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:


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

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 $2$.