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,

$$f_{n}(x)=a_{0}+a_{1}x+a_{2}x^{2}+ ...+a_{n} x^{n}$$ (B.1)

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

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

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

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

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

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

Since the quotient $f_{n-2}(x)$ and the remainder $R(x)$ are obtained by standard synthetic division the co-efficients $b_{i} (i=0...x)$ can be obtained by the following recurrence relation.

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

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

$$b_{i}=a_{i}+rb_{i+1}+sb_{i+2} \quad i=n-2 \quad {to} \quad 0$$ (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 verily 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:

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

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

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

$$\frac{\partial b_{0}}{\partial r}\Delta r+\frac{\partial b_{0}} {\partial s}\Delta s=-b_{0}$$ (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.

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

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

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

for $i=n-2...1$

where

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

$\therefore$ the system of equation (B.7a)-(B.7b) may be written as.

$$c_{2}\quad \Delta r+c_{3}\quad \Delta s=-b_{1}$$ (B.10a)

$$c_{1}\quad \Delta r+c_{2}\quad \Delta s=-b_{0}$$ (B.10b)

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

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

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

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

$$x=\frac{r\pm\sqrt{r^{2}+4s}}{2}$$ (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

   $$x=\frac{-b}{a}$$ (B.13)

   
   

Example:
Find all the roots of the polynomial

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

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

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

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

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

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

$\therefore$ the simultaneous equations for $\Delta r$ and $\Delta s$ are:

$$4.75\quad\Delta r-4\quad\Delta s=+4.125$$

$$0.25\quad\Delta r+4.75\quad\Delta s=1.6875$$

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

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

$$S=-0.5+\Delta s=-0.203580916$$

and

$$\vert\varepsilon_{a,r}\vert=\left\vert\frac{1.1180371}{1.6180371}\right\vert\times 100=69.0983582$$

$$\vert\varepsilon_{a},s\vert=\left\vert\frac{0.296419084}{-0.203580916}\right\vert\times100=145.602585$$

Set iteration=2

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

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

$\therefore$ now we have to solve
$1.26659977\quad\Delta r-1.76392567\quad\Delta s=2.31465483$
$0.0938522071\quad\Delta r+1.26659977\quad\Delta s=0.625537872$

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

$${\vert\varepsilon_{a},r\vert=\left\vert\frac{2.27996969}{3.8900602}\right\vert}\times 100=58.490654$$

$${\vert\varepsilon_{a},s\vert=\left\vert\frac{0.32493115}{0.121350199}\right\vert}\times 100=267.763153$$

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

$$\vert\varepsilon_{a},r\vert\sim 7.95\times10^{-6}<\varepsilon_{s}=0.01$$

$$\vert\varepsilon_{a},s\vert\sim 5.96\times10^{-6}<\varepsilon_{s}=0.01$$

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

So at this point Quotient is a quadratic equation

$$f_{2}(x)=x^{2}-2x+2$$

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

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