## Ac3 one year ago use equation 1 to find a power series representation for f(x)=ln(1-x). What is the radius of convergence?

1. Ac3

$f(x)=\ln(1-x)$

2. Ac3

@Vocaloid Any idea?

3. anonymous

Not sure what you mean by equation (1), but I'll take a wild guess and suppose it's $\frac{1}{1-x}=\sum_{n=0}^\infty x^n\quad\text{for }|x|<1$ Notice that $\frac{d}{dx}\ln(1-x)=-\frac{1}{1-x}=-\sum_{n=0}^\infty x^n$ Integrating, you have $\ln(1-x)=-\sum_{n=0}^\infty \frac{x^{n+1}}{n+1}+C$ If we consider $$x=0$$, you would find that $\ln(1-0)=-\sum_{n=0}^\infty \frac{0^n}{n+1}+C~\implies~C=0$

4. Ac3

Alright so we have $\sum_{0}^{\infty} \frac{ x ^{n+1} }{ n+1 }$

5. Ac3

using that do we now we use that to find the power series of xln(1-x)

6. Ac3

@SithsAndGiggles Would I just multiply the whole thing by x getting. $\sum_{n=0}^{\infty} \frac{ x ^{n+2} }{ n+1 }$

7. Ac3

??

8. anonymous

Right, if the power series for $$f(x)$$ is given by $$S$$, then the power series for $$x f(x)$$ is $$xS$$. You're missing the minus sign, btw.

9. Ac3

THANK YOU!!!

10. anonymous

yw

11. Ac3

hold on though.

12. Ac3

we're on the last part of the entire question so now by putting x=1/2 in your result from part a (that's our first one), express ln2 as te sum of an infintie series.

13. anonymous

Okay, so if $$x=\dfrac{1}{2}$$, then $\ln\left(1-\frac{1}{2}\right)=\ln\frac{1}{2}=\ln2^{-1}=-\ln2$

14. Ac3

That's it?

15. anonymous

Yep! The series representation is simple enough, you're just replacing $$x$$ with $$\dfrac{1}{2}$$. $-\ln2=-\sum_{n=0}^\infty \frac{\left(\frac{1}{2}\right)^{n+1}}{n+1}$which you can rewrite in several ways.

16. Ac3

Thank you dude your freaking awesome!!!

17. anonymous

yw, here to help :)