Help check my derivation of multivariable functions extremum criterion
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I'm trying myself to derive the criterion of extremum of several variables.
Let $f:Bbb R^2toBbb R$, $c=(x_0,y_0)inBbb R^2,~u=(a,b)inBbb R^2$ be a unit vector. Define $r(t)=(x_0+at,y_0+bt)$. Then $(fcirc r)''(0)=[a~~b]begin{bmatrix}f_{xx}(c)&f_{xy}(c)\f_{yx}(c)&f_{yy}(c)end{bmatrix}begin{bmatrix}a\bend{bmatrix}$. So if $f_{xx}(c)$ and the determinant of the Hessian matrix are both positive, then $[a~~b]begin{bmatrix}f_{xx}(c)&f_{xy}(c)\f_{yx}(c)&f_{yy}(c)end{bmatrix}begin{bmatrix}a\bend{bmatrix}$ is always positive, hence $c$ is a relative minimum. Am I all correct?
real-analysis calculus extreme-value-analysis
asked Dec 8 '18 at 14:43
SharaShara
725
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add a comment |
$begingroup$
I'm trying myself to derive the criterion of extremum of several variables.
Let $f:Bbb R^2toBbb R$, $c=(x_0,y_0)inBbb R^2,~u=(a,b)inBbb R^2$ be a unit vector. Define $r(t)=(x_0+at,y_0+bt)$. Then $(fcirc r)''(0)=[a~~b]begin{bmatrix}f_{xx}(c)&f_{xy}(c)\f_{yx}(c)&f_{yy}(c)end{bmatrix}begin{bmatrix}a\bend{bmatrix}$. So if $f_{xx}(c)$ and the determinant of the Hessian matrix are both positive, then $[a~~b]begin{bmatrix}f_{xx}(c)&f_{xy}(c)\f_{yx}(c)&f_{yy}(c)end{bmatrix}begin{bmatrix}a\bend{bmatrix}$ is always positive, hence $c$ is a relative minimum. Am I all correct?
real-analysis calculus extreme-value-analysis
asked Dec 8 '18 at 14:43
SharaShara
725
$endgroup$
$begingroup$
Looks correct to me.
$endgroup$
– jgon
Dec 8 '18 at 15:16
add a comment |
$begingroup$
I'm trying myself to derive the criterion of extremum of several variables.
Let $f:Bbb R^2toBbb R$, $c=(x_0,y_0)inBbb R^2,~u=(a,b)inBbb R^2$ be a unit vector. Define $r(t)=(x_0+at,y_0+bt)$. Then $(fcirc r)''(0)=[a~~b]begin{bmatrix}f_{xx}(c)&f_{xy}(c)\f_{yx}(c)&f_{yy}(c)end{bmatrix}begin{bmatrix}a\bend{bmatrix}$. So if $f_{xx}(c)$ and the determinant of the Hessian matrix are both positive, then $[a~~b]begin{bmatrix}f_{xx}(c)&f_{xy}(c)\f_{yx}(c)&f_{yy}(c)end{bmatrix}begin{bmatrix}a\bend{bmatrix}$ is always positive, hence $c$ is a relative minimum. Am I all correct?
real-analysis calculus extreme-value-analysis
asked Dec 8 '18 at 14:43
SharaShara
725
$endgroup$
I'm trying myself to derive the criterion of extremum of several variables.
Let $f:Bbb R^2toBbb R$, $c=(x_0,y_0)inBbb R^2,~u=(a,b)inBbb R^2$ be a unit vector. Define $r(t)=(x_0+at,y_0+bt)$. Then $(fcirc r)''(0)=[a~~b]begin{bmatrix}f_{xx}(c)&f_{xy}(c)\f_{yx}(c)&f_{yy}(c)end{bmatrix}begin{bmatrix}a\bend{bmatrix}$. So if $f_{xx}(c)$ and the determinant of the Hessian matrix are both positive, then $[a~~b]begin{bmatrix}f_{xx}(c)&f_{xy}(c)\f_{yx}(c)&f_{yy}(c)end{bmatrix}begin{bmatrix}a\bend{bmatrix}$ is always positive, hence $c$ is a relative minimum. Am I all correct?
real-analysis calculus extreme-value-analysis
real-analysis calculus extreme-value-analysis
asked Dec 8 '18 at 14:43
SharaShara
725
asked Dec 8 '18 at 14:43
SharaShara
725
edited Dec 8 '18 at 15:10
Shara
asked Dec 8 '18 at 14:43
SharaShara
725
asked Dec 8 '18 at 14:43
SharaShara
725
asked Dec 8 '18 at 14:43
SharaShara
725
725
$begingroup$
Looks correct to me.
$endgroup$
– jgon
Dec 8 '18 at 15:16
add a comment |
$begingroup$
Looks correct to me.
$endgroup$
– jgon
Dec 8 '18 at 15:16
$begingroup$
Looks correct to me.
$endgroup$
– jgon
Dec 8 '18 at 15:16
$begingroup$
Looks correct to me.
$endgroup$
– jgon
Dec 8 '18 at 15:16
add a comment |
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$begingroup$
Looks correct to me.
$endgroup$
– jgon
Dec 8 '18 at 15:16