Descartes Rule

Descartes Rule

In this section we discuss some bounds for the number of positive roots, number of negative roots and number of nonreal complex roots for a polynomial over ¡ . These bounds can be computed using a powerful tool called “Descartes Rule”.

1. Statement of Descartes Rule

To discuss the rule we first introduce the concept of change of sign in the coefficients of a polynomial.

Consider the polynomial.

2x⁷ - 3x⁶ - 4x⁵ + 5x⁴ + 6x³ - 7x + 8

For this polynomial, let us denote the sign of the coefficients using the symbols ‘ + ’ and ‘ − ’as

+, -, -, +, +, -, +

Note that we have not put any symbol corresponding to x². We further note that 4 changes of sign occurred (at x⁶ , x⁴, x¹ and x⁰ ).

Definition 3.2

A change of sign in the coefficients is said to occur at the  j ^th power of  x  in a polynomial P(x) , if the coefficient of  x ^{j +1}  and the coefficient of  x^j  (or) also coefficient of x ^{j −1}  coefficient of x ^j are of different signs. (For zero coefficient we take the sign of the immediately preceding nonzero coefficient.)

From the number of sign changes, we get some information about the roots of the polynomial using

Descartes Rule. As the proof is beyond the scope of the book, we state the theorem without proof.

Theorem 3.7 (Descartes Rule)

If p is the number of positive zeros of a polynomial P(x) with real coefficients and s is the number of sign changes in coefficients of P(x), then s − p is a nonnegative even integer.

The theorem states that the number of positive roots of a polynomial P(x) cannot be more than the number of sign changes in coefficients of P(x) . Further it says that the difference between the number of sign changes in coefficients of P(x) and the number of positive roots of the polynomial P(x) is even.

As a negative zero of P(x) is a positive zero of P(−x) we may use the theorem and conclude that the number of negative zeros of the polynomial P( x) cannot be more than the number of sign changes in coefficients of P(− x) and the difference between the number of sign changes in coefficients of P(− x) and the number of negative zeros of the polynomial P( x) is even.

As the multiplication of a polynomial by x^k, for some positive integer k , neither changes the number of positive zeros of the polynomial nor the number of sign changes in coefficients, we need not worry about the constant term of the polynomial. Some authors assume further that the constant term of the polynomial must be non zero.

We note that nothing is stated about 0 as a root, in Descartes rule. But from the very sight of the polynomial written in the customary form, one can say whether 0 is a root of the polynomial or not. Now let us verify Descartes rule by means of certain polynomials.

2. Attainment of bounds

2 (a) Bounds for the number of real roots

The polynomial P ( x) = (x +1)(x −1)(x − 2)(x + i)(x − i) has the zeros −1, 1, 2, − i, i . The polynomial, in the customary form is x⁵ − 2x⁴ − x + 2 .This polynomial P(x) has 2 sign changes, namely at fourth and zeroth powers. Moreover,

P(−x) = −x⁵ − 2x⁴ + x + 2

has one sign change. By our Descartes rule, the number of positive zeros of the polynomial P(x) cannot be more than 2; the number of negative zeros of the polynomial P(x) cannot be more than 1. Clearly 1 and 2 are positive zeros, and −1 is the negative zero for the polynomial, x⁵ − 2x⁴ − x + 2 , and hence the bounds 2 for positive zeros and the bound 1 for negative zeros are attained. We note that i and −i are neither positive nor negative.

We know (x + 2)(x + 3)(x + i)(x − i) is a polynomial with roots −2, −3, −i, i . The polynomial, say P(x) , in the customary form is x⁴ + 5x³ + 7x² + 5x + 6 .

This polynomial P(x) has no sign change and P(−x) = x⁴ − 5x³ + 7x² − 5x + 6 has 4 sign changes. By Descartes rule, the polynomial P(x) cannot have more than 0 positive zeros and the number of negative zeros of the polynomial P(x) cannot be more than 4 .

As another example, we consider the polynomial.

xⁿ −ⁿC₁ xⁿ⁻¹ +ⁿC₂ xⁿ⁻² −ⁿC₃ xⁿ⁻³ +….+ (−1)ⁿ⁻¹ n C_(n−1) x + (−1)ⁿ .

This is the expansion of (x −1)ⁿ . This polynomial has n changes in coefficients and P(−x) has no change of sign in coefficients. This shows that the number of positive zeros of the polynomial cannot be more than n and the number of negative zeros of the polynomial cannot be more than 0. The statement on negative zeros gives a very useful information that the polynomial has no negative zeros. But the statement on positive zeros gives no good information about the positive zeros, though there are exactly n positive zeros; in fact, it is well-known that for a polynomial of degree n , the number of zeros cannot be more than n and hence the number of positive zeros cannot be more than n .

2. (b) Bounds for the number of Imaginary (Nonreal Complex)roots

Using the Descartes rule, we can compute a lower bound for the number of imaginary roots. Let m denote the number of sign changes in coefficients of P(x) of degree n; let k denote the number of sign changes in coefficients of P(−x) . Then there are at least n − (m + k ) imaginary roots for the polynomial P(x) . Using the other conclusion of the rule, namely, the difference between the number of roots and the corresponding sign changes is even, we can sharpen the bounds in particular cases.

Example 3.30

Show that the polynomial 9x⁹ + 2x⁵ − x⁴ − 7x² + 2 has at least six imaginary roots.

Solution

Clearly there are 2 sign changes for the given polynomial P(x) and hence number of positive  roots of P ( x) cannot be more than two. Further, as P(-x) = -9x⁹ - 2x⁵ - x⁴ - 7x² + 2, there is one sign change for P(-x) and hence the number of negative roots cannot be more than one. Clearly 0 is not a root. So maximum number of real roots is 3 and hence there are atleast six imaginary roots.

Remark: From the above discussion we note that the Descartes rule gives only upper bounds for the number of positive roots and number of negative roots; the Descartes rule neither gives the exact number of positive roots nor the exact number of negative roots. But we can find the exact number of positive, negative and nonreal roots in certain cases. Also, it does not give any method to find the roots.

Example 3.31

Discuss the nature of the roots of the following polynomials:

(i) x²⁰¹⁸ +1947x¹⁹⁵⁰ +15x⁸ + 26x⁶ + 2019

(ii) x⁵ -19x⁴ + 2x³ + 5x² +11

Solution

Let P(x) be the polynomial under consideration.

(i) The number of sign changes for P(x) and P(-x) are zero and hence it has no positive roots and no negative roots. Clearly zero is not a root. Thus the polynomial has no real roots and hence all roots of the polynomial are imaginary roots.

(ii) The number of sign changes for P(x) and P(-x) are 2 and 1 respectively. Hence it has at most two positive roots and at most one negative root.Since the difference between number of sign changes in coefficients of P(-x) and the number of negative roots is even, we  cannot have zero negative roots. So the number of negative roots is 1. Since the difference between number of sign changes in coefficient of P(x) and the number of positive roots must be even, we must have either zero or two positive roots. But as the sum of the coefficients is zero, 1 is a root. Thus we must have two and only two positive roots. Obviously the other two roots are imaginary numbers.