# College Math Teaching

## June 25, 2015

### Workshop in Geometric Topology: TCU 2015 morning session 1

Filed under: advanced mathematics, conference, editorial, topology — Tags: — collegemathteaching @ 3:59 pm

I’ll be blunt: I’ve been teaching at a 11-12 hour load (mostly 11; one time I had a 9 hour load; 3 courses) since fall, 1991. Though I’ve published, most of what I’ve done has been extremely “bare handed”; it is tough to learn the most advanced techniques (which is a full time job in and of itself)

So, at math conferences, I get to see how much further behind I’ve fallen.

But these things help in the following way:

1. They are an excellent change of pace from the usual routine of teaching calculus.
2. I do learn things, even if it is “looking up” a definition or two; for example I looked up the definition of “pure braid group” in between the 20 minute talks.
3. I have to review my own stuff to see if I am indeed making progress; I don’t want to say something idiotic in front of some very smart, informed people.

But yes, the talks have been given by smart, (mostly) young, energetic people who have been studying the topic that they are talking about very intensely for a long time; frequently it is tough to hang in to the second half of the 20 minute talks. But I can see WHAT is being studied, what tools are being used and, as I said before, find stuff to look up.

The final talk: didn’t understand much beyond the general gist but it was well organized, well presented..exactly what you get when you have a brilliant energetic young researcher working full time in mathematical research.

On one hand, I envy his talent. On the other hand, I am glad that we have some smart humans among us; they benefit all of us.

The trip here The plane was about 2.5 hours late getting in, then there was a long ride to the car rental place and a 35 minute drive to campus, then finding my way around in the dark. So no morning run; I might do a gentle “after the talks” focused walk (5K-ish?).

I talk at 9 am tomorrow and I want to make it worth their while.

## June 19, 2015

### Scientific American article about finite simple groups

Filed under: advanced mathematics, algebra, mathematician — Tags: , — collegemathteaching @ 2:42 pm

For those of you who are a bit rusty: a finite group is a group that has a finite number of elements. A simple group is one that has no proper non-trivial normal subgroups (that is, only the identity and the whole group are normal subgroups).

It is a theorem that if $G$ is a finite simple group then $G$ falls into one of the following categories:

1. Cyclic (of prime order, of course)
2. Alternating (and not isomorphic to $A_4$ of course)
3. A member of a subclass of Lie Groups
4. One of 26 other groups that don’t fall into 1, 2 or 3.

Scientific American has a nice article about this theorem and the effort to get it written down and understood; the problem is that the proof of such a theorem is far from simple; it spans literally hundreds of research articles and would take thousands of pages to be complete. And, those who have an understanding of this result are aging and won’t be with us forever.

Here is a link to the preview of the article; if you don’t subscribe to SA it is probably in your library.

## June 9, 2015

### Volumes of n-balls: why?

Filed under: calculus, geometry — Tags: , , — collegemathteaching @ 11:38 am

I talked about a method of finding the hypervolume of an n-ball in a previous post. To recap: the volume of the n-ball is that “hypervolume” (I’ll be dropping “hyper”) of the region described by $\{(x_1, x_2,...x_n) | x_1^2 + x_2^2 + ...x_n^2 \leq R^2 \}$.

The formula is: $V_n = \frac{\pi^{\frac{n}{2}}}{\Gamma[\frac{n}{2} + 1]} R^n$

Here, we’ll explore this topic further, both giving a different derivation (from Greg Huber’s American Mathematical Monthly paper) and make some informal observations.

Derivation: the argument I present can be tweaked to produce a formal induction argument that, if the volume is $V_n$, it is proportional to the n’th power of the radius $R$.

Now note that if the surface area of the $n-1$ sphere is given by $W_{n-1} = w_{n-1}R^{n-1}$ we have, from the theory of differentials, $\frac{dV_n}{dR} = W_{n-1}$. Think of taking a sphere and adding just a bit $\Delta R$ to its radius; you obtain a shell of thickness $\Delta R$ all the way around the sphere which is roughly equal to the surface area times $\Delta R$

So we can rewrite this as $V_n = \int^R_0 W_{n-1} dr = \int^R_0 w_{n-1}r^{n-1} dr = w_{n-1}\int^R_0r^{n-1} dr$

To see what comes next, we first write this same quantity in two different ways:

$\int \int \int ...\int_{S^{n-1}} \Pi^{n}_{i=1} dx_i = \int^R_0 w_{n-1}r^{n-1} dr = w_{n-1}\int^R_0r^{n-1} dr = \int \int \int ...\int^R_{0} r^{n-1} J(\theta_1, ..\theta_{n-1}) dr$.

The first integral is integral in rectangular coordinates within the boundary of the n-1 sphere. The rightmost integral is the same integral in generalized spherical coordinates (see Blumenson’s excellent Monthly article) where the first iterated integrals are those with angular limits with $J$ being the angular volume element. The middle integral is the volume integral. All are equal to the volume of the n-ball. The key here is that the iterated integrals evaluated over the entire n-1 sphere are equal to $w_{n-1}$.

Now integrate $e^{-r^2}$ over the region bounded by the sphere $r^2 = x_1^2 + x_2^2 + ...x_n^2$, noting that $e^{-r^2} = e^{-x_1^2}e^{-x_2^2}...e^{-x_n^2}$:

$\int \int \int ...\int_{S^{n-1}} \Pi^{n}_{i=1}e^{-x_i^2} dx_i = w_{n-1}\int^R_0e^{-r^2}r^{n-1} dr = \int \int \int ...\int^R_{0} r^{n-1}e^{-r^2}J(\theta_1, ..\theta_{n-1}) dr$

Equality holds between the middle and right integral because in “angle/r” space, the r and angular coordinates are independent. Equality between the leftmost and rightmost integrals holds because this is a mere change of variables.
So we can now drop the rightmost integral. Now take a limit as $R \rightarrow \infty$:

$\int \int \int ...\int_{R^{n}} \Pi^{n}_{i=1}e^{-x_i^2} dx_i = (\int^{\infty}_{-\infty} e^{-x^2} dx)^n = w_{n-1}\int^{\infty}_0e^{-r^2}r^{n-1} dr$

The left integral is just the n-th power of the Gaussian integral and is therefore $\pi^{\frac{n}{2}}$ and a substitution $r = \sqrt{u}$ turns this into $\frac{w_{n-1}}{2}\int^{\infty}_{0} u^{\frac{1}{2} -1}e^{-u}du = w_{n-1} \frac{1}{2}\Gamma[\frac{1}{2}]$ (recall $\Gamma[x] = \int^{\infty}_{0} t^{x-1}e^{-t} dt$ ).

So $w_{n-1} = \frac{2 \pi^{\frac{n}{2}}}{\Gamma[\frac{n}{2}]}$ and hence, by integration, $v_n = \frac{2 \pi^{\frac{n}{2}}}{n\Gamma[\frac{n}{2}]}= \frac{ \pi^{\frac{n}{2}}}{n\Gamma[\frac{n}{2}+1]}$

Now $v_n =V_n$ when $R = 1$. $\frac{v_n}{2^n}$ can be thought of as the percentage of the cube with vertexes $(\pm 1, \pm 1, ... \pm 1)$ that is taken up by the inscribed unit sphere.

Now we set $R = 1$ at look at a graph of hypervolume vs. n:

The first graph is the ratio of the volume taken up by the ball verses the hypervolume of the hyper cube that the ball is inscribed in.

Next we see the the hypervolume peaks at n = 5 (max is between 5 and 6 )and then starts to decline to zero. Of course there has to be an inflection point somewhere; it turns out to be between n = 10 and n = 11.

Now we plot the hyperarea of the n-1 sphere vs. the hypervolume of the ball that it bounds; we see that more and more of the hypervolume of the ball is concentrated near the boundary as the dimension goes up.

For more: see the interesting discussion on Division by Zero.