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

Question

Ezra Adams · Accepted Answer

#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.

Ezekiel Alexander · Answer

undefined To find a power series representation for $ \ln(1-x^2) $, you can use the known Maclaurin series expansion for $ \ln(1+x) $.

1. Maclaurin series for $ \ln(1+x) $ is:
$$ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots $$
for $ |x| < 1 $.

2. Replace $ x $ with $ -x^2 $:
$$ \ln(1-x^2) = -x^2 - \frac{(-x^2)^2}{2} + \frac{(-x^2)^3}{3} - \frac{(-x^2)^4}{4} + \dots $$

3. Simplify each term:
$$ \ln(1-x^2) = -x^2 - \frac{x^4}{2} - \frac{x^6}{3} - \frac{x^8}{4} + \dots $$

This is the power series representation for $ \ln(1-x^2) $.

The radius of convergence $ R $ for this series can be found using the ratio test:

$$ R = \lim_{n \to \infty} \left| \frac{a_{n}}{a_{n+1}} \right| $$

For this series, the general term $ a_n $ is $ \frac{(-1)^n x^{2n}}{n} $.

$$ \lim_{n \to \infty} \left| \frac{\frac{(-1)^n x^{2n}}{n}}{\frac{(-1)^{n+1} x^{2(n+1)}}{n+1}} \right| $$

$$ = \lim_{n \to \infty} \left| \frac{n+1}{n} x^2 \right| $$

$$ = |x^2| $$

Since this limit should be less than 1 for convergence:
$$ |x^2| < 1 $$

Thus, the radius of convergence $ R $ is 1.