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Jemurray3 Group Title

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 n-dimensional case. Anybody who's bored is welcome to find mistakes in my work or make suggestions for more elegance.

  • 2 years ago
  • 2 years ago

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  1. badreferences Group Title
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    This question would be in better company amongst those of http://math.stackexchange.com/.

    • 2 years ago
  2. Jemurray3 Group Title
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    Maybe, but I don't think it's quite advanced enough work for that.

    • 2 years ago
  3. badreferences Group Title
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    You and I haven't been visiting the same site. I see a precalculus question there every day.

    • 2 years ago
  4. Jemurray3 Group Title
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    In 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.

    • 2 years ago
  5. Jemurray3 Group Title
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    I'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 \[ (H)_{ij} = \frac{\partial^2 f}{\partial x_i \partial x_j}\]

    • 2 years ago
  6. Jemurray3 Group Title
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    Oops... 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.

    • 2 years ago
  7. badreferences Group Title
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    Are you attempting a generalized analytic continuation?

    • 2 years ago
  8. Jemurray3 Group Title
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    No, I'm just extending to the case of n variables rather than just 2.

    • 2 years ago
  9. badreferences Group Title
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    Oh right. On your way then. :P

    • 2 years ago
  10. Jemurray3 Group Title
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    If 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.

    • 2 years ago
  11. Jemurray3 Group Title
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    Pulling 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.

    • 2 years ago
  12. badreferences Group Title
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    Got lazy with the TeX, huh?

    • 2 years ago
  13. Jemurray3 Group Title
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    As 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.

    • 2 years ago
  14. Jemurray3 Group Title
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    And no, I just figured that since the only mathematical symbols I had to type were regular QWERTY letters I wouldn't bother.

    • 2 years ago
  15. badreferences Group Title
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    I will reread in more detail after dinner. Something seems weird, but sometimes I'm just dumb.

    • 2 years ago
  16. Jemurray3 Group Title
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    By all means. If I've made a mistake I would like to know :)

    • 2 years ago
  17. badreferences Group Title
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    Sounds solid. I recall a similar problem set in some text. Let's see if I can find it.

    • 2 years ago
  18. badreferences Group Title
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    Not relevant, but it's important to feed mathematicians. http://upload.wikimedia.org/wikipedia/commons/2/20/Roast_duck_fs.JPG

    • 2 years ago
  19. badreferences Group Title
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    Have you taken PDE yet?

    • 2 years ago
  20. badreferences Group Title
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    Like, finished it?

    • 2 years ago
  21. Jemurray3 Group Title
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    Only the standard, linear ones.

    • 2 years ago
  22. badreferences Group Title
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    Well, the text is eluding me for now. I will find it maybe when I review regular differential equations.

    • 2 years ago
  23. badreferences Group Title
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    But otherwise your modest conclusion is solid. :P

    • 2 years ago
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