# Manipulating power series #2

• Apr 9th 2009, 09:05 AM
mollymcf2009
Manipulating power series #2
Find a power series representation of the function:

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

Here is what I did:

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

$\displaystyle 2+x \sum^{\infty}_{n=0} (x)^n$

$\displaystyle = 2 + \sum^{\infty}_{n=0} x^{2n}$

Which is not right. The correct answer is:

$\displaystyle 2 + 3 \sum^{\infty}_{n=1}x^n$

But I don't understand how you would get that answer.
• Apr 9th 2009, 09:20 AM
o_O

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

Now you have to modify your indices so you can add them together:
$\displaystyle = 2\sum_{n = 0}^{\infty} x^n + \sum_{n=0}^{\infty}x^{n+1} = 2\left(1 + \sum_{n = 1}^{\infty} x^n\right) + \sum_{n=1}^{\infty}x^{n} = \cdots$
• Apr 9th 2009, 02:32 PM
Jester
Quote:

Originally Posted by mollymcf2009
Find a power series representation of the function:

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

Here is what I did:

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

$\displaystyle 2+x \sum^{\infty}_{n=0} (x)^n$

$\displaystyle = 2 + \sum^{\infty}_{n=0} x^{2n}$

Which is not right. The correct answer is:

$\displaystyle 2 + 3 \sum^{\infty}_{n=1}x^n$

But I don't understand how you would get that answer.

It's probably easier to notice that

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

use the usual power series for $\displaystyle \frac{1}{1-x}$ multiply by 3 and substract 1.
• Apr 9th 2009, 04:41 PM
mollymcf2009
Quote:

Originally Posted by o_O

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

Now you have to modify your indices so you can add them together:
$\displaystyle = 2\sum_{n = 0}^{\infty} x^n + \sum_{n=0}^{\infty}x^{n+1} = 2\left(1 + \sum_{n = 1}^{\infty} x^n\right) + \sum_{n=1}^{\infty}x^{n} = \cdots$

That definitely helps, but I still don't see how you end up with 2+3..
• Apr 9th 2009, 04:46 PM
mollymcf2009
Quote:

Originally Posted by danny arrigo
It's probably easier to notice that

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

use the usual power series for $\displaystyle \frac{1}{1-x}$ multiply by 3 and substract 1.

THANK YOU!!! That is the simple explanation that I needed!