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It's a rainy day anyway.
Posted:
Jul 9, 1996 8:59 AM


Hello,
My name is Matthew and I need your help. I am struggling with the summation of a double series:
Let alpha_n=(2n+1) pi/a; Let beta_m =(2m+1) pi/b; Both a and b are lengths, so the units of alpha_n and beta_m are m^1. Let gamma^2 be a complex number which real part is zero; the unit of gamma is m^1. C and ab*C_0 have the same dimensions.
The series:
(1) C=64*C_0/(ab) sum of n=0 to infinity sum of m=0 to infinity of
{ (alpha_n)^2 + (beta_m)^2 } divided by { (alpha_n)^2 * (beta_m)^2 }
times 1 over {(alpha_n)^2 +(beta_m)^2gamma^2 }.
It's a real hard one. I am studying for my Ph.D at Twente University in Enschede, the Netherlands, so I went to the mathematics department over here. (I am a physicist myself.) They solved it partially, removing one summation. They got:
(2) C/C_0= ab * tan (gamma a/2) over (gamma a /2) +
16 a^4 gamma^2 over (pi^5) times the sum of n=0 to infinity of
tan{ b/2 sqrt(gamma^2pi^2*(2n+1)^2/a^2) }over
(2n+1)^2 * [ gamma^2 a^2/pi^2  (2n+1)^2 ]^(3/2).
If you manage to obtain these sums in a closed mathematical form, I would be appreciate it. Since I am not into any of the mathematical software packages, I did not try them. It might help to look upon (1) as a Laurent series, with poles at gamma^2=(alpha_n)^2+(beta_m)^2.
I studied two cases: small and large gamma. For gamma small, the series is easy summable and it should say that C=ab*C_0. In the case of large gamma, I obtained:
C=ab*C_0 times [tan(gamma a/2) over (gamma a/2) + tan (gamma b/2) over (gamma b/2) ];
Therefore, it might be that (1) is some weird product of tangenses.
I appreciate your effort. If you succeed, you can email me at M.W.denOtter@ct.utwente.nl . Response guaranteed.
Good luck!
Matthew.



