How do you find a power series representation for $ln(1-x^2)$ and what is the radius of convergence?

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Answer 1 · 2017-05-06T05:06:44

$ln(1-x^2)=sum_(n=0)^oo frac{-(x)^(2n+2)}{n+1}$

The radius of convergence is equal to $1$ by the ratio test.

Remember the MacLaurin series representation for $ln(1+u)$ :
$color(blue)(ln(1+u)=sum_(n=0)^(oo) frac{(-1)^n(u)^(n+1)}{n+1})$

Substitute $u=-x^2$ :

$ln(1-x^2)=sum_(n=0)^(oo) frac{(-1)^n(-x^2)^(n+1)}{n+1}$

$color(white)(ln(1-x^2))= sum_(n=0)^oo frac{(-1)^n(-1)^(n+1)(x)^(2n+2)}{n+1}$

$color(white)(ln(1-x^2))=sum_(n=0)^oo frac{-(x)^(2n+2)}{n+1}$

To find radius of convergence, use ratio test, which states that if $lim_(x->oo) |frac{a_(n+1)}{a_n}|<1$ , then $sum_(n=0)^oo a_n$ converges

$lim_(n->oo)|frac{color(red)(-)(x)^(2n+4)}{n+2}*frac{n+1}{color(red)(-)(x)^(2n+2)}|$

$=lim_(x->oo) |frac{color(red)((x)^(2n))(x)^4(n+1)}{(n+2)color(red)((x)^(2n))(x)^2}|$

$=|x^2|$

In order for the series to converge, set this less than 1:
$x^2<1, color(red)(x^2> -1)$

The radius of convergence is equal to $1$ by the ratio test.

How do you find a power series representation for ln(1-x^2) ln(1-x^2) and what is the radius of convergence?