Monday, 13 August 2018

Fabian Franklin's Beautiful Proof of the Pentagonal Theorem

 *Wik

On Aug 16, 1878 Charles Hermite wrote to J J Sylvester at Johns Hopkins concerned about his accepting a Math Chair in America and questioning the ability of the American people to contribute to research-level mathematics. Only three years later he would be reading the paper of Fabian Franklin, a young assistant mathematics instructor at Johns Hopkins, before the French Academy. The paper was on a short, purely graphic, proof of Euler's theorem on pentagonal numbers. Hans Rademacher called this proof “the first major achievement of American mathematics.”

Some background for students: Pentagonal numbers are named for the ways of arranging dots into pentagons, much like the square numbers or triangular numbers. The true or pure pentagonal numbers are 1, 5, 12, 22, 35, 51, 70, 92,...
 *Wik

You can get them by using the formula $\frac{3n^2-n}{2}$ with n a positive integer.
But for what we are doing today, we need to also include the generalized pentagonal numbers. They are obtained from the formula given above, but with n taking values in the sequence 0, 1, −1, 2, −2, 3, −3, 4..., producing the sequence 0, 1, 2, 5, 7, 12, 15, 22, 26, 35,...

One of the amazing things about this sequence is that it shows up in relation to finding the sum of the divisors of n, and finding the number of partitions of n. As Euler used it for the Pentagonal Number theorem, it was written out as $1 − n − n^2 + n^5 + n^7 − n^{12} − n^{15} + n^{22} + n^{26} − n^{35} − etc$ Notice all these numbers are the same as the ones in the generalized pentagonal numbers, hence, the pentagonal number theorem.

All that is beautiful math, but today I focus on a detail of the theorem that led to the graphic proof. What the terms of the Pentagonal Number theorem really say, and a question about them. If we look, for instance, in the partitions of five, they are {5},{4,1}, {3,2}, {3,1,1}, {2,2,1}, {2,1,1,1,1}, and {1,1,1,1,1} These can be divided into two sets, can you figure out how some are different than the others? Look at the first three. Now look at the last four. Each of the last four have repeats of one or more numbers. The first three are all distinct. Of the three with distinct digits, two of them have an even number of integers in their composition, {4,1} and {3,2}. The other, {5}, has an odd number of integers. And if you look at the exponent of X5 in the theorem, you see that it is positive, and that's what the theorem really says. If you look at all the partitions of any number, this polynomial gives you the the number of even distinct partitions (an even number of integers in it) minus the number of odd distinct partitions. So 5 has one more even than odd, and 7 does as well, but 12 has one more odd than even. but what first aroused Franklin's curiosity, was why there were so many missing exponents, why there were so many like X3 and X4 and X6. These would all have an equal number of odd and even partitions, hence the terms with zero for a coefficient which simply did not appear. what would explain this?

Franklin's insight was that for numbers like 6, the numbers could be matched up into odd and even pairs that offset each other in the count. For example, the distinct partitions of 6 are {6}, {5,1}, {4,2}, {3,2,1} the other seven partitions of 6 all have a repeated value. In his plan, the first two were matched together, and the last two were converted to each other, and he even had a graphic plan to show it always would work.

Here are two partitions of the number 33. The first is a partition into {9,8,7,5,4} The right diagonal which has 3 dots in it, and the bottom row which has 4 dots in it are the key. Since 3 is less than four (and four then, is the smallest number in the partition, we can move the 3 dots in the diagonal to make a new row in the bottom, and reduce the top three rows by one. So the matching partition is {8,7,6,5,4,3} at right. Note that we have changed the odd permutation into an even one. And we can go back by moving the three on the bottom row in the right partition to return to the left. These two always match. But he realized that there might be a situation in which this didn't work. For instance if the number of dots in the diagonal and the number of dots in the bottom row are the same (they share a corner dot) then you couldn't move either one. This can only happen if the bottom row is equal to the diagonal, or if the diagonal is one more than the bottom row. Here are examples of numbers that can't be matched to another. The top row is made up of numbers that have them equal. If you try to shift either one to make and odd partition even, or vice versa it just won't work. And the bottom sets show numbers that can be arranged with a partition that has the lowest row one more than the diagonal. They won't work either.

Now look at these numbers, count the dots, what do you notice. These are all the numbers in the Pentagonal Theorem Polynomial. And try as you may, you can't find another partition of any of these numbers that can't be transformed from odd to even or even to odd. If this partition has an odd number of rows, then the coefficient of that power in the polynomial is positive (5 in top row and 7 in bottom row are both positive coefficients since there is one odd partition that can't be matched to this even partition. The distinct partitions of 7 are {7}, {6,1}. {5,2}, {4,3} and {4,2,1} . Draw the diagrams, 7 can be transformed to {6,1} by dropping the last dot (a diagonal of one) to be a second row. {5,2} and be transformed in the same way to make {4,2,1} but the {4,3} has no match, so 3 evens, 2 odds makes the coefficient one.

I had never seen this before until I read a paper "The Pentagonal Number Theorem and All That" by Dick Koch from 2016. I no longer have the link to it, but still have the PDF (sometimes my reading list gets far longer than my free time) , so if you can't find it online somewhere, drop me a note and I'll send a copy out to you. Lots more really complicated and fascination stuff in it.

tzvi said...

good stuff