I was recently Taylor-expanding ln around $(1,0)$. I noticed that this polynomial will have a range of input that converges between $0$ and $2$ regardless of Taylor order. I then found an expansion that did not seem to have this issue, namely:
$\lim_{n \to +\infty} 2 \cdot \sum\limits_{i=0}^n (\frac{1}{2i + 1} \cdot {(\frac{x-1}{x+1})}^{2i + 1}) = \ln(x)$
My question is: how is this formula derived (as in created, not the derivative) from ln?
On
This is outlined on the wikipedia page for natural logs.
You start with the taylor series about $x=0$ for $\log(1+x) = \sum_n \frac{(-1)^{n+1}}{n} x^n$ for $|x|<1$.
Then, apply a Binomial/Euler transform to get a series for $\log \frac{x}{x-1} = \sum_n \frac{1}{n} x^{-n}$.
Then, substitute $x = \frac{u}{u-1}$ to get the desired series after a shift of indices. note that it doesn't converge for all $x$ (exercise: what $x$ does it converge for?)
On
Replace $x$ by $\frac{1-x}{1+x}$ in
$$\operatorname{arctanh}(x)=-\frac12\ln\left(\frac{1-x}{1+x}\right)=\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}$$
we get
$$-\frac12\ln(x)=\sum_{n=0}^\infty\frac{\left(\frac{1-x}{1+x}\right)^{2n+1}}{2n+1}=-\sum_{n=0}^\infty\frac{\left(\frac{x-1}{1+x}\right)^{2n+1}}{2n+1}$$
or
$$\ln(x)=2\sum_{n=0}^\infty\frac{\left(\frac{x-1}{1+x}\right)^{2n+1}}{2n+1}$$
If you combine the Taylor series for $\log(1+x)$ and $\log(1-x)$ you arrive to a well known series which is $$\log \frac{1+x}{1-x}=2 \sum_{i=0}^\infty \frac{x^{2i+1}}{2i + 1} $$ So now define $y=\frac {1+x}{1-x}$ that is to say $x=\frac{y-1}{y+1}$ and you end with $$\log(y)=\sum_{i=0}^\infty \frac{1}{2i + 1} (\frac{y-1}{y+1})^{2i+1}$$