Show function has no local extremum using Taylor polynomials 
Let $n$ be odd and $f \in C^n(\mathbb{R}^n, \mathbb{R})$. Further, let $x \in \mathbb{R}^n$ with $D^kf(x) = 0$ for $k = 1,...,n-1$ and $D^nf(x) \neq 0$. Show that $f$ has no local extremum in $x$.

$D^nf(x) =: p$ is a homogeneous polynomial of degree $n$ in the increment variables $\xi_1,...,\xi_n$ having their origin in x. $D^nf(x) \neq 0$ means $p$ is not identically $0$. It follows that there is a unit vector $U = (U_1,...,U_n)$ with $p(U) =: q \neq 0$.
For $||\alpha|| := \sum_{i=1}^n \alpha_i$ , $\alpha! := \prod_{i=1}^n\alpha_i!$  , $D^\alpha f := \partial_1^{\alpha_1} \partial_2^{\alpha_2} ... \partial_n^{\alpha_n}f$, $U \subseteq \mathbb{R}^n$ an open set, $f \in C^{k+1}(U)$, $x \in U$ and $\xi \in \mathbb{R}^n$ such that $x + \xi[0,1] \subseteq U$ and $\exists \theta \in (0,1)$ we defined Taylor's theorem as $$f(x + \xi) = \sum_{||\alpha|| \le k} \frac1{\alpha!}D^\alpha f(x)\xi^\alpha + \sum_{||\alpha|| = k+1}\frac1{\alpha!}D^\alpha f(x + \theta \xi)\xi^\alpha.$$
We can write $f$ this way acording to Taylor's theorem with $k := n-1$. As all $D^kf(x) = 0$ for $1 \leq k$ it follows that $$f(x + \xi) - f(x) = p(\xi) + \sum_{||\alpha|| = k+1}\frac1{\alpha!}D^\alpha f(x + \theta \xi)\xi^\alpha.$$
Choosing $\xi := \epsilon U$ we therefore have $$f(x + \epsilon U) - f(x) = \epsilon^n p(U) + \sum_{||\alpha|| = k+1}\frac1{\alpha!}D^\alpha f(x + \theta \epsilon U)\epsilon U^\alpha = \epsilon^n q + \sum_{||\alpha|| = k+1}\frac1{\alpha!}D^\alpha f(x + \theta \epsilon U)\epsilon U^\alpha$$
As $q \neq 0$ and $n$ is odd it follows that $F(x + \xi) - f(x)$ assumes both signs in the immediate neighborhood of $x$, and thus cannot have an extremum at $x$.
 A: A hint:
$D^nf(a)=:p$ is a homogeneous polynomial  of degree $n$ in the increment variables $X_1$, $\ldots$, $X_n$ having their origin at $a$. $D^nf(a)\ne0$ means that  $p$ is not identically $0$. It follows that there is a unit vector $U=(U_1,\ldots, U_n)$ with $p(U)=:\mu\ne0$.
Use this together with Taylor's theorem and the fact that $n$ is odd to produce points $a\pm\epsilon U$ for which $f(a\pm\epsilon U)-f(a)$ have different signs, so $f$ cannot have an extremum at $a$.
Full solution:
According to Taylor's theorem (in its most primitive form) we have
$$f(a+X)=\sum_{k=0}^n {1\over k!} D^kf(a)(X)+o(|X|^n)\qquad(X\to0)\ .$$
As all $D^kf(a)=0$ for $1\leq k\leq n-1$ it follows that
$$f(a+X)-f(a)=p(X)+o(|X|^n)\qquad(X\to0)\ .$$
Choosing $X:=\epsilon\, U$ we therefore have
$$f(a+\epsilon\, U)-f(a)=\epsilon^n \,p(U)+o(|\epsilon|^n)=\epsilon^n\bigl(\mu+o(1)\bigr)\qquad(\epsilon\to0)\ .$$
As $\mu\ne0$ and $n$ is odd it follows that $f(a+X)-f(a)$ assumes both signs in the immediate neighborhood of $a$.
A: It's not true that $D^2 f(x) > 0$ is a necessary condition for $x$ to be a local extremum. Take for example $f(x) = x^4$, where $Df(0) = D^2f(0) = D^3f(0) = 0$. The correct necessary condition for a sufficiently smooth $f$ is that the smallest $n$ with $D^{n+1} f(x) \neq 0$ is even. If the smallest such $n$ is odd, $x$ is a saddlepoint.
The basic idea is that you can approximate $f$ around $x$ by it's Taylor polynomial $$
  f(x + d) \approx \sum_{k=0}^n c_k d^k, \quad d_k = \frac{f^{(k)}(x)}{k!} \text{.}
$$
Since the first $n-1$ deratives are zero, you get $$
  f(x + d) \approx c_0 + c_nd^n \text{,}
$$
and since $x^n$ doesn't have a local extremem at $0$ if $n$ is odd, neither does $f$ at $x$.
To turn this into a formal proof, you need show that the approximation error is too small on some neighborhood of $x$ to intefere. For that, you'll need to use one of the forms of the remainder found here. You'll also need to extend the argument to multiple dimensions, but as far as I can see, nothing changes fundamentally there. Things just get a bit more messy.
