Question

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

Answer 1

\[f(x)=\ln(1-x)\]

Answer 2

@Vocaloid Any idea?

Answer 3

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
\[\ln(1-0)=-\sum_{n=0}^\infty \frac{0^n}{n+1}+C~\implies~C=0\]

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

??

Answer 8

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.

Answer 9

THANK YOU!!!

Answer 10

yw

Answer 11

hold on though.

Answer 12

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.

Answer 13

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

Answer 14

That's it?

Answer 15

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.

Answer 16

Thank you dude your freaking awesome!!!

Answer 17

yw, here to help :)