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.

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

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 $q$