Fourier series solutions V - Computing some Fourier series

#differential_equations

It is a good idea to see computations of some Fourier series of functions that aren't built from sines and cosines. To that end, we consider two simple examples: 1) a "square wave"; and 2) a "triangular wave".

Example: A square wave

Consider the piecewise-defined function $f (t)$ is periodic with period 1, and which on $[0, 1)$ is defined by

f (t) = {\begin{cases} 1, & if 0 \leq t < \frac{1}{2} \\ - 1, & if \frac{1}{2} \leq t < 1 \end{cases}

In other words, this function alternates between the values of $1$ and $- 1$ over intervals of length $\frac{1}{2}$ . Although this function is discontinuous (where it "jumps" between the values of $1$ and $- 1$ ), if we suggestively connect the values across those jumps with vertical lines we get the following visualization of the graph of $f$ :

Based on this image, it's reasonable to call this graph a "rectangular wave", but for some reason the descriptor "square wave" is more popular.

In any case, let's compute the complex Fourier series associated to this function $f$ . We first compute

\hat{f} (0) = \int_{0}^{1} f (t) d t = \int_{0}^{\frac{1}{2}} 1 d t + \int_{\frac{1}{2}}^{1} - 1 d t = 0.

For $n \neq 0$ we compute

\begin{aligned} \hat{f} (n) & = \int_{0}^{1} f (t) e^{- 2 π i n t} d t \\ = \int_{0}^{\frac{1}{2}} e^{- 2 π i n t} d t - \int_{\frac{1}{2}}^{1} e^{- 2 π i n t} d t \\ = - \frac{1}{2 π i n} (e^{- π i n} - 1) + \frac{1}{2 π i n} (e^{- 2 π i n} - e^{- π i n}) \\ = \frac{1}{π i n} (1 - e^{- π i n}) \\ = \frac{1}{π i n} (1 - (- 1)^{n}) \\ = {\begin{cases} 0, & if n even \\ \frac{2}{π i n}, & if n odd \end{cases} \end{aligned}

Note that we computed the integrals using $u$ -substitutions, and simplified some expressions using the facts that $e^{2 π i k} = 1$ whenever $k$ is an integer, while $e^{π i k}$ is $1$ when $k$ is even and $- 1$ when $k$ is odd; i.e., $e^{π i k} = (- 1)^{k}$ .

It follows that that Fourier series associated to our function $f$ is

\begin{aligned} \sum_{n = - \infty}^{\infty} \hat{f} (n) e^{2 π i n t} & = \sum_{n odd} \frac{2}{π i n} e^{2 π i n t} \\ = \dots - \frac{2}{π i \cdot 3} e^{- 6 π i t} - \frac{2}{π i} e^{- 2 π i t} + \frac{2}{π i} e^{2 π i t} + \frac{2}{π i \cdot 3} e^{6 π i t} + \dots \end{aligned}

If we group together the term at index $n$ with the term at index $- n$ and use the identity $e^{2 π i n t} - e^{- 2 π i n t} = 2 i \sin (2 π n t)$ , we can rewrite the above series in the sines-and-cosines form as

\sum_{n odd} \frac{2}{π i n} e^{2 π i n t} = \sum_{odd n > 0} \frac{4}{n π} \sin (2 π n t)

Some Fourier approximations

Let's truncate our Fourier series after various amounts of terms and see how well the corresponding finite Fourier series approximates our function $f$ . If we only use the first term, the finite Fourier series we obtain is

S_{1} (t) = \frac{4}{π} \sin (2 π t),

whose graph is

If we truncate instead at $N = 9$ (i.e., include all terms with $n \leq 9$ ), the finite Fourier series we obtain is

S_{9} (t) = \frac{4}{π} \sin (2 π t) + \frac{4}{3 π} \sin (6 π t) + \frac{4}{5 π} \sin (10 π t) + \frac{4}{7 π} \sin (14 π t) + \frac{4}{9 π} \sin (18 π t),

whose graph is

Finally, if we truncate our series at $N = 99$ , the graph of the corresponding finite Fourier series we obtain is

Although our "Fourier approximations" appear to be converging quickly to the original square wave, you might notice those weird little "peaks" at the corners of each wave. There's something a little surprising going on here, but we'll hold off on investigating further until after our next example.

Example: A triangular wave

Now $f (t)$ denote the "triangle wave" function, which is periodic with period 1 and is defined on $[0, 1)$ by

f (t) = {\begin{cases} t, & 0 \leq t < \frac{1}{2} \\ 1 - t, & \frac{1}{2} \leq t < 1 \end{cases}

As you can see from the graph, it seems reasonable to call this function a "triangle wave":

We first compute

\hat{f} (0) = \int_{0}^{1} f (t) d t = \int_{0}^{1 / 2} t d t + \int_{1 / 2}^{1} (1 - t) d t = \frac{1}{4} .

For $n \neq 0$ , we have

\hat{f} (n) = \int_{0}^{1} f (t) e^{- 2 π i n t} d t = \int_{0}^{1 / 2} t e^{- 2 π i n t} d t + \int_{1 / 2}^{1} (1 - t) e^{- 2 π i n t} d t .

To compute these, we make the substitution $u = - 2 π i n t$ (so $d u = - 2 π i n d t$ ) and use integration by parts, obtaining

\begin{aligned} \hat{f} (n) & = - \frac{1}{4 π n^{2}} \int_{0}^{- π i n} u e^{u} d u - \frac{1}{2 π i n} \int_{- π i n}^{- 2 π i n} e^{u} d u + \frac{1}{4 π^{2} n^{2}} \int_{- π i n}^{- 2 π i n} u e^{u} d u \\ = - \frac{1}{4 π^{2} n^{2}} {[u e^{u} - e^{u}]}_{0}^{- π i n} + \frac{i}{2 π n} {[e^{u}]}_{- π i n}^{- 2 π i n} + \frac{1}{4 π^{2} n^{2}} {[u e^{u} - e^{u}]}_{- π i n}^{- 2 π i n} \\ = - \frac{1}{4 π^{2} n^{2}} (- π i n e^{- π i n} - e^{- π i n} + 1) + \frac{i}{2 π n} (e^{- 2 π i n} - e^{- π i n}) \\ + \frac{1}{4 π^{2} n^{2}} (- 2 π i n e^{- 2 π i n} - e^{- 2 π i n} + π i n e^{- π i n} + e^{- π i n}) \end{aligned}

To simplify this mess, recall that $e^{- 2 π i n} = 1$ for all $n$ , and that $e^{- π i n} = {(e^{- π i})}^{n} = (- 1)^{n}$ . So when $n$ is even, the above expression simplifies to

\begin{aligned} \hat{f} (n) & = - \frac{1}{4 π^{2} n^{2}} (- π i n - 1 + 1) + \frac{i}{2 π n} (1 - 1) + \frac{1}{4 π^{2} n^{2}} (- 2 π i n - 1 + π i n + 1) \\ = 0. \end{aligned}

When $n$ is odd, the expression instead simplifies to

\begin{aligned} \hat{f} (n) & = - \frac{1}{4 π^{2} n^{2}} (π i n + 1 + 1) + \frac{i}{2 π n} (1 + 1) + \frac{1}{4 π^{2} n^{2}} (- 2 π i n - 1 - π i n - 1) \\ = - \frac{1}{π^{2} n^{2}} . \end{aligned}

We've shown that the Fourier series for $f (t)$ is

\begin{aligned} \sum_{n = - \infty}^{\infty} \hat{f} (n) e^{2 π i n t} & = \frac{1}{4} - \frac{1}{π^{2}} \sum_{odd n} \frac{1}{n^{2}} e^{2 π i n t} \\ = \frac{1}{4} - \frac{1}{π^{2}} \sum_{odd n > 0} (\frac{1}{n^{2}} e^{2 π i n t} + \frac{1}{(- n)^{2}} e^{2 π i (- n) t}) \\ = \frac{1}{4} - \frac{1}{π^{2}} \sum_{odd n > 0} \frac{1}{n^{2}} (e^{2 π i n t} + e^{- 2 π i n t}) \\ = \frac{1}{4} - \frac{2}{π^{2}} \sum_{odd n > 0} \frac{1}{n^{2}} \cos (2 π n t) . \end{aligned}

Some Fourier approximations

As with the square wave, we can get a sense for how quickly our Fourier series is converging to the original function by looking at some Fourier approximations with increasingly larger $N$ values (i.e., more and more terms included). Here are the graphs of those approximations when $N = 1, 9$ and $99$ , this time with the original graph overlaid for visual comparison:

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Fourier series solutions VI - Inner product spaces