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## Main Question or Discussion Point

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

- Thread starter Ratzinger
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- #1

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

- #2

AKG

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Correct......

- #3

Hurkyl

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such as for

[tex]

f(x) = \left\{

\begin{array}{ll}

0 & x = 0 \\

e^{-1/x^2} \quad & x \neq 0

\end{array}

[/tex]

at

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