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 + 3 5 = 1 f() = 4 + 6 5 = +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) 1 + 1.5 i.e. 1.5 as the next root Hence f(1.5) = (1.5) + 3(1.5) 5 = 0.315 Since f(1) is negative and f(1.5) is positive, a root lies between x = 1 and x = 1.5 1 + 1.5 i.e. 1.15 Hence f(1.15) = (1.15) + 3(1.15) 5 = 0.359375 Since f(1.15) is negative and f(1.5) is positive, a root lies between x = 1.15 and x = 1.5 1.15 + 1.5 i.e. 1.1875 Hence f(1.1875) = (1.1875) + 3(1.1875) 5 = 0.0734375 385 014, John Bird
Since f(1.1875) is negative and f(1.5) is positive, a root lies between x = 1.1875 and x = 1.5 1.1875 + 1.5 i.e. 1.1875 Hence f(1.1875) = (1.1875) + 3(1.1875) 5 = 0.1416016 Since f(1.1875) is positive and f(1.1875) is negative, a root lies between x = 1.1875 and x = 1.1875 1.1875 + 1.1875 = 1.0315 Hence f(1.0315) = 0.056885 Since f(1.0315) is positive and f(1.1875) is negative, a root lies between x = 1.1875 and x = 1.0315 1.1875 + 1.0315 = 1.195315 Hence f(1.195315) = 0.0147095 Since f(1.195315) is positive and f(1.1875) is negative, a root lies between x = 1.195315 and x = 1.1875 1.195315 + 1.1875 = 1.191406 Hence, f(1.191406) = 0.006334 Since f(1.191406) is negative and f(1.195315) is positive, a root lies between x = 1.191406 and x = 1.195315 1.191406 + 1.195315 = 1.193359 The last two values obtained for the root are 1.1914 and 1.1934. 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 1.19. Using the bisection method solve e x x =, correct to 4 significant figures. Let f(x) = ex x then f(0) = 1 0 = 1 386 014, John Bird
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 1.5 1.5 = 0.98169 Hence, a root lies between x = 1 and x = 1.5 due to a sign change 1 + 1.5 = 1.5 f(1.5) = e1.5 1.5 = 0.40343 Hence, a root lies between x = 1 and x = 1.5 due to a sign change 1 + 1.5 = 1.15 f(1.15) = e1.15 1.15 = 0.04478 Hence, a root lies between x = 1.15 and x = 1.5 due to a sign change f(1.1875) = 0.091374 1.15 + 1.5 = 1.1875 Hence, a root lies between x = 1.1875 and x = 1.15 due to a sign change f(1.1565) = 0.01743 1.1875 + 1.15 = 1.1565 Hence, a root lies between x = 1.1565 and x = 1.15 due to a sign change f(1.14065) = 0.01190 1.1565 + 1.15 = 1.14065 Hence, a root lies between x = 1.14065 and x = 1.1565 due to a sign change 1.14065 + 1.1565 = 1.1484375 f(1.1484375) = 0.00484586 387 014, John Bird
Hence, a root lies between x = 1.1484375 and x = 1.14065 due to a sign change 1.1484375 + 1.14065 = 1.1445315 f(1.1445315) = 0.003566 Hence, a root lies between x = 1.1445315 and x = 1.1484375 due to a sign change f(1.14648) = 0.000694 1.1445315 + 1.1484375 = 1.14648 Hence, a root lies between x = 1.14648 and x = 1.1445315 due to a sign change 1.14648 + 1.1445315 = 1.14551 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 1.146 3. 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) = 1.161 = 1.161 f() = 4 1.6646 = 5.6646 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 + 1.5 = 1.5 f(1.5) = 1.5 4cos1.5 = 0.301 388 014, John Bird
Hence, a root lies between x = 1 and x = 1.5 due to a sign change 1 + 1.5 = 1.15 f(1.15) = 1.15 4cos1.15 = 0.459 Hence, a root lies between x = 1.15 and x = 1.5 due to a sign change f(1.1875) = 0.0858 1.15 + 1.5 = 1.1875 Hence, a root lies between x = 1.1875 and x = 1.5 due to a sign change f(1.165) = 0.0906 1.1875 + 1.5 = 1.165 Hence, a root lies between x = 1.1875 and x = 1.165 due to a sign change f(1.01875) = 0.00065 1.1875 + 1.165 = 1.01875 Hence, a root lies between x = 1.1875 and x = 1.01875 due to a sign change f(1.1946875) = 0.0419 1.1875 + 1.01875 = 1.1946875 Hence, a root lies between x = 1.01875 and x = 1.1946875 due to a sign change f(1.198815) = 0.0199 1.01875 + 1.1946875 = 1.198815 Hence, a root lies between x = 1.01875 and x = 1.198815 due to a sign change 1.01875 + 1.198815 = 1.0008715 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 1.0 389 014, John Bird
4. Solve x ln x = 0, correct to 3 decimal places using the bisection method. Let f(x) = x ln x f() = ln = 0.693 f(3) = 3 ln 3 = 0.0986 f(4) = 4 ln 4 = 0.61371 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 x3 3 3 3 3.15 3.15 3.1565 3.14065 3.14844 3.144535 3.146486 4 3.5 3.5 3.5 3.1875 3.15 3.1565 3.14065 3.14844 3.144535 = 3.5 3.5 3.15 3.1875 3.1565 3.14065 3.14844 3.14535 3.146486 3.1455095 f ( x 3 ) 0.474 0.071345 0.014434 0.0863 0.0068654 0.003797 0.001531 0.001137 0.0001999 Correct to 3 decimal places, the solution of x ln x = 0 is 3.146 5. 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) = 0.416 f() = (sin 4) = 0.346 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 390 014, John Bird
1.5 + = 1.75 f(1.75) = 1.75 (sin 1.75) = 0.186 Hence, a root lies between x = 1.75 and x = due to a sign change 1.75 + = 1.875 f(1.875) = 1.875 (sin 1.875) = 0.054 Hence, a root lies between x = 1.75 and x = 1.875 due to a sign change f(1.815) = 0.079 1.75 + 1.875 = 1.815 Hence, a root lies between x = 1.875 and x = 1.815 due to a sign change f(1.84375) = 0.010 1.875 + 1.815 = 1.84375 Hence, a root lies between x = 1.84375 and x = 1.875 due to a sign change f(1.859375) = 0.01 1.84375 + 1.875 = 1.859375 Hence, a root lies between x = 1.84375 and x = 1.859375 due to a sign change f(1.851565) = 0.0051 1.84375 + 1.859375 = 1.851565 Hence, a root lies between x = 1.851565 and x = 1.84375 due to a sign change f(1.8476565) = 0.0081 1.851565 + 1.84375 = 1.8476565 Hence, a root lies between x = 1.851565 and x = 1.8476565 due to a sign change 391 014, John Bird
f(1.849609375) = 0.00109 1.851565 + 1.8476565 = 1.849609375 Hence, a root lies between x = 1.8476565 and x = 1.849609375 due to a sign change f(1.84863813) = 0.00091 1.8476565 + 1.849609375 = 1.84863813 Hence, a root lies between x = 1.849609375 and x = 1.84863813 due to a sign change 1.849609375 + 1.84863813 = 1.84911094 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 1.849 39 014, John Bird
EXERCISE 105 Page 31 1. 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) = 1 + 5 17 = 11 f() = 1 + 10 17 = 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 ] + 8.5 + 5δ 1 17 0 8.67 + 10.δ 1 + 8.5 + 5δ 1 17 0 15.δ 1 17 8.67 8.5 δ 1 17 8.67 8.5 0.01118 15. Thus x 1.7 0.01118 = 1.6888 Third approximation Let the true value of the root, x 3, be (x + δ ) Let f(x + δ ) = 0, then since x = 1.6888, 3(1.6888 + δ ) + 5(1.6888 + δ ) 17 = 0 Neglecting terms containing products of δ and using the binomial series gives: 393 014, John Bird
3[(1.6888) + (1.6888) δ ] + 8.444 + 5δ 17 0 8.55614 + 10.138δ + 8.444 + 5δ 17 0 15.138δ 17 8.55614 8.444 δ 17 8.55614 8.444 0.0000095 15.138 Thus x 3 (x + δ ) = 1.6888 0.0000095 1.6887 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) = 3 5 17 = 19 f( ) = 1 10 17 = 15 f( 3) = 7 15 17 = 5 f( 4) = 48 0 17 = 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( 3.4 + δ 1 ) + 5( 3.4 + δ 1 ) 17 = 0 Neglecting terms containing products of δ 1 and using the binomial series gives: 3[( 3.4) + ( 3.4) δ 1 ] 17 + 5δ 1 17 0 34.68 0.4δ 1 17 + 5δ 1 17 0 15.4δ 1 17 34.68 + 17 δ 1 17 34.68 17 = 0.044156 15.4 Thus x 3.4 + 0.044156 = 3.3558 Third approximation 394 014, John Bird
Let the true value of the root, x 3, be (x + δ ) Let f(x + δ ) = 0, then since x = 3.3558, 3( 3.3558 + δ ) + 5( 3.3558 + δ ) 17 = 0 Neglecting terms containing products of δ and using the binomial series gives: 3[( 3.3558) + ( 3.3558) δ ] 16.779 + 5δ 17 0 33.78418 0.1348δ 16.779 + 5δ 17 0 15.1348δ 16.779 33.78418 + 17 δ 16.779 33.78418 + 17 0.00034 15.1348 Thus x 3 (x + δ ) = 3.3558 + 0.00034 3.3551 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) = 1 + 14 = 13 f() = 8 4 + 14 = 18 (There are no positive values of x) f( 1) = 1 + + 14 = 15 f( ) = 8 + 4 + 14 = 10 f( 3) = 7 + 6 + 14 = 7 Hence a root lies between x = and x = 3 First approximation Let the first approximation be.6 Second approximation 395 014, John Bird
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 1 + 14 0 17.576 + 0.8δ 1 + 5. δ 1 + 14 0 18.8δ 1 17.576 5. 14 δ 1 17.576 5. 14 0.08884 18.8 Thus x.6 0.08884 =.6888 Third approximation Let the true value of the root, x 3, be (x + δ ) Let f(x + δ ) = 0, then since x =.6888, (.6888 + δ ) 3 (.6888 + δ ) + 14 = 0 Neglecting terms containing products of δ gives: ( ) ( ) 3.6888 + 3.6888 δ + 5.3776 δ + 14 0 19.439 + 1.6888δ + 5.3776 δ + 14 0 19.6888δ 19.439 5.3776 14 δ 19.439 5.3776 14 19.6888 0.003119 Thus x 3 (x + δ ) =.6888 + 0.003119.6857 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, (.6857 + δ 3 ) 3 (.6857 + δ 3 ) + 14 = 0 Neglecting terms containing products of δ 3 gives: ( ) ( ) 3.6857 + 3.6857 δ + 5.3714 δ + 14 0 3 3 396 014, John Bird
19.3719 + 1.63895δ 3 + 5.3714 δ 3 + 14 0 19.63895δ 19.3719 5.3714 14 δ 19.3719 5.3714 14 19.63895 0.0000546 Thus x 4 (x 3 + δ 3 ) =.6857 + 0.00005.6857 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) = 1 3 + 7 5.5 = 0.5 f() = 16 4 + 14 5.5 = 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 ] + 10.5 + 7δ 1 5.5 0 5.065 + 13.5δ 1 10.15 0.5δ 1 + 10.5 + 7δ 1 5.5 0 0.5δ 1 0.065 δ 1 0.065 0.5 0.5 Thus x 1.5 + 0.5 = 1.75 Third approximation 397 014, John Bird
Let the true value of the root, x 3, be (x + δ ) Let f(x + δ ) = 0, then since x = 1.75, (1.75 + δ ) 4 3(1.75 + δ ) 3 + 7(1.75 + δ ) 5.5 = 0 Neglecting terms containing products of δ gives: ( ) ( ) ( ) ( ) 4 3 3 δ δ δ 1.75 + 4 1.75 3[ 1.75 + 3 1.75 ] + 7(1.75 + ) 5.5 0 9.3789 + 1.4375δ 16.07815 7.565δ + 1.5 + 7δ 5.5 0 δ 0.875δ 0.050775 0.050775 0.05803 0.875 Thus x 3 (x + δ ) = 1.75 0.05803 1.69 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, (1.69 + δ 3 ) 4 3(1.69 + δ 3 ) 3 + 7(1.69 + δ 3 ) 5.5 = 0 Neglecting terms containing products of δ 3 gives: ( ) ( ) ( ) ( ) 4 3 3 δ3 δ3 δ3 1.69 + 4 1.69 3[ 1.69 + 3 1.69 ] + 7(1.69 + ) 5.5 0 8.19599 + 19.37586δ 3 14.5318977 5.76578δ 3 + 11.844 + 7δ 3 5.5 0 0.61008δ 3 0.008093 δ 3 0.008093 0.013643 0.61008 Thus x 4 (x 3 + δ 3 ) = 1.69 0.013643 1.6787 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 = 1.6787, (1.6787 + δ 4 ) 4 3(1.6787 + δ 4 ) 3 + 7(1.6787 + δ 4 ) 5.5 = 0 Neglecting terms containing products of δ 4 gives: ( ) ( ) ( ) ( ) 4 3 3 δ4 δ4 δ4 1.6787 + 4 1.6787 3[ 1.6787 + 3 1.6787 ] + 7(1.6787 + ) 5.5 0 7.941314 + 18.95δ 4 14.1919 5.363δ 4 + 11.7509 + 7δ 4 5.5 0 0.560δ 4 0.000314 398 014, John Bird
δ 4 0.000314 0.00056 0.560 Thus x 5 (x 4 + δ 4 ) = 1.6787 0.00056 1.6781 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) = 1 + 3 7 5.5 = 8.5 f( ) = 16 + 4 14 5.5 = 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) = 1 + 1 13 = 0 f() = 16 + 96 13 = 99 Hence, x = 1.000 is a root of the equation There are no further positive roots f( 1) = 1 1 13 = 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 399 014, John Bird
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, ( 1.0 + δ 1 ) 4 + 1( 1.0 + 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 ] 13 0 0 736 691δ 1 0 736 + 5184δ 1 13 0 δ 1 178δ 1 13 13 178 0.0075 Thus x 1.0 0.0075 = 1.0075 Third approximation Let the true value of the root, x 3, be (x + δ ) Let f(x + δ ) = 0, then since x = 1.0075, ( 1.0075 + δ ) 4 + 1( 1.0075 + δ ) 3 13 = 0 Neglecting terms containing products of δ gives: ( ) ( ) δ ( ) ( ) 4 3 3 δ 1.0075 + 4 1.0075 + 1[ 1.0075 + 3 1.0075 ] 13 0 0 787.8886 694.968δ 0 774.904 + 5190.48δ 13 0 δ 1734.486δ 0.0154 0.0154 1734.486 0.000008879 Thus x 3 (x + δ ) = 1.0075 + 0.000008879 1.00749 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 1.000 and 1.01, correct to 4 significant figures 400 014, John Bird
EXERCISE 106 Page 33 1. 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() = 4 4 13 = 13 f(3) = 9 6 13 = 10 f(4) = 14 8 13 = 7 f(5) = 5 10 13 = 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) 13 0.31 r = 4.7 = 4.7 = 4.7419 (4.7) 7.4 f (4.7419) 0.0018156 r3 = 4.7419 = 4.7419 = 4.74166 f '(4.7419) 7.4838 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) = 1 + 13 = 10 f( ) = 4 + 4 13 = 5 f( 3) = 9 + 6 13 = 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) 13 0.31 r =.7 =.7 =.7401 (.7) 7.71 f (.7401) 0.01089 r3 =.7401 =.7401 =.73876 f '(.7401) 7.4804 f (.73876) 0.01674 r4 =.73876 =.73876 =.74166 f '(.73876) 7.4775 f (.74166) 0.000019556 r5 =.74166 =.74166 =.7417 f '(.74166) 7.4833 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) = 3 10 14 = 1 401 014, John Bird
f() = 4 0 14 = 10 f(3) = 81 30 14 = 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(.) 14 4.056 9(.) 10(.) 1.56 r =. =. =.3881 f (.3881).9771577 r3 =.3881 =.3881 =.796 f '(.3881) 7.44619 f (.796) 1.57655 r4 =.796 =.796 =.331 f '(.796) 3.973184 f (.331) 0.797097 r5 =.331 =.331 =.3036 f '(.331) 5.6714 0.3633356 r6 =.3036 =.3183 4.731566 0.196136 r 7 =.3183 =.3105 5.187634 0.10180935 r8 =.3105 =.3146 4.94069 0.0545775 r 9 =.3146 =.314 5.070358 0.0944745 r10 =.314 =.3136 5.0007438 0.01633 r 11 =.3136 =.319 5.0387046 r 1 0.01037987 =.319 =.313 5.0165577 Hence, correct to 4 significant figures, x =.313 3. 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) = 1 3 + 7 1 = 7 f() = 16 4 + 14 1 = 6 f(3) = 81 81 + 1 1 = 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) 40 014, John Bird
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) 1 0.8304 4(.6) 3 9(.6) + 7 16.464 r =.6 =.6 =.65044 f (.65044) 0.0446586 r3 =.65044 =.65044 =.64799 f '(.65044) 18.5095 f (.64799) 0.0000507015 r4 =.64799 =.64799 =.64799 f '(.64799) 18.16586 Hence, correct to 3 decimal places, x =.648 f( 1) = 1 + 3 7 1 = 15 f( ) = 16 + 4 14 1 = 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 7.315 4( 1.5) 3 9( 1.5) + 7 6.75 r = 1.5 = 1.5 = 1.77336 f ( 1.77336).0690068 r3 = 1.77336 = 1.77336 = 1.776 f '( 1.77336) 43.610741 f ( 1.776) 0.0880710 r4 = 1.776 = 1.776 = 1.7057 f '( 1.776) 40.163049 f ( 1.7057) 0.000736118 r5 = 1.7057 = 1.7057 = 1.7056 f '( 1.7057) 40.01784 Hence, correct to 3 decimal places, x = 1.71 i.e. the two roots of x 4 3x 3 + 7x = 1 are.648 and 1.71 4. 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) = 3 4 + 7 1 = 6 f() = 48 3 + 14 1 = 18 Hence, a root lies between x = 1 and x = There are no further positive roots since the x 4 term predominates. f( 1) = 3 + 4 7 1 = 1 f( ) = 48 + 3 14 1 = 54 403 014, John Bird
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 + 7 4 3 ( ) ( ) ( ) ( ) f( r1 ) 3 1.5 4 1.5 + 7(1.5) 1 3.7388 r = r1 = 1.5 = 1.5 = 1.56985 3 f '( r1 ) 1 1.5 1 1.5 + 7 11.6875 1.734046 r 3 = 1.56985 = 1.49715 3.85585 0.195743 r 4 = 1.49715 = 1.49081 0.370909 0.000994417 r 5 = 1.49081 = 1.49076 0.089988 Hence, correct to 3 decimal places, the positive root is: x = 1.491 For the root between x = 1 and x = : Let r 1 = 1. 4 3 ( ) ( ) ( ) ( ) f( r1 ) 3 1. 4 1. + 7( 1.) 1 7.67 r = r1 = 1. = 1. = 1.4343 3 f '( r1 ) 1 1. 1 1. + 7 31.016.4589815 r 3 = 1.4343 = 1.3880 53.094585 0.11488764 r 4 = 1.3880 = 1.3856 48.07045 0.00051387 r 5 = 1.3856 = 1.3856 47.960999 Hence, correct to 3 decimal places, the negative root is: x = 1.386 5. 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) = 0 + 4 5 = 1 f() = 3 ln + 8 5 = 5.0794 Hence, a root lies between x = 1 and x = Let r 1 = 1. 404 014, John Bird
(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.) 5 0.34696467 Hence, r = 1. = 1. = 1.1466 3 /1. + 4 6.5 f (1.1466) 0.003064547 r3 = 1.1466 = 1.1466 = 1.1471 f '(1.1466) 6.61638555 Hence, correct to 3 decimal places, x = 1.147 6. 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 = 3.0807 (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) 0.50683571 3(0.7) + 10sin(1.4) 11.34497 r = 0.7 = 0.7 = 0.7448 f (0.74476) 0.007159769 r3 = 0.74476 = 0.74476 = 0.74414 f '(0.74476) 11.6309913 f (0.74414) 0.00005079 r4 = 0.74414 = 0.74414 = 0.74414 f '(0.74414) 11.67076 Hence, correct to 3 decimal places, x = 0.744 f(0) = 0 5 cos 0 = 5 f( 1) = 1 5 cos( ) = 1.849 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) 0.36600388 r = 0.8 = 0.8 = 0.8453 3( 0.8) + 10sin( 1.6) 8.0757360 405 014, John Bird
f ( 0.8453) 0.00651834 r3 = 0.8453 = 0.8453 = 0.8461 f '( 0.8453) 7.78457567 f ( 0.8461) 0.000181379 r4 = 0.8461 = 0.8461 = 0.8461 f '( 0.8461) 7.77874058 Hence, correct to 3 decimal places, x = 0.846 From above, f(0) = 5 and f( 1) = 1.849 f( ) = 8 5 cos( 4) = 5.5779 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) 0.07900903 3( 1.7) + 10sin( 3.4) 11.54110 r = 1.7 = 1.7 = 1.6996 f ( 1.6996) 0.0007143059 r3 = 1.6996 = 1.6996 = 1.699 f '( 1.6996) 11.0173777 Hence, correct to 3 decimal places, x = 1.693 Thus, the solutions to x 3 = 5 cos x, correct to 3 decimal places, are 0.744, 0.846 and 1.693 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) = 300 6 = 94 f(1) = 35.1 f() = 0.495 f(3) = 3.756 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 + 0.5 f r r = r1 = = =.049 f ( 1) 300 e 4 + 1 6 0.49469 '( r1 ) 600 e 4 + 0.5 10.489383 0.00436387 r3 =.049 =.0497 9.4595774 406 014, John Bird
Hence, correct to 3 significant figures, x =.05 8. 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 + 4 + 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 0.880 When y = 0.88, then π 1 0.88 = t+ + sin t+ sin 3t 4 8 or π 1 t+ + sin t+ sin 3t 0.88 = 0 4 8 Let f(t) = π 1 t+ + sin t+ sin 3t 0.88 4 8 3 Let r 1 = 0.04 f '( t) = 1+ cost+ cos3t 8 π 1 0.04 + + sin 0.04 + sin 3(0.04) 0.88 f( r1 ) 4 8 0.0003515 r = r1 = 0.04 = 0.04 f '( r1 ) 3 1+ cos 0.04 + cos 3(0.04).371503345 8 = 0.03985 0.00000403 r3 = 0.03985 = 0.03985.37159496 Hence, correct to 4 decimal places, t = 0.0399 10. 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. 407 014, John Bird
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. = 4.189 f '( ) e. cos1.6 3.7 sin1.6 18.03833 ( 1) 7.4 e.1 sin1.6.03189 r1.1 + f (4.189) 0.03785 r3 = 4.189 = 4.189 = 4.189 f '(4.189) 180.3134965 (Note, sin 1.6 is sin 1.6 rad) Hence, correct to 3 significant figures, t = 4.19 11. The critical speeds of oscillation, λ, of a loaded beam are given by the equation: λ3 λ λ 3.50 + 0.063 = 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 3 3.50λ + λ 0.063 f λ λ λ '( ) = 3 6.5 + 1 Let r 1 = 3.0 3 ( ) ( ) ( ) ( ) f (3.0) 3.0 3.50 3.0 + 3.0 0.063 0.687 r = 3.0 = 3.0 = 3.0 =.91918 f '(3.0) 3 3.0 6.5 3.0 + 1 8.5 0.0370604 r 3 =.91918 =.91430 7.5901656 0.00015139 r 4 =.91430 =.9148 7.5364835 Hence, correct to 4 decimal places, λ =.9143 408 014, John Bird