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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:
    $$f(x)=\frac{(x-1)^2}{9}\frac{1}{(1-x/3)^2}$$
    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:
    $$(3-x)^2=\left(3\left(1-\frac{x}{3}\right)\right)^2=3^2\left(1-\frac{x}{3}\right)^2=9\left(1-\frac{x}{3}\right)^2$$
    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.

    Chet
     
  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

    jbunniii

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    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

    jbunniii

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    But the original problem statement says "valid for every ##x## with ##|x|< 3##", so that includes ##x=0##.
     
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