Bird - Higher Engineering Mathematics - 5e - Solutions Manual

HIGHER ENGINEERING MATHEMATICS TH

5

EDITION

JOHN BIRD SAMPLE OF WORKED SOLUTIONS TO EXERCISES

© 2006 John Bird. All rights reserved. Published by Elsevier.

INTRODUCTION

In ‘Higher Engineering Mathematics 5th Edition’ are some 1750 further problems arranged regularly throughout the text within 250 Exercises. A sample of solutions for over 1000 of these further problems has been prepared in this document. The reader should be able to cope with the remainder by referring to similar worked problems contained in the text.

CONTENTS Chapter 1 Algebra

Page 1

Chapter 2 Inequalities

13

Chapter 3 Partial fractions

19

Chapter 4 Logarithms and exponential functions

25

Chapter 5 Hyperbolic functions

41

Chapter 6 Arithmetic and geometric progressions

48

Chapter 7 The binomial series

55

Chapter 8 Maclaurin’s series

65

Chapter 9 Solving equations by iterative methods

71

Chapter 10 Computer numbering systems

85

Chapter 11 Boolean algebra and logic circuits

94

Chapter 12 Introduction to trigonometry

110

Chapter 13 Cartesian and polar co-ordinates

131

Chapter 14 The circle and its properties

135

Chapter 15 Trigonometric waveforms

144

Chapter 16 Trigonometric identities and equations

155

Chapter 17 The relationship between trigonometric and hyperbolic functions

163

Chapter 18 Compound angles

168

Chapter 19 Functions and their curves

181

Chapter 20 Irregular areas, volumes and mean values of waveforms

197

© 2006 John Bird. All rights reserved. Published by Elsevier.

ii

Chapter 21 Vectors, phasors and the combination of waveforms

202

Chapter 22 Scalar and vector products

212

Chapter 23 Complex numbers

219

Chapter 24 De Moivre’s theorem

232

Chapter 25 The theory of matrices and determinants

238

Chapter 26 The solution of simultaneous equations by matrices and determinants

246

Chapter 27 Methods of differentiation

257

Chapter 28 Some applications of differentiation

266

Chapter 29 Differentiation of parametric equations

281

Chapter 30 Differentiation of implicit functions

287

Chapter 31 Logarithmic differentiation

291

Chapter 32 Differentiation of hyperbolic functions

295

Chapter 33 Differentiation of inverse trigonometric and hyperbolic functions

297

Chapter 34 Partial differentiation

306

Chapter 35 Total differential, rates of change and small changes

312

Chapter 36 Maxima, minima and saddle points for functions of two variables

319

Chapter 37 Standard integration

327

Chapter 38 Some applications of integration

332

Chapter 39 Integration using algebraic substitutions

350

Chapter 40 Integration using trigonometric and hyperbolic substitutions

356

Chapter 41 Integration using partial fractions

365

Chapter 42 The t = tan θ/2 substitution

372

Chapter 43 Integration by parts

376

Chapter 44 Reduction formulae

384

Chapter 45 Numerical integration

390

Chapter 46 Solution of first order differential equations by separation of variables 398 Chapter 47 Homogeneous first order differential equations

410

Chapter 48 Linear first order differential equations

417

Chapter 49 Numerical methods for first order differential equations

424

Chapter 50 Second order differential equations of the form a

d2 y dy +b + cy = 0 2 dx dx

435

Chapter 51 Second order differential equations of the form a

d2 y dy +b + cy = f (x) dx dx 2

441

Chapter 52 Power series methods of solving ordinary differential equations

© 2006 John Bird. All rights reserved. Published by Elsevier.

458

iii

Chapter 53 An introduction to partial differential equations

474

Chapter 54 Presentation of statistical data

489

Chapter 55 Measures of central tendency and dispersion

497

Chapter 56 Probability

504

Chapter 57 The binomial and Poisson distributions

508

Chapter 58 The normal distribution

513

Chapter 59 Linear correlation

523

Chapter 60 Linear regression

527

Chapter 61 Sampling and estimation theories

533

Chapter 62 Significance testing

543

Chapter 63 Chi-square and distribution-free tests

553

Chapter 64 Introduction to Laplace transforms

566

Chapter 65 Properties of Laplace transforms

569

Chapter 66 Inverse Laplace transforms

575

Chapter 67 The solution of differential equations using Laplace transforms

582

Chapter 68 The solution of simultaneous differential equations using Laplace transforms 590 Chapter 69 Fourier series for periodic functions of period 2π

595

Chapter 70 Fourier series for a non-periodic functions over period 2π

601

Chapter 71 Even and odd functions and half-range Fourier series

608

Chapter 72 Fourier series over any range

616

Chapter 73 A numerical method of harmonic analysis

623

Chapter 74 The complex or exponential form of a Fourier series

627

© 2006 John Bird. All rights reserved. Published by Elsevier.

iv

CHAPTER 1 ALGEBRA EXERCISE 1 Page 2 2. Find the value of 5 pq 2 r 3 when p =

2 , q = -2 and r = -1 5

2 2 3 ⎛2⎞ 5 pq 2 r 3 = 5 ⎜ ⎟ ( −2 ) ( −1) = 5 × × 4 × −1 = -8 5 ⎝5⎠ 5. Simplify ( x 2 y 3 z )( x3 yz 2 ) and evaluate when x =

(x

2

1 , y = 2 and z = 3 2

y 3 z )( x3 yz 2 ) = x 2+3 y 3+1 z1+ 2 = x5 y 4 z 3

1 24 × 33 33 33 27 1 4 3 ⎛1⎞ = = = = 13 When x = , y = 2 and z = 3, x5 y 4 z 3 = ⎜ ⎟ ( 2 ) ( 3) = 5 5− 4 2 2 2 2 2 2 ⎝2⎠ 5

⎛ 32 − 12 ⎞⎛ 12 − 12 ⎞ 6. Evaluate ⎜ a bc ⎟⎜ a b c ⎟ when a = 3, b = 4 and c = 2 ⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞ 1 ⎛ 32 −3 ⎞ ⎛ 12 − 12 ⎞ ⎜ + ⎟ ⎜1− ⎟ −3+1 a2 b 2 2 −2 ⎝ 2 3⎠ ⎝ 2⎠ =a b c = 2 b c ⎜ a bc ⎟ ⎜ a b c ⎟ = a c ⎝ ⎠⎝ ⎠ 1 1

When a = 3, b = 4 and c = 2,

1

1 a 2 b 32 4 9 ( ±2 ) = 2 = =±4 2 2 c 2 4

1 ⎛ 3 12 − 12 ⎞ ⎜ a b c ⎟ ( ab ) 3 ⎠ 8. Simplify ⎝ 3 a bc

(

)

1 ⎛ 3 12 − 12 ⎞ 1 1 1 1 − ⎜ a b c ⎟ ( ab ) 3 ⎛ 1 3⎞ ⎛1 1 1⎞ ⎛ 1 ⎞ ⎛ 18 + 2 − 9 ⎞ 1 3 3 2 2 3 3 − ⎜ 3+ − ⎟ ⎜ + − ⎟ − ⎜ − −1⎟ ⎜ ⎟ a b c a b ⎝ ⎠ 6 ⎝ 3 2⎠ ⎝ 2 3 2⎠ ⎝ 2 ⎠ ⎝ ⎠ 3 2 a b c a b c = = = 3 1 a3 b c a 2b 2c

(

)

11

1

= a 6 b3c

−

3 2

6

or

a 11 3 b c3

© 2006 John Bird. All rights reserved. Published by Elsevier.

1

EXERCISE 2 Page 3 3. Remove the brackets and simplify: 24 p − ⎡⎣ 2 {3 ( 5 p − q ) − 2 ( p + 2q )} + 3q ⎤⎦

24 p − ⎡⎣ 2 {3 ( 5 p − q ) − 2 ( p + 2q )} + 3q ⎤⎦ = 24 p − ⎡⎣ 2 {15 p − 3q − 2 p − 4q} + 3q ⎤⎦ = 24 p − [30 p − 6q − 4 p − 8q + 3q ] = 24 p − [ 26 p − 11q ] = 24p – 26p + 11q = 11q – 2p 6. Simplify 2 y + 4 ÷ 6 y + 3 × 4 − 5 y

2 y + 4 ÷ 6 y + 3× 4 − 5 y = 2 y +

2 4 2 + 3× 4 − 5 y = 2 y + + 12 − 5 y = − 3 y + 12 3y 6y 3y

8. Simplify a 2 − 3ab × 2a ÷ 6b + ab

a 2 − 3ab × 2a ÷ 6b + ab = a 2 − 3ab ×

2a 6a 2b + ab = a 2 − + ab = a 2 − a 2 + ab = ab 6b 6b

© 2006 John Bird. All rights reserved. Published by Elsevier.

2

EXERCISE 3 Page 4 3. Solve the equation:

1 1 + =0 3a − 2 5a + 3

1 1 from which, (5a + 3) = -(3a – 2) =− 3a − 2 5a + 3 i.e.

5a + 3 = -3a + 2

and

5a + 3a = 2 – 3

Thus,

8a = -1

4. Solve the equation:

If

(

6= 3 t

if

If t = 2π

l g

then

t = 2π

µL L + rCR

µL L + rCR

then

t=

then

l g

6 =2 3

and t = 22 = 4

2

l ⎛ t ⎞ and ⎜ ⎟ = g ⎝ 2π ⎠

⎛ t ⎞ l = g⎜ ⎟ ⎝ 2π ⎠

from which,

7. Transpose m =

from which, 6 = 6 t − 3 t = 3 t

l g

6. Make l the subject of t = 2π

If m =

)

3 t = −6 + 6 t

Hence,

1 8

3 t = −6 1− t

3 t = −6 then 3 t = −6 1 − t 1− t

i.e.

and a = -

2

or

l=

gt 2 4π 2

for L

m ( L + rCR ) = µ L

i.e.

mL + mrCR = µ L

© 2006 John Bird. All rights reserved. Published by Elsevier.

3

from which,

mrCR = µ L − mL = L ( µ − m )

8. Make r the subject of the formula

If

x 1+ r2 = y 1− r2

from which, and

Thus,

then

L=

and

mrCR µ−m

x 1+ r2 = y 1− r2

x (1 − r 2 ) = y (1 + r 2 ) x − xr 2 = y + yr 2 x − y = yr 2 + xr 2 = r 2 ( y + x )

r2 =

x− y x+ y

and

⎛ x− r= ⎜ ⎝ x+

y⎞ ⎟ y⎠

© 2006 John Bird. All rights reserved. Published by Elsevier.

4

EXERCISE 4 Page 5

2. Solve the simultaneous equations

5a = 1 − 3b 2b + a + 4 = 0

5a + 3b = 1

(1)

a + 2b = -4

(2)

5 × (2) gives:

5a + 10b = -20

(3)

(1) – (3) gives:

- 7b = 21

from which,

21 = -3 −7

b=

Substituting in (1) gives: 5a + 3(-3) = 1 from which,

5a = 1 + 9 = 10

3. Solve the simultaneous equations

a=

and

10 =2 2

x 2 y 49 + = 5 3 15 3x y 5 − + =0 7 2 7

(1) (2)

15 × (1) gives:

3x + 10y = 49

(3)

14 × (2) gives:

6x – 7y = -10

(4)

6x + 20y = 98

(5)

27y = 108

from which,

2 × (3) gives: (5) – (4) gives: Substituting in (3) gives:

3x + 40 = 49

and

y=

108 =4 27

3x = 49 – 40 = 9

from which,

x=3

4.(b) Solve the quadratic equation by factorisation: 8 x 2 + 2 x − 15 = 0

If

8 x 2 + 2 x − 15 = 0

then

(4x – 5)(2x + 3) = 0 5 4

hence,

4x – 5 = 0

i.e. 4x = 5

i.e.

x=

and

2x + 3 = 0

i.e. 2x = -3

i.e.

x= −

© 2006 John Bird. All rights reserved. Published by Elsevier.

3 2

5

5. Determine the quadratic equation in x whose roots are 2 and -5

If roots are x = 2 and x = -5 then (x – 2)(x + 5) = 0

i.e. x 2 − 2 x + 5 x − 10 = 0

x 2 + 3 x − 10 = 0

i.e.

6.(a) Solve the quadratic equation, correct to 3 decimal places: 2 x 2 + 5 x − 4 = 0

−5 ± ⎡⎣52 − 4(2)(−4) ⎤⎦ −5 ± (25 + 32) −5 ± 57 = = If 2 x + 5 x − 4 = 0 then x = 2(2) 4 4 2

Hence,

x=

−5 + 57 = 0.637 4

or

x=

−5 − 57 = -3.137 4

© 2006 John Bird. All rights reserved. Published by Elsevier.

6

EXERCISE 5 Page 8

(10 x

3. Determine

2

+ 11x − 6 ) ÷ ( 2 x + 3)

5x - 2 2 x + 3 10 x 2 + 11x − 6

10 x 2 + 15 x - 4x - 6 - 4x - 6 Hence,

10 x 2 + 11x − 6 = 5x - 2 2x + 3

5. Divide ( x3 + 3x 2 y + 3xy 2 + y 3 ) by (x + y)

x 2 + 2 xy + y 2 x + y x3 + 3 x 2 y + 3 xy 2 + y 3

x3 + x 2 y

2 x 2 y + 3xy 2 2 x 2 y + 2 xy 2 xy 2 + y 3 xy 2 + y 3

Hence,

x 3 + 3 x 2 y + 3 xy 2 + y 3 = x 2 + 2 xy + y 2 x+ y

Hence,

8 5x2 − x + 4 = 5x + 4 + x −1 x −1

6. Find ( 5 x 2 − x + 4 ) ÷ ( x − 1)

5x + 4 x −1 5x2 − x + 4

5x2 − 5x 4x + 4 4x - 4 8

© 2006 John Bird. All rights reserved. Published by Elsevier.

7

8. Determine ( 5 x 4 + 3x 3 − 2 x + 1) ÷ ( x − 3)

5 x3 + 18 x 2 + 54 x + 160 x − 3 5 x 4 + 3 x3

− 2x + 1

5 x 4 − 15 x 3 18 x 3 18 x 3 − 54 x 2 54 x 2 − 2 x 54 x 2 − 162 x 160x + 1 160x - 480 481 481 5 x 4 + 3x3 − 2 x + 1 Hence, = 5 x 3 + 18 x 2 + 54 x + 160 + x−3 x −3

© 2006 John Bird. All rights reserved. Published by Elsevier.

8

EXERCISE 6 Page 9

2. Use the factor theorem to factorise x 3 + x 2 − 4 x − 4

Let f(x) = x3 + x 2 − 4 x − 4 If x = 1, f(x) = 1 + 1 – 4 – 4 = -6 x = 2, f(x) = 8 + 4 – 8 – 4 = 0

hence, (x – 2) is a factor

x = 3, f(x) = 27 + 9 – 12 – 4 = 20 x = -1, f(x) = -1 + 1 + 4 – 4 = 0 hence, (x + 1) is a factor x = -2, f(x) = -8 + 4 + 8 – 4 = 0 hence, (x + 2) is a factor Thus, x 3 + x 2 − 4 x − 4 = (x + 1)(x + 2)(x – 2)

4. Use the factor theorem to factorise 2 x 3 − x 2 − 16 x + 15

Let f(x) = 2 x 3 − x 2 − 16 x + 15 If x = 1, f(x) = 2 – 1 – 16 + 15 = 0

hence, (x – 1) is a factor

x = 2, f(x) = 16 – 4 – 32 +15 = -5 x = 3, f(x) = 54 – 9 – 48 + 15 = 12 x = -1, f(x) = – 1 – 1 + 16 + 15 = 29 x = -2, f(x) = -16 – 4 + 32 + 15 = 27 x = -3, f(x) = -54 – 9 + 48 + 15 = 0 hence, (x + 3) is a factor 2 x 3 − x 2 − 16 x + 15 2 x 3 − x 2 − 16 x + 15 = ( x − 1)( x + 3) x2 + 2x − 3

2x - 5 x + 2 x − 3 2 x 3 − x 2 − 16 x + 15 2

2 x3 + 4 x 2 − 6 x −5 x 2 − 10 x + 15 −5 x 2 − 10 x + 15

Hence, 2 x 3 − x 2 − 16 x + 15 = (x – 1)(x + 3)(2x – 5) © 2006 John Bird. All rights reserved. Published by Elsevier.

9

6. Solve the equation x3 − 2 x 2 − x + 2 = 0

Let f(x) = x3 − 2 x 2 − x + 2 If x = 1, f(x) = 1 – 2 – 1 + 2 = 0 hence, (x – 1) is a factor x = 2, f(x) = 8 – 8 – 2 + 2 = 0 hence, (x – 2) is a factor x = 3, f(x) = 27 – 18 – 3 + 2 = 8 x = -1, f(x) = -1 – 2 + 1 + 2 = 0 hence, (x + 1) is a factor Hence, x3 − 2 x 2 − x + 2 = (x – 1)(x – 2)(x + 1) If x3 − 2 x 2 − x + 2 = 0 then (x – 1)(x – 2)(x + 1) = 0 from which,

x = 1, x = 2, or x = -1

© 2006 John Bird. All rights reserved. Published by Elsevier.

10

EXERCISE 7 Page 11

2. Determine the remainder when x3 − 6 x 2 + x − 5 is divided by (a) (x + 2) (b) (x – 3)

(a) Remainder is ap 3 + bp 2 + cp + d where a = 1, b = -6, c = 1, d = -5 and p = -2 Hence, remainder = 1(−2)3 − 6(−2) 2 + 1(−2) − 5 = -8 – 24 – 2 – 5 = -39 (b) When p = 3, remainder = 1(3)3 − 6(3) 2 + 1(3) − 5 = 27 – 54 + 3 – 5 = -29

4. Determine the factors of x3 + 7 x 2 + 14 x + 8 and hence solve the cubic equation x3 + 7 x 2 + 14 x + 8 = 0

Remainder is ap 3 + bp 2 + cp + d where a = 1, b = 7, c = 14, d = 8 Let p = 1, then remainder = 1(1)3 + 7(1) 2 + 14(1) + 8 = 30 Let p = -1, then remainder = 1(−1)3 + 7(−1) 2 + 14(−1) + 8 = -1 + 7 – 14 + 8 = 0, hence (x + 1) is a factor Let p = -2, then remainder = 1(−2)3 + 7(−2) 2 + 14(−2) + 8 = -8 + 28 – 28 + 8 = 0, hence (x + 2) is a factor Let p = -3, then remainder = 1(−3)3 + 7(−3) 2 + 14(−3) + 8 = -27 + 63 – 42 + 8 = 2 Let p = -4, then remainder = 1(−4)3 + 7(−4) 2 + 14(−4) + 8 = -64 + 112 – 56 + 8 = 0, hence (x + 4) is a factor Hence, x3 + 7 x 2 + 14 x + 8 = (x + 1)(x + 2)(x + 4) If x3 + 7 x 2 + 14 x + 8 = 0 then (x + 1)(x + 2)(x + 4) = 0 from which,

x = -1, x = -2 or x = -4

© 2006 John Bird. All rights reserved. Published by Elsevier.

11

6. Using the remainder theorem, solve the equation 2 x 3 − x 2 − 7 x + 6 = 0

Remainder is ap 3 + bp 2 + cp + d where a = 2, b = -1, c = -7, d = 6 Let p = 1, then remainder = 2(1)3 + (−1)(1) 2 + (−7)(1) + 6 = 2 - 1 – 7 + 6 = 0, hence (x - 1) is a factor Let p = 2, then remainder = 2(2)3 + (−1)(2) 2 + (−7)(2) + 6 = 16 - 4 – 14 + 6 = 4 Let p = -1, then remainder = 2(−1)3 + (−1)(−1) 2 + (−7)(−1) + 6 = -2 - 1 + 7 + 6 = 10 Let p = -2, then remainder = 2(−2)3 + (−1)(−2) 2 + (−7)(−2) + 6 = -16 - 4 + 14 + 6 = 0, hence (x + 2) is a factor Let p = -3, then remainder = 2(−3)3 + (−1)(−3) 2 + (−7)(−3) + 6 = -54 - 9 + 21 + 6 = -36 The third root can be found by division, i.e.

2 x3 − x 2 − 7 x + 6 2 x3 − x 2 − 7 x + 6 = x2 + x − 2 ( x − 1)( x + 2 ) 2x - 3 x 2 + x − 2 2 x3 − x 2 − 7 x + 6

2 x3 + 2 x 2 − 4 x

−3x 2 − 3x + 6 −3 x 2 − 3 x + 6 Hence, 2 x3 − x 2 − 7 x + 6 = (x – 1)(x + 2)(2x – 3) If 2 x 3 − x 2 − 7 x + 6 = 0 then (x – 1)(x + 2)(2x – 3) = 0 from which,

x = 1, x = -2 or x = 1.5

© 2006 John Bird. All rights reserved. Published by Elsevier.

12

CHAPTER 2 INEQUALITIES EXERCISE 8 Page 13

2. Solve the following inequalities: (a)

(a)

x > 1.5 2

(b) x + 2 ≥ 5

i.e. x > 2(1.5)

i.e. x > 3

i.e. x ≥ 5 – 2

i.e. x ≥ 3

4. Solve the following inequalities: (a)

(a)

x > 1.5 (b) x + 2 ≥ 5 2

7 − 2k ≤1 4

(b) 3z + 2 > z + 3

i.e. 7 – 2k ≤ 4

7 − 2k ≤ 1 (b) 3z + 2 > z + 3 4

i.e. 7 – 4 ≤ 2k

i.e. 3z – z > 3 – 2

i.e. 2z >1

i.e. 3 ≤ 2k and and

z>

k≥

3 2

1 2

5. Solve the following inequalities: (a) 5 – 2y ≤ 9 + y (b) 1 - 6x ≤ 5 + 2x

(a) 5 – 2y ≤ 9 + y

i.e. 5 – 9 ≤ y + 2y

i.e. -4 ≤ 3y

i.e. −

4 ≤y 3

or

y≥ −

(b) 1 - 6x ≤ 5 + 2x

i.e. 1 – 5 ≤ 2x + 6x

i.e. -4 ≤ 8x

i.e. −

4 ≤x 8

or

x≥-

© 2006 John Bird. All rights reserved. Published by Elsevier.

4 3

1 2

13

EXERCISE 9 Page 14 1. Solve the inequality: t + 1 < 4

If t + 1 < 4 then -4 < t + 1 < 4 -4 < t + 1 becomes -5 < t t + 1 < 4 becomes Hence,

i.e. t > -5

t1 t −5

2t + 4 -1>0 t −5

then

t +9 >0 t −5

i.e.

2t + 4 t − 5 − > 0 and t −5 t −5

i.e.

( 2t + 4 ) − ( t − 5) > 0 t −5

Hence, either (i) t + 9 > 0 and t – 5 > 0 or (ii) t + 9 < 0 and t – 5 < 0

(i) t > -9 and t > 5 and both inequalities are true when t > 5 (ii) t < -9 and t < 5 and both inequalities are true when t < -9 2t + 4 > 1 is true when t > 5 or t < -9 t −5

Hence,

3. Solve the inequality:

If

3z − 4 ≤2 z+5

i.e.

then

3z − 4 ≤2 z+5

3z − 4 -2≤0 z+5

3z − 4 − 2 z − 10 ≤0 z +5

Hence,

either (i) z - 14 ≤ 0 or

(ii) z – 14 ≥ 0

i.e. and

i.e.

3z − 4 2( z + 5) − ≤ 0 and ( z + 5) z +5

(3 z − 4) − 2( z + 5) ≤0 z +5

z − 14 ≤0 z +5

z+5>0

and z + 5 < 0

(i) z ≤ 14

and z > -5 i.e. -5 < z ≤ 14

(ii) z ≥ 14

and z ≥ -5 Both of these inequalities are not possible to satisfy.

Hence,

3z − 4 ≤ 2 is true when -5 < z ≤ 14 z+5

© 2006 John Bird. All rights reserved. Published by Elsevier.

15

EXERCISE 11 Page 16

3. Solve the inequality: 2 x 2 ≥ 6

2 x2 ≥ 6

i.e.

x2 ≥ 3

x≥ 3

hence,

or x ≤ - 3

4. Solve the inequality: 3k 2 − 2 ≤ 10

3k 2 − 2 ≤ 10

i.e.

3k 2 ≤ 12

- 4 ≤k≤

Hence,

and

k2 ≤ 4

i.e. -2 ≤ k ≤ 2

4

6. Solve the inequality: ( t − 1) ≥ 36 2

( t − 1)

36

or

(t – 1) ≤ - 36

i.e.

(t – 1) ≥ 6

or

(t – 1) ≤ -6

i.e.

t ≥7

or

t ≤ -5

( 4k + 5 )

>9

2

≥ 36 then (t – 1) ≥

8. Solve the inequality:

( 4k + 5 )

9

or

4k + 5 < - 9

i.e.

4k + 5 > 3

or

4k + 5 < -3

i.e.

4k > -2

or

4k < -8

and

k> −

1 2

or

k < -2

2

>9

2

then

4k + 5 >

© 2006 John Bird. All rights reserved. Published by Elsevier.

16

EXERCISE 12 Page 17

1. Solve the inequality: x 2 − x − 6 > 0

x2 − x − 6 > 0

(x – 3)(x + 2) > 0

thus

Either (i) x – 3 > 0

and

x+2>0

(ii) x – 3 < 0

and

x+2 3

and x > -2

i.e.

x>3

(ii) x < 3

and

x < -2

i.e.

x < -2

3. Solve the inequality: 2 x 2 + 3x − 2 < 0

2 x 2 + 3x − 2 < 0

thus

(2x – 1)(x + 2) < 0

Either (i) 2x – 1 > 0

and

x+20

or

(i) 2x > 1 i.e. x >

1 2

and

x < -2

(ii) 2x < 1 i.e. x <

1 2

and

x > -2

both of which are not possible

thus

-2 < x <

1 2

5. Solve the inequality: z 2 + 4 z + 4 ≤ 4 z2 + 4z + 4 ≤ 4

i.e.

Either (i) z ≤ 0

and

z ≥ -4

and

z ≤ -4

or

(ii) z ≥ 0

z2 + 4z ≤ 0

z(z + 4) ≤ 0

i.e. i.e.

-4 ≤ z ≤ 0

both of which are not possible

7. Solve the inequality: t 2 − 4t − 7 ≥ 0

Since t 2 − 4t − 7 ≥ 0

then

(t − 2)

2

-7–4≥0

© 2006 John Bird. All rights reserved. Published by Elsevier.

17

(t − 2)

i.e. and thus,

t – 2 ≥ 11

(

t ≥ 2 + 11

)

2

≥ 11

or

t – 2 ≤ - 11

or

t ≤ 2 − 11

(

)

© 2006 John Bird. All rights reserved. Published by Elsevier.

18

CHAPTER 3 PARTIAL FRACTIONS EXERCISE 13 Page 20

2. Resolve

Let

4( x − 4) into partial fractions. x2 − 2 x − 3

4( x − 4) 4 x − 16 A B A( x − 3) + B( x + 1) ≡ = + = 2 ( x + 1)( x − 3) x − 2 x − 3 ( x + 1)( x − 3) ( x + 1) ( x − 3)

Hence,

4x – 16 = A(x – 3) + B(x + 1)

If x = -1,

-20 = -4A from which, A = 5

If x = 3,

12 – 16 = 4B

Hence,

4( x − 4) 5 1 = − 2 x − 2 x − 3 ( x + 1) ( x − 3)

4. Resolve

from which, B = -1

3(2 x 2 − 8 x − 1) into partial fractions. ( x + 4)( x + 1)(2 x − 1)

Let 3(2 x 2 − 8 x − 1) A B C A( x + 1)(2 x − 1) + B( x + 4)(2 x − 1) + C ( x + 4)( x + 1) ≡ + + = ( x + 4)( x + 1)(2 x − 1) ( x + 4) ( x + 1) (2 x − 1) ( x + 4)( x + 1)(2 x − 1)

Hence, 6 x 2 − 24 x − 3 = A(x + 1)(2x – 1) + B(x + 4)(2x – 1) + C(x + 4)(x + 1) If x = -4, 96 + 96 -3 = A(-3)(-9) from which, If x = -1,

6 + 24 -3 = B(3)(-3)

from which,

189 = 27A and A = 7 27 = -9B

and B = -3

If x = 0.5, 1.5 - 12 -3 = C(4.5)(1.5) from which, -13.5 = 6.75C and C = -2 3(2 x 2 − 8 x − 1) 7 3 2 = − − ( x + 4)( x + 1)(2 x − 1) ( x + 4) ( x + 1) (2 x − 1)

Hence,

5. Resolve

x2 + 9 x + 8 into partial fractions. x2 + x − 6

Since the numerator is of the same degree as the denominator, division is firstly required. © 2006 John Bird. All rights reserved. Published by Elsevier.

19

1 x2 + x − 6 x2 + 9 x + 8

x2 + x − 6

8x + 14 x + 9x + 8 8 x + 14 = 1+ 2 2 x + x−6 x + x−6 2

Hence, Let

8 x + 14 8 x + 14 A B A( x − 2) + B( x + 3) = ≡ + = 2 ( x + 3)( x − 2) x + x − 6 ( x + 3)( x − 2) ( x + 3) ( x − 2)

Hence,

8x + 14 = A(x – 2) + B(x + 3)

If x = -3, -24 + 14 = -5A form which, -10 = -5A and A = 2 If x = 2,

16 + 14 = 5B from which,

30 = 5B and B = 6

x2 + 9x + 8 2 6 = 1+ + 2 x + x−6 ( x + 3) ( x − 2)

Hence,

7. Resolve

3 x 3 − 2 x 2 − 16 x + 20 into partial fractions. ( x − 2)( x + 2)

3x - 2 x − 4 3 x3 − 2 x 2 − 16 x + 20 2

3x3

− 12 x −2 x 2 − 4 x + 20 −2 x 2 +8

- 4x + 12 Hence, Let

3 x 3 − 2 x 2 − 16 x + 20 12 − 4 x ≡ 3x − 2 + 2 ( x − 2)( x + 2) x −4

12 − 4 x 12 − 4 x A B A( x + 2) + B( x − 2) = ≡ + = 2 ( x − 2)( x + 2) x − 4 ( x − 2)( x + 2) ( x − 2) ( x + 2)

Hence,

12 – 4x = A(x + 2) + B(x - 2)

If x = 2,

4 = 4A from which, A = 1

If x = -2

20 = -4B from which, B = -5

Hence,

3 x 3 − 2 x 2 − 16 x + 20 1 5 ≡ 3x − 2 + − ( x − 2)( x + 2) ( x − 2) ( x + 2)

© 2006 John Bird. All rights reserved. Published by Elsevier.

20

EXERCISE 14 PAGE 22

x2 + 7 x + 3 2. Resolve 2 into partial fractions. x ( x + 3)

x2 + 7 x + 3 A B C A( x)( x + 3) + B ( x + 3) + Cx 2 ≡ + 2+ = Let 2 x ( x + 3) x x ( x + 3) x 2 ( x + 3)

Hence, If x = 0 If x = -3

x 2 + 7x + 3 = A(x)(x + 3) + B (x + 3) + C x 2 3 = 3B

from which,

9 – 21 + 3 = 9C

i.e. -9 = 9C f rom which, C = -1

Equating x 2 coefficients: 1 = A + C Hence,

B=1

from which, A = 2

x2 + 7 x + 3 2 1 1 = + 2− 2 x ( x + 3) x x ( x + 3)

18 + 21x − x 2 4. Resolve into partial fractions. ( x − 5)( x + 2) 2

18 + 21x − x 2 A B C A( x + 2) 2 + B( x − 5)( x + 2) + C ( x − 5) ≡ + + = Let ( x − 5)( x + 2) 2 ( x − 5) ( x + 2) ( x + 2) 2 ( x − 5)( x + 2) 2

Hence, If x = 5 If x = -2

18 + 21x − x 2 = A( x + 2) 2 + B( x − 5)( x + 2) + C ( x − 5) 18 + 105 – 25 = 49A

i.e. 98 = 49A

from which, A = 2

18 – 42 – 4 = -7C

i.e. -28 = -7C

from which, C = 4

Equating x 2 coefficients: -1 = A + B Hence,

from which,

B = -3

18 + 21 x − x 2 2 3 4 = − + 2 ( x − 5)( x + 2) ( x − 5) ( x + 2) ( x + 2)2

© 2006 John Bird. All rights reserved. Published by Elsevier.

21

EXERCISE 15 PAGE 23

x 2 − x − 13 1. Resolve 2 into partial fractions. ( x + 7) ( x − 2)

Let

( Ax + B)( x − 2) + C ( x 2 + 7 ) x 2 − x − 13 Ax + B C ≡ + = ( x 2 + 7 ) ( x − 2 ) ( x2 + 7 ) ( x − 2) ( x 2 + 7 ) ( x − 2)

Hence,

x 2 − x − 13 = ( Ax + B )( x − 2) + C ( x 2 + 7 )

If x = 2,

4 – 2 –13 = 11C

Equating x 2 coefficients:

1=A+C

i.e. -11 = 11C from which,

from which, C = -1 A=2

Equating constant terms: -13 = -2B + 7C = -2B – 7 i.e. 2B = 13 – 7 = 6 from which, B = 3 x 2 − x − 13 2x + 3 1 = 2 − 2 ( x + 7 ) ( x − 2 ) ( x + 7 ) ( x − 2)

Hence,

4. Resolve

x 3 + 4 x 2 + 20 x − 7 into partial fractions. ( x − 1) 2 ( x 2 + 8 )

2 2 2 x 3 + 4 x 2 + 20 x − 7 A B Cx + D A( x − 1) ( x + 8 ) + B ( x + 8 ) + (Cx + D)( x − 1) Let ≡ + + = ( x − 1) ( x − 1) 2 ( x 2 + 8 ) ( x − 1) 2 ( x 2 + 8 ) ( x − 1) 2 ( x 2 + 8 )

x3 + 4 x 2 + 20 x − 7 = A( x − 1) ( x 2 + 8 ) + B ( x 2 + 8 ) + (Cx + D)( x − 1) 2

Hence,

= A( x − 1) ( x 2 + 8 ) + B ( x 2 + 8 ) + (Cx + D)( x 2 − 2 x + 1)

If x = 1,

1 + 4 + 20 – 7 = 9B i.e. 18 = 9B from which, B = 2

Equating x 3 coefficients:

1=A+C

(1’)

Equating x 2 coefficients:

4 = -A + B – 2C + D

(2’)

Equating x coefficients: Since B = 2,

20 = 8A + C – 2D

(3’)

A+C=1

(1)

-A – 2C + D = 2

(2)

© 2006 John Bird. All rights reserved. Published by Elsevier.

22

8A + C – 2D = 20

(3)

2 × (2) gives: -2A – 4C + 2D = 4

(4)

(3) + (4) gives:

6A – 3C = 24

(5)

3 × (1) gives:

3A + 3C = 3

(6)

(5) + (6) gives:

9A

= 27 from which,

A=3

From (1):

3+C=1

from which,

C = -2

From (2):

-3 + 4 + D = 2

from which,

D=1

x 3 + 4 x 2 + 20 x − 7 3 2 1− 2x = + + 2 2 2 2 ( x − 1) ( x − 1) ( x + 8 ) ( x − 1) ( x + 8 )

Hence,

5. When solving the differential equation

d 2θ dθ −6 − 10θ = 20 − e 2t by Laplace transforms, for 2 dt dt

given boundary conditions, the following expression for Λ {θ } results: 39 2 s + 42s − 40 2 Λ {θ } = s ( s − 2 ) ( s 2 − 6s + 10 ) 4s 3 −

Show that the expression can be resolved into partial fractions to give: Λ {θ } =

2 1 5s − 3 − + s 2 ( s − 2 ) 2 ( s 2 − 6s + 10 )

39 2 s + 42s − 40 A B Cs + D 2 ≡ + + 2 2 s ( s − 2 ) ( s − 6s + 10 ) s ( s − 2) ( s − 6s + 10 )

4s 3 −

Let

= Hence, 4 s 3 −

A( s − 2) ( s 2 − 6s + 10 ) + B( s ) ( s 2 − 6 s + 10 ) + (Cs + D)( s )( s − 2) s ( s − 2) ( s 2 − 6s + 10 )

39 2 s + 42 s − 40 = A( s − 2) ( s 2 − 6s + 10 ) + B ( s ) ( s 2 − 6s + 10 ) + (Cs + D)( s )( s − 2) 2 = A ( s 3 − 8s 2 − 2s − 20 ) + B ( s 3 − 6 s 2 + 10s ) + (Cs + D)( s 2 − 2s )

If s = 0, If s = 2,

-40 = A(-20)

from which,

32 – 78 + 84 – 40 = B (8 – 24 + 20)

A=2

i.e. -2 = 4B

from which, B = −

© 2006 John Bird. All rights reserved. Published by Elsevier.

1 2 23

Equating s 3 coefficients:

Equating s 2 coefficients: −

4=A+B+C

i.e.

39 = -8A – 6B – 2C + D 2

4=2-

1 +C 2

i.e.

−

from which,

C=

5 2

39 = -16 + 3 – 5 + D 2 from which, D = −

3 2

39 2 1 5 3 s + 42 s − 40 − s− 2 2 2 + 2 2 ≡ + 2 2 s ( s − 2 ) ( s − 6 s + 10 ) s ( s − 2) ( s − 6s + 10 )

4s 3 −

Hence,

i.e.

Λ {θ } =

2 1 5s − 3 − + 2 s 2 ( s − 2 ) 2 ( s − 6 s + 10 )

© 2006 John Bird. All rights reserved. Published by Elsevier.

24

CHAPTER 4 LOGARITHMS AND EXPONENTIAL FUNCTIONS EXERCISE 16 Page 26

2. Evaluate: log 2 16

then 2 x = 16 = 24

Let x = log 2 16 Hence,

from which,

x=4

log 2 16 = 4

4. Evaluate: log 2

Let x = log 2

1 8

Hence,

1 8

then log 2

2x =

1 1 = = 2−3 8 23

from which, x = -3

1 = -3 8

7. Evaluate: log 4 8

Let x = log 4 8

then

4x = 8

from which, Hence,

2 x

2x = 3 log 4 8 = 1

1 2

then

= 23

and x =

i.e.

2 2 x = 23

3 2

1 2

11. Solve the equation: log 4 x = −2

If log 4 x = −2

(2 )

i.e.

x= 4

−

5 2

=

1 2

1 4

5 2

=

1 4

5

=±

1 1 =± 5 32 2

© 2006 John Bird. All rights reserved. Published by Elsevier.

25

12. Solve the equation: lg x = −2

If lg x = -2 then log10 x = −2 and x = 10−2 =

1 1 = or 0.01 2 100 10

⎛ 16 × 4 5 ⎞ 16. Write in terms of log 2, log 3 and log 5 to any base: log ⎜⎜ ⎟⎟ ⎝ 27 ⎠ ⎛ 4 14 ⎛ 16 × 4 5 ⎞ ⎜ 2 ×5 log ⎜⎜ ⎟⎟ = log ⎜ 3 ⎜ 3 ⎝ 27 ⎠ ⎝

⎞ 1 1 ⎟ = log ⎛ 24 × 5 4 × 3−3 ⎞ = log 24 + log 5 4 + log 3−3 ⎜ ⎟ ⎟ ⎟ ⎝ ⎠ ⎠ 1 = 4 log 2 + log 5 – 3 log 3 4

⎛ 125 × 4 16 ⎞ 17. Write in terms of log 2, log 3 and log 5 to any base: log ⎜⎜ ⎟ 4 813 ⎟⎠ ⎝ ⎛ 125 × 4 16 ⎞ ⎛ 53 × 2 ⎞ −3 −3 3 3 log ⎜⎜ = log ⎟⎟ ⎜ 3 ⎟ = log ( 5 × 2 × 3 ) = log 5 + log 2 + log 3 3 4 81 ⎠ ⎝ 3 ⎠ ⎝

= 3 log 5 + log 2 – 3 log 3

19. Simplify: log 64 + log 32 – log 128

⎛ 2 6 × 25 ⎞ 6+5−7 log 64 + log 32 – log 128 = log 26 + log 25 − log 27 = log ⎜ = log 24 = 4 log 2 ) ⎟ = log ( 2 7 ⎝ 2 ⎠

1 1 log16 − log 8 3 20. Evaluate: 2 log 4

⎛ 22 ⎞ 1 1 1 1 1 1 log 4 3 ⎜ ⎟ log16 − log 8 log16 2 − log 83 log ( 2 ) 2 − log ( 2 ) 3 log 22 − log 2 ⎝ 2 ⎠ = log 2 = 1 2 3 = = = = 2 2 2 log 2 log 2 log 2 log 22 2 log 2 2 log 4

© 2006 John Bird. All rights reserved. Published by Elsevier.

26

22. Solve the equation: log 2t 3 − log t = log16 + log t

log 2t 3 − log t = log16 + log t ⎛ 2t 3 ⎞ i.e. log ⎜ ⎟ = log (16t ) ⎝ t ⎠

from which, Hence,

2t 2 = 16t t=8

i.e.

i.e.

log ( 2t 2 ) = log (16t )

2t 2 − 16t = 0

i.e.

2t(t – 8) = 0

(note that t = 0 is not a valid solution to the equation)

© 2006 John Bird. All rights reserved. Published by Elsevier.

27

EXERCISE 17 Page 27

1. Solve the equation 3x = 6.4 correct to 4 significant figures

If 3x = 6.4

then x log10 3 = log10 6.4 x=

and

log10 6.4 0.80617997... = = 1.689675... = 1.690, correct to 4 significant figures. log10 3 0.47712125...

3. Solve the equation 2 x −1 = 32 x −1 correct to 4 significant figures

If 2 x −1 = 32 x −1

( x − 1) log10 2 = (2 x − 1) log10 3

then

x log10 2 − log10 2 = 2 x log10 3 − log10 3

i.e.

log10 3 − log10 2 = 2 x log10 3 − x log10 2 = x ( 2 log10 3 − log10 2 )

i.e. Hence,

x=

log10 3 − log10 2 = 0.2696, correct to 4 significant figures. 2 log10 3 − log10 2

5. Solve the equation 25.28 = 4.2 x correct to 4 significant figures

If 25.28 = 4.2 x

then

log10 25.28 = x log10 4.2 x=

from which,

log10 25.28 = 2.251, correct to 4 significant figures. log10 4.2

6. Solve the equation 42 x −1 = 5 x + 2 correct to 4 significant figures

If 42 x −1 = 5 x + 2

then

(2 x − 1) log10 4 = ( x + 2) log10 5

2 x log10 4 − log10 4 = x log10 5 + 2 log10 5

i.e. i.e.

2 x log10 4 − x log10 5 = 2 log10 5 + log10 4

i.e.

x(2 log10 4 − log10 5) = 2 log10 5 + log10 4

from which,

x=

2 log10 5 + log10 4 = 3.959, correct to 4 significant figures. 2 log10 4 − log10 5

© 2006 John Bird. All rights reserved. Published by Elsevier.

28

8. Solve the equation 0.027 x = 3.26 correct to 4 significant figures

If 0.027 x = 3.26

then

x log10 0.027 = log10 3.26 x=

from which,

log10 3.26 = -0.3272, correct to 4 significant figures. log10 0.027

⎛P ⎞ 9. The decibel gain n of an amplifier is given by: n = 10 log10 ⎜ 2 ⎟ where P1 is the power input ⎝ P1 ⎠

and P2 is the power output. Find the power gain

When n = 25 then: from which,

Thus,

P2 when n = 25 decibels. P1

⎛P ⎞ 25 = 10 log10 ⎜ 2 ⎟ ⎝ P1 ⎠ ⎛P ⎞ ⎛P ⎞ 25 i.e. 2.5 = log10 ⎜ 2 ⎟ = log10 ⎜ 2 ⎟ 10 ⎝ P1 ⎠ ⎝ P1 ⎠

P2 = 102.5 P1

i.e. power gain,

P2 = 316.2 P1

© 2006 John Bird. All rights reserved. Published by Elsevier.

29

EXERCISE 18 Page 29

5.6823 3. Evaluate, correct to 5 significant figures: (a) −2.1347 e

e 2.1127 − e −2.1127 (b) 2

(c )

4 ( e −1.7295 − 1) e3.6817

Using a calculator: 5.6823 = 48.04106, correct to 5 decimal places. e −2.1347 e 2.1127 − e −2.1127 = 4.07482, correct to 5 decimal places. (b) 2 4 ( e −1.7295 − 1) (c ) = -0.08286, correct to 5 decimal places. e3.6817

(a)

4. The length of a bar, l , at a temperature θ is given by l = l0 eα θ , where l and α are constants.

Evaluate l , correct to 4 significant figures, when l0 = 2.587, θ = 321.7 and α = 1.771×10−4

Using a calculator, l = l0 eα θ = (2.587) e

(321.7×1.771×10 ) = 2.739, correct to 4 significant figures. −4

© 2006 John Bird. All rights reserved. Published by Elsevier.

30

EXERCISE 19 Page 31 2. Use the power series for e x to determine, correct to 4 significant figures, (a) e 2 (b) e −0.3 and

check your result by using a calculator.

(a)

x 2 x3 x 4 + + + .... 2! 3! 4!

ex = 1 + x +

When x = 2, e 2 = 1 + 2 +

2 2 23 2 4 25 2 6 + + + + + ... 2! 3! 4! 5! 6!

= 1 + 2 + 2 + 1.33333 + 0.66666 + 0.26666 + 0.08888 + 0.02540 + 0.00635 + 0.00141 + 0.00028 + 0.00005 + … = 7.389, correct to 4 significant figures, which may be checked with a calculator. (b) When x = -0.3, e −0.3 = 1 − 0.3 +

(−0.3) 2 (−0.3)3 (−0.3) 4 (−0.3)5 + + + + ... 2! 3! 4! 5!

= 1 – 0.3 + 0.04500 – 0.00450 + 0.00034 – 0.00002 + … = 0.7408, correct to 4 significant figures.

3. Expand (1 – 2x) e2 x to six terms.

⎛ ⎞ (2 x) 2 (2 x)3 (2 x) 4 (1 – 2x) e2 x = (1 – 2x) ⎜1 + 2 x + + + + ... ⎟ 2! 3! 4! ⎝ ⎠

4 2 ⎞ ⎛ = (1 – 2x) ⎜1 + 2 x + 2 x 2 + x3 + x 4 ⎟ 3 3 ⎠ ⎝ =1 + 2 x + 2 x2 +

= 1 − 2 x2 −

4 3 2 4 8 x + x − 2 x − 4 x 2 − 4 x3 − x 4 − ... 3 3 3

8 3 x − 2 x4 3

© 2006 John Bird. All rights reserved. Published by Elsevier.

31

EXERCISE 20 Page 32

1. Plot a graph of y = 3 e0.2 x over the range x = -3 to x = 3. Hence determine the value of y when

x = 1.4 and the value of x when y = 4.5

Figure 1

From Figure 1, when x = 1.4, y = 3.95 and when y = 4.5, x = 2.05 4. The rate at which a body cools is given by θ = 250e−0.05 t where the excess temperature of a body

above its surroundings at time t minutes is θ °C . Plot a graph showing the natural decay curve for the first hour of cooling. Hence determine (a) the temperature after 25 minutes, and (b) the time when the temperature is 195 °C

From Figure 2 on page 33, (a) after t = 25 minutes, temperature θ = 70 °C (b) when the temperature is 195 °C , time t = 5 minutes © 2006 John Bird. All rights reserved. Published by Elsevier.

32

Figure 2

© 2006 John Bird. All rights reserved. Published by Elsevier.

33

EXERCISE 21 Page 34

2. Evaluate, correct to 5 significant figures: 2.946 ln e1.76 (a) lg101.41

5e −0.1629 (b) 2 ln 0.00165

(c )

ln 4.8629 − ln 2.4711 5.173

Using a calculator, (a)

2.946 ln e1.76 (2.946)(1.76) = 3.6773, correct to 5 significant figures. = 1.41 (1.41) lg10

5e−0.1629 (b) = -0.33154, correct to 5 significant figures. 2 ln 0.00165 (c )

ln 4.8629 − ln 2.4711 = 0.13087, correct to 5 significant figures. 5.173

4. Solve, correct to 4 significant figures: 7.83 = 2.91e−1.7 x

If 7.83 = 2.91e−1.7 x

i.e.

then

e−1.7 =

⎛ 7.83 ⎞ -1.7x = ln ⎜ ⎟ ⎝ 2.91 ⎠

7.83 2.91

and x = −

and

⎛ 7.83 ⎞ ln e −1.7 x = ln ⎜ ⎟ ⎝ 2.91 ⎠

1 ⎛ 7.83 ⎞ ln ⎜ ⎟ = -0.5822, correct to 4 significant figures. 1.7 ⎝ 2.91 ⎠

t − ⎞ ⎛ 2 5. Solve, correct to 4 significant figures: 16 = 24 ⎜1 − e ⎟ ⎝ ⎠

t − ⎞ ⎛ If 16 = 24 ⎜ 1 − e 2 ⎟ ⎝ ⎠

from which, and

and

then

t − 16 = 1− e 2 24

e

−

t 2

= 1−

16 24

−

t ⎛ 16 ⎞ = ln ⎜ 1 − ⎟ 2 ⎝ 24 ⎠

⎛ 16 ⎞ t = −2 ln ⎜1 − ⎟ = 2.197, correct to 4 significant figures. ⎝ 24 ⎠ © 2006 John Bird. All rights reserved. Published by Elsevier.

34

⎛ 1.59 ⎞ 7. Solve, correct to 4 significant figures: 3.72 ln ⎜ ⎟ = 2.43 ⎝ x ⎠

⎛ 1.59 ⎞ If 3.72 ln ⎜ ⎟ = 2.43 ⎝ x ⎠

then

⎛ 1.59 ⎞ 2.43 ln ⎜ ⎟= ⎝ x ⎠ 3.72 ⎛ 2.43 ⎞

⎜ ⎟ 1.59 = e⎝ 3.72 ⎠ x

from which,

and

x=

1.59 e

⎛ 2.43 ⎞ ⎜ ⎟ ⎝ 3.72 ⎠

= 1.59e

⎛ 2.43 ⎞ −⎜ ⎟ ⎝ 3.72 ⎠

= 0.8274, correct to 4 significant figures.

8. The work done in an isothermal expansion of a gas from pressure p1 to p 2 is given by: ⎛p ⎞ w = w 0 ln ⎜ 1 ⎟ ⎝ p2 ⎠

If the initial pressure p1 = 7.0 kPa, calculate the final pressure p 2 if w = 3 w 0

If w = 3 w 0 then

⎛p ⎞ 3 w 0 = w 0 ln ⎜ 1 ⎟ ⎝ p2 ⎠

i.e.

⎛p ⎞ 3 = ln ⎜ 1 ⎟ ⎝ p2 ⎠

and

e3 =

from which,

p1 7000 = p2 p2

final pressure, p 2 =

7000 = 7000 e −3 = 348.5 Pa 3 e

© 2006 John Bird. All rights reserved. Published by Elsevier.

35

EXERCISE 22 Page 37 2. The voltage drop, v volts, across an inductor L henrys at time t seconds is given by v = 200e

− Rt L

,

where R = 150 Ω and L = 12.5 ×10−3 H. Determine (a) the voltage when t = 160 ×10−6 s, and (b) the time for the voltage to reach 85 V.

(a) Voltage v = 200e

− Rt L

−

= 200 e

(b) When v = 85 V, 85 = 200 e

and

Thus,

−

−

(150 )(160×10−6 ) 12.5×10−3

= 200 e −1.92 = 29.32 volts

150 t 12.5×10−3

150 t

from which,

− 85 −3 = e 12.5×10 200

150 t ⎛ 85 ⎞ = ln ⎜ ⎟ −3 12.5 ×10 ⎝ 200 ⎠ time t = −

12.5 × 10−3 ⎛ 85 ⎞ −6 ln ⎜ ⎟ = 71.31 × 10 s 150 ⎝ 200 ⎠

4. A belt is in contact with a pulley for a sector θ = 1.12 radians and the coefficient of friction

between the two surfaces is µ = 0.26. Determine the tension on the taut side of the belt, T newtons, when tension on the slack side T0 = 22.7 newtons, given that these quantities are related by the law T = T0 e µθ . Determine also the value of θ when T = 28.0 newtons. Tension T = T0 e µθ = 22.7 e( 0.26)(1.12) = 22.7 e0.2912 = 30.4 N When T = 28.0 N,

Thus,

28.0 = 22.7 e0.26θ ⎛ 28.0 ⎞ 0.26θ = ln ⎜ ⎟ ⎝ 22.7 ⎠

from which,

and

θ =

28.0 = e0.26θ 22.7 1 ⎛ 28.0 ⎞ ln ⎜ ⎟ = 0.807 rad 0.26 ⎝ 22.7 ⎠

5. The instantaneous current i at time t is given by: i = 10e

−

t CR

when a capacitor is being charged.

The capacitance C is 7 ×10−6 F and the resistance R is 0.3 ×106 ohms. Determine: (a) the instantaneous current when t is 2.5 seconds, and © 2006 John Bird. All rights reserved. Published by Elsevier.

36

(b) the time for the instantaneous current to fall to 5 amperes. Sketch a curve of current against time from t = 0 to t = 6 seconds.

−

(a) Current, i = 10 e (b) When i = 5,

2.5 7×10−6 × 0.3×106

5 = 10 e

−

= 3.04 A

t 2.1

from which,

t ⎛ 5⎞ ln ⎜ ⎟ = − 2.1 ⎝ 10 ⎠

Thus, Time t

0

1

and 2

3

t − 5 = e 2.1 10

time, t = −(2.1) ln 0.5 = 1.46 s 4

5

6

Current i 10 6.21 3.86 2.40 1.49 0.92 0.57 A graph of current against time is shown in Figure 3.

Figure 3

t − ⎛ ⎞ CR 7. The current i flowing in a capacitor at time t is given by: i = 12.5 ⎜ 1 − e ⎟ ⎝ ⎠ where resistance R is 30 kilohms and the capacitance C is 20 microfarads. Determine:

(a) the current flowing after 0.5 seconds, and (b) the time for the current to reach 10 amperes.

© 2006 John Bird. All rights reserved. Published by Elsevier.

37

0.5 t − ⎛ − ⎛ ⎞ 20×10−6 ×30×103 CR (a) Current, i = 12.5 ⎜1 − e ⎟ = 12.5 ⎜⎜1 − e ⎝ ⎠ ⎝ t − ⎛ ⎞ 10 = 12.5 ⎜1 − e 0.6 ⎟ ⎝ ⎠

(b) When i = 10 A,

Thus,

i.e.

e

−

t 0.6

= 1−

10 12.5

and

⎞ ⎟⎟ = 7.07 A ⎠

from which,

−

t − 10 = 1 − e 0.6 12.5

t 10 ⎞ ⎛ = ln ⎜1 − ⎟ 0.6 ⎝ 12.5 ⎠

10 ⎞ ⎛ time, t = −0.6 ln ⎜1 − ⎟ = 0.966 s ⎝ 12.5 ⎠

© 2006 John Bird. All rights reserved. Published by Elsevier.

38

EXERCISE 23 Page 40 2. At particular times, t minutes, measurements are made of the temperature, θ °C , of a cooling liquid and the following results are obtained: Temperature θ °C Time t minutes

92.2 55.9 33.9 20.6 12.5 10

20

30

40

50

Prove that the quantities follow a law of the form θ = θ 0 e kt , where θ 0 and k are constants, and determine the approximate value of θ 0 and k.

Since θ = θ 0 e kt i.e.

then

ln θ = ln (θ 0 e k t ) = ln θ 0 + ln e k t ln θ = kt + ln θ 0

where gradient = k and intercept on vertical axis = ln θ 0 Using log-linear graph paper, a graph of ln θ against t is shown in Figure 4

Figure 4 Since the graph is a straight line the quantities do follow a law of the form θ = θ 0 e kt © 2006 John Bird. All rights reserved. Published by Elsevier.

39

Gradient, k =

AB ln 92.2 − ln12.5 = = -0.05 BC 10 − 50

At point A in Figure 4, θ = 92.2°C , t = 10 min Substituting in θ = θ 0 e kt gives: from which,

92.2 = θ 0 e( −0.05)(10)

θ0 =

92.2 = 92.2 e0.5 = 152 −0.5 e

© 2006 John Bird. All rights reserved. Published by Elsevier.

40

CHAPTER 5 HYPERBOLIC FUNCTIONS EXERCISE 24 Page 42

1.(a) Evaluate sh 0.64 correct to 4 significant figures.

(a) sh 0.64 =

1 0.64 −0.64 ( e − e ) = 0.6846, correct to 4 significant figures. 2

Alternatively, using a scientific calculator, using hyp, sin 0.64 = 0.6846

2.(b) Evaluate ch 2.4625 correct to 4 significant figures.

(b) ch 2.4625 =

1 2.4625 −2.4625 ( e + e ) = 5.910, correct to 4 significant figures. 2

Alternatively, using a scientific calculator, using hyp, cos 0.72 = 5.910

3.(a) Evaluate th 0.65 correct to 4 significant figures.

(a) th 0.65 =

e0.65 − e −0.65 1.39349505... = = 0.5717, correct to 4 significant figures. e0.65 + e −0.65 2.4375866...

Alternatively, using a scientific calculator, using hyp, tan 0.65 = 0.5717

4.(b) Evaluate cosech 3.12 correct to 4 significant figures.

(b) cosech 3.12 =

1 = 0.08849, correct to 4 significant figures, using a calculator. sh 3.12

5.(a) Evaluate sech 0.39 correct to 4 significant figures.

(a) sech 0.39 =

1 = 0.9285, correct to 4 significant figures, using a calculator. ch 0.39

© 2006 John Bird. All rights reserved. Published by Elsevier.

41

6.(b) Evaluate coth 1.843 correct to 4 significant figures.

(b) coth 1.843 =

1 = 1.051, correct to 4 significant figures, using a calculator. th1.843

7. A telegraph wire hangs so that its shape is described by y = 50 ch

y . Evaluate, correct to 4 50

significant figures, the value of y when x = 25.

When x = 0.25, y = 50 ch

y 25 = 50 ch = 50 ch 0.50 = 56.38, correct to 4 significant figures. 50 50

8. The length l of a heavy cable hanging under gravity is given by l = 2c sh( L / 2c) . Find the value of l when c = 40 and L = 30.

⎛ 30 ⎞ ⎛3⎞ l = 2c sh( L / 2c) = 2(40) sh ⎜ ⎟ = 80 sh ⎜ ⎟ = 30.71 ⎝8⎠ ⎝ 2(40) ⎠

9. V 2 = 0.55 L tanh(6.3d / L) is a formula for velocity V of waves over the bottom of shallow water, where d is the depth, and L is the wavelength. If d = 8.0 and L = 96, calculate the value of V.

⎡ (6.3)(8.0) ⎤ V 2 = 0.55 L tanh(6.3d / L) = 0.55(96) tanh ⎢ ⎥⎦ = 52.8 tanh 0.525 = 25.425829… ⎣ 96 Hence,

V=

25.425829... = 5.042

© 2006 John Bird. All rights reserved. Published by Elsevier.

42

EXERCISE 25 Page 46 2. Prove the following identities: (a) coth x ≡ 2 cosech 2x + th x (b) ch 2θ - 1 ≡ 2 sh 2 θ

(a) R.H.S. = 2 cosech 2x + th x =

2 sh x 2 sh x 1 sh x 1 + sh 2 x + = + = + = sh 2 x ch x 2 sh x ch x ch x sh x ch x ch x sh x ch x =

ch 2 x ch x = = coth x = L.H.S sh x ch x sh x

2

⎛ eθ − e −θ ⎞ 2 θ 1 2θ θ −θ θ −θ −θ −θ −2θ θ (b) R.H.S. = 2 sh θ = 2 ⎜ ⎟ = ( e − e )( e − e ) = ⎡⎣e − e e − e e + e ⎤⎦ 4 2 ⎝ 2 ⎠ 2

=

1 2θ 1 e 2θ e −2θ 2 ⎡⎣e − e0 − e0 + e −2θ ⎦⎤ = ⎣⎡e 2θ − 2 + e−2θ ⎦⎤ = + − 2 2 2 2 2

=

e 2θ + e −2θ − 1 = ch 2θ − 1 = L.H.S. 2

4. Prove the following identities: (a) sh(A + B) ≡ sh A ch B + ch A sh B sh 2 x + ch 2 x − 1 (b) ≡ tanh 4 x 2 2 2ch x coth x

⎛ e A − e− A ⎞ ⎛ e B + e− B ⎞ ⎛ e A + e− A ⎞ ⎛ e B − e− B ⎞ (a) R.H.S. = sh A ch B + ch A sh B = ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟ 2 2 2 2 ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

(b) L.H.S. =

=

1 A+ B ⎡⎣e + e A− B − e− A+ B − e− A− B + e A+ B − e A− B + e − A+ B − e− A− B ⎤⎦ 4

=

( A+ B ) 1 ⎡ A+ B − e −( A+ B ) −( A+ B ) ⎤=e − = sh ( A + B) = L.H.S 2 e 2 e ⎦ 4⎣ 2

sh 2 x + ch 2 x − 1 sh 2 x + sh 2 x = 2ch 2 x coth 2 x ⎛ ch 2 x ⎞ 2ch 2 x ⎜ 2 ⎟ ⎝ sh x ⎠ =

since ch 2 x − 1 = sh 2 x

2 sh 2 x ⎛ sh 2 x ⎞ sh 4 x = tanh 4 x = R.H.S. ⎜ ⎟= 2 ch 2 x ⎝ ch 2 x ⎠ ch 4 x

© 2006 John Bird. All rights reserved. Published by Elsevier.

43

5. Given P e x − Q e − x ≡ 6 ch x − 2 sh x , find P and Q

⎛ e x + e− x P e x − Q e − x ≡ 6 ch x − 2 sh x = 6 ⎜ ⎝ 2

⎞ ⎛ e x − e− x ⎟ − 2⎜ ⎠ ⎝ 2

⎞ x −x x −x ⎟ = 3(e + e ) − (e − e ) ⎠

= 3e x + 3e − x − e x + e − x P e x − Q e − x = 2e x + 4e − x

i.e. from which,

P = 2 and Q = -4

6. If 5 e x − 4 e − x ≡ A sh x + B ch x , find A and B

⎛ e x − e− x 5 e x − 4 e − x ≡ A sh x + B ch x = A ⎜ ⎝ 2

⎞ ⎛ e x + e− x B + ⎟ ⎜ ⎠ ⎝ 2

⎞ A x A −x B x B −x ⎟= e − e + e + e 2 2 2 ⎠ 2

⎛ A B⎞ ⎛ A B⎞ = ⎜ + ⎟ e x − ⎜ − ⎟ e− x ⎝2 2⎠ ⎝2 2⎠ i.e.

⎛ A + B ⎞ x ⎛ A − B ⎞ −x 5 e x − 4 e− x = ⎜ ⎟e −⎜ ⎟e ⎝ 2 ⎠ ⎝ 2 ⎠

Hence,

5=

A+ B 2

i.e.

A + B = 10

(1)

and

4=

A− B 2

i.e.

A–B=8

(2)

(1) + (2) gives: From (1)

2A = 18 from which, A = 9

B=1

© 2006 John Bird. All rights reserved. Published by Elsevier.

44

EXERCISE 26 Page 48

2. Solve 2 ch x = 3, correct to 4 decimal places.

2 ch x = 3 from which, ch x = i.e.

i.e.

Hence,

e x + e− x 3 = 2 2

(e )

x 2

3 2 e x + e− x − 3 = 0

or

+ e − x e x − 3e x = 0

(e )

x 2

or

− 3e x + 1 = 0

2 − ( −3) ± ⎡( −3) − 4 (1)(1) ⎤ 3 ± 5 ⎣ ⎦ = e = = 2.61803 or 0.381966 2 (1) 2 x

Thus,

x = ln 2.61803 or

x = ln 0.381966

i.e.

x = 0.9624

x = -0.9624

or

i.e.

x = ± 0.9624

3. Solve 3.5 sh x + 2.5 ch x = 0, correct to 4 decimal places.

3.5 sh x + 2.5 ch x = 0 i.e.

⎛ e x − e− x 3.5 ⎜ ⎝ 2

i.e.

1.75 e x − 1.75 e − x + 1.25 e x + 1.25 e− x = 0

and

⎞ ⎛ e x + e− x 2.5 + ⎟ ⎜ ⎠ ⎝ 2

3 e x − 0.5 e − x = 0

i.e.

e x 0.5 = e− x 3

Hence,

2x = ln

0.5 3

⎞ ⎟=0 ⎠

3 e x = 0.5 e − x

or i.e.

e2 x =

from which,

0.5 3

x=

1 0.5 ln = -0.8959 2 3

5. Solve 4 th x - 1 = 0, correct to 4 decimal places.

4 th x - 1 = 0

i.e.

⎛ e x − e− x 4 ⎜ x −x ⎝e +e

⎞ ⎟ =1 ⎠

© 2006 John Bird. All rights reserved. Published by Elsevier.

45

4 ( e x − e− x ) = e x + e− x

i.e.

3e x − 5e− x = 0

Hence, Thus,

ex 5 = e− x 3

i.e.

2x = ln

4e x − 4e− x − e x − e− x = 0

and

3e x = 5e− x

and

e2 x =

from which, 5 3

and

x=

5 3

1 5 ln = 0.2554 2 3

6. A chain hangs so that its shape is of the form y = 56 ch ( x / 56) . Determine, correct to 4 significant figures, (a) the value of y when x is 35, and (b) the value of x when y is 62.35

⎛ 35 ⎞ (a) When x = 35, y = 56 ch ( x / 56) = 56 ch ⎜ ⎟ = 67.30, using a calculator. ⎝ 56 ⎠ (b) When, y = 62.35, then

62.35 = 56 ch ( x / 56) x

62.35 x = ch 56 56

Thus, x 56

e +e

i.e.

−

x 56

e 56 + e 2

or

−

x 56

=

62.35 56

⎛ 62.35 ⎞ = 2⎜ ⎟ = 2.22679 ⎝ 56 ⎠

2

Thus,

x ⎛ 56x ⎞ ⎛ 56x ⎞ ⎛ − 56x ⎞ 56 ⎜ e ⎟ + ⎜ e ⎟ ⎜ e ⎟ − 2.22679e = 0 ⎝ ⎠ ⎝ ⎠⎝ ⎠

i.e.

x ⎛ 56x ⎞ 56 ⎜ e ⎟ − 2.22679e + 1 = 0 ⎝ ⎠

2

from which,

e

x 56

2 − ( −2.22679 ) ± ⎡( −2.22679 ) − 4 (1)(1) ⎤ 2.22679 ± 0.95859... ⎣ ⎦ = = 2 (1) 2

= 1.60293 or 0.623857 Hence, i.e.

x = ln1.60293 56

x = ln 0.623857 56

or

x = 56 ln 1.60293 = 26.42

or

x = 56 ln 0.623857 = -26.42, which is not possible.

© 2006 John Bird. All rights reserved. Published by Elsevier.

46

EXERCISE 27 Page 49

3. Expand the following as a power series as far as the term in x 5 : (a) sh 3x (b) ch 2x

(a) sh 3x = ( 3 x )

( 3x ) + 3!

3

( 3x ) +

5

5!

27 3 3 × 3 × 3 × 3 × 3 5 x + x 6 5 × 4 × 3 × 2 ×1

+ ... = 3x + = 3x +

(b)

( 2x ) ch 2x = 1 +

2

2!

(2x) + 4!

9 3 81 5 x + x as far as the term in x5 2 40

4

+ ... = 1 + 2 x 2 +

16 4 x + ... 24 2 4 x as far as the term in x 4 3

= 1 + 2 x2 +

7 31 5 4. Prove the identity: sh 2θ − sh θ ≡ θ + θ 3 + θ as far as the term in θ 5 only. 6 120 3 5 ⎛ ⎞ ⎛ 2θ ) ( 2θ ) ⎞ ( θ3 θ5 L.H.S. = sh 2θ − sh θ ≡ ⎜ 2θ + + + ... ⎟ − ⎜ θ + + + ... ⎟ ⎜ ⎟ ⎝ 3! 5! 3! 5! ⎠ ⎝ ⎠

8 32 5 1 1 5 ⎛ ⎞ ⎛ ⎞ θ + ... ⎟ − ⎜ θ + θ 3 + θ + ... ⎟ = ⎜ 2θ + θ 3 + 6 120 6 120 ⎝ ⎠ ⎝ ⎠ 1 ⎞ 5 ⎛8 1⎞ ⎛ 32 5 = ( 2θ − θ ) + ⎜ − ⎟ θ 3 + ⎜ − ⎟ θ as far as the term in θ only ⎝6 6⎠ ⎝ 120 120 ⎠ 7 31 5 = θ + θ3 + θ = R.H.S. 6 120

5. Prove the identity: 2 sh

θ 2

− ch

θ 2

≡ −1 + θ −

θ2 8

+

θ3

−

θ4

+

θ5

24 384 1920

as far as the term in θ 5 only.

⎛ θ (θ / 2 )3 (θ / 2 )5 ⎞ ⎛ (θ / 2 )2 (θ / 2 )4 ⎞ L.H.S. = 2 sh − ch ≡ 2 ⎜ + + + ... ⎟ − ⎜1 + + + ... ⎟ ⎜2 ⎟ ⎜ ⎟ 2 2 3! 5! 2! 4! ⎝ ⎠ ⎝ ⎠

θ

θ

⎛ ⎞ ⎛ 1 ⎞ 2 2 1 θ 5 ⎟ − ⎜1 + θ 2 + θ4⎟ = ⎜θ + θ 3 + 5 48 2 (120) ⎠ ⎝ 8 (16)(24) ⎠ ⎝ = −1 + θ −

θ2 8

+

θ3 24

+

θ4 384

+

θ5 1920

as far as the term in θ 5 only

© 2006 John Bird. All rights reserved. Published by Elsevier.

47

CHAPTER 6 ARITHMETIC AND GEOMETRIC PROGRESSIONS EXERCISE 28 Page 52 1. Find the 11th term of the series 8, 14, 20, 26, … The 11th term of the series 8, 14, 20, 26,… is given by: a + (n – 1)d

where a = 8, n = 11 and d = 6

Hence, the 11th term is: 8 + (11 – 1)(6) = 8 + 60 = 68

3. The seventh term of a series is 29 and the eleventh term is 54. Determine the sixteenth term.

The n’th term of an arithmetic progression is: a + (n – 1)d The 7th term is:

a + 6d = 29

(1)

The 11th term is:

a + 10d = 54

(2)

(2) – (1) gives:

4d = 25

25 4

from which,

d=

from which,

a = 29 – 37.5 = -8.5

⎛ 25 ⎞ Substituting in (1) gives: a + 6 ⎜ ⎟ = 29 ⎝ 4 ⎠ i.e.

a + 37.5 = 29

Hence, the 16th term is:

⎛ 25 ⎞ -8.5 + (16 – 1) ⎜ ⎟ = -8.5 + 93.75 = 85.25 ⎝ 4 ⎠

5. Determine the number of the term which is 29 in the series 7, 9.2, 11.4, 13.6, …

29 = 7 + (n – 1)(2.2) i.e.

22 =n–1 2.2

i.e.

from which, 10 = n – 1

29 – 7 = 2.2(n – 1)

and n = 11

i.e. 29 is the 11th term of the series.

7. Determine the sum of the series 6.5, 8.0, 9.5, 11.0, …, 32

© 2006 John Bird. All rights reserved. Published by Elsevier.

48

In the series: 6.5, 8.0, 9.5, 11.0, …, 32, a = 6.5 and d = 1.5 The n’th term is 32, hence, a + (n – 1)d = 32

i.e.

from which,

25.5 =n–1 1.5

32 – 6.5 = (n – 1)(1.5)

Sum of series, Sn =

and

6.5 + (n - 1)(1.5) = 32 i.e. 17 = n - 1

and

n = 18

n 18 [ 2a + (n − 1)d ] = [ 2(6.5) + (18 − 1)(1.5)] = 9 [13 + 25.5] = 346.5 2 2

© 2006 John Bird. All rights reserved. Published by Elsevier.

49

EXERCISE 29 Page 53 2. Three numbers are in arithmetic progression. Their sum is 9 and their product is 20

1 . 4

Determine the three numbers.

Let the three numbers be (a – d), a and (a + d) Thus,

(a – d) + a + ( a + d) = 9

i.e. Also,

3a = 9

and

a=3

a(a – d)(a + d) = 20.25

Since, a = 3, then 3 ( 9 − d 2 ) = 20.25

20.25 = 6.75 3

i.e.

9 − d2 =

and

9 – 6.75 = d 2

from which,

d 2 = 2.25

and

d=

2.25 = 1.5

Hence, the three numbers are: (a – d) = 3 – 1.5 = 1.5, a = 3 and (a + d) = 3 + 1.5 = 4.5

4. Find the number of terms of the series 5, 8, 11,… of which the sum is 1025

Sum of n terms is given by:

i.e. i.e. Hence, i.e.

Sn =

1025 =

n [ 2(5) + (n − 1)(3)] 2

2 ×1025 = n [10 + 3(n − 1) ]

2050 = n [10 + 3n − 3] = n [ 7 + 3n ] = 7n + 3n 2 3n 2 + 7n − 2050 = 0

This is a quadratic equation, hence i.e.

n [ 2a + (n − 1)d ] 2

n=

−7 ± 7 2 − 4(3)(−2050) −7 ± 24649 −7 ± 157 = = 2(3) 6 6

number of terms, n = 25 (the negative answer having no meaning)

© 2006 John Bird. All rights reserved. Published by Elsevier.

50

6. The first, tenth and last terms of an arithmetic progression are 9, 40.5 and 425.5 respectively.

Find (a) the number of terms, (b) the sum of all the terms, and (c) the 70th term. (a) a = 9 and the 10th term is: a + (10 – 1)d = 40.5 i.e.

9 + 9d = 40.5 d=

hence

and

9d = 40.5 – 9 = 31.5

31.5 = 3.5 9

Last term is given by: a + (n – 1)d i.e.

9 + (n – 1)(3.5) = 425.5

i.e.

(n – 1)(3.5) = 425.5 – 9 = 416.5

and Hence,

n–1=

416.5 = 119 3.5

the number of terms, n = 120

(b) Sum of all the terms, Sn =

n 120 [ 2a + (n − 1)d ] = [ 2(9) + (120 − 1)(3.5)] = 60 [18 + 416.5] 2 2 = 26070

(c) The 70th term is: a + (n – 1)d = 9 + (70 – 1)(3.5) = 9 + 69(3.5) = 250.5

8. An oil company bores a hole 80 m deep. Estimate the cost of boring if the cost is £30 for drilling

the first metre with an increase in cost of £2 per metre for each succeeding metre.

The series is: 30, 32, 34, … to 80 terms, Thus, total cost, Sn =

i.e. a = 30, d = 2 and n = 80

n 80 [ 2a + (n − 1)d ] = [ 2(30) + (80 − 1)(2)] = 40 [ 60 + 158] = 40(218) = £8720 2 2

© 2006 John Bird. All rights reserved. Published by Elsevier.

51

EXERCISE 30 Page 55 1. Find the 10th term of the series 5, 10, 20, 40, …

The 10th term of the series 5, 10, 20, 40, … is given by: a r n −1 where a = 5, r = 2 and n = 10 i.e. the 10th term = a r n −1 = ( 5 )( 2 )

10 −1

= (5)(2)9 = 5(512) = 2560

3. The first term of a geometric progression is 4 and the 6th term is 128. Determine the 8th and 11th

terms.

The 6th term is given by:

a r 5 = 128

Thus,

r=

Hence, the 8th term is: and the 11th term is:

5

i.e. 4r 5 = 128

and

r5 =

128 = 32 4

32 = 2

a r n −1 = 4(2)8−1 = 4(2)7 = 4(128) = 512 4(2)11−1 = 4(2)10 = 4(1024) = 4096

1 4. Find the sum of the first 7 terms of the series 2, 5, 12 ,… (correct to 4 significant figures). 2

Common ratio, r =

ar 5 = = 2.5 a 2

Sum of 7 terms, Sn =

a ( r n − 1)

( r − 1)

(Also,

=

ar 2 12.5 = = 2.5) ar 5

2 ( 2.57 − 1)

( 2.5 − 1)

=

2 ( 610.35 − 1) = 812.5, correct to 4 significant 1.5 figures.

1 1 5 6. Find the sum to infinity of the series 2 , − 1 , ,..... 2 4 8

The series is a G.P. where r = − Hence, sum to infinity, S∞ =

1.25 = -0.5 and a = 2.5 2.5

2 a 2.5 2.5 5 = = = =1 3 1 − r 1 − (−0.5) 1.5 3

© 2006 John Bird. All rights reserved. Published by Elsevier.

52

EXERCISE 31 Page 57 1. In a geometric progression the 5th term is 9 times the 3rd term, and the sum of the 6th and 7th

terms is 1944. Determine (a) the common ratio, (b) the first term, and (c) the sum of the 4th to 10th terms inclusive.

(a) The 5th term of a geometric progression is: ar 4 and the 3rd term is: ar 2 r4 Hence, ar 4 = 9 ar 2 from which, = 9 i.e. r2 = 9 r2 from which, the common ratio, r = 3 (b) The 6th term is ar 5 and the 7th term is ar 6 ar 5 + ar 6 = 1944

Hence, Since r = 3,

243a + 729a = 1944

i.e.

972a = 1944

and first term, a =

1944 =2 972

(c) Sum of the 4th to 10th terms inclusive is given by: S10 − S3 =

a ( r10 − 1)

( r − 1)

−

a ( r 3 − 1)

( r − 1)

=

2 ( 310 − 1) (3 − 1)

−

2 ( 33 − 1)

( 3 − 1)

= ( 310 − 1) − ( 33 − 1) = 310 − 33 = 59049 − 27 = 59022

4. If the population of Great Britain is 55 million and is decreasing at 2.4% per annum, what will

be the population in 5 years time? G.B. population now = 55 million, population after 1 year = 0.976 × 55 million, population after 2 years = ( 0.976 ) × 55 million 2

Hence, population after 5 years = ( 0.976 ) × 55 = 48.71 million 5

6. If £250 is invested at compound interest of 6% per annum, determine (a) the value after 15 years,

(b) the time, correct to the nearest year, it takes to reach £750. © 2006 John Bird. All rights reserved. Published by Elsevier.

53

(a) First term, a = £250,

common ratio, r = 1.06

Hence, value after 15 years = ar15 = (250) (1.06 ) = £599.14 15

(b) When £750 is reached, 750 = ar n i.e.

750 = 250 (1.06 )

and

750 = 1.06n 250

Taking logarithms gives: from which,

n

i.e.

3 = 1.06n

lg 3 = lg (1.06 ) = n lg1.06 n

n=

lg 3 = 18.85 lg1.06

Hence, it will take 19 years to reach more than £750

7. A drilling machine is to have 8 speeds ranging from 100 rev/min to 1000 rev/min. If the speeds

form a geometric progression determine their values, each correct to the nearest whole number.

First term, a = 100 rev/min The 8th term is given by:

ar 8−1 = 1000

from which,

r7 =

1000 = 10 100

and r = 7 10 = 1.3895

Hence, 1st term is 100 rev/min 2nd term is ar = (100)(1.3895) = 138.95 3rd term is ar 2 = (100)(1.3895) 2 = 193.07 4th term is ar 3 = (100)(1.3895)3 = 268.27 5th term is ar 4 = (100)(1.3895) 4 = 372.76 6th term is ar 5 = (100)(1.3895)5 = 517.96 7th term is ar 6 = (100)(1.3895)6 = 719.70 8th term is ar 7 = (100)(1.3895)7 = 1000 Hence, correct to the nearest whole numbers, the eight speeds are: 100, 139, 193, 268, 373, 518, 720 and 1000 rev/min

© 2006 John Bird. All rights reserved. Published by Elsevier.

54

CHAPTER 7 THE BINOMIAL SERIES EXERCISE 32 Page 59 1. Expand ( 2a + 3b ) using Pascal’s triangle 5

From page 58 of textbook,

(a + x )

5

= a 5 + 5a 4 x + 10a 3 x 2 + 10a 2 x 3 + 5a x 4 + x 5

Thus, replacing a with 2a and x with 3b gives:

( 2a + 3b )

5

= ( 2a ) + 5 ( 2a ) ( 3b ) + 10 ( 2a ) ( 3b ) + 10 ( 2a ) ( 3b ) + 5 ( 2a )( 3b ) + ( 3b ) 5

4

3

2

2

3

4

5

= 32a5 + 240a4b + 720a 3b 2 + 1080a 2b 3 + 810ab 4 + 243b 5

© 2006 John Bird. All rights reserved. Published by Elsevier.

55

EXERCISE 33 Page 61 1. Use the binomial theorem to expand ( a + 2 x )

( a + 2x)

= a 4 + 4a 3 ( 2x ) +

4

( 4 )( 3) a 2 2!

( 2x )

2

+

4

( 4 )( 3)( 2 ) a 3!

( 2x ) + ( 2x ) 3

4

= a4 + 8a 3 x + 24a 2 x 2 + 32a x 3 + 16x 4

3. Expand ( 2 x − 3 y )

( 2x − 3y )

4

4

= ( 2 x ) + 4 ( 2 x ) ( −3 y ) + 4

3

( 4 )( 3) 2!

( 2 x ) ( −3 y ) 2

2

+

( 4 )( 3)( 2 ) 3!

( 2 x )( −3 y ) + ( −3 y ) 3

4

= 16 x 4 − 96 x 3 y + 216 x 2 y 2 − 216 xy 3 + 81 y 4

2⎞ ⎛ 4. Determine the expansion of ⎜ 2 x + ⎟ x⎠ ⎝

5

( 5)( 4 ) 2 x 3 ⎛ 2 ⎞ + ( 5)( 4 )( 3) 2 x 2 ⎛ 2 ⎞ 2⎞ 5 4⎛2⎞ ⎛ ( )⎜ ⎟ ( )⎜ ⎟ ⎜ 2x + ⎟ = ( 2x) + 5 ( 2x ) ⎜ ⎟ + 2! 3! x⎠ ⎝ ⎝ x⎠ ⎝ x⎠ ⎝ x⎠ 5

2

3

( 5)( 4 )( 3)( 2 ) + 4!

= 32 x 5 + 160 x 3 + 320 x +

4

2 2 ( 2 x ) ⎛⎜ ⎞⎟ + ⎛⎜ ⎞⎟ ⎝ x⎠ ⎝ x⎠

5

320 160 32 + 3 + 5 x x x 13

q⎞ ⎛ 6. Determine the sixth term of ⎜ 3 p + ⎟ 3⎠ ⎝ The 6th term of

(a + x)

n

is

n ( n − 1)( n − 2 ) ..... to (r − 1) terms n −( r −1) r −1 a x ( r − 1)! 13

q⎞ ⎛ Hence, the 6 term of ⎜ 3 p + ⎟ is: 3⎠ ⎝ th

(13)(12 )(11)(10 )( 9 ) 5!

(3 p )

13− 5

5

8⎛q⎞ ⎛q⎞ ⎜ ⎟ = 1287 ( 3 p ) ⎜ ⎟ ⎝3⎠ ⎝3⎠

5

= 34749 p8 q 5

© 2006 John Bird. All rights reserved. Published by Elsevier.

56

9. Use the binomial theorem to determine, correct to 5 significant figures:

( 0.98)

(a)

7

(b)

( 2.01)

9

(a)

( 0.98)

7

= (1 − 0.02 ) = 1 + 7 ( −0.02 ) + 7

( 7 )( 6 ) 2!

( −0.02 )

2

+

( 7 )( 6 )( 5) 3!

( −0.02 )

3

+

( 7 )( 6 )( 5)( 4 ) 4!

( −0.02 )

4

= 1 – 0.14 + 0.0084 – 0.00028 + 0.0000056 - … = 0.86813, correct to 5 significant figures. 9

9 ⎛ 0.01 ⎞ 9 (b) ( 2.01) = ( 2 + 0.01) = 2 ⎜1 + ⎟ = 2 (1 + 0.005 ) 2 ⎠ ⎝ 9

9

9

⎡ ( 9 )( 8) (0.005)2 + ( 9 )( 8)( 7 ) (0.005)3 + ...⎤ = 29 ⎢1 + 9(0.005) + ⎥ 2! 3! ⎣ ⎦

= 29 (1 + 0.045 + 0.0009 + 0.0000105 + ...) = 29 (1.0459105) = 535.51, correct to 5 significant figures.

© 2006 John Bird. All rights reserved. Published by Elsevier.

57

+ ..

EXERCISE 34 Page 63

2. Expand

1

(1+ x )

in ascending powers of x as far as the term in x3 , using the binomial theorem.

2

Sate the limits of x for which the series is valid.

1

(1 + x )

= (1 + x ) = 1 − 2 x + −2

2

( −2 )( −3) 2!

( x)

2

+

( −2 )( −3)( −4 ) 3!

= 1 − 2 x + 3 x 2 − 4 x 2 + ...

3. Expand

1

(2 + x)

3

and

( x)

3

+ ...

x µ

The alternative hypothesis, H1 : x = µ z=

x − µ 2050 − 2000 50 = =± = ± 3.54 100 σ 14.142 N 50

The value for a two-tailed test is given in Table 62.1, page 594 of textbook, and at a significance level of 0.01 is ± 2.58. Since the z-value of the sample is outside of this range, the hypothesis is rejected.

3. The internal diameter of a pipe has a mean diameter of 3.0000 cm with a standard deviation of

0.015 cm. A random sample of 30 measurements are taken and the mean of the samples is 3.0078 cm. Test the hypothesis that the mean diameter of the pipe is 3.0000 cm at a level of significance of 0.01. µ = 3.0000 cm, σ = 0.015 cm, N = 30, x = 3.0078cm The null hypothesis,

H 0 : mean diameter = 3.0000 cm

The alternative hypothesis, H1 : mean diameter ≠ 3.0000 cm z=

x − µ 3.0078 − 3.0000 0.0078 = =± = ± 2.85 0.015 σ 0.00274 N 30

The value for a two-tailed test is given in Table 62.1, page 594 of textbook, and at a significance level of 0.01 is ± 2.58. Since the z-value of the sample is outside of this range, the hypothesis is rejected.

© 2006 John Bird. All rights reserved. Published by Elsevier.

546

4. A fishing line has a mean breaking strength of 10.25 kN. Following a special treatment on the

line, the following results are obtained for 20 specimens taken from the line. Breaking strength (kN) 9.8 10.0 10.1 10.2 10.5 10.7 10.8 10.9 11.0 Frequency

1

1

4

5

3

2

2

1

1

Test the hypothesis that special treatment has improved the breaking strength at a level of significance of 0.1. µ = 10.25 kN and N = 20 Sample mean, x =

( 9.8 ×1) + (10.0 ×1) + (10.1× 4 ) + (10.2 × 5) + .... = 207.6 = 10.38 20

20

Sample standard deviation, 1( 9.8 − 10.38 ) + 1(10.0 − 10.38 ) + 4 (10.1 − 10.38 ) + 5 (10.2 − 10.38 ) + ... 2.212 = = 0.33 20 20 2

s=

2

2

2

The null hypothesis is that the sample breaking strength is better than the mean breaking strength. N < 30, therefore a t distribution is used. t =

(x − µ)

( N − 1) (10.38 − 10.25) ( 20 − 1) s

=

0.33

At a level of significance of 0.1, the t value is t ⎛

0.1 ⎞ ⎜1 − ⎟ 2 ⎠ ⎝

= 1.72

i.e. t 0.95 and ν = N – 1 = 20 – 1 = 19, and

from Table 61.1, page 587 of textbook, t 0.95 ν = 19 has a value of 1.73 Since 1.72 is within this range, the hypothesis is accepted.

5. A machine produces ball bearings having a mean diameter of 0.50 cm. A sample of 10 ball

bearings is drawn at random and the sample mean is 0.53 cm with a standard deviation of 0.03 cm. Test the hypothesis that the mean diameter is 0.50 cm at a level of significance of (a) 0.05 and (b) 0.01. µ = 0.50 cm, N = 10, x = 0.53 cm and s = 0.03 cm The null hypothesis is: H 0 : µ = 0.50 cm

© 2006 John Bird. All rights reserved. Published by Elsevier.

547

t =

( x − µ)

( N − 1) ( 0.53 − 0.50 ) (10 − 1) =

s

0.03

(a) From Table 61.1, page 587 of the textbook, t ⎛

=3

0.05 ⎞ ⎜1 − ⎟ 2 ⎠ ⎝

= t 0.975 , ν = 9 has a value of 2.26

Since 3 is outside of this range, the hypothesis is rejected. (b) From Table 61.1, t ⎛

0.01 ⎞ ⎜1 − ⎟ 2 ⎠ ⎝

= t 0.995 , ν = 9 has a value of 3.25

Since 3 is within this range, the hypothesis is accepted.

© 2006 John Bird. All rights reserved. Published by Elsevier.

548

EXERCISE 225 Page 605 1. A comparison is being made between batteries used in calculators. Batteries of type A have a

mean lifetime of 24 hours with a standard deviation of 4 hours, this data being calculated from a sample of 100 of the batteries. A sample of 80 of the type B batteries has a mean lifetime of 40 hours with a standard deviation of 6 hours. Test the hypothesis that the type B batteries have a mean lifetime of at least 15 hours more than those of type A, at a level of significance of 0.05. Battery A: x A = 24 , σ A = 4 and N A = 100 Battery B: x B = 40 , σ B = 6 and N B = 80

The hypothesis is: H: x = x A + 15 Let x = 24 + 15 = 39 z=

x − xB ⎛ σA 2 σB2 ⎞ + ⎜ ⎟ ⎝ NA NB ⎠

=

39 − 40 ⎛ 42 62 ⎞ + ⎟ ⎜ ⎝ 100 80 ⎠

=

−1 = −1.28 0.781025

From Table 62.1, page 594 of textbook, for α = 0.05, one-tailed test, z = 1.645 Since the z-value is within this range, the hypothesis is accepted.

3. Capacitors having a nominal capacitance of 24 µF but produced by two different companies are tested. The values of actual capacitances are: Company 1 21.4 23.6 24.8 22.4 26.3 Company 2 22.4 27.7 23.5 29.1 25.8 Test the hypothesis that the mean capacitance of capacitors produced by company 2 are higher than those produced by company 1 at a level of significance of 0.01. 2 $2 = s N ) (Bessel’s correction is σ N −1

N = 5, x1 =

21.4 + 23.6 + 24.8 + 22.4 + 26.3 = 23.7 5

⎛ ( 21.4 − 23.7 )2 + ( 23.6 − 23.7 )2 + ( 24.8 − 23.7 )2 + ( 22.4 − 23.7 )2 + ( 26.3 − 23.7 )2 ⎞ s1 = ⎜ ⎟ = 1.73 ⎜ ⎟ 5 ⎝ ⎠

© 2006 John Bird. All rights reserved. Published by Elsevier.

549

⎛ N ⎞ ⎛5⎞ σ1 = s1 ⎜ ⎟ = 1.73 ⎜ ⎟ = 1.93 ⎝ N −1 ⎠ ⎝4⎠

x2 =

22.4 + 27.7 + 23.5 + 29.1 + 25.8 = 25.7 5

⎛ ( 22.4 − 25.7 )2 + ( 27.7 − 25.7 )2 + ( 23.5 − 25.7 )2 + ( 29.1 − 25.7 )2 + ( 25.8 − 25.7 )2 ⎞ s2 = ⎜ ⎟ = 2.50 ⎜ ⎟ 5 ⎝ ⎠ ⎛ N ⎞ ⎛5⎞ σ2 = s2 ⎜ ⎟ = 2.50 ⎜ ⎟ = 2.80 ⎝ N −1 ⎠ ⎝4⎠

t =

t⎛

x 2 − x1 ⎛ σx 2 σy 2 ⎞ + ⎜⎜ ⎟ N N y ⎟⎠ x ⎝

0.01 ⎞ ⎜1 − ⎟ 2 ⎠ ⎝

=

25.7 − 23.7 ⎛ 1.932 2.802 ⎞ + ⎜ ⎟ 5 ⎠ ⎝ 5

=

2 = 1.32 1.5208

= t 0.995 and ν = N1 + N 2 − 2 = 5 + 5 − 2 = 8

From Table 61.2, page 587 of textbook, t 0.995 , ν = 8 has a value of 3.36 Since the t value of the difference of the means, i.e. 1.32, is within the range ± 3.36, the hypothesis is accepted.

5. A sample of 12 car engines produced by manufacturer A showed that the mean petrol

consumption over a measured distance was 4.8 litres with a standard deviation of 0.40 litres. Twelve similar engines for manufacturer B were tested over the same distance and the mean petrol consumption was 5.1 litres with a standard deviation of 0.36 litres. Test the hypothesis that the engines produced by manufacturer A are more economical than those produced by manufacturer B at a level of significance of (a) 0.01 and (b) 0.1.

N A = 12, x A = 4.8 litre, σ A = 0.40 litre N B = 12, x B = 5.1 litre, σ B = 0.36 litre

The hypothesis is: H: manufacturer A is more economical than manufacturer B. ⎛ 12 ( 0.40 )2 + 12 ( 0.36 )2 ⎞ ⎛ N A sA 2 + N B sB2 ⎞ 3.4752 σ= ⎜ = 0.397 ⎟= ⎟ = ⎜⎜ ⎟ 12 + 12 − 2 22 ⎝ NA + NB − 2 ⎠ ⎝ ⎠ © 2006 John Bird. All rights reserved. Published by Elsevier.

550

t =

xA − xB ⎛ 1 1 ⎞ σ ⎜ + ⎟ ⎝ NA NB ⎠

4.8 − 5.1

=

⎛1 1⎞ 0.397 ⎜ + ⎟ ⎝ 12 12 ⎠

(a) The t value is t ⎛

0.01 ⎞ ⎜1 − ⎟ 2 ⎠ ⎝

=−

0.3 = −1.85 ( 0.397 )( 0.40825)

i.e. t 0.995 and ν = 12 + 12 – 2 = 22, hence, from Table 61.2, page 587 of

textbook, t 0.995 , ν = 22 has a value of 2.82 Since the t value of the difference of the means is outside the range ± 1.85, the hypothesis is rejected.

(b) From Table 61.2, t ⎛

0.1 ⎞ ⎜1 − ⎟ 2 ⎠ ⎝

i.e. t 0.95 , ν = 22 has a value of 1.72

Since the t value of the difference of the means is within the range ± 1.85, the hypothesis is accepted.

6. Four-star and unleaded petrol is tested in 5 similar cars under identical conditions. For four-star

petrol, the cars covered a mean distance of 21.4 kilometres with a standard deviation of 0.54 kilometres for a given mass of petrol. For the same mass of unleaded petrol the mean distance covered was 22.6 kilometres with a standard deviation of 0.48 kilometres. Test the hypothesis that unleaded petrol gives more kilometres per litre than four-star petrol at a level of significance of 0.1. N = 5, x 4 = 21.4 km, s 4 = 0.54 km x un = 22.6 km, s un = 0.48 km

The hypothesis is: H: x un > x 4 ⎛ 5 ( 0.54 )2 + 5 ( 0.48 )2 ⎞ ⎛ N 4 s 4 2 + N un s un 2 ⎞ 2.61 σ= ⎜ = 0.571 ⎟= ⎟ = ⎜⎜ ⎟ 5 + 5 − 2 8 ⎝ N 4 + N un − 2 ⎠ ⎝ ⎠ t =

x un − x 4 ⎛ 1 1 ⎞ σ ⎜ + ⎟ ⎝ N un N 4 ⎠

=

22.6 − 21.4 ⎛1 1⎞ 0.571 ⎜ + ⎟ ⎝5 5⎠

=

1.2 = 3.32 ( 0.571)( 0.63246 )

© 2006 John Bird. All rights reserved. Published by Elsevier.

551

The t value is t ⎛

0.1 ⎞ ⎜1 − ⎟ 2 ⎠ ⎝

i.e. t 0.95 and ν = 5 + 5 – 2 = 8, hence, from Table 61.2, page 587 of textbook,

t 0.95 , ν = 8 has a value of 1.86. Since the t value of the difference of the means, i.e. 3.32, is outside of the range ± 3.32, the hypothesis is rejected.

© 2006 John Bird. All rights reserved. Published by Elsevier.

552

CHAPTER 63 CHI-SQUARE AND DISTRIBUTION-FREE TESTS EXERCISE 226 Page 607 1. A dice is rolled 240 times and the observed and expected frequencies are as shown.

Face

Observed frequency Expected frequency

1

49

40

2

35

40

3

32

40

4

46

40

5

49

40

6

29

40

Determine the χ2-value for this distribution.

Face

o-e

(o − e)

(o − e)

Observed

Expected

frequency, o

frequency, e

1

49

40

9

81

2.025

2

35

40

-5

25

0.625

3

32

40

-8

64

1.6

4

46

40

6

36

0.9

5

49

40

9

81

2.025

6

29

40

-11

121

3.025

2

2

e

⎧⎪ ( o − e )2 ⎫⎪ χ = ∑⎨ ⎬ = 10.2 ⎩⎪ e ⎭⎪ 2

Hence, the Chi-square value, χ 2 = 10.2

© 2006 John Bird. All rights reserved. Published by Elsevier.

553

2. The numbers of telephone calls received by the switchboard of a company in 200 five-minute

intervals are shown in the distribution below. Number of calls Observed frequency Expected frequency 0

11

16

1

44

42

2

53

52

3

46

42

4

24

26

5

12

14

6

7

6

7

3

2

Calculate the χ2-value for this data.

o-e

(o − e)

(o − e)

Number

Observed

Expected

of calls

frequency, o

frequency, e

0

11

16

-5

25

1.5625

1

44

42

2

4

0.0952

2

53

52

1

1

0.0192

3

46

42

4

16

0.3810

4

24

26

-2

4

0.1538

5

12

14

-2

4

0.2857

6

7

6

1

1

0.1667

7

3

2

1

1

0.5000

2

2

e

⎧⎪ ( o − e )2 ⎫⎪ χ = ∑⎨ ⎬ = 3.16 ⎪⎩ e ⎪⎭ 2

Hence, the Chi-square value, χ 2 = 3.16

© 2006 John Bird. All rights reserved. Published by Elsevier.

554

EXERCISE 227 Page 612 1. Test the null hypothesis that the observed data given below fits a binomial distribution of the

form 250(0.6 + 0.4)7 at a level of significance of 0.05. Observed frequency 8 27 62 79 45 24 5 0 Is the fit of the data ‘too good’ at a level of confidence of 90%?

250 ( 0.6 + 0.4 )

7

(7)(6) (7)(6)(5) 6 5 2 4 3 ⎡ 7 ⎤ ( 0.6 ) ( 0.4 ) ⎢0.6 + 7 ( 0.6 ) ( 0.4 ) + 2! ( 0.6 ) ( 0.4 ) + ⎥ 3! ⎢ ⎥ (7)(6)(5)(4) (7)(6)(5)(4)(3) 3 4 2 5 ⎢ ⎥ = 250 + ( 0.6 ) ( 0.4 ) + ( 0.6 ) ( 0.4 ) ⎢ ⎥ 4! 5! ⎢ ⎥ (7)(6)(5)(4)(3)(2) 6 7 ⎢ + 0.6 )( (0.4 ) + ( 0.4 ) ⎥ ( ⎢⎣ ⎥⎦ 6!

= 250[0.02799 + 0.13064 + 0.26127 + 0.29030 + 0.19354 + 0.07741 + 0.01720 + 0.00164] = 7 + 33 + 65 + 73 + 48 + 19 + 4 + 0 correct to the nearest whole number o-e

(o − e)

(o − e)

Observed

Expected

frequency, o

frequency, e

8

7

1

1

0.14286

27

33

-6

36

1.09091

62

65

-3

9

0.13846

79

73

6

36

0.49315

45

48

-3

9

0.18750

24

19

5

25

1.31579

5

4

1

1

0.25000

0

0

0

0

0

2

2

e

⎧⎪ ( o − e )2 ⎫⎪ χ = ∑⎨ ⎬ = 3.62 e ⎩⎪ ⎭⎪ 2

Degrees of freedom, ν = N – 1 = 8 – 1 = 7 2 For χ 0.95 and ν = 7 from Table 63.1, page 609 of textbook is 14.1

Hence, the hypothesis is accepted, i.e. the observed data fits. 2 χ 0.10 , ν 7 = 2.83, hence the data is not ‘too good’.

© 2006 John Bird. All rights reserved. Published by Elsevier.

555

3. The resistances of a sample of carbon resistors are as shown below.

Resistance (MΩ) 1.28 1.29 1.30 1.31 1.32 1.33 1.34 1.35 1.36 Frequency

7

19

41

50

73

52

28

17

9

Test the null hypothesis that this data corresponds to a normal distribution at a level of significance of 0.05.

x=

(7 × 1.28) + (19 × 1.29) + (41× 1.30) + (50 × 1.31) + .... 390.5 = = 1.32 7 + 19 + 41 + 50 + 73 + 52 + 28 + 17 + 9 296

2 2 2 2 ⎪⎧ 7 (1.28 − 1.32 ) + 19 (1.29 − 1.32 ) + 41(1.30 − 1.32 ) + 50 (1.31 − 1.32 ) + .... ⎫⎪ s= ⎨ ⎬ 296 ⎩⎪ ⎭⎪

=

0.0958 = 0.0180 296

Class

Class

z-value for class

Area from

Area for

Expected

mid-point

boundaries, x

boundary =

0 to z

class

frequency

x − 1.32 0.0180

from Table 58.1, page 561

-2.50

0.4938 0.02

6

0.0561

17

0.121

36

0.1864

55

0.2206

65

0.1864

55

0.121

36

0.0561

17

0.02

6

1.275 1.28 1.285

-1.94

0.4738

1.29 1.295

-1.39

0.4177

1.30 1.305

-0.83

0.2967

1.315

-0.28

0.1103

1.325

0.28

0.1103

1.335

0.83

0.2967

1.345

1.39

0.4177

1.355

1.94

0.4738

1.31 1.32 1.33 1.34 1.35 1.36 1.365

2.5

0.4938

© 2006 John Bird. All rights reserved. Published by Elsevier.

556

Resistance

Observed

Expected

frequency,

frequency, e

(o − e)

o-e

(o − e)

2

2

e

o 1.28

7

6

1

1

0.1667

1.29

19

17

2

4

0.2353

1.30

41

36

5

25

0.6944

1.31

50

55

-5

25

0.4545

1.32

73

65

8

64

0.9846

1.33

52

55

-3

9

0.1636

1.34

28

36

-8

64

1.7778

1.35

17

17

0

0

0

1.36

9

6

3

9

1.5000 ⎧⎪ ( o − e )2 ⎫⎪ χ = ∑⎨ ⎬ = 5.98 ⎪⎩ e ⎪⎭ 2

Degrees of freedom, ν = N – 1 – M = 9 – 1 – 2 = 6 2 For χ 0.95 and ν = 6 from Table 63.1, page 609 of textbook is 12.6

Hence, the null hypothesis is accepted, i.e. the data does correspond to a normal distribution.

5. Test the hypothesis that the maximum load before breaking supported by certain cables produced by a company follows a normal distribution at a level of significance of 0.05, based on the experimental data given below. Also, test to see if the data is ‘too good’ at a level of significance of 0.05. Maximum load (MN) 8.5 9.0 9.5 10.0 10.5 11.0 11.5 12.0 Number of cables x=

2

5

12

17

14

6

3

1

(2 × 8.5) + (5 × 9.0) + (12 × 9.5) + (17 × 10.0) + .... 605.5 = = 10.09 MN 2 + 5 + 12 + 17 + 14 + 6 + 3 + 1 60

⎧⎪ 2 ( 8.5 − 10.09 )2 + 5 ( 9.0 − 10.09 )2 + 12 ( 9.50 − 10.09 )2 + 17 (10.0 − 10.09 )2 + .... ⎫⎪ s= ⎨ ⎬ 60 ⎩⎪ ⎭⎪ =

32.246 = 0.733MN 60

© 2006 John Bird. All rights reserved. Published by Elsevier.

557

Class

Class

z-value for class

Area from

mid-point

boundaries, x

boundary =

0 to z

Area for class

Expected frequency

x − 10.09 0.733 8.25

0.4940

-2.51

8.5 8.75

-1.83

0.4664

9.25

-1.15

0.3749

9.75

-0.46

0.1772

10.25

0.22

0.0871

10.75

0.90

0.3159

11.25

1.58

0.4430

11.75

2.26

0.4881

9.0 9.5 10.0 10.5 11.0 11.5 12.0 12.25

Load

Expected

frequency,

frequency, e

2

0.0915

5

0.1977

12

0.2643

16

0.2288

14

0.1271

8

0.0451

3

0.0103

1

0.4984

2.95

Observed

0.0276

o-e

(o − e)

(o − e)

2

2

e

o 8.5

2

2

0

0

0

9.0

5

5

0

0

0

9.5

12

12

0

0

0

10.0

17

16

1

1

0.0625

10.5

14

14

0

0

0

11.0

6

8

-2

4

0.5000

11.5

3

3

0

0

0

12.0

1

1

0

0

0 ⎧⎪ ( o − e )2 ⎫⎪ χ = ∑⎨ ⎬ = 0.563 e ⎩⎪ ⎭⎪ 2

© 2006 John Bird. All rights reserved. Published by Elsevier.

558

Degrees of freedom, ν = N – 1 – M = 8 – 1 – 2 = 5 2 and ν = 5 from Table 63.1, page 609 of textbook is 11.1 For χ 0.95

Hence, the hypothesis is accepted. 2 χ 0.05 , ν 5 = 1.15 , hence the results are ‘too good to be true’.

© 2006 John Bird. All rights reserved. Published by Elsevier.

559

EXERCISE 228 Page 616 2. In a laboratory experiment, 18 measurements of the coefficient of friction, µ, between metal and leather gave the following results: 0.60 0.57 0.51 0.55 0.66 0.56 0.52 0.59 0.58 0.48 0.59 0.63 0.61 0.69 0.57 0.51 0.58 0.54 Use the sign test at a level of significance of 5% to test the null hypothesis µ = 0.56 against an alternative hypothesis µ ≠ 0.56. Using the procedure for the sign test: (i) Null hypothesis, H 0 : µ = 0.56 Alternative hypothesis, H1 : µ ≠ 0.56 (ii) Significance level, α 2 = 5% (iii) With + sign ≥ 0.56 and - sign < 0.56: + + - - + + - + + - + + + + + - + (iv) There are 12 + signs and 6 – signs, hence, S = 6 (v) From Table 63.3, page 614 of textbook, with n = 18 and α 2 = 5% S ≤ 4 hence, the null hypothesis is accepted.

3. 18 random samples of two types of 9 V batteries are taken and the mean lifetime (in hours) of each are: Type A

8.2

7.0 11.3 13.9

3.6

7.5

9.0 13.8 16.2

6.5 18.0 11.5 13.4

9.8 10.6 16.4 12.7 16.8

9.4

6.9 14.2 12.4

Type B 15.3 15.4 11.2 16.1 18.1 17.1 17.7 7.8

8.6

8.4 13.5

9.9 12.9 14.7

Use the sign test, at a level of significance of 5%, to test the null hypothesis that the two samples come from the same population. Using the procedure for the sign test: (i) Null hypothesis, H 0 : µ A = µ B Alternative hypothesis, H1 : µ A ≠ µ B

© 2006 John Bird. All rights reserved. Published by Elsevier.

560

(ii) Significance level, α 2 = 5% (iii) A - B

-7.1 -8.4 +0.1 -2.2 -9.1 -3.3 -1.5 +0.2 -4.1 -4.2 -2.3 -4.1 +1.6 -1.2 -3.4 -3.0 +1.3 -2.3

(iv) There are 4 + signs and 14 – signs, hence, S = 4 (v) From Table 63.3, page 614 of textbook, with n = 18 and α 2 = 5% S ≤ 4 Since from (iv) S is equal to 4, then the result is significant at α 2 = 5% hence, the alternative

hypothesis H1 is accepted.

© 2006 John Bird. All rights reserved. Published by Elsevier.

561

EXERCISE 229 Page 619 1. The time to repair an electronic instrument is a random variable. The repair times (in hours) for 16 instruments are as follows: 218 275 264 210 161 374 178 265 150 360 185 171 215 100 474 248 Use the Wilcoxon signed-rank test, at a 5% level of significance, to test the hypothesis that the mean repair time is 220 hours Using the procedure for the Wilcoxon signed-rank test: (i) H 0 : t = 220 h H1 : t ≠ 220 h (ii) α 2 = 5% (iii) Taking the time difference between the time taken for repair and 220 h gives: -2 +55 +44 -10 -59 +154 -42 +45 -70 +140 -35 -49 -5 -120 +254 + 28 (iv) Ranking gives: Rank

1 2 3

4

5

6

7

8

9

10 11 12

13

14

15

16

Difference -2 -5 -10 +28 -35 -42 +44 +45 -49 +55 -59 -70 -120 +140 +154 +254 (v) T = 4 + 7 + 8 + 10 + 14 + 15 + 16 = 74 (vi) From Table 63.4, page 617 of textbook, for n = 16, α 2 = 5% , T ≤ 29 Hence, since 74 > 29, the hypothesis H 0 is accepted, i.e. the mean repair time is 220 hours.

3. A paint supplier claims that a new additive will reduce the drying time of their acrylic paint. To test his claim, 12 pieces of wood are painted, one half of each piece with paint containing the regular additive and the other half with paint containing the new additive. The drying time (in hours) were measured as follows: New additive

4.5 5.5 3.9 3.6 4.1 6.3 5.9 6.7 5.1 3.6 4.0 3.0

Regular additive 4.7 5.9 3.9 3.8 4.4 6.5 6.9 6.5 5.3 3.6 3.9 3.9 Use the Wilcoxon signed-rank test at a significance level of 5% to test the hypothesis that there is no difference, on average, in the drying times of the new and regular additive paints.

© 2006 John Bird. All rights reserved. Published by Elsevier.

562

Using the procedure for the Wilcoxon signed-rank test: (i) H 0 : N = R H1 : N ≠ R (ii) α 2 = 5% (iii) Taking the time difference between N and R gives: (N – R)

-0.2 -0.4 0 -0.2 -0.3 -0.2 -1.0 +0.2 -0.2 0 +0.1 -0.9

(iv) Ranking and ignoring the zero’s gives: Rank

1

4

4

4

4

4

7

8

9

10

Difference +0.1 -0.2 -0.2 -0.2 -0.2 +0.2 -0.3 -0.4 -0.9 -1.0 (v) T = 1 + 4 = 5 (vi) From Table 63.4, page 617 of textbook, for n = 10, α 2 = 5% , T ≤ 8 Since from (v) T is less than 8, there is a significant difference in the drying times.

© 2006 John Bird. All rights reserved. Published by Elsevier.

563

EXERCISE 230 Page 624 1. The tar content of two brands of cigarettes (in mg) was measured as follows: Brand P 22.6 4.1 3.9 0.7 3.2 6.1 1.7 2.3 5.6 2.0 Brand Q

3.4 6.2 3.5 4.7 6.3 5.5 3.8 2.1

Use the Mann-Whitney test at a 0.05 level of significance to determine if the tar contents of the two brands are equal. Using the procedure for the Mann-Whitney test: (i)

H 0 : TA = TB H1 : TA ≠ TB

(ii) α 2 = 5% (iii) Brand P

0.7 1.7 2.0 2.3 3.2

Brand Q

2.1

3.9 4.1

3.4 3.5 3.8

5.6 6.1

4.7 5.5

22.6

6.2 6.3

(iv) P P P Q P P Q Q Q P P Q Q P P Q Q P 0 0 0

1 1

4 4

6 6

8

writing the Q’s that precede the P’s

(v) U = 1 + 1 + 4 + 4 + 6 + 6 + 8 = 30 (vi) From Table 63.5, page 622 of textbook, for a sample size of 10 and 8 at α 2 = 5% , U ≤ 17 Hence, H 0 is accepted, i.e. there is no difference between brands P and Q

3. An experiment, designed to compare two preventive methods against corrosion gave the following results for the maximum depths of pits (in mm) in metal strands: Method A 143 106 135 147 139 132 153 140 Method B

98 105 137

94 112 103

Use the Mann-Whitney test, at a level of significance of 0.05, to determine whether the two tests are equally effective. Using the procedure for the Mann-Whitney test: (i)

H0 : A = B H1 : A ≠ B

© 2006 John Bird. All rights reserved. Published by Elsevier.

564

(ii) α 2 = 5% (iii) A B

106 94 98 103 105

132 135

112

139 140 143 147 153 137

(iv) B B B B A B A A B A A A A A 1

3

writing the A’s that precede the B’s

(v) U = 1 + 3 = 4 (vi) From Table 63.5, page 621 of textbook, for a sample size of 8 and 6 at α 2 = 5% , U ≤ 8 Hence, the null hypothesis H 0 is rejected, i.e. the two methods are not equally effective.

4. Repeat problem 3 of Exercise 228, using the Mann-Whitney test.

Using the procedure for the Mann-Whitney test: (i) Null hypothesis, H 0 : µ A = µ B Alternative hypothesis, H1 : µ A ≠ µ B (ii) α 2 = 5% (iii) A 3.6 6.5 6.9 7.0 7.5 8.2 8.6 9.0 9.4 11.3 11.5 12.4 13.4 13.8 13.9 14.2 16.2 18.0 B 7.8 8.4 9.8 9.9 10.6 11.2 12.7 12.9 13.5 14.7 15.3 15.4 16.1 16.4 16.8 17.1 17.7 18.1

(iv) A A A A A B A B A A A B B B B A A A B B A B A A A B B B B A B B B B A B 1

2 2 2

6 6 6

8 9 9 9

13

17

(v) U = 1 + 2 + 2 + 2 + 6 + 6 + 6 + 8 + 9 + 9 + 9 + 13 + 17 = 90 (vi) From Table 63.5, page 622 of textbook, for a sample size of 18 and 18 at α 2 = 5% , U ≤ 99 Hence, the null hypothesis H 0 is rejected and H1 is accepted, i.e. the two means are not

equal.

© 2006 John Bird. All rights reserved. Published by Elsevier.

565

CHAPTER 64 INTRODUCTION TO LAPLACE TRANSFORMS EXERCISE 231 Page 630 1. Determine the Laplace transforms of (a) 2t – 3 (b) 5t 2 + 4t − 3 2 3 ⎛ 1 ⎞ ⎛1⎞ (a) ℒ {2t − 3} = 2 ⎜ 2 ⎟ − 3 ⎜ ⎟ = 2 − s s ⎝s ⎠ ⎝s⎠ 10 4 3 ⎛ 2! ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ (b) ℒ {5t 2 + 4t − 3} = 5 ⎜ 3 ⎟ + 4 ⎜ 2 ⎟ − 3 ⎜ ⎟ = 3 + 2 − s s s ⎝s ⎠ ⎝s ⎠ ⎝s⎠

2. Determine the Laplace transforms of (a)

t3 t5 t2 − 2t 4 + - 3t + 2 (b) 15 2 24

⎧ t3 ⎫ ⎛ 1 ⎞ ⎛ 3! ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ 1 3 2 (a) ℒ ⎨ − 3t + 2 ⎬ = ⎜ ⎟ ⎜ 3+1 ⎟ − 3 ⎜ 2 ⎟ + 2 ⎜ ⎟ = 4 − 2 + 4s s s ⎩ 24 ⎭ ⎝ 24 ⎠ ⎝ s ⎠ ⎝ s ⎠ ⎝ s ⎠ ⎧ t5 8 48 1 t 2 ⎫ ⎛ 1 ⎞⎛ 5! ⎞ ⎛ 4! ⎞ ⎛ 1 ⎞⎛ 2! ⎞ (b) ℒ ⎨ − 2t 4 + ⎬ = ⎜ ⎟⎜ 5+1 ⎟ − 2 ⎜ 4+1 ⎟ + ⎜ ⎟⎜ 3 ⎟ = 6 − 5 + 3 s s s 2 ⎭ ⎝ 15 ⎠⎝ s ⎠ ⎝ s ⎠ ⎝ 2 ⎠⎝ s ⎠ ⎩15

3. Determine the Laplace transform of (a) 5e3t (b) 2e−2t 5 ⎛ 1 ⎞ (a) ℒ {5e3t } = 5 ⎜ ⎟ = s−3 ⎝ s−3⎠ 2 ⎛ 1 ⎞ (b) ℒ {2e −2t } = 2 ⎜ ⎟ = s+2 ⎝s+2⎠

4. Determine the Laplace transform of (a) 4 sin 3t (b) 3 cos 2t 12 ⎛ 3 ⎞ (a) ℒ { 4sin 3t } = 4 ⎜ 2 2 ⎟ = 2 s +9 ⎝s +3 ⎠ 3s ⎛ s ⎞ (b) ℒ { 3cos 2t } = 3 ⎜ 2 = 2 2 ⎟ s +4 ⎝s +2 ⎠

© 2006 John Bird. All rights reserved. Published by Elsevier.

566

5. Determine the Laplace transforms of (a) 7 cosh 2x (b)

1 sinh 3t 3

7s ⎛ s ⎞ (a) ℒ { 7 cosh 2x } = 7 ⎜ 2 = 2 2 ⎟ s −4 ⎝s −2 ⎠ 1 ⎧1 ⎫ 1⎛ 3 ⎞ (b) ℒ ⎨ sinh 3t ⎬ = ⎜ 2 2 ⎟ = 3 s −9 ⎩3 ⎭ 3⎝ s −3 ⎠

6.(a) Determine the Laplace transform of 2 cos 2 t cos 2t = 2 cos 2 t − 1 ℒ { 2 cos t } 2

from which,

cos 2 t =

1 + cos 2t 2

s2 + 4 ) + s2 ( ⎧ ⎛ 1 + cos 2t ⎞ ⎫ 1 s 1 s = + = = ℒ ⎨2 ⎜ ⎟ ⎬ = ℒ {1 + cos 2t} = + 2 2 s s + 22 s s 2 + 4 s ( s2 + 4 ) ⎠⎭ ⎩ ⎝ 2 ( s2 + 2 ) 2s 2 + 4 = = s (s2 + 4) s ( s2 + 4)

7.(b) Determine the Laplace transform of 2sinh 2 2θ cosh 4θ = 1 + 2sinh 2 2θ

from which,

ℒ { 2sinh 2 2θ } = ℒ {cosh 4θ − 1} =

2sinh 2 2θ = cosh 4θ - 1

2 2 16 s 1 s − ( s − 16 ) − = = 2 2 s −4 s s ( s 2 − 16 ) s ( s 2 − 16 )

8. Determine the Laplace transform of 4 sin (at + b), where a and b are constants. ℒ { 4sin(at + b) } = ℒ 4 { sin at cos b + cos at sin b } = ℒ { (4 cos b) sin at + (4sin b) cos at }

⎛ a ⎞ ⎛ s ⎞ + 4sin b ) ⎜ 2 = ( 4 cos b ) ⎜ 2 2 ⎟ ( 2 ⎟ ⎝s +a ⎠ ⎝s +a ⎠

=

4 ( a cos b + s sin b ) s + a2 2

© 2006 John Bird. All rights reserved. Published by Elsevier.

567

10. Show that ℒ ( cos 2 3t − sin 2 3t ) =

s s + 36 2

cos 6t = 2cos 2 3t − 1

from which,

cos 2 3t =

1 + cos 6t 2

and cos 6t = 1 − 2sin 2 3t

from which,

sin 2 3t =

1 − cos 6t 2

⎧⎛ 1 + cos 6t ⎞ ⎛ 1 − cos 6t ⎞ ⎫ ⎧ 1 cos 6t 1 cos 6t ⎫ Hence, ℒ ( cos 2 3t − sin 2 3t ) = ℒ ⎨⎜ − + ⎬ ⎟−⎜ ⎟⎬ = ℒ ⎨ + 2 2 2 2 2 ⎭ ⎠ ⎝ ⎠⎭ ⎩2 ⎩⎝ = ℒ{cos 6t} =

© 2006 John Bird. All rights reserved. Published by Elsevier.

s s + 36 2

568

CHAPTER 65 PROPERTIES OF LAPLACE TRANSFORMS EXERCISE 232 Page 634 1. Determine the Laplace transforms of (a) 2te 2t (b) t 2 e t

(a) ℒ {2te 2t } = 2 ℒ {t1e2t } = (2)

(b) ℒ {t 2 e t } =

2!

( s − 1)

2 +1

=

1!

(s − 2)

1+1

=

2

(s − 2)

2

2

( s − 1)

3

2.(b) Determine the Laplace transform of

1 4 −3t te 2

12 4! ⎞ ⎧1 ⎫ 1⎛ ℒ ⎨ t 4 e −3t ⎬ = ⎜ ⎟ = 5 4 +1 ⎩2 ⎭ 2 ⎜⎝ ( s + 3) ⎟⎠ ( s + 3)

3.(b) Determine the Laplace transform of 3e 2t sin 2t ⎛ ⎞ 6 2 6 ℒ {3e 2t sin 2t} = 3 ⎜ = = ⎟ 2 2 2 ⎜ ( s − 2 ) + 22 ⎟ s − 4s + 4 + 4 s − 4s + 8 ⎝ ⎠

4.(a) Determine the Laplace transform of 5e−2t cos 3t ⎛ ⎞ 5 ( s + 2) 5 (s + 2) s+2 ℒ {5e −2t cos 3t} = 5 ⎜ = = ⎟ ⎜ ( s + 2 )2 + 32 ⎟ s 2 + 4s + 4 + 9 s 2 + 4s + 13 ⎝ ⎠

5.(a) Determine the Laplace transform of 2e t sin 2 t

⎧ ⎛ 1 − cos 2t ⎞ ⎫ 1 s −1 t t ℒ {2e t sin 2 t} = ℒ ⎨2e t ⎜ − ⎟ ⎬ = ℒ {e } - ℒ {e cos 2t} = 2 s − 1 ( s − 1)2 + 22 ⎠⎭ ⎩ ⎝ =

1 s−1 − 2 s − 1 s − 2s + 5

© 2006 John Bird. All rights reserved. Published by Elsevier.

569

6.(b) Determine the Laplace transform of 3e 2t cosh 4t

⎛ ⎞ 3 ( s − 2) s−2 3(s − 2) ℒ {3e 2t cosh 4t} = 3 ⎜ = = ⎟ ⎜ ( s − 2 )2 − 42 ⎟ s 2 − 4s + 4 − 16 s 2 − 4s − 12 ⎝ ⎠

7.(a) Determine the Laplace transform of 2e− t sinh 3t

⎛ ⎞ 6 3 6 = 2 ℒ {2e − t sinh 3t} = 2 ⎜ ⎟= 2 ⎜ ( s + 1) − 32 ⎟ s 2 + 2s + 1 − 9 s + 2s − 8 ⎝ ⎠

8. Determine the Laplace transforms of (a) 2e t ( cos 3t − 3sin 3t )

(b) 3e −2t ( sinh 2t − 2 cosh 2t )

⎛ ⎞ ⎛ ⎞ s −1 3 (a) ℒ {2e t ( cos 3t − 3sin 3t )} = ℒ {2e t cos 3t} - ℒ {6e t sin 3t} = 2 ⎜ 6 − ⎟ ⎜ ⎟ ⎜ ( s − 1)2 + 32 ⎟ ⎜ ( s − 1)2 + 32 ⎟ ⎝ ⎠ ⎝ ⎠ =

2 ( s − 1) 18 2s − 2 − 18 2s − 20 − 2 = 2 = 2 s − 2s + 10 s − 2s + 10 s − 2s + 10 s − 2s + 10 2

=

2 ( s − 10 ) s 2 − 2s + 10

(b) ℒ {3e−2t ( sinh 2t − 2 cosh 2t )} = ℒ {3e−2t sinh 2t} - ℒ {6e −2t cosh 2t}

⎛ ⎞ ⎛ ⎞ 2 s+2 − 6 = 3⎜ ⎟ ⎜ ⎟ 2 2 ⎜ ( s + 2 ) − 22 ⎟ ⎜ ( s + 2 ) − 22 ⎟ ⎝ ⎠ ⎝ ⎠ =

6 (s + 2) 6 (s + 2) 6 6 − = − s 2 + 4s + 4 − 4 s 2 + 4s + 4 − 4 s 2 + 4s s 2 + 4s

=

−6 ( s + 1 ) 6 − 6s − 12 −6s − 6 = = s ( s + 4) s (s + 4) s (s + 4)

© 2006 John Bird. All rights reserved. Published by Elsevier.

570

EXERCISE 233 Page 635 2. Use the Laplace transform of the first derivative to derive the transforms:

(a) ℒ{ eat } =

1 s−a

(b) ℒ{ 3t 2 } =

6 s3

(a) Let f(t) = eat then f ′(t) = a eat and f(0) = 1 From equation (3), page 634 of textbook, ℒ {f '(t)} = s ℒ {f (t)} − f (0) ℒ{ a eat } = sℒ {eat } - 1

Hence,

1 = (s – a)ℒ {eat }

i.e.

ℒ {eat } =

and

1 s−a

(b) Let f(t) = 3t 2 then f ′(t) = 6t and f(0) = 0 Since

ℒ {f '(t)} = s ℒ {f (t)} − f (0)

then,

ℒ{ 6t } = sℒ {3t 2 } + 0

i.e.

6 = sℒ {3t 2 } s2

and

ℒ { 3t 2 } =

6 s3

3. Derive the Laplace transform of the second derivative from the definition of a Laplace

transform. Hence derive the transform ℒ{sin at} =

a s + a2 2

For the derivation, see page 634 of textbook. Let f(t) = sin at, then f ′(t) = a cos at and f ′′(t) = −a 2 sin at , f(0) = 0 and f ′(0) = a From equation (4), page 635 of textbook, Hence,

ℒ {f ''(t)} = s 2 ℒ {f (t)} − sf (0) − f '(0)

ℒ {−a 2 sin at} = s 2 ℒ {sin at} - s(0) - a

© 2006 John Bird. All rights reserved. Published by Elsevier.

571

−a 2 ℒ {sin at} = s 2 ℒ {sin at} - a

i.e.

a = ( s 2 + a 2 ) ℒ {sin at}

Hence,

ℒ {sin at} =

and

a s + a2 2

4.(b) Use the Laplace transform of the second derivative to derive the transform: ℒ{cosh at} =

s s − a2 2

Let f(t) = cosh at then f ′(t) = a sinh at and f ′′(t) = a 2 cosh at , f(0) = 1 and f ′(0) = 0 ℒ {f ''(t)} = s 2 ℒ {f (t)} − sf (0) − f '(0)

Hence,

ℒ {a 2 cosh at} = s 2 ℒ {cosh at} - s(1) - 0

i.e.

a 2 ℒ {cosh at} = s 2 ℒ {cosh at} - s

i.e. and

s = ( s 2 − a 2 ) ℒ {cosh at} ℒ {cosh at} =

s s − a2 2

© 2006 John Bird. All rights reserved. Published by Elsevier.

572

EXERCISE 234 Page 637 1. State the initial value theorem. Verify the theorem for the function (a) 3 – 4 sin t (b) ( t − 4 )

2

and state their initial values. The initial value theorem states: lim it [ f (t) ] = lim it [ s ℒ {f (t)} t →0

s →∞

then ℒ{f(t)} = ℒ{3 – 4 sin t} =

(a) Let f(t) = 3 – 4 sin t

]

3 4 − 2 s s +1

⎡ ⎛3 4 ⎞⎤ lim it [3 − 4sin t ] = lim it ⎢s ⎜ − 2 ⎟ ⎥ t →0 s →∞ ⎣ ⎝ s s + 1 ⎠⎦

Hence,

4s ⎤ ⎡ = lim it ⎢3 − 2 ⎥ s →∞ ⎣ s + 1⎦ i.e.

3 – 4 sin 0 = 3 -

i.e.

3=3

∞ ∞ +1 2

which verifies the theorem.

The initial value is 3

(b) Let f(t) = ( t − 4 ) = t 2 − 8t + 16 2

then

Hence,

ℒ {t 2 − 8t + 16} =

2 8 16 − + s3 s 2 s

⎡ ⎛ 2 8 16 ⎞ ⎤ lim it ⎡⎣ t 2 − 8t + 16 ⎤⎦ = lim it ⎢s ⎜ 3 − 2 + ⎟ ⎥ t →0 s →∞ s ⎠⎦ ⎣ ⎝s s ⎡2 8 ⎤ = lim it ⎢ 2 − + 16 ⎥ s →∞ ⎣s s ⎦

i.e.

16 = 16 which verifies the theorem

The initial value is 16

3. State the final value theorem and state a practical application where it is of use. Verify the

theorem for the function 4 + e−2t (sin t + cos t) representing a displacement and state its final value.

© 2006 John Bird. All rights reserved. Published by Elsevier.

573

The final value theorem states:

lim it [ f (t) ] = lim it [ s ℒ {f (t)} t →∞

s→0

]

The final value theorem is used in investigating the stability of systems such as in automatic aircraft-landing systems. Let f(t) = 4 + e−2t (sin t + cos t) = 4 + e−2t sin t + e−2t cos t then

Hence,

ℒ {f (t)} =

4 1 s+2 + + 2 s ( s + 2 ) + 1 ( s + 2 )2 + 1

⎛4 1 s+2 ⎞ lim it ⎡⎣ 4 + e−2t sin t + e−2t cos t ⎤⎦ = lim it [ s ⎜ + + ⎟ t →∞ s→0 ⎜ s ( s + 2 )2 + 1 ( s + 2 )2 + 1 ⎟ ⎝ ⎠ ⎡ s (s + 2) ⎤ s + = lim it ⎢ 4 + ⎥ 2 2 s→0 ⎢⎣ ( s + 2 ) + 1 ( s + 2 ) + 1 ⎥⎦

i.e.

4+0+0=4+0+0

i.e.

4 = 4 which verifies the theorem

The final value is 4

© 2006 John Bird. All rights reserved. Published by Elsevier.

574

CHAPTER 66 INVERSE LAPLACE TRANSFORMS EXERCISE 235 Page 640

2. (a) Determine the inverse Laplace transform of

3 2s + 1

⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ 3 − 12 t 3 −1 ⎪ 1 ⎪⎪ 1 ⎪ ⎧ 3 ⎫ −1 ⎪ ℒ −1 ⎨ = 3 ℒ = ℒ = e ⎬ ⎨ ⎬ ⎨ ⎬ 1 1 2 2 ⎛ ⎞ ⎛ ⎞ ⎩ 2s + 1 ⎭ ⎪2⎜s + ⎟⎪ ⎪⎜ s + ⎟ ⎪ ⎪⎩ ⎝ 2 ⎠ ⎭⎪ ⎩⎪ ⎝ 2 ⎠ ⎭⎪

3. (a) Determine the inverse Laplace transform of

1 s + 25 2

1 1 −1 ⎧ 5 ⎫ ⎧ 1 ⎫ ℒ −1 ⎨ 2 ⎬ = ℒ ⎨ 2 2 ⎬ = sin 5t 5 5 ⎩ s + 25 ⎭ ⎩s + 5 ⎭

4. (a) Determine the inverse Laplace transform of

5s 2s + 18 2

⎧ ⎫⎪ 5 −1 ⎧ s ⎫ 5 s ⎧ 5s ⎫ −1 ⎪ ℒ −1 ⎨ 2 ⎬ = 5ℒ ⎨ ⎬ = ℒ ⎨ 2 2 ⎬ = cos 3t 2 2 ⎩ 2s + 18 ⎭ ⎩s + 3 ⎭ 2 ⎪⎩ 2 ( s + 9 ) ⎪⎭

5. (a) Determine the inverse Laplace transform of

5 s3

5 ⎧5⎫ ⎧ 2!⎫ 5 ℒ −1 ⎨ 3 ⎬ = ℒ −1 ⎨ 3 ⎬ = t 2 2! ⎩s ⎭ ⎩s ⎭ 2

6. (a) Determine the inverse Laplace transform of

3s 1 2 s −8 2

⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ 3s s ⎧ s ⎫ ℒ −1 ⎨ = 3ℒ −1 ⎨ = 6ℒ −1 ⎨ 2 = 6 cosh 4t ⎬ ⎬ 2⎬ 1 1 − s 4 2 2 ⎩ ⎭ ⎪ s −8⎪ ⎪ ( s − 16 ) ⎪ ⎩2 ⎭ ⎩2 ⎭

© 2006 John Bird. All rights reserved. Published by Elsevier.

575

8. (b) Determine the inverse Laplace transform of

3

( s − 3)

5

⎧⎪ 3 ⎫⎪ ⎧⎪ ⎫⎪ 1 3 −1 ⎧⎪ 4! ⎫⎪ 1 −1 ⎧⎪ 4! ⎫⎪ 1 3t 4 = 3ℒ −1 ⎨ = ℒ ⎨ = ℒ ⎨ = e t ℒ −1 ⎨ 5⎬ 4 +1 ⎬ 4 +1 ⎬ 4 +1 ⎬ 4! 8 − − − − s 3 s 3 s 3 s 3 ( ) ( ) ( ) ( ) ⎩⎪ ⎭⎪ ⎩⎪ ⎭⎪ ⎩⎪ ⎭⎪ ⎩⎪ ⎭⎪ 8

9. (b) Determine the inverse Laplace transform of

3 s + 6s + 13 2

⎧ ⎫⎪ ⎫⎪ 3 −3t 3 1 3 −1 ⎧⎪ 2 ⎧ ⎫ −1 ⎪ ℒ −1 ⎨ 2 = ℒ ⎬ ⎨ ⎬ = e sin 2t ⎬ = 3ℒ ⎨ 2 2 2 2 2 ⎩ s + 6s + 13 ⎭ ⎩⎪ ( s + 3) + 2 ⎭⎪ ⎩⎪ ( s + 3) + 2 ⎭⎪ 2

10. (a) Determine the inverse Laplace transform of

2(s − 3) s − 6s + 13 2

⎧ ⎫⎪ ⎧ 2 ( s − 3) ⎫ s−3 −1 ⎪ 3t ℒ −1 ⎨ 2 ⎬ = 2e cos 2t ⎬ = 2ℒ ⎨ 2 2 ⎩ s − 6s + 13 ⎭ ⎩⎪ ( s − 3) + 2 ⎭⎪

11. Determine the inverse Laplace transforms of (a)

2s + 5 s + 4s − 5 2

(b)

3s + 2 s − 8s + 25 2

⎧ 2 (s + 2) ⎫⎪ 1 ⎧ 2s + 5 ⎫ −1 ⎪ (a) ℒ −1 ⎨ 2 + ⎬ ⎬ =ℒ ⎨ 2 2 2 2 ⎩ s + 4s − 5 ⎭ ⎩⎪ ( s + 2 ) − 3 ( s + 2 ) − 3 ⎭⎪ ⎧⎪ ⎫⎪ 1 −1 ⎧⎪ ⎫⎪ 1 −2t s+2 3 −2t 2e cosh 3t e sinh 3t + = 2ℒ −1 ⎨ ℒ = + ⎬ ⎨ ⎬ 2 2 2 2 3 ⎪⎩ ( s + 2 ) − 3 ⎪⎭ 3 ⎪⎩ ( s + 2 ) − 3 ⎭⎪ ⎧ 3s + 2 ⎫⎪ ⎧ ⎫ ⎧ 3s + 2 ⎫ −1 ⎪ −1 ⎪ 3 ( s − 4 ) + 14 ⎪ = ℒ (b) ℒ −1 ⎨ 2 ⎬ ⎬ =ℒ ⎨ ⎨ ⎬ 2 2 2 2 ⎩ s − 8s + 25 ⎭ ⎪⎩ ( s − 4 ) + 3 ⎪⎭ ⎪⎩ ( s − 4 ) + 3 ⎭⎪ ⎧⎪ ⎫⎪ 14 −1 ⎧⎪ ⎫⎪ 14 4t s−4 3 4t = 3ℒ −1 ⎨ + ℒ ⎬ ⎨ ⎬ = 3e cos 3t + e sin 3t 2 2 2 2 3 ⎪⎩ ( s − 4 ) + 3 ⎪⎭ 3 ⎪⎩ ( s − 4 ) + 3 ⎭⎪

© 2006 John Bird. All rights reserved. Published by Elsevier.

576

EXERCISE 236 Page 642

2. Use partial fractions to find the inverse Laplace transform of:

2s 2 − 9s − 35 4 3 1 = − + ( s + 1)( s − 2 )( s + 3) ( s + 1) ( s − 2 ) ( s + 3)

2s 2 − 9s − 35 ( s + 1)( s − 2 )( s + 3)

from Problem 2, page 19 of textbook

⎧ 4 3 1 ⎪⎫ ⎪⎧ 2s 2 − 9s − 35 ⎪⎫ −1 ⎪ −t 2t −3t − + Hence, ℒ −1 ⎨ ⎬ =ℒ ⎨ ⎬ = 4e − 3e + e + − + + − + s 1 s 2 s 3 s 1 s 2 s 3 ( )( )( ) ( ) ( ) ( ) ⎩⎪ ⎭⎪ ⎩⎪ ⎭⎪

4. Use partial fractions to find the inverse Laplace transform of:

3s 2 + 16s + 15

( s + 3)

3

=

3 2 6 − − 2 ( s + 3 ) ( s + 3 ) ( s + 3 )3

3s 2 + 16s + 15

( s + 3)

3

from Problem 7, page 22 of textbook

⎧⎪ 3s 2 + 16s + 15 ⎫⎪ ⎧ 3 2 6 ⎫⎪ −1 ⎪ Hence, ℒ −1 ⎨ = ℒ − − ⎬ ⎨ 2 3⎬ 3 ⎪⎩ ( s + 3) ⎪⎭ ⎪⎩ ( s + 3) ( s + 3) ( s + 3) ⎪⎭ ⎧ 1 ⎫⎪ ⎧ 1 ⎫⎪ ⎪⎧ 1 ⎪⎫ −1 ⎪ −1 ⎪ = 3ℒ −1 ⎨ 6 ℒ ⎬ - 2ℒ ⎨ ⎬ ⎨ 2 3⎬ ⎪⎩ ( s + 3) ⎭⎪ ⎪⎩ ( s + 3) ⎭⎪ ⎩⎪ ( s + 3) ⎭⎪ ⎧ ⎫⎪ 6 −1 ⎧⎪ 2! ⎫⎪ ⎧⎪ 1 ⎫⎪ 1 −1 ⎪ = 3ℒ −1 ⎨ - ℒ ⎨ ⎬ - 2ℒ ⎨ 1+1 ⎬ 2 +1 ⎬ ⎪⎩ ( s + 3) ⎭⎪ 2! ⎩⎪ ( s + 3) ⎭⎪ ⎩⎪ ( s + 3) ⎭⎪

= 3e −3t − 2e −3t t − 3e −3t t 2

or

e −3t (3 − 2t − 3t 2 )

3 + 6s + 4s 2 − 2s3 6. Use partial fractions to find the inverse Laplace transform of: s 2 ( s 2 + 3) 3 + 6s + 4s 2 − 2s3 2 1 3 − 4s = + 2+ 2 s s s +3 s 2 ( s 2 + 3)

from Problem 9, page 23 of textbook

⎧⎪ 3 + 6s + 4s 2 − 2s3 ⎫⎪ 1 3 − 4s ⎫ −1 ⎧ 2 Hence, ℒ −1 ⎨ ⎬ ⎬ =ℒ ⎨ + 2 + 2 2 2 ⎩s s s +3⎭ ⎩⎪ s ( s + 3) ⎭⎪

© 2006 John Bird. All rights reserved. Published by Elsevier.

577

⎧ 3 − 4s ⎧1 ⎫ −1 ⎧ 1 ⎫ −1 ⎪ = 2ℒ ⎨ ⎬ + ℒ ⎨ 2 ⎬ + ℒ ⎨ ⎩s ⎭ ⎩s ⎭ ⎪ s2 + 3 ⎩ −1

( )

⎫ ⎪ 2⎬ ⎪ ⎭

⎧ ⎫ ⎧ 3 4s ⎪ ⎧1 ⎫ −1 ⎧ 1 ⎫ −1 ⎪ −1 ⎪ = 2ℒ ⎨ ⎬ + ℒ ⎨ 2 ⎬ + ℒ ⎨ -ℒ ⎨ 2⎬ ⎩s ⎭ ⎩s ⎭ ⎪ s2 + 3 ⎪ ⎪ s2 + 3 ⎩ ⎭ ⎩ −1

( )

( )

⎫ ⎪ 2⎬ ⎪ ⎭

⎧ ⎫ ⎧ ⎫ 3 3 s ⎪ ⎪ ⎧1 ⎫ −1 ⎧ 1 ⎫ −1 ⎪ −1 ⎪ = 2ℒ ⎨ ⎬ + ℒ ⎨ 2 ⎬ + ℒ ⎨ - 4ℒ ⎨ 2⎬ 2⎬ 3 ⎩s ⎭ ⎩s ⎭ ⎪ s2 + 3 ⎪ ⎪ s2 + 3 ⎪ ⎩ ⎭ ⎩ ⎭ −1

( )

( )

= 2 + t + 3 sin 3 t − 4cos 3 t

7. Use partial fractions to find the inverse Laplace transform of:

26 − s 2 s ( s 2 + 4s + 13)

A ( s 2 + 4s + 13) + ( Bs + C ) s 26 − s 2 A Bs + C ≡ + 2 = Let s ( s 2 + 4s + 13) s s + 4s + 13 s ( s 2 + 4s + 13)

26 − s 2 = A ( s 2 + 4s + 13) + Bs 2 + Cs

Hence,

When s = 0:

26 = 13A + 0 + 0

i.e. A = 2

Equating s 2 coefficients:

-1 = A + B

i.e. B = -3

Equating s coefficients: Thus,

0 = 4A + C

i.e. C = -8

26 − s 2 2 −3s − 8 ≡ + 2 2 s ( s + 4s + 13) s s + 4s + 13

⎧⎪ ⎫⎪ 3s + 8 ⎫ 26 − s 2 −1 ⎧ 2 ⎫ −1 ⎧ Hence, ℒ −1 ⎨ 2 ⎬ ⎬ =ℒ ⎨ ⎬ -ℒ ⎨ 2 ⎩ s + 4s + 13 ⎭ ⎩s ⎭ ⎩⎪ s ( s + 4s + 13) ⎭⎪

⎧⎪ 3s + 8 ⎫⎪ ⎧1 ⎫ = 2ℒ −1 ⎨ ⎬ - ℒ −1 ⎨ ⎬ 2 2 ⎩s ⎭ ⎩⎪ ( s + 2 ) + 3 ⎭⎪ ⎧⎪ 3 ( s + 2 ) ⎫⎪ ⎧ ⎫⎪ 2 ⎧1 ⎫ −1 ⎪ ℒ = 2ℒ −1 ⎨ ⎬ - ℒ −1 ⎨ ⎬ ⎨ ⎬ 2 2 2 2 ⎩s ⎭ ⎩⎪ ( s + 2 ) + 3 ⎭⎪ ⎩⎪ ( s + 2 ) + 3 ⎭⎪

© 2006 John Bird. All rights reserved. Published by Elsevier.

578

⎧⎪ ( s + 2 ) ⎫⎪ 2 −1 ⎧⎪ ⎫⎪ 3 ⎧1 ⎫ = 2ℒ −1 ⎨ ⎬ - 3ℒ −1 ⎨ ℒ ⎬ ⎨ ⎬ 2 2 2 2 ⎩s ⎭ ⎩⎪ ( s + 2 ) + 3 ⎭⎪ 3 ⎩⎪ ( s + 2 ) + 3 ⎭⎪ 2 = 2 − 3e −2t cos 3t − e −2t sin 3t 3

© 2006 John Bird. All rights reserved. Published by Elsevier.

579

EXERCISE 237 (Page XX)

1. Determine for the transfer function: R(s) =

50 ( s + 4 )

s ( s + 2 ) ( s 2 − 8s + 25 )

(a) the zero and (b) the poles. Show the poles and zeros on a pole-zero diagram.

(a) For the numerator to be zero, (s + 4) = 0 hence, s = -4 is a zero of R(s) (b) For the denominator to be zero, s = 0 or s = -2 or s 2 − 8s + 25 = 0

i.e.

s=

( −8 )

− −8±

2

− 4(1)(25)

2(1)

=

8 ± −36 −8 ± j6 = = 4 ± j3 2 2

Hence, poles occur at s = 0, s = -2, s = 4 + j3 and 4 - j3 A pole-zero diagram is shown below.

3. For the function G(s) =

s −1 determine the poles and zeros and show them on a ( s + 2 ) ( s 2 + 2s + 5 )

pole-zero diagram. For the denominator to be zero, s = -2 or s 2 + 2s + 5 = 0

i.e.

s=

−2 ±

( 2)

2

− 4(1)(5)

2(1)

=

−2 ± −16 −2 ± j4 = = −1 ± j2 2 2

Hence, poles occur at s = -2, s = -1 + j2 and -1 – j2 © 2006 John Bird. All rights reserved. Published by Elsevier.

580

For the numerator to be zero, (s - 1) = 0 hence, s = 1 is a zero of G(s) A pole-zero diagram is shown below.

© 2006 John Bird. All rights reserved. Published by Elsevier.

581

CHAPTER 67 THE SOLUTION OF DIFFERENTIAL EQUATIONS USING LAPLACE TRANSFORMS EXERCISE 238 Page 648 1. A first order differential equation involving current i in a series R-L circuit is given by:

di E + 5i = dt 2

and i = 0 at time t = 0.

Use Laplace transforms to solve for i when (a) E = 20 (b) E = 40e −3t and (c) E = 50 sin 5t.

Taking the Laplace transform of each term of

di E + 5i = gives: dt 2

⎧ di ⎫ ⎧E ⎫ ℒ ⎨ ⎬ + 5ℒ{i} = ℒ ⎨ ⎬ ⎩2⎭ ⎩ dt ⎭ sℒ{i} – i(0) + 5ℒ{i} =

i.e. i = 0 at t = 0, hence,

E/2 s

i(0) = 0 (s + 5)ℒ{i} =

Hence,

ℒ{i} =

i.e.

E/2 s E/2 s(s + 5)

⎧ E / 2 ⎫ E −1 ⎧ 1 ⎫ i = ℒ −1 ⎨ ⎬ ⎬= ℒ ⎨ ⎩ s(s + 5) ⎭ ⎩ s(s + 5) ⎭ 2

and

1 A B A(s + 5) + Bs ≡ + = s(s + 5) s (s + 5) s(s + 5)

Let Hence,

1 = A(s + 5) + Bs

When s = 0:

1 = 5A

i.e.

When s = -5:

1 = -5B

i.e.

Thus,

A=

1 5

B=-

1 5

1 ⎫ ⎧1 E −1 ⎧ 1 ⎫ E −1 ⎪ 5 5 ⎪ = E ⎛ 1 − 1 e −5t ⎞ i= ℒ ⎨ ⎬= ℒ ⎨ − ⎬ ⎜ ⎟ 2 2 2 ⎝5 5 ⎠ ⎩ s(s + 5) ⎭ ⎪ s (s + 5) ⎪ ⎩ ⎭

(a) When E = 20,

i=

20 ⎛ 1 1 −5t ⎞ −5t ⎜ − e ⎟ = 2 1− e 2 ⎝5 5 ⎠

(

)

© 2006 John Bird. All rights reserved. Published by Elsevier.

582

⎧ 40e −3t ⎫ ⎧ di ⎫ ℒ ⎨ ⎬ + 5ℒ{i} = ℒ ⎨ ⎬ ⎩ dt ⎭ ⎩ 2 ⎭

(b) When E = 40e −3t

sℒ{i} – i(0) + 5ℒ{i} =

i.e. i = 0 at t = 0, hence,

20 s+3

i(0) = 0 (s + 5)ℒ{i} =

Hence,

ℒ{i} =

i.e.

20 s+3 20 (s + 3)(s + 5)

and

⎧ ⎫ 20 i = ℒ −1 ⎨ ⎬ ⎩ (s + 3)(s + 5) ⎭

Let

20 A B A(s + 5) + B(s + 3) ≡ + = (s + 3)(s + 5) (s + 3) (s + 5) (s + 3)(s + 5)

Hence,

20 = A(s + 5) + B(s + 3)

When s = -3:

20 = 2A

i.e.

A = 10

When s = -5:

20 = -2B

i.e.

B = -10

Thus,

⎧ ⎫ 20 10 ⎫ −1 ⎧ 10 − i = ℒ −1 ⎨ ⎬ =ℒ ⎨ ⎬ ⎩ (s + 3)(s + 5) ⎭ ⎩ (s + 3) (s + 5) ⎭

i.e.

i = 10e −3t − 10e −5t = 10 ( e−3t − e−5t ) ⎧ di ⎫ ⎧ 50sin 5t ⎫ ℒ ⎨ ⎬ + 5ℒ{i} = ℒ ⎨ ⎬ ⎩ dt ⎭ ⎩ 2 ⎭

(c) When E = 50 sin 5t

sℒ{i} – i(0) + 5ℒ{i} =

i.e. i = 0 at t = 0, hence, Hence, i.e.

and

Let

25(5) s 2 + 52

i(0) = 0 (s + 5)ℒ{i} = ℒ{i} =

125 s + 25 2

125 (s + 5)(s 2 + 25)

⎧ ⎫ 125 i = ℒ −1 ⎨ ⎬ 2 ⎩ (s + 5)(s + 25) ⎭ 125 A Bs + C A(s 2 + 25) + (Bs + C)(s + 5) ≡ + = (s + 5)(s 2 + 25) (s + 5) (s 2 + 25) (s + 5)(s 2 + 25)

© 2006 John Bird. All rights reserved. Published by Elsevier.

583

Hence,

125 = A ( s 2 + 25 ) + ( Bs + C )( s + 5 )

When s = -5:

125 = 50A

Equating s 2 coefficients:

0=A+B

Equating constant terms:

i.e. i.e.

125 = 25A + 5C

from which,

A=

5 2

B=-

i.e. 125 =

5 2

125 + 5C 2

125 25 C= 2 = 5 2

5 25 ⎫ ⎧ 5 − + s ⎪ ⎧ ⎫ 125 −1 2 + 2 2 ⎪ = ℒ i = ℒ −1 ⎨ ⎬ ⎨ ⎬ 2 2 ⎩ (s + 5)(s + 25) ⎭ ⎪ (s + 5) ( s + 25 ) ⎪ ⎩ ⎭

Thus,

⎧ 25 ⎫ ⎧ 5 ⎫ s ⎪ ⎪ ⎪ ⎪ 5 −1 ⎧ 1 ⎫ −1 −1 2 2 = ℒ ⎨ ⎬ +ℒ ⎨ 2 ⎬ -ℒ ⎨ 2 ⎬ 2 ⎩ (s + 5) ⎭ ⎪ ( s + 25 ) ⎪ ⎪ ( s + 25 ) ⎪ ⎩ ⎭ ⎩ ⎭

i.e.

=

⎫⎪ 5 ⎧ ⎫⎪ 5 −1 ⎧ 1 ⎫ 5 −1 ⎧⎪ 5 s −1 ⎪ ℒ ⎨ ⎬+ ℒ ⎨ 2 2 ⎬- ℒ ⎨ 2 2 ⎬ 2 2 ⎩ (s + 5) ⎭ 2 ⎪⎩ ( s + 5 ) ⎪⎭ ⎩⎪ ( s + 5 ) ⎭⎪

=

5 −5t 5 5 e + sin 5t − cos 5t 2 2 2

i=

5 −5t ( e + sin 5t − cos 5t ) 2

d2x 3. Use Laplace transforms to solve the differential equation: 2 + 100x = 0 given x(0) = 2 and dt x′(0) = 0. ⎧ d2x ⎫ ℒ ⎨ 2 ⎬ + ℒ{100x} = ℒ{0} ⎩ dx ⎭

i.e.

s 2 ℒ{x} – s x(0) - x′(0) + 100ℒ{x} = 0

x(0) = 2 and x′(0) = 0, hence i.e.

s 2 ℒ{x} – 2s + 100ℒ{x} = 0 ( s 2 + 100)ℒ{x} = 2s

© 2006 John Bird. All rights reserved. Published by Elsevier.

584

from which,

ℒ{x} =

2s s + 100 2

and

⎧ 2s ⎫ x = ℒ −1 ⎨ 2 ⎬ ⎩ s + 100 ⎭

⎧ s ⎫ x = 2ℒ −1 ⎨ 2 = 2 cos 10t 2⎬ ⎩ s + 10 ⎭

i.e.

4. Use Laplace transforms to solve the differential equation:

d 2i di + 1000 + 250000i = 0 2 dt dt

given i(0) = 0 and i′(0) = 100.

⎧ d 2i ⎫ ⎧ di ⎫ ℒ ⎨ 2 ⎬ + 1000ℒ ⎨ ⎬ + 250000ℒ{i} = ℒ{0} ⎩ dt ⎭ ⎩ dt ⎭

i.e. [ s 2 ℒ{i} – s i(0) - i′(0)] + 1000[sℒ{i} – i(0)] + 250000ℒ{i} = 0 i(0) = 0 and i′(0) = 100, hence s 2 ℒ{i} – 100 + 1000sℒ{i} + 250000ℒ{i} = 0 ( s 2 +1000s + 250000)ℒ{i} = 100

i.e.

ℒ{i} =

from which,

100 s + 1000s + 250000 2

⎧⎪ 100 ⎫⎪ ⎧ ⎫⎪ 1 −1 ⎪ i = ℒ −1 ⎨ = 100 ℒ ⎨ 2⎬ 1+1 ⎬ ⎪⎩ ( s + 500 ) ⎭⎪ ⎪⎩ ( s + 500 ) ⎭⎪

and

i = 100 t e −500t

i.e.

6. Use Laplace transforms to solve the differential equation: y(0) = −

d2 y dy − 2 + y = 3e 4x given 2 dx dx

2 1 and y′(0) = 4 3 3 ⎧ d2 y ⎫ ⎧ dy ⎫ ℒ ⎨ 2 ⎬ - 2ℒ ⎨ ⎬ + ℒ{y} = ℒ {3e 4x } ⎩ dx ⎭ ⎩ dx ⎭

i.e.

[ s 2 ℒ{y} – s y(0) - y′(0)] - 2[sℒ{y} – y(0)] + ℒ{y} =

3 (s − 4)

© 2006 John Bird. All rights reserved. Published by Elsevier.

585

y(0) = −

2 1 and y′(0) = 4 , hence 3 3

3 ⎛ 2 ⎞ 13 ⎛ 2⎞ - 2sℒ{y} + 2 ⎜ − ⎟ + ℒ{y} = s 2 ℒ{y} – s ⎜ − ⎟ (s − 4) ⎝ 3⎠ 3 ⎝ 3⎠ ( s 2 - 2s + 1)ℒ{y} +

i.e.

( s 2 - 2s + 1)ℒ{y} =

from which,

ℒ{y} =

3 2 17 9 − 2s(s − 4) + 17(s − 4) - s+ = (s − 4) 3 3 3(s − 4) 9 − 2s 2 + 8s + 17s − 68 −59 + 25s − 2s 2 = 2 3 ( s − 4 ) ( s 2 − 2s + 1) 3 ( s − 4 )( s − 1)

⎧⎪ −59 + 25s − 2s 2 ⎫⎪ y = ℒ −1 ⎨ 2 ⎬ ⎩⎪ 3 ( s − 4 )( s − 1) ⎭⎪

and

Let

2 13 4 3 s− − = 3 3 3 (s − 4)

1 2 −59 + 25s − 2s 2 ) ( A ( s − 1) + B ( s − 4 )( s − 1) + C ( s − 4 ) A B C 3 ≡ + + = 2 2 ( s − 4 ) ( s − 1) ( s − 1)2 ( s − 4 )( s − 1) ( s − 4 )( s − 1) −

Hence,

59 25 2 2 + s − s 2 = A ( s − 1) + B ( s − 4 )( s − 1) + C ( s − 4 ) 3 3 3 −

When s = 4:

59 100 32 + − = 9A + 0 + 0 3 3 3 −

When s = 1:

i.e. 3 = 9A

59 25 2 + − = 0 + 0 − 3C 3 3 3

Equating s 2 coefficients:

−

2 =A+B 3

i.e. -12 = -3C and i.e.

Hence,

⎧ 1 ⎫ ⎪ ⎪ 1 4 y = ℒ −1 ⎨ 3 − + 2⎬ ⎪ ( s − 4 ) ( s − 1) ( s − 1) ⎪ ⎩ ⎭

i.e.

y=

1 4x x e − e + 4x e x 3

or

and

−

2 1 = + B and 3 3

A=

1 3

C=4 B = -1

1 y = ( 4x − 1) e x + e4x 3

d 2 y dy + − 2y = 3cos 3x − 11sin 3x 8. Use Laplace transforms to solve the differential equation: dx 2 dx given y(0) = 0 and y′(0) = 6.

© 2006 John Bird. All rights reserved. Published by Elsevier.

586

⎧ d2 y ⎫ ⎧ dy ⎫ ℒ ⎨ 2 ⎬ + ℒ ⎨ ⎬ - 2ℒ{y} = ℒ {3cos 3x − 11sin 3x} ⎩ dx ⎭ ⎩ dx ⎭

i.e.

[ s 2 ℒ{y} – s y(0) - y′(0)] + [sℒ{y} – y(0)] - 2ℒ{y} =

3s 33 − 2 s +9 s +9 2

y(0) = 0 and y′(0) = 6, hence s 2 ℒ{y} - 6 + sℒ{y} - 2ℒ{y} =

( s 2 + s - 2)ℒ{y} = 6 +

i.e.

from which,

( s + s - 2)ℒ{y} = 2

6 ( s 2 + 9 ) + 3s − 33

ℒ{y} =

s2 + 9

=

3s − 33 s2 + 9

6s 2 + 3s + 21 s2 + 9

6s 2 + 3s + 21 ( s2 + 9 )( s2 + s − 2 )

⎧⎪ ⎫⎪ 6s 2 + 3s + 21 y = ℒ −1 ⎨ 2 ⎬ ⎪⎩ ( s + 9 ) ( s − 1)( s + 2 ) ⎭⎪

and

Let

3s − 33 s2 + 9

6s 2 + 3s + 21 A B Cs + D ≡ + + 2 2 ( s + 9 ) ( s − 1)( s + 2 ) ( s + 2 ) ( s − 1) ( s + 9 ) =

A ( s − 1) ( s 2 + 9 ) + B ( s + 2 ) ( s 2 + 9 ) + ( Cs + D )( s + 2 )( s − 1)

( s + 2 )( s − 1) ( s 2 + 9 )

6s 2 + 3s + 21 = A ( s − 1) ( s 2 + 9 ) + B ( s + 2 ) ( s 2 + 9 ) + ( Cs + D )( s + 2 )( s − 1)

Hence, When s = 1:

6 + 3 + 21 = B(3)(10)

When s = -2:

24 − 6 + 21 = A(−3)(13)

i.e. 30 = 30B

and

B=1

i.e. 39 = -39A and

A = -1

Equating s3 coefficients:

0 = A+B+C

Equating constant terms:

21 = -9A + 18B – 2D

i.e.

C=0

Hence,

⎧⎪ −1 ⎫⎪ 1 3 y = ℒ −1 ⎨ + + 2 2 ⎬ s + 2 ) ( s − 1) ( s + 3 ) ⎪ ⎩⎪ ( ⎭

i.e.

y = −e−2x + e x + sin 3x

i.e.

21 = 9 + 18 – 2D

and D = 3

or y = ex − e−2x + sin 3x

© 2006 John Bird. All rights reserved. Published by Elsevier.

587

d2 y dy − 2 + 2y = 3e x cos 2x given 9. Use Laplace transforms to solve the differential equation: 2 dx dx y(0) = 2 and y′(0) = 5. ⎧ d2 y ⎫ ⎧ dy ⎫ ℒ ⎨ 2 ⎬ - 2ℒ ⎨ ⎬ + 2ℒ{y} = ℒ {3e x cos 2x} ⎩ dx ⎭ ⎩ dx ⎭

⎛ ⎞ s −1 i.e. [ s 2 ℒ{y} – s y(0) - y′(0)] - 2[sℒ{y} – y(0)] + 2ℒ{y} = 3 ⎜ ⎟ ⎜ ( s − 1)2 + 22 ⎟ ⎝ ⎠ y(0) = 2 and y′(0) = 5, hence ⎛ ⎞ s −1 s 2 ℒ{y} – 2s - 5 - 2sℒ{y} + 4 + 2ℒ{y} = 3 ⎜ ⎟ ⎜ ( s − 1)2 + 22 ⎟ ⎝ ⎠ ⎛ ⎞ s −1 ( s 2 - 2s + 2)ℒ{y} = 2s + 1 + 3 ⎜ ⎟ ⎜ ( s − 1)2 + 22 ⎟ ⎝ ⎠

i.e.

=

=

( 3s − 3) + ( 2s + 1) ( s 2 − 2s + 5)

(s

2

− 2s + 5 )

3s − 3 + 2s3 − 4s 2 + 10s + s 2 − 2s + 5 ( s2 − 2s + 5)

2s3 − 3s 2 + 11s + 2 = ( s2 − 2s + 5) ℒ{y} =

from which,

⎧⎪ 2s3 − 3s 2 + 11s + 2 ⎫⎪ y = ℒ −1 ⎨ 2 ⎬ 2 ⎩⎪ ( s − 2s + 5 )( s − 2s + 2 ) ⎭⎪

and

Let

2s3 − 3s 2 + 11s + 2 As + B Cs + D ≡ 2 + 2 2 2 ( s − 2s + 5)( s − 2s + 2 ) ( s − 2s + 5) ( s − 2s + 2 ) =

Hence,

2s3 − 3s 2 + 11s + 2 ( s2 − 2s + 5)( s2 − 2s + 2 )

( As + B) ( s 2 − 2s + 2 ) + ( Cs + D ) ( s 2 − 2s + 5 )

(s

2

− 2s + 5 )( s 2 − 2s + 2 )

2s3 − 3s 2 + 11s + 2 = ( As + B ) ( s 2 − 2s + 2 ) + ( Cs + D ) ( s 2 − 2s + 5 )

Equating s3 coefficients:

2=A+C

(1)

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588

Equating s 2 coefficients:

-3 = -2A + B – 2C + D

(2)

Equating s coefficients:

11 = 2A – 2B + 5C – 2D

(3)

Equating constant terms:

2 = 2B + 5D

(4)

From (1), C = 2 – A and substituting C = 2 – A in (2):

-3 = -2A + B – 2(2 – A) + D

i.e.

1=B+D

(5)

2 × (5) gives:

2 = 2B + 2D

(6)

(4) – (6) gives:

0 = 3D

i.e. D = 0

From (4), if D = 0, then B = 1 In (2), if B = 1 and D = 0, then i.e.

-3 = -2A + 1 – 2C 4 = 2A + 2C

From (3),

11 = 2A – 2 + 5C

i.e.

13 = 2A + 5C

(8) – (7) gives:

9 = 3C

i.e.

(7)

(8)

C=3

In (1), if C = 3, then A = -1

Hence,

⎧⎪ ⎫⎪ 1− s 3s y = ℒ −1 ⎨ 2 + 2 ⎬ ⎩⎪ ( s − 2s + 5 ) ( s − 2s + 2 ) ⎭⎪

⎧⎪ ⎫⎪ ⎧ ⎫⎪ s −1 3s −1 ⎪ = -ℒ −1 ⎨ +ℒ ⎬ ⎨ ⎬ 2 2 2 2 ⎪⎩ ( s − 1) + 1 ⎭⎪ ⎩⎪ ( s − 1) + 2 ⎭⎪ ⎧⎪ ⎫⎪ ⎧ ⎫ s −1 −1 ⎪ 3 ( s − 1) + 3 ⎪ = -ℒ −1 ⎨ +ℒ ⎬ ⎨ ⎬ 2 2 2 2 ⎪⎩ ( s − 1) + 1 ⎭⎪ ⎩⎪ ( s − 1) + 2 ⎭⎪ ⎧⎪ ⎫⎪ ⎧ ⎫ ⎧ ⎫⎪ s −1 3 −1 ⎪ 3 ( s − 1) ⎪ −1 ⎪ = -ℒ −1 ⎨ +ℒ + ℒ ⎬ ⎨ ⎬ ⎨ ⎬ 2 2 2 2 2 2 ⎪⎩ ( s − 1) + 2 ⎪⎭ ⎪⎩ ( s − 1) + 1 ⎭⎪ ⎪⎩ ( s − 1) + 1 ⎭⎪ i.e.

y = −e x cos 2x + 3e x cos x + 3e x sin x

or

y = 3e x ( cos x + sin x ) − e x cos 2x

© 2006 John Bird. All rights reserved. Published by Elsevier.

589

CHAPTER 68 THE SOLUTION OF SIMULTANEOUS DIFFERENTIAL EQUATIONS USING LAPLACE TRANSFORMS EXERCISE 239 Page 654 2. Solve the following pair of simultaneous differential equations: 2

dy dx −y+x+ − 5sin t = 0 dt dt dy dx 3 + x − y + 2 − et = 0 dt dt

given that when t = 0, x = 0 and y = 0. Taking Laplace transforms of each term in each equation gives: 2[sℒ{y} – y(0)] - ℒ{y} + ℒ{x} + [sℒ{x} – x(0)] -

5 =0 s +1

3[sℒ{y} – y(0)] + ℒ{x} - ℒ{y} + 2[sℒ{x} – x(0)] -

2

1 =0 s −1

y(0) = 0 and x(0) = 0, hence

and

(2s – 1)ℒ{y} + (s + 1)ℒ{x} =

5 s +1

(1)

(3s – 1)ℒ{y} +(2s + 1)ℒ{x} =

1 s −1

(2)

2

(3s – 1) × (1) gives: (3s – 1)(2s – 1)ℒ{y} + (3s – 1)(s + 1)ℒ{x} = (3s – 1)

5 s +1

(3)

1 s −1

(4)

(2s – 1) × (2) gives: (2s – 1)(3s – 1)ℒ{y} + (2s – 1)(2s + 1)ℒ{x} = (2s – 1) (3) – (4) gives:

i.e.

i.e.

⎡( 3s 2 + 2s − 1) − ( 4s 2 − 1) ⎤ ℒ{x} = ⎣ ⎦

( −s

2

+ 2s ) ℒ{x} =

2

5 ( 3s − 1) 2s − 1 − s2 + 1 s −1

(5)

5 ( 3s − 1)( s − 1) − ( 2s − 1) ( s 2 + 1)

( s − 1) ( s 2 + 1)

⎧⎪15s 2 − 20s + 5 − 2s3 − 2s + s 2 + 1 ⎫⎪ ℒ{x} = − ⎨ ⎬ s ( s − 2 )( s − 1) ( s 2 + 1) ⎪⎩ ⎭⎪ ⎧⎪ 2s3 − 16s 2 + 22s − 6 ⎫⎪ = ⎨ ⎬ 2 ⎪⎩ s ( s − 2 )( s − 1) ( s + 1) ⎭⎪

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590

⎧⎪ 2s3 − 16s 2 + 22s − 6 ⎫⎪ x = ℒ −1 ⎨ ⎬ 2 ⎪⎩ s ( s − 2 )( s − 1) ( s + 1) ⎭⎪

and

Let

2s3 − 16s 2 + 22s − 6 A B C Ds + E ≡ + + + 2 2 s ( s − 2 )( s − 1) ( s + 1) s ( s − 2 ) ( s − 1) ( s + 1) =

A ( s − 2 )( s − 1) ( s 2 + 1) + B ( s )( s − 1) ( s 2 + 1) + C ( s )( s − 2 ) ( s 2 + 1) + ( Ds + E )( s )( s − 2 )( s − 1) s ( s − 2 )( s − 1) ( s 2 + 1)

from which, 2s3 − 16s 2 + 22s − 6 = A ( s − 2 )( s − 1) ( s 2 + 1) + B ( s )( s − 1) ( s 2 + 1) + C ( s )( s − 2 ) ( s 2 + 1) + ( Ds + E )( s )( s − 2 )( s − 1) When s = 0: When s = 1: When s = 2:

-6 = A(-2)(-1)(1) 2 – 16 + 22 – 6 = C(1)(-1)(2)

i.e. A = -3 i.e. C = -1

16 – 64 + 44 – 6 = B(2)(1)(5)

i.e. B = -1

Equating s 4 coefficients:

0=A+B+C+D

Equating s3 coefficients:

2 = -3A – B – 2C – 3D + E

i.e.

2 = 9 + 1 + 2 – 15 + E

i.e. D = 5

i.e. E = 5

Hence,

⎧⎪ 3 1 1 5s + 5 ⎫⎪ x = ℒ −1 ⎨− − − + 2 ⎬ ⎪⎩ s ( s − 2 ) ( s − 1) ( s + 1) ⎪⎭

i.e.

x = −3 − e−2t − e t + 5cos t + 5sin t

or

x = 5cos t + 5sin t − e 2t − et − 3

From equations (1) and (2) 5 s +1

(6)

1 s −1

(7)

(2s + 1) × (1) gives: (2s + 1)(2s – 1)ℒ{y} + (2s + 1)(s + 1)ℒ{x} = (2s + 1) (s + 1) × (2) gives:

(6) – (7):

i.e.

and

(s + 1)(3s – 1)ℒ{y} + (s + 1)(2s + 1)ℒ{x} = (s + 1)

⎡( 4s 2 − 1) − ( 3s 2 + 2s − 1) ⎤ ℒ{y} = ⎣ ⎦

( s2 − 2s ) ℒ{y} =

2

2 5 ( 2s + 1) ( s + 1) (10s + 5 )( s − 1) − ( s + 1) ( s + 1) − = s2 + 1 ( s − 1) ( s − 1) ( s 2 + 1)

10s 2 − 5s − 5 − s3 − s − s 2 − 1 ( s − 1) ( s 2 + 1)

−s3 + 9s 2 − 6s − 6 ℒ{y} = s ( s − 2 )( s − 1) ( s 2 + 1)

© 2006 John Bird. All rights reserved. Published by Elsevier.

591

⎧⎪ −s3 + 9s 2 − 6s − 6 ⎫⎪ y = ℒ −1 ⎨ ⎬ 2 ⎪⎩ s ( s − 1)( s − 2 ) ( s + 1) ⎭⎪

and

Let

A B C Ds + E −s3 + 9s 2 − 6s − 6 ≡ + + + 2 2 s ( s − 1)( s − 2 ) ( s + 1) s ( s − 1) ( s − 2 ) ( s + 1) =

from which,

A ( s − 1)( s − 2 ) ( s 2 + 1) + B ( s )( s − 2 ) ( s 2 + 1) + C ( s )( s − 1) ( s 2 + 1) + ( Ds + E )( s )( s − 1)( s − 2 ) s ( s − 1)( s − 2 ) ( s 2 + 1)

−s3 + 9s 2 − 6s − 6 = A ( s − 1)( s − 2 ) ( s 2 + 1) + B ( s )( s − 2 ) ( s 2 + 1) + C ( s )( s − 1) ( s 2 + 1) + ( Ds + E )( s )( s − 1)( s − 2 )

When s = 0: When s = 1: When s = 2:

-6 = A(-1)(-2)(1)

i.e. A = -3

-1 + 9 - 6 – 6 = B(1)(-1)(2)

i.e. B = 2

-8 + 36 - 12 – 6 = C(2)(1)(5)

i.e. C = 1

Equating s 4 coefficients:

0=A+B+C+D

Equating s3 coefficients:

-1 = -3A – 2B – C – 3D + E

i.e.

-1 = 9 - 4 - 1 + 0 + E

i.e. D = 0

i.e. E = -5

Hence,

⎧⎪ 3 2 1 5 ⎫⎪ y = ℒ −1 ⎨− + + − 2 ⎬ ⎪⎩ s ( s − 1) ( s − 2 ) ( s + 1) ⎪⎭

i.e.

y = −3 + 2e t + e 2t − 5sin t

or

y = e2t + 2et − 3 − 5sin t

3. Solve the following pair of simultaneous differential equations: d2x + 2x = y dt 2 d2 y + 2y = x dt 2 given that when t = 0, x = 4, y = 2,

dx dy = 0 and = 0. dt dt

Taking Laplace transforms of each term in each equation gives: [ s 2 ℒ{x} – s x(0) - x′(0)] + 2ℒ{x} = ℒ{y} and

[ s 2 ℒ{y} – s y(0) - y′(0)] + 2ℒ{y} = ℒ{x}

x(0) = 4 and x′(0), hence

[ s 2 ℒ{x} – 4s] + 2ℒ{x} = ℒ{y}

© 2006 John Bird. All rights reserved. Published by Elsevier.

592

y(0) = 2 and y′(0) = 0, hence [ s 2 ℒ{y} – 2s] + 2ℒ{y} = ℒ{x} i.e.

( s 2 +2)ℒ{x} - ℒ{y} = 4s

(1)

and

- ℒ{x} + ( s 2 +2)ℒ{y} = 2s

(2)

( s 2 +2) × (2) gives: - ( s 2 +2)ℒ{x} + ( s 2 + 2) ( s 2 +2)ℒ{y} = 2s( s 2 +2)

(3)

2 2 ⎡ 2 ⎤ ⎢⎣( s + 2 ) − 1⎥⎦ ℒ{y} = 2s( s +2) + 4s

(1) + (3) gives:

(s

i.e.

4

+ 4s 2 + 3) ℒ{y} = 2s3 + 8s ℒ{y} =

and

2s3 + 8s 2s3 + 8s = ( s4 + 4s2 + 3) ( s2 + 3)( s2 + 1)

⎧⎪ 2s3 + 8s ⎫⎪ y = ℒ −1 ⎨ 2 ⎬ 2 ⎪⎩ ( s + 3)( s + 1) ⎭⎪

and

2 2 2s3 + 8s As + B Cs + D ( As + B ) ( s + 1) + ( Cs + D ) ( s + 3) ≡ + = Let ( s2 + 3)( s2 + 1) ( s2 + 3)( s2 + 1) ( s2 + 3) ( s2 + 1)

2s3 + 8s = ( As + B ) ( s 2 + 1) + ( Cs + D ) ( s 2 + 3)

from which,

Equating s3 coefficients:

2=A+C

(4)

Equating s 2 coefficients:

0=B+D

(5)

Equating s coefficients:

8 = A + 3C

(6)

(6) – (4) gives:

6 = 2C

i.e.

C=3

and from (4),

A=2–C

i.e.

A = -1

Equating constant terms:

0 = B + 3D

(7) – (5) gives:

0 = 2D

Hence,

i.e.

(7) i.e.

D=0

⎧ ⎧⎪ −s ⎫⎪ 3s s −1 ⎪ y = ℒ −1 ⎨ 2 + 2 ⎬ = ℒ ⎨− ⎪ s2 + 3 ⎩⎪ ( s + 3) ( s + 1) ⎭⎪ ⎩

(

y = − cos

(

)

3 t + 3cos t

or

and from (5), B = 0

( )) 2

+

y = 3 cos t - cos

(

(

3s s2 + 3t

( )) 2

1

⎫ ⎪ ⎬ ⎪ ⎭

)

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593

If y = 3 cos t - cos

(

3t

)

dy = −3sin t + 3 sin 3 t dt

then

d2 y = −3cos t + 3cos 3 t dt 2

and

Since from one of the original equations, d2 y + 2y = x dt 2 then

−3cos t + 3cos

(

) (

3 t + 2 3cos t − cos

i.e.

x = −3cos t + 3cos

i.e.

x = 3 cos t + cos

(

(

)

(

3t

)) = x

3 t + 6 cos t − 2 cos 3t

(

3t

)

)

© 2006 John Bird. All rights reserved. Published by Elsevier.

594

CHAPTER 69 FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD 2π EXERCISE 240 Page 661 1. Determine the Fourier series for the periodic function: ⎧ − 2, when − π ≤ x ≤ 0 f(x) = ⎨ ⎩ + 2, when 0 ≤ x ≤ π which is periodic outside this range of period 2π. The periodic function is shown in the diagram below.

∞

The Fourier series is given by: f(x) = a 0 + ∑ ( a n cos nx + b n sin nx )

(1)

n =1

a0 =

1 π 1 f (x) dx = ∫ 2π −π 2π

{∫

π

0 −π

} 21π {[−2x]

−2 dx + ∫ 2 dx = 0

= an =

1 π 1 f (x) cos nx dx = ∫ −π π π

{∫

0 −π

+ [ 2x ] 0

π

0 −π

}

1 {[(0) − (2π)] + [(2π) − (0)]} = 0 2π

}

π

−2 cos nx dx + ∫ 2 cos nx dx 0

0 π 1 ⎪⎧ ⎡ 2 ⎤ ⎡2 ⎤ ⎪⎫ = ⎨ ⎢ − sin nx ⎥ + ⎢ sin nx ⎥ ⎬ = 0 π ⎪⎩ ⎣ n ⎦ −π ⎣ n ⎦ 0 ⎪⎭

bn =

1 π 1 f (x) sin nx dx = ∫ −π π π =

0

−π

π

bn =

}

−2sin nx dx + ∫ 2sin nx dx 0

0 π 1 ⎧⎪ ⎡ 2 ⎤ ⎡ 2 ⎤ ⎫⎪ 2 + − cos nx cos nx ⎨ ⎥⎦ ⎢⎣ n ⎥⎦ ⎬ = πn {[ cos 0 − cos n(−π)] − [ cos nπ − cos 0]} π ⎪⎩ ⎢⎣ n −π 0⎪ ⎭

bn =

When n is even, When n is odd,

{∫

2 {[1 − 1] − [1 − 1]} = 0 πn

2 2 8 [1 − −1] − [ −1 − 1]} = ( 4 ) = { πn πn πn

© 2006 John Bird. All rights reserved. Published by Elsevier.

595

Hence,

b1 =

8 , π

b3 =

8 , 3π

8 and so on. 5π

b5 =

Substituting onto equation (1) gives: f(x) = 0 + 0 +

f(x) =

i.e.

8 8 8 sin x + sin 3x + sin 5x + ..... π 3π 5π

8⎛ 1 1 ⎞ sin x + sin 3x + sin 5x + ...... ⎟ ⎜ π⎝ 3 5 ⎠

2. For the Fourier series in problem 1, deduce a series for

When x = i.e.

π , f(x) = 2, hence 2

π π at the point where x = 4 2

2=

8 π 8 3π 8 5π sin + sin + sin + ..... π 2 3π 2 5π 2

2=

8⎛ 1 1 1 ⎞ ⎜1 − + − + .... ⎟ π⎝ 3 5 7 ⎠

and

2π 1 1 1 = 1 − + − + ..... 8 3 5 7

i.e.

π 1 1 1 = 1 − + − + .... 4 3 5 7

5. Find the term representing the third harmonic for the periodic function of period 2π given by: ⎧ 0, when − π ≤ x ≤ 0 f(x) = ⎨ ⎩ 1, when 0 ≤ x ≤ π The periodic function is shown in the diagram below.

π

1 π 1 π 1 ⎡ sin nx ⎤ =0 a n = ∫ f (x) cos nx dx = ∫ 1cos nx dx = ⎢ 0 −π π π π ⎣ n ⎥⎦ 0 bn =

1 π 1 π f (x) sin nx dx = ∫ sin nx dx ∫ π −π π 0

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596

π 1 ⎧⎪ ⎡ cos nx ⎤ ⎫⎪ 1 1 − {1 − cos nπ} ⎨⎢ ⎬ = − {cos nπ − cos 0} = ⎥ π ⎪⎩ ⎣ πn πn n ⎦ 0 ⎪⎭

=

The third harmonic is when n = 3, 1 1 2 {1 − cos 3π} = {1 − −1} = 3π 3π 3π

b3 =

i.e.

∞

Since the Fourier series is given by: f(x) = a 0 + ∑ ( a n cos nx + b n sin nx ) , n =1

2 sin 3x 3π

the 3rd harmonic term is:

6. Determine the Fourier series for the periodic function of period 2π defined by: ⎧ ⎪ 0, when − π ≤ t ≤ 0 ⎪ π ⎪ f(t) = ⎨ 1, when 0 ≤ t ≤ 2 ⎪ π ⎪ ⎪⎩−1, when 2 ≤ t ≤ π The function has a period of 2π. The periodic function is shown in the diagram below.

a0 =

1 π 1 f (t) dt = ∫ −π 2π 2π

{∫

0 −π

0 dt + ∫

π/ 2 0

1dt + ∫

π π/ 2

} 21π {[ t ]

−1dt =

= an =

1 π 1 f (t) cos nt dt = ∫ −π π π

{∫

π/ 2 0

cos nt dt + ∫

π π/ 2

π/2 0

+ [−t] π / 2 π

}

⎤ ⎡ 1 ⎧ ⎡⎛ π ⎞ ⎛ π ⎞⎤ ⎫ ⎨ ⎢⎜ ⎟ − ( 0 ) ⎥ + ⎢( −π ) − ⎜ − ⎟ ⎥ ⎬ = 0 2π ⎩ ⎣⎝ 2 ⎠ ⎝ 2 ⎠⎦ ⎭ ⎦ ⎣

}

− cos nx dx

π/ 2 π 1 ⎧⎪ ⎡ sin nt ⎤ nπ ⎤ ⎫ ⎡ sin nt ⎤ ⎫⎪ 1 ⎧ ⎡ nπ ⎤ ⎡ = ⎨⎢ −⎢ − 0 ⎥ − ⎢sin nπ − sin ⎥ ⎬ ⎬= ⎨ ⎢sin ⎥ ⎥ π ⎪⎩ ⎣ n ⎦ 0 2 2 ⎦⎭ ⎣ n ⎦ π / 2 ⎪⎭ nπ ⎩ ⎣ ⎦ ⎣

© 2006 John Bird. All rights reserved. Published by Elsevier.

597

When n is even,

an = 0

When n = 1,

a1 =

1 ⎧⎡ π ⎤ ⎡ π ⎤⎫ 1 2 ⎨ ⎢sin − 0 ⎥ − ⎢sin π − sin ⎥ ⎬ = {[1 − 0] − [ 0 − 1]} = π ⎩⎣ 2 ⎦ ⎣ 2 ⎦⎭ π π

When n = 3,

a3 =

1 ⎧ ⎡ 3π ⎤ ⎡ 3π ⎤ ⎫ 1 2 ⎨ ⎢sin − 0 ⎥ − ⎢sin 3π − sin ⎥ ⎬ = {[ −1 − 0] − [ 0 − −1]} = − 3π ⎩ ⎣ 2 2 ⎦ ⎭ 3π 3π ⎦ ⎣

When n = 5,

a5 =

1 ⎧ ⎡ 5π ⎤ ⎡ 5π ⎤ ⎫ 1 2 ⎨ ⎢sin − 0 ⎥ − ⎢sin 5π − sin ⎥ ⎬ = {[1 − 0] − [ 0 − 1]} = 5π ⎩ ⎣ 2 2 ⎦ ⎭ 5π 5π ⎦ ⎣

It follows that

a7 = −

bn =

2 , 7π

1 π 1 f (t) sin nt dt = ∫ π −π π

{∫

a9 = π/ 2 0

2 9π

and so on.

sin nt dt + ∫

π π/2

}

− sin nt dt

π/ 2 π 1 ⎧⎪ ⎡ cos nt ⎤ nπ nπ ⎤ ⎫ ⎡ cos nt ⎤ ⎫⎪ 1 ⎧ ⎡ ⎤ ⎡ +⎢ = ⎨⎢− ⎬= ⎨− ⎢ cos − cos 0 ⎥ + ⎢ cos nπ − cos ⎥ ⎬ ⎥ ⎥ n ⎦0 2 2 ⎦⎭ π ⎩⎪ ⎣ ⎣ n ⎦ π / 2 ⎭⎪ πn ⎩ ⎣ ⎦ ⎣

=

1 ⎧⎛ nπ ⎞ ⎛ nπ ⎞ ⎫ 1 ⎧ nπ ⎫ ⎨⎜1 − cos ⎟ + ⎜ cos nπ − cos ⎟ ⎬ = ⎨1 − 2 cos + cos nπ ⎬ πn ⎩⎝ 2 ⎠ ⎝ 2 ⎠ ⎭ πn ⎩ 2 ⎭

When n is odd,

bn =

1 {1 − 0 − 1} = 0 πn

When n is even,

b2 =

1 4 2 {1 − 2(−1) + 1} = = , 2π 2π π

b4 =

1 {1 − 2(1) + 1} = 0 , 4π

b6 =

1 4 2 {1 − 2(−1) + 1} = = 6π 6π 3π

Similarly,

b8 = 0 ,

b10 =

2 , and so on. 5π

∞

Substituting into f(t) = a 0 + ∑ ( a n cos nt + b n sin nt ) n =1

gives:

f(x) = 0 +

2 2 2 2 cos t − cos 3t + cos 5t − cos 7t + ..... π 3π 5π 7π +

i.e. f(x) =

2 2 2 sin 2t + sin 6t + sin10t + .... π 3π 5π

2⎛ 1 1 1 1 ⎞ cos t − cos 3t + cos 5t − ...... + sin 2t + sin 6t + sin10t + ..... ⎟ ⎜ π⎝ 3 5 3 5 ⎠

© 2006 John Bird. All rights reserved. Published by Elsevier.

598

7. Show that the Fourier series for the periodic function of period 2π defined by: when − π ≤ θ ≤ 0 ⎧ 0, f(θ) = ⎨ ⎩ sin θ, when 0 ≤ θ ≤ π is given by:

f ( θ) =

⎞ 2 ⎛ 1 cos 2θ cos 4θ cos 6θ − − − ..... ⎟ ⎜ − π⎝ 2 (3) (3)(5) (5)(7) ⎠

The periodic function is shown in the diagram below.

a0 =

1 π 1 f (θ) dθ = ∫ −π 2π 2π

{∫

0 −π

} 21π {[− cos θ] } = 21π {⎡⎣( − cos π) − ( − cos 0)⎤⎦}

π

0

= an = =

1 π

{∫

−π

0

1 1 1 (1 − cos π ) = (1 − −1) = π 2π 2π

}

π

0

π

0 dθ + ∫ sin θ dθ =

0 cos nθ dθ + ∫ sin θ cos nθ dθ 0

1⎧ π1 ⎫ 1 ⎨ ∫ 0 ⎡⎣sin ( θ + nθ ) + sin ( θ − nθ ) ⎤⎦ ⎬ = π⎩ 2 ⎭ 2π

{∫ sin θ (1 + n ) + sin θ (1 − n ) dθ} from 6, page 398 π

0

of textbook π

θ (1 − n ) ⎤ 1 ⎡ cos θ (1 + n ) 1 ⎡⎛ cos π (1 + n ) cos π (1 − n ) ⎞ ⎛ 1 1 ⎞⎤ − cos − − = ⎢⎜ − ⎟−⎜− ⎢− ⎥ = ⎟⎥ 2π ⎣ 1+ n 1 − n ⎦ 0 2π ⎢⎣⎝ 1+ n 1− n ⎠ ⎝ 1 + n 1 − n ⎠ ⎥⎦ When n is odd, a n =

1 ⎡ 1 1 1 1 ⎤ − − + + =0 ⎢ 2π ⎣ 1 + n 1 − n 1 + n 1 − n ⎥⎦

When n = 2,

a2 =

1 ⎧ cos 3π cos(−π) 1 1 ⎫ 1 ⎧ 1 1 ⎫ 1 ⎧ 4⎫ 2 − + + ⎬= ⎨− ⎨ − 1 + − 1⎬ = ⎨− ⎬ = − 2π ⎩ 3 −1 3 −1 ⎭ 2 π ⎩ 3 3 ⎭ 2π ⎩ 3 ⎭ 3π

When n = 4,

a4 =

1 ⎧ cos 5π cos(−3π) 1 1 ⎫ 1 ⎧ 1 1 1 1 ⎫ − + + ⎬= ⎨− ⎨ − + − ⎬ −3 2π ⎩ 5 5 −3 ⎭ 2 π ⎩ 5 3 5 3 ⎭

=

When n = 6,

a6 =

1 ⎧3 − 5 + 3 − 5 ⎫ 1 ⎧ 4 ⎫ 2 ⎨ ⎬= ⎨− ⎬=− 2π ⎩ (3)(5) ⎭ 2π ⎩ (3)(5) ⎭ π(3)(5) 1 ⎧ cos 7π cos(−5π) 1 1 ⎫ 1 ⎧ 1 1 1 1 ⎫ − + + ⎬= ⎨− ⎨ − + − ⎬ −5 2π ⎩ 7 7 −5 ⎭ 2π ⎩ 7 5 7 5 ⎭

© 2006 John Bird. All rights reserved. Published by Elsevier.

599

= bn =

1 π

{∫

1 ⎧5 − 7 + 5 − 7 ⎫ 1 ⎧ 4 ⎫ 2 ⎨ ⎬= ⎨− ⎬=− 2π ⎩ (5)(7) ⎭ 2π ⎩ (5)(7) ⎭ π(5)(7)

}

π

0 −π

0sin nθ dθ + ∫ sin θ sin nθ dθ 0

π

1⎧ π 1 1 ⎡ sin θ(1 + n) sin θ(1 − n) ⎤ ⎫ = ⎨ ∫ − ⎡⎣cos ( θ + nθ ) − cos ( θ − nθ ) ⎤⎦ ⎬ = − =0 − 0 2 2π ⎣⎢ 1 + n 1 − n ⎦⎥ 0 π⎩ ⎭

from 9, page 398, of textbook

∞

Substituting into f(θ) = a 0 + ∑ ( a n cos nθ + b n sin nθ ) n =1

gives:

f(θ) =

1 2 2 2 − cos 2θ − cos 4θ − cos 6θ − ..... + 0 (3)(5)π (5)(7)π π 3π

i.e.

f(θ) =

⎞ 2 ⎛ 1 cos 2θ cos 4θ cos 6θ − − − ...... ⎟ ⎜ − π⎝2 (3) (3)(5) (5)(7) ⎠

© 2006 John Bird. All rights reserved. Published by Elsevier.

600

CHAPTER 70 FOURIER SERIES FOR A NON-PERIODIC FUNCTION OVER RANGE 2π EXERCISE 241 Page 667 2. Determine the Fourier series for the function defined by:

⎧ 1 − t, when − π ≤ t ≤ 0 f(t) = ⎨ ⎩ 1 + t, when 0 ≤ t ≤ π Draw a graph of the function within and outside of the given range. The periodic function is shown in the diagram below.

1 π 1 a0 = f (t) dt = ∫ 2π −π 2π =

an = =

{∫

0 −π

{

0

}

0 π ⎡ t 2 ⎤ ⎫⎪ 1 ⎧⎪ ⎡ t 2 ⎤ (1 + t) dt = + t+ ⎨ t− ⎬ 2π ⎪ ⎢⎣ 2 ⎥⎦ −π ⎢⎣ 2 ⎥⎦ 0 ⎪ ⎩ ⎭

⎤ ⎫⎪ 1 ⎧⎪ ⎛ ⎛ 1 ⎧⎪ ⎡ π 2 ⎞ ⎤ ⎡⎛ π2 ⎞ π2 ⎞ ⎫⎪ 2π 2π2 π + = 1+ ⎨ ⎢( 0 ) − ⎜ −π − ⎟ ⎥ + ⎢⎜ π + ⎟ − ( 0 ) ⎥ ⎬ = ⎨2 ⎜ π + ⎟ ⎬ = 2π ⎩⎪ ⎣ 2 ⎠ ⎦ ⎣⎝ 2 ⎠ 2 ⎠ ⎭⎪ 2π 4π 2 ⎝ ⎦ ⎭⎪ 2π ⎩⎪ ⎝

1 π 1 f (t) cos nt dt = ∫ π −π π 1 π

(1 − t) dt + ∫

π

{∫

0 −π

}

π

(1 − t) cos nt dt + ∫ (1 + t) cos nt dt 0

}

π

∫ −π ( cos nt − t cos nt ) dt + ∫ 0 ( cos nt + t cos nt ) dt 0

π

0

1 ⎡ sin nt t sin nt cos nt ⎤ ⎡ sin nt t sin nt cos nt ⎤ = ⎢ − − 2 ⎥ +⎢ + + 2 ⎥ by integration by parts n n ⎦ −π ⎣ n n n ⎦0 π⎣ n =

1 ⎧ ⎡⎛ cos 0 ⎞ ⎛ cos(− nπ) ⎞ ⎤ ⎡⎛ cos nπ ⎞ ⎛ cos 0 ⎞ ⎤ ⎫ ⎨ ⎢⎜ 0 − 0 − 2 ⎟ − ⎜ 0 − 0 − ⎟ ⎥ + ⎢⎜ 0 + 0 + ⎟ − ⎜ 0 + 0 + 2 ⎟⎥ ⎬ 2 2 n ⎠ ⎝ n n ⎠ ⎝ n ⎠⎦ ⎭ π ⎩ ⎣⎝ ⎠ ⎦ ⎣⎝

=

1 ⎧ 1 cos(− nπ) cos nπ 1 ⎫ 1 + − 2 ⎬ = 2 ( 2 cos nπ − 2 ) ⎨− 2 + 2 2 π⎩ n n n n ⎭ πn =

When n is even,

since cos(-nπ) = cos nπ

2 ( cos nπ − 1) πn 2

an = 0

© 2006 John Bird. All rights reserved. Published by Elsevier.

601

2 4 −1 − 1) = − 2 ( π(1) π

When n = 1,

a1 ==

When n = 3,

a3 =

2 4 −1 − 1) = − 2 ( π(3) π(3) 2

When n = 5,

a5 =

2 4 −1 − 1) = − 2 ( π(5) π(5) 2

bn =

1 π 1 f (t) sin nt dt = ∫ π −π π

{∫

{∫

0 −π

and so on.

}

π

(1 − t) sin nt dt + ∫ (1 + t) sin nt dt 0

}

π

=

1 π

=

1 ⎡ cos nt t cos nt sin nt ⎤ ⎡ cos nt t cos nt sin nt ⎤ − + − + − − + 2 ⎥ by integration by parts 2 n n n ⎥⎦ −π ⎢⎣ n n n ⎦0 π ⎢⎣

0 −π

( sin nt − t sin nt ) dt + ∫ 0 ( sin nt + t sin nt ) dt 0

π

⎧ ⎡⎛ cos 0 ⎫ ⎞ ⎛ cos(− nπ) π cos(− nπ) ⎞⎤ + 0 − 0⎟ − ⎜ − − − 0 ⎟⎥ ⎪ ⎢⎜ − ⎪ n n n ⎠ ⎝ ⎠⎦ 1 ⎪⎣⎝ ⎪ = ⎨ ⎬ π⎪ ⎡⎛ cos nπ π cos nπ ⎞ ⎛ cos 0 ⎞⎤ ⎪ + ⎢⎜ − − + 0⎟ − ⎜ − + 0 + 0 ⎟⎥ ⎪ n n n ⎠ ⎝ ⎠ ⎦ ⎪⎭ ⎣⎝ ⎩ =

1 ⎧ 1 cos(−nπ) π cos(−nπ) cos nπ π cos nπ 1 ⎫ + − − + ⎬ =0 ⎨− + n n n n n⎭ π⎩ n

since cos nπ = cos(-nπ)

∞

Substituting into f(t) = a 0 + ∑ ( a n cos nt + b n sin nt ) n =1

gives:

f(t) =

π 4 4 4 cos 3t − cos 5t − ..... + 0 + 1 − cos t − 2 π π(3) π(5) 2 2

i.e.

f(t) =

π 4⎛ cos 3t cos 5t ⎞ + 1 − ⎜ cos t + + + ...... ⎟ 2 2 2 3 5 π⎝ ⎠

4. Determine the Fourier series up to and including the third harmonic for the function defined by:

when 0 ≤ x ≤ π ⎧ x, f(x) = ⎨ ⎩ 2π − x, when π ≤ x ≤ 2π Sketch a graph of the function within and outside of the given range, assuming the period is 2π. The periodic function is shown in the diagram below.

© 2006 John Bird. All rights reserved. Published by Elsevier.

602

1 π 1 a0 = f (x) dx = ∫ 2π −π 2π

0

x dx + ∫

2π π

}

1 ⎧⎪ ⎡ x 2 ⎤ ⎡ x2 ⎤ (2π − x) dx = 2 x + π − ⎨ 2π ⎪ ⎢⎣ 2 ⎥⎦ 0 ⎢⎣ 2 ⎥⎦ π ⎩ π

2π

{ }

1 2π 1 f (x) cos nx dx = ∫ 0 π π

{∫

π 0

}

2π

x cos nx dx + ∫ (2π − x) cos nx dx π

1 ⎪⎧ ⎡ x sin nx cos nx ⎤ ⎡ 2π sin nx x sin nx cos nx ⎤ + + − − = ⎨⎢ n 2 ⎥⎦ 0 ⎢⎣ n n n 2 ⎥⎦ π π ⎪⎩ ⎣ n π

2π

⎪⎫ ⎬ by integration by parts ⎪⎭

=

1 ⎧ ⎡⎛ cos nπ ⎞ ⎛ cos 0 ⎞ ⎤ ⎡⎛ cos 2πn ⎞ ⎛ cos nπ ⎞ ⎤ ⎫ − 0 + + 0 − 0 − − 0 − 0 − ⎨ ⎢⎜ 0 + ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎬ n2 ⎠ ⎝ n 2 ⎠ ⎥⎦ ⎢⎣⎝ n2 ⎠ ⎝ n 2 ⎠ ⎥⎦ ⎭ π ⎩ ⎣⎝

=

1 2 cos nπ − 1 − cos 2πn + cos nπ} = 2 ( cos nπ − 1) 2 { πn πn

When n is even,

an = 0

When n = 1,

a1 =

2 4 −1 − 1) = − 2 ( π(1) π

When n = 3,

a3 =

2 4 −1 − 1) = − 2 ( π(3) π(3) 2

When n = 5,

a5 =

2 4 −1 − 1) = − 2 ( π(5) π(5) 2

bn =

1 2π 1 f (x) sin nx dx = ∫ π 0 π

{∫ π

=

⎫⎪ ⎬ ⎪⎭

⎤ ⎡⎛ 2 4π2 ⎞ ⎛ 2 π2 ⎞ ⎤ ⎫⎪ 1 1 ⎧⎪ ⎡⎛ π2 ⎞ π π2 ) = ( ⎨ ⎢⎜ ⎟ − ( 0 ) ⎥ + ⎢ ⎜ 4π − ⎟ − ⎜ 2π − ⎟ ⎥ ⎬ = 2π ⎩⎪ ⎣⎝ 2 ⎠ 2 ⎠ ⎝ 2 ⎠ ⎦ ⎭⎪ 2π 2 ⎦ ⎣⎝

=

an =

{∫

π

0 −π

and so on. π

}

x sin nx dx + ∫ (2π − x) sin nx dx 0

2π

1 ⎡ x cos nx sin nx ⎤ ⎡ 2π cos nx x cos nx sin nx ⎤ − + + − + − n n 2 ⎥⎦ 0 ⎢⎣ n n n 2 ⎥⎦ π π ⎢⎣

by integration by parts

⎧ ⎡⎛ 2π cos 2nπ 2π cos 2nπ ⎤⎫ ⎞ + + 0⎟ ⎪ ⎢⎜ − ⎥⎪ n n 1 ⎪ ⎡⎛ π cos nπ ⎤ ⎢⎝ ⎠ ⎞ ⎥ ⎪⎬ = ⎨ ⎢⎜ − + 0 ⎟ − ( 0 + 0 )⎥ + π ⎪ ⎣⎝ n ⎠ ⎛ 2π cos nπ π cos nπ ⎞⎥ ⎦ ⎢ −⎜− + − 0 ⎟⎥ ⎪ ⎢ n n ⎝ ⎠ ⎦ ⎭⎪ ⎣ ⎩⎪

© 2006 John Bird. All rights reserved. Published by Elsevier.

603

=

1 {−π cos nπ + 2π cos nπ − π cos nπ} = 0 nπ ∞

Substituting into f(x) = a 0 + ∑ ( a n cos nx + b n sin nx ) n =1

gives:

f(x) =

π 4 4 4 cos 3x − cos 5x − ..... + 0 − cos x − 2 π(3) π(5) 2 2 π

i.e.

f(x) =

π 4⎛ cos 3x cos 5x ⎞ − ⎜ cos x + + + ...... ⎟ 2 2 2 π⎝ 3 5 ⎠

5. Expand the function f ( θ ) = θ2 in a Fourier series in the range - π < θ < π

Sketch the function within and outside of the given range. The periodic function is shown in the diagram below.

1 π 1 a0 = f (x) dx = ∫ 2π −π 2π

an =

{

}

π 1 ⎧⎪ ⎡ θ3 ⎤ ⎫⎪ 1 3 2 π3 π 2 3 θ θ = = π − −π = = d { } ⎨ ⎬ ∫ −π 2π ⎪ ⎢⎣ 3 ⎥⎦ −π ⎪ 6π 6π 3 ⎩ ⎭ π

2

1 π 1 f (θ) cos nθ dθ = ∫ π −π π

{∫

π −π

}

θ2 cos nθ dθ π

1 ⎡ θ2 sin nθ 2θ cos nθ 2sin nθ ⎤ + − = ⎢ by integration by parts π⎣ n n2 n 3 ⎥⎦ −π =

1 ⎧ ⎡⎛ 2π cos nπ 2π cos(− nπ) ⎞ ⎛ ⎞ ⎤ ⎫ 4π cos nπ 4 − 0 − 0 − − 0 = 2 cos nπ ⎨ ⎢⎜ 0 + ⎟ ⎜ ⎟⎥ ⎬ = π ⎩ ⎣⎝ πn 2 n2 n2 n ⎠ ⎝ ⎠⎦ ⎭

since cos nπ = cos(-nπ) Hence, bn =

a1 =

4 4 −1) = − 2 , 2 ( (1) 1

1 π 1 f (θ) sin nθ dθ = ∫ −π π π

{∫

π −π

a2 =

4 4 1 = 2, 2 ( ) (2) 2

a3 =

4 4 −1) = − 2 , 2 ( (3) 3

a4 =

4 , and so on. 42

}

θ2 sin nθ dθ π

1 ⎡ θ2 cos nθ 2θ sin nθ 2 cos nθ ⎤ + + = ⎢− by integration by parts π⎣ n n2 n 3 ⎥⎦ −π © 2006 John Bird. All rights reserved. Published by Elsevier.

604

1 ⎧⎪ ⎡⎛ π2 cos nπ 2 cos nπ ⎞ ⎛ π2 cos(− nπ) 2 cos(− nπ) ⎞ ⎤ ⎫⎪ +0+ − − + + 0 = ⎨ ⎢⎜ − ⎟ ⎜ ⎟⎥ ⎬ = 0 n n3 ⎠ ⎝ n n3 π ⎩⎪ ⎣⎝ ⎠ ⎦ ⎭⎪ ∞

Substituting into f(θ) = a 0 + ∑ ( a n cos nθ + b n sin nθ ) n =1

gives:

f(θ) =

π2 4 4 4 4 − 2 cos θ + 2 cos 2θ − 2 cos 3θ + 2 cos 4θ − ..... + 0 2 3 4 3 1

i.e.

f(θ) =

π2 1 1 ⎛ ⎞ − 4 ⎜ cos θ − 2 cos 2θ + 2 cos 3θ − ...... ⎟ 3 2 3 ⎝ ⎠

6. For the Fourier series in problem 5, let θ = π and deduce a series for

∞

1

∑n n =1

2

When θ = π in Problem 5 above, f(θ) = π2 Thus,

π =

π2 1 1 1 ⎛ ⎞ − 4 ⎜ cos π − 2 cos 2π + 2 cos 3π − 2 cos 4π + ...... ⎟ 3 2 3 4 ⎝ ⎠

i.e.

π2 =

π2 4 4 4 4 + + + + + ..... 3 12 22 32 42

2

π2 −

i.e.

π2 ⎛1 1 1 1 ⎞ = 4 ⎜ 2 + 2 + 2 + 2 + ..... ⎟ 3 ⎝1 2 3 4 ⎠

2π 2 ⎛1 1 1 1 ⎞ = 4 ⎜ 2 + 2 + 2 + 2 + ..... ⎟ 3 ⎝1 2 3 4 ⎠

i.e.

and

2π2 1 1 1 1 = + + + + .... 3(4) 12 22 32 42

i.e.

1 1 1 1 π2 .... + + + + = 12 22 32 42 6

i.e.

1 π2 = ∑ 2 6 n =1 n ∞

⎧ 2x ⎪⎪ 1 + π , when − π ≤ x ≤ 0 8. Sketch the waveform defined by: f(x) = ⎨ ⎪ 1 − 2x , when 0 ≤ x ≤ π π ⎩⎪ Determine the Fourier series in this range. The periodic function is shown in the diagram below. © 2006 John Bird. All rights reserved. Published by Elsevier.

605

0 π π ⎡ 1 π 1 ⎧ 0 ⎛ 2x ⎞ 2x x2 ⎤ x 2 ⎤ ⎫⎪ ⎫ 1 ⎧⎪ ⎡ a0 = f (x) dx = ⎨ ⎜1 + ⎨⎢ x + ⎥ + ⎢ x − ⎥ ⎬ ⎟ dx + ∫ 0 (1 − ) dx ⎬ = 2π ∫ −π 2π ⎩ ∫ −π ⎝ π ⎠ π π ⎦ −π ⎣ π ⎦0⎪ ⎭ 2π ⎩⎪ ⎣ ⎭

=

an =

⎤ ⎪⎫ 1 ⎛ 1 ⎧⎪ ⎡ π 2 ⎞ ⎤ ⎡⎛ π2 ⎞ 0 − −π + + π − ( ) {( π − π ) + ( π − π )} = 0 ⎨⎢ ⎜ ⎟ ⎥ ⎢⎜ ⎟ − ( 0 )⎥ ⎬ = 2π ⎪⎩ ⎣ π ⎠ ⎦ ⎣⎝ π⎠ ⎝ ⎦ ⎪⎭ 2π

π 1 π 1 ⎧ 0 ⎛ 2x ⎞ 2x ⎫ f (x) cos nx dx = ⎨ ∫ ⎜1 + cos nx dx + ∫ (1 − ) cos nx dx ⎬ ⎟ ∫ 0 −π −π π π⎩ ⎝ π ⎠ π ⎭ π

0

1 ⎡ sin nx 2 ⎛ x sin nx cos nx ⎞ ⎤ ⎡ sin nx 2 ⎛ x sin nx cos nx ⎞ ⎤ = ⎢ by integration by + ⎜ + − ⎜ + ⎟⎥ + ⎢ ⎟ 2 n ⎠ ⎦ −π ⎣ n n 2 ⎠ ⎥⎦ 0 π⎣ n π⎝ n π⎝ n parts =

1 ⎧⎪ ⎡⎛ 2 ⎛ 1 ⎞⎞ ⎛ 2 ⎛ cos(− nπ) ⎞ ⎞ ⎤ ⎡⎛ 2 ⎛ cos nπ ⎞ ⎞ ⎛ 2 ⎛ 1 ⎞ ⎞ ⎤ ⎫⎪ 0 0 0 0 + − − − − − ⎨ ⎢⎜ 0 + 0 + ⎜ 2 ⎟ ⎟ − ⎜ 0 + 0 + ⎜ ⎟ ⎟ ⎥ ⎢⎜ ⎜ ⎟ ⎜ ⎟ ⎥⎬ n2 π ⎩⎪ ⎣⎝ π ⎝ n ⎠⎠ ⎝ π⎝ π ⎝ n 2 ⎠ ⎟⎠ ⎜⎝ π ⎝ n 2 ⎠ ⎟⎠ ⎦ ⎭⎪ ⎠ ⎠ ⎦ ⎣⎝

=

1 4 2 − 2 cos(− nπ) − 2 cos nπ + 2} = 2 2 (1 − cos nπ ) 2 { πn πn 2

an = 0

When n is even, Hence, bn =

a1 =

since cos nπ = cos(-nπ)

4 8 1 − −1) = 2 , 2 ( π (1) π 2

a3 =

4 8 1 − −1) = 2 2 , 2 ( π (3) π (3) 2

a5 =

8 π (5) 2 2

and so on.

π⎛ 1 π 1 ⎧ 0 ⎛ 2x ⎞ 2x ⎞ ⎫ f (x) sin nx dx = ⎨ ∫ ⎜1 + ⎟ sin nx dx + ∫ 0 ⎜1 − ⎟ sin nx dx ⎬ ∫ −π −π π π⎩ ⎝ π ⎠ π ⎠ ⎝ ⎭

0 π 1 ⎧⎪ ⎡ cos nx 2 ⎛ x cos nx sin nx ⎞ ⎤ ⎡ cos nx 2 ⎛ x cos nx sin nx ⎞ ⎤ ⎫⎪ = ⎨⎢− + ⎜− + + − − ⎜− + ⎟ ⎟ ⎬ by n n n 2 ⎠ ⎥⎦ −π ⎢⎣ n n n 2 ⎠ ⎥⎦ 0 ⎭⎪ π ⎩⎪ ⎣ π⎝ π⎝

integration by parts ⎧ ⎡⎛ cos nπ 2π cos nπ ⎤⎫ ⎞ + 0⎟ ⎪ ⎢⎜ − n + ⎥⎪ n 1 ⎪ ⎡⎛ 1 ⎠ ⎞ ⎛ cos(− nπ) 2π cos(− nπ) ⎞ ⎤ ⎢⎝ ⎥ ⎪⎬ = ⎨ ⎢⎜ − − 0 + 0 ⎟ − ⎜ − + + 0 ⎟⎥ + π ⎪ ⎣⎝ n n n ⎠ ⎝ ⎠⎦ ⎢ ⎛ 1 ⎞⎥ − ⎜ − + 0 − 0 ⎟⎥ ⎪ ⎢ ⎪⎩ ⎝ n ⎠ ⎦ ⎪⎭ ⎣

=

1 ⎧ 1 cos(− nπ) 2π cos(− nπ) cos nπ 2π cos nπ 1 ⎫ − − + + ⎬=0 ⎨− + n n n n n⎭ π⎩ n

© 2006 John Bird. All rights reserved. Published by Elsevier.

606

∞

∞

n =1

n =1

Substituting into f(x) = a 0 + ∑ ( a n cos nx + b n sin nx ) = ∑ a n cos nx

gives:

f(x) =

i.e.

f(x) =

8 8 8 8 cos x + 2 2 cos 3x + 2 2 cos 5x + 2 2 cos 7x + ..... 2 π π (3) π (5) π (7) 8 ⎛ 1 1 1 ⎞ cos x + 2 cos 3x + 2 cos 5x + 2 cos 7x + ...... ⎟ 2 ⎜ π ⎝ 3 5 7 ⎠

π2 9. For the Fourier series of Problem 8, deduce a series for 8 When f(x) = 1 in the series of problem 8 above, x = 0, hence,

i.e.

1=

8 ⎛ 1 1 1 ⎞ cos 0 + 2 cos 0 + 2 cos 0 + 2 cos 0 + ...... ⎟ 2 ⎜ 3 5 7 π ⎝ ⎠

π2 1 1 1 = 1 + 2 + 2 + 2 + ..... 8 3 5 7

© 2006 John Bird. All rights reserved. Published by Elsevier.

607

CHAPTER 71 EVEN AND ODD FUNCTIONS AND HALF-RANGE FOURIER SERIES EXERCISE 242 Page 672 2. Obtain the Fourier series of the function defined by: ⎧ t + π, when − π ≤ t ≤ 0 f(t) = ⎨ ⎩ t − π, when 0 ≤ t ≤ π which is periodic of period 2π. Sketch the given function. The periodic function is shown in the diagram below. Since it is symmetrical about the origin, the ∞

function is odd, and f (t) = ∑ b n sin nt n =1

bn =

1 π 1 f (t) sin nt dt = ∫ π −π π

{∫

0 −π

}

π

(t + π) sin nt dt + ∫ (t − π) sin ntdt 0

π

1 ⎡ t cos nt sin nt π cos nt ⎤ ⎡ t cos nt sin nt π cos nt ⎤ = ⎢− + 2 − + ⎢− + 2 + by integration by parts ⎥ π⎣ n n n ⎦ −π ⎣ n n n ⎥⎦ 0 0

⎧ ⎡⎛ π cos nπ π cos nπ ⎞ ⎤⎫ +0+ ⎪ ⎟ ⎢⎜ − ⎥⎪ n n π ⎞ ⎛ −π cos(− nπ) π cos(− nπ) ⎞ ⎤ ⎢⎝ 1 ⎪ ⎡⎛ ⎠ ⎥ ⎪⎬ = ⎨ ⎢⎜ 0 + 0 − ⎟ − ⎜ − +0− + ⎟⎥ π ⎪ ⎣⎝ n⎠ ⎝ n n π ⎞⎥ ⎠⎦ ⎢ ⎛ − ⎜ 0 + 0 + ⎟⎥ ⎪ ⎢ n ⎠ ⎦ ⎭⎪ ⎝ ⎣ ⎩⎪

=

1 ⎧ π π cos(− nπ) π cos(− nπ) π cos nπ π cos nπ π ⎫ 1 ⎛ 2π ⎞ 2 + − + − ⎬ = ⎜− ⎟ = − ⎨− − π⎩ n n n n n n⎭ π⎝ n ⎠ n

Hence,

b1 = −

2 , 1

b2 = −

2 , 2

b3 = −

2 , 3

b4 = −

2 , and so on. 4

i.e.

2 2 2 2 f(t) = − sin t − sin 2t − sin 3t − sin 4t − ...... 1 2 3 4

i.e.

1 1 1 ⎛ ⎞ f (t) = −2 ⎜ sin t + sin 2t + sin 3t + sin 4t + ..... ⎟ 2 3 4 ⎝ ⎠

© 2006 John Bird. All rights reserved. Published by Elsevier.

608

⎧ 1 − x, when − π ≤ x ≤ 0 3. Determine the Fourier series defined by: f(x) = ⎨ ⎩ 1 + x, when 0 ≤ x ≤ π which is periodic of period 2π. The periodic function is shown in the diagram below. Since it is symmetrical about the f(x) axis, the ∞

function is even, and f (x) = a 0 + ∑ a n cos nx n =1

a0 =

1 π 1 π f (x) dx = ∫ f (x) dx due to symmetry ∫ π 0 2π −π π

⎤ 1 π 1⎡ x2 ⎤ 1 ⎡⎛ π2 ⎞ π = ∫ (1 + x) dx = ⎢ x + ⎥ = ⎢⎜ π + ⎟ − ( 0 ) ⎥ = 1 + π 0 π⎣ 2 ⎦ 0 π ⎣⎝ 2 ⎠ 2 ⎦ an = =

1 π 1 f (x) cos nx dx = ∫ π −π π 1 π

{∫

{∫

0 −π

}

π

(1 − x) cos nx dx + ∫ (1 + x) cos nx dx 0

}

π

( cos nx − x cos nx ) dx + ∫ 0 ( cos nx + x cos nx ) dx −π 0

π

0

1 ⎡ sin nx x sin nx cos nx ⎤ ⎡ sin nx x sin nx cos nx ⎤ by integration by parts = ⎢ − − +⎢ + + 2 ⎥ π⎣ n n n ⎦ −π ⎣ n n n 2 ⎥⎦ 0 =

1 ⎧ ⎡⎛ 1 ⎞ ⎛ cos(−nπ) ⎞ ⎤ ⎡⎛ cos nπ ⎞ ⎛ 1 ⎞⎤ ⎫ ⎨ ⎢⎜ 0 − 0 − 2 ⎟ − ⎜ 0 − 0 − ⎟ ⎥ + ⎢⎜ 0 + 0 + ⎟ − ⎜ 0 + 0 + 2 ⎟⎥ ⎬ 2 2 π ⎩ ⎣⎝ n ⎠ ⎝ n n ⎠ ⎝ n ⎠⎦ ⎭ ⎠ ⎦ ⎣⎝

=

1 ⎧ 1 cos(− nπ) cos nπ 1 ⎫ 2 + − 2 ⎬ = 2 ( cos nπ − 1) ⎨− 2 + 2 2 π⎩ n n n n ⎭ πn

When n is even,

since cos(-nπ) = cos nπ

an = 0

When n = 1,

a1 =

2 4 −1 − 1) = − 2 ( π(1) π

When n = 3,

a3 =

2 4 −1 − 1) = − 2 ( π(3) π(3) 2

When n = 5,

a5 =

2 4 −1 − 1) = − 2 ( π(5) π(5) 2

and so on.

© 2006 John Bird. All rights reserved. Published by Elsevier.

609

∞

f (x) = a 0 + ∑ a n cos nx = n =1

i.e.

f(x) =

π 4 4 4 cos 3x − cos 5x − ..... + 1 − cos x − 2 π π(3) π(5) 2 2

π 4⎛ 1 1 ⎞ + 1 − ⎜ cos x + 2 cos 3x + 2 cos 5x + ...... ⎟ π⎝ 2 3 5 ⎠

4. In the Fourier series of Problem 3, let x = 0 and deduce a series for

π2 8

When x = 0 in the series of Problem 3 above, f(x) = 1, hence,

1=

π 4⎛ cos 0 cos 0 ⎞ + 1 − ⎜ cos 0 + 2 + 2 + ...... ⎟ 2 π⎝ 3 5 ⎠

i.e.

1=

π 4⎛ 1 1 1 ⎞ + 1 − ⎜ 1 + 2 + 2 + 2 + ..... ⎟ 2 π⎝ 3 5 7 ⎠

i.e.

and

−

π 4⎛ 1 1 1 ⎞ = − ⎜1 + 2 + 2 + 2 + ..... ⎟ 2 π⎝ 3 5 7 ⎠

π2 1 1 1 = 1 + 2 + 2 + 2 + .... 8 3 5 7

© 2006 John Bird. All rights reserved. Published by Elsevier.

610

EXERCISE 243 Page 675

π ⎧ ⎪⎪ x, when 0 ≤ x ≤ 2 1. Determine the half-range series for the function defined by: f(x) = ⎨ ⎪ 0, when π ≤ x ≤ π ⎪⎩ 2 The periodic function is shown in the diagram below. Since a half-range sine series is required, the ∞

function is symmetrical about the origin and

f (x) = ∑ b n sin nx n =1

2 π 2 b n = ∫ f (x) sin nx dx = 0 π π

{∫

π/ 2 0

}

π/2

2 ⎡ x cos nx sin nx ⎤ + x sin nx dx = ⎢ − π⎣ n n 2 ⎥⎦ 0

by integration by parts

⎡⎛ π nπ nπ ⎞ ⎤ cos sin ⎟ 2 ⎢⎜ 2 2 + 2 − ( 0)⎥ = ⎢⎜ − ⎥ ⎟ 2 n n ⎟ π ⎢⎜ ⎥ ⎢⎣⎝ ⎥⎦ ⎠

Hence,

⎡⎛ π π π ⎞⎤ cos sin ⎟ ⎥ 2 ⎢⎜ 2 2+ 2 = 2 ⎛0+ 1 ⎞ = 2 , b1 = ⎢⎜ − ⎟⎥ ⎜ ⎟ 2 π ⎢⎜ 1 1 ⎟ ⎥ π ⎝ 12 ⎠ π ⎠ ⎦⎥ ⎣⎢⎝ ⎡⎛ π ⎞⎤ cos π ⎢ ⎜ 2 sin π ⎟ ⎥ 2 ⎛ π ⎞ 2⎛π⎞ b 2 = ⎢⎜ − 2 + 2 ⎟⎥ = ⎜ + 0 ⎟ = ⎜ ⎟ , 2 2 ⎟⎥ π ⎝ 4 π ⎢⎜ ⎠ π⎝ 4⎠ ⎠ ⎦⎥ ⎣⎢⎝ ⎡⎛ π 3π 3π ⎞ ⎤ cos sin ⎢ ⎜ 2 2 + 2 ⎟⎥ = 2 ⎛ 0 − 1 ⎞ = − 2 , b 3 = ⎢⎜ − 2 ⎟⎥ ⎜ ⎟ 2 π ⎢⎜ π(3) 2 3 3 ⎟⎥ π ⎝ 32 ⎠ ⎢⎣⎝ ⎠ ⎥⎦ ⎡⎛ π ⎞⎤ cos 2π 2 ⎢⎜ 2 sin 2π ⎟ ⎥ 2 ⎛ π 2⎛π⎞ ⎞ b 4 = ⎢⎜ − + ⎟ ⎥ = ⎜ − + 0 ⎟ = − ⎜ ⎟ , and so on 2 4 4 ⎟⎥ π ⎝ 8 π ⎢⎜ π⎝8⎠ ⎠ ⎢⎣⎝ ⎠ ⎥⎦ © 2006 John Bird. All rights reserved. Published by Elsevier.

611

∞

Hence,

f (x) = ∑ b n sin nx = n =1

f (x) =

i.e.

2 2⎛π⎞ 2⎛ 1 ⎞ 2⎛π⎞ sin x + ⎜ ⎟ sin 2x − ⎜ 2 ⎟ sin 3x − ⎜ ⎟ sin 4x + ... π π⎝ 4⎠ π⎝3 ⎠ π⎝ 8⎠

2⎛ π 1 π ⎞ ⎜ sin x + sin 2x − sin 3x − sin 4x + ..... ⎟ π⎝ 4 9 8 ⎠

2. Obtain (a) the half-range cosine series and (b) the half-range sine series for the function:

⎧ ⎪⎪ 0, when f(t) = ⎨ ⎪ 1, when ⎪⎩

0≤t ≤

π 2

π ≤t≤π 2

(a) The periodic function is shown in the diagram below. Since a half-range cosine series is ∞

required, the function is symmetrical about the f(t) axis and

f (t) = a 0 + ∑ a n cos nt n =1

a0 =

1 π 1 π 1⎡ π⎤ 1 1dt = [ t ] π / 2 = ⎢ π − ⎥ = ∫ π π/ 2 π π⎣ 2⎦ 2

an =

2 2 ⎡ sin nt ⎤ 1cos nt dt = ⎢ ∫ π / 2 π π ⎣ n ⎥⎦ π / 2

π

π

nπ ⎞ nπ ⎛ sin 2sin ⎜ ⎟ 2 2 =− 2 = ⎜0− ⎟ n ⎟ π⎜ πn ⎝ ⎠

When n is even, a n = 0

and

a1 = −

π 2 =−2 , π π

2sin

∞

a3 = −

Thus, f (t) = a 0 + ∑ a n cos nt = n =1

i.e.

f (t) =

3π 5π 2sin 2 1 ⎛ ⎞ 2 = 2 = − 2 ⎛ 1 ⎞ , and so on. ⎜ ⎟ , a5 = − ⎜ ⎟ π⎝3⎠ π⎝5⎠ 3π 5π

2sin

1 2 2⎛1⎞ 2⎛1⎞ − cos t + ⎜ ⎟ cos 3t − ⎜ ⎟ cos 5t + .... π⎝3⎠ π⎝5⎠ 2 π

1 2⎛ 1 1 ⎞ − ⎜ cos t − cos 3t + cos 5t − ...... ⎟ 2 π⎝ 3 5 ⎠

(b) The periodic function is shown in the diagram below. Since a half-range sine series is required, © 2006 John Bird. All rights reserved. Published by Elsevier.

612

∞

f (x) = ∑ b n sin nx

the function is symmetrical about the origin and

n =1

π

bn =

2 π 2 ⎡ cos nt ⎤ 2 ⎡ nπ ⎤ = − ⎢cos nπ − cos ⎥ 1sin nt dt = ⎢ − ∫ ⎥ π / 2 n ⎦ π/2 nπ ⎣ 2⎦ π π⎣

Hence,

2⎛ π⎞ 2 2 b1 = − ⎜ cos π − cos ⎟ = − ( −1 − 0 ) = , π⎝ π π 2⎠ b2 = −

2 2 2 ( cos 2π − cos π ) = − (1 − −1) = − , 2π 2π π

b3 = −

2 ⎛ 3π ⎞ 2 2 , ⎜ cos 3π − cos ⎟ = − ( −1 − 0 ) = 3π ⎝ 2 ⎠ 3π 3π

b4 = −

2 2 ( cos 4π − cos 2π ) = − ( 0 − 0 ) = 0 , 4π 4π

b5 = −

2 ⎛ 5π ⎞ 2 2 , ⎜ cos 5π − cos ⎟ = − ( −1 − 0 ) = 5π ⎝ 2 ⎠ 5π 5π

b6 = −

2 2 2 ( cos 6π − cos 3π ) = − (1 − −1) = − , and so on. 6π 6π 3π ∞

Thus,

f (t) = ∑ b n sin nt = n =1

i.e.

f (t) =

2 2 2 2 2 sin t − sin 2t + sin 3t + 0 + sin 5t − sin 6t + .... π π 3π 5π 3π

2⎛ 1 1 1 ⎞ ⎜ sin t − sin 2t + sin 3t + sin 5t − sin 6t + ..... ⎟ 3 5 3 π⎝ ⎠

4. Determine the half-range Fourier cosine series in the range x = 0 to x = π for the function

defined by:

⎧ when ⎪⎪ x, f(x) = ⎨ ⎪ ( π − x ) , when ⎪⎩

0 ≤x ≤

π 2

π ≤x≤π 2

The periodic function is shown in the diagram below. Since a half-range cosine series is required, ∞

the function is symmetrical about the f(x) axis and

f (x) = a 0 + ∑ a n cos nx n =1

© 2006 John Bird. All rights reserved. Published by Elsevier.

613

1 a0 = π

{∫

π/ 2 0

}

π/ 2 π ⎡ 1 ⎧⎪ ⎡ x 2 ⎤ x 2 ⎤ ⎫⎪ x dx + ∫ (π − x) dx = ⎨ ⎢ ⎥ + ⎢ πx − ⎥ ⎬ π/2 π ⎪⎣ 2 ⎦ 0 2 ⎦ π/ 2 ⎪ ⎣ ⎩ ⎭ π

1 ⎧⎪ π2 ⎛ 2 π2 ⎞ ⎛ π2 π2 ⎞ ⎫⎪ = ⎨ + ⎜ π − ⎟ − ⎜ − ⎟⎬ π ⎪⎩ 8 ⎝ 2 ⎠ ⎝ 2 8 ⎠ ⎭⎪ = an =

2 π

{∫

π/ 2 0

x cos nx dx + ∫

π π/2

1 ⎛ π 2 π 2 π 2 π 2 ⎞ 1 ⎛ 2π 2 ⎞ π ⎜ + − + ⎟= ⎜ ⎟= π⎝ 8 2 2 8 ⎠ π⎝ 8 ⎠ 4

}

(π − x) cos nx dx

π/ 2 π 2 ⎪⎧ ⎡ x sin nx cos nx ⎤ ⎡ π sin nπ x sin nx cos nx ⎤ ⎪⎫ = ⎨⎢ + +⎢ − − ⎬ π ⎪⎩ ⎣ n n 2 ⎥⎦ 0 n n 2 ⎥⎦ π / 2 ⎪⎭ ⎣ n

⎧⎛ π nπ nπ ⎞ nπ π nπ nπ ⎞ ⎫ ⎛ sin cos π sin sin cos ⎟ ⎜ π 2 ⎪⎪⎜ 2 1 cos n ⎛ ⎞ ⎛ ⎞ 2 + 2 − 0+ 2 −2 2 − 2 ⎟ ⎪⎪ = ⎨⎜ + − − − 0 0 ⎟ ⎜ ⎟⎬ ⎜ ⎟ ⎜ ⎟ π ⎪⎜ n n2 ⎟ ⎝ n2 ⎠ ⎝ n2 ⎠ ⎜ n n n 2 ⎟⎪ ⎠ ⎝ ⎠ ⎭⎪ ⎩⎪⎝ nπ nπ nπ ⎧ ⎫ 2 cos π sin π sin ⎪ 2⎪ 1 cos n π 2 ⎧ nπ ⎫ 2 + 2 − − 2 − = ⎨ − 1 − cos nπ ⎬ ⎬ = 2 ⎨2 cos 2 2 2 n n n n ⎪ πn ⎩ 2 π⎪ n ⎭ ⎩ ⎭

When n is odd, a n = 0 and

a2 =

2 2 8 2 2 cos π − 1 − cos 2π ) = ( −2 − 1 − 1) = − = − , 2 ( π(2) 4π 4π π

a4 =

2 2 2 cos 2π − 1 − cos 4π ) = ( 2 − 1 − 1) = 0 , 2 ( π(4) 16π

a6 =

2 2 8 2 2 cos 3π − 1 − cos 6π ) = =− 2 , ( −2 − 1 − 1) = − 2 ( π(6) 36π 36π (3) π

a8 = 0 , a10 =

2 2 8 2 2 cos 5π − 1 − cos10π ) = = − 2 , and so on. ( −2 − 1 − 1) = − 2 ( π(10) 100π 100π (5) π

© 2006 John Bird. All rights reserved. Published by Elsevier.

614

∞

Thus, f (t) = a 0 + ∑ a n cos nt = n =1

i.e.

f (t) =

π 2 2 2 − cos 2t − 2 cos 6t − 2 cos10t + .... 4 π (3) π (5) π

π 2⎛ 1 1 ⎞ − ⎜ cos 2t + 2 cos 6t + 2 cos10t + ...... ⎟ 4 π⎝ 3 5 ⎠

© 2006 John Bird. All rights reserved. Published by Elsevier.

615

CHAPTER 72 FOURIER SERIES OVER ANY RANGE EXERCISE 244 Page 679 2. Find the Fourier series for f(x) = x in the range x = 0 to x = 5.

The periodic function is shown in the diagram below.

The Fourier series is given by:

∞ ⎡ ⎛ 2πnx ⎞ ⎛ 2πnx ⎞ ⎤ f(x) = a 0 + ∑ ⎢a n cos ⎜ ⎟ + b n sin ⎜ ⎟ ⎥ where L = 5 ⎝ L ⎠ ⎝ L ⎠⎦ n =1 ⎣ 5

1 L 1 5 1 ⎡ x2 ⎤ 1 ⎡ 52 ⎤ 5 a 0 = ∫ f (x) dx = ∫ x dx = ⎢ ⎥ = ⎢ ⎥ = L 0 5 0 5 ⎣ 2 ⎦0 5 ⎣ 2 ⎦ 2 5

⎡ ⎛ 2πnx ⎞ ⎛ 2πnx ⎞ ⎤ x sin ⎜ cos ⎜ ⎢ ⎟ ⎟⎥ 2 L 2 5 2⎢ 5 ⎠ 5 ⎠⎥ ⎛ 2πnx ⎞ ⎛ 2πnx ⎞ ⎝ ⎝ + by parts a n = ∫ f (x) cos ⎜ ⎟ dx = ∫ 0 x cos ⎜ ⎟ dx = ⎢ 2 L 0 5 5 ⎛ 2πn ⎞ ⎝ L ⎠ ⎝ 5 ⎠ ⎛ 2πn ⎞ ⎥ ⎢ ⎜ 5 ⎟ ⎜ ⎟ ⎥ ⎠ ⎝ 5 ⎠ ⎦0 ⎣ ⎝

⎡⎛ ⎞ ⎛ ⎞⎤ ⎢⎜ ⎟ ⎜ ⎟⎥ 2 5sin 2πn cos 2πn ⎟ ⎜ 1 ⎟⎥ = 0 = ⎢⎜ + − 0 + 2 5 ⎢⎜ ⎛ 2πn ⎞ ⎛ 2πn ⎞ 2 ⎟ ⎜ ⎛ 2πn ⎞ ⎟ ⎥ ⎢⎜ ⎜ 5 ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎟⎥ ⎠ ⎝ 5 ⎟⎠ ⎠ ⎝ ⎝ 5 ⎠ ⎠⎦ ⎣⎝ ⎝ 5

⎡ ⎛ 2πnx ⎞ ⎛ 2πnx ⎞ ⎤ x cos ⎜ sin ⎜ ⎢ ⎟ ⎟⎥ 2 L 2 5 2⎢ ⎛ 2πnx ⎞ ⎛ 2πnx ⎞ ⎝ 5 ⎠+ ⎝ 5 ⎠ ⎥ by parts = = − b n = ∫ f (x) sin ⎜ dx x sin dx ⎟ ⎜ ⎟ 2 L 0 5 ∫0 5⎢ ⎛ 2πn ⎞ ⎝ L ⎠ ⎝ L ⎠ ⎛ 2πnx ⎞ ⎥ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎝ 5 ⎠ ⎝ 5 ⎠ ⎦0 ⎣ ⎡⎛ ⎤ ⎞ ⎡ ⎤ ⎢⎜ ⎥ ⎟ ⎢ ⎥ 2 5cos 2πn sin 2πn ⎟ 2 5cos 2πn 5 5 ⎥ = − cos 2πn = − = ⎢⎜ − + − ( 0 + 0 )⎥ = ⎢− 2 ⎥ 5 ⎢ ⎛ 2πn ⎞ ⎥ πn πn 5 ⎢⎜ ⎛ 2πn ⎞ ⎛ 2πn ⎞ ⎟ ⎜ ⎟ ⎥ ⎢⎜ ⎜ 5 ⎟ ⎜ ⎥ ⎟ ⎟ ⎢ ⎠ ⎝ 5 ⎠ ⎠ ⎣ ⎝ 5 ⎠ ⎦ ⎣⎝ ⎝ ⎦

Hence, b1 = − Thus,

5 , π

b2 = −

5 , 2π

b3 = −

5 , 3π

b4 = −

5 , and so on. 4π

∞ ⎡ ⎛ 2πnx ⎞ ⎛ 2πnx ⎞ ⎤ f(x) = a 0 + ∑ ⎢a n cos ⎜ ⎟ + b n sin ⎜ ⎟⎥ ⎝ L ⎠ ⎝ L ⎠⎦ n =1 ⎣

© 2006 John Bird. All rights reserved. Published by Elsevier.

616

i.e.

f(x) =

5 5 ⎛ 2πx ⎞ 5 ⎛ 4πx ⎞ 5 ⎛ 6πx ⎞ − sin ⎜ ⎟ − sin ⎜ ⎟ − sin ⎜ ⎟ − ..... 2 π ⎝ 5 ⎠ 2π ⎝ 5 ⎠ 3π ⎝ 5 ⎠

i.e.

f(x) =

⎤ 5 5 ⎡ ⎛ 2πx ⎞ 1 ⎛ 4 πx ⎞ 1 ⎛ 6πx ⎞ − ⎢sin ⎜ + sin ⎜ + sin ⎜ + ......⎥ ⎟ ⎟ ⎟ 2 π⎣ ⎝ 5 ⎠ 2 ⎝ 5 ⎠ 3 ⎝ 5 ⎠ ⎦

3. A periodic function of period 4 is defined by:

⎧ − 3, when − 2 ≤ x ≤ 0 f(x) = ⎨ ⎩ + 3, when 0 ≤ x ≤ 2

Sketch the function and obtain the Fourier series for the function. The periodic function is shown in the diagram below.

The function is odd since it is symmetrical about the origin, i.e. a n = 0 Thus,

∞ ⎡ ⎛ 2πnx ⎞ ⎤ f(x) = a 0 + ∑ ⎢ b n sin ⎜ ⎟⎥ ⎝ L ⎠⎦ n =1 ⎣

{∫

where L = 4

} 14 {[−3x]

}

1 {( 0 ) − ( 6 ) + ( 6 ) − ( 0 )} = 0 4

a0 =

1 L/2 1 f (x) dx = ∫ L −L / 2 4

bn =

2 2 L/2 2⎧ 0 ⎛ 2πnx ⎞ ⎛ 2πnx ⎞ ⎛ 2πnx ⎞ ⎫ f (x) sin ⎜ dx = ⎨ ∫ −3sin ⎜ dx + ∫ 3sin ⎜ ⎟ ⎟ ⎟ dx ⎬ ∫ 0 L −L / 2 4 ⎩ −2 ⎝ L ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎭

0

−2

2

−3dx + ∫ 3dx = 0

0

−2

+ [3x ] 0 = 2

0 2 ⎧⎡ ⎧⎛ ⎡ ⎞ ⎛ ⎞⎫ ⎛ πnx ⎞ ⎤ ⎛ πnx ⎞ ⎤ ⎫ 3cos ⎜ ⎪ ⎢ 3cos ⎜ ⎪ ⎟ ⎟ ⎪ ⎜ ⎟ ⎜ ⎟⎪ ⎥ ⎢ ⎥ 1⎪ ⎝ 2 ⎠ ⎥ + ⎢− ⎝ 2 ⎠ ⎥ ⎪ = 1 ⎪⎜ 3cos 0 − 3cos(−πn) ⎟ + ⎜ − 3cos πn − − 3cos 0 ⎟ ⎪ = ⎨⎢ ⎬ ⎨ ⎬ 2 ⎪ ⎢ ⎛ πn ⎞ ⎥ ⎛ πn ⎞ ⎥ ⎪ 2 ⎪⎜ ⎛ πn ⎞ ⎛ πn ⎞ ⎟ ⎜ ⎛ πn ⎞ ⎛ πn ⎞ ⎟ ⎪ ⎢ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎟ ⎜ ⎜ ⎟ ⎜ ⎟⎟ ⎢ ⎜ ⎟ ⎥ ⎢ ⎪⎩⎝⎜ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦ 0 ⎭⎪ ⎝ 2 ⎠ ⎠ ⎝ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎠ ⎪⎭ ⎩⎪ ⎣ ⎝ 2 ⎠ ⎦ − 2 ⎣

⎡ ⎤ ⎢ ⎥ 6 1 6 6 = ⎢ − cos πn ⎥ = (1 − cos πn ) 2 ⎢ ⎛ πn ⎞ ⎛ πn ⎞ ⎥ πn ⎢⎣ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎥⎦

When n is even, b n = 0

© 2006 John Bird. All rights reserved. Published by Elsevier.

617

Hence, b1 = Thus,

i.e.

6 12 (1 − −1) = , π π

b3 =

6 12 (1 − −1) = , 3π 3π

b5 =

6 12 (1 − −1) = , and so on. 5π 5π

∞ 12 ⎛ πx ⎞ 12 ⎡ ⎛ 3πx ⎞ 12 ⎛ 5πx ⎞ ⎛ 2πnx ⎞ ⎤ sin ⎜ ⎟ + sin ⎜ f(x) = a 0 + ∑ ⎢ b n sin ⎜ ⎟ + sin ⎜ ⎟ + ...... ⎟⎥ = 0 + π ⎝ 2 ⎠ 3π ⎝ 2 ⎠ 5π ⎝ 2 ⎠ ⎝ L ⎠⎦ n =1 ⎣

f(x) =

⎫ 12 ⎧ ⎛ πx ⎞ 1 ⎛ 3πx ⎞ 1 ⎛ 5πx ⎞ + sin ⎜ + .....⎬ ⎨sin ⎜ ⎟ + sin ⎜ ⎟ ⎟ π ⎩ ⎝ 2 ⎠ 3 ⎝ 2 ⎠ 5 ⎝ 2 ⎠ ⎭

© 2006 John Bird. All rights reserved. Published by Elsevier.

618

EXERCISE 245 Page 681 1. Determine the half-range Fourier cosine series for the function f(x) = x in the range 0 ≤ x ≤ 3.

Sketch the function within and outside of the given range. The periodic function is shown in the diagram below. Since a half-range cosine series is required, the function is symmetrical about the f(x) axis and

1 a0 = L

an =

{

∞ ⎛ nπx ⎞ f (x) = a 0 + ∑ a n cos ⎜ ⎟ ⎝ L ⎠ n =1

}

3 1 3 1 ⎧⎪ ⎡ x 2 ⎤ ⎫⎪ 3 ∫ 0 f (x) dx = 3 ∫ 0 x dx = 3 ⎨⎢⎣ 2 ⎥⎦ ⎬ = 2 ⎪ 0⎭ ⎩⎪ L

2⎧ L ⎛ nπx ⎞ ⎫ 2 3 ⎛ nπx ⎞ ⎨ ∫ 0 f (x) cos ⎜ ⎟ dx ⎬ = ∫ 0 x cos ⎜ ⎟ dx L⎩ ⎝ L ⎠ ⎭ 3 ⎝ 3 ⎠

3 ⎧⎡ ⎡⎛ ⎤ ⎫ ⎞ ⎛ ⎞⎤ n x n x π π ⎛ ⎞ ⎛ ⎞ ⎪ ⎢ x sin ⎜ cos ⎜ ⎢⎜ ⎟ ⎜ ⎟⎥ ⎥ ⎪ ⎟ ⎟ 2 ⎪⎢ 1 ⎟⎥ 3 ⎠ 3 ⎠ ⎥ ⎪ 2 ⎢⎜ 3sin nπ cos nπ ⎟ ⎜ ⎝ ⎝ = ⎨ + + − 0+ ⎬= 2 2 2 3 ⎪ ⎢ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎥ ⎪ 3 ⎢⎜ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎟ ⎜ ⎛ nπ ⎞ ⎟ ⎥ ⎜ ⎟ ⎢⎜ ⎜ 3 ⎟ ⎜ ⎟ ⎟ ⎜ ⎜ ⎟ ⎥ ⎪ ⎜ ⎟ ⎟⎥ ⎪⎢ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎦0⎭ ⎝ 3 ⎠ ⎠⎦ ⎣⎝ ⎝ ⎠ ⎝ 3 ⎠ ⎠ ⎝ ⎩⎣

⎧ ⎫ ⎪ ⎪ 2⎪ cos nπ 1 ⎪ − = ⎨0 + ⎬= 3 ⎪ ⎛ nπ ⎞ 2 ⎛ nπ ⎞ 2 ⎪ ⎪⎩ ⎜⎝ 3 ⎟⎠ ⎜⎝ 3 ⎟⎠ ⎪⎭

by parts

2

2⎛ 3 ⎞ 6 ⎜ ⎟ {cos nπ − 1} = 2 2 ( cos nπ − 1) 3 ⎝ nπ ⎠ n π

When n is even, a n = 0 and

Thus,

i.e.

a1 =

6 12 6 12 −2 ) = − 2 , a 3 = 2 2 ( −2 ) = − 2 2 , 2 ( π (1) π π (3) π (3) 2

a5 = −

12 , and so on. π (5) 2 2

∞ 12 12 ⎛ nπx ⎞ 3 12 ⎛ πx ⎞ ⎛ 3πx ⎞ ⎛ 5πx ⎞ f (x) = a 0 + ∑ a n cos ⎜ ⎟ = − 2 cos ⎜ ⎟ − 2 2 cos ⎜ ⎟ − 2 2 cos ⎜ ⎟ − .... ⎝ L ⎠ 2 π ⎝ 3 ⎠ π (3) ⎝ 3 ⎠ π (5) ⎝ 3 ⎠ n =1

f (x) =

⎫ 3 12 ⎧ ⎛ πx ⎞ 1 ⎛ 3πx ⎞ 1 ⎛ 5πx ⎞ − 2 ⎨cos ⎜ ⎟ + 2 cos ⎜ + 2 cos ⎜ + .....⎬ ⎟ ⎟ 2 π ⎩ ⎝ 3 ⎠ 3 ⎝ 3 ⎠ 5 ⎝ 3 ⎠ ⎭

© 2006 John Bird. All rights reserved. Published by Elsevier.

619

2. Find the half-range Fourier sine series for the function f(x) = x in the range 0 ≤ x ≤ 3.

Sketch the function within and outside of the given range. The periodic function is shown in the diagram below. Since a half-range sine series is required, the function is symmetrical about the origin and

∞ ⎛ nπx ⎞ f (x) = ∑ b n sin ⎜ ⎟ ⎝ L ⎠ n =1

3

⎡ ⎛ nπx ⎞ ⎛ nπx ⎞ ⎤ x cos ⎜ sin ⎜ ⎢ ⎟ ⎟⎥ 2 L 2⎧ 3 3 ⎠ 3 ⎠⎥ ⎛ nπx ⎞ ⎛ nπx ⎞ ⎫ 2 ⎢ ⎝ ⎝ b n = ∫ f (x) sin ⎜ by parts + ⎟ dx = ⎨ ∫ 0 x sin ⎜ ⎟ dx ⎬ = − 2 L 0 3⎩ ⎛ nπ ⎞ ⎝ L ⎠ ⎝ 3 ⎠ ⎭ 3⎢ ⎛ nπ ⎞ ⎥ ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎦0 ⎣ ⎡⎛ ⎤ ⎞ ⎢⎜ ⎥ ⎟ 2 3cos nπ sin nπ ⎟ ⎥ = − 2 cos nπ = − 6 cos nπ 0 0 = ⎢⎜ − + − + ( ) 2 ⎢ ⎥ ⎜ ⎟ 3 nπ ⎛ nπ ⎞ ⎛ n π ⎞ ⎛ nπ ⎞ ⎜ ⎟ ⎢⎜ ⎜ 3 ⎟ ⎜ ⎟ ⎟ ⎥ ⎝ 3 ⎠ ⎣⎝ ⎝ ⎠ ⎝ 3 ⎠ ⎠ ⎦

b1 =

6 , π

b2 = −

6 , 2π

b3 =

6 , 3π

b4 = −

6 , and so on. 4π

∞ ⎛ nπx ⎞ 6 ⎛ πx ⎞ 6 ⎛ 2πx ⎞ 6 ⎛ 3πx ⎞ 6 ⎛ 4πx ⎞ Thus, f (x) = ∑ b n sin ⎜ ⎟ = sin ⎜ ⎟ − sin ⎜ ⎟ + sin ⎜ ⎟ − sin ⎜ ⎟ + .... ⎝ L ⎠ π ⎝ 3 ⎠ 2π ⎝ 3 ⎠ 3π ⎝ 3 ⎠ 4π ⎝ 3 ⎠ n =1

i.e.

f(x) =

⎫ 6 ⎧ ⎛ πx ⎞ 1 ⎛ 2πx ⎞ 1 ⎛ 3πx ⎞ 1 ⎛ 4 πx ⎞ + sin ⎜ − sin ⎜ + .....⎬ ⎨sin ⎜ ⎟ − sin ⎜ ⎟ ⎟ ⎟ π⎩ ⎝ 3 ⎠ 2 ⎝ 3 ⎠ 3 ⎝ 3 ⎠ 4 ⎝ 3 ⎠ ⎭

3. Determine the half-range Fourier sine series for the function defined by:

t, when ⎧⎪ f(t) = ⎨ ⎪⎩ ( 2 − t ) , when

0 ≤t ≤1 1≤ t ≤ 2

© 2006 John Bird. All rights reserved. Published by Elsevier.

620

The periodic function is shown in the diagram below. Since a half-range sine series is required, the function is symmetrical about the origin and

bn =

∞ ⎛ nπt ⎞ f (t) = ∑ b n sin ⎜ ⎟ ⎝ L ⎠ n =1

2 2 L 2⎧ 1 ⎛ nπt ⎞ ⎛ nπt ⎞ ⎛ nπt ⎞ ⎫ f (t) sin ⎜ dt = ⎨ ∫ t sin ⎜ dt + ∫ (2 − t) sin ⎜ ⎟ ⎟ ⎟ dt ⎬ ∫ 1 L 0 2⎩ 0 ⎝ L ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎭

1 2 ⎧⎡ ⎤ ⎡ ⎤ ⎫ n t n t n t n t n t π π π π π ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎪ ⎢ t cos ⎜ ⎟ sin ⎜ ⎟ ⎥ ⎢ 2 cos ⎜ ⎟ t cos ⎜ ⎟ sin ⎜ ⎟⎥ ⎪ ⎪⎢ 2 ⎠ 2 ⎠⎥ ⎢ 2 ⎠ 2 ⎠ 2 ⎠⎥ ⎪ ⎝ ⎝ ⎝ ⎝ ⎝ = ⎨ − + + − + − ⎬ 2 2 n n n π π π ⎢ ⎥ ⎢ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ n π ⎛ ⎞ ⎛ nπ ⎞ ⎥ ⎪ ⎪ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎥ ⎢ ⎜ ⎟ ⎥ ⎪ ⎪ ⎢⎣ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ 2 ⎝ ⎠ ⎝ 2 ⎠ ⎦1 ⎭ ⎦ ⎣ 0 ⎩

by parts

⎧ ⎡⎛ ⎤⎫ ⎞ ⎪ ⎢⎜ ⎥⎪ ⎟ π π π 2 cos n 2 cos n sin n ⎪ ⎢⎜ − ⎥⎪ ⎟ + − 2 ⎪ ⎡⎛ ⎢ ⎥⎪ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎟ ⎤ ⎜ ⎛ nπ ⎞ ⎞ ⎪ ⎢⎜ cos nπ sin ⎜⎛ nπ ⎟⎞ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎪ ⎜ ⎟ ⎥ 2 ⎠ 2 ⎠ ⎜⎝ 2 ⎟⎠ ⎠ ⎝ ⎝ ⎪ ⎢⎜ ⎝ 2 ⎢ ⎥⎪ ⎝ ⎠ 2 + ⎟ − ( 0 + 0)⎥ + =⎨ − 2 ⎢ ⎥⎬ ⎥ ⎞ ⎛ ⎛ nπ ⎞ ⎟ n n n π π π ⎪ ⎢⎢⎜ ⎜⎛ nπ ⎟⎞ ⎪ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎥ ⎢ 2 cos ⎜ ⎟ cos ⎜ ⎟ sin ⎜ ⎟ ⎟ ⎥ ⎪ ⎜ ⎟ ⎟ ⎜ ⎪ ⎣⎝⎜ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎠ ⎦ ⎢ ⎝ 2 ⎠+ ⎝ 2 ⎠− ⎝ 2 ⎠ ⎟⎥ ⎪ ⎪ −⎜− ⎢ ⎥ 2 ⎜ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎪ ⎛ nπ ⎞ ⎟ ⎥ ⎪ ⎢ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎟⎥ ⎪ ⎪ ⎢⎣ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎠⎦ ⎭ ⎝ ⎩ ⎛ nπ ⎞ 2sin ⎜ ⎟ ⎝ 2 ⎠ = 8 sin ⎛ nπ ⎞ = ⎜ ⎟ 2 n 2 π2 ⎝ 2 ⎠ ⎛ nπ ⎞ ⎜ ⎟ ⎝ 2 ⎠

When n is even, b n = 0 b1 =

8 , π2

b3 = −

8 , (3) 2 π2

b5 =

8 , and so on. (5) 2 π2

∞ 8 8 8 ⎛ nπt ⎞ ⎛ πt ⎞ ⎛ 3πt ⎞ ⎛ 5πt ⎞ Thus, f (t) = ∑ b n sin ⎜ ⎟ = 2 sin ⎜ ⎟ − 2 2 sin ⎜ ⎟ + 2 2 sin ⎜ ⎟ − .... ⎝ 2 ⎠ (3) π ⎝ 2 ⎠ (5) π ⎝ 2 ⎠ ⎝ L ⎠ π n =1

i.e.

f(t) =

⎫ 8 ⎧ ⎛ πt ⎞ 1 ⎛ 3πt ⎞ 1 ⎛ 5πt ⎞ sin ⎜ ⎟ − 2 sin ⎜ + 2 sin ⎜ − .....⎬ ⎟ ⎟ 2 ⎨ π ⎩ ⎝ 2⎠ 3 ⎝ 2 ⎠ 5 ⎝ 2 ⎠ ⎭

© 2006 John Bird. All rights reserved. Published by Elsevier.

621

4. Show that the half-range Fourier cosine series for the function f(θ) = θ2 in the range 0 to 4 is

given by: f ( θ ) =

16 64 ⎧ ⎛ πθ ⎞ 1 ⎫ ⎛ 2πθ ⎞ 1 ⎛ 3πθ ⎞ − 2 ⎨cos ⎜ ⎟ − 2 cos ⎜ ⎟ + 2 cos ⎜ ⎟ − .....⎬ 3 π ⎩ ⎝ 4 ⎠ 2 ⎝ 4 ⎠ 3 ⎝ 4 ⎠ ⎭

Sketch the function within and outside of the given range. The periodic function is shown in the diagram below. Since a half-range cosine series is required, the function is symmetrical about the f(θ) axis and

1 a0 = L

an =

{∫

L 0

}

4 1 4 2 1 ⎧⎪ ⎡ θ3 ⎤ ⎫⎪ f (θ) dθ = ∫ θ dθ = ⎨ ⎢ ⎥ ⎬ = 4 0 4 ⎪⎣ 3 ⎦ 0 ⎪ ⎩ ⎭

∞ ⎛ nπθ ⎞ f (θ) = a 0 + ∑ a n cos ⎜ ⎟ ⎝ L ⎠ n =1

1 ⎛ 64 ⎞ 16 ⎜ ⎟= 4⎝ 3 ⎠ 3

2⎧ L ⎛ nπθ ⎞ ⎫ 2 4 2 ⎛ nπθ ⎞ ⎨ ∫ 0 f (θ) cos ⎜ ⎟ dθ ⎬ = ∫ 0 θ cos ⎜ ⎟ dθ L⎩ ⎝ L ⎠ ⎭ 4 ⎝ 4 ⎠

4 ⎧⎡ ⎫ ⎤ n πθ n πθ n πθ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎪ ⎢ θ2 sin ⎜ θ 2 cos 2sin ⎥ ⎟ ⎜ ⎟ ⎜ ⎟ ⎪ 1 ⎪⎢ 4 ⎠ 4 ⎠ 4 ⎠⎥ ⎪ ⎝ ⎝ ⎝ = ⎨ + − ⎬ by parts 2 3 ⎥ ⎪ 2 ⎪⎢ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎜ ⎟ ⎥ ⎪ ⎜ ⎟ ⎜ ⎟ ⎪ ⎢⎣ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎦0⎭ ⎩

⎧⎛ ⎫ ⎞ ⎧ ⎫ ⎪⎜ ⎪ ⎟ ⎪ ⎪ 64 1 ⎪⎜ 16sin nπ 8cos nπ 2sin nπ ⎟ ⎪ 1 ⎪ 8cos nπ ⎪ 1 ⎛ 8(16) ⎞ = ⎨ + − − ( 0 )⎬ = ⎨ = ⎜ 2 2 ⎟ cos nπ = 2 2 cos nπ 2 3 2 ⎬ n π 2 ⎪⎜ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎟ ⎪ 2 ⎪ ⎛ nπ ⎞ ⎪ 2 ⎝ n π ⎠ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎪⎩⎝ ⎝ 4 ⎠ ⎪⎭ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎠ ⎩⎪ ⎝ 4 ⎠ ⎭⎪

and a1 =

64 64 64 64 64 64 −1) = − 2 , a 2 = 2 2 (1) = 2 2 , a 3 = 2 2 (−1) = − 2 2 , and so on. 2 ( π (1) π π (2) π (2) π (3) π (3) 2

Thus,

∞ 64 64 ⎛ nπθ ⎞ 16 64 ⎛ πθ ⎞ ⎛ 2πθ ⎞ ⎛ 3πθ ⎞ f (θ) = a 0 + ∑ a n cos ⎜ ⎟ = − 2 cos ⎜ ⎟ + 2 2 cos ⎜ ⎟ − 2 2 cos ⎜ ⎟ + .... ⎝ L ⎠ 3 π ⎝ 4 ⎠ π (2) ⎝ 4 ⎠ π (3) ⎝ 4 ⎠ n =1

i.e.

f (θ ) =

⎫ 16 64 ⎧ ⎛ πθ ⎞ 1 ⎛ 2πθ ⎞ 1 ⎛ 3πθ ⎞ − 2 ⎨cos ⎜ − 2 cos ⎜ + 2 cos ⎜ − .....⎬ ⎟ ⎟ ⎟ 3 π ⎩ ⎝ 4 ⎠ 2 ⎝ 4 ⎠ 3 ⎝ 4 ⎠ ⎭

© 2006 John Bird. All rights reserved. Published by Elsevier.

622

CHAPTER 73 A NUMERICAL METHOD OF HARMONIC ANALYSIS EXERCISE 246 Page 686 1. Determine the Fourier series to represent the periodic function given by the table of values

below, up to and including the third harmonic and each coefficient correct to 2 decimal places. Use 12 ordinates. Angle θ°

30 60 90 120 150 180 210 240 270 300 330 360

Displacement y 40 43 38 θ°

sin θ

30

y

cosθ

y cos θ

y sin θ

30

40

0.866

34.64

0.5

20

60

43

0.5

21.50

0.866

37.24

23

cos 2θ

17

11

9

10

13

21

y cos 2θ

sin 2θ

y sin 2θ

cos 3θ

0.5

20

0.866

34.64

-0.5

-21.5

0.866

37.24 0

0 1

32

ycos3θ

sin3θ

ycos3θ

0

0

1

40

-1

-43

0

0

0

-1

-38

30

0

0

90

38

0

0

1

38

-1

-38

0

120

30

-0.5

-15.0

0.866

25.98

-0.5

-15

-0.866

-25.98

150

23

-0.866

-19.92

0.5

11.5

0.5

11.5

-0.866

-19.92

0

0

1

23

180

17

-1

-17

0

0

1

17

0

0

-1

-17

0

0

210

11

-0.866

-9.53

-0.5

-5.5

0.5

5.5

0.866

9.53

0

0

-1

-11

240

9

-0.5

-4.5

-0.866

-7.79

-0.5

-4.5

0.866

7.79

1

9

0

0

270

10

0

0

-1

-10

-1

-10

0

0

0

1

10

300

13

0.5

6.5

-0.866

-11.26

-0.5

-6.5

-0.866

-11.26

-1

-13

0

0

330

21

0.866

18.19

-0.5

-10.5

0.5

10.5

-0.866

-18.19

0

0

360

32

1

32

0

0

1

32

0

1

32

12

∑y k =1

k

= 287

12

∑y k =1

k

cos θk = 46.88

12

∑y k =1

k

sin θk = 87.67

12

∑y k =1

k

cos 2θk = 1

0

0

12

∑y k =1

k

sin 2θk = 13.85

12

∑y k =1

k

cos 3θk = −2

-21

-1

0

0

12

∑y k =1

k

sin 3θk = 3

a0 ≈

1 p 1 y k = (287) = 23.92 ∑ p k =1 12

an ≈

2 p ∑ yk cos nx k p k =1

hence, a1 ≈

2 2 2 (46.88) = 7.81 , a 2 ≈ (1) = 0.17 , a 3 ≈ (−2) = −0.33 12 12 12

2 p b n ≈ ∑ y k sin nx k p k =1

hence, b1 ≈

2 2 2 (87.67) = 14.61 , b 2 ≈ (13.85) = 2.31 , b3 ≈ (3) = 0.50 12 12 12 ∞

Substituting these values into the Fourier series: f (x) = a 0 + ∑ ( a n cos nx + b n sin nx ) n =1

gives: y = 23.92 + 7.81 cos θ + 0.17 cos 2θ - 0.33 cos 3θ + …… + 14.61 sin θ + 2.31 sin 2θ + 0.50 sin 3θ or

y = 23.92 + 7.81 cos θ +14.61 sin θ + 0.17 cos 2θ + 2.31 sin 2θ - 0.33 cos 3θ + 0.50 sin 3θ

© 2006 John Bird. All rights reserved. Published by Elsevier.

623

3. Determine the Fourier series to represent the periodic function given by the table of values

below, up to and including the third harmonic and each coefficient correct to 2 decimal places. Use 12 ordinates.

θ°

Angle θ°

30 60

Current i

0 -1.4 -1.8 -1.9 -1.8 -1.3

i

cosθ

30

0

0.866

60

-1.4

0.5

i cos θ

0 -0.7

90

sinθ

120 150 180 210 240 270 300 330 360

i sin θ

cos 2θ

0

i cos 2θ

2.2 3.8 3.9 3.5 2.5 sin 2θ

i sin 2θ

cos 3θ

i cos3θ

sin3θ

i cos3θ

0.5

0

0.866

0

0

0

1

0

0.866

-1.212

-0.5

0.7

0.866

-1.212

-1

1.4

0

0

0

0.5

90

-1.8

0

0

1

-1.8

-1

1.8

0

0

0

0

-1

1.8

120

-1.9

-0.5

0.95

0.866

-1.645

-0.5

0.95

-0.866

1.645

1

-1.9

0

0

150

-1.8

-0.866

1.56

0.5

-0.9

0.5

-0.90

-0.866

1.548

0

0

1

-1.8

180

-1.3

-1

1.3

0

1

-1.3

0

0

-1

1.3

0

0

0

-0.5

0.5

0

0.866

0

0

0

-1

0

-0.866

-1.905

-0.5

-1.1

0.866

1.905

1

2.2

0

0

0

-1

-3.8

-1

-3.8

0

0

0

0

1

3.8

1.95

-0.866

-3.377

-0.5

-1.95

-0.866

-3.377

-1

-3.9

0

0

0.866

3.03

-0.5

-1.75

0.5

1.75

-0.866

-3.031

0

0

-1

-3.5

1

2.5

0

0

1

2.5

0

0

1

2.5

0

0

210

0

-0.866

240

2.2

-0.5

270

3.8

0

300

3.9

0.5

330

3.5

360

2.5

12

∑y k =1

k

= 7.7

0

-1.1

12

∑y k =1

k

cos θk = 9.49

0

12

∑y k =1

k

sin θk = −16.389

12

∑y k =1

k

cos 2θk = −1.35

12

∑y k =1

k

sin 2θk = −2.522

12

∑y k =1

k

cos 3θk = 1.6

12

∑y k =1

k

sin 3θk = 0.3

a0 ≈

1 p 1 y k = (7.7) = 0.64 ∑ p k =1 12

an ≈

2 p 2 2 2 y k cos nx k hence, a1 ≈ (9.49) = 1.58 , a 2 ≈ (−1.35) = −0.23 , a 3 ≈ (1.6) = 0.27 ∑ 12 12 12 p k =1

2 p 2 2 b n ≈ ∑ y k sin nx k hence, b1 ≈ (−16.389) = −2.73 , b 2 ≈ (−2.522) = −0.42 , p k =1 12 12 b3 ≈

2 (0.3) = 0.05 12 ∞

Substituting these values into the Fourier series: f (x) = a 0 + ∑ ( a n cos nx + b n sin nx ) n =1

gives: i = 0.64 + 1.58 cos θ - 0.23 cos 2θ + 0.27 cos 3θ + …… - 2.73 sin θ - 0.42 sin 2θ + 0.05 sin 3θ or

i = 0.64 + 1.58 cos θ - 2.73 sin θ - 0.23 cos 2θ - 0.42 sin 2θ + 0.27 cos 3θ + 0.05 sin 3θ

© 2006 John Bird. All rights reserved. Published by Elsevier.

624

EXERCISE 247 Page 688 2. Analyse the periodic waveform of displacement y against angle θ in the diagram below into its

constituent harmonics as far as and including the third harmonic, by taking 30° intervals.

θ°

y

cosθ

30

10

0.866

8.66

0.5

60

-6

0.5

-3.0

0.866

-5.196

90

-17

0

0

1

-17

-1

120

-17

-0.5

8.5

0.866

-14.72

150

-13

-0.866

0.5

-6.5

180

-4

-1

0

210

10

-0.866

-8.66-

-0.5

240

24

-0.5

12

-0.866

270

33

0

0

-1

300

36

0.5

18

330

33

0.866

28.58

360

24

1

24

12

∑y k =1

k

= 113

k =1

11.26 4

12

∑y

y cos θ

k

cos θk = 79.34

sinθ

cos 2θ

y cos 2θ

sin 2θ

0.5

5

0.866

-0.5

3

0.866

17

0

-0.5

8.5

0.5

-6.5

0

1

-4

-5.0

0.5

5

-20.78

-0.5

-33

-1

-0.866

-31.18

-0.5

-0.5

-16.5

0.5

16.5

0

0

1

24

0

5

12

∑y k =1

y sin θ

k

12

∑y

sin θk

k =1

k

cos 3θ

y cos3θ

sin3θ

y cos3θ

8.66

0

0

1

10

-5.196

-1

6

0

0

0

0

0

-1

17

-0.866

14.722

1

-17

0

0

-0.866

11.258

0

0

1

-13

0

0

-1

4

0

0

0.866

8.66

0

0

-1

-10

-12

0.866

20.784

1

24

0

0

-33

0

0

0

0

1

33

-18

-0.866

-31.176

-1

-36

0

0

-0.866

-28.578

0

0

-1

-33

1

24

0

0

cos 2θk = 5.5

0

12

∑y k =1

y sin 2θ

k

sin 2θk = −0.866

12

∑y k =1

k

cos 3θk = 5

12

∑y k =1

k

sin 3θk = 4

= −144.88

a0 ≈

1 p 1 y k = (113) = 9.4 ∑ p k =1 12

an ≈

2 p 2 2 2 y k cos nx k hence, a1 ≈ (79.34) = 13.2 , a 2 ≈ (5.5) = 0.92 , a 3 ≈ (5) = 0.83 ∑ p k =1 12 12 12

bn ≈

2 p ∑ yk sin nx k p k =1

hence, b1 ≈ b3 ≈

2 2 (−144.88) = −24.1 , b 2 ≈ (−0.866) = −0.14 , 12 12 2 (4) = 0.67 12 ∞

Substituting these values into the Fourier series: f (x) = a 0 + ∑ ( a n cos nx + b n sin nx ) n =1

gives: y = 9.4 + 13.2 cos θ + 0.92 cos 2θ + 0.83 cos 3θ + …… - 24.1 sin θ - 0.14 sin 2θ + 0.67 sin 3θ © 2006 John Bird. All rights reserved. Published by Elsevier.

625

y = 9.4 + 13.2 cos θ - 24.1 sin θ + 0.92 cos 2θ - 0.14 sin 2θ + 0.83 cos 3θ + 0.67 sin 3θ

or

3. For the waveform of current shown below state why only a d.c. component and even cosine

terms will appear in the Fourier series and determine the series, using π/6 intervals, up to and including the sixth harmonic.

The function is even, thus no sine terms will be present. The function repeats itself every half cycle, hence only even terms will be present. Hence, the Fourier series will contains a d.c. component and even cosine terms only. θ°

i

cos 2θ

i cos 2θ

cos 4θ

i cos 4θ

cos 6θ

i cos 6θ

-1

-1.5

30

1.5

0.5

0.75

-0.5

-0.75

60

5.5

-0.5

-2.75

-0.5

-2.75

90

10

-1

-10

1

10

120

5.5

-0.5

-2.75

-0.5

-2.75

1

5.5

150

1.5

0.5

0.75

-0.5

-0.75

-1

-1.5

180

0

1

0

1

0

210

1.5

0.5

0.75

-0.5

-0.75

-1

-1.5

240

5.5

-0.5

-2.75

-0.5

-2.75

1

5.5

270

10

-1

-10

1

10

-1

-10

300

5.5

-0.5

-2.75

-0.5

-2.75

1

5.5

330

1.5

0.5

0.75

-0.5

-0.75

-1

-1.5

360

0

1

0

1

0

12

∑y k =1

k

= 48

12

∑y k =1

k

cos 2θk = −28

12

∑y k =1

k

1

5.5

-1

-10

0

1

0

1

12

∑y

cos 4θk = 6

k =1

k

cos 6θk = −4

a0 ≈

1 p 1 y k = (48) = 4.00 ∑ p k =1 12

an ≈

2 p 2 2 2 y k cos nx k hence, a 2 ≈ (−28) = −4.67 , a 4 ≈ (6) = 1.00 , a 3 ≈ (−4) = −0.66 ∑ 12 12 12 p k =1 ∞

Substituting these values into the Fourier series: f (x) = a 0 + ∑ ( a n cos nx ) n =1

gives:

i = 4.00 – 4.67 cos 2θ + 1.00 cos 4θ - 0.66 cos 6θ + ……

© 2006 John Bird. All rights reserved. Published by Elsevier.

626

CHAPTER 74 THE COMPLEX OR EXPONENTIAL FORM OF A FOURIER SERIES EXERCISE 248 Page 694 1. Determine the complex Fourier series for the function defined by:

⎧ 0, when − π ≤ t ≤ 0 f(t) = ⎨ ⎩ 2, when 0 ≤ t ≤ π The function is periodic outside of this range of period 2π. The periodic function is shown in the diagram below.

f (t) =

The complex Fourier series is given by:

∞

∑

n =−∞

cn e

j

2π n t L

2π n t −j 1 L/2 c n = ∫ f (t) e L dt L −L / 2

where

π

2π n t −j π ⎫ 1 π − jn t 1 ⎡ e − jn t ⎤ 1 ⎧ 0 1 2π ⎡⎣e − jn π − e0 ⎤⎦ cn = 0 dt 2 e dt + = ∫ e = ⎢ =− ⎨ ∫ −π ⎬ ⎥ ∫ 0 0 π ⎣ − jn ⎦ 0 2π ⎩ jπ n ⎭ π

i.e.

= −

=

j j − jn π ⎡⎣e − jnπ − 1⎤⎦ = ( e − 1) πn j πn 2

j j [cos nπ − jsin nπ − 1] = [ cos nπ − 1] πn πn

for all

integer values of n f (t) =

Hence,

∞

∑

n=− ∞

c0 = a 0 = mean value = c1 =

j 2 (−1 − 1) = − j , π π

c3 = − j

2 , 3π

c5 = − j

cn e

j

2πn t L

=

∞

j ( cos nπ − 1) e jn t n= − ∞ n π

∑

2× π =1 2π c2 =

j (1 − 1) = 0 and all even terms will be zero 2π

2 , and so on. 5π

© 2006 John Bird. All rights reserved. Published by Elsevier.

627

c −1 =

j 2 (−2) = j , −π π

c −3 =

j 2 , (−2) = j −3π 3π

c5 = j

2 , and so on. 5π

2 jt 2 2 2 2 2 e − j e j3t − j e j5t − .... + j e− jt + j e− j3t + j e− j5t π π 3π 5π 3π 5π

Thus,

f(t) = 1 − j

i.e.

2⎛ 1 1 2⎛ 1 1 ⎞ ⎞ f (t) = 1 − j ⎜ e jt + e j3t + e j5t + .... ⎟ + j ⎜ e − jt + e − j3t + e − j5t + .... ⎟ π⎝ 3 5 3 5 ⎠ π⎝ ⎠

2. Show that the complex Fourier series for the waveform shown below, that has period 2, may be

represented by:

f (t) = 2 +

∞

j2 ( cos nπ − 1) e jπ n t n = − ∞ πn

∑

(n ≠ 0)

f (t) =

The complex Fourier series is given by:

∞

∑

n=− ∞

cn e

j

2π n t L

2π n t −j 1 L/2 c n = ∫ f (t) e L dt L −L / 2

where

1

i.e.

1 ⎡ e− jπ n t ⎤ 1 ⎧ 1 − j2π n t ⎫ 2 ⎡⎣ e− jπ n − e0 ⎤⎦ c n = ⎨ ∫ 4e 2 ⎬ = 2 ∫ e − jπ n t = 2 ⎢ =− ⎥ 0 0 2⎩ jπ n ⎣ − jπ n ⎦ 0 ⎭

= −

=

2 j2 − jn π ⎡e − jn π − 1⎦⎤ = ( e − 1) ⎣ πn jπ n

j2 j2 [cos nπ − jsin nπ − 1] = [ cos nπ − 1] πn πn

for all

integer values of n c0 = a 0 = mean value =

1 1 1 1 4 dt = [ 4t ] 0 = 2 ∫ 2 0 2

Hence,

f (t) = 2 +

∞

j2

∑ n π ( cos nπ − 1) e

jπ n t

n =−∞ (n ≠ 0)

© 2006 John Bird. All rights reserved. Published by Elsevier.

628

3. Show that the complex Fourier series of Problem 2 is equivalent to:

8⎛ 1 1 ⎞ f (t) = 2 + ⎜ sin πt + sin 3πt + sin 5πt + .... ⎟ π⎝ 3 5 ⎠ j2 4 (−1 − 1) = − j , π(1) π

c2 =

c3 = j

2 4 (−2) = − j , 3π 3π

c5 = − j

c −1 =

j2 4 (−2) = j , −π π

c1 =

Thus,

f (t) = 2 +

c −3 =

j2 (1 − 1) = 0 and all even terms will be zero 2π 4 , and so on. 5π

j2 4 , (−2) = j −3π 3π

c5 = j

4 , and so on. 5π

∞

j2 ( cos nπ − 1) e jπ n t n=− ∞ n π

∑

(n ≠ 0)

4 j πt 4 4 4 4 4 e − j e j3πt − j e j5 πt − .... + j e− jπ t + j e− j3πt + j e− j5πt π π 3π 5π 3π 5π

i.e.

f(t) = 2 − j

i.e.

4⎛ 1 1 1 1 ⎞ 4⎛ ⎞ f (t) = 2 − j ⎜ e j πt + e j3πt + e j5πt + .... ⎟ + j ⎜ e − j πt + e− j3πt + e− j5 πt + .... ⎟ π⎝ 3 5 3 5 ⎠ π⎝ ⎠ 4⎡ 1 1 ⎤ = 2 − j ⎢( e j π t − e − jπ t ) + ( e j3 π t − e− j3 π t ) + ( e j5 πt − e− j5 π t ) + ....⎥ π⎣ 3 5 ⎦ = 2 − j2

i.e.

f (t) = 2 +

8 ⎡⎛ e j π t − e − j π t ⎢⎜ 2j π ⎣⎝

⎞ 1 ⎛ e j3 π t − e− j3 π t ⎟+ ⎜ 2j ⎠ 3⎝

⎞ 1 ⎛ e j5 π t − e − j5 π t ⎟+ ⎜ 2j ⎠ 5⎝

⎤ ⎞ ⎟ + ....⎥ ⎠ ⎦

8⎡ 1 1 ⎤ sin πt + sin 3πt + sin 5πt + ....⎥ ⎢ π⎣ 3 5 ⎦

4. Determine the exponential form of the Fourier series for the function defined by:

f(t) = e2t when – 1 < t < 1 and has period 2. The function is shown in the diagram below.

The complex Fourier series is given by:

f (t) =

∞

∑

n =−∞

cn e

j

2π n t L

© 2006 John Bird. All rights reserved. Published by Elsevier.

629

cn =

where

2π n t −j 1 L/2 L f (t) e dt L ∫ −L / 2

− 2− j π n ) t 2− jπ n ) 2− jπ n ) 2πn t ⎤ ⎤ −j ⎫ 1 1 −e ( 1⎧ 1 1 ⎡e ( 1 ⎡ e( = ⎢ c n = ⎨ ∫ e 2t e 2 dt ⎬ = ∫ e 2t − j π n t dt = ⎢ ⎥ ⎥ −1 2 ⎩ −1 2 ⎣ 2 − jπ n ⎦ −1 2 ⎣ 2 − jπ n ⎭ 2 ⎦ 1

i.e.

∞

Thus,

f(t) =

∑

n =− ∞

cn e

j

2π n t L

=

1 ∞ ⎛ e( 2− j π n) − e − ( 2− j π n) ⎞ j π n t ∑ ⎜ 2 − jπ n ⎟e 2 n=− ∞ ⎝ ⎠

© 2006 John Bird. All rights reserved. Published by Elsevier.

630

EXERCISE 249 Page 698 1. Determine the exponential form of the Fourier series for the periodic function defined by: π ⎧ ⎪ − 2, when − π ≤ x ≤ − 2 ⎪ π π ⎪ f(x) = ⎨ 2, when − ≤ x ≤ + 2 2 ⎪ π ⎪ ⎪ − 2, when + 2 ≤ x ≤ +π ⎩

and which has a period of 2π.

The periodic waveform is shown below. It is an even function and contains no sine terms, hence b n = 0 and between -π and +π, the mean value is zero, hence a 0 = 0 .

cn =

2 L/2 2 π ⎛ 2πnx ⎞ ⎛ 2πnx ⎞ f (x) cos ⎜ dx = f (x) cos ⎜ ⎟ ⎟ dx ∫ ∫ L 0 2π 0 ⎝ L ⎠ ⎝ 2π ⎠ =

1 π

{∫

π/ 2 0

2 cos nx dx + ∫

π π/2

since L = 2π

}

−2 cos nx dx

π/2 π 1 ⎧⎪ ⎡ 2sin nx ⎤ ⎡ 2sin nx ⎤ ⎫⎪ −⎢ = ⎨⎢ ⎬ π ⎩⎪ ⎣ n ⎥⎦ 0 ⎣ n ⎥⎦ π / 2 ⎭⎪

∞

Hence,

f(x) =

∑ce

n =− ∞

n

j

2π n x L

=

1 ⎡⎛ nπ nπ ⎞ ⎤ 4 nπ ⎞ ⎛ − 0 ⎟ − ⎜ 2sin nπ − 2sin ⎟ ⎥ = sin ⎜ 2sin ⎢ 2 2 ⎠ ⎦ πn 2 πn ⎣⎝ ⎠ ⎝

=

⎧ 4 ⎛ nπ ⎞ ⎫ jn x ⎨ sin ⎜ ⎟⎬ e ⎝ 2 ⎠⎭ n = − ∞ ⎩ πn ∞

∑

2. Show that the exponential form of the Fourier series in Problem 1 above is equivalent to:

f (x) =

8⎛ 1 1 1 ⎞ ⎜ cos x − cos 3x + cos 5x − cos 7x + .... ⎟ π⎝ 3 5 7 ⎠

© 2006 John Bird. All rights reserved. Published by Elsevier.

631

Since from Problem 1, c n = c0 = 0 , c1 = c3 =

4 π 4 4 2π = 0 = c 4 = c6 = c −2 = c −4 and so on , sin = , c 2 = sin π 2 π 2π 2

4 3π 4 sin =− , 3π 2 3π

4 −π 4 c −1 = − sin = , π 2 π ∞

f(x) =

∑

n =− ∞

4 nπ sin , then πn 2

cn e

j

2π n x L

c5 =

c −3 = − ∞

=

⎧4

4 −3π 4 4 and so on. sin = − , c −5 = 3π 2 3π 5π ⎛ nπ ⎞ ⎫ jn x ⎟⎬ e 2 ⎠⎭

∑ ⎨⎩ πn sin ⎜⎝

n =− ∞

=

4 4 and so on. , c7 = − 5π 7π

4 jx 4 j3x 4 j5x 4 4 4 e − e + e + ..... + e − jx − e − j3x + e − j5x − ..... π π 3π 5π 3π 5π

4 4 4 ⎛4 ⎞ ⎛ 4 ⎞ ⎛ 4 ⎞ = ⎜ e jx + e − jx ⎟ − ⎜ e j3x + e − j3x ⎟ + ⎜ e j5x + e − j5x ⎟ − ..... π 3π 5π ⎝π ⎠ ⎝ 3π ⎠ ⎝ 5π ⎠

i.e.

=

8 ⎛ e jx + e − jx ⎞ 8 ⎛ e j3x + e − j3x ⎞ 8 ⎛ e j5x + e− j5x ⎞ ⎜ ⎟− ⎜ ⎟+ ⎜ ⎟ − ..... π⎝ 2 2 2 ⎠ 3π ⎝ ⎠ 5π ⎝ ⎠

=

8 8 8 cos x − cos 3x + cos 5x − ..... π 3π 5π

f (x) =

8⎛ 1 1 1 ⎞ ⎜ cos x − cos 3x + cos 5x − cos 7x + ..... ⎟ π⎝ 3 5 7 ⎠

3. Determine the complex Fourier series to represent the function f(t) = 2t in the range - π to + π.

The triangular waveform shown below is an odd function since it is symmetrical about the origin.

The period of the waveform, L = 2π. Thus,

2 L2 ⎛ 2πnt ⎞ c n = − j ∫ f (t) sin ⎜ ⎟ dt 0 L ⎝ L ⎠

= −j

2 π 2 π ⎛ 2πnt ⎞ 2t sin ⎜ dt = − j ∫ t sin nt dt ⎟ ∫ 2π 0 π 0 ⎝ 2π ⎠

© 2006 John Bird. All rights reserved. Published by Elsevier.

632

π

2 ⎡ − t cos nt sin nt ⎤ 2 ⎡⎛ −π cos nπ sin nπ ⎞ ⎤ = −j ⎢ + 2 ⎥ = − j ⎢⎜ + ⎟ − ( 0 + 0 )⎥ 2 n n ⎦0 n n ⎠ π⎣ π ⎣⎝ ⎦ i.e.

by parts

2 c n = j cos nπ n

Hence, the complex Fourier series is given by: ∞

f(t) =

∑c

n =− ∞

n e

j

2 πnt L

∞

=

⎛ j2

⎞

∑ ⎜⎝ n cos nπ ⎟⎠ e

jnt

n=− ∞

4. Show that the complex Fourier series is Problem 3 above is equivalent to:

1 1 1 ⎛ ⎞ f(t) = 4 ⎜ sin t − sin 2t + sin 3t − sin 4t + .... ⎟ 2 3 4 ⎝ ⎠ 2 From Problem 3 above, c n = j cos nπ n When n = 1, c1 = j

2 2 j2 cos π = j ( −1) = − (1) (1) 1

2 2 When n = 2, c 2 = j cos 2π = j 2 2 2 2 j2 When n = 3, c3 = j cos 3π = j ( −1) = − 3 3 3π By similar reasoning, c 4 =

j2 , 4

c5 = −

j2 , and so on. 5

When n = -1, c −1 = j

2 2 j2 cos(−π) = + j ( −1) = (−1) (−1) 1

When n = -2, c −2 = j

2 2 j2 cos(−2π) = j (1) = − (−2) (−2) 2

2 j2 By similar reasoning, c −3 = j , c −4 = − , and so on. 3 4 Since the waveform is odd, c0 = a 0 = 0 ∞

From Problem 3,

f(t) =

∑c

n =−∞

Hence,

f(t) = −

n e

j

2 πnt L

∞

=

⎛ j2

⎞

∑ ⎜⎝ n cos nπ ⎟⎠ e

jn t

n =−∞

j2 jt j2 j2t j2 j3t j2 j4 t j2 − jt j2 − j2 t j2 − j3t j2 − j4t + e − e + ... e + e − e + e − ... + e − e 1 2 3 4 1 2 3 4

j2 j2 j2 ⎛ j2 ⎞ ⎛ j2 ⎞ ⎛ j2 ⎞ = ⎜ − e jt + e − jt ⎟ + ⎜ e j2 t − e− j2 t ⎟ + ⎜ − e j3t + e− j3t ⎟ + .... 1 2 3 ⎝ 1 ⎠ ⎝2 ⎠ ⎝ 3 ⎠

© 2006 John Bird. All rights reserved. Published by Elsevier.

633

⎛ e jt − e − jt = − j4 ⎜ 2 ⎝

⎞ j4 ⎛ e j2t − e − j2 t ⎟+ ⎜ 2 ⎠ 2⎝

⎛ e jt − e − jt = − j2 4 ⎜ ⎝ 2j

⎞ j4 ⎛ e j3t − e − j3t ⎟− ⎜ 2 ⎠ 3⎝

⎞ j2 4 ⎛ e j2t − e − j2 t ⎟+ ⎜ 2j ⎠ 2 ⎝

⎞ ⎟ + .... ⎠

⎞ j2 4 ⎛ e j3t − e− j3t ⎟− ⎜ 2j ⎠ 3 ⎝

⎞ ⎟ + .... by multiplying top and ⎠ bottom by j

4 4 = 4sin t − sin 2t + sin 3t + .... 2 3 i.e.

1 1 1 ⎛ ⎞ f(t) = 4 ⎜ sin t − sin 2t + sin 3t − sin 4t + ... ⎟ 2 3 4 ⎝ ⎠

Hence,

f(t) =

∞

⎛ j2

⎞

∑ ⎜⎝ n cos nπ ⎟⎠ e

n =−∞

jnt

1 1 1 ⎛ ⎞ ≡ 4 ⎜ sin t − sin 2t + sin 3t − sin 4t + ... ⎟ 2 3 4 ⎝ ⎠

© 2006 John Bird. All rights reserved. Published by Elsevier.

634

EXERCISE 250 Page 703 2. Determine the pair of phasors that can represent the harmonic given by:

v = 10 cos 2t – 12 sin 2t v = 10 cos 2t – 12 sin 2t ⎡1 ⎤ ⎡1 ⎤ = 10 ⎢ ( e j2 t + e − j2 t ) ⎥ − 12 ⎢ ( e j2 t − e− j2t ) ⎥ ⎣2 ⎦ ⎣2j ⎦ 6 6 = 5e j2 t + 5e − j2 t − e j2 t + e− j2 t j j = 5e j2 t + 5e − j2 t + 6 je j2 t − 6 je− j2 t

(note:

1 −j = j 1

or

1 j = ) −j 1

i.e. v = ( 5 + j6 ) e j2t + ( 5 − j6 ) e − j2t Hence,

v = 7.81∠0.88 rad, rotating anticlockwise with an angular velocity, ω = 2 rad/s

and v = 7.81∠-0.88 rad, rotating clockwise with an angular velocity, ω = 2 rad/s, as shown in the diagram below.

3. Find the pair of phasors that can represent the fundamental current:

i = 6 sin t + 4 cos t i = 6 sin t + 4 cos t ⎡1 ⎤ ⎡1 ⎤ 3 = 6 ⎢ ( e j t − e − j t ) ⎥ + 4 ⎢ ( e j t + e − jt ) ⎥ = ( e jt − e− jt ) + 2 ( e jt + e− jt ) ⎣2 ⎦ j ⎣2j ⎦ = −3j ( e jt − e − jt ) + 2 ( e jt + e− jt )

i.e. i = ( 2 − j3 ) e j t + ( 2 + j3 ) e − j t

© 2006 John Bird. All rights reserved. Published by Elsevier.

635

Hence,

i = 3.61∠-0.98 rad, rotating anticlockwise with an angular velocity, ω = 1 rad/s

and i = 3.61∠0.98 rad, rotating clockwise with an angular velocity, ω = 1 rad/s, as shown in the diagram below.

© 2006 John Bird. All rights reserved. Published by Elsevier.

636