Descartes' Rule of Signs

Descartes' rule of signs is a criterion which gives an upper bound on the number of positive or negative real roots of a polynomial with real coefficients. The bound is based on the number of sign changes in the sequence of coefficients of the polynomial.

Let $f(x) = a_nx^n + a_{n-1}x^{n-1}+ \cdots+a_0$ be a polynomial with real coefficients. Let $s$ be the number of sign changes in the sequence $a_n,a_{n-1},\ldots,a_0$: that is, delete the terms of the sequence that are $0,$ and let $s$ be the number of pairs of consecutive terms in the remaining sequence that have opposite signs. Let $p$ be the number of positive roots of $f(x)$ (counted with multiplicity). Then $s-p$ is a nonnegative even number.

Let $f(x) = x^3-3x-2.$ Then the sequence of nonzero coefficients is $1,-3,-2,$ which changes sign once. So $s=1,$ and $s-p$ is nonnegative and even, so $p$ has to be $1.$ There is exactly one positive root.

It is possible to get some information about the negative roots as well: note that $f(-x) = (-x)^3-3(-x)-2 = -x^3+3x-2,$ and the sequence of coefficients is $-1,3,-2,$ with two sign changes. So there are either $0$ or $2$ positive roots of $f(-x).$

But the positive roots of $f(-x)$ correspond to negative roots of $f(x).$ So there are $0$ or $2$ of these. Descartes' rule of signs cannot distinguish between these two possibilities.

In fact, $x^3-3x-2 = (x+1)^2(x-2),$ so there is one positive root $2,$ and two negative roots, $-1$ and $-1$ (counted twice because it has multiplicity two).

How many positive and negative roots does $x^3-3x^2+1$ have?

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By Descartes' rule of signs, the number of sign changes is $2,$ so there are zero or two positive roots. And $f(-x) = -x^3-3x^2+1$ has one sign change, so there is exactly one negative root.

To decide whether there are zero or two positive roots, it is a good idea to look at the graph of $y=f(x),$ or to use the intermediate value theorem. In particular, $f(0) = 1$ and $f(1) = -1,$ so there is a root between $0$ and $1.$ So there can't be zero positive roots, so there must be two. (There is also a root between 2 and 3, by a similar IVT argument.) $_\square$

Suppose that $f(x)$ is a polynomial of degree $n$ with real coefficients which is expressible as the sum of at most $\big\lceil \frac n2 \big\rceil$ monomials. Suppose also that $f(0) \ne 0.$ Show that $f(x)$ has at least one non-real root.

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The number of sign changes in the coefficients of $f(x)$ is at most $\big\lceil \frac n2 \big\rceil-1,$ as is the number of sign changes in $f(-x).$ So, by Descartes' rule of signs, the number of positive roots plus the number of negative roots is at most

$2 \left\lceil \frac n2 \right\rceil - 2.$

If $n$ is even, this number is $n-2,$ and if $n$ is odd it is $n-1.$ In both cases it is less than $n.$

The fundamental theorem of algebra says that $f$ has exactly $n$ complex roots, with multiplicity, and the above argument shows that there are fewer than $n$ real roots, so $f$ has at least one non-real root. $_\square$

A specific example to illustrate the above: Suppose $f(x) = ax^5+bx^2+1,$ where $a,b$ are any real numbers and $a\ne 0.$ Then $f(x)$ has at least one non-real root.

There are two parts. The first is to show that $s-p$ is even; the second is to show that it is nonnegative.

Without loss of generality, assume that $f(x)$ has a positive leading coefficient. Now suppose the rightmost coefficient, $f(0),$ is positive as well. Then $s$ is even, since the signs of the coefficients start and end with $+,$ and the graph of $y=f(x)$ crosses the $x$-axis an even number of times as well (since it starts and ends above the $x$-axis). So $s-p$ is even. If $f(0)$ is negative, similar arguments show that both $s$ and $p$ are odd, so $s-p$ is even.

To show that $s\ge p,$ induct on $n.$ The base case $n=1$ is clear. Suppose that all polynomials of degree $n-1$ satisfy $s\ge p,$ and take a polynomial $f(x)$ of degree $n.$ Consider the polynomial $f'(x)$ of degree $n-1.$ The number of sign changes $s'$ of $f'(x)$ is either $s$ or $s-1,$ depending on whether there is a sign change at the rightmost coefficient of $f.$ But the number of positive zeroes $p'$ of $f'(x)$ is at least $p-1$ by Rolle's theorem, since between any two zeroes of $f$ there is a zero of $f'.$

Since $f'(x)$ has degree $n-1,$ the inductive hypothesis implies that $s'\ge p',$ but then

$s \ge s' \ge p' \ge p-1,$

so $s-p \ge -1,$ but $s-p$ is even, so it must be nonnegative. This proves the theorem by induction.