# Thread: Hessian matrix for 2d function

1. ## Hessian matrix for 2d function

Hi to all !

I have this function :

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

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

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

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

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

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

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

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

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

The question :
P2 and P3 are points of saddle ?

2. Originally Posted by Gustavis
I have this function :

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

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

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

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

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

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

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

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

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

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

3. 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 $\displaystyle f_{xx}f_{yy} - (f_{xy})^2$, which in this example is $\displaystyle 2(2-8y) - 64x^2$. That gives the value 4 at the origin (so that is a minimum). At the points $\displaystyle \bigl(\pm\tfrac1{\sqrt8}, \tfrac14\bigr)$ 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:
$\displaystyle f(x,y)=x^2 + y^2 - 4x^2y$

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

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

thanks again !