第四讲 高阶导数与 Taylor 公式
未央软-11 鲁睿
1.高阶微分
称 \(E \subseteq \mathbb{R}^{m}\) 是一个开集,如果任何 \(x \in E\) 都是 \(E\) 的内点。
设 \(f: E \rightarrow \mathbb{R}^{n}\) 在任何 \(x \in E\) 处都可微。则
\[
\mathrm{D}^{1} f: x \mapsto \operatorname{D} f(x), \quad E \rightarrow \mathcal{L}\left(\mathbb{R}^{m} ; \mathbb{R}^{n}\right)
\]
定义了 \(E\) 上一个映射, 其中 \(\mathcal{L}\left(\mathbb{R}^{m} ; \mathbb{R}^{n}\right)\) 是由从 \(\mathbb{R}^{m}\) 到 \(\mathbb{R}^{n}\) 上所有线性映射组成的一个 \(m n\) 维向量空间。
如果 \(\mathrm{D}^{1} f\) 连续, 则称 \(f\) 是 \(\mathscr{C}^{1}\) 的。
如果 \(\mathrm{D}^{1} f\) 可微, 则称 \(f\) 是二阶可微的, 称 \(\mathrm{DD}^{1} f(\mathbf{x})\) 为 \(f\) 的二阶导数, 记
\[
\mathrm{D}^{2}(\mathbf{x})=\mathrm{DD}^{1} f(\mathbf{x}) .
\]
如果 \(\mathrm{D}^{2} f: E \rightarrow \mathcal{L}\left(\mathbb{R}^{m} ; \mathcal{L}\left(\mathbb{R}^{m} ; \mathbb{R}^{n}\right)\right)\) 连续, 则称 \(f\) 是 \(\mathscr{C}^{2}\) 的。
归纳地, 可以定义 \(f\) 的 \(k\) 阶导数
\[
\mathrm{D}^{k} f(\mathbf{x})=\mathrm{DD}^{k-1} f(\mathbf{x}) \in \mathcal{L}\left(\mathbb{R}^{m} ; \ldots ; \mathcal{L}\left(\mathbb{R}^{m} ; \mathbb{R}^{n}\right)\right.
\]
如果 \(\mathrm{D}^{k} f\) 是连续的, 则称 \(f\) 是 \(\mathscr{C}^{k}\) 的。
如果对任意 \(k\) , \(f\) 是 \(\mathscr{C}^{k}\) 的, 则称 \(f\) 是 \(\mathscr{C}^{\infty}\) 的。并有
\[
\mathrm{D}^{k} f(\mathbf{x})\left(\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}\right)=\mathrm{D}\left(\mathrm{D}^{k-1} f(\mathbf{x})\left(\mathbf{v}_{1}, \ldots, \mathbf{v}_{k-1}\right)\right)\left(\mathbf{v}_{k}\right)
\]
例:矩阵求逆 $\mathrm{inv} $ 函数的高阶导数
上节课我们知道求逆映射是可微的,且 \(\operatorname{D \ inv}\left(A_{0}\right)(B)=-A_{0}^{-1} B A_{0}^{-1}\) 。有双重变化
\[
\begin{aligned}
& \operatorname{D inv}\left(A+B_{2}\right)\left(B_{1}\right) \\
=&-\left(A+B_{2}\right)^{-1} B_{1}\left(A+B_{2}\right)^{-1} \\
=&-\left[A^{-1}-A^{-1} B_{2} A^{-1}+o\left(B_{2}\right)\right] B_{1}\left[A^{-1}-A^{-1} B_{2} A^{-1}+o\left(B_{2}\right)\right] \\
=& \operatorname{D inv}(A)\left(B_{1}\right)+A^{-1} B_{2} A^{-1} B_{1} A^{-1}+A^{-1} B_{1} A^{-1} B_{2} A^{-1}+o\left(B_{2}\right)
\end{aligned}
\]
当 \(B_2\to0\) 。所以 \(\mathrm{D}^{2} \operatorname{inv}(A)\left(B_{1}, B_{2}\right)=A^{-1} B_{2} A^{-1} B_{1} A^{-1}+A^{-1} B_{1} A^{-1} B_{2} A^{-1}\) 。 归纳可知,\(\mathrm{inv}\) 是任意阶可微的,有:
\[
\begin{aligned}
&\mathrm{D}^{k} \operatorname{inv}(A)\left(B_{1}, B_{2}, \ldots, B_{k}\right) \\
&=(-1)^{k} \sum_{i_{1}, i_{2}, \ldots, i_{k} \text { 是 } 1,2, \ldots, k \text { 的任意排列 }} A^{-1} B_{i_{1}} A^{-1} B_{i_{2}} \cdots A^{-1} B_{i_{k}} A^{-1}
\end{aligned}
\]
定理:若 \(g \circ f\) 中 \(f, g\) 都是 \(\mathscr{C}^{r}\) 的, 则 \(g \circ f\) 也是 \(\mathscr{C}^{r}\) 的。
证明:数学归纳法和链锁法则。
定理:(逆映射的高阶连续性)若 \(\mathscr{C}^{r}\) 映射 \(g\) 有可微的逆映射 \(g^{-1}\), 则 \(g^{-1}\) 也是 \(\mathscr{C}^{r}\) 的。
证明:由链索法则 \(I=\mathrm{D}\left(g \circ g^{-1}\right)(\mathbf{y})=\mathrm{D} g\left(g^{-1}(\mathbf{y})\right) \circ \mathrm{Dg}^{-1}(\mathbf{y})\) 。 所以
\[
\operatorname{D} g^{-1}(\mathbf{y})=\left(\operatorname{D}g\left(g^{-1}(\mathbf{y})\right)\right)^{-1}=\operatorname{inv}\left(\operatorname{D}g\left(g^{-1}(\mathbf{y})\right)\right),
\]
高阶可微性由 \(\mathrm{inv}\) 的高阶可微性和数学归纳法得到。
2.高阶偏导数
以二阶微分为例,
\[
\begin{aligned}
\mathrm{d}^{2} f\left(x^{1}, \ldots, x^{m}\right) &=\sum_{j=1}^{m} \partial_{j}\left(\sum_{i=1}^{m} \partial_{i} f\left(x^{1}, \ldots, x^{m}\right) \mathrm{d} x^{i}\right) \mathrm{d} x^{j} \\
&=\sum_{1 \leq i, j \leq m} \partial_{j} \partial_{i} f\left(x^{1}, \ldots, x^{m}\right) \mathrm{d} x^{i} \otimes \mathrm{d} x^{j} \\
&=\sum_{1 \leq i, j \leq m} \partial_{j, i}^{2} f\left(x^{1}, \ldots, x^{m}\right) \mathrm{d} x^{i} \otimes \mathrm{d} x^{j}
\end{aligned}
\]
上式中 \(\mathrm{d} x^{i} \otimes \mathrm{d} x^{j}\) 是一个双线性型符号,满足
\[
\mathrm{d} x^{i} \otimes \mathrm{d} x^{j}(\mathbf{v}, \mathbf{w})=\mathrm{d} x^{i}(\mathbf{v}) \mathrm{d} x^{j}(\mathbf{w})
\]
即
\[
\mathrm{d} x^{i} \otimes \mathrm{d} x^{j}\left(\sum_{p=1}^{m} \xi^{p} \mathbf{e}_{p}, \sum_{q=1}^{m} \eta^{q} \mathbf{e}_{q}\right)=\xi^{i} \eta^{j}
\]
一般地,
\[
\mathrm{d}^{k} f(\mathbf{x})=\sum_{1 \leq i_{1}, \ldots, i_{k} \leq m} \partial_{i_{k}, \ldots, i_{1}}^{k} f(\mathbf{x}) \mathrm{d} x^{i_{1}} \otimes \cdots \otimes \mathrm{d} x^{i_{k}}
\]
于是, \(f\) 是 \(\mathscr{C}^{k}\) 的当且仅当 \(f\) 的所有 \(k\) 阶偏导数都是连续函数。
一般而言, 高阶偏导数的求导顺序不能随意交换,但是当 \(f\) 是 \(\mathscr{C}^{k}\) 函数时,其直到 \(k\) 阶的高阶偏导数的值与求导顺序无关。
符号与写法:
\[
\begin{array}{ccc}
\hline \text { 符号 } & \text { 传统符号 } & \text { 含义 } \\
\hline \partial_{i} f & \frac{\partial f}{\partial x^{i}} & \text { 一阶偏导数 } \\
\partial_{\mathbf{v}} f=\partial_{\left(v^{1}, \ldots, v^{m}\right)} f & \frac{\partial f}{\partial \mathbf{v}} & \text { 沿向量 } \mathbf{v} \text { 的导数 } \\
\partial_{i_{k}, \ldots, i_{2}, i_{1}}^{k} f & \frac{\partial^{k} f}{\partial x^{i_{k}} \cdots \partial x^{i_{2}} \partial x^{i_{1}}} & k \text { 阶偏导数 } \\
\partial_{\left[\alpha_{1}, \ldots, \alpha_{m}\right]}^{|\alpha|} f & \frac{\partial^{\alpha_{1}+\cdots+\alpha_{m}} f}{\partial\left(x^{1}\right)^{\alpha_{1}} \cdots \partial\left(x^{m}\right)^{\alpha_{m}}} & |\alpha| \text { 阶偏导数 }
\\ \hline
\end{array}
\]
\(\alpha=\left[\alpha_{1}, \ldots, \alpha_{m}\right]\) 是一组非负整数, $|\alpha|=\alpha_{1}+\cdots+\alpha_{m} $ 。
当同时出现不同坐标系时, 用 \(\partial_{x^{i}} f\) 代替 \(\partial_{i} f ;\) 用 \(\partial_{x^{i_{k}}, \ldots, x^{i_{2}, x^{i_{1}}} f}^{k} f\) 代替 \(\partial_{i_{k}, \ldots, i_{2}, i_{1}}^{k} f\); 用 \(\partial_{\left(x^{1}\right)^{\alpha_{1}}, \ldots,\left(x^{m}\right)^{\alpha_{m}}}^{|\alpha|} f\) 代替 \(\partial_{\left[\alpha_{1}, \ldots, \alpha_{m}\right]}^{|\alpha|} f\) 。
应用:Laplace 方程
Laplace 方程 \(\Delta f=0\) 。Laplace 算子 \(\Delta=\nabla \cdot \nabla, \nabla\) 是梯度算子。求 Laplace 方程在平面直角坐标系 \((x, y)\) 和极坐标系 \((r, \theta)\) 下的形式。
直角坐标系下,\(\nabla=\mathbf{e}_{1} \partial_{x}+\mathbf{e}_{2} \partial_{y}\),因 \(\mathbf{e}_{i} \cdot \mathbf{e}_{j}=\left\{\begin{array}{ll}1, & i=j \\ 0, & i \neq j\end{array} \quad \partial_{x} \mathbf{e}_{i}=\partial_{y} \mathbf{e}_{i}=0\right.\) ,有
\[
\begin{aligned}
\nabla^{2} &=\left(\mathbf{e}_{1} \partial_{x}+\mathbf{e}_{2} \partial_{y}\right) \cdot\left(\mathbf{e}_{1} \partial_{x}+\mathbf{e}_{2} \partial_{y}\right) \\
&=\mathbf{e}_{1} \cdot \mathbf{e}_{1} \partial_{x, x}^{2}+\mathbf{e}_{2} \cdot \mathbf{e}_{2} \partial_{y, y}^{2}=\partial_{x, x}^{2}+\partial_{y, y}^{2}
\end{aligned}
\]
平面极坐标下,有 \(\nabla=\mathbf{e}_{r} \partial_{r}+\dfrac{\mathbf{e}_{\theta}}{r^{2}} \partial_{\theta}\) ,且基向量不再恒定不变:
\[
\partial_{r} \mathbf{e}_{r}=0, \quad \partial_{\theta} \mathbf{e}_{r}=\partial_{r} \mathbf{e}_{\theta}=\frac{\mathbf{e}_{\theta}}{r}, \quad \partial_{\theta} \mathbf{e}_{\theta}=-r \mathbf{e}_{r}
\]
因此
\[
\begin{aligned}
\nabla^{2} f &=\left(\mathbf{e}_{r} \partial_{r}+\frac{\mathbf{e}_{\theta}}{r^{2}} \partial_{\theta}\right) \cdot\left(\mathbf{e}_{r} \partial_{r}+\frac{\mathbf{e}_{\theta}}{r^{2}} \partial_{\theta}\right) f \\
&=\partial_{r, r}^{2} f \mathbf{e}_{r} \cdot \mathbf{e}_{r}+\frac{1}{r^{2}} \partial_{r} f \mathbf{e}_{\theta} \cdot \partial_{\theta} \mathbf{e}_{r}+0+\frac{1}{r^{4}} \partial_{\theta, \theta}^{2} f \mathbf{e}_{\theta} \cdot \mathbf{e}_{\theta} \\
&=\partial_{r, r}^{2} f+\frac{1}{r} \partial_{r} f+\frac{1}{r^{2}} \partial_{\theta, \theta}^{2} f
\end{aligned}
\]
课后练习
问题:(分离变量法)求 Laplace 方程形如 \(u(r, \theta)=U(r) V(\theta)\) 形式的解。
代入公式有 \(\Delta u=V(\theta)(\partial^2_{r,r}U(r)+\dfrac{1}{r}\partial_r{U(r)})+\dfrac{U(r)}{r^2}\partial_{\theta,\theta}{V(\theta)}=0\) ,或
\[
\frac{1}{U(r)}(r^2\partial^2_{r,r}U(r)+r\partial_r{U(r)})+\frac{1}{V(\theta)}\partial_{\theta,\theta}{V(\theta)}=0
\]
由于前两项与后一项没有关联,可以设 \(r^2\partial^2_{r,r}U(r)+r\partial_r{U(r)}=CU(r)\)(这是一个欧拉方程),\(\partial_{\theta,\theta}{V(\theta)}=-CV(\theta)\) ,解得
\[
\begin{gathered}
\begin{gathered}
U(r)=k_{1} \cosh (\sqrt{C} \ln r)+ k_{2} \sinh (\sqrt{C} \ln r),\\V(\theta)=k_3\cos(\sqrt{C}\theta)+k_4\sin(\sqrt{C}\theta);
\end{gathered}\quad \text{when \ }C>0\\
\begin{gathered}
U(r)=k_{1} \cos (\sqrt{C} \ln r)+ k_{2} \sin (\sqrt{C} \ln r),\\V(\theta)=k_3\cosh(\sqrt{C}\theta)+k_4\sinh(\sqrt{C}\theta);
\end{gathered}\quad \text{when \ }C<0\\
U(r)=k_1\ln r+k_2 \ , \ V(\theta)=k_3\theta \ ;\quad \text{when \ }C=0
\end{gathered}
\]
因此,
\[
u(r,\theta)=\left\{\begin{array}{l}(C_1r^C+C_2r^{-C})\cos(C\theta+C_3)\\\cos(C\ln r+C_1)(C_2\mathrm{e}^{C\theta}+C_3\mathrm{e}^{-C\theta})\\
C_1(\ln r+C_2)\theta
\end{array}\right.
\]
问题:试写出在三维柱坐标、球坐标以及任意维数空间中直角坐标系和一般坐标系下Laplace 算子的坐标形式。
由上节课的结论可知,球坐标下的梯度算子:
\[
\nabla =\mathbf{e}_{r}\partial_{r} +\frac{\mathbf{e}_{\theta}}{r^{2}} \partial_{\theta} +\frac{\mathbf{e}_\varphi}{r^{2} \sin ^{2} \varphi} \partial_\varphi
\]
且有
\[
\begin{gathered}
\partial_r\mathbf{e}_r=0,\partial_r\mathbf{e}_\theta=\frac{\mathbf{e}_\theta}{r},\partial_r\mathbf{e}_\varphi=\frac{\mathbf{e}_\varphi}{r}\\ \partial_\theta\mathbf{e}_r=\frac{\mathbf{e}_\theta}{r},\partial_\theta\mathbf{e}_\theta=-r\mathbf{e}_r,\partial_\theta\mathbf{e}_\varphi=\cot\theta\mathbf{e}_\varphi\\ \partial_\varphi\mathbf{e}_r=\frac{\mathbf{e}_\varphi}{r},\partial_\varphi\mathbf{e}_\theta=\cot\theta\mathbf{e}_\varphi,\partial_\varphi\mathbf{e}_\varphi=\begin{bmatrix}-r\sin\theta\cos\varphi\\-r\sin\theta\sin\varphi\\
0\end{bmatrix}
\end{gathered}
\]
故对球坐标系
\[
\begin{aligned}
\Delta&=\nabla\cdot\nabla\\
&=(\mathbf{e}_{r}\partial_{r} +\frac{\mathbf{e}_{\theta}}{r^{2}} \partial_{\theta} +\frac{\mathbf{e}_\varphi}{r^{2} \sin ^{2} \varphi} \partial_\varphi)\cdot(\mathbf{e}_{r}\partial_{r} +\frac{\mathbf{e}_{\theta}}{r^{2}} \partial_{\theta} +\frac{\mathbf{e}_\varphi}{r^{2} \sin ^{2} \varphi} \partial_\varphi)\\
&=\partial_{r,r}+0+0+\frac{1}{r}\partial_r+\frac{1}{r^2}\partial_{\theta,\theta}+0+\frac{1}{r}\partial_r+\frac{\cot\theta}{r^2}\partial_\theta+\frac{1}{r^2\sin^2\varphi}\partial_{\varphi,\varphi}\\
&=\partial_{r,r}+\frac{2}{r}\partial_r+\frac{1}{r^2}\partial_{\theta,\theta}+\frac{\cot\theta}{r^2}\partial_\theta+\frac{1}{r^2\sin^2\varphi}\partial_{\varphi,\varphi}
\end{aligned}
\]
3. Taylor 展开
设 \(f: E \rightarrow \mathbb{R}\) 是一个 \(\mathscr{C}^{k}\) 函数。则 \(g(t)=f\left(\mathbf{x}_{0}+t \mathbf{v}\right)\) 也是 \(\mathscr{C}^{k}\) 函数, 从而
\[
g(t)=g(0)+g^{\prime}(0) t+\cdots+\frac{g^{(k)}(0)}{k !} t^{k}+o\left(t^{k}\right), \quad t \rightarrow 0
\]
由链索法则和数学归纳法可得
\[
\begin{aligned}
g^{(k)}(t) &=d^{k} f\left(\mathbf{x}_{0}+t \mathbf{v}\right)(\underbrace{\mathbf{v}, \ldots, \mathbf{v})}_{k \text { 个 }}\\
&=\sum_{1 \leq i_{1}, \ldots, i_{k} \leq m} \partial_{i_{k}, \ldots, i_{1}}^{k} f\left(\mathbf{x}_{0}+t \mathbf{v}\right) v^{i_{1}} \cdots v^{i_{k}}
\end{aligned}
\]
所以
\[
f\left(\mathbf{x}_{0}+t \mathbf{v}\right)=f\left(\mathbf{x}_{0}\right)+\sum_{r=1}^{k} \frac{t^{r}}{r !} \sum_{1 \leq i_{1}, \ldots, i_{r} \leq m} \partial_{i_{r}, \ldots, i_{1}}^{r} f\left(\mathbf{x}_{0}\right) v^{i_{1}} \cdots v^{i_{r}}+o\left(t^{k}\right)
\]
对 \(\mathrm{x}\), 取 \(v=\frac{\mathrm{x}-\mathrm{x}_{0}}{\left\|\mathrm{x}-\mathrm{x}_{0}\right\|}, t=\left\|\mathrm{x}-\mathrm{x}_{0}\right\|\), 则
\[
\begin{aligned}
f(\mathbf{x})=& f\left(\mathbf{x}_{0}\right) \\
&+\sum_{r=1}^{k} \frac{1}{r !} \sum_{1 \leq i_{1}, \ldots, i_{r} \leq m} \partial_{i_{r}, \ldots, i_{1}}^{r} f\left(\mathbf{x}_{0}\right)\left(x^{i_{1}}-x_{0}^{i_{1}}\right) \cdots\left(x^{i_{r}}-x_{0}^{i_{r}}\right) \\
&+o\left(\left\|\mathbf{x}-\mathbf{x}_{0}\right\|^{k}\right), \quad \mathbf{x} \rightarrow \mathbf{x}_{0}
\end{aligned}
\]
对拉格朗日余项,使用 \(\xi\in(0,1)\) 为余项
\[
\begin{aligned}
f(\mathbf{x})=& f\left(\mathbf{x}_{0}\right)+\sum_{r=1}^{k-1} \frac{1}{r !} \sum_{1 \leq i_{1}, \ldots, i_{r} \leq m} \partial_{i_{r}, \ldots, i_{1}}^{r} f\left(\mathbf{x}_{0}\right) v^{i_{1}} \cdots v^{i_{r}} \\
&+\frac{1}{k !} \sum_{1 \leq i_{1}, \ldots, i_{k} \leq m} \partial_{i_{k}, \ldots, i_{1}}^{k} f\left(\mathbf{x}_{0}+\xi\left(\mathbf{x}-\mathbf{x}_{0}\right)\right) v^{i_{1}} \cdots \cdot v^{i_{k}}
\end{aligned}
\]
使用组合数对求导次数相同的合并在一起
\[
\begin{aligned}
& \frac{1}{k !} \sum_{1 \leq i_{1}, \ldots, i_{k} \leq m} \partial_{i_{k}, \ldots, i_{1}}^{k} f\left(\mathbf{x}_{0}\right) v^{i_{1}} \cdots v^{i_{k}} \\
=& \sum_{\alpha_{1}+\cdots+\alpha_{m}=k} \frac{1}{\alpha_{1} ! \alpha_{2} ! \cdots \alpha_{m} !} \partial_{\left[\alpha_{1}, \ldots, \alpha_{m}\right]}^{k} f\left(\mathbf{x}_{0}\right)\left(v^{1}\right)^{\alpha_{1}} \cdots\left(v^{m}\right)^{\alpha_{m}}
\end{aligned}
\]
对应算符计算方法泰勒展开在 \(k\) 阶处有
\[
\partial ^{k}f=\dfrac{1}{k!}(\displaystyle \sum_{m} \dfrac{\partial }{\partial \mathbf{x^{i}}})^{k}f=\sum_{\sum\alpha_{i}=k}\dfrac{1}{\alpha_{1}!\alpha_2!\cdots\alpha_{m}!}\dfrac{\partial ^{\alpha_1}f}{\partial \mathbf{x^{1}}^{\alpha_1}}\dfrac{\partial ^{\alpha_2}f}{\partial \mathbf{x^{2}}^{\alpha_2}}\cdots\dfrac{\partial ^{\alpha_m}f}{\partial \mathbf{x^{m}}^{\alpha_m}}
\]
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