College Math Teaching

August 31, 2014

The convolution integral: do some examples in Calculus III or not?

Filed under: applications of calculus, applied mathematics, calculus, integrable function, integrals, Uncategorized — Tags: , — collegemathteaching @ 11:31 pm

For us, calculus III is the most rushed of the courses, especially if we start with polar coordinates. Getting to the “three integral theorems” is a real chore. (ok, Green’s, Divergence and Stoke’s theorem is really just \int_{\Omega} d \sigma = \int_{\partial \Omega} \sigma but that is the subject of another post)

But watching this lecture made me wonder: should I say a few words about how to calculate a convolution integral?

Note: I’ve discussed a type of convolution integral with regards to solving differential equations here.

In the context of Fourier Transforms, the convolution integral is defined as it was in analysis class: f*g = \int^{\infty}_{-\infty} f(x-t)g(t) dt . Typically, we insist that the functions be, say, L^1 and note that it is a bit of a chore to show that the convolution of two L^1 functions is L^1 ; one proves this via the Fubini-Tonelli Theorem.

(The straight out product of two L^1 functions need not be L^1 ; e.g, consider f(x) = \frac {1}{\sqrt{x}} for x \in (0,1] and zero elsewhere)

So, assuming that the integral exists, how do we calculate it? Easy, you say? Well, it can be, after practice.

But to test out your skills, let f(x) = g(x) be the function that is 1 for x \in [\frac{-1}{2}, \frac{1}{2}] and zero elsewhere. So, what is f*g ???

So, it is easy to see that f(x-t)g(t) only assumes the value of 1 on a specific region of the (x,t) plane and is zero elsewhere; this is just like doing an iterated integral of a two variable function; at least the first step. This is why it fits well into calculus III.

f(x-t)g(t) = 1 for the following region: (x,t), -\frac{1}{2} \le x-t \le \frac{1}{2}, -\frac{1}{2} \le t \le \frac{1}{2}

This region is the parallelogram with vertices at (-1, -\frac{1}{2}), (0, -\frac{1}{2}), (0 \frac{1}{2}), (1, \frac{1}{2}) .

convolutiondraw

Now we see that we can’t do the integral in one step. So, the function we are integrating f(x-t)f(t) has the following description:

f(x-t)f(t)=\left\{\begin{array}{c} 1,x \in [-1,0], -\frac{1}{2} t \le \frac{1}{2}+x \\ 1 ,x\in [0,1], -\frac{1}{2}+x \le t \le \frac{1}{2} \\ 0 \text{ elsewhere} \end{array}\right.

So the convolution integral is \int^{\frac{1}{2} + x}_{-\frac{1}{2}} dt = 1+x for x \in [-1,0) and \int^{\frac{1}{2}}_{-\frac{1}{2} + x} dt = 1-x for x \in [0,1] .

That is, of course, the tent map that we described here. The graph is shown here:

tentmapgraph

So, it would appear to me that a good time to do a convolution exercise is right when we study iterated integrals; just tell the students that this is a case where one “stops before doing the outside integral”.

August 26, 2014

How some mathematical definitions are made

Filed under: advanced mathematics, applied mathematics, mathematician, mathematics education — Tags: , , — collegemathteaching @ 5:30 pm

I love what Brad Osgood says at 47:37.

The context: one is showing that the Fourier transform of the convolution of two functions is the product of the Fourier transforms (very similar to what happens in the Laplace transform); that is \mathcal{F}(f*g) = F(s)G(s) where f*g = \int^{\infty}_{-\infty} f(x-t)g(t) dt

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