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§1.11 Zeros of Polynomials

Contents

§1.11(i) Division Algorithm

Horner’s Scheme

Let

1.11.1 f(z)=anzn+an-1zn-1++a0.

Then

1.11.2 f(z)=(z-α)(bnzn-1+bn-1zn-2++b1)+b0,

where bn=an,

1.11.3 bk=αbk+1+ak,
k=n-1,n-2,,0,
1.11.4 f(α)=b0.

Extended Horner Scheme

With bk as in (1.11.1)–(1.11.3) let cn=an and

1.11.5 ck=αck+1+bk,
k=n-1,n-2,,1.

Then

1.11.6 f(α)=c1.

More generally, for polynomials f(z) and g(z), there are polynomials q(z) and r(z), found by equating coefficients, such that

1.11.7 f(z)=g(z)q(z)+r(z),

where 0degr(z)<degg(z).

§1.11(ii) Elementary Properties

A polynomial of degree n with real or complex coefficients has exactly n real or complex zeros counting multiplicity. Every monic (coefficient of highest power is one) polynomial of odd degree with real coefficients has at least one real zero with sign opposite to that of the constant term. A monic polynomial of even degree with real coefficients has at least two zeros of opposite signs when the constant term is negative.

Descartes’ Rule of Signs

The number of positive zeros of a polynomial with real coefficients cannot exceed the number of times the coefficients change sign, and the two numbers have same parity. A similar relation holds for the changes in sign of the coefficients of f(-z), and hence for the number of negative zeros of f(z).

Example

1.11.8 f(z) =z8+10z3+z-4,
f(-z) =z8-10z3-z-4.

Both polynomials have one change of sign; hence for each polynomial there is one positive zero, one negative zero, and six complex zeros.

Next, let f(z)=anzn+an-1zn-1++a0. The zeros of znf(1/z)=a0zn+a1zn-1++an are reciprocals of the zeros of f(z).

The discriminant of f(z) is defined by

1.11.9 D=an2n-2j<k(zj-zk)2,

where z1,z2,,zn are the zeros of f(z). The elementary symmetric functions of the zeros are (with an0)

1.11.10 z1+z2++zn =-an-1/an,
1j<knzjzk =an-2/an,
z1z2zn =(-1)na0/an.

§1.11(iii) Polynomials of Degrees Two, Three, and Four

Quadratic Equations

The roots of az2+bz+c=0 are

1.11.11 -b±D2a,
D =b2-4ac.

The sum and product of the roots are respectively -b/a and c/a.

Cubic Equations

Set z=w-13a to reduce f(z)=z3+az2+bz+c to g(w)=w3+pw+q, with p=(3b-a2)/3, q=(2a3-9ab+27c)/27. The discriminant of g(w) is

1.11.12 D=-4p3-27q2.

Let

1.11.13 A =-272q+32-3D3,
B =-3p/A.

The roots of g(w)=0 are

1.11.14 13(A+B),
13(ρA+ρ2B),
13(ρ2A+ρB),

with

1.11.15 ρ =-12+12-3
=2π/3,
ρ2 =-2π/3.

Addition of -13a to each of these roots gives the roots of f(z)=0.

Example

f(z)=z3-6z2+6z-2, g(w)=w3-6w-6, A=343, B=323. Roots of f(z)=0 are 2+43+23, 2+43ρ+23ρ2, 2+43ρ2+23ρ.

For another method see §4.43.

Quartic Equations

Set z=w-14a to reduce f(z)=z4+az3+bz2+cz+d to

1.11.16 g(w) =w4+pw2+qw+r,
p =(-3a2+8b)/8,
q =(a3-4ab+8c)/8,
r =(-3a4+16a2b-64ac+256d)/256.

The discriminant of g(w) is

1.11.17 D=16p4r-4p3q2-128p2r2+144pq2r-27q4+256r3.

For the roots α1,α2,α3,α4 of g(w)=0 and the roots θ1,θ2,θ3 of the resolvent cubic equation

1.11.18 z3-2pz2+(p2-4r)z+q2=0,

we have

1.11.19 2α1 =-θ1+-θ2+-θ3,
2α2 =-θ1--θ2--θ3,
2α3 =--θ1+-θ2--θ3,
2α4 =--θ1--θ2+-θ3.

The square roots are chosen so that

1.11.20 -θ1-θ2-θ3=-q.

Add -14a to the roots of g(w)=0 to get those of f(z)=0.

Example

f(z)=z4-4z3+5z+2,   g(w)=w4-6w2-3w+4. Resolvent cubic is z3+12z2+20z+9=0 with roots θ1=-1, θ2=-12(11+85), θ3=-12(11-85), and -θ1=1, -θ2=12(17+5), -θ3=12(17-5). So 2α1=1+17, 2α2=1-17, 2α3=-1+5, 2α4=-1-5, and the roots of f(z)=0 are 12(3±17), 12(1±5).

§1.11(iv) Roots of Unity and of Other Constants

The roots of

1.11.21 zn-1=(z-1)(zn-1+zn-2++z+1)=0

are 1, 2π/n, 4π/n,,(2n-2)π/n, and of zn+1=0 they are π/n,3π/n,,(2n-1)π/n.

The roots of

1.11.22 zn=a+b,
a,b real,

are

1.11.23 Rn(cos(α+2kπn)+sin(α+2kπn)),

where R=(a2+b2)1/2, α=ph(a+b), with the principal value of phase (§1.9(i)), and k=0,1,,n-1.

§1.11(v) Stable Polynomials

1.11.24 f(z)=a0+a1z++anzn,

with real coefficients, is called stable if the real parts of all the zeros are strictly negative.

Hurwitz Criterion

Let

1.11.25 D1 =a1,
D2 =|a1a3a0a2|,
D3 =|a1a3a5a0a2a40a1a3|,

and

1.11.26 Dk=det[hk(1),hk(3),,hk(2k-1)],

where the column vector hk(m) consists of the first k members of the sequence am,am-1,am-2, with aj=0 if j<0 or j>n.

Then f(z), with an0, is stable iff a00; D2k>0, k=1,,12n; signD2k+1=signa0, k=0,1,,12n-12.