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Local Extrema of Functions Exhibiting Self-Similarity

  1. Oct 5, 2012 #1
    Hi guys. Today was my last day of Calculus I class, and I’ve had this question simmering in my mind for some time now. Perhaps you guys may enlighten me or point me in the right direction?

    Consider the Weierstrass function, which is continuous everywhere but differentiable nowhere. We know that a local extremum, let’s say a local maximum, means that there is a number b, a < b < c, such that f(b) >= f(x) in (a, c). We also know that a local extremum either has a derivative of 0 or is non-differentiable.

    My question is: Is it possible for every point of the Weierstrass function to be a local extremum (considering it is a non-constant function), given its self-similarity? If you look at it from one viewing rectangle it appears as if it the function has a maximum/minimum at a point in a given neighborhood, but then you zoom in and zoom in and zoom in. . . .

    If not, how does one locate the points in which there is a local extremum?

    Sorry if that was a lame question, and thanks in advance!
     
    Last edited: Oct 5, 2012
  2. jcsd
  3. Oct 7, 2012 #2
    You should recall the definition of extremum. It involves a neighborhood. Can every point be an extremum given this definition?
     
  4. Oct 8, 2012 #3
    Yes, it involves a neighborhood. Well, logically the answer would be no, right? But what counts as a "neighborhood" here?
     
  5. Oct 8, 2012 #4
    Well, use whatever definition you were given in your course of Calculus. Anyway it will have more than one point. So every point cannot be an extremum, because its neighborhood must contain other points, which are non-extremal.
     
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