I am reading an article of Etienne Ghys and I got stuck on a "well known" fact.
Let $f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n$ be a (function) polynomial. Define the valuation of $f$ denoted $\nu(P)$ to be the smallest integer $k$ such that $a_k\ne 0.$
Quote from the article "It is well known that the function $f$ changes sign in the neighborhood of $0$ if and only if its valuation is odd."
I am not sure about proving this. I tried on severals exemples, as $4x^3+6x^4+x^5$ to see if there is any pattern I can recognize.
Cleary we have $f(0)=0$ so there is something that ensures that $f$ is monotone over a neighborhood of $0.$
How do tackle this problem ?
The polynomial can be written as
$$x^k(a_k + a_{k+1}x^{1} + \cdots + a_n x^{n-k})$$
Now, let's take a look at that value in parentheses. Think about what happens to it when $x$ is near $0$. You can prove, in one of many ways, that there must exist some $\epsilon$ such that $a_k + a_{k+1}x^{1} + \cdots + a_n x^{n-k}$ has the same sign as $a_k$ whenever $x\in(-\epsilon, \epsilon)$.
(for example, simply from continuity of the above function, its $\epsilon-\delta$ definition, and the fact that it has a value of $a_k\neq 0$ at $x=0$, we can see that there must exist, for $\delta=\frac{|a_k|}{2}$, some $\epsilon$ such that the value of the expression falls into the interval $\left(a_k-\frac{|a_k|}{2}, a_k+\frac{|a_k|}{2}\right)$)
Assuming $a_k$ is positive, the value in parentheses will be positive for small enough values of $x$, while the value $x^k$ will be either always positive (if $k$ is even) or go from positive to negative at $0$ (if $k$ is odd).