Asymptotic Expansion in zero of $\frac{1}{\ln(1+x)}$

Question

On wolfram the expansion is:

$$\frac {1}{x} + \dfrac{1}{2} ...\,.$$

But I don't understand from where it outside comes the $\frac{1}{2}$

thanks

Check what you typed on Worlfram: $$ \ln\frac{1-x}{1+x} = \ln\left((1-x)(1-x+o(x))\right)=\ln(1-2x+o(x))\sim -2x$$ — Clement C., Jan 22 '14 at 20:38
Yes this is the expansion, I've calculated with Mathematica. — Vitasalato, Jan 22 '14 at 20:39
http://www.wolframalpha.com/input/?i=Series%5B+Log%5B%281-x%29%2F%281%2Bx%29%5D+%2C+%7Bx%2C0%2C3%7D%5D — Clement C., Jan 22 '14 at 20:41
I believe I saw the function $\frac{1}{\ln(1+x)}$ first when you posted the question. — Mhenni Benghorbal, Jan 22 '14 at 20:43

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

2

Note that, the function has a pole of order $1$ with residue equals $1$. Now, just write the function as

$$ \frac{1}{\ln(1+x)}=\frac{1}{x}+a_1+a_2 x+\dots \implies a_1= \lim_{x\to 0}\left(\frac{1}{\ln(1+x)}-\frac{1}{x} \right). $$

edited Apr 13 '17 at 12:21

Community

answered Jan 22 '14 at 20:52

Mhenni Benghorbal

1 Answers1