Even Order Derivative of Odd Function Vanishes at Zero

Theorem

Let $X$ be a symmetric subset of $\R$ containing $0$.

Let $n$ be a positive integer.

Let $f:X \to \R$ be an odd function.

Let $f$ be at least $\paren {2 n}$-times differentiable.

Then:

$\map {f^{\paren {2 n} } } 0 = 0$

Proof

From the definition of an odd function, for all $x \in X$ we have:

$\map f x = -\map f {-x}$

Differentiating $2 n$ times, we have, by the Chain Rule for Derivatives:

$\map {f^{\paren {2 n} } } x = -\paren {-1}^{2 n} \map {f^{\paren {2 n} } } {-x} = -\map {f^{\paren {2 n} } } {-x}$

Setting $x = 0$ gives:

$\map {f^{\paren {2 n} } } 0 = -\map {f^{\paren {2 n} } } {- 0}$

That is:

$2 \map {f^{\paren {2 n} } } 0 = 0$

The result follows on division by $2$.

$\blacksquare$