# College Math Teaching

## August 11, 2016

### Post Promotion Summer

Filed under: editorial, topology — Tags: — collegemathteaching @ 12:02 am

This is my first “terminal promotion” summer. And while I have something that I have “sort of” written up…I just don’t like the result; it basically fills in some gaps in a survey article. But I think that my thinking about this article has lead me to something that I can add to the paper so that I’ll actually LIKE what I submit.

Then again, my quandary can be summed up in this tweet:

If I wait until I am absolutely in love with my work before I send it out, it will never get sent out.

Hopefully, I’ll have more material to add to this blog this semester.

What I am working on: equivalence classes of simple closed curves; these are one to one, continuous images of the unit circle in 3-space. The objects that I am studying are so pathological that these curves fail to have a tangent at ANY point. One of these beasts can be constructed by taking the intersection of these nested, solid tori.

## June 15, 2016

### Elementary Math in the news: elections

Filed under: calculus, elementary mathematics, news — Tags: — collegemathteaching @ 9:11 pm

Ok, mostly I am trying to avoid writing up the painful details of a proposed mathematics paper.
But I do follow elections relatively closely. In the California Democratic primary, CNN called the election for Hillary Clinton late on June 7; at the time she lead Bernie Sanders 1,940,588-1,502,043, which is a margin of 438,537 votes. Percentage wise, the lead was 55.8-43.2, or 12.6 percentage points.

But due to mail in balloting and provisional ballot counting, there were still many votes to count. As of this morning, the totals were:

2,360,266-1,887,178 for a numerical lead of 473,088 votes. Percentage wise, the lead was 55.1-44.0, or 11.1 percentage points.

So, the lead grew numerically, but shrunk percentage wise.

“Big deal”, you say? Well, from reading social media, it is not obvious (to some) how a lead can grow numerically but shrink as a percentage.

Conceptually, it is pretty easy to explain: suppose one has an election involving 1100 voters who MUST choose between candidates. Say the first 100 votes that are counted happened to come from a strongly pro-Hillary group, and the tally after 100 was 90 Hillary, 10 Bernie. Then suppose the next 1000 was closer, say 550 for Hillary and 450 for Bernie. Then the lead grew by 100 votes (80 to 180) but the percentage lead shrunk from 80 percentage points to a 16.36 percentage point lead (58.18 to 41.82 percent). And it is easy to see that if the rest of the vote was really 55 percent Hillary, her percent of the vote would asymptotically shrink to close to 55 percent as the number of votes counted went up.

So, how might one have students model it? Let $H(t), B(t)$ be increasing functions of $t$ which represent the number of votes for Hillary and Bernie as a function of time. Assume no mistakes, hence $H(t), B(t)$ can be assumed to be increasing functions. So we want a case there $D(t) = H(t)-B(t)$ is an increasing function but $P(t) = \frac{H(t)}{H(t)+ B(t)}$ decreases with time.

Without calculus: rewrite $P(t) = \frac{1}{1+\frac{B(t)}{H(t)}}$ and note that $P(t)$ decreases as $\frac{B(t)}{H(t)}$ increases; that is, as $B(t)$ outgrows $H(t)$. But $H(t)$ must continue to outgrow $B(t)$. That is, the new ballots must still include more Hillary Bernie ballots, but the ratio of Bernie ballots to Hillary ballots must be going down.

If we use some calculus, we see that $H'(t)$ must exceed $B'(t)$ but to make $P(t)$ decrease, use the quotient rule plus a tiny bit of algebra to conclude that $H'(t)B(t)-B'(t)H(t)$ must be negative, or that $\frac{B'(t)}{B(t)} > \frac{H'(t)}{H(t)}$. That is, the Bernie ballots must be growing at a higher percentage rate than the Hillary ballots are.

None of this is surprising, but it might let the students get a feel of what derivatives are and what proportional change means.

## June 7, 2016

### Pop-math: getting it wrong but being close enough to give the public a feel for it

Space filling curves: for now, we’ll just work on continuous functions $f: [0,1] \rightarrow [0,1] \times [0,1] \subset R^2$.

A curve is typically defined as a continuous function $f: [0,1] \rightarrow M$ where $M$ is, say, a manifold (a 2’nd countable metric space which has neighborhoods either locally homeomorphic to $R^k$ or $R^{k-1})$. Note: though we often think of smooth or piecewise linear curves, we don’t have to do so. Also, we can allow for self-intersections.

However, if we don’t put restrictions such as these, weird things can happen. It can be shown (and the video suggests a construction, which is correct) that there exists a continuous, ONTO function $f: [0,1] \rightarrow [0,1] \times [0,1]$; such a gadget is called a space filling curve.

It follows from elementary topology that such an $f$ cannot be one to one, because if it were, because the domain is compact, $f$ would have to be a homeomorphism. But the respective spaces are not homeomorphic. For example: the closed interval is disconnected by the removal of any non-end point, whereas the closed square has no such separating point.

Therefore, if $f$ is a space filling curve, the inverse image of a points is actually an infinite number of points; the inverse (as a function) cannot be defined.

And THAT is where this article and video goes off of the rails, though, practically speaking, one can approximate the space filling curve as close as one pleases by an embedded curve (one that IS one to one) and therefore snake the curve through any desired number of points (pixels?).

So, enjoy the video which I got from here (and yes, the text of this post has the aforementioned error)

### Infinite dimensional vector subspaces: an accessible example that W-perp-perp isn’t always W

Filed under: integrals, linear albegra — Tags: , — collegemathteaching @ 9:02 pm

This is based on a Mathematics Magazine article by Irving Katz: An Inequality of Orthogonal Complements found in Mathematics Magazine, Vol. 65, No. 4, October 1992 (258-259).

In finite dimensional inner product spaces, we often prove that $(W^{\perp})^{\perp} = W$ My favorite way to do this: I introduce Grahm-Schmidt early and find an orthogonal basis for $W$ and then extend it to an orthogonal basis for the whole space; the basis elements that are not basis elements are automatically the basis for $W^{\perp}$. Then one easily deduces that $(W^{\perp})^{\perp} = W$ (and that any vector can easily be broken into a projection onto $W, W^{\perp}$, etc.

But this sort of construction runs into difficulty when the space is infinite dimensional; one points out that the vector addition operation is defined only for the addition of a finite number of vectors. No, we don’t deal with Hilbert spaces in our first course. 🙂

So what is our example? I won’t belabor the details as they can make good exercises whose solution can be found in the paper I cited.

So here goes: let $V$ be the vector space of all polynomials. Let $W_0$ the subspace of even polynomials (all terms have even degree), $W_1$ the subspace of odd polynomials, and note that $V = W_0 \oplus W_1$

Let the inner product be $\langle p(x), q(x) \rangle = \int^1_{-1}p(x)q(x) dx$. Now it isn’t hard to see that $(W_0)^{\perp} = W_1$ and $(W_1)^{\perp} = W_0$.

Now let $U$ denote the subspace of polynomials whose terms all have degree that are multiples of 4 (e. g. $1 + 3x^4 - 2x^8$ and note that $U^{\perp} \subset W_1$.

To see the reverse inclusion, note that if $p(x) \in U^{\perp}$, $p(x) = p_0 + p_1$ where $p_0 \in W_0, p_1 \in W_1$ and then $\int^1_{-1} (p_1(x))x^{4k} dx = 0$ for any $k \in \{1, 2, ... \}$. So we see that it must be the case that $\int^1_{-1} (p_0(x))x^{4k} dx = 0 = 2\int^1_0 (p_0(x))x^{4k} dx$ as well.

Now we can write: $p_0(x) = c_0 + c_1 x^2 + ...c_n x^{2n}$ and therefore $\int^1_0 p_0(x) x^{4k} dx = c_0\frac{1}{4k+1} + c_1 \frac{1}{2 + 4k+1}...+c_n \frac{1}{2n + 4k+1} = 0$ for $k \in \{0, 1, 2, ...2n+1 \}$

Now I wish I had a more general proof of this. But these equations (for each $k$ leads a system of equations:

$\left( \begin{array}{cccc} 1 & \frac{1}{3} & \frac{1}{5} & ...\frac{1}{2n+1} \\ \frac{1}{5} & \frac{1}{7} & \frac{1}{9}...&\frac{1}{2n+5} \\ ... & ... & ... & ... \\ \frac{1}{4k+1} & \frac{1}{4k+3} & ...& \frac{1}{10n+4} \end{array} \right) \left( \begin{array}{c} c_0 \\ c_1 \\ ... \\ c_n \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ ... \\ 0 \end{array} \right)$

It turns out that the given square matrix is non-singular (see page 92, no. 3 of Polya and Szego: Problems and Theorems in Analysis, Vol. 2, 1976) and so the $c_j = 0$. This means $p_0 = 0$ and so $U^{\perp} = W_1$

Anyway, the conclusion leaves me cold a bit. It seems as if I should be able to prove: let $f$ be some, say…$C^{\infty}$ function over $[0,1]$ where $\int^1_0 x^{2k} f(x) dx = 0$ for all $k \in \{0, 1, ....\}$ then $f = 0$. I haven’t found a proof as yet…perhaps it is false?

## May 20, 2016

### Student integral tricks…

Ok, classes ended last week and my brain is way out of math shape. Right now I am contemplating how to show that the complements of this object

and of the complement of the object depicted in figure 3, are NOT homeomorphic.

I can do this in this very specific case; I am interested in seeing what happens if the “tangle pattern” is changed. Are the complements of these two related objects *always* topologically different? I am reasonably sure yes, but my brain is rebelling at doing the hard work to nail it down.

Anyhow, finals are graded and I am usually treated to one unusual student trick. Here is one for the semester:

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

Now I was hoping that they would say $u = x +1 \rightarrow u-1 = x \rightarrow x^2 = u^2-2u+1$ at which case the integral is translated to: $\int u^{\frac{5}{2}} - 2u^{\frac{3}{2}} + u^{\frac{1}{2}} du$ which is easy to do.

Now those wanting to do it a more difficult (but still sort of standard) way could do two repetitions of integration by parts with the first set up being $x^2 = u, \sqrt{x+1}dx =dv \rightarrow du = 2xdx, v = \frac{2}{3} (x+1)^{\frac{3}{2}}$ and that works just fine.

But I did see this: $x =tan^2(u), dx = 2tan(u)sec^2(u)du, x+1 = tan^2(x)+1 = sec^2(u)$ (ok, there are some domain issues here but never mind that) and we end up with the transformed integral: $2\int tan^5(u)sec^3(u) du$ which can be transformed to $2\int (sec^6(u) - 2 sec^4(u) + sec^2(u)) tan(u)sec(u) du$ by elementary trig identities.

And yes, that leads to an answer of $\frac{2}{7}sec^7(u) +\frac{4}{5}sec^5(u) + \frac{2}{3}sec^3(u) + C$ which, upon using the triangle

Gives you an answer that is exactly in the same form as the desired “rationalization substitution” answer. Yeah, I gave full credit despite the “domain issues” (in the original integral, it is possible for $x \in (-1,0]$ ).

What can I say?

## April 12, 2016

### At long last…

Filed under: academia, editorial — Tags: — collegemathteaching @ 9:17 pm

I’ve been silent on this blog for too long. Part of what is happening: our department is slowly morphing into a “mostly service courses” department due to new regulations on “minimum class size” (set to 10 students for upper division courses). THAT, plus a dearth of “mathematics teaching majors” is hurting our “majors” enrollment.

So it has been “all calculus/all the time” for me lately. Yes, calculus can be fun to teach but after close to 30 years…..zzzzzz….

And it would be unethical for me to try something new just because I am bored.

But I finally have something I want to talk about: next post!

## February 10, 2016

### Vector subspaces: two examples

Filed under: linear albegra, pedagogy — Tags: — collegemathteaching @ 8:41 pm

I am teaching linear algebra our of the book by Fraleigh and Beauregard. We are on “subspaces” (subsets of $R^n$ for now) and a subspace is defined to be a set of vectors that is closed under both vector addition and scalar multiplication. Here are a couple of examples of non-subspaces:

1. $W= \{(x,y)| xy = 0 \}$. Now this space IS closed under scalar multiplication, note that this space IS closed under additive inverses. But it is not closed under addition as $[x,0] + [0,y]=[x,y] \notin W$ for $x \neq 0, y \neq 0$.

2. (this example is in the book): the vectors $\{(n, m) | n, m \in Z \}$ are closed under vector addition but not under scalar multiplication.

## February 9, 2016

### Bedside manner

Filed under: editorial — Tags: — collegemathteaching @ 7:53 pm

One of the things I’ve had to change is how I related to students.

I grew up playing football and wrestling. I went to a service academy and served in the Navy. Then, of course, I got pushed in graduate school.

I cannot treat my students the way that I got treated; they would break down rather than get motivated; in general they have trouble handling even a hint of anger.

### An economist talks about graphs

Filed under: academia, economics, editorial, pedagogy, student learning — Tags: , — collegemathteaching @ 7:49 pm

Paul Krugman is a Nobel Laureate caliber economist (he won whatever they call the economics prize).
Here he discusses the utility of using a graph to understand an economic situation:

Brad DeLong asks a question about which of the various funny diagrams economists love should be taught in Econ 101. I say production possibilities yes, Edgeworth box no — which, strange to say, is how we deal with this issue in Krugman/Wells. But students who go on to major in economics should be exposed to the box — and those who go on to grad school really, really need to have seen it, and in general need more simple general-equilibrium analysis than, as far as I can tell, many of them get these days.

There was, clearly, a time when economics had too many pictures. But now, I suspect, it doesn’t have enough.

OK, this is partly a personal bias. My own mathematical intuition, and a lot of my economic intuition in general, is visual: I tend to start with a picture, then work out both the math and the verbal argument to make sense of that picture. (Sometimes I have to learn the math, as I did on target zones; the picture points me to the math I need.) I know that’s not true for everyone, but it’s true for a fair number of students, who should be given the chance to learn things that way.

Beyond that, pictures are often the best way to convey global insights about the economy — global in the sense of thinking about all possibilities as opposed to small changes, not as in theworldisflat. […]

And it probably doesn’t hurt to remind ourselves that our students are, in general, NOT like us. What comes to us naturally probably does not come to them naturally.

## February 8, 2016

### Where these posts are (often) coming from

Filed under: academia, linear albegra, student learning — Tags: , — collegemathteaching @ 9:57 pm

Yes, my office is messy. Deal with it. 🙂 And yes, some of my professional friends (an accountant and a lawyer) just HAD to send me their office shots…pristine condition, of course.
(all in good fun!)

Note: this semester I teach 3 classes in a row: second semester “business/life science” calculus, second semester “engineering/physical science” calculus and linear algebra. Yes, I love the topics, but there is just enough overlap that I have to really clear my head between lessons. Example: we covered numerical integration in both of my calculus classes, as well as improper integrals. I have to be careful not to throw in $\int^{\infty}_{-\infty} \frac{dx}{1+x^2}$ as an example during my “life science calculus” class. I do the “head clearing” by going up the stairs to my office between classes.

Linear algebra is a bit tricky; we are so accustomed to taking things like “linear independence” for granted that it is easy to forget that this is the first time the students are seeing it. Also, the line between rigor and “computational usefulness” is tricky; for example, how rigorously do we explain “the determinant” of a matrix?

Oh well…back to some admin nonsense.

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