## Ian Cavey on sphere bounded regions

October 10, 2015

I met Ian Cavey, an undergraduate at Boise State, about a year ago. He took my Communication in the mathematical sciences course. It was a pleasure to have him as a student. (You can see the slides of one of his group projects here.)

Last Spring, Ian took my advanced linear algebra class. Also, he told me he was interested in trying some independent research under my guidance. He proved to be quite independent (he chose the problem he wanted to work on) and resourceful (for instance, finding workarounds when his direct approach would not work). The result of this was a paper (Volumes of Sphere-Bounded Regions in High Dimensions) that he presented at the recent MAA centennial meeting (see page 18 of the program). You can see his slides here.

[You can see Norm Richert representing Math Reviews at the meeting in this post.]

At the conference, Ian not only presented his talk. He also competed and won first place in the US National Collegiate Mathematics Championship. Some additional details can be found here.

Ian’s talk is about his approach to estimate the $n$-dimensional volume of the central region of a unit $n$-cube bounded by $n$-spheres centered at the vertices. Ian proves that this volume quickly approaches 1. His slides detail his nice combinatorial argument that circumvents the need for explicit computations of unwieldy iterated integrals.

## Amazing grace

June 29, 2015

I.

(Photo by Chuck Burton)

At least six black churches burned last week.

Bree Newsome has been charged with “defacing monuments on state Capitol grounds”, which holds a fine of up to \$5,000, and “a prison term of up to three years or both.”

(See here.)

II.

So, we moved to Ann Arbor.

We visited the Hands-On Museum (thanks to Zach and Diana, and congratulations!),

and attended the Math Reviews Summer picnic.

I started today.

## Generations

May 13, 2015

My dad teaching Francisco to play chess (March 24, 2015).

## 311 – HW 8

April 23, 2015

HW 8 is due Tuesday, April 28, at the beginning of lecture.

Work in hyperbolic geometry. Given a triangle $\triangle ABC$, recall that its Saccheri quadrilateral $\Box ABB'A'$ based at $\overleftrightarrow{AB}$ is defined as follows: Let $M$ be the midpoint of $\overline{AC}$ and $N$ be the midpoint of $\overline{CB}$. Let $A',B'$ be the feet of the perpendiculars from $A$ and $B$ to $MN,$ respectively.

Continuing with the same notation, suppose now that $G$ is an arbitrary point on $\overleftrightarrow{MN}$, and let $H$ be a point on the ray $\overrightarrow{AG}$ with $GH=AG$. Show that $\Box ABB'A'$ is also the Saccheri quadrilateral of $\triangle ABH$ based at $\overleftrightarrow{AB}$.

## MathReviews

April 22, 2015

I will be taking a leave from BSU this coming academic year, and moving to Ann Arbor, to work as an Associate Editor at MathReviews.

## 403/503 – HW7

April 16, 2015

This exercise is due Tuesday, April 21, at the beginning of lecture.

Find the Singular Value Decomposition of

$\displaystyle M=\left(\begin{array}{ccccc}1&0&0&0&2\\ 0&0&3&0&0\\ 0&0&0&0&0\\ 0&4&0&0&0\end{array}\right).$

(I am not so interested in the specific answer, which can be found online, but rather in the process describing how one arrives to this answer.)

## 311 – HWs 6 and 7

April 15, 2015

HW 6 is due Thursday, April 16 and HW 7 is due Tuesday, April 21, both at the beginning of lecture.

HW6

Work in hyperbolic geometry.

1. Let $\ell$ and $m$ be two parallel lines admitting a common perpendicular: There are points $P\in\ell$ and $Q\in m$ with $\overleftrightarrow{PQ}$ perpendicular to both $\ell$ and $m$. Suppose that $A,B$ are other points in $\ell$ with $P*A*B$, that is, $A$ is between $P$ and $B$. Let $C$ be the foot of the perpendicular from $A$ to $m$, and let $D$ be the foot of the perpendicular from $B$ to $m$.

Show that $PQ. That is, $\ell$ and $m$ drift apart away from their common perpendicular.

(Note that $\Box PACQ$ and $\Box PBDQ$ are Lambert quadrilaterals, and therefore $PQ and $PQ. The problem is to show that $AC.)

As an extra credit problem, show that for any number $r>0$ we can find $B$ (on either side of $P$) such that $BD>r$, that is, $\ell$ and $m$ not just drift apart but they do so unboundedly.

2. Now suppose instead that $\ell$ and $m$ are critical (or limiting) parallel lines, that is, they are parallel, and if $Q\in m$ and $P\in\ell$ is the foot of the perpendicular from $Q$ to $\ell$, then on one of the two sides determined by the line $\overleftrightarrow{PQ}$, any line through $Q$ that forms with $\overleftrightarrow{PQ}$ a smaller angle than $m$ does, cuts $\ell$ at some point.

On the same side as just described, suppose that $C,D$ are points on $m$ with $Q*C*D$, that is, $C$ is between $Q$ and $D$. Let $A$ be the foot of the perpendicular from $C$ to $\ell$, and let $B$ be the foot of the perpendicular from $D$ to $\ell$.

Show that $PQ>AC>BD$. That is, $\ell$ and $m$ approach each other in the appropriate direction.

As an extra credit problem, show that for any $r>0$ we can choose $D$ so that $BD. That is, $\ell$ and $m$ are asymptotically close to one another. Do they drift away unboundedly in the other direction?

HW 7

Show that the critical function $\kappa$ is continuous. Recall that $\kappa:(0,\infty)\to(0,\pi/2)$ measures the critical angle, that is, $\kappa(x)=\theta$ iff there are critical parallel lines $\ell$ and $m$ and a point $Q\in m$ such that if $P$ is the foot of the perpendicular from $Q$ to $\ell$, and $PQ=x$, then $m$ and $\overleftrightarrow{PQ}$ make an angle of measure $\theta$ in the appropriate direction.

(In lecture we verified that $\kappa$ is strictly decreasing. This means that the only possible discontinuities of $\kappa$ are jump discontinuities. We also verified that $\kappa(x)$ approaches $0$ as $x\to\infty$, and approaches $\pi/2$ as $x\to0$. It follows that to show that $\kappa$ has no jump discontinuities, it suffices to verify that it takes all values between $0$ and $\pi/2$, that is, one needs to prove that for any $\theta\in(0,\pi/2)$ there is an $x>0$ such that $\kappa(x)=\theta$.)