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Convulation and integration

  1. Jan 2, 2014 #1
    I was verifying that:

    02cdb0955d8b54c21acde435bcc14991.png

    http://en.wikipedia.org/wiki/Convolution#Integration


    But, I'd like to know if this equality is true:
    [tex]\int f(x)g(x)dx \overset{?}{=} \int f(x)dx \ast \int g(x)dx[/tex]

    Subquestion: this theorem above can be applied in summation too?
    [tex]\sum_{x}\left (f(x)\ast f(x) \right )\overset{?}{=}\left(\sum_{x} f(x)\right)\left (\sum_{x} g(x) \right )[/tex]
    [tex]\sum_{x}f(x)g(x) \overset{?}{=} \sum_{x}f(x) \ast \sum_{x}g(x)[/tex]
    Thanks!
     
  2. jcsd
  3. Jan 2, 2014 #2

    pwsnafu

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    Science Advisor

    The easy way to check is differentiate both sides. What do you know about the derivative of the convolution.
     
  4. Jan 2, 2014 #3
    But ##\int f(x) dx## and ##\int g(x)dx## are just real numbers, no? What is the convolution of two numbers?
     
  5. Jan 2, 2014 #4
    That:
    a63515c305a187bb7a3e71570429c3dc.png
     
  6. Feb 25, 2014 #5
    No! The result is a function!

    Results: f(x) g(x) = f(x) * ∫ g(x) dx

    But, my ask remains! My as ask now is f(x) g(x) is equal to f(x) * ∫ g(x) dx ???
     
    Last edited: Feb 25, 2014
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