nLab power series

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).

As adic completions

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 $k$.

Taylor series

• Taylor series

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.

Laurent series

• Laurent series

Puiseux series

• Puiseux series

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.

Is a principal ideal domain over a field

If $R$ is a field then the nonzero ideals of the power series ring are entirely of the form $x^n R[[x]]$:

Lemma

(ideals of a power series ring over a field) let $R$ be a field and let $I \subseteq R[[x]]$ be a nonzero ideal of the algebra of power-series over $R[[x]]$. There is $n \in \mathbb{N}$ such that $I = x^n \cdot R[[x]]$.

Proof

let $f = \sum_{i =0}^{\infty} a_i x^i$ be a nonzero element of $I$ chosen to satisfy

• $a_n = 1$ where $n = \mathrm{min} \{ i \in \mathbb{N} : a_i \neq 0 \}$.

• for each nonzero $g = \sum_{i = 0}^{\infty} b_i x^i \in I$,

$$\mathrm{ord}(f)\ \leq\ \mathrm{ord}(g)$$

where for a given nonzero element $h = \sum_{i =0}^{\infty} c_i x^i$ in $R[[x]]$,

$$\mathrm{ord}(h) = \mathrm{min} \{ i \in \mathbb{N} : c_i \neq 0 \}$$

Such an element exists since, if $f' = \sum_{i=0}^{\infty} a_i' x^i$ is any nonzero element of $I$ satisfying the second property, then we can set

$$f = \sum_{i = 0}^{\infty} a_{\mathrm{min} \{ j \in \mathbb{N} : a_j' \neq 0 \} }'{}^{-1} \cdot a_i' \cdot x^i$$

Define a sequence $f_k = \sum_{i =0}^{\infty} a_{k,i} x^i$ by induction on $k \in \mathbb{N}$ satisfying the inductive hypothesis that $a_{k,i} = 0$ for $i \in \mathbb{N}$ such that

• $i \leq n + k$

• $i \neq n$

For the case in which $k = 0$, set $f_0 = f$.

For the inductive case, suppose that for some $k \in \mathbb{N}$ we have defined $f_k = \sum_{i = 0}^{\infty} a_{k,i} x^i$ satisfying $a_{k,i} = 0$ for $i \in \mathbb{N}$ such that

• $i \leq n + k$

• $i \neq n$

$$f_{k+1} = \sum_{i = 0}^{\infty} a_{k+1,i} x^i = f_k ( 1 - a_{k,n+k+1} x^{k+1} )$$

Then $a_{k+1,i} = 0$ for $i \in \mathbb{N}$ such that

• $i \leq n + k + 1$

• $i \neq n$

This concludes the construction of the elements

$$f_{k} = \sum_{i = 1}^{\infty} a_{k,i} x^i \in R[[x]]$$

Next consider the sequence

$$g_m = \sum_{j = 0}^{m} b_{m,j} x^j = \Pi_{k = 0}^{m} ( 1 - a_{k,n+k+1} x^{k+1} ) \in R[[x]]$$

and consider that

$$b_{m+\ell, j} = b_{m, j}$$

for $j \leq m+1$. This identity implies that

$$g = \Pi_{k =0}^{\infty} (1 - a_{k,n+k+1} x^{k+1} )$$

is well defined.

For each $m \in \mathbb{N}$, $f \cdot g_m$ agrees with $x^n$ through degree $n + m$. The coefficients of $f \cdot g$ and $x^n$ agree in every degree, so $fg = x^n$.

This shows that $f = x^n u$ for $u = g^{-1}$, hence $x^n \in f \cdot R[[x]]$.

Since $x^n \in I$, $x^n \cdot R[[x]] \subseteq I$.

If $h \in I$, then by minimality of $n$, $n \leq \mathrm{ord}(h)$. Hence $h = x^n \cdot q$ for some $q \in R[[x]]$. So $h \in x^n \cdot R[[x]]$. So $I \subseteq x^n \cdot R[[x]]$.

It follows from this argument that $I = x^n \cdot R[[x]]$ for some $n \in \mathbb{N}$.

It follows from this argument that $R[[x]]$ is a principal ideal domain for any field $R$.

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

ring with infinitesimals	function
dual numbers	differentiable function
Weil ring	smooth function
power series ring	analytic function

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

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

Related concepts

• convergence radius

• asymptotic series, transseries

• Tate algebra

• characteristic series

• power series and their formal spectra

References

• Wikipedia, Formal power series

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

• Álvaro Pelayo, Vladimir Voevodsky, Michael Warren, A preliminary univalent formalization of the p-adic numbers (arXiv:1302.1207)

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

• Tim Gowers’s blog: differentiating-power-series

category: analysis, algebra