Power Series

**cgiulz** · October 5th 2009, 02:29 PM

From the power series,

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

obtain the power series whose sum is

$\ln (\frac{1+x}{1-x}).$

**tonio** · October 5th 2009, 02:53 PM

Originally Posted by cgiulz

From the power series,

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

obtain the power series whose sum is

$\ln (\frac{1+x}{1-x}).$

(1+x)/(1-x) = 1 + 2x/(1-x) ==> simmilarly as the above one we get:

ln(1 - 2x/(1-x)) = 2x/(1-x) - [2x/(1-x)]^2/2 + [2x/(1-x)]^3/3 -....

Tonio

**Krizalid** · October 5th 2009, 03:12 PM

$\ln (1+x)-\ln (1-x)=\ln (1+x)-\ln \big(1+(-x)\big).$

Thread: Power Series

LinkBack

Thread Tools

Display

Power Series

Similar Math Help Forum Discussions

Exploiting geometric series with power series to find Taylors series

question about power series, taylor series, maclaurin series

Power Series&Function Series Questions

Formal power series & Taylor series

Finding Power Series and Evaluating as a Power Series

Search Tags