Exercise 21, § 2

Let $(a_i)_{i \leq n}$ be a finite sequence of $n$ elements of an ordered field K, not all zero; let $(a_{i_k})_{k \leq p}$ be the subsequence of $(a_i)$ consisting of the nonzero $a_i$ ($i_1 < i_2 < \ldots < i_p$); then the number of indices $k \leq p - 1$ such that $a_{i_k}$ and $a_{i_{k-1}}$ have opposite signs is called the number of variations of the sequence $(a_i)$. Let K be a maximal ordered field and let $f$ be a nonzero polynomial in $K[T]$ of degree $n > 0$; for all $a \in K$ let $w(a)$ denote the number of variations of the sequence $(f''(a))_{i \leq n}$. If $v$ is the number of roots of $f$, counted with multiplicity, in the interval ]$a, b$ (a < b), show that $v \leq w(a) - w(b)$ and that $w(a) - w(b) - v$ is even (« Budan-Fourier rule »; split the interval ]$a, b$ by the roots of $f$, and evaluate the amount of variation in $w(x)$ as $x$ passes one of these roots). Deduce that if $f(T) = a_0 + a_1 T + \ldots + a_n T^n$, then the number of roots $> 0$ of $f$, counted with multiplicity, is at most equal to the number of variations of the sequence $(a_i)_{0 \leq i \leq n}$, and that the difference between these two numbers is even (« Descartes' rule »).