# Award Fest 2012

The Math department made out like bandits last week.  Not one, not two, but three OU math faculty won big awards.

David Ross Boyd

Andy Miller received a David Ross Boyd Professorship.  This and the George Lynn Cross professorships are the most prestigious awards given to OU faculty.   According to the University:

To qualify for a David Ross Boyd Professorship a faculty member must have consistently demonstrated outstanding teaching, guidance, and leadership for students in an academic discipline or in an interdisciplinary program within the University.

David Ross Boyd was the first president of OU.  He’s famous (he has his own Wikipedia page!) for planting the thousands of trees at OU, and for being fired by the first governor of Oklahoma for being an “aristocrat, not democratic enough.”

Christian Remling received a Regent’s Award for Research and Creative Activity.  At most three of these are given each year (this year only one was given out!) for excellence in research.  Dr. Remling got it for, of course, his recent paper in the Annals (which we talked about here).  One of the world experts in this area of research called the paper “Remling’s earthquake”!

And Jon Kujawa received the Irene Rothbaum Outstanding Assistant Professor award.  It’s given for:

The nominee should demonstrate outstanding teaching as shown by student-teacher evaluations, chair recommendations and additional information such as student or peer letters of support…. The candidate also should show evidence for good progress in scholarship, such as publication in peer-reviewed outlets and fellowship or grant support.

As you know, our own Ravi Shankar won the award in 2006 and Nikola Petrov won it in 2009.  We reported on Dr. Petrov winning here, but Dr. Shankar won BB (Before Blog).

With Dr. Kujawa winning it in 2012, we fully expect someone in Math to win it in 2015 (we’re looking at you Dr. Pitale!)

# Noetherian Zombies

One of our favorite mathematicians, Emmy Noether, got a shout out in xkcd this week:

No word on zombie Noether sightings. Although Dr. Pitale's and Dr. Özaydin's final exams are likely location for zombie-tired people mumbling about Noetherian rings.

# Pi Day!

If you see Andrew Holmes today, wish him a happy Pi Day!  (He’s a fan of $\pi$ (and pie)).

Those of you who were at Dr. Pitale’s Math Club talk learned many cool things.  We thought we’d share one in honor of Pi Day.

Question:  If you choose two positive integers at random, what is the probability that they will be relatively prime (i.e. only 1 will divide them both evenly)?

Answer: $6/\pi^2 \sim .6079$.  So around 60%.

The good question is how does $\pi$ show up in a question about relatively prime numbers?  Well, when you took Calc III and learned about series, you learned about the series

$\sum_{n=1}^{\infty} \frac{1}{n^2}.$

You probably even learned that this is a p-series with p=2, and that using the integral test it’s easy to see that it converges.

You might have even learned that it converges to $\pi^2/6$ (this was first shown by Euler, but Dr. Pitale told us that Euler only gets partial credit as his answer was right, but his proof was wrong!).  Notice that this number is the reciprocal of the probability from above.  How strange!

We won’t give the full explanation (ask Dr. Pitale for the details, or take his number theory class in the fall!), but let’s at least sketch how it goes:

Say you pick two positive integers at random: call them A and B.  We want to know the probability that they are relatively prime.

Well, they’re relatively prime if they have no common factors.  That is, if for each prime p, p does not divide both A and B.  What’s the probability of that?  It’s easier to compute the probability that p does divide both A and B first.

For example, if p=2, then both A and B have to be even numbers if 2 divides both of them.  The possibilities are: A is even/B is even, A is even/B is odd, A is odd/B is even, A is odd/B is even.  Only 1/4 of the possible outcomes gives both A and B divisible by p=2.

Similarily, if p=3, then 1/9 of the possible outcomes gives both A and B divisible by p=3.  In general, it’s not too hard to see that $1/p^2$ is the fraction of possible outcomes where both A and B are divisible by p.

Therefore, $1-(1/p^2) = (1-p^{-2})$ is the probability that A and B are not both divisible by p.

But the odds that A and B are both divisible by one prime is independent of whether they are both divisible by another prime.  The probability of independent events is the product of the probabilities of each individual event.  So the probability that A and B are relatively prime is equal to

$\Pi_{p \text{ prime}} (1-p^{-2})$.

Let’s consider the reciprocal of this number:

$\frac{1}{\Pi_{p \text{ prime}} (1-(p^{-2}))} = \Pi_{p \text{ prime}} \left(\frac{1}{1-(p^{-2})} \right)$.

Now let $r= p^{-2} = \frac{1}{p^2} < 1$, so if we think about the geometric series

$\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$

we see we can substitute this geometric series into the above product to get

$\Pi_{p \text{ prime}} (1+p^{-2} + (p^{-2})^2+ (p^{-2})^{3}+ \cdots)$.

Now, when you FOIL out this product, you should keep in mind that all but finitely many of the terms should be the 1 (That’s how infinite products are defined).    So what do you get when you multiple out?

By the Fundamental Theorem of Arithmetic every natural number can uniquely be written in this as a product of prime powers!  So on the one hand,

$p_1^{a_1} p_2^{a_2}p_3 \cdots p_r^{a_r} = n$

for some natural number n, and on the other hand, each natural number n appears exactly once as one of these products.

If you look carefully what happens when you FOIL it out, you see using the Fundamental Theorem of Arithmetic that you get exactly

$1^{-2} + 2^{-2}+3^{-2}+4^{-2}+ 5^{-2} + \cdots$

That is, you get exactly

$\sum_{n=1}^{\infty} n^{-2} = \sum_{n=1}^{\infty} \frac{1}{n^2}$.

Therefore, the reciprocal of the probability is exactly the p-series we started with!

It’s absolutely amazing that this well know series computes such a natural probability/number theory question.  And it has the bonus of having $\pi$ showing up, too!

On the other hand, you could just enjoy this $\pi$ video suggested by our own Dr. McCullough:

Or, you could just make some $\pi$ pie.

# The Riemann Hypothesis

Ameya Pitale

Our own Dr. Ameya Pitale will be speaking on

Wednesday, March 2nd at 5pm in PHSC 1105

on the famous Riemann Hypothesis.

The Riemann zeta function is defined for any complex number, s, using an infinite series as:

Well, technically, this is not right since even your friend in Calc III knows (or at least should know!) that this series doesn’t converge if s is less than 1.  The above function is the Riemann Zeta function for s greater than 1, and then you have to use some complex analysis to define it for all other s values.

The Riemann hypothesis is then the conjecture that the only non-stupid zeros of the the function are when the real part of s is 1/2.  According to Wikipedia, 10,000,000,000,000 zeros have so far been found and all of them are where they should be.  If you prove or disprove the Riemann hypothesis, then besides being as famous as Bat Boy (see below), you’ll earn yourself a million bucks.

Here’s a cool video which shows you what we mean:

Dr. Pitale will tell us all about the math involved and why the Riemann Zeta function is so darned important:

I am sure that students have often wondered while studying sequence and series in a Calculus 3 class as to the need and usefulness of the material.  I would like to start with some very simple questions whose answers lead naturally to understanding certain series. These series are special cases of a very famous function in number theory and all of math – the Riemann Zeta function. This function is full of rich information about prime numbers and studying it gives us deep insights into some very difficult questions in number theory. And if you are really smart and very very hard working, then it can also lead you to a million dollars.

– Dr. Pitale’s abstract

Or, xkcd’s take on it: