Proper equilibrium

The following examples 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 tricks of the trade in context, so that people may incorporate them into their knowledge.

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

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

External links

• Generating Functions, Power Indices and Coin Change at cut-the-knot
• Generatingfunctionology (PDF)