What does it mean for a formal power series to be "well-defined"?

by odnerpmocon   Last Updated September 21, 2018 18:20 PM - source

I see the term "well-defined" used in Stanley's Enumerative Combinatorics, but I'm not sure what it means. Is it equivalent to saying that the formal power series converges to a certain function, or it is something else?

EDIT: Picture exemplifying the use of "well-defined" in Stanley now attached: Tags :

What the author means is that apart from a constant term $$1$$ the $$n$$th factor contains terms of degree and higher only. When you (formally) expand a product like $$(1+a_{1,1}x+a_{1,2}x^2+\cdots)(1+a_{2,2}x^2+a_{2,3}x^3+\cdots)(1+a_{3,3}x^3+a_{3,4}x^4+\cdots)\cdots$$ what happens is that you only get a finite number of terms of a given degree. Basically because after the first $$n$$ factors the lowest degree terms have "converged". The remaining factors, even though there are infinitely many of them, no longer alter the coefficients of those low degree terms.
Consider the following example that surprises most when they first hear about it. In the ring of formal power series $$\Bbb{R}[[x]]$$ the series $$e^{x+1}=\sum_{n=0}^\infty\frac1{n!}(x+1)^n$$ does not make sense! In other words, it is not well-defined. Each and every one of those terms has a non-zero constant term. As a formal power series this would make sense in $$\Bbb{Q}[[x+1]]$$ though.
OTOH, in the ring of formal power series we don't need to worry about the series converging w.r.t. some weird topology of the field of coefficients. Therefore $$1+x+4x^2+27x^3+\cdots=\sum_{n=0}^\infty n^nx^n$$ makes perfect sense in $$\Bbb{Z}[[x]]$$ even though it does not converge as a Taylor series for any value of $$x\in\Bbb{R}$$ other than zero. In the ring of formal power series we never give the variable $$x$$ any value - its powers are just placeholders for the coeffcients.
These properties show up in the calculation of inverses in $$K[[x]]$$. If $$n(x)$$ is any formal power series with constant term zero, then the geometric series $$1+n(x)+n(x)^2+n(x)^3+\cdots$$ is a well-defined power element in $$K[[x]]$$ because only the $$n+1$$ first terms contain terms of degrees $$\le n$$ for any $$n\in\Bbb{N}$$. This is why the above sum serves as the reciprocal $$1/(1-n(x))$$ in $$K[[x]].$$