Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My Problem is to expand $f(x)=\dfrac{\ln(1-x)}{1+x}$ into a power series.

My Approach: from looking onto the Graphs of this function, i know, for rising x the y is falling towards Zero, without reaching it. but i don't think there is convergence for the series.

A power series has the scheme: $\sum\limits_{n=0}^{\infty} a_{n}\cdot x^{n}$ but im stuck in trying to convert $f(x)$ into a suitable sequence $a_{n}$.

Any hints?

share|improve this question
1  
We know power series expansion for $ ln(1-x)$ and for $1/(1+x)$ too, just multiply those two power series. –  Girish May 11 '13 at 10:07

1 Answer 1

up vote 2 down vote accepted

You have two series expansions, both for $|x|<1$, namely:

1) $\log(1+x)=\displaystyle\sum_{n=1}^{+\infty}(-1)^{n+1}\frac{x^n}{n}$

2) $\frac{1}{1-x}=\displaystyle\sum_{n=0}^{+\infty}x^n$

Substituting $x$ with $-x$ doesn't change the convergence radius, hence you get

1') $\log(1-x)=-\displaystyle\sum_{n=1}^{+\infty}\frac{x^n}{n}$

2') $\frac{1}{1+x}=\displaystyle\sum_{n=0}^{+\infty}(-1)^nx^n$

Now multiply the two series together

$$\frac{\log(1-x)}{1+x}=(1-x+x^2-x^3+\ldots)\cdot(-x-\frac{x^2}{2}-\frac{x^3}{3}-\ldots)$$ $$=-x+\frac{x^2}{2}-\frac{5x^3}{6}+\frac{7x^4}{12}-\frac{47x^5}{60}+ O(x^6)$$

share|improve this answer
    
how did you multiply that? –  Toralf Westström May 11 '13 at 11:34
1  
@ToralfWestström Same way you multiply polynomials, considering a series as a polynomyal with "infinite degree"" –  Federica Maggioni May 11 '13 at 11:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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