# Maclaurin Series

• November 22nd 2011, 04:55 PM
statmajor
Maclaurin Series
How would I find the maclaurin series for the following function: $f(x) = \frac{3}{(1-x)^2}$.

The first, second, and third derivatives are: $\frac{6}{(1-x)^3}, \frac{18}{(1-x)^4}, \frac{72}{(1-x)^5}$.

$f(x)=\Sigma \frac{f^{(n)}(0)}{n!}x^n = 3 + 6x +\frac{18}{2!}x^2 + \frac{72}{3!}x^3 + ...$.

Here's where I'm stuck.

Any help would be greatly appreciated.
• November 22nd 2011, 05:11 PM
skeeter
Re: Maclaurin Series
$f(x) = \frac{3}{(1-x)^2}$

let $F(x)$ be the antiderivative of $f(x)$ ...

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

$F'(x) = f(x) = \frac{3}{(1-x)^2} = 3(1 + 2x + 3x^2 + ... )$
• November 22nd 2011, 05:13 PM
Prove It
Re: Maclaurin Series
Quote:

Originally Posted by statmajor
How would I find the maclaurin series for the following function: $f(x) = \frac{3}{(1-x)^2}$.

The first, second, and third derivatives are: $\frac{6}{(1-x)^3}, \frac{18}{(1-x)^4}, \frac{72}{(1-x)^5}$.

$f(x)=\Sigma \frac{f^{(n)}(0)}{n!}x^n = 3 + 6x +\frac{18}{2!}x^2 + \frac{72}{3!}x^3 + ...$.

Here's where I'm stuck.

Any help would be greatly appreciated.

The easiest way would be to notice that $\displaystyle 3\,\frac{d}{dx}\left(\frac{1}{1 - x}\right) = \frac{3}{(1 - x)^2}$ and that $\displaystyle \frac{1}{1 - x}$ is the closed form of the geometric series $\displaystyle \sum_{n = 0}^{\infty}x^n$ for $\displaystyle |x| < 1$.