(11C) Bairstow's Method
06-14-2015, 06:27 PM (This post was last modified: 06-15-2017 01:17 PM by Gene.)
Post: #1
 Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
(11C) Bairstow's Method
Bairstow's Method

Wikipedia
MathWorld
Numerical Recipes in Fortran 77 9.5 Roots of Polynomials

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 R0 - R5 with the corresponding values of the equation:

$$\begin{matrix} n & \rightarrow & R_0 \\ \sum x & \rightarrow & R_1 \\ \sum x^2 & \rightarrow & R_2 \\ \sum y & \rightarrow & R_3 \\ \sum y^2 & \rightarrow & R_4 \\ \sum xy & \rightarrow & R_5 \end{matrix}$$

This mapping leaves us with the following linear equation:

$$\begin{bmatrix} R_2 & R_1 \\ R_1 & R_0 \end{bmatrix}\cdot\begin{bmatrix} Y \\ X \end{bmatrix}=\begin{bmatrix} R_5 \\ R_3 \end{bmatrix}$$

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 2
-3 STO 1
5 STO 0
13 STO 5
-21 STO 3
L.R.

Y: 2
X: -3

Registers

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

Program

Code:
001 - 42,21,11   LBL A                             032 - 44,40, 7   STO + 7
002 -   42  32   CLEAR Σ                           033 -       34   x<>y
003 -    45  6   RCL 6                             034 - 44,40, 8   STO + 8
004 -   44  25   STO I                             035 -   42  26   →P
005 - 42,21, 0   LBL 0                             036 -   43  34   RND
006 -    45  3   RCL 3                             037 -   43  30   x≠0
007 -    45  1   RCL 1                             038 -   22  11   GTO A
008 -    44  2   STO 2                             039 -    45  6   RCL 6
009 -    45  8   RCL 8                             040 -        2   2
010 -       20   ×                                 041 -       26   EEX
011 -       30   -                                 042 -        3   3
012 -    45  0   RCL 0                             043 -       16   CHS
013 -    44  1   STO 1                             044 -       30   -
014 -    45  7   RCL 7                             045 -    44  6   STO 6
015 -       20   ×                                 046 -   44  25   STO I
016 -       30   -                                 047 -        0   0
017 -    44  0   STO 0                             048 -       36   ENTER
018 -   45  24   RCL (i)                           049 - 42,21, 1   LBL 1
019 -    45  5   RCL 5                             050 -   45  24   RCL (i)
020 -    45  8   RCL 8                             051 -    45  8   RCL 8
021 -       20   ×                                 052 -   43  33   R↑
022 -       30   -                                 053 -       20   ×
023 -    45  3   RCL 3                             054 -       30   -
024 -    44  5   STO 5                             055 -    45  7   RCL 7
025 -    45  7   RCL 7                             056 -   43  33   R↑
026 -       20   ×                                 057 -       20   ×
027 -       30   -                                 058 -       30   -
028 -    44  3   STO 3                             059 -   44  24   STO (i)
029 -    42  6   ISG                               060 -    42  6   ISG
030 -    22  0   GTO 0                             061 -    22  1   GTO 1
031 -   42  49   L.R.                              062 -   43  32   RTN

Description

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

Polynomial division
Lines 005 - 030
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 - 017
$$(c, {c}′, {c}'') \leftarrow (b - c p - {c}′ q, c, {c}′)$$

Lines 018 - 028
$$(b, {b}′) \leftarrow (a_k - b p - {b}′ q, b)$$

Solve the linear equation
Line 031
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 032 - 034
$$p \leftarrow p + \delta p$$
$$q \leftarrow q + \delta q$$

Stop Criterion
Lines 035 - 037
$$\left | (\delta p, \delta q) \right |< \varepsilon$$
$$\varepsilon = 10^{-n}$$ if display is set to FIX n

Polynomial Division
Lines 039 - 061
We have to repeat the division here since we didn't keep $$b_k$$ in lines 018 - 028.

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

Code:
063 - 42,21,12  LBL B
064 -    45  7  RCL 7
065 -        2  2
066 -       16  CHS
067 -       10  ÷
068 -       36  ENTER
069 -       36  ENTER
070 -    43 11  x^2
071 -    45  8  RCL 8
072 -       30  -
073 -       11  SQRT
074 -       30  -
075 -       34  x<>y
076 -    43 36  LSTx
077 -       40  +
078 -    43 32  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 6
1      STO 7
STO 8

Run program

Code:
GSB A        -52.0000

RCL 7          1.5000
RCL 8         -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 7
STO 8

Run program again

Code:
GSB A         -4.0000

RCL 7         -4.0000
RCL 8         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

$$x_1=1.5$$
$$x_2=2$$
$$x_3=-3$$
$$x_4=2+3i$$
$$x_5=2-3i$$

Attached File(s) bairstow-method.zip (Size: 874 bytes / Downloads: 11)
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