Taylor series of $\ln\frac{1+x}{1-x}$

Question

Let $f(x)=\ln\frac{1+x}{1-x}$ 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?

Answer 1

\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.

Answer 2

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.

Answer 3

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