This question is from the book "Advanced Engineering Mathematics" by Stroud.
I can't seem to get the required answer for this. I've derived the two Fourier transform equations for them.
.
U and V are dummy variables.
I've been told I need to use the convolution theorem by multiplying the functions in the frequency domain and finding the inverse transform of the solution, but I don't know how to proceed with the equations I've got.
Best Answer
The Fourier transform is a linear operator, so
$$ F_1(\omega) = \sum_{n=-\infty}^{\infty} c_n \mathcal{F}(e^{jn\omega_0 t}) = \sum_{n=-\infty}^{\infty} c_n \delta( \omega-n\omega_0) $$
Use the above to multiply the two Fourier transforms. Then see if you can take the take the inverse transform to finish the proof.
EDIT:
To see how to find the product of the two sums note that
$$ \delta( \omega - n \omega_0) \cdot \delta( \omega - m \omega_0) \neq 0 \Leftrightarrow n = m. $$
Furthermore,
$$ m=n \Rightarrow \delta( \omega - n \omega_0) \cdot \delta( \omega - m \omega_0) = \delta( \omega - n \omega_0) = \delta( \omega - m \omega_0). $$
Thus,
$$ d_m \delta( \omega - m\omega_0) \cdot \sum_{n=-\infty}^{\infty} c_n \delta( \omega-n\omega_0) = d_m \delta( \omega - m\omega_0) \cdot c_m \delta( \omega - m\omega_0) = d_m c_m \delta( \omega - m\omega_0) $$
EDIT 2:
We know that
$$ F_1(\omega) = \sum_{n=-\infty}^{\infty} c_n \delta( \omega-n\omega_0) $$
$$ F_2(\omega) = \sum_{m=-\infty}^{\infty} d_m \delta( \omega-m\omega_0) $$
Their product is then
$$ F_1 (\omega) F_2 (\omega) = \left( \sum_{n=-\infty}^{\infty} c_n \delta( \omega-n\omega_0) \right) \left( \sum_{m=-\infty}^{\infty} d_m \delta( \omega-m\omega_0) \right) $$
By the definition of the Dirac delta function, we know that
$$ \delta(\omega - a) \cdot \delta( \omega - b) = \left\{ \begin{array}{lr} \delta(x-a), & a = b \\ 0, & a \neq b \end{array} \right. $$
Thus, if we consider an individual term in the series $F_1 (\omega)$ by fixing $n$ and take its product with $F_2(\omega)$, we get
$$ c_n \delta(\omega - n\omega_0) \left( \sum_{m=-\infty}^{\infty} d_m \delta( \omega-m \omega_0) \right) = c_n d_n \delta(\omega - n\omega_0). $$
Since this is true for all $n$, we have that
$$ F_1 (\omega) F_2 (\omega) = \sum_{n=-\infty}^{\infty} c_n d_n \delta( \omega-n \omega_0). $$