- #1
FranzDiCoccio
- 342
- 41
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
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