New iterative methods

Applied Mathematics and Computation 189 (2007) 1260–1267

www.elsevier.com/locate/amc

New iterative methods

Arif Rafiq a,*, Sifat Hussain b, Farooq Ahmad b, Muhammad Awais a

a Department of Mathematics, COMSATS Institute of Information Technology, Plot No. 30, Sector H-8/1, Islamabad 44000, Pakistanb Centre for Advanced Studies in Pure and Applied Mathematics, Bahaudin Zakariya University, Multan 60800, Pakistan

Abstract

In this paper, we present a new three-step predictor corrector type iterative method for finding simple and real roots of anon-linear equation in one variable, i.e., f(x) = 0. Experiments show that our new method is more efficient than the otherknown methods. A comparison of the proposed method with other methods reveals that the new method performs betteras shown in Table 2.� 2006 Elsevier Inc. All rights reserved.

Keywords: Three-step methods; Non-linear equations; Numerical examples

1. Introduction

Finding the roots of a non-linear equation f(x) = 0 is a common yet important problem in science and engi-neering. Analytical methods for solving such equations are difficult or almost non-existent. Therefore it is onlypossible to obtain approximate solutions by numerical techniques based on iteration procedures.

In this direction, three-step iterative methods have been proposed by Chen and Li [2] and Noor and Ahmad[5]. These are based on a Trapezoidal like method developed earlier by Nedzhibov [3]. In fact, Algorithm 1 in[4] is due to Nedzhibov [1,3]. In the present paper we present a new three-step method and show that it per-forms better than the other known methods.

2. Iterative methods

Let us see first some well known iteration formulae for finding roots of f(x) = 0.

Algorithm 1. (due to Nedzhibov [3]):

Step 1. For initial guess xo, a tolerance e > 0, for iterations n, set k = 0.

0096-3003/$ - see front matter � 2006 Elsevier Inc. All rights reserved.

doi:10.1016/j.amc.2006.12.042

* Corresponding author.E-mail addresses: [email protected] (A. Rafiq), [email protected] (S. Hussain), [email protected] (F. Ahmad),

[email protected] (M. Awais).

mailto:[email protected]




A. Rafiq et al. / Applied Mathematics and Computation 189 (2007) 1260–1267 1261

Step 2. Calculate x1; x2; . . . ; such that

xnþ1 ¼ xn �4f ðxnÞ

f 0ðxnÞ þ 2f 0 xnþxnþ1

2

� �þ f 0ðxnþ1Þ

:

Step 3. For given e > 0, if jxkþ1 � xkj < e, or k > n, then stop.Step 4. Set k ¼ k þ 1 and go to Step 2.

Algorithm 2. Ch (due to Chen and Li [2]):

Step 1. For initial guess xo, a tolerance e > 0, for iterations n, set k ¼ 0.Step 2. Calculate x1; x2; . . . ; such that

xnþ1 ¼ xn exp�f ðxnÞxnf 0ðxnÞ

� �:

Step 3. For given e > 0, if jxkþ1 � xkj < e , or k > n, then stop.Step 4. Set k ¼ k þ 1 and go to Step 2.

Algorithm 3. Nr (due to Noor and Ahmad [5]):


xnþ1 ¼ xn �2f ðxnÞ

f 0ðxnÞ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffif 02ðxnÞ þ 4p3f 3ðxnÞ

p ;

where p 2 R and jpj <1; also p is chosen so that f ðxkÞ and p have the same sign.Step 3. For given e > 0, if jxkþ1 � xkj < e, or k > n, then stop.Step 4. Set k ¼ k þ 1 and go to Step 2.

We suggest here a new three-step method as follows:

Algorithm 4. SAAF:


zn ¼ xn exp�f ðxnÞxnf 0ðxnÞ

� �;

yn ¼ zn �2f ðznÞ

f 0ðznÞ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffif 02ðznÞ þ 4p3f 3ðznÞ

p ;

xnþ1 ¼ yn �4f ðynÞ

f 0ðynÞ þ 2f 0 ynþzn

2

� �þ f 0ðznÞ

;

where p 2 R and jpj <1, also p is chosen so that f ðxnÞ and p have the same sign.Step 3. For given e > 0, if jxkþ1 � xkj < e, or k > n, then stop.Step 4. Set k ¼ k þ 1 and go to Step 2.

3. Convergence analysis

Now, we prove that our three-step method (SAAF) has sixth order of convergence.

Theorem 1. Let a 2 I be a simple zero of a sufficiently differentiable function f : I � R! R on an open interval I.

If x0 is close to a, then the Algorithm 4 has sixth order of convergence.

1262 A. Rafiq et al. / Applied Mathematics and Computation 189 (2007) 1260–1267

Proof. The iterative technique is given by

zn ¼ xn exp�f ðxnÞxnf 0ðxnÞ

� �; ð3:1Þ

yn ¼ zn �2f ðznÞ

f 0ðznÞ þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffif 02ðznÞ þ 4p3f 3ðznÞ

p ; ð3:2Þ

xnþ1 ¼ yn �4f ðynÞ

f 0ðynÞ þ 2f 0 znþyn2

� �þ f 0ðznÞ

: ð3:3Þ

Let a be a simple zero of f. By Taylor’s expansion, we have,

f ðxnÞ ¼ f 0ðaÞ en þ c2e2n þ c3e3

n þ c4e4n þ c5e5

n þ c6e6n þOðe7

nÞ� �

; ð3:4Þf 0ðxnÞ ¼ f 0ðaÞ 1þ 2c2en þ 3c3e2

n þ 4c4e3n þ 5c5e4

n þ 6c6e5n þOðe6

nÞ� �

; ð3:5Þ

where

ck ¼1

k!

� �f ðkÞðaÞf 0ðaÞ ; k ¼ 2; 3; . . . ; and en ¼ xn � a:

From (3.1), (3.4) and (3.5), we have,

zn ¼ aþ c2 þ1

2a

� �e2

n þ 2c3 �2

3a2� 2c2

2 �c2

a

� �e3

n þ 3c4 �2c3

aþ 3c2

2a2þ 7

8a3� 7c2c3 þ

5c22

2aþ 4c3

2

� �e4

n

þ � 13c2

6a3þ 3c3

a2� 3c4

aþ 4c5 þ

9c3c2

a� 4c2

2

a2� 6c3

2

aþ 20c3c2

2 � 10c4c2 � 6c23 � 8c4

2 �17

15a4

� �e5

n

þ 5c6 �4c5

a� 29c3c2

2a2� 31c3c2

2

aþ 13c4c2

a� 17c4c3 þ 28c4c2

2 � 13c5c2 þ 33c23c2 � 52c3c3

2

�

þ 209

144a5þ 73c2

24a4� 13c3

3a3þ 9c4

2a2þ 73c2

2

12a3þ 61c3

2

6a2þ 8c2

3

aþ 14c4

2

aþ 16c5

2

�e6

n

þ 6c7 þ133c3c2

6a3� 5c6

aþ 92c4c3c2 þ

53c3c22

a2� 21c4c2

a2þ 23c4c3

a� 44c4c2

2

aþ 17c5c2

a� 53c2

3c2

a

�

þ 94c3c32

a� 22c5c3 þ 36c5c2

2 � 16c6c2 � 72c4c32 � 126c2

3c22 þ 128c3c4

2 þ73c3

12a4� 13c4

2a3

þ 6c5

a2� 107c2

2

12a4� 97c3

2

6a3� 13c2

3

a2� 25c4

2

a2� 32c5

2

a� 12c2

4 þ 18c33 � 32c6

2 �167c2

40a5� 773

420a6

�e7

n þOðe8nÞ:

ð3:6Þ

Now, by Taylor’s series, we compute

f ðznÞ ¼ f 0ðaÞ c2 þ1

2a

� �e2

n þ 2c3 �2

3a2� 2c2

2 �c2

a

� �e3

n þ 3c4 �2c3

aþ 7c2

4a2þ 7

8a3� 7c2c3 þ

7c22

2aþ 5c3

2

� �e4

n

þ 24c3c22 �

19c22

3a2� 12c4

2 �10c3

2

aþ 11c3c2

a� 17c2

6a3þ 3c3

a2� 3c4

aþ 4c5 � 10c4c2 � 6c2

3 �17

15a4

� �e5

n

þ 5c6 �4c5

a� 221c3c2

12a2� 89c3c2

2

2aþ 16c4c2

a� 17c4c3 þ 34c4c2

2 � 13c5c2 þ 37c23c2 � 73c3c3

2

�

þ 209

144a5þ 157c2

36a4� 101c3

24a3þ 9c4

2a2þ 32c2

2

3a3þ 58c3

2

3a2þ 8c2

3

aþ 27c4

2

aþ 28c5

2

�e6

n


þ 6c7 þ343c3c2

12a3� 5c6

aþ 104c4c3c2 þ

485c3c22

6a2� 28c4c2

a2þ 23c4c3

a� 66c4c2

2

aþ 21c5c2

a� 61c2

3c2

a

�

þ 155c3c32

a� 22c5c3 þ 44c5c2

2 � 16c6c2 � 104c4c32 � 160c2

3c22 þ 206c3c4

2 þ67c3

12a4� 13c4

2a3þ 6c5

a2

� 171c22

10a4� 103c3

2

3a3� 23c2

3

2a2� 166c4

2

3a2� 70c5

2

a� 12c2

4 þ 18c33 � 64c6

2 �259c2

40a5� 773

420a6

�e7

n þOðe8nÞ; ð3:7Þ

and

f 0ðznÞ ¼ f 0ðaÞ 1þ 2c22 þ

c2

a

� �e2

n þ 4c2c3 �4c2

3a2� 4c3

2 �2c2

2

a

� �e3

n þ 6c2c4 �c2c3

aþ 3c2

2

a2þ 7c2

4a3� 11c3c2

2

�

þ 5c32

aþ 8c4

2 þ3c3

4a2

�e4

n þ�c3c2

a2þ 28c3c3

2 þ6c3c2

2

aþ 6c2

3

a� 2c3

a3� 13c2

2

3a3� 6c4c2

aþ 8c5c2 �

8c32

a2

�

� 12c42

a� 20c4c2

2 � 16c52 �

34c2

15a4

�e5

n þ61c3c2

12a3� 16c4c3c2 �

3c3c22

2a2þ 12c4c2

a2þ 9c4c3

aþ 32c4c2

2

a

�

� 8c5c2

a� 29c2

3c2

a� 23c3c3

2

a� 26c5c2

2 þ 10c6c2 þ 60c4c32 � 68c3c4

2 þ95c3

24a4þ c4

2a3þ 73c2

2

12a4þ 73c3

2

6a3

� 14c23

a2þ 61c4

2

3a2þ 28c5

2

aþ 12c3

3 þ 32c62 þ

209c2

72a5

�e6

n þ12c3c5

a� 61c3c2

2

6a3� 15c3c4

a2þ 77c2

3c2

a2� 743c3c2

60a4

�

þ 104c23c2

2

aþ 112c3c4c2

2 � 20c3c5c2 þ4c3c4c2

a� 64c7

2 þ 36c4c23 þ

15c3c32

a2þ 74c3c4

2

a� 68c4c2

2

a2

� 34c5c22

a� 10c2c6

a� 24c2c4

a3þ 12c2c5

a2� 84c3

3c2 �69c3

10a5þ 55c2

3

2a3� 42c3

3

aþ 160c3c5

2 þ 12c2c7 þ 72c5c32

� 32c6c22 � 168c4c4

2 �107c3

2

6a4� 97c4

2

3a3� 50c5

2

a2� 64c6

2

a� 24c2c2

4 �167c2

2

20a5� 773c2

210a6� 2c4

a4

�e7

n þOðe8nÞ:

ð3:8Þ
Using (3.6)–(3.8) , we have
yn ¼ aþ c2

4a2þ c2

2

aþ c3

2

� �e4

n þ 4c3c22 �

7c22

3a2� 4c4

2 �4c3

2

aþ 2c3c2

a� 2c2

3a3

� �e5

n

þ �19c3c2

6a2� 12c3c2

2

aþ 3c4c2

aþ 6c4c2

2 þ 4c23c2 � 20c3c3

2 þ95c2

72a4þ c3

4a3þ 13c2

2

3a3þ 23c3

2

3a2þ 10c4

2

aþ 10c5

2

� �e6

n

þ 11c3c2

3a3þ 12c4c3c2 þ

55c3c22

3a2� 7c4c2

a2� 22c4c2

2

aþ 4c5c2

a� 2c2

3c2

aþ 40c3c3

2

aþ 8c5c2

2 � 32c4c32 � 28c2

3c22

�

þ 60c3c42 �

c3

a4� 431c2

2

60a4� 38c3

2

3a3þ 3c2

3

a2� 52c4

2

3a2� 20c5

2

a� 20c6

2 �23c2

10a5

�e7

n þOðe8nÞ: ð3:9Þ

Again, by Taylor’s series, we compute

f ðynÞ ¼ f 0ðaÞ c2

4a2þ c2

2

aþ c3

2

� �e4

n þ 4c3c22 �

7c22

3a2� 4c4

2 �4c3

2

aþ 2c3c2

a� 2c2

3a3

� �e5

n

þ �19c3c2

6a2� 12c3c2

2

aþ 3c4c2

aþ 6c4c2

2 þ 4c23c2 � 20c3c3

2 þ95c2

72a4þ c3

4a3

�

þ 13c22

3a3þ 23c3

2

3a2þ 10c4

2

aþ 10c5

2

�e6

n þ11c3c2

3a3þ 12c4c3c2 þ

55c3c22

3a2� 7c4c2

a2� 22c4c2

2

a

�

þ 4c5c2

a� 2c2

3c2

aþ 40c3c3

2

aþ 8c5c2

2 � 32c4c32 � 28c2

3c22 þ 60c3c4

2 �c3

a4� 431c2

2

60a4� 38c3

2

3a3

þ 3c23

a2� 52c4

2

3a2� 20c5

2

a� 20c6

2 �23c2

10a5

�e7

n þOðe8nÞ; ð3:10Þ


and

f 0ðynÞ ¼ f 0ðaÞ 1þ c22

2a2þ 2c3

2

aþ 2c4

2

� �e4

n þ 8c3c32 �

14c32

3a2� 8c5

2 �8c4

2

aþ 4c3c2

2

a� 4c2

2

3a3

� �e5

n

þ � 19c3c22

3a2� 24c3c3

2

aþ 6c4c2

2

aþ 12c4c3

2 þ 8c23c2

2 � 40c3c42 þ

95c22

36a4þ c3c2

2a3þ 26c3

2

3a3þ 46c4

2

3a2þ 20c5

2

aþ 20c6

2

� �e6

n

þ 22c3c22

3a3þ 24c3c4c2

2 þ110c3c3

2

3a2� 14c4c2

2

a2� 44c4c3

2

aþ 8c5c2

2

a� 4c2

3c22

aþ 80c3c4

2

aþ 16c5c3

2 � 64c4c42

�

� 56c23c3

2 þ 120c3c52 �

2c3c2

a4� 431c3

2

30a4� 76c4

2

3a3þ 6c2

3c2

a2� 104c5

2

3a2� 40c6

2

a� 40c7

2 �23c2

2

5a5

�e7

n þOðe8nÞ: ð3:11Þ

From (3.6) and (3.9), we have

fzn þ yn

2

� �¼ f 0ðaÞ c2

2þ 1

4a

� �e2

n þ c3 �1

3a2� c2

2 �c2

2a

� �e3

n

þ 3c4

2� c3

aþ 15c2

16a2þ 7

16a3� 7c3c2

2þ 2c2

2

aþ 11c3

2

4

� �e4

n þ 13c3c22 �

15c22

4a2� 7c4

2 �6c3

2

a

�

þ 6c3c2

a� 19c2

19a3þ 3c3

2a2� 3c4

2aþ 2c5 � 5c4c2 � 3c2

3 �17

30a4

�e5

n þ5c6

2� 2c5

a� 317c3c2

32a2

�

� 401c3c22

16aþ 35c4c2

4a� 17c4c3

2þ 37c4c2

2

2� 13c5c2

2þ 39c2

3c2

2� 331c3c3

2

8þ 209

288a5þ 241c2

96a4

� 389c3

192a3þ 9c4

4a2þ 77c2

2

12a3þ 139c3

2

12a2þ 4c2

3

aþ 16c4

2

aþ 33c5

2

2

�e6

n þ 3c7 þ1427c3c2

96a3� 5c6

2aþ 55c4c3c2

�

þ 707c3c22

16a2� 63c4c2

4a2þ 23c4c3

2a� 77c4c2

2

2aþ 23c5c2

2a� 121c2

3c2

4aþ 691c3c3

2

8a� 11c5c3

þ 24c5c22 � 8c6c2 � 60c4c3

2 �345c2

3c22

4þ 469c3c4

2

4þ 119c3

48a4� 13c4

4a3þ 3c5

a2� 2483c2

2

240a4� 61c3

2

3a3

� 77c23

16a2� 32c4

2

a2� 40c5

2

a� 6c2

4 þ 9c33 � 37c6

2 �61c2

16a5� 773

840a6

�e7

n þ Oðe8nÞ; ð3:12Þ

and

f 0zn þ yn

2

� �¼ f 0ðaÞ 1þ c2

2 þc2

2a

� �e2

n þ 2c3c2 �2c2

3a2� 2c3

2 �c2

2

a

� �e3

n þ 3c4c2 �5c3c2

4aþ 7c2

2

4a2þ 7c2

8a3

�

� 25c3c22

4þ 7c3

2

2aþ 5c4

2 þ3c3

16a2

�e4

n þ �3c23c2 þ

5c3c2

4a2þ 21c3c3

2 þ8c3c2

2

aþ 3c2

3

a� c3

2a3� 19c3

2

3a2

�

� 12c52 �

10c42

a� 17c2

2

6a3� 3c4c2

aþ 4c5c2 � 10c4c2

2 �17c2

15a4

�e5

n þ � 11c3c2

24a3� 25c4c3c2

2� 29c3c2

2

3a2

�

þ 39c4c2

8a2þ 9c4c3

4aþ 67c4c2

2

4a� 4c5c2

a� 13c2

3c2

4a� 31c3c3

2

a� 13c5c2

2 þ 5c6c2 þ69c4c3

2

2þ 41c2

3c22

2

� 123c3c42

2þ 95c3

96a4þ c4

16a3þ 157c2

2

a4þ 125c3

2

12a3� 7c2

3

2a2þ 107c4

2

6a2þ 24c5

2

aþ 3c3

3 þ 26c62 þ

209c2

144a5

�e6

n

þ 3c3c5

aþ 97c3c2

2

12a3� 9c3c4

2a2þ 18c2

3c2

a2� 433c3c2

240a4þ 13c2

3c22

2aþ 83c3c4c2

2 � 16c3c5c2

�

þ 19c3c4c2

2a� 52c7

2 þ 9c4c23 þ

233c3c32

6a2þ 92c3c4

2

a� 125c4c2

2

4a2� 141c4c3

2

2aþ 21c5c2

2

a

� 5c2c6

a� 63c2c4

8a3þ 6c2c5

a2� 12c3

3c2 � 82c23c3

2 �69c3

40a5þ 55c2

3

8a3� 21c3

3

2aþ 155c3c5

2

þ 6c2c7 þ 44c5c32 � 16c6c2

2 � 107c4c42 �

161c32

10a4� 173c4

2

6a3� 127c5

2

3a2� 52c6

2

a� 12c2c2

4

� 259c22

40a5� 773c2

420a6� c4

a4

�e7

n þ Oðe8nÞ: ð3:13Þ


Using (3.3), (3.8), (3.9), (3.10), (3.11) and (3.13), we have

Table

No.

1

2

34567

891011

12

1314

15

xnþ1 ¼ aþ c52 þ

3c42

2aþ 3c3

2

4a2þ c2

2

8a3

� �e6

n þ 6c3c42 �

13c42

2a2� 6c6

2 �9c5

2

aþ 6c3c3

2

a� 11c3

2

4a3þ 3c3c2

2

2a2� c2

2

2a4

� �e7

n

þ Oðe8nÞ;

implies

enþ1 ¼ c52 þ

3c42

2aþ 3c3

2

4a2þ c2

2

8a3

� �e6

n þ 6c3c42 �

13c42

2a2� 6c6

2 �9c5

2

aþ 6c3c3

2

a� 11c3

2

4a3þ 3c3c2

2

2a2� c2

2

2a4

� �e7

n þOðe8nÞ:

Thus, we observe that the new three-step method (SAAF) has sixth order of convergence. h

4. Numerical examples

In this section we consider some numerical examples to demonstrate the performance of our newly devel-oped iterative method. We compare Chen’s method (Ch), Noor’s Method (Nr) with the newly developedmethod (SAAF). All the computations for above mentioned methods, are performed using software Maple9, precision 60 digits and e ¼ 10�18 as tolerance, using the following criteria for estimating the zero:

(i) d ¼ jxnþ1 � xnj < e,(ii) jf ðxnÞj < e,

(iii) Maximum numbers of iterations = 500.

We use fifteen examples for numerical testing and the results are given in Table 2.

1

f(x) Root

x10 � 1 1�1

arctan(x) 0�0.25476336284103557e�14�0.440747298854002e�67�0.18138519616e�121

0.27360581313052272e�14ex2þ7x�30 � 1 3ex � 1� cosðxÞ 0.6013467677258198x3 � 2x2 � 5 2.690647448028613751=x� 1 1e1�x � 1 1

0*(which is not a root)x� 0:8� 0:2 sin x .9643338876952226sinð1=x� xÞ �0.8975394612804872sinðxÞ 0ex � 2�x þ 2 cosðxÞ � 6 1.829383601933849

0***

ðx� 2Þ2 � lnðxÞ 3.0571035499947381.412391172023885****

ex � 3x2 0.9100075724887091sinðxÞ � e�x 3.096363932410646

37.69911184307751*

9.424697254738521*****

6.285049273382587x2 � 10 cosðxÞ �1.379364594222031

0*(which is not a root)


** Root is real part of a complex number.*** Not a root.

Table 2Values of f(x) and d ar for our algorithm SAAF

f(x) x0 Ch Nr SAAF f(x) d

1 �1.5 9 10 4 0 3.5e51

0.5 D 11 498** – –0.9 6 6 3 0 2.6e�53

2 12 21 5 0 3.19e�30

4 19 *** 8 0 7.53e�17

2 �5 29 10 3 0 1.37e�66

�1 33 5 3 �1.81e�122 1.60e�24

5 29 10 3 0 1.60e�24

3 0 No result 6 D – –2.8 19 13 9** – –2.9 7 7 3 0 2.59e�28

3.5 12 15 5 0 2.96e�34

5 35 D 17 0 04 �0.1 – 5 3 �1.1e�60 0

1 6 5 2 �1.1e�60 7.1e�21

5 11 29 4 �1.1e�60 4.7e�16

5 3 5 5 2 0.1e�57 0.16e�262 7 6 3** – –0.5 – D 3 0 5.36e�49

�0.1 – D 5 1.0e58 3.58e�29

�1 – D 5 1.0e�58 3.58e�29

6 0.5 5 6 2 0 3.96�15

2.7 6 7 3 0 4.4e�39

8 11 12 5 0 9.8e�26

7 �0.1 0* 7 3 0 03 6 6 3 0 5.6.0e�23

3.1 5 7 3 0 08 0 No result 4 3 0 0

0.9 6 6 3 0 02 12 21 5 0 3.19e�30

4 19 D 8 0 7.53e�17

9 1.2 5 4 2 0 5.8e�51

8 7 19 3 1.0e�60 4.3e�23

�1 4 3 2 0 010 1.5 25 5 1 8.6e�108 1.8e�21

�1 33 5 3 �2.8e�146 3.6e�29

1 33 5 2 2.8e�146 3.6e�29

11 0 1*** 7 7 0 5.7e�46

3.14159265 6 7 2 0.1e-59 0.38e-191.3 7 5 3** – –8 13 112 6 0 0

12 8 5**** 6**** 2**** 6e�60 1.1e�15

2.5 7 6 3** – –8 8 23 3 0 3.7e�15

2 5**** 6**** 2**** 6e�60 1.1e�15

13 0.6 6 5 2 0 9.7e�18

1 4 4 2 �1.0e�59 8.0e�42

2 4 4 2 �1.0e�59 8.0e�42

14 2 6* 5 3***** �5.1e�61 03 4 4 2 9.0e�61 06 4 3 2 �4.5e�60 0

15 �3.1 5 11 2 5.0e�59 3.0e�21

�2 5 6 2 5.0e�59 1.5e�31

0 1* 8 4** – –


5. Conclusion

From Tables 1 and 2, we observe that our new iterative method, namely (SAAF) is comparable with theother known methods cited in the table. Almost in all the cases our new method (SAAF) gives better resultsin terms of number of iterations. It is noted that new method is more successful in cases where the initial guessis away from the root.

References

[1] D.K.R. Babajee, M.Z. Dauhoo, An analysis of the properties of the variants of Newon’s method with third order convergence, Appl.Math. Comput. 183 (1) (2006) 659–684.

[2] J. Chen, W. Li, On new exponential quadratically convergent iterative formulae, Appl. Math. Comput. 180 (2006) 242–246.[3] G. Nedzhibov, On a few iterative methods for solving nonlinear equations, Application of Mathematics in Engineering and Economics,

vol. 28, in: Proceedings of XXVIII Summer School Sozopol’ 002, Heroon press, Sofia, 2002.[4] M.A. Noor, New iterative methods for nonlinear equations, Appl. Math. Comput., in press, doi:10.1016/j.amc.2006.05.146.[5] M.A. Noor, F. Ahmad, Fourth-order convergent iterative method for nonlinear equation, Appl. Math. Comput. 182 (2) (2006) 1149–

1153.

http://dx.doi.org/10.1016/j.amc.2006.05.146

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