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 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 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 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 $(x,y)$) 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 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 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 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 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 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... 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2020 find function from points calculator


# 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: $$3x+2y^2=4x^2+4y^2 \Rightarrow \color{blue}{2y^2=3x-4x^2}$$

Substitution into $(*)$ and simplifying, will give you $0$.


# 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:

$$\prod_{i=1}^n(3i-1)(3i-2)=\frac{(3n)!}{3^nn!}$$

$$\prod_{i=1}^n(4i-1)(4i-3)=\frac{(4n)!}{2^{2n}(2n)!}$$

$$\prod_{i=1}^n(5i-1)(5i-2)(5i-3)(5i-4)=\frac{(5n)!}{5^nn!}$$

$$\prod_{i=1}^n(6i-1)(6i-5)=\frac{(6n)!(n)!}{2^{2n}3^n(2n)!(3n)!}$$

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

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.


# 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.

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.

• 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 $$u_i < u_k$$.

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

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.


# 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.

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;
}

$count++ if$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

$$a_{\mathrm{lcm}(l_1, l_2)}^{l_1 l_2 / \mathrm{lcm}(l_1, l_2)} = a_{\mathrm{lcm}(l_1, l_2)}^{\gcd(l_1, l_2)}.$$

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);

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

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

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.

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 $$n=5$$, $$Z(Q_5) = {\frac {{a_{{1}}}^{25}}{28800}}+{\frac {{a_{{1}}}^{15}{a_{{ 2}}}^{5}}{1440}}+{\frac {{a_{{1}}}^{9}{a_{{2}}}^{8}}{288}}+ {\frac {{a_{{1}}}^{10}{a_{{3}}}^{5}}{720}}+{\frac {{a_{{1}} }^{5}{a_{{2}}}^{10}}{192}}\\+{\frac {{a_{{1}}}^{3}{a_{{2}}}^{ 11}}{96}}+{\frac {a_{{1}}{a_{{2}}}^{12}}{128}}+{\frac {{a_{ {1}}}^{6}{a_{{2}}}^{2}{a_{{3}}}^{3}a_{{6}}}{72}}+{\frac {{a _{{1}}}^{4}{a_{{3}}}^{7}}{72}}\\+{\frac {{a_{{1}}}^{5}{a_{{4} }}^{5}}{480}}+1/24\,{a_{{1}}}^{3}{a_{{2}}}^{3}{a_{{4}}}^{4} +{\frac {{a_{{2}}}^{5}{a_{{3}}}^{5}}{720}}+1/48\,{a_{{1}}}^ {3}a_{{2}}{a_{{4}}}^{5}\\+1/48\,{a_{{1}}}^{2}{a_{{2}}}^{4}a_{ {3}}{a_{{6}}}^{2}+{\frac {{a_{{2}}}^{5}{a_{{3}}}^{3}a_{{6}} }{72}}+1/32\,a_{{1}}{a_{{2}}}^{2}{a_{{4}}}^{5}\\+1/48\,{a_{{2 }}}^{5}a_{{3}}{a_{{6}}}^{2}+1/36\,{a_{{2}}}^{2}{a_{{3}}}^{5 }a_{{6}}+1/12\,{a_{{1}}}^{2}a_{{2}}a_{{3}}{a_{{6}}}^{3}\\+{ \frac {3\,a_{{1}}{a_{{4}}}^{6}}{32}}+{\frac {{a_{{2}}}^{2}{ a_{{3}}}^{3}{a_{{6}}}^{2}}{72}}+1/24\,{a_{{1}}}^{2}a_{{3}}{ a_{{4}}}^{2}a_{{12}}+1/24\,a_{{2}}a_{{3}}{a_{{4}}}^{2}a_{{ 12}}\\+{\frac {13\,{a_{{5}}}^{5}}{600}}+1/8\,a_{{1}}{a_{{8}}} ^{3}+1/12\,a_{{3}}a_{{4}}a_{{6}}a_{{12}}+{\frac {{a_{{5}}}^ {3}a_{{10}}}{60}}+1/30\,{a_{{5}}}^{2}a_{{15}}\\+1/8\,a_{{5}}{ a_{{10}}}^{2}+1/20\,a_{{5}}a_{{20}}+1/30\,a_{{10}}a_{{15}} .$$

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;

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;

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.

• 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.

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

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 := [];

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:

$${z}^{36}+{z}^{35}+2\,{z}^{34}+4\,{z}^{33}+10\,{z}^{32} +20\,{z}^{31}+50\,{z}^{30}+99\,{z}^{29}+227\,{z}^{28} \\ +458\,{z}^{27}+934\,{z}^{26}+ 1711\,{z}^{25}+3057\,{z}^{24}+4889\,{z}^{23}+7400\,{z}^{22} \\ +10071\,{z}^{21}+12755\,{z}^{20}+14524\,{z}^{19} +15331\,{z}^{18}+14524\,{z}^{17}+12755\,{z}^{16} \\+10071\,{z}^{15}+7400\,{z}^{14}+4889\,{z}^{13}+ 3057\,{z}^{12}+1711\,{z}^{11}+934\,{z}^{10} \\ +458\,{z}^{9}+227\,{z}^{8}+99\,{z}^{7}+50\,{z}^{6}+20\,{z}^{5} +10\,{z}^{4}+4\,{z}^{3}+2\,{z}^{2}+z+1.$$

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;

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;


# 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.

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


# 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.

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


$A+B+C=2149$, 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.

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 $k$:

$$k = \frac{2150 - 3n^2 - n}{6}$$

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!

• 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

\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.

• 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.


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.