On series expansions and valid arguments

Question

Hopefully the answer to this question isn't so obvious one way or the other that it ends up just creating more confusion. Briefly, I'm concerned about the potential for subtly illegal moves to creep into arguments that otherwise arrive at the right solution.

Consider an utterly trivial equation such as $$y\left(\frac{1}{1-x}\right) = \frac{1}{1-x} ~\Longrightarrow~ y = \left( \frac{1}{1-x} \right)\left( \frac{1 -x}{1} \right) = 1$$

This is valid for $\textbf{all}$ $x \in \mathbb{R}$ such that $x\neq1$. No other restrictions are required.

$\textit{But}$...we could've gotten cute. Suppose again that $x \neq 1$ is our only restriction. Now argue as follows: $$y\left(\frac{1}{1-x}\right) = 1 + x +x^2 + \dots = $$ $$ =(1 -x)(1 + x +x^2 + \dots) = (1 + x +x^2 + \dots) + (-x -x^2 -x^3 - \dots) = 1$$

Algebraically, this seems perfectly fine. However, the series expansion for $\frac{1}{1-x}$ is valid only for $|x| < 1$, right? So, while the first derivation is $\textit{true at every step}$ for, say, $x = 2$, the second contains a step in the argument where it seems to me we could potentially be implying that $$y = (1 - x)(1 + x + x^2 + \dots) = (1 - 2)(1 + 2 + 2^2 + \dots) = -1 \times \infty = 1$$ despite arriving at the same solution.

My question is this: Even though both prove that $y = 1$, is the second derivation in any sense an invalid argument unless we're specifically told that $x$ can only lie within the radius of convergence of $\frac{1}{1 -x}$? Or is this sort of move fine so long as we are algebraically manipulating unknowns, not actual values?

Answer 1

Spot on. When you ignore the issue of convergence, a power series becomes a formal power series wherein the series itself is considered an algebraic object. You have made one error though: in treating the series as a formal power series and considering $x=2$, you were incorrect to conclude that $y=1$, because it does not converge there. You still have to be explicit with regards to the domain (here, that'd be $|x| \lt 1$). Even then, it is nonstandard to assign a numerical value to a formal power series (not that it's wrong to, per se, but it must converge).

To expand on formal power series, their motivation for existing is to act as a "container/clothesline" for a sequence, the members of which are given by the coefficients; for example, the fifth member of a sequence would be the coefficient of $x^5$ in the series. In this context, they are known as "generating functions" (an admittedly misleading name) and are used chiefly for solving problems in combinatorics and number theory. These aren't the only applications of formal power series, but I'd wager that they're the most common.

Answer 2

Here is a very general way of looking at this question: you specify an identity $$ y(\frac{1}{1-x}) = \frac{1}{1-x} \tag{$\dagger$} $$ required of two variables $x$ and $y$ and ask what algebraic manoeuvers are legitimate when we try and solve for $y$.

What you then actually do (or at least how H. sapiens rex reads what you do) is consider what solutions you can get in two different rings: you first took

$$ \mathbb R(\dagger) := \left\{(x,y) \in \mathbb R \mid y.(\frac{1}{1-x}) = \frac{1}{1-x}\right\} $$ and observe that, provided $a\neq 1$, the pair $(a,1)$ is in $S_{\mathbb R}$: the condition $a\neq 1$ is necessary because $1-a$ must be a unit in $\mathbb R$, so $1-a\neq 0$, but then multiplying both sides by $(1-x)$ we find $y =1$.

One way of saying what you do in your "cute" manipulation is consider solutions in the ring $R_1=\mathbb R[\![t]\!]$ of formal power series in one variable $t$, i.e. consider $$ R_1(\dagger) = \{(x,y) \in R_1^2: y.(\frac{1}{1-x})= \frac{1}{1-x} \} $$ Just as in the case of $\mathbb R$, the fact that (eq.1) involves $\frac{1}{1-x}$ means that if $(x,y) \in R_1^2$ is a pair of elements in $R_1$ satisfying the equation, $1-x$ has to be invertible (i.e. a unit) in $R_1$. But $t \in R_1$ is a unit, because it has inverse $1+t+t^2+\ldots$, so we can take $x=t$ and then see if we can find an element of $s \in R_1$ with $(t,s) \in S_{R_1}$. But just as for $\mathbb R$, once we multiply by $(1-t)$ on both sides, we find the only possibility for $s$ is $1$. Thus $(t,1) \in S_{R_1}$, and in fact in general $S_{R_1} = \{f(t),1): f(t) = \sum_{i=0}^\infty a_it^i: a_0 \neq 1\}$ is the complete set of pair $(r,s) \in R_1$ satisfying (eq.1).

[Aside: This is because a formal power series with coefficients in a field has an inverse precisely when its constant term is nonzero,} so that $1-f$ is a unit precisely when $a_0 \neq 1$.}]

Now suppose that $R$ is any commutative ring with $1$ and consider $$ R(\dagger) = \{(r,s) \in R^2: s(\frac{1}{1-r}) = \frac{1}{1-r}\} $$ Now again because (eq.1) involves the expression $\frac{1}{1-x}$ any pair $(r,s)$ must have $1-r$ a unit in $R$. But if that is the case then whatever its inverse is (a power series or a real number) in the ring $R$ if $s(\frac{1}{1-r}) = \frac{1}{1-r}$ then $s.(\frac{1}{1-r})(1-r)= \frac{1}{1-r}.(1-r) =1$, so that $s=1$. Thus is any commutative ring $R$ we have $$ R(\dagger) = \{(r,1): 1-r \in R^{\times}\}. $$ In other words, the algebraic operations that force $y=1$ hold in any commutative ring (indeed in any associative ring with a multiplicative identity!)

Notice that if I rearrange (eq.1) slightly by cross-multiplying, I get $$ y(1-x)=(1-x) \tag{$\ddagger$} $$ Then if we set $$ R(\ddagger) = \{(r,s)\in R^2: s(1-r)=(1-r)\}, $$ the behaviour of $R(\ddagger)$ is much more varied: if we take $R=\mathbb R$, then $\mathbb R(\ddagger) = \{(r,1): r \in R\}\cup\{(1,s): s\in R\}$.

Indeed $s(1-r)=(1-r)$ is any commutative ring $R$ is and only if $(s-1)(r-1)=0$, thus for example if we take $R = \mathbb Z/6\mathbb Z$ then $$ \mathbb Z/6\mathbb Z(\ddagger) = \{(r,s), (s,r) \in \mathbb Z/6\mathbb Z^2 \mid r=1, s \text{ arbitrary, or }r\in \{3,5\}, s=4\} $$

The idea of studying an equation by considering its solutions in all possible commutative rings is in fact what is known as Grothendieck's "functor of points" approach to schemes!