Infinitely differentiable

[Ratzinger]

infinitely differentiable doesn't care if all the higher derivatives are zeroes (like for polynomials), it only has to be defined...correct?

 

[AKG]

Correct...

 

[Hurkyl]

It's okay for them to all be zero as well!

such as for

$$f(x) = \left\{ \begin{array}{ll} 0 & x = 0 \\ e^{-1/x^2} \quad & x \neq 0 \end{array}$$

at x = 0.