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Bairstow Method
Bairstow Method is an iterative method used to find both the real
and complex roots of a polynomial. It is based on the idea of
synthetic division of the given polynomial by a quadratic function
and can be used to find all the roots of a polynomial. Given a
polynomial say,
![](img262a.gif) |
(B.1) |
Bairstow's method divides the polynomial by a quadratic function.
![$\displaystyle x^{2}-rx-s.$](img263.png) |
(B.2) |
Now the quotient will be a polynomial
![$\displaystyle f_{n-2}(x)=b_{2}+b_{3}x+b_{4}x^{2}+...+b_{n-1}x^{n-3}+b_{n}x^{n-2}$](img265.png) |
(B.3) |
and the remainder is a linear function
, i.e.
![$\displaystyle R(x)=b_{1}(x-r)+b_{0}$](img267.png) |
(B.4) |
Since the quotient
and the remainder
are
obtained by standard synthetic division the co-efficients
can be obtained by the following recurrence relation.
![$\displaystyle b_{n}=a_{n}$](img270.png) |
(B.5a) |
![$\displaystyle b_{n-1}=a_{n-1}+rb_{n}$](img271.png) |
(B.5b) |
for![$\displaystyle \quad i=n-2 \quad {to} \quad 0$](img273.png) |
(B.5c) |
If
![$ x^{2}-rx-s$](img274.png)
is an exact factor of
![$ f_{n}(x)$](img275.png)
then the
remainder
![$ R(x)$](img266.png)
is zero and the real/complex roots of
![$ x^{2}-rx-s$](img274.png)
are the roots of
![$ f_{n}(x)$](img275.png)
. It may be noted
that
![$ x^{2}-rx-s$](img274.png)
is considered based on some guess values for
![$ r,s$](img276.png)
. So Bairstow's method reduces to determining the values of r
and s such that
![$ R(x)$](img266.png)
is zero. For finding such values Bairstow's
method uses a strategy similar to Newton Raphson's method.
Since both
![$ b_{0}$](img277.png)
and
![$ b_{1}$](img278.png)
are functions of r and s we can
have Taylor series expansion of
![$ b_{0}$](img277.png)
,
![$ b_{1}$](img278.png)
as:
![$\displaystyle b_{1}(r+\Delta r,\quad s+\Delta s)=b_{1}+\frac{\partial b_{1}}{\p...
...\Delta r+\frac{\partial b_{1}}{\partial s}\Delta s+O(\Delta r^{2},\Delta s^{2})$](img279.png) |
(B.6a) |
![$\displaystyle b_{0}(r+\Delta r,\quad s+\Delta s)=b_{0}+\frac{\partial b_{0}}{\p...
...elta r+\frac{\partial b_{0}}{\partial s}\Delta s+ O (\Delta r^{2},\Delta s^{2})$](img280.png) |
(B.6b) |
For
,
terms
i.e. second and higher order terms may be
neglected, so that
the improvement over
guess value
may be obtained by equating (B.6a),(B.6b) to
zero i.e.
![$\displaystyle \frac{\partial b_{1}}{\partial r}\Delta r+\frac{\partial b_{1}}{\partial s}\Delta s=-b_{1}$](img286.png) |
(B.7a) |
![$\displaystyle \frac{\partial b_{0}}{\partial r}\Delta r+\frac{\partial b_{0}} {\partial s}\Delta s=-b_{0}$](img287.png) |
(B.7b) |
To solve the system of equations
![$ (B.7a)-(B.7b)$](img288.png)
, we need the
partial derivatives of
![$ b_{0}, b_{1}$](img289.png)
w.r.t. r and s. Bairstow has
shown that these partial derivatives can be obtained by synthetic
division of
![$ f_{n-2}(x)$](img268.png)
, which amounts to using the recurrence
relation
![$ (B.5a)-(B.5c)$](img290.png)
replacing
![$ a_{i}'s$](img291.png)
with
![$ b_{i}'s$](img292.png)
and
![$ b_{i}'s$](img292.png)
with
![$ c_{i}'s$](img293.png)
i.e.
![$\displaystyle c_{n}=b_{n}$](img294.png) |
(B.8a) |
![$\displaystyle c_{n-1}=b_{n-1}+rc_{n}$](img295.png) |
(B.8b) |
![$\displaystyle c_{i}=b_{i}+r c_{i+1}+sc_{i+2}$](img296.png)
|
(B.8c) |
for
where
The system of equations (B.7a)-(B.7b) may be written
as.
![](img300a.gif) |
(B.10a) |
![](img301a.gif) |
(B.10b) |
These equations can be solved for
and turn
be used to improve guess value
to
.
Now we can calculate the percentage of approximate errors in (r,s)
by
![$\displaystyle \vert\varepsilon_{a,r}\vert=\vert\frac{\Delta r}{r}\vert\times 100;\quad\varepsilon_{a,s}=\vert\frac{\Delta s}{s}\vert\times 100$](img303.png) |
(B.11) |
If
or
, where
is the iteration
stopping error, then we repeat the process with the new guess
i.e.
. Otherwise the roots of
can be determined by
![$\displaystyle x=\frac{r\pm\sqrt{r^{2}+4s}}{2}$](img307.png) |
(B.12) |
If we want to find all the roots of
then at this point
we have the following three possibilities:
-
If the quotient polynomial
![$ f_{n-2}(x)$](img268.png)
is a third (or higher)
order polynomial then we can again apply the Bairstow's method to
the quotient polynomial. The previous values of
![$ (r,s)$](img285.png)
can serve
as the starting guesses for this application.
- If the quotient polynomial
is a quadratic
function then use (B.12) to obtain the remaining two roots of
.
- If the quotient polynomial
is a linear function
say
then the remaining single root is given by
Example:
Find all the roots of the polynomial
by Bairstow method . With the initial values
Solution:
Set iteration=1
Using the recurrence relations (B.5a)-(B.5c) and
(B.8a)-(B.8c) we get
the simultaneous equations for
and
are:
on solving we get
and
Set iteration=2
Now on using
![$ \displaystyle{x=\frac{r\pm\sqrt{r^{2}+4s}}{2}}\quad(i.e. \quad
eqn. \quad B.12)$](img338.png)
we get
So at this point Quotient is a quadratic equation
Roots of
are:
Roots
are
i.e
![$ f_{4}(x)=(x-(1-i))(x-(1+i))(x-1)(x-2).$](img346.png)
Exercises:
(1) Use initial approximation
to find a quadratic factor of the form
of the polynomial equation
![](e23.gif)
using Bairstow method and hence find all its roots.
(2) Use initial approximaton
to find a quadratic factor of the form
of the polynomial equation
![](e24.gif)
using Bairstow method and hence find all the roots.
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