Solutions Manual - Contents Chapter 1 Number, Algebra and Geomet...

Solutions Manual (All Chapters) Modern Engineering Mathematics Sixth edition Glyn James Phil Dyke and John Searl Matthew Craven Yinghui Wei 1 / 4

iii © Pearson Education Limited 2020 Contents Chapter 1 Number, Algebra and Geometry 1 Chapter 2 Functions 38 Chapter 3 Complex Numbers 104 Chapter 4 Vector Algebra 153 Chapter 5 Matrix Algebra 181 Chapter 6 An Introduction to Discrete Mathematics 230 Chapter 7 Sequences, Series and Limits 264 Chapter 8 Differentiation and Integration 317 Chapter 9 Further Calculus 447 Chapter 10 Introduction to Ordinary Differential Equation 517 Chapter 11 Introduction to Laplace Transforms 621 Chapter 12 Introduction to Fourier Series 649 Chapter 13 Data Handling and Probability Theory 690 2 / 4

1 © Pearson Education Limited 2020

CHAPTER 1

Number, Algebra and Geometry

1.2.4 Exercises

 1

110110.101

• = 2

5

• 2

4

• 2

2

• 2

1

• 2

−1

• 2

−3

= 54.625

10

 2

16 321 = 2

13

• 2

12

• 2

11

• 2

10

• 2

9

• 2

8

• 2

7

• 2

6

• 2

= 11111111000001

2

16 321 = 3 × 8

4

+ 7 × 8

3

+ 7 × 8

2

• 8

= 37701

8 To convert from binary to octal: take the first three entries immediately to the right and include the 2

term in the binary expansion; this may be considered as a three-digit binary number; convert this number into octal; the resulting octal number is the 8

term of the octal expansion. Now do the same with the next three digits of the binary expansion to get the 8 1 term of the octal expansion, and so on.101

• = 58, 1002 = 48, 0112 = 38, and 12 = 18, so

[1011100101101

2 = 134558]

 3

30 .6 = 2

4

• 2

3

• 2

2

• 2

1

• 2

−1

• 2

−4

• 2

−5

• 2

−8

• 2

−9

• 2

−12

• 2

−13

= 11110.1001100110011 . . .

2

30 .6 = 3 × 8 + 6 × 8

+ 4 × 8

−1

+ 6 × 8

−2

+ 3 × 8

−3

• 8

−4

+ 4 × 8

−5

+ 6 × 8

−6

+ 3 × 8

−7

• 8

−8

+ . . .

= 36.46314631 . . .

8

The rule works in this case as well: 100

• = 48, 1102 = 68, 0112 = 38 and 0012 = 18.

 4(a)

100011.011

2

+ 1011.0012

101110.1002

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James et al., Modern Engineering Mathematics, 6th edition, Solutions Manual 2 © Pearson Education Limited 2020 4(b)

111.10011

2

× 10.111

2

0.11110011

1.1110011

11.110011

+ 1111.0011

10101.11010101

2

 5(a) 2 3 × 2 −4 = 2 3 ÷ 2 4 = 1/2

5(b) 2 3 ÷ 2 −4 = 2 3 × 2 4 = 2 3+4 = 2 7

5(c) (2 3 ) −4

= 1/(2

3 ) 4 = 1/2 12

5(d) 3 1/3 × 3 5/3 = 3

(1/3+5/3)

= 3 2

5(e) 36

−1/2

= 1/(36)

1/2 = 1/6

5(f) 16 3/4 = (16 1/4 ) 3 = 2 3

 6(a)

(21 + ((4 × 3) ÷ 2))

6(b)

(17 − 6

(2+3) )

6(c)

((4 × 2

3

) − ((7 ÷ 6) × 2))

6(d)

((2 × 3) − (6 ÷ 4) + 3

(2 −5 ) )

 7(a) (7 + 3

)25 = (7 + )25(7 +

2 )25

= (7 +

)25(99 + )270

= 7 × 99 + 5 × 70 × 2 + (7 × 70 + 5 × 99)

2

= 1393 +

2985

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