# Differentiable $N$ times, $f'(x_{0})=\cdots=f^{(N-1)}(x_0)=0$, $f^{(N)}(x_0)>0\implies f$ increasing on $x_0$?

Let $I\subseteq \mathbb {R}$ be an open interval and $f:I\rightarrow \mathbb {R}$ is differentiable $N$ times in $x_0\in I$. It's given that:

$$f'(x_{0})=\cdots=f^{(N-1)}(x_0)=0, \qquad f^{(N)}(x_0)>0$$

Prove that exists some $\delta >0$ such that $f$ is strictly monotone increasing in $(x_0-\delta,x_0+\delta)$.

I tried to use Taylor expansion to prove that the first derivative is positive close enough to $x_0$ but I need some help formalize it.

• $f(x)=x^2$; then $f'(0)=0$, $f''(0)=2>0$, but of course $f$ is not increasing in any neighborhood of $0$. You probably also have the condition that $N$ is odd. Mar 3, 2015 at 13:56
• The original question is with $N=2007$. Maybe it's important that $N>2$ ? Mar 3, 2015 at 13:58
• No, it's important that $N$ is odd, I guess. Mar 3, 2015 at 14:02

The expansion reads $$f(x_0+h)=f(x_0)+\left(\frac{f^{n}(x_0)}{n!}+\rho(h)\right)h^n,$$ with $\lim_{h\to 0}\rho(h)=0$. Argue that the expression in parenthesis has the same sign as $f^{n}(x_0)$ and is nonzero when $h$ is small. This implies $f$ strictly monotone in some neighborhood of $x_0$ when $n$ is odd.
• The remainder is usually written $r_n(h)$ with $\lim_\limits{h\to 0}r_n(h)/h^n = 0$; here I wrote $\rho(h) = r_n(h)/h^n$ with $\lim_\limits{h\to 0} \rho(h) =0$ and simplified the expansion for our analysis. Mar 3, 2015 at 14:14
It looks like the statement is false: a counterexample is $f(x)=x^{2n}\sin(1/x)$. If we added the assumption that $f$ had a continuous $n$th derivative, then we could use a Taylor expansion of $f'$ to show that $f'$ is positive in a punctured neighborhood of 0 (assuming $n$ is odd), hence $f$ would be strictly increasing.