Budan's theorem

Remarks

The roots of $p(x + c)$ are the roots of $p$ shifted by $-c$. So Descartes' rule of signs applied to $p(x + l)$ and $p(x + r)$ bounds the roots of $p$ greater than $l$ and greater than $r$. The theorem says that the difference of the two bounds is again a bound, with the same parity rule. Example: take $p = x^2 - 2$, $l = 0$ and $r = 2$. Then $p(x) = x^2 - 2$ has $V_0 = 1$, and $p(x + 2) = x^2 + 4x + 2$ has $V_2 = 0$. So there is exactly one root in $(0, 2]$, namely $\sqrt 2$. Take $l = 0$ and $r$ larger than the real part of every complex root of $p$. Then all coefficients of $p(x + r)$ have the same sign, so $V_r = 0$, and one gets back Descartes' rule.

Going deeper

For a student

The theorem is used in root isolation: an interval with $V_l - V_r = 0$ has no root, and one with $V_l - V_r = 1$ has exactly one. An equivalent form, due to Fourier, counts sign changes in the sequence $p(c), p'(c), \dots, p^{(n)}(c)$. By Taylor's formula, these numbers are the coefficients of $p(x + c)$ up to positive factors. Compared with Sturm's theorem, it is cheaper to compute but gives only a bound.