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I want the geometrical interpretation of the following:

If $f_{xx}f_{yy} < f_{xy}^2$ and $f_{xx}$ has the same sign as $f_{yy}$ at a point, then why is that point a saddle point? Because,in the case that they have the same sign,one would expect that point to be a minimum or maximum point,not a saddle point.

Bear in mind,that i have checked the intuitive explanation that is available in wikipedia,but did not understand it.

Also, i am studying physics and not math,so please don't complicate things with symbolisms that i might not understand.

Also, i want an intuitive answer,because i know the maths behind categorizing critical points.

TheQuantumMan
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  • I guess $f \in C^2(\mathbb{R}^2, \mathbb{R})$? – Keba Aug 06 '15 at 14:21
  • I don't understand this notation.I am reading from the calculus book of Anton,which does not have notation for mathematicians. – TheQuantumMan Aug 06 '15 at 14:23
  • @LandosAdam, Keba meant that $f$ is twice differentiable real function of two variables – Michael Galuza Aug 06 '15 at 14:24
  • Oh,yes it is..Thanks for the clarification – TheQuantumMan Aug 06 '15 at 14:25
  • @LandosAdam, what's your definition of saddle point? – Michael Galuza Aug 06 '15 at 14:26
  • A point that has a minimum in one direction and maximum in another orthogonal direction – TheQuantumMan Aug 06 '15 at 14:27
  • $f_{xx} f_{yy} < f_{xy}^2$ implies that the determinant of the hessian is negative. thus, the two eigenvalues of the hessian have different signs, that's the hessian is indefinite. – user251257 Aug 06 '15 at 14:29
  • Yes,i know this.But i am looking for the geometrical interpretation of this.(what causes the -at first glance- maximum or minimum critical point to be a saddle rather than the max/min?).I think it has to do with fxy being very large with respect to fxx and fyy.But i can't visualize it(create an example of a graph in my mind that has this characteristic) – TheQuantumMan Aug 06 '15 at 14:32
  • @user251257 https://en.wikipedia.org/wiki/Second_partial_derivative_test this might help you understand what i am looking for(the section "Geometric interpretation in the two-variable case").I just don't understand this explanation,although i think that this is the kind of explanation that i need.. – TheQuantumMan Aug 06 '15 at 14:34
  • @LandosAdam okay. say $0$ is a critical point. then, $f(x, 0) \approx f(0) + \frac12 f_{xx} x^2$. the same applies to the $y$ argument. since $f_{xx}$ and $f_{yy}$ have different signs, say for example $f_{xx} > 0$, then $f$ increases in $x$ direction but decreases in $y$ direction. – user251257 Aug 06 '15 at 14:41
  • I am asking about the case when they have the same signs.This is the more trickier part i think. – TheQuantumMan Aug 06 '15 at 14:43
  • @LandosAdam it is basically the same. $f$ increases in the direction of the eigenvector to the positive eigenvalue and decreases in th orthogonal direction (same as the eigenvector to the negative eigenvalue). – user251257 Aug 06 '15 at 14:50
  • But why is that the case???The second derivatives have the same signs.So it should decrease or increase in every direction – TheQuantumMan Aug 06 '15 at 14:51

2 Answers2

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Consider $f:\mathbb R^2 \to \mathbb R$, twice continuously differentiable in a neighborhood of $0$ with $f(0)=0$ and $f'(0)=0$. (We discuss how the value of $f$ changes near the critical point $0$)

Assume the determinant of Hessian $H=f''(0)$ of $f$ at $0$ is negative. Since $f$ is twice continuously differentiable, $H$ is symmetric. Hence, both eigenvalues $\mu_1 \le \mu_2$ of $H$ are real. From $\det H = \mu_1 \mu_2 < 0$ follows that $\mu_1 < 0 < \mu_2$.

Now, let $u$ be the eigenvector to $\mu_1$. By Taylor theorem we have $$ f(u) \approx \frac12 u^T H u = \frac12 \mu_1 u^T u < 0 $$ for $u$ sufficiently small. Similarly, for a sufficiently small eigenvector $v$ to $\mu_2$ we have $$ f(v) \approx \frac12 v^T H v = \frac12 \mu_2 v^T v > 0. $$ Thus, $0$ is a saddle point of $f$.

user251257
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    This is an answer to a very similar question.You describe the mathematical proof of what i am asking. But,i have already stated that i already know the mathematics behind it. What i need is a geometrical interpretation of it. I want to know how to describe it visually/graphycally. Like the explanation in the wiki article in the link en.wikipedia.org/wiki/Second_partial_derivative_test under "Geometric interpretation in the two-variable case". Thank you for your answer, but i am looking for a more visual/intuitive explanation – TheQuantumMan Aug 06 '15 at 18:17
  • the axes of the "saddle" are given by the eigenvectors. I don't see how it could get more graphical. for which part do you want a more graphical interpretation? – user251257 Aug 06 '15 at 18:21
  • In more plain words. Like in the above article. – TheQuantumMan Aug 06 '15 at 18:23
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I found a very good answer at Geometric interpretation of $\frac {\partial^2} {\partial x \partial y} f(x,y)$ Its the last answer by @harvard man
also,a link to see before reading it is: http://www.math.harvard.edu/archive/21a_fall_08/exhibits/fxy/index.html posted by @ user26439
(sorry,i cant tag you guys)

If any of the other users have anything else to say in the lines of the answer in the link,feel free to do so.

Community
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TheQuantumMan
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