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Textbook question on power series

  1. Dec 2, 2005 #1
    My textbook has an example on multiplication of power series.
    " Multiply the geometric series [tex] x^n[/tex] by itself to get a power series for [tex]1/(1-x)^2 [/tex] for |x|<1 "
    from this we get the [tex]c_{n}[/tex]=n+1
    O.K. I get that the coefficients are 1 for all n but why +1.
    Could someone please explain this to me if possible.
  2. jcsd
  3. Dec 2, 2005 #2

    Physics Monkey

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    You want to multiply [tex] (1 + x + x^2 + x^3 + ...)(1 + x + x^2 + x^3 + ...)[/tex], so just try few examples first to see what's going on. The [tex]x^0[/tex] term is just [tex] 1*1 = 1[/tex], the [tex] x^1 [/tex] term is [tex] x*1+1*x = 2x [/tex], the [tex] x^2 [/tex] term is [tex] x^2*1 + x*x + 1*x^2 = 3x^2 [/tex], and the [tex] x^3 [/tex] is [tex] x^3*1 + x^2*x+x*x^2+1*x^3 =4x^3[/tex]. Now, can you begin to see a pattern forming?

    Alternatively, you can just apply the formula for the coeffecients of the product series in terms of the coeffecients of the two original series. It looks something like [tex] c_k = \sum^k_{r=0} a_r b_{k-r}[/tex] where the a's and b's are the coeffecients of the original series, and the c's are coeffecients of the product series.
    Last edited: Dec 2, 2005
  4. Dec 2, 2005 #3


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    Right call Physics Monkey:
    The formula for the Cauchy product of series as it is presented in this link.
    A quick version is:
    Suppose [itex] \sum_{n=0}^{\infty} a_n[/itex] and [itex] \sum_{n=0}^{\infty} b_n[/itex] converge absolutely. Then
    [tex]\left( \sum_{n=0}^{\infty} a_n\right) \left( \sum_{n=0}^{\infty} b_n\right) = \sum_{n=0}^{\infty} \sum_{k=0}^{n} a_{k}b_{n-k}[/tex] also converges absolutely.
    Alternately, look here, under the heading A Variant.
  5. Dec 3, 2005 #4
    O.K. I see what's going on now, thanks. But I still don't get how to get n+1 from [tex] c_k = \sum^k_{r=0} a_r b_{k-r}[/tex] without doing some multiplication. Please explain to me if you can.
  6. Dec 3, 2005 #5


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    the answer to you above question


    [tex]\left( \sum_{n=0}^{\infty} a_n\right) \left( \sum_{n=0}^{\infty} b_n\right) = \sum_{n=0}^{\infty} \sum_{k=0}^{n} a_{k}b_{n-k}[/tex]

    we have

    [tex]\left( \sum_{n=0}^{\infty} x^n\right) \left( \sum_{n=0}^{\infty} x^n\right) = \sum_{n=0}^{\infty} \sum_{k=0}^{n} x^{k}x^{n-k} = \sum_{n=0}^{\infty} \sum_{k=0}^{n} x^{k+(n-k)} = \sum_{n=0}^{\infty} x^{n} \sum_{k=0}^{n} 1 = \sum_{n=0}^{\infty}(n+1) x^{n} [/tex]

    where [itex]\sum_{k=0}^{n} 1 =n+1[/itex] is the answer to you above question.
  7. Dec 3, 2005 #6
    O.K. I getting close to understanding this. What puzzles me is the last sum. Do you put n+1 so the first term isn't 0? I think I'm missing some information to understand this.
  8. Dec 3, 2005 #7

    matt grime

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    We put n+1 because that is what it is... You're adding up the number 1, n+1 times, so the answer is n+1.

    But what's wrong with just multiplying out the power series?

    You can multiply (1+x+x^2+.....)(1+x+x^2+...)

    and count and think and, well, it's just true.... there is nothing clever going on. to end up with x^n in the product you can only get it from multiplying x^r in the first and x^{n-r} and each of those multiplications contributes 1 to the coefficiant of x^n and there is one contributrion from each r as r goes from 0 to n so you get 1 added up n+1 times.
    Last edited: Dec 3, 2005
  9. Dec 3, 2005 #8
    I think I understand this no, just hvae to get some practice solving these kind of problems. Thanks.
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