You want to modify the functions f_n(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 f_n. But not any function will do: For instance, you cannot add a Gaussian exp(-x²), 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.

Let h(x) equal the equation for this hump
Let f(x) = h(x) when -1 <= x <= 1

h(x) = { if x <= -1, 0
             if x >= 1, 0
             if -1 <= x <= 1, f(x) }

In order for h(x) to be infinitely differentiable, f⁽ⁿ⁾(-1) = f⁽ⁿ⁾(1) = 0 for all whole numbers n.

I think it will be useful to think of f(x) in terms of its taylor series.

f⁽ⁿ⁾(x) = Σ(i = 0 to ∞) ( f⁽ⁱ⁺ⁿ⁾(a)*(x-a)^i/i! )
where a is any constant (did I get this equation right?)

Let's set a to 1 and n to zero

f(x) = Σ(i = 0 to ∞) ( f⁽ⁱ⁾(1)*(x-1)^i/i! )
= Σ(i = 0 to ∞) (0*(x-1)^i/i! )
= 0

But this contradicts our assumption that f(0) = 1

Therefore, no such "hump" function exists.

Edited on October 24, 2006, 5:39 pm

Posted by Tristan on 2006-10-24 17:38:50