I would like to express the product $$ \left( \sum_{k \in \mathbb{Z}} a_k \sin(k t) \right) \left( \sum_{k \in \mathbb{Z}} b_k \cos(k t) \right) $$ as $$ \sum_{k \in \mathbb{Z}} c_k \sin(k t). $$ Question : How can I express the coefficients $c_k$ as functions of $a_k$ and $b_k$ ? I know that when multiplying two $\cos$ series, for example, the $c_k$ are standard convolution products. I believe using complex exponentials, Cauchy products and the formulae for Fourier coefficients might be useful, but I can't quite see how to proceed.

  • 3
    $\begingroup$ Use the fact that $2 \sin(kt)\cos(k't) = \sin((k+k')t) + \sin((k-k')t)$ $\endgroup$
    – user10676
    Jun 2, 2014 at 14:47

1 Answer 1


Let's assume without loss of generality the the indices for the sine series are positive and the ones of the cosine series are non-negative:

$$x(t)=\sum_{k=1}^{\infty}a_k\sin(kt),\quad y(t)=\sum_{k=0}^{\infty}b_k\cos(kt)$$

Rewrite $x(t)$ as

$$x(t)=\frac{1}{2i}\sum_{k=1}^{\infty}a_k(e^{ikt}-e^{-ikt})= \frac{1}{2i}\left(\sum_{k=1}^{\infty}a_ke^{ikt}-\sum_{k=-\infty}^{-1}a_{-k}e^{ikt}\right)=\\=\sum_{k=-\infty}^{\infty}d_ke^{ikt},\quad d_k=\begin{cases}\frac{a_k}{2i},&k>0\\0,&k=0\\-\frac{a_{-k}}{2i},&k<0\end{cases}$$

Note that the coefficients $d_k$ are purely imaginary and satisfy $d_k=-d_{-k}$. In a completely analogous way we can write $y(t)$ as

$$y(t)=\sum_{k=-\infty}^{\infty}f_ke^{ikt},\quad f_k=\begin{cases}\frac{b_k}{2},&k>0\\b_0,&k=0\\\frac{b_{-k}}{2},&k<0\end{cases}$$

Now we can express $z(t)=x(t)y(t)$ as




Note that $g_k$ is purely imaginary and satisfies $g_k=-g_{-k}$ and $g_0=0$. So we can rewrite $z(t)$ as


Note that the final coefficients $2ig_k$ are of course real-valued.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .