Examples of generating functions

The following examples of generating functions are in the spirit of George Pólya, who advocated learning mathematics by doing and re-capitulating as many examples and proofs as possible. The purpose of this article is to present common ways of creating generating functions.

Worked example A: basics

New generating functions can be created by extending simpler generating functions. For example, starting with

G(1;x)=\sum _{{n=0}}^{{\infty }}x^{n}={\frac {1}{1-x}}

and replacing $x$ with $ax$ , we obtain

G(1;ax)={\frac {1}{1-ax}}=1+(ax)+(ax)^{2}+\cdots +(ax)^{n}+\cdots =\sum _{{n=0}}^{{\infty }}a^{n}x^{n}=G(a^{n};x).

Bivariate generating functions

One can define generating functions in several variables, for series with several indices. These are often called super generating functions, and for 2 variables are often called bivariate generating functions.

For instance, since $(1+x)^n$ is the generating function for binomial coefficients for a fixed n, one may ask for a bivariate generating function that generates the binomial coefficients ${\binom {n}{k}}$ for all k and n. To do this, consider $(1+x)^n$ as itself a series (in n), and find the generating function in y that has these as coefficients. Since the generating function for $a^{n}$ is just $1/(1-ay)$ , the generating function for the binomial coefficients is:

{\frac {1}{1-(1+x)y}}=1+(1+x)y+(1+x)^{2}y^{2}+\dots ,

and the coefficient on $x^{k}y^{n}$ is the ${\binom {n}{k}}$ binomial coefficient.

Worked example B: Fibonacci numbers

Consider the problem of finding a closed formula for the Fibonacci numbers F_n defined by F₀ = 0, F₁ = 1, and F_n = F_n−1 + F_n−2 for n ≥ 2. We form the ordinary generating function

f=\sum _{{n\geq 0}}F_{n}x^{n}

for this sequence. The generating function for the sequence (F_n−1) is xf and that of (F_n−2) is x²f. From the recurrence relation, we therefore see that the power series xf + x²f agrees with f except for the first two coefficients:

{\begin{array}{rcrcrcrcrcrcr}f&=&F_{0}x^{0}&+&F_{1}x^{1}&+&F_{2}x^{2}&+&\cdots &+&F_{i}x^{i}&+&\cdots \\xf&=&&&F_{0}x^{1}&+&F_{1}x^{2}&+&\cdots &+&F_{{i-1}}x^{i}&+&\cdots \\x^{2}f&=&&&&&F_{0}x^{2}&+&\cdots &+&F_{{i-2}}x^{i}&+&\cdots \\(x+x^{2})f&=&&&F_{0}x^{1}&+&(F_{0}+F_{1})x^{2}&+&\cdots &+&(F_{{i-1}}+F_{{i-2}})x^{i}&+&\cdots \\&=&&&&&F_{2}x^{2}&+&\cdots &+&F_{i}x^{i}&+&\cdots \\\end{array}}

Taking these into account, we find that

f=xf+x^{2}f+x.\,\!

(This is the crucial step; recurrence relations can almost always be translated into equations for the generating functions.) Solving this equation for f, we get

f={\frac {x}{1-x-x^{2}}}.

The denominator can be factored using the golden ratio φ₁ = (1 + √5)/2 and φ₂ = (1 − √5)/2, and the technique of partial fraction decomposition yields

f={\frac {1}{{\sqrt {5}}}}\left({\frac {1}{1-\varphi _{1}x}}-{\frac {1}{1-\varphi _{2}x}}\right).

These two formal power series are known explicitly because they are geometric series; comparing coefficients, we find the explicit formula

F_{n}={\frac {1}{{\sqrt {5}}}}(\varphi _{1}^{n}-\varphi _{2}^{n}).

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