Higher Engineering Mathematics - BS Grewal

Solving equations by iterative methods 83 Neglecting terms containing products of δ3 gives: f (1) = 3(1)3 − 10(1)2 + 4(1) + 7 = 4 2.3759 − 6.1656 δ3 + 4.6242 − 6 δ3 − 7 ≈ 0 f (2) = 3(2)3 − 10(2)2 + 4(2) + 7 = −1 i.e. δ3 ≈ −2.3759 − 4.6242 + 7 −6.1656 − 6 Following the above procedure: ≈ −0.0001 First approximation −12.156 (a) Let the first approximation be such that it divides ≈ +0.00000822 the interval 1 to 2 in the ratio of 4 to −1, i.e. let x1 be 1.8. Thus, x4, the fourth approximation to the root is Second approximation (−0.7707 + 0.00000822), i.e. x4 = −0.7707, cor- (b) Let the true value of the root, x2, be (x1 + δ1). rect to 4 significant figures, and −0.771, correct to (c) Let f (x1 + δ1) = 0, then since x1 = 1.8, 3 significant figures. 3(1.8 + δ1)3 − 10(1.8 + δ1)2 Since the values of the roots are the same on + 4(1.8 + δ1) + 7 = 0 two consecutive approximations, when stated to the required degree of accuracy, then the negative Neglecting terms containing products of δ1 and root of 4x2 − 6x − 7 = 0 is −0.771, correct to 3 using the binomial series gives: significant figures. 3[1.83 + 3(1.8)2 δ1] − 10[1.82 + (2)(1.8) δ1 ] [Checking, using the quadratic formula: + 4(1.8 + δ1) + 7 ≈ 0 −(−6) ± [(−6)2 − (4)(4)(−7)] 3(5.832 + 9.720 δ1) − 32.4 − 36 δ1 x= + 7.2 + 4 δ1 + 7 ≈ 0 (2)(4) 17.496 + 29.16 δ1 − 32.4 − 36 δ1 = 6 ± 12.166 = −0.771 and 2.27, + 7.2 + 4 δ1 + 7 ≈ 0 8 correct to 3 significant figures] [Note on accuracy and errors. Depending on the −17.496 + 32.4 − 7.2 − 7 accuracy of evaluating the f (x + δ) terms, one or two δ1 ≈ 29.16 − 36 + 4 iterations (i.e. successive approximations) might be saved. However, it is not usual to work to more than ≈ − 0.704 ≈ −0.2479 about 4 significant figures accuracy in this type of cal- 2.84 culation. If a small error is made in calculations, the only likely effect is to increase the number of iterations.] Thus x2 ≈ 1.8 −0.2479 =1.5521 Problem 5. Determine the value of the Third approximation smallest positive root of the equation 3x3 − 10x2 + 4x + 7 =0, correct to 3 significant (d) Let the true value of the root, x3, be (x2 + δ2). figures, using an algebraic method of successive (e) Let f (x2 + δ2) = 0, then since x2 = 1.5521, approximations. 3(1.5521 + δ2)3 − 10(1.5521 + δ2)2 The functional notation method is used to find the value of the first approximation. + 4(1.5521 + δ2) + 7 = 0 f (x) = 3x3 − 10x2 + 4x + 7 Neglecting terms containing products of δ2 gives: f (0) = 3(0)3 − 10(0)2 + 4(0) + 7 = 7 11.217 + 21.681 δ2 − 24.090 − 31.042 δ2 + 6.2084 + 4 δ2 + 7 ≈ 0

84 Higher Engineering Mathematics δ2 ≈ −11.217 + 24.090 − 6.2084 − 7 9.4 The Newton-Raphson method 21.681 − 31.042 + 4 ≈ −0.3354 The Newton-Raphson formula, often just referred to as −5.361 Newton’s method, may be stated as follows: ≈ 0.06256 If r1 is the approximate value of a real root of the equation f (x) = 0, then a closer approximation to the Thus x3 ≈ 1.5521 + 0.06256 ≈ 1.6147 root r2 is given by: (f) Values of x4 and x5 are found in a similar way. r2 = r1 − f (r1) f (r1) f (x3 + δ3) = 3(1.6147 + δ3)3 − 10(1.6147 The advantages of Newton’s method over the alge- + δ3)2 + 4(1.6147 + δ3) + 7 = 0 braic method of successive approximations is that it can be used for any type of mathematical equation giving δ3 ≈ 0.003175 and x4 ≈ 1.618, i.e. 1.62 (i.e. ones containing trigonometric, exponential, loga- correct to 3 significant figures. rithmic, hyperbolic and algebraic functions), and it is usually easier to apply than the algebraic method. f (x4 + δ4) = 3(1.618 + δ4)3 − 10(1.618 Problem 6. Use Newton’s method to determine + δ4)2 + 4(1.618 + δ4) + 7 = 0 the positive root of the quadratic equation 5x2 + 11x − 17 =0, correct to 3 significant figures. giving δ4 ≈ 0.0000417, and x5 ≈ 1.62, correct to Check the value of the root by using the quadratic 3 significant figures. Since x4 and x5 are the same when expressed to formula. the required degree of accuracy, then the required root is 1.62, correct to 3 significant figures. The functional notation method is used to determine the first approximation to the root. Now try the following exercise f (x) = 5x2 + 11x − 17 Exercise 36 Further problems on solving f (0) = 5(0)2 + 11(0) − 17 = −17 equations by an algebraic method of f (1) = 5(1)2 + 11(1) − 17 = −1 successive approximations f (2) = 5(2)2 + 11(2) − 17 = 25 Use an algebraic method of successive approx- This shows that the value of the root is close to x = 1. imation to solve the following equations to the accuracy stated. Let the first approximation to the root, r1, be 1. 1. 3x2 + 5x − 17 = 0, correct to 3 significant Newton’s formula states that a closer approximation, figures. [−3.36, 1.69] r2 = r1 − f (r1) f (r1 ) f (x) = 5x2 + 11x − 17, 2. x3 − 2x + 14 =0, correct to 3 decimal places. [−2.686] thus, f (r1) = 5(r1 )2 + 11(r1) − 17 = 5(1)2 + 11(1) − 17 = −1 3. x4 − 3x3 + 7x − 5.5 = 0, correct to 3 signifi- f (x) is the differential coefficient of f (x), cant figures. [−1.53, 1.68] 4. x4 + 12x3 − 13 = 0, correct to 4 significant i.e. f (x) = 10x + 11. figures. [−12.01, 1.000] Thus f (r1 ) = 10(r1) + 11 = 10(1) + 11 = 21