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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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That's exactly convolution of coefficients: $c_k''=\sum_l с_{k-l}c_l'$. – Andrew Jul 4 '11 at 10:50
Have you tried working out what the answer looks like for Fourier series with a finite number of terms? It's the same as multiplication of polynomials in $e^{it}$. – Qiaochu Yuan Jul 4 '11 at 11:09
@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... – Dave Jul 4 '11 at 11:33
up vote 4 down vote accepted

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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thanks for pointing the confusing operator sign. I edited my question. Thanks for your answer... Actually, I realised that I am looking for a finite series approximation (I should have framed my question better)... Is there any way to use the infinite expression to get a useful result with a limited development? – Dave Jul 4 '11 at 11:27
Simply truncate the formula given in my post so that only the finite terms are used. You can do that by first restricting $ n $ how you wish and then making sure you only sum over $ k $ such that both $ n - k $ and $ k $ obey that original restriction. – anon Jul 4 '11 at 11:57
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? – Dave Jul 4 '11 at 15:36

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) – Dave Jul 4 '11 at 11:29

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