College Math Teaching

October 25, 2013

A Laplace Transform of a function of non-exponential order

Many differential equations textbooks (“First course” books) limit themselves to taking Laplace transforms of functions of exponential order. That is a reasonable thing to do. However I’ll present an example of a function NOT of exponential order that has a valid (if not very useful) Laplace transform.

Consider the following function: n \in \{1, 2, 3,...\}

g(t)= \begin{cases}      1,& \text{if } 0 \leq t \leq 1\\      10^n,              & \text{if } n \leq t \leq n+\frac{1}{100^n} \\  0,  & \text{otherwise}  \end{cases}

Now note the following: g is unbounded on [0, \infty) , lim_{t \rightarrow \infty} g(t) does not exist and
\int^{\infty}_0 g(t)dt = 1 + \frac{1}{10} + \frac{1}{100^2} + .... = \frac{1}{1 - \frac{1}{10}} = \frac{10}{9}

One can think of the graph of g as a series of disjoint “rectangles”, each of width \frac{1}{100^n} and height 10^n The rectangles get skinnier and taller as n goes to infinity and there is a LOT of zero height in between the rectangles.

notexponentialorder

Needless to say, the “boxes” would be taller and skinnier.

Note: this is an example can be easily modified to provide an example of a function which is l^2 (square integrable) which is unbounded on [0, \infty) . Hat tip to Ariel who caught the error.

It is easy to compute the Laplace transform of g :

G(s) = \int^{\infty}_0 g(t)e^{-st} dt . The transform exists if, say, s \geq 0 by routine comparison test as |e^{-st}| \leq 1 for that range of s and the calculation is easy:

G(s) = \int^{\infty}_0 g(t)e^{-st} dt = \frac{1}{s} (1-e^{-s}) + \frac{1}{s} \sum^{\infty}_{n=1} (\frac{10}{e^s})^n(1-e^{\frac{-s}{100^n}})

Note: if one wants to, one can see that the given series representation converges for s \geq 0 by using the ratio test and L’Hoptial’s rule.

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Why the sequence cos(n) diverges

We are in the sequences section of our Freshman calculus class. One of the homework problems was to find whether the sequence a_n = cos(\frac{n}{2}) converged or diverged. This sequence diverges, but it isn’t easy for a freshman to see.

I’ll discuss this problem and how one might go about explaining it to a motivated student. To make things a bit simpler, I’ll discuss the sequence a_n = cos(n) instead.

Of course cos(x) is periodic with a fundamental region [0, 2\pi] so we will work with that region. Now we notice the following:

n (mod 2 \pi) is a group with the usual operation of addition.

By n (mod 2 \pi) , I mean the set n + k*2\pi where k \in \{..-2, -1, 0, 1, 2, 3,...\} ; one can think of the analogue of modular arithmetic, or one might see the elements of the group \{ r| r \in [0, 2 \pi), r = n - k 2\pi \} .

Of course, to get additive inverses, we need to include the negative integers, but ultimately that won’t matter. Example: 1, 2, 3, 4, 5, 6 are just equal to themselves mod 2 \pi. 7 = 7 - 2\pi (mod 2\pi), 13 = 13 - 4 \pi (mod 2\pi) , etc. So, I’ll denote the representative of n (mod 2\pi) by [n] .

Now if n \ne m then [n] \ne [m] ; for if [n]=[m] then there would be integers j, k so that n + j2\pi = m +k2\pi which would imply that |m-n| is a multiple of \pi . Therefore there are an infinite number of [n] in [0, 2\pi] which means that the set \{[n]\} has a limit point in the compact set [0, 2\pi] which means that given any positive integer m there is some interval of width \frac{2\pi}{m} that contains two distinct [i], [j] (say, j greater than i .)

This means that [j-i] \in (0, \frac{2\pi}{m}) so there is some integers k_2, k_3, so that k_2[j-i] \in (\frac{2\pi}{m}, \frac{2*2\pi}{m}), k_3[j-i] \in (\frac{2*2\pi}{m}, \frac{3*2\pi}{m})  , etc. Therefore there is some multiple of [j-i] in every interval of width \frac{2\pi}{m} . But m was an arbitrary positive integer; this means that the [n] are dense in [0,2\pi] . It follows that cos([n]) = cos(n) is dense in [-1,1] and hence a_n = cos(n) cannot converge as a sequence.

Frankly, I think that this is a bit tough for most Freshman calculus classes (outside of, say those at MIT, Harvard, Cal Tech, etc.).

October 22, 2013

Training mathematics teachers

Filed under: academia, pedagogy — Tags: — collegemathteaching @ 10:05 pm

The following article from the New York Times caught my eye:

“What I hadn’t realized was that setting a high bar at the beginning of the profession sends a signal to everyone else that you are serious about education and teaching is hard,” Ripley told me. “When you do that, it makes it easier to make the case for paying teachers more, for giving them more autonomy in the classroom. And for kids to buy into the premise of education, it helps if they can tell that the teachers themselves are extremely well educated.”

Once they are admitted, critics say, prospective teachers need more rigorous study, not just of the science and philosophy of education but of the contents, especially in math and the sciences, where America trails the best systems in Asia and Europe. A new study by the Education Policy Center at Michigan State, drawing on data from 17 countries, concluded that while American middle school math teachers may know a lot about teaching, they often don’t know very much about math. Most of them are not required to take the courses in calculus and probability that are mandatory in the best-taught programs.

Now, of course, there is the debate about “content” and “taking a course” (e. g. some states allow for a course to qualify as a “content check off” without actually being a course in the subject).

There will be pushback; I’ve been frequently distressed to hear students complain “why do I have to learn X…all I want to do is to teach math.”

The worst kind of paper to referee

Filed under: academia, advanced mathematics, research — Tags: — collegemathteaching @ 9:59 pm

Of course, refereeing journal articles is an expected duty; I’ve published a few and therefore benefited from the service of referees.

And it is very important that referees do their jobs responsibly.

I’ve refereed a few articles and some were very easy to reject: they either contained gross errors OR contained proofs of items that were already well known…and the existing “known” proofs were simpler (e. g. appeared in widely read textbooks).

But the most difficult articles to referee are those that are both

1. Poorly written and
2. contain some content that might have mathematical value.

These sorts of articles are time-sinks; one has to read them carefully because those ideas might well be worth seeing in print…but my goodness they are painful to read.

October 16, 2013

Convincing calculus students that the symbols MEAN SOMETHING

Filed under: applications of calculus, calculus, integrals — collegemathteaching @ 12:04 am

On a recent exam, the first 5 questions were as follows:

Given the region bounded by x = \frac{1}{2}, x = 1, y = 0, y = \frac{1}{x}

1. Find the AREA enclosed by the region.

2. Find the volume obtained by revolving this area about the y axis (the line x = 0 ).

3. Find the volume obtained by revolving this area about the x axis (the line y = 0 ).

4. Find \bar{x} . (constant density lamina).

5. Find \bar{y} .

Many students did fine, though there were a couple who literally blanked out on how to integrate \frac{1}{x} .

But some…well expect a few errors. But there were some who put a factor of \pi in their answers to 4, 5, and…yes, even 1.

Evidently, I’ll have to give my “these symbols actually have MEANING” speech again.

Note: yes, there were some interesting symmetries here; perhaps some students didn’t believe their answers.

October 15, 2013

What do you mean “you don’t know”???

Filed under: calculus, integrals, student learning — Tags: — collegemathteaching @ 6:38 pm

goatprof

I am grading calculus II exams and the above photo looks a bit like me. I’ll calm down before I hand the exams back to the students.

But I swear: I had one student DURING THE EXAM say “hey, you can’t do \int^1_{\frac{1}{2}} \frac{1}{x} dx  because \frac{1}{x} = x^{-1} and x^{-1+1} \frac{1}{-1+1} is undefined. Yes, this person passed calculus I and yes, we did that integral in calculus I (some schools hold off and develop ln(x) = \int^x_1 \frac{1}{t} dt using the Fundamental Theorem of Calculus). And yes, previously THIS SEMESTER we did integrals like \int \frac{1}{(x-1)(x+1)} dx .

(facepalm).

Hydrostatic force and work problems and …topology?

I just gave my second “calculus two” exam; the final two problems involved the following:

Suppose a trough with semi-circular ends is filled with water (say, length = 40 feet, radius = 5 feet).

1. How much work does one do in pumping all of the water to the top of the tank and out? (work against gravity only)

2. How much hydrostatic force is there against one of the semi-circular end plates?

trough

Assuming water is 62.5 pounds per gallon (I gave that to them):

1. Work = 62.5 \int^5_0 40x \sqrt{25 - x^2} dx ; of course, x is the distance a molecule of water is lifted and 40 \sqrt{25 - x^2} dx is the cross sectional volume of water.

2. Force = 62.5 \int^5_0 x \sqrt{25 - x^2} dx ; of coure, x is the depth of the water (hence 62.5 x is the pressure at that depth) and \sqrt{25 - x^2} dx is the area at depth x that the pressure is applied to.

The student can easily notice that the two answers differ by a factor of 40, which is the length of the trough.

So, what is the lesson here? Well, for one, I always envisioned a wall of a tank holding back a long mass of water. That isn’t correct; ONLY THE DEPTH matters. If the tank were a mile long or, say, an inch long, the pressure on the semi-circular ends would be the same. That runs counter to my intuition (which is clearly bad).

This lead me to think about the following: what if one were to put in a baffle between the two ends and the baffle was the same shape as the semi-circular ends. What would be the force on that plate?

Of course, the net force would be zero; there are two sides of the plate.

Now drop an open cylinder into the tank (think: a can with the top and bottom cut away). Clearly: zero force on the sides, right? Two sides, right?

Now, drop a Mobius band into the tank. A Mobius band has but one side. What is the force on it?

The key here: the Mobius band is one sided, but it is LOCALLY two sided; one can break the surface into tiny rectangles and note that the net force on each rectangle is zero as it is locally two sided. Hence zero total force on the one sided object.

October 9, 2013

Fun with complex numbers and particular solutions

I was fooling around with y''+by'+cy = e^{at}cos(bt) and thought about how to use complex numbers in the case when e^{at}cos(bt) is not a solution to the related homogenous equation. It then hit me: it is really quite simple.

First notes the following: e^{rt}cos(st) = \frac{1}{2}(e^{(r + si)t} + e^{(r - si)t}) and e^{rt}sin(st) = \frac{1}{2i}(e^{(r + si)t} - e^{(r - si)t}).

Then it is a routine exercise to see the following: given that z = r+si, \bar{z} = r-si are NOT solutions to p(m)= m^2 + bm + c = 0 p(m) is the characteristic equation of the differential equation. Then: attempt y_p = Ae^{zt} + Be^{\bar{z}t} Put into the differential equation to see y''_p + by'_p + cy_p = A(z^2+bz+c)e^{zt} + B(\bar{z}^2 + b\bar{z} + c)e^{\bar{z}t} .

Then: if the forcing function is e^{rt}cos(st) , a particular solution is y_p = Ae^{zt} + Be^{\bar{z}t} where A = \frac{1}{2(p(z))}, B = \frac{1}{2(p(\bar{z}))}. If the forcing function is e^{rt}cos(st) , a particular solution is y_p = Ae^{zt} - Be^{\bar{z}t} where A = \frac{1}{2i(p(z))}, B = \frac{1}{2i(p(\bar{z}))}.

That isn’t profound, but it does lead to the charming exercise: if z, \bar{z} are NOT roots to the quadratic with real coefficients p(x), then \frac{1}{p(z)} + \frac{1}{p(\bar{z})} is real as is \frac{i}{p(z)} - \frac{i}{p(\bar{z})} .

Let’s check this out: \frac{1}{p(z)} + \frac{1}{p(\bar{z})} = \frac{p(\bar{z})+p(z)}{p(z)p(\bar{z})}. Now look at the numerator and the denominator separately. The denominator: p(z)p(\bar{z})= (z^2 + bz +c)(\bar{z}^2 + b\bar{z} + c) = (z^2 \bar(z)^2) + b(b (z \bar{z}) + (z(z \bar{z}) +\bar{z}(z \bar{z})) + c (\bar{z} + z) + (z^2 + \bar{z}^2)) + (c^2) Now note that every term inside a parenthesis is real.

The numerator: z^2 + \bar{z}^2 + b (z + \bar{z}) + 2c is clearly real.

What about \frac{i}{p(z)} - \frac{i}{p(\bar{z})} ? We need to only check the numerator: i (z^2 - \bar{z}^2 + b(z - \bar{z}) + c-c) is indeed real.

Yeah, this is elementary but this might appear as an exercise for my next complex variables class.

October 1, 2013

I’ve lost any reason to live…

Filed under: student learning — Tags: , — collegemathteaching @ 7:45 pm

Today, I was grading differential equation papers. Some were really good, others were not in that category.

A calculus 2 student came in (we teach techniques of integration, applications of integration, and infinite series).

She, in a very polite manner, complained that the course was. too. easy.

Too. Easy.

I wish that I drank.

But there is no denying it: we have strong freshmen and some very weak, “take the class multiple times” students in the same class, and heaven help you if your flunk out rate is too high.
I might encourage her to take her complaint to the dean and put it in writing.

#$#@!!!

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