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Expanding Power Series

  1. Nov 30, 2011 #1
    1. The problem statement, all variables and given/known data
    Expand f(x)= (x+x2)/(1-x)3

    2. Relevant equations

    3. The attempt at a solution
    I've tried everything I can think of to simplify this equation: substitution of various other power series, partial fraction decomposition, taking derivatives, multiplying out the denominator. It's driving me nuts. Thanks for any steps in a new direction.
  2. jcsd
  3. Nov 30, 2011 #2
    OK, let's do this in steps. First, can you find the power series expansion of [itex]\frac{1}{(1-x)^3}[/itex]??

    If you can't, try to use that


    and differentiate both sides.
  4. Nov 30, 2011 #3
    I tried to find a power expansion for that term and found it as 2(1-x)^3, which would equal the power series (n^(2)-n)x^(n-2) from n=2. From here, I'm not sure if there is an equation for the numerator, or if there is a way to multiply the numerator into this equation in a way that makes sense.
  5. Nov 30, 2011 #4
    OK, and now you simply can do


    So exchange both terms of [itex]\frac{1}{(1-x)^3}[/itex] by its power series and multiply and add everything.
  6. Nov 30, 2011 #5
    Oh wow. Thanks for your help, I guess my brain is rebelling against obvious steps. :blushing:

    Ok, so now I have the sum of (n^2-n)x^(n-1) + sum of (n^2-n)x^n, both starting from n=2, and both expressions multiplied by 1/2.
    Last edited: Nov 30, 2011
  7. Nov 30, 2011 #6
    Ok, so I have the sums 1/2(n^(2)-n)x^(n-1) + 1/2(n^(2)-n)x^(n-1) from n=2. The next part of the question says I should relate (n^2)/(2^n) to the previous equation. Is there a simplification I'm missing?
  8. Dec 1, 2011 #7
    Well, you will have to put x=2. Try to rewrite the equation a bit to see if you get anything nice.
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