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I Manipulate this summation with Exp

  1. Jul 20, 2017 #1
    I am trying to manipulate this summation such that I have a summation of a function of r only by itself somewhere:


    This could be rewritten:
    [tex]\sum_{r=1}^{\infty}\left(e^{-r}\right)^B[/tex] or [tex]\sum_{r=1}^{\infty}\left(e^{-B}\right)^r[/tex]

    What I would like is:
    Where [itex]g[/itex] depends on [itex]B[/itex] as well.

    I really just want to be able to get some kind of independent [itex]f(r)[/itex] out of this. The biggest problem I've had with this is that while the sum converges, doing something like [itex]ln\left[\sum_{r=1}^{\infty}e^{-B⋅r}\right][/itex] does not converge, so I'm stuck...

    Thank you!
    Last edited: Jul 20, 2017
  2. jcsd
  3. Jul 20, 2017 #2


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    I guess I am missing something. The terms do not involve n. It's the infinite sum of the same constant.
  4. Jul 20, 2017 #3


    Staff: Mentor

    In addition to what @FactChecker said, this doesn't make sense. Neither of the last two summations is equal to the first. Both of the latter two summations start with n = s (what is the value of s) and end with t, so both are finite sums, assuming that s is given a value and t is some finite number.

    Note that ##\sum_{n=1}^{\infty}\left(e^{-B}\right)^r## is a geometric series. Under certain conditions, such a series converges and there is a formula for the value it converges to. This fact might help you reach the form you're looking for.
  5. Jul 20, 2017 #4
    Oops, typos corrected. And actually, this is an oversimplification of the real problem, which is this:


    So, [itex]r[/itex] is not necessarily integral. If it were, this would converge automatically.

    However, I was hoping if I started from that more simple version, I could apply the same principle to the real problem.
  6. Jul 20, 2017 #5


    Staff: Mentor

    Is this problem part of a volume integral?
  7. Jul 20, 2017 #6
    It's part of a volume sum, not integral. The [itex](i,j,k)[/itex] values are integer coordinates in a grid, so they cannot be manipulated, or gaps reduced to [itex]{\Delta}i[/itex] (for example) for purposes of integration.
  8. Jul 20, 2017 #7


    Staff: Mentor

    This seems related to the integral ##\int_0^\infty \int_0^\infty e^{-x^2 - y^2}dx~dy##. As this integral sits, it can't be done by ordinary means, but if you rewrite it in polar form, it's easy.
    For your sum, instead of looking at volume elements that are little cubes, perhaps the trick is to rewrite the sum in polar form. In that case, ##\sqrt{x^2 + y^2 + z^2}=\sqrt r##, or more simply, ##x^2 + y^2 + z^2 = r^2##, and write the sum as r goes from 1 to infinity.
  9. Jul 20, 2017 #8
    The only problem with that is that r is not always an integer (in fact it almost always isn't). But Feynmann's trick is a good one. Too bad it won't work here.
  10. Jul 20, 2017 #9


    Staff: Mentor

    Maybe you can force r to be an integer (in the set ##\{1, 2, 3, \dots \}##) by a suitable choice for a volume element, which will have curved inner and outer surfaces and four slanted sides. Off the top of my head I don't recall the name of this type of solid.

    I don't know Feynmann's trick...
  11. Jul 21, 2017 #10
    Well you do, but you probably didn't know that's what it's called (among students I've known). When you make that substitution to polar coordinates for double (or triple) Gaussian terms in an integral, that's Feynmann's trick. Though I think Griffiths made it popular.
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