All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 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)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: Solution | Comment 2 of 9 |
(In reply to Solution by Tristan)

It is not true that every infinitely differentiable function of a real variable has a convergent Taylor series.  The function that is equal to exp(-1/x) for x>0 and equal to 0 for all other real x is an example.
 Posted by Richard on 2006-10-24 18:28:32

 Search: Search body:
Forums (0)