3
$\begingroup$

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

  1. 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?
  2. 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 $\displaystyle \sum_{n=0}^{+\infty}n^2\frac{x^n}{n!}=x(x+1)e^x$?
$\endgroup$
3
$\begingroup$

(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,

$\endgroup$
  • $\begingroup$ brilliant! thanks a lot :-) $\endgroup$ – xan Mar 22 '12 at 13:46
0
$\begingroup$

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 $\mathrm{e}^z$:

$\begin{align} z \frac{\mathrm{d}}{\mathrm{d} z} \left( z \frac{\mathrm{d}}{\mathrm{d} z} \mathrm{e}^z \right) &= z \frac{\mathrm{d}}{\mathrm{d} z} \left( z \mathrm{e}^z \right) \\ &= z \left( \mathrm{e}^z + z \mathrm{e}^z \right) \\ &= (z + z^2) \mathrm{e}^z \end{align}$

$\endgroup$

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.