CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS

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1 CHAPTER 5 SOLVING EQUATIONS BY ITERATIVE METHODS EXERCISE 104 Page 8 1. Find the positive root of the equation x + 3x 5 = 0, correct to 3 significant figures, using the method of bisection. Let f(x) = x + 3x 5 then, using functional notation: f(0) = 5 f(1) = = 1 f() = = +5 Since there is a change of sign from negative to positive there must be a root of the equation between x = 1 and x = The method of bisection suggests that the root is at 1 + = 1.5, i.e. the interval between 1 and has been bisected Hence f(1.5) = (1.5) + 3(1.5) 5 = 1.75 Since f(1) is negative, f(1.5) is positive, and f() is also positive, a root of the equation must lie between x = 1 and x = 1.5, since a sign change has occurred between f(1) and f(1.5) i.e. 1.5 as the next root Hence f(1.5) = (1.5) + 3(1.5) 5 = Since f(1) is negative and f(1.5) is positive, a root lies between x = 1 and x = i.e Hence f(1.15) = (1.15) + 3(1.15) 5 = Since f(1.15) is negative and f(1.5) is positive, a root lies between x = 1.15 and x = i.e Hence f(1.1875) = (1.1875) + 3(1.1875) 5 = , John Bird

2 Since f(1.1875) is negative and f(1.5) is positive, a root lies between x = and x = i.e Hence f(1.1875) = (1.1875) + 3(1.1875) 5 = Since f(1.1875) is positive and f(1.1875) is negative, a root lies between x = and x = = Hence f(1.0315) = Since f(1.0315) is positive and f(1.1875) is negative, a root lies between x = and x = = Hence f( ) = Since f( ) is positive and f(1.1875) is negative, a root lies between x = and x = = Hence, f( ) = Since f( ) is negative and f( ) is positive, a root lies between x = and x = = The last two values obtained for the root are and The last two values are both 1.19, correct to 3 significant figures. We therefore stop the iterations here. Thus, correct to 3 significant figures, the positive root of x + 3x 5 = 0 is Using the bisection method solve e x x =, correct to 4 significant figures. Let f(x) = ex x then f(0) = 1 0 = , John Bird

3 f(1) = e 1 1 = 0.8 f() = e = 3.38 Hence, a root lies between x = 1 and x = Let the root be x = 1.5 f(1.5) = e = Hence, a root lies between x = 1 and x = 1.5 due to a sign change = 1.5 f(1.5) = e = Hence, a root lies between x = 1 and x = 1.5 due to a sign change = 1.15 f(1.15) = e = Hence, a root lies between x = 1.15 and x = 1.5 due to a sign change f(1.1875) = = Hence, a root lies between x = and x = 1.15 due to a sign change f(1.1565) = = Hence, a root lies between x = and x = 1.15 due to a sign change f( ) = = Hence, a root lies between x = and x = due to a sign change = f( ) = , John Bird

4 Hence, a root lies between x = and x = due to a sign change = f( ) = Hence, a root lies between x = and x = due to a sign change f( ) = = Hence, a root lies between x = and x = due to a sign change = The last two values are both 1.146, correct to 4 significant figures. We therefore stop the iterations here Thus, correct to 4 significant figures, the positive root of ex x = is Determine the positive root of x = 4 cos x, correct to decimal places using the method of bisection. Let f(x) = x 4 cos x then f(0) = 0 4 = 4 f(1) = = f() = = Hence, a root lies between x = 1 and x = Let the root be x = 1.5 f(1.5) = 1.5 4cos1.5 = 1.97 Hence, a root lies between x = 1 and x = 1.5 due to a sign change = 1.5 f(1.5) = 1.5 4cos1.5 = , John Bird

5 Hence, a root lies between x = 1 and x = 1.5 due to a sign change = 1.15 f(1.15) = cos1.15 = Hence, a root lies between x = 1.15 and x = 1.5 due to a sign change f(1.1875) = = Hence, a root lies between x = and x = 1.5 due to a sign change f(1.165) = = Hence, a root lies between x = and x = due to a sign change f( ) = = Hence, a root lies between x = and x = due to a sign change f( ) = = Hence, a root lies between x = and x = due to a sign change f( ) = = Hence, a root lies between x = and x = due to a sign change = The last two values are both 1.0, correct to 4 decimal places. We therefore stop the iterations here Thus, correct to decimal places, the root of x = 4 cos x is , John Bird

6 4. Solve x ln x = 0, correct to 3 decimal places using the bisection method. Let f(x) = x ln x f() = ln = f(3) = 3 ln 3 = f(4) = 4 ln 4 = Hence, the root lies between x = 3 and x = 4 because of the sign change. Table 1 shows the results in tabular form. Table 1 x 1 x x1+ x x = f ( x 3 ) Correct to 3 decimal places, the solution of x ln x = 0 is Solve, correct to 4 significant figures, x sin x = 0 using the bisection method. Let f(x) = x sin x then f(0) = 0 0 = 0 f(1) = 1 (sin 1) = f() = (sin 4) = Hence, a root lies between x = 1 and x = Let the root be x = 1.5 f(1.5) = 1.5 (sin 1.5) = 0.49 Hence, a root lies between x = 1.5 and x = due to a sign change , John Bird

7 1.5 + = 1.75 f(1.75) = 1.75 (sin 1.75) = Hence, a root lies between x = 1.75 and x = due to a sign change = f(1.875) = (sin 1.875) = Hence, a root lies between x = 1.75 and x = due to a sign change f(1.815) = = Hence, a root lies between x = and x = due to a sign change f( ) = = Hence, a root lies between x = and x = due to a sign change f( ) = = Hence, a root lies between x = and x = due to a sign change f( ) = = Hence, a root lies between x = and x = due to a sign change f( ) = = Hence, a root lies between x = and x = due to a sign change , John Bird

8 f( ) = = Hence, a root lies between x = and x = due to a sign change f( ) = = Hence, a root lies between x = and x = due to a sign change = The last two values are both 1.849, correct to 4 significant figures. We therefore stop the iterations here Thus, correct to 4 significant figures, the root of x sin x = 0 is , John Bird

9 EXERCISE 105 Page Use an algebraic method of successive approximation to solve: 3x + 5x 17 = 0, correct to 3 significant figures. Let f(x) = 3x + 5x 17 f(0) = 17 f(1) = = 11 f() = = 5 Hence a root lies between x = 1 and x = Since f(3), f(4), and so on do not produce a change of sign, then there is only one positive root First approximation Let the first approximation be 1.7 Second approximation Let the true value of the root, x, be (x 1 + δ 1 ) Let f(x 1 + δ 1 ) = 0, then since x 1 = 1.7, 3(1.7 + δ 1 ) + 5(1.7 + δ 1 ) 17 = 0 Neglecting terms containing products of δ 1 and using the binomial series gives: 3[(1.7) + (1.7) δ 1 ] δ δ δ δ δ Thus x = Third approximation Let the true value of the root, x 3, be (x + δ ) Let f(x + δ ) = 0, then since x = , 3( δ ) + 5( δ ) 17 = 0 Neglecting terms containing products of δ and using the binomial series gives: , John Bird

10 3[(1.6888) + (1.6888) δ ] δ δ δ δ δ Thus x 3 (x + δ ) = Since x and x 3 are the same when expressed to the required degree of accuracy, then the required positive root is 1.69, correct to 3 significant figures Now for the negative root: f( 1) = = 19 f( ) = = 15 f( 3) = = 5 f( 4) = = 11 Hence a root lies between x = 3 and x = 4 Since f( 5), f( 6), and so on do not produce a change of sign, then there is only one negative root First approximation Let the first approximation be 3.4 Second approximation Let the true value of the root, x, be (x 1 + δ 1 ) Let f(x 1 + δ 1 ) = 0, then since x 1 = 3.4, 3( δ 1 ) + 5( δ 1 ) 17 = 0 Neglecting terms containing products of δ 1 and using the binomial series gives: 3[( 3.4) + ( 3.4) δ 1 ] δ δ δ δ δ = Thus x = Third approximation , John Bird

11 Let the true value of the root, x 3, be (x + δ ) Let f(x + δ ) = 0, then since x = , 3( δ ) + 5( δ ) 17 = 0 Neglecting terms containing products of δ and using the binomial series gives: 3[( ) + ( ) δ ] δ δ δ δ δ Thus x 3 (x + δ ) = Since x and x 3 are the same when expressed to the required degree of accuracy, then the required negative root is 3.36, correct to 3 significant figures Thus, the two solutions of the equation 3x + 5x 17 = 0 are 1.69 and 3.36, correct to 3 significant figures. Use an algebraic method of successive approximation to solve: x 3 x + 14 = 0, correct to 3 decimal places. Let f(x) = x3 x+ 14 f(0) = 14 f(1) = = 13 f() = = 18 (There are no positive values of x) f( 1) = = 15 f( ) = = 10 f( 3) = = 7 Hence a root lies between x = and x = 3 First approximation Let the first approximation be.6 Second approximation , John Bird

12 Let the true value of the root, x, be (x 1 + δ 1 ) Let f(x 1 + δ 1 ) = 0, then since x 1 =.6, (.6 + δ 1 ) 3 (.6 + δ 1 ) + 14 = 0 Neglecting terms containing products of δ 1 and using the binomial series gives: [(.6) 3 + 3(.6) δ 1 ] + 5. δ δ δ δ δ Thus x =.6888 Third approximation Let the true value of the root, x 3, be (x + δ ) Let f(x + δ ) = 0, then since x =.6888, ( δ ) 3 ( δ ) + 14 = 0 Neglecting terms containing products of δ gives: ( ) ( ) δ δ δ δ δ δ Thus x 3 (x + δ ) = Fourth approximation Let the true value of the root, x 4, be (x 3 + δ 3 ) Let f(x 3 + δ 3 ) = 0, then since x 3 =.6857, ( δ 3 ) 3 ( δ 3 ) + 14 = 0 Neglecting terms containing products of δ 3 gives: ( ) ( ) δ δ , John Bird

13 δ δ δ δ Thus x 4 (x 3 + δ 3 ) = Since x 3 and x 4 are the same when expressed to the required degree of accuracy, then the required root is.686, correct to 3 decimal places 3. Use an algebraic method of successive approximation to solve: x 4 3x 3 + 7x 5.5 = 0, correct to 3 significant figures. Let f(x) = x4 3x3+ 7x 5.5 f(0) = 5.5 f(1) = = 0.5 f() = = 0.5 Hence a root lies between x = 1 and x = First approximation Let the first approximation be 1.5 Second approximation Let the true value of the root, x, be (x 1 + δ 1 ) Let f(x 1 + δ 1 ) = 0, then since x 1 = 1.5, (1.5 + δ 1 ) 4 3(1.5 + δ 1 ) 3 + 7(1.5 + δ 1 ) 5.5 = 0 Neglecting terms containing products of δ 1 and using the binomial series gives: [(1.5) 4 + 4(1.5) 3 δ 1 ] 3[(1.5) 3 + 3(1.5) δ 1 ] δ δ δ δ δ δ Thus x = 1.75 Third approximation , John Bird

14 Let the true value of the root, x 3, be (x + δ ) Let f(x + δ ) = 0, then since x = 1.75, ( δ ) 4 3( δ ) 3 + 7( δ ) 5.5 = 0 Neglecting terms containing products of δ gives: ( ) ( ) ( ) ( ) δ δ δ [ ] + 7( ) δ δ δ δ 0.875δ Thus x 3 (x + δ ) = Fourth approximation Let the true value of the root, x 4, be (x 3 + δ 3 ) Let f(x 3 + δ 3 ) = 0, then since x 3 = 1.69, ( δ 3 ) 4 3( δ 3 ) 3 + 7( δ 3 ) 5.5 = 0 Neglecting terms containing products of δ 3 gives: ( ) ( ) ( ) ( ) δ3 δ3 δ [ ] + 7( ) δ δ δ δ δ Thus x 4 (x 3 + δ 3 ) = Fifth approximation Let the true value of the root, x 5, be (x 4 + δ 4 ) Let f(x 4 + δ 4 ) = 0, then since x 4 = , ( δ 4 ) 4 3( δ 4 ) 3 + 7( δ 4 ) 5.5 = 0 Neglecting terms containing products of δ 4 gives: ( ) ( ) ( ) ( ) δ4 δ4 δ [ ] + 7( ) δ δ δ δ , John Bird

15 δ Thus x 5 (x 4 + δ 4 ) = Since x 4 and x 5 are the same when expressed to the required degree of accuracy, then the required root is 1.68, correct to 3 decimal places From earlier, f(x) = x4 3x3+ 7x 5.5 f(0) = 5.5 f( 1) = = 8.5 f( ) = = 0.5 Hence, a root lies between x = 1 and x = This root, x = 1.53, may be found in exactly the same way as for the positive root above 4. Use an algebraic method of successive approximation to solve: x 4 + 1x 3 13 = 0, correct to 4 significant figures. Let f(x) = x 4 + 1x 3 13 f(0) = 13 f(1) = = 0 f() = = 99 Hence, x = is a root of the equation There are no further positive roots f( 1) = = 4 f( ) = 93 f( 3) = 56 f( 4) = 55 f( 5) = 888 f( 6) = 1303 f( 7) = 178 f( 8) = 061 f( 9) = 00 f( 10) = 013 f( 11) = 1344 f( 1) = 13 f( 13) = 184 (Since the sign of f( 14) onwards does not change, there are no further negative roots) Hence a root lies between x = 1 and x = 13 First approximation , John Bird

16 Let the first approximation be 1.0 Second approximation Let the true value of the root, x, be (x 1 + δ 1 ) Let f(x 1 + δ 1 ) = 0, then since x 1 = 1.0, ( δ 1 ) 4 + 1( δ 1 ) 3 13 = 0 Neglecting terms containing products of δ 1 and using the binomial series gives: [( 1.0) 4 + 4( 1.0) 3 δ 1 ] + 1[( 1.0) 3 + 3( 1.0) δ 1 ] δ δ δ 1 178δ Thus x = Third approximation Let the true value of the root, x 3, be (x + δ ) Let f(x + δ ) = 0, then since x = , ( δ ) 4 + 1( δ ) 3 13 = 0 Neglecting terms containing products of δ gives: ( ) ( ) δ ( ) ( ) δ [ ] δ δ 13 0 δ δ Thus x 3 (x + δ ) = Since x and x 3 are the same when expressed to the required degree of accuracy, then the required root is 1.01, correct to 4 significant figures Thus, the two solutions of the equation x 4 + 1x 3 13 = 0 are and 1.01, correct to 4 significant figures , John Bird

17 EXERCISE 106 Page Use Newton s method to solve: x x 13 = 0, correct to 3 decimal places. Let f(x) = x x 13 f(0) = 13 f(1) = 1 13 = 14 f() = = 13 f(3) = = 10 f(4) = = 7 f(5) = = Hence a root lies between x = 4 and x = 5. Let r 1 = 4.7 A better approximation is given by: ( 1) f r r = r1 f f '( r) '( x ) = x 1 Hence, (4.7) (4.7) r = 4.7 = 4.7 = (4.7) 7.4 f (4.7419) r3 = = = f '(4.7419) Hence, correct to 3 decimal places, x = 4.74 Since f(6), f(7), and so on do not change sign, there are no further positive roots f( 1) = = 10 f( ) = = 5 f( 3) = = Hence a root lies between x = and x = 3. Let r 1 =.7 A better approximation is given by: ( 1) f r r = r1 f f '( r) '( x ) = x 1 Hence, (.7) (.7) r =.7 =.7 =.7401 (.7) 7.71 f (.7401) r3 =.7401 =.7401 = f '(.7401) f (.73876) r4 = = = f '(.73876) f (.74166) r5 = = =.7417 f '(.74166) Hence, correct to 3 decimal places, x =.74 i.e. the two roots of x x 13 = 0 are 4.74 and.74. Use Newton s method to solve: 3x 3 10x = 14, correct to 4 significant figures. Let f(x) = 3x3 10x 14 f(0) = 14 f(1) = = , John Bird

18 f() = = 10 f(3) = = 36 Hence, a root lies between x = and x = 3 Let r 1 =. A better approximation is given by: ( 1) f r r = r1 f '( x) = 9x 10x f '( r) 1 Hence, 3(.) 3 10(.) (.) 10(.) 1.56 r =. =. =.3881 f (.3881) r3 =.3881 =.3881 =.796 f '(.3881) f (.796) r4 =.796 =.796 =.331 f '(.796) f (.331) r5 =.331 =.331 =.3036 f '(.331) r6 =.3036 = r 7 =.3183 = r8 =.3105 = r 9 =.3146 = r10 =.314 = r 11 =.3136 = r =.319 = Hence, correct to 4 significant figures, x = Use Newton s method to solve: x 4 3x 3 + 7x = 1, correct to 3 decimal places. Let f(x) = x 4 3x 3 + 7x 1 f(0) = 1 f(1) = = 7 f() = = 6 f(3) = = 9 Hence, a root lies between x = and x = 3 Let r 1 =.6 (Since f(4), f(5), and so on do not change sign, there are no further positive roots) , John Bird

19 A better approximation is given by: ( 1) f r r = r1 f '( x) = 4x 3 9x + 7 f '( r) 1 Hence, (.6) 4 3(.6) 3 + 7(.6) (.6) 3 9(.6) r =.6 =.6 = f (.65044) r3 = = = f '(.65044) f (.64799) r4 = = = f '(.64799) Hence, correct to 3 decimal places, x =.648 f( 1) = = 15 f( ) = = 14 Hence a root lies between x = 1 and x =. Let r 1 = 1.5 (Since f( 3), f( 4), and so on do not change sign, there are no further negative roots). A better approximation is given by: ( 1) f r r = r1 f '( x) = 4x 3 9x + 7 f '( r) 1 Hence, ( 1.5) 4 3( 1.5) 3+ 7( 1.5) ( 1.5) 3 9( 1.5) r = 1.5 = 1.5 = f ( ) r3 = = = f '( ) f ( 1.776) r4 = = = f '( 1.776) f ( ) r5 = = = f '( ) Hence, correct to 3 decimal places, x = 1.71 i.e. the two roots of x 4 3x 3 + 7x = 1 are.648 and Use Newton s method to solve: 3x 4 4x 3 + 7x 1 = 0, correct to 3 decimal places. Let f(x) = 3x4 4x3+ 7x 1 f(0) = 1 f(1) = = 6 f() = = 18 Hence, a root lies between x = 1 and x = There are no further positive roots since the x 4 term predominates. f( 1) = = 1 f( ) = = , John Bird

20 Hence, a root lies between x = 1 and x = There are no further negative roots since, once again, the x 4 term predominates. For the root between x = 1 and x = : Let r 1 = 1.5 f '( x) = 1x3 1x ( ) ( ) ( ) ( ) f( r1 ) (1.5) r = r1 = 1.5 = 1.5 = f '( r1 ) r 3 = = r 4 = = r 5 = = Hence, correct to 3 decimal places, the positive root is: x = For the root between x = 1 and x = : Let r 1 = ( ) ( ) ( ) ( ) f( r1 ) ( 1.) r = r1 = 1. = 1. = f '( r1 ) r 3 = = r 4 = = r 5 = = Hence, correct to 3 decimal places, the negative root is: x = Use Newton s method to solve: 3 ln x + 4x = 5, correct to 3 decimal places. Let f(x) = 3 ln x + 4x 5 f(1) = = 1 f() = 3 ln = Hence, a root lies between x = 1 and x = Let r 1 = , John Bird

21 (Since f(3), f(4), and so on do not change sign, there are no further positive roots) A better approximation is given by: r = r 1 ( 1) f r f '( r) 1 f 3 '( x) = + 4 x 3ln (1.) + 4(1.) Hence, r = 1. = 1. = / f (1.1466) r3 = = = f '(1.1466) Hence, correct to 3 decimal places, x = Use Newton s method to solve: x 3 = 5 cos x, correct to 3 decimal places. Let f(x) = x 3 5 cos x f(0) = 0 5 cos 0 = 5 f(1) = 1 5 cos = (note: cos means cos( rad)) Hence, a root lies between x = 0 and x = 1 Let r 1 = 0.7 (Since f(), f(3), and so on do not change sign, there are no further positive roots) A better approximation is given by: ( 1) f r r = r1 f '( x) = 3x + 10sin x f '( r) 1 Hence, (0.7) 3 5cos(1.4) (0.7) + 10sin(1.4) r = 0.7 = 0.7 = f ( ) r3 = = = f '( ) f ( ) r4 = = = f '( ) Hence, correct to 3 decimal places, x = f(0) = 0 5 cos 0 = 5 f( 1) = 1 5 cos( ) = Hence, a root lies between x = 0 and x = 1 Let r 1 = 0.8 A better approximation is given by: ( 1) f r r = r1 f '( x) = 3x + 10sin x f '( r) 1 Hence, ( 0.8) 3 5cos( 1.6) r = 0.8 = 0.8 = ( 0.8) + 10sin( 1.6) , John Bird

22 f ( ) r3 = = = f '( ) f ( ) r4 = = = f '( ) Hence, correct to 3 decimal places, x = From above, f(0) = 5 and f( 1) = f( ) = 8 5 cos( 4) = Hence, a root lies between x = 1 and x = Let r 1 = 1.7 (Since f( 3), f( 4), and so on do not change sign, there are no further roots) A better approximation is given by: ( 1) f r r = r1 f '( x) = 3x + 10sin x f '( r) 1 Hence, ( 1.7) 3 5cos( 3.4) ( 1.7) + 10sin( 3.4) r = 1.7 = 1.7 = f ( ) r3 = = = f '( ) Hence, correct to 3 decimal places, x = Thus, the solutions to x 3 = 5 cos x, correct to 3 decimal places, are 0.744, and θ 7. Use Newton s method to solve: 300e θ + = 6, correct to 3 significant figures. θ Let f(θ) = 300e θ + 6 f(0) = = 94 f(1) = 35.1 f() = f(3) = Hence, a root lies between x = and x = 3, very close to x = There are no further positive roots since the 300e θ term predominates. There are no negative roots since f(x) will always be positive Let r 1 = f '( x) = 600e θ f r r = r1 = = =.049 f ( 1) 300 e '( r1 ) 600 e r3 =.049 = , John Bird

23 Hence, correct to 3 significant figures, x = Solve the equations in Problems 1 to 5, Exercise 104, page 8 and Problems 1 to 4, Exercise 105, page 31 using Newton s method This is left as further practise at using Newton s method. 9. A Fourier analysis of the instantaneous value of a waveform can be represented by y = π t sin t + 1 sin 3t 8 Use Newton s method to determine the value of t near to 0.04, correct to 4 decimal places, when the amplitude, y, is When y = 0.88, then π = t+ + sin t+ sin 3t 4 8 or π 1 t+ + sin t+ sin 3t 0.88 = Let f(t) = π 1 t+ + sin t+ sin 3t Let r 1 = 0.04 f '( t) = 1+ cost+ cos3t 8 π sin sin 3(0.04) 0.88 f( r1 ) r = r1 = 0.04 = 0.04 f '( r1 ) 3 1+ cos cos 3(0.04) = r3 = = Hence, correct to 4 decimal places, t = A damped oscillation of a system is given by the equation: y = 7.4e 0.5t sin 3t. Determine the value of t near to 4., correct to 3 significant figures, when the magnitude y of the oscillation is zero , John Bird

24 Let f(t) = 7.4e0.5t sin 3t ( )( ) ( )( ) f '( t) = 7.4e0.5 t 3cos3t + sin 3t 3.7 e0.5 t =.e0.5 t cos3t 3.7 e0.5 t sin 3t Let r 1 = 4. ( ) = e0.5t (.cos3t+ 3.7sin 3t) f r r = r1 = 4. = 4. = f '( ) e. cos sin ( 1) 7.4 e.1 sin r1.1 + f (4.189) r3 = = = f '(4.189) (Note, sin 1.6 is sin 1.6 rad) Hence, correct to 3 significant figures, t = The critical speeds of oscillation, λ, of a loaded beam are given by the equation: λ3 λ λ = 0 Determine the value of λ which is approximately equal to 3.0 by Newton s method, correct to 4 decimal places. Let f(λ) = λ λ + λ f λ λ λ '( ) = Let r 1 = ( ) ( ) ( ) ( ) f (3.0) r = 3.0 = 3.0 = 3.0 = f '(3.0) r 3 = = r 4 = = Hence, correct to 4 decimal places, λ = , John Bird

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