Mathematics Pocket Book for Engineers and Scientists 9781000726800, 9780367266530, 9780367266523, 9780429294402, 1000726800

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Why is knowledge of science and mathematics important in engineering? A career in any engineering field will require both basic and advanced mathematics and science. Without mathematics and science to determine principles, calculate dimensions and limits, explore variations, prove concepts, and so on, there would be no mobile telephones, televisions, stereo systems, video games, microwave ovens, computers, or virtually anything electronic. There would be no bridges, tunnels, roads, skyscrapers, automobiles, ships, planes, rockets or most things mechanical. There would be no metals beyond the common ones, such as iron and copper, no plastics, no synthetics. In fact, society would most certainly be less advanced without the use of mathematics and science throughout the centuries and into the future. Electrical engineers require mathematics and science to design, develop, test, or supervise the manufacturing and installation of electrical equipment, components, or systems for commercial, industrial, military, or scientific use. Mechanical engineers require mathematics and science to perform engineering duties in planning and designing tools, engines, machines, and other mechanically functioning equipment; they oversee installation, operation, maintenance, and repair of such equipment as centralised heat, gas, water, and steam systems. Aerospace engineers require mathematics and science to perform a variety of engineering work in designing, constructing, and testing aircraft, missiles, and spacecraft; they conduct basic and applied research to evaluate adaptability of materials and equipment to aircraft design and manufacture and recommend improvements in testing equipment and techniques. Nuclear engineers require mathematics and science to conduct research on nuclear engineering problems or apply principles and theory of nuclear science to problems concerned with release, control, and utilisation of nuclear energy and nuclear waste disposal. Petroleum engineers require mathematics and science to devise methods to improve oil and gas well production and determine the need for new or modified tool designs; they oversee drilling and offer technical advice to achieve economical and satisfactory progress.

Industrial engineers require mathematics and science to design, develop, test, and evaluate integrated systems for managing industrial production processes, including human work factors, quality control, inventory control, logistics and material flow, cost analysis, and production coordination. Environmental engineers require mathematics and science to design, plan, or perform engineering duties in the prevention, control, and remediation of environmental health hazards, using various engineering disciplines; their work may include waste treatment, site remediation, or pollution control technology. Civil engineers require mathematics and science in all levels in civil engineering – structural engineering, hydraulics and geotechnical engineering are all fields that employ mathematical tools such as differential equations, tensor analysis, field theory, numerical methods and operations research. Architects require knowledge of algebra, geometry, trigonometry and calculus. They use mathematics for several reasons, leaving aside the necessary use of mathematics in the engineering of buildings. Architects use geometry because it defines the spatial form of a building, and they use mathematics to design forms that are considered beautiful or harmonious. The front cover of this text shows a modern London architecture and the financial district, all of which at some stage required in its design a knowledge of mathematics. Knowledge of mathematics and science is clearly needed by each of the disciplines listed above. It is intended that this text – Mathematics Pocket Book for Engineers and Scientists – will provide a step by step, helpful reference, to essential mathematics topics needed by engineers and scientists.

Mathematics Pocket Book for Engineers and Scientists Fifth Edition

John Bird

Fifth edition published 2020 by Routledge 2 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN and by Routledge 52 Vanderbilt Avenue, New York, NY 10017 Routledge is an imprint of the Taylor & Francis Group, an informa business © 2020 John Bird The right of John Bird to be identified as author of this work has been asserted by him in accordance with sections 77 and 78 of the Copyright, Designs and Patents Act 1988. All rights reserved. No part of this book may be reprinted or reproduced or utilised in any form or by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying and recording, or in any information storage or retrieval system, without permission in writing from the publishers. Trademark notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. First edition published as Newnes Mathematics for Engineers Pocket Book by Newnes 1983 Fourth edition published as Engineering Mathematics Pocket Book by Routledge 2008 British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloging-in-Publication Data A catalog record has been requested for this book ISBN: 978-0-367-26653-0 (hbk) ISBN: 978-0-367-26652-3 (pbk) ISBN: 978-0-429-29440-2 (ebk) Typeset in Frutiger 45 Light by Servis FIlmsetting Ltd, Stockport, Cheshire.

Mathematics Pocket Book for Engineers and Scientists

John Bird is the former Head of Applied Electronics in the Faculty of Technology at Highbury College, Portsmouth, UK. More recently, he has combined freelance lecturing at the University of Portsmouth, with examiner responsibilities for Advanced Mathematics with City and Guilds and examining for International Baccalaureate. He has over 45 years’ experience of successfully teaching, lecturing, instructing, training, educating and planning trainee engineers study programmes. He is the author of 140 textbooks on engineering, science and mathematical subjects, with worldwide sales of over one million copies. He is a chartered engineer, a chartered mathematician, a chartered scientist and a Fellow of three professional institutions. He is currently lecturing at the Defence College of Marine Engineering in the Defence College of Technical Training at H.M.S. Sultan, Gosport, Hampshire, UK, one of the largest technical training establishments in Europe.

Contents Preface

xiii

Section 1 Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5

Engineering conversions, constants and symbols General conversions and the Greek alphabet Basic SI units, derived units and common prefixes Some physical and mathematical constants Recommended mathematical symbols Symbols for physical quantities

1 2 3 5 7 9

Section 2 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21

Some algebra topics Introduction to algebra Polynomial division The factor theorem The remainder theorem Continued fractions Solving simple equations Transposing formulae Solving simultaneous equations Solving quadratic equations by factorising Solving quadratic equations by completing the square Solving quadratic equations by formula Logarithms Exponential functions Napierian logarithms Hyperbolic functions Partial fractions

17 18 22 23 25 25 27 32 36 42 45 47 49 53 55 58 63

Section 3 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29

Some number topics Simple number sequences Arithmetic progressions Geometric progressions Inequalities The binomial series Maclaurin’s theorem Limiting values – L’Hopital’s rule Solving equations by iterative methods (1) – the bisection method Solving equations by iterative methods (2) – an algebraic method of successive approximations Solving equations by iterative methods (3) – the Newton-Raphson method Computer numbering systems

67 68 70 71 73 80 84 87

Chapter 30 Chapter 31 Chapter 32

88 90 92 93

viii

Contents

Section 4 Chapter 33 Chapter 34 Chapter 35 Chapter 36 Chapter 37 Chapter 38 Chapter 39

Areas and volumes Area of plane figures Circles Volumes and surface areas of regular solids Volumes and surface areas of frusta of pyramids and cones The frustum and zone of a sphere Areas and volumes of irregular figures and solids The mean or average value of a waveform

101 103 106 111 116 119 122 127

Section 5 Chapter 40 Chapter 41 Chapter 42 Chapter 43 Chapter 44 Chapter 45 Chapter 46 Chapter 47 Chapter 48 Chapter 49 Chapter 50 Chapter 51 Chapter 52 Chapter 53

131 133 133 135 136 137 138 139 142 144 149 149 152 157

Chapter 54

Geometry and trigonometry Types and properties of angles Properties of triangles The theorem of Pythagoras Trigonometric ratios of acute angles Evaluating trigonometric ratios Fractional and surd forms of trigonometric ratios Solution of right-angled triangles Cartesian and polar co-ordinates Sine and cosine rules and areas of any triangle Graphs of trigonometric functions Angles of any magnitude Sine and cosine waveforms Trigonometric identities and equations The relationship between trigonometric and hyperbolic functions Compound angles

Section 6 Chapter 55 Chapter 56 Chapter 57 Chapter 58 Chapter 59 Chapter 60 Chapter 61 Chapter 62 Chapter 63

Graphs The straight-line graph Determination of law Graphs with logarithmic scales Graphical solution of simultaneous equations Quadratic graphs Graphical solution of cubic equations Polar curves The ellipse and hyperbola Graphical functions

170 172 174 179 183 184 189 190 196 197

Section 7 Chapter 64 Chapter 65 Chapter 66 Chapter 67 Chapter 68 Chapter 69

Complex numbers General complex number formulae Cartesian form of a complex number Polar form of a complex number Applications of complex numbers De Moivre’s theorem Exponential form of a complex number

204 205 205 208 210 212 214

Section 8 Chapter 70 Chapter 71 Chapter 72 Chapter 73 Chapter 74 Chapter 75

Vectors Scalars and vectors Vector addition Resolution of vectors Vector subtraction Relative velocity i, j, k notation

216 217 217 219 220 222 224

161 163

Contents

ix

Chapter 76 Chapter 77 Chapter 78

Combination of two periodic functions The scalar product of two vectors Vector products

225 227 230

Section 9 Chapter 79 Chapter 80 Chapter 81 Chapter 82 Chapter 83 Chapter 84 Chapter 85

233 234 235 236 238 239 242

Chapter 87

Matrices and determinants Addition, subtraction and multiplication of matrices The determinant and inverse of a 2 by 2 matrix The determinant of a 3 by 3 matrix The inverse of a 3 by 3 matrix Solution of simultaneous equations by matrices Solution of simultaneous equations by determinants Solution of simultaneous equations using Cramer’s rule Solution of simultaneous equations using Gaussian elimination Eigenvalues and eigenvectors

Section 10 Chapter 88 Chapter 89 Chapter 90 Chapter 91 Chapter 92 Chapter 93 Chapter 94

Boolean algebra and logic circuits Boolean algebra and switching circuits Simplifying Boolean expressions Laws and rules of Boolean algebra De Morgan’s laws Karnaugh maps Logic circuits and gates Universal logic gates

253 254 257 258 259 261 266 269

Section 11 Chapter 95 Chapter 96 Chapter 97 Chapter 98 Chapter 99 Chapter 100 Chapter 101 Chapter 102 Chapter 103 Chapter 104 Chapter 105 Chapter 106 Chapter 107 Chapter 108 Chapter 109 Chapter 110 Chapter 111 Chapter 112 Chapter 113 Chapter 114

Differential calculus and its applications Common standard derivatives Products and quotients Function of a function Successive differentiation Differentiation of hyperbolic functions Rates of change using differentiation Velocity and acceleration Turning points Tangents and normals Small changes using differentiation Parametric equations Differentiating implicit functions Differentiation of logarithmic functions Differentiation of inverse trigonometric functions Differentiation of inverse hyperbolic functions Partial differentiation Total differential Rates of change using partial differentiation Small changes using partial differentiation Maxima, minima and saddle points of functions of two variables

273 275 276 277 278 279 280 281 282 285 287 288 291 293 295 298 302 305 306 307

Integral calculus and its applications Standard integrals Non-standard integrals Integration using algebraic substitutions

315 317 320 321

Chapter 86

Section 12 Chapter 115 Chapter 116 Chapter 117

245 247 248

308

x

Contents

Chapter 118 Chapter 119 Chapter 120 Chapter 121 Chapter 122 Chapter 123 Chapter 124 Chapter 125 Chapter 126 Chapter 127 Chapter 128 Chapter 129 Chapter 130 Chapter 131 Section 13 Chapter 132 Chapter 133 Chapter 134 Chapter 135 Chapter 136 Chapter 137 Chapter 138 Chapter 139 Chapter 140

Chapter 141

Chapter 142 Chapter 143 Chapter 144 Chapter 145 Chapter 146 Chapter 147

Integration using trigonometric and hyperbolic substitutions Integration using partial fractions θ The t = tan substitution 2 Integration by parts Reduction formulae Double and triple integrals Numerical integration Area under and between curves Mean or average values Root mean square values Volumes of solids of revolution Centroids Theorem of Pappus Second moments of area Differential equations dy The solution of equations of the form  f(x) dx dy  f(y) dx dy  f(x).f(y) The solution of equations of the form dx Homogeneous first order differential equations Linear first order differential equations Numerical methods for first order differential equations (1) – Euler’s method Numerical methods for first order differential equations (2) – Euler-Cauchy method Numerical methods for first order differential equations (3) – Runge-Kutta method Second order differential equations of the form d2y dy a 2 b  cy  0 dx dx Second order differential equations of the form d2y dy a 2 b  cy  f(x) dx dx Power series methods of solving ordinary differential equations (1) – Leibniz theorem Power series methods of solving ordinary differential equations (2) – Leibniz-Maclaurin method Power series methods of solving ordinary differential equations (3) – Frobenius method Power series methods of solving ordinary differential equations (4) – Bessel’s equation Power series methods of solving ordinary differential equations (5) – Legendre’s equation and Legendre’s polynomials Power series methods of solving ordinary differential equations (6) – Rodrigue’s formula The solution of equations of the form

323 329 331 334 337 342 346 349 355 357 359 361 365 369 375 377 377 379 381 383 384 386 388 391

395 400 401 403 406 407 408

Contents

Chapter 148 Chapter 149 Chapter 150 Chapter 151 Section 14 Chapter 152 Chapter 153 Chapter 154 Chapter 155 Chapter 156 Chapter 157 Chapter 158

Solution of partial differential equations (1) – by direct integration Solution of partial differential equations (2) – the wave equation Solution of partial differential equations (3) – the heat conduction equation Solution of partial differential equations (4) – Laplace’s equation

xi

409 410 414 415

Laplace transforms Standard Laplace transforms The initial and final value theorems Inverse Laplace transforms Poles and zeros The Laplace transform of the Heaviside function Solving differential equations using Laplace transforms Solving simultaneous differential equations using Laplace transforms

418 419 423 425 428 431 439

Section 15 Chapter 159 Chapter 160 Chapter 161 Chapter 162

Z-transforms Sequences Properties of z-transforms Inverse z-transforms Using z-transforms to solve difference equations

447 448 451 456 460

Section 16 Chapter 163 Chapter 164 Chapter 165 Chapter 166 Chapter 167 Chapter 168

464 466 469 471 474 476

Chapter 169 Chapter 170 Chapter 171

Fourier series Fourier series for periodic functions of period 2π Fourier series for a non-periodic function over period 2π Even and odd functions Half range Fourier series Expansion of a periodic function of period L Half-range Fourier series for functions defined over range L The complex or exponential form of a Fourier series A numerical method of harmonic analysis Complex waveform considerations

Section 17 Chapter 172 Chapter 173 Chapter 174 Chapter 175 Chapter 176 Chapter 177 Chapter 178 Chapter 179 Chapter 180 Chapter 181 Chapter 182 Chapter 183 Chapter 184 Chapter 185 Chapter 186

Statistics and probability Presentation of ungrouped data Presentation of grouped data Measures of central tendency Quartiles, deciles and percentiles Probability Permutations and combinations Bayes’ theorem The binomial distribution The Poisson distribution The normal distribution Linear correlation Linear regression Sampling and estimation theories Chi-square values The sign test

494 497 500 504 507 509 511 513 514 516 517 522 524 525 532 535

442

479 482 487 490

xii

Contents

Chapter 187 Chapter 188 Index

Wilcoxon signed-rank test The Mann-Whitney test

537 540 547

Preface Mathematics Pocket Book for Engineers and Scientists 5th Edition is intended to provide students, technicians, scientists and engineers with a readily available reference to the essential engineering mathematics formulae, definitions, tables and general information needed during their work situation and/or studies – a handy book to have on the bookshelf to delve into as the need arises. In this 5th edition, the text has been re-designed to make information easier to access. The importance of why each mathematical topic is needed in engineering and science is explained at the beginning of each section. Essential theory, formulae, definitions, laws and procedures are stated clearly at the beginning of each chapter, and then it is demonstrated how to use such information in practice. The text is divided, for convenience of reference, into seventeen main sections embracing engineering conversions, constants and symbols, some algebra topics, some number topics, areas and volumes, geometry and trigonometry, graphs, complex numbers, vectors, matrices and determinants, Boolean algebra and logic circuits, differential and integral calculus and their applications, differential equations, Laplace transforms, z-transforms, Fourier series and statistics and probability. To aid understanding, over 675 application examples have been included, together with some 300 line diagrams. The text assumes little previous knowledge and is suitable for a wide range of disciplines and/or courses of study. It will be particularly useful as a reference for those in industry involved in engineering and science and/or for students studying mathematics within Engineering and Science Degree courses, as well as for National and Higher National Technician Certificates and Diplomas, GCSE and A levels. JOHN BIRD BSc(Hons), CEng, CSi, CMath, FIET, FIMA, FCollP Royal Naval Defence College of Marine Engineering, HMS Sultan,formerly University of Portsmouth and Highbury College, Portsmouth

Section 1 Engineering conversions, constants and symbols Why are engineering conversions, constants and symbols important? In engineering there are many different quantities to get used to, and hence many units to become familiar with. For example, force is measured in Newtons, electric current is measured in amperes and pressure is measured in Pascals. Sometimes the units of these quantities are either very large or very small and hence prefixes are used. For example, 1000 Pascals may be written as 103 Pa which is written as 1 kPa in prefix form, the k being accepted as a symbol to represent 1000 or 103. Studying, or working, in an engineering and science discipline, you very quickly become familiar with the standard units of measurement, the prefixes used and engineering notation. An electronic calculator is extremely helpful with engineering notation. Unit conversion is very important because the rest of the world other than three countries uses the metric system. So, converting units is important in science and engineering because it uses the metric system. Without the ability to measure, it would be difficult for scientists to conduct experiments or form theories. Not only is measurement important in science and engineering, it is also essential in farming, construction, manufacturing, commerce, and numerous other occupations and activities. Measurement provides a standard for everyday things and processes. Examples include weight, temperature, length, currency and time, and all play a very important role in our lives.

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Mathematics Pocket Book for Engineers and Scientists

Chapter 1

General conversions and Greek alphabet

General conversions Length (metric)

1 kilometre (km)  1000 metres (m) 1 metre (m)  100 centimetres (cm) 1 metre (m)  1000 millimetres (mm) 1 cm  102 m 1 mm  103 m 1 micron ()  106 m 1 angstrom (A)  1010 m

Length (imperial)

1 inch (in)  2.540 cm or 1 cm  0.3937 in 1 foot (ft)  30.48 cm 1 mile (mi)  1.609 km or 1 km  0.6214 mi 1 cm  0.3937 in 1 m  39.37 in  3.2808 ft  1.0936 yd 1 km  0.6214 mile 1 nautical mile  1.15 mile

Area (metric)

1 m2  106 mm2 1 mm2  106 m2 1 m2  104 cm2 1 cm2  104 m2 1 hectare (ha)  104 m2

Area (imperial)

1 m2  10.764 ft2  1.1960 yd2 1 ft2  929 cm2 1 mile2  640 acres 1 acre  43560 ft2  4840 yd2 1 ha  2.4711 acre  11960 yd2  107639 ft2

Volume

1 litre (l)  1000 cm3 1 litre  1.057 quart (qt)  1.7598 pint (pt)  0.21997 gal 1 m3  1000 l 1 British gallon  4 qt  4.545 l  1.201 US gallon 1 US gallon  3.785 l

Mass

1 kilogram (kg)  1000 g  2.2046 pounds (lb) 1 lb  16 oz  453.6 g 1 tonne (t)  1000 kg  0.9842 ton

Speed

1 km/h  0.2778 m/s  0.6214 m.p.h. 1 m.p.h.  1.609 km/h  0.4470 m/s 1 rad/s  9.5493 rev/min 1 knot  1 nautical mile per hour  1.852 km/h  1.15 m.p.h. 1 km/h  0.540 knots 1 m.p.h.  0.870 knots

Angular measure

1 rad  57.296°

Engineering conversions, constants and symbols

Greek alphabet Letter Name

Upper Case

Lower Case

Alpha Beta Gamma Delta Epsilon Zeta Eta Theta Iota Kappa Lambda Mu Nu Xi Omicron Pi Rho Sigma Tau Upsilon Phi Chi Psi

A B   E Z H θ l K  M N  O  P  T Y  X 

    e   θ       o π     φ  

Omega





Chapter 2

Basic SI units, derived units and common prefixes Basic SI units

Quantity

Unit

Length Mass Time Electric current Thermodynamic temperature Luminous intensity Amount of substance

metre, m kilogram, kg second, s ampere, A kelvin, K candela, cd mole, mol

3

4

Mathematics Pocket Book for Engineers and Scientists

SI supplementary units Plane angle

radian, rad

Solid angle

steradian, sr

Derived units Quantity

Unit

Electric capacitance Electric charge Electric conductance Electric potential difference Electrical resistance Energy Force Frequency Illuminance Inductance Luminous flux Magnetic flux Magnetic flux density Power Pressure

farad, F coulomb, C siemens, S volts, V ohm,  joule, J Newton, N hertz, Hz lux, lx henry, H lumen, lm weber, Wb tesla, T watt, W pascal, Pa

Some other derived units not having special names Quantity

Unit

Acceleration Angular velocity Area Current density Density Dynamic viscosity Electric charge density Electric field strength Energy density Heat capacity

metre per second squared, m/s2 radian per second, rad/s square metre, m2 ampere per metre squared, A/m2 kilogram per cubic metre, kg/m3 pascal second, Pa s coulomb per cubic metre, C/m3 volt per metre, V/m joule per cubic metre, J/m3 joule per Kelvin, J/K

Engineering conversions, constants and symbols

Quantity

Unit

Heat flux density Kinematic viscosity Luminance Magnetic field strength Moment of force Permeability Permittivity Specific volume Surface tension Thermal conductivity Velocity Volume

watt per square metre, W/m3 square metre per second, m2/s candela per square metre, cd/m2 ampere per metre, A/m newton metre, Nm henry per metre, H/m farad per metre, F/m cubic metre per kilogram, m3/kg newton per metre, N/m watt per metre Kelvin, W/(mK) metre per second, m/s2 cubic metre, m3

5

Common prefixes Prefix

Name

Meaning

Y Z E P T G M k m  n p f a z y

yotta zeta exa peta tera giga mega kilo milli micro nano pico femto atto zepto yocto

multiply by 1024 multiply by 1021 multiply by 1018 multiply by 1015 multiply by 1012 multiply by 109 multiply by 106 multiply by 103 multiply by 103 multiply by 106 multiply by 109 multiply by 1012 multiply by 1015 multiply by 1018 multiply by 1021 multiply by 1024

Chapter 3

Some physical and mathematical constants

Below are listed some physical and mathematical constants, each stated correct to 4 decimal places, where appropriate.

6

Mathematics Pocket Book for Engineers and Scientists

Quantity

Symbol

Value

Speed of light in a vacuum Permeability of free space Permittivity of free space Elementary charge Planck constant

c 0 e0 e h

2.9979  108 m/s 4π  107 H/m 8.8542  1012 F/m 1.6022  1019 C 6.6261  1034 J s

Reduced Planck constant

h=

h 2π

1.0546  1034 J s

Fine structure constant

=

e2 4πe0hc

7.2974  103

Coulomb force constant Gravitational constant Atomic mass unit Rest mass of electron Rest mass of proton Rest mass of neutron Bohr radius Compton wavelength of electron Avogadro constant Boltzmann constant Stefan-Boltzmann constant Bohr constant Nuclear magnetron Triple point temperature Molar gas constant Micron Characteristic impedance of vacuum

ke G u me mp mn a0 C

8.9875  109 Nm2/C2 6.6726  1011 m3/kg s2 1.6605  1027 kg 9.1094  1031 kg 1.6726  1027 kg 1.6749  1027 kg 5.2918  1011 m 2.4263  1012 m

NA k 

6.0221  1023/mol 1.3807  1023 J/K 5.6705  108 W /m2K4

B N Tt

9.2740  1024 J/T 5.0506  1027 J/T 273.16 K

R m Zo

8.3145 J/K mol 106 m 376.7303 

Astronomical constants Mass of earth Radius of earth Gravity of earth’s surface Mass of sun Radius of sun Solar effective temperature Luminosity of sun Astronomical unit Parsec Jansky

mE RE g Me Re Te Le AU pc Jy

5.976  1024 kg 6.378  106 m 9.8067 m/s2 1.989  1030 kg 6.9599  108 m 5800 K 3.826  1026 W 1.496  1011 m 3.086  1016 m 1026 W/m2HZ

Engineering conversions, constants and symbols

Tropical year Standard atmosphere

3.1557  107 s 101325 Pa

atm Mathematical constants

Pi (Archimedes’ constant) Exponential constant Apery’s constant Catalan’s constant Euler’s constant Feigenbaum’s 1st constant Feigenbaum’s 2nd constant Gibb’s constant Golden mean Khintchine’s constant

Chapter 4

π e (3) G    G φ K

3.1416 2.7183 1.2021 0.9160 0.5772 4.6692 2.5029 1.8519 1.6180 2.6855

Recommended mathematical symbols

equal to not equal to identically equal to

  

corresponds to approximately equal to approaches proportional to

=  → 

infinity smaller than larger than smaller than or equal to larger than or equal to much smaller than

    

much larger than

 

plus minus plus or minus minus or plus

  

a multiplied by b

ab or a  b or a ? b

a divided by b

a or a/b or ab-1 b

magnitude of a

|a|

a raised to power n

an

square root of a

7

1

a or a 2

7

8

Mathematics Pocket Book for Engineers and Scientists

n’th root of a

n

1

a or a n or a1/n

mean value of a

a

factorial of a

a!

sum



function of x

f(x)

limit to which f(x) tends as x approaches a

lim f(x)

x →a

finite increment of x

Dx

variation of x

x

differential coefficient of f(x) with respect to x

df or df/dy or f′(x) dx

differential coefficient of order n of f(x)

dnf or dnf/dx 2 or fn (x) dxn

partial differential coefficient of f(x, y, …) w.r.t. x when y, … are held constant

 ∂f  ∂ f(x,y,...) or   or fx  ∂ x  ∂x y

total differential of f indefinite integral of f(x) with respect to x

df

∫ f(x) dx

definite integral of f(x) from x  a to x  b

∫

logarithm to the base a of x common logarithm of x exponential of x natural logarithm of x sine of x cosine of x tangent of x secant of x cosecant of x cotangent of x inverse sine of x inverse cosine of x inverse tangent of x inverse secant of x inverse cosecant of x inverse cotangent of x hyperbolic sine of x hyperbolic cosine of x hyperbolic tangent of x hyperbolic secant of x hyperbolic cosecant of x

loga x lg x or log10 x ex or exp x ln x or loge x sin x cos x tan x sec x cosec x cot x sin1 x or arcsin x cos1 x or arccos x tan1 x or arctan x sec1 x or arcsec x cosec1 x or arccosec x cot1 x or arccot x sinh x cosh x tanh x sech x cosech x

b a

f(x) dx

Engineering conversions, constants and symbols

hyperbolic cotangent of x inverse hyperbolic sine of x inverse hyperbolic cosine of x inverse hyperbolic tangent of x inverse hyperbolic secant of x inverse hyperbolic cosecant of x inverse hyperbolic cotangent of x complex operator modulus of z argument of z complex conjugate of z transpose of matrix A determinant of matrix A vector magnitude of vector A scalar product of vectors A and B vector product of vectors A and B

Chapter 5

coth x sinh1 x or arsinh x cosh1 x or arcosh x tanh1 x or artanh x sech1 x or arsech x cosech1 x or arcosech x coth1 x or arcoth x i, j |z| arg z z* AT |A| → A or A |A| A•B AB

Symbols for physical quantities

(a) Space and time angle (plane angle) solid angle length breadth height thickness radius diameter distance along path rectangular co-ordinates cylindrical co-ordinates spherical co-ordinates area volume time dθ angular speed, dt angular acceleration, speed,

ds dt

d dt

, , , θ, φ, etc. ,  l b h d,  r d s, L x, y, z r, φ, z r, θ, φ A V t   u, v, w

9

10

Mathematics Pocket Book for Engineers and Scientists

acceleration,

du dt

acceleration of free fall speed of light in a vacuum Mach number

a g c Ma

(b) Periodic and related phenomena period frequency rotational frequency circular frequency wavelength damping coefficient attenuation coefficient phase coefficient propagation coefficient

T f n      

(c) Mechanics mass density relative density specific volume momentum moment of inertia second moment of area second polar moment of area force bending moment torque; moment of couple pressure normal stress shear stress linear strain shear strain volume strain Young’s modulus shear modulus bulk modulus Poisson ratio compressibility section modulus coefficient of friction viscosity fluidity kinematic viscosity diffusion coefficient surface tension angle of contact work

m  d v p I, J Ia Ip F M T p, P   e, e  θ E G K ,   Z, W   φ  D ,  θ W

Engineering conversions, constants and symbols

energy potential energy kinetic energy power gravitational constant Reynold’s number

E, W Ep, V,  Ek, T, K P G Re

(d) Thermodynamics thermodynamic temperature common temperature linear expansivity cubic expansivity heat; quantity of heat work; quantity of work heat flow rate thermal conductivity heat capacity specific heat capacity entropy internal energy enthalpy Helmholtz function Planck function specific entropy specific internal energy specific enthalpy specific Helmholz function

T,  t, θ ,  ,  Q, q W, w , q , k C c S U, E H A, F Y s u, e h a, f

(e) Electricity and magnetism Electric charge; quantity of electricity electric current charge density surface charge density electric field strength electric potential electric potential difference electromotive force electric displacement electric flux capacitance permittivity permittivity of a vacuum relative permittivity electric current density magnetic field strength magnetomotive force magnetic flux magnetic flux density self inductance

Q I   E V, φ U, V E D  C e e0 er J, j H Fm  B L

11

12

Mathematics Pocket Book for Engineers and Scientists

mutual inductance coupling coefficient leakage coefficient permeability permeability of a vacuum relative permeability magnetic moment resistance resistivity conductivity reluctance permeance number of turns number of phases number of pairs of poles loss angle phase displacement impedance reactance resistance quality factor admittance susceptance conductance power, active power, reactive power, apparent

M k   0 r m R  ,  Rm, S  N m p  φ Z X R Q Y B G P Q S

(f) Light and related electromagnetic radiations radiant energy radiant flux, radiant power radiant intensity radiance radiant exitance irradiance emissivity quantity of light luminous flux luminous intensity luminance luminous exitance illuminance light exposure luminous efficacy absorption factor, absorptance reflexion factor, reflectance transmission factor, transmittance linear extinction coefficient linear absorption coefficient refractive index

Q, Qe , e, P I, Ie L, Le M, Me E, Ee e Q, Qv , v I, Iv L, Lv M, Mv E, Ev H K     a n

Engineering conversions, constants and symbols

refraction angle of optical rotation

R 

(g) Acoustics speed of sound speed of longitudinal waves speed of transverse waves group speed sound energy flux sound intensity reflexion coefficient acoustic absorption coefficient transmission coefficient dissipation coefficient loudness level

c cl ct cg P I, J  , a   LN

(h) Physical chemistry atomic weight molecular weight amount of substance molar mass molar volume molar internal energy molar enthalpy molar heat capacity molar entropy molar Helmholtz function molar Gibbs function (molar) gas constant compression factor mole fraction of substance B mass fraction of substance B volume fraction of substance B molality of solute B amount of substance concentration of solute B chemical potential of substance B absolute activity of substance B partial pressure of substance B in a gas mixture fugacity of substance B in a gas mixture relative activity of substance B activity coefficient (mole fraction basis) activity coefficient (molality basis) activity coefficient (concentration basis) osmotic coefficient osmotic pressure surface concentration electromotive force Faraday constant charge number of ion i ionic strength velocity of ion i

Ar Mr n M Vm Um Hm Cm Sm Am Gm R Z xB wB φB mB cB B B pB fB B fB B yB φ, g   E F zi I vi

13

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Mathematics Pocket Book for Engineers and Scientists

electric mobility of ion i electrolytic conductivity molar conductance of electrolyte transport number of ion i molar conductance of ion i overpotential exchange current density electrokinetic potential intensity of light transmittance absorbance (linear) absorption coefficient molar (linear) absorption coefficient angle of optical rotation specific optical rotatory power molar optical rotatory power molar refraction stoiciometric coefficient of molecules B extent of reaction affinity of a reaction equilibrium constant degree of dissociation rate of reaction rate constant of a reaction activation energy of a reaction

ui   ti i  j0  I T A a e  m n Rm B  A K  , J k E

(i) Molecular physics Avogadro constant number of molecules number density of molecules molecular mass molecular velocity molecular position molecular momentum average velocity average speed most probable speed mean free path molecular attraction energy interaction energy between molecules i and j distribution function of speeds Boltzmann function generalized co-ordinate generalized momentum volume in phase space Boltzmann constant partition function grand partition function statistical weight symmetrical number dipole moment of molecule

L, NA N n m c, u r p c , u , c 0 , u0 c , u , c, u ˆ uˆ c, l,  e φij, Vij f(c) H q p  k Q, Z

 g , s p, 

Engineering conversions, constants and symbols

quadrupole moment of molecule polarizability of molecule Planck constant characteristic temperature Debye temperature Einstein temperature rotational temperature vibrational temperature Stefan-Boltzmann constant first radiation constant second radiation constant rotational quantum number vibrational quantum number

  h  D E r v  c1 c2 J, K v

(j) Atomic and nuclear physics nucleon number; mass number atomic number; proton number neutron number (rest) mass of atom unified atomic mass constant (rest) mass of electron (rest) mass of proton (rest) mass of neutron elementary charge (of protons) Planck constant Planck constant divided by 2π Bohr radius Rydberg constant magnetic moment of particle Bohr magneton Bohr magneton number, nuclear magneton nuclear gyromagnetic ratio g-factor Larmor (angular) frequency nuclear angular precession frequency cyclotron angular frequency of electron nuclear quadrupole moment nuclear radius orbital angular momentum quantum number spin angular momentum quantum number total angular momentum quantum number nuclear spin quantum number hyperfine structure quantum number principal quantum number magnetic quantum number fine structure constant electron radius Compton wavelength mass excess packing fraction

A Z N ma mu me mp mn e h h a0 R  B N  g L N c Q R L, l1 S, s1 J, j1 I, J F n, n1 M, m1  re C  f

15

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Mathematics Pocket Book for Engineers and Scientists

mean life level width activity specific activity decay constant half-life

  A a  T1 , t 1

disintegration energy spin-lattice relaxation time spin-spin relaxation time indirect spin-spin coupling

Q T1 T2 J

2

2

(k) Nuclear reactions and ionising radiations reaction energy cross-section macroscopic cross-section impact parameter scattering angle internal conversion coefficient linear attenuation coefficient atomic attenuation coefficient mass attenuation coefficient linear stopping power atomic stopping power linear range recombination coefficient

Q   b θ, φ  , 1  m S, S1 Sa R, R1 

Section 2 Some algebra topics

Why is Algebra important? Algebra is one of the most fundamental tools for engineers and scientists because it allows them to determine the value of something (such as length, material constant, temperature, mass, and so on) given values that they do know (possibly other length, material properties, mass). Although the types of problems that mechanical, chemical, civil, environmental, electrical engineers deal with vary, all engineers use algebra to solve problems. An example where algebra is frequently used is in simple electrical circuits, where the resistance is proportional to voltage. Using Ohm’s Law, V = I × R, an engineer or scientist simply multiplies the current in a circuit by the resistance to determine the voltage across the circuit. Engineers and scientists use algebra in many ways, and so frequently that they don’t even stop to think about it. Algebra lays the foundation for the mathematics needed to become an engineer or scientist. A basic form of mathematics, algebra is nevertheless among the most commonly used forms of mathematics in the workplace. Although relatively simple, algebra is a powerful problem-solving tool used in many fields of engineering and science. For example, in designing a rocket to go to the moon, engineers and scientists must use algebra to solve for flight trajectory, how long to burn each thruster and at what intensity, and at what angle to lift off. Engineers and scientists use mathematics all the time – and, in particular, algebra; becoming familiar with algebra will make all engineering mathematics studies so much easier.

Mathematics Pocket Book for Engineers and Scientists

18

Chapter 6

Introduction to algebra

Some rules of algebra Algebra merely uses letters to represent numbers. If, say, a, b, c and d represent any four numbers then in algebra: (i)

a + a + a + a = 4a

(ii) 5b means 5 × b (iii) 2a + 3b + a – 2b = 2a + a + 3b – 2b = 3a + b Only similar terms can be combined in algebra. (iv) 4abcd = 4 × a × b × c × d (v) (a)(c)(d) means a × c × d Brackets are often used instead of multiplication signs. (vi) ab = ba (vii) b2 = b × b (viii) a3 = a × a × a

Application: Simplify 4x2 – 2x – 3y + 5x + 7y

Re-ordering 4x2 – 2x – 3y + 5x + 7y gives 4x2 + 5x – 2x + 7y – 3y = 4x2 + 3x + 4y

Application: Simplify 3xy – 7x + 5xy + 3x

Re-ordering 3xy – 7x + 5xy + 3x gives 3xy + 5xy + 3x – 7x = 8xy – 4x

Application: Simplify ab × b2c × a

ab × b2c × a = a × b × b × b × c × a = a × a × b × b × b × c = a2× b3 × c = a2b3c

Some algebra topics

19

Laws of indices The laws of indices in algebraic terms are as follows: am × an = am+n am (ii) = am–n an (iii) (am)n = amn

For example, a3 × a4 = a3+4 = a7 c5 For example, 2 = c5–2 = c3 c For example, (d2)3 = d2×3 = d6

(i)

m

(iv) a n =

4

n

3

For example, x 3 = x 4

am

(v) a–n = 1 an (vi) a0 = 1

For example, 3-2 = 1 = 1 9 32 For example, 170 = 1

Application: Simplify a2b3c × ab2c5 a2b3c × ab2c5 = a2 × b3 × c × a × b2 × c5 = a2 × b3 × c1 × a1 × b2 × c5 Grouping together like terms gives: a2 × a1 × b3 × b2 × c1 × c5 a2+1 × b3+2 × c1+5 = a3 × b5 × c6

Using law (i) of indices gives:

a2b3c × ab2c5 = a3b5c6

i.e.

Application: Simplify

x5y 2z x 2y z3 5

2

x y z x5y 2z x5 × y 2 × z = 2 = 2 × 1 × 3 = x5–2 × y2–1 × z1–3 by law (ii) of indices 2 3 x yz x × y × z3 x y z x3 y by law (v) of indices = x3 × y1 × z–2 × x3yz–2 or z2 Application: Simplify (p3)2(q2)4 Using law (iii) of indices gives: (p3)2(q2)4 = p3 × 2 × q2 × 4 = p6q8

Brackets With algebra

(i) 2(a + b) = 2a + 2b (ii)

(a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd

Application: Determine 2b(a – 5b) 2b(a – 5b) = 2b × a + 2b × – 5b = 2ba – 10b2 = 2ab – 10b2 (Note that 2ba is the same as 2ab)

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Mathematics Pocket Book for Engineers and Scientists

Application: Determine (3x + 4y)(x – y) (3x + 4y)(x – y) = 3x(x – y) + 4y(x – y) = 3x2 – 3xy + 4yx – 4y2 = 3x2 – 3xy + 4xy – 4y2 = 3x2 + xy – 4y2 Application: Simplify 3(2x – 3y) – (3x – 5y) 3(2x – 3y) – (3x – 5y) = 3 × 2x – 3 × 3y – 3x – – 5y = 6x – 9y – 3x + 5y = 6x – 3x + 5y – 9y = 3x – 4y Application: Remove the brackets from the expression and simplify: 2[x2 – 3x (y + x) + 4xy] 2[x2 – 3x (y + x) + 4xy] = 2[x2 – 3xy – 3x2 + 4xy] Whenever more than one type of brackets are involved, always start with the inner brackets = 2[–2x2 + xy] = – 4x2 + 2xy = 2xy – 4x2

Factorisation The factors of 8 are 1, 2, 4, and 8, because 8 may be divided by 1, 2, 4, and 8 The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24 because 24 may be divided by 1, 2, 3, 4, 6, 8, 12 and 24 The common factors of 8 and 24 are 1, 2, 4 and 8 since 1, 2, 4 and 8 are factors of both 8 and 24. The highest common factor (HCF) is the largest number that divides into two or more terms. Hence, the HCF of 8 and 28 is 4 When two or more terms in an algebraic expression contain a common factor, then this factor can be shown outside of a bracket. For example,

df + dg = d(f + g)

which is just the reverse of

d(f + g) = df + dg

This process is called factorisation

Application: Factorise ab – 6ac ‘a’ is common to both of the terms ab and – 6ac. ‘a’ is therefore taken outside of the bracket. What goes inside the bracket?

Some algebra topics

21

(i) What multiplies ‘a’ to make ab? Answer: b (ii) What multiplies ‘a’ to make – 6ac? Answer: – 6c Hence, b – 6c appears in the bracket. Thus,

ab – 6ac = a(b – 6c)

Application: Factorise 2x2 + 14xy3 For the numbers 2 and 14, the highest common factor (HCF) is 2 (i.e. 2 is the largest number that divides into both 2 and 14) For the x terms, x2 and x, the HCF is x Thus, the HCF of 2x2 and 14xy3 is 2x 2x is therefore taken outside of the bracket. What goes inside the bracket? (i) What multiplies 2x to make 2x2? Answer: x (ii) What multiplies 2x to make 14xy3? Answer: 7y3 Hence x + 7y3 appears in the bracket Thus,

2x2+ 14xy3 = 2x(x+7y3)

Laws of precedence With the laws of precedence the order is Brackets Order (or pOwer) Division Multiplication Addition Subtraction The first letter of each word spells BODMAS

Application: Simplify 5x + 3x × 4x – x 5x + 3x × 4x – x = 5x + 12x2 – x

(M)

= 5x – x + 12x2 = 4x + 12x2 or 4x( 1 + 3x) by factorising

(S)

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Mathematics Pocket Book for Engineers and Scientists

Application: Simplify (y + 4y) × 2y – 5y (y + 4y) × 2y – 5y = 5y × 2y – 5y = 10y2 – 5y

(B) (M)

or 5y(2y – 1) by factorising Application: Simplify 2y ÷ (8y + 3y – 5y) 2y ÷ (8y + 3y – 5y) = 2y ÷ 6y =

(B)

2y 6y

(D)

= 1 by cancelling 3

Chapter 7

Polynomial division

A polynomial is an expression of the form: f(x) = a + bx + cx2 + dx3 + ….. and polynomial division is sometimes required when resolving a fraction into partial fractions. Application: Divide 2x2  x  3 by x  1 2x2  x  3 is called the dividend and x  1 the divisor. The usual layout is shown below with the dividend and divisor both arranged in descending powers of the symbols.

x

)

2x

3

1 2x 2 x 3 2x 2 − 2x 3x 3 3x 3 . .

Dividing the first term of the dividend by the first term of the divisor, i.e. 2x 2 /x gives 2x, which is placed above the first term of the dividend as shown. The divisor is then multiplied by 2x, i.e. 2x(x  1)  2x2  2x, which is placed under the dividend as shown. Subtracting gives 3x  3. The process is then repeated, i.e. the first term of the divisor, x, is divided into 3x, giving 3, which is placed above the dividend as shown. Then 3(x  1)  3x  3 which is placed under the 3x  3. The remainder, on subtraction, is zero, which completes the process. Thus, (2x2  x  3) ÷ (x  1)  (2x  3) The answer, (2x + 3), is called the quotient.

Some algebra topics

23

Application: Divide (x2  3x  2) by (x  2)

)

x+5

x − 2 x 2 + 3x − 2 x 2 − 2x 5x − 2 5x − 10 8 Hence

8 x 2  3x - 2  x 5 x-2 x-2

Application: Divide (3x3  x2  3x + 5) by (x + 1)

)

(1) 3x 2

(4) (7) 2x 5

x + 1 3x 3 x 2 + 3x + 5 3x 3 + 3x 2 – 2x 2 + 3x + 5 −2x 2 − 2x 5x + 5 5x + 5 . (1) (2) (3) (4) (5) (6) (7) (8) (9)

3

2

2

.

3

x into 3x goes 3x . Put 3x above 3x 3x2(x + 1) = 3x3 + 3x2 Subtract x into –2x2 goes –2x. Put –2x above the dividend –2x(x+1) = –2x2– 2x Subtract x into 5x goes 5. Put 5 above the dividend 5(x+1) = 5x+5 Substract

Thus

3x 3 + x 2 + 3x + 5 = 3x2 – 2x +5 x +1

Chapter 8

The factor theorem

A factor of (x  a) in an equation corresponds to a root of x  a If x  a is a root of the equation f(x)  0, then (x  a) is a factor of f(x) Application: Factorise x3  7x  6 and use it to solve the cubic equation x3  7x  6  0

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Mathematics Pocket Book for Engineers and Scientists

Let f(x)  x3  7x  6 If x  1, then f(1)  13  7(1)  6  12 If x  2, then f(2)  23  7(2)  6  12 If x  3, then f(3)  33  7(3)  6  0 If f(3)  0, then (x  3) is a factor – from the factor theorem. We have a choice now. We can divide x3  7x  6 by (x  3) or we could continue our ‘trial and error’ by substituting further values for x in the given expression – and hope to arrive at f(x)  0. Let us do both ways. Firstly, dividing out gives:

)

x2

3x

2

x − 3 x 3 0 − 7x − 6 x 3 − 3x 2 3x 2 − 7x − 6 3x 2 − 9x 2x − 6 2x − 6 . Hence, i.e.

.

x 3 - 7x - 6  x 2  3x  2 x-3

x3  7x  6  (x  3)(x2  3x  2)

x2  3x  2 factorises ‘on sight’ as (x  1)(x  2) Therefore, x3  7x  6  (x  3)(x  1)(x  2) A second method is to continue to substitute values of x into f(x). Our expression for f(3) was 33  7(3)  6. We can see that if we continue with positive values of x the first term will predominate such that f(x) will not be zero. Therefore let us try some negative values for x. f(1)  (1)3  7(1)  6  0; hence (x  1) is a factor (as shown above). Also f(2)  (2)3  7(2)  6  0; hence (x  2) is a factor. To solve

x3  7x  6  0, we substitute the factors, i.e. (x - 3)(x  1)(x  2)  0

from which, x  3, x  1 and x  2 Note that the values of x, i.e. 3, 1 and 2, are all factors of the constant term, i.e. the 6. This can give us a clue as to what values of x we should consider.

Some algebra topics

Chapter 9

25

The remainder theorem

If (ax2  bx  c) is divided by (x  p), the remainder will be ap2  bp  c If (ax3  bx2  cx  d) is divided by (x  p), the remainder will be ap3  bp2  cp  d Application: When (3x2  4x  5) is divided by (x  2) find the remainder ap2  bp  c, (where a  3, b  4, c  5 and p  2), hence the remainder is 3(2)2  (4)(2)  5  12  8  5  9 We can check this by dividing (3x2  4x  5) by (x  2) by long division:

)

3x

2

x − 2 3x 2 − 4x + 5 3x 2 − 6x 2x 5 2x − 4 9 Application: When (2x2  x  3) is divided by (x  1), find the remainder ap2  bp  c, (where a  2, b  1, c  3 and p  1), hence the remainder is 2(1)2  1(1)  3  0, which means that (x  1) is a factor of (2x2  x  3). In this case, the other factor is (2x  3), i.e. (2x2  x  3)  (x  1)(2x  3) Application: When (3x3  2x2  x  4) is divided by (x  1), find the remainder The remainder is ap3  bp2  cp  d (where a  3, b  2, c  1, d  4 and p  1), i.e. the remainder is: 3(1)3  2(1)2  (1)(1)  4  3  2  1  4  8

Chapter 10

Continued fractions

A continued fraction is an expression obtained through an iterative process of representing a number as the sum of its integer part and the reciprocal of another number, then writing this other number as the sum of its integer part and another reciprocal, and so on. These approximations to fractions are used to obtain practical

26

Mathematics Pocket Book for Engineers and Scientists

ratios for gearwheels or for a dividing head (used to give a required angular displacement). Any fraction may be expressed in the form shown below for the fraction 26/55: 26 1 1 1 1 1 = = = = = 55 3 1 1 1 55 2+ 2+ 2+ 2+ 26 2 1 26 26 8+ 8+ 3 3 3 2 1

=

1

2+

1

8+

1+

1 2

The latter factor can be expressed as: 1 A B C  D Comparisons show that A, B, C and D are 2, 8, 1 and 2 respectively. A fraction written in the general form is called a continued fraction and the integers A, B, C and D are called the quotients of the continued fraction. The quotients may be used to obtain closer and closer approximations, called convergents. A tabular method may be used to determine the convergents of a fraction: 1 a

 bp b   bq

0 1

2

3

4

5

2

8

1

2

1 2

8 17

9 19

26 55

The quotients 2, 8, 1 and 2 are written in cells a2, a3, a4 and a5 with cell a1 being left empty. The fraction

0 1

is always written in cell b1.

The reciprocal of the quotient in cell a2 is always written in cell b2, i.e. The fraction in cell b3 is given by

(a3 × b2p)  b1p , (a3 × b2q)  b1q

1 2

in this case.

Some algebra topics

i.e.

(8 × 1)  0 8  (8 × 2)  1 17

The fraction in cell b4 is given by

i.e.

27

(a4 × b3p)  b2p , (a4 × b3q)  b2q

(1× 8)  1 9 , and so on.  (1× 17)  2 19

1 Hence the convergents of 26 are , 8 , 9 and 26 , each value approximating 2 17 19 55 55 26 . closer and closer to 55

Chapter 11

Solving simple equations

Introduction 3x – 4 is an example of an algebraic expression. 3x – 4 = 2 is an example of an algebraic equation (i.e. it contains an ‘=’ sign) An equation is simply a statement that two expressions are equal. Hence, A = πr2 (where A is the area of a circle of radius r) 9 F = C + 32 (which relates Fahrenheit and Celsius temperatures) 5 and y = 3x + 2 (which is the equation of a straight line graph) are all examples of equations.

Solving equations To ‘solve an equation’ means ‘to find the value of the unknown’. Application: Solve the equation: 4x = 24 Dividing each side of the equation by 4 gives:

x = 6 by cancelling

i.e. which is the solution to the equation 4x = 24

Application: Solve the equation:

4x 24 = 4 4

2x =4 5

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Mathematics Pocket Book for Engineers and Scientists

 2x  5   = 5(4)  5 

Multiplying both sides by 5 gives: Cancelling and removing brackets gives:

2x = 20 20 2x Dividing both sides of the equation by 2 gives: = 2 2 x = 10

Cancelling gives: Application: Solve the equation: x + 3 = 8

Subtracting 3 from both sides gives: x + 3 – 3 = 8 – 3 i.e.

x=8–3

i.e.

x=5

Application: Solve the equation: 7x + 1 = 3x + 9 In such equations the terms containing x are grouped on one side of the equation and the remaining terms grouped on the other side of the equation. Changing from one side of an equation to the other must be accompanied by a change of sign. Since

7x + 1 = 3x + 9

then

7x – 3x = 9 – 1

i.e.

4x = 8

Dividing both sides by 4 gives: Cancelling gives:

8 4x = 4 4 x=2

Application: Solve the equation: 3(x – 2) = 12 Removing the bracket gives:

3x – 6 = 12

Rearranging gives:

3x = 12 + 6

i.e.

3x = 18

Dividing both sides by 3 gives:

x=6

Application: Solve the equation: 4(2y – 3) – 2(y – 4) = 3(y – 3) – 1 Removing brackets gives:

8y – 12 – 2y + 8 = 3y – 9 – 1

Rearranging gives: i.e. Dividing both sides by 3 gives:

8y – 2y – 3y = – 9 – 1 + 12 – 8 3y = – 6 y=

-6 =–2 3

Some algebra topics

Application: Solve the equation:

29

2 4 = 5 t

The lowest common multiple (LCM) of the denominators, i.e. the lowest algebraic expression that both t and 5 will divide into, is 5t 4  2 Multiplying both sides by 5t gives: 5t   = 5t    t   5  Cancelling gives:

5(4) = t(2)

i.e.

20 = 2t

(1)

20 2t = 2 2 10 = t or t = 10

Dividing both sides by 2 gives: Cancelling gives:

When there is just one fraction on each side of the equation as in this example, there is a quick way to arrive at equation (1) without needing to find the LCM of the denominators.  4  2 We can move from   = to: 4 × 5 = 2 × t  t  5 by what is called ‘cross-multiplication’. a c = then: ad = bc b d We can use cross-multiplication when there is one fraction only on each side of the equation. In general, if

Application: Solve the equation: ‘Cross-multiplication’ gives:

3 4 = x-2 3x + 4

3(3x + 4) = 4(x – 2)

Removing brackets gives:

9x + 12 = 4x – 8

Rearranging gives:

9x – 4x = – 8 – 12

i.e.

5x = – 20 x=

Dividing both sides by 5 gives:

–20 =–4 5

Application: Solve the equation: 2 d = 6 Whenever square roots are involved in an equation, the square root term needs to be isolated on its own before squaring both sides ‘Cross-multiplying’ gives:

6 d = 2

Cancelling gives:

d =3

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Mathematics Pocket Book for Engineers and Scientists

( d)

2

Squaring both sides gives:

2

= ( 3)

d9

i.e. Application: Solve the equation:   

2 15  3 4 t2

We need to rearrange the equation to get the t2 term on its own. 15(3)  2(4t2)

‘Cross-multiplying’ gives:

45  8t2

i.e. Dividing both sides by 8 gives: By cancelling: or

45 8t 2 = 8 8 5.625  t2 t2  5.625

Taking the square root of both sides gives: t 2 = 5.625 i.e.

t  ± 2.372, correct to 4 significant figures,

Practical problems involving simple equations Application: Applying the principle of moments to a beam results in the following equation: F × 3  (7.5 – F) × 2 where F is the force in Newtons. Determine the value of F. Removing brackets gives: Rearranging gives:

3 F  15 – 2 F 3 F + 2 F  15

i.e.

5 F  15 5F 15 = Dividing both sides by 5 gives: 5 5 from which,

force, F  3 N

Application: PV = mRT is the characteristic gas equation. Find the value of gas constant R when pressure, P = 3 × 106Pa, volume, V = 0.90m3, mass, m = 2.81kg and temperature, T = 231 K.

Some algebra topics

31

Dividing both sides of PV = mRT by mT gives: PV mRT = mT mT Cancelling gives:

PV =R mT

Substituting values gives: R =

(3 × 106 ) ( 0.90) ( 2.81)( 231)

Using a calculator, gas constant, R = 4160 J/(kg K), correct to 4 significant figures. Application:A formula relating initial and final states of pressures, P1 and P2, volumes V1 and V2, and absolute temperatures, T1 and T2, of an ideal PV PV gas is: 1 1 = 2 2 T2 T1 Find the value of P2 given P1 = 100 × 103 Pa, V1 = 1.0 m3, V2 = 0.266 m3 , T1 = 423 K and T2 = 293 K

P1V1

Since

T1

=

P2V2 T2

then

P2 (0.266) (100 × 103 )(1.0) = 423 293

‘Cross-multiplying’ gives: (100 × 103)(1.0)(293) = P2(0.266)(423) 3 P2 = (100 × 10 )(1.0)(293) (0.266)(423)

P2 = 260 × 103 Pa or 2.6 × 105 Pa

Hence,

Application: The stress f in a material of a thick cylinder can be obtained D  008f1+ pf   =  +1800 the stress, given that D = 21.5, d = 10.75 and from:   Calculate d  008f1pf   -1800 p = 1800

21.5   f+ pf  then +1800 = 0081   10.75 =   008f1 pf   -1800

 f + 1800     f - 1800 

i.e.

2=

 f + 1800     f - 1800 

Squaring both sides gives:

4=

Since

D d

f + 1800 f - 1800

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Mathematics Pocket Book for Engineers and Scientists

‘Cross-multiplying’ gives:

4(f – 1800) = f + 1800 4f – 7200 = f + 1800 4f – f = 1800 + 7200 3f = 9000 f = 9000 = 3000 3 stress, f = 3000

Hence,

Chapter 12

Transposing formulae

Introduction to transposing formulae In the formula I = V, I is called the subject of the formula. R Similarly, in the formula y = mx + c, y is the subject of the formula. When a symbol other than the subject is required to be the subject, then the formula needs to be rearranged to make a new subject. This rearranging process is called transposing the formula or transposition. There are no new rules for transposing formulae. The same rules as were used for simple equations in chapter 11 are used, i.e. the balance of an equation must be maintained. Whatever is done to one side of an equation must be done to the other. Application: If a + b = z – x + y, express x as the subject A term can be moved from one side of an equation to the other side but with a change of sign. Hence, rearranging gives:

x=z+y–a–b

Application: Transpose v = f  to make  the subject v = f  relates velocity v, frequency f and wavelength  Rearranging gives:

f=v

Dividing both sides by f gives: f  = v f f Cancelling gives: λ =v f

Some algebra topics

33

Application: When a body falls freely through a height h, the velocity v is given by v2 = 2gh. Express this formula with h as the subject. Rearranging gives:

2gh = v2

2 Dividing both sides by 2g gives: 2gh = v 2g 2g

Cancelling gives:

h=

v2 2g

Application: If I = V , rearrange to make V the subject R I = V is Ohm’s law, where I is the current, V is the voltage and R is the resistance. R V=I Rearranging gives: R  0081 + fV+f 1800   R   = R(I)  Multiplying both sides by R gives:  0081 fR-f 1800  V = IR

Cancelling gives:

Application: Transpose y = mx + c to make m the subject y = mx + c is the equation of a straight-line graph, where y is the vertical axis variable, x is the horizontal axis variable, m is the gradient of the graph and c is the y-axis intercept. Subtracting c from both sides gives: or

y – c = mx mx = y – c

Dividing both sides by x gives:

m=

Application: Transpose the formula v = u + v=u+

y–c x

Ft , to make t the subject m

Ft relates final velocity v, initial velocity u, force F, mass m and time t. m

F is acceleration ‘a’) m

(

Rearranging gives: and Multiplying each side by m gives: Cancelling gives: Dividing both sides by F gives:

Ft =v m Ft =v–u m  0081+  fFt + f 1800     = m(v – u)   00m f81f 1800 m u+

Ft = m(v – u) Ft m(v – u) = F F

Mathematics Pocket Book for Engineers and Scientists

34

t = m(v – u) or t = m (v – u) F F This shows two ways of expressing the answer. There is often more than one way of expressing a transposed answer.

Cancelling gives:

In this case, both equations for t are equivalent; neither one is more correct than the other. Application: In a right-angled triangle having sides x, y and hypotenuse z, Pythagoras’ theorem states z2 = x2 + y2. Transpose the formula to find y. x2 + y2 = z2

Rearranging gives: and

y2 = z2 – x2

Taking the square root of both sides gives:

y=

Application: Given t = 2π

z2 – x 2

l , find g in terms of t, l and π g

Whenever the prospective new subject is within a square root sign, it is best to isolate that term on the LHS and then to square both sides of the equation. 2π

Rearranging gives:

l =t g

l t = g 2π

Dividing both sides by 2π gives:

2

t2 l  t  =   = 2 g  2π  4π

Squaring both sides gives:

Cross-multiplying, (i.e. multiplying each term by 4π2g), gives: 4π2l = gt2 or

gt2 = 4π2l

Dividing both sides by t2 gives:

gt2 4π2l = 2 t2 t

Cancelling gives:

g=

4π2l t2

Application: The impedance Z of an a.c. circuit is given by: Z = where R is the resistance. Make the reactance, X , the subject. Rearranging gives:

Z= Z R2 +– X R22

Z R22 R2 +– X

Some algebra topics

R2 + X2 = Z2

Squaring both sides gives:

X2 = Z2 – R2

Rearranging gives:

Taking the square root of both sides gives: X =

Z2 – R2

x-y

Application: Make b the subject of the formula a =

x-y

Rearranging gives:

bd + be

Multiplying both sides by

bd + be gives:

or

a

Dividing both sides by a gives:

bd + be

=a

x–y=a

bd + be

bd + be = x – y bd + be =

x–y a 2

   a 

Squaring both sides gives:

x - y  bd + be = 

Factorising the LHS gives:

 x - y   b(d + e) =   a 

Dividing both sides by (d + e) gives:

 x − y    b =   a 

2

2

or b =

( d + e)

Application: If a =

(x – y)2 a2(d + e)

b make b the subject of the formula. 1+b

Rearranging gives:

b =a 1+b

Multiplying both sides by (1 + b) gives:

b = a(1 + b)

Removing the bracket gives:

b = a + ab

Rearranging to obtain terms in b on the LHS gives: Factorising the LHS gives: Dividing both sides by (1 – a) gives:

b – ab = a b(1 – a) = a b=

a 1- a

35

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Mathematics Pocket Book for Engineers and Scientists

Application: Given that

D  008 1  1800 f  express +p =  f + p in terms of D, d and f    1 d  008 f -p 1800 f  

 008f1+ p  D 1800   + f  =  008f1pf  d -1800

Rearranging gives:

 0081f + pf1800  D2 =     d2  0081f - pf1800

Squaring both sides gives:

Cross-multiplying, i.e. multiplying each term by d2(f – p), gives: d2(f + p) = D2(f – p) d2f + d2p = D2f – D2p

Removing brackets gives:

Rearranging, to obtain terms in p on the LHS gives: d2p + D2p = D2f – d2f p(d2 + D2) = f(D2 – d2)

Factorising gives: Dividing both sides by (d2 + D2) gives:

2 2 p = f (D - d ) ( d2 + D 2 )

Chapter 13

Solving simultaneous equations

Introduction When an equation contains two unknown quantities it has an infinite number of solutions. When two equations are available connecting the same two unknown values then a unique solution is possible. Equations which have to be solved together to find the unique values of the unknown quantities, which are true for each of the equations, are called simultaneous equations. Two methods of solving simultaneous equations analytically are: (a) by substitution, and (b) by elimination. Further methods of solving simultaneous equations are explained in Chapters 83 to 86.

Some algebra topics

37

Solving simultaneous equations in two unknowns Application: Solve the following equations for x and y, (a) by substitution, and (b) by elimination: x + 2y = – 1

(1)

4x – 3y = 18

(2)

(a) By substitution From equation (1):

x = – 1 – 2y

Substituting this expression for x into equation (2) gives: 4(– 1 – 2y) – 3y = 18 This is now a simple equation in y. Removing the bracket gives: – 4 – 8y – 3y = 18 – 11y = 18 + 4 = 22 y = 22 = – 2 –11 Substituting y = – 2 into equation (1) gives: x + 2(– 2) = – 1 x–4=–1 x=–1+4=3 Thus, x = 3 and y = – 2 is the solution to the simultaneous equations (b) By elimination x + 2y = – 1

(1)

4x – 3y = 18

(2)

If equation (1) is multiplied throughout by 4 the coefficient of x will be the same as in equation (2), giving: 4x + 8y = – 4

(3)

Subtracting equation (3) from equation (2) gives: 4x – 3y = 18

(2)

4x + 8y = – 4

(3)

0 – 11y = 22 22 =–2 Hence, y = – 11

(Note, in the above subtraction, 18 – – 4 = 18 + 4 = 22)

Mathematics Pocket Book for Engineers and Scientists

38

Application: Solve, by a substitution method, the simultaneous equations 3x – 2y = 12

(1)

x + 3y = – 7

(2)

From equation (2), x = – 7 – 3y Substituting for x in equation (1) gives: 3(– 7 – 3y) – 2y = 12 i.e.

– 21 – 9y – 2y = 12 – 11y = 12 + 21 = 33

33 =–3 – 11 Substituting y = – 3 in equation (2) gives:

Hence,

y=

x + 3(– 3) = – 7 i.e. Hence

x–9=–7 x=–7+9=2

Thus, x = 2, y = – 3 is the solution of the simultaneous equations. Application: Solve 3p = 2q 4p + q + 11 = 0

(1) (2)

Rearranging gives: 3p – 2q = 0

(3)

4p + q = – 11

(4)

Multiplying equation (4) by 2 gives: 8p + 2q = – 22 Adding equations (3) and (5) gives: 11p + 0 = – 22 p = –22 = – 2 11 Substituting p = – 2 into equation (1) gives: 3(– 2) = 2q – 6 = 2q q = –6 = – 3 2

(5)

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39

Checking, by substituting p = – 2 and q = – 3 into equation (2) gives: LHS = 4(– 2) + (– 3) + 11 = – 8 – 3 + 11 = 0 = RHS Hence, the solution is p = – 2, q = – 3

Practical problems involving simultaneous equations There are several situations in engineering and science where the solution of simultaneous equations is required.

Application: The law connecting friction F and load L for an experiment is of the form F = aL + b, where a and b are constants. When F = 5.6 N, L = 8.0 N and when F = 4.4 N, L = 2.0 N. Find the values of a and b and the value of F when L = 6.5 N Substituting F = 5.6, L = 8.0 into F = aL + b gives: 5.6 = 8.0a + b

(1)

Substituting F = 4.4, L = 2.0 into F = aL + b gives: 4.4 = 2.0a + b

(2)

Subtracting equation (2) from equation (1) gives: 1.2 = 6.0 a a = 1.2 = 1 or 0.2 5 6.0 Substituting a = 1 into equation (1) gives: 5  0081+1 f+ f 1800   + b  5.6 = 08.0 081-f5f-1800  5.6 = 1.6 + b 5.6 – 1.6 = b i.e.

b=4

Hence, a = 1 and b = 4 5 When L = 6.5, F = aL + b = 1 (6.5) + 4 = 1.3 + 4 i.e. F = 5.3 N 5 Application: When Kirchhoff’s laws are applied to the electrical circuit shown in Figure 13.1 the currents I1 and I2 are connected by the equations: 27 = 1.5I1 + 8(I1 – I2) – 26 = 2I2 – 8(I1 – I2) Solve the equations to find the values of currents I1 and I2

(1) (2)

Mathematics Pocket Book for Engineers and Scientists

40

l1

l2 (l1  l 2)

27 V

26 V

8 2

1.5 

Figure 13.1

Removing the brackets from equation (1) gives: 27 = 1.5I1 + 8I1 – 8I2 Rearranging gives:

9.5I1 – 8I2 = 27

(3)

Removing the brackets from equation (2) gives: – 26 = 2I2 – 8I1 + 8I2 Rearranging gives:

– 8I1 + 10I2 = – 26

(4)

Multiplying equation (3) by 5 gives: 47.5 I1 – 40I2 = 135

(5)

Multiplying equation (4) by 4 gives: – 32I1 + 40I2 = – 104

(6)

Adding equations (5) and (6) gives: 15.5I1 + 0 = 31 I1 = 31 = 2 15.5 Substituting I1 = 2 into equation (3) gives: 9.5(2) – 8I1 = 27 19 – 8I2 = 27 19 – 27 = 8I2 – 8 = 8I2 and

I2 = – 1

Hence, the solution is I1 = 2 and I2 = – 1 Application: The resistance R Ω of a length of wire at toC is given by: R = R0(1 + αt), where R0 is the resistance at 0oC and α is the temperature coefficient of resistance in /oC. Find the values of α and R0 if R = 30 Ω at 50oC, and R = 35 Ω at 100oC

Some algebra topics

41

Substituting R = 30, t = 50 into R = R0(1 + αt) gives: 30 = R0(1 + 50α)

(1)

Substituting R = 35, t = 100 into R = R0(1 + αt) gives: 35 = R0(1 + 100α)

(2)

Although these equations may be solved by the conventional substitution method, an easier way is to eliminate R0 by division. Thus, dividing equation (1) by equation (2) gives: 30 R0 (1 + 50α ) 1 + 50α = = 35 R0 (1 + 50α ) 1 + 100α ‘Cross-multiplying’ gives: 30(1 + 100α) = 35(1 + 50α) 30 + 3000α = 35 + 1750α 3000α – 1750α = 35 – 30 1250α = 5 α = 5 = 1 or 0.004 1250 250

i.e.

1 Substituting  = 250 into equation (1) gives:





 00f81 + 1800 f   1+  } 30 = R0 {1 + (50)  00f81 1800 -f   250 30 = R0 (1.2)

30 = 25 R0 = 1.2 Thus the solution is: α = 0.004/oC and R0 = 25Ω

Solving simultaneous equations in three unknowns Application: Solve the simultaneous equations: x+y+z=4

(1)

2x – 3y + 4z = 33

(2)

3x – 2y – 2z = 2

(3)

There are a number of ways of solving these equations. One method is shown below. The initial object is to produce two equations with two unknowns. For example, multiplying equation (1) by 4 and then subtracting this new equation from equation (2) will produce an equation with only x and y involved.

Mathematics Pocket Book for Engineers and Scientists

42

Multiplying equation (1) by 4 gives:

4x + 4y + 4z = 16

(4)

Equation (2) – equation (4) gives:

– 2x – 7y

(5)

= 17

Similarly, multiplying equation (3) by 2 and then adding this new equation to equation (2) will produce another equation with only x and y involved. Multiplying equation (3) by 2 gives:

6x – 4y – 4z = 4

(6)

Equation (2) + equation (6) gives:

8x – 7y

= 37

(7)

– 2x – 7y

= 17

(5)

Rewriting equation (5) gives:

Now we can use the previous method for solving simultaneous equations in two unknowns. Equation (7) – equation (5) gives:

10x = 20 x=2

from which, (note that 8x – – 2x = 8x + 2x = 10x) Substituting x = 2 into equation (5) gives: from which,

– 4 – 7y = 17

– 7y = 17 + 4 = 21 y=–3

and

Substituting x = 2 and y = – 3 into equation (1) gives: 2–3+z=4 z=5

from which,

Hence, the solution of the simultaneous equations is: x = 2, y = – 3 and z = 5

Chapter 14

Solving quadratic equations by factorising

Introduction A quadratic equation is one in which the highest power of the unknown quantity is 2. For example, x2 – 3x + 1 = 0 is a quadratic equation.

Factorisation Multiplying out (x + 1)(x – 3) gives x2 – 3x + x – 3 i.e. x2 – 2x – 3 The reverse process of moving from x2 – 2x – 3 to (x + 1)(x – 3) is called factorising. For example, if x2 – 2x – 3 = 0, then, by factorising: (x + 1)(x – 3) = 0

Some algebra topics

Hence either (x + 1) = 0

i.e. x = – 1

or

i.e. x = 3

(x – 3) = 0

43

Hence, x = – 1 and x = 3 are the roots of the quadratic equation x2 – 2x – 3 = 0 The technique of factorising is often one of ‘trial and error’.

Application: Solve the equation x2 + x – 6 = 0 by factorisation. The factors of x2 are: x and x. These are placed in brackets: (x

)(x

)

The factors of – 6 are: +6 and – 1, or – 6 and +1, or +3 and – 2, or – 3 and +2. The only combination to give a middle term of +x is +3 and – 2 i.e.

x2 + x – 6 = (x + 3)(x – 2)

The quadratic equation, x2 + x – 6 = 0 thus becomes (x + 3)(x – 2) = 0 Since the only way that this can be true is for either the first or the second, or both factors, to be zero, then either (x + 3) = 0 i.e. x = – 3 or

(x – 2) = 0

i.e. x = 2

Hence, the roots of x2 + x – 6 = 0 are x = – 3 and x = 2 Application: Solve the equation x2 + 2x – 8 = 0 by factorisation The factors of x2 are: x and x. These are placed in brackets thus: (x

)(x

)

The factors of – 8 are: +8 and – 1, or – 8 and +1, or +4 and – 2, or – 4 and +2. The only combination to give a middle term of +2x is +4 and – 2, x2 + 2x – 8 = (x + 4)(x – 2)

i.e.

(Note that the product of the two inner terms, 4x, added to the product of the two outer terms, – 2x, must equal the middle term, +2x in this case.) The quadratic equation x2 + 2x – 8 = 0 thus becomes (x + 4)(x – 2) = 0 Since the only way that this can be true is for either the first or the second, or both factors to be zero, then either (x + 4) = 0 i.e. x = – 4 or

(x – 2) = 0

i.e. x = 2

2

Hence, the roots of x + 2x – 8 = 0 are x = – 4 and x = 2 Application: Determine the roots of x2 – 8x + 16 = 0 by factorisation.

Mathematics Pocket Book for Engineers and Scientists

44

x2 – 8x + 16 = (x – 4)(x – 4)

i.e. (x – 4)2 = 0

The left-hand side is known as a perfect square. Hence, x = 4 is the only root of the equation x2 – x + 16 = 0 Application: Solve the equation: x2 – 5x = 0 Factorising gives: If

x (x – 5) = 0

x (x – 5) = 0

then either x = 0 or x – 5 = 0 x=0

i.e.

or x = 5

These are the two roots of the given equation. Answers can always be checked by substitution into the original equation.

Application: Determine the roots of 4x2 – 25 = 0 by factorisation. The left-hand side of 4x2 – 25 = 0 is the difference of two squares, (2x)2 and (5)2 By factorising, Hence,

4x2 – 25 = (2x + 5)(2x – 5)

either (2x + 5) = 0 i.e. or

Application:

(2x – 5) = 0 i.e.

i.e.

(2x + 5)(2x – 5) = 0

x = – 5 = – 2.5 2 x = 5 = 2.5 2

Solve the equation 3x2 – 11x – 4 = 0 by factorisation.

The factors of 3x2 are: 3x and x. These are placed in brackets: (3x

) (x

)

The factors of – 4 are: – 4 and +1, or +4 and – 1, or – 2 and 2 Remembering that the product of the two inner terms added to the product of the two outer terms must equal – 11x, the only combination to give this is +1 and – 4 3x2 – 11x – 4 = (3x + 1)(x – 4)

i.e.

The quadratic equation 3x2 – 11x – 4 = 0 thus becomes (3x + 1)(x – 4) = 0 Hence,

either or

(3x + 1) = 0

i.e.

x=– 1 3

(x – 4) = 0

i.e.

x=4

and both solutions may be checked in the original equation. Application: Solve the quadratic equation 15x2 + 2x – 8 = 0 by factorising The factors of 15x2 are: 15x and x

or

5x and 3x.

Some algebra topics

45

The factors of – 8 are: – 4 are +2, or 4 and – 2, or – 8 and +1, or 8 and – 1. By trial and error the only combination that works is 15x2 + 2x – 8 = (5x + 4)(3x – 2) Hence (5x + 4)(3x – 2) = 0 from which either

5x + 4 = 0

or

3x – 2 = 0 x=– 4 5 or

Hence,

x= 2 3

which may be checked in the original equation.

Chapter 15

Solving quadratic equations by completing the square

An expression such as x2 or (x + 2)2 or (x – 3)2 is called a perfect square. If x2 = 3 then x = ± 3 If (x + 2)2 = 5 then x + 2 = ±

5 and x = – 2 ±

If (x – 3)2 = 8 then x – 3 = ±

8 and x = 3 ±

5 8

Hence, if a quadratic equation can be rearranged so that one side of the equation is a perfect square and the other side of the equation is a number, then the solution of the equation is readily obtained by taking the square root of each side as in the above examples. The process of rearranging one side of a quadratic equation into a perfect square before solving is called ‘completing the square’. (x + a)2 = x2 + 2ax + a2 Thus in order to make the quadratic expression x2 + 2ax into a perfect square it  2 is necessary to add (half the coefficient of x)2 i.e.  2a  or a2  2   3 2 For example, x2 + 3x becomes a perfect square by adding   , i.e.  2   3 2 x2 + 3x +   =  2 

 2  x + 3    2 

The method of ‘completing the square’ is demonstrated in the following Applications.

46

Mathematics Pocket Book for Engineers and Scientists

Application: Solve 2x2 + 5x = 3 by ‘completing the square’ The procedure is as follows: 1. Rearrange the equation so that all terms are on the same side of the equals sign (and the coefficient of the x2 term is positive). Hence,

2x2 + 5x – 3 = 0

2. Make the coefficient of the x2 term unity. In this case this is achieved by dividing throughout by 2. 2x 2 5x 3 Hence, + – =0 2 2 2 i.e.

x2 +

3 5 x– =0 2 2

3. Rearrange the equations so that the x2 and x terms are on one side of the equals sign and the constant is on the other side. Hence, x2 +

3 5 x= 2 2

4. Add to both sides of the equation (half the coefficient of x)2. In this case the 5 coefficient of x is . 2  5 2 Half the coefficient squared is therefore    4   5 2 3  5 2   = +    4  2  4  The LHS is now a perfect square, i.e.

Thus,

x2 +

5 x+ 2

 2  2  x + 5  = 3 +  5    4   2 4  5. Evaluate the RHS. Thus  2  x + 5  = 3 + 25 = 24 + 25 = 49   2 16 16 16 4  6. Take the square root of both sides of the equation (remembering that the square root of a number gives a ± answer). Thus  2  x + 5  =   4 

 49     16 

x+ 5 = ± 7 4 4 7. Solve the simple equation, i.e. x=– 5 ± 7 4 4 i.e.

i.e.

1 2 x=– 5 + 7 = = or 0.5 2 4 4 4

Some algebra topics

47

x = – 5 – 7 = – 12 = – 3 4 4 4

and

Hence, x = 0.5 or x = – 3 i.e. the roots of the equation 2x2 + 5x = 3 are 0.5 and – 3 Application: Solve 2x2 + 9x + 8 = 0, correct to 3 significant figures, by ‘completing the square’ Making the coefficient of x2 unity gives: x2 + and rearranging gives:

9 x+4=0 2 9 x2 + x = – 4 2

Adding to both sides (half the coefficient of x)2 gives: x2 +

9 x+ 2

 9 2  9 2   =   – 4  4   4 

The LHS is now a perfect square, thus 81 64 81 17  2  x + 9  = –4= =  16 16 16 16  4 Taking the square root of both sides gives: x+ Hence, i.e.

9 = 4

 17     16  = ± 1.031

9 ± 1.031 4 x = – 1.22 or – 3.28, correct to 3 significant figures

Chapter 16

x=–

Solution of quadratic equations by formula

If ax2  bx  c  0 then x =

-b ± b2 - 4ac 2a

This is known as the quadratic formula. Application: Solve 3x2  11x  4  0 by using the quadratic formula Comparing 3x2  11x  4  0 with ax2  bx  c  0 gives a  3, b  11 and c  4

Mathematics Pocket Book for Engineers and Scientists

48

-(-11) ± (-11)2 - 4(3)(-4) 11 ± 121  48  2(3) 6 11 ± 169 11 ± 13   6 6

Hence, x 



Hence, x 

11  13 11 - 13 or 6 6 -2 1 24  4 or 6 6 3

Application: Solve 4x2  7x  2  0 giving the roots correct to 2 decimal places Comparing 4x2  7x  2  0 with ax2  bx  c gives a  4, b  7 and c  2 Hence, x  

7  72  4(4)(2) 7  17  2(4) 8 7  4.123 7  4.123 7  4.123  or 8 8 8

Hence, x  0.36

or

1.39, correct to 2 decimal places.

Application: The height s metres of a mass projected vertically upwards at time t seconds is s  ut 

1 2

gt 2 . Determine how long the mass will take after

being projected to reach a height of 16 m (a) on the ascent and (b) on the descent, when u  30 m/s and g  9.81 m/s2

When height s  16 m, 16  30t  i.e.

1 (9.81)t 2 2

4.905t2  30t  16  0

Using the quadratic formula: t 

(30)  (30)2  4(4.905)(16) 2(4.905) 30  586.1 30  24.21   5.53 or 0.59 9.81 9.81

Hence the mass will reach a height of 16 m after 0.59 s on the ascent and after 5.53 s on the descent.

Some algebra topics

49

Application: A shed is 4.0 m long and 2.0 m wide. A concrete path of constant width is laid all the way around the shed and the area of the path is 9.50 m2. Calculate its width, to the nearest centimetre Figure 16.1 shows a plan view of the shed with its surrounding path of width t metres t t

2.0 m

4.0 m

(4.0  2t )

SHED

Figure 16.1

Area of path  2(2.0  t)  2t(4.0  2t) i.e. or Hence

9.50  4.0t  8.0t  4t 2 4t 2  12.0t  9.50  0 t 

(12.0)  (12.0)2  4(4)(9.50) 2(4) 12.0  296.0 12.0  17.20465  8 8

Hence, t  0.6506 m or 3.65058 m Neglecting the negative result which is meaningless, the width of the path, t  0.651 m or 65 cm, correct to the nearest centimetre. In this chapter and in the previous two, three methods of solving quadratic equations have been shown. However, scientific notation electronic calculators with an equation mode are also able to solve quadratic equations – and much more quickly!

Chapter 17

Logarithms

Introduction to logarithms The theory of logarithms is important, for there are several scientific and engineering laws that involve the rules of logarithms.

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Mathematics Pocket Book for Engineers and Scientists

If y = ax then x = loga y

Definition of a logarithm:

Logarithms having a base of 10 are called common logarithms and log10 is usually abbreviated to lg. Logarithms having a base of e (where ‘e’ is a mathematical constant approximately equal to 2.7183) are called hyperbolic, Napierian or natural logarithms, and loge is usually abbreviated to ln. Application: Evaluate log3 9 Let x = log3 9 then 3x = 9

from the definition of a logarithm,

3x = 32 from which, x = 2

i.e.

log3 9 = 2

Hence,

Application: Evaluate log16 8 Let x = log16 8 then 16x = 8 from the definition of a logarithm, (24)x = 23

i.e.

24x = 23

i.e.

from the laws of indices,

4x = 3 and x = 3 4 log16 8 = 3 4

from which, Hence,

Application: Evaluate lg 0.001 Let x = lg 0.001 = log10 0.001

then 10x = 0.001 10x = 10–3

i.e.

from which, x = – 3

lg 0.001 = – 3 (which may be checked by a calculator)

Hence,

Application: Evaluate ln e Let x = ln e = logee

then

ex = e ex = e1

i.e.

ln e = 1

Hence,

Application: Evaluate log3

Let x = log3 Hence,

1 81

then

3x =

from which, x = 1

(which may be checked by a calculator)

1 81 1 1 = 4 3 81

log3 1 = – 4 81

= 3–4

from which, x = –4

Some algebra topics

51

Application: Solve the equation: lg x = 4 If lg x = 4 then log10 x = 4 x = 104

and

x = 10,000

i.e.

Application: Solve the equation: log5 x = – 2 If log5 x = – 2

then

x = 5–2 =

1 52

=

1 25

Laws of logarithms There are three laws of logarithms, which apply to any base: (i)

log (A × B) = log A + log B

(ii)

A log   = log A – log B  B 

(iii)

log An = n log A

Application: Write log 3 + log 8 as the logarithm of a single number log 3 + log 8 = log (3 × 8) by the first law of logarithms = log 24 Application: Write log 15 – log 3 as the logarithm of a single number  15  log 15 – log 3 = log   by the second law of logarithms  3  = log 5 Application: Write 2 log 5 as the logarithm of a single number 2 log 5 = log 52

by the third law of logarithms

= log 25 Application: Solve the equation: log(x – 1) + log(x + 8) = 2 log(x + 2) LHS = log(x – 1) + log(x + 8) = log(x – 1)(x + 8)

from the first law of logarithms

= log(x2 + 7x – 8) RHS = 2 log(x + 2) = log(x + 2)2 from the third law of logarithms = log(x2 + 4x + 4)

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Mathematics Pocket Book for Engineers and Scientists

log(x2 + 7x – 8) = log(x2 + 4x + 4 )

Hence,

x2 + 7x – 8 = x2 + 4x + 4

from which, i.e.

7x – 8 = 4x + 4

i.e.

3x = 12

and

x=4

Indicial equations The laws of logarithms may be used to solve certain equations involving powers – called indicial equations. Application: Solve 3x = 27 Logarithms to a base of 10 are taken of both sides, log10 3x = log10 27

i.e. and

x log10 3 = log10 27

Rearranging gives:

x =

(Note,

log10 27 log10 3

=

by the third law of logarithms

1.43136 ... 0.47712 ...

= 3

which may be readily checked.

27 log 27 is not equal to log ) 3 log 3

Application: Solve the equation: 2x = 5, correct to 4 significant figures. Taking logarithms to base 10 of both sides of 2x = 5 gives: log10 2x = log10 5 i.e.

x log10 2 = log10 5

Rearranging gives: x =

log10 5 log10 2

=

by the third law of logarithms 0.6989700 .. = 2.322, correct to 4 significant 0.3010299 .. figures.

Application: Solve the equation: x2.7 = 34.68, correct to 4 significant figures. Taking logarithms to base 10 of both sides gives: log10 x2.7 = log10 34.68 2.7 log10 x = log10 34.68 Hence, Thus,

log10 x =

log10 34.68

= 0.57040 2.7 x = antilog 0.57040 = 100.57040 = 3.719, correct to 4 significant figures.

Some algebra topics

53

Graphs of logarithmic functions A graph of y  log10 x is shown in Figure 17.1 and a graph of y  loge x is shown in Figure 17.2. Both are seen to be of similar shape; in fact, the same general shape occurs for a logarithm to any base. y

0.5 0 1 x y  log10x

0.5

2 3

x

3 2

1

0.5

0.2

0.1

0.48 0.30 0 0.30 0.70 1.0

1.0

Figure 17.1 y 2 1

0

1

2

3

4

5

6

x

1

2

x 6 5 4 3 2 1 0.5 0.2 0.1 y  logex 1.79 1.61 1.39 1.10 0.69 0 0.69 1.61 2.30

Figure 17.2

In general, with a logarithm to any base a, it is noted that: 1. loga 1  0 2. loga a  1 3. loga 0 → 

Chapter 18 Exponential functions Introduction to exponential functions An exponential function is one which contains ex, e being a constant called the exponent and having an approximate value of 2.7183. The exponent arises from the natural laws of growth and decay and is used as a base for natural or Napierian logarithms.

Mathematics Pocket Book for Engineers and Scientists

54

Application: The instantaneous voltage v in a capacitive circuit is related to time t CR

-

t by the equation: v = Ve

where V, C and R are constants. Determine v,

correct to 4 significant figures, when t = 50 ms, C = 10 μF, R = 47 kΩ and V = 300 volts. -

v = Ve

t CR

-

= 300e

50 × 10-3 10× 10-6 × 47× 103

Using a calculator, v = 300e– 0.1063829... = 300(0.89908025...) = 269.7 volts The power series for ex is: ex  1 x 

x2 x3 x 4 ...    2! 3! 4!

(where 3!  3  21 and is called ‘factorial 3’) The series is valid for all values of x.

Graphs of exponential functions Figure 18.1 shows graphs of y  ex and y  ex y 20 y  ex 16

y  ex

12 8

4

3

2

1

0

1

2

3

x

Figure 18.1

Application: The decay of voltage, v volts, across a capacitor at time t seconds is given by v  250et/3. Draw a graph showing the natural decay curve over the first 6 seconds. Determine (a) the voltage after 3.4 s, and (b) the time when the voltage is 150 volts A table of values is drawn up as shown below. t e

0 t/3

v  250et/3

1.00

1

2

0.7165 0.5134

250.0 179.1

128.4

3

4

5

6

0.3679 0.2636 0.1889 0.1353 91.97

65.90

47.22

33.83

Some algebra topics

55

The natural decay curve of v  250et/3 is shown in Figure 18.2. 250

v = 250e −t /3

Voltage v (volts)

200 150 100 80 50

0

1 1.5 2

3 3.4 4

5

6

Time t (seconds)

Figure 18.2

From the graph, (a) when time t  3.4 s, voltage v  80 volts (b) when voltage v  150 volts, time t  1.5 seconds

Chapter 19

Napierian logarithms

Introduction to Napierian logarithms Logarithms having a base of ‘e’ are called hyperbolic, Napierian or natural logarithms and the Napierian logarithm of x is written as loge x, or more commonly as ln x. Logarithms were invented by John Napier, a Scotsman (1550–1617). Check using your calculator,

ln 1.812 = 0.59443, correct to 5 significant figures

ln 1 = 0, ln e3 = 3 and ln e1 = 1 From the last two examples we can conclude that: loge ex = x This is useful when solving equations involving exponential functions. Application: Solve e3x  8

Taking Napierian logarithms of both sides, gives ln e3x = ln 8 ie from which

3x = ln 8 x = 1 ln 8 = 0.6931, correct to 4 decimal places 3

Mathematics Pocket Book for Engineers and Scientists

56

t

Application: Given 32 = 70(1 – e– 2 ) determine the value of t, correct to 3 significant figures -

Rearranging 32 = 70 (1 – e

t 2

32 = 1 – e– 2t 70

) gives:

-

e

and

t 2

38 = 1 – 32= 70 70 t

e 2 = 70

Taking the reciprocal of both sides gives:

38

t 2

t

 70    38 

Taking Napierian logarithms of both sides gives: ln e e=2 ln = 

t 2t  70  =   2 e= ln  38 

i.e.

 70   = 1.22, correct to 3 significant figures  38 

t

t =e22 ln = 

from which,

Application: The work done in an isothermal expansion of a gas from pressure p1 to p2 is given by:  p  w  w 0 ln  1   p2  If the initial pressure p1  7.0 kPa, calculate the final pressure p2 if w  3w0  p   3 w 0  w 0 ln  1   p2 

If w  3w0 then

i.e.

 p  3  ln  1   p2 

and

e3 

from which,

p1 p2



7000 p2

final pressure, p2 = 7000 = 7000e–3 = 348.5 Pa e3

Laws of growth and decay

The laws of exponential growth and decay are of the form y  Aekx and y  A(1  ekx), where A and k are constants. When plotted, the form of each of these equations is as shown in Figure 19.1. The laws occur frequently in engineering and science and examples of quantities related by a natural law include (i) Linear expansion

l  l0 eθ

(ii) Change in electrical resistance with temperature

Rθ  R0 e θ

(iii) Tension in belts

T1  T0 e θ

Some algebra topics y A

57

y A

y  Aekx y  A(1  ekx )

x

0

x

0

Figure 19.1

(iv) Newton’s law of cooling

θ  θ0 ekt

(v) Biological growth

y  y0 ekt

(vi) Discharge of a capacitor

q  Q et/CR

(vii) Atmospheric pressure

p  p0 eh/c

(viii) Radioactive decay

N  N0 et

(ix) Decay of current in an inductive circuit

i  I eRt/L

(x) Growth of current in a capacitive circuit

i  I(1  et/CR)

Application: In an experiment involving Newton’s law of cooling, the temperature θ(°C) is given by θ  θ0ekt. Find the value of constant k when θ0  56.6°C, θ  16.5°C and t  83.0 seconds Transposing θ  θ0 ekt gives θ0 θ



1 e ekt

θ  ekt from which, θ0

kt

Taking Napierian logarithms of both sides gives: ln from which, kk  fromwhich,

θ0 θ

 kt

 56.6  1 θ0 1   1 (1.2326486 ..) ln ln   t θ 83.0  16.5  83.0  1.485  102

Application: The current i amperes flowing in a capacitor at time t seconds is given by i  8.0(1  et/CR), where the circuit resistance R is 25 k and capacitance C is 16 F. Determine (a) the current i after 0.5 seconds and (b) the time, to the nearest ms, for the current to reach 6.0 A 6 )(25103 )

(a) Current i  8.0(1  et/CR )  8.0 [1  e0.5/ (1610  8.0(1  e1.25 )

 8.0(1  0.2865047..)  8.0(0.7134952..)  5.71 amperes

]

Mathematics Pocket Book for Engineers and Scientists

58

(b) Transposing i  8.0(1  et/CR) gives: from which, et/CR  1 

i  1  et/CR 80

i 8.0  i  8.0 8.0

Taking the reciprocal of both sides gives: et/CR 

8.0 8.0  i

Taking Napierian logarithms of both sides gives:  8.0  t   ln   8.0  i  CR  8.0   Hence t  CR ln   8.0  i 

i.e.

 8.0   when i  6.0 amperes,  (16  106 )(25  103 ) ln   8.0  6.0   8.0    0.4 ln 4.0 t  0.40 ln   2.0   0.4(1.3862943..)  0.5545 s  555 ms, to the nearest millisecond.

A graph of current against time is shown in Figure 19.2. 8 i(A) 6 5.71 4

i  8.0 (1  et /CR )

2 0

Chapter 20

0.5 1.0 0.555

1.5

t(s)

Figure 19.2

Hyperbolic functions

sinh x 

e x  ex 2

cosech x 

cosh x 

e x  ex 2

sech x 

1 2  x cosh x e  ex

tanh x 

sinh x e x  ex  x cosh x e  ex

coth x 

1 e x  ex  x tanh x e  ex

1 2  x sinh x e  ex

Some algebra topics

cosh x  1 

x2 x 4 ..  + (which is valid for all values of x) 2! 4!

sinh x  x 

x3 x 5 ..   (which is valid for all values of x) 3! 5!

59

Graphs of hyperbolic functions A graph of y  sinh x is shown in Figure 20.1. Since the graph is symmetrical about the origin, sinh x is an odd function. y 10 8 6

y  sinh x

4

y

2 3 2 1 0 1 2

2

3

10

x

4

6

6

4 2

8 10

y  cosh x

8

3 2 1 0

Figure 20.1

1 2 3

x

Figure 20.2

A graph of y  cosh x is shown in Figure 20.2. Since the graph is symmetrical about the y-axis, cosh x is an even function. The shape of y  cosh x is that of a heavy rope or chain hanging freely under gravity and is called a catenary. Examples include transmission lines, a telegraph wire or a fisherman’s line, and are used in the design of roofs and arches. Graphs of y  tanh x, y  coth x, y  cosech x and y  sech x are shown in Figures 20.3 and 20.4. y 3 2

y  coth x

1 y

y  tanh x

1

1 2 3

x

1 0 1 2 3 x

3 2 1

3 2 1 0

1 (a)

y  coth x 2 3 (b)

Figure 20.3

60

Mathematics Pocket Book for Engineers and Scientists y 3 2 y  cosech x 1 y 0 1

3 2 1

2

3

x 1

1 y  cosech x

2

y  sech x 0 1

3 2 1

2

3

x

3 (a)

(b)

Hyperbolic identities Trigonometric identity

Corresponding hyperbolic identity

cos2x  sin2x  1

ch2x  sh2x  1

1  tan2x  sec2x

1  th2x  sech2x

cot 2x  1  cosec2x

coth2x  1  cosech2x

Compound angle formulae

sin(A  B)  sin A cos B  cos A sin B

sh(A  B)  sh A ch B  ch A sh B

cos(A  B)  cos A cos B 7 sin A sin B

ch(A  B)  ch A ch B  sh A sh B

tan(A  B) 

tan A  tan B 1 7 tan A tan B

tan(A  B) 

th A  th B 1  th A th B

Double angles

sin 2x  2 sin x cos x

sh 2x  2 sh x ch x

cos 2x  cos2x  sin2x

ch 2x  ch2x  sh2x

 2 cos2x  1

 2 ch2x  1

 1  2 sin2x

 1  2 sh2x

tan 2x 

2 tan x 1  tan2x

th 2x 

2 th x 1  th2x

Figure 20.4

Some algebra topics

61

Solving equations involving hyperbolic functions Equations of the form a ch x  b sh x  c, where a, b and c are constants may be solved either by: (a) plotting graphs of y  a ch x  b sh x and y  c and noting the points of intersection, or more accurately, (b) by adopting the following procedure: x   e x  ex   x  and ch x to  e  e  1. Change sh x to        2 2 2. Rearrange the equation into the form pex  qex  r  0, where p, q and r are constants. 3. Multiply each term by ex, which produces an equation of the form p(ex)2  rex  q  0 (since (ex)(ex)  e0  1) 4. Solve the quadratic equation p(ex)2  rex  q  0 for ex by factorising or by using the quadratic formula. 5. Given ex  a constant (obtained by solving the equation in 4), take Napierian logarithms of both sides to give x  ln(constant) Application: Solve the equation sh x  3, correct to 4 significant figures Following the above procedure:  e x  ex    3 1. sh x     2 2. ex  ex  6, i.e. ex  ex  6  0 3. (ex)2  (ex)(ex)  6ex  0, i.e. (ex)2  6ex  1  0 4. e x 

(6)  [(6)2  4(1)(1)] 6  40 6  6.3246   2(1) 2 2

Hence, ex  6.1623 or 0.1623 5. x  ln 6.1623 or x  ln(0.1623) which has no solution since it is not possible in real terms to find the logarithm of a negative number. Hence x  ln 6.1623  1.818, correct to 4 significant figures. The above solution may be obtained much quicker with a calculator. Using a calculator: (i) Press hyp

(ii) Choose 4, which is sinh–1

(iii) Type in 3

(iv) Close bracket

(v) Press = and the answer is 1.818448459 i.e. the solution of sh x = 3 is: x = 1.818, correct to 4 significant figures, as above.

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Mathematics Pocket Book for Engineers and Scientists

Application: A chain hangs in the form given by y  40 ch

x . Determine, 40

correct to 4 significant figures, (a) the value of y when x is 25 and (b) the value of x when y  54.30 (a)

y  40 ch

x and when x  25, 40

y  40 ch

 e0.625  e0.625  25   40 ch 0.625  40    40 2  20(1.8682  0.5353)  48.07

(b) When y  54.30, 54.30  40 ch ch

x , from which 40

54.30 x  1.3575  40 40

Following the above procedure: e x/ 40  ex/ 40  1.3575 2 2. ex/40  ex/40  2.715 i.e. 1.

x/40 2

3. (e

x/40

)  1  2.715 e

0

ex/40  ex/40  2.715  0 i.e.

(ex/40)2  2.715 ex/40  1  0

2 4. e x/ 40  (2.715)  [(2.715)  4(1)(1)] 2(1)

2.715  (3.3712) 2.715  1.8361  2 2 Hence ex/40  2.2756 or 0.43945 

5.

x x  ln 2.2756 or  ln(0.43945) 40 40 x x  0.8222 or  0.8222 40 40 Hence, x  40(0.8222) or x  40(0.8222) Hence,

i.e. x  32.89, correct to 4 significant figures.

Some algebra topics

Chapter 21

63

Partial fractions

Provided that the numerator f(x) is of less degree than the relevant denominator, the following identities are typical examples of the form of partial fraction used: Linear factors

f(x) A B C    (x  a)(x  b)(x  c)  (x  a) (x  b) (x  c) Repeated linear factors f(x) A B C ≡   (x  a) (x  a) 3 (x  a) 2 (x  a) 3

Quadratic factors f(x) Ax  B C ≡  (x  d) (ax 2  bx  c)(x  d) (ax 2  bx  c)

11  3x Application: Resolve x 2  2x  3 into partial fractions

The denominator factorises as (x  1)(x  3) and the numerator is of less degree than the denominator. Thus Let

11  3x may be resolved into partial fractions. x 2  2x  3

11  3x 11  3x A B where A and B are constants    (x  1)(x  3) (x  1) (x  3) x 2  2x  3

to be determined, i.e.

11  3x A(x  3)  B(x  1)  by algebraic addition (x  1)(x  3) (x  1)(x  3)

Since the denominators are the same on each side of the identity then the numerators are equal to each other. Thus,

11  3x  A(x  3)  B(x  1)

To determine constants A and B, values of x are chosen to make the term in A or B equal to zero. When x  1, then 11  3(1)  A(1  3)  B(0) i.e. i.e.

8  4A A2

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Mathematics Pocket Book for Engineers and Scientists

When x   3, then 11  3(3)  A(0)  B(3  1) i.e.

20  4B

i.e.

B  5 11  3x 5 2 5 2     (x  1) (x  3) (x  1) (x  3) x 2  2x  3

Thus

 2 5 2(x  3)  5(x  1) 11  3x   Check:    2  (x 1) (x 3) (x 1)(x 3)     x  2x  3  

Application: Express

x 3  2x 2  4x  4 in partial fractions x2  x  2

The numerator is of higher degree than the denominator. Thus dividing out gives: x

x2

x

)

2 x3 x3

3 2x 2 4x x 2 2x 3x 2 2x 3x 2 3x x

Thus

4 4 6 10

x 3  2x 2  4x  4 x  10  x 3 2 x2  x  2 x x 2  x 3

Let

x  10 (x  2)(x  1)

x  10 A B A(x  1)  B(x  2)    (x  2)(x  1) (x  2) (x  1) (x  2)(x  1)

Equating the numerators gives: x  10  A(x  1)  B(x  2) Let x  2, then i.e.

A4

Let x  1, then i.e.

12  3A

9  3B B  3

Hence

x  10 4 3   (x  2)(x  1) (x  2) (x  1)

Thus

4 3 x 3  2x 2  4x  4  x 3  ( x  2) ( x  1) x2  x  2

Some algebra topics

Application: Express

65

5x 2  2x  19 as the sum of three partial fractions (x  3)(x  1)2

The denominator is a combination of a linear factor and a repeated linear factor. Let 5x 2  2x  19 A B C    (x  3) (x  1) (x  1)2 (x  3)(x  1)2 

A(x  1)2  B(x  3)(x  1)  C(x  3) by algebraic algebraic addition by (x  3)(x  1)2 addition

Equating the numerators gives: 5x 2  2x  19  A(x  1)2  B(x  3)(x  1)  C(x  3)

(1)

Let x  3, then 5(3)2  2(3)  19  A(4)2  B(0)(4)  C(0) i.e.

32  16A

i.e.

A2

Let x  1, then

5(1)2  2(1)  19  A(0)2  B(4)(0)  C(4)

i.e.

16  4C

i.e.

C  4

Without expanding the RHS of equation (1) it can be seen that equating the coefficients of x2 gives: 5  A  B, and since A  2, B  3 Hence

5x 2  2x  19 2 3 4 ≡   (x  2) (x  1) (x  3)(x  1) 2 (x  1) 2

Application: Resolve

3  6x  4x 2  2x 3 into partial fractions x 2 (x 2  3)

Terms such as x2 may be treated as (x  0)2, i.e. they are repeated linear factors. (x2  3) is a quadratic factor which does not factorise without containing surds and imaginary terms. Let

3  6x  4x 2  2x 3 A B Cx  D   2  2 2 2 x x (x  3) x (x  3) Ax (x 2  3)  B (x 2  3)  (Cx  D) x 2  x 2 (x 2  3)

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Mathematics Pocket Book for Engineers and Scientists

Equating the numerators gives: 3  6x  4x 2  2x 3  Ax(x 2  3)  B(x 2  3)  (Cx  D)x 2  Ax 3  3Ax  Bx 2  3B  Cx 3  Dx 2 Let x  0, then i.e.

3  3B B1

Equating the coefficients of x3 terms gives: Equating the coefficients of x2 terms gives:

2  A  C 4 BD

Since B  1, D  3 Equating the coefficients of x terms gives:

6  3A

i.e.

A2

From equation (1), since A  2, C  4 Hence

2 1 3  6x  4x 2  2x 3 4x  3   2  2 x x x 2 (x 2  3) x 3 

2 1 3  4x  2  2 x x x 3

(1)

Section 3 Some number topics

Why are number topics important? Number sequences are widely used in engineering and scientific applications including computer data structure and sorting algorithms, financial engineering, audio compression, and architectural engineering. Thanks to engineers, robots have migrated from factory shop floors – as industrial manipulators, to outer space – as interplanetary explorers, hospitals – as minimally invasive surgical assistants, homes – as vacuum cleaners and lawn mowers, and battlefields – as unmanned air, underwater, and ground vehicles. Arithmetic progressions are used in simulation engineering and in the reproductive cycle of bacteria. Some uses of A.P.s in daily life include uniform increase in the speed at regular intervals, completing patterns of objects, calculating simple interest, speed of an aircraft, increase or decrease in the costs of goods, sales and production, and so on. Geometric progressions are used in compound interest and the range of speeds on a drilling machine. In fact, G.P.s are used throughout mathematics, and they have many important applications in physics, engineering, biology, economics, computer science, queuing theory, and finance. There are many, many different types of equations used in every branch of engineering and science. There are straight forwardmethods for solving simple, quadratic and simultaneous equations. However, there are many other types of equations than these three. Great progress has been made in the engineering and scientific disciplines regarding the use of iterative methods for linear systems. In engineering it is important that we can solve any equation; iterative methods help us do that. There are infinite ways to represent a number. The four commonly associated with modern computers and digital electronics are decimal, binary, octal, and hexadecimal. All four number systems are equally capable of representing any number. Furthermore, a number can be perfectly converted between the various number systems without any loss of numeric value. At a first look, it seems like using any number system other than decimal is complicated and unnecessary. However, since the job of electrical and software engineers is to work with digital circuits, engineers and scientists require number systems that can best transfer information between the human world and the digital circuit world. Thus, the way in which a number is represented can make it easier for the engineer to perceive the meaning of the number as it applies to a digital circuit, i.e. the appropriate number system can actually make things less complicated.

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Mathematics Pocket Book for Engineers and Scientists

Chapter 22

Simple number sequences

Simple sequences A set of numbers which are connected by a definite law is called a series or a sequence of numbers. Each of the numbers in the series is called a term of the series. For example, 1, 3, 5, 7, .. is a series obtained by adding 2 to the previous term, and 2, 8, 32, 128, .. is a sequence obtained by multiplying the previous term by 4

Application: Determine the next two terms in the series: 3, 6, 9, 12, … We notice that the sequence 3, 6, 9, 12, … progressively increases by 3, thus the next two terms will be 15 and 18 Application: Find the next three terms in the series: 9, 5, 1, … We notice that each term in the series 9, 5, 1, … progressively decreases by 4, thus the next two terms will be 1 – 4, i.e. – 3 and – 3 – 4, i.e. – 7 Application: Determine the next two terms in the series: 2, 6, 18, 54, … We notice that the second term, 6, is three times the first term, the third term, 18, is three times the second term, and that the fourth term, 54, is three times the third term. Hence the fifth term will be 3 x 54 = 162, and the sixth term will be 3 x 162 = 486

The n’th term of a series If a series is represented by a general expression, say, 2n + 1, where n is an integer (i.e. a whole number), then by substituting n = 1, 2, 3, … the terms of the series can be determined; in this example, the first three terms will be: 2(1) + 1,

2(2) + 1,

2(3) + 1,… , i.e.

3, 5, 7, …

What is the n’th term of the sequence 1, 3, 5, 7, …? Firstly, we notice that the gap between each term is 2, hence the law relating the numbers is: ‘2n + something’ The second term, 3 = 2n + something, hence when n = 2 (i.e. the second term of the series), then

Some number topics

69

3 = 4 + something, and the ‘something’ must be – 1 Thus, the n’th term of 1, 3, 5, 7, … is 2n – 1 Hence the fifth term is given by 2(5) – 1 = 9, and the twentieth term is 2(20) – 1 = 39, and so on.

Application: The n’th term of a sequence is given by 3n + 1. Write down the first four terms.

The first four terms of the series 3n + 1 will be: 3(1) + 1, 3(2) + 1, 3(3) + 1 and 3(4) + 1 i.e. 4, 7, 10 and 13

Application: The n’th term of a series is given by 4n – 1. Write down the first four terms. The first four terms of the series 4n – 1 will be: 4(1) – 1, 4(2) – 1, 4(3) – 1 and 4(4) – 1 i.e. 3, 7, 11 and 15 Application: Find the n’th term of the series: 1, 4, 7, … We notice that the gap between each of the given three terms is 3, hence the law relating the numbers is: ‘3n + something’ The second term,

4 = 3n + something,

so when n = 2, then

4 = 6 + something,

so the ‘something’ must be – 2 (from simple equations) Thus, the n’th term of the series 1, 4, 7, … is: 3n – 2 Application: Find the n’th term of the sequence: 3, 9, 15, 21, … Hence determine the 15th term of the series. We notice that the gap between each of the given four terms is 6, hence the law relating the numbers is: ‘6n + something’ The second term,

9 = 6n + something,

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Mathematics Pocket Book for Engineers and Scientists

so when n = 2, then

9 = 12 + something,

so the ‘something’ must be – 3 Thus, the n’th term of the series 3, 9, 15, 21, … is: 6n – 3 The 15th term of the series is given by 6n – 3 when n = 15 Hence, the 15th term of the series 3, 9, 15, 21, … is: 6(15) – 3 = 87 Application: Find the n’th term of the series: 1, 4, 9, 16, 25, .. This is a special series and does not follow the pattern of the previous examples. Each of the terms in the given series are square numbers, i.e. 1, 4, 9, 16, 25, … ≡ 12, 22, 32, 42, 52, .. Hence the n’th term is: n2

Chapter 23

Arithmetic progressions

When a sequence has a constant difference between successive terms it is called an arithmetic progression (often abbreviated to AP). If a = first term, d = common difference and n = number of terms, then the arithmetic progression is: a, a  d, a  2d, .... The n’th term is: a  (n  1)d The sum of n terms, Sn 

n [2a  (n  1)d] 2

Application: Find the sum of the first 7 terms of the series 1, 4, 7, 10, 13, . . . The sum of the first 7 terms is given by S7  

7 [2(1)  (7  1)3] 2

since a  1 and d  3

7 7 [2  18]  [20]  70 2 2

Application: Determine (a) the ninth, and (b) the sixteenth term of the series 2, 7, 12, 17, . . .

Some number topics

71

2, 7, 12, 17, ..... is an arithmetic progression with a common difference, d, of 5 (a) The n’th term of an AP is given by a  (n  1)d Since the first term a  2, d  5 and n  9 then the 9th term is: 2  (9  1)5  2  (8)(5)  2  40  42 (b) The 16th term is: 2  (16  1)5  2  (15)(5)  2  75  77 Application: Find the sum of the first 12 terms of the series 5, 9, 13, 17, ..... 5, 9, 13, 17, ..... is an AP where a  5 and d  4 The sum of n terms of an AP, Sn 

n [2a  (n  1)d] 2 12 [2(5)  (12  1)4] 2  6[10  44]  6(54)

Hence the sum of the first 12 terms, S12 

 324

Application: 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,

n

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

80 

   Sn = 2a + (n - 1)d = 2(30) + (80 - 1)(2) = 40 60 + 158 2 2  = 40(218) = £8720

Chapter 24

Geometric progressions

When a sequence has a constant ratio between successive terms it is called a geometric progression (often abbreviated to GP). If a = first term, r = common ratio, and n = number of terms, then the geometric progression is: a, ar, ar 2 , ar 3 , .... The n’th term is:

ar n1

The sum of n terms, Sn 

a( 1  r n ) (1  r )

which is valid when r < 1

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Mathematics Pocket Book for Engineers and Scientists

or Sn 

a(r n  1) (r  1)

If 1 < r 1 a (1  r)

Application: Find the sum of the first 8 terms of the GP 1, 2, 4, 8, 16, .... The sum of the first 8 terms is given by

i.e.

S8 

1(28  1) (2  1)

S8 

1(256  1)  255 1

since a  1 and r  2

Application: Determine the tenth term of the series 3, 6, 12, 24, .... 3, 6, 12, 24, .... is a geometric progression with a common ratio r of 2. The n’th term of a GP is arn1, where a is the first term. Hence the 10th term is: (3)(2)101  (3)(2)9  3(512)  1536 Application: A tool hire firm finds that their net return from hiring tools is decreasing by 10% per annum. Their net gain on a certain tool this year is £400. Find the possible total of all future profits from this tool (assuming the tool lasts forever) The net gain forms a series: £400  £400  0.9  £400  0.92  ....., which is a GP with a  400 and r  0.9 The sum to infinity, S 

a 400   £4000  total future profits (1  r) (1  0.9)

Application: A drilling machine is to have 6 speeds ranging from 50 rev/min to 750 rev/min. Determine their values, each correct to the nearest whole number, if the speeds form a geometric progression Let the GP of n terms be given by a, ar, ar2, .... arn1 The first term a  50 rev/min The 6th term is given by ar61, which is 750 rev/min, i.e.

ar5  750

Some number topics

r5 

from which

73

750 750   15 a 50

r  5 15  1.7188

Thus the common ratio, The first term is a  50 rev/min

the second term is ar  (50)(1.7188)  85.94, the third term is ar2  (50)(1.7188)2  147.71, the fourth term is ar3  (50)(1.7188)3  253.89, the fifth term is ar4  (50)(1.7188)4  436.39, the sixth term is ar5  (50)(1.7188)5  750.06 Hence, correct to the nearest whole number, the 6 speeds of the drilling machine are: 50, 86, 148, 254, 436 and 750 rev/min

Chapter 25

Inequalities

Introduction to inequalities An inequality is any expression involving one of the symbols < , > , < _ or > _ p < q means p is less than q

p > q means p is greater than q

p< _ q means p is less than or equal to q

p> _ q means p is greater than or equal to q

Some simple rules (i) When a quantity is added or subtracted to both sides of an inequality, the inequality still remains. For example, if p < 3 then p + 2 < 3 + 2 (adding 2 to both sides) and

p – 2 < 3 – 2 (subtracting 2 from both sides)

(ii) When multiplying or dividing both sides of an inequality by a positive quantity, say 5, the inequality remains the same. For example, if p > 4 p 4 > 5 5 (iii) When multiplying or dividing both sides of an inequality by a negative quantity, say –3, the inequality is reversed. For example, then 5p > 20 and

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Mathematics Pocket Book for Engineers and Scientists

if p > 1 then –3p < –3 example)

and

p 4 < –3 –3

(Note > has changed to < in each

To solve an inequality means finding all the values of the variable for which the inequality is true.

Simple inequalities Application: Solve the following inequalities: (a) (c) z – 2 > _5

(d)

3 + x > 7

(b)

3t < 6

p < _2 3

(a) Subtracting 3 from both sides of the inequality: 3 + x > 7 gives: 3 + x – 3 > 7 – 3 i.e. x > 4 Hence, all values of x greater than 4 satisfy the inequality. (b) Dividing both sides of the inequality: 3t < 6 by 3 gives: Hence, all values of t less than 2 satisfy the inequality.

3t 6 i.e. t < 2 < 3 3

(c) Adding 2 to both sides of the inequality: z – 2 > _ 5 gives: z – 2 + 2 > _5+2 i.e. z > _7 Hence, all values of z equal to or greater than 7 satisfy the inequality. (d) Multiplying both sides of the inequality:

p < _ 2 by 3 gives: 3

(3)

p < _ (3)2 3

i.e. p < _6 Hence, all values of p equal to or less than 6 satisfy the inequality. Application: Solve the inequality: 4x + 1 > x + 5 Subtracting 1 from both sides of the inequality: 4x + 1 > x + 5 gives: 4x > x + 4 Subtracting x from both sides of the inequality: 4x > x + 4 gives: 3x > 4 4 3 4 satisfy the inequality: 4x + 1 > x + 5 Hence all values of x greater than 3 Dividing both sides of the inequality: 3x > 4 by 3 gives: x >

Application: Solve the inequality: 3 – 4t < _8+t

Some number topics

75

Subtracting 3 from both sides of the inequality: 3 – 4t < _ 8 + t gives: – 4t < _5+t Subtracting t from both sides of the inequality: –4t < _ 5 + t gives:

–5t < _5

Dividing both sides of the inequality: –5t < _ 5 by –5 gives: t > _ –1 (remembering to reverse the inequality) Hence, all values of t equal to or greater than –1 satisfy the inequality.

Inequalities involving a modulus The modulus of a number is the size of the number, regardless of sign. Vertical lines enclosing the number denote a modulus. For example,  4  = 4 negative)

and  –4  = 4

(the modulus of a number is never

The inequality:  t  < 1 means that all numbers whose actual size, regardless of sign, is less than 1, i.e. any value between –1 and +1. Thus  t  < 1 means – 1 < t < 1 Similarly,  x  > 3 means all numbers whose actual size, regardless of sign, is greater than 3, i.e. any value greater than 3 and any value less than – 3. Thus  x  > 3 means x > 3 and x < – 3

Application: Solve the following inequality:  3x + 1  < 4 Since  3x + 1  < 4 then – 4 < 3x + 1 < 4 Now – 4 < 3x + 1 becomes

– 5 < 3x

i.e. – 5 < x and 3x + 1 < 4 becomes 3

3x < 3 i.e. x < 1 Hence, these two results together become – 5 < x < 1 and mean that the 3 5 inequality  3x + 1  < 4 is satisfied for any value of x greater than – but less 3 than 1.

Application: Solve the inequality: | 1+ 2t | < _ 5 Since  1+ 2t  < _ 5 then –5 < _ 1 + 2t < _5 Now – 5 < _ 1 + 2t becomes – 6 < _ 2t i.e. – 3 < _ t and 1 + 2t < _ 5 becomes 2t < _ 4 i.e. t < _2 Hence, these two results together become: – 3 < _t< _2

Mathematics Pocket Book for Engineers and Scientists

76

Inequalities involving quotients p p > 0 then must be a positive value. q q

If

For

p to be positive, either p is positive and q is positive q

or

p is negative

and q is negative. + = + +

i.e.

and

– = + –

p p < 0 then must be a negative value. q q

If

For

p to be negative, either p is positive and q is negative or p is negative q

and q is positive. i.e.

+ = – and –

– = – +

Application: Solve the inequality:

t+1 >0 3t – 6

t+1 t+1 > 0 then must be positive. 3t – 6 3t – 6 t+1 For to be positive, either (i) t + 1 > 0 and 3t – 6 > 0 3t – 6 and 3t – 6 < 0 Since

or (ii) t + 1 < 0

(i) If t + 1 > 0 then t > – 1 and if 3t – 6 > 0 then 3t > 6 and t > 2 Both of the inequalities t > – 1 and t > 2 are only true when t > 2, i.e. the fraction

t+1 is positive when t > 2 3t – 6

(ii) If t + 1 < 0 then t < –1 and if 3t – 6 < 0 then 3t < 6 and t < 2 Both of the inequalities t < – 1 and t < 2 are only true when t < – 1, i.e. the fraction Summarising,

t+1 is positive when t < – 1 3t – 6

t+1 > 0 when t > 2 or t < – 1 3t – 6

Application: Solve the inequality:

2x + 3 < _1 x+2

Some number topics

2x + 3 2x + 3 < _ 1 then –1< _ 0 x+2 x+2 2x + 3 x + 2 2x + 3 – (x + 2) < _ 0 or – < _ 0 i.e. i.e. x+2 x+2 x+2 x+1 to be negative or zero, either (i) x + 1 < _0 For x+2 or (ii) x + 1 > _0

77

Since

x+1 < _ 0 x+2 and x + 2 > 0 and x + 2 < 0

(i) If x + 1 < _ 0 then x < _ – 1 and if x + 2 > 0 then x > –2 (Note that > is used for the denominator, not > _ ; a zero denominator gives a value for the fraction which is impossible to evaluate.) x+1 < _ 0 Hence, the inequality x+2 or equal to – 1,

is true when x is greater than – 2 and less than

which may be written as – 2 < x < _–1 (ii) If x + 1 > _ 0 then x > _ – 1 and if x + 2 < 0 then x < – 2 It is not possible to satisfy both x > _ – 1 and x < – 2 satisfies (ii). Summarising,

thus no values of x

2x + 3 < _ 1 when – 2 < x < _–1 x+2

Inequalities involving square functions The following two general rules apply when inequalities involve square functions: (i) if x2 > k

then x >

(ii) if x2 < k

then

–

k

or x < – k

k 9

Since t2 > 9 then t2– 9 > 0

i.e.

(t + 3)(t – 3) > 0 by factorising

For (t + 3)(t – 3) to be positive, either (i) (t + 3) > 0 and (t – 3) > 0 or (ii) (t + 3) < 0 and (t – 3) < 0 (i) If (t + 3) > 0 then t > – 3 and if (t – 3) > 0 then t > 3 Both of these are true only when t > 3 (ii) If (t + 3) < 0 then t < – 3 and if (t – 3) < 0 then t < 3 Both of these are true only when t < – 3

(1) (2)

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Mathematics Pocket Book for Engineers and Scientists

Summarising, t2 > 9 when t > 3 or t < – 3 This demonstrates the general rule: if x2 > k then x >

or x < – k

k

(1)

Application: Solve the inequality: ×2 > 4

From the general rule stated above in equation (1):

4

if x2 > 4 then x >

or x < – 4

i.e. the inequality: x2 > 4 is satisfied when

x > 2 or x < – 2

Application: Solve the inequality: (2z + 1)2 > 9

From equation (1), if (2z + 1)2 > 9

then

2z + 1 >

9

or 2z + 1 < – 9

i.e.

2z + 1 > 3

or

2z + 1 < – 3

i.e.

2z > 2

or

2z < –4

i.e.

z > 1 or z < – 2

Application: Solve the inequality: t2 < 9 Since t2 < 9

then t2 – 9 < 0

i.e.

(t + 3)(t – 3) < 0 by factorising.

For (t + 3)(t – 3) to be negative, either (i) (t + 3) > 0 and (t – 3) < 0 or (ii) (t + 3) < 0 and (t – 3) > 0 (i) If (t + 3) > 0 then t > – 3 and if (t – 3) < 0 then t < 3 Hence (i) is satisfied when –3 0 then t > 3 It is not possible to satisfy both t < – 3 and t > 3, thus no values of t satisfies (ii). Summarising, t2 < 9 when – 3 < t < 3 which means that all values of t between – 3 and +3 will satisfy the inequality. This demonstrates the general rule: if x2 < k

then

– k 0 Since x2 + 2x – 3 > 0 then (x – 1)(x + 3) > 0 by factorising. For the product (x – 1)(x + 3) to be positive, either (i) (x – 1) > 0 and (x + 3) > 0 or (ii) (x – 1) < 0 and (x + 3) < 0 (i) Since (x – 1) > 0 then x > 1

and since (x + 3) > 0 then x > – 3

Both of these inequalities are satisfied only when x > 1 (ii) Since (x – 1) < 0 then x < 1 and since (x + 3) < 0 then x < – 3 Both of these inequalities are satisfied only when x < – 3

Summarising, x2+ 2x – 3 > 0 is satisfied when either x > 1 or x < – 3 Application: Solve the inequality: t2– 2t – 8 < 0 Since t2– 2t – 8 < 0 then (t – 4)(t + 2) < 0 by factorising. For the product (t – 4)(t + 2) to be negative, either (i) (t – 4) > 0 and (t + 2) < 0 or (ii) (t – 4) < 0 and (t + 2) > 0 (i) Since (t – 4) > 0 then t > 4 and since (t + 2) < 0 then t < – 2 It is not possible to satisfy both t > 4 and t < – 2, thus no values of t satisfies the inequality (i)

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Mathematics Pocket Book for Engineers and Scientists

(ii) Since (t – 4) < 0 then t < 4 and since (t + 2) > 0 then t > – 2 Hence, (ii) is satisfied when – 2 < t < 4 Summarising, t2– 2t – 8 < 0 is satisfied when – 2 < t < 4 Application: Solve the inequality: x2+ 6x + 3 < 0

x2 + 6x + 3 does not factorise; completing the square gives: x2+ 6x + 3 = (x + 3)2 + 3 – 32 = (x + 3)2 – 6 The inequality thus becomes: (x + 3)2 – 6 < 0 From equation 2,

– 6 < (x + 3)
1

4. {k2} = {0, 1, 4, 9, ...}

z (z + 1) (z - 1)3

for |z| > 1

5. {k3} = {0, 1, 8, 27, ...}

z (z 2 + 4z + 1) (z - 1)4

for |z| > 1

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Mathematics Pocket Book for Engineers and Scientists

Table 159.1 Continued Sequence

Transform F(z)

6. {ak} = {1, a, a2, a3, ...}

for |z| > |a|

z z -a

7. {kak} = {0, a, 2a2, 3a3, ...}

az

for |z| > |a| 2

(z - a ) 8. {k2ak} = {0, a, 4a2, 9a3, ...}

az (z + a ) 3

(z - a ) 9. {e-ak} = {e-a, e-2a, e-3a, ...}

z z - e -a

10. sin ak = {sin a, sin 2a, ...}

z sin a z 2 - 2z cos a + 1

11. cos ak = {cos a, cos 2a, ...}

z (z - cos a ) z 2 - 2z cos a + 1

12. e-ak sin bk = {e-a sinb, e-2a sin 2b, ...}

z2

ze -a sin b - 2ze -a cos b + e -2a

z2

z 2 - ze -a cos b - 2ze -a cos b + e -2a

13. e–ak cos bk = {e-a cosb, e-2a cos 2b, ...}

Application: Determine the z-transform of 5k2

z (z + 1) (z - 1)3 5 z ( z + 1) Z {5k 2 } = 5Z {k 2 } = ( z − 1)3

{ }

From 4 in Table 159.1, Z k 2 = Hence,

Application: Determine the z-transform of (a) 3k (b) (–3)k

z z -a z Z { 3k } = z −3

{ }

(a) From 6 in Table 159.1, Z a k = If a = 3, then

z z -a z z Z { (−3)k } = = z − −3 z +3

{ }

(b) From 6 in Table 159.1, Z a k = If a = - 3, then

Application: Determine the z-transform of 2e-3k

for |z| > |a|

Z-transforms

{

}

From 9 in Table 159.1, Z e -ak =

z z - e -a

Z { 2e −3k } = 2Z {e −3k } =

Hence,

451

2z z − e −3

Application: Determine Z{cos3k} From 11 in Table 159.1, Z {cos ak } = Z { cos 3k } =

Hence, since a = 3,

z2

z (z - cos a ) - 2z cos a + 1

z2

z ( z − cos 3) − 2z cos 3 + 1

Application: Determine Z{e–2kcos4k}

{

}

z2

z 2 - ze -a cos b - 2ze -a cos b + e -2a

{

}

z2

z 2 − ze −2 cos 4 − 2ze −2 cos 4 + e −4

Z e -ak cos bk =

From 13 in Table 159.1,

Hence, since a = 2 and b = 4, Z e −2k cos 4k =

Chapter 160

Some properties of z-transforms

(a) Linearity property The z-transform is a linear transform, i.e.

Z (a { xk } + b { yk }) = aZ { xk } + bZ { yk }

(1) where a and b are constants

Application: Determine the z-transform of 2{k} – 3{e–2k}

Now, and since

  z  from 4 in Table 159.1 2Z {k} = 2  2  (z - 1)  z Z {e–ak} = from 9 in Table 159.1, z - e -a

z z - e -2 Hence, Z (2{k} -3 {e–2k}) = 2Z {k} -3Z {e–2k}

then

Z {e–2k} =

    z z  = 2  - 3   2  z - e   (z - 1)2   2z   3z  =  -   (z - 1)2   z - e -2 

from equation (1)

Mathematics Pocket Book for Engineers and Scientists

452

2

(

) = 2 ( z - 1) ( z - e -2 ) 2z 2 - 2ze -2 - 3z ( z 2 - 2z + 1) = 2 ( z - 1) ( z - e -2 ) 2z z - e -2 - 3z ( z - 1)

=

i.e.

2z 2 - 2ze -2 - 3z 3 + 6z 2 - 3z 2

( z - 1) ( z - e -2 )

2Z{k} − 3Z{e–2k} =

(

−3z 3 + 8 z 2 − z 2e −2 + 3 2

( z − 1) ( z

− e −2

)

)

(b) First shift theorem (shifting to the left) It may be shown by the first shift theorem (shifting to the left), that if

Z{xk} = F(z)

then

Z{xk+m} = zmF(z) – [zmx0 + zm−1x1 + .... + zxm−1]

(2)

is the z-transform of the sequence that has been shifted by m places to the left. This theorem is often needed when solving difference equations (see Chapter 162).

Application: Determine Z{3k+2} Since from equation (2), Z{xk+m} = zmF(z) – [zmx0 + zm–1x1 + .... + zxm–1] then

Z{3k+2} = z2Z{3k} – [z230 + z31]

z z thus Z{3k} = From 6 of Table 159.1, Z{a } = z -a z -3 Hence, substituting in equation (3),  z    -  z 2 + 3z  Z{3k+2} = z 2    z - 3   k

z3 -  z 2 + 3z   z - 3  3 2  z - ( z - 3)  z + 3z    = z -3 z 3 -  z 3 + 3z 2 - 3z 2 - 9z    = z -3 z 3 -  z 3 - 9z    = z -3 =

(3)

Z-transforms

453

9z z −3 This is the z-transform of the sequence {9, 27, 81,…} by shifting the sequence {1, 3, 9, 27,…} two places to the left and losing the first two terms. Z{3k+2} =

i.e.

Application: Determine Z{k + 1} From 3 of Table 159.1,

Z{k} =

z 2

( z - 1)

Since from equation (2), Z{xk + m} = zmF(z) – [zmx0 + zm–1x1 + .... + zxm–1] Z{k + 1} = z1Z(k) – [z1x0]

then

    z  -  z 1 × 0  from 3 of Table 159.1 = z 1  2      ( z - 1)  Z{k + 1} =

i.e.

z2 2

( z − 1)

(c) Second shift theorem (shifting to the right) It may be shown by the second shift theorem (shifting to the right), that if

Z{xk} = F(z)

then

Z{xk–m} = z –m F(z)

(4)

is the z-transform of the sequence that has been shifted by m places to the right. Application: Determine Z{xk–2} Since from equation (4), Z{xk–m} = z–mF(z) then

Z{xk–2} = z–2F(z)  z   since Z{xk} = z from 2 of = z -2   z - 1 z -1 Table 159.1

i.e.

 z −1  1  = Z{xk–2} =   z − 1 z ( z − 1)

This is the z-transform of the sequence {0, 0, 1, 1, 1, ...} by shifting the sequence {1, 1, 1, 1, ...} two places to the right and defining the first two terms as zeros.

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Mathematics Pocket Book for Engineers and Scientists

Application: Determine Z{ak–1} Since from equation (4), z{xk–m} = z–mF(z) Z{ak–1} = z–1F(z)  z   since Z{ak} = z from 6 of = z -1   z - a  z -a Table 159.1

then

 z -1 × z  1  = Z{ak–1} =   z - a  (z − a ) which is the z-transform of {ak} shifted one place to the right. i.e.

(d) Translation If the sequence {xk} has the z-transform Z{akxk} = F(z), then the sequence {akxk} has the z-transform Z{akxk} = F(a–1z) Application: Determine Z{3kk}

z from 3 of Table 159.1 (z - 1)2 then by the translation property,

Since

Z(k) =

Z{3kk} = F(3–1z) =

3-1z 3-1z = 2 2 - 1)     3-1  z - 1     3-1  

(3-1z

=

i.e.

Z{3kk} =

3-1z 2

3-2 ( z - 3)

=

32 z 2

31 ( z - 3)

3z 2

( z − 3)

(e) Final value theorem For the sequence {xk} with the z-transform F(z),   z - 1  F ( z )  provided that Lim xk exists Lim x k = Lim   k →∞    z →1  z  k →∞ 

Z-transforms

455

 k     1  Application: Determine Lim     3   k →∞       

F{xk} =

Now

i.e.

 k  z z 3z 1 = and    = z -1 1 3z - 1  3   z  3  k      z - 1  1  F (z )  = Lim  z - 1 3z  Lim    = Lim             z → 1 z → 1 k →∞   z  3z 1  z    3      3(z - 1)   = 0 = Lim  z →1  3z - 1     k     1  Lim    = 0  3   k →∞       

   33z 2 - 25z   Application: Determine Lim  2 k →∞   ( z - 1)( 3z - 1)       33z 2 - 25z     = Lim  z - 1 F (z )   By the final value theorem, Lim     2   k →∞   z → 1 z   ( z - 1)( 3z - 1)         z - 1 33z 2 - 25z  = Lim    2 z →1   z  ( z - 1)( 3z - 1)       33z - 25   = Lim  2 z →1  ( 3z - 1)    33 - 25 8 = = =2 4 22

(f) The initial value theorem For the sequence {xk} with the z-transform F(z), {F(z)} initial value, x0 = zLim →∞

{ }

ak Application: Determine zLim →∞

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Mathematics Pocket Book for Engineers and Scientists

z from 6 of Table 159.1 z -a  z   Lim {F (z ) } = Lim  z →∞ z →∞  z - a     d  (z ) dz     dz   = Lim  1 = by L’Hopital’s rule = Lim    z →∞  d (z - a ) dz  z →∞  1 (see Chapter 28)  dz  F(z) = F{ak } =

and

Lim {ak} = 1

i.e.

z→∞

(g) The derivative of the transform Z{xk} = F(z)

If

–zF’(z) = Z{kxk}

then

Application: Determine the derivative of Z{kak}

z from 6 of Table 159.1 z -a Z{kxk} = –zF’(z) from above   z    = -z  (z - a )(1) - z (1)  using the quotient rule Z {kx k } = -zF ’    2  z - a  (z - a )  

F(z) = F{ak} = Now i.e.

 -a  az    ( z - a )2  = ( z - a )2   az k i.e. the derivative of Z{ka } = –zF’(z) = which confirms result 7 in ( z − a )2 Table 159.1 z -a -z = -z  2  (z - a )

Chapter 161

  = -z  

Inverse z-transforms

If the sequence {xk} has a Z transform Z{xk} = F(z), then the inverse z-transform is defined as:

Z–1F(z) = {xk}

In the following Applications, some inverses may be determined directly from Table 159.1, page 449–450, albeit with a little manipulation; others sometimes require the use of partial fractions – just as with inverse Laplace transforms.

Z-transforms

Application: Determine the inverse z-transform of F(z) =

{ }

From 6 in Table 159.1, Z a k = Comparing Thus,

z z +5

 z  z  = ak hence Lim Z-1   z - a  z →∞ z -a

z z shows that a = - 5 with z +5 z -a  z  Z −1   = (-5)k  z + 5 

Application: Determine the inverse z-transform of F(z) =

2z 2z + 1

2z 2z z = =   1 2z + 1 1 2  z +  z +  2  2  z  z  = ak hence Lim Z–1  z →∞  z - a  z -a   z z 1 Comparing shows that a = – with 1 z -a 2 z+ 2

{ }

From 6 in Table 159.1, Z a k =

    k  2z   z   1    = Z -1  Thus, Z -1   = -    2z + 1 1   2   z +  2   3z determine Z–1F(z) 3z - 1

Application: If F(z) =

3z 3z z = =   1 3z – 1 1 3  z –  z –  3  3

{ }

From 6 in Table 159.1, Z a k =

z

z z -a



z   = ak a  

hence Lim Z–1 

z →∞  z

z 1 shows that a = z -a 3    z   1 k  3z  − 1  =      Thus, Z −1  = Z    3z – 1  1   3   z –  3   Comparing

1 z3

with

Application: Determine the inverse z-transform of F(z) =

z z - e2

457

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Mathematics Pocket Book for Engineers and Scientists

{

}

From 9 in Table 159.1, Z e -ak =

 z  z  = e -ak hence Z–1 a a  z - e  z -e  

Comparing

z z with shows that a = - 2 z - e2 z - e -a

Thus,

 z  Z −1   = e −−2k = e 2k  z - e 2 

Application: If F(z) =

z determine Z–1 F(z) z2 +1

In Table 159.1, results 10 and 11 have z2 in their denominators. If the numerator is to be z in result 10, then sin a has to equal 1, i.e. a = From 10 in Table 159.1, Z { sin ak } =

z sin a z 2 - 2z cos a + 1

π 2

  z sin a  = sin ak hence Z -1  2  z - 2z cos a + 1 When a =

π z sin a = , 2 z 2 - 2z cos a + 1

 z  π Thus, Z −1   = sin k  z 2 + 1 2

z sin

π 2

π z 2 - 2z cos + 1 2

=

z z2 +1

Application: Determine the inverse z-transform of F(z) =

Using partial fractions, let

z z A B = = + (z - 4)(z - 3) (z - 4) (z - 3) z 2 - 7z + 12 =

from which,

z = A(z – 3) + B(z – 4)

Letting z = 4 gives:

4=A

Letting z = 3 gives: Hence,

z z 2 - 7z + 12

A(z - 3) + B (z - 4) (z - 4)(z - 3)

3 = – B i.e. B = – 3 z 4 3 = F(z) = 2 (z - 4) (z - 3) z - 7z + 12

The nearest transform in Table 159.1 to either of these partial fractions is z Z{ak} = z -a 4 3 4 3 z z - × = × Rearranging gives: F(z) = (z - 4) (z - 3) z (z - 4) z (z - 3) = 4 × z–1Z{4k} – 3 × z–1Z{3k}

Z-transforms

Hence,

459

Z–1F(z) = 4 × {4k–1} – 3 × {3k–1} by the second shift theorem = {4k} – {3k} = {4k – 3k}

i.e.

the sequence is xk = 4k – 3k

z being z, there is an alternative, and z2 - 7z + 12 more straight-forward method of determining the inverse transform, F (z ) 1 = 2 i.e. by initially rearranging as: z - 7z + 12 z With the denominator of F(z) =

Using partial fractions,

z2

1 1 A B = = + ( 4)( z z - 3) (z - 4) (z - 3) - 7z + 12 =

from which,

1 = A(z – 3) + B(z – 4)

Letting z = 4 gives:

1=A

Letting z = 3 gives:

1 = – B i.e. B = – 1

F (z )

Hence,

z

and

=

F(z) =

1 1 1 = (z - 4) (z - 3) z 2 - 7z + 12

z z (z - 4) (z - 3)

Z-1F(z ) = {4k} – {3k} = {4k – 3k}

and

A(z - 3) + B (z - 4) (z - 4)(z - 3)

from 6 in Table 159.1

Application: Determine the inverse z-transform of F(z) =

Since F(z) =

z then z 2 - 3z + 2

Using partial fractions, let

z z 2 - 3z + 2

F (z ) 1 = 2 z z - 3z + 2

1 11 A 1 B A B = = + (z 2- 1)(z - 2) = ((zz - 1)( 1) z -(z2)-=2)(z - 1) + (z - 2) z 2 - 3z + 2 z - 3z + 2 A(z - 2) + B (z -A1)(z - 2) + B (z - 1) = (z - 1)(z - = 2) (z - 1)(z - 2)

from which,

1 = A(z – 2) + B(z – 1)

Letting z = 1 gives:

1=–A

Letting z = 2 gives:

1=B

i.e. A = – 1

Hence,

-1 F (z ) 1 1 1 = + = z (z - 1) (z - 2) (z - 2) (z - 1)

and

F(z) =

z z (z - 2) (z - 1)

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Mathematics Pocket Book for Engineers and Scientists

 Z-1F(z) = Z -1   (z  = Z -1   (z

Thus,

Z–1F(z) = (2)k – (1)k = (2)k – 1

From 6 in Table 159.1,

Chapter 162

z z   - 2) (z - 1)   z  z   - Z -1     - 2)   (z - 1) 

Using z-transforms to solve difference equations

In Chapter 157, Laplace transforms were used to solve differential equations; in this section, the solution of difference equations using z-transforms is demonstrated. Difference equations arise in several different ways – sometimes from the direct modelling of systems in discrete time, or as an approximation to a differential equation describing the behaviour of a system modelled as a continuous-time system. The z-transform method is based on the first shift theorem, (see earlier, page 452), and the method of solution is explained through the following applications. Application: Solve the difference equation xk+1 – 2xk = 0 given the initial condition that x0 = 3 Taking the z-transform of each term gives: Z{xk+1} – 2Z{xk} = Z{0} Since from equation (2), Chapter 160, page 452 Z{xk+m} = zmF(z) – [zmx0 + zm–1x1 + .... + zxm–1] then

(z1Z{k} – [ z1(3)

i.e.

z Z{xk} – 3z – 2Z{xk} = 0

([

i.e.

–2Z{xk} = 0

(z – 2)Z{xk} = 3z

3z z -2  z   3z  Taking the inverse z-transform gives: {xk} = Z -1   = 3Z -1    z - 2   z - 2  and

i.e.

Z {xk } =

{xk} = 3(2k) from 6 of Table 159.1

Application: Solve the difference equation: xk+2 – 3xk+1 + 2xk = 1 given that x0 = 0 and x1 = 2

Z-transforms

461

Taking the z-transform of each term gives: Z{xk+2} – 3Z{xk+1} + 2Z{xk} = Z{1} Since from equation (2), Chapter 160, page 452 Z{xk+m} = zmF(z) – [zmx0 + zm-1x1 +…+ zxm-1] (z2Z{xk} – [z2(0) + z1(2)]) – 3(z1Z{xk} – [z1(0)]) + 2Z{xk} =

z 2Z { x k } - 2z - 3zZ { x k } + 2Z { x k } =

i.e. and

z z -1

z z -1

z z + 2z (z - 1) 2z 2 - z z (2z - 1) + 2z = = = z -1 z -1 z -1 z -1 z (2z - 1) z (2z - 1) Z {xk } = = (z - 1)(z - 2)(z - 1) (z - 1)(z 2 - 3z + 2)

(z 2 - 3z + 2)Z { x k } =

from which,

Z {xk } (2z - 1) (2z - 1) = = z (z - 1)(z - 2)(z - 1) (z - 1)2 (z - 2)

or

Using partial fractions, let

(2z - 1) A B C = + + (z - 1) (z - 2) (z - 1)2 (z - 2) (z - 1)2

A(z - 1)(z - 2) + B (z - 2) + C (z - 1)2 (z - 1)2 (z - 2) 2z – 1 = A(z-1)(z-2) + B(z-2) + C(z-1)2 =

and Letting z = 1 gives:

1=-B

Letting z = 2 gives:

3=C

Equating z2 coefficients gives:

0 = A + C i.e. A = - 3

Hence, Therefore,

i.e. B = - 1

(2z - 1) -3 -1 3 Z {xk } = = + + (z - 1) (z - 2) (z - 1)2 (z - 2) (z - 1)2 z    z  z  - 3  z  Z { x k } = 3     (z - 1)  (z - 1)2  (z - 2) 

Taking the inverse z-transform gives:      z    - 3Z -1  z  - Z -1  z { x k } = 3Z -1   (z - 1)2   (z - 1)   (z - 2)  = 3(2)k – 3(1)k – k i.e.

from 6 and 3 of Table 159.1

{xk} = 3(2k) – 3 – k

Application: Solve the difference equation: xk+2 – xk = 1 given that x0 = 0 and x1 = – 1 Taking the z-transform of each term gives: Z{xk+2} – Z{xk} = Z{1}

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Mathematics Pocket Book for Engineers and Scientists

Since from equation (2), Chapter 160, page 452 Z{xk+m} = zmF(z) – [zmx0 + zm–1x1 + .... + zxm–1] z z2Z{xk} – [z2(0) + z1(–1)] – Z{xk} = z -1 z i.e. z2Z{xk} + z – Z{xk} = z -1 z z - z (z - 1) 2z - z 2 and (z 2 - 1)Z { x k } = -z = = z -1 z -1 z -1 2z - z 2 2z - z 2 2z - z 2 = = Z {xk } = from which, 2 (z - 1)(z - 1)(z + 1) (z - 1)2 (z + 1) (z - 1) z - 1

Z {xk }

(

)

2-z (z - 1)2 (z + 1) 2-z A B C = + + Using partial fractions, let (z - 1) (z + 1) (z - 1)2 (z + 1) (z - 1)2 and

z

=

=

A(z - 1)(z + 1) + B (z + 1) + C (z - 1)2 (z - 1)2 (z + 1)

2 – z = A(z–1)(z+1) + B(z+1) + C(z–1)2

and Letting z = 1 gives:

1 = 2B

Letting z = – 1 gives:

3 = 4C

Equating z2 coefficients gives:

Z {xk }

Hence,

z

i.e. C = 3/4

0 = A + C i.e. A = – 3/4 =

2-z -3 / 4 1/ 2 3/4 + + = (z - 1)2 (z + 1) (z - 1) (z + 1) (z - 1)2

Z{xk} = -

Therefore,

i.e. B = 1/2

 3  z  3  z  1  z   -    +  2  4  z - 1 2  (z - 1)  4  z + 1

Taking the inverse z-transform gives: {xk} = -

i.e.

{xk} = −

   3 3 -1  z  1 -1  z  - Z -1  z  + Z  Z    2    z + 1  (z - 1)  4  (z - 2)  2 4

1 3 k 3 2 + k − (−1)k from 6 and 3 of Table 159.1 4 2 4

( )

Application: Solve the difference equation: xk+2 – 3xk+1 + 2xk = 1 given that x0 = 0 and x1 = 1 Taking the z-transform of both sides of the equation gives: Z{xk+2 – 3xK+1 + 2xk} = Z{1} i.e.

Z{xk+2} – 3Z{xK+1} + 2Z{xk} = Z{1}

Z-transforms

463

Using the first shift theorem and Z {xk} = F(z) gives:

( z 2F (z ) - z 2x 0 - zx1) - 3 ( zF (z ) - zx 0 ) + 2F (z ) = z z- 1 x0 = 0 and x1 = 1, hence

( z 2F (z ) - z 2 (0) - z (1)) - 3 ( zF (z ) - z (0)) + 2F (z ) = z z- 1

i.e.

z 2F (z ) - z - 3zF (z ) + 2F (z ) =

and

( z 2 - 3z + 2) F (z ) = z z- 1 + z z z z + z (z - 1) z + z2 - z z2 + = = = z -1 1 z -1 z -1 z -1

= F(z) =

Hence,

z z -1

z2

=

( z 2 - 3z + 2) (z - 1)

z2 z2 = (z - 2)(z - 1)(z - 1) (z - 2)(z - 1)2

F (z ) z = z (z - 2)(z - 1)2

and

Using partial fractions, let

z A B C = + + (z - 2) (z - 1) (z - 2)(z - 1)2 (z - 1)2 =

A(z - 1)2 + B (z - 2)(z - 1) + C (z - 2) (z - 2)(z - 1)2

z = A(z - 1)2 + B(z - 2)(z - 1) + C(z - 2)

from which, Letting z = 2 gives:

2 = A(1)2

Letting z = 1 gives:

1 = C(- 1) i.e.

2

Equating z coefficients gives: 0 = A + B Therefore, or

i.e. A = 2

i.e.

C=-1 B=-2

F (z ) 2 2 1 = z (z - 2) (z - 1) (z - 1)2 F(z) =

2z 2z z (z - 2) (z - 1) (z - 1)2

Taking the inverse z-transform of F(z) gives:  z       - 2Z =1  z  - Z -1  z  Z -1 F (z ) = 2Z -1   (z - 1)   (z - 1)2   (z - 2)  = 2(2k) - 2(1) -k from 2, 6 and 7 of Table 159.1 i.e.

{xk} = 2k+1 – 2 – k

Section 16 Fourier series

Why are Fourier series important? A Fourier series changes a periodic function into an infinite expansion of a function in terms of sines and cosines. In engineering and physics, expanding functions in terms of sines and cosines is useful because it makes it possible to more easily manipulate functions that are just too difficult to represent analytically. The fields of electronics, quantum mechanics and electrodynamics all make great use of Fourier series. The Fourier series has become one of the most widely used and useful mathematical tools available to any scientist. There are many practical uses of Fourier series in science and engineering. The technique has practical applications in the resolution of sound waves into their different frequencies, for example, in an MP3 player, in telecommunications and Wi-Fi, in computer graphics and image processing, in climate variation, in water waves, and much more. Any field of physical science that uses sinusoidal signals, such as engineering, applied mathematics, and chemistry will make use of Fourier series. Applications are found in electrical engineering, such as in determining the harmonic components in ac waveforms, in vibration analysis, acoustics, optics, signal processing, image processing and in quantum mechanics. If it can be found ‘on sight’ that a function is even or odd, then determining the Fourier series becomes an easier exercise. In communications, Fourier series are essential to understanding how a signal behaves when it passes through filters, amplifiers and communications channels. In astronomy, radar and digital signal processing, Fourier analysis is used to map the planet. In geology, seismic research uses Fourier analysis, and in optics, Fourier analysis is used in light diffraction. In music, if a note has frequency f, integer multiples of that frequency, 2f, 3f, 4f, and so on, are known as harmonics. As a result, the mathematical study of overlapping waves is called harmonic analysis; this analysis is a diverse field and may be used to produce a Fourier series. Signal processing, medical imaging,

Fourier series

465

astronomy, optics, and quantum mechanics are some of the fields that use harmonic analysis extensively. A Fourier series may be represented not only as a sum of sines and cosines, but as a sum of complex exponentials. The complex exponentials provide a more convenient and compact way of expressing the Fourier series than the trigonometric form. It also allows the magnitude and phase spectra to be easily calculated. This form is widely used by engineers, for example, in circuit theory and in control theory.

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Mathematics Pocket Book for Engineers and Scientists

Chapter 163

Fourier series for periodic functions of period 2π

The basis of a Fourier series is that all functions of practical significance which are defined in the interval π  x π can be expressed in terms of a convergent trigonometric series of the form: f(x)  a0  a1 cos x  a2 cos 2x  a3 cos 3x  ….  b1 sin x  b2 sin 2x  b3 sin 3x  .. when a0, a1, a2, … b1, b2, … are real constants, i.e. f(x)  a0 

∞

∑ ( an cos nx  bn sin nx )

(1)

n1

where for the range π to π:

and

a0 

1 2π

an 

1 π

∫π f(x)cosnx

bn 

1 π

∫π f(x)sin nx

π

∫π f(x) dx π

π

dx (n  1, 2, 3, … )

dx (n  1, 2, 3, … )

Fourier series provides a method of analysing periodic functions into their constituent components. Alternating currents and voltages, displacement, velocity and acceleration of slider-crank mechanisms and acoustic waves are typical practical examples in engineering and science where periodic functions are involved and often require analysis. For an exact representation of a complex wave, an infinite number of terms are, in general, required. In many practical cases, however, it is sufficient to take the first few terms only. Application: Obtain a Fourier series for the periodic function f(x) defined as: k, when  π 〈 x 〈 0 f(x)   k, when 0 〈 x 〈 π (The function is periodic outside of this range with period 2π) The square wave function defined is shown in Figure 163.1. Since f(x) is given by two different expressions in the two halves of the range the integration is performed in two parts, one from π to 0 and the other from 0 to π.

Fourier series

467

f(x) k

π

0

π

2π

x

k

Figure 163.1

From above:

a0 

1 

π

1 2π

π

0

∫π f(x) dx  2π  ∫π k dx  ∫0 1 0  [kx]π [kx] π 0 2π 0 

{

 k dx  

}

[a0 is in fact the mean value of the waveform over a complete period of 2π and this could have been deduced on sight from Figure 163.1] an 

1 π

π

∫π f(x) cos nx

π  1  0 k cos nx dx  k cos nx dx   0 π  π  0 π     k sin nx   1 k sin nx           n   π   n    0   π  0

∫

dx 

∫

Hence a1, a2, a3, … are all zero (since sin 0  sin(nπ)  sin nπ  0), and therefore no cosine terms will appear in the Fourier series. bn 

1 π

1 

π

0

 k sin nx dx   π   k cos nx         n   0  π

∫π f(x) sin nx dx  π  ∫π k sin nx dx  ∫0 

0  1   k cos nx    π   n π 

When n is odd: bn 

k π

       1   1    1   1   k  2 2  4 k                      n   n    n   n   π  n n  nπ      

Hence, b1 

4k 4k 4k , b3  , b5  , and so on 3π 5π π

When n is even: bn 

k π

  1 1   1  1               0   n n   n  n      

468

Mathematics Pocket Book for Engineers and Scientists

Hence, from equation (1), the Fourier series for the function shown in Figure 163.2 is given by: f(x)  a0 

i.e.

f(x) 





n1

n1

∑ (an cos nx  bn sin nx)  0  ∑ (0  bn sin nx) 4k 4k 4k sin x  sin 3x  sin 5x  .. 3π 5π π

    4k4k   1111 1111 … … ...... f(x) sisn sin xx sisin x3+x +sin ssin isn x5x  4 4sin xx 5x   i.e. f(fx()xf(x) ) in sn3x in33x in55x  sin   3333 5555 ππ    If k  π in the above Fourier series then:   1 1 f(x)  4  sin x  sin 3x + sin 5x  …   3 5 4 sin x is termed the first partial sum of the Fourier series of f(x),    4 sin x  4 sin 3x  is termed the second partial sum of the Fourier series, and   3     4 sin x  4 sin 3x  4 sin 5x   is termed the third partial sum, and so on.  3 5    4 Let P1  4 sin x, P2   4 sin x  sin 3x  and  3    4 4 P3   4 sin x  sin 3x  sin 5x  .   3 5 Graphs of P1, P2 and P3, obtained by drawing up tables of values, and adding waveforms, are shown in Figures 163.2(a) to (c) and they show that the series is convergent, i.e. continually approximating towards a definite limit as more and more partial sums are taken, and in the limit will have the sum f(x)  π. Even with just three partial sums, the waveform is starting to approach the rectangular wave the Fourier series is representing. Thus, a rectangular wave is comprised of a fundamental and an infinite number of odd harmonics.

Fourier series f (x)

P1

f (x)

4 

f (x)

f (x) 

P2

P1 

0

/2



/2

x 

0

/2

469

/2



x

4/3 sin 3x 

 4

(b)

(a) P2

f(x)

f (x)

π P3

π/2 π

0

π/2

π

x

4/5 sin 5x π (c)

Figure 163.2

Chapter 164

Fourier series for a non-periodic function over range 2π

If a function f(x) is not periodic then it cannot be expanded in a Fourier series for all values of x. However, it is possible to determine a Fourier series to represent the function over any range of width 2π. For determining a Fourier series of a non-periodic function over a range 2π, exactly the same formulae for the Fourier coefficients are used as in equation (1), page 466.

Application: Determine the Fourier series to represent the function f(x)  2x in the range π to π The function f(x)  2x is not periodic. The function is shown in the range π to π in Figure 164.1 and is then constructed outside of that range so that it is periodic of period 2π (see broken lines) with the resulting saw-tooth waveform.

470

Mathematics Pocket Book for Engineers and Scientists f(x) f(x)  2x 2 0

2 



2

3 x

2

Figure 164.1 ∞

For a Fourier series: f(x)  a0  π

∑ (an cos nx  bn sin nx) n1

π

1 1 1  2 π f(x) dx  2x dx  x 0 2π π 2π π 2π  π 1 π 1 π an  f(x) cos nx dx  2x cos nx dx π  π π π 2  x sin nx sin nx     dx  by parts (see Chapter 121) n π n π

∫

a0 

∫

∫

∫

∫



2 π

bn 

1 π



2 π



2 π



2 π



2 π

π  x sin nx cos nx  2  cos nπ   cos n(π)      0    0    0   n  2 2  π  n π n   n2   π 1 π f(x) sin nx dx  2x sin nx dx π π π π  x cox nx   cos nx      dx   by parts     n n π  π  x cos nx sin nx     n n2 π   π cos nπ sin nπ   (π) cos n(π) sin n(π )        n n n2   n2    π cos nπ π cos(nπ )  4    cos nπ since cos nπ  cos (nπ)   n n n  

∫

When n is odd, bn 

∫

∫

4 4 4 . Thus b1  4, b3  , b5  , and so on. 3 5 n

4 4 4 4 When n is even, bn   . Thus b2   , b4   , b6   , and so on. 4 6 n 2 4 4 4 4 Thus , f(x)  2x  4 sin x  sin 2x  sin 3x  sin 4x  sin 5x 2 3 4 5 4  sin 6x  .. 6

Fourier series

i.e.

2x  4 ( sin x 

471

1 1 1 sin 2x  sin 3x  sin 4x 2 3 4 1 1  sin 5x  sin 6x  ...) 5 6

for values of f(x) between π and π.

Chapter 165

Even and odd functions

A function y  f(x) is said to be even if f(x)  f(x) for all values of x. Graphs of even functions are always symmetrical about the y-axis (i.e. a mirror image). Two examples of even functions are y  x2 and y  cos x as shown in Figure 63.2, page 198. A function y  f(x) is said to be odd if f(x)  f(x) for all values of x. Graphs of odd functions are always symmetrical about the origin. Two examples of odd functions are y  x3 and y  sin x as shown in Figure 63.3, page 198. Many functions are neither even nor odd, two such examples being y  ln x and y  ex.

Fourier cosine series The Fourier series of an even periodic function f(x) having period 2π contains cosine terms only (i.e. contains no sine terms) and may contain a constant term. Hence f(x)  a0 



∑ an cos nx

(1)

n1

where

a0 

1 2π

and

an 

1 π

π

1

π

∫π f(x) dx  π ∫ 0 π

f(x) dx (due to symmetry) 2

π

∫π f(x) cos nx dx  π ∫ 0

f(x) cos nx dx

Fourier sine series The Fourier series of an odd periodic function f(x) having period 2π contains sine terms only (i.e. contains no constant term and no cosine terms). Hence



f(x) 

∑ bn sin nx

(2)

n1

where

bn 

1 π

π

2

π

∫π f(x) sin nx dx  π ∫ 0

f(x) sin nx dx

Mathematics Pocket Book for Engineers and Scientists

472

Application: Determine 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  〈x〈π 2, when 2 

and has a period of 2π

The square wave shown in Figure 165.1 is an even function since it is symmetrical about the f(x) axis. f(x) 2

0

π π/2

3π/2

π/2

3π/2 2π x

π

2

Figure 165.1

Hence from equation (1), the Fourier series is given by: 

∑ an cos nx

f(x)  a0 

a0 

an 

2 π

(i.e. the series contains no sine terms).

n1

π

∫0

1 π

π

∫0

f(x) cos nx dx 

π  1  π / 2 2 dx  2 dx    π/2 π  0 π   1    π / 2     2x   2x     π / 2  π    0 1   (π)   (2π)  (π)    π 0

f(x) dx 

2   π 

π/2

∫0

∫

2 cos nx dx 

∫

π

   π/2   π     sin nx     sin nx    4  sin(π / 2)n  0        n    n  π / 2  π  n   0     sin(π /2)n    0    n  4  2 sin(π /2)n  8  nπ     sin    2  π  n  πn 



4 π



∫π / 2 2 cos nx dx 

Fourier series

473

When n is even, an  0 When n is odd, an  an 

and Hence, a1 

8 for n  1, 5, 9,… πn 8 for n  3, 7, 11,… πn

8 8 8 ,a  ,a  , and so on π 3 3π 5 5π

Hence the Fourier series for the waveform of Figure 165.1 is given by:   π1 π  2 π 11  π / 2 81  π / 1 1 1 π ...  f(x) dx cos 7x 2 dx f(xdx )  a0 c os x2 cosdx3x a0  f(x) dx   2 dx  cos 5x2   π5  0 π/2 ππ  0 π 0 3 7 π/2 π 0  / 2 π π π/2 π   1   1     2x   2x     2x  0  2x  π / 2   series  π / 2  for Application: Obtain π  the square waveshown in Figure π   the 0 Fourier 165.2. 1 1    (π)   (2π)  (π)    (π)   (2π)  (π)        π π f(x) 0 0

∫

∫

∫

∫

∫

∫

2 π

π

0

2π

3π x

2

Figure 165.2

The square wave is an odd function since it is symmetrical about the origin. Hence, from equation (2), the Fourier series is given by: 

f(x) 

∑ bn sin nx n1

2, when  π 〈 x 〈 0 The function is defined by: f(x)    2, when 0 〈 x 〈 π π π π nx  2 ππ 4   cos  nx dx  4   cos nx  f(x) sin nx b dx  2 2 sinsin nxnx dxdx  2 f(x) 2 sin  0n  n  0 π 00 ππ n π π   0 0 4   cos nπ   1 4  4cos nπ   1  4        (1  cos  nπ)   (1  cos nπ)      n  πn n n π  πn n π   

bn 

2 π

∫

π

∫

When n is even, bn  0. When n is odd, bn  Hence, b1 

∫

4 8 [1  (1)]  πn πn

8 8 8 ,b  ,b  , and so on π 3 3π 5 5π

474

Mathematics Pocket Book for Engineers and Scientists

Hence the Fourier series is: f(x) 

Chapter 166

8 1 1 1 (sin x  sin 3x  sin 5x  sin 7x  ...) π 3 5 7

Half range Fourier series

When a function is defined over the range say 0 to π instead of from 0 to 2π it may be expanded in a series of sine terms only or of cosine terms only. The series produced is called a half-range Fourier series. When a half range cosine series is required then: 

f(x)  a0  ∑ an cos nx n1

where a0 

1 π

π

∫0

f(x) dx and an 

2 π

π

∫0

f(x) cos nx dx

If a half-range cosine series is required for the function f(x)  x in the range 0 to π then an even periodic function is required. In Figure 166.1, f(x)  x is shown plotted from x  0 to x  π. Since an even function is symmetrical about the f(x) axis the line AB is constructed as shown. If the triangular waveform produced is assumed to be periodic of period 2π outside of this range then the waveform is as shown in Figure 166.1. f(x) B

π

f(x)  x

A 2π

0

π

π

2π x

Figure 166.1

When a half-range sine series is required then the Fourier coefficient bn is calculated as earlier, i.e. 

f(x) 

∑ bn sin nx n1

where

bn 

2 π

π

∫0

f(x) sin nx dx

If a half-range sine series is required for the function f(x)  x in the range 0 to π then an odd periodic function is required. In Figure 166.2, f(x)  x is shown

Fourier series

f(x) π

475

f(x)  x

C 0

2π π D

π

3π x

2π

π

Figure 166.2

plotted from x  0 to x  π. Since an odd function is symmetrical about the origin the line CD is constructed as shown. If the sawtooth waveform produced is assumed to be periodic of period 2π outside of this range, then the waveform is as shown in Figure 166.2

Application: Determine the half-range Fourier cosine series to represent the function f(x)  x in the range 0  x π The function is shown in Figure 166.1.

an 

π

1 π

When f(x)  x, a0 

2 π

∫0

π

∫0

f(x) dx 

1 π

π

∫0

f(x) cos nx dx  π

x dx  2 π

π

∫0

1 π

 x2  π   π  2 2   0

x cos nx dx

 x sin nx cos nx    by parts  n n2  0  2  π sin nπ cos nπ   cos 0       0  2   2  π  n n   n  2 cos nπ cos 0  2   0   2  (cos nπ  1)   n2 n  π n2 

2 π

When n is even, an  0 When n is odd, an 

2 4 (1  1)  2 π n2 πn

Hence, a1  4 , a3  4 , a5  4 , and so on π π32 π52 Hence, the half-range Fourier cosine series is given by: f(x)  x 

 4 1 1 π   cos x  2 cos 3x  2 cos 5x  ...  2 π  3 5

Application: Determine the half-range Fourier sine series to represent the function f(x)  x in the range 0  x π The function is shown in Figure 166.2.

476

Mathematics Pocket Book for Engineers and Scientists

When f(x)  x, bn 

2 π

π

∫0

f(x) sin nx dx 

2 π

π

∫0

x sin nx dx

π  x cos nx sin nx    2  by parts  n n 0   2  π cos nπ sin nπ  2       (0  0)    cos nπ 2  π  n n n  



2 π

When n is odd, bn 

2 2 2 2 . Hence, b1  , b3  , b5  and so on. 3 5 n 1

When n is even, bn  

2 . Hence b   2 , b   2 , b   2 and so on 2 4 6 n 2 4 6

Hence the half-range Fourier sine series is given by:   1 1 1 1 f(x)  x  2  sin x  sin 2x  sin 3x  sin 4x  sin 5x  ...   2 3 4 5

Chapter 167

Expansion of a periodic function of period L

If f(x) is a function of period L, then its Fourier series is given by:    2πnx   2πnx  f(x)  a0  ∑  an cos    bn sin     L   L   n1 

where, in the range 

a0 

1 L/2 f(x)dx, L ∫L/2

and

bn 

2 L

L/2

(1)

L L to  : 2 2 an 

2 L

L/2

 2πnx   dx L 

∫L/2 f(x)cos 

 2πnx   dx L 

∫L/2 f(x)sin 

(The limits of integration may be replaced by any interval of length L, such as from 0 to L)

Fourier series

477

Application: The voltage from a square wave generator is of the form:  0, 4 〈 t 〈 0 v(t)   10, 0 〈 t 〈 4

and has a period of 8 ms.

Find the Fourier series for this periodic function f(x) 10 0

4

8

4

8

12 t (ms)

Figure 167.1

Period L  8 ms

The square wave is shown in Figure 167.1. From above, the Fourier series is of the form:  nt   2πnt    bn sin  2π   L   L   n1  L /2 4 0   1 1 1 a0  v(t)dt  v(t)dt   0 dt  L L /2 8 4 8  4 4 1  10 t   5 0 8 

v(t)  a0 



∑  an cos 

∫

an 

2 L

∫

∫

 2πnt  2  dt  L  8

L /2

∫L /2 v(t) cos 



1 4

4



∫0 10 dt 

 2πnt   dt 8 

4

∫4 v(t) cos     

0

 πnt   dt  4 

∫4 0 cos 

4

 πnt    dt  4  

∫0 10 cos 

4     10 sin  πnt     4   1   10 [sin πn  sin 0]  0 for n  1, 2, 3, ...     πn  πn 4       for n 1, 2, 3,...  4    0

 2πnt   dt L 

bn 

2 L

∫L / 2 v(t) sin 



2 8

∫4 v(t) sin 



1   4 

L /2

 2πnt   dt 8 

4

0

 πnt   dt  4 

∫4 0 sin 

4

 πnt    dt  4  

∫0 10 sin 

4     10 cos  πnt    4   1   10 [cos πn  cos 0]     πn  πn 4         4    0

When n is even, bn  0

478

Mathematics Pocket Book for Engineers and Scientists

When n is odd, b1  b3 

20 -10 , (1  1)  π π

10 20 20 and so on (1  1)  ,b  , 3π 3π 5 5π

Thus the Fourier series for the function v(t) is given by:

v(t)  5 

20 π

   πt  1      sin    sin  3πt   1 sin  5πt   ...       4  3      4   4  5    

Application: Obtain the Fourier series for the function defined by: 0, when  2 〈 x 〈1  f(x)  5, when  1 〈 x 〈 1  0, when 1 〈 x 〈 2 The function is periodic outside of this range of period 4 The function f(x) is shown in Figure 167.2 where period, L  4. Since the function is symmetrical about the f(x) axis it is an even function and the Fourier series contains no sine terms (i.e. bn  0) f(x) 5

5 4 3 2 1 0

1

2

3

4

5

x

L4

Figure 167.2

 2πnx   L 



Thus, from equation (1), f(x)  a0 

a0 

1 L

1

L/2

∑ an cos  n1

2

∫L / 2 f(x) dx  4 ∫2 f(x) dx

 0 dx   1 10 1 5   [5x]11  [(5)  (5)]  4 4 4 2 

1   4 

1

1

2

∫2 0 dx  ∫15 dx ∫1

 2πnx   dx L 

an 

2 L

∫L / 2 f(x) cos 

=

2 4

∫2 f(x) cos 



1   2 

L/2

 2πnx   dx 4 

2

1

 πnx   dx  2 

∫2 0 cos 

1

 πnx   dx  2 

∫15 cos 

∫1

2

 πnx    dx  0 cos   2   

L

∫L / 2 f(x) cos 

=

2 4

∫2 f(x) cos 



1   2 

an 

 dx 

L

 2πnx   dx 4 

2

1

 πnx   dx  2 

∫2 0 cos 

  sin πnx 5 2   2  πn  2 

Fourier series

 πnx   dx  2 

1

∫15 cos 

∫1

2

1   πn   5   πn   sin    sin      2   πn   2    1

479

 πnx    dx  0 cos   2   

When n is even, an  0 When n is odd, a1  a5 

5 10 5 10 , a3  , (1  1)  (1  1)  π π 3π 3π

5 10 (1  1)  , and so on 5π 5π

Hence the Fourier series for the function f(x) is given by: f(x) 

        πx  1 5 10    cos  3πx   1 cos  5πx   1 cos  7πx   … cos             2   2  5  2  7  2  3 2 π  

Chapter 168

Half-range Fourier series for functions defined over range L

A half-range cosine series in the range 0 to L can be expanded as:   nπx  f(x)  a0  ∑ an cos    L  n1

(1)

where

a0 

1 L f(x) dx L ∫0

an 

and

2 L

L

∫0

 nπx  f(x)cos   dx  L 

A half-range sine series in the range 0 to L can be expanded as:

f(x) 

 nπx   L 



∑ bn sin  n1

where

bn 

2 L

L

∫0

 nπx  f(x)sin   dx  L 

(2)

Mathematics Pocket Book for Engineers and Scientists

480

Application: Determine the half-range Fourier cosine series for the function f(x)  x in the range 0  x  2 A half-range Fourier cosine series indicates an even function. Thus the graph of f(x)  x in the range 0 to 2 is shown in Figure 168.1 and is extended outside of this range so as to be symmetrical about the f(x) axis as shown by the broken lines. f(x)

f(x)  x

2 4

2

0

2

4

6

x

Figure 168.1 

For a half-range cosine series: f(x)  a0 

a0 

1 L

∫0

an 

2 L

∫0 f(x) cos 



2 2

L

f(x) dx 

L

∫

1 2

2

∫0

 nπx   L 

∑ an cos  n1

from equation (1)

2

x dx 

1  x 2  1 2  2  0

 nπx   dx L 

2   nπx    nπx       cos   x sin   2  nπx   2    2    dx     x cos    nπ   2  0  nπ 2         2   2    0

                    2 sin nπ cos nπ     cos 0  cos nπ 1         0          nπ 2   nπ 2    nπ 2  nπ 2     nπ           2                2    2     2   2           2  2     (cos nπ  1)  πn  When n is even, an  0, a1 

8 8 , a3  , a  8 , and so on. 5 π2 π2 32 π252

Hence the half-range Fourier cosine series for f(x) in the range 0 to 2 is given by: f(x)  1 

8 π2

         cos  πx   1 cos  3πx   1 cos  5πx   ...     2   2  5 2  2  32  

Fourier series

481

Application: Determine the half-range Fourier sine series for the function f(x)  x in the range 0  x  2 A half-range Fourier sine series indicates an odd function. Thus the graph of f(x)  x in the range 0 to 2 is shown in Figure 168.2 and is extended outside of this range so as to be symmetrical about the origin, as shown by the broken lines. f(x)

f(x)  x

2 4

0

2

2

4

6

x

2

Figure 168.2 

For a half-range sine series: f(x) 

n1

bn 

2 L

L

 nπx   dx  2 L  2

∫0 f(x) sin 

 nπx   from equation (2) L 

∑ bn sin  2

∫0

 nπx   dx x sin   L 

  nπx     x cos   2       nπ        2  

2  nπx     sin    2     nπ 2     2   0

 2 cos nπ 4 sin nπ   sin 0  2 cos nπ     cos nπ      0  2   nπ nπ  nπ 2   nπ      nπ                  2   2   2    2      4 4 4 4 4 4 , b3  , and so on (1)  , b2  (1)  (1)  π π 2π 2π 3π 3π  3πx   2πx  1 4   πx  1  series  in sthe  sinsine f(x)  sin  in  range 0 to 2 is given by: Thus the half-range  Fourier  2   3  2  π   2  2    4πx      πx  1  1 4   ...    sin  2πx   1 sin  3πx  sin  f(                                                                                  x)   sin      2   2  4  2  3 π   2  2    4πx  1   ...                                                                                    sin   4  2   Hence, b1 

482

Mathematics Pocket Book for Engineers and Scientists

Chapter 169

The complex or exponential form of a Fourier series

The form used for the Fourier series considered previously consisted of cosine and sine terms. However, there is another form that is commonly used – one that directly gives the amplitude terms in the frequency spectrum and relates to phasor notation. This form involves the use of complex numbers (see Chapter 69). It is called the exponential or complex form of a Fourier series. e jθ  cos θ  j sin θ e jθ  ejθ  2 cos θ

and

cos  

from which,

ejθ  cos θ  j sin θ e j  ej 2

e j  ej 2j The complex or exponential form of the Fourier series. e jθ  ejθ  2j sin θ

sin  

from which,



f(x) 

∑

n

where

cn 

1 L

L 2 L  2

∫

c ne

j

f(x) e

j

2πnx L

2πnx L

(1) (2)

(3)

dx

(4)

Care needs to be taken when determining c0. If n appears in the denominator of an expression the expansion can be invalid when n  0. In such circumstances it is usually simpler to evaluate c0 by using the relationship: c 0  a0 

1 L

L

∫2L f(x) dx

(5)

2

Application: Determine the complex Fourier series for the function defined by:

0, when  2  x  1  f( x )  5, when  1  x  1  0, when 1  x  2  The function is periodic outside this range of period 4. This is the same Application Problem as on page 478 and we can use this to demonstrate that the two forms of Fourier series are equivalent. The function f(x) was shown in Figure 167.2, where the period, L  4. From equation (3), the complex Fourier series is given by: 

f(x) 

∑ cn e

n

j

2πnx L

Fourier series L

1 L

where cn is given by: cn 

j

∫2L f(x) e

2πnx L dx

483

(from equation (4))

2

With reference to Figure 167.2, when L  4, cn 

 1   4   

1

1

j

∫2 0 dx  ∫15 e

1 4



5   jπ2n e j2πn 



πn 5 sin πn 2



∫15 e



2

∫1

 0 dx   

 jπnx 1 1    5   jπ2nx  5 e 2      e  4  jπn  j2πn  1     2 1 πn   j πn j   jπn   5  e 2  e 2  2   e     πn   2j  

jπnx 2 dx



1

2πnx 4 dx

(from equation (2))

Hence, from equation (3), the complex form of the Fourier series is given by: 

f(x) 

∑

n

cn e

j

2πnx L





5 πn j sin e n 2 π n

∑

πnx 2

(6)

Let us show how this result is equivalent to the result involving sine and cosine terms determined on page 479. From equation (5), L 2 L  2

c 0  a0 

1 L

Since cn 

5 n 5 π 5 , then c1  sin  sin π π πn 2 2

∫

f(x) dx 

1 4

1

5

5

5

∫15 dx  4 [x]11  4 [1 1]  2

5 sin π  0 (in fact, all even terms will be zero since sin nπ  0) 2π 5 πn 5 3π 5 sin sin c3    2 3π 2 3π πn c2 

By similar substitution, c 5 

Similarly, c1 

5 5 , c7   , and so on. 5π 7π

5 π 5 sin  π 2 π c2  

2π 5  0  c4  c6 , and so on sin 2π 2

c3  

3π 5 5  sin 3π 3π 2

c5  

5 5π 5  sin , and so on. 5π 5π 2

484

2π 5 Mathematics Pocket c  Book  for  0and  Scientists c c sinEngineers 2

2π

4

2

6 ,

c3  

3π 5 5  sin 3π 3π 2

c5  

5 5π 5  sin , and so on. 5π 5π 2

and so on

Hence, the extended complex form of the Fourier series shown in equation (6) becomes: f(x) 

5 5 j π2x 5 j 32πx 5 j 5π2x 5 j 7π2x ....  e  e  e  e  2 π 3π 5π 7π

5 j π2x 5 j 32πx 5 j 5π2x 5 j 7π2x .... e  e  e  e  3π 5π 7π π πx  3πx  5πx  j j j   5  j 5πx  5 5  j πx 5  j 3πx    e 2  e 2   e 2  e 2     e 2  e 2   ....   3π   5π   2 π  πx  3πx 3πx   j πx  j j    2  j 2 5 5 e  e 2  5 e  e 2  (2)    (2)        2 π 2 2  3π      

5πx  j 5πx j  2 e 2 5 e              (2)   5π 2  5 x   3πx  10  πx  10 5 10  + cos  π   ...   cos    cos   2      2 π  2  5π  2  3π

 i.e.

f(x) 

 

    ....   

(from equation 1)

 

      πx  1 5 10    cos  3πx   1 cos  5πx   ....   cos    2   2  5  2  3 2 π   which is the same as obtained on page 479.

Hence,



∑

n

5 nπ j sin e 2 πn

πnx 2

is equivalent to:

  πx  1  3πx  1  5πx  5 10   cos    cos    cos    ....      2  3  2  5  2  2 π  

Symmetry relationships If even or odd symmetry is noted in a function, then time can be saved in determining coefficients. The Fourier coefficients present in the complex Fourier series form are affected by symmetry.

Fourier series

485

For even symmetry: cn 

an 2



2 L

∫0

L 2

 2πnx   dx f(x)cos   L 

(7)

For odd symmetry: cn 

jbn 2

 j

2 L

∫0

L 2

 2πnx   dx f(x) sin   L 

(8)

For example, in the Application Problem on page 482, the function f(x) is even, since the waveform is symmetrical about the f(x) axis. Thus equation (7) could have been used, giving: cn 

2 L

L

 2πnx   dx L 

∫02 f(x) cos 

   2πnx   1  dx  1  5 cos  πnx  dx  f(x) cos      4  0 2   2  1   πnx      sin    5   2   5 nπ 5  2   nπ    0  sin      sin  πn π 2 π n 2 2 n 2        2   0

  

2 4

2

∫0

∫

2

∫1

 0 dx   

which is the same answer as on page 483; however, a knowledge of even functions has produced the coefficient more quickly.

Application: Obtain the Fourier series, in complex form, for the square wave shown in Figure 169.1 f(x) 2

π

π

0

2π

3π x

2

Figure 169.1

The square wave shown in Figure 169.1 is an odd function since it is symmetrical about the origin. The period of the waveform, L  2π. Thus, using equation (8): cn  j

2 L

L

 2πnx   dx L 

∫02 f(x) sin 

486

Mathematics Pocket Book for Engineers and Scientists

2 2π 2  j π  j

π

∫0 π

∫0

 2πnx   dx 2 sin   2π 

sin nx dx π

2   cos nx   n π  0 2  j ( cos πn)  ( cos 0) πn  j

(

i.e.

c n  j

)

2  1  cos πn   πn 

(9)

From equation (3), the complex Fourier series is given by: 

f(x) 

∑ cn e

j

2πnx L





2

∑ j nπ (1  cos nπ)e jnx

(10)

n

n

This is the same as that obtained on page 474, i.e. f(x) 

 8  1 1 1  sin x  sin 3x  sin 5x  sin 7x  ...  π  3 5 7

which is demonstrated below. From equation (9), cn  j

2 (1 cos nπ ) nπ

When n  1, c1  j

2 2 j4 (1 cos π )  j π (1 1)   π (1)π

When n  2, c2  j

2 (1 cos 2π )  0 ; in fact, all even values of cn will be zero. 2π

When n  3, c3  j

2 2 j4 (1 cos 3π )  j 3π (1 1)   3π 3π

By similar reasoning, c5   When n  1, c1  j When n  3, c3  j

j4 j4 , c   , and so on. 5π 7 7π

2 2 j4 1  cos (π )  j (1  1)   (1)π π π

(

)

2 2 j4 (1 cos(3π))  j 3π (1 1)  3π (3)π

By similar reasoning, c5  

j4 j4 , and so on. , c  5π 7 7π

Since the waveform is odd, c0  a0  0 

From equation (10), f(x) 

2

∑ j nπ (1 cos nπ ) e jnx

n

Fourier series

487

Hence, j4 jx j4 j3x j4 j5x j4 j7x e  e  e  e … 3π 5π 7π π j4 j4 j3x j4 j5x j4 j7x                                   ejx    … e e e 3π 5π 7π π

f(x)  

 j4   j4 j4 j4 3x  e3x  e    e jx  e jx       π 3π π   3π  j4 j4 5x  .... e5x  e                                                                  5π 5π j4 jx j4 3x j4 5x e  ejx  e  e3x  e  e5x  .... π 3π 5π 4 jx 4 4 5x (e  ejx )  (e3x  e3x )  (e  e5x )  ....  jπ j3π j5π by multiplying top and bottom by j



i.e.

(

)

(

)

(

)



8  e jx  ejx  8  e j3x  ej3  8  e j5x  ej5x  ....            5π   3π  2j 2j 2j π 



8 8 8 sin x  sin 3x  sin 5x  .... π 3π 3x

f(x) 

from equation 2, page 482

 8  1 1 1  sin x  sin 3x  sin 5x  sin 7x  ...  π  3 5 7 

Hence, f(x) 

by rearranging

2

∑ j nπ (1  cos nπ) e jnx

n

≡

 8  1 1 1  sin x  sin 3x  sin 5x  sin 7x  ...  3 5 7 π 

Chapter 170

A numerical method of harmonic analysis

Many practical waveforms can be represented by simple mathematical expressions, and, by using Fourier series, the magnitude of their harmonic components determined, as above. For waveforms not in this category, analysis may be achieved by numerical methods. Harmonic analysis is the process of resolving a periodic, non-sinusoidal quantity into a series of sinusoidal components of ascending order of frequency. The trapezoidal rule can be used to evaluate the Fourier coefficients, which are given by: a0 ≈

1 p

p

∑ yk k 1

(1)

488

Mathematics Pocket Book for Engineers and Scientists

2 P

an ≈ bn ≈

2 P

p

∑ yk cos nxk

(2)

k 1 p

∑ yk k 1

(3)

sin nx k

Application: A graph of voltage V against angle θ is shown in Figure 170.1. Determine a Fourier series to represent the graph.

Voltage (volts)

y10 80 60 40 y 1 y2 20

0 20 40 60 80

90

180

y3 y4 y5 y6

y8

y9

y11 y12

270 y7

360  degrees

Figure 170.1

The values of the ordinates y1, y2, y3, …. are 62, 35, 38, 64, 63, 52, 28, 24, 80, 96, 90 and 70, the 12 equal intervals each being of width 30°. (If a larger number of intervals are used, results having a greater accuracy are achieved). The voltage may be analysed into its first three constituent components as follows: The data is tabulated in the proforma shown in Table 170.1. p

From equation (1), a0 

1 p

∑ yk  12 (212)  17.67

From equation (2), an 

2 p

∑ yk cos nxk

hence a1 

(since p  12)

p

k1

2 (417.94)  69.66 12 a2 

2 (39)  6.50 12

From equation (3), bn  hence

1

k1

2 p

and

a3 

2 (49)  8.17 12

and

b3 

2 (55)  9.17 12

p

∑ yk sin nxk k1

2 b1  (278.53)  46.42 12 b2 

2 (29.43)  4.91 12

1

0.866

0.5

180 52

210 28

24

80

96

90

70

240

270

300

330

360

Y6

Y7

Y8

Y9

Y10

Y11

y12

k=1

∑ yk  212

12

0.866

150 63

Y5

70

77.94

48

0

12

24.25

52

54.56

k =1

 417.94

∑ yk cos θk

12

1

0.866

0.5

0

0.5

120 64

Y4

32

0

53.69

0

0.866

V cos 

90 38

Y3

35

62

cos 

17.5

60

Y2

V

0.5

30

Y1

Ordinates 

Table 170.1

0

k =1

 278.53

0

45

83.14

80

20.78

∑ yk sin θk

12

0

0.5

0.866

1

0.866

14

0

0.5

31.5

55.42

38

30.31

31

V sin 

0.5

0.866

1

0.866

0.5

sin  31

70

45

48

80

12

14

52

31.5

32

38

17.5

k =1

 39

∑ yk cos 2θk

12

1

0.5

0.5

1

0.5

0.5

1

0.5

0.5

1

0.5

0.5

cos 2 V cos 2

0

77.94

83.14

0

20.78

24.25

0

54.56

55.42

0

30.31

53.69

V sin 2

k =1

 29.43

∑ yk sin 2θk

12

0

0.866

0.866

0

0.866

0.866

0

0.866

0.866

0

0.866

0.866

sin 2 0

1

70

0

96

0

0

0

12

0

1

0

1

0

1

0

90

0

80

0

28

0

63

0 1

38

0

62

1

0

1

sin 3 Vsin 3

k =1

 49

k =1

 55

∑ yk cos 3θk ∑ yk sin 3θk

12

0

1

0

24

0 1

52 0

0

1

64

0

0

35

1

0

1

0

cos 3 V cos 3

Fourier series 489

490

Mathematics Pocket Book for Engineers and Scientists

Substituting these values into the Fourier series: 

f(x)  a0 

∑ (an cos nx  bn sin nx) n1

gives: v  17.67  69.66 cos   6.50 cos 2  8.17 cos 3  … 46.42 sin   4.91 sin 2  9.17 sin 3  ...

(4)

Note that in equation (4), (46.42 sinθ  69.66 cosθ) comprises the fundamental, (4.91 sin 2θ  6.50 cos 2θ) comprises the second harmonic and (9.17 sin 3θ  8.17 cos 3θ) comprises the third harmonic. It is shown in Chapter 54 that: a sin t  b cos t  R sin(t  ) where a  R cos , b  R sin , R  a2  b2

and   tan1 b a

For the fundamental, R  (46.42)2  (69.66)2  83.71 If a  R cos , then cos  

a -46.42  R 83.71

which is negative,

b 69.66 which is positive.  R 83.71 The only quadrant where cos  is negative and sin  is positive is the second quadrant. and if b  R sin , then sin  

Hence,   tan1

b 69.66 2.l6 rad8  tan1  123.68 a 46.42

Thus, (46.42 sin θ  69.66 cos θ)  83.71 sin(θ  2.16) By a similar method it may be shown that the second harmonic (4.91 sin 2θ  6.50 cos 2θ)  8.15 sin(2θ  0.92) and the third harmonic (9.17 sin 3θ  8.17 cos 3θ)  12.28 sin(3θ  0.73) Hence equation (4) may be re-written as: v  17.67  83.71 sin(   2.16 )  8.15 sin(2  0.92)  12.28 sin(3  0.73) volts which is the form normally used with complex waveforms.

Chapter 171

Complex waveform considerations

It is sometimes possible to predict the harmonic content of a waveform on inspection of particular waveform characteristics. 1. If a periodic waveform is such that the area above the horizontal axis is equal to the area below then the mean value is zero. Hence a0  0 (see Figure 171.1(a)).

Fourier series

491

2. An even function is symmetrical about the vertical axis and contains no sine terms (see Figure 171.1(b)). 3. An odd function is symmetrical about the origin and contains no cosine terms (see Figure 171.1(c)). 4. f(x)  f(x  π) represents a waveform which repeats after half a cycle and only even harmonics are present (see Figure 171.1(d)). 5. f(x)  f(x  π) represents a waveform for which the positive and negative cycles are identical in shape and only odd harmonics are present (see Figure 171.1(e)). f(x)

f (x)

π

0

2π x

π

a0  0

(a)

2π x

(b) Contains no sine terms

f(x)

2π π

π

0

f(x) π

0

2π x

2ππ 0

π

2π x

(d) Contains only even harmonics

(c) Contains no cosine terms f(x)

π

π

0

2π

x

(e) Contains only odd harmonics

Figure 171.1

Application: An alternating current i amperes is shown in Figure 171.2. Analyse the waveform into its constituent harmonics as far as and including the fifth harmonic, taking 30° intervals. 10 i y5

5 y1 y2 y3 y4

180 120 60

180

150 90 30 0 30 60 90120150 5 10 Figure 171.2

240

300 °

210 270 y7 y8 y9

330 360 y11 y10

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Mathematics Pocket Book for Engineers and Scientists

With reference to Figure 171.2, the following characteristics are noted: (i) The mean value is zero since the area above the θ axis is equal to the area below it. Thus the constant term, or d.c. component, a0  0 (ii) Since the waveform is symmetrical about the origin the function i is odd, which means that there are no cosine terms present in the Fourier series. (iii) The waveform is of the form f(θ)  f(θ  π) which means that only odd harmonics are present. Investigating waveform characteristics has thus saved unnecessary calculations and in this case the Fourier series has only odd sine terms present, i.e. i  b1 sin θ  b3 sin 3θ  b5 sin 5θ  .. Table 171.1 

i

sin 

Y1

30

2

0.5

Y2

60

7

0.866

Y3

90

10

Y4

120

7

0.866

Y5

150

2

0.5

Y6

180

0

Y7

210

Y8

240

Y9

Ordinate

i sin 

sin 3

1

i sin 3θ

1

i sin 5

2

0.5

1

0.866

6.06

0

0

1

10

6.06

0

0

0.866

6.06

1

1

2

0.5

1

0

0

0

0

0

0

2

0.5

1

1

2

0.5

7

0.866

6.06

0

0

1

10

0

0

1

6.06

sin 5

10

10

1

10

1

0.866

6.06

270 10

1

Y10

300

7

0.866

6.06

Y11

330

2

0.5

1

1

2

0.5

1

Y12

360

0

0

0

0

0

0

0

12

12

∑ y sin θ k

∑ y sin 3θ

k

k

k =1

k

k =1

 48.24

10

1 0.866

6.06

12

∑ y sin 5θ k

k

k =1

 12

 0.24

A proforma, similar to Table 170.1, page 489, but without the ‘cosine terms’ columns and without the ‘even sine terms’ columns is shown in Table 171.1 up to, and including, the fifth harmonic, from which the Fourier coefficients b1, b3 and b5 can be determined. Twelve co-ordinates are chosen and labelled y1, y2, y3, .. y12 as shown in Figure 171.2.

From equation (3), bn  Hence, b1 

2 p

p

∑ ik sin nθk

where p  12

k1

2 2 (48.24)  8.04 , b3  (12)  2.00 12 12

and

Fourier series

2 (0.24)  0.04 12 Thus the Fourier series for current i is given by: b5 

i  8.04 sin   2.00 sin 3  0.04 sin 5

493

Section 17

Statistics and probability

Why are statistics and probability important? Statistics is the study of the collection, organisation, analysis, and interpretation of data. It deals with all aspects of this, including the planning of data collection in terms of the design of surveys and experiments. Statistics is applicable to a wide variety of academic disciplines, including natural and social sciences, engineering, government, and business. Statistical methods can be used for summarising or describing a collection of data. Engineering statistics combines engineering and statistics. Design of experiments is a methodology for formulating scientific and engineering problems using statistical models. Quality control and process control use statistics as a tool to manage conformance to specifications of manufacturing processes and their products. Time and methods engineering use statistics to study repetitive operations in manufacturing in order to set standards and find optimum manufacturing procedures. Reliability engineering measures the ability of a system to perform for its intended function (and time) and has tools for improving performance. Probabilistic design involves the use of probability in product and system design. System identification uses statistical methods to build mathematical models of dynamical systems from measured data. System identification also includes the optimal design of experiments for efficiently generating informative data for fitting such models. In many real-life situations, it is helpful to describe data by a single number that is most representative of the entire collection of numbers. Such a number is called a measure of central tendency; the most commonly used measures are mean, median, mode and standard deviation, the latter being the average distance between the actual data and the mean. Statistics is important in the field of engineering since it provides tools to analyse collected data. For example, a chemical engineer may wish to analyse temperature measurements from a mixing tank. Statistical methods can be used to determine how reliable and reproducible the temperature measurements are, how much the temperature varies within the data set, what future temperatures of the tank may be, and how confident the engineer can be in the temperature measurements made. When performing statistical analysis on a set of data, the mean, median, mode, and standard deviation are all helpful values to calculate.

Statistics and probability

495

Engineers deal with uncertainty in their work, often with precision and analysis, and probability theory is widely used to model systems in engineering and scientific applications. There are several examples where probability is used in engineering. For example, with electronic circuits, scaling down the power and energy of such circuits reduces the reliability and predictability of many individual elements, but the circuits must nevertheless be engineered so that the overall circuit is reliable. Centres for disease control need to decide whether to institute massive vaccination or other preventative measures in the face of globally threatening, possibly mutating diseases in humans and animals. System designers must weigh the costs and benefits of measures for reliability and security, such as levels of backups and firewalls, in the face of uncertainty about threats from equipment failures or malicious attackers. Models incorporating probability theory have been developed and are continuously being improved for understanding the brain, gene pools within populations, weather and climate forecasts, microelectronic devices, and imaging systems such as computer aided tomography (CAT) scan and radar. The electric power grid, including power generating stations, transmission lines, and consumers, is a complex system with many redundancies; however, breakdowns occur, and guidance for investment comes from modelling the most likely sequences of events that could cause outage. Similar planning and analysis are done for communication networks, transportation networks, water, and other infrastructure. Probabilities, permutations and combinations are used daily in many different fields that range from gambling and games, to mechanical or structural failure rates, to rates of detection in medical screening. Uncertainty is clearly all around us, in our daily lives and in many professions. Use of standard deviation is widely used when results of opinion polls are described. The language of probability theory lets people break down complex problems, and to argue about pieces of them with each other, and then aggregate information about subsystems to analyse a whole system. The binomial distribution is used only when both of two conditions are met – the test has only two possible outcomes, and the sample must be random. If both conditions are met, then this distribution may be used to predict the probability of a desired result. For example, a binomial distribution may be used in determining whether a new drug being tested has or has not contributed to alleviating symptoms of a disease. Common applications of this distribution range from scientific and engineering applications to military and medical ones, in quality assurance, genetics and in experimental design. A Poisson distribution has several applications and is essentially a derived limiting case of the binomial distribution. It is most applicably relevant to a situation in which the total number of successes is known, but the number of trials is not. An example of such a situation would be if the mean expected number of cancer cells present per sample is known and it was required to determine the probability of finding 1.5 times that number of cells in any given sample; this is an example when the Poisson distribution would be used. The Poisson distribution has widespread applications in analysing traffic flow, in fault prediction on electric cables, in the prediction of randomly occurring accidents, and in reliability engineering. A normal distribution is a very important statistical data distribution pattern occurring in many natural phenomena, such as height, blood pressure, lengths

496

Mathematics Pocket Book for Engineers and Scientists

of objects produced by machines, marks in a test, errors in measurements, and so on. In general, when data is gathered, we expect to see a particular pattern to the data, called a normal distribution. This is a distribution where the data is evenly distributed around the mean in a very regular way, which when plotted as a histogram will result in a bell curve. The normal distribution is the most important of all probability distributions; it is applied directly to many practical problems in every engineering discipline. There are two principal applications of the normal distribution to engineering and reliability. One application deals with the analysis of items which exhibit failure due to wear, such as mechanical devices frequently the wear-out failure distribution is sufficiently close to normal that the use of this distribution for predicting or assessing reliability is valid. Another application is in the analysis of manufactured items and their ability to meet specifications. No two parts made to the same specification are exactly alike; the variability of parts leads to a variability in systems composed of those parts. The design must take this variability into account, otherwise the system may not meet the specification requirement due to the combined effect of part variability. Correlation coefficients measure the strength of association between two variables. The most common correlation coefficient, called the product-moment correlation coefficient, measures the strength of the linear association between variables. A positive value indicates a positive correlation and the higher the value, the stronger the correlation. Similarly, a negative value indicates a negative correlation and the lower the value the stronger the correlation. The general process of fitting data to a linear combination of basic functions is termed linear regression. Linear least squares regression is by far the most widely used modelling method; it is what most people mean when they say they have used ‘regression’, ‘linear regression’ or ‘least squares’ to fit a model to their data. Not only is linear least squares regression the most widely used modelling method, but it has been adapted to a broad range of situations that are outside its direct scope. It plays a strong underlying role in many other modelling methods. Estimation theory is a branch of statistics and signal processing that deals with estimating the values of parameters based on measured/empirical data that has a random component. Estimation theory can be found at the heart of many electronic signal processing systems designed to extract information; these systems include radar, sonar, speech, image, communications, control and seismology. In statistical testing, a result is called statistically significant if it is unlikely to have occurred by chance, and hence provides enough evidence to reject the hypothesis of ‘no effect’. The tests involve comparing the observed values with theoretical values. The tests establish whether there is a relationship between the variables, or whether pure chance could produce the observed results. For most scientific research, a statistical significance test eliminates the possibility that the results arose by chance, allowing a rejection of the null hypothesis. Chi-square and distribution-free tests are used in science and engineering. Chi-square is a statistical test commonly used to compare observed data with data we would expect to obtain according to a specific hypothesis. Distributionfree methods do not rely on assumptions that the data are drawn from a given probability distribution. Non-parametric methods are widely used for studying populations that take on a ranked order.

Statistics and probability

Chapter 172

497

Presentation of ungrouped data

Ungrouped data can be presented diagrammatically by: (a) pictograms, in which pictorial symbols are used to represent quantities, (b) horizontal bar charts, having data represented by equally spaced horizontal rectangles, (c) vertical bar charts, in which data are represented by equally spaced vertical rectangles, (d) percentage component bar chart, where rectangles are subdivided into values corresponding to the percentage relative frequencies of the members, and (e) pie diagrams, where the area of a circle represents the whole, and the areas of the sectors of the circle are made proportional to the parts that make up the whole. Application: The number of television sets repaired in a workshop by a technician in six, one-month periods is as shown below. Month

January

February

March

April May June

Number repaired

11

6

15

9

13

8

Present the data in a pictogram This data is represented as a pictogram as shown in Figure 172.1 where each symbol represents two television sets repaired. Thus, in January, 5 1 symbols are used to 2 represent the 11 sets repaired, in February, 3 symbols are used to represent the 6 sets repaired, and so on. Month

Number of TV sets repaired

2 sets

January February March April May June

Figure 172.1

Application: The distance in miles travelled by four salesmen in a week are as shown below. Salesmen P Distance travelled (miles) 413

Q R 264 597

Represent the data by a horizontal bar chart

S 143

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Mathematics Pocket Book for Engineers and Scientists

Salesmen

To represent these data diagrammatically by a horizontal bar chart, equally spaced horizontal rectangles of any width, but whose length is proportional to the distance travelled, are used. Thus, the length of the rectangle for salesman P is proportional to 413 miles, and so on. The horizontal bar chart depicting these data is shown in Figure 172.2.

S R Q P 0

100

200

300

400

500

600

Distance travelled, miles

Figure 172.2

Application: The number of issues of tools or materials from a store in a factory is observed for seven, one-hour periods in a day, and the results of the survey are as follows: Period Number of issues

1 2 3 4 5 6 34 17 9 5 27 13

7 6

Represent the data by a vertical bar chart

Number of issues

In a vertical bar chart, equally spaced vertical rectangles of any width, but whose height is proportional to the quantity being represented, are used. Thus the height of the rectangle for period 1 is proportional to 34 units, and so on. The vertical bar chart depicting these data is shown in Figure 172.3.

40 30 20 10 1

2

3 4 5 Periods

6

7

Figure 172.3

Application: The numbers of various types of dwellings sold by a company annually over a three-year period are as shown below. 4-roomed bungalows 5-roomed bungalows 4-roomed houses

Year 1 24 38 44

Year 2 17 71 50

Year 3 7 118 53

Statistics and probability

5-roomed houses 6-roomed houses

64 30

82 30

499

147 25

Draw a percentage component bar chart to represent the above data To draw a percentage component bar chart to present these data, a table of percentage relative frequency values, correct to the nearest 1%, is the first requirefrequency of member  100 then for ment. Since, percentage relative frequency  total frequency 4-roomed bungalows in year 1: percentage relative frequency 

24  100  12% 24  38  44  64  30

The percentage relative frequencies of the other types of dwellings for each of the three years are similarly calculated and the results are as shown in the table below. 4-roomed bungalows 5-roomed bungalows 4-roomed houses 5-roomed houses 6-roomed houses

Year 1 12% 19% 22% 32% 15%

Year 2 7% 28% 20% 33% 12%

Year 3 2% 34% 15% 42% 7%

The percentage component bar chart is produced by constructing three equally spaced rectangles of any width, corresponding to the three years. The heights of the rectangles correspond to 100% relative frequency, and are subdivided into the values in the table of percentages shown above. A key is used (different types of shading or different colour schemes) to indicate corresponding percentage values in the rows of the table of percentages. The percentage component bar chart is shown in Figure 172.4.

Percentage relative frequency

Key 100 90

6-roomed houses 5-roomed houses

80

4-roomed houses

70

5-roomed bungalows

60

4-roomed bungalows

50 40 30 20 10 1

2 Year

3

Figure 172.4

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Mathematics Pocket Book for Engineers and Scientists

Application: The retail price of a product costing £2 is made up as follows: materials 10p, labour 20p, research and development 40p, overheads 70p, profit 60p. Present this data on a pie diagram To present these data on a pie diagram, a circle of any radius is drawn, and the area of the circle represents the whole, which in this case is £2. The circle is subdivided into sectors so that the areas of the sectors are proportional to the parts, i.e. the parts that make up the total retail price. For the area of a sector to be proportional to a part, the angle at the centre of the circle must be proportional to that part. The whole, £2 or 200p, corresponds to 360°. Therefore, 10p corresponds to 360 

10 degrees, i.e. 188 200

20p corresponds to 360 

20 degrees, i.e. 368 200

and so on, giving the angles at the centre of the circle for the parts of the retail price as: 18°, 36°, 72°, 126° and 108°, respectively. The pie diagram is shown in Figure 172.5.

Research and development Labour 72° 36° 18° Materials 126° 108° Overheads Profit lp

Chapter 173

1.8°

Figure 172.5

Presentation of grouped data

Grouped data can be presented diagrammatically by: (a) a histogram, in which the areas of vertical, adjacent rectangles are made proportional to frequencies of the classes, (b) a frequency polygon, which is the graph produced by plotting frequency against class mid-point values and joining the co-ordinates with straight lines, (c) a cumulative frequency distribution, which is a table showing the cumulative frequency for each value of upper class boundary, and

Statistics and probability

501

(d) an ogive or a cumulative frequency distribution curve, which is a curve obtained by joining the co-ordinates of cumulative frequency (vertically) against upper class boundary (horizontally).

Application: The masses of 50 ingots, in kilograms, are measured correct to the nearest 0.1 kg and the results are as shown below. 8.0 8.3 7.7 8.1 7.4

8.6 7.1 8.4 7.4 8.2

8.2 8.1 7.9 8.8 8.4

7.5 8.3 8.8 8.0 7.7

8.0 8.7 7.2 8.4 8.3

9.1 7.8 8.1 8.5 8.2

8.1 8.7 7.8 8.1 7.9

7.6 8.5 8.2 7.3 8.5

8.2 8.4 7.7 9.0 7.9

7.8 8.5 7.5 8.6 8.0

Produce for this data (a) a frequency distribution for 7 classes, (b) a frequency polygon, (c) a histogram, (d) a cumulative frequency distribution, and (e) an ogive. (a) The range of the data is the member having the largest value minus the member having the smallest value. Inspection of the set of data shows that: range  9.1  7.1  2.0 The size of each class is given approximately by range number of classes If about seven classes are required, the size of each class is 2.0/7, that is approximately 0.3, and thus the class limits are selected as 7.1 to 7.3, 7.4 to 7.6, 7.7 to 7.9, and so on. The class mid-point for the 7.1 to 7.3 class is 7.35  7.05 , i.e. 7.2, for the 7.4 2 7.65  7.35 to 7.6 class is i.e. 7.5, and so on. 2 To assist with accurately determining the number in each class, a tally diagram is produced as shown in Table 173.1. This is obtained by listing the classes in the left-hand column and then inspecting each of the 50 members of the set of data in turn and allocating it to the appropriate class by putting a ‘1’ in the appropriate row. Each fifth ‘1’ allocated to a particular row is marked as an oblique line to help with final counting. A frequency distribution for the data is shown in Table 173.2 and lists classes and their corresponding frequencies. Class mid-points are also shown in this table, since they are used when constructing the frequency polygon and histogram. (b) A frequency polygon is shown in Figure 173.1, the co-ordinates corresponding to the class mid-point/frequency values, given in Table 173.2. The co-ordinates are joined by straight lines and the polygon is ‘anchored-down’ at each end by joining to the next class mid-point value and zero frequency. (c) A histogram is shown in Figure 173.2, the width of a rectangle corresponding to (upper class boundary value – lower class boundary value) and height corresponding to the class frequency. The easiest way to draw a histogram is to mark class

Mathematics Pocket Book for Engineers and Scientists

mid-point values on the horizontal scale and to draw the rectangles symmetrically about the appropriate class mid-point values and touching one another. A histogram for the data given in Table 173.2 is shown in Figure 173.2. Table 173.1 Class

Tally

7.1 to 7.3 7.4 to 7.6 7.7 to 7.9 8.0 to 8.2 8.3 to 8.5 8.6 to 8.8 8.9 to 9.1

111 1111 1111 1111 1111 1111 1111 1111 1111 1 1111 1 11

Table 173.2 Class

Class mid-point

Frequency

7.1 to 7.3 7.4 to 7.6 7.7 to 7.9 8.0 to 8.2 8.3 to 8.5 8.6 to 8.8 8.9 to 9.1

14 12 10 8 6 4 2 0

Frequency

7.2 7.5 7.8 8.1 8.4 8.7 9.0

3 5 9 14 11 6 2

Frequency polygon

7.2

7.5

7.8

8.4

8.1

8.7

9.0

Class mid-point values

Figure 173.1

8.4

Class mid-point values

8.7

9.0

9.15

8.1

8.85

7.8

8.55

7.5

8.25

7.2

7.95

Histogram

7.65

14 12 10 8 6 4 2 0

7.35

Frequency

502

Figure 173.2

Statistics and probability

503

(d) A cumulative frequency distribution is a table giving values of cumulative frequency for the values of upper class boundaries, and is shown in Table 173.3. Columns 1 and 2 show the classes and their frequencies. Column 3 lists the upper class boundary values for the classes given in column 1. Column 4 gives the cumulative frequency values for all frequencies less than the upper class boundary values given in column 3. Thus, for example, for the 7.7 to 7.9 class shown in row 3, the cumulative frequency value is the sum of all frequencies having values of less than 7.95, i.e. 3  5  9  17, and so on. (e) The ogive for the cumulative frequency distribution given in Table 173.3 is shown in Figure 173.3. The co-ordinates corresponding to each upper class boundary/ cumulative frequency value are plotted and the co-ordinates are joined by straight lines (– not the best curve drawn through the co-ordinates as in experimental work). The ogive is ‘anchored’ at its start by adding the co-ordinate (7.05, 0). Table 173.3 1 Class

2 Frequency

7.1–7.3 7.4–7.6 7.7–7.9 8.0–8.2 8.3–8.5 8.6–8.8 8.9–9.1

3 5 9 14 11 6 2

3 Upper class boundary

4 Cumulative frequency

Less than 7.35 7.65 7.95 8.25 8.55 8.85 9.15

3 8 17 31 42 48 50

Cumulative frequency

50 40 30 20 10 7.05 7.35 7.65 7.95 8.25 8.55 8.85 9.15 Upper class boundary values in kilograms

Figure 173.3

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Mathematics Pocket Book for Engineers and Scientists

Chapter 174

Measures of central tendency

(a) Discrete data mean value, x 

∑x

n the median is the middle term of a ranked set of data, the mode is the most commonly occurring value in a set of data, and  standard deviation, σ    

∑ ( x -x )2  n

 

Application: Find the median of the set {7, 5, 74, 10} The set: {7, 5, 74, 10} is ranked as {5, 7, 10, 74}, and since it contains an even number of members (four in this case), the mean of 7 and 10 is taken, giving a median value of 8.5 Application: Find the median of the set {3, 81, 15, 7, 14} The set: {3, 81, 15, 7, 14} is ranked as {3, 7, 14, 15, 81} and the median value is the value of the middle member, i.e. 14 Application: Find the modal value of the set {5, 6, 8, 2, 5, 4, 6, 5, 3} The set: {5, 6, 8, 2, 5, 4, 6, 5, 3} has a modal value of 5, since the member having a value of 5 occurs three times. Application: Find the mean, median and modal values for the set {2, 3, 7, 5, 5, 13, 1, 7, 4, 8, 3, 4, 3} For the set {2, 3, 7, 5, 5, 13, 1, 7, 4, 8, 3, 4, 3} mean value, _ 2  3  7  5  5  13  1  7  4  8  3  4  3 65 x  5 13 13 To obtain the median value the set is ranked, that is, placed in ascending order of magnitude, and since the set contains an odd number of members the value of the middle member is the median value. Ranking the set gives: {1, 2, 3, 3, 3, 4, 4, 5, 5, 7, 7, 8, 13} The middle term is the seventh member, i.e. 4, thus the median value is 4. The modal value is the value of the most commonly occurring member and is 3, which occurs three times, all other members only occurring once or twice.

Statistics and probability

505

Application: Determine the standard deviation from the mean of the set of numbers: {5, 6, 8, 4, 10, 3}, correct to 4 significant figures

The arithmetic mean, x 

∑ x  5  6  8  4  10  3  6 n

     

Standard deviation,

6

∑ (x  x)2  n

 

The (x  x )2 values are: (5  6)2, (6  6)2, (8  6)2, (4  6)2, (10  6)2 and (3  6)2 The sum of the (x  x )2 values,

∑ (x  x)2  1  0  4  4  16  9  34 i (x  x)2 34 and ∑   5.6 since there are 6 members in the set.

i.e.

n

6

Hence, standard deviation,

∑

i  ( x -x )2  σ   = 5. 6   n    2.380, correct to 4 significant figures. = 2.380, correct to 4 significant figures.

(b) Grouped data mean value, x 

∑ (fx) ∑f

  standard deviation, σ    

∑ {f(x - x )2 }  ∑ f 

Application: Find (a) the mean value, and (b) the standard deviation for the following values of resistance, in ohms, of 48 resistors: 20.5–20.9 3, 22.0–22.4 13,

21.0–21.4 10, 22.5–22.9 9,

21.5–21.9 11, 23.0–23.4 2

(a) The class mid-point/frequency values are: 20.7 3,

21.2 10,

21.7 11,

22.2 13,

22.7 9 and 23.2 2

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Mathematics Pocket Book for Engineers and Scientists

For grouped data, the mean value is given by: x 

∑ (f x) ∑f

where f is the class frequency and x is the class mid-point value. Hence (3  20.7)  (10  21.2)  (11  21.7)  (13  22.2)  (9  22.7)  (2  23.2) mean value, x  48 1052.1   21.919.. 48 i.e. the mean value is 21.9 ohms, correct to 3 significant figures. (b) From part (a), mean value, x  21.92 , correct to 4 significant figures. The ‘x-values’ are the class mid-point values, i.e. 20.7, 21.2, 21.7, …. Thus the (x  x )2 values are (20.7 – 21.92)2, (21.2 – 21.92)2, (21.7 – 21.92)2, …, and the f(x  x )2 values are 3(20.7 – 21.92)2, 10(21.2 – 21.92)2, 11(21.7 – 21.92)2, …. The ^ f(x  x)2 values are 4.4652  5.1840  0.5324  1.0192  5.4756  3.2768  19.9532

∑ {f(x  x)2 }  19.9532  0.41569 48 ∑f and standard deviation,

∑{

}

 f(x  x)2    0.41569      f    0.645, correct to 3 significant figures

∑

Application: The time taken in minutes to assemble a device is measured 50 times and the results are as shown below: 14.5–15.5 20.5–21.5

5, 12,

16.5–17.5 8, 22.5–23.5 6,

18.5–19.5 16, 24.5–25.5 3

Determine the mean, median and modal values of the distribution by depicting the data on a histogram The histogram is shown in Figure 174.1. The mean value lies at the centroid of the histogram. With reference to any arbitrary axis, say YY shown at a time of 14 minutes, the position of the horizontal value of the centroid can be obtained from the relationship AM  ^(am), where A is the area of the histogram, M is the horizontal distance of the centroid from the axis YY, a is the area of a rectangle of the histogram and m is the distance of the centroid of the rectangle from YY. The areas of the individual

Statistics and probability

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507

Mean Median Mode A B

Y

Frequency

14 12

5.6

10 8 6

16

4 2

C D

10

24

32 12

6 E F 14 15 16 17 18 19 20 21 22 23 24 25 26 27 Time in minutes

Figure 174.1

rectangles are shown circled on the histogram giving a total area of 100 square units. The positions, m, of the centroids of the individual rectangles are 1, 3, 5, … units from YY. Thus 100 M  (10  1)  (16  3)  (32  5)  (24  7)  (12  9)  (6  11) 560  5.6 units from YY 100 Thus the position of the mean with reference to the time scale is 14  5.6, i.e. 19.6 minutes. i.e.

M

The median is the value of time corresponding to a vertical line dividing the total area of the histogram into two equal parts. The total area is 100 square units; hence the vertical line must be drawn to give 50 units of area on each side. To achieve this with reference to Figure 174.1, rectangle ABFE must be split so that 50  (10  16) units of area lie on one side and 50  (24  12  6) units of area lie on the other. This shows that the area of ABFE is split so that 24 units of area lie to the left of the line and 8 units of area lie to the right, i.e. the vertical line must pass through 19.5 minutes. Thus the median value of the distribution is 19.5 minutes. The mode is obtained by dividing the line AB, which is the height of the highest rectangle, proportionally to the heights of the adjacent rectangles. With reference to Figure 174.1, this is done by joining AC and BD and drawing a vertical line through the point of intersection of these two lines. This gives the mode of the distribution and is 19.3 minutes.

Chapter 175

Quartiles, deciles and percentiles

The quartile values of a set of discrete data are obtained by selecting the values of members which divide the set into four equal parts.

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When a set contains a large number of members, the set can be split into ten parts, each containing an equal number of members; these ten parts are then called deciles. For sets containing a very large number of members, the set may be split into one hundred parts, each containing an equal number of members; one of these parts is called a percentile.

Application: The frequency distribution given below refers to the overtime worked by a group of craftsmen during each of 48 working weeks in a year. 25–29 5, 45–49 12,

30–34 4, 50–54 8,

35–39 7, 55–59 1

40–44

11,

Draw an ogive for this data and hence determine the quartiles values The cumulative frequency distribution (i.e. upper class boundary/cumulative frequency values) is: 29.5 5, 34.5 9, 39.5 16, 49.5 39, 54.5 47, 59.5 48

44.5 27,

The ogive is formed by plotting these values on a graph, as shown in Figure 175.1.

Cumulative frequency

50 40 30 20 10 25

30

35 Q1 40 Q2 45 Q3 50

55

Upper class boundary values, hours

60

Figure 175.1

The total frequency is divided into four equal parts, each having a range of 48/4, i.e. 12. This gives cumulative frequency values of 0 to 12 corresponding to the first quartile, 12 to 24 corresponding to the second quartile, 24 to 36 corresponding to the third quartile and 36 to 48 corresponding to the fourth quartile of the distribution, i.e. the distribution is divided into four equal parts. The quartile values are those of the variable corresponding to cumulative frequency values of 12, 24 and 36, marked Q1, Q2 and Q3 in Figure 175.1. These values, correct to the nearest hour, are 37 hours, 43 hours and 48 hours, respectively. The Q2 value is also equal to the median value of the distribution. One measure of the dispersion of a distribution is called the semi-interquartile range and is given by (Q3 – Q1)/2, and is (48 – 37)/2 in this case, i.e. 5½ hours.

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Application: Determine the numbers contained in the (a) 41st to 50th percentile group, and (b) 8th decile group of the set of numbers shown below: 14 24

22 17

17 20

21 22

30 27

28 19

37 26

7 21

23 15

32 29

The set is ranked, giving: 7 14 15 17 17 19 20 21 21 22 22 23 24 26 27 28 29 30 32 37 (a) There are 20 numbers in the set, hence the first 10% will be the two numbers 7 and 14, the second 10% will be 15 and 17, and so on Thus the 41st to 50th percentile group will be the numbers 21 and 22 (b) The first decile group is obtained by splitting the ranked set into 10 equal groups and selecting the first group, i.e. the numbers 7 and 14. The second decile group are the numbers 15 and 17, and so on. Thus the 8th decile group contains the numbers 27 and 28

Chapter 176

Probability

The probability of events A or B or C or …. . N happening is given by p A  pB  pC  ....  pN The probability of events A and B and C and … N happening is given by p A  pB  pC  ....  pN

Application: Determine the probability of selecting at random the winning horse in a race in which 10 horses are running Since only one of the ten horses can win, the probability of selecting at random the 1 number of winners winning horse is , i.e. or 0.10 10 number of horses Application: Determine the probability of selecting at random the winning horses in both the first and second races if there are 10 horses in each race The probability of selecting the winning horse in the first race is 1 10 The probability of selecting the winning horse in the second race is 1 10 The probability of selecting the winning horses in the first and second race is given by the multiplication law of probability, i.e.

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probability 

1 1 1   or 0.01 10 10 100

Application: The probability of a component failing in one year due to exces1 1 , due to excessive vibration is 25 and due to excessive sive temperature is 20 1 . Determine the probabilities that during a one year period a humidity is 50 component: (a) fails due to excessive temperature and excessive vibration, (b) fails due to excessive vibration or excessive humidity, and (c) will not fail due to excessive temperature and excessive humidity Let pA be the probability of failure due to excessive temperature, then pA 

1 19 and pA  20 20

(where pA is the probability of not failing)

Let pB be the probability of failure due to excessive vibration, then pB 

1 25

and

pB 

24 25

Let pC be the probability of failure due to excessive humidity, then pC 

1 50

and

pC 

49 50

(a) The probability of a component failing due to excessive temperature and excessive vibration is given by: pA  pB 

1 1 1   or 0.002 20 25 500

(b) The probability of a component failing due to excessive vibration or excessive humidity is: pB  pC 

1 1 3   or 0.06 25 50 50

(c) The probability that a component will not fail due to excessive temperature and will not fail due to excess humidity is: pA  pC 

19 49 931   or 0.931 20 50 1000

Application: A batch of 40 components contains 5 which are defective. If a component is drawn at random from the batch and tested and then a second component is drawn at random, calculate the probability of having one defective component, both with and without replacement.

The probability of having one defective component can be achieved in two ways. If p is the probability of drawing a defective component and q is the probability

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511

of drawing a satisfactory component, then the probability of having one defective component is given by drawing a satisfactory component and then a defective component or by drawing a defective component and then a satisfactory one, i.e. by qppq With replacement: 5 35 1 7  and q   40 8 40 8 Hence, probbility of having one defective component is: p

1 7 7 1 7 7 7    , i.e.   or 0.2188 8 8 8 8 64 64 32 Without replacement: 7 1 p1  and q1  on the first of the two draws. The batch number is now 39 for 8 8 the second draw, thus, p2  5 and q2  35 39 39 p1q2  q1p2 

Chapter 177

1 35 7 5 35 + 35 70      or 0.2244 8 39 8 39 312 312

Permutations and combinations

Permutations If n different objects are available, they can be arranged in different orders of selection. Each different ordered arrangement is called a permutation. For example, permutations of the three letters X, Y and Z taken together are: XYZ, XZY, YXZ, YZX, ZXY and ZYX This can be expressed as 3P3 = 6, the upper 3 denoting the number of items from which the arrangements are made, and the lower 3 indicating the number of items used in each arrangement. If we take the same three letters XYZ two at a time the permutations XY, YZ, XZ, ZX, YZ, ZY can be found, and denoted by 3P2 = 6 (Note that the order of the letters matters in permutations, i.e. YX is a different permutation from XY). n! n In general, nPr = n(n1) (n2) (nr1) or Pr = (n − r)! For example, 5P4 = 5(4)(3)(2) = 120 or 5P4 =

5! 5! = = (5)(4)(3)(2) = 120 (5 - 4)! 1!

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Mathematics Pocket Book for Engineers and Scientists

Also, 3P3 = 6 from above; using

nP r

=

3! 6 n! = gives 3P3 = . (3 3)! 0! (n - r)!

Since this must equal 6, then 0! = 1 (check this with your calculator).

Combinations If selections of the three letters X, Y, Z are made without regard to the order of the letters in each group, i.e. XY is now the same as YX for example, then each group is called a combination. The number of possible combinations is denoted by nCr where n is the total number of items and r is the number in each selection. n! r!(n − r )!

In general, nCr =

For example, 5 C4 =

5! 5! 5 × 4 × 3 × 2 ×1 = = =5 4 !(5 - 4)! 4! 4 × 3 × 2 ×1

Application: Calculate the number of permutations there are of: (a) 5 distinct objects taken 2 at a time, (b) 4 distinct objects taken 2 at a time. (a) 5P2 =

5! 5! 5 × 4 × 3 × 2 = = = 20 (5 - 2)! 3! 3×2

(b) 4P2 =

4! 4! = = 12 (4 - 2)! 2!

Application: Calculate the number of combinations there are of: (a) 5 distinct objects taken 2 at a time, (b) 4 distinct objects taken 2 at a time. (a)

5C 2

=

5! 5! 5 × 4 × 3 × 2 ×1 = = = 10 2!(5 - 2)! 2!3! (2 × 1)(3 × 2 × 1)

(b)

4C 2

=

4! 4! = =6 2!(4 - 2)! 2!2!

Application: A class has 24 students. 4 can represent the class at an exam board. How many combinations are possible when choosing this group? Number of combinations possible, nCr = i.e.

24

C4 =

n! r !(n - r)!

24 ! 24 ! = = 10626 4 !(24 - 4)! 4 !20!

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Application: In how many ways can a team of eleven be picked from sixteen possible players? Number of ways = nCr = 16C11 =

Chapter 178

16! 16! = = 4368 11!(16 – 11)! 11!5!

Bayes’ theorem

Bayes’ theorem is one of probability theory (originally stated by the Reverend Thomas Bayes), and may be seen as a way of understanding how the probability that a theory is true is affected by a new piece of evidence. The theorem has been used in a wide variety of contexts, ranging from marine biology to the development of ‘Bayesian’ spam blockers for email systems; in science, it has been used to try to clarify the relationship between theory and evidence. Insights in the philosophy of science involving confirmation, falsification and other topics can be made more precise, and sometimes extended or corrected, by using Bayes’ theorem. Bayes’ theorem may be stated mathematically as:

(

)

P A1 B =

(

) P (B A ) P ( A ) + P (B A ) P ( A ) + .... P B A1 P ( A1 )

1

or

(

)

P Ai B =

(

1

2

2

)

P B Ai P ( Ai ) n

∑ P (B A j ) P ( A j )

(i = 1, 2, ..., n)

j =1

where

P(A  B) is the probability of A given B, i.e. after B is observed P(A) and P(B) are the probabilities of A and B without regard to each other

and

P(B  A) is the probability of observing event B given that A is true

In the Bayes theorem formula, ‘A’ represents a theory or hypothesis that is to be tested, and ‘B’ represents a new piece of evidence that seems to confirm or disprove the theory.

Application: An outdoor degree ceremony is taking place tomorrow, 5 July, in the hot climate of Dubai. In recent years it has rained only 2 days in the fourmonth period June to September. However, the weather forecaster has predicted rain for tomorrow. When it actually rains, the weatherman correctly forecasts rain 85% of the time. When it doesn’t rain, he incorrectly forecasts rain 15% of the time. Determine the probability that it will rain tomorrow.

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Mathematics Pocket Book for Engineers and Scientists

There are two possible mutually-exclusive events occurring here – it either rains or it does not rain. Also, a third event occurs when the weatherman predicts rain. Let the notation for these events be:

Event A1

It rains at the ceremony

Event A2 It does not rain at the ceremony Event B

The weatherman predicts rain

The probability values are: 2 1 P ( A1) = = (i.e. it rains 2 days in the months June to 30 + 31 + 31 + 30 61 September) 120 60 P ( A2 ) = = (i.e. it does not rain for 120 of the 122 days 30 + 31 + 31 + 30 61 in the months June to September) P(B|A1) = 0.85 (i.e. when it rains, the weatherman predicts rain 85% of the time) P(B|A2) = 0.15 (i.e. when it does not rain, the weatherman predicts rain 15% of the time) Using Bayes’ theorem to determine the probability that it will rain tomorrow, given the forecast of rain by the weatherman:

(

)

P A1 B =

(

)

(

)

P B A1 P ( A1)

(

)

P B A1 P ( A1) + P B A2 P ( A2 )

 1 (0.85)    61 0.0139344 = 0.0863 or 8.63% = = 1 60 0.1614754 + 0.15 × 0.85 × 61 61 Even when the weatherman predicts rain, it rains only between 8% and 9% of the time. Hence, there is a good chance it will not rain tomorrow in Dubai for the degree ceremony.

Chapter 179 The binomial distribution

If p is the probability that an event will happen and q is the probability that the event will not happen, then the probabilities that the event will happen 0, 1, 2, 3,…, n times in n trials are given by the successive terms of the expansion of (q  p)n, taken from left to right, i.e.

q n , nqn1p,

n(n  1) n2 2 n(n  1)(n  2) n3 3 q p , q p , ... 2! 3!

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Industrial inspection The probabilities that 0, 1, 2, 3, … , n components are defective in a sample of n components, drawn at random from a large batch of components, are given by the successive terms of the expansion of (q  p)n, taken from left to right. Application: A die is rolled 9 times. Find the probabilities of having a 4 upwards (a) 3 times and (b) less than 4 times Let p be the probability of having a 4 upwards. Then p  1/6, since dice have six sides. Let q be the probability of not having a 4 upwards. Then q  5/6. The probabilities of having a 4 upwards 0, 1, 2.. n times are given by the successive terms of the expansion of (q  p)n, taken from left to right. From the binomial expansion (see Chapter 26): (q  q)9  q9  9q8p  36q7p2  84q6p3  ... The probability of having a 4 upwards no times is q9  (5/6)9  0.1938 The probability of having a 4 upwards once is 9q8p  9(5/6)8 (1/6)  0.3489 The probability of having a 4 upwards twice is 36q7p2  36(5/6)7(1/6)2  0.2791 (a) The probability of having a 4 upwards 3 times is 84q6p3  84(5/6)6 (1/6)3  0.1302 (b) The probability of having a 4 upwards less than 4 times is the sum of the probabilities of having a 4 upwards 0,1, 2, and 3 times, i.e. 0.1938  0.3489  0.2791  0.1302  0.9520 Application: A package contains 50 similar components and inspection shows that four have been damaged during transit. If six components are drawn at random from the contents of the package, determine the probabilities that in this sample (a) one and (b) less than three are damaged The probability of a component being damaged, p, is 4 in 50, i.e. 0.08 per unit. Thus, the probability of a component not being damaged, q, is 1  0.08, i.e. 0.92 The probability of there being 0, 1, 2,…, 6 damaged components is given by the successive terms of (q  p)6, taken from left to right. (q  p)6  q6  6q5p  15q4p2  20q3p3   ... (a) The probability of one damaged component is 6q5p  6  0.925  0.08  0.3164 (b) The probability of less than three damaged components is given by the sum of the probabilities of 0, 1 and 2 damaged components, i.e.

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Mathematics Pocket Book for Engineers and Scientists

q6  6q5p  15q4p2  0.926  6  0.925  0.08  15  0.924  0.082  0.6064  0.3164  0.0688  0.9916

Chapter 180

The Poisson distribution

If  is the expectation of the occurrence of an event then the probability of 0, 1, 2, 3, …. occurrences is given by: eλ , λeλ , λ 2

eλ eλ , λ3 , ... 2! 3!

Application: If 3% of the gearwheels produced by a company are defective, determine the probabilities that in a sample of 80 gearwheels (a) two and (b) more than two will be defective The sample number, n, is large, the probability of a defective gearwheel, p, is small and the product np is 80  0.03, i.e. 2.4, which is less than 5. Hence a Poisson approximation to a binomial distribution may be used. The expectation of a defective gearwheel,   np  2.4 The probabilities of 0, 1, 2,… defective gearwheels are given by the successive  2 3  terms of the expression e 1 +     ... taken from left to right, i.e. by   2! 3! 2  e, e,  e , .. 2! The probability of no defective gearwheels is e  e2.4  0.0907 The probability of 1 defective gearwheel is e  2.4e2.4  0.2177 (a) the probability of 2 defective gearwheels is

 2e 2.42 e2.4  2! 2 1  0.2613  0.2613

(b) The probability of having more than 2 defective gearwheels is 1 – (the sum of the probabilities of having 0, 1, and 2 defective gearwheels), i.e. 1  (0.0907  0.2177  0.2613), that is, 0.4303 Application: A production department has 35 similar milling machines. The number of breakdowns on each machine averages 0.06 per week. Determine the probabilities of having (a) one, and (b) less than three machines breaking down in any week

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Since the average occurrence of a breakdown is known but the number of times when a machine did not break down is unknown, a Poisson distribution must be used. The expectation of a breakdown for 35 machines is 35  0.06,i.e. 2.1 breakdowns per week. The probabilities of a breakdown occurring 0, 1, 2, … times are given by  2  3 ... the successive terms of the expression e 1       , taken from left   2! 3! to right. Hence the probability of no breakdowns e  e2.1  0.1225 (a) The probability of 1 breakdown is e  2.1e2.1  0.2572 (b) The probability of 2 breakdowns is

 2e 2.12 e2.1   0.2700 2! 2 1

The probability of less than 3 breakdowns per week is the sum of the probabilities of 0, 1 and 2 breakdowns per week, i.e.

0.1225  0.2572  0.2700  0.6497

Chapter 181

The normal distribution

A table of partial areas under the standardised normal curve is shown in Table 181.1. Application: The mean height of 500 people is 170 cm and the standard deviation is 9 cm. Assuming the heights are normally distributed, determine (a) the number of people likely to have heights between 150 cm and 195 cm, (b) the number of people likely to have heights of less than 165 cm, and (c) the number of people likely to have heights of more than 194 cm _

(a) The mean value, x , is 170 cm and corresponds to a normal standard variate value, z, of zero on the standardised normal curve. A height of 150 cm has a xx z-value given by z  standard deviations, i.e. 150  170 or 2.22 standard  9 deviations. Using a table of partial areas beneath the standardised normal curve (see Table 181.1), a z-value of 2.22 corresponds to an area of 0.4868 between the mean value and the ordinate z  2.22. The negative z-value shows that it lies to the left of the z  0 ordinate. This area is shown shaded in Figure 181.1(a). Similarly, 195 cm has a z-value of 195  170 that is 2.78 standard deviations. From Table 181.1, this value of z 9

z

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3

xx σ

0

0.0000 0.0398 0.0793 0.1179 0.1554 0.1915 0.2257 0.2580 0.2881 0.3159 0.3413 0.3643 0.3849 0.4032

0

z

0.0040 0.0438 0.0832 0.1217 0.1591 0.1950 0.2291 0.2611 0.2910 0.3186 0.3438 0.3665 0.3869 0.4049

1 0.0080 0.0478 0.0871 0.1255 0.1628 0.1985 0.2324 0.2642 0.2939 0.3212 0.3451 0.3686 0.3888 0.4066

2 0.0120 0.0517 0.0910 0.1293 0.1664 0.2019 0.2357 0.2673 0.2967 0.3238 0.3485 0.3708 0.3907 0.4082

3 0.0159 0.0557 0.0948 0.1331 0.1700 0.2054 0.2389 0.2704 0.2995 0.3264 0.3508 0.3729 0.3925 0.4099

4

Table 181.1 Partial areas under the standardised normal curve

0.0199 0.0596 0.0987 0.1388 0.1736 0.2086 0.2422 0.2734 0.3023 0.3289 0.3531 0.3749 0.3944 0.4115

5 0.0239 0.0636 0.1026 0.1406 0.1772 0.2123 0.2454 0.2760 0.3051 0.3315 0.3554 0.3770 0.3962 0.4131

6 0.0279 0.0678 0.1064 0.1443 0.1808 0.2157 0.2486 0.2794 0.3078 0.3340 0.3577 0.3790 0.3980 0.4147

7

0.0319 0.0714 0.1103 0.1480 0.1844 0.2190 0.2517 0.2823 0.3106 0.3365 0.3599 0.3810 0.3997 0.4162

8

0.0359 0.0753 0.1141 0.1517 0.1879 0.2224 0.2549 0.2852 0.3133 0.3389 0.3621 0.3830 0.4015 0.4177

9

1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 `2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9

0.4192 0.4332 0.4452 0.4554 0.4641 0.4713 0.4772 0.4821 0.4861 0.4893 0.4918 0.4938 0.4953 0.4965 0.4974 0.4981 0.4987 0.4990 0.4993 0.4995 0.4997 0.4998 0.4998 0.4999 0.4999 0.5000

0.4207 0.4345 0.4463 0.4564 0.4649 0.4719 0.4778 0.4826 0.4864 0.4896 0.4920 0.4940 0.4955 0.4966 0.4975 0.4982 0.4987 0.4991 0.4993 0.4995 0.4997 0.4998 0.4998 0.4999 0.4999 0.5000

0.4222 0.4357 0.4474 0.4573 0.4656 0.4726 0.4783 0.4830 0.4868 0.4898 0.4922 0.4941 0.4956 0.4967 0.4076 0.4982 0.4987 0.4991 0.4994 0.4995 0.4997 0.4998 0.4999 0.4999 0.4999 0.5000

0.4236 0.4370 0.4484 0.4582 0.4664 0.4732 0.4785 0.4834 0.4871 0.4901 0.4925 0.4943 0.4957 0.4968 0.4977 0.4983 0.4988 0.4991 0.4994 0.4996 0.4997 0.4998 0.4999 0.4999 0.4999 0.5000

0.4251 0.4382 0.4495 0.4591 0.4671 0.4738 0.4793 0.4838 0.4875 0.4904 0.4927 0.4945 0.4959 0.4969 0.4977 0.4984 0.4988 0.4992 0.4994 0.4996 0.4997 0.4998 0.4999 0.4999 0.4999 0.5000

0.4265 0.4394 0.4505 0.4599 0.4678 0.4744 0.4798 0.4842 0.4878 0.4906 0.4929 0.4946 0.4960 0.4970 0.4978 0.4984 0.4989 0.4992 0.4994 0.4996 0.4997 0.4998 0.4999 0.4999 0.4999 0.5000

0.4279 0.4406 0.4515 0.4608 0.4686 0.4750 0.4803 0.4846 0.4881 0.4909 0.4931 0.4948 0.4961 0.4971 0.4979 0.4985 0.4989 0.4992 0.4994 0.4996 0.4997 0.4998 0.4999 0.4999 0.4999 0.5000

0.4292 0.4418 0.4525 0.4616 0.4693 0.4756 0.4808 0.4850 0.4884 0.4911 0.4932 0.4949 0.4962 0.4972 0.4980 0.4985 0.4989 0.4992 0.4995 0.4996 0.4997 0.4998 0.4999 0.4999 0.4999 0.5000

0.4306 0.4430 0.4535 0.4625 0.4699 0.4762 0.4812 0.4854 0.4887 0.4913 0.4934 0.4951 0.4963 0.4973 0.4980 0.4986 0.4990 0.4993 0.4995 0.4996 0.4997 0.4998 0.4999 0.4999 0.4999 0.5000

0.4319 0.4441 0.4545 0.4633 0.4706 0.4767 0.4817 0.4857 0.4890 0.4916 0.4936 0.4952 0.4964 0.4974 0.4981 0.4986 0.4990 0.4993 0.4995 0.4997 0.4998 0.4998 0.4999 0.4999 0.4999 0.5000

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Mathematics Pocket Book for Engineers and Scientists

corresponds to an area of 0.4973, the positive value of z showing that it lies to the right of the z  0 ordinate. This area is shown shaded in Figure 181.1(b). The total area shaded in Figures 181.1(a) and (b) is shown in Figure 181.1(c) and is 0.4868  0.4973, i.e. 0.9841 of the total area beneath the curve. However, the area is directly proportional to probability. Thus, the probability that a person will have a height of between 150 and 195 cm is 0.9841. For a group of 500 people, 500  0.9841, i.e. 492 people are likely to have heights in this range.

2.22

0 (a)

z-value

0 (b)

0

2.22

2.78 z-value

2.78 z-value

(c)

Figure 181.1

(b) A height of 165 cm corresponds to 165 - 170 , i.e. 0.56 standard deviations. 9

The area between z  0 and z  0.56 (from Table 181.1) is 0.2123, shown shaded in Figure 181.2(a). The total area under the standardised normal curve is unity and since the curve is symmetrical, it follows that the total area to the left of the z  0 ordinate is 0.5000. Thus the area to the left of the z  0.56 ordinate (‘left’ means ‘less than’, ‘right’ means ‘more than’) is 0.5000  0.2123, i.e. 0.2877 of the total area, which is shown shaded in Figure 181.2(b). The area is directly proportional to probability and since the total area beneath the standardised normal curve is unity, the probability of a person’s height being less than 165 cm is 0.2877. For a group of 500 people, 500  0.2877, i.e. 144 people are likely to have heights of less than 165 cm.

0.56 0

z-value

(a)

0.56 0 (b)

Figure 181.2

z-value

Statistics and probability

521

194  170 that is, 2.67 standard deviations. 9 From Table 181.1, the area between z  0, z  2.67 and the standardised normal curve is 0.4962, shown shaded in Figure 181.3(a). Since the standardised normal curve is symmetrical, the total area to the right of the z  0 ordinate is 0.5000, hence the shaded area shown in Figure 181.3(b) is 0.5000  0.4962, i.e. 0.0038. This area represents the probability of a person having a height of more than 194 cm, and for 500 people, the number of people likely to have a height of more than 194 cm is 0.0038  500, i.e. 2 people.

(c) 194 cm corresponds to a z-value of

0 (a)

2.67

0 (b)

z-value

2.67

z-value

Figure 181.3

Testing for a normal distribution Application: Use normal probability paper to determine whether the data given below, which refers to the masses of 50 copper ingots, is approximately normally distributed. If the data is normally distributed, determine the mean and standard deviation of the data from the graph drawn. Class mid-point value (kg) 29.5 30.5 31.5 32.5 33.5 34.5 35.5 36.5 37.5 38.5 Frequency 2

4

6

8

9

8

6

4

2

1

To test the normality of a distribution, the upper class boundary/percentage cumulative frequency values are plotted on normal probability paper. The upper class boundary values are: 30, 31, 32,…, 38, 39. The corresponding cumulative frequency values (for ‘less than’ the upper class boundary values) are: 2, (4  2)  6, (6  42)  12, 20, 29, 37, 43, 47, 49 and 50. The corresponding percentage 2 6 cumulative frequency values are  100  4,  100  12, 24, 40, 58, 74, 50 50 86, 94, 98 and 100% The co-ordinates of upper class boundary/percentage cumulative frequency values are plotted as shown in Figure 181.4. When plotting these values, it will always be found that the co-ordinate for the 100% cumulative frequency value cannot be plotted, since the maximum value on the probability scale is 99.99. Since the points plotted in Figure 181.4 lie very nearly in a straight line, the data is approximately normally distributed. The mean value and standard deviation can be determined from Figure 181.4. Since a normal curve is symmetrical, the mean value is the value of the variable corre-

522

Mathematics Pocket Book for Engineers and Scientists 99.99 99.9 99.8

Percentage cumulative frequency

99 98 95 90 Q

80 70 60 50 40 30

P

20

R

10 5 2 1 0.5 0.2 0.1 0.05 0.01

30

32 34 36 38 Upper class boundary

40

42

Figure 181.4

sponding to a 50% cumulative frequency value, shown as point P on the graph. This shows that the mean value is 33.6 kg. The standard deviation is determined using the 84% and 16% cumulative frequency values, shown as Q and R in Figure 181.4. The variable values for Q and R are 35.7 and 31.4 respectively; thus two standard deviations correspond to 35.7  31.4, i.e. 4.3, showing that the standard deviation of the distribution is approximately 4.3 i.e. 2.15 standard deviations. 2

Chapter 182 Linear correlation The Pearson product-moment formula for determining the linear correlation coefficient states, coefficient of correlation, r 

∑ xy {( ∑ x2 )( ∑ y2 )}

where x  (X  X ) and y  (Y  Y )

Statistics and probability

523

Application: In an experiment to determine the relationship between force on a wire and the resulting extension, the following data is obtained: Force (N)

10

20

30

40

50

60

70

Extension (mm)

0.22

0.40

0.61

0.85

1.20

1.45

1.70

Determine the linear coefficient of correlation for this data Let X be the variable force values and Y be the dependent variable extension values, respectively. Using a tabular method to determine the quantities of this formula gives: x2

y2

x  (X  X )

y  (Y  Y )

xy

0.22

30

0.699

20.97

900

0.489

0.40

20

0.519

10.38

400

0.269

30

0.61

10

0.309

3.09

100

0.095

40

0.85

0

0.069

0

0

0.005

50

1.20

10

0.281

2.81

100

0.079

60

1.45

20

0.531

10.62

400

0.282

70

1.70

30

0.781

23.43

900

0.610

∑ X  280 280 X 7  40

∑ Y  6.43 6.43 Y 7  0.919

^ xy  71.30

^ x2  2800

^ y2  1.829

X

Y

10 20

Thus, coefficient of correlation, r

∑ xy  {( ∑ x2 )( ∑ y2 )}

71.3 [2800  1.829]

 0.996

This shows that a very good direct correlation exists between the values of force and extension.

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Mathematics Pocket Book for Engineers and Scientists

Chapter 183 Linear regression

The least-squares regression lines If the equation of the least-squares regression line is of the form: Y  a0  a1X the values of regression coefficient a0 and a1 are obtained from the equations:

∑ Y  a0N  a1∑ X

(1)

∑ (XY)  a0 ∑ X  a1∑ X2

(2)

If the equation of the regression line is of the form: X  b0  b1Y the values of regression coefficient b0 and b1 are obtained from the equations:

∑ X  b0N  b1∑ Y

(3)

∑ (XY)  b0 ∑ Y  b1∑ Y2

(4)

Application: The experimental values relating centripetal force and radius, for a mass travelling at constant velocity in a circle, are as shown: Force (N) Radius (cm)

5

10

15

20

25

30

35

40

55

30

16

12

11

9

7

5

Determine the equations of (a) the regression line of force on radius and (b) the regression line of radius on force. Hence, calculate the force at a radius of 40 cm and the radius corresponding to a force of 32 N (a) Let the radius be the independent variable X, and the force be the dependent variable Y. The equation of the regression line of force on radius is of the form Y  a0  a1X Using a tabular approach to determine the values of the summations gives: Radius, X 55 30 16 12 11 9 7 5

^ X  145

Force, Y 5 10 15 20 25 30 35 40

^ Y  180

X2 3025 900 256 144 121 81 49 25

^ X2  4601

XY 275 300 240 240 275 270 245 200

^ XY  2045

Y2 25 100 225 400 625 900 1225 1600

^ Y2  5100

Statistics and probability

525

Thus, from equations (1) and (2), 180  8a0  145a1 and 2045  145a0  4601a1 Solving these simultaneous equations gives a0  33.7 and a1  0.617, correct to 3 significant figures. Thus the equation of the regression line of force on radius is: Y  33.7  0.617 X Thus the force, Y, at a radius of 40 cm, is: Y  33.7  0.617(40)  9.02 i.e.

the force at a radius of 40 cm is 9.02 N

(b) The equation of the regression line of radius on force is of the form X  b0  b1Y From equations (3) and (4), 145  8b0  180b1 and 2045  180b0  5100b1 Solving these simultaneous equations gives b0  44.2 and b1  1.16, correct to 3 significant figures. Thus the equation of the regression line of radius on force is: X  44.2  1.16Y Thus, the radius, X, when the force is 32 N is: X  44.2  1.16(32)  7.08, i.e. the radius when the force is 32 N is 7.08 cm

Chapter 184

Sampling and estimation theories

Theorem 1 If all possible samples of size N are drawn from a finite population, Np, without replacement, and the standard deviation of the mean values of the sampling distribution of means is determined, then: standard error of the means, σ x =

σ N

 N - N   p     Np - 1 

(1)

where  x is the standard deviation of the sampling distribution of means and  is the standard deviation of the population For an infinite population and/or for sampling with replacement: σx =

σ N

(2)

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Mathematics Pocket Book for Engineers and Scientists

Theorem 2 If all possible samples of size N are drawn from a population of size Np and the mean value of the sampling distribution of means  x is determined then

µx  µ

(3)

where  is the mean value of the population Application: The heights of 3000 people are normally distributed with a mean of 175 cm and a standard deviation of 8 cm. If random samples are taken of 40 people, predict the standard deviation and the mean of the sampling distribution of means if sampling is done (a) with replacement, and (b) without replacement For the population: number of members, Np  3000; standard deviation,   8 cm; mean,   175 cm For the samples:

number in each sample, N  40

(a) When sampling is done with replacement, the total number of possible samples (two or more can be the same) is infinite. Hence, from equation (2) the standard error of the mean (i.e. the standard deviation of the sampling distribution of means) x 

 N



8 40

 1.265 cm

From equation (3), the mean of the sampling distribution µ x    175 cm (b) When sampling is done without replacement, the total number of possible samples is finite and hence equation (1) applies. Thus the standard error of the means, x 

 N

 N  N   3000  40   p    8     N 1    p 40  3000  1   (1.265)(0.9935)  1.257 cm

Provided the sample size is large, the mean of the sampling distribution of means is the same for both finite and infinite populations. Hence, from equation (3), µ x  175 cm

Statistics and probability

527

The estimation of population parameters based on a large sample size Table 184.1 Confidence levels Confidence level, %

99

98

96

95

90

80

50

Confidence coefficient, zC

2.58

2.33

2.05

1.96

1.645

1.28

0.6745

Application: Determine the confidence coefficient corresponding to a confidence level of 98.5% 98.5% is equivalent to a per unit value of 0.9850. This indicates that the area under the standardised normal curve between –zC and zC, i.e. corresponding to 2zC, is 0.9850 of the total area. Hence the area between the mean value and zC is 0.9850/2 i.e. 0.4925 of the total area. The z-value corresponding to a partial area of 0.4925 is 2.43 standard deviations from Table 181.1 on page 518. Thus, the confidence coefficient corresponding to a confidence level of 98.5% is 2.43

Estimating the mean of a population when the standard deviation of the population is known The confidence limits of the mean of a population are:

x

zC  N

 N  N   p   Np  1 

(4)

for a finite population of size Np The confidence limits for the mean of the population are:

x

zC  N

(5)

for an infinite population.

Application: It is found that the standard deviation of the diameters of rivets produces by a certain machine over a long period of time is 0.018 cm. The diameters of a random sample of 100 rivets produced by this machine in a day have a mean value of 0.476 cm. If the machine produces 2500 rivets a day, determine (a) the 90% confidence limits, and (b) the 97% confidence limits for an estimate of the mean diameter of all the rivets produced by the machine in a day For the population:

standard deviation,   0.018 cm number in the population, Np  2500

For the sample:

number in the sample, N  100 mean, x  0 . 476 cm

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Mathematics Pocket Book for Engineers and Scientists

There is a finite population and the standard deviation of the population is known, hence expression (4) is used. (a) For a 90% confidence level, the value of zC, the confidence coefficient, is 1.645 from Table 184.1. Hence, the estimate of the confidence limits of the population mean, , is:

 (1 .645)(0 . 018)   0 .476     100 i.e.

 2500  100     2500  1 

0 . 476  (0 . 00296)(0 . 9800)  0 . 476  0 . 0029 cm

Thus, the 90% confidence limits are 0.473 cm and 0.479 cm This indicates that if the mean diameter of a sample of 100 rivets is 0.476 cm, then it is predicted that the mean diameter of all the rivets will be between 0.473 cm and 0.479 cm and this prediction is made with confidence that it will be correct nine times out of ten. (b) For a 97% confidence level, the value of zC has to be determined from a table of partial areas under the standardised normal curve given in Table 181.1, as it is not one of the values given in Table 184.1. The total area between ordinates drawn at zC and  zC has to be 0.9700. Because the standardised normal curve is symmetrical, the area between zC  0 and zC is 0 . 9700 / 2, i.e. 0.4850. From Table 181.1 an area of 0.4850 corresponds to a zC value of 2.17. Hence, the estimated value of the confidence limits of the population mean is between

x

zC  N

 N  N     p   0 . 476   (2 . 17)(0 . 018)   2500  100     2500  1     Np  1  100  0 . 476  (0 . 0039)(0 . 9800)  0 . 4 7 6  0 . 0038

Thus, the 97% confidence limits are 0.472 cm and 0.480 cm It can be seen that the higher value of confidence level required in part (b) results in a larger confidence interval.

Estimating the mean and standard deviation of a population from sample data The confidence limits of the mean value of the population, , are given by:

µ x  zC  x

(6)

If s is the standard deviation of a sample, then the confidence limits of the standard deviation of the population are given by:

s  zC s

(7)

Statistics and probability

529

Application: Several samples of 50 fuses selected at random from a large batch are tested when operating at a 10% overload current and the mean time of the sampling distribution before the fuses failed is 16.50 minutes. The standard error of the means is 1.4 minutes. Determine the estimated mean time to failure of the batch of fuses for a confidence level of 90% For the sampling distribution: the mean,  x  16 . 50, the standard error of the means,

 x  1.4

The estimated mean of the population is based on sampling distribution data only and so expression (6) is used. For an 90% confidence level, zC  1.645 (from Table 184.1), thus  x  zC  x  16 . 50  (1 . 645)(1 . 4 )  16 . 50  2 . 30 minutes. Thus, the 90% confidence level of the mean time to failure is from 14.20 minutes to 18.80 minutes.

Estimating the mean of a population based on a small sample size Table 184.2 Percentile values (tp) for Student’s t distribution with  degrees of freedom (shaded area  p)

tp



t0.995

t0.99

t0.975

t0.95

t0.90

t0.80

t0.75

t0.70

t0.60

t0.55

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

63.66 9.92 5.84 4.60 4.03 3.71 3.50 3.36 3.25 3.17 3.11 3.06 3.01 2.98 2.95 2.92 2.90

31.82 6.96 4.54 3.75 3.36 3.14 3.00 2.90 2.82 2.76 2.72 2.68 2.65 2.62 2.60 2.58 2.57

12.71 4.30 3.18 2.78 2.57 2.45 2.36 2.31 2.26 2.23 2.20 2.18 2.16 2.14 2.13 2.12 2.11

6.31 2.92 2.35 2.13 2.02 1.94 1.90 1.86 1.83 1.81 1.80 1.78 1.77 1.76 1.75 1.75 1.74

3.08 1.89 1.64 1.53 1.48 1.44 1.42 1.40 1.38 1.37 1.36 1.36 1.35 1.34 1.34 1.34 1.33

1.376 1.061 0.978 0.941 0.920 0.906 0.896 0.889 0.883 0.879 0.876 0.873 0.870 0.868 0.866 0.865 0.863

1.000 0.816 0.765 0.741 0.727 0.718 0.711 0.706 0.703 0.700 0.697 0.695 0.694 0.692 0.691 0.690 0.689

0.727 0.617 0.584 0.569 0.559 0.553 0.549 0.546 0.543 0.542 0.540 0.539 0.538 0.537 0.536 0.535 0.534

0.325 0.289 0.277 0.271 0.267 0.265 0.263 0.262 0.261 0.260 0.260 0.259 0.259 0.258 0.258 0.258 0.257

0.158 0.142 0.137 0.134 0.132 0.131 0.130 0.130 0.129 0.129 0.129 0.128 0.128 0.128 0.128 0.128 0.128

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Mathematics Pocket Book for Engineers and Scientists

Table 184.2 Continued 

t0.995

t0.99

t0.975

t0.95

t0.90

t0.80

t0.75

t0.70

t0.60

t0.55

18 19 20 21 22 23 24 25 26 27 28 29 30 40 60 120 

2.88 2.86 2.84 2.83 2.82 2.81 2.80 2.79 2.78 2.77 2.76 2.76 2.75 2.70 2.66 2.62 2.58

2.55 2.54 2.53 2.52 2.51 2.50 2.49 2.48 2.48 2.47 2.47 2.46 2.46 2.42 2.39 2.36 2.33

2.10 2.09 2.09 2.08 2.07 2.07 2.06 2.06 2.06 2.05 2.05 2.04 2.04 2.02 2.00 1.98 1.96

1.73 1.73 1.72 1.72 1.72 1.71 1.71 1.71 1.71 1.70 1.70 1.70 1.70 1.68 1.67 1.66 1.645

1.33 1.33 1.32 1.32 1.32 1.32 1.32 1.32 1.32 1.31 1.31 1.31 1.31 1.30 1.30 1.29 1.28

0.862 0.861 0.860 0.859 0.858 0.858 0.857 0.856 0.856 0.855 0.855 0.854 0.854 0.851 0.848 0.845 0.842

0.688 0.688 0.687 0.686 0.686 0.685 0.685 0.684 0.684 0.684 0.683 0.683 0.683 0.681 0.679 0.677 0.674

0.534 0.533 0.533 0.532 0.532 0.532 0.531 0.531 0.531 0.531 0.530 0.530 0.530 0.529 0.527 0.526 0.524

0.257 0.257 0.257 0.257 0.256 0.256 0.256 0.256 0.256 0.256 0.256 0.256 0.256 0.255 0.254 0.254 0.253

0.127 0.127 0.127 0.127 0.127 0.127 0.127 0.127 0.127 0.127 0.127 0.127 0.127 0.126 0.126 0.126 0.126

The confidence limits of the mean value of a population based on a small sample drawn at random from the population are given by

x

tCs (N  1)

(8)

Application: A sample of 12 measurements of the diameter of a bar are made and the mean of the sample is 1.850 cm.The standard deviation of the samples is 0.16 mm.Determine (a) the 90% confidence limits and (b) the 70% confidence limits for an estimate of the actual diameter of the bar For the sample: the sample size, N  12; mean, x  1.850 cm; standard deviation, s  0.16 mm 0.016 cm Since the sample number is less than 30, the small sample estimate as given in expression (8) must be used. The number of degrees of freedom, i.e. sample size minus the number of estimations of population parameters to be made, is 12  1, i.e. 11 (a) The confidence coefficient value corresponding to a percentile value of t0.90 and a degree of freedom value of   11 can be found by using Table 184.2, and is 1.36, i.e. tC  1.36. The estimated value of the mean of the population is given by:

x

tC s (1.36)(0.016)  1.850  (N  1) 11  1.850  0.0066 cm

Statistics and probability

531

Thus, the 90% confidence limits are 1.843 cm and 1.857 cm This indicates that the actual diameter is likely to lie between 1.843 cm and 1.857 cm and that this prediction stands a 90% chance of being correct. (b) The confidence coefficient value corresponding to t0.70 and to   11 is obtained from Table 184.2, and is 0.540, i.e. tC  0.540. The estimated value of the 70% confidence limits is given by:

x

tC s (0.540)(0.016)  1.850  (N  1) 11  1.850  0.0026 cm

Thus, the 70% confidence limits are 1.847 cm and 1.853 cm, i.e. the actual diameter of the bar is between 1.847 cm and 1.853 cm and this result has a 70% probability of being correct.

Chi-square distribution

Chi-square values

1 2 3 4 5 6 7 8 9 10 11 12



7.88 10.6 12.8 14.9 16.7 18.5 20.3 22.0 23.6 25.2 26.8 28.3

χ 20.995

6.63 9.21 11.3 13.3 15.1 16.8 18.5 20.1 21.7 23.2 24.7 26.2

χ 20.99

5.02 7.38 9.35 11.1 12.8 14.4 16.0 17.5 19.0 20.5 21.9 23.3

χ 20.975

3.84 5.99 7.81 9.49 11.1 12.6 14.1 15.5 16.9 18.3 19.7 21.0

χ 20.95 2.71 4.61 6.25 7.78 9.24 10.6 12.0 13.4 14.7 16.0 17.3 18.5

χ 20.90 1.32 2.77 4.11 5.39 6.63 7.84 9.04 10.2 11.4 12.5 13.7 14.8

χ 20.75 0.455 1.39 2.37 3.36 4.35 5.35 6.35 7.34 8.34 9.34 10.3 11.3

χ 20.50 0.102 0.575 1.21 1.92 2.67 3.45 4.25 5.07 5.90 6.74 7.58 8.44

χ 20.25

Percentile values ( χ 2ρ ) for the Chi-square distribution with  degrees of freedom

χp2

Table 185.1

Chapter 185

0.0158 0.211 0.584 1.06 1.61 2.20 2.83 3.49 4.17 4.87 5.58 6.30

χ 20.10

0.0039 0.103 0.352 0.711 1.15 1.64 2.17 2.73 3.33 3.94 4.57 5.23

χ 20.05

0.0010 0.0506 0.216 0.484 0.831 1.24 1.69 2.18 2.70 3.25 3.82 4.40

χ 20.025

0.0002 0.0201 0.115 0.297 0.554 0.872 1.24 1.65 2.09 2.56 3.05 3.57

χ 20.001

0.0000 0.0100 0.072 0.207 0.412 0.676 0.989 1.34 1.73 2.16 2.60 3.07

χ 20.005

13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 40 50 60 70 80 90 100

29.8 31.3 32.8 34.3 35.7 37.2 38.6 40.0 41.4 42.8 44.2 45.6 46.9 48.3 49.6 51.0 52.3 53.7 66.8 79.5 92.0 104.2 116.3 128.3 140.2

27.7 29.1 30.6 32.0 33.4 34.8 36.2 37.6 38.9 40.3 41.6 43.0 44.3 45.9 47.0 48.3 49.6 50.9 63.7 76.2 88.4 100.4 112.3 124.1 135.8

24.7 26.1 27.5 28.8 30.2 31.5 32.9 34.4 35.5 36.8 38.1 39.4 40.6 41.9 43.2 44.5 45.7 47.7 59.3 71.4 83.3 95.0 106.6 118.1 129.6

22.4 23.7 25.0 26.3 27.6 28.9 30.1 31.4 32.7 33.9 35.2 36.4 37.7 38.9 40.1 41.3 42.6 43.8 55.8 67.5 79.1 90.5 101.9 113.1 124.3

19.8 16.0 21.1 17.1 22.3 18.2 23.5 19.4 24.8 20.5 26.0 21.6 27.2 22.7 28.4 23.8 29.6 24.9 30.8 26.0 32.0 27.1 33.2 28.2 34.4 29.3 35.6 30.4 36.7 31.5 37.9 32.6 39.1 33.7 40.3 34.8 51.8 45.6 63.2 56.3 74.4 67.0 85.5 77.6 96.6 88.1 107.6 98.6 118.5 109.1

12.3 13.3 14.3 15.3 16.3 17.3 18.3 19.3 20.3 21.3 22.3 23.3 24.3 25.3 26.3 27.3 28.3 29.3 39.3 49.3 59.3 69.3 79.3 89.3 99.3

9.30 10.2 11.0 11.9 12.8 13.7 14.6 15.5 16.3 17.2 18.1 19.0 19.9 20.8 21.7 22.7 23.6 24.5 33.7 42.9 52.3 61.7 71.1 80.6 90.1

7.04 7.79 8.55 9.31 10.1 10.9 11.7 12.4 13.2 14.0 14.8 15.7 16.5 17.3 18.1 18.9 19.8 20.6 29.1 37.7 46.5 55.3 64.3 73.3 82.4

5.89 6.57 7.26 7.96 8.67 9.39 10.1 10.9 11.6 12.3 13.1 13.8 14.6 15.4 16.2 16.9 17.7 18.5 26.5 34.8 43.2 51.7 60.4 69.1 77.9

5.01 5.63 6.26 6.91 7.56 8.23 8.91 9.59 10.3 11.0 11.7 12.4 13.1 13.8 14.6 15.3 16.0 16.8 24.4 32.4 40.5 48.8 57.2 65.6 74.2

4.11 4.66 5.23 5.81 6.41 7.01 7.63 8.26 8.90 9.54 10.2 10.9 11.5 12.2 12.9 13.6 14.3 15.0 22.2 29.7 37.5 45.4 53.5 61.8 70.1

3.57 4.07 4.60 5.14 5.70 6.26 6.84 7.43 8.03 8.64 9.26 9.89 10.5 11.2 11.8 12.5 13.1 13.8 20.7 28.0 35.5 43.3 51.2 59.2 67.3

534

Mathematics Pocket Book for Engineers and Scientists

Application: As a result of a survey carried out of 200 families, each with five children, the distribution shown below was produced. Test the null hypothesis that the observed frequencies are consistent with male and female births being equally probable, assuming a binomial distribution, a level of significance of 0.05 and a ‘too good to be true’ fit at a confidence level of 95% Number of boys (B) and girls (G)

5B,OG

4B,1G

3B,2G

2B,3G

1B,4G,

0B,5G

Number of families

11

35

69

55

25

5

To determine the expected frequencies Using the usual binomial distribution symbols, let p be the probability of a male birth and q  1  p be the probability of a female birth. The probabilities of having 5 boys, 4 boys,.., 0 boys are given by the successive terms of the expansion of (q  p)n. Since there are 5 children in each family, n  5, and (q  p)5  q5  5q4 p  10q3p2  10q2p3  5qp4  p5 When q  p0.5, the probabilities of 5 boys, 4 boys,…, 0 boys are 0.03125, 0.15625, 0.3125, 0.3125, 0.15625 and 0.03125 For 200 families, the expected frequencies, rounded off to the nearest whole number are: 6, 31, 63, 63, 31 and 6 respectively. To determine the 2-value Using a tabular approach, the 2-value is calculated using 2 

Number of boys(B) and girls(G) 5B, 0G 4B, 1G 3B, 2G 2B, 3G 1B, 4G 0B, 5G

Observed frequency, o

 (ο  e)2    e  

∑ 

Expected frequency, e

11 35 69 55 25 5

6 31 63 63 31 6

oe

(o  e)2

5 4 6 8 6 1

χ2 

25 16 36 64 36 1

(o  e) 2 e 4.167 0.516 0.571 1.016 1.161 0.167

 (ο  e)2    7.598  e  

∑ 

To test the significance of the 2-value The number of degrees of freedom is given by   N  1 where N is the number of rows in the table above, thus   6  1  5. For a level of significance of 0.05, the 2 confidence level is 95%, i.e. 0.95 per unit. From Table 185.1, for the 0.95 ,   5 2 2  value, the percentile value p is 11.1. Since the calculated value of  is less than

Statistics and probability

535

p2 the null hypothesis that the observed frequencies are consistent with

male and female births being equally probable is accepted. 2 For a confidence level of 95%, the 0.05 ,   5 value from Table 185.1 is 1.15 and 2 because the calculated value of  (i.e. 7.598) is greater than this value, the fit is not so good as to be unbelievable.

Chapter 186

The sign test

Table 186.1 Critical values for the sign test α1  5%

2 21 %

1%

n

α2  10%

1 2

5%

2%

1%

1 2 3 4 5 6 7 8 9 10 11 12 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

— — — — 0 0 0 1 1 1 2 2 2 3 3 3 4 4 5 5 5 6 6 7 7 7 8 8 9 9 10 10

— — — — — 0 0 0 1 1 1 2 2 2 2 3 3 4 4 4 5 5 5 6 6 7 7 7 8 8 9 9

— — — — — — 0 0 0 0 1 1 1 1 2 2 2 3 3 4 4 4 5 5 5 6 6 7 7 7 8 8

%

— — — — — — — 0 0 0 0 1 1 1 1 2 2 2 3 3 3 4 4 4 5 5 6 6 6 7 7 7

536

Mathematics Pocket Book for Engineers and Scientists

Table 186.1 Continued α1  5%

2 21 %

1%

n

α2  10%

1 2

5%

2%

1%

32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

10 11 11 12 12 13 13 13 14 14 15 15 16 16 16 17 17 18 18

8 9 9 10 10 10 11 11 12 12 13 13 13 14 14 15 15 15 16

8 8 9 9 9 10 10 11 11 11 12 12 13 13 13 14 14 15 15

9 10 10 11 11 12 12 12 13 13 14 14 15 15 15 16 16 17 17

%

Procedure for sign test 1. State for the data the null and alternative hypotheses, H0 and H1 2. Know whether the stated significance level,  , is for a one-tailed or a twotailed test. Let, for example, H0: x  φ, then if H1: xφ then a two-tailed test is suggested because x could be less than or more than φ (thus use 2 in Table 186.1), but if say H1: x  φ or H1: x  φ then a one-tailed test is suggested (thus use 1 in Table 186.1) 3. Assign plus or minus signs to each piece of data – compared with φ or assign plus and minus signs to the difference for paired observations 4. Sum either the number of plus signs or the number of minus signs. For the two-tailed test, whichever is the smallest is taken; for a one-tailed test, the one which would be expected to have the smaller value when H1is true is used. The sum decided upon is denoted by S 5. Use Table 186.1 for given values of n, and 1 or 2 to read the critical region of S. For example, if, say, n  16 and 1  5%, then from Table 186.1, S  4. Thus if S in part (iv) is greater than 4 we accept the null hypothesis H0 and if S is less than or equal to 4 we accept the alternative hypothesis H1

Statistics and probability

537

Application: A manager of a manufacturer is concerned about suspected slow progress in dealing with orders. He wants at least half of the orders received to be processed within a working day (i.e. 7 hours). A little later he decides to time 17 orders selected at random, to check if his request had been met. The times spent by the 17 orders being processed were as follows:

3 3 1 1 1 3 3 h 9 h 15 h 11h 8 h 6 h 9 h 8 h 10 h 4 4 2 4 2 4 4 1 1 1 3 3 1 3 h 8 h 9 h 15 h 13 h 8 h 7 h 6 h 2 2 2 4 4 4 4

Use the sign test at a significance level of 5% to check if the managers request for quicker processing is being met Using the above procedure: 1. The hypotheses are H0: t  7 h and H1: t > 7 h, where t is time. 2. Since H1 is t  7 h, a one-tail test is assumed, i.e. α1  5% 3. In the sign test each value of data is assigned a  or  sign. For the above data let us assign a  for times greater than 7 hours and a – for less than 7 hours. This gives the following pattern:

                 4. The test statistic, S, in this case is the number of minus signs (if H0 were true there would be an equal number of  and  signs). Table 186.1 gives critical values for the sign test and is given in terms of small values; hence in this case S is the number of  signs, i.e. S  4 5. From Table 186.1, with a sample size n  17, for a significance level of 1  5%, S ≤ 4. Since S  4 in our data, the result is significant at 1  5%, i.e. the alternative hypothesis is accepted – it appears that the managers request for quicker processing of orders is not being met.

Chapter 187

Wilcoxon signed-rank test

Table 187.1 Critical values for the Wilcoxon signed-rank test

n 1 2 3

α1  5%

2 21 %

1%

α2  10%

1 2

5%

2%

1%

— — —

— — —

— — —

— — —

%

538

Mathematics Pocket Book for Engineers and Scientists

Table 187.1 Continued α1  5%

2 21 %

1%

α2  10%

1 2

5%

2%

1%

4 5

— 0

— —

— —

— —

6 7 8 9 10

2 3 5 8 10

0 2 3 5 8

— 0 1 3 5

— — 0 1 3

11 12 13 14 15

13 17 21 25 30

10 13 17 21 25

7 9 12 15 19

5 7 9 12 15

16 17 18 19 20

35 41 47 53 60

29 34 40 46 52

23 27 32 37 43

19 23 27 32 37

21 22 23 24 25

67 75 83 91 100

58 65 73 81 89

49 55 62 69 76

42 48 54 61 68

26 27 28 29 30

110 119 130 140 151

98 107 116 126 137

84 92 101 110 120

75 83 91 100 109

31 32 33 34 35

163 175 187 200 213

147 159 170 182 195

130 140 151 162 173

118 128 138 148 159

36 37 38 39 40

227 241 256 271 286

208 221 235 249 264

185 198 211 224 238

171 182 194 207 220

41 42 43 44 45

302 319 336 353 371

279 294 310 327 343

252 266 281 296 312

233 247 261 276 291

n

%

Statistics and probability

539

Table 187.1 Continued α1  5%

2 21 %

1%

n

α2  10%

1 2

5%

2%

1%

46 47 48 49 50

389 407 426 446 466

361 378 396 415 434

328 345 362 379 397

307 322 339 355 373

%

Procedure for the Wilcoxon signed-rank test 1. State for the data the null and alternative hypotheses, H0 and H1 2. Know whether the stated significance level,  , is for a one-tailed or a twotailed test (see 2. in the procedure for the sign test on page 536) 3. Find the difference of each piece of data compared with the null hypothesis or assign plus and minus signs to the difference for paired observations 4. Rank the differences, ignoring whether they are positive or negative 5. The Wilcoxon signed-rank statistic T is calculated as the sum of the ranks of either the positive differences or the negative differences – whichever is the smaller for a two-tailed test, and the one which would be expected to have the smaller value when H1 is true for a one-tailed test 6. Use Table 187.1 for given values of n, and 1 or 2 to read the critical region of T. For example, if, say, n  16 and 1  5%, then from Table 187.1, t  35. Thus if T in part 5 is greater than 35 we accept the null hypothesis H0 and if T is less than or equal to 35 we accept the alternative hypothesis H1

Application: The following data represents the number of hours that a portable car vacuum cleaner operates before recharging is required. Operating time (h) 1.4 2.3 0.8 1.4 1.8 1.5 1.9 1.4 2.1 1.1 1.6 Use the Wilcoxon signed-rank test to test the hypothesis, at a 5% level of significance, that this particular vacuum cleaner operates, on average, 1.7 hours before needing a recharge

Using the above procedure: 1. H0: t  1.7 h and H1: t  1.7 h 2. Significance level, α2  5% (since this is a two-tailed test) 3. Taking the difference between each operating time and 1.7 h gives: 0.3 h 0.2 h

0.6 h 0.3 h

0.9 h 0.4 h

0.3 h 0.6 h

0.1 h 0.1 h

0.2 h

4. These differences may now be ranked from 1 to 11 (ignoring whether they are positive or negative).

540

Mathematics Pocket Book for Engineers and Scientists

Some of the differences are equal to each other. For example, there are two 0.1’s (ignoring signs) that would occupy positions 1 and 2 when ordered. We average these as far as rankings are concerned i.e. each is assigned a ranking of

12 2

i.e. 1.5. Similarly the two 0.2 values in positions 3 and 4 when ordered are each assigned rankings of

34 2

i.e. 3.5, and the three 0.3 values in positions 5, 6,

and 7 are each assigned a ranking of therefore:

i.e. 6, and so on. The rankings are

567 3

Rank

1.5

1.5

3.5

3.5

6

6

Difference

0.1

0.1

0.2

0.2

0.3

0.3

Rank

6

8

9.5

9.5

11

Difference

0.3

0.4

0.6

0.6

0.9

5. There are 4 positive terms and 7 negative terms. Taking the smaller number, the four positive terms have rankings of 1.5, 3.5, 8 and 9.5. Summing the positive ranks gives: T  1.5  3.5  89.5  22.5 6. From Table 187.1, when n  11 and 2  5%, T ≤ 10 Since T  22.5 falls in the acceptance region (i.e. in this case is greater than 10), the null hypothesis is accepted, i.e. the average operating time is not significantly different from 1.7 h [Note that if, say, a piece of the given data was 1.7 h, such that the difference was zero, that data is ignored and n would be 10 instead of 11 in this case.]

Chapter 188

The Mann-Whitney test

Table 188.1 Critical values for the Mann-Whitney test α1  5%

2 21 %

1%

1 2

%

n1

n2

α2  10%

5%

2%

1%

2 2 2 2 2 2 2 2 2 2 2 2

2 3 4 5 6 7 8 9 10 11 12 13

— — — 0 0 0 1 1 1 1 2 2

— — — — — — 0 0 0 0 1 1

— — — — — — — — — — — 0

— — — — — — — — — — — —

Statistics and probability

541

Table 188.1 Continued α1  5%

2 21 %

1%

5%

2%

1 2

%

n1

n2

α2  10%

2 2 2 2 2 2 2

14 15 16 17 18 19 20

3 3 3 3 4 4 4

1 1 1 2 2 2 2

0 0 0 0 0 1 1

— — — — — 0 0

3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

0 0 1 2 2 3 4 4 5 5 6 7 7 8 9 9 10 11

— — 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8

— — — — 0 0 1 1 1 2 2 2 3 3 4 4 4 5

— — — — — — 0 0 0 1 1 1 2 2 2 2 3 3

4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4

4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

1 2 3 4 5 6 7 8 9 10 11 12 14 15 16 17 18

0 1 2 3 4 4 5 6 7 8 9 10 11 11 12 13 14

— 0 1 1 2 3 3 4 5 5 6 7 7 8 9 9 10

— — 0 0 1 1 2 2 3 3 4 5 5 6 6 7 8

5 5 5

5 6 7

4 5 6

2 3 5

1 2 3

0 1 1

1%

542

Mathematics Pocket Book for Engineers and Scientists

Table 188.1 Continued α1  5%

2 21 %

1%

5%

2%

1 2

%

n1

n2

α2  10%

5 5 5 5 5 5 5 5 5 5 5 5 5

8 9 10 11 12 13 14 15 16 17 18 19 20

8 9 11 12 13 15 16 18 19 20 22 23 25

6 7 8 9 11 12 13 14 15 17 18 19 20

4 5 6 7 8 9 10 11 12 13 14 15 16

2 3 4 5 6 7 7 8 9 10 11 12 13

6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

7 8 10 12 14 16 17 19 21 23 25 26 28 30 32

5 6 8 10 11 13 14 16 17 19 21 22 24 25 27

3 4 6 7 8 9 11 12 13 15 16 18 19 20 22

2 3 4 5 6 7 9 10 11 12 13 15 16 17 18

7 7 7 7 7 7 7 7 7 7 7 7 7 7

7 8 9 10 11 12 13 14 15 16 17 18 19 20

11 13 15 17 19 21 24 26 28 30 33 35 37 39

8 10 12 14 16 18 20 22 24 26 28 30 32 34

6 7 9 11 12 14 16 17 19 21 23 24 26 28

4 6 7 9 10 12 13 15 16 18 19 21 22 24

8 8 8

8 9 10

15 18 20

13 15 17

9 11 13

7 9 11

1%

Statistics and probability

543

Table 188.1 Continued α1  5%

2 21 %

1%

5%

2%

1 2

%

n1

n2

α2  10%

8 8 8 8 8 8 8 8 8 8

11 12 13 14 15 16 17 18 19 20

23 26 28 31 33 36 39 41 44 47

19 22 24 26 29 31 34 36 38 41

15 17 20 22 24 26 28 30 32 34

13 15 17 18 20 22 24 26 28 30

9 9 9 9 9 9 9 9 9 9 9 9

9 10 11 12 13 14 15 16 17 18 19 20

21 24 27 30 33 36 39 42 45 48 51 54

17 20 23 26 28 31 34 37 39 42 45 48

14 16 18 21 23 26 28 31 33 36 38 40

11 13 16 18 20 22 24 27 29 31 33 36

10 10 10 10 10 10 10 10 10 10 10

10 11 12 13 14 15 16 17 18 19 20

27 31 34 37 41 44 48 51 55 58 62

23 26 29 33 36 39 42 45 48 52 55

19 22 24 27 30 33 36 38 41 44 47

16 18 21 24 26 29 31 34 37 39 42

11 11 11 11 11 11 11 11 11 11 12 12

11 12 13 14 15 16 17 18 19 20 12 13

34 38 42 46 50 54 57 61 65 69 42 47

30 33 37 40 44 47 51 55 58 62 37 41

25 28 31 34 37 41 44 47 50 53 31 35

21 24 27 30 33 36 39 42 45 48 27 31

1%

544

Mathematics Pocket Book for Engineers and Scientists

Table 188.1 Continued α1  5%

2 21 %

1%

5%

2%

1 2

%

n1

n2

α2  10%

12 12 12 12 12 12 12

14 15 16 17 18 19 20

51 55 60 64 68 72 77

45 49 53 57 61 65 69

38 42 46 49 53 56 60

34 37 41 44 47 51 54

13 13 13 13 13 13 13 13

13 14 15 16 17 18 19 20

51 56 61 65 70 75 80 84

45 50 54 59 63 67 72 76

39 43 47 51 55 59 63 67

34 38 42 45 49 53 57 60

14 14 14 14 14 14 14

14 15 16 17 18 19 20

61 66 71 77 82 87 92

55 59 64 69 74 78 83

47 51 56 60 65 69 73

42 46 50 54 58 63 67

15 15 15 15 15 15

15 16 17 18 19 20

72 77 83 88 94 100

64 70 75 80 85 90

56 61 66 70 75 80

51 55 60 64 69 73

16 16 16 16 16

16 17 18 19 20

83 89 95 101 107

75 81 86 92 98

66 71 76 82 87

60 65 70 74 79

17 17 17 17

17 18 19 20

96 102 109 115

87 92 99 105

77 82 88 93

70 75 81 86

18 18 18

18 19 20

109 116 123

99 106 112

88 94 100

81 87 92

19 19

19 20

123 130

112 119

101 107

93 99

20

20

138

127

114

105

1%

Statistics and probability

545

Procedure for the Mann-Whitney test 1. State for the data the null and alternative hypotheses, H0 and H1 2. Know whether the stated significance level, , is for a one-tailed or a twotailed test (see 2. in the procedure for the sign test on page 536) 3. Arrange all the data in ascending order whilst retaining their separate identities 4. If the data is now a mixture of, say, A’s and B’s, write under each letter A the number of B’s that precede it in the sequence (or vice-versa) 5. Add together the numbers obtained from 4 and denote total by U. U is defined as whichever type of count would be expected to be smallest when H1 is true 6. Use Table 188.1 for given values of n1 and n2 , and 1 or 2 to read the critical region of U. For example, if, say, n1  10 and n2  16 and 2  5%, then from Table 188.1, U  42. If U in part 5 is greater than 42 we accept the null hypothesis H0, and if U is equal or less than 42, we accept the alternative hypothesis H1 Application: 10 British cars and 8 non-British cars are compared for faults during their first 10000 miles of use. The percentage of cars of each type developing faults were as follows: Non-British cars,

P

5

8

14

10

15

7

12

4

British cars,

Q

18

9

25

6

21

20

28

11

16

34

Use the Mann-Whitney test, at a level of significance of 1%, to test whether non-British cars have better average reliability than British models Using the above procedure: 1. The hypotheses are: H0: Equal proportions of British and non-British cars have breakdowns H1: A higher proportion of British cars have breakdowns 2. Level of significance 1  1% 3. Let the sizes of the samples be nP and nQ, where nP  8 and nQ  10 The Mann-Whitney test compares every item in sample P in turn with every item in sample Q, a record being kept of the number of times, say, that the item from P is greater than Q, or vice-versa. In this case there are nPnQ, i.e. (8)(10)  80 comparisons to be made. All the data is arranged into ascending order whilst retaining their separate identities – an easy way is to arrange a linear scale as shown in Figure 188.1.

Sample P 4

5 6

Sample Q 0

7 8 10 12 14 15 9 11 16 18 20 21

10

20

Figure 188.1

25

28

30

34

546

Mathematics Pocket Book for Engineers and Scientists

From Figure 188.1, a list of P’s and Q’s can be ranked giving: PPQPPQPQPPPQQQQQQQ 4. Write under each letter P the number of Q’s that precede it in the sequence, giving: P PQPPQPQPPPQQQQQQQ 00 11 2 333 5. Add together these 8 numbers, denoting the sum by U, i.e. U  0  0  1  1  2  3  3  3  13 6. The critical regions are of the form U  critical region From Table 188.1, for a sample size 8 and 10 at significance level 1  1% the critical regions is U ≤ 13 The value of U in our case, from 5, is 13 which is significant at 1% significance level. The Mann-Whitney test has therefore confirmed that there is evidence that the non-British cars have better reliability than the British cars in the first 10,000 miles, i.e. the alternative hypothesis applies.

Index

Acute angle, 133 Acute angled triangle, 133 Adding alternating waveforms, 225 Addition law of probability, 509 of matrices, 234 of two periodic functions, 225 of vectors, 217 by calculation, 217 Adjoint of matrix, 238 Alarm indicator, 272 Algebra, 17, 18 rules of, 18 topics, 17 Algebraic equation, 27 expression, 27 method of successive approximations, 90 substitution, integration, 321 Alternate angles, 133 Amplitude, 153 And-function, 254 And-gate, 266 Angle, double, 60, 166 lagging and leading, 153 of any magnitude, 149 of depression, 140 of elevation, 140 Angles of any magnitude, 152 Angular measurement, 2 velocity, 155 Annulus, 107 Applications of complex numbers, 210 differentiation, 280 rates of change, 280 small changes, 287 tangents and normals, 285 turning points, 282 velocity and acceleration, 281 Applications of integration, 349 areas, 349 centroids, 361 mean value, 355

r.m.s. value, 357 second moment of area, 369 volumes, 359 Arc, 107 Arc length, 107 Arch design, 59 Area between curves, 349 in Imperial units, 2 conversions, 2 in metric units, 2 conversions, 2 of circle, 106 irregular figures, 122 plane figures, 103 sector, 107 similar shapes, 106 triangle, 103, 145 under a curve, 349 Areas and volumes, 101 Argand diagram, 207, 208 Argument, 208 Arithmetic progression, 70 Astroid, 288 Astronomical constants, 6 Asymptotes, 201 Auxiliary equation, 392 Average value of waveform, 127, 355 Axes, 172 Bar charts, 497 Base, 50 Basic algebra, 19 SI units, 3 Bayes’ theorem, 513 Bell curve, 496 Bessel functions, 407 Bessel’s equation, 406 Binary addition, 95 numbers, 93 to decimal, 93 to hexadecimal, 100

548

Mathematics Pocket Book for Engineers and Scientists Index

Binomial distribution, 495, 514 series/theorem, 80 practical problems, 83 Bisection method, 88 BODMAS with algebra, 21 Boiler, 114 Boolean algebra, 253 laws and rules of, 256 Boolean expressions, 254 Boundary conditions, 377 Brackets, 19 By parts, integration, 324 Calculator, 137 Calculus, 273, 315 Cardioid, 194, 288 Cartesian complex numbers, 205 co-ordinates, 142 Catenary, 59 Centroids, 361 Chain rule, 277 Change of limits, integration, 322 Characteristic determinant, 248 equation, 248 Chi-square distribution, 496 tests, 496 values, 532 Circle, 100, 106 area, 106 equation of, 110 Classes, 501 Class interval, 501 limits, 501 mid-point, 501 Coefficient of correlation, 523 Cofactor of matrix, 237 Combinational logic networks, 266 Combination of two periodic functions, 225 Combinations, 512 Common factors, 20 prefixes, 5 ratio, 71 shapes, 103 Complementary angles, 133 function, 395 Completing the square, 45, 79 Complex numbers, 204 addition and subtraction, 205 applications of, 210 Cartesian form, 205 conjugate, 206 De Moivre’s theorem, 212

exponential form, 214 form of Fourier series, 482 general formulae, 205 multiplication and division, 206, 207 polar form, 208 powers of, 212 roots of, 212 use of calculator, 210 Complex wave considerations, 490 Compound angles, 60, 163 Computer numbering systems, 93 Cone, 111 frustum of, 116 Confidence coefficients, 527 limits, 527 Congruent triangles, 134 Conjugate, complex, 206 Continued fractions, 25, 26 Continuous function, 197 Contour map, 309 Convergents, 26 Conversion of a sin ωt + b cos ωt into R sin(ωt + α), 163 Cooling tower, 119 Correlation, coefficients, 496 linear, 522 Corresponding angles, 133 Cosecant, 137 Cosh, 58 Coshec, 58 Cosine, 136 curves, 152 rule, 144 wave, 152 Cotangent, 137 Coth, 58 Cramer’s rule, 245 Crane jib, 146 Crank mechanism, 147 Cross multiplication, 29 product, 230 Cubic equations, 189 graphs, 189 Cuboid, 111 Cumulative frequency curve, 501 distribution, 500, 503 Cycle, 179 of log graph paper, 179 Cycloid, 288 Cylinder, 111 Deciles, 508 Decimal to binary, 94

Statistics and probability Index

via octal, 96 to hexadecimal, 99 Definite integrals, 319 De Moivre’s theorem, 212 De Morgan’s laws, 259 Depression, angle of, 140 Derivatives, Laplace transforms of, 439 of the z-transform, 456 standard list, 275 Derived units, 4 Determinant, 2 by 2, 235 3 by 3, 236 Determinants to solve simultaneous equations, 242 Determination of law, 174 involving logarithms, 175 Difference equations, 460 Difference of two squares, 44 Differential calculus, 273 coefficient, 276 total, 305 Differential equations, 375 d2 x dy a b  cy  0 type, 391 dx dy 2 d2 x dy b  cy  f(x) type, 395 dx dy 2 dy  f(x) type, 377 dx dy  f(y) type, 377 dx dy  f(x).f(y) type, 379 dx dy  Py  Q type, 383 dx first order, separation of variables, 379 homogeneous first order, 381 linear first order, 383 numerical methods, 384–390 dy P  Q , 381 dx partial, 409–417 power series methods, 400–409 simultaneous, using Laplace transforms, 442 using Laplace transforms, 439 Differentiation, 273 function of a function rule, 277 implicit, 291 in parameters, 288 inverse hyperbolic function, 298 trigonometric function, 295 a

logarithmic, 293 methods of, 275 of axn, 275 of common functions, 275 of eax and ln ax, 275 of [f(x)]x, 294 of hyperbolic functions, 279 of parametric equations, 288 of sine and cosine functions, 275 partial, 302 first order, 302 second order, 303 product rule, 276 quotient rule, 276 rates of change, 280 successive, 278 Direction cosines, 229, 230 Discontinuous function, 197 Discrete data, 504 standard deviation of, 504 Distribution free tests, 496 Dividend, 22 Dividing head, 26 Divisor, 22 D-operator form, 392 Dot product, 227 Double angles, 60, 166 integrals, 342 Drilling machine speeds, 72 Eigenvalues, 248 Eigenvectors, 248 Elastic string, 411 Elevation, angle of, 140 Ellipse, 103, 196, 288 Engineering constants, 1 conversions, 1 notation, 5 symbols, 1 Equations, 27 Bessel’s, 406 heat conduction, 414 hyperbolic, 61 indicial, 52, 404 Laplace, 415 Legendre’s, 407 Newton-Raphson, 92 normal, 285 of circle, 110 quadratic, 42 simple, 27 simultaneous, 36 solving by iterative methods, 88–92

549

550

Mathematics Pocket Book for Engineers and Scientists Index

straight line graph, 172 tangent, 285 trigonometric, 157 wave, 410 Equilateral triangle, 134 Estimation theory, 496 Euler-Cauchy method, 386 Euler’s method, 384 Evaluation of trigonometric ratios, 137 Even function, 59, 198, 471 symmetry, Fourier series, 485 Exponent, 53 Exponential form of complex number, 214 Fourier series, 482 Exponential functions, 53 graphs of, 54 power series, 54 Exterior angle of triangle, 134 Extrapolation, 173 Factorisation, 20, 79 to solve quadratic equations, 42 Factors, 20, 23 Factor theorem, 23 Final value theorem, 423, 454 Finite discontinuities, 197 First order differential equations, 377–390 partial derivatives, 303 First shift theorem, 452 Fisherman’s line, 59 Floodlit area, 109 Formulae, transposition of, 32 Fourier series, 464 cosine, 471 exponential form, 482 half range, 474, 479 non-periodic over range 2π, 469 over any range, 476 periodic of period 2π, 466 sine, 471 Fractional form of trigonometric ratios, 138 Fractions, continued, 25 partial, 63 Frequency, 155 distribution, 501 polygon, 500, 501 Frobenius method, 403 Frustum of pyramids and cones, 116 sphere, 119 Full-wave rectified waveform, 128 Function of a function rule, 277 Functions of two variables, 302

Gamma function, 407 Garden spray area, 109 Gaussian elimination, 247 Gear wheels, 26 General solution of a differential equation, 377 Geometric progression, 71 Geometry and trigonometry, 131 Gradient of a curve, 276 of a straight-line graph, 172 Graphical functions, 197 solution of cubic equation, 189 quadratic, 184, 186 simultaneous, 183 Graphs, 170 of exponential functions, 54 hyperbolic functions, 59 logarithmic functions, 53 reducing non-linear to linear form, 174 sine and cosine, 149 straight line, 172 trigonometric functions, 149 y = abx, 175 y = aekx, 175 y = axn, 175 Graphs with logarithmic scales, 179 Greek alphabet, 3 Grouped data, 505 mean, median and mode, 505 standard deviation, 505 Growth and decay laws, 56 Half range cosine series, 474 Fourier series, 474 sine series, 474 Half-wave rectified waveform, 128 Harmonic analysis, 487 H.C.F., 20 Heat conduction equation, 414 Heaviside unit step function, 431 inverse Laplace transform of, 437 Laplace transform of, 431 Heptagon, 103 Hexadecimal number, 98 to binary, 100 to decimal, 99 Hexagon, 103 Histogram, 500, 501 Hollow shaft, 107 Homogeneous first order differential equations, 381 Hooke’s law, 170

Statistics and probability Index

Horizontal bar chart, 497 component, 219 Hyperbola, 196, 288 rectangular, 196, 288 Hyperbolic functions, 58 differentiation of, 279 graphs of, 59 identities, 60 inverse, 298 logarithms, 55 solving equations, 61 substitutions, integration, 323 Identities, hyperbolic, 60 trigonometric, 157 i, j, k notation, 224 Imaginary part of a complex number, 205 Impedance, 34 Implicit differentiation, 291 function, 291 Inclined jib of crane, 146 Indefinite integrals, 317 Indices, 19 laws of, 19 Indicial equations, 52, 404 Industrial inspection, 515 Inequalities, 73 Initial value theorem, 423, 455 Integral calculus, 315 Integrals, definite, 319 double, 342 standard, 317 triple, 344 Integrating factor, 383 Integration, algebraic substitution, 321 applications of, areas, 349 centroids, 361 mean value, 355 r.m.s. value, 357 second moment of area, 369 volumes, 359 by partial fractions, 329 by parts, 334 change of limits, 322 cosh θ substitution, 323 definite, 318 hyperbolic substitutions, 323 numerical, 384–391 of axn, 317 reduction formulae, 337 sine θ substitution, 323

551

sinh θ substitution, 323 standard, 317 t = tan θ/2 substitution, 331 tan θ substitution, 323 trigonometric substitutions, 323 Intercept, y-axis, 172 Interior angles of a triangle, 133, 134 Interpolation, 173 Inverse functions, 199, 295 hyperbolic, 298 differentiation of, 298 trigonometric, 200 differentiation of, 298 z-transforms, 456 Inverse Laplace transforms, 425 of Heaviside unit step function, 437 using partial fractions, 427 Inverse matrix, 2 by 2, 235 3 by 3, 238 Inverse z-transforms, 456 Invert-gate, 266 Irregular areas, 122 volumes, 124 Isosceles triangle, 134 Iterative methods, 88–93 Karnaugh maps, 261 Kirchhoff laws, 39 Lagging angle, 153 Lampshade, area of, 118 Laplace’s equation, 415 Laplace transforms, 418 common notations, 420 definition, 419 derivatives, 439 for differential equations, 439 for simultaneous differential equations, 442 inverse, 425 using partial fractions, 427 of elementary functions, 419 of Heaviside step function, 431 Laws of algebra, 18 Boolean algebra, 258 growth and decay, 53, 56 indices, 19 logarithms, 51, 293 precedence, 21 probability, 509 Leading angle, 153 Least-squares regression lines, 524 Legendre polynomials, 408

552

Mathematics Pocket Book for Engineers and Scientists Index

Legendre’s equation, 407 Leibniz theorem, 400 Leibniz-Maclaurin method, 401 L’Hopital’s rule, 87 Length in Imperial units, 2 conversions, 2 Length in metric units, 2 conversions, 2 Limiting values, 87 Linear correlation, 522 extrapolation, 173 first order differential equation, 383 interpolation, 173 regression, 496, 524 Linearity property, z-transform, 451 Logarithmic differentiation, 293 forms of inverse hyperbolic functions, 301 function, 301 scales, 179 Logarithms, 49 graphs of, 53 laws of, 51, 293 Logic circuits, 266 universal, 269 Log-linear graph paper, 183 Log-log graph paper, 180 Maclaurin’s series/theorem, 84 numerical integration, 86 Major axis of ellipse, 196 Mann-Whitney test, 540 Mass, 2 Mathematical constants, 7 symbols, 7 Matrices, 233 to solve simultaneous equations, 239 Matrix, adjoint, 238 determinant of, 235, 236 inverse, 238 transpose, 238 unit, 236 Maxima, minima and saddle points, 308 Maximum point, 309 Mean value, 355, 504 of waveform, 127 Measures of central tendency, 504 Median, 504 Metric conversions, 2 Metric to Imperial conversions, 2 Mid-ordinate rule, 122 Minimum point, 309

Minor axis of ellipse, 196 Minor of an element, 236 Mode, 504 Modulus, 75 inequalities involving, 75 Moment of a force, 231 Nand-gate, 267 Napier, John, 55 Napierian logarithms, 55 Natural laws of growth and decay, 53 logarithms, 55 Newton-Raphson method, 92 Newton’s law of cooling, 57 method, 92 Non-right-angled triangles, 144 Non-standard integrals, 320 Nor-gate, 267 Norm, 228 Normal, 285 Normal curve, 517–519 distribution, 495, 517 equations of, 285 probability paper, 521 Nose-to-tail method, 217 Not-function, 254 Not-gate, 266 n’th term of a series, 69 a G.P., 71 an A.P., 70 Number sequences, 68 topics, 67 Numerical integration, 86, 346 method of harmonic analysis, 487 methods, first order differential equations, 384–391 Obtuse angle, 133 Obtuse-angled triangle, 134 Octagon, 103 Octal numbers, 96 Odd function, 59, 198, 471 symmetry, Fourier series, 485 Ogive, 501, 503 Ohms law, 33 Or-function, 254 Or-gate, 266 Oscillating mechanism, 156 Pappus’ theorem, 365 Parabola, 288 Parallel axis theorem, 370 lines, 133

Statistics and probability Index

Parallelogram, 103 method, 218 Parametric equations, 288 Partial differential equations, 409–417 differentiation, 302 rates of change, 306 small changes, 307 fractions, 63 inverse Laplace transforms, 427 integration, using, 329 linear factors, 63, 329 quadratic factors, 63, 330 repeated linear factors, 63, 330 Particular integral, 395 solution of differential equation, 377, 392 Pearson product-moment formula, 522 Pentagon, 103 Percentage component bar chart, 497 Percentile, 508 Perfect square, 44, 45 Perimeter, 134 Period, 153, 197 Periodic function, 153, 197 Periodic functions, combination of, 225 Periodic time, 155 Permutations, 511 Perpendicular axis theorem, 370 Phasor, 146, 226 Physical constants, 5 quantities, 9 Pictograms, 497 Pie diagram, 497 Plotting periodic functions, 225 Poisson distribution, 495, 516 Polar complex numbers, 208 co-ordinates, 142 curves, 190 form, 208 second moment of area, 373 Poles, 428 Pole-zero diagram, 429 Pol/Rec function on calculator, 210 Polygon, 103 frequency, 500, 501 Polynomial, 22 division, 22 Legendre’s, 408 Power of complex number, 212 Power series for ex, 54 methods of solving differential equations, 400–408 by Bessel, 406 by Frobenius’s method, 403

553

by Leibniz-Maclaurin method, 401 by Leibniz theorem, 400 Powers of complex numbers, 212 Practical problems, binomial series, 83 quadratic equations, 48, 49 simple equations, 30 simultaneous equations, 39 straight line graphs, 173 trigonometry, 145–148 Precedence, 21 Prefixes, 5 Presentation of grouped data, 500 ungrouped data, 497 Principal value, 166 Principle of moments, 30 Prism, 111 Prismoidal rule, 126 Probability, 509 laws of, 509 Product-moment formula, 522 Product rule of differentiation, 276 Properties of triangles, 133 z-transforms, 451 Pyramid, 111 volume and surface area of frustum, 116 Pythagoras, theorem of, 135 Quadratic equations, 42 by completing the square, 45 factorisation, 42 formula, 47 graphically, 184 practical problems, 48, 49 Quadratic formula, 47 graphs, 184 inequalities, 79 Quadrilaterals, 103 Quantities and their units, 3–5 Quartiles, 507 Quotient, 22 rule of differentiation, 296 Quotients, 26 inequalities involving, 76 Radian, 106 Radius of curvature, 290 of gyration, 369 Range, 501 Rates of change, 280 using partial differentiation, 306 Real part of a complex number, 205 Reciprocal ratios, 137

554

Mathematics Pocket Book for Engineers and Scientists Index

Recommended mathematical symbols, 7 Rectangle, 103 Rectangular complex numbers, 205 hyperbola, 288 prism, 111 Recurrence formula, 401 relation, 401 Reduction formulae, 337 of non-linear laws to linear form, 174 Reflex angle, 133 Regression, coefficients, 524 linear, 524 Relationship between trigonometric and hyperbolic functions, 161 Relative velocity, 222 Remainder theorem, 25 Resolution of vectors, 219 Right angle, 133 Right-angled triangle, 133 solution of, 139 Rivet, 114 R.m.s. values, 357 Rodrigue’s formula, 408 Roof design, 59 span, 145 Roots of a complex numbers, 212 Rules of algebra, 18 Runge-Kutta method, 388 Saddle point, 309 Sampling and estimation theories, 525 Scalar product, 227 application of, 229 quantity, 217 Scalene triangle, 134 Secant, 136 Sech, 58 Second moment of area, 369 order differential equations, 391, 395 partial derivatives, 307 shift theorem, 453 Sector, 107 area of, 107 Semi-interquartile range, 508 Separation of variables, 379 Sequences, 68, 448 Series, binomial, 80 exponential, 54 Maclaurin’s, 84 sinh and cosh, 59 Shift theorem, z-transforms, 452, 453 Sign test, 535

Similar shapes, areas of, 106 triangles, 134 Simple equations, 27 practical problems, 30 inequalities, 74 sequences, 68 Simplifying Boolean expressions, 257 de Morgan’s rule, 259 Karnaugh maps, 261 laws and rules of Boolean algebra, 258 Simpson’s rule, 123, 124 Simultaneous differential equations by Laplace transforms, 442 Simultaneous equations, 36 by Cramers rule, 245 by determinants, 242 by elimination, 36, 37 by Gaussian elimination, 247 by matrices, 239 by substitution, 36, 37 graphical solution, 183 in three unknowns, 41 in two unknowns, 37, 183 practical problems, 39 SI units, 3, 4 Sine, 136 curves, 152 rule, 144 series, Fourier, 471 wave, 128 Sine θ substitution, 324 Sinh, 59 series, 59 Sinh θ substitution, 324 Sinusoidal form, A sin(ωt ± α), 153 Small changes, using differentiation, 287 using partial differentiation, 307 Solid of revolution, 261 Solution of any triangle, 152 differential equations, 375–417 using Laplace transforms, 439 right-angled triangles, 139 simple equations, 27 simultaneous equations, 36, 41 Space diagram, 223 Speed, 2 Sphere, 112 frustum of, 119 Spherical storage tank, 121 Square functions, inequalities involving, 77 Standard derivatives, 275 deviation, 504

Statistics and probability Index

discrete data, 504 grouped data, 505 error of the means, 525 integration, 317 Laplace transforms, 419 Stationary points, 282 Statistical data, presentation of, 494 table of normal curve, 518, 519 Statistics and probability, 494 Straight line, equation of, 172 Straight line graphs, 172 practical problems, 173 Stress in thick cylinder, 31 Student’s t distribution, 529 Subject of formulae, 32 Subtraction of matrices, 234 of vectors, 220 Successive approximations, 90 differentiation, 278 Sum of n terms of a G.P., 71 an A.P., 70 Sum to infinity of GP, 72 Supplementary angles, 133 Surd form, 138 Surface areas of frusta of pyramids and cones, 116 solids, 111 zone of a sphere, 119 Surface tension, 290 Switching circuits, 254 Symbols, for physical quantities, 9 Symmetry relationships, Fourier series, 484 Table of normal curve, 518–519 z-transforms, 449–450 Tally diagram, 501 Tangent, 136 equation of, 285 graph of, 149 Tangential velocity, 232 Tanh, 58 Tan θ substitution, 324 Tan θ/2 substitution, 331 Tapered groove, 109 Telegraph wire, 59 Testing for a normal distribution, 521 Theorems: binomial, 80 Maclaurin’s, 84 Pappus, 365 Parallel axis, 370 Perpendicular axis, 370 Pythagoras, 135

555

Total differential, 305 Transfer function, 429 Translation, z-transform, 454 Transmission lines, 59 Transpose of matrix, 238 Transposing formulae, 32 Transposition, 32 Transversal, 133 Trapezium, 103 Trapezoidal rule, 122, 487 Triangle, 103 area of, 103, 145 congruent, 134 properties of, 133 similar, 134 Trigonometric and hyperbolic substitutions, integration, 323 equations, 157 identities, 157 inverse function, 200, 295 ratios, 136 evaluation of, 137 waveforms, 149 Trigonometry, 131 practical situations, 145–148 Triple integrals, 344 Truth tables, 254, 255 t = tan θ/2 substitution, 331 Turning points, 282 Types and properties of angles, 133 Ungrouped data, 497 Unit matrix, 236 step function, 431 Universal logic gates, 269 Upper class boundary value, 503 Vector addition, 217 nose-to-tail method, 217 parallelogram method, 218 Vector drawing, 217, 218 products, 230 applications of, 231, 232 quantities, 217 subtraction, 220 Vectors, 216 addition of, 217 by calculation, 217 by drawing, 217, 218 by horizontal and vertical components, 219 Velocity and acceleration, 281

556

Mathematics Pocket Book for Engineers and Scientists Index

Velocity, angular, 155 relative, 222 Vertical-axis intercept, 172 Vertical bar chart, 497 component, 219 Vertically opposite angles, 133 Volume, 2 Volumes in metric units, 2 conversions, 2 Volumes of frusta of pyramids and cones, 116 irregular solids, 124 pyramids, 111 regular solids, 111 similar shapes, 115 solids of revolution, 359 zone of a sphere, 119 Wallis’s formula, 340 Wave equation, 410

Waveform addition, 225 Wavelength, 32 Wilcoxon signed-rank test, 537 Work done, 229 Xor-gate, 267 Xnor-gate, 267 y-axis intercept, 172 Zeros (and poles), 429 Zone of a sphere, 119 Z-transform pair, 448 Z-transforms, 447 definition of, 448 inverse, 456 properties of, 451 table of, 449–450 to solve difference equations, 460