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Super-smooth hump (Posted on 2006-10-24) Difficulty: 5 of 5

You want to modify the functions fn(x)=xn (n = 1, 2, 3, ...) in the following way: Outside the interval [-1,1] the functions should remain completely unchanged. At zero, the new functions should be 1 (instead of 0). And the most important requirement: The functions should remain continuously differentiable infinitely many times, everywhere.

You can achieve this by adding a "hump" function to fn. But not any function will do: For instance, you cannot add a Gaussian exp(-x2), because that would change fn outside the interval [-1,1]. You also cannot add a sine hump, (0.5 + 0.5*sin(π x)) for x in [-1,1] and 0 otherwise, because the second derivative would become discontinuous at -1 and 1. Find a hump function which will do the trick.

No Solution Yet Submitted by vswitchs    
Rating: 4.3333 (3 votes)

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re(2): My Way | Comment 6 of 9 |
(In reply to re: My Way by Tristan)

While it doesn't answer your exact question, the Wikipedia article on the subject Non-analytic smooth function does fairly adequately address your concerns. Using a graphing calculator on exp(-1/x) for 0<x<2, say, you willl see that the graph practically coincides with the x-axis for 0<x<1/16, lending credence to the assertion that at x=0, the function s is utterly flat.  A rigorous proof that the kth derivative of s at x=0 is zero is somewhat involved, but depends on the fact that the kth derivative of s for very small positive x always contains the factor exp(-1/x) so that the right-hand derivative is zero just as the left-hand derivative clearly is.

It is homework problem 1 of paragraph 16 of chapter 4 of the well-known book "Analysis on Manifolds" by James R. Munkres to prove that s is of class C-infinity (has infinitely many, necessarily continuous, derivatives) on the real line.

Edited on October 25, 2006, 11:59 pm
  Posted by Richard on 2006-10-25 23:54:37

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