# College Math Teaching

## December 28, 2016

### Commentary: our changing landscape and challenges

Filed under: calculus, editorial — collegemathteaching @ 10:34 pm

Yes, I haven’t written anything of substance in a while; I hope to remedy that in upcoming weeks. I am teaching differential equations this next semester and that is usually good for a multitude of examples.

Our university is undergoing changes; this includes admitting students who are nominally STEM majors but who are not ready for even college algebra.

Our provost wants us to reduce college algebra class sizes…even though we are down faculty lines and we cannot find enough bodies to cover courses. Our wonderful administrators didn’t believe us when we explained that it is difficult to find “masters and above” part time faculty for mathematics courses.

And so: with the same size freshmen class, we have a wider variation of student abilities: those who are ready for calculus III, and those who cannot even add simple fractions (yes, one of these was admitted as a computer science major!). Upshot: we need more people to teach freshmen courses, and we are down faculty lines!

Then there is the pressure from the bean-counters in our business office. They note that many students are avoiding our calculus courses and taking them at community colleges. So, obviously, we are horrible teachers!

Here is what the administrators will NOT face up to: students frequently say that passing those courses at a junior college is much easier; they don’t have to study nearly as much. Yes, engineering tells us that students with JC calculus don’t do any worse than those who take it from the mathematics department.

What I think is going on: at universities like ours (I am NOT talking about MIT or Stanford!), the mathematics required in undergraduate engineering courses has gone down; we are teaching more mathematics “than is necessary” for the engineering curriculum, at least the one here.

So some students (not all) see the extra studying required to learn “more than they need” as wasted effort and they resent it.

The way we get these students back: lower the mathematical demands in our calculus courses, or at least lower the demands on studying the more abstract stuff (“abstract”, by calculus standards).

Anyhow, that is where we are. We don’t have the resources to offer both a “mathematical calculus” course and one that teaches “just what you need to know”.

## November 29, 2016

### Facebook data for a statistics class

Filed under: statistics — Tags: , , , — collegemathteaching @ 6:04 pm

I have to admit that teaching statistics has kind of ruined me. I find myself seeking patterns and data sets everywhere.

Now a national election does give me some data to play with; I used 2012 data for those purposes a few years ago.

But now I have Facebook. And I have a very curious Facebook friendship (I won’t embarrass the person by naming the person).

She became my FB friend in January of 2014. Lately, we’ve been talking a lot, mostly about the 2016 general election. But we went a long time without conversing via “private message”.

I noticed in the first 560 days of our FB “friendship” we exchanged 30 private messages. Then we started to talk more and more. $t$ is time in days since we started to talk (March 2014) and $NMSG$ is the cumulative number of private messages that we exchanged:

So I figured: this has to be an example of an exponential situation, so I ran a regression $r^2 \geq 0.99$ and got: $N = .1248e^{.010835 t}$ where $N$ is the number of messages and $t$ is the time in days.

Of course, practically speaking, this can’t continue but this “virtually zero” for a long time followed by an “explosion” is a classical exponential phenomenon.

## November 1, 2016

### A test for the independence of random variables

Filed under: algebra, probability, statistics — Tags: , — collegemathteaching @ 10:36 pm

We are using Mathematical Statistics with Applications (7’th Ed.) by Wackerly, Mendenhall and Scheaffer for our calculus based probability and statistics course.

They present the following Theorem (5.5 in this edition)

Let $Y_1$ and $Y_2$ have a joint density $f(y_1, y_2)$ that is positive if and only if $a \leq y_1 \leq b$ and $c \leq y_2 \leq d$ for constants $a, b, c, d$ and $f(y_1, y_2)=0$ otherwise. Then $Y_1, Y_2$ are independent random variables if and only if $f(y_1, y_2) = g(y_1)h(y_2)$ where $g(y_1), h(y_2)$ are non-negative functions of $y_1, y_2$ alone (respectively).

Ok, that is fine as it goes, but then they apply the above theorem to the joint density function: $f(y_1, y_2) = 2y_1$ for $(y_1,y_2) \in [0,1] \times [0,1]$ and 0 otherwise. Do you see the problem? Technically speaking, the theorem doesn’t apply as $f(y_1, y_2)$ is NOT positive if and only if $(y_1, y_2)$ is in some closed rectangle.

It isn’t that hard to fix, I don’t think.

Now there is the density function $f(y_1, y_2) = y_1 + y_2$ on $[0,1] \times [0,1]$ and zero elsewhere. Here, $Y_1, Y_2$ are not independent.

But how does one KNOW that $y_1 + y_2 \neq g(y_1)h(y_2)$?

I played around a bit and came up with the following:

Statement: $\sum^{n}_{i=1} a_i(x_i)^{r_i} \neq f_1(x_1)f_2(x_2).....f_n(x_n)$ (note: assume $r_i \in \{1,2,3,....\}, a_i \neq 0$

Proof of the statement: substitute $x_2 =x_3 = x_4....=x_n = 0$ into both sides to obtain $a_1 x_1^{r_1} = f_1(x_1)(f_2(0)f_3(0)...f_n(0))$ Now none of the $f_k(0) = 0$ else function equality would be impossible. The same argument shows that $a_2 x_2^{r_2} = f_2(x_2)f_1(0)f_3(0)f_4(0)...f_n(0)$ with none of the $f_k(0) = 0$.

Now substitute $x_1=x_2 =x_3 = x_4....=x_n = 0$ into both sides and get $0 = f_1(0)f_2(0)f_3(0)f_4(0)...f_n(0)$ but no factor on the right hand side can be zero.

This is hardly profound but I admit that I’ve been negligent in pointing this out to classes.

## October 12, 2016

### P-values and precision of language

Filed under: media, popular mathematics — Tags: , — collegemathteaching @ 2:00 am

I read yet another paper proclaiming that it is “now time to do away with p-values.” And yes, I can recommend reading the article.

From my point of view, one of the troubles with p-values is that there is a misunderstanding as to what they actually mean.

So here goes: the p-value is the probability that, given the null hypothesis is true, one obtains an observation as extreme (or greater) than the given observation. That is, if $Y$ is a random variable with a probability distribution as given by the null hypothesis, and $Y^*$ is the observation, $P(Y \geq Y^*) = p$.

Example: suppose you assume that a coin is fair (the null hypothesis), and you toss it 100 times and observe 65 heads. It can be shown that $P(Y \geq 65) = 0.00175882086148504$. So that is the p-value of that particular experiment. That is, IF the coin really were fair, you’d expect to 65 or more heads .1716 percent of the time.

That seems clear enough, statistically speaking.

But when one gets down to the science, one wants to determine whether there is evidence enough to believe one thing or another thing. So, is this coin biased or did this result happen “just by chance”? And strictly speaking, we don’t really know. For example, it could be that we did a precision scientific measurement on the coin and found it to be fair before doing the above experiment. Or it could be that this was just some coin we came across, or it could be that we were asked to examine this coin because of previous suspicious results. This information matters.

And think of it this way: suppose the above experiment was repeated, say, 100,000 times with a coin known to be fair. Then we’d expect to see the above result about 176 times and ALL of those “positives” would be “due to chance”.

Upshot: when it comes to scientific experiments, we still need replication.

## October 11, 2016

### The bias we have toward the rational numbers

Filed under: analysis, Measure Theory — Tags: , , — collegemathteaching @ 5:39 pm

A brilliant scientist (full tenure at the University of Chicago) has a website called “Why Evolution is True”. He wrote an article titled “why is pi irrational” and seemed to be under the impression that being “irrational” was somehow special or unusual.

That is an easy impression to have; after all, almost every example we use rationals or sometimes special irrationals (e. g. multiples of $pi$, $e^1$, square roots, etc.

We even condition our students to think that way. Time and time again, I’ve seen questions such as “if $f(.9) = .94, f(.95) = .9790, f(1.01) = 1.043$ then it is reasonable to conclude that $f(1) =$. It is as if we want students to think that functions take integers to integers.

The reality is that the set of rationals has measure zero on the real line, so if one were to randomly select a number from the real line and the selection was truly random, the probability of the number being rational would be zero!

So, it would be far, far stranger had “pi” turned out to be rational. But that just sounds so strange.

So, why do the rationals have measure zero? I dealt with that in a more rigorous way elsewhere (and it is basic analysis) but I’ll give a simplified proof.

The set of rationals are countable so one can label all of them as $q(n), n \in \{0, 1, 2, ... \}$ Now consider the following covering of the rational numbers: $U_n = (q(n) - \frac{1}{2^{n+1}}, q(n) + \frac{1}{2^{n+1}})$. The length of each open interval is $\frac{1}{2^n}$. Of course there will be overlapping intervals but that isn’t important. What is important is that if one sums the lengths one gets $\sum^{\infty}_{n = 0} \frac{1}{2^n} = \frac{1}{1-\frac{1}{2}} = 2$. So the rationals can be covered by a collection of open sets whose total length is less than or equal to 2.

But there is nothing special about 2; one can then find new coverings: $U_n = (q(n) - \frac{\epsilon}{2^{n+1}}, q(n) + \frac{\epsilon}{2^{n+1}})$ and the total length is now less than or equal to $2 \epsilon$ where $\epsilon$ is any real number. Since there is no positive lower bound as to how small $\epsilon$ can be, the set of rationals can be said to have measure zero.

## October 7, 2016

### Now what is a linear transformation anyway?

Filed under: linear albegra, pedagogy — Tags: , — collegemathteaching @ 9:43 pm

Yes, I know, a linear transformation $L: V \rightarrow W$ is a function between vector spaces such that $L(V \oplus W) = L(V) \oplus L(W)$ and $L(a \odot V) = a \odot L(V)$ where the vector space operations of vector addition and scalar multiplication occur in their respective spaces.

Consider the set $R^+ = \{x| x > 0 \}$ endowed with the “vector addition” $x \oplus y = xy$ where $xy$ represents ordinary real number multiplication and “scalar multiplication $r \odot x = x^r$ where $r \in R$ and $x^r$ is ordinary exponentiation. It is clear that $\{R^+, R | \oplus, \odot \}$ is a vector space with $1$ being the vector “additive” identity and $0$ playing the role of the scalar zero and $1$ playing the multiplicative identity. Verifying the various vector space axioms is a fun, if trivial exercise.

Then $L(x) = ln(x)$ is a vector space isomophism between $R^+$ and $R$ (the usual addition and scalar multiplication) and of course, $L^{-1}(x) = exp(x)$.

Can we expand this concept any further?

Question: (I have no idea if this has been answered or not): given any, say, non-compact, connected subset of $R$, is it possible to come up with vector space operations (vector addition, scalar multiplication) so as to make a given, say, real valued, continuous one to one function into a linear transformation?

The answer in some cases is “yes.”

Consider $L(x): R^+ \rightarrow R^+$ by $L(x) = x^r$, $r$ any real number.

Exercise 1: $L$ is a linear transformation.

Exercise 2: If we have ANY linear transformation $L: R^+ \rightarrow R^+$, let $L(e) = e^a$.
Then $L(x) = L(e^{ln(x)}) = L(e)^{ln(x)} = (e^a)^{ln(x)} = x^a$.

Exercise 3: we know that all linear transformations $L: R \rightarrow R$ are of the form $L(x) = ax$. These can be factored through:

$x \rightarrow e^x \rightarrow (e^x)^a = e^{ax} \rightarrow ln(e^{ax}) = ax$.

So this isn’t exactly anything profound, but it is fun! And perhaps it might be a way to introduce commutative diagrams.

## October 4, 2016

### Linear Transformation or not? The vector space operations matter.

Filed under: calculus, class room experiment, linear albegra, pedagogy — collegemathteaching @ 3:31 pm

This is nothing new; it is an example for undergraduates.

Consider the set $R^+ = \{x| x > 0 \}$ endowed with the “vector addition” $x \oplus y = xy$ where $xy$ represents ordinary real number multiplication and “scalar multiplication $r \odot x = x^r$ where $r \in R$ and $x^r$ is ordinary exponentiation. It is clear that $\{R^+, R | \oplus, \odot \}$ is a vector space with $1$ being the vector “additive” identity and $0$ playing the role of the scalar zero and $1$ playing the multiplicative identity. Verifying the various vector space axioms is a fun, if trivial exercise.

Now consider the function $L(x) = ln(x)$ with domain $R^+$. (here: $ln(x)$ is the natural logarithm function). Now $ln(xy) = ln(x) + ln(y)$ and $ln(x^a) = aln(x)$. This shows that $L:R^+ \rightarrow R$ (the range has the usual vector space structure) is a linear transformation.

What is even better: $ker(L) =\{x|ln(x) = 0 \}$ which shows that $ker(L) = \{1 \}$ so $L$ is one to one (of course, we know that from calculus).

And, given $z \in R, ln(e^z) = z$ so $L$ is also onto (we knew that from calculus or precalculus).

So, $R^+ = \{x| x > 0 \}$ is isomorphic to $R$ with the usual vector operations, and of course the inverse linear transformation is $L^{-1}(y) = e^y$.

Upshot: when one asks “is F a linear transformation or not”, one needs information about not only the domain set but also the vector space operations.

## October 3, 2016

### Lagrange Polynomials and Linear Algebra

Filed under: algebra, linear albegra — Tags: — collegemathteaching @ 9:24 pm

We are discussing abstract vector spaces in linear algebra class. So, I decided to do an application.

Let $P_n$ denote the polynomials of degree $n$ or less; the coefficients will be real numbers. Clearly $P_n$ is $n+1$ dimensional and $\{1, x, x^2, ...x^n \}$ constitutes a basis.

Now there are many reasons why we might want to find a degree $n$ polynomial that takes on certain values for certain values of $x$. So, choose $x_0, x_1, x_2, ..., x_{n-1}$. So, let’s construct an alternate basis as follows: $L_0 = \frac{(x-x_1)(x-x_2)(x-x_3)..(x-x_{n})}{(x_0 - x_1)(x_0-x-x_2)..(x_0 - x_{n})}, L_1 = \frac{(x-x_0)(x-x_2)(x-x_3)..(x-x_{n})}{(x_1 - x_0)(x_1-x-x_2)..(x_1 - x_{n})}, ...L_k = \frac{(x-x_0)(x-x_1)(x-x_2)..(x-x_{k-1})(x-x_{k+1})...(x-x_{n})}{(x_k - x_1)(x_k-x-x_2)..(x_k - x_{k-1})(x_k - x_{k+1})...(x_k - x_{n})}.$ $....L_{n} = \frac{(x-x_0)(x-x_1)(x-x_2)..(x-x_{n-1})}{(x_{n}- x_1)(x_{n}-x-x_2)..(x_{n} - x_{n})}$

This is a blizzard of subscripts but the idea is pretty simple. Note that $L_k(x_k) = 1$ and $L_k(x_j) = 0$ if $j \neq k$.

But let’s look at a simple example: suppose we want to form a new basis for $P_2$ and we are interested in fixing $x$ values of $-1, 0, 1$.

So $L_0 = \frac{(x)(x-1)}{(-1-0)(-1-1)} = \frac{(x)(x-1)}{2}, L_1 = \frac{(x+1)(x-1)}{(0+1)(0-1)} = -(x+1)(x-1),$
$L_2 = \frac{(x+1)x}{(1+1)(1-0)} = \frac{(x+1)(x)}{2}$. Then we note that

$L_0(-1) = 1, L_0(0) =0, L_0(1) =0, L_1(-1)=0, L_1(0) = 1, L_1(1) = 0, L_2(-1)=0, L_2(0) =0, L_2(1)=1$

Now, we claim that the $L_k$ are linearly independent. This is why:

Suppose $a_0 L_0 + a_1 L_1 + ....a_n L_n =0$ as a vector. We can now solve for the $a_i$ Substitute $x_i$ into the right hand side of the equation to get $a_iL_i(x_i) = 0$ (note: $L_k(x_i) = 0$ for $i \neq k$). So $L_0, L_1, ...L_n$ are $n+1$ linearly independent vectors in $P_n$ and therefore constitute a basis.

Example: suppose we want to have a degree two polynomial $p(x)$ where $p(-1) =5, p(0) =3, p(1) = 17.$. We use our new basis to obtain:

$p(x) = 5L_0(x) + 3 L_1(x) + 17L_2(x) = \frac{5}{2}(x)(x-1) -3(x+1)(x-1) + \frac{17}{2}x(x+1)$. It is easy to check that $p(-1) = 5, p(0) =3, p(1) = 17$

## September 23, 2016

### Carmichael Numbers: “not quite” primes…

Filed under: algebra, elementary number theory, number theory, recreational mathematics — collegemathteaching @ 9:49 pm

We had a fun mathematics seminar yesterday.

Andrew Shallue gave a talk about the Carmichael numbers and gave a glimpse into his research. Along the way he mentioned the work of another mathematician…one that I met during my ultramarathon/marathon walking adventures! Talk about a small world..

So, to kick start my brain cells, I’ll say a few words about these.

First of all, prime numbers are very important in encryption schemes and it is a great benefit to be able to find them. However, for very large numbers, it can be difficult to determine whether a number is prime or not.

So one can take short cuts in determining whether a number is *likely* prime or not: one can say “ok, prime numbers have property P and if this number doesn’t have property P, it is not a prime. But if it DOES have property P, we hare X percent sure that it really is a prime.

If this said property is relatively “easy” to implement (via a computer), we might be able to live with the small amount of errors that our test generates.

One such test is to see if this given number satisfies “Fermat’s Little Theorem” which is as follows:

Let $a$ be a positive integer and $p$ be a prime, and suppose $a \neq kp$, that is $a \neq 0 (mod p)$ Then $a^{p-1} = 1 (mod p)$

If you forgotten how this works, recall that $Z_p$ is a field if $p$ is a prime, so $a \in Z_p, a \neq 0 (mod p)$ means that the set $\{a, 2a, 3a, ...(p-1)a \}$ consists of $\{1, 2, 3, ...(p-1) \}$. So take the product $(a)(2a)(3a)...((p-1)a)) = 1(2)(3)..(p-1)a^{p-1} = 1(2)(3)...(p-1) (mod p)$. Now note that we are working in a field, so we can cancel the $(1)(2)...(p-1)$ factor on both sides to get $a^{p-1} = 1 (mod p)$.

So one way to check to see if a number $q$ might be a prime is to check all $a^{q-1}$ for all $a \leq q$ and see if $a^{q-1} = 1 mod q$.
Now this is NOT a perfect check; there are non-prime numbers for which $a^{q-1} = 1 mod q$ for all $a \leq q$; these are called the Carmichael numbers. The 3 smallest such numbers are 561, 41041 and 825265.

The talk was about much more than this, but this was interesting.

## August 19, 2016

### A fun question concerning projections

Filed under: geometry, popular mathematics — Tags: — collegemathteaching @ 11:34 am

The semester is about to start. I decided to have some fun on Facebook. I took some office shots (yes, my office is messy) and decided to retouch the photos by adding joke photos from Facebook; I wanted to see which friends noticed (without being specifically notified). One did.

Of course, the color alone will give away which “on the wall” photos are genuine and which were put in (via Paint).

But a fun question is: if those photos were to be genuine, what size and shape would they be in real life (to appear the way that they do in those shots). One would give the genuine dimensions of the genuine photos on the wall to help the student solve the problem.

A more sophisticated 3-d version of the problem can be obtained from this cool video:

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