Odd Order Derivative of Even 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 even function.

Let $f$ be at least $\paren{2 n + 1}$-times differentiable.

Then:

$\map {f^{\paren {2 n + 1} } } 0 = 0$

Proof

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

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

Differentiating $2 n + 1$ times, we have, by the Chain Rule for Derivatives:

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

Setting $x = 0$ gives:

$\map {f^{\paren {2 n + 1} } } 0 = -\map {f^{\paren {2 n + 1} } } {-0}$

That is:

$2 \map {f^{\paren {2 n + 1 } } } 0 = 0$

The result follows on division by $2$.

$\blacksquare$