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Sums of Cubes

Good news, everyone! I finally found the geometrical interpretation of this fact from Triangular Numbers! (The sums are over k from 1 to n.)

sum ( k3 ) = [ sum ( k ) ]2

And, yes, the interpretation is four-dimensional!

For anyone interested in the untidy process of discovery as well as the tidy final result, here's what happened. I was trying once again to visualize the whole thing in four dimensions. The left side is clearly a stack of n cubes, while the right is some kind of product of planar triangles. I've never been able to visualize the true product, but I thought I'd just ignore that aspect and see if I could pack individual triangles into the stack. For n = 2 it worked, but for n = 3 it didn't, and in fact I could prove it didn't only one triangle point can fit into the 111 cube at the top, so only two cells of the 222 cube can be covered. I'd been through all of that before, more or less, but this time I had one new idea: if I definitely can't pack one into the other, maybe I can pack both into some larger object? I knew the closed form for the sum of cubes was (1/4) n2(n+1)2, so I guessed a rectangular block might work. When I went back to n = 2 and tried a 2233 block, it didn't exactly work, but it did give me the first construction below, and the rest followed.

I'd like to think I've discovered something genuinely new, but probably I haven't. The fact itself is very old, known as Nicomachus' Theorem, and apparently there have been many different proofs of it over the years. No matter there may be many proofs, but I think what I have here is immediately recognizable as the true explanation; and since it was new to me, clearly it needs to be more widely known. Also it gives me an excuse to show you some pretty pictures!

Let's start with a simple fact. If we have a normal two-dimensional square, we can wrap four linear pieces around it to make a square that's two units larger.

As an equation, that's this.

n2 + 4(n+1) = n2 + 4n + 4 = (n+2)2

If we multiply the whole thing by (n+1)2, we get this.

n2(n+1)2 + 4(n+1)3 = (n+1)2(n+2)2

But, in geometrical terms, all the multiplication does is extend the original figure uniformly into another two dimensions. So, if we have a four-dimensional rectangular block, we can wrap four cubes around the short sides to make a rectangular block that's one unit larger.

By induction, that proves this.

n2(n+1)2 = 4 sum ( k3 )

Of course we also need something to start the induction, but it's easy enough to check the case n = 0, and maybe also the case n = 1 since it's not entirely obvious that we can wrap cubes around a block of size zero.

Now we know how to pack cubes into a rectangular block, but what about the other side of the equation, the triangles? It turns out that's easy. In Triangular Numbers, to explain the formula for T(n) I talked about combining two copies of a triangle to get a square with the diagonal covered twice but I only said that because it went with the next point there about matrices. Normally the way I think about T(n) is, we can combine two triangles of side n to get an n(n+1) rectangle.

If we multiply that two-dimensional figure by the same figure in two orthogonal directions, we see that the four-dimensional rectangular block we've been talking about is composed of four doubly-triangular wedges (which are those things I can't visualize).

n2(n+1)2 = 4 [ sum ( k ) ]2

And there we have it! An nn(n+1)(n+1) rectangular block is on the one hand a sum of cubes and on the other a sum of doubly-triangular wedges. Putting the last two equations together and canceling the 4s gives us the desired result.

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@ February (2011)