How to I find the Taylor series of $\ln {\frac{|1-x|}{1+x^2}}$?

Question

The expression is $$\ln {\frac{|1-x|}{1+x^2}}$$

I'm told there's an easy way to do it to get the first 2 non-zero term but I ended up differentiating this answer several times and got a very long answer that is not correct.

What I did in specific:

1. split up the ln expressions.

2. Differentiate once, so I get 1/|1-x| and 2x/(1+x^2).

3. Differentiate another time, and get an even longer expression.

   What is the easy way to do this that I do not see?

Answer 1

Note that $f(x) =\ln|1-x|-\ln(1+x^2)$. but we know that $$ (\ln|1-x| )' =-\frac{1}{1-x} = -\sum_{n=1}^{\infty} x^n ~~~~\mbox{only for}~~0\le |x|<1\\\implies \ln|1-x| = -\sum_{n=1}^{\infty}\frac{x^n}{n} ~~~~\mbox{only for}~~0\le |x|<1$$ also from since $-\ln(1-h) = \sum_{n=1}^{\infty}\frac{h^n}{n} ~~~~\mbox{only for}~~0\le h<1$ we have $$\ln(1+x^2) = -\sum_{n=1}^{\infty}(-1)^n\frac{x^{2n}}{n} ~~~~\mbox{only for}~~0\le |x|<1$$ hence

$$f(x) = \sum_{n=1}^{\infty}\frac{ (-1)^{n}x^{2n}-x^n }{n} ~~~~\mbox{only for}~~0\le |x|<1$$

Answer 2

So you have $$f(x)=\ln {\frac{|1-x|}{1+x^2}},$$ that means $$f'(x)=-\frac1{1-x}-\frac{2x}{1+x^2}=(-1-x-x^2-\ldots)-(2x-2x^3\pm\ldots)=-1-3x-x^2+\ldots$$ Integration gives $$f(x)=-x-\frac32x^2-\frac13x^3+\ldots,$$ since $f(0)=0$.

Answer 3

If you mean the Taylor series at $x=0$, then you can rewrite your function as $$ \ln(1-x)-\ln(1+x^2)= -\sum_{n\ge1}\frac{x^n}{n}-\sum_{n\ge1}\frac{(-1)^nx^{2n}}{n} =\sum_{n\ge1}a_nx^n $$ where $$ a_n=\begin{cases} -\dfrac{1}{(n-1)/2} & \text{$n$ odd} \\[8px] -\dfrac{1}{n}-\dfrac{(-1)^{n/2}}{n/2} & \text{$n$ even} \end{cases} $$