Power series

**LeoBloom.** · June 15th 2008, 05:01 PM

Find a power series representation for the function

f(x) = x^3 / (x - 2)^2

**Mathstud28** · June 15th 2008, 05:42 PM

Originally Posted by LeoBloom.

Find a power series representation for the function

f(x) = x^3 / (x - 2)^2

Ok first we note that

$\frac{d}{dx}\bigg[\frac{1}{x-2}\bigg]=\frac{-1}{(x-2)^2}$

But we also know that

$\frac{1}{x-2}=\frac{-1}{2}\frac{1}{1-\frac{x}{2}}=-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{x}{2}\ri ght)^n=-\sum_{n=0}^{\infty}\frac{x^n}{2^{n+1}}$

So we have

$-x^3\cdot\bigg[-\sum_{n=0}^{\infty}\frac{x^n}{2^{n+1}}\bigg]=x^3\cdot\sum_{n=1}^{\infty}\frac{nx^{n-1}}{2^{n+1}}=\sum_{n=0}^{\infty}\frac{nx^{n+2}}{2^ {n+1}}$

$\therefore\frac{x^3}{(x-2)^2}=\sum_{n=0}^{\infty}\frac{nx^{n+2}}{2^{n+1}}$

**LeoBloom.** · June 15th 2008, 05:55 PM

Ok, so when the n becomes 1 in the Sigma, that means that you took the derivative? And can you only multiply the x^3 after you do that?

**Mathstud28** · June 15th 2008, 06:01 PM

Originally Posted by LeoBloom.

Ok, so when the n becomes 1 in the Sigma, that means that you took the derivative? And can you only multiply the x^3 after you do that?

What do you mean? I multiplied after I took the derivative because if I would have done so before hand I would be saying the following is true

$f(x)\cdot{g'(x)}=\bigg(f(x)\cdot{g(x)}\bigg)'$

Which I hope is not true, or I got an A+ in calculus when I shouldnt have

**LeoBloom.** · June 15th 2008, 06:35 PM

Good man, you've answered my question, you deserve your A+

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