power series representation

**john52302** · September 3rd 2009, 02:21 PM

Hi,

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

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

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

Thanks!

**Calculus26** · September 3rd 2009, 02:49 PM

For

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

**skeeter** · September 3rd 2009, 02:53 PM

Originally Posted by john52302

Hi,

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

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

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

Thanks!

$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...<br />$

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

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

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