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Let $$f(x)=a^{(1)}_0/2+\sum_{n=1}^{\infty}a^{(1)}_n\cos(nx)+b^{(1)}_n\sin(nx) ~~~(1)$$ and $$g(x)=a^{(2)}_0/2+\sum_{n=1}^{\infty}a^{(2)}_n\cos(nx)+b^{(2)}_n\sin(nx)~~~ (2) $$, where the conditions $$\sum_{n=1}^{\infty}| a^{(1)}_n|+ |b^{(1)}_n|<+\infty,$$ $$\sum_{n=1}^{\infty}| a^{(2)}_n|+ |b^{(2)}_n|<+\infty$$ hold true and $x \in (-\pi,\pi)$. Find the Fourier trigonometric series
$$a^{(3)}_0/2+\sum_{n=1}^{\infty}a^{(3)}_n\cos(nx)+b^{(3)}_n\sin(nx)$$

of the product $f\times g$ on the interval $(-\pi,\pi)$ by using (1)-(2).

George
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1 Answers1

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The expressions are complicated. For instance $$\begin{align} a_0^{(3)}&=\frac1\pi\int_{-\pi}^\pi \Bigl(\frac{a_0^{(1)}}{2}+\sum_{n=1}^{\infty}a^{(1)}_n\cos(nx)+b^{(1)}_n\sin(nx)\,\Bigr)\Bigl(\frac{a_0^{(2)}}{2}+\sum_{n=1}^{\infty}a^{(2)}_n\cos(nx)+b^{(2)}_n\sin(nx)\Bigr)\,dx\\ &=\frac{a_0^{(1)}a_0^{(2)}}{2}+\sum_{n=1}^{\infty}\bigl(a_n^{(1)}a_n^{(2)}+b_n^{(1)}b_n^{(2)}\bigr). \end{align}$$ The interchange of integration and summation is justified by the absolute suability of the coefficients.

The computations will be easier if you rewrite the series as complex Fourier series: $$ f(x)=\sum_{k\in\mathbb{Z}}c_k^{(1)}e^{ikx},\quad g(x)=\sum_{k\in\mathbb{Z}}c_k^{(2)}e^{ikx}. $$ Then $$ f(x)\,g(x)=\sum_{k\in\mathbb{Z}}\Bigl(\sum_{j\in\mathbb{Z}}c_j^{(1)}c_{k-j}^{(2)}\Bigr)e^{ikx}. $$

Julián Aguirre
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