Descartes' Rule of Signs

Descartes' rule of signsis a criterion which gives an upper bound on the number of positive or negative realrootsof 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多重性). 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 $2$这些。笛卡尔的规则不能distingu迹象ish 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 theintermediate 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 $n$与实际系数表达ssible 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 $n.$

Thefundamental theorem of algebrasays 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 $p-1$byRolle'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 byinduction.