# Hessian matrix for 2d function

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• Sep 27th 2009, 06:26 AM
Gustavis
Hessian matrix for 2d function
Hi to all ! (Happy)

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 ? (Crying)
• Sep 27th 2009, 08:44 AM
Opalg
Quote:

Originally Posted by Gustavis
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 ? (Crying)

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.
• Sep 27th 2009, 11:24 AM
Gustavis
Quote:

Originally Posted by Opalg
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 ! :D

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 ! :D