# College Math Teaching

## 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 23, 2014

### Asking: DOES THIS MAKE SENSE…

Filed under: academia, linear regression — Tags: , , — oldgote @ 6:17 pm

I am teaching two first semester calculus classes and one numerical analysis class. I doubt that I’ll ask for 2 preparations ever again; I’ll get into those reasons later.

I gave a take home test; they had 10 days to do it. I decided to ask an easy warm up question to give them some confidence:

Yes, I covered “linear regression” in class and assigned a homework problem. Data: each student was assigned ONE of these data sets. I didn’t tell them where these came from, but these were the human population of: the earth or China or India or the US (in billions or millions respectively).

And much to my disgust: 2 of the students didn’t do this problem at all, and one who did GOT A NEGATIVE SLOPE for the “best fit line”. He duly plotted the negatively sloped line with the data….

And to think that this student is a graduate student in engineering! Unbelievable. Remember: this was a TAKE HOME test.

Moral: even students that you think are “advanced” have to be TOLD to do a “does this make sense” analysis ofter they obtained a result.

Evidently, this problem is not an isolated one.

And it is unlikely to get better. The article I linked to talks about “punching the button”. Well, they won’t even have to do that.

### Being part of the problem…

Filed under: academia, basic algebra, editorial — Tags: , — collegemathteaching @ 10:55 am

An eight-month probe has estimated that the “shadow curriculum” that existed at the University of North Carolina from 1993 to 2011 offered a grade-point boost from phony coursework to more than 3,100 students, including a disproportionately high percentage of student-athletes.[…]

A number of the students’ papers included brief introductions and closings with only “fluff” in between, according to the report.

“Hundreds” of the courses, the report reads, were independent-study. Yet when the university tightened standards on the amount of independent study a student could undertake, Crowder altered her program, creating courses she identified as lecture courses, but which mirrored independent study in that lectures never happened. The Wainstein report found 188 such courses between 1999 and 2011, in which 47.4 percent of the enrollments in these “paper classes” were student-athletes, who generally comprise 4 percent of the student population. Once Crowder retired in 2009, Nyang’oro sustained the practices for two more years until his retirement in 2011, albeit less voluminously.

In all the “paper classes,” the report found an average issued grade of 3.62, set against 3.28 for the regular classes in the department. Among student-athletes, though, the report found an average issued grade of 3.55, further above the 2.84 among student-athletes in the regular department classes.

The grade was based on one paper, which wasn’t merely scanned. But note the following: the average issued grade IN THE DEPARTMENT was 3.28???

Now this is far from the first public scandal involving university athletes, but university athletes aren’t the focus of this post.

I am thinking about what *I* have been complicit in over the years.

No, I’ve never given credit for incorrect work. And no, I’ve never falsified a grade.

But….well, let me start with a story from my youth.

When I was young, my dad wanted me to learn how to hit a baseball. So, he’d toss a ball at me when I was holding a bat. BUT, he’d often..deliberately hit my bat with the ball, so I got the impression that I was hitting the ball.

And, at times, it appears that I do that with my exams; I hit their bats.

If I gave what I considered *honest* exams in calculus and assigned partial credit honestly, I’d say that 70 percent of my “C” students would get D’s or F’s. About 50 percent of my B students would make C’s and perhaps 50 percent of my A students could get B’s. And I’d get called on the carpet…get complaints and accused of “not being student centered”…of “lacking compassion.”

What is going on?

Well, though we have a smaller percentage of students going to college now than in, say, 2009, check out the data:

When I was an undergraduate, about 50 percent of high school graduates went to college. When I started teaching college, it was about 60 percent…it rose to roughly 70 percent and is now about 65 percent. I think that we are seeing at least a little bit of “regression to the mean” effects.

Perhaps as a result, we are seeing things like this:

Algebra is an onerous stumbling block for all kinds of students: disadvantaged and affluent, black and white. In New Mexico, 43 percent of white students fell below “proficient,” along with 39 percent in Tennessee. Even well-endowed schools have otherwise talented students who are impeded by algebra, to say nothing of calculus and trigonometry.

California’s two university systems, for instance, consider applications only from students who have taken three years of mathematics and in that way exclude many applicants who might excel in fields like art or history. Community college students face an equally prohibitive mathematics wall. A study of two-year schools found that fewer than a quarter of their entrants passed the algebra classes they were required to take.

“There are students taking these courses three, four, five times,” says Barbara Bonham of Appalachian State University. While some ultimately pass, she adds, “many drop out.”

ALGEBRA? (and no, I am not talking about group theory or ring theory!)

Yes, that algebra.

So there are times when I wonder if I am participating in a fraud of a type.

## October 3, 2014

### There are semesters and classes like this

Filed under: editorial — Tags: , — collegemathteaching @ 1:46 pm

This semester: I am ok with the classes that I have to teach. No, my students aren’t perfect, but neither am I.

This semester is what I signed up for when I took the job here.

But there ARE semesters when one ends up with…well…a class full of people who have already failed once and have a low probability of success and…

it is a bit like this:

### 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.

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.

## October 1, 2014

### Osgood’s uniqueness theorem for differential equations

I am teaching a numerical analysis class this semester and we just started the section on differential equations. I want them to understand when we can expect to have a solution and when a solution satisfying a given initial condition is going to be unique.

We had gone over the “existence” theorem, which basically says: given $y' = f(x,y)$ and initial condition $y(x_0) = y_0$ where $(x_0,y_0) \in int(R)$ where $R$ is some rectangle in the $x,y$ plane, if $f(x,y)$ is a continuous function over $R$, then we are guaranteed to have at least one solution to the differential equation which is guaranteed to be valid so long as $(x, y(x)$ stays in $R$.

I might post a proof of this theorem later; however an outline of how a proof goes will be useful here. With no loss of generality, assume that $x_0 = 0$ and the rectangle has the lines $x = -a, x = a$ as vertical boundaries. Let $\phi_0 = f(0, y_0)x$, the line of slope $f(0, y_0)$. Now partition the interval $[-a, a]$ into $-a, -\frac{a}{2}, 0, \frac{a}{2}, a$ and create a polygonal path as follows: use slope $f(0, y_0)$ at $(0, y_0)$, slope $f(\frac{a}{2}, y_0 + \frac{a}{2}f(0, y_0))$ at $(\frac{a}{2}, y_0 + \frac{a}{2}f(0, y_0))$ and so on to the right; reverse this process going left. The idea: we are using Euler’s differential equation approximation method to obtain an initial piecewise approximation. Then do this again for step size $\frac{a}{4},$

In this way, we obtain an infinite family of continuous approximation curves. Because $f(x,y)$ is continuous over $R$, it is also bounded, hence the curves have slopes whose magnitude are bounded by some $M$. Hence this family is equicontinuous (for any given $\epsilon$ one can use $\delta = \frac{\epsilon}{M}$ in continuity arguments, no matter which curve in the family we are talking about. Of course, these curves are uniformly bounded, hence by the Arzela-Ascoli Theorem (not difficult) we can extract a subsequence of these curves which converges to a limit function.

Seeing that this limit function satisfies the differential equation isn’t that hard; if one chooses $t, s \in (-a.a)$ close enough, one shows that $| \frac{\phi_k(t) - \phi_k(s)}{(t-s)} - f(t, \phi(t))| 0$ where $|f(x,y_1)-f(x,y_2)| \le K|y_1-y_2|$ then the differential equation $y'=f(x,y)$ has exactly one solution where $\phi(0) = y_0$ which is valid so long as the graph $(x, \phi(x) )$ remains in $R$.

Here is the proof: $K > 0$ where $|f(x,y_1)-f(x,y_2)| \le K|y_1-y_2| < 2K|y_1-y_2|$. This is clear but perhaps a strange step.
But now suppose that there are two solutions, say $y_1(x)$ and $y_2(x)$ where $y_1(0) = y_2(0)$. So set $z(x) = y_1(x) -y_2(x)$ and note the following: $z'(x) = y_1(x) - y_2(x) = f(x,y_1)-f(x,y_2)$ and $|z'(x)| = |f(x,y_1)-f(x,y_2)| 0$. A Mean Value Theorem argument applied to $z$ means that we can assume that we can select our $x_1$ so that $z' > 0$ on that interval (since $z(0) = 0$).

So, on this selected interval about $x_1$ we have $z'(x) < 2Kz$ (we can remove the absolute value signs.).

Now we set up the differential equation: $Y' = 2KY, Y(x_1) = z(x_1)$ which has a unique solution $Y=z(x_1)e^{2K(x-x_1)}$ whose graph is always positive; $Y(0) = z(x_1)e^{-2Kx_1}$. Note that the graphs of $z(x), Y(x)$ meet at $(x_1, z(x_1))$. But $z'(x) 0$ where $z(x_1 - \delta) > Y(x_1 - \delta)$.

But since $z(0) = 0 z'(x)$ on that interval.

So, no such point $x_1$ can exist.

Note that we used the fact that the solution to $Y' = 2KY, Y(x_1) > 0$ is always positive. Though this is an easy differential equation to solve, note the key fact that if we tried to separate the variables, we’d calculate $\int_0^y \frac{1}{Kt} dt$ and find that this is an improper integral which diverges to positive $\infty$ hence its primitive cannot change sign nor reach zero. So, if we had $Y' =2g(Y)$ where $\int_0^y \frac{1}{g(t)} dt$ is an infinite improper integral and $g(t) > 0$, we would get exactly the same result for exactly the same reason.

Hence we can recover Osgood’s Uniqueness Theorem which states:

If $f(x,y)$ is continuous on $R$ and for all $(x, y_1), (x, y_2) \in R$ we have a $K > 0$ where $|f(x,y_1)-f(x,y_2)| \le g(|y_1-y_2|)$ where $g$ is a positive function and $\int_0^y \frac{1}{g(t)} dt$ diverges to $\infty$ at $y=0$ then the differential equation $y'=f(x,y)$ has exactly one solution where $\phi(0) = y_0$ which is valid so long as the graph $(x, \phi(x) )$ remains in $R$.