CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODSdocuments.routledge-interactive.s3.amazonaws.com/9780415662840/... · chapter 25 solving equations by iterative methods . ... = 4

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CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS

EXERCISE 104 Page 228

1. Find the positive root of the equation x 2 + 3x – 5 = 0, correct to 3 significant figures, using the method of bisection. Let f(x) = 2 3 5x x+ − then, using functional notation:

f(0) = –5

f(1) = 1 + 3 – 5 = –1

f(2) = 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 = 2

The method of bisection suggests that the root is at 1 22+ = 1.5, i.e. the interval between 1 and 2 has been

bisected

Hence f(1.5) = (1.5) 2 + 3(1.5) – 5 = 1.75

Since f(1) is negative, f(1.5) is positive, and f(2) 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)

Bisecting this interval gives 1 1.52+ i.e. 1.25 as the next root

Hence f(1.25) = (1.25) 2 + 3(1.25) – 5 = 0.3125

Since f(1) is negative and f(1.25) is positive, a root lies between x = 1 and x = 1.25

Bisecting this interval gives 1 1.252

+ i.e. 1.125

Hence f(1.125) = (1.125) 2 + 3(1.125) – 5 = –0.359375

Since f(1.125) is negative and f(1.25) is positive, a root lies between x = 1.125 and x = 1.25

Bisecting this interval gives 1.125 1.252+ i.e. 1.1875

Hence f(1.1875) = (1.1875) 2 + 3(1.1875) – 5 = –0.02734375

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Since f(1.1875) is negative and f(1.25) is positive, a root lies between x = 1.1875 and x = 1.25

Bisecting this interval gives 1.1875 1.252+ i.e. 1.21875

Hence f(1.21875) = (1.21875) 2 + 3(1.21875) – 5 = 0.1416016

Since f(1.21875) is positive and f(1.1875) is negative, a root lies between x = 1.1875 and x = 1.21875

Bisecting this interval gives 1.1875 1.218752+ = 1.203125

Hence f(1.203125) = 0.056885

Since f(1.203125) is positive and f(1.1875) is negative, a root lies between x = 1.1875 and x = 1.203125


Hence f(1.1953125) = 0.0147095

Since f(1.1953125) is positive and f(1.1875) is negative, a root lies between x = 1.1953125 and

x = 1.1875


Hence, f(1.191406) = –0.006334

Since f(1.191406) is negative and f(1.1953125) is positive, a root lies between x = 1.191406 and

x = 1.1953125


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 2 + 3x – 5 = 0 is 1.19

2. Using the bisection method solve e x – x = 2, correct to 4 significant figures. Let f(x) = e 2x x− − then f(0) = 1 – 0 – 2 = –1

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f(1) = 1e 1 2 0.28− − = − f(2) = 2e 2 2 3.38− − = Hence, a root lies between x = 1 and x = 2 Let the root be x = 1.5 f(1.5) = 1.5e 1.5 2 0.98169− − = Hence, a root lies between x = 1 and x = 1.5 due to a sign change

Bisecting this interval gives 1 1.52+ = 1.25

f(1.25) = 1.25e 1.25 2 0.240343− − = Hence, a root lies between x = 1 and x = 1.25 due to a sign change


+ = 1.125

f(1.125) = 1.125e 1.125 2 0.04478− − = − Hence, a root lies between x = 1.125 and x = 1.25 due to a sign change


f(1.1875) = 0.091374 Hence, a root lies between x = 1.1875 and x = 1.125 due to a sign change




f(1.140625) = –0.011902 Hence, a root lies between x = 1.140625 and x = 1.15625 due to a sign change


f(1.1484375) = 0.004824586

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Hence, a root lies between x = 1.1484375 and x = 1.140625 due to a sign change





Bisecting this interval gives 1.14648 1.144531252

+ = 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 e 2x x− = is 1.146

3. Determine the positive root of x 2 = 4 cos x, correct to 2 decimal places using the method of bisection. Let f(x) = x 2 – 4 cos x then f(0) = 0 – 4 = –4 f(1) = 1 – 2.161 = –1.161 f(2) = 4 – 1.6646 = 5.6646 Hence, a root lies between x = 1 and x = 2 Let the root be x = 1.5 f(1.5) = 1.5 2 4cos1.5 1.97− = Hence, a root lies between x = 1 and x = 1.5 due to a sign change

Bisecting this interval gives 1 1.52+ = 1.25

f(1.25) = 1.25 2 4cos1.25 0.301− =

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Hence, a root lies between x = 1 and x = 1.25 due to a sign change


+ = 1.125

f(1.125) = 1.125 2 4cos1.125 0.459− = − Hence, a root lies between x = 1.125 and x = 1.25 due to a sign change












The last two values are both 1.20, correct to 4 decimal places. We therefore stop the iterations here

Thus, correct to 2 decimal places, the root of x 2 = 4 cos x is 1.20

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4. Solve x – 2 – ln x = 0, correct to 3 decimal places using the bisection method. Let f(x) = x – 2 – ln x f(2) = 2 – 2 – ln 2 = –0.693 f(3) = 3 – 2 – ln 3 = –0.0986 f(4) = 4 – 2 – 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

1x 2x 1 23

2x xx + =

( )3f x

3 3 3 3.125 3.125 3.15625 3.140625 3.14844 3.1445325 3.146486

4 3.5 3.25 3.25 3.1875 3.125 3.15625 3.140625 3.14844 3.1445325

3.5 3.25 3.125 3.1875 3.15625 3.140625 3.14844 3.145325 3.146486 3.14550925

0.24724 0.071345 –0.014434 0.028263 0.0068654 –0.003797 0.0015312 –0.0011327 0.0001999

Correct to 3 decimal places, the solution of x – 2 – ln x = 0 is 3.146 5. Solve, correct to 4 significant figures, x – 2 sin 2 x = 0 using the bisection method. Let f(x) = x – 2 sin 2 x then f(0) = 0 – 0 = 0 f(1) = 1 – 2 (sin 1) 2 = –0.416 f(2) = 2 – 2 (sin 4) 2 = 0.346 Hence, a root lies between x = 1 and x = 2 Let the root be x = 1.5 f(1.5) = 1.5 – 2 (sin 1.5) 2 = –0.49 Hence, a root lies between x = 1.5 and x = 2 due to a sign change

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Bisecting this interval gives 1.5 22+ = 1.75

f(1.75) = 1.75 – 2 (sin 1.75) 2 = –0.186 Hence, a root lies between x = 1.75 and x = 2 due to a sign change

Bisecting this interval gives 1.75 22+ = 1.875

f(1.875) = 1.875 – 2 (sin 1.875) 2 = 0.054 Hence, a root lies between x = 1.75 and x = 1.875 due to a sign change











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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 – 2 sin 2 x = 0 is 1.849

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1. Use an algebraic method of successive approximation to solve: 3x2 + 5x – 17 = 0, correct to 3

significant figures.

Let f(x) = 3x2 + 5x – 17

f(0) = –17

f(1) = 1 + 5 – 17 = –11

f(2) = 12 + 10 – 17 = 5 Hence a root lies between x = 1 and x = 2 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, x2 , be (x1 + δ1) Let f(x1 + δ1) = 0, then since x1 = 1.7, 3(1.7 + δ1)2 + 5(1.7 + δ1) – 17 = 0 Neglecting terms containing products of δ1 and using the binomial series gives: 3[(1.7)2 + 2(1.7) δ1] + 8.5 + 5δ1 – 17 ≈ 0 8.67 + 10.2δ1 + 8.5 + 5δ1 – 17 ≈ 0 15.2δ1 ≈ 17– 8.67 – 8.5

δ1 ≈ 17 8.67 8.5 0.0111815.2

− −≈ −

Thus x2 ≈ 1.7 – 0.01118 = 1.6888 Third approximation Let the true value of the root, x3 , be (x2 + δ2) Let f(x2 + δ2) = 0, then since x2 = 1.6888, 3(1.6888 + δ2)2 + 5(1.6888 + δ2) – 17 = 0 Neglecting terms containing products of δ2 and using the binomial series gives:

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3[(1.6888)2 + 2(1.6888) δ2] + 8.444 + 5δ2 – 17 ≈ 0 8.55614 + 10.1328δ2 + 8.444 + 5δ2 – 17 ≈ 0 15.1328δ2 ≈ 17– 8.55614 – 8.444

δ2 ≈ 17 8.55614 8.444 0.0000092515.1328

− −≈ −

Thus x3 ≈ (x2 + δ2) = 1.6888 – 0.00000925 ≈ 1.6887 Since x2 and x3 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(– 2) = 12 – 10 – 17 = –15

f(– 3) = 27 – 15 – 17 = –5

f(– 4) = 48 – 20 – 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, x2 , be (x1 + δ1) Let f(x1 + δ1) = 0, then since x1 = – 3.4, 3(– 3.4 + δ1)2 + 5(– 3.4 + δ1) – 17 = 0 Neglecting terms containing products of δ1 and using the binomial series gives: 3[(– 3.4)2 + 2(– 3.4) δ1] – 17 + 5δ1 – 17 ≈ 0 34.68 – 20.4δ1 – 17 + 5δ1 – 17 ≈ 0 –15.4δ1 ≈ 17– 34.68 + 17

δ1 ≈ 17 34.68 17 0.04415615.4

− −=

−

Thus x2 ≈ –3.4 + 0.044156 = –3.3558 Third approximation

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Let the true value of the root, x3 , be (x2 + δ2) Let f(x2 + δ2) = 0, then since x2 = – 3.3558, 3(–3.3558 + δ2)2 + 5(–3.3558 + δ2) – 17 = 0 Neglecting terms containing products of δ2 and using the binomial series gives: 3[(–3.3558)2 + 2(–3.3558) δ2] – 16.779 + 5δ2 – 17 ≈ 0 33.78418 – 20.1348δ2 – 16.779 + 5δ2 – 17 ≈ 0 – 15.1348δ2 ≈ 16.779 – 33.78418 + 17

δ2 ≈ 16.779 33.78418 17 0.00034215.1348− +

≈−

Thus x3 ≈ (x2 + δ2) = –3.3558 + 0.000342 ≈ –3.3551 Since x2 and x3 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 3x2 + 5x – 17 = 0 are 1.69 and –3.36, correct to 3 significant

figures

2. Use an algebraic method of successive approximation to solve: x3 – 2x + 14 = 0, correct to 3 decimal

places.

Let f(x) = 3 2 14x x− +

f(0) = 14

f(1) = 1 – 2 + 14 = 13

f(2) = 8 – 4 + 14 = 18 (There are no positive values of x)

f(–1) = –1 + 2 + 14 = 15

f(–2) = –8 + 4 + 14 = 10

f(–3) = –27 + 6 + 14 = –7 Hence a root lies between x = –2 and x = –3 First approximation Let the first approximation be –2.6 Second approximation

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Let the true value of the root, x2 , be (x1 + δ1) Let f(x1 + δ1) = 0, then since x1 = –2.6, (–2.6 + δ1)3 – 2(–2.6 + δ1) + 14 = 0 Neglecting terms containing products of δ1 and using the binomial series gives: [(–2.6)3 + 3(–2.6)2 δ1] + 5.2 – 2δ1 + 14 ≈ 0 –17.576 + 20.28δ1 + 5.2 – 2δ1 + 14 ≈ 0 18.28δ1 ≈ 17.576 – 5.2 – 14

δ1 ≈ 17.576 5.2 14 0.0888418.28− −

≈ −

Thus x2 ≈ –2.6 – 0.08884 = –2.6888 Third approximation Let the true value of the root, x3 , be (x2 + δ2) Let f(x2 + δ2) = 0, then since x2 = –2.6888, (–2.6888 + δ2)3 – 2(–2.6888 + δ2) + 14 = 0 Neglecting terms containing products of δ2 gives: ( ) ( )3 2

2 22.6888 3 2.6888 5.3776 2 14 0δ δ− + − + − + ≈ –19.439 + 21.6888δ2 + 5.3776 – 2δ2 + 14 ≈ 0 19.6888δ2 ≈ 19.439 – 5.3776 – 14

δ2 ≈ 19.439 5.3776 1419.6888− − ≈ 0.003119

Thus x3 ≈ (x2 + δ2) = –2.6888 + 0.003119 ≈ –2.6857 Fourth approximation Let the true value of the root, x4 , be (x3 + δ3) Let f(x3 + δ3) = 0, then since x3 = –2.6857, (–2.6857 + δ3)3 – 2(–2.6857 + δ3) + 14 = 0 Neglecting terms containing products of δ3 gives: ( ) ( )3 2

3 32.6857 3 2.6857 5.3714 2 14 0δ δ− + − + − + ≈

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–19.3719 + 21.63895δ3 + 5.3714 – 2δ3 + 14 ≈ 0 19.63895δ2 ≈ 19.3719 – 5.3714 – 14

δ2 ≈ 19.3719 5.3714 1419.63895− − ≈ 0.00002546

Thus x4 ≈ (x3 + δ3) = –2.6857 + 0.000025 ≈ –2.6857 Since x3 and x4 are the same when expressed to the required degree of accuracy, then the required root is –2.686, correct to 3 decimal places 3. Use an algebraic method of successive approximation to solve: x4 – 3x3 + 7x – 5.5 = 0, correct to 3


Let f(x) = 4 33 7 5.5x x x− + −

f(0) = –5.5

f(1) = 1 – 3 + 7 – 5.5 = –0.5

f(2) = 16 – 24 + 14 – 5.5 = 0.5

Hence a root lies between x = 1 and x = 2 First approximation Let the first approximation be 1.5 Second approximation Let the true value of the root, x2 , be (x1 + δ1)

Let f(x1 + δ1) = 0, then since x1 = 1.5,

(1.5 + δ1) 4 – 3(1.5 + δ1) 3 + 7(1.5 + δ1) – 5.5 = 0

Neglecting terms containing products of δ1 and using the binomial series gives:

[(1.5) 4 + 4(1.5) 3δ1] – 3[(1.5) 3 + 3(1.5) 2 δ1] + 10.5 + 7δ1 – 5.5 ≈ 0

5.0625 + 13.5δ1 – 10.125 – 20.25δ1 + 10.5 + 7δ1 – 5.5 ≈ 0

0.25δ1 ≈ 0.0625

δ1 ≈ 0.0625 0.250.25

≈

Thus x2 ≈ 1.5 + 0.25 = 1.75 Third approximation

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Let the true value of the root, x3 , be (x2 + δ2) Let f(x2 + δ2) = 0, then since x2 = 1.75,

(1.75 + δ2) 4 – 3(1.75 + δ2)3 + 7(1.75 + δ2) – 5.5 = 0

Neglecting terms containing products of δ2 gives:

( ) ( ) ( ) ( )4 3 3 22 2 21.75 4 1.75 3[ 1.75 3 1.75 ] 7(1.75 ) 5.5 0δ δ δ+ − + + + − ≈

9.3789 + 21.4375δ2 – 16.078125 – 27.5625δ2 + 12.25 + 7δ2 – 5.5 ≈ 0

0.875δ2 ≈ –0.050775

δ2 ≈ 0.050775 0.058030.875

−≈ −

Thus x3 ≈ (x2 + δ2) = 1.75 – 0.05803 ≈ 1.692 Fourth approximation Let the true value of the root, x4 , be (x3 + δ3) Let f(x3 + δ3) = 0, then since x3 = 1.692, (1.692 + δ3) 4 – 3(1.692 + δ3)3 + 7(1.692 + δ3) – 5.5 = 0 Neglecting terms containing products of δ3 gives: ( ) ( ) ( ) ( )4 3 3 2

3 3 31.692 4 1.692 3[ 1.692 3 1.692 ] 7(1.692 ) 5.5 0δ δ δ+ − + + + − ≈ 8.19599 + 19.37586δ3 – 14.5318977 – 25.76578δ3 + 11.844 + 7δ3 – 5.5 ≈ 0 0.61008δ3 ≈ –0.0080923

δ3 ≈ 0.0080923 0.01326430.61008

−≈ −

Thus x4 ≈ (x3 + δ3) = 1.692 – 0.0132643 ≈ 1.6787 Fifth approximation Let the true value of the root, x5 , be (x4 + δ4) Let f(x4 + δ4) = 0, then since x4 = 1.6787, (1.6787 + δ4) 4 – 3(1.6787 + δ4)3 + 7(1.6787 + δ4) – 5.5 = 0 Neglecting terms containing products of δ4 gives: ( ) ( ) ( ) ( )4 3 3 2

4 4 41.6787 4 1.6787 3[ 1.6787 3 1.6787 ] 7(1.6787 ) 5.5 0δ δ δ+ − + + + − ≈ 7.941314 + 18.9225δ4 – 14.1919 – 25.3623δ4 + 11.7509 + 7δ4 – 5.5 ≈ 0 0.5602δ4 ≈ –0.000314

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δ4 ≈ 0.000314 0.000560.5602

−≈ −

Thus x5 ≈ (x4 + δ4) = 1.6787 – 0.00056 ≈ 1.6781 Since x4 and x5 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) = 4 33 7 5.5x x x− + − f(0) = –5.5 f(–1) = 1 + 3 – 7 – 5.5 = –8.5

f(–2) = 16 + 24 – 14 – 5.5 = 20.5 Hence, a root lies between x = –1 and x = –2

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: x4 + 12x3 – 13 = 0, correct to 4


Let f(x) = x4 + 12x3 – 13

f(0) = –13

f(1) = 1 + 12 – 13 = 0 Hence, x = 1.000 is a root of the equation

f(2) = 16 + 96 – 13 = 99 There are no further positive roots

f(–1) = 1 – 12 – 13 = –24

f(–2) = –93

f(–3) = –256

f(–4) = –525

f(–5) = –888

f(–6) = –1303

f(–7) = –1728

f(–8) = –2061

f(–9) = –2200

f(–10) = –2013

f(–11) = –1344

f(–12) = –13

f(–13) = 2184 (Since the sign of f(–14) onwards does not change, there are no further

negative roots)

Hence a root lies between x = –12 and x = –13 First approximation

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Let the first approximation be –12.0 Second approximation Let the true value of the root, x2 , be (x1 + δ1)

Let f(x1 + δ1) = 0, then since x1 = – 12.0,

(–12.0 + δ1) 4 + 12(–12.0 + δ1) 3 – 13 = 0

Neglecting terms containing products of δ1 and using the binomial series gives:

[(–12.0) 4 + 4(–12.0) 3δ1] + 12[(–12.0) 3 + 3(–12.0) 2 δ1] – 13 ≈ 0

20 736 – 6912δ1 – 20 736 + 5184δ1 – 13 ≈ 0

– 1728δ1 ≈ 13

δ1 ≈ 13 0.007521728

≈ −−

Thus x2 ≈ – 12.0 – 0.00752 = –12.0075 Third approximation Let the true value of the root, x3 , be (x2 + δ2) Let f(x2 + δ2) = 0, then since x2 = – 12.0075,

(–12.0075 + δ2) 4 + 12(–12.0075 + δ2)3 – 13 = 0

Neglecting terms containing products of δ2 gives:

( ) ( ) ( ) ( )4 3 3 22 212.0075 4 12.0075 12[ 12.0075 3 12.0075 ] 13 0δ δ− + − + − + − − ≈

20 787.8886 – 6924.968δ2 – 20 774.904 + 5190.48δ2 – 13 ≈ 0

– 1734.486δ2 ≈ 0.0154

δ2 ≈ 0.0154 0.0000088791734.486

≈−

Thus x3 ≈ (x2 + δ2) = – 12.0075 + 0.000008879 ≈ – 12.00749 Since x2 and x3 are the same when expressed to the required degree of accuracy, then the required root is –12.01, correct to 4 significant figures Thus, the two solutions of the equation x4 + 12x3 – 13 = 0 are 1.000 and –12.01, correct to 4 significant

figures

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1. Use Newton’s method to solve: x2 – 2x – 13 = 0, correct to 3 decimal places. Let f(x) = x2 – 2x – 13 f(0) = –13 f(1) = 1 – 2 – 13 = –14 f(2) = 4 – 4 – 13 = –13 f(3) = 9 – 6 – 13 = –10 f(4) = 14 – 8 – 13 = –7 f(5) = 25 – 10 – 13 = 2 Hence a root lies between x = 4 and x = 5. Let 1r = 4.7

A better approximation is given by: ( )12 1

1' ( )f r

r rf r

= − ' ( )f x = 2x – 2

Hence, 2

2(4.7) 2(4.7) 13 0.314.7 4.7 4.7419

2(4.7) 2 7.4r − − −= − = − =

−

3(4.7419) 0.00181564.7419 4.7419 4.74166'(4.7419) 7.4838

frf

= − = − =

Hence, correct to 3 decimal places, x = 4.742 Since f(6), f(7), and so on do not change sign, there are no further positive roots f(–1) = 1 + 2 – 13 = –10 f(–2) = 4 + 4 – 13 = –5 f(–3) = 9 + 6 – 13 = 2 Hence a root lies between x = –2 and x = –3. Let 1r = –2.7


1' ( )f r

r rf r

= − ' ( )f x = 2x – 2

Hence, 2

2( 2.7) 2( 2.7) 13 0.312.7 2.7 2.74021

2( 2.7) 2 7.71r − − − − −= − − = − − = −

− − −

3( 2.74021) 0.0108292.74021 2.74021 2.73876'( 2.74021) 7.48042

frf−

= − − = − − = −− −

4( 2.73876) 0.0216742.73876 2.73876 2.74166'( 2.73876) 7.47752

frf− −

= − − = − − = −− −

5( 2.74166) 0.0000195562.74166 2.74166 2.7417'( 2.74166) 7.48332

frf−

= − − = − − = −− −

Hence, correct to 3 decimal places, x = –2.742 i.e. the two roots of x2 – 2x – 13 = 0 are 4.742 and –2.742 2. Use Newton’s method to solve: 3x3 – 10x = 14, correct to 4 significant figures. Let f(x) = 33 10 14x x− − f(0) = –14 f(1) = 3 – 10 – 14 = –21

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f(2) = 24 – 20 – 14 = –10 f(3) = 81 – 30 – 14 = 36 Hence, a root lies between x = 2 and x = 3 Let 1 2.2r =


1' ( )f r

r rf r

= − 2' ( ) 9 10f x x x= −

Hence, 3

22

3(2.2) 10(2.2) 14 4.0562.2 2.2 2.38819(2.2) 10(2.2) 21.56

r − − −= − = − =

−

3(2.3881) 2.97715772.3881 2.3881 2.2796'(2.3881) 27.44619

frf

= − = − =

4(2.2796) 1.2576552.2796 2.2796 2.3321'(2.2796) 23.973184

frf

−= − = − =

5(2.3321) 0.72970972.3321 2.3321 2.3036'(2.3321) 25.627214

frf

= − = − =

60.36333562.3036 2.3183

24.7231566r −= − =

70.19621362.3183 2.310525.187634

r = − =

80.101809352.3105 2.314624.94069

r −= − =

90.054527752.3146 2.312425.070358

r = − =

100.029447452.3124 2.3136

25.0007438r −

= − =

110.01633222.3136 2.312925.0387046

r = − =

120.010379872.3129 2.313

25.0165577r −

= − =

Hence, correct to 4 significant figures, x = 2.313 3. Use Newton’s method to solve: x4 – 3x3 + 7x = 12, correct to 3 decimal places . Let f(x) = x4 – 3x3 + 7x – 12 f(0) = –12 f(1) = 1 – 3 + 7 – 12 = –7 f(2) = 16 – 24 + 14 – 12 = –6 f(3) = 81 – 81 + 21 – 12 = 9 Hence, a root lies between x = 2 and x = 3 Let 1r = 2.6 (Since f(4), f(5), and so on do not change sign, there are no further positive roots)

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1' ( )f r

r rf r

= − 3 2' ( ) 4 9 7f x x x= − +

Hence, 4 3

23 2

(2.6) 3(2.6) 7(2.6) 12 0.83042.6 2.6 2.650444(2.6) 9(2.6) 7 16.464

r − + − −= − = − =

− +

3(2.65044) 0.04465862.65044 2.65044 2.64799'(2.65044) 18.252095

frf

= − = − =

4(2.64799) 0.00005070152.64799 2.64799 2.64799'(2.64799) 18.162586

frf

= − = − =

Hence, correct to 3 decimal places, x = 2.648 f(– 1) = 1 + 3 – 7 – 12 = – 15 f(– 2) = 16 + 24 – 14 – 12 = 14 Hence a root lies between x = –1 and x = –2. Let 1r = –1.5 (Since f(– 3), f(– 4), and so on do not change sign, there are no further negative roots).


1' ( )f r

r rf r

= − 3 2' ( ) 4 9 7f x x x= − +

Hence, 4 3

23 2

( 1.5) 3( 1.5) 7( 1.5) 12 7.31251.5 1.5 1.773364( 1.5) 9( 1.5) 7 26.75

r − − − + − − −= − − = − − = −

− − − + −

3( 1.77336) 2.206900681.77336 1.77336 1.72276'( 1.77336) 43.610741

frf−

= − − = − − = −− −

4( 1.72276) 0.08807101.72276 1.72276 1.72057'( 1.72276) 40.163049

frf−

= − − = − − = −− −

5( 1.72057) 0.00027361181.72057 1.72057 1.72056'( 1.72057) 40.017284

frf−

= − − = − − = −− −

Hence, correct to 3 decimal places, x = –1.721 i.e. the two roots of x4 – 3x3 + 7x = 12 are 2.648 and –1.721 4. Use Newton’s method to solve: 3x4 – 4x3 + 7x – 12 = 0, correct to 3 decimal places. Let f(x) = 4 33 4 7 12x x x− + − f(0) = –12 f(1) = 3 – 4 + 7 – 12 = –6 f(2) = 48 – 32 + 14 – 12 = 18 Hence, a root lies between x = 1 and x = 2 There are no further positive roots since the 4x term predominates. f(–1) = 3 + 4 – 7 – 12 = –12 f(–2) = 48 + 32 – 14 – 12 = 54

© 2014, John Bird

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Hence, a root lies between x = –1 and x = –2 There are no further negative roots since, once again, the 4x term predominates. For the root between x = 1 and x = 2: Let 1 1.25r = 3 2' ( ) 12 12 7f x x x= − +

( ) ( )( ) ( )

4 31

2 1 3 21

3 1.25 4 1.25 7(1.25) 12( ) 3.738281.25 1.25 1.56985'( ) 11.687512 1.25 12 1.25 7

f rr rf r

− + − −= − = − = − =

− +

31.7340461.56985 1.49715

23.8522585r = − =

40.129257431.49715 1.4908120.3720909

r = − =

50.0009944171.49081 1.49076

20.089988r = − =

Hence, correct to 3 decimal places, the positive root is: x = 1.491 For the root between x = –1 and x = –2: Let 1 1.2r = −

( ) ( )( ) ( )

4 31

2 1 3 21

3 1.2 4 1.2 7( 1.2) 12( ) 7.26721.2 1.2 1.4343'( ) 31.01612 1.2 12 1.2 7

f rr rf r

− − − + − − −= − = − − = − − = −

−− − − +

32.45898151.4343 1.388053.094585

r = − − = −−

40.114887641.3880 1.3856

48.207045r = − − = −

−

50.000513871.3856 1.3856

47.960999r = − − = −

−

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(2) = 3 ln 2 + 8 – 5 = 5.0794 Hence, a root lies between x = 1 and x = 2 Let 1r = 1.2

© 2014, John Bird

405

(Since f(3), f(4), and so on do not change sign, there are no further positive roots)


1' ( )f r

r rf r

= − 3'( ) 4f xx

= +

Hence, 23ln (1.2) 4(1.2) 5 0.346964671.2 1.2 1.14662

3 /1.2 4 6.5r + −= − = − =

+

3(1.14662) 0.0030645471.14662 1.14662 1.1471'(1.14662) 6.61638555

frf

−= − = − =

Hence, correct to 3 decimal places, x = 1.147 6. Use Newton’s method to solve: x3 = 5 cos 2x, correct to 3 decimal places. Let f(x) = x3 – 5 cos 2x f(0) = 0 – 5 cos 0 = –5 f(1) = 1 – 5 cos 2 = 3.0807 (note: cos 2 means cos(2 rad)) Hence, a root lies between x = 0 and x = 1 Let 1r = 0.7 (Since f(2), f(3), and so on do not change sign, there are no further positive roots)


1' ( )f r

r rf r

= − 2' ( ) 3 10sin 2f x x x= +

Hence, 3

22

(0.7) 5cos(1.4) 0.506835710.7 0.7 0.74483(0.7) 10sin(1.4) 11.324497

r − −= − = − =

+

3(0.74476) 0.0071597690.74476 0.74476 0.74414'(0.74476) 11.6309913

frf

= − = − =

4(0.74414) 0.00005027290.74414 0.74414 0.74414'(0.74414) 11.6272076

frf

−= − = − =

Hence, correct to 3 decimal places, x = 0.744 f(0) = 0 – 5 cos 0 = –5 f(– 1) = – 1 – 5 cos(– 2) = 1.8249 Hence, a root lies between x = 0 and x = –1 Let 1r = –0.8


1' ( )f r

r rf r

= − 2' ( ) 3 10sin 2f x x x= +

Hence, 3

22

( 0.8) 5cos( 1.6) 0.3660023880.8 0.8 0.845323( 0.8) 10sin( 1.6) 8.0757360

r − − − −= − − = − − = −

− + − −

© 2014, John Bird

406

3( 0.84532) 0.00625183240.84532 0.84532 0.8461'( 0.84532) 7.78457567

frf− −

= − − = − − = −− −

4( 0.8461) 0.000182137290.8461 0.8461 0.8461'( 0.8461) 7.77874058

frf− −

= − − = − − = −− −

Hence, correct to 3 decimal places, x = –0.846 From above, f(0) = –5 and f(–1) = 1.8249 f(–2) = –8 – 5 cos(–4) = –5.5779 Hence, a root lies between x = –1 and x = –2 Let 1r = –1.7 (Since f(– 3), f(– 4), and so on do not change sign, there are no further roots)


1' ( )f r

r rf r

= − 2' ( ) 3 10sin 2f x x x= +

Hence, 3

22

( 1.7) 5cos( 3.4) 0.079009031.7 1.7 1.692963( 1.7) 10sin( 3.4) 11.2254110

r − − − −= − − = − − = −

− + −

3( 1.69296) 0.00071430591.69296 1.69296 1.6929'( 1.69296) 11.01737772

frf− −

= − − = − − = −−

Hence, correct to 3 decimal places, x = –1.693 Thus, the solutions to x3 = 5 cos 2x, correct to 3 decimal places, are 0.744, –0.846 and –1.693

7. Use Newton’s method to solve: 300e 2θ− + 2θ = 6, correct to 3 significant figures.

Let f(θ) = 2300e 62

θθ

− + − f(0) = 300 – 6 = 294 f(1) = 35.1 f(2) = 0.495

f(3) = –3.756 Hence, a root lies between x = 2 and x = 3, very close to x = 2 There are no further positive roots since the 2300e θ− term predominates. There are no negative roots since f(x) will always be positive Let 1 2r = 2' ( ) 600e 0.5f x θ−= − +

41

2 141

( ) 300e 1 6 0.4946922 2 2.0492'( ) 600e 0.5 10.489383

f rr rf r

−

−

+ −= − = − = − =

− + −

30.004363872.0492 2.04979.45952774

r −= − =

−

© 2014, John Bird

407

Hence, correct to 3 significant figures, x = 2.05 8. Solve the equations in Problems 1 to 5, Exercise 104, page 228 and Problems 1 to 4, Exercise 105, page 231 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 = 4

t π +

+ sin t + 18

sin 3t

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 10.88 sin sin 34 8

t t tπ = + + +

or 1sin sin 3 0.88 04 8

t t tπ + + + − =

Let f(t) = 1sin sin 3 0.884 8

t t tπ + + + −

Let 1r = 0.04 3'( ) 1 cos cos38

f t t t= + +

12 1

1

10.04 sin 0.04 sin 3(0.04) 0.88( ) 0.00035154 80.04 0.043'( ) 2.3715033451 cos 0.04 cos3(0.04)8

f rr rf r

π+ + + −

= − = − = −+ +

= 0.03985

30.0000042030.03985 0.03985

2.371529496r −= − =

Hence, correct to 4 decimal places, t = 0.0399 10. A damped oscillation of a system is given by the equation: y = –7.4e0.5t sin 3t. Determine the value of t near to 4.2, correct to 3 significant figures, when the magnitude y of the oscillation is zero.

© 2014, John Bird

408

Let f(t) = 0.57.4e sin 3t t− ( )( ) ( )( )0.5 0.5 0.5 0.5' ( ) 7.4e 3cos3 sin 3 3.7e 22.2e cos3 3.7e sin 3t t t tf t t t t t= − + − = − − = ( )0.5e 22.2cos3 3.7sin 3t t t− + Let 1 4.2r =

( )

2.112 1

2.11

( ) 7.4e sin12.6 2.031829224.2 4.2 4.189'( ) e 22.2cos12.6 3.7sin12.6 182.2023833

f rr rf r

− −= − = − = − =

− + −

(Note, sin 12.6 is sin 12.6 rad)

3(4.189) 0.0378254.189 4.189 4.189'(4.189) 180.3134965

frf

−= − = − =

−

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:

3 23.250 0.063 0λ λ λ− + − =

Determine the value of λ which is approximately equal to 3.0 by Newton’s method, correct to

4 decimal places.

Let f(λ) = 3 23.250 0.063λ λ λ− + − 2' ( ) 3 6.5 1f λ λ λ= − + Let 1 3.0r =

( ) ( )( ) ( )

3 2

2 2

3.0 3.250 3.0 3.0 0.063(3.0) 0.6873.0 3.0 3.0 2.91918'(3.0) 8.53 3.0 6.5 3.0 1

frf

− + −= − = − = − =

− +

30.03706042.91918 2.914307.5901656

r = − =

40.000151392.91430 2.914287.5364835

r = − =

Hence, correct to 4 decimal places, λ = 2.9143

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