# What does it mean for a formal power series to be “well-defined”?

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:

• I'm sure "well-defined" means "well-defined". Could you give us an example of its use in Stanley's book? – Lord Shark the Unknown Sep 21 '18 at 17:30
• @LordSharktheUnknown Are you sure that "well-defined" means "well-defined?" Are you sure that you are sure? ;-)) – Mark Viola Sep 21 '18 at 17:33
• You can also look here: en.wikipedia.org/wiki/Well-defined – georg Sep 21 '18 at 17:40
• Added one such example in the question description. – odnerpmocon Sep 21 '18 at 17:41
• See an old answer of mine for an example of an infinite product of power series that is well-defined for the same reason - the lower degree terms in the product stabilize when the factors coming later no longer change them. – Jyrki Lahtonen Sep 22 '18 at 17:47

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.

Remember that when handling formal power series we are not analysts, and hence we are incapable of making sense of the sum of infinitely many terms of the same degree.

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 $$K[[x]].$$

• The argument showing that the inverse of $1/(1-n(x))$ is a well-defined formal power series generalize to show that given another formal power series $A(x)=\sum_ka_kx^x$ we can make the substitution $$A(n(x))=\sum_k a_k n(x)^k$$ and this is a well-defined power series as long as $n(x)$ has no constant term. See here. Don't forget to check out Bill Dubuque's link. – Jyrki Lahtonen Sep 21 '18 at 18:14

Your assumption is correct. A formal power series $$F(x)$$ is well-defined if it is an element in $$\mathbb{C}[[x]]$$, the ring of formal power series.

Here we have an infinite product \begin{align*} F(x)=\prod_{n\geq 1}(1-x^n)^{-\mu/n} \end{align*} and in order to show that $$F(x)\in \mathbb{C}[[x]]$$ we have to assure that it converges in the standard topology of formal power series. This means we have to show that according to proposition 1.1.9 in Stanley's book:

• The infinite product $$\prod_{n\geq 1}(1+F_n(x))$$, where $$F_n(0)=0$$, converges if and only if $$\lim_{n\to\infty}\mathrm{deg} F_n(x)=\infty$$.