1
$\begingroup$

Let $f$ and $g$ be $2\pi$-periodic, piece-wise smooth functions having Fourier series $f(x)=\sum_n\alpha_ne^{inx}$ and $g(x)=\sum_n\beta_ne^{inx}$, and define the convolution of $f$ and $g$ to be $f\star g(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)g(x-t)dt$. Show that the complex form of the Fourier series for $f\star g$ is $$f\star g(x)=\sum_{n=-\infty}^{\infty}\alpha_n\beta_ne^{inx}.$$

I have been trying different approaches to this for a while but I haven't been able to figure it out. I think that I am supposed to use the fact that $$\frac{1}{2\pi}\langle f,g\rangle=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\overline{g(t)}dt=\sum_{n=-\infty}^{\infty}\alpha_n\overline{\beta_n}$$ but I can't figure out how to manipulate the algebra to show the correct result. I may be forgetting some key property of the inner product or something.

$\endgroup$
  • 1
    $\begingroup$ Try convolving the Fourier series expressions instead. Note that the convolution operator is linear, so you just have to convolve two exponentials. $\endgroup$ – Cameron Williams Feb 2 '16 at 2:39
  • $\begingroup$ I have, but that gets me $\frac{1}{2\pi}\int_{-\pi}^{\pi}\sum_n\alpha_ne^{int}\sum_n\beta_ne^{in(x-t)}dt$ and I don't know how to simplify that $\endgroup$ – Matt G Feb 2 '16 at 2:40
  • $\begingroup$ Bring out the sums (and use different indices on the sums!). $\endgroup$ – Cameron Williams Feb 2 '16 at 2:41
  • $\begingroup$ Bring them outside the integral? $\endgroup$ – Matt G Feb 2 '16 at 2:53
  • 1
    $\begingroup$ Matt. You need only express one of the functions as a Fourier series. I posted a way forward that avoids the "messiness" of the double sum and ensuing manipulation thereof. - Mark $\endgroup$ – Mark Viola Feb 2 '16 at 3:55
3
$\begingroup$

Let the Fourier Series representation for $f$ and $g$ be written respectively by

$$f(x)=\sum_{n=-\infty}^\infty \alpha_ne^{inx}$$

$$g(x)=\sum_{n=-\infty}^\infty \beta_ne^{inx}$$

with coefficients $\alpha_n$ and $\beta_n$ given by

$$\alpha_n=\frac1{2\pi}\int_{-\pi}^\pi f(t)e^{-int}\,dt$$

$$\beta_n=\frac1{2\pi}\int_{-\pi}^\pi g(t)e^{-int}\,dt$$

The convolution of $f$ and $g$ is defined as

$$\begin{align} (f*g)(x)&\equiv \frac1{2\pi}\int_{-\pi}^\pi f(t)g(x-t)\,dt\\\\ &=\frac1{2\pi}\int_{-\pi}^\pi f(t)\left(\sum_{n=-\infty}^\infty \beta_n e^{in(x-t)}\right)\,dt\\\\ &=\sum_{n=-\infty}^\infty \beta_n e^{inx}\left(\frac1{2\pi}\,\int_{-\pi}^\pi f(t)e^{-int}\,dt\right)\\\\ &=\sum_{n=-\infty}^\infty \alpha_n\,\beta_n e^{inx} \end{align}$$

and we are done!

$\endgroup$
  • $\begingroup$ Nice answer :) I haven't seen this approach before. $\endgroup$ – Cameron Williams Feb 2 '16 at 4:56
  • $\begingroup$ @CameronWilliams Thank you! And glad to see you're back Cam. - Mark $\endgroup$ – Mark Viola Feb 2 '16 at 5:24
  • $\begingroup$ Yeah I've been in a bit of a rut lately so I've been pretty scarce around here. But it has gotten better now :) $\endgroup$ – Cameron Williams Feb 2 '16 at 6:02
3
$\begingroup$

Since Dr. MV already posted an answer, I'll post the one I alluded to in my comment:

Convolving $f$ with $g$ gives

\begin{align} (f*g)(x) &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \left(\sum_m \alpha_m e^{imt}\right) \left(\sum_n \beta_n e^{in(x-t)}\right)\,dt \\ &= \frac{1}{2\pi} \sum_m\sum_n \alpha_m\beta_n e^{inx} \int_{-\pi}^{\pi} e^{i(m-n)t}\,dt. \end{align}

This integral evaluates to $2\pi$ if $m=n$ and $0$ otherwise. (Check this yourself by breaking it into these two cases.) Meaning that the double sum is zero if $m\neq n$ so really you can drop down to one sum to get

$$(f*g)(x) = \sum_n \alpha_n \beta_n e^{inx}.$$

$\endgroup$
  • $\begingroup$ How can you combine the two sums like that? $\endgroup$ – Matt G Feb 2 '16 at 17:53
  • 1
    $\begingroup$ Try working it out with finite sums to see why it works. $\endgroup$ – Cameron Williams Feb 2 '16 at 17:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.