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Tricky Euler-Maclaurin sum

  1. Mar 12, 2015 #1
    Hi everybody,

    I'm facing a tricky summation problem. The problem is (both mathematically and physically) related to this thread I started a while ago. I'm starting a new thread because the functions are not exactly the same, and I have (perhaps) made some progress, using Euler-Maclaurin sum formula. However, there is something that still really puzzles me.

    So, my goal would be to evaluate a function

    [tex]
    F(b,\ell) = \sinh(2 \ell\, b)- \sum_{k=1}^{\ell-1} e^{2 \ell\, b\, c_k} \left(2 \ell\, b\, s_k^2-c_k \right)
    [/tex]
    where
    [tex]
    f_k(b,\ell) = e^{2 \ell\, b\, c_k} \left(2 \ell\, b\, s_k^2-c_k \right)= -\frac{\ell}{\pi}\frac{d}{dk} \left(s_k e^{2 \ell\, b\, c_k}\right)
    [/tex]

    [tex]
    c_k = \cos \left(\frac{\pi}{\ell} k\right)
    , \quad s_k = \sin \left(\frac{\pi}{\ell} k\right)
    [/tex]


    and [itex]\ell[/itex] is a positive integer. Actually, I'd like to find out the result for large [itex]\ell[/itex] .

    I thought of using the Euler-Maclaurin sum formula (EMSF) to evaluate the sum.

    The result seems to be that the above sum exactly cancels the hyperbolic sin everywhere.
    Specifically, as it is easy to check, the integral term in the EMSF vanishes. The term in the EMSF involving the value of [itex]f_k[/itex] at [itex]k=0[/itex] and [itex]k=\ell[/itex] is exactly the sinh term.

    Thus [itex]F[b,L][/itex] should be given by the series containing the Bernoulli numbers and the (odd) derivatives of [itex]f_k[/itex] evaluated at [itex]k=0[/itex] and [itex]k=\ell[/itex].
    The problem is that it seems to me that all of these derivatives contain an overall factor [itex]s_k[/itex], which is exactly zero at both "endpoints". Several tests using Mathematica confirm this.

    But this cannot be right. I know (e.g. numerically) that [itex]F[b,\ell][/itex] is not zero at all. I expect that [itex]F[b,\ell]\sim C b^{2\ell-1} [/itex] for small [itex]b [/itex].

    I do not understand what's wrong.

    Thanks a lot for your interest.
    F
     
  2. jcsd
  3. Mar 12, 2015 #2
    How sure are you about the scaling of ##F[b,l]##? Expanding around ## b = 0## seems to give zero to second order.
     
  4. Mar 12, 2015 #3
    Hi Strum,

    thanks for your interest in my problem. I hope you can help me.

    As to your comment, I'm not sure I understand it. Are you saying that you find a nonzero contribution to the third order? Or that you find zero up to the second?

    Because it seems to me that the latter is compatible with what I expect.
    If a function vanishes as [itex]b^{2\ell-1}[/itex], it is zero to second order, provided that [itex]\ell>1[/itex], which in my case is (as I say, I'd like to find an analytic expression for [itex]F[/itex] in the limit [itex]\ell \gg 1[/itex]).
    I mean, the first nonzero contribution is of order [itex]{2\ell-1}[/itex].

    I expect that behavior because [itex]F(b,\ell) b^{1-2\ell}[/itex] is a known constant for [itex]b=0[/itex]. I can check this behavior numerically, although for very small [itex]b[/itex] the calculations become very imprecise due to roundoff (or poor programming).

    Thanks again
    -F
     
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