Question

ln(1-x^2) power series

Answer 1

\[\Large\rm \ln(1-x^2)=\ln\left[(1-x)(1+x)\right]=\ln(1-x)+\ln(1+x)\]From there we can try to do something fancy to get a power series, hmm let's see..

Answer 2

Since,\[\Large\rm \frac{d}{dx}\ln(1+x)=\frac{1}{1+x}\]This implies,\[\Large\rm \ln(1+x)=\int\limits \frac{1}{1+x}dx\]If we make restrictions on our x, we can write this as a geometric series, yes?\[\Large\rm =\int\limits \sum_{n=0}^{\infty} (-x)^n dx,\qquad \qquad |x|<1\]

Answer 3

We can pass the integral into the sum,\[\Large\rm =\sum_{n=0}^{\infty} (-1)^n\int\limits x^n\;dx\]And then apply power rule to integrate our x.
The other term ln(1-x) should work pretty similarly I think....

Is this making any sense? :o