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.
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