Solutions Manual Modern Engineering Mathematics 5th Edition By Glyn James

Solutions Manual
Modern Engineering
Mathematics
Fifth edition

Glyn James
and
David Burley
Dick Clements
Phil Dyke
John Searl
Jerry Wright

, Contents
Chapter 1 Numbers, Algebra and Geometry 1
Chapter 2 Functions 37
Chapter 3 Complex Numbers 103
Chapter 4 Vector Algebra 152
Chapter 5 Matrix Algebra 180
Chapter 6 An Introduction to Discrete Mathematics 229
Chapter 7 Sequences, Series and Limits 263
Chapter 8 Differentiation and Integration 316
Chapter 9 Further Calculus 446
Chapter 10 Introduction to Ordinary Differential Equation 516
Chapter 11 Introduction to Laplace Transforms 620
Chapter 12 Introduction to Fourier Series 648
Chapter 13 Data Handling and Probability Theory 689




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, CHAPTER 1


Numbers, Algebra and Geometry

1.2.4 Exercises
„ 1
110110.1012 = 25 + 24 + 22 + 21 + 2−1 + 2−3
= 54.62510

„ 2
16 321 = 213 + 212 + 211 + 210 + 29 + 28 + 27 + 26 + 20
= 111111110000012
16 321 = 3 × 84 + 7 × 83 + 7 × 82 + 80
= 377018

To convert from binary to octal: take the first three entries immediately to the right and include the
20 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 80 term of the octal expansion. Now do the same
with the next three digits of the binary expansion to get the 81 term of the octal expansion, and so on.

1012 = 58, 1002 = 48, 0112 = 38, and 12 = 18 so
[10111001011012 = 134558]

„ 3
30.6 = 24 + 23 + 22 + 21 + 2−1 + 2−4 + 2−5 + 2−8 + 2−9 + 2−12 + 2−13
= 11110.1001100110011 . . . 2
30.6 = 3 × 8 + 6 × 80 + 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: 1002 = 48, 1102 = 68, 0112 = 38 and 0012 = 18.

„ 4(a)
100011.0112
+ 1011.0012
101110.1002




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© Pearson Education Limited 2015

, th
James, Burley, Clements, Dyke, Searl and Wright, Modern Engineering Mathematics, 5 Edition,
Solutions Manual on the Web

4(b)
111.100112
× 10.1112
0.11110011
1.1110011
11.110011
+ 1111.0011
10101.110101012

„ 5(a)
23 × 2−4 = 23 ÷ 24 = 1/2

5(b)
23 ÷ 2−4 = 23 × 24 = 23+4 = 27

5(c)
(23)−4 = 1/(23)4 = 1/212

5(d)
31/3 × 35/3 = 3(1/3+5/3) = 32

5(e)
36−1/2 = 1/(36)1/2 = 1/6

5(f)
163/4 = (161/4)3 = 23

„ 6(a)
(21 + ((4 × 3) ÷ 2))

6(b)
(17 − 6(2+3))

6(c)
((4 × 23) − ((7 ÷ 6) × 2))

6(d)
−5
((2 × 3) − (6 ÷ 4) + 3(2 ))

„ 7(a)
(7 + 5 2 )3 = (7 + 5 2 ) (7 + 5 2 ) 2
= (7 + 5 2 ) (99 + 70 2 )
= 7 × 99 + 5 × 70 × 2 + (7 × 70 + 5 × 99) 2
= 1393 + 985 2




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