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Convolution product between x² and x³

  1. Apr 12, 2014 #1
    I tryied make the convolution product between x² and x³ by ##\int_{- \infty}^{+ \infty} f(u) g(x-u) du## and the result is an indeterminate form, however, by defintion ##\int_{0}^{x} f(u) g(x-u) du##, the result is 1/60 x6. So, [tex]\int_{- \infty}^{+ \infty} f(u) g(x-u) du \overset{?}{=} \frac{1}{60} x^6[/tex] ?
     
  2. jcsd
  3. Apr 12, 2014 #2

    Mark44

    Staff: Mentor

    Show us what you did, please.
     
  4. Apr 14, 2014 #3
    imagem.png

    BTW, exist a symmetric algebraic definition for convolution where the graphic of the function f(x) is moved oppositely to g(x) and vice-versa (different of the standard definition, where a function is fixed and the another is moved over the first)? I tryied ##\int_{- \infty}^{+ \infty} f(\frac{1}{2}x+u) g(\frac{1}{2}x-u) du##, but this expression isn't equal to the convolution of f and g...
     
  5. Apr 14, 2014 #4
    OK, so far.

    According to you, how does this follow exactly:

     
  6. Apr 14, 2014 #5
    This equation isn't an affirmation, is a question. In acctually, I'm asking if ##\int_{- \infty}^{+\infty} f(u)g(x−u)du## is equal to ##\int_{0}^{x} f(u)g(x−u)du##.
     
  7. Apr 14, 2014 #6
    I know it's a question. But you're not going to learn much if we solve the question for you. So again, why do you think it's equal? Try to give the proof you're thinking of.
     
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