Jtdylktuy

Let $f(x)=lnfrac{1+x}{1-x}$ for $x$ in $(-1,1)$. Calculate the Taylor series of $f$ at $x_0=0$

I determined some derivatives:

$f'(x)=frac{2}{1-x^2}$; $f''(x)=frac{4x}{(1-x^2)^2}$; $f^{(3)}(x)=frac{4(3x^2+1)}{(1-x^2)^3}$; $f^{(4)}(x)=frac{48x(x^2+1)}{(1-x^2)^4}$; $f^{(5)}(x)=frac{48(5x^2+10x^2+1)}{(1-x^2)^5}$

and their values at $x_0=0$:

$f(0)=0$; $f'(0)=2$; $f''(0)=0$; $f^{(3)}(0)=4=2^2$

$f^{(4)}(0)=0$; $f^{(5)}(0)=48=2^4.3$; $f^{(7)}(0)=1440=2^5.3^2.5$

I can just see that for $n$ even, $f^{(n)}(0)=0$, but how can I generalize the entire series?

begin{align}
ln frac{1+x}{1-x} &= ln (1+x) - ln (1-x) \
&= sum_{n=0}^infty frac{(-1)^n}{n+1}x^{n+1} - sum_{n=0}^infty frac{(-1)^n}{n+1}(-x)^{n+1}\
&= sum_{n=0}^infty frac{(-1)^n}{n+1}x^{n+1} + sum_{n=0}^infty frac{1}{n+1}x^{n+1}\
&= sum_{n=0}^infty frac{(-1)^n+1}{n+1}x^{n+1}\
&=sum_{n=1}^infty frac{(-1)^{n-1}+1}{n}x^n\
&=sum_{n=1}^infty frac{2}{2n-1}x^{2n-1}\
end{align}

Remark: Your observation that all the even terms vanishes is due to this is an odd function.

The derivative is
$$biggl(lnfrac{1+x}{1-x}biggr)'=frac{1-x}{1+x},frac 2{(1-x)^2}=frac 2{1-x^2}$$
Now
$$frac 2{1-x^2}=2sum_{n=0}^infty x^{2n},enspacetext{so}quadlnfrac{1+x}{1-x}=2sum_{n=0}^infty frac{x^{2n+1}}{2n+1}, $$
taking into account that both sides are $0$ for $x=0$.

Beginning with

$$frac{1}{1 - x} = sum_{k = 0}^{infty} x^{k}, $$

we can integrate to get $$log(1 - x) = -sum_{k = 1}^{infty} frac{x^{k}}{k}.$$

Also, if we plug in $(-x)$ for $x$ into the equation for $frac{1}{1 - x}$, we get

$$frac{1}{1 + x} = sum_{k = 0}^{infty}(-x)^{k}.$$

Integrating, we get

$$log(1 + x) = sum_{k = 1}^{infty} frac{(-x)^{k}}{k}.$$

Now, by properties of $log$, $frac{log(1 + x)}{log(1 - x)} = log(1 + x) - log(1 - x)$. So,

$$frac{log(1 + x)}{log(1 - x)} = sum_{k = 1}^{infty} frac{(-x)^{k}}{k} + sum_{k = 1}^{infty} frac{x^{k}}{k}$$

$$= sum_{k = 1}^{infty} frac{(-x)^{k} + x^{k}}{k}.$$

Note that when $k$ is odd, the terms vanish.