Integral of exponentiated Fourier expansion


by quelarion
Tags: expansion, exponentiated, fourier, integral
quelarion
quelarion is offline
#1
Nov11-09, 08:44 AM
P: 9
Hello everyone,
I need to solve an integral, opfully extracting Fourier coefficients, and I don't have a clue.

It is just

[tex]
\[
\int^{2\pi}_0 d\sigma \text{exp}{A(\sigma)}
\]
[/tex]

where

[tex]
\[
A(\sigma) = \sum_n a_n e^{-i n \sigma }
\]
[/tex]

I tried to work with complex coordinates [tex]
\[
z = e^{-i \sigma}
\]
[/tex] and look for the poles, but since there are poles at every order I can't go through that...

Any idea? I don't even know how to look on the net for something like this...
Thank you guys!
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LCKurtz
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#2
Nov13-09, 01:14 PM
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PF Gold
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I don't know if I can help you or not, but your question is not clear to me. What do you mean by "poles at every order"? Is [itex]A(\sigma)[/itex] a given periodic function for which you seek the complex Fourier coefficients? Does it satisfy the Dirichlet conditions? More details please.
quelarion
quelarion is offline
#3
Nov13-09, 01:36 PM
P: 9
well, [tex] \[ A(\sigma) \] [/tex] should be a [tex] \[ C^\infty \] [/tex] function (all derivatives exist and are differentiable).

I don't know if the problem is really well defined mathematically, for me [tex] \[ A(\sigma) \] [/tex] is a field for which I know the expansion in creation and annihilation operators. And everything is normal ordered. But I would like to avoid all the complications of considering it as a quantum operator and go straight to the mathematical problem, if it is solvable ;)

As for the "poles at every order" i meant that i can define [tex] \[ z = e^{-i \sigma} \] [/tex], and write the integral as

[tex] i \int_\gamma d z \frac{1}{z} \text{exp}\left( \sum_{n>0}( f_n z^n + f_{-n} z^{-n}) + f_0 \right)[/tex]

where [tex] \[ \gamma \] [/tex] is a closed contour around the origin in the complex plane.
I could try now to compute the residues at the poles, expanding the exponential for the negative n.

[tex] i \int_\gamma d z \frac{e^{f_0}}{z} \sum_{k>0} \left( \sum_{m>0} f_{-m} z^{-m}\right)^k \text{exp}\left( \sum_{n>0} f_n z^n \right)[/tex]

but even assuming that I can compute the k power of this series i would get

[tex] i \int_\gamma d z e^{f_0} \sum_{k>0} \sum_{m>0} F_{m,k} z^{-mk-1} \text{exp}\left( \sum_{n>0} f_n z^n \right)[/tex]

where F_{m,k} are some combinations of the f's.
At this point there are poles at every order, from 1 to infinity, so one should take an arbitrary high derivative of the exponential left (that depends on z) and then compute it for z=0 to get the residue...
It doesn't seems a good way to go through this problem, does it?

Thank you anyway ;)

LCKurtz
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#4
Nov13-09, 01:57 PM
HW Helper
Thanks
PF Gold
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P: 7,198

Integral of exponentiated Fourier expansion


Well, that clarifies enough for me to tell that I'm not the guy to help you with it. Sorry, but it's a bit far afield for me.
quelarion
quelarion is offline
#5
Nov13-09, 02:22 PM
P: 9
hehehe, thank you anyway ;)
Mute
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#6
Nov13-09, 04:22 PM
HW Helper
P: 1,391
You might try the following:

[tex]\exp\left[ \sum_n a_n e^{-in\sigma} \right] = \prod_n \exp\left[a_n e^{-in\sigma}\right] = \prod_n \exp\left[a_n \cos(n\sigma)\right]\exp\left[-ia_n \sin(n\sigma)\right],[/tex]

then use the identities

[tex] e^{iz \sin\phi} = \sum_{m=-\infty}^{\infty} J_m(z) e^{im\phi} [/tex]

and


[tex] e^{z \cos\phi} = I_0(z) + 2\sum_{m=1}^{\infty} I_m(z) \cos{m\phi} [/tex]

where J and I are bessel functions and modified bessel functions, respectively.

http://en.wikipedia.org/wiki/Bessel_function

This will, however, place your coefficients [itex]a_n[/itex] inside the bessel functions, which may not be so useful in being able to extract them. You will also have very many products of sums, which may be quite difficult, if not impossible, to deal with.
quelarion
quelarion is offline
#7
Nov13-09, 04:50 PM
P: 9
thanks, it seems to be a good solution ;)

I just checked these functions and since they contain just powers of my creation/annihilation operators they would work perfectly. Anyway tomorrow I will check in detail and I will let you know ;)

thanks again!


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