proof that Even powers of an odd function's taylor polynomial vanish

Question

Let $f$ be a $k$ times continuously differentiable function defined on a neighborhood of $0 \in \mathbb{R}$. Show that if $f(-x) = -f(x) \forall x \in \mathbb{R}^n$, then the coefficients of the taylor series expanded about $0$ has all even coefficients equal $0$.

So this is an odd function, it makes sense. let's write a $k$-degree taylor polynomial:

$$\sum\limits_{i=0}^k \frac{f^{i}(0)}{i!}x^i = f(0) + f'(0)x+\frac{f''(0)x^2}{2} + . . .$$

$f(0) = 0$ we can easily conclude since $f$ is odd, but how can we say anything about any other terms? (since they all involve derivatives)

Answer 1

Here is a simple proof:

We know that the derivative of an odd function is even and the derivative of an even function is odd. Obviously, an odd function evaluated at $0$ must vanish, otherwise we would get a contradiction. Consequently, if $f(x)$ is odd, then $f^{(n)}(x)$ is either even or odd. In particular, if $n$ is even, then $f^{(n)}(x)$ is odd and as a result $f^{(n)}(0) = 0$. Hence, the Maclaurin series expansion of $f(x)$ can consists only of the odd degree terms.

Answer 2

First, show that $f\left(0\right)=0$ by continuity.

Next, show that $f''\left(0\right)=0$. This follows because

$$f''\left(0\right)=\lim_{x\to0}\left[\frac{f\left(0+x\right)-2f\left(0\right)+f\left(0-x\right)}{x^{2}}\right]=\lim_{x\to0}\left[\frac{f\left(x\right)-f\left(-x\right)}{2}\right]=0.$$

Next, do the same for all even power differentials, in a similar way.