BEST 2014, First Announcement

March 26, 2014

BOISE EXTRAVAGANZA IN SET THEORY (BEST)
http://diamond.boisestate.edu/~best/
June 18 – 20, 2014
University of California, Riverside

The 21-st meeting of BEST will be hosted at University of California, Riverside, as a symposium of the 95th annual meeting of the American Association for the Advancement of Science – Pacific Division (AAAS-PD). Contributed and invited talks at BEST will be held on Wednesday, Thursday and Friday.

In addition to four invited speakers, the conference program has reserved speaking slots for students, post docs and pre-tenure tenure track faculty. NSF supported funding to assist up to eight student speakers, four post-doc speakers and two pre-tenure tenure track faculty speakers is available. For details on applying to the BEST program committee for these, please visit the conference website. In addition, the AAAS-PD provides up to \$150 in travel funding for students. Please see the BEST conference website for more details on these also. There are a number of deadlines associated with applications for a travel grant.

The four invited speakers at BEST 2014 are:

Dr. Joel Hamkins, CUNY Graduate Center
Dr. Dima Sinapova, University of Illinois at Chicago
Dr. Nam Trang, Carnegie Mellon University
Dr. Andrew Marks, Caltech University

Special features of BEST 2014 include:

The BEST 2014 conference marks the end of three weeks of intensive set theory meetings in California. In addition to the BEST symposium there are several other symposia and workshops of interest offered at the AAAS-PD annual meeting. On Thursday, June 19, the AAAS-PD hosts a banquet at which awards of excellence are given to student speakers selected by a panel of judges. On Friday, June 20, Dr. Joel Hamkins will deliver a plenary lecture from 12:15 to 1:15 about our field to the general audience of the AAAS-PD annual meeting.

DEADLINE 1: REGISTRATION: Please consult http://diamond.boisestate.edu/~best/ for registration costs and deadlines. Registration fees depend on date of registration. We kindly request that tenure track mathematicians planning to participate in BEST 2014 consider acting as judges for the student presentations. The registration form has a place where willingness to act as a judge can be indicated. There are also a number of excursions available that can be indicated on the registration form. Also consider attending the award banquet in support of our student speakers – meal choices are available on the registration form.

DEADLINE 2: ABSTRACTS: Atlas Conferences, Inc. is providing abstract services for BEST 2014. Abstracts submitted by the deadline will appear in the proceedings of the annual conference of the AAAS-PD. The deadline for submitting an abstract is APRIL 16. The url for the abstract submission is available at the BEST 2014 website.

Organized by Liljana Babinkostova, Andres Caicedo, Sam Coskey and Marion Scheepers.

403 – Some references on Geršgorin’s theorem

March 24, 2014

I am posting here some references on Geršgorin’s theorem.

(Marsli and Hall have published recently a series of papers further exploring extensions of the theorem. Varga’s book is highly recommended.) If you want to practice some of the topics we have been covered, work through some of the exercises in the posted chapter, and turn them in for some extra credit, by April 8.

Related, though of a different nature, is the following. Geršgorin’s theorem is discussed in section 6:

Here I briefly review the result:

Theorem (Geršgorin, 1931). Let $A=(a_{ij})_{i,j=1}^n$ be a complex-valued matrix. For $1\le i\le n$ let $r_i(A)=\sum_{j\ne i}|a_{ij}|$. If $\lambda$ is an eigenvalue of $A$, then ${}|\lambda-a_{ii}|\le r_i(A)$ for at least one $i$.

To prove this, let $x$ be an eigenvector of $A$ with eigenvalue $\lambda$, say $x=(x_1,\dots,x_n)^T$, and let $i$ be such that $|x_i|=\max\{|x_k|\mid 1\le k\le n\}$. Since $Ax=\lambda x$, we have that $\sum_j a_{ij}x_j=\lambda x_i$, so $(\lambda-a_{ii})x_i=\sum_{j\ne i}a_{ij}x_j$. From this, the triangle inequality gives us that

$|\lambda-a_{ii}||x_i|\le\sum_{j\ne i}|a_ij||x_j|\le\sum_{j\ne i}|a_{ij}||x_i|,$

and the result follows since $x_i\ne0$.

March 19, 2014

This is a (somewhat expanding) list of suggested additional references. Some cover topics discussed in lecture, others add new material that complements what we covered. The level varies: Some are basic, others are more advanced and portions of them may require knowledge beyond this course.

For the group project: Choose one of these articles. Inform me by email, to make sure it has not already been chosen. Feel free to suggest different papers or other topics, I’ll see whether we can use them.

Write (type) a note on the topic discussed in the paper you have chosen, include details of some of the results discussed there. Make sure the proofs you include contain all needed details (typically proofs in articles are more sketchy than what we are aiming for through the course), and that the write up is your own, even if modeled on the arguments in the paper. Include references as usual. Turn this in by Thursday, May 15, at 10:30 am. Feel free to turn it in earlier, of course. I encourage you, as you work through the paper, to share your progress with me during office hours, so I can give you some feedback before your final submission.

Groups:

• Booker Ahl, Dorthee Berman, and Stephanie Potter: Russ’s translation of Bolzano’s paper.
• Tim Deidrick, Justin Durflinger, and Ariel Farber: Calkin-Wilf and Malter-Schleicher-Zagier on enumerating the rationals.
• Carrie Smith, and Jordan Wilson: Fleron’s note on the history of the Cantor set and function.
• Caleb Richards, and Chris VanDerhoff: McShane’s paper on the Henstock–Kurzweil integral.
• Kenny Ballou, Sarah Devore, and Luke Warren: Nitecki’s paper on subseries.
• Farrghun Abdulrahim, and Kenneth Coiteux: Burns and Hasselblatt’s paper on Sharkovsky’s theorem.
• Tyler Clark: Niven’s paper on formal power series.
• Joe Magdaleno, and Piper Gutridge: Bruckner and Bruckner-Leonard  on derivatives.

314 – On √n

March 10, 2014

Let’s prove that if $n\in\mathbb N$, then either $\sqrt n$ is an integer, or else it is irrational. (Cf. Abbott, Understanding analysis, Exercise 1.2.1.) There are many proofs of this fact. I present three.

1.

The standard proof of this fact uses the prime factorization of $n$: There is a unique way of writing $n$ as $\prod_{i=1}^k p_i^{\alpha_i}$, where the $p_i$ are distinct primes numbers, and the $\alpha_i$ are positive integers (the number $n=1$ corresponds to the empty product, but since $1$ is a square, we may as well assume in what follows that $n>1$).

We show that if $\sqrt{n}$ is rational, then in fact each $\alpha_i$ is even, so $\sqrt n$ is actually an integer. Write $\sqrt n=a/b$ where $a,b$ are integers that we may assume relatively prime. This gives us that $b^2n=a$.

Consider any of the primes $p=p_i$ in the factorization of $n$. Let $p^\beta$ and $p^\gamma$ be the largest powers of $p$ that divide $a$ and $b$, respectively, say $a=p^\beta c$ and $b=p^\gamma d$ where $p$ does not divide either of $c$ and $d$. Similarly, write $n=p^{\alpha}m$, where $p$ does not divide $m$ ($\alpha$ is what we called $\alpha_i$ above). We have

$p^{2\gamma} p^{\alpha}d^2m=p^{2\beta}c^2.$

The point is that since $p$ is prime, it does not divide $c^2$ or $d^2m$: If $q$ is a prime and $q$ divides a  product $hj$ (where $h,j$ are integers), then $q$ divides $h$ or it divides $j$.

This means that either $\alpha$ is even (as we wanted to show), so that $2\gamma+\alpha=2\beta$, or else (upon dividing both sides of the displayed equation by the smaller of $p^{2\gamma+\alpha}$ and $p^{2\beta}$), $p$ divides one of the two sides of the resulting equation, but not the other, a contradiction.

2.

The above is the standard proof, but there are other arguments that do not rely on prime factorizations. One I particularly like uses Bézout theorem: If $c$ is the greatest common divisor of the positive integers $a$ and $b$, then there are integers $x,y$ such that $ax+by=c$.

Suppose $\sqrt n=a/b$. We may assume that $a,b$ are relatively prime, and therefore there are integers $x,y$ such that $ax+by=1$. The key observation is that $\sqrt n=n/\sqrt n=nb/a$. This, coupled with elementary algebra, verifies that

$\displaystyle \sqrt n= \frac ab=\frac{nb}a=\frac {ay+nbx}{by+ax},$

but the latter is an integer, and we are done.

3.

Another nice way of arguing, again by contradiction, is as follows: Suppose that $\sqrt n$ is not an integer, but it is rational. There is a unique integer $m$ with $m<\sqrt n, so $0<\sqrt n -m<1$. Let $a$ be the least positive integer such that $(\sqrt n-m)a$ is an integer, call it $b$. Note that $0, which gives us a contradiction if $({\sqrt n}-m)b$ is again an integer. But this can be verified by a direct computation:

$(\sqrt n-m)b=(\sqrt n-m)^2a=(n+m^2)a-2m\sqrt n a$ $=(n-m^2)a-2m(\sqrt n-m)a$.

4.

As a closing remark, the three arguments above generalize to show that $\root k\of n$ is either an integer or irrational, for all positive integers $n,k$. Similarly, if $\displaystyle\root k\of {\frac ab}$ is rational for some positive integers $a,b$, then both $a,b$ are $k$th powers. (It is a useful exercise to see precisely how these generalizations go.)

403 – HW 2 – Linear algebra over F_2, and orthogonality

March 3, 2014

This set is due Tuesday, March 18, at the beginning of lecture.