# College Math Teaching

## June 11, 2012

### Well, what do you mean by…..

Filed under: class room experiment, mathematics education, statistics, well posed problem — collegemathteaching @ 12:09 pm

Often seemingly simple questions don’t have simple answers; in fact, a seemingly simple question can be ambiguous.

I’ll give two examples:

1. Next week, Peoria, IL has the Steamboat 4 mile/15 km running race. So one question is: which race is more competitive?
The answer is: “it depends on what you mean by “more competitive”.”

On one hand, the 4 mile race offers prize money and attracts Olympic caliber runners, current world record holders in the marathon and the like. Typical winning times for males is under 18 minutes and the first woman sometimes breaks 20 minutes. There are also a large number of university runners chasing them. So, at the very front of the pack, the 4 mile race is much more competitive.

But the “typical” 15 Km runner is far more serious than the “typical” 4 mile runner. Here is what I mean:

(2011 statistics) 4 mile race had 3346 finishers, median runner (half faster, half slower) was 39:58 (9:59.5 minutes per mile). The 15K race had 836 finishers; the median time was 1:23:25 (8:57 minutes per mile) and that was LONGER and on a much more difficult course (4 mile course is pancake flat).

If you wonder about the mix of men and women, I went ahead and compared the male and female age groups (50-54; my group):
4 mile men: 138 finishers, median time 37:05, median pace: 9:16
15K men: 45 finishers, median time 1:19:50, median pace: 8:34

4 mile women: 128 finishers median time 46:10, median pace: 11:32
15K women: 27 finishers, median time: 1:28:41, median pace: 9:32

That is, the typical 15 km runner will run a course that is over twice as long and much, much, much hillier at a faster pace than the typical 4 mile runner. So in this sense, the 15 km race is far more competitive.

In other words, I’ll be faster than the median pace for my age group if I ran the 4 mile but will be slower (much) in the 15K.

So, for this question “which race is more competitive”, the answer depends on “what you mean by “more competitive””.

Example two: this is the Bertrand paradox: Inscribe an equilateral triangle into a circle. Now pick a random chord in the circle (a line from one point on the circle to some other point). What is the probability that the length of this chord is longer than the length of one of the sides of the triangle?

Answer: it depends on what you mean by “randomly pick”!

Method 1. If you just pick some point “p” on the circle and then some second random point “q” on the circle then:
you can arrange for a vertex of the triangle to coincide with “p”. Then the chord will be longer if the chord pq lies in that 60 degree angle at the vertex; hence the probability of that happening is 1/3.

Method 2. Pick the line as follows: draw a random radius (segment from the center to the edge) and then randomly pick some point on the radius and construct a perpendicular to that. Arrange for the inscribed triangle to have one angle bisector to overlap the radius. Now the chord will be longer than the side of the triangle if the second point is between the center and the opposite edge of the triangle. Since the side bisects the radius, the probability is 1/2. Method 3. Choose a point anywhere in the circle and let that be the midpoint of the random chord. Then the chord is longer than a side of the inscribed triangle if and only if the point happens to lie inside the circle that is inscribed INSIDE the equilateral triangle. Since that area is 1/4’th of the area inside the circle, the probability is 1/4’th.

For more on what is the “best method”: read the article:

In his 1973 paper The Well-Posed Problem, Edwin Jaynes proposed a solution to Bertrand’s paradox, based on the principle of “maximum ignorance”—that we should not use any information that is not given in the statement of the problem. Jaynes pointed out that Bertrand’s problem does not specify the position or size of the circle, and argued that therefore any definite and objective solution must be “indifferent” to size and position. In other words: the solution must be both scale invariant and translation invariant.

It turns out that method 2 is both scale and translation invariant.

## June 5, 2012

### Quantum Mechanics, Hermitian Operators and Square Integrable Functions

In one dimensional quantum mechanics, the state vectors are taken from the Hilbert space of complex valued “square integrable” functions, and the observables correspond to the so-called “Hermitian operators”. That is, if we let the state vectors be represented by $\psi(x) = f(x) + ig(x)$ and we say $\psi \cdot \phi = \int^{\infty}_{-\infty} \overline{\psi} \phi dx$ where the overline decoration denotes complex conjugation.

The state vectors are said to be “square integrable” which means, strictly speaking, that $\int^{\infty}_{-\infty} \overline{\psi}\psi dx$ is finite.
However, there is another hidden assumption beyond the integral existing and being defined and finite. See if you can spot the assumption in the following remarks:

Suppose we wish to show that the operator $\frac{d^2}{dx^2}$ is Hermitian. To do that we’d have to show that: $\int^{\infty}_{-\infty} \overline{\frac{d^2}{dx^2}\phi} \psi dx = \int^{\infty}_{-\infty} \overline{\phi}\frac{d^2}{dx^2}\psi dx$. This doesn’t seem too hard to do at first, if we use integration by parts: $\int^{\infty}_{-\infty} \overline{\frac{d^2}{dx^2}\phi} \psi dx = [\overline{\frac{d}{dx}\phi} \psi]^{\infty}_{-\infty} - \int^{\infty}_{-\infty}\overline{\frac{d}{dx}\phi} \frac{d}{dx}\psi dx$. Now because the functions are square integrable, the $[\overline{\frac{d}{dx}\phi} \psi]^{\infty}_{-\infty}$ term is zero (the functions must go to zero as $x$ tends to infinity) and so we have: $\int^{\infty}_{-\infty} \overline{\frac{d^2}{dx^2}\phi} \psi dx = - \int^{\infty}_{-\infty}\overline{\frac{d}{dx}\phi} \frac{d}{dx}\psi dx$. Now we use integration by parts again: $- \int^{\infty}_{-\infty}\overline{\frac{d}{dx}\phi} \frac{d}{dx}\psi dx = -[\overline{\phi} \frac{d}{dx}\psi]^{\infty}_{-\infty} + \int^{\infty}_{-\infty} \overline{\phi}\frac{d^2}{dx^2} \psi dx$ which is what we wanted to show.

Now did you catch the “hidden assumption”?

Here it is: it is possible for a function $\psi$ to be square integrable but to be unbounded!

If you wish to work this out for yourself, here is a hint: imagine a rectangle with height $2^{k}$ and base of width $\frac{1}{2^{3k}}$. Let $f$ be a function whose graph is a constant function of height $2^{k}$ for $x \in [k - \frac{1}{2^{3k+1}}, k + \frac{1}{2^{3k+1}}]$ for all positive integers $k$ and zero elsewhere. Then $f^2$ has height $2^{2k}$ over all of those intervals which means that the area enclosed by each rectangle (tall, but thin rectangles) is $\frac{1}{2^k}$. Hence $\int^{\infty}_{-\infty} f^2 dx = \frac{1}{2} + \frac{1}{4} + ...\frac{1}{2^k} +.... = \frac{1}{1-\frac{1}{2}} - 1 = 1$. $f$ is certainly square integrable but is unbounded!

It is easy to make $f$ into a continuous function; merely smooth by a bump function whose graph stays in the tall, thin rectangles. Hence $f$ can be made to be as smooth as desired.

So, mathematically speaking, to make these sorts of results work, we must make the assumption that $lim_{x \rightarrow \infty} \psi(x) = 0$ and add that to the “square integrable” assumption.