03 March 2008

From products of infinite sums to subtraction and division

In some of the research I'm doing, I have occasion to write expressions like

A + ABA + ABABA + ABABABA + \dots = (A^{-1} - B)^{-1}

with no consideration of convergence.

Although in my setting I have (multiplicative) inverses and (additive) negatives, it nonetheless becomes interesting to ask what sorts of operations we can do with just plus and times, provided that we can do such things infinitely.

For example, with just positive integers and some questionable summations, we can get all negative fractions:

-1 = 1 + 2 + 4 + 8 + 16 + \dots
-m/n = m + (n+1)\times m + (n+1)^2\times m + \dots

and some irrationals:

\sqrt{2\pi} = 1 \times 2 \times 3 \times 4 \times 5 \times \dot

The first two sums are from the fact that 1+r+r^2+\dots=1/(1-r). Indeed, this comes from multiplying and shifting. If, e.g., we add 1 to the first equation, and then telescope the sum, we have

1+1 + 2 + 4 + 8 + 16 + \dots = 2+2 + 4 + 8 + 16 + \dots = 4+4 + 8 + 16 + \dots = 8+8 + 16 + \dots = etc.

The product is from \zeta-function regularization: we recognize the right hand side as the exponential of d\zeta(s)/ds at s=0.

Since we can multiply power series, we can get all fractions; for any given fraction, there are lots of ways of writing it as a series. For example,

1 = (-1)^2 = (1+2+4+8+\dots)^2 = 1 + (2+2) + (4+4+4) + (8+8+8+8) + \dots
= \sum (n+1)2^n.

Since \sum 2^n = -1, we clearly have \sum n2^{n-1} = 1 as well, which can be checked directly (it is d[\sum 2^n]/d2 = d[1/(1-2)]/d2 = 1/(1-2)^2.).

How about inverses of variables? Especially if you're not sure of the 2-adic sum, perhaps you'd rather use

-1/X = 1 + (1+X) + (1+X)^2 + (1+X)^3 + \dots

to get your negative and inverse Xs. This formula is hard to verify, since it still requires the same shifting tricks:

1 + ( X + X(1+X) + X(1+X)^2 + \dots ) = 1 + X + X(1+X) + X(1+X)^2 + \dots
= 1+X + X(1+X) + X(1+X)^2 + \dots
= (1+X)(1+X) + X(1+X)^2 + \dots
= (1+X)(1+X)^2 + \dots
= etc.

See, infinite sums are not associative --- the distance between terms matters --- but these manipulations are justified since infinite sums are finitely associative.

We get, either by squaring or differentiating,

1/X^2 = 1 + 2(1+X) + 3(1+X)^2 + 4(1+X)^3 + \dots

and hence all powers of X. Well, we get \pm 1/X^n; for, say, -X, the shortest is to use X^2(-1/X) = X^2 + X^2(1+X) + X^2(1+X)^2 + X^2(1+X)^3 + \dots, which telescopes in the same way.

These manipulations require that there be a number 1 which multiplies with X to X and which we can add to X. Which is true is X is, say, a matrix, or a CW complex, but not if X is a two-zero-tensor. If X is such a tensor, then X^{-1} is the tensor which convolves on both sides to the identity matrix. This is the setting of my original line (A is two-zero, and B is zero-two). In this situation, I don't have a good construction of X^{-1}.

2 comments:

Urs Schreiber said...

Hi,

I arrived here by following the link provided with a comment you posted over at the n-Category Cafe.

You are probably aware of the following, but since you don't mention it in the above entry, let me say it anyway:

that "groupoidification" program which John Baez talks about is at least closely related to an idea where one wants to see how various kinds of numbers (natural, negative, rational, irrational, complex) arise as "cardinalities" of things that may be thought of as categories.

In this context, one finds oneself confronted with precisely the kind of considerations that you mention above.

A good place to start reading about this is From finite sets to Feynman diagrams.

Sorry if you know all this already. I just thought I'd mention it.

Theo said...

Hi Urs,

Indeed, like many of y'all over at the Cafe, I'm trying to understand Feynman diagrams, and prove consistency, etc., results.

As you can see, I haven't written here on Orange Juice Files for months (in part because I started the blog in undergrad, whereas now at Berkeley there are enough other graduate students to tell my ideas to that I don't have as much need to telegraph them to the world, and in part because teaching takes up sufficiently much time that I don't have a lot of research ideas to post about). But, anyway, I think that the reason I was interested in A + ABA + ABABA + \dots = (A^{-1} - B)^{-1} is that I was trying to show combinatorially the usual integration rules (u-substitution, fubini theorem) for formal Feynman-diagrammatic integration, and these can lead to this type of expression, exactly with the 0,2 and 2,0 tensors. And famously there's the problem in Feynman integration of inverting the kinetic energy, hence the final comment.