Define a 1-periodic function on $\mathbb{R}$ by:

$f(x) :=$ $\left\{\begin{matrix} 1 & if & 0<x<\frac{1}{10}\\ 0 & if & \frac{1}{10}<x<1 \end{matrix}\right.$

with Fourier series $f(x) = \sum_{-\infty}^{\infty}c_k e^{2\pi ikx}$.

I'm trying to find the graphs of the following two Fourier series:

$\sum_{-\infty}^{\infty}(c_k)^2 e^{2\pi ikx}$ and $\sum_{-\infty}^{\infty}c_{3k} e^{2\pi ikx}$.

Basically my strategy with the others has been to find a way to get these Fourier series into the form $a\sum_{-\infty}^{\infty}c_k e^{2\pi ik(bx)}$ for $a,b\in\mathbb{C}$. But I haven't been able to do it with these, hopefully someone can help, thanks.


2 Answers 2


This doesn't require any complicated computations! Look up some tables of identities about the Fourier Series. Which operation would you have to apply to $f$ so that its Fourier series would become $\sum_{-\infty}^{\infty}(c_k)^2 e^{2\pi ikx}$?

For example, we know that derivating in the time domain is equivalent to multiplying the Fourier series coefficients by $i k$ ; so we know that the graph of the Fourier series $\sum_{-\infty}^{\infty} ik (c_k) e^{2\pi ikx}$ is the graph of the derivative of $f$.

  • $\begingroup$ Ok so I found an identity which says that $F(x)=\int_0^1 f(t)f(x-t)dt$ has coefficients $(c_k)^2$, and I think F(x) comes out to be a 1-periodic function zero everywhere but for $\frac{1}{10}<x<\frac{2}{10}$ where it is given by the function $\frac{-3}{2}x + \frac{2}{10}$. Does this look correct? $\endgroup$ Apr 9, 2012 at 0:10
  • $\begingroup$ Right approach but the $F(x)$ you find (which is the convolution of $f$ by itself) is not correct. Maybe you have forgotten to take into account that $f$ is periodic when computing your integral. The interpretation of your integral is: $F(a)$ is the area, over on period, of the area common to the graph of $f$ and the graph of $f$ reflected along the $x = \frac{a}{2}$ axis. Visualize $f$ and its reflected version sliding on top of each other, and how the area of the intersection evolves. First no overlap, then a maximum overlap when $a = \frac{1}{10}$, then no overlap. $\endgroup$ Apr 9, 2012 at 0:28
  • $\begingroup$ What about for the other one, c_3k, how should I approach that one? $\endgroup$ Apr 9, 2012 at 14:25
  • $\begingroup$ Start from the expression giving $c_k$ as an integral of $f(x)$ and a complex exponential. Which transformation do you have to apply to $f$ to get the $c_{3k}$ coefficient instead? $\endgroup$ Apr 9, 2012 at 14:38
  • $\begingroup$ Well we have $c_k = \int_0^{\frac{1}{10}}e^{-2\pi ikx}dx$. Thus what I would like to do is multiply $f(x)$ by $e^{-4\pi ikx}$, but I can't include $k$ in my transformation of $f(x)$, thus I'm not sure what to do.. $\endgroup$ Apr 9, 2012 at 16:32

Is it cheating to inverse the Fourier transform to obtain an expression for the $c_k$?

  1. Find the graph of the inverse function (compute the inverse Fourier transform)

  2. Transform the inverse function according to the transformation of the $c_k$

  3. Compute the Fourier transform of this result, yielding the same expression.

So, if you could find the graph of $c_{3k}$ and $(c_k)^2$, you could find the graph of those expressions by taking the Fourier transform.

My immediate instinct was to identify the time-domain transformation and to analytically work out how this changes the frequencies and the amplitudes. Clearly the $c_{3k}$ option is going to see a shift in the frequencies, and no change in amplitude where $(c_k)^2$ is going to see a shift in the amplitude but no shift in the frequencies.

  • $\begingroup$ Ya I think it's cheating to do that, supposedly this is doable without knowing c_k explicitly. $\endgroup$ Apr 8, 2012 at 22:28

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