# College Math Teaching

## 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.

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 $| \frac{\phi_k(t) - \phi_k(s)}{(t-s)} - f(t, \phi(t))| < \epsilon$ whenever $k$ is large enough and $s$ is sufficiently close to $t$. This is not hard to do because the segments of the functions are getting smaller and the differential equation is satisfied exactly at the nodes and, because $f(x,y)$ is continuous, $f(x,y)$ cannot vary very much if $s, t$ are close together.

But a careful proof of the above is for another time.

So, once we establish that there is at least one solution satisfying a given initial condition, how do we establish that there is only one?

First, let’s establish the domain: I’ll assume that the point in question is $(0, y_0)$ and I’ll first start with a rectangle of the form: $-a \le x \le a, -b \le y \le b$ and $(0, y_0)$ in the interior of this rectangle $R$. Now we state a proposed condition (often called “Lipschitz in y”:

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 $|f(x,y_1)-f(x,y_2)| \le K|y_1-y_2|$ 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 $|z'(x)| = |f(x,y_1)-f(x,y_2)| < 2K|y_1-y_2| =2K|z|$ on $R$. Now because $y_1$ and $y_2$ are different functions on $R$, there is some $x_1$ where, say, $y_1(x_1) - y_2(x_1) > 0$. 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 $z'(x) < Y'(x)$ hence there is some $\delta > 0$ where $z(x_1 - \delta) > Y(x_1 - \delta)$.

But since $z(0) = 0 < Y(0)$, the graphs of $z, Y$ must meet at some point in the interval $(0, x_1)$, let $x = s$ be the first such point. Then we have $Y(s) = z(s)$ AND $Y(x_1) = z(x_1)$. Then apply Rolle’s Theorem to the function $Y(x) - z(x)$ find a point $c$ between $s, x_1$ where $Y'(c)-z'(c) = 0$ which is impossible as $Y'(x) > z'(x)$ 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 $R$.

## September 23, 2014

### Ok, what do you see here? (why we don’t blindly trust software)

I had Dfield8 from MATLAB propose solutions to $y' = t(y-2)^{\frac{4}{5}}$ meeting the following initial conditions:

$y(0) = 0, y(0) = 3, y(0) = 2$.

Now, of course, one of these solutions is non-unique. But, of all of the solutions drawn: do you trust ANY of them? Why or why not?

Note: you really don’t have to do much calculus to see what is wrong with at least one of these. But, if you must know, the general solution is given by $y(t) = (\frac{t^2}{10} +C)^5 + 2$ (and, of course, the equilibrium solution $y = 2$). But that really doesn’t provide more information that the differential equation does.

By the way, here are some “correct” plots of the solutions, (up to uniqueness)

## September 19, 2014

### Freshman calculus: they don’t always know the basics….

Filed under: basic algebra, calculus — Tags: , — collegemathteaching @ 5:43 pm

Example one: many students don’t know that $\frac{\frac{a}{b}}{c} \ne \frac{a}{\frac{b}{c}}$ (of course, assume that $a, b, c, \ne 0$. ) Where this came up: when we computed $lim_{h \rightarrow 0} \frac{\frac{1}{1+h} - 1}{h}$ we obtained $lim_{h \rightarrow 0} \frac{\frac{-h}{1+h}}{h}$ and a student didn’t understand why this was equal to $lim_{h \rightarrow 0} \frac{-1}{1+h}$

I ended up asking the student to simplify $\frac{\frac{2}{3}}{2} =$ and ….asking: ok, “if I have 2/3’rds of a pie and I give two people an equal piece of that, how much pie does each person get?

Example two: I gave “find the domain of $\frac{1}{\sqrt{x^2 - 9}}$ and two students didn’t understand why an answer of $-3 > x > 3$ was logically impossible. One of them told me: “my calculus teacher in high school told me to do it this way”: I am 99.99 percent that this isn’t true, but, well, I stayed with it until the student understood why such a statement was logically impossible. Oh yes, this same “I had calculus in high school” student was sure that I was wrong when I told him that “the derivative of a constant function is zero”; he was SURE that it is “1”.

## September 9, 2014

### Chebyshev polynomials: a topological viewpoint

Chebyshev (or Tchebycheff) polynomials are a class of mutually orthogonal polynomials (with respect to the inner product: $f \cdot g = \int^1_{-1} \frac{1}{\sqrt{1 - x^2}} f(x)g(x) dx$) defined on the interval $[-1, 1]$. Yes, I realize that this is an improper integral, but it does converge in our setting.

These are used in approximation theory; here are a couple of uses:

1. The roots of the Chebyshev polynomial can be used to find the values of $x_0, x_1, x_2, ...x_k \in [-1,1]$ that minimize the maximum of $|(x-x_0)(x-x_1)(x-x_2)...(x-x_k)|$ over the interval $[-1,1]$. This is important in minimizing the error of the Lagrange interpolation polynomial.

2. The Chebyshev polynomial can be used to adjust an approximating Taylor polynomial $P_n$ to increase its accuracy (away from the center of expansion) without increasing its degree.

The purpose of this note isn’t to discuss the utility but rather to discuss an interesting property that these polynomials have. The Wiki article on these polynomials is reasonably good for that purpose.

Let’s discuss the polynomials themselves. They are defined for all positive integers $n$ as follows:

$T_n = cos(n acos(x))$. Now, it is an interesting exercise in trig identities to discover that these ARE polynomials to begin with; one shows this to be true for, say, $n \in \{0, 1, 2\}$ by using angle addition formulas and the standard calculus resolution of things like $sin(acos(x))$. Then one discovers a relation: $T_{n+1} =2xT_n - T_{n-1}$ to calculate the rest.

The $cos(n acos(x))$ definition allows for some properties to be calculated with ease: the zeros occur when $acos(x) = \frac{\pi}{2n} + \frac{k \pi}{n}$ and the first derivative has zeros where $arcos(x) = \frac{k \pi}{n}$; these ALL correspond to either an endpoint max/min at $x=1, x = -1$ or local max and mins whose $y$ values are also $\pm 1$. Here are the graphs of $T_4(x), T_5 (x)$

Now here is a key observation: the graph of a $T_n$ forms $n$ spanning arcs in the square $[-1, 1] \times [-1,1]$ and separates the square into $n+1$ regions. So, if there is some other function $f$ whose graph is a connected, piecewise smooth arc that is transverse to the graph of $T_n$ that both spans the square from $x = -1$ to $x = 1$ and that stays within the square, that graph must have $n$ points of intersection with the graph of $T_n$.

Now suppose that $f$ is the graph of a polynomial of degree $n$ whose leading coefficient is $2^{n-1}$ and whose graph stays completely in the square $[-1, 1] \times [-1,1]$. Then the polynomial $Q(x) = T_n(x) - f(x)$ has degree $n-1$ (because the leading terms cancel via the subtraction) but has $n$ roots (the places where the graphs cross). That is clearly impossible; hence the only such polynomial is $f(x) = T_n(x)$.

This result is usually stated in the following way: $T_n(x)$ is normalized to be monic (have leading coefficient 1) by dividing the polynomial by $2^{n-1}$ and then it is pointed out that the normalized $T_n(x)$ is the unique monic polynomial over $[-1,1]$ that stays within $[-\frac{1}{2^{n-1}}, \frac{1}{2^{n-1}}]$ for all $x \in [-1,1]$. All other monic polynomials have a graph that leaves that box at some point over $[-1,1]$.

Of course, one can easily cook up analytic functions which don’t leave the box but these are not monic polynomials of degree $n$.

## August 31, 2014

### The convolution integral: do some examples in Calculus III or not?

For us, calculus III is the most rushed of the courses, especially if we start with polar coordinates. Getting to the “three integral theorems” is a real chore. (ok, Green’s, Divergence and Stoke’s theorem is really just $\int_{\Omega} d \sigma = \int_{\partial \Omega} \sigma$ but that is the subject of another post)

But watching this lecture made me wonder: should I say a few words about how to calculate a convolution integral?

Note: I’ve discussed a type of convolution integral with regards to solving differential equations here.

In the context of Fourier Transforms, the convolution integral is defined as it was in analysis class: $f*g = \int^{\infty}_{-\infty} f(x-t)g(t) dt$. Typically, we insist that the functions be, say, $L^1$ and note that it is a bit of a chore to show that the convolution of two $L^1$ functions is $L^1$; one proves this via the Fubini-Tonelli Theorem.

(The straight out product of two $L^1$ functions need not be $L^1$; e.g, consider $f(x) = \frac {1}{\sqrt{x}}$ for $x \in (0,1]$ and zero elsewhere)

So, assuming that the integral exists, how do we calculate it? Easy, you say? Well, it can be, after practice.

But to test out your skills, let $f(x) = g(x)$ be the function that is $1$ for $x \in [\frac{-1}{2}, \frac{1}{2}]$ and zero elsewhere. So, what is $f*g$???

So, it is easy to see that $f(x-t)g(t)$ only assumes the value of $1$ on a specific region of the $(x,t)$ plane and is zero elsewhere; this is just like doing an iterated integral of a two variable function; at least the first step. This is why it fits well into calculus III.

$f(x-t)g(t) = 1$ for the following region: $(x,t), -\frac{1}{2} \le x-t \le \frac{1}{2}, -\frac{1}{2} \le t \le \frac{1}{2}$

This region is the parallelogram with vertices at $(-1, -\frac{1}{2}), (0, -\frac{1}{2}), (0 \frac{1}{2}), (1, \frac{1}{2})$.

Now we see that we can’t do the integral in one step. So, the function we are integrating $f(x-t)f(t)$ has the following description:

$f(x-t)f(t)=\left\{\begin{array}{c} 1,x \in [-1,0], -\frac{1}{2} t \le \frac{1}{2}+x \\ 1 ,x\in [0,1], -\frac{1}{2}+x \le t \le \frac{1}{2} \\ 0 \text{ elsewhere} \end{array}\right.$

So the convolution integral is $\int^{\frac{1}{2} + x}_{-\frac{1}{2}} dt = 1+x$ for $x \in [-1,0)$ and $\int^{\frac{1}{2}}_{-\frac{1}{2} + x} dt = 1-x$ for $x \in [0,1]$.

That is, of course, the tent map that we described here. The graph is shown here:

So, it would appear to me that a good time to do a convolution exercise is right when we study iterated integrals; just tell the students that this is a case where one “stops before doing the outside integral”.

## August 27, 2014

### Nice collection of Math GIFs

Filed under: basic algebra, calculus, elementary mathematics, pedagogy — Tags: — collegemathteaching @ 12:25 am

No, these GIF won’t explain better than an instructor, but I found many of them to be fun.

Via: IFLS

## August 25, 2014

### How to succeed at calculus, and why it is worth it!

Filed under: calculus, student learning — Tags: , — collegemathteaching @ 2:06 pm

This post is intended to help the student who is willing to put time and effort into succeeding in a college calculus class.

Part One: How to Study

The first thing to remember is that most students will have to study outside of class in order to learn the material. There are those who pick things up right away, but these students tend to be the rare exception.

Think of it this way: suppose you want to learn to play the piano. A teacher can help show you how to play it and provide a practice schedule. But you won’t be any good if you don’t practice.

Suppose you want to run a marathon. A coach can help you with running form, provide workout schedules and provide feedback. But if you don’t run those workouts, you won’t build up the necessary speed and endurance for success.

The same principle applies for college mathematics classes; you really learn the material when you study it and do the homework exercises.

Here are some specific tips on how to study:

1. It is optimal if you can spend a few minutes scanning the text for the upcoming lesson. If you do this, you’ll be alert for the new concepts as they are presented and the concepts might sink in quicker.

2. There is some research that indicates:
a. It is better to have several shorter study sessions rather than one long one and
b. There is an optimal time delay between study sessions and the associated lecture.

Look at it this way: if you wait too long after the lesson to study it, you would have forgotten much of what was presented. If you study right away, then you really have, in essence, a longer class room session. It is probably best to hit the material right when the initial memory starts to fade; this time interval will vary from individual to individual. For more on this and for more on learning for long term recall, see this article.

3. Learn the basic derivative formulas inside and out; that is, know what the derivatives of functions like $sin(x), cos(x), tan(x), sec(x), arctan(x), arcsin(x), exp(x), ln(x)$ are on sight; you shouldn’t have to think about them. The same goes for the basic trig identities such as $\sin ^{2}(x)+\cos ^{2}(x)=1$ and $\tan^{2}(x)+1 = \sec^{2}(x)$

Why is this? The reason is that much of calculus (though not all!) boils down to pattern recognition.

For example, suppose you need to calculate:

$\int \dfrac{(\arctan (x))^{5}}{1+x^{2}}dx=$

If you don’t know your differentiation formulas, this problem is all but impossible. On the other hand, if you do know your differentiation formulas, then you’ll immediately recognize the $arctan(x)$ and it’s derivative $\dfrac{1}{1+x^{2}}$ and you’ll see that this problem is really the very easy problem $\int u^{5}du$.

But this all starts with having “automatic” knowledge of the derivative formulas.

Note: this learning is something your professor or TA cannot do for you!

4. Be sure to do some study problems with your notes and your book closed. If you keep flipping to your notes and book to do the homework problems, you won’t be ready for the exams. You have to kick up the training wheels.
Try this; the difference will surprise you. There is also evidence that forcing yourself to recall the material FROM YOUR OWN BRAIN helps you learn the material! Give yourself frequent quizzes on what you are learning.

5. When reviewing for an exam, study the problems in mixed fashion. For example, get some note cards and write problems from the various sections on them (say, some from 3.1, some from 3.2, some from 3.3, and so on), mix the cards, then try the problems. If you just review section by section, you’ll go into each problem knowing what technique to use each time right from the start. Many times, half of the battle is knowing which technique to use with each problem; that is part of the course! Do the problems in mixed order.

If you find yourself whining complaining “I don’t know where to start” it means that you don’t know the material well enough. Remember that a trained monkey can repeat specific actions; you have to be a bit better than that!

6. Read the book, S L O W L Y, with pen and paper nearby. Make sure that you work through the examples in the text and that you understand the reasons for each step.

7. For the “more theoretical” topics, know some specific examples for specific theorems. Here is what I am talking about:

a. Intermediate value theorem: recall that if $f(x)=\frac{1}{x}$, then $f(-1)=-1,f(1)=1$ but there is no $x$ such that $f(x) = 0$. Why does this not violate the intermediate value theorem?

b. Mean value theorem: note also that there is no $c$ such that $f'(c) = \frac{f(1)-f(-1)}{2} = 0$. Why does this NOT violate the Mean Value Theorem?

c. Series: it is useful to know basic series such as those for $exp(x), sin(x), cos(x)$. It is also good to know some basic examples such as the geometric series, the divergent harmonic series $\sum \frac{1}{k}$ and the conditionally convergent series $\sum (-1)^{k}\frac{1}{k}$.

d. Limit definition of derivative: be able to work a few basic examples of the derivative via the limit definition: $f(x) = x^{n}, f(x) = \frac{1}{x}, f(x)=\sqrt{x}$ and know why the derivative of $f(x) = |x|$ and $f(x) = x^{1/3}$ do not exist at $x = 0$.

Part II: Attitude
Your attitude will be very important.

1. Remember that your effort will be essential! Again, you can’t learn to run a marathon without getting off of the couch and making your muscles sore. Learning mathematics involves some frustration and, yes, at times, some tedium. Learning is fun OVERALL but it isn’t always fun at all times. You will encounter discomfort and unpleasantness at times.

2. Remember that winners look for ways to succeed; losers and whiners look for excuses for failure. You can always find those who will be willing to enable your underachievement. Instead, seek out those who bring out your best.

3. Success is NOT guaranteed; that is what makes success rewarding! Think of how good you’ll feel about yourself if you mastered something that seemed impossible to master at first. And yes, anyone who has achieved anything that is remotely difficult has taken some lumps and bruises along the way. You will NOT be spared these.

Remember that if you duck the calculus challenge, you are, in essence, slamming many doors of opportunity shut right from the get-go.

4. On the other hand, remember that Calculus (the first two semesters anyway) is a Freshman level class; exceptional mathematical talent is not a prerequisite for success. True, calculus is easy for some but that isn’t the point. Most reasonably intelligent people can have success, if they are willing to put forth the proper effort in the proper manner.

Just think of how good it will feel to succeed in an area that isn’t your strong suit!

### Dinette set on calculus…

Filed under: calculus, media — Tags: , — collegemathteaching @ 12:47 pm

Note: if you haven’t followed Julie Larson’s comic strip Dinette Set, the characters featured in it are not, well, the world’s most intellectually minded characters (with the exception of Patty). :-)

Ironically, I see such attitudes displayed by people…posting their thoughts on the internet via a computer or smart phone. The irony doesn’t even occur to them.

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