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Help with manipulating power series.

  1. Aug 16, 2015 #1
    So i am given (1+x)/(1-x)^2 and I have to put it into a power series. I know that 1/(1-x)= 1+x+x^2+x^3+...=∑x^n from 0 to infinity. I am having problems factoring series.

    I differentiate 1/(1-x).

    I get, 1/(1-x)^2= 1+2x+3x^2+...= ∑nX^(n-1) the sum from 1 to infinity.

    rewriting this equation.

    1/(1-x)^2=∑(n+1)X^n. The sum from 0 to infinity.

    multiply both sides by (1+x)

    I get (x+1)/(1-x)^2 = (1+x)∑(n+1)X^(n) , from 0 to infinity.

    Then I distribute 1+x.

    ∑(n+1)X^(n) + ∑(n+1)X^(n+1) both sums have an index of 0.

    My problem is that I have having trouble factoring using summation notation and I am not sure how my book factored these 2 sums into 1 to get the answer

    ∑(2n+1)X^n with the index of 0.



    When I factored my previous work I got. ∑(n+1)(X^(n)[1+X]

    which I cannot seem to to put into the form the book has it.
     
  2. jcsd
  3. Aug 16, 2015 #2

    FactChecker

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    Pick an exponent n and look at the terms with xn. From the first and second summation, respectively, they are (n+1)xn and nxn. Sum them to get the book answer.
     
  4. Aug 16, 2015 #3
    hmm. Do you mean by finding a general formula for the partial sums? Ie. n=0 +n=1+n=3+....+n= whatever? from here look at a pattern and construct the formula?

    What i am thinking: Is it kind of the way of how we play around with a telescopic series?
     
  5. Aug 16, 2015 #4

    micromass

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    Make the exponents of the ##x## equal and add.
     
  6. Aug 16, 2015 #5

    FactChecker

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    No. Nothing fancy. Just find the term in each summation that has the same xn and add them together. Combine like terms.
     
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