College Math Teaching

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.

crosssections

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 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

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. :-)

March 13, 2015

Moving from “young Turk” to “old f***”

Filed under: calculus, class room experiment, editorial, pedagogy — Tags: , , — collegemathteaching @ 9:09 pm

Today, one of our hot “young” (meaning: new here) mathematicians came to me and wanted to inquire about a course switch. He noted that his two course load included two different courses (two preparations) and that I was teaching different sections of the same two courses…was I interested in doing a course swap so that he had only one preparation (he is teaching 8 hours) and I’d only have two?

I said: “when I was your age, I minimized the number of preparations. But at my age, teaching two sections of the same low level course makes me want to bash my head against the wall”. That is, by my second lesson of the same course in the same day; I just want to be just about anywhere else on campus; I have no interest, no enthusiasm, etc.

I specifically REQUESTED 3 preparations to keep myself from getting bored; that is what 24 years of teaching this stuff does to you.

COMMENTARY
Every so often, someone has the grand idea to REFORM the teaching of (whatever) and the “reformers” usually get at least a few departments to go along with it.

The common thing said is that it gets professors to reexamine their teaching of (whatever).

But I wonder if many try these things….just out of pure boredom. Seriously, read the buzzwords of the “reform paper” I linked to; there is really nothing new there.

January 23, 2015

Making a math professor happy…

Filed under: calculus, class room experiment, elementary mathematics — Tags: , — collegemathteaching @ 10:28 pm

Calculus III: we are talking about polar curves. I give the usual lesson about how to graph r = sin(2 \theta) and r = sin(3 \theta) and give the usual “if n is even, the graph of r = sin (n \theta) has 2n petals and if n is odd, it has n petals.

Question: “does that mean it is impossible to have a graph with 6 petals then”? :-)

Yes, one can have intersecting petals and one try: r = |sin(3 \theta) | . But you aren’t going to get it without a trick of some sort.

6petals

January 9, 2015

Bad notation drove me nuts….(and still does)

Filed under: advanced mathematics, calculus, topology — Tags: , — collegemathteaching @ 8:36 pm

I remember one of my first classes in algebraic topology. The professor was talking about how to prove that \pi_1(S^1) = Z . For those who might be rusty: I am talking about the fundamental group of the circle, which is a group structure put on the set of continuous maps of the circle into the circle, where the maps all contain a set “base point” and two maps are equivalent if there is a homotopy (continuous deformation) between the two.

He remarked that he hoped “it was clear” that the circle was NOT simply connected.

That confused the heck out of me, because I had fallen into the trap of confusing the circle with a disk bounded by the circle!

Remember, for years, I had heard things like “the area of a circle is”…when in fact, the circle has area zero. The disk in the plane bounded by the circle has an area though.

So, when I teach, I try to point out bad or inconsistent notation. Example: sin^2(x) means (sin(x))^2 rather than sin(sin(x)) as the notation f^2 might suggest. But sin^{-1}(x) means arcsin(x) and NOT csc(x) = \frac{1}{sin(x)} . But…\frac{d^2 y}{dx^2} means \frac{d}{dx}(\frac{dy}{dx}) .

And please, don’t even get me started about dx that appears in integrals. I remember a student asking me about that when we did “integration by substitution”: “we never used the dx for anything up until now!” he said…correctly.

What got me thinking about this
This blog describes many of the things that I am thinking about at the moment. Currently, I am thinking about “wild knots”, which are embeddings of the circle into 3 space which cannot be deformed (by a deformation of space) into a smooth embedding of the circle.

Here are two examples of knots that can’t be deformed into a smooth knot:

Now the term “knot” implies that an embedding is present; the space that is being embedded is a circle. Of course, one might confuse a particular embedding with the equivalence class of equivalent embeddings; some old time authors distinguished the two concepts. Most modern ones (myself included) don’t.

Now I am interested in knots that are formed by the embedding of two “arcs”, each of which is non-wildly embedded (not wild is called “tame”).

In the case of arcs, authors sometimes mean “the arc itself” and in other cases mean “the embedding of an arc” (e. g. “arcs in 3 space”). Yes, there are some arcs that are so pathologically embedded that there is no deformation of space that takes the arc to a smooth arc. Unfortunately, the term “arc” can mean “the underlying space” or “the embedding”.

This will be one focus of my research in 2015: I hope to show that a knot that has one wild point (roughly speaking: one point that can never be assigned a tangent vector) that is the union of two tamely embedded arcs is never determined by its compliment. That might sound like gibberish, but in 2014 I proved that a knot that is an infinite product of knots (which are converging to a single wild point) has a complement which is homeomorphic to the complement of a knot that is wild at ALL of its points.

Of course THINKING that I can prove something and proving it are two different things. I remember spending two years trying to prove something that was false (I published the counter example) and, for part of my Ph.D. thesis, I attempted to prove something that turned out to be false; of course the counterexample came over 20 years after my attempt.

January 6, 2015

Quick Diversion: a rotating circle of dots within a circle..

Filed under: calculus — Tags: , — collegemathteaching @ 3:37 am

Since grading final exams, I’ve been travelling a bit. I am doing some admin duties but should have some time to do some research prior to….SEARCH COMMITTEE. That is such a time suck.

But here is a bit of fun:

Check out this video

Now here is a challenge (that I will take; feel free to beat me to it): find a set of equations that describes the motion of the centers of these disks.

My idea: I might start with a helix (of the type x = cos(t), y = sin(t), z = t and then have this helix change its center as it “moves up”; perhaps something like x = 4sos(t) - cos(t), y =4sin(t) - sin(t), z = t . Then intersect this with planes of the following type: (cylindrical coordinates: r = \theta and then, perhaps the points might described by the intersection of the helix with these planes? I’ll have to check it out.

November 1, 2014

Ok Graduate Student, do you want a pure math Ph. D.???

Filed under: academia, calculus, editorial, research — collegemathteaching @ 2:19 am

appliedvspuremath

This slide made me chuckle (click to see a larger version). But here is the point of it: it is very, very difficult to earn your living by researching in pure mathematics.

Is it a reasonable expectation for you?

Ask yourself this: look at your advisor. Is your advisor considerably smarter than you are, or even moderately smarter than you are? If so, then forget about earning your living as a research professor in pure math. It. Is. NOT. Going. To. Happen.

Yeah, you might get a post-doc. You might even manage to get one of those “tenure track with little hope for tenure” jobs at a D-I research university…maybe (perhaps unlikely?).

I’ve been on search committees. I’ve seen the letters for those who didn’t get tenure; often these folks had decent publication records but didn’t get large enough external grants.

It is brutal out there.

If you get a pure math Ph. D. and you aren’t your advisor’s intellectual equal, about your only hope for a tenured academic job is at the “teaching intensive” universities; basically you’ll spend the vast majority of your time attempting to teach calculus to students of very average ability; after all, most of the teaching load in mathematics is teaching service courses rather than majors courses.

It does have its charm at times, but after 20+ years, it gets very, very old. I’ll discuss how to alleviate the boredom in a responsible way in another post. (e. g., it is probably a bad idea to, say, spice it up by teaching integration via hyperbolic trig functions or to try to teach residue integrals).

So, ask yourself: is your passion research and discovery? Or, is it teaching average students? If it is the latter: well, go ahead and get that theoretical math Ph. D.; after all, there ARE jobs out there, and we’ve hired a couple of people last year and might hire some more in the next couple of years.

IF your passion is research and mathematical discovery and you aren’t your advisor’s intellectual equal, either switch to applied mathematics (more demand for such research) OR enhance your education with sellable skills such as computer programming/modeling, software engineering or perhaps picking up a masters in statistics. Make yourself more marketable to industry.

October 29, 2014

Hyperbolic Trig Functions and integration…

In college calculus courses, I’ve always wrestled with “how much to cover in the hyperbolic trig functions” section.

On one hand, the hyperbolic trig functions make some integrals much easer. On the other hand: well, it isn’t as if our classes are populated with the highest caliber student (I don’t teach at MIT); many struggle with the standard trig functions. There is only so much that the average young mind can absorb.

In case your memory is rusty:

cosh(x) =\frac{e^x + e^{-x}}{2}, sinh(x) = \frac{e^x -e^{-x}}{2} and then it is immediate that the standard “half/double angle formulas hold; we do remember that \frac{d}{dx}cosh(x) = sinh(x), \frac{d}{dx} = cosh(x).

What is less immediate is the following: sinh^{-1}(x)  = ln(x+\sqrt{x^2+1}), cosh^{-1}(x) = ln(x + \sqrt{x^2 -1}) (x \ge 1).

Exercise: prove these formulas. Hint: if sinh(y) = x then e^{y} - 2x- e^{-y} =0 so multiply both sides by e^{y} to obtain e^{2y} -2x e^y - 1 =0 now use the quadratic formula to solve for e^y and keep in mind that e^y is positive.

For the other formula: same procedure, and remember that we are using the x \ge 0 branch of cosh(x) and that cosh(x) \ge 1

The following follows easily: \frac{d}{dx} sinh^{-1} (x) = \frac{1}{\sqrt{x^2 + 1}} (just set up sinh(y) = x and use implicit differentiation followed by noting cosh^2(x) -sinh^2(x) = 1 . ) and \frac{d}{dx} cosh^{-1}(x) = \frac{1}{\sqrt{x^2-1}} (similar derivation).

Now, we are off and running.

Example: \int \sqrt{x^2 + 1} dx =

We can make the substitution x =sinh(u), dx = cosh(u) du and obtain \int cosh^2(u) du = \int \frac{1}{2} (cosh(2u) + 1)du = \frac{1}{4}sinh(2u) + \frac{1}{2} u + C . Now use sinh(2u) = 2 sinh(u)cosh(u) and we obtain:

\frac{1}{2}sinh(u)cosh(u) + \frac{u}{2} + C . The back substitution isn’t that hard if we recognize cosh(u) = \sqrt{sinh^2(u) + 1} so we have \frac{1}{2} sinh(u) \sqrt{sinh^2(u) + 1} + \frac{u}{2} + C . Back substitution is now easy:

\frac{1}{2} x \sqrt{x^2+1} + \frac{1}{2} ln(x + \sqrt{x^2 + 1}) + C . No integration by parts is required and the dreaded \int sec^3(x) dx integral is avoided. Ok, I was a bit loose about the domains here; we can make this valid for negative values of x by using an absolute value with the ln(x + \sqrt{x^2 + 1}) term.

October 3, 2014

Gaps in my mathematics education

Filed under: calculus, editorial, elementary mathematics — Tags: , , — collegemathteaching @ 1:19 pm

I’ve spoken about the many gaps in my mathematics education; I’ve written about a few. But in these cases, I was writing about the gaps at, say, the senior undergraduate to beginning graduate level.

I admit that I’ve enjoyed filling in some of these.

But, I also…have…elementary level gaps that I frequently overlook.

In my case: I never learned trigonometry all that well; I had forgotten about the laws of cosines and sines. And I had forgotten how to derive the following types of formulae: sin(A+B) = sin(A)cos(B) + cos(A)sin(b), cos(A+B) = cos(A)cos(B) - sin(A)sin(B) .

So, I spent a few minutes going over these old facts.

lawofcosines

They aren’t hard but I am a bit surprised that I let my basic ignorance continue on this long.

October 2, 2014

ARGH!!! I got stuck at the board…

Filed under: calculus, elementary mathematics, pedagogy — Tags: , , — collegemathteaching @ 5:51 pm

Related rate problem that required the “law of cosines”, which…is a trig rule that I never bothered to learn and couldn’t derive on the spot.

ARRRRGGGHHHH!!!!!!!!! (even after 20+ years, even AFTER preparing, things like this happen from time to time).

Now, of course, I won’t rest until I’ve learned those stupid rules. :-)

I nailed the rest of them though.

Note: a student pulled out the manual and, given the diagram, finished it while I worked on another problem. He showed me the answer and I gave him a fist bump.

Older Posts »

The WordPress Classic Theme. Blog at WordPress.com.

Follow

Get every new post delivered to your Inbox.

Join 610 other followers