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In the Wikipedia we can find two formulas using power series to $\ln(x)$,

but I would like a formula that holds on the interval $x \geq 1$ (or is possible to calculate $\ln(x)$ to $x \geq 1$ with the same formula).

Is there one?

In other words, a formula like,

$\ln(x)=\sum_{i=0}^{\infty}a_i x^i$

The wikipedia formulas are:

$\ln(x+1)=\sum_{i=1}^{\infty}\frac{(-1)^{i-1}x^i}{i}\text{ for }1 < x \leq1$

and

$\ln\left(\frac{x}{x-1}\right)=\sum_{i=1}^{\infty}\frac{1}{i x^i}$

PS.: Is important to me to use power series, but, any help are welcome.

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    @AlexBecker Sush! Shut the non-believer!2012-06-14

3 Answers 3

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Another alternative, that works for $x > -\dfrac 1 2 $ is

$\log (1+x) =\sum_{n>0} \frac 1 n\left( \frac{x}{x+1}\right)^n$

This is trivially obtained by setting $u=\dfrac{x}{x+1}$ in

$-\log(1-u)=\sum_{n >0}\frac {u^n} n $

or by showing

$\int_0^x \left( \frac{t}{t+1}\right)^{n+1}\frac{dt}t=\log(1+x)-\sum_{k \leq n} \frac 1 k\left( \frac{x}{x+1}\right)^k$

Then since

$\left|\log(1+x)-\sum_{k \leq n} \frac 1 k\left( \frac{x}{x+1}\right)^k\right|=\left| \int_0^x \left( \frac{t}{t+1}\right)^{n+1}\frac{dt}t \right|<\epsilon$

for sufficiently large $n$ and $x > -\dfrac 1 2 $ the equality follows.

5

To add to what was written by Gerry Myerson, if $|w|<1$, we have by the usual power series expansion for $\log(1+w)$, $\log(1+w)=w-\frac{w^2}{2}+\frac{w^3}{3}-\frac{w^4}{4}+\cdots.\tag{$1$}$ Replacing $w$ by $-w$, we have $\log(1-w)=-w-\frac{w^2}{2}-\frac{w^3}{3}-\frac{w^4}{4}-\cdots.\tag{$2$}$ Subtracting $(2)$ from $(1)$, we obtain $\log(1+w)-\log(1-w)=\log\left(\frac{1+w}{1-w}\right)=2w+2\frac{w^3}{3}+2\frac{w^5}{5}+\cdots.\tag{$3$}$ For any $x>1$, there is a $w$ between $0$ and $1$ such that $x=\frac{1+w}{1-w}$. We can solve for $w$ explicitly in terms of $x$, obtaining $w=\frac{x-1}{x+1}$.

Thus we get an expansion for $\log x$ in terms of powers of $\frac{x-1}{x+1}$. This is not a power series expansion of $\ln x$, but it can be useful. For large $x$, the number $w$ is close to $1$, so the convergence is relatively slow. But for smallish $x$, there is usefully fast convergence. Euler, among others, used the series $(3)$ for computations.

Take for example $x=2$. The ordinary series for $\log(1+x)$ converges when $x=1$, but too slowly for practical use. However, in this case $w$ turns out to be $1/3$, and the series $(3)$ converges quite fast. For $x=3$, we get $w=1/2$, and again we get usefully fast convergence.

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At http://www.efunda.com/math/exp_log/series_exp.cfm you will find $\log x=2\sum_1^{\infty}{1\over2n-1}\left({x-1\over x+1}\right)^{2n-1}$ claimed to be valid for $x\gt0$, and several other formulas you might enjoy.

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    I like the similar form for all $x\in\mathbb{R}$: $ \log|x|=\sum_{n=0}^\infty\frac{1}{2n+1}\left(\frac{x^2-1}{x^2+1}\right)^{2n+1} $2012-04-12