Monday, November 25, 2013

On Leaving Academia (Twice)

Moments of decision draw our attention because of the way they lead us into realms of psychological experience. A reporter asked me recently at what point I realized I was going to leave academia to pursue work in education policy. Hers was a novelist’s question, and it has prompted this rather novelistic answer.

As in novels, I begin with a preface. Regular readers of this blog will have realized that the recent years have been marked, to a degree greater than in my previous life, by sleeplessness and overwork. (See herehere, and the comments section here.) I would date this period to the moment when I became a parent: young children and theoretical physicists turn out to have very different opinions about what counts as “early morning,” while somehow managing to agree on what counts as a late night. And then there were the labors of delivering the math standards, a large process that I’ve spoken about often (including here and here).

My work on the standards was reaching high intensity during a period that also saw the birth of my second child and the death of my mother in 2009. Indeed, until the death of my father in 2011, I was in charge of my parents’ long-term care, having moved them from the Detroit area to a nursing home in North Bennington in 2007.

Inevitably perhaps, given all this, my memories seem weak...at least relative to such things as the ability to quote standards passages verbatim. But then by nature I am not a person who looks backwards very much.

***

As challenging as the years prior to my departure from Bennington College were, they were also an exciting period of rapid intellectual growth—growth that I attribute in large part to my membership in Bennington’s unusually vital intellectual community. My academic output during this time was varied, and my course designs were novel. I participated in the founding conversations that ultimately led to the College’s new Center for the Advancement of Public Action. But teaching is a job that requires total presence, and at a residential liberal arts college, this total presence must be sustained for fourteen weeks at a time. My life had become logistically and spiritually incompatible with that. My students weren't getting my best anymore. Neither, for that matter, was my own family; as heroic as I can make my own efforts sound, it was my wife's strength and not mine that carried us through those years.  

***

Fall 2010 wasn’t the first time I left academia. As a doctoral student at Berkeley in 1999, I came to the conclusion that I needed to abandon physics in order to support my family. My father was 72 years old at the time, working 60 to 70 hours per week in a factory. And if he never complained about it, I could stand it no longer. I looked at hedge funds and other things that quantitatively skilled people do to earn money, but I chose a riskier path instead by moving to the East Coast and starting a company. Soon after moving to New York City, I arranged a better job for my father with his plant supervisor. My father would transition from the floor to the tool crib, a job where he could sit during his shift in an air conditioned environment. This was a major improvement, especially as my father’s Parkinsons symptoms had become pronounced by this time. Those symptoms progressed quickly, and it wasn’t long before he was confined to a bed in his home.

***

How does one 'come to a conclusion' like the one I drew in 1999? Mysteriously, I think. At first, there is no locus of decision—it’s as if the teeming cells of the body are jostling with one another to cast their votes. Eventually the hindbrain is informed. Later, the frontal cortex is brought in to make a formal announcement. Until that moment, you suffer existential agitation.

Leaving academia the second time was much the same. I was on a leave of absence from the College, but the schedules for classes and committees were taking shape for the coming term, making a decision necessary. The deciding “cellular votes” in this case probably came from my skeleton; I knew especially in my bones that I was too tired to return to the intensity of a Bennington semester, on top of all the standards-related work that remained to be done.

***

I don’t remember now why I phoned the President of the College, rather than meeting with her face to face. Liz might have been travelling. Or I might have been acting on a sudden burst of courage, one that might fade if subjected to the delay of getting on her official calendar. In any case, I remember that I called from my kitchen in Pownal. Liz probably knew what I was going to say as soon as she saw my number on her phone. After I got to my point, it was clear that she was relieved by my decision.

Liz and I have been friends for a long time. She was my President as a faculty member, and for me she probably played an even greater role during those years—years when my own mother, a similarly strong and intelligent woman, was slowly dying. Liz was relieved because she could see that I had pulled myself to the breaking point, and she knew I wasn’t bodily and spiritually safe as long as it continued. I expressed my most important concern, that I wasn't capable of doing a good enough job in the coming semester. This also was clear to her.

***

In some ways, leaving academia (again) has been a shift (again). But as I view my life in perspective, I also see a continuity: I’ve been an educator for my entire working life. But why did I take up K-12 education again, after so many years in higher education? One straightforward reason: I saw a problem, and I wanted to solve it. Physicists are problem solvers above all, as I noted in a faculty lecture I gave in 2007; that is one reason why you find physicists in all kinds of fields.


It is true that in many ways, I was flourishing at Bennington College. I was widening my intellectual horizons and deepening my understanding of what it means to teach. I worked on varied problems that interested me. I published the textbook I’d long envisioned writing.

But even with all that, I felt restless. A clue as to why might be found in an experience that I had in 2008.


Pownal is a small town in southern Vermont. A brief term as Chair of the town’s planning commission in 2008 reawakened me to public service. For several years, Pownal had been having a difficult time passing new zoning regulations. Leading the process to a successful conclusion that year was something I was proud of, because it made my town a better place to live for thousands of people. This episode—as minor as minor could be on the scale of national events—has nevertheless grown in personal significance as I’ve reflected on it these past few years.

I left academia twice, because twice in my life I came to realize that there was a specific, important thing that I was capable of doing and that I ought to be doing. I left once to serve my parents. I left a second time to serve my country, agreeing with others that mathematics is important to the prosperity of a nation and to the soul of an educated person.

***

People often ask me if I miss academia. Actually, I had a sharp reminder of my earlier life only a few days ago. It was after dinner, and my daughters were sitting at their desks doing extra math problems in the workbooks they use at home. As they filled up their pages, I drifted quietly from one desk to the other, getting one or the other of them unstuck at times and checking over the answers on each page. I fell so naturally into my old professorial rhythms that the experience would have been poignant, if not for the sense of privilege I felt to be participating so deeply in my children's learning. It may have been as brief as ten minutes. But for those ten minutes, there was only the hum of quiet concentration in concert, set to the ticking of the living room clock. I felt the opposite of restless.

Friday, September 20, 2013

Number/Syllable

The puzzles in my last post led to great answers in the comments, and also some great thinking in email threads with friends. A variety of puzzle answers and open questions are below.

First, however, for further reading on the topic of numbers, or even just a gift idea, you might check out The Book of Numbers by John Conway and Richard Guy. I haven't read it (I saw the citation while browsing here), but Conway is a famous mathematician with a flair for games and what he writes is usually well worth reading. He spoke to my college Linear Algebra class one morning, and it was an amusing experience.

Without further ado:

1) Amy subtracted a three-syllable number from a three-syllable number and obtained a thirty-seven syllable number. What could her numbers have been?

For this puzzle, a reader posted the following answer (which was my answer too):

    12,000,000,000,000 - 23 = 11,999,999,999,977.

To devise this puzzle, I first thought about the difference "1 trillion minus one," which would have led to a puzzle of the form [3] - [1] = [30]. But then it seemed more intriguing for the first two numbers to have the same (small) number of syllables---a decision that also allowed me to pump up the syllable count of the difference in two ways:  choosing the subtrahend 23 (which leads to a 77 in the difference) and upping the minuend to 12 trillion (which leads to an 11 in the difference). That resulted in a puzzle of the form [3] - [3] = [37].

We can find some amusing new kinds of answers if we relax the whole-number constraint, an idea suggested by the first comment on the original post. Then we can find cases like this one:

    (0.000000000003) - (8 + i) = -7.999999997 - i.

In words,

    "three trillionths minus eight plus i equals negative seven and nine hundred ninety-nine billion, nine hundred ninety-nine million, nine hundred ninety-nine thousand, nine hundred ninety-seven trillionths minus i"

or in terms of syllable counts, [3] - [3] = [42].


2) What is the smallest number you can find with nineteen syllables?

A reader posted this answer, which I believe is optimal:

    177,777.

He also observed that Spanish beats English in this context, because one can get to 19 syllables with a smaller number. The smallest I have found is

   144,444.

I don't know if this is optimal for Spanish. Nor do I know whether another language would do even better. (German might be worth a look, but Esperanto doesn't look promising.) For those interested in pursuing the Spanish question, this tool is helpful.

Relaxing the whole number constraint once again, note that the words "decillionths" and "one hundred" both have three syllables. So if we just delete the 1 from 177,777 and append "decillionths," then we find a very small, 19-syllable (rational) number:

    77,777 decillionths = 0.000000000000000000000000000077777.


3) What is the largest number you can find for which the number of syllables equals the sum of the digits?

One reader found several examples for this one. First,

    721,721,721,721,721,721,721,341,341,341,341,141

or "seven hundred twenty-one decillion, etc. etc." (Number of syllables = 108 = sum of digits.)

Second, working again in the decillions, he found the improved answer

    999,999,992,111,111,111,111,111,111,111,111,111

(number of syllables = 101 = sum of digits).

This was my own answer as well. I suspect it is optimal relative to the decillions-and-below nomenclature that I linked to in the original post. Nomenclature for higher numbers can be found here and here.

My own initial strategy for this puzzle was just to futz around until I found a 36-digit example that began with 999. Then I said, Can I add another 9? If I do, then my partial digit sum will be 4*9 = 36, while my partial syllable count will be 12 ("nine hundred ninety-nine decillion, nine hundred..."). That puts the digit sum 24 points ahead of the syllable count already. If I want to recover parity, then the digits I add from here on out will have to give me a lot of net syllables. Adding groups of 111 does so, because each 111 adds eight or nine to the syllable count, but only adds three to the digit sum. (Each group of 111 buys me five or six points net.) In fact, I can add enough 111's at the back of the number to buy a lot more 9's at the front. Finally, when the 9's and the 1's met up in the middle, I knitted them together with a 2, the digit that made the number of syllables and the number of digits equal.

Applying this proto-algorithm to numbers in the trillions and below leads to

    998,111,111,111,111

which I suspect is optimal for 15 digits.

The reader's third number, a clever tongue-in-cheek suggestion, refers to a naming convention that I didn't even know. It is

   999,999,992,111,111,111,111,111,111,111,111,111 googolplexian 6.


One more note on puzzle #3: in the email thread, an interesting question was raised as to whether there is a largest whole number for which the syllable count equals the digit sum. Unfortunately, none of the conventions I've seen so far for assigning words to numbers has been spelled out to infinity, so the question would have to be made precise in some fashion. For example, consider any naming convention for which there exists a value of k such that all period names from the kth onward have greater than 27 syllables. In such a naming system, is the digit sum eventually doomed to fall behind the syllable count forever? If so, then relative to that system, there will exist a largest whole number for which the syllable count equals the digit sum.  (Assuming there is at least one such number in the naming system!)


4) What is the largest prime number you can find for which the number of syllables equals the sum of the digits?

Here a reader found

    102,701,

and my mathematical computing platform helped me in finding a pretty big example. But before I give it, let's first note the smallest prime with this property, namely

    103.

This is the 27th prime number, which makes puzzle #4 a little more difficult than I would have liked, because you have to go a fair ways up through the primes before the first example can be found. (Puzzle #3 was more accessible, because both 1 and 10 are solutions to it.)

Working in the decillions, the largest prime I could find with syllable count equal to digit sum was

    999,999,991,111,111,111,121,111,111,111,111,111.

(Number of syllables = 101 = sum of digits.)

I don't think there are any larger examples in the decillions. I actually found this number by permuting digits of the answer to puzzle #3 and just hoping that the result would turn out to be a prime number with a syllable count of 101. (Scientific, huh?)


Finally, in case interesting, here is a scatter plot that shows, in a randomly sampled way, the relationship between the number of digits in a number (horizontal axis) and the number of syllables in the number (vertical axis).



To make the plot, I randomly selected ~200,000 exponents uniformly in the range from 0 to 36; raised 10 to the power of each; and rounded the result. For each such integer, I created an (x, y) data point by counting the number of digits and the number of syllables. Unfortunately, because this data set is large and lives on a lattice, a simple scatter plot is deceptive: some data markers would represent hundreds of data points, while other data markers might represent only one data point---and there would be no way to see the difference. I usually deal with this problem by adding a little random jitter to the data when I plot it. But here that didn't suffice. So instead I gave each data marker a color commensurate with the number of data points that live on the lattice point it occupies. The resulting graph better reveals the "main sequence" within the data (darker blue dots).

I'll end this post with a similar plot for the digit sum vs. the number of syllables - n.b., this one uses the same general technique but a different color function.


Thanks everyone for your creative answers and ideas!

Saturday, September 7, 2013

Numbers of Syllables / Syllables of Numbers


When spoken aloud, the number 11 has three syllables, the number 105 has four syllables, and the number 1,200 has six syllables. Some puzzles on this general topic are below...feel free to put answers in the comments---or add more puzzles!  


1) Amy subtracted a three-syllable number from a three-syllable number and obtained a thirty-seven syllable number. What could her numbers have been?


2) What is the smallest number you can find with nineteen syllables?

(You should be able to do better than 13,012,321,012,001.)


3) What is the largest number you can find for which the number of syllables equals the sum of the digits?


4) What is the largest prime number you can find for which the number of syllables equals the sum of the digits?



Wednesday, January 30, 2013

Curious Cuboids, Part 2


Many of the problems in my previous post were instances of a general question:

Given three positive numbers V, S, and E, does there exist a cuboid with volume V in cubic inches, surface area S in square inches, and edge sum E in inches? 

When I considered this question, I found that such a cuboid exists if and only if


   E^2 S^2 - 2 S^3 - 16 E^3 V + 36 E S V - 108 V^2 >= 0.


The inequality was easy to obtain as soon as I realized that the desired cuboid exists if and only if the following polynomial in p has three positive roots:


 p^3 - E p^2  + (S/2) p -V.

(The left-hand side of the inequality is the discriminant of the polynomial.) Proof sketches are appended at the end.

***

The VSE inequality can be expressed in terms of two variables, the scaled surface area s = S/(2E^2) and scaled volume v = V/(E^3). In terms of these parameters, the space of all possible cuboids looks like so:


Each light-blue or dark-blue point corresponds to a cuboid (except for the dashed line along the bottom, which is s = 0). The two curved boundaries are where the discriminant vanishes, indicating a repeated root. Along the left boundary, there is a double root and its value is smaller than the other root; these are the rod-shaped cuboids. Along the right boundary, there is a double root and its value is larger than the other root; these are the tile-shaped cuboids. The right and left boundaries meet at a point right at the top; this cuboid is both rod-shaped and tile-shaped, i.e., it is a cube.

I put a little tour of the space up on YouTube here.

***



Well, it turns out that a cuboid can't have V, S, and E numerically equal when measured in the same system of units. Seems a shame, really. Well, there's a desperate example in one of my notebooks with edge lengths 2, i, and -i.

But to keep it real, let's open things up to other shapes. Define the diameter of a 3D shape to be the greatest possible distance between two of its points. Can you find a shape for which the volume, surface area, and diameter are all numerically equal when measured in the same system of units? (A few examples are on my whiteboard right now.)


***

It is nice to have an if-and-only-if condition, but the inequality is bit complicated. A few simpler, "only-if" inequalities (necessary conditions) are easy consequences of the inequality relating arithmetic and geometric means:  Given n nonnegative numbers c_1, ..., c_n,

   (1/n)(c_1+c_2 + ...+c_n)  >=  (c_1 c_2 ... c_n)^(1/n),

with equality if and only if all the c_i are equal. (This inequality played a role in one of our previous puzzles.)

If we choose the c_i to be the edge lengths x, y, and z, then we find

   E >= 3V^(1/3),

with equality if and only if the cuboid is a cube.

This inequality verifies our intuition that a cuboid can't enclose a given volume using arbitrarily little edge length. For example, there is no cuboid with a volume of a million cubic inches and an edge sum of 1 inch.

The inequality can also be written

   (E/3)^3 >= V

again with equality if and only if the cuboid is a cube.

The two forms of the inequality correspond to two optimal properties of the cube:  Of all cuboids with a given volume, the cube has the least edge sum; and of all cuboids with a given edge sum, the cube has greatest volume.

(Proof of the second property: Let C be a cuboid with edge sum E. Suppose K is any cuboid with edge sum E that is not a cube. Denote the volumes by V_C and V_K. Applying the inequality to K, we have (E/3)^3 > V_K. But of course, (E/3)^3 = V_C. Therefore V_C > V_K.)

Now take the c_i to be xy, yz, zx. This gives

   S >= 6V^(2/3)


again with equality if and only if the cuboid is a cube.


This again establishes something intuitively obvious, which is that a cuboid can't enclose a given volume using arbitrarily little surface area. For example, there is no cuboid with a volume of 1 cubic inch and a surface area of a millionth of an inch.

The inequality can also be written

   (S/6)^(3/2) >= V


again with equality if and only if the cuboid is a cube.


The two forms again correspond to two optimal properties of the cube:  Of all cuboids with a given volume, the cube has the least surface area; and of all cuboids with a given surface area, the cube has greatest volume.

The last two inequalities imply a third:

   ES >= 18V.

This can also be found by applying the inequality of arithmetic and geometric means, taking the c_i to be the nine expressions xyzxyzxyzx^2yxy^2, x^2zxz^2, y^2z, and yz^2.

For a cuboid with measures V, S, and E to exist, it is necessary but not sufficient that V, S, and E satisfy these simpler inequalities. For example, the numbers E = 8, S = 78, and V = 9 satisfy all of the "only-if" inequalities, although in fact there is no cuboid with these measures.

Another "only-if" inequality follows from applying the Cauchy-Schwarz inequality to the vectors (x, y, z) and (z, x, y). The result is (1/2)S <= x^2 + y^2 + z^2. But x^2 + y^2 + z^2 = E^2 - S, so we have

   S <= (2/3)E^2


again with equality if and only if the cuboid is a cube.


This inequality can be used to show two more properties of the cube:  Of all cuboids with a given surface area, the cube has the least edge sum; and of all cuboids with a given edge sum, the cube has greatest surface area.

***

By the way, we still need the if-and-only-if VSE inequality. None of the "only-if" inequalities is strong enough to rule out the existence of a cuboid with a volume of 80 cubic inches, a surface area of 122 square inches, and an edge sum of 16 inches (though in fact no such cuboid exists).

***

Proof sketch:

(1) Given that a cuboid exists with measures V, S, and E, to show that the polynomial p^3 - E p^2  + (S/2) p -has three positive roots. Idea: the positive roots are none other than the edge lengths.

Denote the cuboid's edge lengths by (positive) x, y, and z. Then since x + y + z = E, 2(xy + yz + zx = S, and xyz = V, the polynomial is p^3 - (x + y + z)p^2 + (xy + yz + zx)p - xyz. This however factors as (p-x)(p-y)(p-z), hence the polynomial has three positive roots.

(2) Given that the polynomial p^3 - E p^2  + (S/2) p -V has three positive roots, to show that a cuboid with measures V, S, and E exists. Idea: the edge lengths are none other than the positive roots.

Denote the positive roots by r1, r2, and r3. Since the polynomial is a monic cubic, the coefficients can be expressed in terms of the roots as symmetric functions, specifically as E = r1+r2+r3, S/2 = r1r2+r1r3+r2r3, and V = r1r2r3. But these functions are also the formulas for the measures of a cuboid. So create a cuboid with edge lengths r1, r2, and r3. The cuboid then has the desired measures.

(3) To arrive finally at the desired inequality.

It is obvious by inspection that every real root of  p^3 - E p^2  + (S/2) p -V  is positive. (Try substituting a negative number or zero for p.) Hence by (1) and (2), a cuboid exists having the specified measures if and only if the polynomial has three real roots. It now follows that the discriminant of the polynomial provides a necessary and sufficient condition for the existence of a cuboid with measures V, S, and E.

***

I would guess that some of this generalizes to higher dimensions. The kth coefficient of an nth-degree monic polynomial is going to be an elementary symmetric polynomial function of the roots...and meanwhile, the k-dimensional measure M_k of an n-dimensional cuboid is going to be the same elementary symmetric polynomial function of the edge lengths, multiplied by an overall combinatorial factor C_nk that depends on the dimension of the measure and the dimension of the space. So if we define the reduced measures of a cuboid by m_k = M_k/C_nk, then there will exist an n-dimensional cuboid with measures M_1, ..., M_n if and only if the polynomial


p^n - m_1 p^(n-1) + m_2 p^(n-2) - ... + m_n 

has n real roots.

Even supposing this principle does generalize, however, the discriminant for higher polynomials doesn't tell the real/complex nature of the roots in the simple way that it does for n = 2 and n = 3. So for n >= 4, I don't think the discriminant will give an immediately usable condition on the coefficients m_i.

***

Usual disclaimers...I assume everything here that is correct is well-known. There may be some interesting related readings out there - if anybody knows of good ones, please put them in the comments!