Confused on Exponential Generating function

**oldguynewstudent** · July 24th 2010, 08:27 AM

Given the following notation: $\left[\left[f(x)\right]\right]_{x^{k}}$ = the coefficient of $x^{k}$ in f(x).

Also for an exponential generating function $\left[\left[e^{x}\right]\right]_{\frac{x^{k}}{k!}}$ = $\left[\left[\sum_{k\geq0}\frac{x^{k}}{k!}\right]\right]_{\frac{x^{k}}{k!}}=1$ .

We have an example in Combinatorics A Guided Tour by David R. Mazur p. 129-130 as follows:

Find a formula for the n-th term of the sequence $\{d_{n}\}_{n\geq0}$ defined by $d_{0}=1\,\, d_{n}=nd_{n-1}+1\,\,\, n\geq1$ .

After a short derivation we have $d_{n}=\left[\left[e^{x}*\frac{1}{1-x}\right]\right]_{\frac{x^{n}}{n!}}$ = $\sum_{j=0}^{n}\left({n\atop j}\right)1*(n-j)!$ = $\sum_{j=0}^{n}\left({n\atop j}\right)(n-j)!$

I am familiar with $(1+x)^{n}=\sum_{k\geq0}\left({n\atop k}\right)x^{k}$ . Also I know $\frac{1}{1-x}=\sum_{k\geq0}k!\frac{x^{k}}{k!}$ . But I just do not see how the binomial fits into the above summation.

I have been working almost nonstop on this course and applied statistics this summer. Any help with how the binomial gets into the above summation would be greatly appreciated. By the way, this is the most fascinating course I have ever taken even though I seem to be struggling just a little.

**Renji Rodrigo** · July 26th 2010, 10:16 AM

In this case you can use the cauchy product of the series

Given two series $\sum\limits^{\infty}_{k=0}a_{k}$

and $\sum\limits^{\infty}_{k=0}b_{k}$ , the Cauchy product is

$\sum\limits^{\infty}_{k=0}c_{k}$

where

$c_{n}= \sum\limits^{n}_{k=0}a_{k}b_{n-k}\;\;n \in N.$

**awkward** · July 26th 2010, 05:21 PM

Hi oldguynewstudent,

I think the theorem you need is that if
$f(x) = \sum_{n=0}^{\infty} \frac{a_n}{n!} x^n$
and
$g(x) = \sum_{n=0}^{\infty} \frac{b_n}{n!} x^n$
then the coefficient of $\frac{x^n}{n!}$ in $f(x) \cdot g(x)$
is
$\sum_r \binom{n}{r} a_r b_{n-r}$

**oldguynewstudent** · July 27th 2010, 09:25 AM

Originally Posted by awkward

Hi oldguynewstudent,

I think the theorem you need is that if
$f(x) = \sum_{n=0}^{\infty} \frac{a_n}{n!} x^n$
and
$g(x) = \sum_{n=0}^{\infty} \frac{b_n}{n!} x^n$
then the coefficient of $\frac{x^n}{n!}$ in $f(x) \cdot g(x)$
is
$\sum_r \binom{n}{r} a_r b_{n-r}$

Yes, that is what I was looking for!

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