College Math Teaching

April 5, 2018

A talk at University of South Alabama

Filed under: advanced mathematics, knot theory, topology — Tags: — collegemathteaching @ 3:27 pm

My slides (in order, more or less), can be found here.

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March 12, 2018

And I embarrass myself….integrate right over a couple of poles…

Filed under: advanced mathematics, analysis, calculus, complex variables, integrals — Tags: — collegemathteaching @ 9:43 pm

I didn’t have the best day Thursday; I was very sick (felt as if I had been in a boxing match..chills, aches, etc.) but was good to go on Friday (no cough, etc.)

So I walk into my complex variables class seriously under prepared for the lesson but decide to tackle the integral

\int^{\pi}_0 \frac{1}{1+sin^2(t)} dt

Of course, you know the easy way to do this, right?

\int^{\pi}_0 \frac{1}{1+sin^2(t)} dt =\frac{1}{2}  \int^{2\pi}_0 \frac{1}{1+sin^2(t)} dt and evaluate the latter integral as follows:

sin(t) = \frac{1}{2i}(z-\frac{1}{z}), dt = \frac{dz}{iz} (this follows from restricting z to the unit circle |z| =1 and setting z = e^{it} \rightarrow dz = ie^{it}dt and then obtaining a rational function of z which has isolated poles inside (and off of) the unit circle and then using the residue theorem to evaluate.

So 1+sin^2(t) \rightarrow 1+\frac{-1}{4}(z^2 -2 + \frac{1}{z^2}) = \frac{1}{4}(-z^2 + 6 -\frac{1}{z^2}) And then the integral is transformed to:

\frac{1}{2}\frac{1}{i}(-4)\int_{|z|=1}\frac{dz}{z^3 -6z +\frac{1}{z}} =2i \int_{|z|=1}\frac{zdz}{z^4 -6z^2 +1}

Now the denominator factors: (z^2 -3)^2 -8  which means z^2 = 3 - \sqrt{8}, z^2 = 3+ \sqrt{8} but only the roots z = \pm \sqrt{3 - \sqrt{8}} lie inside the unit circle.
Let w =  \sqrt{3 - \sqrt{8}}

Write: \frac{z}{z^4 -6z^2 +1} = \frac{\frac{z}{((z^2 -(3 + \sqrt{8})}}{(z-w)(z+w)}

Now calculate: \frac{\frac{w}{((w^2 -(3 + \sqrt{8})}}{(2w)} = \frac{1}{2} \frac{-1}{2 \sqrt{8}} and \frac{\frac{-w}{((w^2 -(3 + \sqrt{8})}}{(-2w)} = \frac{1}{2} \frac{-1}{2 \sqrt{8}}

Adding we get \frac{-1}{2 \sqrt{8}} so by Cauchy’s theorem 2i \int_{|z|=1}\frac{zdz}{z^4 -6z^2 +1} = 2i 2 \pi i \frac{-1}{2 \sqrt{8}} = \frac{2 \pi}{\sqrt{8}}=\frac{\pi}{\sqrt{2}}

Ok…that is fine as far as it goes and correct. But what stumped me: suppose I did not evaluate \int^{2\pi}_0 \frac{1}{1+sin^2(t)} dt and divide by two but instead just went with:

$latex \int^{\pi}_0 \frac{1}{1+sin^2(t)} dt \rightarrow i \int_{\gamma}\frac{zdz}{z^4 -6z^2 +1} where \gamma is the upper half of |z| = 1 ? Well, \frac{z}{z^4 -6z^2 +1} has a primitive away from those poles so isn’t this just i \int^{-1}_{1}\frac{zdz}{z^4 -6z^2 +1} , right?

So why not just integrate along the x-axis to obtain i \int^{-1}_{1}\frac{xdx}{x^4 -6x^2 +1} = 0 because the integrand is an odd function?

This drove me crazy. Until I realized…the poles….were…on…the…real…axis. ….my goodness, how stupid could I possibly be???

To the student who might not have followed my point: let \gamma be the upper half of the circle |z|=1 taken in the standard direction and \int_{\gamma} \frac{1}{z} dz = i \pi if you do this property (hint: set z(t) = e^{it}, dz = ie^{it}, t \in [0, \pi] . Now attempt to integrate from 1 to -1 along the real axis. What goes wrong? What goes wrong is exactly what I missed in the above example.

February 22, 2018

What is going on here: sum of cos(nx)…

Filed under: analysis, derivatives, Fourier Series, pedagogy, sequences of functions, series, uniform convergence — collegemathteaching @ 9:58 pm

This started innocently enough; I was attempting to explain why we have to be so careful when we attempt to differentiate a power series term by term; that when one talks about infinite sums, the “sum of the derivatives” might fail to exist if the sum is infinite.

Anyone who is familiar with Fourier Series and the square wave understands this well:

\frac{4}{\pi} \sum^{\infty}_{k=1} \frac{1}{2k-1}sin((2k-1)x)  = (\frac{4}{\pi})( sin(x) + \frac{1}{3}sin(3x) + \frac{1}{5}sin(5x) +.....) yields the “square wave” function (plus zero at the jump discontinuities)

Here I graphed to 2k-1 = 21

Now the resulting function fails to even be continuous. But the resulting function is differentiable except for the points at the jump discontinuities and the derivative is zero for all but a discrete set of points.

(recall: here we have pointwise convergence; to get a differentiable limit, we need other conditions such as uniform convergence together with uniform convergence of the derivatives).

But, just for the heck of it, let’s differentiate term by term and see what we get:

(\frac{4}{\pi})\sum^{\infty}_{k=1} cos((2k-1)x) = (\frac{4}{\pi})(cos(x) + cos(3x) + cos(5x) + cos(7x) +.....)...

It is easy to see that this result doesn’t even converge to a function of any sort.

Example: let’s see what happens at x = \frac{\pi}{4}: cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}

cos(\frac{\pi}{4}) + cos(3\frac{\pi}{4}) =0

cos(\frac{\pi}{4}) + cos(3\frac{\pi}{4}) + cos(5\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}

cos(\frac{\pi}{4}) + cos(3\frac{\pi}{4}) + cos(5\frac{\pi}{4}) + cos(7\frac{\pi}{4}) = 0

And this repeats over and over again; no limit is possible.

Something similar happens for x = \frac{p}{q}\pi where p, q are relatively prime positive integers.

But something weird is going on with this sum. I plotted the terms with 2k-1 \in \{1, 3, ...35 \}

(and yes, I am using \frac{\pi}{4} csc(x) as a type of “envelope function”)

BUT…if one, say, looks at cos(29x) + cos(31x) + cos(33x) + cos(35x)

we really aren’t getting a convergence (even at irrational multiples of \pi ). But SOMETHING is going on!

I decided to plot to cos(61x)

Something is going on, though it isn’t convergence. Note: by accident, I found that the pattern falls apart when I skipped one of the terms.

This is something to think about.

I wonder: for all x \in (0, \pi), sup_{n \in \{1, 3, 5, 7....\}}|\sum^{n}_{k \in \{1,3,..\}}cos(kx)| \leq |csc(x)| and we can somehow get close to csc(x) for given values of x by allowing enough terms…but the value of x is determined by how many terms we are using (not always the same value of x ).

February 11, 2018

Posting went way down in 2017

Filed under: advanced mathematics, complex variables, editorial — collegemathteaching @ 12:05 am

I only posted 3 times in 2017. There are many reasons for this; one reason is the teaching load, the type of classes I was teaching, etc.

I spent some of the year creating a new course for the Business College; this is one that replaced the traditional “business calculus” class.

The downside: there is a lot of variation in that course; for example, one of my sections has 1/3 of the class having a math ACT score of under 20! And we have many who are one standard deviation higher than that.

But I am writing. Most of what I write this semester can be found at the class blog for our complex variables class.

Our class does not have analysis as a prerequisite so it is a challenge to make it a truly mathematical class while getting to the computationally useful stuff. I want the students to understand that this class is NOT merely “calculus with z instead of x” but I don’t want to blow them away with proofs that are too detailed for them.

The book I am using does a first pass at integration prior to getting to derivatives.

August 28, 2017

Integration by parts: why the choice of “v” from “dv” might matter…

We all know the integration by parts formula: \int u dv = uv - \int v du though, of course, there is some choice in what v is; any anti-derivative will do. Well, sort of.

I thought about this as I’ve been roped into teaching an actuarial mathematics class (and no, I have zero training in this area…grrr…)

So here is the set up: let F_x(t) = P(0 \leq T_x \leq t) where T_x is the random variable that denotes the number of years longer a person aged x will live. Of course, F_x is a probability distribution function with density function f and if we assume that F is smooth and T_x has a finite expected value we can do the following: E(T_x) = \int^{\infty}_0 t f_x(t) dt and, in principle this integral can be done by parts….but…if we use u = t, dv = f_x(t), du = dt, v = F_x we have:

\

t(F_x(t))|^{\infty}_0 -\int^{\infty}_0 F_x(t) dt which is a big problem on many levels. For one, lim_{t \rightarrow \infty}F_x(t) = 1 and so the new integral does not converge..and the first term doesn’t either.

But if, for v = -(1-F_x(t)) we note that (1-F_x(t)) = S_x(t) is the survival function whose limit does go to zero, and there is usually the assumption that tS_x(t) \rightarrow 0 as t \rightarrow \infty

So we now have: -(S_x(t) t)|^{\infty}_0 + \int^{\infty}_0 S_x(t) dt = \int^{\infty}_0 S_x(t) dt = E(T_x) which is one of the more important formulas.

August 1, 2017

Numerical solutions to differential equations: I wish that I had heard this talk first

The MAA Mathfest in Chicago was a success for me. I talked about some other talks I went to; my favorite was probably the one given by Douglas Arnold. I wish I had had this talk prior to teaching numerical analysis for the fist time.

Confession: my research specialty is knot theory (a subset of 3-manifold topology); all of my graduate program classes have been in pure mathematics. I last took numerical analysis as an undergraduate in 1980 and as a “part time, not taking things seriously” masters student in 1981 (at UTSA of all places).

In each course…I. Made. A. “C”.

Needless to say, I didn’t learn a damned thing, even though both professors gave decent courses. The fault was mine.

But…I was what my department had, and away I went to teach the course. The first couple of times, I studied hard and stayed maybe 2 weeks ahead of the class.
Nevertheless, I found the material fascinating.

When it came to understanding how to find a numerical approximation to an ordinary differential equation (say, first order), you have: y' = f(t,y) with some initial value for both y'(0), y(0) . All of the techniques use some sort of “linearization of the function” technique to: given a step size, approximate the value of the function at the end of the next step. One chooses a step size, and some sort of schemes to approximate an “average slope” (e. g. Runga-Kutta is one of the best known).

This is a lot like numerical integration, but in integration, one knows y'(t) for all values; here you have to infer y'(t) from previous approximations of %latex y(t) $. And there are things like error (often calculated by using some sort of approximation to y(t) such as, say, the Taylor polynomial, and error terms which are based on things like the second derivative.

And yes, I faithfully taught all that. But what was unknown to me is WHY one might choose one method over another..and much of this is based on the type of problem that one is attempting to solve.

And this is the idea: take something like the Euler method, where one estimates y(t+h) \approx y(t) + y'(t)h . You repeat this process a bunch of times thereby obtaining a sequence of approximations for y(t) . Hopefully, you get something close to the “true solution” (unknown to you) (and yes, the Euler method is fine for existence theorems and for teaching, but it is too crude for most applications).

But the Euler method DOES yield a piecewise linear approximation to SOME f(t) which might be close to y(t)  (a good approximation) or possibly far away from it (a bad approximation). And this f(t) that you actually get from the Euler (or other method) is important.

It turns out that some implicit methods (using an approximation to obtain y(t+h) and then using THAT to refine your approximation can lead to a more stable system of f(t) (the solution that you actually obtain…not the one that you are seeking to obtain) in that this system of “actual functions” might not have a source or a sink…and therefore never spiral out of control. But this comes from the mathematics of the type of equations that you are seeking to obtain an approximation for. This type of example was presented in the talk that I went to.

In other words, we need a large toolbox of approximations to use because some methods work better with certain types of problems.

I wish that I had known that before…but I know it now. 🙂

Big lesson that many overlook: math is hard

Filed under: advanced mathematics, conference, editorial, mathematician, mathematics education — Tags: — collegemathteaching @ 11:43 am

First of all, it has been a very long time since I’ve posted something here. There are many reasons that I allowed myself to get distracted. I can say that I’ll try to post more but do not know if I will get it done; I am finishing up a paper and teaching a course that I created (at the request of the Business College), and we have a record enrollment..many of the new students are very unprepared.

Back to the main topic of the post.

I just got back from MAA Mathfest and I admit that is one of my favorite mathematics conferences. Sure, the contributed paper sessions give you a tiny amount of time to present, but the main talks (and many of the simple talks) are geared toward those of us who teach mathematics for a living and do some research on the side; there are some mainstream “basic” subjects that I have not seen in 30 years!

That doesn’t mean that they don’t get excellent people for the main speaker; they do. This time, the main speaker was Dusa McDuff: someone who was a member of the National Academy of Sciences. (a very elite level!)

Her talk was on the basics of symplectec geometry (introductory paper can be found here) and the subject is, well, HARD. But she did an excellent job of giving the flavor of it.

I also enjoyed Erica Flapan’s talk on graph theory and chemistry. One of my papers (done with a friend) referenced her work.

I’ll talk about Douglas Arnold’s talk on “when computational math meets geometry”; let’s just say that I wish I had seen this lecture prior to teaching the “numerical solutions for differential equations” section of numerical analysis.

Well, it looks as if I have digressed yet again.

There were many talks, and some were related to the movie Hidden Figures. And the cheery “I did it and so can you” talks were extremely well attended…applause, celebration, etc.

The talks on sympletec geometry: not so well attended toward the end. Again, that stuff is hard.

And that is one thing I think that we miss when we encourage prospective math students: we neglect to tell them that research level mathematics is difficult stuff and, while some have much more talent for it than others, everyone has to think hard, has to work hard, and almost all of us will fail, quite a bit.

I remember trying to spend over a decade trying to prove something, only to fail and to see a better mathematician get the result. One other time I spent 2 years trying to “prove” something…and I couldn’t “seal the deal”. Good thing too, as what I was trying to prove was false..and happily I was able to publish the counterexample.

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:

fbmessages

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.
carmenpmgraph

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.

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