Two questions about generating functions I'm very interested in the topic of generating functions, so I have two questions:


*

*I just realized that when I have exponential generating function for example $F(x)=e^{e^x-1}$, I can take n-th derivative and count the value in $x=0$ to get n-th element of sequence that this function represents. It's very useful I think :-) but are the ordinary generating functions that useful too? Is there any operation on ordinary generating function that can help me count the n-th element of sequence that this function represents?

*Here: http://en.wikipedia.org/wiki/Generating_function#Examples we have got exponential generating function for sequence $a_n=n^2$. It's simple to find the ordinary generating function for this sequence (taking derivatives and subtracting something) but how can I deduce that $\displaystyle \sum_{n=0}^{+\infty}n^2\frac{x^n}{n!}=x(x+1)e^x$?

 A: (1) Just do the same and divide by $n!$ then.
(2) We have
\begin{align*}
   \sum_{n=0}^{\infty} n^2 \frac{x^n}{n!} 
   &= \sum_{n=1}^\infty n \frac{x^n}{(n-1)!}\\
   &= x \sum_{n=1}^\infty n\frac{x^{n-1}}{(n-1)!}\\
   &= x \sum_{n=1}^\infty \bigl((n-1)+1\bigr)\frac{x^{n-1}}{(n-1)!}\\
   &= x\sum_{n=2}^\infty \frac{x^{n-1}}{(n-2)!} + x\exp(x)\\
   &= x^2\exp(x) + x\exp(x)\\
   &= x(x+1)\exp(x).
\end{align*}
HTH, AB,
A: Another way. Note that for any sequence $a_n$, with $A(z) = \sum_{n \ge 0} a_n \frac{z^n}{n!}$
$\begin{align}
z \frac{\mathrm{d}}{\mathrm{d} z} A(z)
  &=z \sum_{n \ge 0} a_n \frac{z^{n - 1}}{(n - 1)!} \\
  &= \sum_{n \ge 0} n a_n \frac{z^n}{n!}
\end{align}$
Thus you get what you want by doing the above twice to $\mathrm{e}^z$:
$\begin{align}
  z \frac{\mathrm{d}}{\mathrm{d} z}
    \left( z \frac{\mathrm{d}}{\mathrm{d} z} \mathrm{e}^z \right) 
    &= z \frac{\mathrm{d}}{\mathrm{d} z}
         \left( z \mathrm{e}^z \right) \\
    &= z \left( \mathrm{e}^z + z \mathrm{e}^z \right) \\
    &= (z + z^2) \mathrm{e}^z
\end{align}$
