# maclaurin series

• Apr 26th 2010, 02:27 PM
Brotha
maclaurin series
differentiate the maclaurin series for 1/(1-x) twice to find the maclaurin series of 1/(1-x)^3. Thanks
• Apr 26th 2010, 04:34 PM
$\frac{1}{1-x} = \sum^{\infty}_{n=0} x^n \text{ for }|x| < 1$

If you differentiate $\frac{1}{1-x}$ twice you get $\frac{2}{(1-x)^3}$

You just differentiate each term in the sum above and divide by 2. After a few terms you should see a pattern then just turn it into an infinite sum.
• Apr 26th 2010, 06:05 PM
Brotha
Thanks, but what would the maclaurin series be? Is it something like
sum_(n=0)^infinity1/2 n (n-1) x^(n-2). How do you get that?
• Apr 26th 2010, 06:21 PM
Quote:

Originally Posted by Brotha
Thanks, but what would the maclaurin series be? Is it something like
sum_(n=0)^infinity1/2 n (n-1) x^(n-2). How do you get that?

$\frac{1}{(1-x)} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 +$ etc...

So differentiate both sides...

$-\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + 5 x^4 + 6x^5 +$ etc...

differentiate again...

$\frac{2}{(1-x)^3} = 2 + 6x + 12x^2 + 20x^3 + 30x^4 +$ etc...

This is a series of the form...

$\sum_{n=0}^{\infty} (n+1)(n+2)x^n$.

How did I get this? Just differentiate $x^n$ twice. You don't even need to all that stuff I did above. Just it's good to see it done so you know what's going on.

Differentiating $x^n$ twice gives you $n(n-1)x^{n-2}$. I.e. $(n+2)(n+1)x^n$

So summarizing...

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

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