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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,

(B.1)

Bairstow's method divides the polynomial by a quadratic function.

$\displaystyle x^{2}-rx-s.$ (B.2)

Now the quotient will be a polynomial $ f_{n-2}(x), i.e.$

$\displaystyle f_{n-2}(x)=b_{2}+b_{3}x+b_{4}x^{2}+...+b_{n-1}x^{n-3}+b_{n}x^{n-2}$ (B.3)

and the remainder is a linear function $ R(x)$, i.e.

$\displaystyle R(x)=b_{1}(x-r)+b_{0}$ (B.4)

Since the quotient $ f_{n-2}(x)$ and the remainder $ R(x)$ are obtained by standard synthetic division the co-efficients can be obtained by the following recurrence relation.

$\displaystyle b_{n}=a_{n}$ (B.5a)

$\displaystyle b_{n-1}=a_{n-1}+rb_{n}$ (B.5b)

$\displaystyle b_{i}=a_{i}+rb_{i+1}+sb_{i+2}$   for$\displaystyle \quad i=n-2 \quad {to} \quad 0$ (B.5c)

If $ x^{2}-rx-s$ is an exact factor of $ f_{n}(x)$ then the remainder $ R(x)$ is zero and the real/complex roots of $ x^{2}-rx-s$ are the roots of $ f_{n}(x)$. It may be noted that $ x^{2}-rx-s$ is considered based on some guess values for $ r,s$. So Bairstow's method reduces to determining the values of r and s such that $ R(x)$ is zero. For finding such values Bairstow's method uses a strategy similar to Newton Raphson's method.

Since both $ b_{0}$ and $ b_{1}$ are functions of r and s we can have Taylor series expansion of $ b_{0}$, $ b_{1}$ as:

$\displaystyle b_{1}(r+\Delta r,\quad s+\Delta s)=b_{1}+\frac{\partial b_{1}}{\p... ...\Delta r+\frac{\partial b_{1}}{\partial s}\Delta s+O(\Delta r^{2},\Delta s^{2})$ (B.6a)

$\displaystyle b_{0}(r+\Delta r,\quad s+\Delta s)=b_{0}+\frac{\partial b_{0}}{\p... ...elta r+\frac{\partial b_{0}}{\partial s}\Delta s+ O (\Delta r^{2},\Delta s^{2})$ (B.6b)

For $ \Delta s, \Delta r << 1$, $ O(\Delta r^{2},\Delta s^{2})$ terms $ \approx 0$ i.e. second and higher order terms may be neglected, so that $ (\Delta r, \Delta s)$ the improvement over guess value $ (r,s)$ may be obtained by equating (B.6a),(B.6b) to zero i.e.

$\displaystyle \frac{\partial b_{1}}{\partial r}\Delta r+\frac{\partial b_{1}}{\partial s}\Delta s=-b_{1}$ (B.7a)

$\displaystyle \frac{\partial b_{0}}{\partial r}\Delta r+\frac{\partial b_{0}} {\partial s}\Delta s=-b_{0}$ (B.7b)

To solve the system of equations $ (B.7a)-(B.7b)$, we need the partial derivatives of $ b_{0}, b_{1}$ w.r.t. r and s. Bairstow has shown that these partial derivatives can be obtained by synthetic division of $ f_{n-2}(x)$, which amounts to using the recurrence relation $ (B.5a)-(B.5c)$ replacing $ a_{i}'s$ with $ b_{i}'s$ and $ b_{i}'s$ with $ c_{i}'s$ i.e.

$\displaystyle c_{n}=b_{n}$ (B.8a)

$\displaystyle c_{n-1}=b_{n-1}+rc_{n}$ (B.8b)

 

$\displaystyle c_{i}=b_{i}+r c_{i+1}+sc_{i+2}$
(B.8c)

for

where

$\displaystyle \frac{\partial b_{0}}{\partial r}=c_{1},\quad \frac{\partial b_{o... ...b_{1}}{\partial r}=c_{2}\quad and \quad \frac{\partial b_{1}}{\partial s}=c_{3}$
 (B.9)

% latex2html id marker 3781 $ \therefore$ The system of equations (B.7a)-(B.7b) may be written as.

(B.10a)

(B.10b)

These equations can be solved for $ (\Delta r, \Delta s)$ and turn be used to improve guess value $ (r,s)$ to .

Now we can calculate the percentage of approximate errors in (r,s) by

$\displaystyle \vert\varepsilon_{a,r}\vert=\vert\frac{\Delta r}{r}\vert\times 100;\quad\varepsilon_{a,s}=\vert\frac{\Delta s}{s}\vert\times 100$ (B.11)

If or , where $ \varepsilon_{s}$ is the iteration stopping error, then we repeat the process with the new guess i.e. . Otherwise the roots of $ f_{n}(x)$ can be determined by

$\displaystyle x=\frac{r\pm\sqrt{r^{2}+4s}}{2}$ (B.12)

If we want to find all the roots of $ f_{n}(x)$ then at this point we have the following three possibilities:
  1. If the quotient polynomial $ f_{n-2}(x)$ 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)$ can serve as the starting guesses for this application.
  2. If the quotient polynomial $ f_{n-2}(x)$ is a quadratic function then use (B.12) to obtain the remaining two roots of $ f_{n}(x)$.
  3. If the quotient polynomial $ f_{n-2}(x)$ is a linear function say $ ax+b=0$ then the remaining single root is given by

Example:
Find all the roots of the polynomial

$\displaystyle f_{4}(x)=x^{4}-5x^{3}+10x^{2}-10x+4$

by Bairstow method . With the initial values $ r=0.5,\quad s=-0.5 \quad and \quad \varepsilon_{s}=0.01.$

Solution:
Set iteration=1

$\displaystyle a_{0}=4,\quad a_{1}=-10,\quad a_{2}=10,\quad a_{3}=-5 \quad a_{4}=1$

Using the recurrence relations (B.5a)-(B.5c) and (B.8a)-(B.8c) we get

$\displaystyle b_{4}=1,\quad b_{3}=-4.5, \quad b_{2}=7.25, \quad b_{1}=-4.125,\quad b_{0}=-1.6875$

$\displaystyle c_{4}=1,\quad c_{3}=-4\quad c_{2}=4.75,\quad c_{1}=0.25$

% latex2html id marker 3833 $ \therefore$ the simultaneous equations for $ \Delta r$ and $ \Delta s$ are:

on solving we get $ \Delta r= 1.1180371,\quad \Delta s=0.296419084$

$\displaystyle r=0.5+\Delta r =1.6180371$

and

Set iteration=2

$\displaystyle b_{4}=1.0,\quad b_{3}=-3.38196278,\quad b_{2}=4.32427788,\quad b_{1}=-2.31465483,\quad b_{0}=-0.625537872$

$\displaystyle c_{4}=1.0,\quad c_{3}=-1.76392567,\quad c_{2}=1.26659977,\quad c_{1}=0.0938522071$

% latex2html id marker 3857 $ \therefore$ now we have to solve





On solving we get $ \Delta r = 2.27996969,\quad\Delta s=0.324931115$
% latex2html id marker 3865 $ \therefore r=1.6180371+\Delta r=3.89800692$
$ \quad s=-0.203580916+\Delta s=0.121350199$





Now proceeding in the above manner in about ten iteration we get $ r=3,\quad s=-2$ with

Now on using $ \displaystyle{x=\frac{r\pm\sqrt{r^{2}+4s}}{2}}\quad(i.e. \quad eqn. \quad B.12)$ we get $ \displaystyle{x=\frac{3\pm\sqrt{9-8}}{2}=2,1}$

So at this point Quotient is a quadratic equation

Roots of $ f_{2}(x)$ are: $ x=1-i,\quad1+i$

% latex2html id marker 3889 $ \therefore$ Roots $ f_{4}(x)$ are $ =1-i,\quad 1+i,\quad 1,\quad 2$
i.e $ f_{4}(x)=(x-(1-i))(x-(1+i))(x-1)(x-2).$

Exercises:
(1) Use initial approximation to find a quadratic factor of the form of the polynomial equation

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

using Bairstow method and hence find all the roots.


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