How smooth?

Suppose f: R -> R is a smooth (=infinitely differentiable) function such that f(x) = 0 for all x < 0. Does it follow that f(x) = 0 for all x?

Originally posted by Acolyte
Suppose f: R -> R is a smooth (=infinitely differentiable) function such that f(x) = 0 for all x < 0. Does it follow that f(x) = 0 for all x?

i don't think it follows thus.
how about f(x) = x^n for n in Z+? or is that not smooth by your definition?
i know that f(x) couldn't be soemthing like sin x since f'(x) is not differentiable at x=0, nor exp x for the same reason.
here is how i think it works for f(x) = x^n:
start with n=1. then f'(x) = 0 for x&lt;0 and 1 elsewise. this function, although it contains a disconinuity, is differentiable everywhere.
for n=2, it becomes 0 and 2 at f''(x). and so forth.
it thus appears that there are infinitely many functions that meet the criterion.
am i right?

Originally posted by BarefootChessPlayer
i don't think it follows thus.
how about f(x) = x^n for n in [b]Z+? or is that not smooth by your definition?
i know that f(x) couldn't be soemthing like sin x since f'(x) is not differentiable at x=0, nor exp x for the same reason.
here is how i think it works for f(x) = x^n:
start with n=1. then f'(x) = 0 for x<0 and 1 elsewi ...[text shortened]...
it thus appears that there are infinitely many functions that meet the criterion.
am i right?[/b]

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 &gt; 0, and 0 for x &lt; 0, so it's not n times differentiable at x = 0. I haven't thought of the answer to this myself yet, but it would have to be a function whose every derivative is 0 at x = 0.

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 in Z+ 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 in R.
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&lt;d) zero and the right half that function.
would that work?

It cant be polynomial. cause the value of the nth derivitive in 0 equals (the coefficient of x^n)*n! . If that equals 0 then all coefficients equal 0.

Secondly it cant be any other function. Cause every everywhere-infinite-differentiable function can be writen as a infinite-Taylor polynomial.

Originally posted by Fiathahel
It cant be polynomial. cause the value of the nth derivitive in 0 equals (the coefficient of x^n)*n! . If that equals 0 then all coefficients equal 0.

Secondly it cant be any other function. Cause every everywhere-infinite-differentiable function can be writen as a infinite-Taylor polynomial.

Actually, BCP is right. The function he gave, exp(-1/x^2), is an example of a smooth function for which a Taylor series isn't very useful, since the error term in its Taylor series about zero is equal to the function itself.

Also, not all power series are polynomials, but in this case, it is true that the only power series which is 0 for all negative x is the zero function. From this it follows that if a power series is zero on an interval of non-zero length, it's zero everywhere; in fact, it has at most countably many zeroes, since if a dense subset of an open interval was composed of zeroes, the whole interval would be zeroes by continuity. But the same cannot be said of an arbitrary smooth function!

Yes, I see BCP is right. When I was writing my messages his wasn't there yet.