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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.