Table of Contents

  1. 微分法
    1. 微分方法
    2. 复合函数的微分
    3. 逆函数的微分法
    4. 指数函数和对数函数
    5. 高阶微分法
    6. 偏微分
    7. 微分可能性 全微分
    8. 微分的顺序
    9. 高阶的全微分
    10. Taylor公式
    11. 极大极小
    12. 切线和曲率

微分法

微分方法

定理 连续性是微分可能性的必要条件

但是它不是充分条件

例子,$ f(x) = x \sin{\frac{1}{x}}, f(0) = 0 $的f(x)在0时的区间上连续,x = 0时却不能微分。实际上

$ \frac{f(h) - f(0)}{h} = \sin{\frac{1}{h}} $

当 $ h \to 0 $时,极限不存在

上述函数在x = 0这一点是特异点,Weierstras(1872)作出了区间各点都不能微分的连续函数的例子时,震惊了当时的数学界

复合函数的微分

定理 f(x) $ \varphi(t) $微分存在,则 $ F(t) = f(\varphi(t)) $微分存在

$ F^{\prime} (t) = f^{\prime}(x) \cdot \varphi^{\prime}(t) $

$ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} $

这是复合函数的微分法

证明 t的变动为 $ \Delta t $,x的变动为 $ \Delta x $,这样y的变动为 $ \Delta y $

$ \frac{\Delta y }{\Delta t} = \frac{\Delta y}{\Delta x} \cdot \frac{\Delta x}{\Delta t} $

当 $ \Delta t \to 0 $时

$ \frac{\Delta x}{\Delta t} \to \frac{dx}{dt} $

同时当 $ \Delta x \to 0 $时

$ \frac{\Delta y}{\Delta x} \to \frac{dy}{dx} $

所以

$ \frac{\Delta y}{\Delta t} \to \frac{dy}{dx} \cdot \frac{dx}{dt} $

这样

$ F^{\prime}(t) = f^{\prime}(x)\varphi^{\prime}(t) $

注意上述证明方法的问题

x为独立变量,$ \Delta x $任意的话需要 $ \Delta x \ne 0 $,上面的情况是x是t的函数值,$ \Delta t $的值引起 $ \Delta x = 0$的情况是有的。这样的话,之前的证明写法有不合理的地方,变量 $ \Delta x, \Delta y $的写法要做修改如下:

$ \Delta y = f^{\prime}(x) \Delta x + \epsilon \Delta x, \quad \Delta x = \varphi^{\prime}(t)\Delta t + \epsilon^{\prime} \Delta t $

当 $ \Delta t \to 0 $时, $ \epsilon^{\prime} \to 0 $,有 $ \Delta x \to 0 $,但 $ \Delta t \ne 0 $时,$ \Delta x = 0 $的情况也有。$ \Delta x = 0 $时 $ \epsilon = 0 $定义为 $ \Delta t \to 0 $时 $ \epsilon \to 0 $,这样

$ \begin{aligned} \Delta t &= (f^{\prime}(x) + \epsilon)(\varphi^{\prime}(t) + \epsilon^{\prime}) \Delta t \\ &= f^{\prime}(x)\varphi^{\prime}(t) \cdot \Delta t + [\epsilon \varphi^{\prime}(t) + \epsilon^{\prime} f^{\prime}(x) + \epsilon \epsilon^{\prime}] \Delta t \end{aligned} $

右边括弧中的用 $ \epsilon^{\prime \prime} $表达,则

$ \Delta y = f^{\prime}(x) \varphi^{\prime}(t) \Delta t + \epsilon^{\prime \prime} \Delta t, \quad \epsilon^{\prime \prime} = \epsilon \varphi^{\prime}(t) + \epsilon^{\prime} f^{\prime}(x) + \epsilon \epsilon^{\prime} $

当 $ \Delta t \to 0 $时 $ \epsilon^{\prime \prime} \to 0 $,则

$ dy = f^{\prime}(x) \varphi^{\prime}(t) dt $

这样结果为

$ dy = f^{\prime}(x)dx $

同样,y为x的函数,x为t的函数,t为u的函数时,微分为

$ \frac{dy}{du} = \frac{dy}{dx} \frac{dx}{dt} \frac{dt}{du} $

逆函数的微分法

假设有一个 $ a \le x \le b $区间上的连续函数y = f(x)。如果y在该区间的最大值和最小值为p和q,则y可取遍 $ p \le y \le q $区间上的任意值。但是,只有当y = f(x)是单调的,才能通过x值唯一确定一个y值

如果f(x)不是单调的,$ x_ {1} < x_ {2} < x_ {3} $对应 $ y_ {1} < y_ {2} < y_ {3} $及$ y_ {1} > y_ {2} > y_ {3} $都有可能。如果 $ y_ {1} < y_ {2}, y_ {2} > y_ {3}, y_ {2} \eta > Max(y_ {1}, y_ {3}) $,区间 $ (x_ {1}, x_ {2}) $及 $ (x_ {2}, x_ {3}), \eta = f(x) $的x的值最少会出现一次

单调的时候,区间 $ p \le y \le q $中y=f(x)中各种值唯一对应。对应 $ x = \varphi(y), \varphi $是f的逆函数。这样,f是 $ \varphi $的逆函数,f和 $ \varphi $互为逆函数

定理18 关于x的区间的函数y单调且连续,则y的变动区间中x是其反函数。反函数也连续单调。如果y可微则x也可微

$ \frac{dy}{dx} \cdot \frac{dy}{dx} = 1 $

证明:我们用y = f(x), $ x = \varphi(y) $表示两个函数。$ x = \xi $对应 $ y = \eta $。$ \{y_ {n}\} $为收敛到 $ \eta $的任意单调数列,对应 $ \{x_ {n} \} $单调有界,极限值收敛到 $ \lambda $。由于f(x)的连续性,$ f(\lambda) = \eta $,则有 $ \lambda = \varphi(\eta) = \xi $。则 $ y_ {n} \to \eta $的同时 $ x_ {n} \to \xi $,即 $ \varphi(y_ {n}) \to \varphi(\eta) $,这样反函数 $ \varphi(y) $连续

img

$ \frac{\Delta x}{\Delta y} = 1 / \frac{\Delta y}{\Delta x} $

当 $ \Delta y \to 0 $时,有$ \Delta x \to 0 $,同时 $ \lim{\frac{\Delta x}{\Delta y}} = 1 / \lim{\frac{\Delta y}{\Delta x}} $,即 $ \frac{d x}{d y} = 1 / \frac{d y}{d x} $

img

但 $ \frac{d y}{d x} = 0 $的情况要排除

取三角函数的逆函数的例子

(1) $ \arcsin{x} $

y = sinx 在区间 $ - \frac{\pi}{2} \le x \le \frac{\pi}{2} $,一般化区间为

$ (2n - 1)\frac{\pi}{2} \le x \le (2n + 1)\frac{\pi}{2}, \qquad (n = 0, \pm 1, \pm 2, \cdots) $

内单调,y在区间 $ -1 \le y \le 1 $之间取值。则函数y = sinx的逆函数,即 $ x = \arcsin{y} $中y在 $ -1 \le y \le 1 $之间,所以x必须限定为上述区间中的一个

这样从y = sinx中有

$ \frac{d \sin{x}}{d x} = \cos{x}, \qquad \frac{d \arcsin{y}}{d y} = \frac{1}{\cos{x}} = \pm \frac{1}{\sqrt{1 - y^{2}}} $

由于我们选取 $ - \frac{\pi}{2} \le x \le \frac{\pi}{2} $区间,则 $ \cos{x} \ge 0 $。则这里 $ \pm $应该为 +,这样变量x,y置换可得

$ d \arcsin{x} = \frac{1}{\sqrt{1 - x^{2}}} $

(2) $ \arctan{x} $

$ y = \tan{x} $在区间 $ - \frac{\pi}{2} < x < \frac{\pi}{2} $内从 $ -\infty $ 到 $ + \infty $单调递增。这样arctan定义为

$ y = \arctan{x}, \qquad - \frac{\pi}{2} < y < \frac{\pi}{2} $

例如:

$ \arctan{0} = 0, \qquad \arctan{\pm 1} = \pm \frac{\pi}{4} \\ \arctan{\pm \infty} = \lim_ {x \to \pm \infty} \arctan{x} = \pm \frac{\pi}{2} $

从 $ y = \tan{x}, \frac{dy}{dx} = \frac{1}{\cos^{2}{x}} = 1 + y^{2} $,通过记号变换,有

$ \operatorname{D}{\arctan{x}} = \frac{1}{1 + x^{2}} $

例1,$ y = \arcsin{\sqrt{1 - x^{2}}} $意味着 $ \sqrt{1 - x^{2}} = \sin{y} $。从而 $ x^{2} = \cos^{2}{y}, x = \pm \cos{y} $。$ \arcsin $的主值在ABC如下图,y的主值微分为

img

$ \frac{dy}{dx} = \frac{1}{\sqrt{1 - (1 - x^{2})}} \frac{-x}{\sqrt{1 - x^{2}}} = \frac{1}{| x |} \frac{-x}{\sqrt{1 - x^{2}}} = \mp \frac{1}{\sqrt{1 - x^{2}}} \qquad (x \ge 0) $

当x = 0时 $ \operatorname{D^{+}}{y} = -1, \operatorname{D^{-}}{y} = +1 $

指数函数和对数函数

当a > 0时,对 $ a^{x} $微分有

$ \frac{d(a^{x})}{dx} = \lim_ {h \to 0}{\frac{a^{x+h} - a^{x}}{h}} = a^{x} \lim_ {h \to 0}\frac{a^{h} - 1}{h} $

因为h > 0,则 $ a^{h} > 1 $,则设 $ a^{h} = 1 + \frac{1}{t}, t > 0 $。当 $ h \to 0, a^{h} \to 1, t \to \infty $

从 $ h = \log_ {a}{(1 + \frac{1}{t})} $中得到

$ \frac{a^{h} - 1}{h} = \frac{\frac{1}{t}}{\log_ {a}{(1 + \frac{1}{t})}} = \frac{1}{\log_ {a}{(1 + \frac{1}{t})^{t}}} $

当 $ h \to 0, t \to \infty $有 $ (1 + \frac{1}{t})^{t} \to e $,因为 $ \log_ {a} $是连续函数,$ h \to 0 $时 $ \log_ {a}{(1 + \frac{1}{t})^{t}} \to \log_ {a}{e} $。则

$ \lim_ {h \to 0}{\frac{a^{h} - 1}{h}} = \frac{1}{\log_ {a}{e}} = \log_ {e}{a} $

当h < 0时,把h用-h带入,得:

$ \frac{a^{-h} - 1}{-h} = \frac{a^{h} - 1}{h} \cdot \frac{1}{a^{h}}, \qquad (h > 0) $

当 $ h \to 0 $时, $ a^{h} \to 1 $,则有

$ \frac{a^{-h} - 1}{-h} \to \log_ {e}{a} $

所以

$ \frac{d(a^{x})}{dx} = a^{x} \log_ {e}{a} $

对数微分法 u, v, w为x的函数,u, v, w在x = 0时,$ \log{| uvw |} $微分为

$ \operatorname{D}\log{ | uvw | } = \operatorname{D}{(\log{| u |} + \log{| v |} + \log{| w |})} = \frac{u^{\prime}}{u} + \frac{v^{\prime}}{v} + \frac{w^{\prime}}{w} $

但是又

$ \operatorname{D}{\log{ | uvw | }} = \frac{(uvw)^{\prime}}{uvw} $

$ \frac{(uvw)^{\prime}}{uvw} = \frac{u^{\prime}}{u} + \frac{v^{\prime}}{v} + \frac{w^{\prime}}{w}, \qquad (u \ne 0, v \ne 0, w \ne 0) $

同样的

$ (\frac{u}{v})^{\prime} / \frac{u}{v} = \frac{u^{\prime}}{u} - \frac{v^{\prime}}{v} $

高阶微分法

y = f(x)的导数是 $ f^{\prime}(x) $,$ f^{\prime}(x) $的导数是 $ f^{\prime \prime}(x) $。第n阶导数是 $ f^{(n)}(x) $。$ f^{\prime \prime}(x) $可写为

$ \frac{d}{dx} (\frac{dy}{dx}) \text{ 或 } \frac{d^{2} y}{dx^{2}} $

同样的

$ \frac{d^{n}y}{dx^{n}} = f^{(n)}(x) $

上述记号中,$ dx^{2} $是 $ (dx)^{2} $,$ d^{2}y $是 $ d(dy) $,微分记号可写为

$ dy = y^{\prime}_ {x} dx $

两边微分,$ d(dy), d(dx) $表示 $ d^{2}y, d^{2} x $,得

$ d^{2}y = y^{\prime \prime}_ {x}(dx)^{2} + y^{\prime}_ {x}d^{2}x $

这里x为独立变量,$ dx = \delta x $,$ d^{2}x = d(\delta x) = 0 $,则

$ d^{2}y = y^{\prime \prime}_ {x}dx^{2} $

则 $ \frac{d^{2}y}{dx^{2}} = f^{\prime \prime}(x) $。而 $ x = \varphi(t) $,有 $ d^{2}x = x^{\prime \prime}_ {t}dt^{2} $,则

$ d^{2}y = y^{\prime \prime}_ {x}x^{\prime 2}_ {t} dt^{2} + y^{\prime}_ {x}x^{\prime \prime}_ {t}dt^{2} $

则有

$ \frac{d^{2}}{dt^{2}} f(\varphi(t)) = f^{\prime \prime}(\varphi(t)) \varphi^{\prime}(t)^{2} + f^{\prime}(\varphi(t))\varphi^{\prime \prime}(t) $

u, v为x的函数,则根据Leibniz法则有

$ \frac{d^{n}(uv)}{dx^{n}} = u^{(n)}v + {n \choose 1}u^{(n-1)}v^{\prime} + \cdots + {n \choose k}u^{(n-k)}v^{(k)} + \cdots + uv^{(n)} $

而 $ u / v $的高阶导函数没有简单的函数表示

偏微分

两个以上的变量,根据一个变量变动,相关的微分叫做偏微分。例如z = f(x, y)

$ \frac{\partial z}{\partial x} = \lim_ {\Delta x \to 0} \frac{f(x + \Delta x, y) - f(x,y)}{\Delta x}, \qquad \frac{\partial z}{\partial y} = \lim_ {\Delta y \to 0} \frac{f(x, y+\Delta y) - f(x, y)}{\Delta y} $

当区域内各点关于 $ \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} $存在时,这样的x, y函数

$ \frac{\partial z}{\partial x} = f_ {x}(x, y) = D_ {x} f(x, y), \quad \frac{\partial z}{\partial y} = f_ {y}(x, y) = D_ {y}f(x, y) $

同样对高阶微分

$ \frac{\partial}{\partial x} (\frac{\partial z}{\partial x}) = \frac{\partial^{2} z}{\partial x^{2}} = f_ {xx}(x, y) $

$ \frac{\partial}{\partial y} (\frac{\partial z}{\partial x}) = \frac{\partial^{2} z}{\partial x \partial y} = f_ {xy}(x, y), \quad \frac{\partial}{\partial x} (\frac{\partial z}{\partial y}) = \frac{\partial^{2} z}{\partial y \partial x} = f_ {yx}(x, y) $

$ \frac{\partial}{\partial y} (\frac{\partial z}{\partial y}) = \frac{\partial^{2} z}{\partial y^{2}} = f_ {yy}(x, y) $

三个以上的变量也一样

例1 $ \sqrt{x^{2} + y^{2}} = r $

$ f(x, y) = \log{r} = \frac{1}{2} \log{x^{2} + y^{2}} $

然而

$ f_ {x} = \frac{x}{x^{2} + y^{2}} = \frac{x}{r^{2}}, \qquad f_ {y} = \frac{y}{r^{2}} $

$ f_ {xx} = \frac{1}{x^{2} + y^{2}} - \frac{2x^{2}}{(x^{2}+y^{2})^{2}} = \frac{1}{r^{2}} - \frac{2x^{2}}{r^{4}} $

$ f_ {xy} = - \frac{2xy}{(x^{2} + y^{2})^{2}} = - \frac{2xy}{r^{4}} = f_ {yx} $

$ f_ {yy} = \frac{1}{r^{2}} - \frac{2y^{2}}{r^{4}} $

$ \frac{\partial^{2}f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} $用 $ \Delta f $表示,则

$ \Delta f = \frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} = \frac{2}{r^{2}} - \frac{2(x^{2} + y^{2})}{r^{4}} = 0 $

例2 $ \sqrt{x^{2} + y^{2} + z^{2}} = r, f(x, y, z) = \frac{1}{r} = (x^{2} + y^{2} + z^{2})^{- \frac{1}{2}} $

$ f_ {x} = - \frac{x}{(x^{2} + y^{2} + z^{2})^{\frac{3}{2}}} = - \frac{x}{r^{3}}, \qquad f_ {y} = - \frac{y}{r^{3}}, \qquad f_ {z} = - \frac{z}{r^{3}} $

$ f_ {xx} = - \frac{1}{r^{3}} - x\{-3 \frac{1}{r^{4}} \cdot \frac{\partial r}{\partial x} \} = - \frac{1}{r^{3}} + \frac{3x}{r^{4}} \cdot \frac{x}{r} = - \frac{1}{r^{3}} + \frac{3x^{2}}{r^{5}} $

$ f_ {yy} = - \frac{1}{r^{3}} + \frac{3y^{2}}{r^{5}} $

$ f_ {zz} = - \frac{1}{r^{3}} + \frac{3z^{2}}{r^{5}} $

$ \Delta f = \frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} + \frac{\partial^{2} f}{\partial z^{2}} = - \frac{3}{r^{3}} + \frac{3(x^{2} + y^{2} + z^{2})}{r^{5}} = - \frac{3}{r^{3}} + \frac{3r^{2}}{r^{5}} = 0 $

$ f_ {xy} = -x(-3 \frac{1}{r^{4}} \cdot \frac{\partial r}{\partial y}) = 3 \frac{xy}{r^{5}} = f_ {yx} $

微分可能性 全微分

函数z = f(x, y)的一点P = (x, y)。$ \Delta z = f(x + \Delta x, y + \Delta y) - f(x, y) $,则

$ \Delta z = A \Delta x + B \Delta y + \varepsilon \rho $

A, B是跟 $ \Delta x, \Delta y $无关的系数,点(x, y)的值有的话,$ \rho $是定点(x,y)和动点 $ (x + \Delta x, y + \Delta y) $的距离 $ (\rho = \sqrt{(\Delta x)^{2} + (\Delta y)^{2}} ) $,$ \varepsilon $跟 $ \Delta x, \Delta y $相关,$ \rho \to 0 $时,$ \varepsilon \to 0 $。用之前页的记号有 $ \varepsilon \rho = o \rho $。这样函数z在点(x, y)微分是可能的

上面的等式成立的话,$ \Delta y = 0, \rho = | \Delta x | $时,

$ \frac{\Delta z}{\Delta x} = A \pm \varepsilon $

这时,$ \Delta x \to 0, \varepsilon \to 0 $的话,(x, y)的 $ \frac{\partial z}{\partial x} $存在,为A。同样的,$ \frac{\partial z}{\partial y} $存在,为B。点 $(x + \Delta x, y + \Delta y) $从点(x, y)一定方向上收敛时,$ \alpha $确定,$ \Delta x = \rho \cos{\alpha}, \Delta y = \rho \sin{\alpha} $时,

$ \frac{\Delta z}{\rho} = A \cos{\alpha} + B \sin{\alpha} + \varepsilon $

$ \lim_ {\rho \to 0} \frac{\Delta z}{\rho} = A \cos{\alpha} + B \sin{\alpha} = \frac{\partial z}{\partial x} \cos{\alpha} + \frac{\partial z}{\partial y} \sin{\alpha} $

这种情况下,$ \lim_ {\rho \to 0} \frac{\Delta z}{\rho} $是 $ \Delta x = \rho \cos{\alpha}, \Delta y = \rho \sin{\alpha} $方向上的偏微分商

之前的等式成立的话各方向上的偏微分商存在,即为上式

z可微分的话,$ \Delta z $的主要部分 $ \Delta x, \Delta y $的一次方程式 $ \frac{\partial z}{\partial x} \Delta x + \frac{\partial z}{\partial y} \Delta y $是z的全微分,用dz表示。特别地,z = x, z = y时 $ dx = \Delta x, dy = \Delta y $,这样全微分为

$ dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy $

z可微分时 $ dz = \frac{\partial z}{\partial x} \Delta x + \frac{\partial z}{\partial y} \Delta y $为(x, y)关于曲面z = f(x, y)相接的平面。这样的平面上坐标为X, Y, Z的话,dx, dy, dz为 $ X - x, Y - y, Z - z $,接平面的方程式为

$ Z - z = \frac{\partial z}{\partial x}(X - x) + \frac{\partial z}{\partial y}(Y - y) $

这是接平面的定义

z = f(x, y)有领域的各点可微分时,这样的领域可微分。这样的情况下f(x, y)的领域是连续的

定理 $ \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} $领域存在且连续的时候,z的领域可微分

证明 $ \Delta x, \Delta y $用h, k带入得

$ \begin{aligned} \Delta z &= f(x + h, y + k) - f(x, y) \\ &= \{ f(x + h, y + k) - f(x, y + k)\} + \{ f(x, y + k) - f(x, y) \} \end{aligned} $

使用x的平均值定理

$ f(x + h, y + k) - f(x, y + k) = hf_ {x}(x + \theta h, y + k), \qquad 0 < \theta < 1 $

假设 $ f_ {x} $ 连续

$ f_ {x}(x + \theta h, y + k) = f_ {x}(x, y) + \varepsilon $

如果 $ h \to 0, k \to 0 $时 $ \varepsilon \to 0 $

y的偏微分

$ f(x, y + k) - f(x, y) = kf_ {y}(x, y) - \varepsilon^{\prime} k $

$ k \to 0 $时 $ \varepsilon^{\prime} \to 0 $,则

$ \Delta z = hf_ {x}(x, y) + k f_ {y}(x, y) + h \varepsilon + k \varepsilon^{\prime} $

$ | h | \le \rho, | k | \le \rho, (\rho = \sqrt{h^{2} + k^{2}}) $,所以 $ | h \varepsilon + k \varepsilon^{\prime} | \le (| \varepsilon | + | \varepsilon^{\prime} |) \rho $

$ \Delta z = hf_ {x}(x, y) + kf_ {y}(x, y) + o \rho $

即z是可微分的

注意 该定理的假设过大,上述证明用了 $ f_ {x} $的连续性。领域内 $ z_ {x}, z_ {y} $关于点(x, y)存在,这样如果连续的话,该点的z可微分

微分的顺序

f(x, y)关于x微分得导函数$ f_ {x}(x, y) $,$ f_ {x}(x, y) $关于y微分的导函数为 $ f_ {xy} $。这样 $ f_ {xy}, f_ {yx} $指代不同的事物。然而在一定条件下,$ f_ {xy}, f_ {yx} $为同一函数。现在采用这些条件中最容易获得的条件:

定理 在某个区域 $ f_ {xy}, f_ {yx} $连续的话这样的区域 $ f_ {xy} = f_ {yx} $

证明 对区域的任意一点(a, b)的附近考察

$ \Delta = f(a + h, b + k) - f(a + h, b) - f(a, b + k) + f(a, b) $

用图的记号有

$ \Delta = f(P_ {3}) - f(P_ {1}) - f(P_ {2}) + f(P) $

img

写成

$ \varphi(x) = f(x, b + k) - f(x, b) $

$ \varphi(a) = f(P_ {2}) - f(P), \varphi(a + h) = f(P_ {3}) - f(P_ {1}) $

$ \Delta = \varphi(a + h) - \varphi(a) $

假设(a, b)附近 $ f_ {x} $存在,则

$ \varphi^{\prime}(x) = f_ {x}(x, b + k) - f_ {x}(x, b) $

这样x = a 和 x = a + h之间,用 $ \varphi(x) $的平均值定理

$ \varphi(a + h) - \varphi(a) = h\varphi^{\prime}(a + \theta h) \qquad (0 < \theta < 1) $

综合得

$ \Delta = h\{ f_ {x}(a + \theta h, b + k) - f_ {x}(a + \theta h, b) \} $

假设(a, b)附近 $ f_ {xy} $存在,则右边y = b和y = b + k间,用平均值定理得

$ \Delta = hkf_ {xy}(a + \theta h, b + \theta^{\prime} k) \qquad (0 < \theta^{\prime} < 1) $

假设 $ f_ {xy} $在(a, b)点连续,则

$ \lim_ {(h, k) \to (0, 0)} \frac{\Delta}{hk} = f_ {xy}(a, b) $

x,h和y,k交换可得

$ \lim_ {h, k) \to (0, 0)} \frac{\Delta}{hk} = f_ {yx}(a, b) $

这样,区域内各点有

$ f_ {xy} = f_ {yx} $

在 $ f_ {x}, f_ {y}, f_ {xy} $存在的区域,该区域内一点 $ f_ {xy} $连续,这样的点 $ f_ {yx} $存在,且 $ f_ {xy} = f_ {yx} $(Schwarz定理)

如果区域内 $ f_ {x}, f_ {y} $存在,区域内某点可微,则在该点 $ f_ {xy} = f_ {yx} $(Young定理)

这里可用之前的式子证明,假定h = k且 $ f_ {x} $可微

$ f_ {x}(a + \theta h, b + h) = f_ {x}(a, b) + \theta h f_ {xy}(a, b) + h f_ {xy}(a, b) + oh $

$ f_ {x}(a + \theta h, b) = f_ {x}(a, b) + \theta h f_ {xx}(a, b) + oh $

这样 $ \Delta = h^{2} f_ {xy}(a, b) + oh^{2} $,有

$ \lim_ {h \to 0} \frac{\Delta}{h^{2}} = f_ {xy}(a, b) $

假设x, y对称,有$ f_ {xy}(a, b) = f_ {yx}(a, b) $

不能无条件的认为 $ f_ {xy} = f_ {yx} $,这个很重要,现在举一个例子

$ \left\{ \begin{array}{ll} f(x,y) = xy \frac{x^{2} - y^{2}}{x^{2} + y^{2}}, & (x, y) \ne (0, 0), \\ f(0, 0) = 0 \end{array} \right. $

$ (x, y) \ne (0, 0) $时计算得

$ f_ {x}(x, y) = \frac{3x^{2} - y^{3}}{x^{2} + y^{2}} - \frac{2x^{2}y(x^{2} - y^{2})}{(x^{2} + y^{2}x)^{2}} $

x,y 交换得 $ f_ {y} $,然而

$ f_ {xy}(x, y) = \frac{x^{2} - y^{2}}{x^{2} + y^{2}} + \frac{8x^{2}y^{2}(x^{2} - y^{2})}{(x^{2} + y^{2})^{3}} $

$ (x, y) \ne (0, 0) $时连续,$ f_ {yx} $相等,然而,$ f_ {xy}(0, y)= -1, (y \ne 0), \lim_ {y \to 0} f_ {xy}(0, y) = -1 $。然而,$ f_ {x}(0, y) = -y, y \ne 0 $,$ f_ {x}(0, 0) = 0 $时$ f_ {x}(0, y) $在y = 0时连续,这样 $ f_ {xy}(0, 0) = -1 $。同样的,$ f_ {yx}(x, 0) = 1 $时 $ f_ {yx}(0, 0) = 1 $

三阶以上,导函数连续的话,可以改变微分顺序。例如 $ f_ {xxy} = (f_ {x})_ {xy} = (f_ {x})_ {yx} = f_ {xyx} $,相邻的顺序进行交换,这样

$ f_ {xyz} = f_ {xzy} = f_ {zxy} = f_ {zyx} = f_ {yzx} = f_ {yxz} $

$ f_ {xxyy} = f_ {xyxy} = f_ {xyyx} = f_ {yxxy} = f_ {yxyx} = f_ {yyxx} $

这样二阶导函数

$ f_ {x^{2}} = \frac{\partial^{2} f}{\partial x^{2}}, \qquad f_ {xy} = \frac{\partial^{2} f}{\partial x \partial y}, \qquad f_ {y^{2}} = \frac{\partial^{2} f}{\partial y^{2}} $

第n阶为

$ f_ {x^{r}y^{s}} = \frac{\partial^{n} f}{\partial x^{r} \partial y^{s}} \qquad \qquad (r + s = n, s = 0, 1, 2, \ldots, n) $

高阶的全微分

u = f(x, y)的第一阶全微分

$ du = \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy $

$ \frac{\partial u}{\partial x} $可微分,x, y关于du的全微分为 $ d^{2} u $

即x, y为独立变量(h = dx, k = dy)

$ \begin{aligned} d^{2}u = d(du) &= \frac{\partial}{\partial x}(\frac{\partial u}{\partial x} h + \frac{\partial u}{\partial y} k)h + \frac{\partial}{\partial y}(\frac{\partial u}{\partial x} h + \frac{\partial u}{\partial y}k)k \\ &= \frac{\partial^{2} u}{\partial x^{2}}h^{2} + 2 \frac{\partial^{2}u}{\partial x \partial y} hk + \frac{\partial^{2} u}{\partial y^{2}} k^{2} \\ &= \frac{\partial^{2} u}{\partial x^{2}}d x^{2} + 2 \frac{\partial^{2} u}{\partial x \partial y}dxdy + \frac{\partial^{2} u}{\partial y^{2}} dy^{2} \end{aligned} $

同样的

$ d^{3}u = \frac{\partial^{3} u}{\partial x^{3}} dx^{3} + 3 \frac{\partial^{3}u}{\partial x^{2} \partial y} dx^{2}dy + 3 \frac{\partial^{3} u}{\partial x \partial y^{2}}dxdy^{2} + \frac{\partial^{3} u}{\partial y^{3}}dy^{3} $

一般的

$ d^{n}u = \frac{\partial^{n} u}{\partial x^{n}}dx^{n} + \cdots + {n \choose k} \frac{\partial^{n} u}{\partial x^{k} \partial y^{n-k}} dx^{k}dy^{n-k} + \cdots + \frac{\partial^{n} u}{\partial y^{n}}dy^{n} $

这样

$ d^{n} u = (\frac{\partial}{\partial x}dx + \frac{\partial}{\partial y}dy)^{n} u $

三个以上的变量的话,同样的

$ \begin{aligned} d^{2} u &= \frac{\partial^{2} u}{\partial x^{2}} dx^{2} + \frac{\partial^{2} u}{\partial y^{2}}dy^{2} + \frac{\partial^{2}u}{\partial z^{2}}dz^{2} + \cdots &+ 2 \frac{\partial^{2} u}{\partial x \partial y}dxdy + 2 \frac{\partial^{2} u}{\partial x \partial y} + 2 \frac{\partial^{2} u}{\partial x \partial z}dxdz + 2 \frac{\partial^{2} u}{\partial y \partial z}dydz + \cdots , \end{aligned} $

一般的

$ d^{n}u = (\frac{\partial}{\partial x}dx + \frac{\partial}{\partial y}dy + \frac{\partial}{\partial z}dz + \cdots)^{n} u $

合成函数 u为x, y的函数,x, y为t的函数,u为t的函数

u跟x, y相关,x, y跟t相关连续(可微分)的话,u连续(可微分)

可微分的时候 $ \frac{du}{dt} $得

$ \Delta u = u_ {x} \Delta x + u_ {y} \Delta y + o(\sqrt{\Delta x^{2} + \Delta y^{2}}) $

$ \Delta x, \Delta y $使得 $ \Delta t $对应变化

$ \Delta x = x^{\prime} \Delta t + o(\Delta t), \qquad \Delta y = y^{\prime} \Delta t + o(\Delta t) $

综合得

$ \Delta u = (u_ {x}x^{\prime} + u_ {y}y^{\prime}) \Delta t + o(\Delta t) $

$ \lim_ {\Delta t \to 0} \frac{\Delta u}{\Delta t} $存在,这样

$ \frac{du}{dt} = u_ {x}x^{\prime} + u_ {y}y^{\prime}. \qquad (x^{\prime} = \frac{dx}{dt}, y^{\prime} = \frac{dy}{dt}) $

第二阶以上可微分的话,有

$ \begin{aligned} \frac{d^{2}u}{dt^{2}} &= \frac{du_ {x}}{dt}x^{\prime} + u_ {x}x^{\prime \prime} + \frac{du_ {y}}{dt}y^{\prime} + u_ {y}y^{\prime \prime} \\ &= (u_ {xx}x^{\prime} +u_ {xy}y^{\prime})x^{\prime} + (u_ {xy}x^{\prime} + u_ {yy}y^{\prime})y^{\prime} + u_ {x}x^{\prime \prime} + u_ {y}y^{\prime \prime} \\ &= u_ {xx}x^{\prime 2} + 2u_ {xy}x^{\prime}y^{\prime} + u_ {yy}y^{\prime 2} + u_ {x}x^{\prime \prime} + u_ {y}y^{\prime \prime} \end{aligned} $

x、y是两个变量 $ \xi, \eta $的函数,u也是 $ \xi, \eta $的函数。这样

$ u_ {\xi} = u_ {x} x_ {\xi} + u_ {y}y_ {\xi}, \qquad u_ {\eta} = u_ {x}u_ {\eta} + u_ {y}y_ {\eta} $

同样的

$ u_ {\xi \xi} = u_ {xx}x^{2}_ {\xi} + 2u_ {xy}x_ {\xi}y_ {\xi} + u_ {yy}y^{2}_ {\xi} + u_ {x}x_ {\xi \xi} + u_ {y}y_ {\xi \xi} $

$ u_ {\xi \eta} = u_ {xx}x_ {\xi}x_ {\eta} + u_ {xy}(x_ {\xi}y_ {\eta} + x_ {\eta}y_ {\xi}) + u_ {yy}y_ {\xi}y_ {\eta} + u_ {x}x_ {\xi \eta} + u_ {y}y_ {\xi \eta} $

$ u_ {\eta \eta} = u_ {xx}x^{2}_ {\eta} + 2u_ {xy}x_ {\eta}y_ {\eta} + u_ {yy}y^{2}_ {\eta} + u_ {x}x_ {\eta \eta} + u_ {y}y_ {\eta \eta} $

例子 u = u(x, y),x,y 相关偏微分商的极坐标变换

$ x = r \cos{\theta}, \qquad y = r \sin{\theta} $

$ r = \sqrt{x^{2} + y^{2}}, \qquad \theta = \arctan{\frac{y}{x}} $

u是r, $ \theta $的函数,$ r, \theta $是x, y的函数,计算得

$ r_ {x} = \frac{x}{r} = \cos{\theta}, \qquad r_ {y} = \frac{y}{r} = \sin{\theta} $

$ \theta_ {x} = - \frac{y}{r^{2}} = - \frac{\sin{\theta}}{r}, \qquad \theta_ {y} = \frac{x}{r^{2}} = \frac{\cos{\theta}}{r} $

$ u_ {x} = u_ {r}r_ {x} + u_ {\theta}\theta_ {x} = u_ {r}\cos{\theta} - \frac{u_ {\theta}}{r} \sin{\theta} $

$ u_ {y} = u_ {r}r_ {y} + u_ {\theta}\theta_ {y} = u_ {r} \sin{\theta} + \frac{u_ {\theta}}{r} \cos{\theta} $

$ u_ {xx} = u_ {rr}\cos^{2}{\theta} + \frac{u_ {\theta \theta}}{r^{2}} \sin^{2}{\theta} - 2 \frac{u_ {r\theta}}{r} \cos{\theta} \sin{\theta} + \frac{u_ {r}}{r} \sin^{2}{\theta} + 2 \frac{u_ {\theta}}{r^{2}} \cos{\theta} \sin{\theta} $

$ u_ {yy} = u_ {rr}\sin^{2}{\theta} + \frac{u_ {\theta \theta}}{r^{2}} \cos^{2}{\theta} + 2 \frac{u_ {r\theta}}{r} \cos{\theta} \sin{\theta} + \frac{u_ {r}}{r} \cos^{2}{\theta} - 2 \frac{u_ {\theta}}{r^{2}} \cos{\theta} \sin{\theta} $

这样

$ \frac{\partial^{2}u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} = \frac{\partial^{2} u}{\partial r^{2}} + \frac{1}{r^{2}} \frac{\partial^{2} u}{\partial \theta^{2}} + \frac{1}{r} \frac{\partial u}{\partial r} $

Taylor公式

固定差异 对于y = f(x),x一定的增量 $ \Delta x(\Delta x \ge 0) $,对于y的增量

$ \Delta y = \Delta f(x) = f(x + \Delta x) - f(x) $

为y的定差的差分。$ \Delta y $是x的函数,对应 $ \Delta x $的增加,这样的固定差异为y的第二阶定差,记为 $ \Delta^{2} y $

$ \begin{aligned} \Delta^{2} y &= \Delta f(x + \Delta x) - \Delta f(x) \\ &= \{f(x + 2 \Delta x) - f(x + \Delta x)\} - \{f(x + \Delta x) - f(x)\} \\ &= f(x + 2 \Delta x) - 2f(x + \Delta x) + f(x) \end{aligned} $

同样的,第n阶的定差

$ \begin{aligned} \Delta^{n} y &= \Delta^{n-1}f(x + \Delta x) - \Delta^{n-1}f(x) \\ &= f(x + n \Delta x) - {n \choose 1} f(x + (n-1) \Delta x) + {n \choose 2}f(x + (n-2) \Delta x) + \cdots + (-1)^{n} f(x) \end{aligned} $

例如,$ g(x) = ax^{n} + \cdots $为n的多项式,$ \Delta x = h $

$ \Delta g(x) = nahx^{n-1} + \cdots, \qquad \Delta^{2}g(x) = n(n-1)ah^{2}x^{n-2} + \cdots $

$ \Delta^{n} g(x) = n! ah^{n} \qquad \Delta^{n+1} g(x) = 0 $

则对十分小的 $ \Delta x $

$ f(x + k \Delta x) = \sum^{n}_ {\nu = 0}(k \Delta x)^{\nu} \frac{f^{(\nu)}(x)}{\nu !} + o(\Delta x)^{n} $

这样

$ \begin{aligned} \Delta^{n} y &= \sum^{n}_ {k, \nu = 0}(-1)^{n-k} {n \choose k} k^{\nu} \Delta x^{\nu} \frac{f^{(\nu)}(x)}{\nu !} + o(\Delta x)^{n} \\ &= \sum^{n}_ {\nu = 0} \Delta x^{\nu} \frac{f^{(\nu)}(x)}{\nu !} (\sum^{n}_ {k=0} (-1)^{n-k} {n \choose k}k^{\nu}) + o(\Delta x)^{n} \end{aligned} $

现在

$ \sum^{n}_ {k=0}(-1)^{k}{n \choose k}k^{\nu} = \left\{ \begin{array}{ll} 0, & (\nu = 0, 1, \ldots, n-1) \\ (-1)^{n} n! & (\nu = n) \end{array} \right. $

这样 $ y = x^{n} $,从以上的式子及 $ \Delta^{n} y = n! \Delta x^{n} $有

$ \Delta^{n} y = (\Delta x)^{n} f^{(n)}(x) + o(\Delta x)^{n} $

从而

$ \lim_ {\Delta x \to 0} \frac{\Delta^{n} y}{\Delta x^{n}} = f^{(n)}(x) $

二元以上的Taylor公式 f(x, y) n次可微分时A = (x, y)区域内一点,取很小的 $ | h |, | k | $,点B = (x+h, y+k)的线分AB有

$ F(t) = f(x + ht, y + kt) $

$ 0 \le t \le 1 $区间里t的函数,有

$ F^{\prime}(t) = (h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y}) f(x + ht, y + kt), \ldots, $

$ F^{n}(t) = (h \frac{\partial}{\partial x} + k \frac{\partial}{\partial y})^{n} f(x+ht, y+kt) $

$ F(t) = F(0) + tF^{\prime}(0) + \cdots + \frac{t^{n-1}}{(n-1)!} F^{(n-1)}(0) + \frac{t^{n}}{n!} F^{(n)}(\theta t), \qquad 0 < \theta < 1 $

t = 1时

$ \begin{aligned} f(x+h, y+k) &= f(x,y) + df(x, y) + \frac{1}{2} d^{2} f(x, y) + \cdots \\ &+ \frac{1}{(n-1)!} d^{n-1} f(x, y) + \frac{1}{n!} d^{n} f(x + \theta h, y + \theta k) \end{aligned} $

用 $ d^{\nu} $简记为

$ df(x, y) = hf_ {x}(x, y) + kf_ {y}(x, y) $

$ d^{2} f(x, y) = h^{2} f_ {xx}(x,y) + 2hkf_ {xy}(x, y) + k^{2}f_ {yy}(x, y), \ldots $

最后的剩余项变量x, y用 $ x + \theta h, y + \theta k $带入

当n = 1时

$ f(x + h, y + k) - f(x, y) = hf_ {x}(x + \theta h, y + \theta k) + kf_ {y}(x + \theta h, y + \theta k) $

$ (x + \theta h, y + \theta k) $为线分AB上的点。这是二元的平均值定理

对点A = (x, y)第n阶微分,假定可能性

$ f(x + h, y + k) = f(x, y) + df(x, y) + \frac{d^{2} f(x, y)}{2!} + \cdots + \frac{d^{n} f(x, y)}{n!} + o \rho^{n} $

$ \rho = \sqrt{h^{2} + k^{2}} $

这样证明

$ F(t) = f(x + \frac{h}{\rho} t, y + \frac{k}{\rho} t) $

$ F(t) = F(0) + tF^{\prime}(0) + \cdots + \frac{t^{n} F^{(n)}(0)}{n!} + ot^{n} $

$ t = \rho $,这样 $ ot^{n} / t^{n} $AB线分的方向无关收敛

Taylor级数 如果f(x)各阶微分存在的话,区间内关于x

$ lim_ {n \to \infty}R_ {n} = 0 $

$ f(x) = lim_ {n \to \infty} \sum^{n}_ {\nu = 0}(x - a)^{\nu} \frac{f^{(\nu)}(a)}{\nu !} $

右边用无限级数写成为

$ f(x) = f(a) + (x - a)\frac{f^{\prime}(a)}{1!} + (x - a)^{2} \frac{f^{\prime \prime}(a)}{2!} + \cdots + (x - a)^{n} \frac{f^{(n)}(a)}{n!} + \cdots $

这样的形式是f(x)的Taylor级数。特别地a = 0时被称为Maclaurin级数

例子 $ f(x) = e^{x} $

a = 0且 $ f^{(n)} = 1 $则

$ R_ {n} = \frac{x^{n}}{n!} e^{\theta x}, \qquad 0 < \theta < 1 $

$ | R_ {n} | < \frac{| x |^{n}}{n!} e^{| x |} $

固定x则 $ \lim \frac{| x |^{n}}{n!} = 0 $,则 $ - \infty < x < \infty $

$ e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \cdots + \frac{x^{n}}{n!} + … $

特别当x = 1时,剩余项写入得

$ e = 1 + \frac{1}{1!} + \frac{1}{2!} + \cdots + \frac{1}{n!} + R_ {n+1}, \qquad R_ {n+1} = \frac{e^{\theta}}{(n+1)!} < \frac{3}{(n+1)!} $

e的计算 e是 $ lim{(1 + \frac{1}{n})^{n}} $,现在用 $ \frac{1}{n!} $小数第7位数计算,到n = 10有

$ \begin{aligned} 1 + \frac{1}{1!} + \frac{1}{2!} &= 2.5 \\ \frac{1}{3!} &= 0.1666666 \\ \frac{1}{4!} &= 0.0416666 \\ \frac{1}{5!} &= 0.0083333 \\ \frac{1}{6!} &= 0.0013888 \\ \frac{1}{7!} &= 0.0001984 \\ \frac{1}{8!} &= 0.0000248 \\ \frac{1}{9!} &= 0.0000027 \\ \frac{1}{10!} &= 0.0000002 \\ e &= 2.7182814 \end{aligned} $

这样得到e的近似值,$ n \ge 3 $有8项误差大于 $ \frac{1}{10^{7}} $,剩余项

$ R_ {11} < \frac{3}{11!} = \frac{1}{10!} \frac{3}{11} < \frac{1}{10^{7}} $

误差小于 $ \frac{1}{10^{6}} $,实际上 $ e = 2.718281828 \cdots $

e为无理数的证明 假设e是有理数,$ e = \frac{m}{n} $,m, n为整数,则 $ n! e $为整数,从而

$ n! R_ {n+1} = \frac{e^{\theta}}{n + 1} > 0 \qquad (0 < \theta < 1) $

必须为整数。从而

$ 1 \le \frac{e^{\theta}}{n+1} < \frac{3}{n+1} $

则 n + 1 < 3, n < 2, n = 1。然而e = m,则e必须是整数,而由于2 < e < 3,则矛盾

例子 $ \sin{x} = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - + \cdots $

$ \cos{x} = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} -+ \cdots $

x任意,这样 $ | R_ {n} | \to 0 $

极大极小

函数f(x)在点 $ x = x_ {0} $取值为 $ f(x_ {0}) $,其是 $ x_ {0} $的附近的值大(小),则 $ f(x_ {0}) $为最大值(最小值),$ x_ {0} $为f(x)的极大点(极小点),最大值、最小值统称为极值,这样 $ x_ {0} $为极值点

$ x_ {0} $为f(x)的极小点,则

$ 0 < | x - x_ {0} | < \delta \text{ 时 } f(x) - f(x_ {0}) > 0 $

如果把上式中的>换成 $ \ge $,则 $ f(x_ {0}) $ 为弱极小值

定理 点 $ x_ {0} $为函数f(x)定义域内一点

(1) f(x)在 $ x_ {0} $可微分时,$ x = x_ {0} $如果是f(x)的极值,则 $ f^{\prime}(x_ {0}) = 0 $。$ f^{\prime}(x_ {0}) $存在是极值的必要条件

(2) f(x)在 $ x_ {0} $处连续,$ x_ {0} $附近可微分,如果 $ f^{\prime}(x) $在 $ x = x_ {0} $处符号改变,$ f(x_ {0}) $为极值。详细地说,x逐渐增大地通过 $ x_ {0} $,$ f^{\prime} $符号从+变成-,则 $ f(x_ {0}) $为极大值,如果是从-变成+则是极小值

(3) f(x)在 $ x_ {0} $附近可微分,$ f^{\prime \prime}(x_ {0}) $存在,$ f^{\prime}(x_ {0}) = 0, f^{\prime \prime}(x_ {0}) > 0 $则 $ f(x_ {0}) $为极小值,$ f^{\prime}(x_ {0}) = 0, f^{\prime \prime}(x_ {0}) < 0 $则 $ f(x_ {0}) $为极大值

例子 一个平面两侧的亮点A, B,动点P在这个平面的两侧以一定的速度 $ c_ {1}, c_ {2} $运动,P从A到B最短的运动时间是什么?

解答 问题的要求是P需要在平面的两侧沿直线运动,问题简化如下

平面的直角坐标x轴上侧和下侧的点 $ A = (0, h_ {1}), B = (a, -h_ {2}), a > 0 $。P = (x, 0),则求使 $ \frac{AP}{c_ {1}} + \frac{BP}{c_ {2}} $值最小的P点位置

img

则可认为是区间为 $ 0 \le x \le a $的

$ f(x) = \frac{\sqrt{h^{2}_ {1} + x^{2}}}{c_ {1}} + \frac{\sqrt{h^{2}_ {2} + (a-x)^{2}}}{c_ {2}} $

的最小值

上述区间内,f(x)可微分,计算得

$ f^{\prime}(x) = \frac{1}{c_ {1}} \cdot \frac{x}{\sqrt{h^{2}_ {1} + x^{2}}} - \frac{1}{c_ {2}} \cdot \frac{a - x}{\sqrt{h^{2}_ {2} + (a-x)^{2}}} $

考察 $ f^{\prime}(x) = 0 $

$ f^{\prime}(x) $的第一项

$ \frac{1}{c_ {1}} \frac{x}{\sqrt{h^{2}_ {1} + x^{2}}} = \frac{\sin{\alpha}}{c_ {1}} $

这里$ [0, a] $区间x增大则单调增大,而第二项

$ \frac{1}{c_ {2}} \frac{a-x}{\sqrt{h^{2}_ {2} + (a - x)^{2}}} = \frac{\sin{\beta}}{c_ {2}} $

x增大则单调减小

则点 $ x_ {0} $应该是

$ \frac{\sin{\alpha}}{\sin{\beta}} = \frac{c_ {1}}{c_ {2}} \qquad \text{光的折射率} $

多变量的极值定义 两变量的 $ P_ {0} = (x_ {0}, y_ {0}) $的附近,$ P_ {0} $之外的点P = (x, y)

$ f(P) < f(P_ {0}) \qquad f(P) > f(P_ {0}) $

则 $ f(P_ {0}) $为极大(极小)值

这样的定义,$ f(x_ {0}, y_ {0}) $的极值,有

$ f_ {x}(x_ {0}, y_ {0}) = 0, \qquad f_ {y}(x_ {0}, y_ {0}) = 0 $

这是极值的必要条件

这样的条件下

$ f(x, y) - f(x_ {0}, y_ {0}) = \frac{1}{2} \{ a(x - x_ {0})^{2} + 2b(x - x_ {0})(y - y_ {0}) + c(y - y_ {0})^{2} \} + o \rho^{2} $

$ a = f_ {xx}(x_ {0}, y_ {0}), b = f_ {xy}(x_ {0}, y_ {0}), c = f_ {yy}(x_ {0}, y_ {0}), \rho = \sqrt{(x - x_ {0})^{2} + (y - y_ {0})^{2}} $

$ \rho $非常小,上式右边的符号决定了二次式

$ aX^{2} + 2bXY + cY^{2} \qquad (X = x - x_ {0}, Y = y - y_ {0}) $

这样有三种情况

(1) $ ac - b^{2} > 0 $。由二次式的符号,a和c的正负符号相同,a为正则是极小值,否则为极大值

(2) $ ac - b^{2} < 0 $。$ P_ {0} $的附近 $ f(P) > f(P_ {0}), f(P) < f(P_ {0}) $都有,$ f(P_ {0}) $不是极值

(3) $ ac - b^{2} = 0 $。这种情况下,需要考虑三阶及以上微分,但是用一般的理论相当复杂,现在为了简单起见,坐标进行变换,$ (x_ {0}, y_ {0}) = (0, 0) $,这样的情况是一个完全平方形,用a = 0, b = 0, c = 1做二三个例子

例1 $ z = y^{2} $。在(0, 0)点是z的最小值0,(x, 0)时z = 0,这是极小值

例2 $ z = y^{2} + x^{4} $。(0, 0)点极小值

例3 $ z = y^{2} - x^{3} $。(0, 0)点不是极值点,(x, y)下图左图,阴影部分z < 0,其他部分则z > 0

例4 $ z = (y - x^{2})(y - 2x^{2})。同上,下图右图

img

有三个以上变量的情况,$ f(x_ {1}, x_ {2}, \cdots, x_ {n}) $关于点 $ A = (a_ {1}, a_ {2}, \ldots, a_ {n}) $为极值的话,这样的点

$ f_ {x_ {1}} = 0, f_ {x_ {2}} = 0, \cdots, f_ {x_ {n}} = 0 $

为必要的,这样的话关于A $ f_ {x_ {i}x_ {j}}(A) = a_ {ij} $

$ f(a_ {1} + \xi_ {1}, \cdots, a_ {n} + \xi_ {n}) - f(a_ {1}, \cdots, a_ {n}) = \frac{1}{2} Q(\xi_ {1}, \xi_ {2}, \cdots, \xi_ {n}) + o \rho^{2} $

$ Q(\xi_ {1}, \xi_ {2}, \cdots, \xi_ {n}) = \sum^{n}_ {i, j = 1}a_ {ij}\xi_ {i}\xi_ {j} $

$ \rho = \sqrt{\sum^{n}_ {i=1} \xi^{2}_ {i}} $

$ a_ {ij} $的行列式为

$ D = \left | \begin{array}{cccc} a_ {11} & a_ {12} & \cdots & a_ {1n} \\ a_ {21} & a_ {22} & \cdots & a_ {2n} \\ \cdots & \cdots & \cdots & \cdots \\ a_ {n1} & a_ {n2} & \cdots & a_ {nn} \end{array} \right | \ne 0 \qquad (a_ {ij} = a_ {ji}) $

这样我们可以对A是否有极值点做断言

如果二次形式Q符号固定,则为正时极小,负时极大。它的判别法可通过D的前导行列式的符号获得。$ D_ {k} $全为正时极小,$ D_ {k} $的符号为 $ (-1)^{k} $时极大

二次形式Q符号不固定时,f(A)没有极值。$ D \ne 0 $时,$ D_ {k} $不符合上述条件,就可能会这样

D = 0时,只有二阶微分时不能断言

例子 求给定周长的三角形的最大面积

解 周长为2p,边为x, y, z,面积为S,S用f(x, y)带入得

$ f(x, y) = p(p - x)(p - y)(p -z), \qquad z = 2p - x - y $

独立变量为x,y,其变动范围为

$ 0 < x < p, \qquad, 0 < y < p, \qquad p < x + y $

这样固定三角形的一条边y,则从几何学上我们可知最大面积为x = z时。这样 $ f_ {x} = 0 $,同样 $ f_ {y} = 0 $时y = z。这样

$ x = y = z = \frac{2}{3} p $

img

从这个关系我们得出一个正的直角三角形,但这是不可能的。我们知道“如果有一个最大值,它是一个等边三角形“,由于f(x, y)是连续的,但它不是闭区间,不能保证最大值的存在,这是问题的关键

幸好根据Weierstrass定理,我们可以越过这个障碍,我们构造一个封闭区域

$ 0 \le x \le p, \qquad 0 \le y \le p \qquad p \le x + y $

在三角形的极端情况下,它只不过是一条双线段,因此即使是面积为0的线段也参与了最大值的竞争。现在在这个封闭区间中有最大值。然而,在区间的边界处f = 0,在区域内f > 0。因此最大值在区域内部,则 $ x = y = z = \frac{2}{3} p $时是最大值

我们计算这里的二阶导数,我们使用上一节的符号

$ a = c = - \frac{2}{3} p^{2}, \qquad b = - \frac{1}{3} p^{2}, \qquad ac - b^{2} > 0 $

则这是极大值点,但极大值不是最大值,所以这个不能解决问题

例子 行列式的最大值(Hadamard定理)。例如,n阶行列式

$ D = \left| \begin{array}{cccc} a_ {1} & b_ {1} & \cdots & l_ {1} \\ a_ {2} & b_ {2} & \cdots & l_ {2} \\ \cdots & \cdots & \cdots & \cdots \\ a_ {n} & b_ {n} & \cdots & l_ {n} \end{array} \right| $

的绝对值的最大值

$ a^{2}_ {i} + b^{2}_ {i} + \cdots + l^{2}_ {i} = s^{2}_ {i} \qquad (i = 1, 2, \cdots, n) $

要求这个条件下。(但是 $ s_ {i} $是正数),目标有以下关系表达式

$ | D | \le s_ {1} s_ {2} \cdots s_ {n} $

D是 $ n^{2} $个变量 $ a_ {1}, \cdots, l_ {n} $的多项式。这种条件下独立变量有n(n - 1)个。现在考察n次元球面上的点 $ P_ {i} $

$ P = (P_ {1}, P_ {2}, \cdots, P_ {n}) $

的组合P的函数的行列式,D = D(P)是关于P连续,P变量变动的闭区间,这样这样的点都是内点。D的最大值、最小值存在,这样可从哪个D的极值中找到

$ D = a_ {i}A_ {i} + b_ {i}B_ {i} + \cdots + l_ {i}L_ {i} $

$ A_ {i}, B_ {i}, \cdots, L_ {i} $是 $ a_ {i}, b_ {i}, \cdots, l_ {i} $的余因子,这样D的第i行以外的组成的多项式,D的极值的必要条件是

$ \frac{\partial D}{\partial a_ {i}} = A_ {i} + L_ {i} \frac{\partial l_ {i}}{\partial a_ {i}} = A_ {i} - L_ {i}\frac{a_ {i}}{l_ {i}} = 0 $

$ b_ {i}, c_ {i}, \cdots $相关也相同,则

$ \frac{A_ {i}}{a_ {i}} = \frac{B_ {i}}{b_ {i}} = \cdots = \frac{L_ {i}}{l_ {i}} $

$ i \ne k $有

$ a_ {k}A_ {i} + b_ {k}B_ {i} + \cdots + l_ {k}L_ {i} = 0 $

则有

$ a_ {i}a_ {k} + b_ {i}b_ {k} + \cdots + l_ {i}l_ {k} = 0 $

由以上的式子的值,D的绝对值已确定。这样

$ D^{2} = \left| \begin{array}{cccc} s^{2}_ {1} & 0 & \cdots & 0 \\ 0 & s^{2}_ {2} & \cdots & 0 \\ \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & \cdots & s^{2}_ {n} \end{array} \right| = (s_ {1}s_ {2} \cdots s_ {n})^{2} $

$ D = \pm s_ {1} s_ {2} \cdots s_ {n} $

这样D的最大值为 $ s_ {1}s_ {2} \cdots s_ {n} $(最小值为 $ - s_ {1}s_ {2}\cdots s_ {n} $ ),这是它的极大极小值

切线和曲率

本章的最后我们讨论切线和曲率,这是微分法的起源。为了简化描述,使用矢量法的记号,矢量作为具有一定大小和方向的量是已知的。从直角坐标原点O到点P = (x, y, z)的线段OP所表示的向量用 $ \upsilon = (x, y, z) $表示,$ | \upsilon | = \sqrt{x^{2} + y^{2} + z^{2}} $

两个向量 $ u = (x_ {1}, y_ {1}, z_ {1}), \upsilon = (x_ {2}, y_ {2}, z_ {2}) $,定义两种乘法

(1) 标量乘法

$ u \upsilon = x_ {1}x_ {2} + y_ {1}y_ {2} + z_ {1}z_ {2} $

$ u, \upsilon $不为零向量时,它们的方向余弦

$ \frac{x_ {1}}{| u |}, \frac{y_ {1}}{| u |}, \frac{z_ {1}}{| u |}, \frac{x_ {2}}{| \upsilon |}, \frac{y_ {2}}{| \upsilon |}, \frac{z_ {2}}{| \upsilon |} $

$ u, \upsilon $间的夹角为 $ \theta $

$ \cos{\theta} = \frac{x_ {1}x_ {2} + y_ {1}y_ {2} + z_ {1}z_ {2}}{| u | | \upsilon |} $

$ u \upsilon = | u | | \upsilon | \cos{\theta} $

这是标量乘积几何上的意义。$ u, \upsilon $互相垂直时

$ u \upsilon = 0 $

$ uu = x^{2}_ {1} + y^{2}_ {1} + z^{2}_ {1} = | u |^{2} $

对于标量积,交换律和分配律成立

$ u \upsilon = \upsilon u, \qquad (u_ {1} + u_ {2}) \upsilon = u_ {1} \upsilon + u_ {2} \upsilon $

x,y,z为变量t的函数时,向量

$ OP = u = (x, y, z) $

这时,$ t + \Delta t $对应的向量

$ OP^{\prime} = u + \Delta u = (x + \Delta x, y + \Delta y, z + \Delta z) $

则 $ \Delta u $是向量 $ PP^{\prime} $

$ \Delta u = (\Delta x, \Delta y, \Delta z) $

img

x,y,z可微分的话,$ \Delta t \to 0 $时向量

$ \frac{\Delta u}{\Delta t} = (\frac{\Delta x}{\Delta t}, \frac{\Delta y}{\Delta t}, \frac{\Delta z}{\Delta t}) $

的极限值是一个常量。这样的极限值 $ \dot{u} = \frac{du}{dt} $写为

$ \dot{u} = (\dot{x}, \dot{y}, \dot{z}) $

由向量积的定义,有

$ \frac{d}{dt} (u \upsilon) = \dot{u} \upsilon + u \dot{\upsilon} $

特别地,u常常是单位向量( $ | u | = 1 $),这样当只有方向改变时,uu = 1,$ u \dot{u} = 0 $,这里 $ \dot{u} \ne 0 $时,u和 $ \dot{u} $相互垂直

(2) 向量积 $ u \times \upsilon $,两个向量

$ u = (x_ {1}, y_ {1}, z_ {1}), \upsilon = (x_ {2}, y_ {2}, z_ {2}) $

坐标的行列

$ \left( \begin{array}{ccc} x_ {1} & y_ {1} & z_ {1} \\ x_ {2} & y_ {2} & z_ {2} \end{array} \right) $

坐标为三个行列式的向量

$ \omega = (y_ {1}z_ {2} - y_ {2}z_ {1}, z_ {1}x_ {2} - z_ {2}x_ {1}, x_ {1}y_ {2} - x_ {2}y_ {1}) $

这个是 $ u, \upsilon $的向量积,写为 $ u \times \upsilon $。其几何学的意义如下

现在简单的 $ \omega = (x, y, z) $写为

$ xx_ {1} + yy_ {1} + zz_ {1} = 0, \qquad xx_ {2} + yy_ {2} + zz_ {2} = 0 $

$ \omega u = 0, \qquad \omega \upsilon = 0 $

则 $ \omega $ 与u 和 $ \upsilon $相互垂直

$ \left | \begin{array}{ccc} x & y & z \\ x_ {1} & y_ {1} & z_ {1} \\ x_ {2} & y_ {2} & z_ {2} \end{array} \right | = x^{2} + y^{2} + z^{2} = | \omega |^{2} $

是一个平行六面体的体积,它的三条边为向量u, $ \upsilon, \omega $,因为它是正的,所以$ u, \upsilon, \omega $是一个与坐标轴一致的系统(右手螺旋),由于它的体积等于 $ | \omega |^{2} $,$ | \omega | $为 $ u, \upsilon $为边的平行四边形的面积

img

特别地当 $ u, \upsilon $方向一致时

$ u \times \upsilon = 0 $

根据上面的定义,向量积的交换律不成立

$ u \times \upsilon = - \upsilon \times u $

有分配律

$ (u_ {1} + u_ {2}) \times \upsilon = u_ {1} \times \upsilon + u_ {2} \times \upsilon $

$ \upsilon \times (u_ {1} + u_ {2}) = \upsilon \times u_ {1} + \upsilon \times u_ {2} $

$ u, \upsilon $为t的函数时

$ \frac{d}{dt} (u \times \upsilon) = \dot{u} \times \upsilon + u \times \dot{\upsilon} $

(3) 此外,三个向量

$ u = (x_ {1}, y_ {1}, z_ {1}), \qquad \upsilon = (x_ {2}, y_ {2}, z_ {2}), \qquad \omega = (x_ {3}, y_ {3}, z_ {3}) $

平行六面体的体积由 $ (u, \upsilon, \omega) $来表示

$ (u, \upsilon, \omega) = \left | \begin{array}{ccc} x_ {1} & y_ {1} & z_ {1} \\ x_ {2} & y_ {2} & z_ {2} \\ x_ {3} & y_ {3} & z_ {3} \end{array} \right | $

如果三个单位向量i, j, k(两两互相垂直)是右旋螺旋,则它们构成一个单位系统,在这种情况下

$ i^{2} = j^{2} = k^{2} = 1, \qquad i \times i = j \times j = k \times k = 0 $

$ ij = ji = 0, \qquad i \times j = k = -j \times i $

$ jk = kj = 0, \qquad j \times k = i = -k \times j $

$ ki = ik = 0, \qquad k \times i = j = - i \times k $

$ (i, j, k) = 1 $

现在曲线C用变量t来表示,C上的点P = (x, y, z)的坐标x, y, z是t的函数,向量OP用 $ \upsilon $表示,现在曲线上 $ t + \Delta t $对应的点 $ P^{\prime} = (x + \delta x, y + \delta y, z + \delta z) $,则向量 $ OP^{\prime} $为 $ \upsilon + \delta \upsilon $,则 $ \delta \upsilon $为向量 $ PP^{\prime} $

$ \delta \upsilon = (\delta x, \delta y, \delta z) $

现在x, y, z三阶可微分的话,则泰勒公式

$ \delta x = \dot{x} \delta t + \ddot{x} \frac{\delta t^{2}}{2} + \dddot{x}\frac{\delta t^{3}}{6} + o \delta t^{3} $

$ \delta y, \delta z $也一样,这样可写为

$ \delta \upsilon = \dot{\upsilon} \delta t + \ddot{\upsilon} \frac{\delta t^{2}}{2} + \dddot{\upsilon} \frac{\delta t^{3}}{6} + o \delta t^{3} $

$ \dot{\upsilon} = (\dot{x}, \dot{y}, \dot{z}) $,$ \ddot{\upsilon}, \dddot{\upsilon} $也一样,$ \delta t \to 0 $时 $ | o | \to 0 $,$ \dot{\upsilon}, \ddot{\upsilon}, \dddot{\upsilon} $三个向量关于曲线上点P在几何学上有重要的性质

对向量 $ \dot{\upsilon} = (\dot{x}, \dot{y}, \dot{z}) $,

$ | \dot{\upsilon} | = \sqrt{\dot{x}^{2} + \dot{y}^{2} + \dot{z}^{2}} $

点P在曲线C的切线的方向余弦为

$ \frac{\cos{\alpha}}{\dot{x}} = \frac{\cos{\beta}}{\dot{y}} = \frac{\cos{\gamma}}{\dot{z}} = \frac{1}{| \dot{\upsilon} |} $

但是 $ \dot{\upsilon} = 0 $时,要排除 $ \dot{x}, \dot{y}, \dot{z} $同时为0的点(特异点)

如果以曲线C的一个固定点计算的弧长s代替t作为参赛,则结果很简明

$ \delta \upsilon = \upsilon^{\prime} \delta s + \upsilon^{\prime \prime} \frac{\delta s^{2}}{2} + \upsilon^{\prime \prime \prime} \frac{\delta s^{3}}{6} + o \delta s^{3} $

弧长理论将在后面介绍,这里弧 $ PP^{\prime} $和弦 $ PP^{\prime} $的比在距离 $ PP^{\prime} \to 0 $时收敛到1,即

$ \frac{\delta x^{2} + \delta y^{2} + \delta z^{2}}{\delta s^{2}} \to 1 $

$ \frac{dx}{ds} = \cos{\alpha}, \qquad \frac{dy}{ds} = \cos{\beta}, \qquad \frac{dz}{ds} = \cos{\gamma} $

则有

$ x^{\prime 2} + y^{\prime 2} + z^{\prime 2} = 1 \Rightarrow | \upsilon^{\prime} | = 1 $

s为变量,$ \upsilon^{\prime} $切线上关于s的增加方向上取的单位向量。$ | \upsilon^{\prime} | = 1 $的话,假定$ \upsilon^{\prime \prime} \ne 0 $,则 $ \upsilon^{\prime \prime} $和 $ \upsilon^{\prime} $垂直。P上关于 $ \upsilon^{\prime \prime} $平行的直线是曲线C的主法线,含有$ \upsilon^{\prime}, \upsilon^{\prime \prime} $的平面是接触平面

通过P点的任意平面的方程式的标准形为

$ l(X - x) + m(Y - y) + n(Z - z) = 0 \qquad (l^{2} + m^{2} + n^{2} = 1) $

p = (l, m, n)是平面的法线上的单位向量。曲线上的点 $ P^{\prime} = (x + \delta x, y + \delta y, z + \delta z) $到平面的距离为

$ l\delta x + m \delta y + n \delta z $

等于 $ p \cdot \delta \upsilon $的标量积。但 $ \delta \upsilon = (\delta x, \delta y, \delta z) $,这样

$ p \upsilon^{\prime} \delta s + p \upsilon^{\prime \prime} \frac{\delta s^{2}}{2} + o \delta s^{2} $

p和 $ \upsilon^{\prime}, \upsilon^{\prime \prime} $垂直,即含有$ \upsilon^{\prime}, \upsilon^{\prime \prime} $的平面时$ \delta \upsilon $为高位的小数。这就意味着是接触平面

设P关于$ P^{\prime} $切线间的夹角为 $ \alpha $,$ \delta \alpha $为向量 $ \upsilon^{\prime}, \upsilon^{\prime} + \delta \upsilon^{\prime} $间的夹角,$ \upsilon^{\prime} $的长为常量1时,$ \delta s \to 0 $

$ \frac{\delta \alpha}{| \delta \upsilon^{\prime} |} \to 1 $

img

然而

$ \frac{\delta \upsilon^{\prime}}{\delta s} \to \upsilon^{\prime \prime}, \qquad \frac{| \delta \upsilon^{\prime} |}{\delta s} \to | \upsilon^{\prime \prime} | $

则有

$ \frac{\delta \alpha}{\delta s} \to | \upsilon^{\prime \prime} | \Rightarrow \frac{d \alpha}{d s} = | \upsilon^{\prime \prime} | $

这里 $ \frac{d \alpha}{ds} $为C的切线方向随弧长变动的改变率,这里点P上的曲率,它的倒数 $ \rho $为曲率半径。即

$ \frac{1}{\rho} = \frac{d \alpha}{ds} = | \upsilon^{\prime \prime} | = \sqrt{(\frac{d^{2}x}{ds^{2}})^{2} + (\frac{d^{2}y}{ds^{2}})^{2} + (\frac{d^{2}z}{ds^{2}})^{2}} $

这是P上接触平面的垂线即陪法线。现在,切线、主法线、陪法线上单位用i,j,k,记为

$ \left. \begin{array}{l} i = \upsilon^{\prime}, \\ j = \rho \upsilon^{\prime \prime}, \\ k = i \times j. \end{array} \right\} $

然而

$ k^{\prime} = i^{\prime} \times j + i \times j^{\prime} $

有$ i^{\prime} = \upsilon^{\prime \prime} $与j平行,则 $ i^{\prime} \times j = 0 $,从而

$ k^{\prime} = i \times j^{\prime} $

则 $ k^{\prime} $与i垂直。而 $ | k | = 1 $得 $ k^{\prime} $ 与k垂直,从而 $ k^{\prime} $与j平行。(这里假定 $ k^{\prime} \ne 0 $)跟 $ | \upsilon^{\prime \prime} | $一样,$ | k^{\prime} | $是s随陪法线方向的变动率,即接触平面绕切线旋转的角度的变动率。由于将s的增加方向定为切线的正方向,因此这种回旋可以定正负。$ k^{\prime} $与j平行,从$ | j | = 1 $得到 $ k^{\prime} = \pm | k^{\prime} | j $,现在

$ k^{\prime} = - \frac{1}{\tau} j $

$ \frac{1}{\tau} $为切线C的第二曲率即扭率,它的倒数为扭率半径

点P在曲线C上移动时,单位系统(i,j,k)以右手还是左手变动取决于 $ \tau $的正负

任意向量用i,j,k组合表达为ai+bj+ck,现在 $ j^{\prime} $考察

$ j = k \times i, j^{\prime} = k^{\prime} \times i + k \times i^{\prime} $

$ i^{\prime} = \upsilon^{\prime \prime} = \frac{1}{\rho} j, k^{\prime} = - \frac{1}{\tau} j $,则 $ j \times i = -k, k \times j = -i $

$ j^{\prime} = - \frac{1}{\rho} i + \frac{1}{\tau} k $

上述 $ i^{\prime}, j^{\prime}, k^{\prime} $一起写为

$ \left\{ \begin{array}{ccccc} i^{\prime} & = & & \frac{1}{\rho}j, & \\ j^{\prime} & = & - \frac{1}{\rho}i & & + \frac{1}{\tau} k, \\ k^{\prime} & = & & -\frac{1}{\tau} j & \end{array} \right. $

这是Frenet公式,i, j, k的常量部分是切线、主法线、陪法线的方向余弦,$ \prime $是弧长s的微分表示,利用 $ \upsilon^{\prime \prime} = \frac{1}{\rho} j $

$ \begin{aligned} \upsilon^{\prime \prime \prime} &= - \frac{\rho^{\prime}}{\rho^{2}} j + \frac{1}{\rho} j^{\prime} \\ &= - \frac{1}{\rho^{2}} i - \frac{\rho^{\prime}}{\rho^{2}} j + \frac{1}{\rho \tau} k \end{aligned} $

从而

这样作为行列式

$ \begin{aligned} (\upsilon^{\prime}, \upsilon^{\prime \prime}, \upsilon^{\prime \prime \prime}) &= (i, \frac{1}{\rho} j, - \frac{1}{\rho^{2}} i - \frac{\rho^{\prime}}{\rho^{2}}j + \frac{1}{\rho \tau} k) \\ &= (i, \frac{1}{\rho} j, \frac{1}{\rho \tau} k) \\ &= (i, \frac{1}{\rho}j, \frac{1}{\rho \tau}k) = \frac{1}{\rho^{2} \tau}(i, j, k) = \frac{1}{\rho^{2} \tau} \end{aligned} $

从而

$ \frac{1}{\tau} = \rho^{2} (\upsilon^{\prime}, \upsilon^{\prime \prime}, \upsilon^{\prime \prime \prime}) = \frac{\left| \begin{array}{ccc} x^{\prime} & y^{\prime} & z^{\prime} \\ x^{\prime \prime} & y^{\prime \prime} & z^{\prime \prime} \\ x^{\prime \prime \prime} & y^{\prime \prime \prime} & z^{\prime \prime \prime} \end{array} \right |}{x^{\prime \prime 2} + y^{\prime \prime 2} + z^{\prime \prime 2}} $

这是s作为变量的扭率

任意变量t相关 $ \rho, \tau $计算,t的微分用 $ \dot{} $表示

$ \dot{\upsilon} = \upsilon^{\prime} \frac{ds}{dt} $

$ \ddot{\upsilon} = \upsilon^{\prime \prime} (\frac{ds}{dt})^{2} + \upsilon^{\prime} \frac{d^{2}s}{dt^{2}} $

$ \dddot{\upsilon} = \upsilon^{\prime \prime \prime} (\frac{ds}{dt})^{3} + 3 \upsilon^{\prime \prime} \frac{ds}{dt} \frac{d^{2}s}{dt^{2}} + \upsilon^{\prime} \frac{d^{3}s}{dt^{3}} $

$ \upsilon^{\prime}, \upsilon{\prime \prime} $互相垂直,上面第二式 $ \ddot{\upsilon} $是两个互相垂直的向量的分解的和。$ \upsilon^{\prime} \frac{d^{2}s}{dt^{2}} $这部分,和C的切线平行,它的大小为 $ \frac{d^{2}s}{dt^{2}} $( $ | \upsilon^{\prime} | = 1 $),另一部分 $ \upsilon^{\prime \prime}(\frac{ds}{dt})^{2} $,与C的主法线平行,大小为 $ \frac{1}{\rho} (\frac{ds}{dt})^{2} $($ | \upsilon^{\prime \prime} | = \frac{1}{\rho} $)。t为时间,加速度 $ \ddot{\upsilon} $这样的两成分分解得

$ | \ddot{\upsilon} |^{2} = \ddot{x}^{2} + \ddot{y}^{2} + \ddot{z}^{2} = \frac{1}{\rho} (\frac{ds}{dt})^{4} + (\frac{d^{2}s}{dt^{2}})^{2} $

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$ (\frac{ds}{dt})^{2} = \dot{x}^{2} + \dot{y}^{2} + \dot{z}^{2} $微分得

$ \frac{ds}{dt} \frac{d^{2}s}{dt^{2}} = \dot{x}\ddot{x} + \dot{y}\ddot{y} + \dot{z}\ddot{z} $

从而

$ \begin{aligned} \frac{1}{\rho^{2}} (\frac{ds}{dt})^{6} = | \ddot{\upsilon} |^{2} (\frac{ds}{dt})^{2} - (\frac{ds}{dt})^{2} (\frac{d^{2}s}{dt^{2}})^{2} &= (\dot{x}^{2} + \dot{y}^{2} + \dot{z}^{2})(\ddot{x}^{2} + \ddot{y}^{2} + \ddot{z}^{2}) - (\dot{x} \ddot{x} + \dot{y}\ddot{y} + \dot{z}\ddot{z})^{2} \\ &= \left | \begin{array}{cc} \dot{y} & \dot{z} \\ \ddot{y} & \ddot{z} \end{array} \right |^{2} + \left | \begin{array}{cc} \dot{z} & \dot{x} \\ \ddot{z} & \ddot{x} \end{array} \right |^{2} + \left | \begin{array}{cc} \dot{x} & \dot{y} \\ \ddot{x} & \ddot{y} \end{array} \right |^{2} \end{aligned} $

$ \frac{1}{\rho} = \frac{| \dot{\upsilon} \times \ddot{\upsilon} |}{| \dot{\upsilon} |^{3}} $

$ (\dot{\upsilon}, \ddot{\upsilon}, \dddot{\upsilon}) = (\frac{ds}{dt})^{6}(\upsilon^{\prime}, \upsilon^{\prime \prime}, \upsilon^{\prime \prime \prime}) $

$ \frac{1}{\tau} = \frac{(\dot{\upsilon}, \ddot{\upsilon}, \dddot{\upsilon})}{| \dot{\upsilon} \times \ddot{\upsilon} |^{2}} $

例子 螺旋线

$ x = a \cos{t}, \qquad y = a\sin{t}, \qquad z = ht \qquad (a > 0) $

$ \dot{x} = -a \sin{t}, \qquad \dot{y} = a \cos{t}, \qquad \dot{z} = h $

$ \ddot{x} = -a \cos{t}, \qquad \ddot{y} - -a \sin{t}, \qquad \ddot{z} = 0 $

$ \dddot{x} = a \sin{t}, \qquad \dddot{y} = -a \cos{t}, \qquad \dddot{z} = 0 $

$ | \dot{\upsilon} | = \sqrt{\dot{x}^{2} + \dot{y}^{2} + \dot{z}^{2}} = \sqrt{a^{2} + h^{2}} $

$ | \dot{\upsilon} \times \ddot{\upsilon} | = \sqrt{a^{2}h^{2} \sin^{2}{t} + a^{2}h^{2} \cos^{2}{t} + a^{4}} = a \sqrt{a^{2} + h^{2}} $

$ (\dot{\upsilon}, \ddot{\upsilon}, \dddot{\upsilon}) = a^{2}h $

$ \frac{1}{\rho} = \frac{a}{a^{2} + h^{2}}, \qquad \frac{1}{\tau} = \frac{h}{a^{2} + h^{2}} $

即曲率和扭率是一定的,$ \tau, h $同符号,正负号区分右旋螺纹和左旋螺纹

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平面曲线z = 0时,泰勒展开取二次项

$ \delta \upsilon = \upsilon^{\prime} \delta s + \upsilon^{\prime \prime} \frac{\delta s^{2}}{2} + o \delta s^{2} $

这时 $ | \upsilon^{\prime} | = 1 $,从而 $ \theta $是x轴的正方向,切线的正方向的角

$ | \upsilon^{\prime \prime} = | \frac{d \theta}{ds} | $

弧长s对应切线方向的变动率,即曲率

$ \frac{1}{\rho} = \frac{d \theta}{ds} $

$ \rho $为曲率半径,然而

$ \upsilon^{\prime} = (x^{\prime}, y^{\prime}) = (\cos{\theta}, \sin{\theta}) $

微分s得

$ \upsilon^{\prime \prime} = (x^{\prime \prime}, y^{\prime \prime}) = (- \sin{\theta \frac{d \theta}{ds}}, \cos{\theta \frac{d \theta}{ds}}) = \frac{1}{\rho} (- \sin{\theta}, \cos{\theta}) = \frac{1}{\rho} (-y^{\prime}, x^{\prime}) $

$ \frac{1}{\rho} = \frac{- x^{\prime \prime}}{y^{\prime}} = \frac{y^{\prime \prime}}{x^{\prime}} $

用一般的变量t表示为

$ \left | \begin{array}{cc} \dot{x} & \dot{y} \\ \ddot{x} & \ddot{y} \end{array} \right | = \left | \begin{array}{cc} x^{\prime} & y^{\prime} \\ x^{\prime \prime} & y^{\prime \prime} \end{array} \right | (\frac{ds}{dt})^{3} = \left | \begin{array}{cc} x^{\prime} & y^{\prime} \\ - \frac{y^{\prime}}{\rho} & \frac{x^{\prime}}{\rho} \end{array} \right | (\frac{ds}{dt})^{3} = \frac{1}{\rho} (\frac{ds}{dt})^{3} $

从而

$ \frac{1}{\rho} = \frac{\dot{x}\ddot{y} - \ddot{x}\dot{y}}{\dot{s}^{3}} $

用之前的记号

$ \frac{1}{\rho} = \frac{(\dot{\upsilon}, \ddot{\upsilon})}{| \dot{\upsilon} |^{3}} $

总之,用独立变量无关的微分记号得

$ \frac{1}{\rho} = \frac{dxd^{2}y - d^{2}xdy}{(dx^{2}+dy^{2})^{\frac{3}{2}}} $

特别地曲线为y = f(x),有

$ \frac{1}{\rho} = \frac{\frac{d^{2}y}{dx^{2}}}{(1 + (\frac{dy}{dx})^{2})^{\frac{3}{2}}} $

点P与曲线C相交,与C相对于切线在同一侧,半径等于 $ | \rho | $的圆为曲率圆,其中心 $ (\xi, \eta) $为曲率的中心

$ \xi = x - \rho \sin{\theta}, \qquad \eta = y + \rho \cos{\theta} $

如果$ \rho $和 $ \frac{d^{2}y}{dx^{2}} $同符号,下图右图曲率中心在曲线的凹侧

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曲线C的曲率中心 $ (\xi, \eta) $的轨迹是曲线E的话,特别地C的弧长s作为变量,E为

$ \xi = x - \rho y^{\prime}, \qquad \eta = y + \rho x^{\prime} $

对s的微分的话

$ \xi^{\prime} = x^{\prime} - \rho^{\prime}y^{\prime} - \rho y^{\prime \prime} = - \rho^{\prime}y^{\prime} $

$ \eta^{\prime} = y^{\prime} + \rho^{\prime}x^{\prime} + \rho x^{\prime \prime} = \rho^{\prime}x^{\prime} $

从而

$ \xi^{\prime} x^{\prime} + \eta^{\prime}y^{\prime} = 0 $

即源曲线C的切线,和对应的点的E的切线互相垂直。则C的法线与E的曲率中心相接,E为源曲线C的法线的包络线

现在E的弧长为 $ \sigma $的话

$ (\frac{d \sigma}{ds})^{2} = (\frac{d \xi}{ds})^{2} + (\frac{d \eta}{ds})^{2} = \rho^{\prime 2} (x^{\prime 2} + y^{\prime 2}) = \rho^{\prime 2} $

即 $ \sigma^{\prime} = \pm \rho^{\prime} $。则在适当的方向上测量E的弧长,在 $ \rho^{\prime} \ne 0 $的各范围内,$ \sigma^{\prime} = \rho^{\prime} $,$ \sigma_ {0} $对应 $ \rho_ {0} $,有 $ \sigma - \sigma_ {0} = \rho - \rho_ {0} $。这样的条件下,E的两点间弧长对应C的亮点曲率半径的差

如果在E上缠绕一根线,在拉紧的同时松开P端,使线不下垂,P将画出C。因此,C是E的渐开线。反之,E是C的闭合线。C给定的时候,其渐开线E是固定的,但给定的渐开线E其C有无穷个

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例子 当一个圆在另一个圆的圆周上或直线上滚动而不滑动,粘在运动圆上的一点的轨道是广义上称为摆线的曲线。最简单的应用是圆周上一点画的曲线为一个在恒定直线上滚动的圆,这是一条法线(狭义的)摆线(所谓的抓线)。如果半径为a,旋转角度为t,则直线为x轴,t = 0处圆圆周上的不动点P为坐标原点,抓线表示如下,以t作为变量

$ x = a(t - \sin{t}), \qquad y = a(1 - \cos{t}) $

$ dx = a(1 - \cos{t})dt, \qquad dy = a \sin{t}dt $

$ ds = \sqrt{dx^{2} + dy^{2}} = \sqrt{2a^{2}(1 - \cos{t})}dt = 2a | \sin{\frac{t}{2}} | dt $

$ d^{2}x = a \sin{t}dt^{2}, \qquad d^{2}y = a \cos{t}dt^{2} $

$ dxd^{2}y - dyd^{2}x = a^{2} \left | \begin{array}{cc} 1 - \cos{t} & \sin{t} \\ \sin{t} & \cos{t} \end{array} \right | dt^{3} = a^{2} (\cos{t} - 1)dt^{3} = -2a^{2} \sin^{2}{\frac{t}{2}} dt^{3} $

$ \rho = \frac{ds^{3}}{dxd^{2}y - dyd^{2}x} = -4a | \sin{\frac{t}{2}} | $

$ \xi = x - \rho \frac{dy}{ds} = a(t+ \sin{t}) $

$ \eta = y + \rho \frac{dx}{ds} = a(-1 + \cos{t}) $

因此闭合线和原曲线全等,具体来说,闭合线的弧 $ AB^{\prime}, B^{\prime}C $和原曲线的弧BC, AB全等。$ t = 0, t = \pi $对应 $ \rho = 0, \rho = -4a $,弧 $ AB^{\prime} $ 长为4a,从而摆线ABC的全长为8a

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例子 椭圆的闭合线。椭圆

$ x = a \cos{t}, \qquad y = b \sin{t} $

其计算

$ \rho = \frac{(a^{2} \sin^{2}{t} + b^{2} \cos^{2}{t})^{\frac{3}{2}}}{ab} $

$ \xi = \frac{a^{2} - b^{2}}{a} \cos^{3}{t}, \qquad \eta = - \frac{a^{2} - b^{2}}{b} \sin^{3}{t} $

消掉t的闭合线方程式为

$ (a \xi)^{\frac{2}{3}} + (b \eta)^{\frac{2}{3}} = (a^{2} - b^{2})^{\frac{2}{3}} $

如下图中星形。原曲线曲率的极大极小点为闭合线的尖点

椭圆的四个法线是从E里面的点画的,两条法线是从E外面的点画的

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例子 圆的渐开线,半径为1

img

$ \xi = \cos{t}, \qquad \eta = \sin{t} $

则渐开线为

$ x = \cos{t} + t \sin{t}, \qquad y = \sin{t} - t \cos{t} $