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Topic: an elementary calculus question
Replies: 2   Last Post: Sep 9, 2011 1:20 PM

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Stephen J. Herschkorn

Posts: 2,297
Registered: 1/29/05
Re: an elementary calculus question
Posted: Sep 8, 2011 10:40 PM
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rge11x wrote:

>Take the following two-variable function f(x,y) = (1-x^2).cos(w.x-y) in the rectangle -1<x<1 and 0<y<2pi, and let w = 10000pi, so that there are many thousands of cycles of the sinusiod along the x-axis within (-1,1).
>
>Now calculate the critical (stationary) points of f(x,y) (points where the partial derivatives with respect to both x and y are zero) by taking partial derivatives.
>
>The partial derivative with respect to y is @f/@y = (1-x^2).sin(w.x-y) and that can only be zero in -1<x<1 when w.x-y=k.pi.
>
>Now @f/@x = -w.(1-x^2).sin(w.x-y) - 2.x.cos(w.x-y), so if w.x-y=k.pi then @f/@x=-2.x.(-1)^k, and for this to be zero we must have x=0. From w.x-y=k.pi we also get y=0 or pi, thus the critical points are (0,0) and (0,pi).
>
>Question: what has happened to the thousands of other local minima and maxima of this function that can be easily seen by plotting it but not found by these partial derivatives?
>
>


I haven't thought this through, but is it the case that there are
thousands of local minima if you fix x, but not so many minima in the
open rectangle? I.e., at your suggested local "minima," the function
may decrease as x varies.

--
Stephen J. Herschkorn sjherschko@netscape.net
Math Tutor on the Internet and in Central New Jersey and Manhattan




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