Mathematical ancestors of Penn math faculty, from October 1999. (This was the department's 100th anniversary.) This lists the advisor's advisor's advisor's... until historical data that was (easily?) available at that time gave out. The longest-ago person listed on this page is Otto Mencke (Ph. D. 1665, 12 generations from Penn professor Stephen Shatz); most chains die out in the 19th century.

As of right now, the math genealogy project claims to know that my advisor's 26th-generation advisor is Elissaeus Judaeus (who was a student in the 1380s). Most mentions of Judaeus on the Internet seem to be by other people who have discovered this (Judaeus has 77000 or so mathematical descendents). But this post from the person who added him to the database gives some background -- he was for the most part a philosopher, it seems. He is described as "a mysterious figure who may or may not have been a Jew". His student Gemistus Pletho seems a little better understood; Wikipedia says "He was one of the chief pioneers of the revival of Greek learning in Western Europe." It seems that in that time a lot more data has been collected for the 14th through 17th centuries.

(As for me, hopefully in a few weeks it'll be possible to add me to the mathematical genealogy project. I defend on April 15.)

## 25 March 2010

## 18 March 2010

### One down, six to go

Perelman has been awarded the Millennium Prize. Press release from the Clay Math Institute.

As Peter Woit points out, no mention of the fate of the million dollars.

As Peter Woit points out, no mention of the fate of the million dollars.

### March Math Madness

From quomodocumque: what would the NCAA tournament look like if every game were won by the college or university with the better math department? (Berkeley -- excuse me, "California", as they're usually called in athletic contexts -- wins.)

Rather interestingly, 20 out of 32 first-round "games", and 37 out of 63 "games" overall -- more than half -- are won by the team that actually has the better seed in the basketball tournament. I suspect this is because quality of math departments and basketball teams are both correlated with the size of the school. This is especially true because ties were broken by asking how many people they could name at the school., which clearly has a bias towards

Rather interestingly, 20 out of 32 first-round "games", and 37 out of 63 "games" overall -- more than half -- are won by the team that actually has the better seed in the basketball tournament. I suspect this is because quality of math departments and basketball teams are both correlated with the size of the school. This is especially true because ties were broken by asking how many people they could name at the school., which clearly has a bias towards

*larger*departments.## 17 March 2010

### A puzzle about polynomials

I have a polynomial, P, with nonnegative integer coefficients. You want to know what it is. For any algebraic number x, you're allowed to ask me what P(x) is. How many questions do you have to ask me to be sure that you know what P is?

## 16 March 2010

### McDonalds: more than 10 served

Megabus.com adds Philadelphia-D.C. line, from the Daily Pennsylvanian.

We learn from a Megabus spokesperson that their vehicles use "less than a pint of fuel per passenger mile".

For those of you who don't have the misfortune of knowing this, there are eight pints in a gallon. So these busses get better than eight passenger-miles to the gallon!

Since most cars in the US get at least 20 miles or more to the gallon, this is really nothing to be proud of.

(I'm guessing that busses are actually more fuel-efficient than cars, at least if they run sufficiently full.)

We learn from a Megabus spokesperson that their vehicles use "less than a pint of fuel per passenger mile".

For those of you who don't have the misfortune of knowing this, there are eight pints in a gallon. So these busses get better than eight passenger-miles to the gallon!

Since most cars in the US get at least 20 miles or more to the gallon, this is really nothing to be proud of.

(I'm guessing that busses are actually more fuel-efficient than cars, at least if they run sufficiently full.)

Labels:
measurement,
transportation

## 14 March 2010

### The Calkin-Wilf shirt

You can actually buy a shirt with the Calkin-Wilf tree on it. I probably should buy it, if only so I can wear it when I inevitably give a talk on this subject again, either as a stand-alone talk (I've done it twice) or when I teach it in a class.

This is an infinite binary tree of rational numbers. It has 1/1 at the root, and the children of r/s are r/(r+s) and (r+s)/s. It turns out that this tree contains every positive rational number exactly once, giving an explicit enumeration of the rational numbers.

Also -- and this is what's really surprising to me -- let f(n) be the number of ways to write n as a sum of powers of 2, where no power of 2 can be used more than twice. Then (f(0), f(1), f(2), ...) = (1,1,2,1,3,2,3,1,4,3,5,2,5,3,4,...). The sequence of numbers f(n)/f(n+1) are the rational numbers in that tree, in the order which they appear if you traverse each level from left to right.

I learned about the shirt from this mathoverflow thread, where Richard Stanley gives the theorem of the previous paragraph as an example of a theorem with an "unexpected conclusion".

See the original paper by Calkin and Wilf. I've mentioned it before here and here. I think I first heard of this paper from Brent Yorgey's series of posts expanding upon the paper, or perhaps from Mark Dominus' blog. (Somewhat coincidentally, Yorgey and Dominus both live in Philadelphia. I claim this is coincidence, even though Wilf is at Penn, because I don't think either of them heard about the paper from Wilf.)

And can anything nice be said about the number of ways to write n as a sum of powers of 3, where no power of 3 can be used more than three times?

This is an infinite binary tree of rational numbers. It has 1/1 at the root, and the children of r/s are r/(r+s) and (r+s)/s. It turns out that this tree contains every positive rational number exactly once, giving an explicit enumeration of the rational numbers.

Also -- and this is what's really surprising to me -- let f(n) be the number of ways to write n as a sum of powers of 2, where no power of 2 can be used more than twice. Then (f(0), f(1), f(2), ...) = (1,1,2,1,3,2,3,1,4,3,5,2,5,3,4,...). The sequence of numbers f(n)/f(n+1) are the rational numbers in that tree, in the order which they appear if you traverse each level from left to right.

I learned about the shirt from this mathoverflow thread, where Richard Stanley gives the theorem of the previous paragraph as an example of a theorem with an "unexpected conclusion".

See the original paper by Calkin and Wilf. I've mentioned it before here and here. I think I first heard of this paper from Brent Yorgey's series of posts expanding upon the paper, or perhaps from Mark Dominus' blog. (Somewhat coincidentally, Yorgey and Dominus both live in Philadelphia. I claim this is coincidence, even though Wilf is at Penn, because I don't think either of them heard about the paper from Wilf.)

And can anything nice be said about the number of ways to write n as a sum of powers of 3, where no power of 3 can be used more than three times?

Labels:
Calkin-Wilf tree,
combinatorics,
number theory

## 09 March 2010

### People round their incomes to the nearest $5,000?

Here's something interesting: lots of people, when asked by the US Census Bureau "how much money do you make?", round to the nearest five thousand dollars.

See the data tables from the 2006 census. These give the number of people whose personal income is in each interval of the form [2500N, 2500N+2499], for integer N.

One sees, for instance, that the number of people making between $27,500 and $29,999 (which is near the mode of the distribution) is less than

Surprisingly, the effect occurs even at very low levels of earnings. If you make $87,714 in a year I can see rounding to $90,000 -- but is the person who makes $7,714 in a year really rounding to $10,000?

(I found this while trying to answer a question at Metafilter: How many people in the United States make more than $10,000,000 per year?. I seem to recall reading somewhere that personal income roughly follows a power law in the tails, but can't actually find a reference for this.)

There also seems to be a preference for multiples of $10,000 over multiples of $5,000 that are not multiples of $10,000. But I have work to do, so I'm not going to do the statistics.

See the data tables from the 2006 census. These give the number of people whose personal income is in each interval of the form [2500N, 2500N+2499], for integer N.

One sees, for instance, that the number of people making between $27,500 and $29,999 (which is near the mode of the distribution) is less than

*both*those making $25,000 to $27,499 and those making $30,000 to $32,499. Something similar occurs at all income levels -- the number of people making between 2500N and 2500(N+1)-1 dollars is smaller if N is odd (and thus this interval doesn't contain a multiple of 5000) than if N is even (and so it does).Surprisingly, the effect occurs even at very low levels of earnings. If you make $87,714 in a year I can see rounding to $90,000 -- but is the person who makes $7,714 in a year really rounding to $10,000?

(I found this while trying to answer a question at Metafilter: How many people in the United States make more than $10,000,000 per year?. I seem to recall reading somewhere that personal income roughly follows a power law in the tails, but can't actually find a reference for this.)

There also seems to be a preference for multiples of $10,000 over multiples of $5,000 that are not multiples of $10,000. But I have work to do, so I'm not going to do the statistics.

## 03 March 2010

### Turning clocks upside down

Apparently this is a puzzle blog now.

This morning at 9:30 I picked up my watch and turned it upside down. It appeared to read 4:00. The hour hand was actually in the position it would be at at 3:30, of course. But the minute hand was pointing straight up, so the time must be on the hour. Since I could easily tell that the hour hand wasn't pointing directly to the right, I suppose my brain interpreted it as 4:00 instead of 3:00. Of course I did not think any of this consciously, but only reconstructed it after the fact; my thought process was more like "hey, 9:30 upside down looks like 4:00. that's weird."

Is it ever possible to interpret the hands of a clock, turned upside-down (i. e. rotated by 180 degrees), unambiguously as a time? (Fudging like my brain apparently did this morning does not count.)

This morning at 9:30 I picked up my watch and turned it upside down. It appeared to read 4:00. The hour hand was actually in the position it would be at at 3:30, of course. But the minute hand was pointing straight up, so the time must be on the hour. Since I could easily tell that the hour hand wasn't pointing directly to the right, I suppose my brain interpreted it as 4:00 instead of 3:00. Of course I did not think any of this consciously, but only reconstructed it after the fact; my thought process was more like "hey, 9:30 upside down looks like 4:00. that's weird."

Is it ever possible to interpret the hands of a clock, turned upside-down (i. e. rotated by 180 degrees), unambiguously as a time? (Fudging like my brain apparently did this morning does not count.)

## 01 March 2010

### A puzzle with lights

Since I seem to be getting this question a lot lately: yes, I'm still alive! But I am writing a dissertation. (And waiting to hear back from places where I applied for jobs, but that is not an excuse because those applications are already out there.)

Here's a puzzle I heard a couple weeks ago. You have a 10-by-10 grid of lights. Some of them are on, some are off. You are allowed to make the following moves:

(a): you pick one of the lights which is the center of some 3-by-3 square (i. e. it is not on the edge of the grid) and switch

(b): like in (a), but with a 5-by-5 square.

Is it possible to get from an arbitrary starting position to the all-off position?

Here's a puzzle I heard a couple weeks ago. You have a 10-by-10 grid of lights. Some of them are on, some are off. You are allowed to make the following moves:

(a): you pick one of the lights which is the center of some 3-by-3 square (i. e. it is not on the edge of the grid) and switch

*all*the lights in that 3-by-3 square (on becomes off, off becomes on).(b): like in (a), but with a 5-by-5 square.

Is it possible to get from an arbitrary starting position to the all-off position?

Subscribe to:
Posts (Atom)