# College Math Teaching

## August 21, 2014

### Calculation of the Fourier Transform of a tent map, with a calculus tip….

I’ve been following these excellent lectures by Professor Brad Osgood of Stanford. As an aside: yes, he is dynamite in the classroom, but there is probably a reason that Stanford is featuring him. 🙂

And yes, his style is good for obtaining a feeling of comradery that is absent in my classroom; at least in the lower division “service” classes.

This lecture takes us from Fourier Series to Fourier Transforms. Of course, he admits that the transition here is really a heuristic trick with symbolism; it isn’t a bad way to initiate an intuitive feel for the subject though.

However, the point of this post is to offer a “algebra of calculus trick” for dealing with the sort of calculations that one might encounter.

By the way, if you say “hey, just use a calculator” you will be BANNED from this blog!!!! (just kidding…sort of. 🙂 )

So here is the deal: let $f(x)$ represent the tent map: the support of $f$ is $[-1,1]$ and it has the following graph: The formula is: $f(x)=\left\{\begin{array}{c} x+1,x \in [-1,0) \\ 1-x ,x\in [0,1] \\ 0 \text{ elsewhere} \end{array}\right.$

So, the Fourier Transform is $F(f) = \int^{\infty}_{-\infty} e^{-2 \pi i st}f(t)dt = \int^0_{-1} e^{-2 \pi i st}(1+t)dt + \int^1_0e^{-2 \pi i st}(1-t)dt$

Now, this is an easy integral to do, conceptually, but there is the issue of carrying constants around and being tempted to make “on the fly” simplifications along the way, thereby leading to irritating algebraic errors.

So my tip: just let $a = -2 \pi i s$ and do the integrals: $\int^0_{-1} e^{at}(1+t)dt + \int^1_0e^{at}(1-t)dt$ and substitute and simplify later:

Now the integrals become: $\int^{1}_{-1} e^{at}dt + \int^0_{-1}te^{at}dt - \int^1_0 te^{at} dt.$
These are easy to do; the first is merely $\frac{1}{a}(e^a - e^{-a})$ and the next two have the same anti-derivative which can be obtained by a “integration by parts” calculation: $\frac{t}{a}e^{at} -\frac{1}{a^2}e^{at}$; evaluating the limits yields: $-\frac{1}{a^2}-(\frac{-1}{a}e^{-a} -\frac{1}{a^2}e^{-a}) - (\frac{1}{a}e^{a} -\frac{1}{a^2}e^a)+ (-\frac{1}{a^2})$

Add the first integral and simplify and we get: $-\frac{1}{a^2}(2 - (e^{-a} -e^{a})$. NOW use $a = -2\pi i s$ and we have the integral is $\frac{1}{4 \pi^2 s^2}(2 -(e^{2 \pi i s} -e^{-2 \pi i s}) = \frac{1}{4 \pi^2 s^2}(2 - cos(2 \pi s))$ by Euler’s formula.

Now we need some trig to get this into a form that is “engineering/scientist” friendly; here we turn to the formula: $sin^2(x) = \frac{1}{2}(1-cos(2x))$ so $2 - cos(2 \pi s) = 4sin^2(\pi s)$ so our answer is $\frac{sin^2( \pi s)}{(\pi s)^2} = (\frac{sin(\pi s)}{\pi s})^2$ which is often denoted as $(sinc(s))^2$ as the “normalized” $sinc(x)$ function is given by $\frac{sinc(\pi x)}{\pi x}$ (as we want the function to have zeros at integers and to “equal” one at $x = 0$ (remember that famous limit!)

So, the point is that using $a$ made the algebra a whole lot easier.

Now, if you are shaking your head and muttering about how this calculation was crude that that one usually uses “convolution” instead: this post is probably too elementary for you. 🙂