A textbook of Computer Based Numerical and Statiscal Techniques part 10 ppt

Since x3 = x4 hence the required root is 2.798 correct to three places of decimal.. Find the real root of the equation x = e –x using the Newton-Raphson’s method... Find the four places

First approximation:

The first approximation is given by

x1 = x0 – 0 0 0

cos

+

x1 = π – sin cos

cos

π π = 2.8233

Similarly successive iterations are x2 = 2.7986, x3 = 2.7984, x4 = 2.7984

Since x3 = x4 hence the required root is 2.798 correct to three places of decimal

Example 9 Find the real root of the equation x = e –x using the Newton-Raphson’s method.

Sol We have f(x) = xe x – 1 then f ´(x) = (1 + x)e x

First approximation:

x1 = 1 – 1 1 1 1

e

−

    = 0.6839397

Now, f(x1) = 0.3553424 and f′(x1) = 3.337012

So that,

Second approximation:

x2 = 0.6839397 – 0.3553424

3.337012

  = 0.5774545

Third approximation:

x3 = 0.5672297 Similarly, x4 = 0.5671433

Hence the required root is 0.5671 correct to 4 decimal places.’

Example 10 Using the starting value 2(1 + i), solve x 4 – 5x 3 – 20x 2 – 40x + 60 = 0 by Newton-Raphson’s method given that all the roots of the equation are complex.

Sol Let f(x) = x4 – 5x3 + 20x2 – 40x + 60

So that f´(x) = 4x3 – 15x2 + 40x – 40

Therefore Newton-Raphson method gives,

x n+1 = x n – ( )

( )n n

f x

f x′

x n+1 = x n –

Put n = 0, take x0 = 2 (1 + i) by trial, we get

x1 = 1.92(1 + i)

x2 = 1.915 + 1.908i

Since, imaginary roots occur in conjugate pairs roots are 1.915 ± 1.908i upto 3 places of

decimal Assuming other pair of roots to be α ± iβ, then

Sum = 1.915 1.908

1.915 1.908

α + β + α − β

i i

= 2α + 3.83 = 5

Also, products of roots are (α2 + β2) [(1.915)2 + (1.908)2] = 60

Hence other two roots are 0.585 ± 2.805i.

Example 11 Apply Newton’s formula to prove that the recurrence formula for finding the nth roots

−

i 1 n i

nx Hence, evaluate (240) 1/5

Sol Let x = a 1/n ⇒ x n = a or x n – a = 0

Let f(x) = x n – a = 0

Now, by Newtons’s-Raphson method, we have

x i+1 = x i – ( )

( )i i

f x

f′ x

1

1 n i

n i

n x a

nx −

− +

(1) Now to find the value of (240)1/5

We know that (243)1/5 = (35)1/5 = 3

x i+1 =

5 4

5

i i

x x

+

(2) First approximation:

Let i = 0, x i = x0 = 2.9 (say), then from above equation (2), we get

( )

5 5

0

0

x x

+

= 4 205.111( ) 240 1060.444

2.99

×

Second approximation:

Let i = 1, x i = x1 = 2.99 (say), then from above equation (2), we get

5 1

1

x x

= 4 238.977( ) 240 399.627

+ = 2.9925 Hence, the required value of (240)1/5 correct to three places of decimal is 2.993

Example 12 Determine the value of p and q so that rate of convergence of the iterative method

x n +1 = px n + q 2

n

N

x for computing N 1/3 becomes as high as possible.

Sol We have x3 = N therefore f(x) = x3 – N.

Let α be the exact root, we have α3 = N.

Substituting x n = α + e n , x n + 1 = α + e n+1, N = α3 in x n+1 = px n + q 2

n

N

x , we have

α + e n+1 = p(α + e n ) + q ( )

3 2

n

a e

α +

= p (α + e n ) + q

3 2

2 1

α

α

n e

= p(α + e n ) + qα

2

−

 + 

 α

= p(α + e n ) + qα

2

1 2e n 3 e n

= p(α + e n ) + qα – 2qe n + 3q

2 α

n e

–

⇒ e n+1 = (p + q – 1) α + (p – 2q) e n + 0 (e n)2 +

Now for the method to become of order as high as possible i.e., of order 2, we must have

p + q = 1 and p – 2q = 0 so that p = 2/3 and q = 1/3.

PROBLEM SET 2.4

1 Use Newton-Raphson method to find a root of the equation x3 – 3x – 5 = 0.

(U.P.T.U 2005) [Ans 2.279]

2 Find the four places of decimal, the smallest root of the equation e –x = sin x [Ans 0.5885]

4 Show that the square roots of N = AB is given by N S4 N

S

≈ + , where S = A + B.

5 Use Newton-Raphson method to obtain a root, correct to three decimal places of following equations:

(a) sin x =

2

x

[Ans 1.896]

6 Using N–R method, obtain formula for N and find 20 correct to two decimal places.

[Ans 4.47]

7 Find cube root of 3 correct to three decimal places by Newton’s iterative method

[Ans 1.442]

8 Find the positive root of the equation ex = 1 + x +

+ e 0.3x correct to 6 decimal places

[Ans 2.363376]

9 Apply Newton’s formula to find the values of (30)1/5 [Ans 1.973]

10 Prove the Chebyshev formula x1 = x0 – ( )

2

0

3

1

f x

′′

−

′

  for the roots of the

equation f(x) = 0.

2.8 SECANT METHOD

The Secant method is similar to the Regula-Falsi method, except for the fact that we drop the

condition that f(x) should have opposite signs at the two points used to generate the next

approximation Instead, we always retain the last two points to generate the next Thus, if x n−1

and x n are two approximations to the root, then the next approximation x n+1 to root is given by

+ −

−

−

1

1

( ), 1, 2, 3,

Geometrically, in Secant method we replace the function f x( ) by a straight line passing through the points (x n, (f x n)) and (x n−1, (f x n−1)) and take the point of intersection of the straight line with the x-axis as the next approximation to the root In contrast to the Regula-Falsi method, the Secant iteration does not bracket the root and it is not even necessary to bracket the root to start the iteration Hence, it is obvious that the iteration may not always coverage on the other hand, it generally converges faster Thus, by dropping the necessity of bracketing the root, we improve the rate of convergence, however, in some cases, the iteration may not converge at all

2.8.1 Procedure for Secant Method to Find the Root of f x( ) 0=

Step 1: Choose the interval [x x0, 1] in which f x( )=0 has a root, where x1>x0 Step 2: Find the next approximation x2 of the required root using the formula

2 1 1 0 1

( )

f x f x

−

Step 3: Find the successive approximations of the required root using the formula

+ −

−

−

− 1 1

1

( ), 1, 2, 3,

Step 4: Stop the process when the prescribed accuracy is obtained

2.8.2 Rate or Order of Convergence of Secant Method

On substituting x n= +î ån, etc in (1) we obtain the error equation as

å å å å î å

− +

−

+ −1 + 1

1

f

(2)

On expanding f(î å )+ n and f(î å+ n−1) in Taylor’s series about the point î in (2), and using f( )ξ = 0, we obtain

1

2 1

1

2 2 1

1

2 1

2

n

−

− +

−

On simplifying, and neglecting higher powers of εn, we get

εn+1 = ε εc n n−1 (3) where,

( ) 1

f c f

′′ ξ

=

′ ξ

Now, we seek a relation of the form

εn+1 = εk p n (4)

where constants k and p are to be determined.

We have from equation (4), ε = εn k p n−1 or εn−1 =k−1/pε1/n p (5) Substituting εn+1 from (4), and εn−1from (5), into the error equation (3), we get

(1 )/ (1 )/

n ck− + n+

On comparing the powers of εn, we have

1 p

p p

+

=

or p2 – p – 1 = 0

Roots of above equation are 1 5

2

+

and 1 5

2

−

Taking, p = 1 5

2

+

= 1.618 and neglecting

the other, we obtain from equation (4), the rate of convergence for the Secant method as p = 1.618 The constant k is determined from (6), and it is given by k = c p/(p+1)

The convergence of the secant method is superlinear

The purpose of this document is to show the following theorem:

Theorem 2.1: Let { }x k k∞ be the sequence produced by the secant method Assume the sequence converges to a root of f (x) = 0, i.e., x k → x∞, f x( ∞) 0= Moreover, assume the root x∞ is regular:

f x∞ ≠ Denote the error in the kth step by E k = x k – x∞ Under these assumptions, we have (1 5)/ 2 1.618

1

E+ ≈CE + ≈CE , for some constant C. (1) The theorem is implied by three lemmas

Lemma 2.1: Under the assumptions and notations of the theorem:

+ ∞ −

∞

′′

1

f x

Proof Using the definition of x k+1, we find

+ + ∞ − ∞

−

−

− 1

1

k k

We can replace x k + 1 by x k + E k and x k by x k – 1 + E k – 1 , so that

+ ∞ ∞ ∞ ∞ − ∞

1

1

To simplify this expression, we apply the Taylor expansion of f(x∞+E k) and f x( ∞+E k−1)

about x∞:

1 2 3

2

f x∞+E = f x∞ + f x E′ ∞ + f x E′′ ∞ +O E , (5)

1 1 2 1 31

1

2

f x∞+E − = f x∞ + f x E′ ∞ − + f x E′′ ∞ − +O E − . (6) Subtracting f x( ∞+E k−1) fromf x( ∞+E k):

∞+ − − ∞+ − = ′ ∞ − − + ′′ ∞ 2− 2 − + 3 − 3 −

1

2

1

( k) ( k )

O E −O E− is of a smaller order than E k and E k–1 we omit this term Using

1

−

−

k k

E E = (E k – E k–1 ) (E k + Ek–1), we organize the above expression as

∞+ − ∞+ −1 ≈ − −1 ′ ∞ + ′′ ∞ + −1

The left of (8) appears at the right of (4), we derive the following expression:

+ ∞ −

−

1

Using a Taylor expansion for f x( ∞ +E k) about x∞ (recall f x( ∞)=0) we have

+

1

1

1

1

2

k

Now we put everything on the same denominator:

+ ∞ ∞ − ∞ ∞

≈

1 1

1

1

2

.(11) which can be simplified as

∞ − +

′′

≈

1 1

1

1

2 1

2

k

Because E k→0 as → ∞,1 ′′( ∞)( + −1)

k f x E E is negligible compared to f x′( ∞), so we omit the second term in the denominator, to find the estimate

+ ∞ −

∞

′′

1

f x

Lemma 2.2: There exists a positive real number r such that:

1 1/

E+ ≈CE−E ⇒E+ ≈KE for some constants C and K. (14)

Proof Assuming the convergence rate is r, there exists some constant A, so we can write

E k+1≈AE k r and E k≈AE r k−1 or   ≈ −

1/

1

1

r

Now we can replace the expressions for E k and E k–1 in the left hand side of (14):

+ ≈    ≈ +

 

1/

1

1

r

Together with the assumption thatE k+1≈AE r k, we obtain 1 1/r k r k

A

B

+ ≈ So, we set K A

B

=

Lemma 2.3: For the r of Lemma 2.1, we have

2

Proof r satisfies the following equation:

The roots of n2 – r – 1 =0 are r = 1 5

2

±

We take the positive value for r.

The constant r = 1 5 1.618 is the golden ratio

2

Example 1 A real root of the equation f(x) = x 3 – 5x + 1 = 0 lies in the interval (0.1) Perform four iterations of the Secant method.

Sol We have x0 = 0, x1 = 1, f(x0) = 1, f(x1) = – 3

By Secant method

First approximation:

First approximation is given by

x2 = x1 – ( )11 0( )0

  f(x1) = 0.25

f(x2) = – 0.234375

Second approximation:

Second approximation is given by

x3 = x2 – ( )22 1( )1

  f(x2) = 0.186441

f(x3) = 0.074276 Third approximation:

Third approximation is given by

x4 = x3 –

( )33 2( )2

  f(x3) = 0.201736

f(x4) = – 0.000470 Fourth approximation:

Fourth approximation is given by

x5 = x4 –

( )44 3( )3

  f(x4) = 0.201640

Example 2 Compute root of the equation x 2 e –x/2 = 1 in the interval [0.2] using Secant method The root should be correct to three decimal places.

Sol We have, x0 = 1.42, x1 = 1.43, f(x0) = – 0.0086, f(x1) = 0.00034

By Secant method,

First approximation:

First approximation is given by

x2 = x1 –

( )11 0( )0

  f(x1)

x2 = 1.43 – 1.43 1.42

0.00034 0.0086

f(x2) = – 0.000011 Second approximation: Second approximation is given by

x3 = x2 – ( )22 1( )1

x3 = 1.4296 – 1.4296 1.42

0.000011 0.00034

Since x2 and x3 agree up to three decimal places hence the required root is 1.429

Example 3 Find the root of the equation x 3 – 5x 2 – 17x + 20 = 0 by Secant method.

Sol Taking initial approximations as, x0 = 0, x1 = 1 and f(x0) = 20, f(x1) = – 1, then by Secant method the next approximation is given by

x2 = x1 –

( )11 0( )0

x2 = 1 – 1 0

1 20

−

 (–1) = 1 – 0.04762 = 0.95238

Hence, f(x2) = 0.13824

Now the next approximation can be obtained by using x1 and x2 in Secant method Similarly, other approximations can be obtained by using two recent approximations in Secant method

These are x3 = 0.95816, f(x3) = 0.00059

x4 = 0.95818, f(x3) = 0.00011

x5 = 0.95818

Thus the approximate root can be taken as 0.65818, which is correct up to five decimals

Example 4 Find the root of the equation x 3 – 2x – 5 by Regula-Falsi and Secant method.

Sol Solution by Regula-Falsi method:

Here f(x) = x3 – 2x – 5 then f(2) = – 1, f(3) = 16 and f(2) f(3) < 0

Therefore initial approximations are taken as x0 = 2, x1 = 3 and f (x0) = – 1, f (x1) = 16 Then by Regula-Falsi method the next approximation is given by

x2 = x1 – ( )11 0( )0

  f(x1)

x2 = 3 – ( )

(16 13 2)

− + 16 = 3 – 0.9412

x2 = 2.0588

Hence, f(x2) = – 0.3911 and f(x1) f (x2) < 0, therefore the next approximation to the root is

obtained by using the values of x1 and x2 in Regula-Falsi method as

x3 = x2 – ( ) ( )22 1 1

  f (x1)

x3 = 2.0588 – ( )

( 2.0588 30.3911 16)

−

− − (–0.3911) = 2.0588 + 0.0225

x3 = 2.0813

Hence, f(x3) = – 0.1468 and f(x1) f (x3) < 0, therefore the next approximation is obtained by

using the values of x1 and x3 in Regula-Falsi method

Proceed in similar way to obtain the iterations as follows

x4 = 2.0899, f(x4) < 0

x5 = 2.0928, f(x5) < 0

x6 = 2.0939, f(x6) < 0

x7 = 2.0943, f(x7) < 0

x8 = 2.0945, f(x8) < 0

x9 = 2.0945

Thus, the root can be taken as 2.0945 correct up to four decimals

Solution by Secant method: Taking initial approximations as x0 = 2, x1 = 3 and

f(x0) = –1, f(x1) = 16, then by Secant method, the next approximation is given by

x2 = x1 – ( ) ( )

−

−

x2 = 3 – ( )

( −+ )

3 2

16 1 16 = 3 – 0.9412

Hence, x2 = 2.0588, f(x2) = – 0.3911

Now, the next approximation can be obtained by using the values of x1 and x2 in Secant method Similarly, other approximations can be obtained by using two recent values of approximations These are

x3 = 2.0813, f(x3) = – 0.1468

x4 = 2.0948, f(x4) = – 0.0028

x5 = 2.0945, f(x5) = – 0.0006

x6 = 2.0945

Thus, the root can be taken as 2.0945 correct to four decimals

Example 5 Find the root of the equation f(x) = 4 sin x + x 2 = 0 by Secant method.

Sol In this method we neglect the condition f(x n ) f(x n−1) < 0 Initially, take x0 = – 1, x1 = –

2 and f(x0) = – 2.36588, f(x1) = 0.36281, the next approximation to the root by Secant method is given by

x2 = x1 – ( ) ( )

−

−

x2 = (–2) – ( 2 1 0.36281) ( )

0.36281 2.36588

− + +

x2 = – 2 + 0.13296 = – 1.86704 Hence, x2 = – 1.86704, f(x2) = – 0.33992

Now, the next approximation x3 can be obtained by using the values of x1 and x2 in Secant method, which is given by,

x3 = x2 – ( ) ( )

−

−

x3 = (–1.86704) – ( 1.86704 2) ( 0.33992)

0.33992 0.36281

x3 = – 1.86704 – 0.06431 = – 1.93135 Hence, x3 = – 1.93135, f(x3) = – 0.01269

Now, the next approximation x4 can be obtained by using the values of x2 and x3 in Secant method Continuing this process and using two recent approximations, to get next approximation,

in Secant method, we get

x4 = – 1.93384, f(x4) = 0.00045

x5 = – 1.93375, f(x5) = – 0.00002