Zero functions on open interval

Question

Are there non-constant differentiable functions that are zero on an open interval of real line? I've tried using the product integral: $$ f(x) = \exp(\int_0^1 \log(x-u) \mathrm{d}u ) = \frac{x^x (x-1)^{x-1}}{e} $$ (result from Wolfram), however $$ f(1/2)= \sqrt{\frac{1}{2}} \sqrt{-2} \frac{1}{e} = \frac{i}{e} $$ which is not $0$. Why?

Answer 1

I think the standard example of the kind of function you're looking for is: $$f\left(x\right) = \exp\left(-1/x^2\right)$$

This function is infinitely differentiable on the whole real line, and all of its derivatives are zero at $x=0$ (that is, $f'\left(0\right)=f''\left(0\right)=\cdots=0$). (of course, if you're pedant enough, you need to define explicitly $f\left(0\right)=0$).

You can use this function to "glue" together pieces of functions in an "infinitely differentiable" way.