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?


2 Answers 2


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 $$\log(y)=\sum_{i=0}^\infty \frac{1}{2i + 1} (\frac{y-1}{y+1})^{2i+1}$$

  • $\begingroup$ I question; why was this necessary? Should this not be the direct output of the taylor algorithm? Is there a specific method to finding taylor expansions of functions that are not expanded so easily as cos or sin (like ln)? Also; what is the intuition behind this method? I understand that substitution is in place, but is it because 1-x and 1+x allow Maclaurin series? $\endgroup$ Apr 18, 2014 at 5:01
  • $\begingroup$ I am not sure to answer your question in the comment. A Taylor expansion gives you a polynomial in the rhs. Then, specific changes of variable allow this kind of manipulation. $\endgroup$ Apr 18, 2014 at 5:05
  • $\begingroup$ @BourgondAries This approach is necessary due to the singularity of $\log(x)$ at $x=0$. For well-behaved functions like $\sin, \cos, \exp$, it is not necessary. $\endgroup$
    – AlexR
    Apr 18, 2014 at 5:05
  • $\begingroup$ $f(x) = \ln x$ is actually easily expanded: its derivatives ares $f^{(n)}(x) = (-1)^{n-1} (n-1)! x^{-n}$, so around any point $a$ its series is $$\ln x = \ln a + \sum_{k=1}^{\infty} (-1)^{k-1} \frac{(x-a)^{n}}{a^n n} $$ Of course, this could just has been easily computed by plugging in $x \mapsto x/a$ into the series for $\ln x$ centered at $1$. $\endgroup$
    – user14972
    Apr 18, 2014 at 5:39
  • $\begingroup$ @AlexR: Does this imply there a theorem stating that in order to approximate any function, it must be expanded from its origin? $\endgroup$ Apr 18, 2014 at 6:04

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?)

  • $\begingroup$ In the sentence mentioning the Binomial/Euler transform, is the log part of the formula supposed to be part of the equation? $\endgroup$ Apr 18, 2014 at 3:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.