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Jemurray3
 3 years ago
Not a question: I spent most of the day exploring questions that popped up in my head recently, some basic and some a little bit more subtle. One of them was the extension of the "2nd Derivative Test" for functions of 2 variables to the ndimensional case. Anybody who's bored is welcome to find mistakes in my work or make suggestions for more elegance.
Jemurray3
 3 years ago
Not a question: I spent most of the day exploring questions that popped up in my head recently, some basic and some a little bit more subtle. One of them was the extension of the "2nd Derivative Test" for functions of 2 variables to the ndimensional case. Anybody who's bored is welcome to find mistakes in my work or make suggestions for more elegance.

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badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0This question would be in better company amongst those of http://math.stackexchange.com/.

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0Maybe, but I don't think it's quite advanced enough work for that.

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0You and I haven't been visiting the same site. I see a precalculus question there every day.

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0In truth I rarely visit there, but since I'm bored and there are some good mathematicians here too, I figured I'd just put it up for the hell of it.

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0I'll denote a function of n variables, in general, by \[ f = f(\mathbf{x}) \] where \[\mathbf{x} = x_1,x_2,x_3 ...\] the first order differential of such a function would be \[ df = \sum_{i=1}^n dx_i\frac{\partial f}{\partial x_i}\] the second order differential would be \[ d^2 f = \left(\sum_{i = 1}^n dx_i \frac{\partial}{\partial x_i}\right)^2f(\mathbf{x})\] \[ = \sum_{i = 1}^n \sum_{j = }^n dx_idx_j \frac{\partial^2f}{\partial x_i \partial x_j} = \sum_{i = 1}^n dx_i \sum_{j = }^n dx_j \frac{\partial^2f}{\partial x_i \partial x_j}\] which can be rewritten as \[d^2f(\mathbf{x}) = \mathbf{x}^TH\mathbf{x} \] where H is the Hessian matrix of f defined by \[ (H)_{ij} = \frac{\partial^2 f}{\partial x_i \partial x_j}\]

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0Oops... I should have really said \[d^2f = d\mathbf{x}^THd\mathbf{x} \] anyway... Assuming the first order variation df vanishes, the sign of the second order variation is the determining factor in whether we are at a minimum, maximum, or something else. At this point, we'll switch gears and move into some matrix theory, though I'll obviously keep the same notation for clarity. If we assume that the function f is at least twice differentiable on the region in question, we can use the equality of mixed partial derivatives to make note that the Hessian matrix is symmetric, i.e. \[H = H^T\] or equivalently \[(H)_{ij} = (H)_{ji} \] From Linear Algebra, we have the fact that n dimensional symmetric matrices have a set of orthonormal eigenvectors (not necessarily having distinct eigenvalues) that span R^n. Therefore, we can write any arbitrary vector dx as a sum of these eigenvectors, which I'll call u > \[ \mathbf{x} = \sum_{i=1}^n c_i \mathbf{u}_i \] This yields \[ d\mathbf{x}^THd\mathbf{x} = \left(\sum_{i=1}^nc_i\mathbf{u}^T_i\right)H\left(\sum_{j=1}^nc_j\mathbf{u}^T_j\right)\] since the eigenvectors are orthonormal, this reduces to \[ d\mathbf{x}^THd\mathbf{x} = \sum_{i=1}^n(c_i)^2\lambda_i\] where the lambdas are the eigenvalues.

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0Are you attempting a generalized analytic continuation?

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0No, I'm just extending to the case of n variables rather than just 2.

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0Oh right. On your way then. :P

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0If we're looking for a local maximum, then it's clear that \[d^2f >0\] for any direction we choose, corresponding to any possible choice in the coefficients above, so \[c_i \text{ are completely arbitrary} \] Let's say eigenvalue k is not positive. Then, it's clear that if we chose \[c_i = \delta_{i,k} \] then our condition would not be satisfied. Therefore, for a point to be a local maximum, the eigenvalues of the Hessian matrix must all be positive. Since the determinant of any matrix is the product of its eigenvalues, obviously \[det(H) > 0 \] Similarly, if we seek a local minimum, all of the eigenvalues must be negative. In that case, \[det(H) >0 \text{ if the space has even dimension, and }det(H)<0 \text{ if the space}\] \[\text{ has odd dimension} \] if we have a saddle point, then d^2f changes sign depending on our direction. That means that some of the eigenvalues must be positive and some must be negative.

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0Pulling back to the 2 dimensional case, both local maxima and local minima are characterized by det(H)>0. det(H) < 0 implies that the eigenvalues have opposite sign, i.e. you have a saddle point. det(H) = 0 implies that one of the eigenvalues is zero, which means that you are at a maximum in one direction but movement along the other direction yields no change  as though you were walking through a pringles tube.

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0Got lazy with the TeX, huh?

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0As far as extending the test to higher dimensions, I conclude that a determinant equal to zero implies that there is neither a local minimum or a local maximum as before, but the other cases are more complicated because we can't deduce the sign of ALL the eigenvalues based purely on the sign of the determinant. Analysis of the eigenvalues of the Hessian is the only recourse.

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0And no, I just figured that since the only mathematical symbols I had to type were regular QWERTY letters I wouldn't bother.

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0I will reread in more detail after dinner. Something seems weird, but sometimes I'm just dumb.

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0By all means. If I've made a mistake I would like to know :)

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0Sounds solid. I recall a similar problem set in some text. Let's see if I can find it.

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0Not relevant, but it's important to feed mathematicians. http://upload.wikimedia.org/wikipedia/commons/2/20/Roast_duck_fs.JPG

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0Have you taken PDE yet?

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0Like, finished it?

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.0Only the standard, linear ones.

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0Well, the text is eluding me for now. I will find it maybe when I review regular differential equations.

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0But otherwise your modest conclusion is solid. :P
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