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Strange resummation

  1. Sep 16, 2006 #1
    According to Hardy's book "divergent series" the sum:

    [tex] g(x)=1/2+cos(x)+cos(2x)+cos(3x)+....... [/tex] has sum equal to 0 (¡¡¡¡¡¡) then if we integrate in the sense:

    [tex] \int_{0}^{b}f(x)\sum_{n=0}^{\infty}cos(nx) dx \rightarrow (-1/2)\int_{0}^{b}f(x)dx [/tex] :rolleyes: :rolleyes: since the sum "regularized" has the value 0..but is this true?...can you manipulate divergent series giving them a "sum" although they diverge and even in this case that is clearly 0?..
  2. jcsd
  3. Sep 16, 2006 #2


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    You're omitting a lot of details; if you paid attention to them you might have your answer already. :wink:

    I'll assume Hardy knows what he's talking about and that you have faithfully represented him... and I'll try and guess at what the missing details are.

    The sum of functions

    [tex]\sum_{n = 0}^{+\infty} \cos nx[/tex]

    certainly doesn't exist. But that doesn't mean you can't do some other operation to this series to get a number. I'll select a certain class of "test functions", and define a "regularizaion" operation (which I'll call [itex]\mathcal{R}\sum[/itex]) on a sequence [itex]f_n[/itex] of functions to be:

    [tex]\mathcal{R}\sum_{n = 0}^{+\infty} f_n(x) = g(x)[/tex]

    if and only if

    [tex]\sum_{n = 0}^{+\infty} \int_0^b k(x) f_n(x) \, dx = \int_0^b k(x) g(x) \, dx[/tex]

    for every "test function" [itex]k[/itex].

    The other way I could interpret this is due to an annoying notation often used in physics that is rarely explicitly explained. If a physicist wrote what you had written, I would expect that they meant [itex]\cos nx[/itex] does not refer to the cosine function, but instead to the operator that maps:

    [tex]k(x) \rightarrow \int_0^b k(x) \cos nx \, dx[/tex]

    Then, I imagine for a certain class of "test functions", the infinite sum of these operators does, in fact, converge to the operator [itex]-1/2[/itex] that maps

    [tex]k(x) \rightarrow \int_0^b k(x) (-1/2) \, dx[/tex]
    Last edited: Sep 16, 2006
  4. Sep 17, 2006 #3
    I was refering to a sum over "cosines"...what i have done comes from the fact that "Poisson sum formula"..is:

    [tex] \sum_{n=-\infty}^{\infty}f(n)= \int_{-\infty}^{\infty}f(x)(1+ \sum_{n=0}^{\infty}Cos(nx)) [/tex]

    So from this you could "Suppose"(¿?) that the sum of cosines tends to -1/2 .. in fact i think you can deduce this identity by taking the "Laplace discrete trasnform:

    [tex] \sum_{n=0}^{\infty}cos(na)e^{-sn} [/tex] and take the limit s-->0+ (tends to 0 by the right side) to get this identity...my objective in proposing that is to get a "resummation" method for divergent integrals that appear in QFT theory (Quantum mehcanics) of the form:

    [tex] \int_{0}^{\infty}dx x^{m} [/tex] m>0 or m=0 so you can use "Poisson sum formula" to get a relation between these integrals and the sum:

    [tex] 1+2^{m}+3^{m}+................. =\zeta (-m) [/tex] m>0
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