How to obtain explicit formula of coefficients for generating functions?

110 Views Asked by At

Is there any function in python or sage to obtain explicit formula of coefficients for generating functions. For example, Catalan generating function is given as

$${\displaystyle c(x)={\dfrac {1-{\sqrt {1-4x}}}{2x}}}$$ And I need to obtain general formula for $n$th coefficient:

$${\displaystyle C_{n}={\dfrac {1}{2n+1}}{2n+1 \choose n}}$$

1

There are 1 best solutions below

0
On

In this case you want the generalized binomial theorem which is

$(1+x)^a =\sum_{n=0}^{\infty} \binom{a}{n}x^n $ where $\binom{a}{n} =\dfrac{\prod_{k=0}^{n-1}(a-k)}{n!} $.

For your case $a=\dfrac12$ so, for $n \ge 1$,

$\begin{array}\\ \binom{\frac12}{n} &=\dfrac{\prod_{k=0}^{n-1}(\frac12-k)}{n!}\\ &=\dfrac{\prod_{k=0}^{n-1}(\frac12(1-2k))}{n!}\\ &=\dfrac{(-1)^n\prod_{k=0}^{n-1}(2k-1)}{2^nn!}\\ &=\dfrac{(-1)^{n+1}\prod_{k=1}^{n-1}(2k-1)}{2^nn!}\\ &=\dfrac{(-1)^{n+1}\prod_{k=1}^{n-1}(2k-1)\prod_{k=1}^{n-1}(2k)}{2^nn!\prod_{k=1}^{n-1}(2k)}\\ &=\dfrac{(-1)^{n+1}\prod_{k=1}^{2n-2}k}{2^nn!2^{n-1}(n-1)!}\\ &=\dfrac{(-1)^{n+1}(2n-2)!}{2^{2n-1}n!(n-1)!}\\ \end{array} $

so

$\begin{array}\\ \sqrt {1-4x} &=\sum_{n=0}^{\infty}\binom{\frac12}{n}(-4x)^n\\ &=1+\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}(2n-2)!}{2^{2n-1}n!(n-1)!}(-1)^n4^nx^n\\ &=1-\sum_{n=1}^{\infty}\dfrac{2(2n-2)!}{n!(n-1)!}x^n\\ \end{array} $

so

$\begin{array}\\ \dfrac {1-\sqrt {1-4x}}{2x} &=\dfrac{2\sum_{n=1}^{\infty}\dfrac{(2n-2)!}{n!(n-1)!}x^n}{2x}\\ &=\sum_{n=1}^{\infty}\dfrac{(2n-2)!}{n!(n-1)!}x^{n-1}\\ &=\sum_{n=0}^{\infty}\dfrac{(2n)!}{n!(n+1)!}x^{n}\\ &=\sum_{n=0}^{\infty}\dfrac{(2n+1)!}{(2n+1)n!(n+1)!}x^{n}\\ &=\sum_{n=0}^{\infty}\dfrac1{2n+1}\binom{2n+1}{n}x^{n}\\ \end{array} $

I'll be glad to explain any of this you do not understand since it uses a number of standard manipulations you should get familiar with.