# power series representation

• Sep 3rd 2009, 02:21 PM
john52302
power series representation
Hi,

I was trying to find a power series for function $\displaystyle ln((1+x)/(1-x))$.

I separated this into $\displaystyle ln(1+x)-ln(1-x)$ and used differentiation & integration to get $\displaystyle \sum_{n=0}^\infty\frac{(-1)^n(x)^{n+1}}{n+1} + \sum_{n=0}^\infty\frac{x^{n+1}}{n+1}$

The answer says $\displaystyle \sum_{n=0}^\infty\frac{2x^{2n+1}}{2n+1}$. I have no idea how this answer was derived. Can anyone explain to me how to get this answer?

Thanks!
• Sep 3rd 2009, 02:49 PM
Calculus26
For

http://www.mathhelpforum.com/math-he...18a5afe8-1.gif

Note if n is even (or 0 ) the terms add and if n is odd they cancel so you have only odd powers and all are positive
• Sep 3rd 2009, 02:53 PM
skeeter
Quote:

Originally Posted by john52302
Hi,

I was trying to find a power series for function $\displaystyle ln((1+x)/(1-x))$.

I separated this into $\displaystyle ln(1+x)-ln(1-x)$ and used differentiation & integration to get $\displaystyle \sum_{n=0}^\infty\frac{(-1)^n(x)^{n+1}}{n+1} + \sum_{n=0}^\infty\frac{x^{n+1}}{n+1}$

The answer says $\displaystyle \sum_{n=0}^\infty\frac{2x^{2n+1}}{2n+1}$. I have no idea how this answer was derived. Can anyone explain to me how to get this answer?

Thanks!

$\displaystyle \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...$

$\displaystyle \ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - ...$

$\displaystyle \ln(1+x) - \ln(1-x) = 2\left(x + \frac{x^3}{3} + \frac{x^5}{5} + ... \right)$