A textbook of Computer Based Numerical and Statiscal Techniques part 12 pdf

Using Bairstow’s method we find the values of p and q: First approximation: Let p0 and q0 be the initial approximations, then the first approximation can be obtained p1 = p0 + ∆p and q1

Now to obtain the values of b i and C i we use the following procedure:

1.5 1.5 6.75 15.375 6.1875

1 4.5 10.25 4.125 0.4375 1.5 1.5 4.5 7.125 9

1 3 4.75 6 3.8125

−

−

Here from the table b3 = –4.125, b4 = –0.4375, C1 = –3, C2 = 4.75, C3 = 6 therefore after

substituting the values of b i and C i in equations (1) and (2) we get

22.56 3(10.125)

+ +

∆p = –0.3949

∆q = – ( 39.6875)

(52.935)

−

∆q = 0.74974

Therefore the first approximation are given by

p1 = p0 + ∆p = –1.5 + (–0.3949) = –1.8949

q1 = q0 + ∆q = 1 + 0.74974 = 1.74974

Second approximation: Using p1 = –1.8949 and q1 = 1.74974 for second approximation, then

p2 = p1 + ∆p and q2 = q1 + ∆q.

Now to obtain the values of b i and C i we use the following procedure:

−

−

1.8949 1.8949 7.7787 16.0526 1.4487 1.74974 1.74974 7.1828 14.8229

1 4.1051 8.4715 0.7645 0.2716 1.8949 1.8949 4.188 4.801 15.4044 1.74974 1.74974 3.8673 4.433

1 2.2102 2.5336 7.9038 10.6996

i

i

i

a

b

C

After substituting the values of b i and C i in equations (1) and (2) we get

∆p = 2.53722752

25.57781

∆p = –0.09919

∆q = ( 5.93879)

25.57781

−

∆q = 0.23218

Therefore the second approximations are given by

p2 = p1 + ∆p = –1.99409

q2 = q1 + ∆q = 1.98192

Third approximation: Using p2 = –1.99409 and q2 = 1.98192 for third approximation, then

p3 = p2 + ∆p and q3 = q2 + ∆q.

Now to obtain the values of b i and C i we use the following procedure:

−

−

1.99409 1.99409 7.988 16.0124 0.0962 1.98192 1.98192 7.9394 15.9146

1 4.0059 8.0299 0.048226 0.01082 1.99409 1.99409 4.01173 4.06046 15.952 1.98192 1.98192 3.9873 4.0357

1 2.0118 0.03625 7.9995 11.9

i

i

i

a

b

After substituting the values of b i and C i in equations (1) and (2) we get

p

∆ = 0.11997 20.3367

p

∆ = – 0.005899

q

∆ = – ( 0.36607)

20.3367

−

q

∆ = 0.01800 Therefore the third approximation are given by

p3 = p2 + ∆p = –1.9882

q3 = q2 + ∆q = 1.9999

Thus, we obtain p = –1.9882 and q = 1.9999 Hence quadratic factor of the given equation

is x2 – 1.9882x + 1.9999 = 0 Now if root of the quadratic factor is α ± β, theni

2α = 1.9882 ⇒ α = 0.9941, α2 + β2 = 1.9999 ⇒ β = 1.0058 Hence a pair of roots is 0.9941±1.0058i

Other roots can be obtained by using default polynomial

Default polynomial is given by b0x2 + b1x + b2 = 0, where b i are given by the same procedure,

when p and q are of required accuracy.

1 1 6 18 24 16 1.9882 1.9882 7.97626 15.95299 0.047343 1.9999 1.9999 8.023198 16.04687

1 4.0118 8.02384 0.023812 0.094213

i

a

b

Thus b0 = 1, b1 = –4.0118, b2 = 8.02384 and thus the polynomial becomes 1x2 – 4.0118x

+ 8.02384 = 0, whose roots are γ = 2.0059, δ = 2.00005

Hence the pair of roots is 2.0059 2.00005i±

Example 10 Solve x 4 – 5x 3 + 20x 2 – 40x + 60 = 0 given that all the roots are complex, by using Lin-Bairstow method Take the values as p 0 = –4, q 0 = 8.

Sol Let the quadratic factor of the equation be x2 + px + q Using Bairstow’s method we find the values of p and q:

First approximation: Let p0 and q0 be the initial approximations, then the first approximation

can be obtained p1 = p0 + ∆p and q1 = q0 + ∆q Because given equation is of the degree four then

2

−

b C b C

2

b C b b C

C C C b

− −

Now to obtain the values of b i and C i we use the following procedure:

1 5 20 40 60

i

i

i

a

b

C

−

−

Here from the table b3 = 0, b4 = –4, C1 = 3, C2 = 12, C3 = 24 therefore after substituting the

values of b i and C i in equations (1) and (2), we get

∆p = – ( 4) 3 0 (12)2

(12) 3 (24 0)

− × − ×

− × −

∆p = 0.166666

0 (24 0) ( 4) 12 (12) 3 (24 0)

× − − − ×

− × −

∆q = –0.666666

Therefore the first approximation are given by

p1 = p0 + ∆p = –4 + 0.166666 = –3.833334

q1 = q0 + ∆q = 8 – 0.666666 = 7.333334

Second approximation: Using p1 = –3.833334 and q1 = 7.333334 for second approximation,

then p2 = p1 + ∆p and q2 = q1 + ∆q.

Now to obtain the values of b i and C i we use the following procedure:

3.833334 3.833334 4.472220 31.412048 0.124203 7.333334 7.333334 8.555551 60.092609

1 1.166666 8.194446 0.032401 0.216812 3.833334 3.8333334 10.222229 42.486147 87.7763681 7.333334 7.333334 19.5555

i

i

a

b

1 2.666668 11.083341 22.898179 6.2817151

i

C

−

After substituting the values of b i and C i in equations (1) and (2), we get

∆p = –0.015192

∆q = –0.026908

Therefore the second approximation are given by

p2 = p1 + ∆p = –3.833334 – 0.015192 = –3.848526

q2 = q1 + ∆q = 7.333334 – 0.026908 = 7.306426

Third approximation: Using p2 = – 3.848526 and q2 = 7.306426 for third approximation,

then p3 = p2 + ∆p and q3 = q2 + ∆q.

Now to obtain the values of b i and C i we use the following procedure

3.848526 3.848526 4.431477 31.796895 0.808398 7.306426 7.306426 8.413159 60.366400

1 1.151474 8.262097 0.210054 0.441998 3.848526 3.848526 10.379674 43.624369 92.859594 7.306426 7.306426 19.705810 82

i

i

a

b

−

−

1 2.697052 11.335345 24.128613 10.480733

i

C

After substituting the values of b i and C i in equations (1) and (2), we get

∆p = 0.018582

∆q = – 0.0002186

Therefore the third approximation are given by

p3 = p2 + ∆p= –3.848526 + 0.018582 = – 3.829944

q3 = q2 + ∆q= 7.306426 – 0.000218 = 7.306208

Thus, we obtain p = – 3.83 and q = 7.3062 Hence quadratic factor of the given equation is

x2 – 3.83x + 7.3062 Now if root of the quadratic factor is α ± iβ2, then

2α = 3.83 ⇒ α = 1.915, α2 + β2 = 7.3062 ⇒ β = 1.9081

Hence a pair of roots is 1.915 ± 1.908li

Other roots can be obtained by using default polynomial

Default polynomial is given by b0x2 + b1x + b2 = 0, where b i are given by the same procedure,

when p and q are of required accuracy.

−

−

3.829944 8.829944 4.481248 31.453583 0.008636 7.306208 7.306208 8.548672 60.002554

1 1.170056 8.212544 0.002255 0.00608185

i

a

b

Thus b0 = 1, b1 = – 1.17, b2 = 8.2125 and thus the polynomial becomes 1x2 – 1.17x +

8.2125 = 0, whose roots are γ = 0.585, δ = 2.8054

Hence the pair of roots is 0.585 ± 2.8054i.

This is a general method to obtain the approximate roots of the polynomial equations The procedure

is quite general and is illustrated here with a cubic polynomial Let the given cubic equation be

and let x1, x2 and x3 be its root such that 0 < x1 < x2 < x3

The roots can be obtained, directly by considering the transformed equation

Whose roots are the reciprocals of those of (1)

1

i i i

x

a x a x a x a

∞

=

So that (a3x3 + a2x2 +a1x + a0) (α0 + α1x + α2x2 + )=1 (3)

Comparing the coefficients of like powers of x on both sides of (3), we get

α0 =

0

1

a , α1 = – 12

0

a

a , α2 =

2

− +

0

α

α = – 10

a a

q1(2) = 2

1

α

α =

− 2

0 1

a a a

a a

And so, ∆1(1) = q1(2) – q(1)= 2

1

a

a , ∆2(0) = 3

2

a a

In general, ∆m (m) = m 1

m

a a

+ , m = 1, 2, 3, (n – 1)

q m (1–m) = 0, m = 2, 3, , n

That is q1(0), q2 (–1) , q3(–2), , top q’s are 0.

We also set ∆0(k) =∆n (k) = 0, for all k [i.e., first and last columns of Q-d table are zero].

The Quotient Difference table for a Cubic Equation

Remarks:

(1) If an ∆ element is at the top of the rhombus, then the product of one pair is equal to that

of the other pair For example, in the rhombus

∆1(1)

q1(2) q2(1) We have ∆(1)1.q2(1)= ∆(2)1 .q1(2)

∆1(2)

From which ∆1(2) can be computed since the other quantities are known

(2) If a q-element is at the top, then the sum of one pair is equal to that of the other pair For

example, in the rhombus

q2(0)

∆(1)1 ∆2(0) We have q2(0) + ∆(0) 2 = q2(1)+ ∆(1)1

q2(1)

From which q2(1) can be computed when q2(0), ∆1(1), ∆2(0) are known

As the building up of table proceeds, the quantities q1(i) , q2(i) , q3(i) tend to roots of cubic equations The disadvantage of this method is that additional computation is also necessary This method can be applied to find the complex roots and multiple roots of polynomials and also for determining the Eigen values of a matrix

An important feature of the method is that it gives approximate values of all the roots simultaneously and this fact enables one to use this method to obtain the first approximations of all roots and then apply a rapidly convergent method such as the generalized Newton method

to obtain the roots to the desired accuracy

Example 11 Solve the following equation by using quotient-difference method x3 – 6x2 +

11x – 6 = 0.

Sol To obtain the roots directly, we consider the transformed equation

–6x3 + 11x2 – 6x + 1 = 0

Here a3 = – 6, a2 = 11, a1 = – 6 and a0 = 1

Therefore, we have q1(1) = – 1

0

a

a = 6

q1(2) = –

2

0 1

a a a

a a

− = 11 36 6

−

− = 4.167

∆1(1) = q1(2) – q1(1) = 2

1

a

a = – 1.833 Also q2(0) = 0, q3(–1) = 0 and ∆2(0) = 3

2

a

a = ` 116 = – 0.5454 First two rows containing starting values of

−

The succeeding rows can be constructed as below:

It is evident that q1, q2, q3 are gradually converging to the roots 3, 2 and 1 respectively

PROBLEM SET 2.5

1 Use Secant method to determine the root of the equation cos x – xe x = 0

[Ans 0.5177573637]

2 Using Secant method, find the root of x–e –x = 0 correct to three decimal places by taking

3 Apply Muller’s method to obtain the root of the equation cos x – xe x = 0 which lies between

4 Solve by Muller’s method x3 + 2x2 + 10x –20 = 0 by taking x = 0, x = 1, x = 2 as initial

5 Find a quadratic factor of the polynomial x4 + 5x3 + 3x2 – 5x – 9 = 0 Starting with p0 = 3,

q0 = – 5 by using Bairstow’s Method [Ans x2 + 2.90255x – 4.91759]

6 Solve the equation x4 – 8x3 + 39x2 – 62x + 50 = 0 Starting with p = 0, q = 0.

7 Find the real roots of the equation x3 – 7x2 + 10x – 2 = 0 by using Quotient difference

8 Solve the following equation x3 – 8x2 + 17x – 10 = 0 by using Quotient difference method.

[Ans 5, 2.001, 0.9995]

9 Find all the roots of the equation x3 – 5x2 – 17x + 20 = 0 by using Quotient difference

GGG

CHAPTER 3

Calculus of Finite Differences

Finite differences: The calculus of finite differences deals with the changes that take place in the value of the dependent variable due to finite changes in the independent variable from this we study the relations that exist between the values, which can be assumed by function, whenever the independent variable changes by finite jumps whether equal or unequal

The study of finite difference calculus has become very important due to its wide variety

of application in routine life It has been originated by Sir Issac Newton It has been of great use for Mathematicians as well as Computer Scientists for solution of the Scientific, business and engineering problems There it helps in reducing complex mathematical expressions like trigonometric functions in terms of simple arithmetic operations

Numerical methods are very important tools to provide practical methods for calculating the solution of problems to applied mathematics for a desired degree of accuracy

If f is a function from x into y for a ≤ x ≤ b such that y = f(x), this means that one or more

values of y = f(x) exist corresponding to every value of x in the given range However if the function f is not known, the value of y can be obtained, when a set of values of x is given The

method to find out such values is based on principle of finite differences provided the function

is continuous

If y = f(x) be a function assumes the values f(a), f(a + h), f(a + 2h), corresponding to the values

of x then each value of x is called argument and its corresponding values of y is called entry.

Let y = f(x) be a function tabulated for the equally spaced values or argument a = x0, x0 + h, x0 + 2h, , x0 + 2h, x0 + nh, where h is the increment given to the independent variable of function y = f(x) To determine the values of function y = f(x) for given intermediate or argument values of x, three types of differences are useful:

104

3.4.1 Forward or Leading Differences

If we subtract from each value of y except y0, the previous value of y, we have

y1 – y0, y2 – y1, y3 – y2, , y n – y n–1 These differences are called the first forward differences

of y and is denoted by ∆ y The symbol ∆ denotes the forward difference operator That is,

∆y0 = y1 – y0

∆y1 = y2 – y1

∆y2 = y3 – y2

.

∆y n = y n + 1 – y n

Also it can be written as,

∆f(x) = f(x+h) – f(x)

where h is the interval of differencing.

Similarly for second and higher order differences,

∆2y0 = ∆y1– ∆y0

∆2y1 = ∆y2– ∆y1

∆2y n–1 = ∆y n– ∆y n–1

or ∆3y0 = ∆2y1– ∆2y0

∆3y1 = ∆2y2– ∆2y1

∆3y n–1 = ∆2y n– ∆2y n–1

In general, nth forward difference are given by

∆n y r = ∆n–1 y r–1– ∆n–1 y r, or

∆n f(x) = ∆n–1 f(x + h)– ∆n–1 f(x)

Forward Difference Table:

0 2

3

3

2

3

2

3

4

x y

x y

y

y

x h y

∆

∆ +

where x0 + h = x1, x0 + 2h = x2, , x0 + nh = x n