66 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Putting i = 2 in (1) x 3 = 0.02439 Therefore reciprocal of 41 is 0.0244. Example 8. Find the square root of 20 correct to 3 decimal places by using recursion formula x i+1 = + i i 120 x 2x . Sol. Put i = 0, x 1 = 0 0 120 2 x x + Let x 0 = 4.5 ∴ x 1 = 120 4.5 24. 5 + = 4.47 Put i = 1, x 2 = 120 4.47 4.4 72 24.47 += Put i = 2, x 3 = 120 4.472 4.47 21 2 4.472 += . Therefore 2 0 ≈ 4.472 correct to three decimal places. Example 9. Show that the following rearrangement of equation x 3 + 6x 2 + 10x – 20 = 0 does not yield a convergent sequence of successive approximations by iteration method near x = 1, x = (20 – 6x 2 – x 3 )/10. Sol. Here, x = () 23 20 6 10 xx f x −− = Hence, f ′ (x)= 2 12 3 10 xx−− Clearly, f ’(x) < – 1 in nbd of x = 1 Hence () ′ fx > 1 and therefore the method and hence the sequence < x n > does not converge. Example 10. Find the smallest root of the equation () () () () !!!! 2345 2222 xxxx 1x 0. 2345 −+−+−+= Sol. Written the given equation as x = 1 + () () () () 2345 222 2 2! 3! 4! 5! xxxx −+− + = φ(x) Omitting x 2 and higher powers of x, we get x = 1 approximately. Taking x 0 = 1, we obtain x 1 = φ(x 0 ) = 1 + () () () () −+− 2222 1111 2! 3! 4! 5! + = 1.2239 x 2 = φ(x 1 ) = 1 + () () () () −+− 2345 2222 (1.2239) (1.2239) (1.2239) (1.2239) 2! 3! 4! 5! + = 1.3263 ALGEBRAIC AND TRANSCENDENTAL EQUATION 67 Similarly, x 3 = φ(x 2 ) = 1.38 x 4 = φ(x 3 ) = 1.409 x 5 = φ(x 4 ) = 1.425 x 6 = φ(x 5 ) = 1.434 x 7 = φ(x 6 ) = 1.439 x 8 = φ(x 7 ) = 1.442 Values of x 7 and x 8 indicate that the root is 1.44 correct to two decimal places. PROBLEM SET 2.3 1. Use the method of Iteration to find a positive root between 0 and 1 of the equation xe x – 1. [Ans. 0.5671477] 2. Find the Iterative method, the real root of the equation 3x – log 10 x = 6 correct to four significant figures. [Ans. 2.108] 3. Solve by Iteration method: (a) x 3 + x + 1 = 0 [Ans. –0.682327803] (b) sin x = 1 1 x x + − [Ans. –0.420365] (c) x 3 – 2x 2 – 4 = 0 [Ans. 2.5943] 4. By Iteration method, find 3 0 .[Ans. 5.477225575] 5. If f(x) is sufficiently differentiable and the iteration x n+1 = F (x n ) converges, prove that the order of convergence is a positive integer. 6. The equation f(x) = 0, where f(x) = 0.1 – x + () () () 234 222 2! 3! 4! xxx −+ – has one root in the interval (0,1). Calculate this root correct to 5 decimal places. [Ans. 0.10260] 7. The equation x 2 + ax + b = 0 has two real roots α and β. Show that the iteration method x n+1 = – 2 n x b a + is convergent near x = α if 2 aα< +β . 2.7 NEWTON–RAPHSON METHOD (OR NEWTON’S METHOD) This method can be derived from Taylor’s series as follows: Let f(x) = 0 be the equation for which we are assuming x 0 be the initial approximation and h be a small corrections to x 0 , so that f(x 0 + h)= 0 Expanding it by Taylor’s series, we get f(x 0 + h) = f(x 0 ) + hf ′(x 0 ) + 2 2! h f ′(x 0 ) + = 0 Since h is small, we can neglect second and higher degree terms in h and therefore, we get f(x 0 ) + hf ′ (x 0 )= 0 68 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES From which we have, h = – () () 0 0 fx fx ′ , where [f ′(x 0 )¹ 0]. Hence, if x 0 be the initial approximation, then next (or first) approximation x 1 is given by x 1 = x 0 + h = x 0 – () () 0 0 fx fx ′ The next and second approximation x 2 is given by x 2 = x 1 – () () ′ 1 1 fx fx In general, x n+1 = x n – () () ′ n n fx fx This formula is well known as Newton-Raphson formula. The iterative procedure terminates when the relative error for two successive approximations becomes less than or equal to the prescribed tolerance. 2.7.1 Procedure for Newton Raphson Method to Find the Root of the Equation f ( x ) = 0 Step 1: Take a trial solution (initial approximation) as x 0 . Find f(x 0 ) and f ′(x 0 ). Step 2: Find next (first) approximation x 1 by using the formula x 1 = x 0 – () () ′ 0 0 fx fx Step 3: Follow the above procedure to find the successive approximations x n +1 using the formula x n+1 = x n – () () n n fx fx ′ , where n = 1, 2, 3, Step 4: Stop the process when 1n n xx + − < ε, where ε is the prescribed accuracy. 2.7.2 Order (or Rate) of Convergence of Newton-Raphson Method Let α be the actual root of equation f(x) = 0 i.e., f(a) = 0. Let x n and x n+1 be two successive approximations to the actual root α. If e n and e n+1 are the corresponding errors we have, x n = α + e n and x n+1 = α + e n+1 . By Newton’s-Raphson formula, e n+1 = e n – () () n n fe fe α+ ′ α+ e n+1 = e n – () () () () () () 2 2 2! 2! ′′′ α+ α+ α+ ′ ′′ ′′′ α+ α+ α+ n n n n e fef f e fef f (By Taylor’s expansion) e n+1 = e n – () () () () () 2 2 2! 2! ′′′ α+ α+ ′ ′′ ′′′ α+ α+ α+ n n n n e ef f e fef f [3 f (α) = 0] ALGEBRAIC AND TRANSCENDENTAL EQUATION 69 e n+1 = () () () ′′ α ′′′ α+ α 2 2 n n ef fef (On neglecting high powers of e n ) = () () () () ′′ α ′′ α ′ α+ ′ α 2 2 1 n n f e f fe f = () () () () − ′′ ′′ αα + ′′ αα 1 2 1 2 n n ff e e ff = 2 () () 1 2() () n n ffe e ff αα αα ′′ ′′ −+ ′′ = () () () () ′′ ′′ αα −+ ′ αα 2 23 2' 2 nn ff ee ff or 2 1 2 () () 1 2()2 () nn n ff ee ff e + ′′ ′′ αα =− + ′′ αα () () ′′ α ≈ ′ α2 f f (Neglecting terms containing powers of e n ) Hence by definition, the order of convergence of Newton-Raphson method is 2 i.e., Newton- Raphson method is quadratic convergent. This also shows that subsequent error at each step is proportional to the square of the previous error and as such the convergence is quadratic. Example 1. Find the real root of the equation x 2 – 5x + 2 = 0 between 4 and 5 by Newton- Raphson’s method. Sol. Let f(x)= x 2 – 5x + 2 (1) Now, f(4) = 4 2 – 5 × 4 + 2 = – 2 and f(5) = 5 2 – 5 × 5 + 2 = 2 Therefore, the root lies between 4 and 5. From (1), we get f ′(x) = 2x – 5 (2) Now, Newton-Raphson’s method becomes x n+1 = x n – () () n n fx fx ′ = x n – 2 5 2 25 nn n xx x −+ − or x n+1 = 2 2 2 5 n n x x − − n = 0, 1, 2, (3) Let us take x 0 = 4 to obtain the approximation to the root by putting n = 0, 1, 2 into (3), we get First approximation: () −− == == −− 22 0 1 0 242 14 4.66 67 252453 x x x 70 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Second approximation: The root is given by () () 2 2 1 2 1 4.6667 2 2 19.7781 4.5641 2 5 2 4.6667 5 4.3334 x x x − − == = = −− Third approximation: The root is given by () () 2 2 2 3 2 4.5641 2 2 18.8310 4.5616 2 5 2 4.5641 5 4.1282 x x x −− == == −− Fourth approximation: The root is given by () () 2 2 3 4 3 4.5616 2 2 18.8082 4.5616 2 5 2 4.5616 5 4.1232 x x x − − == == −− Since x 3 = x 4 , hence the root of the equation is 4.5616 correct to four decimal places. Example 2. Solve x 3 + 2x 2 + 10x – 20 = 0 by Newton-Raphson’s method Sol. Let f(x)= x 3 + 2x 2 + 10x – 20 ⇒ f ′(x)= 3x 2 + 4x + 10 Now, by Newton-Raphson method, we have x n +1 = x n – () () n n fx fx ′ = x n – 32 2 21020 3410 nn n nn xx x xx ++ − ++ or x n+1 = () 32 2 210 3410 nn nn xx xx ++ ++ (1) Clearly, f(1) = – 7 < 0 and f(2) = 16 > 0 Therefore root lies between 1 and 2. Let x 0 = 1.2 be the initial approximation then First approximation: x 1 = () 32 00 2 00 210 3410 xx xx ++ ++ = () () () () 32 2 2( 1.2 1.2 10) 26.336 19.12 31.2 41.2 10 ++ = ++ or x 1 = 1.3774059 Second approximation: x 2 = () 32 11 2 11 210 3410 xx xx ++ ++ = ()() () 32 2 2 1.3774059 1.3774059 10 29.021052 2.201364 3 1.3774059 1.3774059 10 ++ = ++ or x 2 = 1.3688295 ALGEBRAIC AND TRANSCENDENTAL EQUATION 71 Third approximation: x 3 = 32 22 2 22 2( 1 0) 341 0 xx xx ++ ++ = 32 2 2(1.3688295) (1.3688295) 10 28.87692 4 21.0964 3(1.3688295) 4(1.3688295) 10 ++ = ++ or x 3 = 1.3688081 Fourth approximation: x 4 = () 32 33 2 33 21 0 341 0 xx xx ++ ++ = ()() ()() 32 2 2 1.3688081 1.3688081 10 28.87656 7 21.0961 4 3 1.3688081 4 1.3688081 10 ++ = ++ or x 4 = 1.3688081 Hence the required root is 1.3688081. Example 3. Find the real root of the equation x log 10 x = 1.2 by Newton-Raphson’s method. Sol. Let f(x)= x log 10 x – 1.2 = 0 (1) Then f(1) = – 1.2 f(2) = 2 log 10 2 – 1.2 = – 0.5979 f(3) = 3 log 10 3 – 1.2 = 0.2314 Therefore root lies between 2 and 3. Let us take x 0 = 2, then from (1) f´(x) = log 10 x + 1 x . x log 10 e = log 10 x + 0.4343 (2) Now, by Newton’s-Raphson method, we have x n+1 = x n – () () n n fx fx ′ = x n – 10 10 log 1 .2 log 0.43 43 nn n xx x − + or x n+1 = 10 0.4343 1.2 log 0.4343 + + n n x x , n = 0, 1, 2, 3 (3) Putting n = 0 in (3), we get first approximation First approximation: x 1 = 0 10 0 0.4343 1.2 log 0.434 3 x x + + = () () 10 0.4343 2 1.2 2.06 86 log 2 0.4343 0.73 53 + = + or x 1 = 2.8133 Putting n = 1 in (3), we get second approximation Second approximation: x 2 = 1 10 1 0.4343 1.2 log 0.43 43 x x + + 72 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES = () () 10 0.4343 2.8133 1.2 2.4128 log 2.8133 0.4343 0.8835 + = + or x 2 = 2.7411 Putting n = 2 in (3), we get third approximation Third approximation: x 3 = 2 10 2 0.4343 1.2 log 0.4343 x x + + = () () 10 0.4343 2.7411 1.2 2.3905 log 2.7411 0.4343 0.8722 + = + or x 3 = 2.7408 Putting n = 3 in (3), we get fourth approximation Fourth approximation: x 4 = 3 10 3 0.4343 1.2 log 0.4343 x x + + = () () 10 0.4343 2.7408 1.2 2.3903 log 2.7408 0.4343 0.8721 + = + or x 4 = 2.7408 Since x 3 = x 4 , hence the root of the equation is 2.7408 correct to four decimal places. Example 4. Find the real root of the equation 3x = cos x + 1 by Newton’s method. Sol. Let f(x)= 3x – cos x – 1 = 0 (1) So that f(0) = – 2 f(1) = 3 – cos 1 – 1 = 1.4597 So the root lies between 0 and 1. Let us take x 0 = 0.6 From (1) f´(x) = 3 + sin x (2) Therefore the Newton’s method gives x n+1 = x n – () () n n fx fx ′ or x n+1 = sin cos 1 3sin nn n n xx x x ++ + (3) First approximation: Putting n = 0, in (3) we get first approximation x 1 = 00 0 0 sin cos 1 3sin xx x x ++ + = () () () () 0.6 sin 0.6 cos 0.6 1 3sin0.6 ++ + = ()( ) 0.6 0.5646 0.8253 1 2.16406 3 0.5646 3.5646 ++ = + or x 1 = 0.6071 ALGEBRAIC AND TRANSCENDENTAL EQUATION 73 Second approximation: Putting n = 1, in (3) we get second approximation x 2 = ++ + 11 1 1 sin cos 1 3sin xx x x = ()()() () ++ + 0.6071 sin 0.6071 cos 0.6071 1 3 sin 0.6071 = ()() ++ = + 0.6071 0.5705 0.8213 1 2.1677 3 0.5705 3.5705 or x 2 = 0.6071 Since x 1 = x 2 Therefore the root as 0.6071 correct to four decimal places. Example 5. Find the real root of the equation log x – cos x = 0 correct to three places of decimal by Newton-Raphson’s method. Sol. Let f(x) = log x – cos x = 0 (1) So that f(1) = – 0.5403 f(2) = 1.1092 ∴ The root lies between 1 and 2. Also, f(1.1) = log 1.1 – cos 1.1 = 0.0953 – 0.4535 = – 0.3582 f(1.2) = log 1.2 – cos 1.2 = – 0.18 f(1.3) = log 1.3 – cos 1.3 = – 0.0051 f(1.4) = log1.4 – cos 1.4 = 0.1665 Thus the root lies between 1.3 and 1.4. From (1) f´(x)= 1 x + sin x (2) Then by Newton’s-Raphson method, we get x n +1 = x n – () () ′ n n fx fx Using (1) and (2) x n +1 = x n – − + log cos 1 sin nn n n xx x x or x n +1 = +−+ + 2 sin log cos 1sin nn nn nn n nn xx xx xx x xx (3) Let us take x 0 = 1.3 Now putting n = 0 into (3), we get first approximation First approximation: x 1 = +−+ + 2 00 00 00 0 00 sin log cos 1sin xx xx xx x xx = ()() ()() ()() () () () +−+ + 2 1.3 1.3 sin 1.3 1.3 log 1.3 1.3 cos 1.3 1 1.3 sin 1.3 74 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES = () 1.3 1 1.2526 0.2623 0.2674 1 1.2526 +− + + or x 1 = 2.93501 2.2526 = 1.3029 Now putting n = 1 into (3), we get second approximation Second approximation: x 2 = 2 11 11 11 1 11 sin log cos 1sin xx xx xx x xx +−+ + = ()()()() ()() ()() +− + + 2 1.3029 1.3029 sin 1.3029 1.3029 log 1.3029 1.3029 cos 1.3029 1 1.3029 sin 1.3029 = 1.3029 (1 1.2526 0.2645 0.2647) 1 1.2564 +−+ + or x 2 = 2.9401 2.2564 = 1.3030 Hence the required root is 1.303 correct to three decimal places. Example 6. Evaluate 12 to four decimal places by Newton’s iterative method. Sol. Let x = 12 ⇒ x 2 – 12 = 0 (1) Therefore Newton’s Iterative formula gives, x n+1 = x n – () () n n fx fx ′ x n+1 = x n – 2 12 112 22 n n nn x x xx − =+ (2) Now since f(3) = – 3 (–ve) and f(4) = 4 (+ve) Therefore the root lies between 3 and 4. Take x 0 = 3.5, equation (2) gives x 1 = 0 0 112 2 x x + x 1 = 112 3.5 23.5 + = 3.4643 x 2 = 112 3.4643 2 3.4643 + = 3.4641 x 3 = 112 3.4641 2 3.4641 + = 3.4641 Since, x 2 = x 3 up to four decimal places. So we have 1 2 = 3.4641. ALGEBRAIC AND TRANSCENDENTAL EQUATION 75 Example 7. Find a positive root of (17) 1/3 correct to four decimal places by Newton-Raphson’s method. Sol. The iterative formula for the given equation is x n+1 = 2 1 2 3 n n k x x + (1) Here k = 17. Take x 0 = 2.5 because ! & = 2, ! % = 3 Putting n = 0 in (1), we get First approximation: x 1 = 0 2 0 117 2 3 x x + x 1 = 117 5 36.25 + = 2.5733 Putting n = 1 in (1), we get Second approximation: x 2 = 1 2 1 117 2 3 x x + x 2 = 117 5.1466 3 6.6220 + = 2.5713 Putting n = 2 in (1), we get Third approximation: x 3 = 2 2 2 117 2 3 x x + x 3 = 117 5.1426 3 6.61158 + = 2.57128 Putting n = 3 in (1), we get Fourth approximation: x 4 = 3 2 3 117 2 3 x x + x 1 = 117 5.14256 3 6.61148 + = 2.57138 Since x 3 and x 4 are accurate to four decimal places hence the required root is 2.5713. Example 8. Using Newton’s iterative method, find the real root of x sin + cos x = 0, which is near x = π correct to 3 decimal places. Sol. Given, f(x)= x sin x + cos x = 0 therefore f ′(x) = x cos x The iteration formula is, x n+1 = x n – sin cos cos nn n nn xx x xx + With x 0 = π . 66 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Putting i = 2 in (1) x 3 = 0.024 39 Therefore reciprocal of 41 is 0.0244. Example 8. Find the square root of 20 correct to 3 decimal places. 0 Since h is small, we can neglect second and higher degree terms in h and therefore, we get f(x 0 ) + hf ′ (x 0 )= 0 68 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES From which we have, h =. 1.3688 295 ALGEBRAIC AND TRANSCENDENTAL EQUATION 71 Third approximation: x 3 = 32 22 2 22 2( 1 0) 341 0 xx xx ++ ++ = 32 2 2(1.3688 295 ) (1.3688 295 ) 10 28.87 692 4 21. 096 4 3(1.3688 295 ) 4(1.3688 295 )