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Math Help - Hessian matrix for 2d function

  1. #1
    Newbie Gustavis's Avatar
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    Hessian matrix for 2d function

    Hi to all !

    I have this function :

    f(x,y)=x^2 + y^2 - 4xy

    I should find the critical point.
    http://en.wikipedia.org/wiki/Critica..._(mathematics)

    the system of first derivative :
    \left\{\begin{matrix} 2x-8xy=0 \\ 2y-4x^2=0 \end{matrix}\right. \quad

    solutions : \quad p1(0,0) \quad p2(\frac{1}{\sqrt{8}}, \frac{1}{4}) \quad p3(\frac{-1}{\sqrt{8}},\frac{1}{4})

    then I should study the hessian matrix to understand the type of points
    The determinant of Hessian is :

    H(x,y)=(2-8y)*2+8x \quad
    (where f''_{xx}(x,y)=2-8y )

    in p1(0,0) \quad  f''_{xx}(0,0)=2 \quad H(0,0)=4 \quad this is a point of Minimun

    p2(\frac{1}{\sqrt{8}}, \frac{1}{4}) \quad f''_{xx}(\frac{1}{\sqrt{8}}, \frac{1}{4})=0 \quad H(\frac{1}{\sqrt{8}}, \frac{1}{4})=\frac{8}{\sqrt{8}}

    p3(\frac{-1}{\sqrt{8}},\frac{1}{4}) \quad f''_{xx}(\frac{-1}{\sqrt{8}},\frac{1}{4})=0 \quad H(\frac{-1}{\sqrt{8}},\frac{1}{4})=\frac{-1}{\sqrt{8}}

    The question :
    P2 and P3 are points of saddle ?
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Gustavis View Post
    I have this function :

    f(x,y)=x^2 + y^2 - 4x^{\color{red}2}y

    I should find the critical point.
    http://en.wikipedia.org/wiki/Critical_point_(mathematics)

    the system of first derivative :
    \left\{\begin{matrix} 2x-8xy=0 \\ 2y-4x^2=0 \end{matrix}\right. \quad

    solutions : \quad p1(0,0) \quad p2(\frac{1}{\sqrt{8}}, \frac{1}{4}) \quad p3(\frac{-1}{\sqrt{8}},\frac{1}{4})

    then I should study the hessian matrix to understand the type of points
    The determinant of Hessian is :

    H(x,y)=(2-8y)*2+8x \quad
    (where f''_{xx}(x,y)=2-8y )

    in p1(0,0) \quad  f''_{xx}(0,0)=2 \quad H(0,0)=4 \quad this is a point of Minimun

    p2(\frac{1}{\sqrt{8}}, \frac{1}{4}) \quad f''_{xx}(\frac{1}{\sqrt{8}}, \frac{1}{4})=0 \quad H(\frac{1}{\sqrt{8}}, \frac{1}{4})=\frac{8}{\sqrt{8}}

    p3(\frac{-1}{\sqrt{8}},\frac{1}{4}) \quad f''_{xx}(\frac{-1}{\sqrt{8}},\frac{1}{4})=0 \quad H(\frac{-1}{\sqrt{8}},\frac{1}{4})=\frac{-1}{\sqrt{8}}

    The question :
    P2 and P3 are points of saddle ?
    Either f(x,y) should be modified as in red above, or the differentiation is wrong. I'll assume the former.

    The Hessian determinant is f_{xx}f_{yy} - (f_{xy})^2, which in this example is 2(2-8y) - 64x^2. That gives the value 4 at the origin (so that is a minimum). At the points \bigl(\pm\tfrac1{\sqrt8}, \tfrac14\bigr) it works out as –8 (for both points). This is negative, so both those points are saddle points.
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  3. #3
    Newbie Gustavis's Avatar
    Joined
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    Quote Originally Posted by Opalg View Post
    Either f(x,y) should be modified as in red above, or the differentiation is wrong. I'll assume the former.

    The Hessian determinant is f_{xx}f_{yy} - (f_{xy})^2, which in this example is 2(2-8y) - 64x^2. That gives the value 4 at the origin (so that is a minimum). At the points \bigl(\pm\tfrac1{\sqrt8}, \tfrac14\bigr) it works out as 8 (for both points). This is negative, so both those points are saddle points.
    Thanks of all !

    exactly the equation was:
    f(x,y)=x^2 + y^2 - 4x^2y

    and determinant of hessian
    H(x,y)=(2-8y)*2 - (8x)^2

    yes I have lost some raise to a power... poor me

    thanks again !
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