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Stationary points and chain rule

  1. Dec 26, 2005 #1
    Hi, I would like some help verifying the nature of a stationary point of the following function of two variables.

    f\left( {x,y} \right) = \sin \left( x \right) + \sin \left( y \right) + \sin \left( {x + y} \right)

    Ok so I equated grad(f) to zero and solved for x and y. I got three points, in particular I found (x,y) = (pi,pi). As I found it, the Hessian matrix is the zero matrix so the determinant is also zero. So in terms of the "second derivatives test" the test is inconclusive for the point (pi,pi). I would like to know if there are any other ways of determining the nature (max, min, saddle) of the stationary point at (x,y) = (pi,pi).

    My other question relates to the chain rule. I have been looking through a solution to a problem.

    ...conjectures that [tex]\mathop x\limits^ \to \bullet \nabla f\left( {\mathop x\limits^ \to } \right) = \nu f\left( {\mathop x\limits^ \to } \right)[/tex] for [tex]f:U \subset R^n \to R[/tex] provided that [tex]f\left( {a\mathop x\limits^ \to } \right) = a^\nu f\left( {\mathop x\limits^ \to } \right),a > 0[/tex].

    The question is to prove the conjecture. As part of the solution there is a step which goes along the lines of let [tex]g\left( a \right) = f\left( {a\mathop x\limits^ \to } \right)[/tex] then [tex]g'\left( a \right) = \nabla f\left( {a\mathop x\limits^ \to } \right) \bullet \mathop x\limits^ \to [/tex].

    I'm not sure how the chain rule has been implemented here. The LHS (g'(a)) is just the usual single variable derivative whereas the RHS involves the gradient and hence partial derivatives. The only time when I've seen a LHS with a derivative with respect to a single variable while the RHS involves partial and/or 'usual' derivatives is when I've had something like g = g(x,y) where x = x(t) and y = y(t) in which case I have:

    \frac{{dg}}{{dt}} = \frac{{\partial g}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial g}}{{\partial y}}\frac{{dy}}{{dt}}

    However in the above example I'm at a loss as to how the chain rule has been used. Can someone help me out with either of my questions? Thanks.
  2. jcsd
  3. Dec 26, 2005 #2


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    Try writing the gradient as a Taylor's series and keep only the first twp non-zero terms. Since [itex]sin(\pi)= 0[itex], sin'(x)= cos(x), [itex]sin'(\pi)= -1[/itex], sin"(x)= -sin(x) so [itex]sin"(\pi)= 0[/itex], and sin"'(x)= -cos(x) so [itex]sin"'(\pi)= 1, the third degree Taylor's polynomial for sin(x)+ sin(y)+ sin(x+y) about [itex](\pi,\pi)[/itex] is [itex](-x+ \frac{1}{3}x^3)+ (-y+ /frac{1}{3}y^3)+ (x+ y- /frac{1}{3}(x+y)^3[/itex] (the Taylor's Polynomial about [itex]\pi[/itex] in x+y is the Taylor's polynomial for x about [itex]2\pi[/itex] with x+y substituted for x). Multiply that out and see what it looks like close to [itex](\pi,\pi)[/itex].
  4. Dec 26, 2005 #3
    Thanks HallsofIvy, I'll try a Taylor series and see where that leads.

    I'm also still working on the chain rule problem.:biggrin:
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