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 Super-smooth hump (Posted on 2006-10-24)

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)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: My Way | Comment 5 of 9 |
(In reply to My Way by Richard)

Can you prove that the function s(x) is infinitely smooth?  It just doesn't seem obvious to me.
 Posted by Tristan on 2006-10-25 21:06:24

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