Is the double fourier series just the product of the single series?

Question

Let $f(x)=\begin{cases}x & 0 \le x \le 1 \\ 2-x & 1<x\le 2 \end{cases}$.

Find the double Fourier series of $f(x)f(y)$ on $R_{2,2}$

To find the double Fourier series can I just multiply the Fourier sine series of $f(x)$ and $f(y)$?

Answer 1

As Dr. MV noted, not every double Fourier series has the product form: for example, $1+e^{\pi i(x+y)}$ does not factor.

But if you are expanding a function of the form $F(x,y)=f(x)g(y)$, then the formal multiplication of Fourier series is indeed what you want. The reason is that the double integral defining the Fourier coefficients can be factored: $$ \widehat{F}(m,n) = \int_0^2 \int_0^2 F(x,y)e^{-\pi im x}e^{-\pi i n y}\,dx\,dy = \int_0^2 e^{-\pi im x} f(x)\,dx \int_0^2 g(y)e^{-\pi i n y}\,dy = \hat f(m)\hat g(n) $$ (and similarly for other types of Fourier series).