The n-Category Café

A proof by pictures is given on page 58 of Conway and Guy’s The Book of Numbers. If I have a little more time today, I might write out what the pictures say.

There are several visual proofs linked as references from the Wikipedia article on this equation.

One of my favorite Theomathematicians.

http://www.britannica.com/EBchecked/topic/414433/Nicomachus-of-Gerasa

“Neo-Pythagorean philosopher and mathematician who wrote Arithmētikē eisagōgē (Introduction to Arithmetic), an influential treatise on number theory. Considered a standard authority for 1,000 years, the book sets out the elementary theory and properties of numbers and contains the earliest-known Greek multiplication table…. He was not interested, however, in abstract theorems on whole numbers and their proofs, as found in Books VII–IX of Euclid’s Elements; contrary to Euclid’s approach, Nicomachus would merely give specific numerical examples. A Latin translation of the Arithmētikē by Lucius Apuleius (c. 124–170) is lost, but a version by Ancius Boethius (c. 470–524) survived and was used in schools up to the Renaissance. Nicomachus also wrote Encheiridion Harmonikēs (‘Handbook of Harmony’) on the Pythagorean theory of music and the two-volume Theologoumena arithmetikēs (‘The Theology of Numbers’) on the mystic properties of numbers; only fragments of the latter survive.”

“Introduction to Arithmetic” was originally written using Arabic numbers, and the Arabs preserved and transformed Greek ideas during the Middle Ages.

This book is chock full of references to Plato, and so why does the Wikipedia entry on Nicomachus say that he was highly influenced by Aristotle?

Plato totally rules compared to Aristotle, and Plato would beat Aristotle on the football (soccer) field any day!

Dylan Thurston told me this one. I think it is due to Dylan, Walter Neumann, and David Bayer. The sum of cubes can be thought of as the cone on a cube. The square of triangles is that: a triangle times a triangle.

The cone on the cube can be decomposed as the union of two cones on two prisms. These cones on prisms can be re-glued as the triangle times a triangle.

These considerations provoked the works displayed here, and if that link is too slow to download then look here. Specifically, the pieces that have triangle in their names are relevant.

We were discussing this in Hanoi, and Dror Bar-Natan pointed out that in the continuous version you could conceive of the problem like this: A 3-cube consists of 3 un-ordered points in the interval. The cone on a cube is four points in the interval in which one is always to the left of the others. A triangle is an ordered pair of points in the interval, and the triangle cross the triangle is two ordered pairs of points in the interval in which the two pairs of points have no further relations. Let the two pairs of points be named (a,b) and (x,y). There are the following possible configurations: type A: (abxy), (axby), (axyb), and type B: (axyb) (xayb), (xyab). Those of type A form the cone on the prism b(xy) — with “a” being the cone point. Those of type B form the cone on the prism a(xy) with “b” as the cone point.

Meanwhile, the cone on the cube can be thought of as a{xyz} where the {xyz} go through the 6 permutations. Three of these (xyz, xzy, zxy) form one prism that is coned to a, and the other three (yxz,yzx,zyx) form the other prism.

The problem is related to categorification through John and Jeff’s musings about Falhauber and also with regards to some properadic considerations about brackets and parentheses.

The way I think about it (very close to Scott’s discussion, above) is to say the following. $$f(n) = 1+2+\dots+n = \frac{n(n+1)}{2}$$ is the area of a right-angle triangle, of sides, $n$ and $n+1$.

The product of two such triangles is a 4-dimensional figure, with volume $$f(n)^2 = \frac{n^2 (n+1)^2}{4}$$ Consider the difference in volume, when increasing $n$ by 1: $$f(n)^2 - f(n-1)^2 = \frac{n^2[(n+1)^2-(n-1)^2]}{4} = n^3$$

By induction on $n$, we obtain Nichomacus’s formula.

What! no mention of the Pythagorean comma (musical)?

Very nice replies, all!

Here is Gavin’s solution. Remember, I’d said “It would be nice to exhibit Nicomachus of Gerasa’s formula as a way of taking the balls in a 4-dimensional stack of cannonballs with cubical layers, and rearranging them to form a 4-dimensional shape that’s a product of two triangles.”

Here’s how he did it:

Here is a bijection of which Nicomachus’s formula is the shadow. Let {$n$} denote the set of integers from $1$ to $n$ inclusive. Let us denote by $S(k,n)$ the coproduct of ${m}^k$ for $m$ running from $1$ to $n$. More concretely:

$S(3,n) = \{(t,x,y,z) | 1 \le t \le n, 1 \le x \le t, 1 \le y \le t, 1 \le z \le t\}$

$S(1,n) = \{(t,x) | 1 \le t \le n, 1 \le x \le t \}$

Define two functions $u,v : S(3,n) \to S(1,n)$ by:

$u(t,x,y,z) = (t,y)$ if $x \gt z$ else $(z,x)$

$v(t,x,y,z) = (t-z+1,t-x+1)$ if $x \gt z$ else $(t,y)$.

Then the induced function

$\langle u,v\rangle : S(3,n) \to S(1,n)^2$

is a bijection, I claim. This is about the simplest I can cook up. It has a simple geometrical meaning, more laborious to describe in words than with a picture. Are you acquainted with Dynnakov’s formula for braid ordering? I do not claim any connection but I am reminded by the little language: linear arithmetic extended by the .. if $\langle$cond$\rangle$ else .. operator.

I replied:

It looks like you’re slicing the ‘hyperpyramid’ $S(3,n)$ into two pieces along the hyperplane $x = z$, and then rearranging these to form $S(1,n)^2$. If so, I need to grok what these pieces look like.

He replied:

A ------ B -- C
|        |    |
|        |    |
D ------ E -- F
|        |    |
G ------ H -- J

$t$ = AC = AG

$z$ = AB

$t-z+1$ = FJ

Rectangles ABHG and DFJG fit together to make a square, the $z$th horizontal slice of the $t$th cube.

I think this idea of slicing a hyperpyramid into two pieces and fitting them together to form a product of two triangles is also the basis of Dror Bar-Natan’s idea as described by Scott Carter. But, I don’t know whether this solution is secretly ‘the same’ as Gavin’s.

The “derivative” of the continuous version is really easy to visualise.

Think of a hyperpyramid of dimension $L$ as a movie of a cube of side length $t$ that runs from $t=0$ to $t=L$. If you increase $L$ you don’t need to do anything except run the movie longer; the “derivative” wrt $L$ is just “the last frame”, the cube of side $L$.

Now, think of a product of triangles of dimension $L$ as a movie of a right triangular prism of height $t$ and a base that’s half of an $L\times L$ square, running from $t=0$ to $t=L$.

When you change $L$, there are now two distinct three-dimensional regions in the “derivative”. One is “the last frame”, the triangular prism that is half of an $L\times L\times L$ cube. The other is the three-dimensional region you get by following the top triangular face of the growing prism, as it grows from a height of $0$ to a height of $L$. But the second of these is identical to the first, so they fit together into a single $L\times L\times L$ cube.

I know this discussion is getting old, but Greg’s pictures are really nice, and it reminded me of a couple of things. In a talk I gave in Tampa, I had a drawing of Dylan-Walter-Dave’s observation. This should give clues to Gavin’s and Greg’s version.

At the beginning of that talk, I asked, “If the integral of x to the third is one quarter, what is it a quarter of?” The answer is, of course, that it is a quarter of the 4d volume of a hyper-cube. Also, it is just as obvious that the integral of x to the nth takes up 1/(n+1) of the hyper volume of an n+1 cube, since the integral is measuring the hypervolume of the cone on an n-cube. Abhijit Champanerkar and I wrote down a proof here , but there were previous discoveries of this idea. Mathematica fans may enjoy playing with the code in the back of the paper — it has to be rewritten for the current version, but you get a nice rotating hyper-pyramid.

I still wonder what the Bernoulli numbers are measuring in these contexts.

Neat. I haven’t read all the comments so this might be redundant, but this reminds me of a binary tree:

• Heads, my wealth increases by a factor of 2.
• Tails, my wealth becomes the sum of my previous wealth and what my wealth would have been if we had flipped heads, which is (x+2*x) = 3x.

I wish I could play this game because my wealth would always increase! :)