第九周。
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14多变量函数的微分学.tex
259
14多变量函数的微分学.tex
@@ -288,4 +288,261 @@ J_y \bvec{F} = \begin{bmatrix}
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\item 对$\bvec{x} \in B_\delta (\bvec{x}_0)$,$\bvec{y} = \bvec{f}(\bvec{x})$,有
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\[J\bvec{f}(\bvec{x}) = -(J_y \bvec{F}(\bvec{x}, \bvec{y}))^{-1} J_x \bvec{F}(\bvec{x}, \bvec{y})\eqper\]
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\end{enumerate}
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\end{theorem}
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\end{theorem}
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\section{逆映射定理}
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给定$\boldf: D \to \ndreal$,$D \subset \ndreal$。考察\boldf 的反函数及其性质:$\bvec{y} = \boldf^{-1}(\bvec{x})$。
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考虑应用隐函数定理。$\bvec{y} = \boldf^{-1}(\bvec{x})$意味着$\bvec{x} = \boldf(\bvec{x})$。因此定义$\bvec{F}:\tilde{D} \to \ndreal$,满足
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\[\bvec{F}(\bvec{x}, \bvec{y}) = \bvec{x} - \boldf(\bvec{y}), (\bvec{x}, \bvec{y}) \in \ndreal \times D = \tilde{D} \subset \realnum^{2n}\]
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再取$\bvec{y}_0 \in D\interior$,$\bvec{x}_0 = \boldf(\bvec{y}_0)$,那么$(\bvec{x}_0, \bvec{y}_0) \in \tilde{D}\interior$,$\bvec{F}(\bvec{x}_0, \bvec{y}_0) = \bvec{0}$。
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如果我们假设$\bvec{f} \in C^1$,那么
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\[J_{\bvec{y}} \bvec{F}(\bvec{x}, \bvec{y}) = -J \bvec{f}(\bvec{y})\]
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且
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\[J_{\bvec{x}} \bvec{F}(\bvec{x}, \bvec{y}) = J\bvec{x} = \bvec{I}_n \eqper\]
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\begin{theorem}[逆映射定理(局部)]
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设$\bvec{f} \in C^1(D, \ndreal), D \subset \ndreal, \bvec{y}_0 \in D\interior$满足$\det(J\bvec{f}(\bvec{y}_0)) \neq 0$,那么存在$\delta, \eta > 0$以及函数$\bvec{g}: B_\delta (\bvec{x}_0) \to B_\eta(\bvec{y}_0)$,其中$\bvec{x}_0 = \bvec{f}(\bvec{y}_0)$满足以下性质:
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\begin{enumerate}
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\item 对任意的$\bvec{x}$满足$\norm{\bvec{x} - \bvec{x}_0} < \delta$,$\bvec{g}(\bvec{x}_0) = \bvec{y}_0$,$\bvec{f}(\bvec{g}(\bvec{x})) = \bvec{x}$;
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\item $\bvec{g} \in C^1(B_\delta(\bvec{x}_0), \ndreal)$;
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\item $J\bvec{g}(\bvec{x}) = [J\bvec{f}(\bvec{y})]^{-1}$,其中$\bvec{y} = \bvec{g}(\bvec{x})$。
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\end{enumerate}
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\end{theorem}
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\begin{theorem}[逆映射定理]
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设$\bvec{f} \in C^1(D, \ndreal), D \subset\ndreal$为开集,且$\bvec{f}:D \to \ndreal$为单射,对任意的$\bvec{y}$,$\det(J\bvec{f}(\bvec{y})) \neq 0$。则记$\Omega = \bvec{f}(D)$,存在$\bvec{f}$的反函数$\bvec{f}^{-1} \in C^1 (\Omega, \ndreal)$且对任意的$\bvec{x} \in \Omega$,
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\[J\bvec{f}^{-1}(\bvec{x}) = (J\bvec{f}(\bvec{y}))^{-1}, \bvec{y} = \bvec{f}^{-1}(\bvec{x})\eqper\]
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\end{theorem}
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\section{高阶偏导数}
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\begin{definition}
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设在开集$D$上的每一点,函数$f$存在偏导数
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\[D_i f(\bvec{x}) = \frac{\partial f}{\partial x_i}(\bvec{x}), i = 1, 2, \dots, n\]
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称他们为$f$的一阶偏导数,如果对这些偏导函数又可取偏导数,得出的就是$f$的二阶偏导函数,仿照此可以定义三阶偏导函数乃至更高阶的偏导数。我们将以一阶偏导数$\dfrac{\partial f}{\partial x_j}$再对$x_i$求偏导数时,把$\dfrac{\partial}{\partial x_i}\left(\dfrac{\partial f}{\partial x_j}\right)$记作$\dfrac{\partial^2 f}{\partial x_i \partial x_j}$,如果$i = j$,那么把$\dfrac{\partial^2 f}{\partial x_i \partial x_i}$记作$\dfrac{\partial^2 f}{\partial x_i^2}$。
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\end{definition}
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\begin{theorem}[Clairaut定理]
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设$f:D \to \realnum$,$D \subset \realnum^2$是开集,$P = (x_0, y_0) \in D$。若$\dfrac{\partial^2 f}{\partial x \partial y}$和$\dfrac{\partial^2 f}{\partial y \partial x}$在$D$内存在且在$P$点连续,那么二者在该点相等。
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\end{theorem}
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\begin{corollary}
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设$f: D \to \realnum$,$D \subset \ndreal$是开集。若$f$在$D$内所有$k$阶偏导数都存在且连续,则$k$阶偏导数的值与关于自变量的求导次序无关。
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\end{corollary}
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\section{拟微分平均值定理}
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这个下面这个定理主要阐述的是多元数值函数的中值定理。
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\begin{theorem}
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设定义在凸区域$D \subset \ndreal$上的函数$f$可微,则对任何两点$\bvec{a}, \bvec{b} \in D$,在由$\bvec{a}$与$\bvec{b}$确定的直线段上有一点$\bvec{\xi}$使得
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\[f(\bvec{b}) - f(\bvec{a}) = Jf(\bvec{\xi})(\bvec{b} - \bvec{a})\eqper\]
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\end{theorem}
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对于向量值函数,中值定理不一定成立,而有下面的拟微分平均值定理:
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\begin{theorem}[拟微分平均值定理]
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设凸区域$D \subset \ndreal$且$\boldf: D \to \realnum^m$,映射\boldf 在$D$上可微,则对于任何$\bvec{a}, \bvec{b} \in D$,在由$\bvec{a}, \bvec{b}$确定的线段上必有一点$\bvec{\xi}$使得
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\[\norm{\boldf(\bvec{b}) - \boldf(\bvec{a})} \leq \norm{J\boldf(\bvec{\xi})} \norm{\bvec{b} - \bvec{a}}\eqper\]
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\end{theorem}
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\begin{corollary}
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设区域$D \subset \ndreal$,$\bvec{f}:D \to \realnum^m$,如果$J\bvec{f} = \bvec{0}$在$D$上成立,则\boldf 在$D$上为一常向量。
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\end{corollary}
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\section{Taylor公式}
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考虑用多项式近似一个多元函数,即对在$\bvec{a}$点有$m + 1$阶连续偏导数的$n$元函数$f(x)$,是否有$m$次多项式$P_m(\bvec{x})$,使得
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\[f(\bvec{a} + \Delta \bvec{x}) = P_m (\Delta \bvec{x}) + o(\norm{\Delta \bvec{x}}^m)\]
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成立?
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我们设$f \in C^{m + 1}(B_r(\bvec{a})), \bvec{a} \in \ndreal, r > 0$,那么对任意满足$\norm{\Delta \bvec{x}} < r$的$\Delta \bvec{x}$,定义
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\[\varphi(t) = f(\bvec{a} + t\Delta \bvec{x}) \in C^{m + 1}[0, 1]\]
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应用一元函数的Taylor公式:对任意的$t \in [0, 1]$,都存在$\theta \in (0, 1)$满足
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\[\varphi(t) = \sum_{k = 1}^m \frac{\varphi^{(k)}(0)}{k!}t^k + \frac{\varphi^{(m + 1)}(\theta t)}{(m + 1)!} t^{m + 1}\]
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特别取$t = 1$得到
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\[\varphi(1) = \sum_{k = 1}^m \frac{\varphi^{(k)}(0)}{k!} + \frac{\varphi^{(m + 1)}(\theta)}{(m + 1)!}\]
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将引入的一元函数的表达式带入(特别注意求导项的带入):
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计算
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\begin{align*}
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& \deriv{\varphi} (t) = \sum_{i = 1}^n \frac{\partial f(\bvec{a} + t\Delta \bvec{x})}{\partial x_i} \Delta x_i, \deriv{\varphi}(0) = \sum_{i = 1}^n \frac{\partial f(\bvec{a})}{\partial x_i}\Delta x_i\\
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& \varphi^{\prime \prime} (t) = \sum_{i, j = 1}^n \frac{\partial^2 f(\bvec{a} + t\Delta \bvec{x})}{\partial x_i \partial x_j} \Delta x_i \Delta x_j, \varphi^{\prime \prime} (0) = \sum_{i, j = 1}^n \frac{\partial ^2 f(\bvec{a})}{\partial x_i \partial x_j}\Delta x_i \Delta x_j\\
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& \dots \dots
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\end{align*}
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我们引入记号$\alpha = (\alpha_1, \dots, \alpha_n)$,其中每个$\alpha_i$都是非负整数,记
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\[\abs{\alpha} = \alpha_1 + \dots + \alpha_n, \alpha! = \alpha_1! \dots \alpha_n!\]
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如果$\bvec{x} = (x_1, \dots, x_n)$,那么记$x^\alpha = x_1^{\alpha_1} \dots x_n^{\alpha_n}$。
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对于多重指标$\alpha = (\alpha_1, \dots, \alpha_n)$,我们还引进记号
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\[D^\alpha f(\bvec{a}) = \frac{\partial^{\alpha_1 + \dots + \alpha_n}f}{\partial x_1^{\alpha_1} \dots \partial x_n^{\alpha_n}} (\bvec{a})\eqper\]
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于是我们可以叙述多元函数Taylor公式:
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\begin{theorem}[多元函数Taylor公式]\label{多元函数Taylor公式}
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设$D \subset \ndreal$是一个凸区域,$f \in C^{m + 1}(D)$。$\bvec{a} = (a_1, \dots, a_n)$和$\bvec{a} + \bvec{h} = (a_1 + h_1, \dots, a_n + h_n)$是$D$中两点,则必存在$\theta \in (0, 1)$使得
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\[f(\bvec{a} + \bvec{h}) = \sum_{k = 0}^m \sum_{\abs{\alpha} = k} \frac{D^\alpha f(\bvec{a})}{\alpha!} h^\alpha + R_m\]
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其中
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\[R_m = \sum_{\abs{\alpha} = m + 1} \frac{D^\alpha f(\bvec{a} + \theta \bvec{h})}{\alpha!} h^\alpha\]
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称为Lagrange余项。
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\end{theorem}
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这个定理中的和式的意思是,对每个$\alpha$,都求对$f$求$\alpha$次偏导的导函数在$\bvec{a}$处的值,并且对$x_i$求了$\alpha_i$次偏导就要在后面乘上$\dfrac{x_i^{\alpha_i}}{\alpha_i!}$。
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我们再引入一个高阶微分的记号。对$\bvec{h} = (h_1, h_2, \dots, h_3)$,我们定义
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\begin{align*}
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& \qquad\left(h_1 \frac{\partial}{\partial x_1} + \dots + h_n \frac{\partial}{\partial x_n}\right)^k f(\bvec{a})\\
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& = \sum_{\abs{\alpha} = k} \frac{k!}{\alpha!} \frac{\partial^{\alpha_1}}{\partial x_1^{\alpha_1}}\dots \frac{\partial^{\alpha_n}}{\partial x_n^{\alpha_n}}f(\bvec{a}) \bvec{h}^\alpha\\
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& = \sum_{\abs{\alpha} = k} \frac{k!}{\alpha!}D^\alpha f(\bvec{a}) h^\alpha
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\end{align*}
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特别考虑二元函数的情况。如果设$x, y$的改变量为$h, k$,那么
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\begin{align*}
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\left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y}\right) f(x, y) & = \frac{\partial f(x, y)}{\partial x}h + \frac{\partial f(x, y)}{\partial y}k\\
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\left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y}\right)^2 f(x, y) & = \frac{\partial^2 f(x, y)}{\partial x^2}h^2 + 2 \frac{\partial^2 f(x, y)}{\partial x \partial y}hk + \frac{\partial^2 f(x, y)}{\partial y^2}k^2\\
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\left(h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y}\right)^m f(x, y) & = \sum_{i = 0}^m \binom{m}{i} \frac{\partial^m f(x, y)}{\partial x^ \partial y^{m - i}} h^i k^{m - i}, m = 1, \dots, n + 1
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\end{align*}
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在一般的应用中,特别重要的是Taylor公式的前三项。把他们具体写出来:
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\[f(\bvec{a} + \bvec{h}) = f(\bvec{a}) + \sum_{i = 1}^n \frac{\partial f}{\partial x_i}(\bvec{a})h_i + \frac{1}{2} \sum_{i, j = 1}^n \frac{\partial^2 f}{\partial x_i \partial x_j} (\bvec{a}) h_i h_j + \cdots\]
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如果记
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\[Hf(\bvec{a}) = \begin{bmatrix}
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\dfrac{\partial^2 f}{\partial x_1^2}(\bvec{a}) & \cdots & \dfrac{\partial^2 f}{\partial x_1 \partial x_n}(\bvec{a})\\[1em]
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\vdots & \ddots & \vdots\\[1ex]
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\dfrac{\partial^2 f}{\partial x_n \partial x_1}(\bvec{a}) & \cdots & \dfrac{\partial^2 f}{\partial x_n^2}(\bvec{a})
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\end{bmatrix}\]
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进一步简记为
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\[Hf(\bvec{a}) = \begin{bmatrix}
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D_{11} f(\bvec{x}) & \cdots & D_{1n} f(\bvec{x})\\
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\vdots & \ddots & \vdots\\
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D_{n1} f(\bvec{x}) & \cdots & D_{nn} f(\bvec{x})
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\end{bmatrix}\eqper\]
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这$Hf$称为$f$的Hessian,它是一个$n$阶对称方阵。
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那么前面的Taylor公式可以写成
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\[f(\bvec{a} + \bvec{h}) = f(\bvec{a}) + Jf(\bvec{a})\bvec{h} + \frac{1}{2} \bvec{h}^{\mathrm{T}} Hf(\bvec{a}) \bvec{h} + \cdots\]
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\begin{theorem}
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在定理\ref{多元函数Taylor公式}的条件下,
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\[R_m = O(\norm{\bvec{h}}^{m + 1})\eqper\]
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于是我们可以把Taylor公式写成Peano余项的形式:
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\[f(\bvec{a} + \bvec{h}) = f(\bvec{a}) + Jf(\bvec{a}) \bvec{h} + \frac{1}{2} \bvec{h}^{\mathrm{T}} Hf(\bvec{a}) \bvec{h} + o(\norm{\bvec{h}}^2)\]
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\end{theorem}
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\section{极值}
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\begin{definition}
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设$D \subset \ndreal$,函数$f:D \to \realnum$,点$\bvec{p}_0 \in D\interior$,如果存在一个球$B_r(\bvec{p}_0) \subset D\interior$,使得$f(\bvec{p}) \geq f(\bvec{p}_0)$($f(\bvec{p}) > f(\bvec{p}_0)$)对一切$\bvec{p} \in B_r(\hat{\bvec{p}}_0)$成立,那么$\bvec{p}_0$称为$f$的一个(严格)极小值点,而$f(\bvec{p}_0)$称为函数$f$的一个(严格)极小值。
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同样地可以定义(严格)极大值点和(严格)极大值。极小值和极大值统称极值。
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\end{definition}
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类似于Fermat引理,我们可以得到极值点的必要条件
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\begin{theorem}
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设$n$元函数$f$在$\bvec{p}_0$取得极值,且$Jf(\bvec{p}_0)$存在,那么必须有$Jf(\bvec{p}_0) = \bvec{0}$。
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设$\bvec{u}$是任意方向向量,则$D_{\bvec{u}} f(\bvec{a}) = 0$。
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\end{theorem}
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\begin{definition}
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$D$中使得$Jf(\bvec{p}) = \bvec{0}$的一切内点称为函数函数$f$的驻点。极值点一定是煮点,而驻点未必是极值点。
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\end{definition}
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\begin{definition}
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设
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\(A = \begin{bmatrix}
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a_{ij}
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\end{bmatrix}\)
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是一个$n$阶对称方阵。设
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\[\bvec{x} = \begin{bmatrix}
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x_1\\ x_2\\ \vdots\\ x_n
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\end{bmatrix}\]
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称
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\[Q(\bvec{x}) = \bvec{x}^\mathrm{T} A \bvec{x} = \sum_{i, j = 1}^n a_{ij}x_i x_j\]
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为$x_1, x_2, \dots, x_n$的一个二次型,方阵$A$称为二次型$Q$的系数方阵。
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如果对任意$\bvec{x} \neq \bvec{0}$都有$Q(\bvec{x}) \geq 0$($\leq 0$),则称二次型$Q$是正(负)定的,其系数方阵$A$相应地称为正(负)定方阵。
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如果对任意$\bvec{x} \neq \bvec{0}$都有$Q(\bvec{x}) > 0$($< 0$),则称二次型$Q$是严格正(负)定的,其系数方阵$A$相应地称为严格正(负)定方阵。
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如果总存在$\bvec{p}, \bvec{q} \in \ndreal$,使得$Q(\bvec{p}) < 0 < Q(\bvec{q})$,旧称二次型$Q$是不定的,其系数方阵$A$相应地称为不定方阵。
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\end{definition}
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\begin{theorem}
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设
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\(A = \begin{bmatrix}
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a_{ij}
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\end{bmatrix}\)
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是一个$n$阶方阵。方阵$A$为严格正定的一个必要充分条件是它的各级顺序主子式均大于零。
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\end{theorem}
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欲证明一个方阵$A$负定,只需证明$-A$是正定的即可。
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\begin{theorem}
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设二阶对称方阵
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\[A = \begin{bmatrix}
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a_{11} & a_{12}\\
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a_{21} & a_{22}
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\end{bmatrix}\]
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$A$为严格正(负)定的一个必要充分条件是
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\[a_{11} > 0(a_{11} < 0),\]
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\[\begin{vmatrix}
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a_{11} & a_{12}\\
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a_{21} & a_{22}
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\end{vmatrix} > 0\]
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$A$为不定矩阵的一个充分必要条件是
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\[\begin{vmatrix}
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a_{11} & a_{12}\\
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a_{21} & a_{22}
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\end{vmatrix} < 0\eqper\]
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\end{theorem}
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\begin{theorem}
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设$\bvec{x}_0$是函数$f$的一个驻点,函数$f$在$\bvec{x}_0$的某一临域内有连续的二阶偏导数。
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\begin{enumerate}
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\item 如果Hessian $Hf(\bvec{x}_0)$是严格正定(负)方阵,那么$\bvec{x}_0$是$f$的一个严格极小(大)值点。
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\item 如果Hessian $Hf(\bvec{x}_0)$是不定方阵,那么$\bvec{x}_0$不是$f$的极值点。
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\end{enumerate}
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\end{theorem}
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\section{条件极值}
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首先我们引入一个问题。设$f: D \to \realnum$,$\Phi: D \to \realnum^m$,$D \subset \realnum^{n + m}$是开集,$(\bvec{x}, \bvec{y}) \in \realnum^{n + m}$。求满足$\Phi(\bvec{x}, \bvec{y}) = \bvec{0}$的条件下$f(\bvec{x}, \bvec{y})$的最大/最小值,记为
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\[\begin{cases}
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\max f(\bvec{x}, \bvec{y})\\
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\Phi(\bvec{x}, \bvec{y}) = 0
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\end{cases}
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\text{或}
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\begin{cases}
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\min f(\bvec{x}, \bvec{y})\\
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\Phi(\bvec{x}, \bvec{y}) = 0
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\end{cases}\]
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其中$f(\bvec{x}, \bvec{y})$称为目标函数,$\Phi(\bvec{x}, \bvec{y}) = \bvec{0}$称为约束条件。
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设在约束$\varphi(\bvec{x}, y) = 0$下$f(\bvec{x}, y)$在$P = (\bvec{x}_0, y_0) \in D$点达到极值,且$\dfrac{\partial \varphi}{\partial y}(\bvec{x}_0, y_0) \neq 0$,应用隐函数定理:在$P$点附近,方程$\varphi(\bvec{x}, y) = \bvec{0}$确定了隐函数$y = y(\bvec{x})$。函数$F(\bvec{x}) = f(\bvec{x}, y(\bvec{x}))$在$\bvec{x}_0$达到极值,因此$J F(\bvec{x}_0) = \bvec{0}$。
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因此
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\[\frac{\partial F}{\partial x_i} (\bvec{x}_0) = \frac{\partial f}{\partial x_i}(\bvec{x}_0, y_0) + \frac{\partial f}{\partial y} \frac{\partial y}{\partial x_i}(\bvec{x}_0) = 0, i = 1, 2, \cdots, n\]
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而根据隐函数定理,
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\[\frac{\partial y}{\partial x_i}(\bvec{x}_0) = -\frac{\dfrac{\partial \varphi}{\partial x_i}(\bvec{x}_0, y_0)}{\dfrac{\partial \varphi}{\partial y}(\bvec{x}_0, y_0)}\]
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将它带入上式,
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\[\frac{\partial f}{\partial x_i}(\bvec{x}_0, y_0) - \frac{\dfrac{\partial f}{\partial y}{(\bvec{x}_0, y_0)} \dfrac{\partial \varphi}{\partial x_i}(\bvec{x}_0, y_0)}{\dfrac{\partial \varphi}{\partial y}(\bvec{x}_0, y_0)} = 0\]
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引入参数
|
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\[\lambda = -\frac{\dfrac{\partial f}{\partial y}(\bvec{x}_0, y_0)}{\dfrac{\partial \varphi}{\partial y}(\bvec{x}_0, y_0)}\]
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那么
|
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\[\frac{\partial f}{\partial x_i}(\bvec{x}_0, y_0) + \lambda \frac{\partial \varphi}{\partial x_i}(\bvec{x}_0, y_0) = 0, i = 1, 2, \cdots, n\]
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因此我们得到
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\[Jf(\bvec{x}_0, y_0) + \lambda J\varphi(\bvec{x}_0, y_0) = \bvec{0}\]
|
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是条件极值的一个必要条件。
|
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\begin{theorem}[条件极值的必要条件]
|
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设$f \in C^1(D)$,$\Phi \in C^1 (D, \realnum^m)$,$D \subset \realnum^{n + m}$是开集。记$P = (\bvec{x}_0, \bvec{y}_0) \in D$,又设$\det(J_{\bvec{y}}\Phi(\bvec{x}_0, \bvec{y}_0)) \neq 0$。如果$f(\bvec{x}, \bvec{y})$在约束$\Phi(\bvec{x}, \bvec{y}) = \bvec{0}$下在$P$点达到极值,则存在$\bvec{\Lambda} = (\lambda_1, \dots, \lambda_m)$使得
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\[Jf(\bvec{x}_0, \bvec{y}_0) + \bvec{\Lambda} J \Phi(\bvec{x}_0, \bvec{y}_0) = \bvec{0}\eqper\]
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这也就是说$(\bvec{x}_0, \bvec{y}_0)$满足方程组
|
||||
\begin{align*}
|
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& \frac{\partial f}{\partial x_k}(\bvec{x}_0, \bvec{y}_0) + \sum_{i = 1}^m \lambda_i \frac{\partial \Phi_i}{\partial x_k}(\bvec{x}_0, \bvec{y}_0) = 0, k = 1, \cdots, n\\
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& \frac{\partial f}{\partial y_j}(\bvec{x}_0, \bvec{y}_0) + \sum_{i = 1}^m \lambda_i \frac{\partial \Phi_i}{\partial y_j}(\bvec{x}_0, \bvec{y}_0) = 0, j = 1, \cdots, m\\
|
||||
\end{align*}
|
||||
\end{theorem}
|
||||
|
||||
根据这个定理,我们可以引入Lagrange乘数法。定义函数$L: D \times \realnum^m \to \realnum$,
|
||||
\[L(\bvec{z}, \bvec{\Lambda}) = f(\bvec{z}) + \bvec{\lambda} \Phi(\bvec{z}), (\bvec{z}, \bvec{\Lambda}) \in D \times \realnum^m\]
|
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$L$称为条件极值问题的Lagrange函数,$\bvec{\Lambda}$称为Lagrange乘数/乘子。根据条件极值的表要条件,在条件极值点$\bvec{z}_0 \in D$,存在$\bvec{\Lambda} \in \realnum^m$满足
|
||||
\[J_{\bvec{z}} L (\bvec{z}_0, \bvec{\Lambda}) = J_{\bvec{z}} f(\bvec{z}_0) + \bvec{\Lambda} J_{\bvec{z}}\Phi(\bvec{z}_0) = \bvec{0}\]
|
||||
此外
|
||||
\[J_{\bvec{\Lambda}} L(\bvec{z}_0, \bvec{\Lambda}) = \Phi(\bvec{z}_0) = \bvec{0}\]
|
||||
因此
|
||||
\[JL(\bvec{z}, \bvec{\Lambda}) = (J_{\bvec{z}}, J_{\bvec{\Lambda}}) = \bvec{0} \eqper\]
|
||||
Reference in New Issue
Block a user