Calculus - Interval of Convergence and Power Series

**dark_knight_307** · May 3rd 2008, 09:42 AM

I was having a bit of trouble with this:

Use the power series ${1\over{1+x}}=\sum_{n=0}^{\infty}{(-1)^nx^n}$ to determine a power series, centered at 0, for the function:
$f(x) = ln(1-x^2) = \int{{1\over{1+x}}dx} - \int{{1\over{1-x}}dx}$

**Moo** · May 3rd 2008, 09:51 AM

Hello,

Originally Posted by dark_knight_307

I was having a bit of trouble with these:

1. Use the power series ${1\over{1+x}}=\sum_{n=0}^{\infty}{(-1)^nx^n}$ to determine a power series, centered at 0, for the function:
$f(x) = ln(1-x^2) = \int{{1\over{1+x}}dx} - \int{{1\over{1-x}}dx}$

Now continue this way

$\begin{aligned} f(x) & =\int \frac{1}{1+x} dx-\int \frac{1}{1+(-x)} dx \\ & =\int \sum_{n=0}^\infty (-1)^n x^n \ dx - \int \sum_{n=0}^\infty (-1)^n (-x)^n \ dx \\ & =\sum \int (-1)^n x^n dx - \sum \int x^n dx \\ & = \sum (-1)^n \frac{x^{n+1}}{n+1} - \sum \frac{x^{n+1}}{n+1} \\ & = \dots \end{aligned}$

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