A textbook of Computer Based Numerical and Statiscal Techniques part 11 pdf

We now consider methods for determining complex roots of non-linear equations.. Even if all coefficients of a non-linear equation are real, the equation can have complex roots.. The iter

x6 = – 1.93375.

Thus, the approximation value to the root is –1.93375, correct up to five decimals

We now consider methods for determining complex roots of non-linear equations Even if all coefficients of a non-linear equation are real, the equation can have complex roots The iterative methods like the Secant method or the Newton-Raphson method are applicable to complex roots also, provided complex arithmetic is used Starting with the complex initial approximation, if the iteration converges to a complex root, then the asymptotic convergence rate is the same as that for a real root

The problem of finding a complex root of f(z) = 0, where z is a complex variable, is equivalent

to finding real values x and y, such that

f(z) = f (x + iy) = u (x, y) + iv (x, y) = 0 Where u and v are real functions.

This problem is equivalent to solving a system of two non-linear equations in two real

unknowns x and y,

u (x, y) = 0, v (x, y) = 0

Which can be solved using the methods discussed in previous section

Example 6 Find all roots of the equation f(x) = x 3 + 2x 2 – x + 5 using Newton-Raphson method Use initial approximations x 0 = – 3 for real root and x 0 = 1 + i for complex root.

Sol Given f(x) = x3 + 2x2 – x + 5

f´(x) = 3x2 + 4x – 1

Newton-Raphson formula is given by

x n+1 = x n – ( )

( )

′

n n

f x

f x

For real root: Taken initial approximation as x0 = – 3

First approximation: x1 = x0 – ( )

( )

′ 00

f x

f x

x1 = – 3 – ( )−1

14 = – 2.928571429

Second approximation: x2 = x1 – ( )

( )

′ 11

f x

f x

x2 = – 2.928571429 + (0.035349848)

13.01530612

x2 = – 2.925855408

Third approximation: x3 = x2 – ( )

( )

′ 22

f x

f x

x3 = – 2.925855408 + (0.000050045)

12.97846797

x3 = – 2.925851552 Since the second and third approximations are same for five decimals hence the real root is –2.92585 correct up to five decimals

For complex root: Initial approximation is x0 = 1 + i

First approximation: x1 = x0 – ( )

( )00

f x

f x′

x1 = 1 + i – ( ) ( ) ( )

x1 = 53 ( )114

109

i

+

= 0.486238 + i (1.045871) Thus 0.486238 + i (1.045871) is the first approximation value of the root Proceeding similarly,

we get next iterations as

x2 = 0.448139 + i (1.23665)

x3 = 0.462720 + i (1.22242)

x4 = 0.462925 + i (1.22253)

x5 = 0.462925 + i (1.22253) Since the last two iterations are similar, we take 0.462925 + i(1.22253) as the value of the

complex root

This method can also used to determine the both real and complex root of equation

f(x) = 0 Let x i–2 , x i–1 , x i be three distinct approximations to a root of f(x) = 0 and let y i–2 , y i–1 , y i

be the corresponding value of y = f(x).

1

p x A x x B x x y is a parabola passing through the points

2 (x i− ,y i− ), (x i− ,y i− ), ( , ),x y i i we have

y i−1=A x( i−1−x i)2+B x( i−1−x i)+y i (1) − = − − 2+ − − +

Equations (1) and (2) can be written as

2

( i i) ( i i) i i

A x− −x +B x− −x =y− −y

( i i) ( i i) i i

Therefore 2 1 1 2

A

=

− − − −

=

B

From A and B, the quadratic equation p x( )=A x x( − i)2+B x x( − i)+y i =0 gives the next approximation

2 1

4 2

i

A

But direct solution from (3) lead to loss of accuracy therefore for maximum accuracy, equation (3) can be written as

1 2

2 4

i

i

y

Note: If B > 0, we use +ve sign with square root of the equation and if B < 0, we use

–ve sign with square root of the equation

Example 7.Find the root of the equation y(x)=x 3−x 2− − =x 1 0 Muller’s method, taking initial approximations as x 0 = 0, x 1 = 1, x 2 = 2

Sol Let x i – 2 = 0, x i – 1 = 1, x i = 2

Then y i−2= −1,y i−1= −2,y i =1

2 1 1 2

A

=

A=(0 2)( 2 1) (1 2)( 1 1)− (1 0)(1 2)(0 2)− − − − − −

A = 2

− − − −

−

=

2

B

(0 2) ( 2 1) (1 2) ( 1 1) (0 1) (1 2) (0 2)

B= − − − − − − −

B = 5

The next approximation to the desired root is

1 2

2 4

i

y

1

2 1 2

5 25 4 2 1

i

x+ = − ×

± − × × (taking +ve sign) 1

2

5 4.123106

i

+

The procedure can now be repeated with three approximations as 1, 2, 1.780776

Let x i−2 =1,x i−1=2,x i =1.780776

Then y i−2= −2,y i−1=1,y i = −0.304808

A

=

A = − + − − − +

(1 1.780776)(1 0.304808) (2 1.780776)( 2 0.304808)

(2 1)(2 1.780776)(1 1.780776)

A = 3.780773

2

(1 1.780776) (1 0.304808) (2 1.780776) ( 2 0.304808)

(1 2)(2 1.780776)(1 1.780776)

B = 5.123098

The next approximation to the desired root is 1

2

2 4

i

i

y

x x

B B Ay

1 2

2 ( 0.304808) 1.780776

5.123098 (5.123098) 4 3.780773 ( 0.304808)

i

x i+1=1.837867

The procedure can repeated with three approximations as 2, 1.780776, 1.837867

Let x i−2 =2,x i−1=1.780776,x i=1.837867

Then y i−2=1,y i−1= −0.304808,y i= −0.007757

(2 1.837867)( 0.304808 0.007757) (1.780776 1.837867)(1 0.007757)

(1.780776 2)(1.780776 1.837867)(2 1.837867)

A = 4.619024

(2 1.837867) ( 0.304808 0.007757) (1.780776 1.837867) (1 0.007757)

(2 1.780776)(1.780776 1.837867)(2 1.837867)

B = 5.467225

The next approximation to the desired root is

1 2

2 4

i

i

y

x x

B B Ay

+

× −

2 ( 0.007757) 1.837867

5.467225 (5.467225) 4 4.619024 (0.007757)

i x

x i+1 = 1.839284 The procedure can now be repeated with three approximations as 1.780776, 1.837867, and 1.839284

Let x i−2 =1.780776,x i−1=1.837867,x i=1.839284

Then y i-2 = – 0.304808, y i-1 = – 0.007757, y i = – 0.000015

(1.780776 1.839284)( 0.007757 0.000015) (1.837867 1.839284)( 0.304808 0.000015)

(1.837867 1.78.776)(1.837867 1.839284)(1.780776 1.839284)

A = 4.20000

B

=

(1.780776 1.839284) ( 0.007757 0.000015) (1.837867 1.839284) ( 0.304808 0.000015)

(1.780776 1.837867)(1.837867 1.839284)(1.780776 – 1.839284)

B = 5.20000

The next approximation to the desired root is

x i+1 = x i –

2

2 4

i i

y

B± B − Ay

x i+1 = 1.839284 – ( )

2 0.000015 5.2 5.2 4 4.2 0.000015

× −

x i+1 = 1.839287 Hence the required root is 1.839287

Example 8 Using Muller’s method, find the root of the equation y(x) = x 3 – 2x – 5 = 0, which lies between 2 and 3.

Sol Let x i–2 = 1.9, x i–1 = 2, x i = 2.1

Then y i–2 = – 1.941, y i–1 = – 1, y i = 0.061

x x y y x x y y

x x x x x x

0.2 1.061 0.1 2.002 0.1 0.1 0.2

A = 0.2122 0.2002 6

0.002

1

( ) ( )( )

0.2 1.061 0.1 2.002 0.1 0.1 0.2

B = 0.04244 0.02002

0.002

The next approximation to the desired root is

xi+1 = xi –

2

2 4

i i

y

B± B − Ay

x i+1 = 2.1 – ( )

2 0.061 11.21 11.21 24 0.061

×

x i+1 = 2.1 – 0.122

11.21 11.1445+ = 2.094542

The procedure can now be repeated with three approximations as 2, 2.1 and 2.094542

Let x i–2 = 2, x i–1 = 2.1, x i = 2.094542

Then y i–2 = – 1, y i–1 = 0.061, y i = – 0.0001058

2 2.094542 0.061 0.0001058 2.1 2.094542 1 0.0001058

2.1 2 2.1 2.094542 2 2.094542

A = 6.194492

0.094542 0.0611058 0.005458 ( 0.9998942)

0.1 0.005458 0.094542

B = 11.161799

The next approximation to the desired root is

x i+ 1 = x i –

2

2 4

i i

y

B± B − Ay

2 0.0001058 11.161799 11.161799 4 6.194492 0.0001058

× −

xi+1= 2.094542 + 0.0002116

11.161799 11.161916+ = 2.094551

Hence the required root is 2.0945 correct up to 4 decimal places

Let the polynomial equation be

p n (x) = a0x n + a1x n–1 + a2x n–2 + a3x n–3 + + a n–1 x + a n = 0 (1) where a0 ≠ 0 and all ai′s are real

For polynomials, if the coefficients are all real valued then the complex roots occurs in conjugate pair Therefore we extract the quadratic factors that are the products of the pairs of

complex roots, and then complex arithmetic can be avoided because such quadratic factors have real coefficients

This method extracts a quadratic factor from polynomial given by equation (1), which gives

a pair of complex roots or a pair of real roots

Let us divide the given polynomial p n (x) by a quadratic factor x2 + px + q, we obtain a quotient polynomial Q n–2 (x) of degree (n–2) and a linear remainder of the form Rx + S Therefore

p n (x) = (x2 + px + q)Q n – 2 (x) + (Rx + S) (2) where Q n–2 (x) = b0x n–2 + b1x n–3 + b2x n–4 + +b n–3 x + b n–2

If (x2 + px + q) is a factor of equation (1) then the remainder terms must vanish, therefore the problem is then to find p and q such that

If we regularly change the values of p and q, we can make the remainder zero or at least

make its coefficient smaller, however this equation (3) will normally not be so, for the approximated

values of p and q.

Since R and S are both functions of the two parameters p and q then the improved values

are given by

S(p + ∆p, q + ∆q) = 0

Expand equation (4) by Taylor’s series for a function of two variables, were the second and higher order terms are neglected We get

R(p, q) + R p R

∂ ∂ ∆q = 0

S(p, q) + S p S

On solving equation (5), we get

∆p = –

−

∂  ∂  − ∂  ∂ 

∂  ∂  ∂  ∂ 

∆q = –

−

∂  ∂  − ∂  ∂ 

∂  ∂  ∂  ∂ 

(6)

Now the coefficients of b i ’s, R and S can be obtained by comparing the like powers of x in (2) i.e.,

a0x n + a1x n–1 + a2x n–2 + a3x n–3 + a n–2 x2 + a n–1 x + a n = (x2 + px + q)

× (b0x n–2 + b1x n–3 + b2x n–4 + b n–4 x2 + b n–3 x + b n–2 ) + (Rx + S)

⇒ a0 = b0

a1 = b1 + pb0

a2 = b2 + pb1 + qb0

a3 = b3 + pb2 + qb1

a r = b r + pb r–1 + qb r–2

a n–1 = pb n–2 + qb n–3 + R

a n = S + qb n–2

Hence, b0 = a0

b1 = a1 – pb0

b2 = a2 – pb1 – qb0

b3 = a3 – pb2 – qb1

.

b r = a r – pb r-1 – qb r–2

R = a n–1 – pb n–2 – qb n–3

Using b r = a r – pb r–1 – qb r–2 , r = 1, 2, 3, n (8) where b0 = a0, b– 1 = 0, Also from (8)

b n–1 = a n–1 – pb n–2 – qb n–3

Therefore a n–1 = b n–1 + pb n–2 + qb n–3

and b n = a n – pb n–1 – qb n–2

Therefore a n = b n – pb n–1 – qb n–2

So equation (7) becomes R = b n–1 and S = b n + pb n–1 (9)

For partial derivaties of R and S equation (8) can be used i.e., differentiate equation (8) with respect to ‘p’ and ‘q’.

– b r

p

∂

∂ = b r–1 + p b r 1 q b r 2

or b0 b 1

−

=

– b r

q

∂

∂ = b r–2 + p b r 1 q b r 2

or b0 b1

−

∂ ∂ = 0; where r = 1, 2, 3, 4, (11)

Now, an equation (10) and (11) shows that ∂ + 1=∂

Now set, – b r

p

∂

∂ = C r–1 so that – b r

p

∂

∂ = C r–2 ; r = 1, 2, 3, (13)

Thus from (10) b r

p

∂

−

r

C r–1 = b r–1 – pC r–2 – qC r–3

( )

b

From equations (9) and (15),

R p

∂

∂ = (b n 1) R C n 2

R q

∂

∂ = ∂ ( −1)⇒∂ = − −3

R

S p

∂

⇒ ∂ = − −1+ ∂ −1 + −1 ∂ ( )

n

b S

= −C n−10−pC n−2+b n−1

S q

∂

n

b

pb

= – C n–2 – pC n–3

On substituting these values in equation (6), we get

p

Since R = b n–1 and S = b n + pbn–1, therefore

p

p

−

−

2

Similarly ∆q = ( )

−

2

b C b b C

C C C b

Then the improved values of p and q are given by

p1 = p + ∆p and q1 = q + ∆q

The method for computation of b r and C r can be given by

The quotient polynomial Q n−2( )x = p x n( ) /(x2+px q+ )

= b x0 n−2+b x1 n−3+ +b n−3

can be obtained when p and q have been determined to the desired accuracy This polynomial is

called the defaulted polynomial Another quadratic factor is of obtained using this default polynomial

If the initial approximation of p and q are not known then the last three terms of given

polynomial a n – 2 x2 + a n–1 x + a n = 0 can be used to get approximations as p0 = 1

2

n n

a a

−

−

, q0 =

2

n n

a

a−

Example 9 Find the quadratic factor of the equation x 4 – 6x 3 + 18x 2 – 24x + 16 = 0 using Bairstow’s method where p 0 = – 1.5 and q 0 = 1 Also, find all the roots of the equation.

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

p

2

q

2

b C b b C