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