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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Entirely different solution Comment 9 of 9 |
(In reply to re: Entirely different solution by Richard)

Sorry Richard, seems I misunderstood your post. Functional composition is of course a much easier way to get the job done. I was thinking to complicated when reading your post...
  Posted by JLo on 2006-10-31 12:33:07

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