Convolution: Is There an Exception to Gaussian?

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Homework Help Overview

The discussion revolves around the properties of convolution, particularly whether there are exceptions to the notion that repeated convolution of functions leads to Gaussian-like results. Participants explore the implications of the convolution theorem and the behavior of various functions under convolution.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants question the conditions under which convolution leads to Gaussian functions, discussing specific cases such as constant functions and functions with narrow peaks. They also explore the implications of raising functions to high powers versus repeated convolution.

Discussion Status

The discussion is ongoing, with participants providing insights and questioning assumptions about the behavior of different functions under convolution. There is no explicit consensus, but various interpretations and lines of reasoning are being explored.

Contextual Notes

Some participants note the importance of defining what is meant by "approaching" a Gaussian and the need for clarity regarding the intervals over which convolution is considered.

indigojoker
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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?
 
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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 -\infty to \infty, of two Gaussians is a Gaussian so this may be a "fixed point" theorem.
 
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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
 
indigojoker said:
rasing any well defined function to a high integer power gives something increasingly similar to a Gaussian.

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.
 
I don't think indigojoker meant "high power". I think he meant applying the convolution a large number of times.
 
yep, that's what i mean
 
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.
 
"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?
 
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?
 
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indigojoker said:
so my original question was what types of functions will not end up with a gaussian?

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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