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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)

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 Entirely different solution | Comment 7 of 9 |
Interesting problem vswitchs!

Actually my solution is identical to Richard's, but I don't think one needs the concept of convolutions to solve the problem. I also started with exp(-1/x). With school math it is easy to show that exp(-1/x)->0 for x->0+0. In fact it is even exp(-1/x)/x^n->0 for x->0+0 for any integer n, which makes it an excellent candidate for our solution as this means it has a zero of "infinite order" at x=0. Now in order to make the whole thing symmetrical, one quickly comes up with s(x)=exp(-1/x^2) which fullfills s(x)/x^n->0 for x->0 (x approaching from left or right). Since we need our zeroes at x=-1 and x=1, not at x=0, you adjust the denominator a bit and get t(x)=exp(1/(x^2-1)). Add a factor of 1/e to have function value 1 at x=0 and you'll get u(x)=exp(x^2/(x^2-1).

Edited on October 31, 2006, 8:09 am
 Posted by JLo on 2006-10-31 08:06:22

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