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.
Step I: Make a bump with sharp corners. Let f(x) be defined as
x for x>0 and as 0 for all other real x. Then f(x+1) is f
shifted left by 1 and f(1-x) is f reversed and shifted right by
1. Thus g(x) defined as f(x+1)f(1-x) is equal to 1-x² for
|x|<1 and is zero for all other real x. g(x) is almost what we seek,
but it has a pair of fatal flaws in that it has sharp corners when
|x|=1 and therefore has no continuous derivatives there.
Step II: Fix the the flaws. A way to do this is to apply a nice
smooth function to the flawed g(x) . The nice smooth function s
is defined by setting s(x) equal to exp(-1/x) for x>0 and equal
to 0 for all other real x. It isn't hard to see that the result
of following g by s is nice and smooth -- the corners of g where |x|=1
have been smoothed out by applying s. All that remains is to
multiply by e to adjust the value at x=0 so it equals 1. The
final result for the hump funtion h in terms of x is
h(x)=exp(-x²/(1-x²)) for |x|<1 and 0 otherwise.
Edited on October 25, 2006, 7:54 pm
Posted by Richard
on 2006-10-25 19:29:38