Chapter 4

Roots of Polynomials

Objectives• Understand the importance of finding polynomial

roots in engineering applications• Know the conventional method concept• Know the Muller’s method concept• Know the Bairstow’s method

Content

• Polynomials in engineering and science• Conventional method• Muller’s method• Bairstow’s method• Conclusions

Polynomials in… (1)General solutions of linear ODE

Solve for general solution

Change to characteristic equations:

The results can be :-

Polynomials in…(2)General solutions of linear ODE

Polynomials in…(3)Problem :

nnon xaxaxaaxf 2

21)(Follow these rules:1.For an nth order equation, there are

n real or complex roots.2.If n is odd, there is at least one real

root.3.If complex roots exist, they will be

in conjugate pairs (that is, l+mi and l-mi), where i=sqrt(-1).

Conventional…Only real roots exist However,

Finding good initial guesses complicates both the open and bracketing methods, also the open methods could be susceptible to divergence.

Real and complex roots of polynomials – Müller and Bairstow methods.

Müller method (1) Like Secant, Müller’s method

obtains a root estimate by projecting a parabola to the x axis through three function values.

Müller method (2)

,3,2,1)()(

)(

)()()(

1

11

1

1

ixfxf

xxxfxx

xxxfxfxf

ii

iiiii

ii

iii

Secant method (linear approximation)

Müller method (3)

Müller method(Parabola or 2nd order

approximation)

Must use three points to approximate function

Müller method (4)Müller methodology derivation

cxxbxxaxf )()()( 22

22

Write the equation in a convenient form at point x2:

We then have three eqns now (from x0, x1, and x2)

cxxbxxaxf

cxxbxxaxf

cxxbxxaxf ooo

)()()(

)()()(

)()()(

222

222

212

211

22

2

Müller method (5)Step III Reduce to two eqns

)()()()(

)()()()(

212

2121

22

22

xxbxxaxfxf

xxbxxaxfxf ooo

Right now u can solve for a and b from

When u know a, b, c you are ready toestimate root from

cxxbxxaxf )()()( 22

22

Müller method (6)Step IV Here the new estimated root is

acbb

cxx4

2223

Error can be derived from

%1003

23

xxx

a

Two roots, but which one ?

Müller method (7)Summary of algorithm

Start with 3 points [x0,f(x0)] [x1,f(x1)] and [x2,f(x2)]

)(

)()()()(x-xhx-xh If

2111

1

1

121

1

121o1o

xfcahbhh

a

hxfxf

hxfxf

o

o

o

oo

Calculate a, b, and c from

Müller method (8)Summary of algorithm (cont’d)

acbb

cxx4

2223

Calculate new root from

%1003

23

xxx

a

Calculate error

Check whethersa

new xi-1 = old xi

Bairstow’s method (1)An iterative approach loosely related

to both Müller and Newton Raphson methods.

Based on dividing a polynomial by a factor x-t:

0 to11

nitbabab

iii

nn

nnon xaxaxaaxf 2

21)(Start with

123211 )(

nnn xbxbxbbxf

Dividing with x-t yields

and a remainder R=b0 The coefficients of polynomial are

Bairstow’s method (2)To permit the evaluation of complex

roots, Bairstow’s method divides the polynomial by a quadratic factor x2-rx-s:

02

)()(

21

11

1

231322

ton-isbrbabrbab

ab

brxbRxbxbxbbxf

iiii

nn-n-

nn

o

nn

nnn

For the remainder to be zero, bo and b1 must be zero. However, it is unlikely that our initial guesses at the values of r and s will lead to this result, so we do this…

Bairstow’s method (3)Using a similar approach to Newton Raphson method, both bo and b1 can be expanded as function of both r and s in Taylor series.

),(),(

),(),(

22

221111

srOssbr

rbbssrrb

srOssbr

rbbssrrb

oooo

ooo bssbr

rb

bssbr

rb

111

Neglect higher-order termsWe estimate Δr and Δs from

How can we find these partial derivatives ???

Bairstow’s method (4)Partial derivatives can be obtained by a synthetic division of the b’s in a similar fashion the b’s themselves are derived:

31

21

1

21

11

12

csbc

rb

sbc

rb

toniscrcbcrcbc

bc

oo

iiii

nnn

nn

where

then

obscrcbscrc

21

132

Bairstow’s method (5)At each step, the error can be estimated as

%100

%100

,

,

ssrr

sa

ra

242 srrx

The roots can be determined from

Bairstow’s method (6)• At this point three possibilities exist:

1.The quotient is a third-order polynomial or greater. The previous values of r and s serve as initial guesses and Bairstow’s method is applied to the quotient to evaluate new r and s values.

2.The quotient is quadratic. The remaining two roots are evaluated directly, using

3.The quotient is a 1st order polynomial. The remaining single root can be evaluated simply as x=-s/r.

242 srrx