Taylor Expansion from Universal Properties

 

Taylor Expansion as a Quotient Projection

There is a way of reading Taylor's formula which is less computational and more structural.

The usual finite Taylor expansion

$$f(t)\mapsto \sum _{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}{x}^k\mathrm{mod}{x}^n$$

is not merely a formula obtained by differentiating repeatedly. It is the canonical map forced by the universal property of the free $C^{\mathrm{\infty }}$$C^\infty$-ring $C^{\mathrm{\infty }}(\mathbb{R})$$C^\infty(\mathbb R)$.

In this view, finite Taylor expansion is a quotient projection, and formal Taylor expansion is the inverse limit of these quotient projections.

The free $C^{\mathrm{\infty }}$$C^\infty$-ring on one generator

The ring $C^{\mathrm{\infty }}(\mathbb{R})$$C^\infty(\mathbb R)$ is the free $C^{\mathrm{\infty }}$$C^\infty$-ring on one generator.

Let $t\in C^{\mathrm{\infty }}(\mathbb{R})$$t\in C^\infty(\mathbb R)$ denote the identity function

$$t:\mathbb{R}\to \mathbb{R},t(a)=a.$$

Then for every $C^{\mathrm{\infty }}$$C^\infty$-ring $A$$A$, evaluation at the generator gives a natural bijection

$${\mathrm{Hom}}_{C^{\mathrm{\infty }}\mathrm{Ring}}(C^{\mathrm{\infty }}(\mathbb{R}),A)\cong A,$$

where a morphism

$$\varphi :C^{\mathrm{\infty }}(\mathbb{R})\to A$$

corresponds to the element

$$\varphi (t)\in A.$$

Equivalently, to define a $C^{\mathrm{\infty }}$$C^\infty$-ring morphism out of $C^{\mathrm{\infty }}(\mathbb{R})$$C^\infty(\mathbb R)$, it is enough to say where the coordinate function $t$$t$ goes.

This is the smooth analogue of the usual universal property of a polynomial ring. The ordinary polynomial ring $\mathbb{R}[x]$$\mathbb R[x]$ is the free commutative $\mathbb{R}$$\mathbb R$-algebra on one generator, while $C^{\mathrm{\infty }}(\mathbb{R})$$C^\infty(\mathbb R)$ is the free $C^{\mathrm{\infty }}$$C^\infty$-ring on one generator.

The distinction matters: a $C^{\mathrm{\infty }}$$C^\infty$-ring morphism must preserve not only addition and multiplication, but every smooth operation.

The canonical $C^{\mathrm{\infty }}$$C^\infty$-structure on $\mathbb{R}[x]/(x^n)$$\mathbb R[x]/(x^n)$

Consider the truncated polynomial algebra

$$D_n=\mathbb{R}[x]/(x^n).$$

Write

$$\epsilon =[x]\in D_n.$$

Then

$${\epsilon }^n=0.$$

The algebra $D_n$$D_n$ has a canonical $C^{\mathrm{\infty }}$$C^\infty$-ring structure. If

$$u=a+\eta \in D_n,$$

where

$$a\in \mathbb{R},\eta \in (\epsilon ),$$

then for a smooth function

$$f:\mathbb{R}\to \mathbb{R}$$

we define

$$f_{D_n}(u)=\sum _{k=0}^{n-1}\frac{f^{(k)}(a)}{k!}{\eta }^k.$$

This is finite because ${\eta }^n=0$$\eta^n=0$.

In particular, if we evaluate at the nilpotent element $\epsilon$$\varepsilon$, then $a=0$$a=0$ and $\eta =\epsilon$$\eta=\varepsilon$, so

$$f_{D_n}(\epsilon )=\sum _{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}{\epsilon }^k.$$

Thus the usual Taylor polynomial appears as the value of the smooth operation $f$$f$ on the nilpotent element $\epsilon$$\varepsilon$.

The Taylor map is determined by the free property

By the free $C^{\mathrm{\infty }}$$C^\infty$-ring property of $C^{\mathrm{\infty }}(\mathbb{R})$$C^\infty(\mathbb R)$, the choice of the element

$$\epsilon \in D_n$$

determines a unique $C^{\mathrm{\infty }}$$C^\infty$-ring morphism

$$T_{n-1}:C^{\mathrm{\infty }}(\mathbb{R})\to D_n$$

such that

$$T_{n-1}(t)=\epsilon .$$

For any smooth function $f\in C^{\mathrm{\infty }}(\mathbb{R})$$f\in C^\infty(\mathbb R)$, this morphism must satisfy

$$T_{n-1}(f)=f_{D_n}(\epsilon ).$$

Therefore

$$T_{n-1}(f)=\sum _{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}{\epsilon }^k.$$

So the finite Taylor expansion map

$$C^{\mathrm{\infty }}(\mathbb{R})\to \mathbb{R}[\epsilon ]/({\epsilon }^n)$$

is not an arbitrary construction. It is the unique $C^{\mathrm{\infty }}$$C^\infty$-ring morphism determined by

$$t\mapsto \epsilon .$$

This is the key point:

$$\boxed{{\displaystyle \text{Finite Taylor expansion is the canonical map forced by freeness.}}}$$

Its kernel is $(t^n)$$(t^n)$

Now let us compute the kernel of

$$T_{n-1}:C^{\mathrm{\infty }}(\mathbb{R})\to \mathbb{R}[\epsilon ]/({\epsilon }^n).$$

By the formula above,

$$T_{n-1}(f)=0$$

if and only if

$$f(0)=f^{\prime }(0)=\cdots =f^{(n-1)}(0)=0.$$

By Taylor's formula with smooth remainder,

$$f(t)=\sum _{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}{t}^k+t^{n}g(t)$$

for some smooth function

$$g\in C^{\mathrm{\infty }}(\mathbb{R}).$$

Thus, if the first $n$$n$ Taylor coefficients vanish, then

$$f(t)=t^{n}g(t).$$

Hence

$$f\in (t^n).$$

Conversely, if

$$f=t^{n}g,$$

then $f$$f$ vanishes to order at least $n$$n$ at $0$$0$, so

$$T_{n-1}(f)=0.$$

Therefore

$$\ker T_{n-1}=(t^n).$$

By the first isomorphism theorem,

$$C^{\mathrm{\infty }}(\mathbb{R})/(t^n)\cong \mathbb{R}[\epsilon ]/({\epsilon }^n).$$

This is not just an isomorphism of ordinary rings. It is an isomorphism of $C^{\mathrm{\infty }}$$C^\infty$-rings.

Finite Taylor expansion is the quotient projection

Let

$$q_n:C^{\mathrm{\infty }}(\mathbb{R})\to C^{\mathrm{\infty }}(\mathbb{R})/(t^n)$$

be the quotient projection.

Since

$$\ker T_{n-1}=(t^n),$$

the Taylor map factors uniquely through the quotient:

$$C^{\mathrm{\infty }}(\mathbb{R})\stackrel{q_n}{\to }{C}^{\mathrm{\infty }}(\mathbb{R})/(t^n)\stackrel{\sim }{\to }\mathbb{R}[\epsilon ]/({\epsilon }^n).$$

Under the isomorphism

$$C^{\mathrm{\infty }}(\mathbb{R})/(t^n)\cong \mathbb{R}[\epsilon ]/({\epsilon }^n),$$

the quotient projection becomes

$$f\mapsto \sum _{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}{\epsilon }^k.$$

Therefore:

$$\boxed{{\displaystyle \text{Finite Taylor expansion is the quotient map modulo }(t^n),\text{ written in polynomial coordinates.}}}$$

This is the structural content of Taylor's formula.

The formula says that modulo $(t^n)$$(t^n)$, every smooth function has a unique representative of the form

$$a_0+a_{1}t+\cdots +a_{n-1}{t}^{n-1}.$$

The coefficients are exactly

$$a_k=\frac{f^{(k)}(0)}{k!}.$$

The inverse system of infinitesimal neighborhoods

The quotients

$$\mathbb{R}[\epsilon ]/({\epsilon }^n)$$

form an inverse system:

$$\cdots ⟶\mathbb{R}[\epsilon ]/({\epsilon }^{n+1})⟶\mathbb{R}[\epsilon ]/({\epsilon }^n)⟶\cdots ⟶\mathbb{R}[\epsilon ]/(\epsilon ).$$

The transition maps are the obvious truncation maps:

$$p(\epsilon )\mathrm{mod}{\epsilon }^{n+1}\mapsto p(\epsilon )\mathrm{mod}{\epsilon }^n.$$

Equivalently, the smooth quotients

$$C^{\mathrm{\infty }}(\mathbb{R})/(t^n)$$

also form an inverse system:

$$\cdots ⟶C^{\mathrm{\infty }}(\mathbb{R})/(t^{n+1})⟶C^{\mathrm{\infty }}(\mathbb{R})/(t^n)⟶\cdots ⟶C^{\mathrm{\infty }}(\mathbb{R})/(t).$$

Taking the inverse limit gives the $(t)$$(t)$-adic completion of $C^{\mathrm{\infty }}(\mathbb{R})$$C^\infty(\mathbb R)$ at $0$$0$:

$${\widehat{C^{\mathrm{\infty }}(\mathbb{R})}}_{(t)}=\underset{n}{\underleftarrow{\lim }}{C}^{\mathrm{\infty }}(\mathbb{R})/(t^n).$$

Using the finite Taylor identifications, we get

$$\underset{n}{\underleftarrow{\lim }}{C}^{\mathrm{\infty }}(\mathbb{R})/(t^n)\cong \underset{n}{\underleftarrow{\lim }}\mathbb{R}[\epsilon ]/({\epsilon }^n).$$

But

$$\underset{n}{\underleftarrow{\lim }}\mathbb{R}[\epsilon ]/({\epsilon }^n)\cong \mathbb{R}[[\epsilon ]].$$

Indeed, giving a compatible family

$$(p_n{)}_{n\ge 1},p_n\in \mathbb{R}[\epsilon ]/({\epsilon }^n),$$

is the same as giving infinitely many coefficients

$$a_0,a_1,a_2,\dots$$

such that

$$p_n=a_0+a_1\epsilon +\cdots +a_{n-1}{\epsilon }^{n-1}\mathrm{mod}{\epsilon }^n.$$

This is exactly the same data as a formal power series

$$\sum _{k\ge 0}{a}_k{\epsilon }^k.$$

Therefore

$$\boxed{{\displaystyle \underset{n}{\underleftarrow{\lim }}\mathbb{R}[\epsilon ]/({\epsilon }^n)\cong \mathbb{R}[[\epsilon ]].}}$$

Formal Taylor expansion by the universal property of the inverse limit

For each $n$$n$, we have the finite Taylor projection

$$T_{n-1}:C^{\mathrm{\infty }}(\mathbb{R})\to \mathbb{R}[\epsilon ]/({\epsilon }^n).$$

These maps are compatible with truncation. That is, the diagram

$$\begin{array}{ccc}{C}^{\mathrm{\infty }}(\mathbb{R})& \stackrel{T_n}{\to }& \mathbb{R}[\epsilon ]/({\epsilon }^{n+1})\\ & {\searrow }_{T_{n-1}}& \downarrow \\ & & \mathbb{R}[\epsilon ]/({\epsilon }^n)\end{array}$$

commutes.

Hence, by the universal property of the inverse limit, there is a unique ring homomorphism

$$\hat{T}:C^{\mathrm{\infty }}(\mathbb{R})\to \underset{n}{\underleftarrow{\lim }}\mathbb{R}[\epsilon ]/({\epsilon }^n)\cong \mathbb{R}[[\epsilon ]]$$

such that for every $n$$n$,

$${\pi }_n\circ \hat{T}=T_{n-1}.$$

This unique map is

$$\hat{T}(f)=\sum _{k\ge 0}\frac{f^{(k)}(0)}{k!}{\epsilon }^k.$$

So the formal Taylor expansion is not defined by first writing down an infinite analytic series. Instead, it is defined by a universal property:

$$\boxed{{\displaystyle \text{The formal Taylor expansion is the unique map induced by the compatible family of finite Taylor quotient maps.}}}$$

Equivalently,

$$\boxed{{\displaystyle \text{Formal Taylor expansion is the completion map}{C}^{\mathrm{\infty }}(\mathbb{R})\to {\widehat{C^{\mathrm{\infty }}(\mathbb{R})}}_{(t)}\cong \mathbb{R}[[\epsilon ]].}}$$

The crucial warning: smooth functions are not analytic functions

The map

$$\hat{T}:C^{\mathrm{\infty }}(\mathbb{R})\to \mathbb{R}[[\epsilon ]]$$

is not injective.

Its kernel is

$$\ker \hat{T}=\bigcap _{n\ge 1}(t^n).$$

This consists of all smooth functions which are flat at $0$$0$, meaning

$$f^{(k)}(0)=0\text{for all }k\ge 0.$$

For example,

$$\begin{array}{c}f(t)=\{\begin{array}{ll}{e}^{-1/t^2},& t\ne 0,\\ 0,& t=0\end{array}\end{array}$$

is a nonzero smooth function whose Taylor series at $0$$0$ is zero.

Thus the formal Taylor expansion does not embed $C^{\mathrm{\infty }}(\mathbb{R})$$C^\infty(\mathbb R)$ into formal power series. It only records the infinite jet at $0$$0$.

This is exactly what should happen: a smooth function is not determined by its Taylor series, but its formal jet is.