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Strange relation

  1. May 31, 2008 #1
    [tex]\sum\ (m^2+n^2) = 2\pi\int^{\infty}_{1} (\frac{1}{r^2} r dr) [/tex]

    The sum goes over [tex]\mathbb{N}\times \mathbb{N}[/tex]
     
    Last edited: May 31, 2008
  2. jcsd
  3. May 31, 2008 #2
    Wait, that is what I mean.

    [tex]\sum\ (m^2+n^2)^{-1} \sim 2 \pi\int^{\infty}_{1} (\frac{1}{r^2} r dr) [/tex]

    The sum goes over [tex]\mathbb{N}\times \mathbb{N}[/tex]

    I can make no sense of it.

    Edit: Now everything fixed.
     
    Last edited: May 31, 2008
  4. May 31, 2008 #3

    Dick

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    The sum (m^2+n^2) clearly diverges. You must mean 1/(m^2+n^2). Then it's just an integral estimate. And it still diverges.
     
  5. May 31, 2008 #4

    HallsofIvy

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    And the right side makes no sense. In [itex]2\pi\int^{\infty}_{1} (\frac{1}{r^2} r dr) [/itex], "r" is clearly intended to be a vector (otherwise it reduces to [itex]\int 1/r dr= ln(r)[/itex]) in which case integrating from 1 to infinity doesn't make sense. My guess is that the right side is supposed to be integrated over all of R2.
     
  6. May 31, 2008 #5
    Right, that's what I meant. I'm still fighting with latex.

    What you mean by 'just an integral estimate'. Why are both expressions [tex]\sim[/tex]?

    And, yes my book says too, it diverges.
     
  7. May 31, 2008 #6

    tiny-tim

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    Hi Icosahedron! :smile:

    Looks like you're supposed to integrate something over x and y from 1 to ∞ which has poles "of integral 1/r²" whenever x and y are both whole numbers, and then convert to polar coordinates. :smile:
     
  8. May 31, 2008 #7

    Dick

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    The right side is the integral of 1/(x^2+y^2) over the plane (excluding some stuff near zero) in polar coordinates. It should be a reasonable estimate for the integer sum (also excluding (0,0), I hope). Just like the integral test for infinite series.
     
    Last edited: May 31, 2008
  9. Jun 1, 2008 #8
    Thanks!
     
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