# College Math Teaching

## 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.

## February 5, 2016

### More fun with selective sums of divergent series

Just a reminder: if $\sum_{k=1}^{\infty} a_k$ is a series and $c_1, c_2, ...c_n ,,$ is some sequence consisting of 0’s and 1’s then a selective sum of the series is $\sum_{k=1}^{\infty} c_k a_k$. The selective sum concept is discussed in the MAA book Real Infinite Series (MAA Textbooks) by Bonar and Khoury (2006) and I was introduced to the concept by Ferdinands’s article Selective Sums of an Infinite Series in the June 2015 edition of Mathematics Magazine (Vol. 88, 179-185).

There is much of interest there, especially if one considers convergent series or alternating series.

This post will be about divergent series of positive terms for which $lim_{n \rightarrow \infty} a_n = 0$ and $a_{n+1} < a_n$ for all $n$.

The first fun result is this one: any selected $x > 0$ is a selective sum of such a series. The proof of this isn’t that bad. Since $lim_{n \rightarrow \infty} a_n = 0$ we can find a smallest $n$ such that $a_n \leq x$. Clearly if $a_n = x$ we are done: our selective sum has $c_n = 1$ and the rest of the $c_k = 0$.

If not, set $n_1 = n$ and note that because the series diverges, there is a largest $m_1$ so that $\sum_{k=n_1}^{m_1} a_k \leq x$. Now if $\sum_{k=n_1}^{m_1} a_k = x$ we are done, else let $\epsilon_1 = x - \sum_{k=n_1}^{m_1} a_k$ and note $\epsilon_1 < a_{m_1+1}$. Now because the $a_k$ tend to zero, there is some first $n_2$ so that $a_{n_2} \leq \epsilon_1$. If this is equality then the required sum is $a_{n_2} + \sum_{k=n_1}^{m_1} a_k$, else we can find the largest $m_2$ so that $\sum_{k=n_1}^{m_1} a_k + \sum_{k=n_2}^{m_2} a_k \leq x$

This procedure can be continued indefinitely. So if we label $\sum_{k=n_j}^{m_{j}} a_k = s_j$ we see that $s_1 + s_2 + ...s_{n} = t_{n}$ form an increasing, bounded sequence which converges to the least upper bound of its range, and it isn’t hard to see that the least upper bound is $x$ because $x-t_{n} =\epsilon_n < a_{m_n+1}$

So now that we can obtain any positive real number as the selective sum of such a series, what can we say about the set of all selective sums for which almost all of the $c_k = 0$ (that is, all but a finite number of the $c_k$ are zero).

Answer: the set of all such selective sums are dense in the real line, and this isn’t that hard to see, given our above construction. Let $(a,b)$ be any open interval in the real line and let $a < x < b$. Then one can find some $N$ such that for all $n > N$ we have $x - a_n > a$. Now consider our construction and choose $m$ large enough such that $x - t_m > x - a_n > a$. Then the $t_m$ represents the finite selected sum that lies in the interval $(a,b)$.

We can be even more specific if we now look at a specific series, such as the harmonic series $\sum_{k=1}^{\infty} \frac{1}{k}$. We know that the set of finite selected sums forms a dense subset of the real line. But it turns out that the set of select sums is the rationals. I’ll give a slightly different proof than one finds in Bonar and Khoury.

First we prove that every rational in $(0,1]$ is a finite select sum. Clearly 1 is a finite select sum. Otherwise: Given $\frac{p}{q}$ we can find the minimum $n$ so that $\frac{1}{n} \leq \frac{p}{q} < \frac{1}{n-1}$. If $\frac{p}{q} = \frac{1}{n}$ we are done. Otherwise: the strict inequality shows that $pn-p < q$ which means $pn-q < p$. Then note $\frac{p}{q} - \frac{1}{n} = \frac{pn-q}{qn}$ and this fraction has a strictly smaller numerator than $p$. So we can repeat our process with this new rational number. And this process must eventually terminate because the numerators generated from this process form a strictly decreasing sequence of positive integers. The process can only terminate when the new faction has a numerator of 1. Hence the original fraction is some sum of fractions with numerator 1.

Now if the rational number $r$ in question is greater than one, one finds $n_1$ so that $\sum^{n_1}_{k=1} \frac{1}{k} \leq r$ but $\sum^{n_1+1}_{k=1} \frac{1}{k} > r$. Then write $r-\sum^{n_1+1}_{k=1} \frac{1}{k}$ and note that its magnitude is less than $\frac{1}{n_1+1}$. We then use the procedure for numbers in $(0,1)$ noting that our starting point excludes the previously used terms of the harmonic series.

There is more we can do, but I’ll stop here for now.