What is the Taylor series for $\ln(1-x^2)$ around x=0 point?

Question

As you can see from the title I need Taylor series for $$\ln(1-x^2)$$. I am not sure which Taylor series to use: $$\ln(1-x)=-\sum \limits_{n=0}^{\infty}\frac{x^n}{n}$$ or $$\ln(x)=\sum \limits_{n=0}^{\infty}\frac{x^n}{n}$$ Also I am not sure if this has any effect on question or answer but I am also given that I should setup my series around x=0 point. I would really appreciate any help as it is my exam few days later.

Answer 1

$$\ln(1-x)=-\sum \limits_{n=0}^{\infty}\frac{x^n}{n}$$ let $x\rightarrow x^2$ $$\ln(1-x^2)=-\sum \limits_{n=0}^{\infty}\frac{x^{2n}}{n}$$

Answer 2

The first series converges for $|x|<1$ so changing $x$ to $x^{2}$ in it gives you the answer: $$\ln (1-x^{2})=-\sum\limits_{n=1}^{\infty} \dfrac {x^{2n}} n.$$

Answer 3

Consider the derivatives of $\log(1+x)$, evaluated at $x=0$.

$$\log(1+x)\to0$$

$$(\log(1+x))'=(1+x)^{-1}\to1$$ $$\left((1+x)^{-1}\right)'=-(1+x)^{-2}\to-1$$ $$\left(-(1+x)^{-2}\right)'=2(1+x)^{-3}\to2$$ $$\cdots$$ $$(-1)^nn!(1+x)^{-n-1}\to(-1)^nn!$$

From this we get the Taylor development

$$\log(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots$$

Now we can either directly substitute $x$ for $-x^2$,

$$\log(1-x^2)=-x^2-\frac{x^4}2-\frac{x^6}3-\frac{x^8}4-\cdots$$

or add and simplify

$$\log(1+x)+\log(1-x)=x-x-\frac{x^2}2-\frac{x^2}2+\frac{x^3}3-\frac{x^3}3-\frac{x^4}4-\frac{x^4}4+\cdots \\=-x^2-\frac{x^4}2-\cdots$$