# nLab power series

### Context

#### Algebra

higher algebra

universal algebra

# Contents

## Definition

A power series in a variable $X$ and with coefficients in a ring $R$ is a series of the form

$\sum_{n = 0}^\infty a_n X^n$

where $a_n$ is in $R$ for each $n\ge 0$. Given that there are no additional convergence conditions, a power series is also termed emphatically as a formal power series. If $R$ is commutative, then the collection of formal power series in a variable $X$ with coefficients in $R$ forms a commutative ring denoted by $R [ [ X ] ]$.

More generally, a power series in $k$ commuting variables $X_1,\ldots, X_k$ with coefficients in a ring $R$ has the form $\sum_{n_1=0,n_2=0,\ldots, n_k = 0}^\infty a_{n_1\ldots n_k} X_1^{n_1} X_2^{n_2}\cdots X_k^{n_k}$. If $R$ is commutative, then the collection of formal power series in $k$ commuting variables $X_1,\ldots, X_k$ form a formal power series ring denoted by $R [ [ X_1,\ldots, X_k ] ]$.

More generally, we can consider noncommutative (associative unital) ring $R$ and words in noncommutative variables $X_1,\ldots, X_k$ of the form

$w = X_{i_1}\cdots X_{i_m}$

(where $m$ has nothing to do with $k$) and with coefficient $a_w \in R$ (here $w$ is a word of any length, not a multiindex in the previous sense). Thus the power sum is of the form $\sum_w a_w X_w$ and they form a formal power series ring in variables $X_1,\ldots, X_k$ denoted by $R\langle \langle X_1,\ldots, X_k \rangle\rangle$. Furthermore, $R$ can be even a noncommutative semiring in which case the words belong to the free monoid on the set $S = \{ X_1,\ldots, X_k\}$, the partial sums are then belong to a monoid semiring $R\langle S\rangle$. The formal power series then also form a semiring, by the multiplication rule

$\sum_{r} a_r X_r \cdot \sum b_s X_s = \sum_w \sum_{u,v; w = u v} a_u b_v X_w$

Of course, this implies that in a specialization, $b$-s commute with variables $X_{i_k}$; what is usually generalized to take some endomorphisms into an account (like at noncommutative polynomial level of partial sums where we get skew-polynomial rings, i.e. iterated Ore extensions).

Let $R$ be a commutative ring, and let $R[X]$ be the polynomial ring on one indeterminant $X$. Then $(X)$ is a maximal ideal in $R[X]$, and results in an adic topology on $R[X]$. The ring of formal power series in $R$ is the adic completion of the limit of the quotient of $R[X]$ by powers of $(X)$:

$R[[X]] \coloneqq \underset{\leftarrow}\lim R/(X)^n$

The ring of formal power series for multiple indeterminants $X_i$ is constructed iteratively: because $R[[X_1, X_2, \ldots X_n]]$ is a commutative ring, one could construct the polynomial ring $R[[X_1, X_2, \ldots X_n]][X_{n+1}]$ on the indeterminant $X_{n+1}$. As above, $(X_{n+1})$ is a maximal ideal in $R[[X_1, X_2, \ldots X_n]][X_{n+1}]$ with a corresponding adic topology, and one can then take the adic completion

$R[[X_1, X_2, \ldots X_n]][[X_{n+1}]] \coloneqq \underset{\leftarrow}\lim R[[X_1, X_2, \ldots X_n]]/(X_{n+1})^m$

The resulting commutative ring is usually just written as $R[[X_1, X_2, \ldots X_n, X_{n+1}]]$.

## Examples

### Polynomials

For a natural number $k$, a power series $\sum_{n=0}^\infty a_n X^n$ such that $a_n = 0$ for all $n \gt k$ is a polynomial of degree at most $k$.

### MacLaurin series

For $f \in C^\infty(\mathbb{R})$ a smooth function on the real line, and for $f^{(n)} \in C^\infty(\mathbb{R})$ denoting its $n$th derivative its MacLaurin series (its Taylor series at $0$) is the power series

$\sum_{n = 0}^\infty \frac{1}{n!} f^{(n)}(0) x^n \,.$

If this power series converges to $f$, then we say that $f$ is analytic.

## Properties

• An element $a = a_0 + a_1 x + a_2 x^2 + \ldots$ in $R[ [x] ]$ is (multiplicatively) invertible iff $a_0$ is invertible.

This follows easily from the observation that we can invert $1 + x b$ for any power series $b$ by forming $1 - x b + x^2 b^2 - \ldots$ and collecting only finitely many terms in each degree. As a simple corollary,

• If $R$ is a local ring, then the power series ring $R[ [X] ]$ is also a local ring.

### Functional substitution and inversion

###### Proposition

$R[ [x_1, \ldots, x_n] ]$ equipped with the ideal $(x_1, \ldots, x_n)$ is the free adic $R$-algebra on $n$ generators, in the sense that it is the value of the left adjoint $Pow$ to the forgetful functor

$Ideal: AdicRAlg \to Set: (A, I) \mapsto I$

as applied to the set $\{x_1, \ldots, x_n\}$.

###### Proof

The idea is that for each adic $R$-algebra $(S, I)$ and element $(s_1, \ldots s_n) \in I^n$, there is a unique adic algebra map $R[ [x_1, \ldots, x_n] ] \to S$ that sends $x_i$ to $s_i$; this adic algebra map sends a power series $\sum a_{k_1, \ldots, k_n} x_1^{k_1} x_n^{k_n}$ to the sequence of truncations

$\left(\sum_{k_1 + \ldots + k_n \lt k} a_{k_1, \ldots, k_n} s_1^{k_1} \ldots s_n^{k_n} \mod I^k\right)_k$

belonging to $\underset{\longleftarrow}{\lim}_k S/I^k \cong S$.

It follows that we may define a clone or cartesian operad as follows: the $n^{th}$ component is the set $I_n = (x_1, \ldots, x_n) \subset R[ [x_1, \ldots, x_n] ]$ which is the monad value $Ideal Pow(\{x_1, \ldots, x_n\})$. Letting $M$ denote the monad $Ideal \circ Pow$, with monad multiplication $\mu$, and $[n]$ the set $\{x_1, \ldots, x_n\}$, the clone multiplication

$I_n \times I_k^n \to I_k$

is the composition of the maps

$M(n) \times M(k)^n \cong M(n) \times \hom([n], M(k)) \stackrel{1 \times func}{\to} M(n) \times \hom(M(n), M M(k)) \stackrel{eval}{\to} M M(k) \stackrel{\mu(k)}{\to} M(k)$

The clone multiplication thus defined is called substitution of power series; it takes a tuple consisting of $p(x_1, \ldots, x_n) \in I_n, q_1(x_1, \ldots x_k) \in I_k, \ldots q_n(x_1, \ldots, x_k) \in I_k)$ to a power series denoted as

$p(q_1(x_1, \ldots, x_k), \ldots q_n(x_1, \ldots, x_k)).$

The resulting clone or operad yields, in the particular case $k = n = 1$, an associative substitution operation

$x R[ [x] ] \times x R[ [x] ] \stackrel{sub}{\to} x R[ [x] ]$

with $sub(p, q) = p \circ q$ the power series $p(q(x))$.

###### Proposition

The group of invertible elements in the substitution monoid $x R[ [x] ]$ consists of power series of the form $a_1 x + a_2 x^2 + \ldots$ where $a_1$ is multiplicatively invertible in the ring $R$.

In other words, we can functionally invert a power series provided that the linear coefficient $a_1$ is invertible in $R$.

###### Proof

Given power series $a = a_1 x + a_2 x^2 + \ldots$ and $b = b_1 x + b_2 x^2 + \ldots$, we may read off coefficients of the composite $a \circ b$ as

$(a \circ b)_k = \sum_{n \geq 1} a_n \sum_{k = k_1 + \ldots + k_n} b_{k_1} b_{k_2} \ldots b_{k_n}$

where in particular $(a \circ b)_1 = a_1 b_1$. Now $a$ is the left functional inverse of $b$, or $b$ is the right inverse of $a$, if $(a \circ b)(x) = x$, i.e., if $(a \circ b)_k = 1$ if $k = 1$ and $0$ otherwise. The first equation says simply $(a \circ b)_1 = a_1 b_1 = 1$ which implies $a_1$ is invertible. Conversely, if $a_1$ is multiplicatively invertible and $b_1 = a_1^{-1}$, then the equations

$\array{ \sum_{n \geq 1} a_n \sum_{k = k_1 + \ldots + k_n} b_{k_1} b_{k_2} \ldots b_{k_n} & = 1\; if\; k = 1 \\ & = 0\; if\; k \neq 1 }$

may be uniquely solved for the remaining $a_i$'s given the $b_j$'s, and uniquely solved for the remaining $b_j$'s given the $a_i$'s, by an inductive procedure: for $k \neq 1$ we have

$a_1 b_k + a_k b_1^k + \; terms\; a_n b_{k_1} \ldots b_{k_n} = 0$

and this allows us to solve for $b_k$,

$b_k = -a_1^{-1}(a_k b_1^k + \; terms\; a_n b_{k_1} \ldots b_{k_n})$

given the values $a_1, \ldots, a_k$ and earlier $b$-values $b_{k_j}$ for $k_j \lt k$ given by inductive hypothesis. Similarly we can solve for $a_k$ in terms of given coefficients $b_1, \ldots, b_k$ and earlier $a$-values $a_n$, $n \lt k$. Thus every power series $a$ has a right inverse if $a_1^{-1}$ exists, and $b$ has a left inverse if $b_1^{-1}$ exists, and this completes the proof.

### Formal differentiation

One way to define the formal differentiation operator, as a function $\frac{\partial}{\partial X}:R[[X]] \to R[[X]]$, is via the usual formula

$\frac{\partial}{\partial X}\left(\sum_{n = 0}^\infty a_n X^n\right) \coloneqq \sum_{n = 0}^\infty a_{n + 1} (n + 1) X^n.$

Then $\frac{\partial}{\partial X}$ is an $R$-linear function on $R[[X]]$ which satisfies the Leibniz rule, meaning that it is a derivation and $R[[X]]$ is a differential algebra.

Here is a conceptual story underlying the formalism. Let $D = R[\varepsilon]/(\varepsilon^2)$ be the representing object for derivations (the “ring of dual numbers”). Let $\delta: R[ [X] ] \to R[ [X] ] \otimes_R D \cong R[ [X] ][\varepsilon]/(\varepsilon^2)$ be the unique topological $R$-algebra map (under the $(X)$-adic topologies described above) that sends $X$ to $X + \varepsilon$. (If it helps, think $\delta(q) = q(X + \varepsilon)$.)

###### Definition

For $p \in R[ [X] ]$, the derivative $p'$ is the unique element of $R[ [X] ]$ satisfying

$\delta(p) = p(X) + p'(X)\varepsilon.$

We leave as an exercise the proof that the two definitions of derivative match:

$p'(X) = \frac{\partial}{\partial X} p(X).$

(Hint: the restriction of $p \mapsto p'$ to $R[X]$ is by construction a derivation such that $X' = 1$, and $(X^k)' = k X^{k-1}$ by induction. This induces derivations on quotient algebras $R[X]/(X^n)$, satisfying the same formula. Then pass to the inverse limit.)

###### Proposition

(Chain rule) For $p \in R[ [X] ]$ and $q \in x R[ [X ] ]$,

$(p \circ q)' = (p' \circ q) \cdot q'.$

See here for a conceptual proof, using the universal property of adic completion.

Relatedly but in a slightly different direction, we can consider differentiation in coalgebraic terms. Suppose the commutative ring $R$ is a commutative algebra over $\mathbb{Q}$ (thus permitting division by nonzero integers). Then the set $R[ [X]]$ may be identified with the terminal coalgebra $R^\mathbb{N}$ of the endofunctor $R \times - \colon Set \to Set$ via the map

$R[ [X]] \to R^\mathbb{N}\; : \; \sum_{n \geq 0} \frac{a_n X^n}{n!} \mapsto (a_0, a_1, a_2, \ldots)$

whereby the coalgebra structure on $R^\mathbb{N}$,

$R^\mathbb{N} \to R \times R^\mathbb{N}\; \colon \; (a_n)_{n \geq 0} \mapsto \langle a_0, (a_{n+1})_{n \geq 0} \rangle,$

corresponds to

$R[ [X]] \to R \times R[ [X]]\; \colon\; f(X) \mapsto \langle f(0), f'(X) \rangle.$

One may then apply coinductive techniques to prove various facts. One illustration is given here, where coinduction on power series is used to prove the general binomial theorem

$(1 + x)^r \coloneqq \exp(r \log(1 + x)) = \sum_{k \geq 0} \frac{r^\underline{k} x^k}{k!}$

where, remarkably, $r$ is an arbitrary element of $R$.

## Ring of power series as rings with infinitesimals

### Purely real and purely infinitesimal elements

Suppose that $K$ is a Archimedean ordered field and $K[[\epsilon]]$ is the ring of power series in $K$. Since $K[[\epsilon]]$ is a local ring, the quotient of $K[[\epsilon]]$ by its ideal of non-invertible elements $\epsilon K[[\epsilon]]$ is the residue field $K$ itself, and the canonical function used in defining the quotient is the function $\Re:K[[\epsilon]] \to K$ which takes a number $a \in K[[\epsilon]]$ to its purely real component $\Re(a) \in K$ and takes $\Re(\epsilon) = 0$. Since $K[[\epsilon]]$ is an ordered $K$-algebra, there is a strictly monotone ring homomorphism $h:K \to K[[\epsilon]]$. An element $a \in K[[\epsilon]]$ is purely real if $h(\Re(a)) = a$, and an element $a \in K[[\epsilon]]$ is purely infinitesimal if it is in the fiber of $\Re$ at $0 \in K$. Zero is the only element in $K[[\epsilon]]$ which is both purely real and purely infinitesimal.

### Analytic functions

Suppose that $K$ is a sequentially Cauchy complete Archimedean ordered field with lattice structure, and $K[[\epsilon]]$ is the ring of power series of $K$. Then analytic functions are each definable on $K$ using the algebraic, order, metric, and convergence structure on $K$.

The ring homomorphism $h:K \to K[[\epsilon]]$ preserves analytic functions: given a natural number $n \in \mathbb{N}$ and a purely infinitesimal element $\eta \in \epsilon K[[\epsilon]]$, then for every analytic function $f \in C^\infty(K)$, there is a function $f_{K[[\epsilon]]}:K[[\epsilon]] \to K[[\epsilon]]$ such that for all elements $x \in K$, $f_{K[[\epsilon]]}(h(x)) = h(f(x))$ and

$f_{K[[\epsilon]]}(h(x) + \eta) = \sum_{i = 0}^{\infty} \frac{1}{i!} h\left(\frac{d^i f}{d x^i}(x)\right) \eta^i$

A formalization in homotopy type theory and there in Coq is discussed in section 4 of

The discussion of the differentiation of a converging power series term by term is at

category: analysis, algebra