College Math Teaching

April 18, 2012

Construction of the antiderivative of sqrt(r^2 – x^2) without trig substitution

Filed under: calculus, integrals, integration by substitution, pedagogy, popular mathematics — collegemathteaching @ 1:15 am

The above figure shows the graph of y = \sqrt{r^2 - x^2} for r = 1 .
Let A(t) = \int^t_0 \sqrt{r^2 - x^2} dx This gives the area bounded by the graphs of y=0, x = t, x =0, y = \sqrt{r^2 - x^2} .

We break this down into the wedge shaped area and the triangle.
Wedge shaped area: has area equal to \frac{1}{2} \theta r^2 where \theta is the angle made by the y axis and the radius.
Triangle shaped area: has area equal to \frac{1}{2}xy = \frac{1}{2}t\sqrt{r^2 - t^2}
But we can resolve \theta = arcsin(\frac{t}{r}) so the wedge shaped area is \frac{1}{2}r^2arcsin(\frac{t}{r}) hence:
A(t) = \int^t_0 \sqrt{r^2 - x^2} dx = \frac{1}{2}r^2arcsin(\frac{t}{r}) + \frac{1}{2}t\sqrt{r^2 - t^2}

We can verify this by computing A'(t) = \frac{1}{2} \frac{1}{r}\frac{r^2}{\sqrt{ 1- (\frac{t}{r})^2}} + \frac{1}{2}\sqrt{r^2 - t^2}-t^2\frac{1}{2}\frac{1}{\sqrt{ r^2- t^2}}
=  \frac{1}{2} \frac{r^2}{\sqrt{ r^2- t^2}}-t^2\frac{1}{2}\frac{1}{\sqrt{ r^2- t^2}} + \frac{1}{2}\sqrt{r^2 - t^2}
= \frac{1}{2} \frac{r^2 - t^2}{\sqrt{ r^2- t^2}} + \frac{1}{2}\sqrt{r^2 - t^2} = \sqrt{r^2 - t^2} as required.

April 17, 2012

Pointwise versus Uniform convergence of sequences of continuous functions: Part II

Filed under: analysis, calculus, complex variables, uniform convergence — collegemathteaching @ 12:48 am

In my complex analysis class I was grading a problem of the following type:
given K \subset C where K is compact and given a sequence of continuous functions f_n which converges to 0 pointwise on K and if |f_1(z)| > |f_2(z)|>...|f_k(z)|... for all z \in K show that the convergence is uniform.

In a previous post, I talked about why it was important that each f_k(z) be continuous.

Now what about the |f_1(z)| > |f_2(z)|>...|f_k(z)|... hypothesis? Can it be dispensed with?

Answer: well, no.

Let’s look at an example in real variables:

Let f_n(x) = sin(\frac{e \pi}{2}e^{-nx})sin(\frac{n \pi}{2} x) with x \in [0,1] . f_n(0) = 0 for all n . To see that f_n converges to zero pointwise, note that lim_{n \rightarrow \infty}e^{-nx} = 0 for all x > 0 , hence lim_{n \rightarrow \infty}sin(\frac{e \pi}{2}e^{-nx}) = 0 which implies that f_n \rightarrow 0 by the squeeze theorem. But f_n does not converge to 0 uniformly as for t = \frac{1}{n} we have f_n(t) = 1

Here is a graph of the functions for n = 5, 10, 20, 40

April 12, 2012

Pointwise vs. Uniform convergence for functions: Importance of being continuous

In my complex analysis class I was grading a problem of the following type:
given K \subset C where K is compact and given a sequence of continuous functions f_n which converges to 0 pointwise on K and if |f_1(z)| > |f_2(z)|>...|f_k(z)|... for all z \in K show that the convergence is uniform.

The proof is easy enough to do; my favorite way is to pick \epsilon > 0 for a given z \in K and n such that |f_n(z)| < \epsilon find a “delta disk” about z so that for all w in that disk, |f_n(w)| < \epsilon also. Then cover K by these open “delta disks” and then one can select a finite number of such disks, each with an associated n and then let M be the maximum of this finite collection of n .

But we used the fact that f_n is continuous in our proof.

Here is what can happen if the f_n in question are NOT continuous:

Let’s work on the real interval [0,1] . Define g(x) = q if x = \frac{p}{q} in lowest terms, and let g(x) = 0 if x is irrational.

Now let f_n(x) = \frac{g(x)}{n} . Clearly f_n converges to 0 pointwise and the f_n have the decreasing function property. Nevertheless, it is easy to see that the convergence is far from uniform; in fact for each n, f_n is unbounded!

Of course, we can also come up with a sequence of bounded functions that converge to 0 pointwise but fail to converge uniformly.

For this example, choose as our domain [0,1] and let h(x) = \frac{q-1}{q} if x = \frac{p}{q} in lowest terms, and let h(x) = 0 if x is irrational. Now let our sequence f_n(x) = h(x)^n . Clearly f_n converges to zero pointwise. To see that this convergence is not uniform: let \epsilon > 0 be given and if (\frac{q-1}{q})^n < \epsilon, n > \frac{ln(\epsilon)}{ln(\frac{q-1}{q})} and the right hand side of the inequality varies with q and is, in fact, unbounded. Given a fixed n and \epsilon one can always find a q to exceed the fixed n . Hence n varies with q

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