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Convolution question

  1. Feb 25, 2007 #1
    I just realized that the convolution of any function with itself many times will ultimately give a gaussian. I was just wondering if there was a function that was an exception to this?
     
  2. jcsd
  3. Feb 25, 2007 #2

    HallsofIvy

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    I'm not sure I can make sense out of your question. Over what interval are you taking the convolution? What happens if f(x) is a constant or f(x)= x?

    It is true that the convolution, over [itex]-\infty[/itex] to [itex]\infty[/itex], of two Gaussians is a Gaussian so this may be a "fixed point" theorem.
     
    Last edited: Feb 25, 2007
  4. Feb 25, 2007 #3
    Well, because the fourier transform of a convolution is a product and because the Fourier transform of a Gaussian is a Gaussian, the convolution theorem means that rasing any well defined function to a high integer power gives something increasingly similar to a Gaussian. I was wondering if there was an actual counterexample to this
     
  5. Feb 25, 2007 #4

    StatusX

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    What makes you think this? In raising a function to a very high power, the parts with |f(x)|>1 get increasingly larger while those with |f(x)|<1 vanish. So, for example, if the function has several narrow peaks whose height is greater than one, its very high powers will have spikes where each of those peaks were and vanish everywhere else.
     
  6. Feb 25, 2007 #5

    HallsofIvy

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    I don't think indigojoker meant "high power". I think he meant applying the convolution a large number of times.
     
  7. Feb 25, 2007 #6
    yep, that's what i mean
     
  8. Feb 25, 2007 #7

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    Right, but then you noted that a convolution in, say, the time domain corresponds to a product in the frequency domain, so convolving a function with itself n times corresponds to raising its fourier transform to the nth power. This is correct, but it doesn't follow that you'll always end up with gaussian like functions.
     
  9. Feb 25, 2007 #8
    "This is correct, but it doesn't follow that you'll always end up with gaussian like functions."

    so my original question was what types of functions will not end up with a gaussian?
     
  10. Feb 25, 2007 #9

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    I don't see why anything that isn't a guassian should "approach" a gaussian. And this is impossible to tell anyway unless you say exactly what you mean by "approach", ie, how do you measure how similar two functions are?
     
  11. Feb 26, 2007 #10

    AlephZero

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    Think about what convolution means in the frequency domain, and it's clear that almost any function will not end up as a gaussian.
     
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