Originally posted by Acolyte
Well, there are either an uncountable number of suitable functions, or there are none. The function you've given (x^n for x non-negative) would be an example, had I said n-1 times differentiable. But its (n-1)th derivative is n!x for x > ...[text shortened]... would have to be a function whose every derivative is 0 at x = 0.
something liek that, i'd say.
the problem with a step function (f(x) = 0 for x=0 and n! for n
otherwise) is that both sides have derivative 0, and the limit at the discontinuity is 0, but you'd never be able to get the original function back if you tried to integrate the derivative!
by the way, i think the family of fucntions we're talking about here (x^n) is countable unless you allow a*x^n for all a
my calculus book gave a function which is infinitely differentiable which has derivative zero at one point: f(x) = exp (-1/x^2) and all variants thereof (a* exp (b/(x-d)^2), all in R
). i don't remember the proof; it was over thirty years ago that i did this stuff!
so maybe the solution is the make the left half (x<d
) zero and the right half that function.
would that work?