# Formal power series: identity problem

• Feb 16th 2011, 09:58 AM
Greg98
Formal power series: identity problem
Hello,
the problem is to show by calculating that:
$\displaystyle \displaystyle\frac{x(1+x)}{(1-x)^3}=\sum_{i=0}^{\infty}k^2x^k$.

I know the solution has to do something formal power series. I found following pieces: link 1 and link 2 , but even those couldn't get me any further. Also WolframAlpha didn't gave me any clue, though there's some information. I'm completely stuck, which means any help is welcome. Thank you!
• Feb 16th 2011, 10:20 AM
TheEmptySet
Quote:

Originally Posted by Greg98
Hello,
the problem is to show by calculating that:
$\displaystyle \displaystyle\frac{x(1+x)}{(1-x)^3}=\sum_{i=0}^{\infty}k^2x^k$.

I know the solution has to do something formal power series. I found following pieces: link 1 and link 2 , but even those couldn't get me any further. Also WolframAlpha didn't gave me any clue, though there's some information. I'm completely stuck, which means any help is welcome. Thank you!

Very formally we have

$\displaystyle \displaystyle \frac{1}{1-x}=\sum_{k=0}^{\infty}x^k$

Now consider

What is

$\displaystyle \displaystyle x\frac{d}{dx}\left( x\frac{d}{dx}x^k\right)=?$

Why does this help?
• Feb 16th 2011, 01:29 PM
chisigma
Quote:

Originally Posted by Greg98
Hello,
the problem is to show by calculating that:
$\displaystyle \displaystyle\frac{x(1+x)}{(1-x)^3}=\sum_{i=0}^{\infty}k^2x^k$.

I know the solution has to do something formal power series. I found following pieces: link 1 and link 2 , but even those couldn't get me any further. Also WolframAlpha didn't gave me any clue, though there's some information. I'm completely stuck, which means any help is welcome. Thank you!

Proceeding [tediously...] 'step by step' we have...

$\displaystyle \displaystyle \sum_{k=0}^{\infty} k^{2}\ x^{k} = x\ \sum_{k=0}^{\infty} k^{2}\ x^{k-1} = x\ \frac{d}{dx} \sum_{k=0}^{\infty} k\ x^{k}= x\ \frac{d}{dx} (x\ \sum_{k=0}^{\infty} k\ x^{k-1})=$

$\displaystyle \displaystyle = x\ \frac{d}{dx} (x\ \frac{d}{dx} \frac{1}{1-x}) = x\ \frac{d}{dx} \frac{x}{(1-x)^{2}}= x\ \frac{(1-x)^{2} +2 x (1-x)}{(1-x)^{4}} =$

$\displaystyle \displaystyle = x\ \frac{1-x+2 x}{(1-x)^{3}} = x\ \frac{1+x}{(1-x)^{3}}$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$