Identifying the product of two Fourier series with a third?

Question

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)

Answer 1

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 $.

Answer 2

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