• Ac3

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

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  • Ac3

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

Mathematics
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  • Ac3
\[f(x)=\ln(1-x)\]
  • Ac3
@Vocaloid Any idea?
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 \[\ln(1-0)=-\sum_{n=0}^\infty \frac{0^n}{n+1}+C~\implies~C=0\]

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  • Ac3
Alright so we have \[\sum_{0}^{\infty} \frac{ x ^{n+1} }{ n+1 }\]
  • Ac3
using that do we now we use that to find the power series of xln(1-x)
  • Ac3
@SithsAndGiggles Would I just multiply the whole thing by x getting. \[\sum_{n=0}^{\infty} \frac{ x ^{n+2} }{ n+1 }\]
  • Ac3
??
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.
  • Ac3
THANK YOU!!!
yw
  • Ac3
hold on though.
  • 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.
Okay, so if \(x=\dfrac{1}{2}\), then \[\ln\left(1-\frac{1}{2}\right)=\ln\frac{1}{2}=\ln2^{-1}=-\ln2\]
  • Ac3
That's it?
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.
  • Ac3
Thank you dude your freaking awesome!!!
yw, here to help :)

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