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Homework Help: Convolution of a polynomial with itself

  1. May 11, 2010 #1

    Char. Limit

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    Gold Member

    Again, in my quest to learn things I won't use in a class for at least a year, I've been looking at convolutions. Specifically, after finishing the multiple choice section of an AP Chemistry test 50 minutes early, I looked at the convolution of a polynomial with itself. I'm confused about one thing, though.

    Here's what I have:

    [tex]t^m \star t^m = t^{2m+1} \sum_{n=1}^{m+1} \frac{\binom{m}{n}}{n+m}[/tex]

    However, I don't like that sum in the solution, I don't think I computed it right, and I want it gone. Is there any way to do that?
     
  2. jcsd
  3. May 11, 2010 #2

    Mark44

    Staff: Mentor

    That doesn't look like the convolution I learned quite a few years ago, which was this:
    [tex](f \star g)(t) = \int_{-\infty}^{\infty} f(\tau)g(t - \tau)d\tau[/tex]
     
  4. May 12, 2010 #3

    Char. Limit

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    Hmm... I used the same formula, but the bounds I learned were from 0 to t. Using that, I came up with the OP formula for the convolution of a polynomial (really a monomial, now that I think about it, but I can't change the title) with itself.
     
  5. May 12, 2010 #4

    Mark44

    Staff: Mentor

    I got the formula I posted from Wikipedia, but it's the same formula that I remember from when I studied convolution way back when. If you can show us how you got your summation formula from the definition it would be helpful.
     
  6. May 12, 2010 #5

    Char. Limit

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    It was just the pattern I noticed. For example...

    For t:

    [tex]t^3(\frac{1}{2}-\frac{1}{3})[/tex]

    For t^2:

    [tex]t^5(\frac{1}{3}-\frac{2}{4}+\frac{1}{5})[/tex]

    For t^3:

    [tex]t^7(\frac{1}{4}-\frac{3}{5}+\frac{3}{6}-\frac{1}{7})[/tex]

    For t^4:

    [tex]t^9(\frac{1}{5}-\frac{4}{6}+\frac{6}{7}-\frac{4}{8}+\frac{1}{9})[/tex]

    The summation was my attempt at phrasing this pattern, which of course I forgot to put in a (-1)^(n+1). I just noticed that the denominators for t^m's convolution with itself were in consecutive integers from m+1 to 2m+1, and the numberators were each a row of Pascal's triangle, which if I remember right is also the binomial coefficients.
     
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