(15C) Bairstow's Method

[Thomas Klemm]

Bairstow's Method

Links

Wikipedia
MathWorld

Algorithm

We are looking for a quadratic factor T(x) of the polynomial P(x). In general there will be a linear remainder R(x) after the division:

\(P(x) = Q(x) \cdot T(x) + R(x)\)

\(P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0\)

\(T(x) = x^2 + p x + q\)

\(Q(x) = b_n x^{n-2} + b_{n-1} x^{n-3} + \cdots + b_4 x^2 + b_3 x + b_2\)

\(R(x) = b_1 ( x + p ) + b_0\)

If we set \(b_{n+1}\equiv b_{n+2}\equiv 0\), then \(\forall k\leq n\):

\(b_k = a_k - b_{k+1} p - b_{k+2} q\)

The two coefficients of the remainder R(x) depend on p and q:

\(b_0 = b_0(p, q)\)

\(b_1 = b_1(p, q)\)

To make T(x) a factor of P(x) the remainder R(x) must vanish. Let's assume that we already have p and q close to the solution. We want to know how to change these values to get a better solution. Therefore we set the first order Taylor series expansion to 0:

\(b_0(p + \delta p, q + \delta q) \approx b_0(p, q) + \frac{\partial b_0}{\partial p} \delta p + \frac{\partial b_0}{\partial q} \delta q = 0\)

\(b_1(p + \delta p, q + \delta q) \approx b_1(p, q) + \frac{\partial b_1}{\partial p} \delta p + \frac{\partial b_1}{\partial q} \delta q = 0 \)

It turns out that the partial derivatives of \(b_k\) with respect to p and q can be calculated with a similar recurrence formula:

\(c_{k+1} = - \frac{\partial b_k}{\partial p}\)

\(c_{k+2} = - \frac{\partial b_k}{\partial q}\)

\(c_k = b_k - c_{k+1} p - c_{k+2} q\)

Thus we have to solve:

\(\begin{matrix}
c_1 \delta p + c_2 \delta q &= b_0 \\
c_2 \delta p + c_3 \delta q &= b_1
\end{matrix}\)


Using Linear Regression

We have to solve a 2-dimensional linear equation:

\(\begin{bmatrix}
c_1 & c_2 \\
c_2 & c_3
\end{bmatrix}\cdot\begin{bmatrix}
\delta p \\
\delta q
\end{bmatrix}=\begin{bmatrix}
b_0 \\
b_1
\end{bmatrix}\)

But since the matrix is symmetric we can use the built-in function L.R. for linear regression.

We know that the following equation is solved for A and B:

\(\begin{bmatrix}
\sum x^2 & \sum x \\
\sum x & n
\end{bmatrix}\cdot\begin{bmatrix}
A \\
B
\end{bmatrix}=\begin{bmatrix}
\sum xy \\
\sum y
\end{bmatrix}\)

Cramer's Rule gives the solutions:

\(A=\frac{\begin{vmatrix}
\sum xy & \sum x \\
\sum y & n
\end{vmatrix}}{\begin{vmatrix}
\sum x^2 & \sum x \\
\sum x & n
\end{vmatrix}}=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}\)

\(B=\frac{\begin{vmatrix}
\sum x^2 & \sum xy \\
\sum x & \sum y
\end{vmatrix}}{\begin{vmatrix}
\sum x^2 & \sum x \\
\sum x & n
\end{vmatrix}}=\frac{\sum y\sum x^2-\sum x\sum xy}{n\sum x^2-(\sum x)^2}\)

For this to work we have to fill the registers R₂ - R₇ with the corresponding values of the equation:

\(\begin{matrix}
n & \rightarrow & R_2 \\
\sum x & \rightarrow & R_3 \\
\sum x^2 & \rightarrow & R_4 \\
\sum y & \rightarrow & R_5 \\
\sum y^2 & \rightarrow & R_6 \\
\sum xy & \rightarrow & R_7
\end{matrix}\)

This mapping leaves us with the following linear equation:

\(\begin{bmatrix}
R_4 & R_3 \\
R_3 & R_2
\end{bmatrix}\cdot\begin{bmatrix}
Y \\
X
\end{bmatrix}=\begin{bmatrix}
R_7 \\
R_5
\end{bmatrix}\)

Example:

\(\begin{bmatrix}
2 & -3 \\
-3 & 5
\end{bmatrix}\cdot\begin{bmatrix}
2 \\
-3
\end{bmatrix}=\begin{bmatrix}
13 \\
-21
\end{bmatrix}\)

CLEAR Σ
2 STO 4
-3 STO 3
5 STO 2
13 STO 7
-21 STO 5
L.R.

Y: 2
X: -3


Registers

\(\begin{matrix}
R_0 & p \\
R_1 & q \\
R_2 & c = c_j \\
R_3 & {c}′ = c_{j+1} \\
R_4 & {c}'' = c_{j+2} \\
R_5 & b = b_i \\
R_6 & \\
R_7 & {b}′ = b_{i+1} \\
R_8 & \text{index} = 9.fff \\
R_9 & a_n \\
R_{.0} & a_{n-1} \\
R_{.1} & a_{n-2} \\
R_{.2} & \cdots \\
R_{.3} & \cdots \\
\end{matrix}\)

Program

Code:

001 {    42 21 11 } f LBL A
002 {       42 32 } f ∑
003 {       45  8 } RCL 8
004 {       44 25 } STO I
005 {    42 21  0 } f LBL 0
006 {       45  5 } RCL 5
007 {       45  3 } RCL 3
008 {       44  4 } STO 4
009 {    45 20  1 } RCL × 1
010 {          30 } −
011 {       45  2 } RCL 2
012 {       44  3 } STO 3
013 {    45 20  0 } RCL × 0
014 {          30 } −
015 {       44  2 } STO 2
016 {       45 24 } RCL (i)
017 {       45  7 } RCL 7
018 {    45 20  1 } RCL × 1
019 {          30 } −
020 {       45  5 } RCL 5
021 {       44  7 } STO 7
022 {    45 20  0 } RCL × 0
023 {          30 } −
024 {       44  5 } STO 5
025 {    42  6 25 } f ISG I
026 {       22  0 } GTO 0
027 {       42 49 } f L.R.
028 {    44 40  0 } STO + 0
029 {          34 } x↔y
030 {    44 40  1 } STO + 1
031 {       43  1 } g →P
032 {       43 34 } g RND
033 {    43 30  0 } g TEST x≠0
034 {       22 11 } GTO A
035 {       45  8 } RCL 8
036 {           2 } 2
037 {          26 } EEX
038 {           3 } 3
039 {          16 } CHS
040 {          30 } −
041 {       44  8 } STO 8
042 {       44 25 } STO I
043 {           0 } 0
044 {          36 } ENTER
045 {    42 21  1 } f LBL 1
046 {       45 24 } RCL (i)
047 {       45  1 } RCL 1
048 {       43 33 } g R⬆
049 {          20 } ×
050 {          30 } −
051 {       45  0 } RCL 0
052 {       43 33 } g R⬆
053 {          20 } ×
054 {          30 } −
055 {       44 24 } STO (i)
056 {    42  6 25 } f ISG I
057 {       22  1 } GTO 1
058 {       43 32 } g RTN

Description

Initialization k = n
Lines 002 - 004
\((b, {b}′, c, {c}′, {c}'') \leftarrow (0, 0, 0, 0, 0)\)
\(I \leftarrow\) index

Partial Derivatives
Lines 005 - 026
At the end of the loop \(b = b_0\) but \(c = c_1\)!
That's why the second division is performed before the first.

Iteration k+1 → k
Lines 006 - 015
\((c, {c}′, {c}'') \leftarrow (b - c p - {c}′ q, c, {c}′)\)

Lines 016 - 024
\((b, {b}′) \leftarrow (a_k - b p - {b}′ q, b)\)

Solve the linear equation
Line 027
The trick using linear regression only works since the matrix is symmetric.
\(\begin{bmatrix}
c & {c}' \\
{c}' & {c}''
\end{bmatrix}\cdot\begin{bmatrix}
\delta p \\
\delta q
\end{bmatrix}=\begin{bmatrix}
b \\
{b}'
\end{bmatrix}\)

Find improved values for p and q
Lines 028 - 030
\(p \leftarrow p + \delta p\)
\(q \leftarrow q + \delta q\)

Stop Criterion
Lines 031 - 033
\(\left | (\delta p, \delta q) \right |< \varepsilon\)
\(\varepsilon = 10^{-n}\) if display is set to FIX n

Polynomial Division
Lines 035 - 057
We have to repeat the division here since we didn't keep \(b_k\) in lines 016 - 024.


Quadratic Solver

This program solves the quadratic equation: \(T(x)=x^2+px+q=0\)

Code:

059 {    42 21 12 } f LBL B
060 {       45  0 } RCL 0
061 {           2 } 2
062 {          16 } CHS
063 {          10 } ÷
064 {          36 } ENTER
065 {          36 } ENTER
066 {       43 11 } g x²
067 {    45 30  1 } RCL − 1
068 {          11 } √x̅
069 {          30 } −
070 {          34 } x↔y
071 {       43 36 } g LSTx
072 {          40 } +
073 {       43 32 } g RTN

Just be aware that this program can't find complex roots. Instead an Error 0 will be displayed.
However it's easy to find the complex solutions. Just use:

CHS
\(\sqrt{x}\)

The solutions then are: \(Y \pm iX\)


Example

\(P(x)=2x^5-9x^4+15x^3+65x^2-267x+234=0\)

Insert coefficients

Code:

2      STO 9
-9     STO .0
15     STO .1
65     STO .2
-267   STO .3
234    STO .4

Initialization

Code:

9.014  STO 8
1      STO 0
       STO 1

Alternatively use:

Code:

MATRIX 1

Run program

Code:

GSB A        -52.0000

RCL 0          1.5000
RCL 1         -4.5000

RCL 9          2.0000
RCL .0       -12.0000
RCL .1        42.0000
RCL .2       -52.0000

Code:

GSB B          1.5000
x<>y          -3.0000

Conclusion

\(2x^5-9x^4+15x^3+65x^2-267x+234=\)
\((x^2+1.5x-4.5)(2x^3-12x^2+42x-52)\)

Solutions

For \(x^2+1.5x-4.5=0\):
\(x_1=1.5\)
\(x_2=-3\)

Initialize guess

Code:

1      STO 0
       STO 1

Alternatively use:

Code:

MATRIX 1

Run program again

Code:

GSB A         -4.0000

RCL 0         -4.0000
RCL 1         13.0000

RCL 9          2.0000
RCL .0        -4.0000

Code:

GSB B          Error 0
←             -9.0000
CHS            9.0000
√x             3.0000
x<>y           2.0000

Code:

RCL .0        -4.0000
RCL 9          2.0000
÷             -2.0000
CHS            2.0000

Conclusion

\(2x^3-12x^2+42x-52=\)
\((x^2-4x+13)(2x-4)\)

Solutions

For \(x^2-4x+13=0\):
\(x_3=2+3i\)
\(x_4=2-3i\)

For \(2x-4=0\):
\(x_5=2\)

Summary

Factors

\(2x^5-9x^4+15x^3+65x^2-267x+234=\)
\((x^2+1.5x-4.5)(x^2-4x+13)(2x-4)=\)
\((x-1.5)(x+3)(x^2-4x+13)2(x-2)=\)
\((2x-3)(x-2)(x+3)(x^2-4x+13)\)

Solutions

\(x_1=1.5\)
\(x_2=2\)
\(x_3=-3\)
\(x_4=2+3i\)
\(x_5=2-3i\)