BAIRSTOW METHODMULLIER METHOD

DANIEL FERNANDO RODRIGUEZ

ROOTS OF POLYNOMIALS

Bairstow MethodA method for calculating roots of polynomials

can calculate peer (conjugated in the case of complex roots).

Unlike Newton, calculate complex roots without having to make calculations with complex numbers.

 It is based on the synthetic division of the polynomial Pn (x) by the quadratic (x2 - rx - s).

Bairstow Method

The synthetic division can be extended to quadratic factors:

and even by multiplying the coefficients is obtained:

01 ...)( brxbresiduo

RxQsrxxxP nn )()()( 22

residuobxbxbxbsrxx nn

nn

233

122 ...

2100

3211

1233

122

11

2100

3211

1233

122

11

1

::

sbrbab

sbrbab

sbrbab

sbrbab

rbab

ab

sbrbba

sbrbba

sbrbba

sbrbba

rbba

ba

nnnn

nnnn

nnn

nn

nnnn

nnnn

nnn

nn

Bairstow MethodWe want to find the values of r and s that make b1

and b0 equal to zero since, in this case, the factor divided exactly quadratic polynomial.

 The first method works by taking an initial approximation (r0, s0) and generate approximations (rk, sk) getting better using an iterative procedure until the remainder of division by the quadratic polynomial (x2 - rkx - sk) is zero.

 The iterative procedure of calculation is based on the fact that both b1 and b0 are functions of r and s.

Bairstow MethodIn developing b1 (rk, sk) and b0 (rk, sk) in Taylor

series around the point (r *, s *), we obtain:

It takes (r *, s *) as the point where the residue is zero and Δr = r * - rk, Δs = s * - sk. Then:

...)*()*(),(*)*,(

...)*()*(),(*)*,(

0000

1111

kkkk

kkkk

sss

brr

r

bsrbsrb

sss

brr

r

bsrbsrb

ss

br

r

bbsrb

ss

br

r

bbsrb

0000

1111

0*)*,(

0*)*,(

Bairstow Method

Bairstow showed that the required partial derivatives can be obtained from the bi by a second synthetic division between factor (x2 - r0x - s0) in the same way that the bi are obtained from the ai.

 The calculation is:

)2()1(

122

11

:

knknknkn

nnnn

nnn

nn

scrcbc

scrcbc

rcbc

bc

Bairstow Method

Thus, the system of equations can be written

021

132

bscrc

bscrc

22

200

12

110

33

321

23

221

cs

bsb

s

br

s

bc

r

bsb

r

br

r

b

cs

bsb

s

br

s

bc

r

bsb

r

br

r

b

Bairstow Method

Calculation of approximate error:

When tolerance is reached estimated coefficients

r and s is used to calculate the roots:

%100.%100. ,, s

s

r

rsara

2

42 srrx

Bairstow Method

Then:When the resulting polynomial is of third order or

more, the Bairstow method should be applied to obtain a resultant function of order 2.

When the result is quadratic polynomial, defines two of the roots using the quadratic equation.

When the final function is first order root is determined from the clearance of the equation.

a

acbbx

2

42

Muller Method

Is based on the layout of a polynomial function specifically a parable with three initial values.

Is to have the coefficients of a parabola passing through three points. These points are substituted into the quadratic formula to get the value where the parabola intersects the x-axis, ie, the approximate root.

The approach is facilitated by writing the parable in a convenient form:

the parabola must pass through three points. These are evaluated as follows:

cxxbxxaxP ii )()()( 12

1

cxxbxxaxf

cxxbxxaxf

cxxbxxaxf

iiiii

iiiii

iiiii

)()()(

)()()(

)()()(

112

111

12

1

112

111

You can find the three unknown coefficients a, b, c and to two terms of the last equation are zero, f (x +1) = c, resulting in two equations with two unknowns

)()()()(

)()()()(

12

11

112

1111

iiiiii

iiiiii

xxbxxaxfxf

xxbxxaxfxf

An algebraic manipulation allows you to find the remaining coefficients a, b. how to do this is to define the differences

ii

iii

ii

iii

iii

iii

xx

xfxf

xx

xfxf

xxh

xxh

1

1

1

11

1

11

)()(

)()(

These are replaced in the above equations and result:

Where a and b are cleared and get:

iiii

iiiiiiii

hahbh

hhahhbhh

2

112

11 )()(

)( 1

1

1

i

ii

ii

ii

xfc

ahb

hha

Already known evaluate the quadratic coefficients:

Evaluated to determine the sign:

If D1 is developed further with the + quadratic, but is solved with the - sign

acbb

cxx ii

4

2212

acbbD

acbbD

4

4

22

21

EXAMPLE:Use the muller method with initial values:

To determine the root of the equation:

5.41 ix 5.5ix

51 ix

1213)( 3 xxxf

First evaluate the function to baseline:

That are used to calculate:

1213)( 3 xxxf

48)5(

875.82)5.5(

625.20)5.4(

f

f

f

75.695.55

875.8248

25.625.45.5

625.20875.82

5.05.55

15.45.5

1

1

i

i

i

i

h

h

These values are replaced in turn to find the values of a, b, c

Then we find the term D major to determine the sign of the quadratic

28

75.6275.69)5.0(15

1515.0

25.6275.69

c

b

a

22.3048*15*425.6225.62

79.9348*15*425.6225.622

2

21

D

D

As D1> D2 solve the quadratic with a positive sign.

After the error is calculated accordingly to the Xi +1 xi +2 with the new variables:

976487.354451.3125.62

)48(252

ix

%1002

12

Xi

XiXia

74.25%100976487.3

023513.1

a

Now for the new iteration

Xi-1 = Xi previous

Xi +1 = Xi previous

Xi +2 = Xi +1 calculated

BIBLIOGRAPHY CHAPRA, Steven C. y CANALE, Raymond P.:

Métodos Numéricos para Ingenieros. McGraw Hill 2002.

http://ocw.mit.edu/OcwWeb/Mathematics

BURDEN, Richard L. y Faires J.: Análisis Numérico. Séptima Edición.