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Given the product of two functions defined explicitly through their Fourier coefficients (of unknown undeveloped form):

\sum_k{c_k e^{i k t}} \cdot \sum_k{c'_k e^{i k t}}

Is there any way to express it as a Fourier series? (Edit: approximated using a finite number of terms of the original)

That is: \sum_k{c''_k e^{i k t}} where each c''_k could be explicitly defined from a finite sum of $c$ and c'.

I feel the convolution theorem should be of some help here, but I can't see how for the life of me...

(probably not relevant, but my goal is to use this product's equality with a third Fourier series and use coefficient identity in order to extract a set of optimisation constraints based on the terms of all three original series)

Edit: since I am trying to identify coefficients, what I'm really hoping for is an approximated expression of the product, based on a limited number of terms... In the absence of any particular properties of $c$ and c' that would simplify the convolution, is there any way to achieve this?

(thanks a lot to people who already answered and made me realise the issue with my original formulation)

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    @Qiaochu: actually, after reading the two answers below, I realise that *this* is what I am really looking for (a product based on limited Fourier developments). I edited my question accordingly (hope that's OK). Thanks for the tip on multiplication of polynomials: sounds like it might get me there...2011-07-04

2 Answers 2

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I'd use the notation $ \times $ rather than $ * $ because the latter is used for convolutions in this sort of context (Fourier analysis). In any case, you can explicitly calculate the coefficients of the product's Fourier series via c''_n = \sum_{k=-\infty}^{\infty} c_{n-k} c'_k Note that this can be related to convolutions in the sense that c''_n = (c * c')_n .

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    you mean for example $\sum_{k=-5}^{5}c_kc'_{-k}$ is a proper approximation of $c''_0$? Is there any method I should go about to bound the error in that case?2011-07-04
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I'll assume from the text that you're using the asterisk to denote multiplication. This is a bit confusing since in the context of convolutions it is usually used to denote convolution, so I'll use an asterisk to denote convolution and a dot to denote multiplication.

You're right, this can be expressed as a convolution, as follows:

\begin{eqnarray} \sum_k{c_k \mathrm e^{\mathrm i k t}} \cdot \sum_{k'}{c'_{k'} \mathrm e^{\mathrm i {k'} t}} &=& \sum_{k,k'}{c_k c'_{k'}\mathrm e^{\mathrm i k t}} \mathrm e^{\mathrm i {k'} t} \\ &=& \sum_{k,k'}{c_k c'_{k'}\mathrm e^{\mathrm i (k+k') t}} \\ &=& \sum_{k,k''}{c_k c'_{k''-k}\mathrm e^{\mathrm i k'' t}} \\ &=& \sum_{k''}\left(\sum_kc_k c'_{k''-k}\right)\mathrm e^{\mathrm i k'' t} \\ &=& \sum_{k''}\left(c*c'\right)_{k''}\mathrm e^{\mathrm i k'' t} \\ &=& \sum_{k''}c''_{k''}\mathrm e^{\mathrm i k'' t} \end{eqnarray}

with c''_{k''}=\left(c*c'\right)_{k''}.

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    thanks a lot for the very clear and detailed answer! Unfortunately it also made me realise that I am looking for something a little stronger than the straight convolution product :-/ (see my comment above)2011-07-04