# question on indices of a generating function

• May 29th 2013, 05:04 PM
Yuuki
question on indices of a generating function
I want to find the generating function for an=n(n-1).

$\sum_{n=0}^{\infty }n(n-1)\frac{x^n}{n!}$
$=\sum_{n=0}^{\infty }\frac{x^n}{(n-2)!}$
$=x^2\sum_{n=0}^{\infty }\frac{x^{n-2}}{(n-2)!}$
$=x^2\sum_{k=-2}^{\infty }\frac{x^{k}}{k!}$
$=x^2\sum_{k=0}^{\infty }\frac{x^{k}}{k!}$
$=x^2e^x$

Am I justified in the second to last step, ignoring the summands for k=-1 and k=-2?
If so, is there a reason why I can ignore the first two terms?
• May 29th 2013, 05:05 PM
Prove It
Re: question on indices of a generating function
Is there such a thing as (-2)! or (-1)!?
• May 29th 2013, 05:08 PM
Yuuki
Re: question on indices of a generating function
No, but I wasn't sure if (-2)! and (-1)! not existing allowed me to ignore them.
So because (-2)! and (-1)! don't exist, we can't evaluate them, hence it's okay to ignore them?
• May 29th 2013, 05:10 PM
Prove It
Re: question on indices of a generating function
Only in this case, because you can see if you plug n=0 and n=1 into your original sum that those terms are 0 anyway.

Really what you should have written is

\displaystyle \begin{align*} \sum_{n = 0}^{\infty} n(n-1)\frac{x^n}{n!} &= 0 + 0 + \sum_{n = 2}^{\infty} n(n-1)\frac{x^n}{n!} \\ &= \sum_{n = 2}^{\infty} n(n-1)\frac{x^n}{n!} \\ &= \sum_{n = 2}^{\infty} \frac{x^n}{(n-2)!} \\ &= x^2 \sum_{n=2}^{\infty} \frac{x^{n-2}}{(n-2)!} \\ &= x^2\sum_{k=0}^{\infty}\frac{x^k}{k!} \\ &= x^2\,e^x \end{align*}

It doesn't make sense to put into your counter values which will give undefined terms.
• May 29th 2013, 05:23 PM
Yuuki
Re: question on indices of a generating function
Thank you very much, it now makes a lot of sense.