【第九章 习题31】
设在点a∈R2\mathbf{a} \in R^{2}a∈R2的某邻域内，fff三阶连续可微，fff的梯度在a\mathbf{a}a点是0\mathbf{0}0，但fff的诸二阶导数在aaa点不全为零。说明这时从fff在a\mathbf{a}a的（二次）Taylor多项式怎样判断，fff在a\mathbf{a}a点有局部极大或局部极小，或既没有局部极大又没有局部极小。
把对于R2R^{2}R2的这个结果推广到RnR^{n}Rn。
【解】
a. 首先对于二元函数，由于fff的梯度在a\mathbf{a}a点是0\mathbf{0}0，所以
f(a+x)−f(a)=12!(D11f)(a)x12+12!(D22f)(a)x22+12!(D12)(a)x1x2+12!(D21)(a)x1x2+r(x){f\left( \mathbf{a + x} \right) - f\left( \mathbf{a} \right) }{= \frac{1}{2!}\left( D_{11}f \right)\left( \mathbf{a} \right)x_{1}^{2} + \frac{1}{2!}\left( D_{22}f \right)\left( \mathbf{a} \right)x_{2}^{2} + \frac{1}{2!}\left( D_{12} \right)\left( \mathbf{a} \right)x_{1}x_{2} + \frac{1}{2!}\left( D_{21} \right)\left( \mathbf{a} \right)x_{1}x_{2} + r(\mathbf{x})}f(a+x)−f(a)=2!1(D11f)(a)x12+2!1(D22f)(a)x22+2!1(D12)(a)x1x2+2!1(D21)(a)x1x2+r(x)
两边同时除以∣x∣2|x|^{2}∣x∣2得
f(a+x)−f(a)∣x∣2=12(D11f)(a)x12∣x∣2+(D22f)(a)x22∣x∣2+(D12)(a)x1x2∣x∣2+(D21)(a)x1x2∣x∣2+r(x)∣x∣2\frac{f\left( \mathbf{a + x} \right) - f\left( \mathbf{a} \right)}{\left| \mathbf{x} \right|^{2}} = \frac{1}{2}\left\lbrack \left( D_{11}f \right)\left( \mathbf{a} \right)\frac{x_{1}^{2}}{\left| \mathbf{x} \right|^{2}} + \left( D_{22}f \right)\left( \mathbf{a} \right)\frac{x_{2}^{2}}{\left| \mathbf{x} \right|^{2}} + \left( D_{12} \right)\left( \mathbf{a} \right)\frac{x_{1}x_{2}}{\left| \mathbf{x} \right|^{2}} + \left( D_{21} \right)\left( \mathbf{a} \right)\frac{x_{1}x_{2}}{\left| \mathbf{x} \right|^{2}} \right\rbrack + \frac{r\left( \mathbf{x} \right)}{\left| \mathbf{x} \right|^{2}}∣x∣2f(a+x)−f(a)=21(D11f)(a)∣x∣2x12+(D22f)(a)∣x∣2x22+(D12)(a)∣x∣2x1x2+(D21)(a)∣x∣2x1x2+∣x∣2r(x)
设
g(x)=(D11f)(a)x12∣x∣2+(D22f)(a)x22∣x∣2+(D12)(a)x1x2∣x∣2+(D21)(a)x1x2∣x∣2g\left( \mathbf{x} \right) = \left( D_{11}f \right)\left( \mathbf{a} \right)\frac{x_{1}^{2}}{\left| \mathbf{x} \right|^{2}} + \left( D_{22}f \right)\left( \mathbf{a} \right)\frac{x_{2}^{2}}{\left| \mathbf{x} \right|^{2}} + \left( D_{12} \right)\left( \mathbf{a} \right)\frac{x_{1}x_{2}}{\left| \mathbf{x} \right|^{2}} + \left( D_{21} \right)\left( \mathbf{a} \right)\frac{x_{1}x_{2}}{\left| \mathbf{x} \right|^{2}}g(x)=(D11f)(a)∣x∣2x12+(D22f)(a)∣x∣2x22+(D12)(a)∣x∣2x1x2+(D21)(a)∣x∣2x1x2
观察可知
g(kx)=g(x)g\left( k\mathbf{x} \right) = g(\mathbf{x})g(kx)=g(x)
只要限定xxx在平面圆就可以求出g(x)g(x)g(x)的值域。由于平面圆是紧集、连通集，所以g(x)g\left( \mathbf{x} \right)g(x)的值域是紧集、闭集、闭区间。所以g(x)g\left( \mathbf{x} \right)g(x)有最大最小值，g(x)g\left( \mathbf{x} \right)g(x)取最大值、最小值时，∣x∣\left| \mathbf{x} \right|∣x∣可以是任意值。
所以，当∣x∣→0\left| \mathbf{x} \right| \rightarrow 0∣x∣→0时，r(x)∣x∣2→0\frac{r\left( \mathbf{x} \right)}{\left| \mathbf{x} \right|^{2}} \rightarrow 0∣x∣2r(x)→0，此时
f(a+x)−f(a)∣x∣2≈12g(x)\frac{f\left( \mathbf{a + x} \right) - f\left( \mathbf{a} \right)}{\left| \mathbf{x} \right|^{2}} \approx \frac{1}{2}g\left( \mathbf{x} \right)∣x∣2f(a+x)−f(a)≈21g(x)
若g(x)g\left( \mathbf{x} \right)g(x)恒正，也就是g(x)g\left( \mathbf{x} \right)g(x)最小值为正，那么在a\mathbf{a}a点的某邻域内，f(a+x)>f(a)f\left( \mathbf{a + x} \right) > f\left( \mathbf{a} \right)f(a+x)>f(a)，f(x)f(\mathbf{x})f(x)在a\mathbf{a}a点有局部极小值；
若g(x)g\left( \mathbf{x} \right)g(x)恒负，也就是g(x)g\left( \mathbf{x} \right)g(x)最大值为负，那么在a\mathbf{a}a点的某邻域内，f(a+x)<f(a)f\left( \mathbf{a + x} \right) < f\left( \mathbf{a} \right)f(a+x)<f(a)，f(x)f(\mathbf{x})f(x)在a\mathbf{a}a点有局部极大值；
若g(x)g\left( \mathbf{x} \right)g(x)有正有负，也就是g(x)g\left( \mathbf{x} \right)g(x)最小值为负，最大值为正，那么在a\mathbf{a}a点的任何邻域内，f(a+x)>f(a)、f(a+x)<f(a)f\left( \mathbf{a + x} \right) > f\left( \mathbf{a} \right)、f\left( \mathbf{a + x} \right) < f\left( \mathbf{a} \right)f(a+x)>f(a)、f(a+x)<f(a)都有可能，f(x)f(x)f(x)在a\mathbf{a}a点必为鞍点；
若g(x)g\left( \mathbf{x} \right)g(x)最小值为零或者最大值为零，就无法判断。因为g(x)=0g\left( \mathbf{x} \right) = 0g(x)=0时，有r(x)∣x∣2\frac{r\left( \mathbf{x} \right)}{\left| \mathbf{x} \right|^{2}}∣x∣2r(x)的干扰。
由于g(x)g\left( \mathbf{x} \right)g(x)可以写成
g(x)=(x1∣x∣,x2∣x∣)(D11f)(a)(D21f)(a)(D12f)(a)(D22f)(a)(x1∣x∣x2∣x∣)g\left( \mathbf{x} \right) = \left( \frac{x_{1}}{\left| \mathbf{x} \right|},\frac{x_{2}}{\left| \mathbf{x} \right|} \right)\begin{bmatrix} \left( D_{11}f \right)\left( \mathbf{a} \right) & \left( D_{21}f \right)\left( \mathbf{a} \right) \\ \left( D_{12}f \right)\left( \mathbf{a} \right) & \left( D_{22}f \right)\left( \mathbf{a} \right) \end{bmatrix}\binom{\frac{x_{1}}{\left| \mathbf{x} \right|}}{\frac{x_{2}}{\left| \mathbf{x} \right|}}g(x)=(∣x∣x1,∣x∣x2)(D11f)(a)(D12f)(a)(D21f)(a)(D22f)(a)(∣x∣x2∣x∣x1)
中间的矩阵若是正定，则为情况1，取极小值；若是负定，则为情况2，取极大值；若是半正定、半负定，则为情况4，无法判定；其他情况是有正有负，则为鞍点。
b. 当为一般的多元函数时
f(a+x)−f(a)∣x∣2=12!∑i1,i2(Di1Di2f)(a)xi1xi2∣x∣2+r(x)∣x∣2{\frac{f\left( \mathbf{a + x} \right) - f\left( \mathbf{a} \right)}{\left| \mathbf{x} \right|^{2}} }{= \frac{1}{2!}\left\lbrack \sum_{i_{1},i_{2}}^{}{\left( D_{i_{1}}D_{i_{2}}f \right)\left( \mathbf{a} \right)\frac{x_{i_{1}}x_{i_{2}}}{\left| \mathbf{x} \right|^{2}}} \right\rbrack + \frac{r\left( \mathbf{x} \right)}{\left| \mathbf{x} \right|^{2}}}∣x∣2f(a+x)−f(a)=2!1i1,i2∑(Di1Di2f)(a)∣x∣2xi1xi2+∣x∣2r(x)
设
g(x)=(x1∣x∣,x2∣x∣...xn∣x∣)(D11f)(a)(D21f)(a)⋯(Dn1f)(a)(D12f)(a)(D22f)(a)⋯(Dn1f)(a)⋮⋮⋱⋮(D1nf)(a)(D2nf)(a)⋯(Dnnf)(a)(x1∣x∣x2∣x∣⋮xn∣x∣)g\left( \mathbf{x} \right) = \left( \frac{x_{1}}{\left| \mathbf{x} \right|},\frac{x_{2}}{\left| \mathbf{x} \right|}\ldots\frac{x_{n}}{\left| \mathbf{x} \right|} \right)\begin{bmatrix} \left( D_{11}f \right)\left( \mathbf{a} \right) & \left( D_{21}f \right)\left( \mathbf{a} \right) & \cdots & \left( D_{n1}f \right)\left( \mathbf{a} \right) \\ \left( D_{12}f \right)\left( \mathbf{a} \right) & \left( D_{22}f \right)\left( \mathbf{a} \right) & \cdots & \left( D_{n1}f \right)\left( \mathbf{a} \right) \\ \vdots & \vdots & \ddots & \vdots \\ \left( D_{1n}f \right)\left( \mathbf{a} \right) & \left( D_{2n}f \right)\left( \mathbf{a} \right) & \cdots & \left( D_{nn}f \right)\left( \mathbf{a} \right) \end{bmatrix}\begin{pmatrix} \frac{x_{1}}{\left| \mathbf{x} \right|} \\ \frac{x_{2}}{\left| \mathbf{x} \right|} \\ \vdots \\ \frac{x_{n}}{\left| \mathbf{x} \right|} \end{pmatrix}g(x)=(∣x∣x1,∣x∣x2...∣x∣xn) (D11f)(a)(D12f)(a)⋮(D1nf)(a)(D21f)(a)(D22f)(a)⋮(D2nf)(a)⋯⋯⋱⋯(Dn1f)(a)(Dn1f)(a)⋮(Dnnf)(a) ∣x∣x1∣x∣x2⋮∣x∣xn
观察可知
g(kx)=g(x)g\left( k\mathbf{x} \right) = g(\mathbf{x})g(kx)=g(x)
只要限定x\mathbf{x}x在球面就可以求出g(x)g(\mathbf{x})g(x)的值域。由于球面是紧集、连通集，所以g(x)g\left( \mathbf{x} \right)g(x)的值域是紧集、闭集、闭区间。所以g(x)g\left( \mathbf{x} \right)g(x)有最大最小值，g(x)g\left( \mathbf{x} \right)g(x)取最大值、最小值时，∣x∣\left| \mathbf{x} \right|∣x∣可以是任意值。
当∣x∣→0\left| \mathbf{x} \right| \rightarrow 0∣x∣→0时，r(x)∣x∣2→0\frac{r\left( \mathbf{x} \right)}{\left| \mathbf{x} \right|^{2}} \rightarrow 0∣x∣2r(x)→0，此时
f(a+x)−f(a)∣x∣2≈12g(x)\frac{f\left( \mathbf{a + x} \right) - f\left( \mathbf{a} \right)}{\left| \mathbf{x} \right|^{2}} \approx \frac{1}{2}g\left( \mathbf{x} \right)∣x∣2f(a+x)−f(a)≈21g(x)
矩阵若是正定，也就是g(x)g\left( \mathbf{x} \right)g(x)最小值为正，那么在a\mathbf{a}a点的某邻域内，f(a+x)>f(a)f\left( \mathbf{a + x} \right) > f\left( \mathbf{a} \right)f(a+x)>f(a)，f(x)f(\mathbf{x})f(x)在a\mathbf{a}a点有局部极小值；
矩阵若是负定，也就是g(x)g\left( \mathbf{x} \right)g(x)最大值为负，那么在a\mathbf{a}a点的某邻域内，f(a+x)<f(a)f\left( \mathbf{a + x} \right) < f\left( \mathbf{a} \right)f(a+x)<f(a)，f(x)f(\mathbf{x})f(x)在a\mathbf{a}a点有局部极大值；
若是半正定、半负定，因为g(x)=0g\left( \mathbf{x} \right) = 0g(x)=0时，有r(x)∣x∣2\frac{r\left( \mathbf{x} \right)}{\left| \mathbf{x} \right|^{2}}∣x∣2r(x)的干扰，无法判定，需要更高阶判定；
其他情况是有正有负，即不定，也就是g(x)g\left( \mathbf{x} \right)g(x)最小值为负，最大值为正，那么在a\mathbf{a}a点的任何邻域内，f(a+x)>f(a)、f(a+x)<f(a)f\left( \mathbf{a + x} \right) > f\left( \mathbf{a} \right)、f\left( \mathbf{a + x} \right) < f\left( \mathbf{a} \right)f(a+x)>f(a)、f(a+x)<f(a)都有可能，f(x)f(x)f(x)在a\mathbf{a}a点必为鞍点。