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Gaussian functions?

  1. May 24, 2007 #1
    Is the summation of an infinite number of different (not just different by constants ) Gaussian functions still a gaussian function?

    I think not because you can becuase you can just pull an e^(ax^2) from the series where a is any value as small as needed.
    Last edited: May 24, 2007
  2. jcsd
  3. May 24, 2007 #2


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    You're asking if whatever A_n, mu_n and sigma_n,


    for some A, mu and sigma. Is that correct?
  4. May 25, 2007 #3
    Yeah. Although I think this result dosen't hold?
  5. May 25, 2007 #4

    matt grime

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    If you sum two gaussians with mean X and Y (think of them as p.d.f.s) then you get one with mean X+Y. So an infinite sum will, in general, have 'infinite mean'. Similarly, infinite standard deviation. So, no, it is not a Gaussian, and in general won't even exist.
  6. May 29, 2007 #5
    But if you sum a finite number of different Gaussian than the sum will still be a gaussian? If so how? how can you get e^a+e^b=e^c where a,b,c are functions that satisfy the exponentials being a gaussian.
  7. May 29, 2007 #6

    matt grime

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    So, you're arguing that if property P holds for a finite number of things (terms, whatever), then it holds for an infinite number of things? Please, back off and think about that for a while. And I'm not arguing that e^a+e^b=e^c. What are a,b,c, for a start?
  8. May 29, 2007 #7
    What I am saying or rather asking is that I don't even see how it would work for the finite case. That is for any two different Gaussians

    How do you add them and produce another gaussian?

    or even if you take e^2+e^3. Can you make it into e^a for any number a?
  9. May 29, 2007 #8
    Well, [itex]e^{2} + e^{3} = e^{a} \rightarrow a = ln(e^{2} + e^{3})[/itex].
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