# College Math Teaching

## May 31, 2015

### And a Fields Medalist makes me feel better

Filed under: calculus, editorial, elementary mathematics, popular mathematics, topology — Tags: — collegemathteaching @ 10:30 pm

I have subscribed to Terence Tao’s blog.

$\frac{d^{k+1}}{dx^{k+1}}(1+x^2)^{\frac{k}{2}}$ for $k \in \{1, 2, 3, ... \}$. Try this yourself and surf to his post to see the “slick, 3 line proof”.

But that really isn’t the point of this post.

This is the point: I often delight in finding something “fun” and “new to me” about an established area. I thought “well, that is because I am too dumb to do the really hard stuff.” (Yes, I’ve published, but my results are not Annals of Mathematics caliber stuff. 🙂 )

But I see that even the smartest, most accomplished among us can delight in the fun, simple things.

That makes me feel better.

Side note: I haven’t published much on this blog lately, mostly because I’ve been busy updating this one. It is a blog giving notes for my undergraduate topology class. That class was time consuming, but I had the teaching time of my life. I hope that my students enjoyed it too.

## May 11, 2015

### The hypervolume of the n-ball enclosed by a standard n-1 sphere

I am always looking for interesting calculus problems to demonstrate various concepts and perhaps generate some interest in pure mathematics.
And yes, I like to “blow off some steam” by spending some time having some non-technical mathematical fun with elementary mathematics.

This post uses only:

1. Integration by parts and basic reduction formulas.
2. Trig substitution.
3. Calculation of volumes (and hyper volumes) by the method of cross sections.
4. Induction
5. Elementary arithmetic involving factorials.

The quest: find a formula that finds the (hyper)volume of the region $\{(x_1, x_2, x_3,....x_k) | \sum_{i=1}^k x_i^2 \leq R^2 \} \subset R^k$

We will assume that the usual tools of calculus work as advertised.

Start. If we done the (hyper)volume of the k-ball by $V_k$ we will start with the assumption that $V_1 = 2R$; that is, the distance between the endpoints of $[-R,R]$ is $2R$.

Step 1: we show, via induction, that $V_k =c_kR^k$ where $c_k$ is a constant and $R$ is the radius.

Our proof will be inefficient for instructional purposes.

We know that $V_1 =2R$ hence the induction hypothesis holds for the first case and $c_1 = 2$. We now go to show the second case because, for the beginner, the technique will be easier to follow further along if we do the $k = 2$ case.

Yes, I know that you know that $V_2 = \pi R^2$ and you’ve seen many demonstrations of this fact. Here is another: let’s calculate this using the method of “area by cross sections”. Here is $x^2 + y^2 = R^2$ with some $y = c$ cross sections drawn in.

Now do the calculation by integrals: we will use symmetry and only do the upper half and multiply our result by 2. At each $y = y_c$ level, call the radius from the center line to the circle $R(y)$ so the total length of the “y is constant” level is $2R(y)$ and we “multiply by thickness “dy” to obtain $V_2 = 4 \int^{y=R}_{y=0} R(y) dy$.

But remember that the curve in question is $x^2 + y^2 = R^2$ and so if we set $x = R(y)$ we have $R(y) = \sqrt{R^2 -y^2}$ and so our integral is $4 \int^{y=R}_{y=0}\sqrt{R^2 -y^2} dy$

Now this integral is no big deal. But HOW we solve it will help us down the road. So here, we use the change of variable (aka “trigonometric substitution”): $y = Rsin(t), dy =Rcos(t)$ to change the integral to:

$4 \int^{\frac{\pi}{2}}_0 R^2 cos^2(t) dt = 4R^2 \int^{\frac{\pi}{2}}_0 cos^2(t) dt$ therefore

$V_2 = c_2 R^2$ where:

$c_2 = 4\int^{\frac{\pi}{2}}_0 cos^2(t)$

Yes, I know that this is an easy integral to solve, but I first presented the result this way in order to make a point.

Of course, $c_2 = 4\int^{\frac{\pi}{2}}_0 cos^2(t) = 4\int^{\frac{\pi}{2}}_0 \frac{1}{2} + \frac{1}{2}cos(2t) dt = \pi$

Therefore, $V_2 =\pi R^2$ as expected.

Exercise for those seeing this for the first time: compute $c_3$ and $V_3$ by using the above methods.

Inductive step: Assume $V_k = c_kR^k$ Now calculate using the method of cross sections above (and here we move away from x-y coordinates to more general labeling):

$V_{k+1} = 2\int^R_0 V_k dy = 2 \int^R_0 c_k (R(x_{k+1})^k dx_{k+1} =c_k 2\int^R_0 (R(x_{k+1}))^k dx_{k+1}$

Now we do the substitutions: first of all, we note that $x_1^2 + x_2^2 + ...x_{k}^2 + x_{k+1}^2 = R^2$ and so

$x_1^2 + x_2^2 ....+x_k^2 = R^2 - x_{k+1}^2$. Now for the key observation: $x_1^2 + x_2^2 ..+x_k^2 =R^2(x_{k+1})$ and so $R(x_{k+1}) = \sqrt{R^2 - x_{k+1}^2}$

Now use the induction hypothesis to note:

$V_{k+1} = c_k 2\int^R_0 (R^2 - x_{k+1}^2)^{\frac{k}{2}} dx_{k+1}$

Now do the substitution $x_{k+1} = Rsin(t), dx_{k+1} = Rcos(t)dt$ and the integral is now:

$V_{k+1} = c_k 2\int^{\frac{\pi}{2}}_0 R^{k+1} cos^{k+1}(t) dt = c_k(2 \int^{\frac{\pi}{2}}_0 cos^{k+1}(t) dt)R^{k+1}$ which is what we needed to show.

In fact, we have shown a bit more. We’ve shown that $c_1 = 2 =2 \int^{\frac{\pi}{2}}_0(cos(t))dt, c_2 = 2 \cdot 2\int^{\frac{\pi}{2}}_0 cos^2(t) dt = c_1 2\int^{\frac{\pi}{2}}_0 cos^2(t) dt$ and, in general,

$c_{k+1} = c_{k}c_{k-1}c_{k-2} ....c_1(2 \int^{\frac{\pi}{2}}_0 cos^{k+1}(t) dt) = 2^{k+1} \int^{\frac{\pi}{2}}_0(cos^{k+1}(t))dt \int^{\frac{\pi}{2}}_0(cos^{k}(t))dt \int^{\frac{\pi}{2}}_0(cos^{k-1}(t))dt .....\int^{\frac{\pi}{2}}_0(cos(t))dt$

Finishing the formula

We now need to calculate these easy calculus integrals: in this case the reduction formula:

$\int cos^n(x) dx = \frac{1}{n}cos^{n-1}sin(x) + \frac{n-1}{n} \int cos^{n-2}(x) dx$ is useful (it is merely integration by parts). Now use the limits and elementary calculation to obtain:

$\int^{\frac{\pi}{2}}_0 cos^n(x) dx = \frac{n-1}{n} \int^{\frac{\pi}{2}}_0 cos^{n-2}(x)dx$ to obtain:

$\int^{\frac{\pi}{2}}_0 cos^n(x) dx = (\frac{n-1}{n})(\frac{n-3}{n-2})......(\frac{3}{4})\frac{\pi}{4}$ if $n$ is even and:
$\int^{\frac{\pi}{2}}_0 cos^n(x) dx = (\frac{n-1}{n})(\frac{n-3}{n-2})......(\frac{4}{5})\frac{2}{3}$ if $n$ is odd.

Now to come up with something resembling a closed formula let’s experiment and do some calculation:

Note that $c_1 = 2, c_2 = \pi, c_3 = \frac{4 \pi}{3}, c_4 = \frac{(\pi)^2}{2}, c_5 = \frac{2^3 (\pi)^2)}{3 \cdot 5} = \frac{8 \pi^2}{15}, c_6 = \frac{\pi^3}{3 \cdot 2} = \frac{\pi^3}{6}$.

So we can make the inductive conjecture that $c_{2k} = \frac{\pi^k}{k!}$ and see how it holds up: $c_{2k+2} = 2^2 \int^{\frac{\pi}{2}}_0(cos^{2k+2}(t))dt \int^{\frac{\pi}{2}}_0(cos^{2k+1}(t))dt \frac{\pi^k}{k!}$

$= 2^2 ((\frac{2k+1}{2k+2})(\frac{2k-1}{2k})......(\frac{3}{4})\frac{\pi}{4})((\frac{2k}{2k+1})(\frac{2k-2}{2k-1})......\frac{2}{3})\frac{\pi^k}{k!}$

Now notice the telescoping effect of the fractions from the $c_{2k+1}$ factor. All factors cancel except for the $(2k+2)$ in the first denominator and the 2 in the first numerator, as well as the $\frac{\pi}{4}$ factor. This leads to:

$c_{2k+2} = 2^2(\frac{\pi}{4})\frac{2}{2k+2} \frac{\pi^k}{k!} = \frac{\pi^{k+1}}{(k+1)!}$ as required.

Now we need to calculate $c_{2k+1} = 2\int^{\frac{\pi}{2}}_0(cos^{2k+1}(t))dt c_{2k} = 2\int^{\frac{\pi}{2}}_0(cos^{2k+1}(t))dt \frac{\pi^k}{k!}$

$= 2 (\frac{2k}{2k+1})(\frac{2k-2}{2k-1})......(\frac{4}{5})\frac{2}{3}\frac{\pi^k}{k!} = 2 (\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(2k-1)...(5)(3)} \frac{\pi^k}{k!}$

To simplify this further: split up the factors of the $k!$ in the denominator and put one between each denominator factor:

$= 2 (\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(k)(2k-1)(k-1)...(3)(5)(2)(3)(1)} \pi^k$ Now multiply the denominator by $2^k$ and put one factor with each $k-m$ factor in the denominator; also multiply by $2^k$ in the numerator to obtain:

$(2) 2^k (\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(2k)(2k-1)(2k-2)...(6)(5)(4)(3)(2)} \pi^k$ Now gather each factor of 2 in the numerator product of the 2k, 2k-2…

$= (2) 2^k 2^k \pi^k \frac{k!}{(2k+1)!} = 2 \frac{(4 \pi)^k k!}{(2k+1)!}$ which is the required formula.

So to summarize:

$V_{2k} = \frac{\pi^k}{k!} R^{2k}$

$V_{2k+1}= \frac{2 k! (4 \pi)^k}{(2k+1)!}R^{2k+1}$

Note the following: $lim_{k \rightarrow \infty} c_{k} = 0$. If this seems strange at first, think of it this way: imagine the n-ball being “inscribed” in an n-cube which has hyper volume $(2R)^n$. Then consider the ratio $\frac{2^n R^n}{c_n R^n} = 2^n \frac{1}{c_n}$; that is, the n-ball holds a smaller and smaller percentage of the hyper volume of the n-cube that it is inscribed in; note the $2^n$ corresponds to the number of corners in the n-cube. One might see that the rounding gets more severe as the number of dimensions increases.

One also notes that for fixed radius R, $lim_{n \rightarrow \infty} V_n = 0$ as well.

There are other interesting aspects to this limit: for what dimension $n$ does the maximum hypervolume occur? As you might expect: this depends on the radius involved; a quick glance at the hyper volume formulas will show why. For more on this topic, including an interesting discussion on this limit itself, see Dave Richardson’s blog Division by Zero. Note: his approach to finding the hyper volume formula is also elementary but uses polar coordinate integration as opposed to the method of cross sections.

## May 4, 2015

### Teaching evaluations ….

Filed under: academia, editorial — Tags: , , — collegemathteaching @ 4:48 pm

When I was in grade school, I was evaluated by people with undergraduate degrees.
When I was in high school, I was evaluated by people with undergraduate degrees, and occasionally by someone with a masters degree.
When I was an undergraduate, I was evaluated by Ph. D. holders and an occasional masters degree holder.
When I was a graduate student, Ph. D’s with impressive research credentials evaluated me.

Now..as a college professor…I am evaluated by those with high school diplomas ….things have come full circle, huh? 🙂

In all honesty, at least my department does “peer classroom visits” at least with non-tenured faculty, and on occasion, with some tenured faculty (the latter is mostly voluntary).

### Hitting the bat with the ball….the vector calculus integral theorems….

Filed under: calculus, editorial, vector calculus — Tags: , — collegemathteaching @ 4:43 pm

When I was a small kid, my dad would play baseball with me. He’d pitch the ball and try to hit my bat with the ball so I could think I was actually hitting the ball.

Well, fast forward 50 years to my vector calculus final exam; we are covering the “big integral” theorems.

Yeah, I know; it is $\int_{\partial \Omega} \sigma = \int_{\Omega} d \sigma$ but, let’s just say that we aren’t up to differential forms as yet. 🙂

And so I am giving them classical Green’s Theorem, Stokes’ Theorem and Divergence Theorem problems….and everything in sight basically boils down to integrating a constant over a rectangle, box, sphere, ball or disk.

I am hitting their bats with the ball; I wonder how many will notice. 🙂