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 \(\vec{u}\).
(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.
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.
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).
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.)
(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 \(E^{\prime}=E^{n}\left(C^{n}\right).\)
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
\[\begin{aligned} R_{m} &=\frac{1}{m !} \int_{0}^{1} h^{(m+1)}(t) \cdot(1-t)^{m} d t \\ &=\frac{1}{m !} \int_{0}^{1} d^{m+1} f(\vec{p}+t \vec{u} ; \vec{u}) \cdot(1-t)^{m} d t. \end{aligned}\]
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!\(\quad\square\)
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.
(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 \(G_{\vec{p}}\).