same is true for him, whereas you both can’t have positive expectation as you win as much as he loses. Then what is wrong with the argument?
14. kla Says:
Even though I know this wasn’t supposed to be a discussion on the solution to the paradox, I can’t help it.
I do not know a lot about probability, but I think this paradox can be resolved very simply. It does not matter how (by which distribution) the sums in the envelopes have been determined. All we know is that we are given the information “there are two envelopes, one contains amount A and the other contains amount 2A”. This is something which sounds quite similar to, but is very different from the information “you may open one envelope. If you find amount A in it, you ‘ll be given the possibility to open a second one which will contain amount A/2 or 2A, each with probability 1/2”. Let us call these situations (1) and (2). In situation (1) (the one we are really considering) you open an envelope. It contains amount X and we don’t know if X=A or X=2A. Both are possible with equal probability 1/2. If we change, with probability 1/2 we go from A to 2A, or from 2A to A. No gain or less is to be expected and obviously, the same holds if we do not change. | {
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On the other hand if we are in situation (2), where the amount in the next envelope actually depends on the amount in the first one (that’s the whole point: it does not in situation (1)!!), then naturally one has to go for the second one. If initially one has found amount x, then by iterating this procedure, our obtained amount will do a random walk on x.2^Z (Z={integers}).
15. mabs Says:
Regarding the paradox presented, i believe its solution is, at least in this case, quite straightforward: One envelope contains \$x\$ money, the other \$2x\$. We can initially choose each with probability 1/2, therefore our expected gain is \$x+x/2=3x/2\$. Afterwards, when we consider switching envelopes, the reasoning proceeds thus: I don’t know how much money is in my current envelope, so the other one can either have \$x\$ or \$2x\$, each with probability 1/2. In the first case, i’m bound to lose an amount \$x/2\$ relative to my current expected value (\$3x/2\$); in the second case, i’ll win \$x/2\$. The expected gain, therefore, is \$1/2*(-x/2)+1/2*(x/2)=0\$, as intuition predicts.
Reworking the example to have amounts \$x\$ and \$kx\$ in the envelopes, with \$k\$ any constant other than 2, works as well.
Now, why does the analysis in the problem statement appear paradoxical? The analysis is certainly correct (refer to the calculation in the 3rd paragraph of the post); it’s just that it’s an analysis for a different problem. Consider:
Suppose i have 10 dollars. A friend offers me the following proposition: I give him my 10 dollars, and in return he will flip a coin. If it comes out heads, he’ll give me 5 dollars; tails, and he’ll give me 20 dollars (the \$x/2\$ and \$2x\$ envelopes, respectively). Now, should i accept his offer? Of course! As the calculation in the post shows, my expected gain is 5/4*10 dollars, minus my original 10 dollars, so i come out winning 2.5 dollars on average. | {
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In conclusion, the paradox stems from applying the incorrect analysis to the problem. As we mostly try to find problems in the analysis itself (that is, we neglect to see if the analysis is really meant to solve THIS problem), but the development of the analysis is indeed correct, we get the “What the hell?” feeling.
16. mabs Says:
Well, apparently the LaTeX thingy doesn’t work as i thought. Never tried to use it before, and couldn’t be bothered to read the instructions. So, sorry… In my post above, the \$…\$ things are supposed to be formulae in LaTeX syntax.
Another thing, this time regarding the paradox proposed in the reply above by mmm: It is assumed that, by ‘symmetry’, both participants have the same chance to win, thus the paradox. The assumption is, of course, wrong. You would only have true symmetry if you both had the same amount of money in your respective wallets at the start.
The relative amounts of money each player has is critical to deciding who wins, and it is implicit in the problem statement that, lacking knowledge of how much more (or less) money i have than the other person, i can assume that i occupy a central (‘symmetric’) position in the overall distribution of money contained in people’s wallets in general.
Now, that assumption is obviously at fault. If i have no money at all, i’m guaranteed never to lose. And if only few people carry more than 100 dollars at once in their wallets but i happen to have 150 right now, i’m more than likely to lose.
So, your expected gain depends on how much money you’re carrying in your wallet right now, and you are certainly aware of that amount. And even if you had no way of knowing the general distribution of money inside people’s wallets, you couldn’t claim an expected win based on ‘symmetry’. You’d have to say that your expected gain is indeterminate.
17. m Says:
not sure it would be useful but
there is a nice recent book by Szekely on paradoxes in probability theory | {
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Szekely G.J. Pcaradoxes in probability theory and mathematical statistics,
it is a large listing of paradoxes with references to literature.
18. nicolas Says:
Ok so I know the point of this was not the paradox itself but to hell with it.
The crux of the paradox lies for me in that x itself is a random variable, and we dont recognize it as such when supposedly taking expectancy to compare outcomes : after taking unconditional expectancy, which is supposed to be a number, we end up with x, which is a random variable. This is not possible, and we have been fooled.
The way it is presented, we are considering the case where \$x=2a\$, and say the other envelope has a 1/2 proba of having \$2x=4a\$. this case do not exist and we know so, hence it should not appear in our computation…
So we take an envelope, call its value \$x\$. the other envelope contains either \$2x\$ if x=a, or \$x/2\$ if \$x=2a\$
the expectancy of it value is \$3/2a\$. the expectancy of the other envelope is \$3/2a\$, there is no incentive to switch.
19. M-2: Two Probability Paradoxes « Concrete Nonsense Says:
[…] original post […]
20. Sohbet Says:
Thanks…
Not sure if anyone posted it above, but I think St. Petersburg paradox is a great one to do…. | {
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Not sure if anyone posted it above, but I think St. Petersburg paradox is a great one to do….
Here’s a lesser known problem; superficially similar to the two-envelope paradox, but is actually not. Two people each pick a number (whether randomly, haphazardly, or otherwise); each writes it down on a piece of paper which is then placed face down. Person A, say, (who only knows their own choice, but not that of the other) has to guess whether the other number, which, to remind you, is hidden from them, is larger (or not) than theirs. The numbers, by the way, can be any two numbers; not necessarily integers, and may be positive or negative etc. The question: is there a strategy whereby the player (A) can guess correctly with probability greater than 0.5 ? In other words, is there a strategy whereby this is a game with positive expectation? Surprisingly, there is such a strategy !
23. bigmoneyoutofpolitics Says:
I hope this thread is still live. Referring to the argument that “your expectation if you switch is 5/4”, you write:
“Of course, the second argument is incorrect, but the reasons are somewhat subtle.”
That is wrong.
You can see this easily by listing the possibilities. The problem is not as simple as you seem to think, but it also has nothing to do with possible prior probability distributions — something that can be easily seen by simply choosing numbers.
There are actually two parts to the problem. Smullyan gives a clear statement of one part, which involves only actual amounts (not expected values), and their modes of designation (see, for example, “Satan, Cantor, and Infinity, pp 189-192). He does not resolve the problem, but stating it clearly is the most important step.
The second part of the problem does involve expectation values. This corresponds to your “second argument”. The last part of Smullyan’s account, on p 192 gives a compelling statement similar to yours, but again does not resolve the problem. | {
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I don’t know whether Smullyan ever figured out the solutions but, in my view, his work on the problem is the most important contribution in print because he shows how the problem splits into two parts.
By the way, your system forces me to use the name of an old website of mine, instead of my real name. You use some extremely intrusive software, at the level of Google. | {
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# Is there a fundamental reason to expect $e$ to appear in this probability question?
I came across the following question, whose answer is $e$. I was sort of amazed, since I didn't see a reason why $e$ should be making an appearance. So, phrasing the main question in the title another way: should it have been possible to predict that at least my answer should involve $e$ in some nontrivial way?
Question: Suppose you draw random variables $x_i \in [0,1]$, where the $x_i$ are uniformly distributed. If $x_i > x_{i-1}$, we draw again, and otherwise, we stop. What's the expected number of draws before we stop?
Solution: The probability that we take $n$ draws is given by $\frac{n-1}{n!}$ since there are $n!$ ways to order $x_1,\ldots, x_n$, and exactly $n-1$ choices for the placement of $x_{n-1}$ in the ordering $x_1 < x_1 < \cdots < x_{\widehat{n-1}} < x_n$. That is, for it to take precisely $n$ draws, we need $x_1 < x_2, x_2 < x_3, \ldots, x_{n-2} < x_{n-1}$ but then $x_n < x_{n-1}$. Thus, the expected value is given by $$E(\text{number of draws}) = \sum_{n = 2}^\infty \frac{1}{(n-2)!} = \fbox{e}$$
P.S. It's also possible I simply made a mistake, and if that's the case please point it out and I can edit or delete the question accordingly. | {
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• See also this question. – John Bentin Jul 25 '17 at 20:14
• The number of months in a year is very close to $12\dfrac1e~.~$ Coincidence ? – Lucian Jul 25 '17 at 20:16
• Another combinatorial problem where $e$ pops up is the number of derangements in a card deck, en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle – An aedonist Jul 25 '17 at 20:18
• Does it surprise you that $\sum \frac1{n!}$ comes up in a probability question? If not then it shouldn't surprise you that $e$ comes up. What should surprise you is that $e=\sum \frac1{n!}$ in the first place. – Gregory Grant Jul 25 '17 at 20:25
• @DerekAllums Yeah if you take $\sum \frac1{n!}$ to be the definition of $e$ then it shouldn't surprise you that $e=\sum \frac1{n!}$ – Gregory Grant Jul 25 '17 at 20:40
First: I would take a slightly different route, but that's a matter of taste: Use that $$E[X]=\sum_{n=0}^\infty nP(X=n)=\sum_{n=0}^\infty P(X> n).$$ Similar to your reasoning, $X> n$ occurs iff (up to almost impossible equalities) $x_1<x_2<\ldots <x_n$, which happens with probability $\frac 1{n!}$. Hence, (again) $$\tag1 E[X]=\sum_{n=0}^\infty\frac1{n!}=e.$$
But regarding your main question: Could we expect $e$ to raise its head? Well, perhaps. The problem is about random order, hence about permutations, and (hand-wave) $e$ very often occurs in the context of permutations - which is of course owed to it being the sum if reciprocals of factorials. Then again, before one follows this thought further, one has already written down $(1)$ ...
I'll give a different way of approaching the problem that shows another way that $e$ can arise.
As with Hagen von Eitzen's answer, let $X$ be the random variable that represents the number of draws.
Consider the closely related problem where $x_i \in \{1, 2, \ldots, N\}$ instead. At most $N + 1$ draws can happen in this case. We can calculate the expectation using the binomial theorem. | {
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$$E[X] = \sum_{k=0}^N{P(X > k)} =\sum_{k=0}^{N}\frac{N\choose k}{N^k} = \left(1 + \frac1{N}\right)^N$$
Here, $N \choose k$ is the number of strictly increasing ways of choosing $x_1, \ldots, x_k$ which is then divided by the total number of ways of choosing them to get the probability.
Of course then, the expectation approaches $e$ in the limit as $N \to \infty$. (I don't think this is a particularly good way to solve the problem, as one still has to justify how this limit is applicable, but hopefully it provides some insight into the appearance of $e$ regardless.)
As for the fundamental question, I don't really see a way of knowing that $e$ will appear until it has more or less appeared, but $e$ does seem to arise here quite naturally, so maybe it should not be so surprising in general when $e$ arises in a combinatorial context.
• Very cool - and I suppose this further illustrates the point that I shouldn't be surprised by the appearance of $e$, if my definition of $e$ is, for example, $\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n$. – Derek Allums Jul 26 '17 at 13:04 | {
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