# College Math Teaching

## 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 $x$).

## August 1, 2014

### Yes, engineers DO care about that stuff…..

I took a break and watched a 45 minute video on Fourier Transforms:

A few take away points for college mathematics instructors:

1. When one talks about the Laplace Transform, one should distinguish between the one sided and two sided transforms (e. g., the latter integrates over the full real line, instead of 0 to $\infty$.

2. Engineers care about being able to take limits (e. g., using L’Hopitals rule and about problems such as $lim_{x \rightarrow 0} \frac{sin(2x)}{x}$ )

3. Engineers care about DOMAINS; they matter a great deal.

4. Sometimes the dabble in taking limits of sequences of functions (in an informal sense); here the Dirac Delta (a generalized function or distribution) is developed (informally) as a limit of Fourier transforms of a pulse function of height 1 and increasing width.

5. Even students at MIT have to be goaded into issuing answers.

6. They care about doing algebra, especially in the case of a change of variable.

So, I am teaching two sections of first semester calculus. I will emphasize things that students (and sometimes, faculty members of other departments) complain about.

## June 12, 2013

### A couple of instances of math in action

Filed under: advanced mathematics, applied mathematics, Fourier Series, physics, popular mathematics — Tags: — collegemathteaching @ 9:02 pm

Via Jerry Coyne’s website; you’ll see some great comments there.

Watch standing waves in action:

Here is what is going on; the particles collect at the “stationary” points.
This is an excellent reason to take a course that deals with Fourier Series!

Here is an example of a projection, and what happens when you take the image and move it a little.

## February 10, 2013

### Just for fun: no professors allowed!

Filed under: basic algebra, calculus, Fourier Series, media — collegemathteaching @ 2:32 pm

Ok, here is the quiz:

Aluminum medal:

What are these formulas?

Bronze medal:

Derive one of these formulas

Silver medal:

Derive two of these formulas (hint: one way to derive one of these involves a change of variables and polar coordinates.

Gold medal:

Assuming you have a piecewise continuous function (ok, make it piecewise smooth if you wish) and is periodic over $[-l,l]$ derive the middle formula.

You win: all of the money I made writing this note. 😉

## August 2, 2012

### MAA Mathfest Madison Day 1, 2 August 2012

I am sitting in the main ballroom waiting for the large public talks to start. I should be busy most of the day; it looks as if there will be some interesting all day long.

I like this conference not only for the variety but also for the timing; it gives me some momentum going into the academic year.

I regret not taking my camera; downtown Madison is scenic and we are close to the water. The conference venue is just a short walk away from the hotel; I see some possibilities for tomorrow’s run. Today: just weights and maybe a bit of treadmill in the afternoon.

The Talks
The opening lecture was the MAA-AMS joint talk by David Mumford of Brown University. This guy’s credentials are beyond stellar: Fields Medal, member of the National Academy of Science, etc.

His talk was about applied and pure mathematics and how there really shouldn’t be that much of a separation between the two, though there is. For one thing: pure mathematics prestige is measured by the depth of the result; applied mathematical prestige is mostly measured by the utility of the produced model. Pure mathematicians tend to see applied mathematics as shallow and simple and they resent the fact that applied math…gets a lot more funding.

He talked a bit about education and how the educational establishment ought to solicit input from pure areas; he also talked about computer science education (in secondary schools) and mentioned that there should be more emphasis on coding (I agree).

He mentioned that he tended to learn better when he had a concrete example to start from (I am the same way).

What amused me: his FIRST example was on PDE (partial differential equations) model of neutron flux through nuclear reactors used for submarines; note that these reactors were light water, thermal reactors (in that the fission reaction became self sustaining via the absorption of neutrons whose energy levels had been lowered by a moderator (the neutrons lose energy when they collide with atoms that aren’t too much heavier).

Of course, in nuclear power school, we studied the PDEs of the situation after the design had been developed; these people had to come up with an optimal geometry to begin with.

Note that they didn’t have modern digital computers; they used analogue computers modeled after simple voltage drops across resistors!

About the PDE: you had two neutron populations: “fast” neutrons (ones at high energy levels) and “slow” neutrons (ones at lower energy levels). The fast neutrons are slowed down to become thermal neutrons. But thermal neutrons in turn cause more fissions thereby increasing the fast neutron flux; hence you have two linked PDEs. Of course there is leakage, absorption by control rods, etc., and the classical PDEs can’t be solved in closed form.

Another thing I didn’t know: Clairaut (from the “symmetry of mixed partial derivatives” fame) actually came up with the idea of the Fourier series before Fourier did; he did this in an applied setting.

Next talk Amie Wilkinson of Northwestern (soon to be University of Chicago) gave a talk about dynamical systems. She is one of those who has publication in the finest journals that mathematics has to offer (stellar).

The whole talk was pretty good. Highlights: she mentioned Henri Poincare and how he worked on the 3-body problem (one massive body, one medium body, and one tiny body that didn’t exert gravitational force on the other bodies). This creates a 3-dimensional system whose dynamics live in 3-space (the system space is, of course, has much higher dimension). Now consider a closed 2 dimensional manifold in that space and a point on that manifold. Now study the orbit of that point under the dynamical system action. Eventually, that orbit intersects the 2 dimensional manifold again. The action of moving from the first point to the first intersection point actually describes a motion ON THE TWO MANIFOLD and if we look at ALL intersections, we get a the orbit of that point, considered as an action on the two dimensional manifold.

So, in some sense, this two manifold has an “inherited” action on it. Now if we look at, say, a square on that 2-dimensional manifold, it was proved that this square comes back in a “folded” fashion: this is the famed “Smale Horseshoe map“:

Other things: she mentioned that there are dynamical systems that are stable with respect to perturbations that have unstable orbits (with respect to initial conditions) and that these instabilities cannot be perturbed away; they are inherent to the system. There are other dynamical systems (with less stability) that have this property as well.

There is, of course, much more. I’ll link to the lecture materials when I find them.

Last morning Talk
Bernd Sturmfels on Tropical Mathematics
Ok, quickly, if you have a semi-ring (no additive inverses) with the following operations:
$x \oplus y =$ min $(x,y)$ and $x \otimes y = x + y$ (check that the operations distribute), what good would it be? Why would you care about such a beast?

Answer: many reasons. This sort of object lends itself well to things like matrix operations and is used for things such as “least path” problems (dynamic programming) and “tree metrics” in biology.

Think of it this way: if one is considering, say, an “order n” technique in numerical analysis, then the products of the error terms adds to the order, and the sum of the errors gives the, ok, maximum of the two summands (very similar).

The PDF of the slides in today’s lecture can be found here.

## August 19, 2011

### Partial Differential Equations, Differential Equations and the Eigenvalue/Eigenfunction problem

Suppose we are trying to solve the following partial differential equation:
$\frac{\partial \psi}{\partial t} = 3 \frac{\partial ^2 \phi}{\partial x^2}$ subject to boundary conditions:
$\psi(0) = \psi(\pi) = 0, \psi(x,0) = x(x-\pi)$

It turns out that we will be using techniques from ordinary differential equations and concepts from linear algebra; these might be confusing at first.

The first thing to note is that this differential equation (the so-called heat equation) is known to satisfy a “uniqueness property” in that if one obtains a solution that meets the boundary criteria, the solution is unique. Hence we can attempt to find a solution in any way we choose; if we find it, we don’t have to wonder if there is another one lurking out there.

So one technique that is often useful is to try: let $\psi = XT$ where $X$ is a function of $x$ alone and $T$ is a function of $t$ alone. Then when we substitute into the partial differential equation we obtain:
$XT^{\prime} = 3X^{\prime\prime}T$ which leads to $\frac{T^{\prime}}{T} = 3\frac{X^{\prime\prime}}{X}$

The next step is to note that the left hand side does NOT depend on $x$; it is a function of $t$ alone. The right hand side does not depend on $t$ as it is a function of $x$ alone. But the two sides are equal; hence neither side can depend on $x$ or $t$; they must be constant.

Hence we have $\frac{T^{\prime}}{T} = 3\frac{X^{\prime\prime}}{X} = \lambda$

So far, so good. But then you are told that $\lambda$ is an eigenvalue. What is that about?

The thing to notice is that $T^{\prime} - \lambda T = 0$ and $X^{\prime\prime} - \frac{\lambda}{3}X = 0$
First, the equation in $T$ can be written as $D(T) = \lambda T$ with the operator $D$ denoting the first derivative. Then the second can be written as $D^2(X) = 3\lambda X$ where $D^2$ denotes the second derivative operator. Recall from linear algebra that these operators meet the requirements for a linear transformation if the vector space is the set of all functions that are “differentiable enough”. So what we are doing, in effect, are trying to find eigenvectors for these operators.

So in this sense, solving a homogeneous differential equation is really solving an eigenvector problem; often this is termed the “eigenfucntion” problem.

Note that the differential equations are not difficult to solve:
$T = a exp(\lambda T)$ $X = b exp(\sqrt{\frac{\lambda}{3}} x) + cexp(-\sqrt{\frac{\lambda}{3}} x)$; the real valued form of the equation in $x$ depends on whether $\lambda$ is positive, zero or negative.

But the point is that we are merely solving a constant coefficient differential equation just as we did in our elementary differential equations course with one important difference: we don’t know what the constant (the eigenvalue) is.

Now if we turn to the boundary conditions on $x$ we see that a solution of the form $A e^{bx} + Be^{-bx}$ cannot meet the zero at the boundaries conditions; we can rule out the $\lambda = 0$ condition as well.
Hence we know that $\lambda$ is negative and we get $X = a cos(\sqrt{\frac{\lambda}{3}} x) + b sin(\sqrt{\frac{\lambda}{3}} x)$ solution and then $T = d e^{\lambda t }$ solution.

But now we notice that these solutions have a $\lambda$ in them; this is what makes these ordinary differential equations into an “eigenvalue/eigenfucntion” problem.

So what values of $\lambda$ will work? We know it is negative so we say $\lambda = -w^2$ If we look at the end conditions and note that $T$ is never zero, we see that the cosine term must vanish ($a = 0$ ) and we can ensure that $\sqrt{\frac{w}{3}}\pi = k \pi$ which implies that $w = 3k^2$ So we get a whole host of functions: $\psi_k = a_k e^{-3k^2 t}sin(kx)$.

Now we still need to meet the last condition (set at $t = 0$ ) and that is where Fourier analysis comes in. Because the equation was linear, we can add the solutions and get another solution; hence the $X$ term is just obtained by taking the Fourier expansion for the function $x(x-\pi)$ in terms of sines.

The coefficients are $b_k = \frac{1}{\pi} \int^{\pi}_{-\pi} (x)(x-\pi) sin(kx) dx$ and the solution is:
$\psi(x,t) = \sum_{k=1}^{\infty} e^{-3k^2 t} b_k sin(kx)$

## July 19, 2011

### Quantum Mechanics and Undergraduate Mathematics IV: measuring an observable (example)

Ok, we have to relate the observables to the state of the system. We know that the only possible “values” of the observable are the eigenvalues of the operator and the relation of the operator to the state vector provides the density function. But what does this measurement do to the state? That is, immediately after a measurement is taken, what is the state?

True, the system undergoes a "time evolution" but once an observable is measured, an immediate (termed "successive") measurement will yield the same value; a "repeated" measurement (one made giving the system to undergo a time evolution) might give a different value.

So we get:

Postulate 4 A measurement of an observable generally (?) causes a drastic, uncontrollable alteration in the state vector of the system; immediately after the measurement it will coincide with the eigenvector corresponding to the eigenvalue obtained in the measurement.

Note: we assume that our observable operators have distinct eigenvalues; that is, no two distinct eigenvectors have the same eigenvalue.

That is, if we measure an observable with operator $A$ and obtain measurement $a_i$ then the new system eigenvector is $\alpha_i$ regardless of what $\psi$ was prior to measurement. Of course, this eigenvector can (and usually will) evolve with time.

Roughly speaking, here is what is going on:
Say the system is in state $\psi$. We measure and observable with operator $A$. We can only obtain one of the eigenvalues $\alpha_k$ as a measurement. Recall: remember all of those “orbitals” from chemistry class? Those were the energy levels of the electrons and the orbital level was a permissible energy state that we could obtain by a measurement.

Now if we get $\alpha_k$ as a measurement, the new state vector is $\alpha_k$. One might say that we started with a probability density function (given the state and the observable), we made a measurement, and now, for a brief instant anyway, our density function “collapsed” to the density function $P(A = a_k) = 1$.

This situation (brief) coincides with our classical intuition of an observable “having a value”.

For the purposes of this example, we’ll set our Hilbert space to the the square integrable piecewise smooth functions on $[-\pi, \pi]$ and let our “state vector” $\psi(x) =\left\{ \begin{array}{c}1/\sqrt{\pi}, 0 < x \leq \pi \\ 0,-\pi \leq x \leq 0 \end{array}\right.$

Now suppose our observable corresponds to the eigenfunctions mentioned in this post, and we measure “-4” for our observable. This is the eigenvalue for $(1/\sqrt{\pi})sin(2x)$ so our new state vector is $(1/\sqrt{\pi})sin(2x)$.

So what happens if a different observable is measured IMMEDIATELY (e. g., no chance for a time evolution to take place).

Example We’ll still use the space of square integrable functions over $[-\pi, \pi]$
One might recall the Legendre polynomials which are eigenfucntions of the following operator:
$d/dt((1-t^2) dP_n/dt) = -(n)(n+1) P_n(t)$. These polynomials obey the orthogonality relation $\int^{1}_{-1} P_m(t)P_n(t)dt = 2/(2n+1) \delta_{m,n}$ hence $\int^{1}_{-1} P_m(t)P_m(t)dt = 2/(2m+1)$.
The first few of these are $P_0 = 1, P_1 =t, P_2 = (1/2)(3t^2-1), P_3 = (1/2)(5t^3 - 3t), ..$

We can adjust these polynomials by the change of variable $t =x/\pi$ and multiply each polynomial $P_m$ by the factor $sqrt{2/(\pi (2m+1) }$ to obtain an orthonormal eigenbasis. Of course, one has to adjust the operator by the chain rule.

So for this example, let $P_n$ denote the adjusted Legendre polynomial with eigenvalue $-n(n+1)$.

Now back to our original state vector which was changed to state function $(1/\sqrt{\pi})sin(2x)$.

Now suppose eigenvalue $-6 = -2(3)$ is observed as an observable with the Lengendre operator; this corresponds to eigenvector $\sqrt{(2/5)(1/\pi)}(1/2)(3(x/\pi)^2 -1)$ which is now the new state vector.

Now if we were to do an immediate measurement of the first observable, we’d have to a Fourier like expansion of our new state vector; hence the probability density function for the observables changes from the initial measurement. Bottom line: the order in which the observations are taken matters….in general.

The case in which the order wouldn’t matter: if the second observable had the state vector (from the first measurement) as an element of its eigenbasis.

We will state this as a general principle in our next post.

## July 15, 2011

### Quantum Mechanics and Undergraduate Mathematics III: an example of a state function

I feel bad that I haven’t given a demonstrative example, so I’ll “cheat” a bit and give one:

For the purposes of this example, we’ll set our Hilbert space to the the square integrable piecewise smooth functions on $[-\pi, \pi]$ and let our “state vector” $\psi(x) =\left\{ \begin{array}{c}1/\sqrt{\pi}, 0 < x \leq \pi \\ 0,-\pi \leq x \leq 0 \end{array}\right.$

Now consider a (bogus) state operator $d^2/dx^2$ which has an eigenbasis $(1/\sqrt{\pi})cos(kx), (1/\sqrt{\pi})sin(kx), k \in {, 1, 2, 3,...}$ and $1/\sqrt{2\pi}$ with eigenvalues $0, -1, -4, -9,......$ (note: I know that this is a degenerate case in which some eigenvalues share two eigenfunctions).

Note also that the eigenfunctions are almost the functions used in the usual Fourier expansion; the difference is that I have scaled the functions so that $\int^{\pi}_{-\pi} (sin(kx)/\sqrt{\pi})^2 dx = 1$ as required for an orthonormal basis with this inner product.

Now we can write $\psi = 1/(2 \sqrt{\pi}) + 4/(\pi^{3/2})(sin(x) + (1/3)sin(3x) + (1/5)sin(5x) +..)$
(yes, I am abusing the equal sign here)
This means that $b_0 = 1/\sqrt{2}, b_k = 2/(k \pi), k \in {1,3,5,7...}$

Now the only possible measurements of the operator are 0, -1, -4, -9, …. and the probability density function is: $p(A = 0) = 1/2, P(A = -1) = 4/(\pi^2), P(A = -3) = 4/(9 \pi^2),...P(A = -(2k-1))= 4/(((2k-1)\pi)^2)..$

One can check that $1/2 + (4/(\pi^2))(1 + 1/9 + 1/25 + 1/49 + 1/81....) = 1.$

Here is a plot of the state function (blue line at the top) along with some of the eigenfunctions multiplied by their respective $b_k$.