You want to modify the functions f_{n}(x)=x^{n} (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 f_{n}.
But not any function will do: For instance, you cannot add a Gaussian exp(x^{2}),
because that would change f_{n} 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.
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^21)). Add a factor of 1/e to have function value 1 at x=0 and you'll get u(x)=exp(x^2/(x^21).
Edited on October 31, 2006, 8:09 am

Posted by JLo
on 20061031 08:06:22 