# Talk:Generating function

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Field: Discrete mathematics

The information here is really not enough... it didn't give me any idea how to calculate the generating function coefficients. It's algebra and series, but the article should list the most used tricks: binomial theorem, infinite geometric series, convolution products, etc. -Iopq 19:59, 18 October 2005 (UTC)

## Definition

I am not an expert on the field, so I will not dare to introduce the following definition myself. But if somebody does agree, please include under "Definitions" the following:

"A generation function is a transformation that converts a given sequence, S = {an}, into a continous function, f(x), through a series expantion whose coeficients are the elements an of the sequence S."

or something similar you find more appropiate.

Well, I don't think that's very clear. The powers of a variable are really place-olders, here. There is no necessary connection to continuity. Charles Matthews 12:19, 16 November 2005 (UTC)
I agree. Many useful generating functions are not continuous or even convergent. Any definition must stress the formal nature of the series. --Zero 22:52, 16 November 2005 (UTC)
Absolutely. (No pun intended.) To call these things "continuous" is absurd. Michael Hardy 20:13, 17 November 2005 (UTC)

I just noticed that the german version is not liked here it's called "Erzeugende Funktionen", url is here: http://de.wikipedia.org/wiki/Erzeugende_Funktion

In mathematics a generating function is a formal power series whose coefficients encode information about a sequence an that is indexed by the natural numbers. For example,... (a nice easy example or two, please!)

This article is fairly typical of current Wikipedia mathematics articles: it dives headlong into a mass of detail without first explaining the basics. This is supposed to be an online encyclopedia, not a maths textbook!

Education is a process of diminishing deception. Start off with the simple stuff; the ifs and buts come later.

--84.9.78.198 14:14, 27 November 2006 (UTC)

If you read on past the Definitions section you will find an Examples section with four examples of different types of generating function for the sequence of square numbers, and also an extended example showing how the ordinary generating function for the Fibonacci numbers is derived. If Examples came before Definitions the article would be more difficult to follow, as you would not know what the Examples were meant to be illustrating. Gandalf61 14:41, 27 November 2006 (UTC)

## Uniqueness of F

I made a change to the article, dropping a condition (something being an integral domain) on the explanation of the uniqueness of the inverse of (1-X). If F is any ring with a unit, not necessarily commutative or an integral domain, then the only power series $f(X) \in F[[X]]$ such that $1=(1-X)f(X)$ is $f(X)=1+X+X^2+\dots$. To see this, let $f(X) = f_0 + f_1 X + f_2 X^2 + \dots$. Then $(1-X)f(X) = f_0 + (f_1 - f_0) X + (f_2 - f_1) X^2 + ...$. Solving $1 = (1-X)f(X)$ means that $f_0 = 1, (f_1 - f_0) = 0, (f_2 - f_0) = 0, \dots$ and therefore $1 = f_0 = f_1 = f_2 = \dots$. The only multiplication in the ring F is used in this proof is multiplication by 1 and -1 in F. Therefore neither general commutativity of the ring F nor F being an integral domain is required. Indeed multiplication in F need not even be associative. (So, the result holds if F is the octonions, for example.) It is only required that multiplication in F have an identity element 1. Multiplication by -1, and its necessary properties, is then implied by F being a ring. Of course, multiplication by X in $F[[X]]$ has been used, and commutativity of this operation , that is, $Xa = aX$ has been used, as has the fact if $Xa=0$ then $a=0$. In general, though if the ring of coefficient is not an integral domain or commutative, then neither is the resulting power series ring. —Preceding unsigned comment added by DRLB (talkcontribs) 15:26, 17 October 2008 (UTC)

That's a nice extension of the given statement. All the article actually uses is that formal power series with coefficients in any ring form a ring -- two-sided inverses are unique in any ring. --Charleyc (talk) 16:23, 18 October 2008 (UTC)
Good point about uniqueness of two-sided inverses, probably worth saying in the article. Instead of saying this is unique, say this is a two-sided inverse, and thus it is unique. (I'm not sure if two-sided inverses are unique in non-associative rings, but I think that's out-of-scope for the article.) DRLB (talk) 14:50, 20 October 2008 (UTC)

## Formulae

I noticed that all summation formulae on the page look like this: for each natural n sum a_i*x^n or something. I believe this should be fixed, because 1/(1-x) is not x+x^2+x^3+... but 1+x+x^2+x^3... —Preceding unsigned comment added by 85.187.35.160 (talk) 13:47, 20 August 2009 (UTC)

Fixed - I have replaced $\sum_{n\in\mathbf{N}}$ with $\sum_{n=0}^{\infty}$, which was clearly what was intended in each case. Gandalf61 (talk) 15:55, 20 August 2009 (UTC)

## "Generating series" terminology

The current version of the article indicates that "generating series" is "more correct" than "generating function." While I agree that generating functions aren't really functions (for instance, because their evaluation at specific points isn't what they're about), I worry that they aren't really series either (for instance, because whether or not they converge isn't what they're about). Given that there is now a citation (which I haven't checked!) to show that "generating series" is also in use, might we simply say that it is an "alternative" rather than "more correct"? Quantling (talk) 16:00, 27 May 2010 (UTC)

The "series" in "generating series" refers to formal power series, where convergence is not much of an issue either (the term "generating formal power series" would be a bit heavy). Series are not necessarily about convergence, so I don't think this is much of a problem. Marc van Leeuwen (talk) 10:35, 28 May 2010 (UTC)

## Is this a generating function?

$\pi(\cot (\pi(c+z))-2\cot (2\pi(c-z))-2\cot (2\pi(c+z))+\cot (\pi(c-z))) = -2 \left ( \sum_{k=0}^\infty z^{2k} \sum_{x=1}^\infty \frac{1}{(x+c-1/2)^{2k+1}} - \frac{1}{(x-c-1/2)^{2k+1}} \right )$


taking multiple derivatives with respect to z closed form sums can be obtained such as:

$\pi^{3}(8(\cot(2c\pi)+\cot^{3}(2c\pi))-(\cot(c\pi)+\cot^{3}(c\pi))) =\sum_{x=1}^\infty \frac{1}{(x+c-1/2)^{3}} - \frac{1}{(x-c-1/2)^{3}}$


$= -\frac{\pi^{3}\sin(c\pi)}{\cos^{3}(c\pi)}$

$\cot(\pi(c+z)) \approx \cot(c\pi)-z\pi(1+\cot^{2}(c\pi))+z^{2}\pi^{2}(\cot(c\pi)+\cot^{3}(c\pi))$

http://iamned.com/math —Preceding unsigned comment added by 67.161.40.148 (talk) 11:02, 30 June 2010 (UTC)