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!

Homework Help: Finding a Power Series for a function

  1. Mar 18, 2014 #1
    1. The problem statement, all variables and given/known data

    Find a power series representation for the function [itex] f(x) = \frac{(x-1)}{(3-x)^2}^2[/itex], valid for every [itex]x[/itex] with [itex]|x|<3[/itex]

    2. Relevant equations

    The equation that I think would be useful is [itex]\frac{1}{1-x} = \sum_{n=0}^\infty x^n[/itex]

    3. The attempt at a solution

    I began by just looking at the term [itex]\frac{1}{(3-x)^2}[/itex]

    [itex]\frac{1}{1-x} = \sum_{n=0}^\infty x^n[/itex]

    differentiating both sides:

    [itex]\frac{1}{(1-x)^2}= \sum_{n=1}^\infty nx^{(n-1)}[/itex]

    now I'm stuck here, my train of thought is that if I can turn [itex]\frac{1}{(3-x)^2}[/itex] into someform of [itex]\frac{1}{(1-x)^2}[/itex] to obtain its summation formula, I can then multiply the polynomial [itex]{(x-1)}^2[/itex] to that summation and go from there.

    Usually the assignment questions involve functions in the likes of [itex]\frac{x}{(1-x^2)}[/itex] or [itex]\frac{x^2}{(1+x)^3}[/itex], which I can solve, but the [itex] 3 [/itex] in front of the x in [itex]\frac{1}{(3-x)^2}[/itex] is really troubling me, any help? Thanks!
  2. jcsd
  3. Mar 18, 2014 #2
    Hi aerospacev!

    You can factor out the 3 from the denominator and write it as following:
    Can you proceed now?
  4. Mar 18, 2014 #3
    I wondered about that, maybe I need to review my basic arithmetic, am I allowed to simply divide each number individually in [itex]{(3-x)^2}[/itex] by 3 and multiply by [itex]3^2[/itex]? Even though the entire term is squared?
  5. Mar 18, 2014 #4
    I am not sure what you mean there. What I did is the following:
    I hope that helps.
    Last edited: Mar 18, 2014
  6. Mar 18, 2014 #5
    Hi Panav-Arora.

    I think you want to correct those last two versions of the right hand sides, to have an exponent of 2.

  7. Mar 18, 2014 #6
    Woops, thanks Chet! :D
  8. Mar 18, 2014 #7
    Oh ok thanks, I think there was just some confusion on my part.

    I got the correct answer, thanks again!
  9. Mar 18, 2014 #8
    How about
    ##f(x) = (\frac{x-1}{3-x})^2 = (\frac{x}{3-x}-\frac{1}{3-x})^2##
    ## = (\frac{-1}{1-(\frac{3}{x})}-\frac{(\frac{1}{3})}{1-(\frac{x}{3})})^2 ##
    ##so, f(x) = [-\sum\limits_{n=0}^∞ (\frac{3}{x})^n - \frac{1}{3}\sum\limits_{n=0}^∞ (\frac{x}{3})^n]^2 ## where ##-3 < x < 3##
    Last edited: Mar 18, 2014
  10. Mar 18, 2014 #9
    I don't think so, wouldn't you then have to divide by zero?
  11. Mar 18, 2014 #10


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Also, even if the calculation is correct,
    $$[-\sum\limits_{n=0}^∞ (\frac{3}{x})^n - \frac{1}{3}\sum\limits_{n=0}^∞ (\frac{x}{3})^n]^2$$
    is not a power series as written. It is a expression involving power series, but that's not the same thing.
  12. Mar 18, 2014 #11
    Divide what by zero? The index of summation n = 0 means that we would first have a zero as an exponent. This series isn't convergent, an error on my part, but there is nothing being divided by zero in this series; x ≠ 0.
  13. Mar 18, 2014 #12


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    But the original problem statement says "valid for every ##x## with ##|x|< 3##", so that includes ##x=0##.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted