Why Must the Hessian Matrix Be Symmetric at a Critical Point?

AndreTheGiant
Messages
26
Reaction score
0

Homework Statement



Given a function f: R^2 -> R of class C^3 with a critical point c.

Why CANNOT the hessian matrix of f at point c be given by:

1 -2
2 3


Homework Equations





The Attempt at a Solution



So first i want to clarify this.

When it says f: R^2 -> R, that means the function is of two variables (x and y)?

And when it says class C^3 that means the third derivative of the function exists and is continuous. So would a function be x^3 or x^4? the third derivative would be 24x for x^4 and is continuous. The third derivative of x^3 would be 6.

I'm not sure about the answer..
 
Physics news on Phys.org
AndreTheGiant said:

Homework Statement



Given a function f: R^2 -> R of class C^3 with a critical point c.

Why CANNOT the hessian matrix of f at point c be given by:

1 -2
2 3


Homework Equations





The Attempt at a Solution



So first i want to clarify this.

When it says f: R^2 -> R, that means the function is of two variables (x and y)?

And when it says class C^3 that means the third derivative of the function exists and is continuous. So would a function be x^3 or x^4? the third derivative would be 24x for x^4 and is continuous. The third derivative of x^3 would be 6.

I'm not sure about the answer..

IF your matrix A above was a Hessian, what would the number a(2,2) = -2 represent? What would the number a(2,1) = +2 represent?

RGV
 
Ah ok. I think i got it. In the hessian which is given by

fxx fxy

fyx fyy

fxy is not equal to fyx which should be the case for mixed partials?
 
AndreTheGiant said:
Ah ok. I think i got it. In the hessian which is given by

fxx fxy

fyx fyy

fxy is not equal to fyx which should be the case for mixed partials?

Yes, exactly.

RGV
 
I have a question considering the applicability of Hessian matrix.

So, Can I use Hessian to prove that x^y > y^x whenever y > x >= e.

At first I start by multiplying by ln() => y*ln(x) > x*ln(y)

Is it enough, if I take g(x,y) such that g(x,y) = y*ln(x) - x*ln(y) > 0 and show det(H(g)) < 0 whenever y > x >= e?

My purpose with this is to show that there are no real local or global critical points in g(x,y) when y > x >= e, and conclude that x^y - y^x diverges. I am not sure if I can use Hessian to draw that kind of conclusion.
 
Hi Viliperi, welcome to PF, please start a new thread if you have a question as opposed to resurrecting an old one. You're more likely to get an anser that way as well - thanks
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top