Return to Answer

For all $f$ and $g$ in $A[[x]]$, $|f+g|_x \leq \max(|f|_x,|g|_x)$ and $|fg|_x \leq |f|_x|g|_x$. If $A$ is an integral domain, such as $A = \mathbf Z_p$ or $\mathbf Q_p$, then $|fg|_x = |f|_x|g|_x$. From $|\cdot|_x$ we get a topology on $A[[x]]$ in which $x^n \to 0$ as $n \to \infty$, and $|f|_x < 1$ if and only if $f$ has constant term $0$. The $x$-adic distance between $f$ and $g$ is defined to be $|f-g|_x$, which is a non-Archimedean metric on $A[[x]]$ that makes it complete. Just as in the $p$-adic numbers, an sequence $\{f_n\}$ in $A[[x]]$ is $x$-adically convergent if and and only if $|f_{n+1} - f_n|_x \to 0$ as $n \to \infty$, so an infinite series of elements in $A[[x]]$ is $x$-adically convergent if and only if its general term tends to $0$.

For all $f$ and $g$ in $A[[x]]$, $|f+g|_x \leq \max(|f|_x,|g|_x)$ and $|fg|_x \leq |f|_x|g|_x$. If $A$ is an integral domain, such as $A = \mathbf Z_p$ or $\mathbf Q_p$, then $|fg|_x = |f|_x|g|_x$. From $|\cdot|_x$ we get a topology on $A[[x]]$ in which $x^n \to 0$ as $n \to \infty$, and $|f|_x < 1$ if and only if $f$ has constant term $0$. The $x$-adic distance between $f$ and $g$ is defined to be $|f-g|_x$, which is a non-Archimedean metric on $A[[x]]$ that makes it complete. Just as in the $p$-adic numbers, a sequence $\{f_n\}$ in $A[[x]]$ is $x$-adically convergent if and and only if $|f_{n+1} - f_n|_x \to 0$ as $n \to \infty$, so an infinite series of elements in $A[[x]]$ is $x$-adically convergent if and only if its general term tends to $0$.

I call this the combinatorial method because I have seen combinatorialists introduce formal power series in this way, which is purely algebraic and emphasizes the finitely many calculations needed to work out any particular coefficient. It seems like a clunky way of setting up formal power series if you want to prove properties of them beyond the simplest ones, but it has the advantage of being intuitive and getting to the point of things in a direct way. To prove things from this point of view you typically need to work them out degree-by-degree with induction.

Remark 1. Armed with the $x$-adic topology, we gain access to Hensel's lemma and topological arguments that can replace tedious degree-by-degree arguments in the combinatorial method. For example, the formal derivative on $A[[x]]$ is $x$-adically uniformly continuous (since $|f'|_x \leq (1/2)|f|_x$), so to prove the usual rules of formal derivatives are valid on $A[[x]]$, by its continuity and the denseness of $A[[x]]$ we are reduced to verifying them on polynomials, which is "already known". There is no need to actually prove the identities using actual power series, but you may need to be more careful in the case of the chain rule to reduce to the polynomial case.

I call this the combinatorial method because I have seen combinatorialists describe computations with formal power series in this way, which is purely algebraic and emphasizes the finitely many calculations needed to work out any particular coefficient. It seems like a clunky way of setting up formal power series if you want to prove properties of them beyond the simplest ones, but it has the advantage of being intuitive and getting to the point of things in a direct way. To prove things from this point of view you typically need to work them out degree-by-degree with induction. See, for example, how proofs look in the Amer. Math. Monthly article on formal power series by Ivan Niven here.

Remark 1. Armed with the $x$-adic topology, we gain access to Hensel's lemma and topological arguments that can replace tedious degree-by-degree arguments in the combinatorial method. (It is not lost on me that the proof of Hensel's lemma itself is based on a degree-by-degree argument, so my point of comparison between Methods 1 and 2 is that Method 1 uses degree-by-degree arguments very often, while Method 2 can often avoid them with simple continuity arguments.) For example, the formal derivative on $A[[x]]$ is $x$-adically uniformly continuous (since $|f'|_x \leq (1/2)|f|_x$), so to prove that the usual rules of formal derivatives are valid on $A[[x]]$, by its continuity and the denseness of $A[[x]]$ we are reduced to verifying the rules on polynomials, where the rules are "already known". There is no need to prove the identities on actual power series, but you may need to be more careful in the case of proving the chain rule to reduce to the polynomial case.

Method 1: the combinatorial method (using finitely many terms at a time). If $f(x) = a_0 + a_1x + a_2x^2 + \cdots$ is an element of $A[[x]]$, where $A$ is a commutative ring, and $g(x) = b_1x + b_2x^2 + \cdots$ is another element of $A[[x]]$ with constant term $0$, then $f(g(x))$ makes sense as $$ a_0 + a_1g(x) + a_2g(x)^2 + \cdots $$ because $g(x)^n = b_1^nx^n + \cdots$ has no terms in degree below $n$, so to find the coefficient of $x^d$ in $f(g(x))$ you only need to work with the finite sum $a_0 + a_1g(x) + \cdots + a_{d-1}g(x)^{d-1}$ and extract the coefficient of $x^d$ in that sum. It "doesn't make sense" to compose $f(g(x))$ if $g(x) = b_0 + b_1x + b_2x^2 + \cdots$ has a nonzero constant term because the constant term of $f(g(x))$ would then be $f(g(0)) = f(b_0) = a_0 + a_1b_0 + a_2b_0^2 + \cdots$ and this is an infinite series that doesn't make algebraic sense unless $b_0$ is nilpotent or some other reason causes all but finitely many terms to be $0$.

Writing $r \in R = A[[x]]$ as $g(x)$, saying $|r|_x < 1$ means $g(x)$ has constant term $0$. This shows $f(g(x))$ converges $x$-adically in $A[[x]]$ when $g(x)$ has constant term $0$. It's left to you to check, using the $x$-adic uniform continuity estimate $|f(g(x)) - f(h(x))|_x \leq |g(x) - h(x)|_x$ when $g(x)$ and $h(x)$ have constant term $0$, that the calculation of the coefficients of $f(g(x))$ as an $x$-adic limit (finding each of its coefficients as an element of $A[[x]]$) can be done by the combinatorial method ("finitely many terms at a time") because $g(x) \equiv g_d(x) \bmod x^d$ where $g_d(x)$ is the truncation of $g(x)$ to the sum of its first $d$ terms (those below degree $d$).

Method 1: the combinatorial method (using finitely many terms at a time). If $f(x) = a_0 + a_1x + a_2x^2 + \cdots$ is an element of $A[[x]]$, where $A$ is a commutative ring, and $g(x) = b_1x + b_2x^2 + \cdots$ is another element of $A[[x]]$ with constant term $0$, then $f(g(x))$ makes sense as $$ a_0 + a_1g(x) + a_2g(x)^2 + \cdots $$ because $g(x)^n = b_1^nx^n + \cdots$ has no terms in degree below $n$, so to find the coefficient of $x^d$ in $f(g(x))$ you only need to work with the finite sum $a_0 + a_1g(x) + \cdots + a_{d}g(x)^{d}$ and extract the coefficient of $x^d$ in that sum. It "doesn't make sense" to compose $f(g(x))$ if $g(x) = b_0 + b_1x + b_2x^2 + \cdots$ has a nonzero constant term because the constant term of $f(g(x))$ would then be $f(g(0)) = f(b_0) = a_0 + a_1b_0 + a_2b_0^2 + \cdots$ and this is an infinite series that doesn't make algebraic sense unless $b_0$ is nilpotent or some other reason causes all but finitely many terms to be $0$.

Writing $r \in R = A[[x]]$ as $g(x)$, saying $|r|_x < 1$ means $g(x)$ has constant term $0$. This shows $f(g(x))$ converges $x$-adically in $A[[x]]$ when $g(x)$ has constant term $0$. It's left to you to check, using the $x$-adic uniform continuity estimate $|f(g(x)) - f(h(x))|_x \leq |g(x) - h(x)|_x$ when $g(x)$ and $h(x)$ have constant term $0$, that the calculation of the coefficients of $f(g(x))$ as an $x$-adic limit (finding each of its coefficients as an element of $A[[x]]$) can be done by the combinatorial method ("finitely many terms at a time") because $g(x) \equiv g_d(x) \bmod x^{d+1}$ where $g_d(x)$ is the truncation of $g(x)$ to the sum of its terms up through degree $d$.