0
$\begingroup$

I'm working on some problems that involve generating functions, and I've been able to understand when exponential vs. ordinary generating functions are used and how to write a generating function to solve a given problem. Where I get stuck is finding the coefficient of $x_n$ for exponential generating functions. Here is an example of what I have tried, but I don't know where to go from there. Any advice would be appreciated.

$$G(x) = (1 + \frac{x^2}{2!} + \frac{x^4}{4!} + ...) (1 + x + \frac{x^2}{2!} +\frac{x^3}{3!} + ...) (1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...) (1+x) $$ $$= \frac{e^x + e^{-x}}{2} e^x e^x (1+x)$$ $$= \frac{1}{2} (e^{3x} +e^x)(1+x) $$ $$= \frac{1}{2} (\sum_{n=0}^{\infty} 3^n \frac{x^n}{n!} + \sum_{n=0}^{\infty} \frac{x^n}{n!}) (1+x) = \frac{1}{2} (\sum_{n=0}^{\infty} (3^n + 1) \frac{x^n}{n!}) (1+x)$$

$\endgroup$
  • $\begingroup$ $$\ldots=\frac 12\sum_{n=0}^\infty\frac{3^n+1}{n!}(x^n+x^{n+1})=\frac 12\left(\frac{3^0+1}{0!}+\sum_{n=1}^\infty\frac{3^n+1}{n!}x^n+\sum_{n=0}^\infty\frac{3^n+1}{n!}x^{n+1}\right)=\frac 12\left(\frac{3^0+1}{0!}+\sum_{n=1}^\infty\frac{3^n+1}{n!}x^n+\sum_{n=1}^\infty\frac{3^{n-1}+1}{(n-1)!}x^n\right)=\ldots~\\~$$ Can you proceed from here? $\endgroup$ – Prasun Biswas Sep 6 '17 at 7:01
1
$\begingroup$

In this example, the coefficient of $x^n$ is $$\frac12\left(\frac{3^n+1}{n!}+\frac{3^{n-1}+1}{(n-1)!}\right).$$

$\endgroup$
  • $\begingroup$ Notice that for $n=0$, the expression becomes undefined. You need to add that this is valid only for $n\geq 1$ and the coefficient of $x^0$, i.e., the constant term is given by $\frac 12\times\frac{3^0+1}{0!}=1$ $\endgroup$ – Prasun Biswas Sep 6 '17 at 7:00
1
$\begingroup$

You're very close. Let $a_{-1}=0$. Then:

$$(1+x)\sum_{n=0}^\infty a_nx^n=\sum_{n=0}a_nx^n+\sum_{n=0}^\infty a_{n-1} x^{n}=\sum_{n=0}^\infty (a_n+a_{n-1})x^n$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.