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Dirac,s delta properties

  1. Aug 17, 2004 #1
    Dirac,s delta properties....

    Let be d(x-a) the Delta function centered at x=a then could this function be approached by using a Fourier series on the interval (-pi,pi) with -Pi<a<Pi

    Anothe question.let be w(x)=sum(1<n<Infinite)d(x-n) then has the function
    z(x)=1/w(x) any sense and in teh case w(x)f(x)=g(x) could we do f(x)=g(x)/w(x)?

    inf we take instead some of the approximation function to delta for example

    sen(nx)/x or nexp(-n^^2x^2)/sqrt(pi) would be the statements above be true?..
     
  2. jcsd
  3. Aug 17, 2004 #2

    matt grime

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    "Let be d(x-a) the Delta function centered at x=a then could this function be approached by using a Fourier series on the interval (-pi,pi) with -Pi<a<Pi"

    well known

    "Anothe question.let be w(x)=sum(1<n<Infinite)d(x-n) then has the function
    z(x)=1/w(x) any sense and in teh case w(x)f(x)=g(x) could we do f(x)=g(x)/w(x)"

    a "function" that is infinity at all points except the integers? could be less than useful.
     
  4. Aug 17, 2004 #3
    then according to you a function that is 0 elsewhere but a point x=a (Dirac,s delta) is useless.
    the point is let be w(x)f(x)=g(x) is legitimated to do f(x)=g(x)/w(x)?for example let suppose we take the fourier transforms W(p)F(p)=G(p) the to recover f(x) could we do f(x)=F^^--1[G(p)/W(p)]
     
  5. Aug 17, 2004 #4

    matt grime

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    No, that is not what I said, I said a "function" that is undefined at all but a countable number of nowhere dense points maybe less than useful. Note that dirac's delta is actually a distribution, not even a function. What are you claiming this reciprocal of it should do? Even the infinite sum of deltas only makes sense when used with functions whose values at integers are summable. So when should one use the reciprocal of that? What do you think it means? what is it even? The dirac delta is not infinite at 0 (or whichever point) in the same sense that this 1/w is since at uncountably many points (all points except the integers) it is not defined, and you offer no indication of what it ought to mean, and at the integers it is not defined per se, but we may assume you mean it is zero there.

    It is not clear what 1/w should be taken to mean at all these points where it is not defined, ok? You've just divided by zero at all those points, without offering some explanation of what's happening there.

    The delta function can be interpreted as the limit of those functions you mention, can be integrated, can be treated as a derivative, and this formal treatment gives it the properties you see there.

    So you need do decide what the analogue of these results is going to be for your notional z there, you need to explain what it is going to be the limiti of perhaps, or do.
     
  6. Aug 24, 2004 #5
    and its legitimate to calculate the Laplace transform of a function defined as

    Sum(1<n<infinity)d(x-n)f(x) where the d(x-n) means the Dirac,s delta function,if we define w(x)=Sum(1<n<infinity)d(x-n) then Int(a-1/2,a+1/2)w(x)f(x)=f(a) is that true?.
     
  7. Aug 24, 2004 #6

    matt grime

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    That would entirely depend on what f is, since you're about to create an infinite possibly divergent series when you do the integral over the whole of R.

    Your second paragraph is trivial.
     
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