1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Expressing series in terms of a Power Series

  1. Jul 18, 2017 #1
    Hello and thank you for trying to help.

    In spite of the fact that this seems a very simple problem, I do not find myself able to get a solution. Here it goes:

    Let $$f(x)=\displaystyle \sum_{k=3}^\infty a_k \frac{x^k}{k(k-1)(k-2)}$$ and $$g(x)=\displaystyle \sum_{k=0}^\infty a_k x^k$$. Express f(x) in terms of g(x).

    My attempt has been to express f(x) as
    $$\displaystyle\sum_{k=3}^{\infty}a_k\frac{x^k}{k(k-1)(k-2)}=\frac{1}{2}\sum_{k=3}^{\infty}\frac{a_k x^k}{k}-\sum_{k=3}^{\infty}\frac{a_k x^k}{k-1}+\frac{1}{2}\sum_{k=3}^{\infty}\frac{a_k x^k}{k-2}$$

    And then

    $$f(x)=\frac{1}{2}\sum_{k=3}^{\infty}\frac{a_k x^k}{k}-x\sum_{k=2}^{\infty}\frac{a_k x^k}{k}+\frac{x^2}{2}\sum_{k=1}^{\infty}\frac{a_k x^k}{k}$$

    What should I do next? I suppose that I have to integrate those series, but I have already tried that and I have not been able to get any relationship with g(x).

    Any help will be appreciated.
     
    Last edited: Jul 18, 2017
  2. jcsd
  3. Jul 18, 2017 #2

    scottdave

    User Avatar
    Homework Helper
    Gold Member

    Your LaTex is not displaying correctly. I have tried 2 different browsers. It looks like you have f(x) = Sum of (k=3 to infinity) (ak xk) / (k(k-1)(k-2)),
    then g(x) = sum of (k=0 to infinity) (ak xk), then I get confused trying to follow your workings, after that.
     
  4. Jul 18, 2017 #3
    Corrected, thank you
     
  5. Jul 18, 2017 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Using summation index ##j = k-2## your sum becomes
    $$f(x) = \sum_{j=1}^{\infty} a_{j+2} \frac{x^{j+2}}{j(j+1)(j+2)}$$
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Expressing series in terms of a Power Series
Loading...