Difference between revisions of "Generating function"

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The idea behind generating functions is to represent a [[combinatorics|combinatorial]] [[function]] <math>A(k)</math> in terms of a [[polynomial function]] which is equivalent for all purposes. This function is <math>A(0)+A(1)x+A(2)x^2+A(3)x^3+\cdots</math>, where the coefficient <math>A(k)</math> of <math>x^k</math> is the number of ways an event <math>\displaystyle{k}</math> can occur.
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The idea behind generating functions is to represent a [[combinatorics|combinatorial]] [[function]] <math>A(k)</math> in terms of a [[polynomial]] [[function]] which is equivalent for all purposes. This function is <math>A(0)+A(1)x+A(2)x^2+A(3)x^3+\cdots</math>, where the coefficient <math>A(k)</math> of <math>x^k</math> is the number of ways an event <math>\displaystyle{k}</math> can occur.
  
 
== Simple Example ==
 
== Simple Example ==

Revision as of 11:33, 7 July 2006

The idea behind generating functions is to represent a combinatorial function $A(k)$ in terms of a polynomial function which is equivalent for all purposes. This function is $A(0)+A(1)x+A(2)x^2+A(3)x^3+\cdots$, where the coefficient $A(k)$ of $x^k$ is the number of ways an event $\displaystyle{k}$ can occur.

Simple Example

If we let $A(k)={n \choose k}$, then we have ${n \choose 0}+{n \choose 1}x + {n \choose 2}x^2+\cdots+$${n \choose n}x^n$.

This function can be described as the number of ways we can get $\displaystyle{k}$ heads when flipping $n$ different coins.

The reason to go to such lengths is that our above polynomial is equal to $(1+x)^n$ (which is clearly seen due to the Binomial Theorem). By using this equation, we can rapidly uncover identities such as ${n \choose 0}+{n \choose 1}+...+{n \choose n}=2^n$(let ${x}=1$), also ${n \choose 1}+{n \choose 3}+\cdots={n \choose 0}+{n \choose 2}+\cdots$.

See also