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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: Entirely different solution | Comment 8 of 9 |
(In reply to Entirely different solution by JLo)

There is no convolution in my solution.  s is applied to g by functional composition.  Convolution with a smooth function is also a means for smoothing that works works more generally than functional composition which only smooths corners that are specially situated. My comment was intended to provoke some thought on the chain rule of calculus, not the subject of convolution.
  Posted by Richard on 2006-10-31 11:39:19

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