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Givenafunctionf(x,y)=x^3+y^3-3xy,where[x,y]^T∈R^2.a)Determinethegradientvector▽f(x,y)andtheHessianmatrix▽^2f(x,y)off(x,y).b)UsingTaylor'sTheorem,approximatef(x,y)byaquadraticfunctionatthepoint[x,y]^T=[2,2]^T.c)Show
题目详情
Given a function f(x,y)=x^3+y^3-3xy,where [x,y]^T∈R^2.
a) Determine the gradient vector ▽f(x,y) and the Hessian matrix ▽^2f(x,y) of f(x,y).
b) Using Taylor's Theorem,approximate f(x,y) by a quadratic function at the point [x,y]^T=[2,2]^T.
c) Show that ▽^2f(1,1) is positive definite by definition.
d) Show that ▽^2f(-2,-2) is negative definite by the principle minor test.
e) Show that ▽^2f(0,0) is indefinite by the eigenvalue test.
我还有两个10财富值的相同问题在外语分类中,请您也务必把那20分拿走,
a) Determine the gradient vector ▽f(x,y) and the Hessian matrix ▽^2f(x,y) of f(x,y).
b) Using Taylor's Theorem,approximate f(x,y) by a quadratic function at the point [x,y]^T=[2,2]^T.
c) Show that ▽^2f(1,1) is positive definite by definition.
d) Show that ▽^2f(-2,-2) is negative definite by the principle minor test.
e) Show that ▽^2f(0,0) is indefinite by the eigenvalue test.
我还有两个10财富值的相同问题在外语分类中,请您也务必把那20分拿走,
▼优质解答
答案和解析
A=Hessian matrix =[6*x,0;0,6y]
▽^2f(1,1)=[6,0;0,6],矩阵正定
▽^2f(-2,-2)=[-12,0;0,-12],矩阵负定
▽^2f(0,0)=[0,0;0,0],矩阵为不定型
▽^2f(1,1)=[6,0;0,6],矩阵正定
▽^2f(-2,-2)=[-12,0;0,-12],矩阵负定
▽^2f(0,0)=[0,0;0,0],矩阵为不定型
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