Where do the powers go?

Question

I have this weird identity between power series. It's kind of like the relation between a geometric series and $\frac{1}{1-x}$. I was wondering if there was some theory developed along these lines I could read up on.

Consider $$F(x) = (1+x^{1/2})(1+x^{1/4})\cdots$$ then $$F(x^2) = (1+x)F(x)$$ so $$\frac{F(x^2)}{1+x} = F(x)$$

and finally, $$F(x) = \frac{1}{1+x}\frac{1}{1+x^2}\frac{1}{1+x^4}\cdots = 1-x.$$

The main thing I'm having trouble wrapping my head around is where all the fractional powers "go"? In an expression like: $$(1+x)(1+x^2)\cdots = 1+ x + x^2 + \ldots = \frac{1}{1-x}$$ everything still makes sense to me, all the powers still appear. But in the above expression you're summing $x$ to the power of all dyadic rationals and somehow getting a $-x$ term.

Edit: I think I have a formulation where this makes sense. As a Hahn-series where formal infinite products are defined analogously. The support will be well-ordered as we are only taking finite sums. But this would mean that Hahn series are not unique, the same Hahn series can have two different representations. Either I'm still making some sort of mistake or the above is true.

https://en.wikipedia.org/wiki/Hahn_series

Answer 1

If you start with $F(x)=1-x= (1-x^{1/2^k}) \prod_{n=1}^k (1+x^{1/2^n})$ then yes everything is fine in $\bigcup_{n\ge 1}\Bbb{C}[[x^{1/2^n}]]$, but you can't let $k\to \infty$ as you did, ie. $\lim_{k\to \infty} \prod_{n=1}^k (1+x^{1/2^n})$ doesn't converge in any topological integral domain where $x\ne 0$

(except in $\overline{\Bbb{F}}_2$ with the discrete topology and $x=1$ which gives $F=0=1-x$ by the way)

Answer 2

The last line of your argument is invalid; you've hidden an argument in the use of "$\dots$" that, if you expand it out, does not actually work at all. You've written

$$F(x) = \frac{1}{1 + x} \frac{1}{1 + x^2} \frac{1}{1 + x^4} \dots$$

when what you mean is

$$F(x) = \frac{1}{1 + x} F(x^2) = \frac{1}{1 + x} \frac{1}{1 + x^2} F(x^4) = \dots.$$

This does not imply that $F(x) = 1 - x$, because all of the power series $F(x^{2^n})$, if expanded out formally, continue to have nonzero terms with arbitrarily small exponents; the sequence of formal power series $F(x^{2^n})$ simply does not in any reasonable sense converge to $1$, as it would if $F$ were an ordinary formal power series with constant term $1$.

It is in fact possible to talk about formal power series with non-negative real exponents; this is a valuation ring. For multiplication to be well-defined we need to only consider formal power series $\sum_{r \in \mathbb{R}_{\ge 0}} c_r x^r$ where the support $\{ r : c_r \neq 0 \}$ is well-ordered. If we expand out the infinite product $F(x)$ formally (ignoring the question of whether or not it makes sense in this valuation ring) we get that $c_r \ge 1$ for all $r \in [0, 1]$ (each such $r$ has at least one infinite binary expansion, and some of them have two) so the support contains $[0, 1]$ and therefore can't be well-ordered. So this infinite product can't exist in the valuation ring. (Concretely, even if it might sort of make sense to expand $F(x)$ out formally, there's no corresponding formal expansion for $F(x)^2$; computing the coefficients of this would involve adding up uncountably many terms which are bounded below!)

I don't know what the "standard" choice of topology on a non-discrete valuation ring is, but this probably means that the infinite product doesn't converge.