In the name of God
Lin-Bairstow Method
Compiled by Naser Bagheri Student ID : 9016393
Lin-Bairstow Method
http://math.fullerton.edu/mathews /n2003/bairstowmetho...
Module for The Th e Lin-Bai Bairsto rstow w Met Method hod
Quadrati Quad ratic c Synt Syntheti hetic c Divi Division sion
Let the polynomial
of degree n have coefficien coefficients ts
. Then
has the familiar
form
(1)
.
Let
be a fix fixed q qu uadratic te term. Then
(2)
,
where . Here
is th the re remainder wh when is a polynomial of degree
(3)
If we we s se et
can be expressed expressed as
and can be represented by
.
and
, then
(4)
where (5)
and equati equation on (4) can can be writ writte ten n
,
is di divided by by
Lin-Bairstow Method
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The terms in (6) can be expanded so that
is represented in powers of x.
(7)
The numbers
are found by comparing the coefficients of
(7). The coefficients
(8)
Set
of
in equations (1) and
and are computed recursively.
and
,
,
and then
for
.
Proof Lin-Bairstow Method Lin-Bairstow Method
Example 1. Use quadratic synthetic division to divide by
.
Solution 1.
Heuristics
In the days when "hand computations" were necessary, the quadratic synthetic division tableau (or table) was used. The coefficients
of the polynomial are entered on the
first row in descending order, the second and third rows are reserved for the intermediate computation steps ( ,
and
and
) and the bottom row contains the coefficients .
Lin-Bairstow Method
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Example 2. Use the "quadratic synthetic division tableau" to divide by
.
Solution 2.
Using vector coefficients
As mentioned above, it is efficient to store the coefficients of degree n in the vector index for
and the polynomial
of a polynomial
. Notice that this is a shift of the
is written in the form
.
Given the quadratic
, the quotient
and remainder
are
of
are
and .
The recursive formulas for computing the coefficients
, and
, and then
Lin-Bairstow Method
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Example 3. Use the vector form of quadratic synthetic division to divide by
.
Solution 3.
The Lin-Bairstow Method
We now build on the previous idea and develop the Lin -Bairstow's method for finding a quadratic factor
of
. Suppose that we start with the initial guess
(9)
and that
can be expressed as
.
When
u
and
v
are small, the quadratic (9) is close to a factor of
new values
so that
(10)
is closer to a factor of Observe that
u
and
than the quadratic (9). v
are functions of r and
, and .
The new values
satisfy the relations
s,
that is
. We want to find
Lin-Bairstow Method
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.
The differentials of the functions
u
and
v
are used to produce the approximations
and
The new values
are to satisfy
, and .
When the quantities
are small, we replace the above approximations with
equations and obtain the linear system:
(11)
All we need to do is find the values of the partial derivatives and
and then use Cramer's rule to compute
,
,
. Let us announce that the
values of the partial derivatives are
where the coefficients
are built upon the coefficients
calculated recursively using the formulas
(12)
Set
,
and
given in (8) and are
Lin-Bairstow Method
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for
.
The formulas in (12) use the coefficients
in (8). Since
, and , and
the linear system in (11) can be written as
Cramer's rule can be used to solve this linear system. The required determinants are
, and
,
and the new values
.
are computed using the formulas
,
and .
Proof Lin-Bairstow Method Lin-Bairstow Method
The iterative process is continued until good approximations to found. If the initial guesses
r
and
s
have been
are chosen small, the iteration does not tend to
wander for a long time before converging. When
, the larger powers of x can be
neglected in equation (1) and we have the approximation
.
He
th initial
fo
uld be
and
vided that
Lin-Bairstow Method
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If hand calculations are done, then the quadratic synthetic division tableau can be extended to form an easy way to calculate the coefficients
.
Bairstow's method is a special case of Newton's method in two dimensions.
Algorithm (Lin-Bairstow Iteration). To find a quadratic factor of approximation
.
Computer Programs Lin-Bairstow Method Lin-Bairstow Method Mathematica Subroutine (Lin-Bairstow Iteration).
given an initial
Lin-Bairstow Method
Example 4. Given
http://math.fullerton.edu/mathews /n2003/bairstowmetho...
. Start with
Lin-Bairstow method to find a quadratic factor of
and
and use the
.
Solution 4.
Research Experience for Undergraduates Lin-Bairstow Method Lin-Bairstow Method Internet hyperlinks to web sites and a bibliography of articles.
Lin-Bairstow Method
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Example 1. Use quadratic synthetic division to divide
by
Solution 1. First, construct the polynomial
Second, given
.
construct the polynomials
.
.
Lin-Bairstow Method
Third, verify that
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.
Lin-Bairstow Method
http://math.fullerton.edu/mathews/n2003/bairstowmetho...
We are done.
Aside. We can have Mathematica compute the quotient and remainder using the built in procedures PolynomialQuotient and PolynomialRemainder. This is just for fun!
Lin-Bairstow Method
Example
http://math.fullerton.edu/mathews /n2003/ bairstowmetho...
2. Use
the "quadratic synthetic division tableau" to divide by
Solution
.
2.
Use the "quadratic synthetic division tableau"
Since
Then simplify and get.
Thus we have
and
we use
.
Lin-Bairstow Method
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This agrees with our previous computation in Example
We are done.
Aside. We can let Mathematica verify the result .
1.
Lin-Bairstow Method
Example
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3.
Use the vector form of quadratic synthetic division to divide by
Solution
.
3.
First, construct the polynomial
Second, given
.
construct the polynomials
.
Lin-Bairstow Method
Third, verify that
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.
Lin-Bairstow Method
Example
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4. Given
.
Start with
Lin-Bairstow method to find a quadratic factor of Solution
and
and use the
.
4.
Enter the coefficients of the polynomial.
Enter the starting values
and
Verify that a quadratic factor has been found.
and call the subroutine Bairstow.
Lin-Bairstow Method
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We are done.
Aside. We can let Mathematica find the factors too. This is just for fun.
Aside. We can let Mathematica find the roots too. This is just for fun.
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Sun 20 Oct 2013 11:52:32 PM IRST
//********************************************** // Programmer : Naser . Bagheri (9016393) //********************************************** #include #include u si ng n am es pa ce std ; #define ESP 0.001 #define F(x) ( x)*( x)*(x) + ( x) + 10 #define a3 1 #define a2 0 #define a1 1 #define a0 10 //#define c3 0void main() int main() { double u,v,u1,v1 ,u2 ,v2 ,b3,b2 ,p,b1 ,b0 ,c2,c1,c0 ,U,V; int i=1; float c3=0; co ut <<"\nEnter the value of u: " ; sca nf ("%lf",&u); co ut <<"\nEnter the value of v: " ; sca nf ("%lf",&v); b3 =a3 ; b2 =a2 +u*b3 ; b1 =a1 +u*b2 +v*b3; b0 =a0 +u*b1 +v*b2; c2 =b3 ; c1 =b2 +u*c2 +v*c3; c0 =b1 +u*c1 +v*c2; p=c1 *c1 -c0 *c2; U=((-(b1*c1-b0 *c2 ))/( p)); V=((-(b0*c1-c0 *b1 ))/( p)); u1 =u+U; v1 =v+V; co ut <<"\n\n b0 = %lf" <
do { u =u1 ; v =v1 ; b3 =a3; b2 =a2+u*b3; b1 =a1+u*b2+v*b3 ; b0 =a0+u*b1+v*b2 ; c2 =b3; c1 =b2+u*c2+v*c3 ; c0 =b1+u*c1+v*c2 ; p =c1 *c1-c0*c2 ; U =((-(b1 *c1 -b0*c2))/(p)); - 1 -
/home/naser/Darss/Analize-adadi/bros/Bairsto.cpp Page 2 of 3 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116
V =((-(b0 *c1 -c0*b1))/(p)); u2 =u+U; v2 =v+V; cou t << "\n\n b0 = %lf" << b0 ; co ut <<"\n\n b1 = %lf" <
if (fabs(u1 - u2 ) < ESP && fabs(v1 -v2) < ESP) { cou t <<"\n\nREAL ROOT = %.3lf" << u2; cou t <<"\n\nREAL ROOT = %.3lf" << v2; i=0; } else { u1 = u2 ; v1 = v2 ; } } while (i!= 0); } /* ---------------------------------OUT PUT ----------------------------------Enter the value of u: 1.8 Enter the value of v: -4 -----------------------------------------------b0 = 3.232000 b1 = 0.240000 b2 = 1.800000 b3 = 1.000000 c0 c1 c2 c3
= = = =
2.720000 3.600000 1.000000 0.000000
u = 2.031250 v = -5.072500 b0 b1 b2 b3
= = = =
-0.194891 0.053477 2.031250 1.000000
c0 c1 c2 c3
= = = =
3.656953 4.271250 1.000000 0.000000 - 2 -
Sun 20 Oct 2013 11:52:32 PM IRST
/home/naser/Darss/Analize-adadi/bros/Bairsto.cpp Page 3 of 3 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150
u = 2.002230 v = -5.002025 b0 b1 b2 b3
= = = =
-0.001389 0.006900 2.002230 1.000000
c0 c1 c2 c3
= = = =
6.121717 5.552230 1.000000 0.000000
u = 2.000623 v = -5.000003 b0 b1 b2 b3
= = = =
0.001859 0.002490 2.000623 1.000000
c0 c1 c2 c3
= = = =
11.224619 8.108540 1.000000 0.000000
u = 2.000287 v = -4.999767 REAL ROOT = 2.000 REAL ROOT = -5.000 */
- 3 -
Sun 20 Oct 2013 11:52:32 PM IRST
Bairstow 's Method
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A procedure for finding the quadratic factors for the complex conjugate roots of a polynomial
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REFERENCES: Wolfram Web Resources » 13,192 entries Last updated: Tue Oct 8 2013
Created, developed, and nurtured by Eric Weisstein at Wolfram Research
Press, W. H.; Flannery, B. P.; Teukolsky, S. A.; and Vetterling, W. T. Numerical Recipes in FORTRAN: The Art of Scientific Computing, 2nd ed. Cambridge, England: Cambridge University Press, pp. 277 and 283-284, 1989.
Referenced on Wolfram|Alpha: Bairstow's Method CITE THIS AS: Weisstein, Eric W. "Bairstow's Method." From MathWorld --A Wolfram Web Resource. http://mathworld.wolfram.com /BairstowsMethod.html
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