Ac3
  • Ac3
use equation 1 to find a power series representation for f(x)=ln(1-x). What is the radius of convergence?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Ac3
  • Ac3
\[f(x)=\ln(1-x)\]
Ac3
  • Ac3
@Vocaloid Any idea?
anonymous
  • 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 \[\ln(1-0)=-\sum_{n=0}^\infty \frac{0^n}{n+1}+C~\implies~C=0\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.