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Consider two piecewise continuous, twice integrable functions $f, g: [-\pi, \pi] \rightarrow \mathbb{R}$, and suppose they have the following convergent Fourier series expansions:

$$ \begin{aligned} f(x) & = \frac{a_0}{2} + \sum_{n = 1}^{\infty}\left(a_n \cos(n x) + b_n \sin(n x)\right) \\ g(x) & = \frac{\alpha_0}{2} + \sum_{n = 1}^{\infty}\left(\alpha_n \cos(n x) + \beta_n \sin(n x)\right) \end{aligned} $$

Define $h := fg$ and suppose $h$ is again twice integrable and has the following convergent Fourier series expansion:

$$ h(x) = \frac{A_0}{2} + \sum_{n = 1}^{\infty}\left(A_n \cos(n x) + B_n \sin(n x)\right) $$

Can $h$'s coefficients be expressed in terms of $f$ and $g$'s coefficients? If the general question cannot be easily answered, how about the case $g = \sin$ or $g = \cos$?

Evan Aad
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    The algebraic product of series for $f$ and $g$ could be reduced to the form of Fourier series using trigonometric identities. – Karolis Juodelė Apr 27 '14 at 09:06
  • @KarolisJuodelė: Thanks. From your comment I gather that the terms of the Fourier expansion of the product can be identified with the terms of the product of the two series. Is this correct? – Evan Aad Apr 27 '14 at 09:12
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    http://math.stackexchange.com/questions/49387/identifying-the-product-of-two-fourier-series-with-a-third It is the same as with fourier transformation. Product changes to convolution. – tom Apr 27 '14 at 09:13
  • @tom: Thanks. However, the post you linked to discusses the case of a complex Fourier series. My question pertains to a real Fourier series. I can't figure out how to translate the results of the complex case to those of the real case. – Evan Aad Apr 27 '14 at 09:34
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    @EvanAad If you know how to express $\sin,\cos$ with complex exponential, than you know how the connection between complex and real fourier coefficients. It is relatively simple exercise try it out yourself. – tom Apr 27 '14 at 10:14

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Here's how I would motivate the solution:

First, the problem will be VASTLY simpler if you replace $\cos(nx)$ with one half the sum of a positive and negative complex exponential and $\sin(nx)$ as one half the difference between the positive and negative complex exponential, and then once the two series are expressed in terms of complex exponentials, then multiplication of Fourier terms becomes easy.

That's because the coefficient of the $m$-th term in the product will contain terms from the two multiplier series whose indices add to $m$, such as $(m-2,2)$, $(m-1,1)$, $(m,0)$, $(m+1,-1)$, $(m+2,-2)$, and so on, where the first index in each pair refers to the first series and the second index refers to the second series, and this is where the convolution on indices comes from.

Gilles Bonnet
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craig s.
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