
# 6.5: Repeated Differentiation. Taylor’s Theorem


In §1 we defined $$\vec{u}$$-directed derived functions, $$D_{\vec{u}} f$$ for any $$f : E^{\prime} \rightarrow E$$ and any $$\vec{u} \neq \overrightarrow{0}$$ in $$E^{\prime}.$$

Thus given a sequence $$\left\{\vec{u}_{i}\right\} \subseteq E^{\prime}-\{\overrightarrow{0}\},$$ we can first form $$D_{\vec{u}_{1}} f,$$ then $$D_{\vec{u}_{2}}\left(D_{\vec{u}_{1}} f\right)$$ (the $$\vec{u}_{2}$$-directed derived function of $$D_{\vec{u}_{1}} f ),$$ then the $$\vec{u}_{3}$$-directed derived function of $$D_{\vec{u}_{2}}\left(D_{\vec{u}_{1}} f\right),$$ and so on. We call all functions so formed the higher-order directional derived functions of $$f.$$

If at each step the limit postulated in Definition 1 of §1 exists for all $$\vec{p}$$ in a set $$B \subseteq E^{\prime},$$ we call them the higher-order directional derivatives of $$f$$ (on $$B$$).

If all $$\vec{u}_{i}$$ are basic unit vectors in $$E^{n}\left(C^{n}\right),$$ we say "partial' instead of "directional."

We also define $$D_{\vec{u}}^{1} f=D_{\vec{u}} f$$ and

$D_{\vec{u}_{1} \vec{u}_{2} \ldots \vec{u}_{k}}^{k} f=D_{\vec{u}_{k}}\left(D_{\vec{u}_{1} \overline{u}_{2} \ldots \vec{u}_{k-1}}^{k-1} f\right), \quad k=2,3, \ldots,$

and call $$D_{\vec{u}_{1} \vec{u}_{2} \ldots \vec{u}_{k}}^{k} f$$ a directional derived function of order $$k.$$ (Some authors denote it by $$D_{\vec{u}_{k} \vec{u}_{k-1} \ldots \vec{u}_{1}} f.)$$

If all $$\vec{u}_{i}$$ equal $$\vec{u},$$ we write $$D_{\vec{u}}^{k} f$$ instead.

For partially derived functions, we simplify this notation, writing $$1 2 \ldots$$ for $$\vec{e}_{1} \vec{e}_{2} \ldots$$ and omitting the "$$k$$" in $$D^{k}$$ (except in classical notation):

$D_{12} f=D_{\vec{e}_{1} \vec{e}_{2}}^{2} f=\frac{\partial^{2} f}{\partial x_{1} \partial x_{2}}, \quad D_{11} f=D_{\vec{e}_{1} \vec{e}_{1}}^{2} f=\frac{\partial^{2} f}{\partial x_{1}^{2}}, \text { etc.}$

We also set $$D_{\vec{u}}^{0} f=f$$ for any vector $$\vec{u}$$.

Example

(A) Define $$f : E^{2} \rightarrow E^{1}$$ by

$f(0,0)=0, \quad f(x, y)=\frac{x y\left(x^{2}-y^{2}\right)}{x^{2}+y^{2}}.$

Then

$\frac{\partial f}{\partial x}=D_{1} f(x, y)=\frac{y\left(x^{4}+4 x^{2} y^{2}-y^{4}\right)}{\left(x^{2}+y^{2}\right)^{2}},$

whence $$D_{1} f(0, y)=-y$$ if $$y \neq 0;$$ and also

$D_{1} f(0,0)=\lim _{x \rightarrow 0} \frac{f(x, 0)-f(0,0)}{x}=0. \quad \text {(Verify!)}$

Thus $$D_{1} f(0, y)=-y$$ always, and so $$D_{12} f(0, y)=-1; D_{12} f(0,0)=-1$$ Similarly,

$D_{2} f(x, y)=\frac{x\left(x^{4}-4 x^{2} y^{2}-y^{4}\right)}{\left(x^{2}+y^{2}\right)^{2}}$

if $$x \neq 0$$ and $$D_{2} f(0,0)=0.$$ Thus $$(\forall x) D_{2} f(x, 0)=x$$ and so

$D_{21} f(x, 0)=1 \text { and } D_{21} f(0,0)=1 \neq D_{12} f(0,0)=-1.$

The previous example shows that we may well have $$D_{12} f \neq D_{21} f,$$ or more generally, $$D_{\vec{u} \vec{v}}^{2} f \neq D_{\vec{v} \vec{u}}^{2} f.$$ However, we obtain the following theorem.

Theorem $$\PageIndex{1}$$

Given nonzero vectors $$\vec{u}$$ and $$\vec{v}$$ in $$E^{\prime},$$ suppose $$f : E^{\prime} \rightarrow E$$ has the derivatives

$D_{\vec{u}} f, D_{\vec{v}} f, \text { and } D_{\vec{u} \vec{v}}^{2} f$

on an open set $$A \subseteq E^{\prime}$$.

If $$D_{\vec{u} \vec{v}}^{2} f$$ is continuous at some $$\vec{p} \in A,$$ then the derivative $$D_{\vec{v} \vec{u}}^{2} f(\vec{p})$$ also exists and equals $$D_{\vec{u} \vec{v}}^{2} f(\vec{p})$$.

Proof

By Corollary 1 in §1, all reduces to the case $$|\vec{u}|=1=|\vec{v}|.$$ (Why?)

Given $$\varepsilon>0,$$ fix $$\delta>0$$ so small that $$G=G_{\vec{p}}(\delta) \subseteq A$$ and simultaneously

$\sup _{\vec{x} \in G}\left|D_{\vec{u} \vec{v}}^{2} f(\vec{x})-D_{\vec{u} \vec{v}}^{2} f(\vec{p})\right| \leq \varepsilon$

(by the continuity of $$D_{\vec{u} \vec{v}}^{2} f$$ at $$\vec{p})$$.

Now $$\left(\forall s, t \in E^{1}\right)$$ define $$H_{t} : E^{1} \rightarrow E$$ by

$H_{t}(s)=D_{\vec{u}} f(\vec{p}+t \vec{u}+s \vec{v}).$

Let

$I=\left(-\frac{\delta}{2}, \frac{\delta}{2}\right).$

If $$s, t \in I,$$ the point $$\vec{x}=\vec{p}+t \vec{u}+s \vec{v}$$ is in $$G_{\vec{p}}(\delta) \subseteq A,$$ since

$|\vec{x}-\vec{p}|=|t \vec{u}+s \vec{v}|<\frac{\delta}{2}+\frac{\delta}{2}=\delta.$

Thus by assumption, the derivative $$D_{\vec{u} \vec{v}}^{2} f(\vec{p})$$ exists. Also,

\begin{aligned} H_{t}^{\prime}(s) &=\lim _{\Delta s \rightarrow 0} \frac{1}{\Delta s}\left[H_{t}(s+\Delta s)-H_{t}(s)\right] \\ &=\lim _{\Delta s \rightarrow 0} \frac{1}{\Delta s}\left[D_{\vec{u}} f(\vec{x}+\Delta s \cdot \vec{v})-D_{\vec{u}} f(\vec{x})\right]. \end{aligned}

But the last limit is $$D_{\vec{u} \vec{v}}^{2} f(\vec{x}),$$ by definition. Thus, setting

$h_{t}(s)=H_{t}(s)-s D_{\vec{u} \vec{v}}^{2} f(\vec{p}),$

we get

\begin{aligned} h_{t}^{\prime}(s) &=H_{t}^{\prime}(s)-D_{\vec{u} \vec{v}}^{2} f(\vec{p}) \\ &=D_{\vec{u} \vec{v}}^{2} f(\vec{x})-D_{\vec{u} \vec{v}}^{2} f(\vec{p}). \end{aligned}

We see that $$h_{t}$$ is differentiable on $$I,$$ and by (2),

$\sup _{s \in I}\left|h_{t}^{\prime}(s)\right| \leq \sup _{\vec{x} \in G}\left|D_{\vec{u} \vec{v}}^{2} f(\vec{x})-D_{\vec{u} \vec{v}}^{2} f(\vec{p})\right| \leq \varepsilon$

for all $$t \in I.$$ Hence by Corollary 1 of Chapter 5, §4,

$\left|h_{t}(s)-h_{t}(0) \leq\right| s\left|\sup _{\sigma \in I}\right| h_{t}^{\prime}(\sigma)| \leq| s | \varepsilon.$

But by definition,

$h_{t}(s)=D_{\vec{u}} f(\vec{p}+t \vec{u}+s \vec{v})-s D_{\vec{u} \vec{v}}^{2} f(\vec{p})$

and

$h_{t}(0)=D_{\vec{u}} f(\vec{p}+t \vec{u}).$

Thus

$\left|D_{\vec{u}} f(\vec{p}+t \vec{u}+s \vec{v})-D_{\vec{u}} f(\vec{p}+t \vec{u})-s D_{\vec{u} \vec{v}}^{2} f(\vec{p})\right| \leq|s| \varepsilon$

for all $$s, t \in I$$.

Next, set

$G_{s}(t)=f(\vec{p}+t \vec{u}+s \vec{v})-f(\vec{p}+t \vec{u})$

and

$g_{s}(t)=G_{s}(t)-s t \cdot D_{\vec{u} \vec{v}}^{2} f(\vec{p}).$

As before, one finds that $$(\forall s \in I) g_{s}$$ is differentiable on $$I$$ and that

$g_{s}^{\prime}(t)=D_{\vec{u}} f(\vec{p}+t \vec{u}+s \vec{v})-D_{\vec{u}} f(\vec{p}+t \vec{u})-s D_{\vec{u} \vec{v}}^{2} f(\vec{p})$

for $$s, t \in I.$$ (Verify!)

Hence by (3),

$\sup _{t \in I}\left|g_{s}^{\prime}(t)\right| \leq|s| \varepsilon.$

Again, by Corollary 1 of Chapter 5, §4,

$\left|g_{s}(t)-g_{s}(0)\right| \leq|s t| \varepsilon,$

or by the definition of $$g_{s}$$ (assuming $$s, t \in I-\{0\}$$ and dividing by $$s t )$$,

$\left|\frac{1}{s t}[f(\vec{p}+t \vec{u}+s \vec{v})-f(\vec{p}+t \vec{u})]-D_{\vec{u} \vec{v}}^{2} f(\vec{p})-\frac{1}{s t}[f(\vec{p}+s \vec{v})-f(\vec{p})]\right| \leq \varepsilon.$

(Verify!) Making $$s \rightarrow 0$$ (with $$t$$ fixed), we get, by the definition of $$D_{\vec{v}} f$$,

$\left|\frac{1}{t} D_{\vec{v}} f(\vec{p}+t \vec{u})-\frac{1}{t} D_{\vec{v}} f(\vec{p})-D_{\vec{u} \vec{v}}^{2} f(\vec{p})\right| \leq \varepsilon$

whenever $$0<|t|<\delta / 2$$.

As $$\varepsilon$$ is arbitrary, we have

$D_{\vec{u} \vec{v}}^{2} f(\vec{p})=\lim _{t \rightarrow 0} \frac{1}{t}\left[D_{v} f(\vec{p}+t \vec{u})-D_{\vec{v}} f(\vec{p})\right].$

But by definition, this limit is the derivative $$D_{\vec{v} \vec{u}}^{2} f(\vec{p}).$$ Thus all is proved.$$\quad \square$$

Note 1. By induction, the theorem extends to derivatives of order $$>2.$$ Thus the derivative $$D_{\vec{u}_{1} \vec{u}_{2} \ldots \vec{u}_{k}} f$$ is independent of the order in which the $$\vec{u}_{i}$$ follow each other if it exists and is continuous on an open set $$A \subseteq E^{\prime},$$ along with appropriate derivatives of order $$<k$$.

If $$E^{\prime}=E^{n}\left(C^{n}\right),$$ this applies to partials as a special case.

For $$E^{n}$$ and $$C^{n}$$ only, we also formulate the following definition.

Definition 1

Let $$E^{\prime}=E^{n}\left(C^{n}\right).$$ We say that $$f : E^{\prime} \rightarrow E$$ is $$m$$ times differentiable at $$\vec{p} \in E^{\prime}$$ iff $$f$$ and all its partials of order $$<m$$ are differentiable at $$\vec{p}$$.

If this holds for all $$\vec{p}$$ in a set $$B \subseteq E^{\prime},$$ we say that $$f$$ is $$m$$ times differentiable on $$B$$.

If, in addition, all partials of order $$m$$ are continuous at $$\vec{p}$$ (on $$B),$$ we say that $$f$$ is of class $$C D^{m},$$ or continuously differentiable $$m$$ times there, and write $$f \in C D^{m}$$ at $$\vec{p}$$ (on $$B).$$

Finally, if this holds for all natural $$m,$$ we write $$f \in C D^{\infty}$$ at $$\vec{p}$$ (on $$B,$$ respectively).

Definition 2

Given the space $$E^{\prime}=E^{n}\left(C^{n}\right),$$ the function $$f : E^{\prime} \rightarrow E,$$ and a point $$\vec{p} \in E^{\prime},$$ we define the mappings

$d^{m} f(\vec{p} ; \cdot), \quad m=1,2, \ldots,$

from $$E^{\prime}$$ to $$E$$ by setting for every $$\vec{t}=\left(t_{1}, \ldots, t_{n}\right)$$

\begin{aligned} d^{1} f(\vec{p} ; \vec{t}) &=\sum_{i=1}^{n} D_{i} f(\vec{p}) \cdot t_{i}, \\ d^{2} f(\vec{p} ; \vec{t}) &=\sum_{j=1}^{n} \sum_{i=1}^{n} D_{i j} f(\vec{p}) \cdot t_{i} t_{j}, \\ d^{3} f(\vec{p} ; \vec{t}) &=\sum_{k=1}^{n} \sum_{j=1}^{n} \sum_{i=1}^{n} D_{i j k} f(\vec{p}) \cdot t_{i} t_{j} t_{k}, \quad \text {and so on. } \end{aligned}

We call $$d^{m} f(\vec{p} ; \cdot)$$ the $$m t h$$ differential (or differential of order $$m)$$ of $$f$$ at $$\vec{p}$$. By our conventions, it is always defined on $$E^{n}\left(C^{n}\right)$$ as are the partially derived functions involved.

If $$f$$ is differentiable at $$\vec{p}$$ (but not otherwise), then $$d^{1} f(\vec{p} ; \vec{t})=d f(\vec{p} ; \vec{t})$$ by Corollary 1 in §3; $$d^{1} f(\vec{p} ; \cdot)$$ is linear and continuous (why?) but need not satisfy Definition 1 in §3.

In classical notation, we write $$d x_{i}$$ for $$t_{i};$$ e.g.,

$d^{2} f=\sum_{j=1}^{n} \sum_{i=1}^{n} \frac{\partial^{2} f}{\partial x_{i} \partial x_{j}} d x_{i} d x_{j}.$

Note 2. Classical analysis tends to define differentials as above in terms of partials. Formula (4) for $$d^{m} f$$ is often written symbolically:

$d^{m} f=\left(\frac{\partial}{\partial x_{1}} d x_{1}+\frac{\partial}{\partial x_{2}} d x_{2}+\cdots+\frac{\partial}{\partial x_{n}} d x_{n}\right)^{m} f, \quad m=1,2, \ldots$

Indeed, raising the bracketed expression to the $$m$$th "power" as in algebra (removing brackets, without collecting "similar" terms) and then "multiplying" by $$f,$$ we obtain sums that agree with (4). (Of course, this is not genuine multiplication but only a convenient memorizing device.)

Example

(B) Define $$f : E^{2} \rightarrow E^{1}$$ by

$f(x, y)=x \sin y.$

Take any $$\vec{p}=(x, y) \in E^{2}.$$ Then

$D_{1} f(x, y)=\sin y \text { and } D_{2} f(x, y)=x \cos y;$

$D_{12} f(x, y)=D_{21} f(x, y)=\cos y,$

$D_{11} f(x, y)=0, \text { and } D_{22} f(x, y)=-x \sin y;$

$D_{111} f(x, y)=D_{112} f(x, y)=D_{121} f(x, y)=D_{211} f(x, y)=0,$

$D_{221} f(x, y)=D_{212} f(x, y)=D_{122} f(x, y)=-\sin y, \text { and}$

$D_{222} f(x, y)=-x \cos y; \text { etc.}$

As is easily seen, $$f$$ has continuous partials of all orders; so $$f \in C D^{\infty}$$ on all of $$E^{2}.$$ Also,

\begin{aligned} d f(\vec{p} ; \vec{t}) &=t_{1} D_{1} f(\vec{p})+t_{2} D_{2} f(\vec{p}) \\ &=t_{1} \sin y+t_{2} x \cos y. \end{aligned}

In classical notation,

\begin{aligned} d f &=d^{1} f=\frac{\partial f}{\partial x} d x+\frac{\partial f}{\partial y} d y \\ &=\sin y d x+x \cos y d y; \\ d^{2} f &=\frac{\partial^{2} f}{\partial x^{2}} d x^{2}+2 \frac{\partial^{2} f}{\partial x \partial y} d x d y+\frac{\partial^{2} f}{\partial y^{2}} d y^{2} \\ &=2 \cos y d x d y-x \sin y d y^{2}; \\ d^{3} f &=-3 \sin y d x d y^{2}-x \cos y d y^{3}; \end{aligned}

and so on. (Verify!)

We can now extend Taylor's theorem (Theorem 1 in Chapter 5, §6) to the case $$E^{\prime}=E^{n}\left(C^{n}\right).$$

Theorem $$\PageIndex{2}$$ (Taylor)

Let $$\vec{u}=\vec{x}-\vec{p} \neq \overrightarrow{0}$$ in $$E^{\prime}=E^{n}\left(C^{n}\right)$$.

If $$f : E^{\prime} \rightarrow E$$ is $$m+1$$ times differentiable on the line segment

$I=L[\vec{p}, \vec{x}] \subset E^{\prime}$

then

$f(\vec{x})-f(\vec{p})=\sum_{i=1}^{m} \frac{1}{i !} d^{i} f(\vec{p} ; \vec{u})+R_{m},$

with

$\left|R_{m}\right| \leq \frac{K_{m}}{(m+1) !}, K_{m} \in E^{1},$

and

$0 \leq K_{m} \leq \sup _{\vec{s} \in I}\left|d^{m+1} f(\vec{s} ; \vec{u})\right|.$

Proof

Define $$g : E^{1} \rightarrow E^{\prime}$$ and $$h : E^{1} \rightarrow E$$ by $$g(t)=\vec{p}+t \vec{u}$$ and $$h=f \circ g$$.

As $$E^{\prime}=E^{n}\left(C^{n}\right),$$ we may consider the components of $$g$$,

$g_{k}(t)=p_{k}+t u_{k}, \quad k \leq n.$

Clearly, $$g_{k}$$ is differentiable, $$g_{k}^{\prime}(t)=u_{k}$$.

By assumption, so is $$f$$ on $$I=L[\vec{p}, \vec{x}].$$ Thus, by the chain rule, $$h=f \circ g$$ is differentiable on the interval $$J=[0,1] \subset E^{1};$$ for, by definition,

$\vec{p}+t \vec{u} \in L[\vec{p}, \vec{x}] \text { iff } t \in[0,1].$

By Theorem 2 in §4,

$h^{\prime}(t)=\sum_{k=1}^{n} D_{k} f(\vec{p}+t \vec{u}) \cdot u_{k}=d f(\vec{p}+t \vec{u} ; \vec{u}), \quad t \in J.$

(Explain!)

By assumption (and Definition 1), the $$D_{k} f$$ are differentiable on $$I.$$ Hence, by (7), $$h^{\prime}$$ is differentiable on $$J.$$ Reapplying Theorem 2 in §4, we obtain

\begin{aligned} h^{\prime \prime}(t) &=\sum_{j=1}^{n} \sum_{k=1}^{n} D_{k j} f(\vec{p}+t \vec{u}) \cdot u_{k} u_{j} \\ &=d^{2} f(\vec{p}+t \vec{u} ; \vec{u}), \quad t \in J. \end{aligned}

By induction, $$h$$ is $$m+1$$ times differentiable on $$J,$$ and

$h^{(i)}(t)=d^{i} f(\vec{p}+t \vec{u} ; \vec{u}), \quad t \in J, i=1,2, \ldots, m+1.$

The differentiability of $$h^{(i)}(i \leq m)$$ implies its continuity on $$J=[0,1]$$.

Thus $$h$$ satisfies Theorem 1 of Chapter 5, §6 (with $$x=1, p=0,$$ and $$Q=\emptyset);$$ hence

\begin{aligned} h(1)-h(0) &=\sum_{i=1}^{m} \frac{h^{(i)}(0)}{i !}+R_{m}, \\\left|R_{m}\right| & \leq \frac{K_{m}}{(m+1) !}, \quad K_{m} \in E^{1}, \\ K_{m} & \leq \sup _{t \in J}\left|h^{(m+1)}(t)\right|. \end{aligned}

By construction,

$h(t)=f(g(t))=f(\vec{p}+t \vec{u});$

so

$h(1)=f(\vec{p}+\vec{u})=f(\vec{x}) \text { and } h(0)=f(\vec{p}).$

Thus using (8) also, we see that (9) implies (6), indeed.$$\quad \square$$

Note 3. Formula (3') of Chapter 5, §6, combined with (8), also yields

\begin{aligned} R_{m} &=\frac{1}{m !} \int_{0}^{1} h^{(m+1)}(t) \cdot(1-t)^{m} d t \\ &=\frac{1}{m !} \int_{0}^{1} d^{m+1} f(\vec{p}+t \vec{u} ; \vec{u}) \cdot(1-t)^{m} d t. \end{aligned}

Corollary $$\PageIndex{1}$$ (the Lagrange form of $$R_{m})$$

If $$E=E^{1}$$ in Theorem 2, then

$R_{m}=\frac{1}{(m+1) !} d^{m+1} f(\vec{s} ; \vec{u})$

for some $$\vec{s} \in L(\vec{p}, \vec{x})$$.

Proof

Here the function $$h$$ defined in the proof of Theorem 2 is real; so Theorem 1' and formula (3') of Chapter 5, §6 apply. This yields (10). Explain!$$\quad\square$$

Corollary $$\PageIndex{2}$$

If $$f : E^{n}\left(C^{n}\right) \rightarrow E$$ is $$m$$ times differentiable at $$\vec{p}$$ and if $$\vec{u} \neq \overrightarrow{0}$$ $$\left(\vec{p}, \vec{u} \in E^{n}\left(C^{n}\right)\right),$$ then the derivative $$D_{\vec{u}}^{m} f(\vec{p})$$ exists and equals $$d^{m} f(\vec{p} ; \vec{u})$$.

This follows as in the proof of Theorem 2 (with $$t=0).$$ For by definition,

\begin{aligned} D_{\vec{u}} f(\vec{p}) &=\lim _{s \rightarrow 0} \frac{1}{s}[f(\vec{p}+s \vec{u})-f(\vec{p})] \\ &=\lim \frac{1}{s}[h(s)-h(0)] \\ &=h^{\prime}(0)=d f(\vec{p} ; \vec{u}) \end{aligned}

by (7). Induction yields

$D_{\vec{u}}^{m} f(\vec{p})=h^{(m)}(0)=d^{m}(\vec{p} ; \vec{u})$

by (8). (See Problem 3.

Example

(C) Continuing Example (B), fix

$\vec{p}=(1,0);$

thus replace $$(x, y)$$ by $$(1,0)$$ there. Instead, write $$(x, y)$$ for $$\vec{x}$$ in Theorem 2. Then

$\vec{u}=\vec{x}-\vec{p}=(x-1, y);$

so

$u_{1}=x-1=d x \text { and } u_{2}=y=d y,$

and we obtain

\begin{aligned} d f(\vec{p} ; \vec{u}) &=D_{1} f(1,0) \cdot(x-1)+D_{2} f(1,0) \cdot y \\ &=(\sin 0) \cdot(x-1)+(1 \cdot \cos 0) \cdot y \\ &=y; \\ d^{2} f(\vec{p} ; \vec{u}) &=D_{11} f(1,0) \cdot(x-1)^{2}+2 D_{12} f(1,0) \cdot(x-1) y \\ &+D_{22} f(1,0) \cdot y^{2} \\ &=(0) \cdot(x-1)^{2}+2(\cos 0) \cdot(x-1) y-(1 \cdot \sin 0) \cdot y^{2} \\ &=2(x-1) y; \end{aligned}

and for all $$\vec{s}=\left(s_{1}, s_{2}\right) \in I$$,

\begin{aligned} d^{3} f(\vec{s} ; \vec{u})=& D_{111} f\left(s_{1}, s_{2}\right) \cdot(x-1)^{3}+3 D_{112} f\left(s_{1}, s_{2}\right) \cdot(x-1)^{2} y \\ &+3 D_{122} f\left(s_{1}, s_{2}\right) \cdot(x-1) y^{2}+D_{222} f\left(s_{1}, s_{2}\right) \cdot y^{3} \\=&-3 \sin s_{2} \cdot(x-1) y^{2}-s_{1} \cos s_{2} \cdot y^{3}. \end{aligned}

Hence by (6) and Corollary 1 (with $$m=2),$$ noting that $$f(\vec{p})=f(1,0)=0,$$ we get

\begin{aligned} f(x, y) &=x \cdot \sin y \\ &=y+(x-1) y+R_{2}, \end{aligned}

where for some $$\vec{s} \in I$$,

$R_{2}=\frac{1}{3 !} d^{3} f(\vec{s} ; \vec{u})=\frac{1}{6}\left[-3 \sin s_{2} \cdot(x-1) y^{2}-s_{1} \cos s_{2} \cdot y^{3}\right].$

As $$\vec{s} \in L(\vec{p}, \vec{x}),$$ where $$\vec{p}=(1,0)$$ and $$\vec{x}=(x, y), s_{1}$$ is between 1 and $$x;$$ so

$\left|s_{1}\right| \leq \max (|x|, 1) \leq|x|+1.$

Finally, since $$\left|\sin s_{2}\right| \leq 1$$ and $$\left|\cos s_{2}\right| \leq 1,$$ we obtain

$\left|R_{2}\right| \leq \frac{1}{6}[3|x-1|+(|x|+1)|y|] y^{2}.$

This bounds the maximum error that arises if we use (11) to express $$x \sin y$$ as a second-degree polynomial in $$(x-1)$$ and $$y.$$ (See also Problem 4 and Note 4 below.)

Note 4. Formula (6), briefly

$\Delta f=\sum_{i=1}^{m} \frac{d^{i} f}{i !}+R_{2},$

generalizes formula (2) in Chapter 5, §6.

As in Chapter 5, §6, we set

$P_{m}(\vec{x})=f(\vec{p})+\sum_{i=1}^{m} \frac{1}{i !} d^{i} f(\vec{p} ; \vec{x}-\vec{p})$

and call $$P_{m}$$ the $$m$$ th Taylor polynomial for $$f$$ about $$\vec{p},$$ treating it as a function of $$n$$ variables $$x_{k},$$ with $$\vec{x}=\left(x_{1}, \ldots, x_{n}\right)$$.

When expanded as in Example (C), formula (6) expresses $$f(\vec{x})$$ in powers of

$u_{k}=x_{k}-p_{k}, \quad k=1, \ldots, n,$

plus the remainder term $$R_{m}$$.

If $$f \in C D^{\infty}$$ on some $$G_{\vec{p}}$$ and if $$R_{m} \rightarrow 0$$ as $$m \rightarrow \infty,$$ we can express $$f(\vec{x})$$ as a convergent power series

$f(\vec{x})=\lim _{m \rightarrow \infty} P_{m}(\vec{x})=f(\vec{p})+\sum_{i=1}^{\infty} \frac{1}{i !} d^{i} f(\vec{p} ; \vec{x}-\vec{p}).$

We then say that $$f$$ admits a Taylor series about $$\vec{p},$$ on $$G_{\vec{p}}$$.