The Hessian of a scalar valued function f:Rn®R is the n×n matrix of second order partial derivatives of f. In MATLAB we can obtain the Hessian of f by computing the Jacobian of the Jacobian of f. Consider once again the function f(x,y)=(4x2-1)e-x2-y2.

>> syms x y real
>> Hf=jacobian(jacobian(f));
>> Hf=simple(Hf)
Hf =
[2*exp(-x^2-y^2)*(2*x+1)*(2*x-1)*(2*x^2-5),  4*x*y*exp(-x^2-y^2)*(-5+4*x^2)]
[4*x*y*exp(-x^2-y^2)*(-5+4*x^2), 2*exp(-x^2-y^2)*(-1+2*y^2)*(2*x+1)*(2*x-1)]

2# pertain

谢谢。我试试之后回来反馈

我没运行出来。。==

直接在MATLAB的面板上输入

jacobian(jacobian((4*x^2-1)*exp(-x^2-y^2)))

得到结果

ans =

[ 8*exp(-x^2-y^2)-32*x^2*exp(-x^2-y^2)-2*(4*x^2-1)*exp(-x^2-y^2)+4*(4*x^2-1)*x^2*exp(-x^2-y^2),                                          -16*x*y*exp(-x^2-y^2)+4*(4*x^2-1)*x*y*exp(-x^2-y^2)]
[                                          -16*x*y*exp(-x^2-y^2)+4*(4*x^2-1)*x*y*exp(-x^2-y^2),                                     -2*(4*x^2-1)*exp(-x^2-y^2)+4*(4*x^2-1)*y^2*exp(-x^2-y^2)]

似乎不是矩阵。不过数目应该是对的了·~~~

下面一个问题是能apply subs 命令吗？

后面应该有变量吧。
jacobian(jacobian((4*x^2-1)*exp(-x^2-y^2), [x y ]), [x y])