A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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?

  • This Question is Closed
  1. Ac3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. Ac3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Vocaloid Any idea?

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  5. Ac3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  6. Ac3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  7. Ac3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ??

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    THANK YOU!!!

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yw

  11. Ac3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hold on though.

  12. Ac3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  14. Ac3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    That's it?

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you dude your freaking awesome!!!

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yw, here to help :)

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.