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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?

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2 Answers 2

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

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  • $\begingroup$ 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
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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?)

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  • $\begingroup$ 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

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