Engineering Math Review

Engineering Mathematics Dr Colin Turner October 15, 2009

i Copyright Notice The contents of this document are protected by a Creative Commons License, that allows copying and modification with attribution, but not commercial reuse. Please contact me for more details.

http://creativecommons.org/licenses/by-nc/2.0/uk/ Queries Should you have any queries about these notes, you should approach your lecturer or your tutor as soon as possible. Don’t be afraid to ask questions, it is possible you may have found an error, and if you have not, your questions will help your lecturer / tutor understand the problems you are experiencing. As mathematics is cumulative, it will be very hard to continue the module with outstanding problems from the start, a bit of work at this point will make the rest much easier going. Practice Mathematics requires practice. No matter how simple a procedure may look when demonstrated in a lecture or a tutorial you can have no idea how well you can perform it until you try. As there is very little opportunity for practice in the university environment it is vitally important that you attempt the questions provided in the tutorial, preferably before attending the relevant tutorial class. Your time with your tutor will be best spent when you arrive at the class with a list of problems you are unable to tackle, the more specific the better. If you find the questions too hard before the tutorial, do not become discouraged, the mere act of thinking about the problem will have a positive affect on your understanding of the problem once explained to you in the tutorial. Contact Details My contact details are as follows Name Dr Colin Turner Room 5F10 Phone 68084 (+44-28-9036-8084 externally) Email [email protected] WWW http://newton.engj.ulst.ac.uk/crt/

Contents 1 Preliminaries 1.1 Introduction . . . . . . . . . 1.2 Notation . . . . . . . . . . . 1.3 Arithmetic . . . . . . . . . . 1.3.1 The law of signs . . . 1.3.2 Order of precedence . 1.4 Decimal Places & Significant 1.4.1 Decimal Places . . . 1.4.2 Significant Figures . 1.5 Standard Form . . . . . . . 1.5.1 Standard prefixes . . 2 Number Systems 2.1 Natural numbers . 2.2 Prime numbers . . 2.3 Integers . . . . . . 2.4 Real numbers . . . 2.5 Rational numbers . 2.6 Irrational Numbers

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3 Basic Algebra 3.1 Rearranging Equations . . . 3.1.1 Example . . . . . . . 3.1.2 Order of Rearranging 3.1.3 Example . . . . . . . 3.1.4 Example . . . . . . . 3.1.5 Example . . . . . . . 3.2 Function Notation . . . . .

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1 1 1 2 2 3 3 3 4 5 5

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11 11 11 12 13 14 15 15

CONTENTS 3.3

Expansion of Brackets . . . . . . . . . . 3.3.1 Examples . . . . . . . . . . . . . 3.3.2 Brackets upon Brackets . . . . . . 3.3.3 Examples . . . . . . . . . . . . . 3.4 Factorization . . . . . . . . . . . . . . . 3.4.1 Examples . . . . . . . . . . . . . 3.5 Laws of Indices . . . . . . . . . . . . . . 3.5.1 Example “proofs” . . . . . . . . . 3.5.2 Examples . . . . . . . . . . . . . 3.6 Laws of Surds . . . . . . . . . . . . . . . 3.6.1 Examples . . . . . . . . . . . . . 3.7 Quadratic Equations . . . . . . . . . . . 3.7.1 Examples . . . . . . . . . . . . . 3.7.2 Graphical interpretation . . . . . 3.7.3 Factorization . . . . . . . . . . . 3.7.4 Quadratic solution formula . . . . 3.7.5 The discriminant . . . . . . . . . 3.7.6 Examples . . . . . . . . . . . . . 3.7.7 Special cases . . . . . . . . . . . . 3.8 Notation . . . . . . . . . . . . . . . . . . 3.8.1 Modulus or absolute value . . . . 3.8.2 Sigma notation . . . . . . . . . . 3.8.3 Factorials . . . . . . . . . . . . . 3.9 Exponential and Logarithmic functions . 3.9.1 Exponential functions . . . . . . 3.9.2 Logarithmic functions . . . . . . 3.9.3 Logarithms to solve equations . . 3.9.4 Examples . . . . . . . . . . . . . 3.9.5 Anti-logging . . . . . . . . . . . . 3.9.6 Examples . . . . . . . . . . . . . 3.10 Binomial Expansion . . . . . . . . . . . . 3.10.1 Theory . . . . . . . . . . . . . . . 3.10.2 Example . . . . . . . . . . . . . . 3.10.3 Examples . . . . . . . . . . . . . 3.10.4 High values of n . . . . . . . . . . 3.11 Arithmetic Progressions . . . . . . . . . 3.11.1 Examples . . . . . . . . . . . . . 3.11.2 Sum of an arithmetic progression 3.11.3 Example . . . . . . . . . . . . . . 3.11.4 Example . . . . . . . . . . . . . . 3.12 Geometric Progressions . . . . . . . . . .

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17 18 18 19 20 21 21 22 23 23 24 24 24 25 26 27 28 28 29 30 30 31 32 32 32 33 34 35 36 37 38 38 40 40 41 42 42 42 43 44 45

CONTENTS 3.12.1 3.12.2 3.12.3 3.12.4 3.12.5

iv Examples . . . . . . . . . . . . Sum of a geometric progression Sum to infinity . . . . . . . . . Example . . . . . . . . . . . . . Example . . . . . . . . . . . . .

4 Trigonometry 4.1 Right-angled triangles . . . . . . . . 4.1.1 Labelling . . . . . . . . . . . 4.1.2 Pythagoras’ Theorem . . . . . 4.1.3 Basic trigonometric functions 4.1.4 Procedure . . . . . . . . . . . 4.1.5 Example . . . . . . . . . . . . 4.2 Notation . . . . . . . . . . . . . . . . 4.2.1 Example . . . . . . . . . . . . 4.3 Table of values . . . . . . . . . . . . 4.4 Graphs of Functions . . . . . . . . . 4.5 Multiple Solutions . . . . . . . . . . 4.5.1 CAST diagram . . . . . . . . 4.5.2 Procedure . . . . . . . . . . . 4.5.3 Example . . . . . . . . . . . . 4.6 Scalene triangles . . . . . . . . . . . 4.6.1 Labelling . . . . . . . . . . . 4.6.2 Scalene trigonmetry . . . . . . 4.6.3 Sine Rule . . . . . . . . . . . 4.6.4 Cosine Rule . . . . . . . . . . 4.6.5 Example . . . . . . . . . . . . 4.7 Radian Measure . . . . . . . . . . . . 4.7.1 Conversion . . . . . . . . . . . 4.7.2 Length of Arc . . . . . . . . . 4.7.3 Area of Sector . . . . . . . . . 4.8 Identities . . . . . . . . . . . . . . . . 4.8.1 Basic identities . . . . . . . . 4.8.2 Compound angle identities . . 4.8.3 Double angle identities . . . . 4.9 Trigonmetric equations . . . . . . . . 4.9.1 Example . . . . . . . . . . . . 4.9.2 Example . . . . . . . . . . . . 4.9.3 Example . . . . . . . . . . . .

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49 49 49 50 50 51 51 52 53 53 53 54 55 56 56 58 58 58 58 59 60 61 61 61 62 63 64 64 64 65 65 65 66

CONTENTS

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5 Complex Numbers 5.1 Basic Principle . . . . . . . . . . . . . . 5.1.1 Imaginary and Complex Numbers 5.2 Examples . . . . . . . . . . . . . . . . . 5.3 Argand Diagram Representation . . . . . 5.4 Algebra of Complex Numbers . . . . . . 5.4.1 Addition . . . . . . . . . . . . . . 5.4.2 Subtraction . . . . . . . . . . . . 5.4.3 Multiplication . . . . . . . . . . . 5.4.4 Division . . . . . . . . . . . . . . 5.4.5 Examples . . . . . . . . . . . . . 5.5 Definitions . . . . . . . . . . . . . . . . . 5.5.1 Modulus . . . . . . . . . . . . . . 5.5.2 Conjugate . . . . . . . . . . . . . 5.5.3 Real part . . . . . . . . . . . . . 5.5.4 Imaginary part . . . . . . . . . . 5.6 Representation . . . . . . . . . . . . . . 5.6.1 Cartesian form . . . . . . . . . . 5.6.2 Polar form . . . . . . . . . . . . . 5.6.3 Exponential form . . . . . . . . . 5.6.4 Examples . . . . . . . . . . . . . 5.6.5 Examples . . . . . . . . . . . . . 5.7 De Moivre’s Theorem . . . . . . . . . . . 5.7.1 Examples . . . . . . . . . . . . . 5.7.2 Roots of Unity . . . . . . . . . . 5.7.3 Roots of other numbers . . . . . . 5.8 Trigonometric functions . . . . . . . . . 6 Vectors & Matrices 6.1 Vectors . . . . . . . . . . . . . 6.1.1 Modulus . . . . . . . . 6.1.2 Unit Vector . . . . . . 6.1.3 Cartesian unit vectors 6.1.4 Examples . . . . . . . 6.1.5 Signs of vectors . . . . 6.1.6 Addition . . . . . . . . 6.1.7 Subtraction . . . . . . 6.1.8 Zero vector . . . . . . 6.1.9 Scalar Product . . . . 6.1.10 Example . . . . . . . .

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CONTENTS

6.2

6.3

6.4

6.5

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6.7

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6.1.11 Example . . . . . . . . . . 6.1.12 Vector Product . . . . . . 6.1.13 Example . . . . . . . . . . Matrices . . . . . . . . . . . . . . 6.2.1 Square matrices . . . . . . 6.2.2 Row and Column vectors . 6.2.3 Examples . . . . . . . . . 6.2.4 Zero and Identity . . . . . Matrix Arithmetic . . . . . . . . 6.3.1 Addition . . . . . . . . . . 6.3.2 Examples . . . . . . . . . 6.3.3 Subtraction . . . . . . . . 6.3.4 Examples . . . . . . . . . 6.3.5 Multiplication by a scalar 6.3.6 Examples . . . . . . . . . 6.3.7 Domino Rule . . . . . . . 6.3.8 Multiplication . . . . . . . 6.3.9 Examples . . . . . . . . . 6.3.10 Exercise . . . . . . . . . . Determinant of a matrix . . . . . 6.4.1 Examples . . . . . . . . . 6.4.2 Sign rule for matrices . . . 6.4.3 Order 3 . . . . . . . . . . 6.4.4 Examples . . . . . . . . . 6.4.5 Order 4 . . . . . . . . . . Inverse of a matrix . . . . . . . . 6.5.1 Order 2 . . . . . . . . . . 6.5.2 Examples . . . . . . . . . 6.5.3 Other orders . . . . . . . . 6.5.4 Exercise . . . . . . . . . . Matrix algebra . . . . . . . . . . 6.6.1 Addition . . . . . . . . . . 6.6.2 Multiplication . . . . . . . 6.6.3 Mixed . . . . . . . . . . . Solving equations . . . . . . . . . 6.7.1 Example . . . . . . . . . . 6.7.2 Example . . . . . . . . . . 6.7.3 Row reduction . . . . . . . Row Operations . . . . . . . . . . 6.8.1 Determinants . . . . . . . 6.8.2 Example . . . . . . . . . .

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6.10 6.11

6.12

6.13

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Solving systems of equations . . . . . . . . 6.9.1 Gaussian Elimination . . . . . . . . 6.9.2 Example . . . . . . . . . . . . . . . 6.9.3 Example . . . . . . . . . . . . . . . 6.9.4 Example . . . . . . . . . . . . . . . 6.9.5 Number of equations vs. unknowns Inversion by Row operations . . . . . . . . Rank . . . . . . . . . . . . . . . . . . . . . 6.11.1 Example . . . . . . . . . . . . . . . 6.11.2 Systems of equations . . . . . . . . 6.11.3 Example . . . . . . . . . . . . . . . 6.11.4 Example . . . . . . . . . . . . . . . 6.11.5 Example . . . . . . . . . . . . . . . 6.11.6 Summary . . . . . . . . . . . . . . 6.11.7 Exercise . . . . . . . . . . . . . . . Eigenvalues and Eigenvectors . . . . . . . 6.12.1 Finding Eigenvalues . . . . . . . . 6.12.2 Example . . . . . . . . . . . . . . . 6.12.3 Finding eigenvectors . . . . . . . . 6.12.4 Example . . . . . . . . . . . . . . . 6.12.5 Example . . . . . . . . . . . . . . . 6.12.6 Other orders . . . . . . . . . . . . . Diagonalisation . . . . . . . . . . . . . . . 6.13.1 Powers of diagonal matrices . . . . 6.13.2 Example . . . . . . . . . . . . . . . 6.13.3 Powers of other matrices . . . . . . 6.13.4 Example . . . . . . . . . . . . . . .

7 Graphs of Functions 7.1 Simple graph plotting . . . . . . . 7.1.1 Example . . . . . . . . . . 7.1.2 Example . . . . . . . . . . 7.1.3 Example . . . . . . . . . . 7.2 Important functions . . . . . . . . 7.2.1 Direct Proportion . . . . . 7.2.2 Inverse Proportion . . . . 7.2.3 Inverse Square Proportion 7.2.4 Exponential Functions . . 7.2.5 Logarithmic Functions . . 7.3 Transformations on graphs . . . .

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CONTENTS

7.4

7.3.1 7.3.2 7.3.3 7.3.4 Even 7.4.1 7.4.2 7.4.3 7.4.4

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Addition or Subtraction . . . Multiplication or Division . . Adding to or Subtracting from Multiplying or Dividing x . . and Odd functions . . . . . . . Even functions . . . . . . . . Odd functions . . . . . . . . . Combinations of functions . . Examples . . . . . . . . . . .

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8 Coordinate geometry 8.1 Elementary concepts . . . . . . . . . 8.1.1 Distance between two points . 8.1.2 Example . . . . . . . . . . . . 8.1.3 Example . . . . . . . . . . . . 8.1.4 Midpoint of two points . . . . 8.1.5 Example . . . . . . . . . . . . 8.1.6 Example . . . . . . . . . . . . 8.1.7 Gradient . . . . . . . . . . . . 8.1.8 Example . . . . . . . . . . . . 8.1.9 Example . . . . . . . . . . . . 8.2 Equation of a straight line . . . . . . 8.2.1 Meaning of equation of line . 8.2.2 Finding the equation of a line 8.2.3 Example . . . . . . . . . . . . 8.2.4 Example . . . . . . . . . . . .

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9 Differential Calculus 9.1 Concept . . . . . . . . . . . . . . 9.2 Notation . . . . . . . . . . . . . . 9.3 Rules & Techniques . . . . . . . . 9.3.1 Power Rule . . . . . . . . 9.3.2 Addition and Subtraction 9.3.3 Constants upon functions 9.3.4 Chain Rule . . . . . . . . 9.3.5 Product Rule . . . . . . . 9.3.6 Quotient Rule . . . . . . . 9.3.7 Trigonometric Rules . . . 9.3.8 Exponential Rules . . . . 9.3.9 Logarithmic Rules . . . .

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149 149 149 150 150 151 151 151 151 152 152 152 152

CONTENTS 9.4 9.5

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Examples . . . . . . . . . . . . . . . . 9.4.1 Solutions . . . . . . . . . . . . . Tangents . . . . . . . . . . . . . . . . . 9.5.1 Example . . . . . . . . . . . . . 9.5.2 Example . . . . . . . . . . . . . Turning Points . . . . . . . . . . . . . 9.6.1 Types of turning point . . . . . 9.6.2 Finding turning points . . . . . 9.6.3 Classification of turning points . 9.6.4 Example . . . . . . . . . . . . . 9.6.5 Example . . . . . . . . . . . . . 9.6.6 Example . . . . . . . . . . . . . Newton Rhapson . . . . . . . . . . . . 9.7.1 Example . . . . . . . . . . . . . 9.7.2 Example . . . . . . . . . . . . . Partial Differentiation . . . . . . . . . 9.8.1 Example . . . . . . . . . . . . . Small Changes . . . . . . . . . . . . . 9.9.1 Example . . . . . . . . . . . . . 9.9.2 Example . . . . . . . . . . . . .

10 Integral Calculus 10.1 Concept . . . . . . . . . . . . . . . 10.1.1 Constant of Integration . . . 10.2 Rules & Techniques . . . . . . . . . 10.2.1 Power Rule . . . . . . . . . 10.2.2 Addition & Subtraction . . 10.2.3 Multiplication by a constant 10.2.4 Substitution . . . . . . . . . 10.2.5 Limited Chain Rule . . . . . 10.2.6 Logarithm rule . . . . . . . 10.2.7 Partial Fractions . . . . . . 10.2.8 Integration by Parts . . . . 10.2.9 Other rules . . . . . . . . . 10.3 Examples . . . . . . . . . . . . . . 10.3.1 Examples . . . . . . . . . . 10.3.2 Examples . . . . . . . . . . 10.3.3 Example . . . . . . . . . . . 10.3.4 Example . . . . . . . . . . . 10.3.5 Examples . . . . . . . . . .

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152 153 156 157 157 158 158 159 160 161 162 163 163 164 165 166 167 168 168 169

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172 172 172 173 173 173 174 174 174 175 176 177 178 179 179 180 182 183 184

CONTENTS

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10.4 Definite Integration . . . . . . 10.4.1 Notation . . . . . . . . 10.4.2 Concept . . . . . . . . 10.4.3 Areas . . . . . . . . . 10.4.4 Example . . . . . . . . 10.4.5 Example . . . . . . . . 10.4.6 Volumes of Revolution 10.4.7 Example . . . . . . . . 10.4.8 Mean Values . . . . . . 10.4.9 Example . . . . . . . . 10.4.10 Example . . . . . . . . 10.4.11 RMS Values . . . . . . 10.4.12 Example . . . . . . . . 10.4.13 Example . . . . . . . . 10.5 Numerical Integration . . . . 10.5.1 Simpson’s rule . . . . . 10.5.2 Example . . . . . . . . 11 Power Series 11.1 Definition . . . . . . . 11.1.1 Convergence . . 11.2 Maclaurin’s Expansion 11.2.1 Odd and Even . 11.2.2 Example . . . . 11.2.3 Exercise . . . . 11.2.4 Example . . . . 11.2.5 Exercise . . . . 11.2.6 Example . . . . 11.3 Taylor’s Expansion . . 11.3.1 Example . . . . 11.3.2 Identification of

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12 Differential Equations 12.1 Concept . . . . . . . . . 12.2 Exact D.E.s . . . . . . . 12.2.1 Example . . . . . 12.2.2 Example . . . . . 12.2.3 Example . . . . . 12.3 Variables separable D.E.s 12.3.1 Example . . . . .

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185 186 186 186 186 186 187 187 188 188 188 189 189 190 191 192 192

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200 200 200 201 201 202 203 203

CONTENTS 12.3.2 Example . . . . . 12.4 First order linear D.E.s . 12.4.1 Example . . . . . 12.4.2 Example . . . . . 12.5 Second order D.E.s . . . 12.5.1 Homogenous D.E. 12.5.2 Example . . . . . 12.5.3 Example . . . . . 12.5.4 Example . . . . . 12.5.5 Example . . . . . 12.5.6 Example . . . . .

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13 Differentiation in several variables 13.1 Partial Differentiation . . . . . . 13.1.1 Procedure . . . . . . . . . 13.1.2 Examples . . . . . . . . . 13.1.3 Notation . . . . . . . . . . 13.1.4 Higher Derivatives . . . . 13.1.5 Example . . . . . . . . . . 13.2 Taylor’s Theorem . . . . . . . . . 13.3 Stationary Points . . . . . . . . . 13.3.1 Types of points . . . . . . 13.3.2 Finding points . . . . . . . 13.3.3 Classifying points . . . . . 13.3.4 Summary . . . . . . . . . 13.3.5 Example . . . . . . . . . . 13.4 Implicit functions . . . . . . . . . 13.5 Lagrange Multipliers . . . . . . . 13.5.1 Example . . . . . . . . . . 13.6 Jacobians . . . . . . . . . . . . . 13.6.1 Differential . . . . . . . . 13.7 Parametric functions . . . . . . . 13.7.1 Example . . . . . . . . . . 13.8 Chain Rule . . . . . . . . . . . . 14 Integration in several variables 14.1 Double integrals . . . . . . . . 14.1.1 Example . . . . . . . . 14.2 Change of order . . . . . . . . 14.3 Examples . . . . . . . . . . .

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204 206 207 208 209 209 211 211 212 212 213

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14.3.1 Example . . . . . . . . . . . . 14.3.2 Example . . . . . . . . . . . . 14.4 Triple integrals . . . . . . . . . . . . 14.4.1 Example . . . . . . . . . . . . 14.5 Change of variable . . . . . . . . . . 14.5.1 Polar coordinates . . . . . . . 14.5.2 Example . . . . . . . . . . . . 14.5.3 Cylindrical Polar Coordinates 14.5.4 Spherical Polar Coordinates . 15 Fourier Series 15.1 Periodic functions . . . . . . . . . 15.1.1 Example . . . . . . . . . . 15.1.2 Example . . . . . . . . . . 15.2 Sets of functions . . . . . . . . . . 15.2.1 Orthogonal functions . . . 15.2.2 Orthonormal functions . . 15.2.3 Norm of a function . . . . 15.3 Fourier concepts . . . . . . . . . . 15.3.1 Fourier coefficents . . . . . 15.3.2 Fourier series . . . . . . . 15.3.3 Convergence . . . . . . . . 15.4 Important functions . . . . . . . . 15.4.1 Trigonometric system . . . 15.4.2 Exponential system . . . . 15.5 Trigonometric expansions . . . . . 15.5.1 Even functions . . . . . . 15.5.2 Odd functions . . . . . . . 15.5.3 Other Ranges . . . . . . . 15.6 Harmonics . . . . . . . . . . . . . 15.6.1 Odd and Even Harmonics 15.6.2 Trigonometric system . . . 15.6.3 Exponential system . . . . 15.6.4 Percentage harmonic . . . 15.7 Examples . . . . . . . . . . . . . 15.7.1 Example . . . . . . . . . . 15.8 Exponential Series . . . . . . . .

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CONTENTS 16 Laplace transforms 16.1 Definition . . . . . . . . . . . . . . . 16.1.1 Example . . . . . . . . . . . . 16.1.2 Example . . . . . . . . . . . . 16.1.3 Example . . . . . . . . . . . . 16.1.4 Inverse Transform . . . . . . . 16.1.5 Elementary properties . . . . 16.1.6 Example . . . . . . . . . . . . 16.2 Important Transforms . . . . . . . . 16.2.1 First shifting property . . . . 16.2.2 Further Laplace transforms . 16.3 Transforming derivatives . . . . . . . 16.3.1 First derivative . . . . . . . . 16.3.2 Second derivative . . . . . . . 16.3.3 Higher derivatives . . . . . . . 16.4 Transforming integrals . . . . . . . . 16.5 Differential Equations . . . . . . . . . 16.5.1 Example . . . . . . . . . . . . 16.5.2 Example . . . . . . . . . . . . 16.5.3 Example . . . . . . . . . . . . 16.5.4 Example . . . . . . . . . . . . 16.5.5 Exercise . . . . . . . . . . . . 16.5.6 Example . . . . . . . . . . . . 16.5.7 Example . . . . . . . . . . . . 16.6 Other theorems . . . . . . . . . . . . 16.6.1 Change of Scale . . . . . . . . 16.6.2 Derivative of the transform . . 16.6.3 Convolution Theorem . . . . . 16.6.4 Example . . . . . . . . . . . . 16.6.5 Example . . . . . . . . . . . . 16.6.6 Example . . . . . . . . . . . . 16.7 Heaviside unit step function . . . . . 16.7.1 Laplace transform of u(t − c) 16.7.2 Example . . . . . . . . . . . . 16.7.3 Example . . . . . . . . . . . . 16.7.4 Delayed functions . . . . . . . 16.7.5 Example . . . . . . . . . . . . 16.8 The Dirac Delta . . . . . . . . . . . . 16.8.1 Delayed impulse . . . . . . . . 16.8.2 Example . . . . . . . . . . . . 16.9 Transfer Functions . . . . . . . . . .

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248 248 248 249 249 250 250 250 251 251 253 254 254 254 254 254 255 255 256 258 260 261 261 262 263 263 263 264 264 264 265 265 268 268 270 270 271 272 273 274 274

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16.9.1 Impulse Response . . . . . . . . . . . . . . . . . . . . . 276 16.9.2 Initial value theorem . . . . . . . . . . . . . . . . . . . 277 16.9.3 Final value theorem . . . . . . . . . . . . . . . . . . . . 277 17 Z-transform 17.1 Concept . . . . . . . . . . . . . 17.2 Important Z-transforms . . . . 17.2.1 Unit step function . . . 17.2.2 Linear function . . . . . 17.2.3 Exponential function . . 17.2.4 Elementary properties . 17.2.5 Real translation theorem 18 Statistics 18.1 Sigma Notation . . . . . . . . . 18.1.1 Example . . . . . . . . . 18.2 Populations and Samples . . . . 18.2.1 Sampling . . . . . . . . 18.3 Parameters and Statistics . . . 18.4 Frequency . . . . . . . . . . . . 18.5 Measures of Location . . . . . . 18.5.1 Arithmetic Mean . . . . 18.5.2 Mode . . . . . . . . . . . 18.5.3 Median . . . . . . . . . 18.5.4 Example . . . . . . . . . 18.5.5 Example . . . . . . . . . 18.6 Measures of Dispersion . . . . . 18.6.1 Range . . . . . . . . . . 18.6.2 Standard deviation . . . 18.6.3 Inter-quartile range . . . 18.7 Frequency Distributions . . . . 18.7.1 Class intervals . . . . . . 18.8 Cumulative frequency . . . . . . 18.8.1 Calculating the median . 18.8.2 Calculating quartiles . . 18.8.3 Calculating other ranges 18.9 Skew . . . . . . . . . . . . . . . 18.10Correlation . . . . . . . . . . . 18.10.1 Linear regression . . . . 18.10.2 Correlation coefficient .

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279 . 279 . 281 . 281 . 282 . 283 . 283 . 283

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285 . 285 . 285 . 287 . 287 . 288 . 288 . 288 . 289 . 289 . 289 . 289 . 290 . 290 . 291 . 291 . 292 . 292 . 292 . 293 . 293 . 294 . 294 . 294 . 294 . 295 . 295

CONTENTS

xv

19 Probability 19.1 Events . . . . . . . . . . . . . . . . . . . . 19.1.1 Probability of an Event . . . . . . . 19.1.2 Exhaustive lists . . . . . . . . . . . 19.2 Multiple Events . . . . . . . . . . . . . . . 19.2.1 Notation . . . . . . . . . . . . . . . 19.2.2 Relations between events . . . . . . 19.3 Probability Laws . . . . . . . . . . . . . . 19.3.1 A or B (mutually exclusive events) 19.3.2 not A . . . . . . . . . . . . . . . . 19.3.3 1 event of N . . . . . . . . . . . . . 19.3.4 n events of N . . . . . . . . . . . . 19.3.5 Examples . . . . . . . . . . . . . . 19.3.6 A and B (independent events) . . . 19.3.7 Example . . . . . . . . . . . . . . . 19.3.8 A or B or C or ... . . . . . . . . . . 19.3.9 A and B and C and ... . . . . . . . 19.3.10 Example . . . . . . . . . . . . . . . 19.3.11 A or B revisited . . . . . . . . . . . 19.3.12 Example . . . . . . . . . . . . . . . 19.3.13 A and B revisited . . . . . . . . . . 19.3.14 Conditional probability . . . . . . . 19.3.15 Example . . . . . . . . . . . . . . . 19.3.16 Bayes Theorem . . . . . . . . . . . 19.4 Discrete Random Variables . . . . . . . . . 19.4.1 Notation . . . . . . . . . . . . . . . 19.4.2 Expected Value . . . . . . . . . . . 19.4.3 Variance . . . . . . . . . . . . . . . 19.4.4 Example . . . . . . . . . . . . . . . 19.5 Continuous Random Variables . . . . . . . 19.5.1 Definition . . . . . . . . . . . . . . 19.5.2 Probability Density Function . . . 20 The Normal Distribution 20.1 Definition . . . . . . . . . . . 20.2 Standard normal distribution 20.2.1 Transforming variables 20.2.2 Calculation of areas . . 20.2.3 Example . . . . . . . . 20.2.4 Confidence limits . . .

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297 297 297 297 298 298 298 299 299 299 300 300 300 300 301 301 301 302 302 302 303 303 304 304 304 304 305 306 307 308 309 309

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310 . 310 . 311 . 311 . 312 . 312 . 313

CONTENTS 20.2.5 Sampling distribution . . . . . 20.3 The central limit theorem . . . . . . 20.4 Finding the Population mean . . . . 20.5 Hypothesis Testing . . . . . . . . . . 20.5.1 Two tailed tests . . . . . . . . 20.6 Difference of two normal distributions

xvi . . . . . .

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313 314 315 315 316 317

A Statistical Tables

318

B Greek Alphabet

321

List of Tables 1.1 1.2 1.3 1.4 1.5

Basic notation . . . . . . . . . The law of signs . . . . . . . . Order of precedence . . . . . . SI prefixes for large numbers . SI prefixes for small numbers .

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2 2 3 6 7

3.1 3.2 3.3 3.4 3.5

The laws of indices . . . . . . . . The laws of surds . . . . . . . . . Examples of quadratic equations . Laws of Logarithms . . . . . . . . Pascal’s Triangle . . . . . . . . .

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21 23 25 34 39

4.1 4.2

Table of trigonometric values . . . . . . . . . . . . . . . . . . . 54 Conversion between degrees and radians . . . . . . . . . . . . 62

5.1

Powers of j . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

6.1 6.2 6.3

Matrix algebra - Addition . . . . . . . . . . . . . . . . . . . . 104 Matrix algebra - Multiplication . . . . . . . . . . . . . . . . . 105 Matrix algebra - Mixed operations . . . . . . . . . . . . . . . . 105

7.1 7.2

Adding and Subtracting even and odd functions . . . . . . . . 139 Multiplying even and odd functions . . . . . . . . . . . . . . . 140

9.1 9.2

Second derivative test . . . . . . . . . . . . . . . . . . . . . . . 160 First derivative turning point classification . . . . . . . . . . . 161

15.1 Symmetry in Fourier Series . . . . . . . . . . . . . . . . . . . . 244 16.1 Common Laplace transforms . . . . . . . . . . . . . . . . . . . 252 16.2 Further Laplace transforms . . . . . . . . . . . . . . . . . . . . 253 16.3 Examples of transfer function denominators . . . . . . . . . . 276

LIST OF TABLES 18.1 An example of class intervals . . . . . . . . . . . . . . . . . . . 293 19.1 Probabilities for total of two rolled dice . . . . . . . . . . . . . 305 19.2 Calculating E(X) and var (X) for two rolled dice. . . . . . . . 308 A.1 Table of Φ(x) (Normal Distribution) . . . . . . . . . . . . . . 319 A.2 Table of χ2 distribution (Part I) . . . . . . . . . . . . . . . . . 320 A.3 Table of χ2 distribution (Part II) . . . . . . . . . . . . . . . . 320

xviii

List of Figures 2.1

The real number line . . . . . . . . . . . . . . . . . . . . . . .

3.1

The quadratic equation . . . . . . . . . . . . . . . . . . . . . . 25

4.1 4.2 4.3 4.4 4.5 4.6

Labelling right-angled triangles . . . Generating the trigonometric graphs The graphs of sin θ and cos θ . . . . . The “CAST” diagram . . . . . . . . Labelling a scalene triangle . . . . . . Length of Arc, Area of Sector . . . .

5.1 5.2

The Argand diagram . . . . . . . . . . . . . . . . . . . . . . . 71 Polar representation of a complex number . . . . . . . . . . . 77

6.1 6.2

Vector Addition . . . . . . . . . . . . . . . . . . . . . . . . . . 88 Vector Subtraction . . . . . . . . . . . . . . . . . . . . . . . . 89

7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12

The graph of x2 + 2x − 3 . . The graph of 2x + 3 . . . . The graph of x1 . . . . . . . The graph of x12 . . . . . . . The graph of ex . . . . . . . The graph of ln x . . . . . . Closeup of graph of ln x . . Graph of sin x . . . . . . . . Graph of sin x + 1 . . . . . . Graph of 2 sin x . . . . . . . Graph of sin(x + 90) . . . . Closeup of graph of sin(2x) .

8.1

Relationships between two points . . . . . . . . . . . . . . . . 142

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129 130 132 133 134 135 136 137 138 139 140 141

LIST OF FIGURES 9.1 9.2

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Types of turning point . . . . . . . . . . . . . . . . . . . . . . 159 The Newton-Rhapson method . . . . . . . . . . . . . . . . . . 171

13.1 The graph of f (x, y) = xy 2 + 1 . . . . . . . . . . . . . . . . . . 215 13.2 The graph of f (x, y) = x3 (sin xy + 3x + y + 3) . . . . . . . . . 216 13.3 A graph with four turning points . . . . . . . . . . . . . . . . 222 14.1 Double integration over the simple region R. . . . . . . . . . . 230 14.2 Double integration over x then y. . . . . . . . . . . . . . . . . 231 14.3 Double integration over y then x. . . . . . . . . . . . . . . . . 232 16.1 16.2 16.3 16.4 16.5 16.6 16.7 16.8 16.9

An L and R curcuit. . . . . . . . . . . . . . . . . . . . An L, C, and R curcuit. . . . . . . . . . . . . . . . . . The unit step function u(t). . . . . . . . . . . . . . . . The displaced unit step function u(t − c). . . . . . . . . Building functions that are on and off when we please. A positive waveform built from steps. . . . . . . . . . . A waveform built from steps. . . . . . . . . . . . . . . A waveform built from delayed linear functions. . . . . An impulse train built from Dirac deltas. . . . . . . . .

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261 262 266 266 267 269 270 272 274

17.1 A continuous (analog) function . . . . . . . . . . . . . . . . . 279 17.2 Sampling the function . . . . . . . . . . . . . . . . . . . . . . 280 17.3 The digital view . . . . . . . . . . . . . . . . . . . . . . . . . . 280 20.1 The Normal Distribution . . . . . . . . . . . . . . . . . . . . . 310 20.2 Close up of the Normal Distribution . . . . . . . . . . . . . . . 311

Chapter 1 Preliminaries 1.1

Introduction

We shall start the course, by recapping many definitions and results that may already be well known. As mathematics is a cumulative subject, it is necessary however, to ensure that all the basics are in place before we can go on. We assume the reader is familiar with the elementary arithmetic of numbers positive, negative and zero. We also assume the reader is familiar with the decimal representation of numbers, and that they can evaluate simple expressions, including fractional arithmetic.

1.2

Notation

We now list some mathematical notation that we may use in the course, or may be encountered elsewhere, this is shown in table 1.1. Another important bit of notation is “. . . ”, which is used as a sort of mathematician’s etcetera. For example • 1, 2, 3, . . . , 10 is short hand for 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. • 1, 2, 3, . . . is short hand for 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, etc. It’s probably worth noting that in algebra when we use letters to represent numbers, then

1.3 Arithmetic

2

=

equal to

6=

not equal to

<

less than

≤

less than or equal to

>

greater than

≥

greater than or equal to

≡

equivalent to

≈

approximately equal to

⇒

implies

∞ infinity

Σ

sum of what follows

Π

product of what follows

Table 1.1: Basic notation • 3a is a shorthand for 3 × a • a is a shorthand for 1 × a • −a is a shorthand for −1 × a

1.3

Arithmetic

Two often forgotten pieces of arithmetic are:

1.3.1

The law of signs

When we combine signs, either by multiplying two numbers, or by subtracting a negative number for example, we use table 1.2 to determine the sign of the outcome. Put simply, a “−” sign reverses our direction, and so two of them take us back to the “+” direction and so on. +

+

→

+

−

−

→

+

−

+

→

−

+

−

→

−

Table 1.2: The law of signs

1.4 Decimal Places & Significant Figures

1.3.2

Order of precedence

We are familiar with the fact that expressions inside brackets must be evaluated first, that is what the bracket signifies. However, without brackets there is still an inherent order in which operations must be done. Consider this simple calculation 2+3×4 opinion is usually split as to whether the answer is 20, or 14. The reason is that multiplication should be performed before addition, and so the 3 × 4 segment should be calculated first. Be aware that not all calculators understand this, test yours with this calculuation. Calculations should be performed in the order shown in table 1.3. B

Brackets first - they override all priority

O

Order (Powers, roots)

D

Division

M

Multiplication

A

Addition

S

Subtraction Table 1.3: Order of precedence

We note that the table provides us with a handy reminder, BODMAS.

1.4

Decimal Places & Significant Figures

Often we are required to produce answers to a specific degree of accuracy. The most well known way to do this is with the number of decimal places.

1.4.1

Decimal Places

The number of decimal places is a measure of how to truncate answers to a given accuracy. If four decimal places are required then we look at the fifth decimal place and beyond, if it is 5 or greater, we round up the last decimal places that is written, otherwise we simply leave it alone. Let us take an

3

1.4 Decimal Places & Significant Figures example. π is a mathematical constant that continues infinitely through all its decimal places, never repeating its pattern. π = 3.141592653589793 . . . rounded to five decimal places we obtain π ≈ 3.14159 since the next digit is 2, wheras rounding to four decimal places give π ≈ 3.1416 since the next digit is 9 which is clearly bigger than 5. It is good to use enough decimal places to obtain an accurate answer, but one must always remember the context of the answer. There is little point in calculating that the length of a piece of metal should be 2.328745 cm if all we will have to measure it with is a ruler accurate to 1 mm.

1.4.2

Significant Figures

Sometimes decimal places are not the most appropriate way to define accuracy. There is no specific number of decimal places that suit all situations. For example, if we quote the radius of the Earth in metres, then probably no number of decimal places are appropriate for most purposes, as the answer will not be that accurate, and there will be so many other figures before it, they are unlikely to be significant. An alternative often used is to specify a number of significant figures. This is essential the number of non-zero numbers that should be displayed. Suppose that we specify four significant figures. Then the speed of light in m/s is written as: c = 2, 997, 992, 458 ≈ 2, 998, 000, 000 m/s which can be written more simply again in standard form (see below). The issue here is that the other figures are less likely to have any real impact on the answer of a problem. Similarly the standard atomic mass of Uranium is 238.02891 g/mol ≈ 238.0 g/mol since we only have four significant figures, we round after the zero. Note that writing the zero helps indicate the precision of the answer.

4

1.5 Standard Form

1.5

5

Standard Form

In science, large and small numbers are often represented by standard form. This takes the form a.bcd × 10n if we are using four significant figures. For example, we saw above that to four significant figures c = 2, 998, 000, 000 m/s = 2, 998 × 1, 000, 000 m/s or, working a bit more, we move the decimal place each time to the left, (which divides the left hand number by ten), and multiply by another ten on the right to compensate. = 2.998 × 1, 000, 000, 000 m/s now all that remains to do, is to write the number on the right as a power of ten. We count the zeros, there are nine, and so c ≈ 2.998 × 109 m/s. The same applies for small numbers. The light emitted by a Helium-Neon Laser has a wavelength of λ = 0.000, 000, 632, 8 m but this is clearly rather unwieldy to write down. This time we move the decimal place to the right until we get to after the first non-zero digit. Each time we do this we essentially multiply by 10, and so to compensate we have to divide by ten. This can be represented by increasingly large negative values of the power.1 So here, we need to move the decimal place seven times to the right, and so we will multiply by 10−7 . λ = 6.328 × 10−7 m.

1.5.1

Standard prefixes

There are a number of prefixes applied to large and small numbers to allow us to write them more meaningfully. You will have met many of them before. The prefixes for large numbers are shown in table 1.5.1. 1

We will have to wait a while, until 3.5 to see exactly why this is.

1.5 Standard Form

6

Name Prefix In English

Power of Ten

deca

da

tens

101

hecto

h

hundreds

102

kilo

k

thousands

103

Mega

M

millions

106

Giga

G

billions

109

Tera

T

trillions

1012

Peta

P

quadrillions 1015

Exa

E

quintillions

1018

Table 1.4: SI prefixes for large numbers Note that in the past there was a difference between billions as used in British English and American English. The English billion was one million million, wheras the American billion is one thousand million. The latter has won out now, and most references to a billion are to the American one. Also, there is a slight disparity between “normal” quantities, and the bytes used in computing. Since computing is based on binary, and therefore powers of 2, a kilobyte (kB) is not 1000 bytes, but 1024 bytes.2 So in computing, 1024 is used rather than thousands to build up such quantities. The prefixes for large numbers are shown in table 1.5.1. Because of these prefixes, it is normal within engineering to adapt standard scientific form to get the power of ten to be a multiple of three. Let us revisit our wavelength example: λ = 6.328 × 10−7 m. So we would prefer to tweak the power of ten here. We could do this λ = .6328 × 10−6 m = 0.6238 µm, but this is pretty ugly to have a fractional number. It would be more normal to write λ ≈ 632.8 × 10−9 m = 623.8 nm, 2

1024 = 210 and so is the closest power of two

1.5 Standard Form

7

Name Prefix In English

Power of Ten

deci

d

tenths

10−1

centi

c

hundredths

10−2

milli

m

thousandths

10−3

micro

µ

millionths

10−6

nano

n

billionths

10−9

pico

p

trillionths

10−12

femto

f

quadrillionths 10−15

atto

a

quintillionths

10−18

Table 1.5: SI prefixes for small numbers

Chapter 2 Number Systems We remind ourselves of different sets of numbers that we will refer to later.

2.1

Natural numbers

The set of all numbers 1, 2, 3, . . . is known as the set of positive integers, (or natural numbers, or whole numbers, or counting numbers and is denoted by N.

2.2

Prime numbers

A prime number is a positive integer which has exactly two factors 1 , namely itself and one. Thus 2 is the first prime number, and the only even prime number. So the prime numbers are given by 2, 3, 5, 7, 11, 13, 17, 19, . . . 1

Recall that a factor of a number x is one that divides into x with no remainder.

2.3 Integers

2.3

9

Integers

The set of all numbers given by . . . , −3, −2, −1, 0, 1, 2, 3, . . . is known as the set of integers and is denoted by the symbol Z.

2.4

Real numbers

The collection of all numbers in ordinary arithmetic, i.e. including fractions, integers, zero, positive and negative numbers etc. is called the set of real numbers and is denoted R. The set of real numbers can be visualised as a line, called the real number line or simply the real line. Each point on the lines represents a unique real number, and every number, including exotic examples such as π is represented by a unique point on the line.

Figure 2.1: The real number line

2.5

Rational numbers

where m and n are integers, The set of all numbers that can be written m n is known as the set of rational numbers and is noted by Q. (Note that n cannot by zero, as division by zero is not permitted). 165 , −2 = −2 , 0 = 10 are all rational numbers. For example, − 23 , 2096 1 (Note that although division by zero is not peemitted, dividing zero by another number is, and as no other number can fit into zero at all, the result is zero). It turns out that if you add, subtract, divide or multiply any two rational numbers together, you still get a rational number.

2.6 Irrational Numbers

10

There’s an easy way of working out whether a given number is rational or not. Simply expand it in it’s decimal form. Rational numbers always have a decimal expansion that ends, or repeats itself every so many digits. For example 2 − = −0.6666666 . . . 3 3436.234523452345 . . . 34.68 are all rational.

2.6

Irrational Numbers

Of course, not all real numbers are rational, and in fact many numbers you will already have met are not. These numbers are called irrational numbers. √ √ √ Examples are 2, 3, and in fact p where p is prime. When written in their decimal forms, irrationals are never ending and non-repeating. This means that irrational numbers can never be written down exactly. , suggesting that π Although in secondary schools we often write π = 22 7 is rational, this is only a simple (and not very accurate) approximation. In fact π is also irrational, as is Euler’s constant e.

Chapter 3 Basic Algebra Algebra is perhaps the most important part of mathematics to master. If you do not, you will find problems in all the areas you study, all caused by the underlying weakness in your algebra skills. In many ways, algebra is representative of mathematics in that it deals with forming an easy problem out of a difficult one.

3.1

Rearranging Equations

Imagine an equation as a pair of balanced weighing scales. What will happen if we add 2kg on both sides? The scales will remain in balance. If we multiply the weights by three on both sides? The scales will remain in balance. In fact, even if we take the sine of both weights, the scales remain in balance. The leads to the fundamental result you must remember. You can do anything to both sides of an equation and you will obtain an equivalent equation.

3.1.1

Example

We shall look at an extremely trivial example of this concept in use. In our learning of algebraic manipulation we are often told that we can take things across the equals sign and change the sign. Rarely are we told why this works. Let’s examine it.

3.1 Rearranging Equations

12

x+4=9 Well, it is simple to see what value x has in this case. However, we wish to show how rearranging works in these very simple cases. We wish to find x, and this is really saying we want to manipulate the equation into the form: x =?? Where ?? represents the answer. Therefore, we wish to have an x on it’s own, on one side of the equation, with everything else on the other side of the equals sign. To that end, we start to look at what is attached to x, how it is attached, and how we should remove it. In our example, 4 is attached to the x by the process of addition. Now, how do you get rid of a 4 that has been added? Of course, the answer is to subtract it, but we must not simply do this on one side, rather in accordance with 3.1 we must do it on both sides of the equation to maintain its validity. So we obtain x+4−4=9−4 Now the +4 − 4 on the L.H.S. cancel, leaving zero, and this step wouldn’t be written normally. So we finally obtain x=9−4=5 If you observe that it appears that the +4 crossed the equals sign to become a −4 on the R.H.S.. However, now we know what has actually happened.

3.1.2

Order of Rearranging

Of course, in most examples, more than one thing is attached to the x, and usually by a combination of operations. It may be equally correct to rearrange by removing these in any order, but some ways will almost certainly be easier than others. We have already noted in 1.3.2 that some operations are naturally done before others, and so when we see an expression such as: 3x2 − 4 = 8 it really means

3.1 Rearranging Equations

13

((3(x2 )) − 4) = 8 where the brackets serve simply to underline the order in which things are done. The effect is somewhat similar to an onion with the x in the very centre. We could peel the onion from the outside in for the most tidy approach. That is, we remove things in the reverse order to the way they were attached in the first place. So in our simple example, the 4 is subtracted last, so remove it first, (adding 4 on both sides). 3x2 − 4 + 4 = 8 + 4 ⇒ 3x2 = 12 Now the x still has two things attached, the three, which is multiplied on, and the 2 which is a power. Powers are done before multiplication, so we remove in the reverse order again. Therefore we divide by 3 on both sides. 3 2 12 x = ⇒ x2 = 4 3 3 Now we only have one thing “stuck” to the x, and that is the power of 2. To remove this we simply take the square root on both sides: x = ±2. Recall that -2 squared is also 4.

3.1.3

Example

Rearrange the following expression for x. 4x + 6 = 2x − 3 Solution We still wish to rearrange to get x =, but we must notice here that x occurs in two places. We could remove the 3 and 6 on the LHS to obtain 2x − 9 4 (try it as an exercise), but this is not very helpful, as x is now defined in terms of itself, so we still don’t know it’s value. x=

3.1 Rearranging Equations

14

Instead we first gather all the x terms together, and we do this by performing the same operation on both sides. For example, we don’t want the 2x on the RHS, it is positive and so it present by addition. We subtract it on both sides. 4x + 6 − 2x = 2x − 3 − 2x ⇒ 2x + 6 = −3 So we now have a simpler equation, with x only on one side. We can proceed as before now to remove things from the x in the LHS. Subtract 6 on both sides. 2x + 6 − 6 = −3 − 6 ⇒ 2x = −9 Finally divide by 2 on both sides. x=−

3.1.4

9 2

Example

Rearrange the following expression for x: 3(2x + 3) − 6 = 0 Solution In this case we find that x is encased in brackets. To get at x so we can rearrange for it we could multiply out the bracket and rearrange from there. This is left as an exercise for the reader. Another way to deal with it is to think of the bracket as an “onion within an onion” to continue the analogy we began above. Begin by taking things off this bracket, rather than x directly. We add 6 both sides 3(2x + 3) − 6 + 6 = 0 + 6 ⇒ 3(2x + 3) = 6. Now divide by 3 on both sides to obtain 6 3(2x + 3) = ⇒ 2x + 3 = 2. 3 3 Note the brackets can “fall off” at this point naturally. Now we start with out new “onion”, subtracting 3 both sides. 2x + 3 − 3 = 2 − 3 ⇒ 2x = −1

3.2 Function Notation

15

and finally divide by 2 both sides 2x 1 1 =− ⇒x=− 2 2 2

3.1.5

Example

Rearrange the following expression for x: −3 =

10 − 5. x

Solution We have a more serious problem here, namely that x is on the bottom line. We begin by removing the −5 to clarify the equation, by adding 5 on both sides of course. 10 10 −5+5⇒2= x x Now, there’s not much attached to x, but the x is still on the bottom line. That means the x has been divided into something (the 10 in this case). To cancel the division by x, we multiply x on both sides. −3 + 5 =

10x ⇒ 2x = 10 x which simplifies our equation quite a lot. We can now divide by two on both sides to finish. 2x =

10 2x = ⇒x=5 2 2

3.2

Function Notation

Very often when we wish to analyse the behaviour of an expression, we make it a function. A function can be thought of as a box, into which goes a value and out of which comes a, usually different, value. You will have seen functions written as formulae before, for example A = πr2

3.2 Function Notation

16

is a function for calculating the area of a circle. A value goes in (the radius of the circle) and a value comes out (the area of the circle). At times we may use notation such as f (x) to represent a function. For example f (x) = 2x − 3 is a very simple function. For different values we put in (x), we will get different values out f (x). The notation f (x) simply means the value of the output of the function. This notation is very useful when we want to consider specific values that we insert. For example, we write f (2) to mean “find the output value of the function f , when the input value for x is 2”. You can see that we have simply replaced the x by a 2. In our example stated above we get f (2) = 2(2) − 3 = 1 f (−3) = 2(−3) − 3 = −9 f (0) = 2(0) − 3 = −3 f (w) = 2(w) − 3 In the last example, we had to replace x by w, but we can’t work out anything further, so we stop there. Very often we want to be able to undo the result of our function. For example, we need to be able to reverse multiplication with division in order to rearrange equations, or reverse a square with a square root. To do this we use an inverse function. An inverse function can be thought of a complementary box to our original one, so that when we plug the output of the first function into it’s input we get the original value. For example, with our simple f (x) above, we inserted 2 and got 1. Our inverse will have to take 1 and give us 2. To find the inverse function, it is usually easier to give f (x) a letter, like y. In our example we obtain y = 2x − 3 Now we rearrange the equation for x, using the rules described above. We obtain x=

y+3 2

3.3 Expansion of Brackets

17

this is left as an exercise for the reader. We’re pretty much done, but it is usual to label our inverse of f (x) with the notation f −1 (x) and have our function in terms of x, not y. So, we swap x and y to obtain x+3 2 and now use our inverse function formula y=

x+3 2 Recall that with our simple example for f (x), that f −1 (x) =

f (2) = 2(2) − 3 = 1. If we now feed this output into the input of the inverse, we should get back to our starting position (2). 1+3 = 2. 2 Just as before we insert the value in the brackets into x throughout the expression for the inverse function, and you can see that indeed the inverse function here has taken us back to the start. We will meet other examples of inverse functions throughout this module. It’s not always possible to do this, and not all functions have inverses unfortunately. f −1 (1) =

3.3

Expansion of Brackets

When we have to multiply something by a bracketed expression, we use the so called distributive law. a(b + c) = ab + ac(a + b)c = ac + bc We can easily show that this can be extended. a(b + c + d + · · · ) = ab + ac + ad + · · · There are some simple things worth remembering • The abscence of a number before an expression is the same as multiplying by 1.

3.3 Expansion of Brackets

18

• The sign preceding a number belongs to that number and must be included in the multiplication. • In particular, a minus sign before a bracket means −1 multiplied by that bracket.

3.3.1

Examples

Here are some examples. 1. −2(3x − 5y + z) = −6x + 10y − 2z 2. 3(2x − y) − (x + 2y) = 6x − 3y − x − 2y = 5x − 5y 3. 4x(y − z + 2(x − y)) = 4x(y − z + 2x − 2y) = 4x(2x − y − z) = 8x2 − 4xy − 4xz 4. 2y(3x − 4z(x + z)) = 2y(3x − 4xz − 4z 2 ) = 6xy − 4xyz − 8yz 2

3.3.2

Brackets upon Brackets

When we encounter a bracketed expression multiplied by another bracketed expression we can apply the same technique, although it appears more complicated. Consider (a + b)(c + d) For the moment, we shall call z = (a + b). Then our expression appears simpler. z(c + d) = zc + zd Now we reinsert the true value of z. = (a + b)c + (a + b)d = ac + bc + ad + bd So we reduce the whole problem to two separate expansions of the type we have already met. There are a number of rules of thumb to make this technique rather simpler, but many depend on multiplying only two brackets

3.3 Expansion of Brackets

19

together, each of which with exactly two terms. We shall examine a general technique without this shortcoming. Consider once more (a + b)(c + d) Pick any bracket, for the sake of demonstration, we shall pick the first. Now take the first term in it (which is a). We now multiply this term on each term of the other bracket in turn, adding all the results. = ac + ad + · · · When we reach the end of the other bracket, we return to the first bracket and move onto the next term, which is now b and do the same again, adding to our existing terms. = ac + ad + bc + bd + · · · Now we return to the first bracket, and move to the next term. We find we have actually exhausted our supply of terms, and so our expansion is really complete. (a + b)(c + d) = ac + ad + bc + bd To multiply several brackets together at once we should multiply two only at a time. For example (a + b)(c + d)(e + f ). We begin my multiplying one pair together, let us say the first two, to obtain: = (ac + ad + bc + bd)(e + f ). We may then complete the expansion, it is left to the reader as an exercise to confirm that the full expansion will be: = ace + ade + bce + bde + acf + adf + bcf + bdf.

3.3.3

Examples

Here are some examples. 1. (x + 1)(x + 2) = x2 + 2x + x + 2 = x2 + 3x + 2

3.4 Factorization

20

2. (x + y)2 = (x + y)(x + y) = x2 + xy + xy + y 2 = x2 + 2xy + y 2 (See binomial expansions later). 3. (x + y)(x − y) = x2 − xy + xy − y 2 = x2 − y 2 (This is called the difference of two squares). 4. (3x − 2y)(x + 3) = 3x2 + 9x − 2xy − 6y 5. (2x − y)(x + 2y) = 2x2 + 4xy − xy − 2y 2 = 2x2 + 3xy − 2y 2

3.4

Factorization

Factorization is the opposite of expansion, we often prefer to condense and simplify expressions rather than expand them. Indeed, even in the expansion examples above we simplified the expressions along the way to make life easier for ourselves. Factorization is recognizing that an expression like this 4x + 2y could be written as 4x + 2y = 2(2x + y) simply because we can clearly see that expanding the result gives us the original. A factor common to all terms is observed - in this case the number 2 clearly divides into all the terms. The factor is divided into the expression and written outside the result which is bracketed. The factor may often be some algebra, and not just a number. In the expression x2 − 3x we see that x divides into both terms. We can thus write x2 − 3x = x(x − 3). The ability to spot factors does not come easily, but with a great deal of practice.

3.5 Laws of Indices

3.4.1

21

Examples

Let us look at some examples. Remember that in each case, expanding the end result should give us our original expression, and this allows you to check and follow the logic. 1. 3x + 12y 2 − 6z = 3(x + 4y 2 − 2z) 2. x3 + 3x2 + 4x = x(x2 + 3x + 4) 3. x3 + 3x2 = x2 (x + 3) 4. 2x2 + 4xy + 8x2 z = 2x(x + 2y + 4xz) We can also factorise expressions into two or more brackets multiplied together, but this is more difficult and we shall examine it later.

3.5

Laws of Indices

The term index is a formal term for a power, such as squaring, cubing etc, and the plural of index is indices. There are some simple laws of indices, which are shown in table 3.1. 1 xa × xb

= xa+b

2 xa ÷ xb

= xa−b

3

(xa )b

=

xab

4

x0

=

1

5

x−b

=

1 xb

6

xb

7

xb

1

a

= =

√ b x √ b xa

Table 3.1: The laws of indices

3.5 Laws of Indices

3.5.1

22

Example “proofs”

We shall attempt to show how a selection of these results “work”, but such demonstrations are for understanding and are not examinable. Let us consider the first law, with a concrete example: x3 × x2 We don’t know what the number x is, but all that is important is that the “base” values of each number are the same. We recall that powers mean a string of the same thing multiplied together, so that: × x} . x3 × x2 = x x × x} × x | {z | × {z x3

x2

Clearly there is no difference between the “×” inside the braced section and between them. In otherwords, this is just x3 × x2 = x x × x} × x × x} = x x × x × x} = x5 | × {z | {z | × x × {z x3

x2

x5

a string of five xs multiplied together, exactly the definition of x5 , and the 3 and 2 add to make 5. We shall show how one other result works, using the more general a and b. Consider (xa )b . By definition, this is just a a a x {z · · · × x} | ×x × b times

a

with b of these x terms. (Recall, x6 just means 6 of the x terms multiplied together). Now each xa terms is itself a collection of a xs multiplied together. So we can expand further b groups

z }| { x × · · · × x × x × · · · × x × · · · × x × · · · × x | {z } | {z } | {z } . a times

a times

a times

Each collection of x terms with a brace below is an expanded xa , so each contains a xs multiplied together. There were b xa terms, and so have b

3.6 Laws of Surds

23

braces, each containing a xs. Therefore we have a long string of xs multiplied together, which are a × b in number, exactly the definition of xa×b . It is worth going through this argument in a concrete case, for example 2 3 (x ) to help follow the logic.

3.5.2

Examples

Here are some examples 1. 25 × 23 = 25+3 = 28 (= 256) 2. 28 ÷ 23 = 28−3 = 25 (= 32) 3. (32 )2 = 32×2 = 34 (= 81) 4.

1

16 4 = 5. 15−3 = 6. 2

27− 3 =

3.6

√ 4

16(= 2)

1 1 (= ) 3 15 3375

1 27

2 3

= √ 3

1 272

=

1 9

Laws of Surds

A surd is technically a square root which has an irrational value, but we often talk about surds whenever we manipulate square roots. We have two main results for manipulating square roots, as shown in table 3.2 1 2

√ √ a b √ a √ b

= =

√ ab pa b

Table 3.2: The laws of surds

3.7 Quadratic Equations

3.6.1

Examples

The second law of surds is most often used to work out the square roots of fractions. 1. r √ 1 1 1 =√ = 4 2 4 2. r √ 4 4 2 =√ =√ 7 7 7 The first law was often used to split large square roots into smaller ones, by attempting to divide the original number by a perfect square. 1. √ √ √ √ √ 12 = 4 × 3 = 4 × 3 = 2 3 2.

√ √ √ √ √ 80 = 16 × 5 = 16 × 5 = 4 5

A very important application of this to come later is that of complex numbers.

3.7

Quadratic Equations

A quadratic equation in x is an equation of the form ax2 + bx + c = 0, a 6= 0 where a, b and c are constants (that is, they do not change value as x does. It’s vital that a is not zero, or the equation collapses into that of a straight line. However, it is quite allowable to have b or c (or both) zero.

3.7.1

Examples

Here are some examples of equations which are, and which are not quadratic equations, shown in table 3.3. Equations 1,2 and 3 are genuine quadratics, even though terms are missing in 2 and 3. Equation 4 is the equation of a straight line, or if you like a = 0 which is not permitted. Equation 5 contains an x3 term, and so is a cubic equation and not a quadratic whose highest term must be x2 .

24

3.7 Quadratic Equations

25

Equation

Quadratic?

1 2x2 + 3x − 4 = 0 YES

a

b

c

2

3

-4

2

x2 + 2x = 0

YES

1

2

0

3

x2 + 3 = 0

YES

1

0

3

4

4x + 2 = 0

NO

n/a n/a n/a

5

x3 + 3x − 2 = 0

NO

n/a n/a n/a

Table 3.3: Examples of quadratic equations

3.7.2

Graphical interpretation

To examine the solutions of this equation, it is helpful to consider the graph of the function given by y = ax2 + bx + c. When this graph is plotted it gives a characteristic “U” shaped curve, called parabola. It has many interesting properties, but the most important is that it is symmetrical about a vertical line through the maximum, or minimum point. Our equation corresponds to the above function when y = 0, or to put it another way, the solutions of our equation occur when the curve cuts the x-axis (where y is zero).

Figure 3.1: The quadratic equation This gives us our first problem, which is that we cannot even be sure that the equation has solutions. If this seems strange or confusing, recall that we never claimed that all equations could be solved. Our problem falls into three categories.

3.7 Quadratic Equations

26

1. The curve cuts the x-axis twice; 2. The curve cuts the x-axis once only (just touching, no more); 3. The curve does not cut the x-axis at all. This situation is shown in figure 3.1. In this figure all the curves have been drawn as “maximum” parabolas, many quadratics show “minimum” parabolas but the cases are just the same. The y-axis has been ommitted as it is not relevant to the problem at hand which is determining the number of solutions of the corresponding equation.

3.7.3

Factorization

One approach to try and solve a quadratic is the so called factorization or sum and product method of solution. We imagine that we can write our equation in the form (x − α)(x − β) = 0. Why have we written it this way? Well, for a product of two things to be zero, one or other, or both must be zero. Therefore in this case, if the left-hand bracket is zero we see that x = α, and if the right-hand bracket is zero we see that x = β. Therefore α and β are our two solutions. If we expand these brackets out, we obtain x2 − αx − βx + αβ = x2 − (α + β) + αβ = 0 Now we try to compare this to our quadratic equation, but before we do so, we “standardize” our equation, by dividing it by a: c b ax2 + bx + c = 0 ⇒ x2 + x + = 0. a a We now compare the coefficients of the x terms in each equation (assuming they are in fact two forms of the same equation). (A coefficient is a number multiplied upon a term). Comparing we obtain x2 term 1=1 b x term = −(α + β) a c constant term a = αβ

This is why we “standardize”;

3.7 Quadratic Equations The two significant equations are this b c − = α + β; = αβ a a which tell us that the sum of the solutions is − ab and that the product of the solutions is ac . If we can guess two numbers with these properties, we have solved the quadratic equation. This method has the advantages that the theory is simple and straight forward, but has two crippling disadvantages. Firstly, we have to guess two numbers, and even in simple cases this may be very difficult, in a hard example next to impossible. Secondly, and more importantly, before we start we have no way of knowing how many solutions the equation has. If it has two, this theory works well, if it has one, then α and β have the same value (we say the solution is repeated). However, if there are no solutions, it will be impossible to guess our two numbers, as they do not exist, but this may not be obvious from the start.

3.7.4

Quadratic solution formula

Fortunately, we do have a more flexible method of solution, but its proof is difficult. The proof is given here, but is not required. First of all consider the problem, we cannot simply take a square root to remove the square, as the other x term gets in the way. Consequently, we try to absorb the x term and the x2 term into one term. This is a difficult procedure known as completing the square. Consider (x + y)2 = (x + y)(x + y) = x2 + 2xy + y 2 We wish to let this be our x2 + ab x terms. For this to be true, we must b . Note that this does not correspond to exactly what we want: let y = 2a  2 b b b2 x+ = x2 + x + 2 2a a 4a which is our quadratic, except that the ac term is missing, and the last term above is extra, so we add and subtract these two terms respectively. b c x2 + x + = 0 a a  2 b c b2 ⇒ x+ + = 2 2a a 4a

27

3.7 Quadratic Equations

28

2  b2 b2 − 4ac b c = 2− = ⇒ x+ 2a 4a a 4a2 r   b b2 − 4ac =± ⇒ x+ 2a 4a2 which using the laws of surds (see 3.5), yields our final result. √ −b ± b2 − 4ac x= 2a

3.7.5

The discriminant

As well as giving a fool-proof method of solving a quadratic, the solution formula has a small section which tells us some useful information. The section b2 − 4ac which was the contents of the square root is known as the discriminant of the equation, as it tells us how many solutions the equation has. b2 − 4ac > 0 Two real solutions Case 1 above b2 − 4ac = 0 One real solution Case 2 above b2 − 4ac < 0 No real solutions Case 3 above

3.7.6

Examples

Here are some examples of quadratic equations and their solutions. 1. x2 + 2x − 3 = 0 2. x2 − 4x + 4 = 0 3. x2 + x + 1 = 0 Solutions 1. In this case a = 1, b = 2, c = −3. We plug this into the solution formula to obtain

3.7 Quadratic Equations

29

p 22 − 4(1)(−3) x= 2 which when calculated out yields answers of 1 and −3. 2. In this case a = 1, b = −4, c = 4. The solution formula yields p +4 ± (−4)2 − 4(1)(4) x= 2 which when calculated out yields answers of 2 and 2. The repeated solution is nothing unusual, and indicated that this quadratic has only one solution. 3. In this case a = 1, b = 1, c = 1. We shall follow this case in more detail. p −1 ± 12 − 4(1)(1) x= 2 √ −1 ± 1 − 4 = 2 √ −1 ± −3 = 2 Note that we have a negative square root here. We cannot calculate this, suppose that the answer was a positive number, then when squared we get a positive number, not −3. We also get a positive number when we square a negative √ number, and we get zero when we square zero. Thus no number can be −3. There are no solutions to this quadratic equation. −2 ±

3.7.7

Special cases

Observe that if b = 0, or if c = 0 in the quadratic, the equation can be solved more directly. We show how for completeness, although the formula may still be used in these cases. b=0 We have here ax2 + c = 0 ⇒ ax2 = −c ⇒ x2 = − Therefore

c a

3.8 Notation

30

r c x=± − . a Note that − ac must be positive, and this will be the case only if a and c have different signs, otherwise there are no solutions. Note also the ± in the formula, solving directly often leads to forgetting the solution coming from the negative branch of the square root. c=0 In this case we have ax2 + bx = 0 ⇒ x(ax + b) = 0 from which we obtain, that either x=0 which is one solution, or ax + b = 0 ⇒ x = −

3.8

b a

Notation

We now introduce two more items of notation.

3.8.1

Modulus or absolute value

A very useful idea in mathematics is the notion of absolute value of a real number. If x is a real number, we define the modulus of x, or mod x, denoted |x| as follows. |x| = larger of x and −x Note that this is not the same as the mod operation in computing.

3.8 Notation

31

Examples Here are some examples |2| = larger of 2 and −2 = 2; | − 2| = larger of −2 and 2 = 2; |0| = larger of 0 and −0 = 0. In other words, the modulus function simply strips off any leading minus sign on the number. Alternative definitions There are some alternative, but equivalent definitions of |x|:  x if x ≥ 0 |x| = −x if x < 0 and |x| =

√

x2

where we adopt the convention that the square root takes the positive branch only, unless we include ±, which is commonly accepted.

3.8.2

Sigma notation

Suppose that f (k) is some expression involving k (a function of k formally speaking). For example, f (k) could be 2k or 2k + 1 etc. Then, the notation n X

f (k)

k=m

is a shorthand for the expression f (m) + f (m + 1) + · · · + f (n) In other words, we insert the value at the bottom of the sigma into the f expression, then insert that value plus one, plus two, etc., until we reach the value on top of the sigma, and add all the results together. We will use the symbol ∞ to denote the lack of an endpoint in the summation. The ranges above and below the sigma are sometimes ommitted when it is clear what is being summed. This may be easier to understand after some simple examples.

3.9 Exponential and Logarithmic functions

32

Examples Here are some examples of sigma notation 1. 4 X 2k = 20 + 21 + 22 + 23 + 24 = 31 k=0

2.

∞ X

k = 4 + 5 + 6 + 7 + 8 + ···

k=4

3. P5

= =

3.8.3

k+1 1 k=2 (−1) 3k

5 (−1)3
312 +
(−1)4
313 + (−1)
1 1 1 1 − 32 + 33 − 34 + 35




1 34




+ (−1)6

1 35




Factorials

The notation n!, where n is an integer greater or equal to 0, is spoken “n factorial” and is a shorthand for n × (n − 1) × (n − 2) × · · · × 2 × 1. So for example 5! = 5 × 4 × 3 × 2 × 1 = 120. It should be clear that this time of expressions gets very large, very quickly, and indeed 69! is so large that most calculators are unable to represent it. We note that by convention we accept 1! = 1 and 0! = 1.

3.9

Exponential and Logarithmic functions

We now consider another, slightly more complex type of function.

3.9.1

Exponential functions

In quadratics, we had x terms with specific constant powers, such as x2 . A very powerful function is formed when x is in the exponent (yet another name for “power” or “index”). An exponential function is one of the form

3.9 Exponential and Logarithmic functions

y = kx where k is some positive constant. These functions are important due to their extraordinary ability to climb or decrease, as we shall see later when we examine their graphs.

3.9.2

Logarithmic functions

For a positive number n, the logarithm or log of n to base b, written logb n is the power to which b must be raised to give n. To put this in mathematics: logb n = x ⇔ bx = n The antilogarithm or antilog of n to the base b is bn . There are two main “types” of logarithms in use today: • log10 , often written simply as log; • loge , often written1 as ln. Examples Logs are often used to condense a very large range of numbers to a more managable one. Examine how in the following table the logs of a large range of values from one thousandth, to one thousand are contracted to a range from -3 to 3. Verifying that there logarithms are correct is left as a simple exercise to the reader. Value Log10

0.001 0.01 0.1 1 -3 -2 -1 0

10 100 1000 1 2 3

The scales of pH (chemical scale of acidity) and the decibel range, the Richter scale of earthquake intensity are all logarithmic,2 base 10. Therefore a pH of 6 is 10 times more acidic than the neutral 7, and 5 is 100 times more acidic than 7, etc. 1

The Scottish mathematician John Napier (1550 - 1617) did a great deal of work

on methods of computation and published his Mirifici logarithmorum canonis descripto (Discription of the marvellous rule of logarithms) in 1614. In his honour logs base e are often called Naperian logarithms. Logs were first used to help perform large calculations. 2 A lesser known example is the warpspeed scale

33

3.9 Exponential and Logarithmic functions Warnings Just as we have problems with the square roots of negative numbers, so we cannot take the logarithms of negative numbers, or of zero. Laws of Logarithms Logarithms are so useful because they exhibit the following properties, known as the laws of logarithms. These are closely related to the laws of indices and are shown in table 3.4. 1 logn (x × y)

=

logn (x) + logn (y)

2 logn (x ÷ y)

=

logn (x) − logn (y)

3

logn (xy )

= y logn (x)

4

logn n =

1

5

logn 1

0

=

Table 3.4: Laws of Logarithms These laws easily account for their use in computation. To multiply two large numbers, one would find their logarithms from tables, add these two values, and then by the first law, this is the logarithm of the product, so one merely antilogs this number to obtain the answer. The other major application ot these is in solving equations where the variable is in the index, i.e. some sort of exponential expression.

3.9.3

Logarithms to solve equations

When we obtain equations like this ax = b it is sometimes possible to guess the power to which we must raise a to obtain b. However, more often than not, a and b are not simple integers, and therefore almost impossible to guess x. In this case, take logarithms on both sides of the equation.

34

3.9 Exponential and Logarithmic functions ⇒ ⇒ ⇒

3.9.4

logn ax = logn b x logn a = logn b (using laws of logs 3) logn b x = log a n

Examples

Solve the following equations for x. 1. 3x = 27 2. 3x = 30 3. 22x − 5(2x ) + 6 = 0 Solutions 1. Although it is clear here that the answer is x = 3 we use logs to illustrate the point. It really doesn’t matter what base of logs we use, provided we are consistent. Take logs both sides log 3x = log 27 ⇒ x log 3 = log 27 using the third law of logs (see 3.9.2). We now can perform ordinary rearranging. x=

log 27 =3 log 3

2. This is only slightly more complicated, although it is not possible to guess the answer easily. log 3x = log 30 ⇒ x log 3 = log 30 x=

log 30 = 3.0959 log 3

to four decimal places. 3. This problem looks very complicated. Taking logs immediately will get us into trouble. Partly because we have no law to split logs over addition (note this carefully, a common mistake), and that the log of zero is a problem (it’s something like −∞). We might observe that the equation roughly resembles a quadratic (see 3.7, and this is the key. Let us assign u = 2x , then clearly

35

3.9 Exponential and Logarithmic functions

u = 2x ⇒ u2 = (2x )2 = 22x by the laws of indices (see 3.5). Placing these into the equation yields u2 − 5u + 6 = 0 which is now a plain quadratic. Solving by whatever method yields u = 2 and u = 3 as solutions. Having dealt with the quadratic, we now deal with the exponential problem, recall that u = 2x . Consider the u = 2 solution, this means 2x = 2 so that clearly x = 1, with no logarithms required. The u = 3 solution proceeds as follows 2x = 3 and following the procedure above we obtain x=

log 3 = 1.5850 log 2

to four decimal places. So we have two solutions x = 1, x = 1.5850.

3.9.5

Anti-logging

Logarithms are used to undo expressions in which the x we wish to obtain is in the power. Similarly, exponential functions may be used to undo logarithms. Recall our definition. logb n = x ⇔ bx = n this means that blogb n = n To put it another way, we can remove a logarithm by taking the base number to the power of the logarithm. Let’s use some numbers to illustrate the point. Example We know that log10 (1000) = −3

36

3.9 Exponential and Logarithmic functions so suppose we were asked to solve the equation log10 x = −3 we know that x = 1000, but we might not notice this in a more difficult problem, or one we had not previously seen. To remove the log, we take the base number (in this case we are using base 10) and raise it to both sides: 10log10 x = 10−3 . Now we know that the log and the anti-log (the process of raising the base to this power) cancel each other out on the left, so we obtain x = 10−3 = 1000 and our equation is solved.

3.9.6

Examples

Solve the following equations for x: 1. log x2 = 2 2. 4 ln x = 70 Solutions 1. In the absence of a specific base, we assume base 10, as noted above. So we may antilog both sides directly here, raising 10 to both sides. 2

10log x = 102 On the LHS, the anti-log and log cancel, leaving x2 = 102 = 100 ⇒ x = ±10 2. In this case, the base of the logarithm is e. We could immediately take anti-logs on both sides, but the 4 in front of the ln x makes it messy. It is easier to rearrange to ln x first (onion within onion as before). ln x =

70 4

37

3.10 Binomial Expansion

38

Now we anti-log, base e, note that exp(· · · ) is a another notation for e··· .     70 70 exp(ln x) = exp ⇒ x = exp ⇒ x = 39, 824, 784.4 4 4

3.10

Binomial Expansion

The binomial theorem or expansion has two main purposes. It allows us to easily expand out high powers of brackets which contain two terms. It is also highly suited to calculating probabilities associated with a simple repeated experiment with a fixed probability of success each time. We shall concentrate on the first application here.

3.10.1

Theory

Consider (p + q)n for all numbers, and in particular positive integers n. We now examine the expansion of powers of (p + q), and let us start by considering some examples3 : (p + q)2 = p2 + 2pq + q 2 ; (p + q)3 = p3 + 3p2 q + 3pq 2 + q 3 ; (p + q)4 = p4 + 4p3 q + 6p2 q 2 + 4pq 3 + q 4 . If you examine this you will note some facts: Patterns Binomial expansions follow these rules: • the powers of each term add to the power of the expansion; • the powers of p begin with n and decrement each time until they reach 0; 3

These were produced by considering (p+q)(p+q) and expanding, and then multiplying

by another (p + q) each time.

3.10 Binomial Expansion

39

• the powers of q begin with 0 and increment each time until they reach n. Armed with these facts we can write out the expansion for any value of n, except that we as yet can’t work out the coefficients. However, we can note something about these too: Binomial coefficients follow these rules: • the coefficients on the first and last term are both 1; • the coefficients on the second and one but last term are both n; • if the coefficients are laid out in a triangle, then any one can be found by adding the two above. This triangle layout is shown in table 3.5 as far as n = 8 and is known as Pascal’s triangle 4 . 1 11 121 1331 14641 1 5 10 10 5 1 1 6 15 20 16 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1

Table 3.5: Pascal’s Triangle We can now work out simple expansions, and this can be done in a two step process for beginners. 4

Named after Blaise Pascal (1623-1662) the noted French mathematician, physicist and

philosopher. Due to his work in hydrostatics the SI unit of pressure takes his name, and a computer language is named after him for his production of a calculating machine at age 19. Pascal, along with Fermat and De Moivre was a pioneer in the mathematics of probability

3.10 Binomial Expansion

3.10.2

40

Example

Write out the expansion of (p + q)8 . Start out by writing the terms, we will start with p8 and end with q 8 ; each term in between will lose 1 in power from p and gain 1 in power on q. p8 + p7 q + p6 q 2 + p5 q 3 + p4 q 4 + p3 q 5 + p2 q 6 + pq 7 + q 8 This still lacks the coefficients, we can place these in front of the terms by taking them straight from Pascal’s triangle, the correct row is the one whose second number matches n = 8. This gives us p8 + 8p7 q + 28p6 q 2 + 56p5 q 3 + 70p4 q 4 + 56p3 q 5 + 28p2 q 6 + 8pq 7 + q 8

3.10.3

Examples

We can deal with more difficult problems in the same way. Expand the following expressions fully 1. 4  1 x− x 2. (2a + b)5 Solutions 1. It is usually easier to do the simple expansion first. That is, expand out (a + b)4 which is a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 and now let a = x and let b = − y1 (note that the minus sign is included in b itself). Now insert these into the expansion, inserting brackets for safety.    2  3  4 1 1 1 1 2 + 6(x) − + 4(x) − + − (x) + 4(x) − x x x x 4

3

Which, with careful working out using the laws of signs gives us    2  3  4 1 1 1 1 4 3 2 = x − 4x + 6x − 4x + x x x x

3.10 Binomial Expansion

41

and finally we have 4 1 + 4 2 x x 2. Similarly, begin with (x + y)5 (I choose x and y because they are not present in the original problem). = x4 − 4x2 + 6 −

x5 + 5x4 y + 10x3 y 2 + 10x2 y 3 + 5xy 4 + y 5 Now we let x = 2a and y = b to obtain (2a)5 + 5(2a)4 (b) + 10(2a)3 (b)2 + 10(2a)2 (b)3 + 5(2a)(b)4 + (b)5 Now the brackets around the b terms are not required, but it’s a good habit, as they are certainly required around the 2a terms. Note that (2a)2 = 2a × 2a = 4a2 6= 2a2 . In otherwords without the brackets the power only acts on the a and not the 2 as well, this is a common error. Expanding all these powers of 2a gives us = 32a5 + 5(16a4 )b + 10(8a3 )b2 + 10(4a2 )b3 + 5(2a)b4 + b5 which finally leaves us with = 32a5 + 80a4 b + 80a3 b2 + 40a2 b3 + 10ab4 + b5

3.10.4

High values of n

This is not examinable and is included for completeness. For high powers of n working out the powers of the terms is not difficult, but calculating the coefficients is more difficult, for this we use the following formula. The number of combinations of r objects picked from n is given by n! r!(n − r)! This formula gives us the terms from Pascal’s triangle precisely, but note that r must be counted from 0. So we could in fact use this to define the expansion. The binomial expansion for (p + q)n is n Cr

=

n X (n Cr pn−r q r ). r=0

Remember that the summation starts with r = 0.

3.11 Arithmetic Progressions

3.11

Arithmetic Progressions

An arithmetic progression is a sequence of numbers of the following form: a, a + d, a + 2d, a + 3d, . . . , a + (n − 1)d, . . . where a and b are constants. Note that a + (n − 1)d is the nth term in the progression, rather than a + nd as the first term has no d. In this context, we call a the initial term; d the common difference. So in an arithmetic progression, (we abbreviate this to A.P.) each term except the first is obtained by adding the common difference d to the its predecessor.

3.11.1

Examples

Here are some examples of A.P.s: 1. 1, 2, 3, 4, 5, . . . is an A.P. with a = 1 and d = 1. 2. 50, 100, 150, 200, . . . is an A.P. with a = 50 and d = 50. 3. 100, 50, 0, −50, −100, . . . is an A.P. with a = 100 and d = −50

3.11.2

Sum of an arithmetic progression

Let a, a + d, a + 2d, . . . be an A.P.. Then the sum of the first n terms of the A.P., denoted Sn is given by Sn =

n (2a + (n − 1)d) 2

42

3.11 Arithmetic Progressions Although it is not examinable, a simple proof is given below. Clearly we can write Sn = a + (a + d) + (a + 2d) + · · · + (a + (n − 2)d) + (a + (n − 1)d) Now, we right the same R.H.S., but “back to front”. Sn = (a + (n − 1)d) + (a + (n − 2)d) + · · · + (a + 2d) + (a + d) + a Now, obviously there’s the same number of terms in the R.H.S. of both expressions, remember there are n terms. We add them together from left to right to obtain: 2Sn = (2a+(n−1)d)+(2a+(n−1)d)+(2a+(n−1)d)+· · ·+(2a+(n−1)d)+(2a+(n−1)d). Now, as pointed out earlier, there are n terms, and so this becomes 2Sn = n(2a + (n − 1)d) dividing by 2 on both sides gives the result. (The mathematician Karl Gauss was allegedly able to perform this trick in his head in primary school, where is mathematical abilities were first observed. Many believe Gauss to be the greatest mathematician ever.) Note The following notation is often used in texts: n (2a + n − 1d) 2 in this context, the line over n − 1 acts as a bracket. Sn =

3.11.3

Example

The 4th term of an A.P. is − 12 and the 8th term is 32 . Find the initial term, the 3rd term, and sum of the first 100 terms.

43

3.11 Arithmetic Progressions

44

Solution Suppose the A.P. has initial term a and common difference d as before. Then we obtain a + 3d = − 21 a + 7d = 32

(i) (ii)

(as the 4th term is − 21 ) (as the 8th term is 32 )

So here we have two equations in two unknowns, so we solve by subtracting (i) from (ii) to obtain 4d = 2 and so we have found that d = 12 . Now that we’ve found d we can use it to find a by simply inserting the value for d into either equation. Inserting it into (ii) yields: 1 3 a + 7( ) = ⇒ a = −2. 2 2 We have found a and d and so we can now go on to finish off the question. The 3rd term will be a + 2d and so is −2 + 2( 12 ) = −2 + 1 = −1. and the sum of the first 100 terms is given by S100

3.11.4

= = = =

100 (2(−2) 2

+ (100 − 1) 12 ) 50(−4 + 99( 12 )) 50(45.5) 2275

Example

The sum of the first ten terms of an A.P. is 10, the sum of the first hundred terms is −8900. Find the initial term and common difference of the progression. Solution First of all we formulate the problem in mathematics rather than words. (2a + 9d) = 10 S10 = 10 2 ⇒ 5(2a + 9d) = 10 ⇒ 2a + 9d = 2 (i)

3.12 Geometric Progressions S100 = 100 (2a + 99d) = −8900 2 ⇒ 50(2a + 99d) = −8900 ⇒ 2a + 99d = −178 (ii) We now have two equations in two unknowns whch we can subtract as they are to yield 90d = −180 ⇒ d = −2. We can now insert this value for d into equation (i) giving 2a + 9(−2) = 2 ⇒ 2a − 18 − 2 ⇒ 2a = 20 ⇒ a = 10.

3.12

Geometric Progressions

A geometric progression (or G.P. for short) is a sequence of numbers of the form a, ar, ar2 , ar3 , . . . , arn−1 , . . . Note then the nth term is arn−1 , and not arn , similarly to A.P.s because there is no r in the first term. In this context, we call a the initial term; r the common ratio.

3.12.1

Examples

Here are some examples of G.P.s 1. 1, 2, 4, 8, 16, . . . is a G.P. with a = 1 and r = 2. 2. 1 1 1 1 1, , , , , . . . 2 4 8 16 1 is a G.P. with a = 1 and r = 2 .

45

3.12 Geometric Progressions

3.12.2

46

Sum of a geometric progression

Let a, ar, ar2 , ar3 , . . . , arn−1 , . . . be a G.P.. Then the sum of the first n terms of the G.P., denoted Sn is given by Sn =

a(1 − rn ) 1−r

provided that r 6= 1. Although it is not examinable, a simple proof is presented below: We can write Sn = a + ar + ar2 + · · · + arn−2 + arn−1 and multiplying both sides by r we obtain rSn = ar + ar2 + · · · + arn−1 + arn just as in the proof for the sum for A.P.s, we combine these two equations, this time by subtraction to obtain Sn − rSn = a − arn ⇒ (1 − r)Sn = a(1 − rn ) and dividing by 1−r on both sides now gives us the result. Of course, the r 6= 1 restriction comes from this devision, ensuring that we are not dividing by zero.

3.12.3

Sum to infinity

Sometimes, we may be asked to find the “sum to infinity” of a G.P., that is, the sum of all the terms added together. This is not a relevant question with A.P.s by the way, as we keep on adding a sizeable chunk each time. The only way the sum to infinity can exist is if the amount we add on with each new term shrinks to negligible levels as n tends to infinity. For example, consider that |r| < 1 in a given G.P., then every subsequent power of r gets smaller and smaller numerically speaking. The sum to infinity is given by S∞ =

a ; |r| < 1 1−r

3.12 Geometric Progressions

3.12.4

47

Example

The 6th term of a geometric progression is 32, and the 11th term is 1. Determine 1. the common ratio of the progression; 2. the initial term of the progression; 3. the sum to infinity of the progression. Solution 1. We begin by forumlating the given information as equations. ar5 = 32 (i) ar10 = 1 (ii)

(6th term is 32) (11th term is 1)

This gives us two equations in two unknowns. There are two main methods of solving these, you could for example rearrange each equation to give a and then put those two expression equal to each other, and solve for r. We will solve by simply dividing equation (i) into (ii) to obtain 1 1 ar10 = ⇒ r5 = 5 ar 32 32 You can see that the as cancel, and the rs combine according to the laws of indices (see 3.5). We find r by taking the 5th root on both sides, showing that r = 21 . 2. We now have to obtain a, and we can do this by inserting our discovered value for r into one of our equations. If we pick (i) we obtain 1 a a( )5 = 32 ⇒ = 32 ⇒ a = 32 × 32 = 1024. 2 32 3. We have now found the two numbers which characterise this progression, and we are hence in a position to answer almost any subsequent question on it. The sum to infinity exists if |r| < 1, and in our example this is certainly the case. We can then plug in our values S∞ =

1024 = 2048. 1 − 21

3.12 Geometric Progressions

3.12.5

Example

Here is a slight reworking of a classical example. Take a chess board (which consists of an 8 by 8 grid of squares). Place a single 10 pence piece on the first square, two on the second, four on the third and so on, doubling for all 64 squares. How much money is there on the last square, and how much money is there on the entire board? Solution This is a geometric progression, as we are multiplying by a constant (2) each time. Therefore r = 2, and the first term is 1 (for 1 coin) so a = 1. The number of coins on the 64th square is given by ar6 3 which is 1 × 26 3 which is 9.22337 × 1016 pounds, or in more usual notation, an incredible 92 233 700 000 000 000 pounds. The quantity of money on the entire board is the sum of the first 64 terms, given by 1 − 264 1(1 − 264 ) = = 264 − 1 1−2 −1 which is as close to 264 as makes no difference (the error in this case is 10p). So the money on the board is 1.84467 × 1017 pounds, or 184 467 000 000 000 000 pounds. This example demonstrates the incredible ability of geometric progressions to grow at speed.

48

Chapter 4 Trigonometry Trigonometry is an essential skill for performing many real life mathematical calculations, including the analysis of simple harmonic motion and waves. There is a strong link between trigonometry and complex numbers as evidenced by the polar and exponential form of complex numbers. We recall the most elementary principles quickly.

4.1

Right-angled triangles

We first recall the trigonometry of simple right-angled triangles.

4.1.1

Labelling

In a right-angled triangle, we call the side opposite the right angle (that is, the side that does not touch the right angle) the hypotenuse. Given any other angle (which clearly cannot be other than acute), we label sides relative to that angle. Thus the side opposite this acute angle is called simply the opposite and the remaining side becomes the adjacent. It should be clear that the opposite and adjacent will switch, depending on what acute angle is considered, while the hypotenuse remains the same side at all times.

4.1 Right-angled triangles

50

Figure 4.1: Labelling right-angled triangles

4.1.2

Pythagoras’ Theorem

We remember this most fundamental of theorems concerning the triangle, proved before the birth of Christ and known in principle in ancient Egypt. The proof is usually attributed to Pythagoras 1 . Given a right-angled triangle with hypotenuse length c and other sides of length a and b. a2 + b 2 = c 2 Thus, this relation may be used when we have two sides of a right-angled triangle and wish to find the third. With less sides known, we must use angles to solve the triangle completely.

4.1.3

Basic trigonometric functions

Recall that sin θ =

A O O ; cos θ = ; tan θ = H H A

where O is the oppositive relative to θ, A is the adjacent relative to θ, and H is the hypotenuse. 1

Pythagoras (c. 560-480 BC) was a Greek methematician and mystic who founded a

cult in which astronomy, geometry and especially numbers were central. Pythagoras was exiled to Metapontum around 500 BC for the political ambitions of the cult.

4.1 Right-angled triangles The “words” SOH, CAH andTOA are often used as mnemonics for these equations. There’s a variety of ways of remembering these, and it’s important you are able to do so. We define some other simple short-cut functions relative to these We shall define sec θ =

4.1.4

1 1 1 ; csc θ = ; cot θ = cos θ sin θ tan θ

Procedure

The trigonometry of right-angled triangles is particularly simple. The method we follow tends to follow these steps 1. Identify 3 things, 2 you know and 1 you want; 2. Determine the correct equation to use (perhaps using a mnemonic); 3. Insert what you know into the equation; 4. Rearrange to find what you want. In step 1, you might for example have two sides and want the third side this is accomplished using Pythagoras’ theorem and nothing else. Generally however, two of the things will be sides and one will be an angle (other than the right angle itself). For example, we might know one angle and a side and use this to find another side. We might have two sides and use this to find an angle.

4.1.5

Example

A plane takes off and maintains an angle of 15◦ to the horizontal while it covers 1000m as measured along the ground. Calculate (a) the distance it has travelled in the air; (b) the height it has achieved; by this time.

51

4.2 Notation

52

Solution If we draw the triangle which reflects this situation then the path of the plane is the hypotenuse, the length of 1000m along the ground is the adjacent and the height is the opposite. (a) We know the angle, and the adjacent, we want the hypotenuse. Looking above (see 4.1.3) we see we need cos, so write the equation, filling the blanks.

cos 15◦ =

1000 1000 ⇒ H cos 15◦ = 1000 ⇒ H = = 1035.276m H cos 15◦

(b) We could now use Pythgoras’ theorem (see 4.1.2) for this, but we shall use trigonometry to demonstrate. We have the angle, adjacent and we want the opposite. We see that tan is the function that associates these together. O ⇒ O = 1000 tan 15◦ ⇒ O = 267.949m 1000 All figures accurate to 3 decimal places. tan 15◦ =

4.2

Notation

We now introduce a short-hand notation in common use. If you have a specific trig function, which we shall denote as f n then: f nn (x) = (f n(x))n For example sin2 x = (sin x)2 We only use this notation when the index n is a positive integer. This is to prevent confusion when using −1. The notation f n−1 x represents the correct inverse function, for example sin−1 x is the function which takes a sine, and returns the angle. Please note that

4.3 Table of values

53

1 . sin x Remember that all trig function work on angles, and they are meaningless without them.√ Therefore sin 20 does not mean sin ×20, rather it is similar to √ the notation 2 which does not mean × 2. sin−1 6=

4.2.1

Example

A ramp which travels 2m along the ground, and finishes 1m up is constructed. At what angle is this ramp to the ground? Solution We know the opposite (1m) and the adjacent (2m) and we want the angle. The tan function relates this things together. 1 2 We now need to obtain θ so we take the inverse tan function on both sides to obtain tan−1 (tan(θ)) = tan−1 (0.5) tan θ =

where the brackets are added for emphasis. Now the tan−1 and tan functions cancel each other (that’s the whole point) so we obtain θ = tan−1 (0.5) = 26.565◦

4.3

Table of values

The exact values of the sine, cosine and tan functions at important angles are given in table 4.1. Note that tan 90 is not defined.

4.4

Graphs of Functions

Consider figure 4.2. A point P moves around the circumference of a circle of radius 1, and its angle from the postive x axis is taken to be θ.

4.5 Multiple Solutions

54

Angle θ

sin θ

cos θ

tan θ

0

0

1

0

√

√

30

1 2 √

3 2 √ 2 2

3 3

60

2 2 √ 3 2

90

1

0

∞

180

0

−1

0

45

1 2

1 √

3

Table 4.1: Table of trigonometric values Observe that x y y sin θ = ; cos θ = ; tan θ = . 1 1 x So to summarise, in this case we have y x As we let θ move, even to angles larger than 90◦ we can work out the values of the major trigonometric functions and plot their graphs, and these can be seen in figure 4.3. We can see that sin and cos are bounded between −1 and 1, while tan takes extreme values as cos θ = x = 0. sin θ = y; cos θ = x; tan θ =

4.5

Multiple Solutions

Examining the graphs above shows an interesting situation. If we try to solve the following equation 1 2 This corresponds to where the line y = 12 cuts the graph of y = sin θ, but this clearly happens at several points. Of course, the trig functions repeat themselves every 360◦ and so we need only look for solutions in that range. sin θ =

4.5 Multiple Solutions

Figure 4.2: Generating the trigonometric graphs To aid us we use a diagram often known as a CAST diagram. This diagram is a helpful memory aid for which functions are positive and which are negative for different angles. Armed with this and the symmetry of trigonomtric functions we can find an easy way to locate multiple solutions.

4.5.1

CAST diagram

We use a type of diagram called a “CAST” diagram (see figure 4.4 on page 57 to remind us of the behaviour of the three basic trig functions. The name of the diagram comes from the single letter to be found in each quadrant. The significance is as follows • A – All functions are positive in this quadrant; • S – Only sin is positive, cos and tan are negative; • T – Only tan is positive, sin and cos are negative; • C – Only cos is positive, sin and tan are negative. The use of the diagram is detailed below. Note that the diagram here is labelled in the range 0◦ to 360◦ , but it is as easy to label in the range −180◦ to 180◦ .

55

4.5 Multiple Solutions

56

Figure 4.3: The graphs of sin θ and cos θ

4.5.2

Procedure

We can follow this step-by-step method to find multiple solutions, given a principal solution from a calculator or tables. 1. Draw the principal solution on the CAST diagram; 2. Work out the angle this solution is to the horizontal; 3. Draw the other three lines with the same angle to the horizontal; 4. Examine the sign of the RHS of our equation, that is whether our target number is positive or negative. 5. Select only those solutions whose quadrants are the right sign for the function you are using.

4.5.3

Example

Solve the equations 1. sin θ =

1 2

2. sin θ = − in the range −180 < θ ≤ 180.

1 3

4.5 Multiple Solutions

Figure 4.4: The “CAST” diagram Solution 1. From a calculator or some other source we can get a principal solution of 30◦ , and from our CAST diagram our other solution is 150◦ . This is because sin is positive in the two upper quadrants, and our equation is sin θ = + 12 . 2. Our principal solution is −19.471◦ , and from our CAST diagram our other solution is −160.529 (all to 3 decimal places). This is because sin is negative in the two lower quadrants, and our equation is sin θ = − 13 .

57

4.6 Scalene triangles Other ranges If we had been asked to solve in the range 0◦ ≤ θ < 360◦ the solutions of part one would be unchanged. The solutions for two would be read from a CAST diagram labelled appropriately to obtain 199.471◦ and 340.529◦ .

4.6

Scalene triangles

Given an equilateral, or isosceles triangle, it is often possible to split it down the middle to form two right angled triangles. In this way it is usually not necessary to employ sophisticated trigonometry to such triangles. However, we need more powerful techniques to deal with scalene or irregular triangles, with no two sides the same length.

4.6.1

Labelling

For non-right-angled triangles, another naming convention and set of equations are required. We name the sides a, b and c in an arbitrary order. We then name the angle opposite side a with A, the angle opposite side b becomes labelled B and similarly for c and C. It should be noted that there is a close relationship between angles and their opposite sides. In particular the larger the angle, the larger the opposite side. This labelling is shown in figure 4.5 on page 59.

4.6.2

Scalene trigonmetry

Under this naming system, we have the following useful laws.

4.6.3

Sine Rule

In a suitably labelled triangle sin B sin C sin A = = a b c This rule can be used when we know one side and its corresponding angle. Then, given another side we can find its corresponding angle, or conversely given the angle we can find the other side.

58

4.6 Scalene triangles

59

Figure 4.5: Labelling a scalene triangle When finding angles we will need to solve an equation of the form sin θ = x, and in this case we must consider the possibility of multiple solutions (see 4.5). This is because there will be a solution in the range 0◦ to 90◦ as usual, but also one in the range 90◦ to 180◦ , and because the triangle is not right angled, such a large angle may be possible.

4.6.4

Cosine Rule

In a suitably labelled triangle a2 = b2 + c2 − 2bc cos A This could also be written b2 = a2 + c2 − 2ac cos B or c2 = a2 + b2 − 2ab cos C This rule is very useful when we know two sides and the angle between them. Then, we can calculate the remaining side’s length.

4.6 Scalene triangles

60

We can also rearrange each equation (left as an exercise for the reader) to obtain b 2 + c 2 − a2 2bc allowing us to obtain an angle given all three sides. For technical reasons there is no ambiguous case for the cosine rule (the second solution will lie in the range of 270◦ to 360◦ and thus cannot appear in a triangle). cos A =

4.6.5

Example

A conventially labelled triangle has A = 63◦ , a = 12 and b = 13. Solve it completely (that is, find all its angles and sides). Solution We have one side and its corresponding angle, and another side. This is a case for the sine rule. sin B sin 63◦ sin B 13 sin 63◦ sin A = ⇒ = ⇒ sin B = = 0.965 a b 12 13 12 Using the inverse function sin−1 we may show that B = 74.853◦ , but this is no right angled triangle, as we must check for multiple solutions. Using the method shown above (see 4.5) we can show that B = 105.147◦ , and there seems to be no reason why this could not be correct also. We have no choice but to accept that this is an ambiguous question, and find both possible triangles, splitting it into two cases. (a) Consider the case B = 74.853◦ , then as all the angles add to 180◦ we obtain that C = 42.147◦ . All that remains to be found is c. We could use the sine rule for this, but we shall use the cosine rule for variety. c2 = a2 + b2 − 2ab cos C ⇒ c2 = 122 + 132 − 2(12)(13) cos 42.147◦ = 81.675. Thus we obtain c = 9.037. This complete solution is A = 63◦ , B = 74.853◦ , C = 42.147◦ a = 12, b = 13, c = 9.037

4.7 Radian Measure

61

(b) Consider the case B = 105.147◦ . Then C = 11.853◦ , and using the cosine rule shows that c2 = 122 + 132 − 2(12)(13) cos 11.853◦ = 7.652◦ so that c = 2.766 and the complete solution in this case is A = 63◦ , B = 105.147◦ , C = 11.853◦ a = 12, b = 13, c = 2.766

4.7

Radian Measure

There is nothing special about the degree. The fact that we have 360◦ degrees in a circle is a historical accident, and has no mathematical significance. There is a “natural” unit of angle, which is mathematically simple. This unit is called the radian. There are 2π radians in a circle, so the radian is quite “big”, there being only slightly over 6 in the whole circle.

4.7.1

Conversion

To convert from degrees to radians we divide by 360 and multiply by 2π, or π alternatively multiply by 180 . 180 We multiply by π to convert from radians to degrees. Some significant angle conversions are given in table 4.2.

4.7.2

Length of Arc

In a circle with radius r and an arc 2 subtended by an angle θ measured in radians (see figure 4.6), the arc length s is given by s = rθ 2

An unbroken section of the circumference

4.7 Radian Measure

62 Degrees Radians 0

0

30

π 6

45

π 4

60

π 3

90

π 2

180

π

360

2π

Table 4.2: Conversion between degrees and radians Proof We present a short, non-examinable, proof. The ratio that the angle θ is of the whole angle in the circle is the same as the ratio of the arc length to the whole circumference. That is s 2πrθ θ = ⇒s= ⇒ s = rθ 2π 2πr 2π Exercise It is left as an exercise for the reader to show that if the angle was φ in degrees the formula would be s=

πrφ 180

which is clearly less “natural”.

4.7.3

Area of Sector

In a circle with radius r and a sector 3 subtended by an angle θ measured in radians, the sector area A is given by 1 A = r2 θ 2 3

A region bounded by two radii and an arc

4.8 Identities

63

Figure 4.6: Length of Arc, Area of Sector Proof We present a short, non-examinable, proof. The ratio that the angle θ is of the whole angle in the circle is the same as the ratio of the sector area to the whole area. That is A πr2 θ 1 θ = 2 ⇒A= ⇒ A = r2 θ 2π πr 2π 2 Exercise It is left as an exercise for the reader to show that if the angle was φ in degrees the formula would be A=

πr2 φ 360

which is clearly less “natural”.

4.8

Identities

Trigonometry is rich in identities. Identities are equations, but with a special significance. For example x+2=4 is an algebraic equation. It is true for some values of x (in this case true for one value), and false for others. Now this equation

4.8 Identities

64

x + x = 2x is always true, regardless of the value of x. This is an identity, an equation that holds true for all values of the variables in it. We may use trigonometric identities to change the form of expressions to make them easier to deal with.

4.8.1

Basic identities tan θ =

sin θ cos θ

sin2 θ + cos2 θ = 1 1 + tan2 θ = sec2 θ cot2 θ + 1 = csc2 θ

4.8.2

Compound angle identities sin(θ ± φ) = sin θ cos φ ± sin φ cos θ cos(θ ± φ) = cos θ cos φ ∓ sin θ sin φ tan(θ ± φ) =

4.8.3

tan θ ± tan φ 1 ∓ tan θ tan φ

Double angle identities sin(2θ) = 2 sin θ cos θ cos(2θ) = cos2 θ − sin2 θ tan(2θ) =

2 tan θ 1 − tan2 θ

4.9 Trigonmetric equations

4.9

65

Trigonmetric equations

In some applications trigonometry leads an almost independent life from angles. For the example the expression A sin(ωt + ) is often used to represent a waveform with angular frequency ω and a so called phase shift . When we deal with equations like this we need to use all we have learned so far of multiple solutions and/or trigonemetric identities.

4.9.1

Example

Solve the following trigonemtric equation for values of θ in the range 0◦ ≤ θ < 360◦ : sin θ = cos θ Solution This is a simple example. Before learning about identities this would have presented us with a severe problem, we might have been able to solve it by graphical means, but exact solutions would be hard to find. In this case I will first divide by cos θ on both sides. Of course that is a little dangerous if that can be zero so we need to be careful, but when cos theta is zero sin θ = ±1 so the two are not equal and there are no solutions like this. Therefore we can exclude this possibility and divide. This obtains cos θ sin θ = ⇒ tan θ = 1 cos θ cos θ (see 4.8.1) which is a much simpler, routine equation. A calculator will yield a principal solution of 45◦ and using our CAST diagram (see 4.5), we can find the other solution in the range to be 225◦ . So θ = 45◦ ; θ = 225◦

4.9.2

Example

Write the following trigonometric equation in terms of sin θ only. Hence or otherwise solve the equation for values of θ in the range −180◦ < θ ≤ 180◦ :

4.9 Trigonmetric equations

6 cos2 θ + sin θ = 5 Solution The first problem consists of getting rid of the cos2 θ term which is not in terms of sin θ. We recall from 4.8.1 that sin2 θ + cos2 θ = 1 ⇒ cos2 θ = 1 − sin2 θ so inserting this into our equation, we obtain 6(1 − sin2 θ) + sin θ = 5 ⇒ 6 − 6 sin2 θ + sin θ = 5 ⇒ 6 sin2 θ − sin θ − 1 = 0 which is actually a quadratic equation (see 3.7) which can be made more clear by letting s = sin θ ⇒ s2 = sin2 θ so we obtain 1 1 6s2 − s − 1 = 0 ⇒ (2s − 1)(3s + 1) = 0 ⇒ s = ; s = − 2 3 which we could have solved using the solution formula of course. Now let us not forget that s was a temporary place holder, we must now solve the equations 1 1 sin θ = ; sin θ = − 2 3 which we solved previously in example 4.5.3 so we can write down our solutions as θ = 30◦ ; θ = 150◦ ; θ = −19.471◦ ; θ = −160.529◦

4.9.3

Example

Suppose that A1 = sin(ωt), A2 = 2cos(ωt) are two waves, what is the resultant wave if they are superimposed?

66

4.9 Trigonmetric equations

67

Solution The principle of superposition says that we can algebraicly add these two signals to find the resultant, but we want to find the resulting wave exactly. When we add the signals together we get sin(ωt) + 2cos(ωt) and let us suppose the resulting signal is of the form r sin(ωt + ), then4 by the results in 4.8.2 we can expand this out to obtain r(sin(ωt) cos φ + sin φ cos(ωt)) Comparing these two equations gives us r cos φ = 1 (i) r sin φ = 2 (ii) If we divide equation (ii) by equation (i), then using the results from 4.8.1 we obtain tan φ = 2 ⇒ φ = 63.435◦ Either using this to find r directly, or by squaring (i) and (ii) and adding together we obtain r2 cos2 φ + r2 sin2 φ = 5 ⇒ r2 (cos2 φ + sin2 φ) = 5 ⇒ r2 = 5 ⇒ r = So the resultant wave is A=

4

√ 5 sin(ωt + 63.435◦ )

We could have just as easily used r cos(ωt + )

√

5

Chapter 5 Complex Numbers We saw in 3.5 that we have a difficulty with certain square roots, namely the square roots of negative numbers. We introduce the set of Complex Numbers to address this situation.

5.1

Basic Principle

Rather than dealing with the square roots of all negative numbers, we can reduce the problem to one such square root. For example, we can use the laws of surds detailed in 3.5 to split negative square roots up. √ √

−9 =

−25 =

√

√ √ −1 9 = 3 × −1

√

√ √ −1 25 = 5 × −1

In general, if we take a to be some positive real number √ √ √ −a = −1 a so in all cases the problem comes down to the square root of −1. Now we know that no real number can have this value, but we shall assign this special square root a specific symbol, namely j. Please note that i is used throughout the mathematical world itself, both in textbooks and in calculators, while engineers often prefer j to represent this square root. You should be aware of this for background reading. Thus we now say that

5.2 Examples

69 √ √ √ √ −9 = 3j; −25 = 5j; −2 = 2j.

Remember, none of these are real numbers. We also sometimes define j by the equivalent equation j 2 = −1.

5.1.1

Imaginary and Complex Numbers

We shall call any number of the form yj an imaginary number where y is a real number. For many people, this terminology produces much of their mistrust and difficulty with this number system. We have invented these numbers in order to solve a specific type of equation, that is true, but it is not the first time. The number zero, and all negative numbers were “invented” in order to solve equations which couldn’t be handled with the positive numbers alone. Numbers of the form x + yj are called complex numbers, where x and y are real numbers. Observe that complex numbers are in general hybrid numbers, having a real section or component and an imaginary component. All real numbers can be thought of as being complex, as for example, we can write π =π+0×j and similarly all imaginary numbers are complex, as for example 3j = 0 + 3 × j so that the complex numbers encompass all the numbers we have used before, and add more besides.

5.2

Examples

Here are some examples of equations that have complex number solutions, and which we could not solve previously. In particular, we now see that quadratic equations (see 3.7) which have “no real solutions” in fact have two complex solutions. 1. x2 + x + 1 = 0

5.3 Argand Diagram Representation Recall that we showed this to be √ √ −1 ± −3 −1 ± 3j = . 2 2 So our two solutions are

√ √ −1 + 3j −1 − 3j x= ;x = 2 2

2. x2 − 6x + 13 = 0 Using the quadratic solution formula once more we obtain p √ 6 ± (−6)2 − 4(1)(13) 6 ± −16 6 ± 4j = = . 2 2 2 So again we have two solutions, which are x = 3 + 2j; x = 3 − 2j

5.3

Argand Diagram Representation

The real numbers are frequently represented by a line often called the number line or simply the real line. Complex numbers are inherently two dimensional and cannot be presented on a line, instead we represent these on a plane. We often talk of the complex plane as an interchangable name for the set of complex numbers, and diagrams in the form of two dimensional graphs are called Argand diagrams. We take the horizontal axis to be, essentially, the real line itself, and we call it the real axis. We call the vertical axis the imaginary axis. Then, given a complex number z = x + yj we represent it as the point (x, y) on the diagram. Note that every point on the diagram corresponds to a unique complex number, and every complex number has a unique point on the diagram. Figure 5.1 shows a typical argand diagrams with two complex numbers, z and w, z = 3 + 5j and w = 4 − 2j.

5.4

Algebra of Complex Numbers

It is surprisingly easy to perform algebraic operations with complex numbers, if we simply do not overly concern ourselves with the origin of j.

70

5.4 Algebra of Complex Numbers

Figure 5.1: The Argand diagram In the remainder of this section, we shall assume that z = a + bj and w = c + dj and that a, b, c and d are all real numbers.

5.4.1

Addition

Adding two complex numbers is extremely elementary. z + w = (a + bj) + (c + dj) = (a + c) + (b + d)j Essentially we treat the “j”s as any other algebraic expression and manipulate it normally. Note that the result of the addition is also a complex number.

71

5.4 Algebra of Complex Numbers Examples Here are some concrete examples involving specific numbers. 1. (2 + 3j) + (−4 + j) = −2 + 4j 2. (j − 4) + (−2 − 3j) = −6 − 2j 3. (1 + 2j) + (2 + 3j) + (−3 − 5j) = 0

5.4.2

Subtraction

Subtraction is really no different than the addition of a negative number. z − w = (a + bj) − (c + dj) = (a − c) + (b − d)j Examples Here are some concrete examples involving specific numbers. 1. (2 + 3j) − (−4 + j) = 6 + 2j 2. (j − 4) − (−2 − 3j) = −2 + 4j 3. (3 + j) − (3 + 4j) = −3j

5.4.3

Multiplication

Multiplication of complex numbers usually requires the expansion of brackets. Additionally it often gives rise to a term containing j 2 , at first glance we might consider such a term imaginary, but recall that j 2 = −1, so this term is real. Here is the process. z × w = (a + bj)(c + dj) = ac + adj + bcj + bdj 2 = ac + (ad + bc)j + (−1)bd = (ac − bd) + (ad + bc)j Because of the confusion caused by the power of j, it is always a good idea to reduce such powers as quickly as possible.

72

5.4 Algebra of Complex Numbers

73

Powers of j We can deal with powers of j as a special case. j1

=

j2

=

−1

j3

=

j × j2

j4

= j2 × j2

j5

=

=

j × j4

j

= −1 =

−1 × j

= −1 × −1 =

j×1

=

−j

=

1

=

j

Table 5.1: Powers of j

We have shown how the powers may be calculated in table 5.1. Note that by j 5 we are repeating ourselves, we can go round and round this table for any power of j. Examples Here are some concrete examples involving specific numbers. 1. (2 + 3j) × (−4 + j) = −8 − 12j + 2j + 3j 2 = −8 − 3 − 10j = −11 − 10j 2. (2 + 3j)2 = (2 + 3j)(2 + 3j) = 4 + 12j + 9j 2 = −5 + 12j 3. (2 + 3j)(2 − 3j) = 4 + 6j − 6j − 9j 2 = 4 + 9 = 13 4. (2 + 3j)3 = (−5 + 12j)(2 + 3j) = −10 + 24j − 15j + 36j 2 = −46 + 9j We could use the binomial expansion (see 3.10) to work out high powers of complex numbers, but it is quite tedious to work out all the powers of j. It is easier to use a technique still to come, De Moivre’s theorem (see 5.7).

5.4 Algebra of Complex Numbers

5.4.4

Division

Division is perhaps the least routine of all the arithmetic operations. We begin by multiplying top and bottom by another, specific, complex number. This leads to two multiplications - it’s a more drawn out process. Remember that multiplying a fraction by a number top and bottom does not change its value, just its form. z a + bj (a + bj)(c − dj) = = w c + dj (c + dj)(c − dj) ac + bcj − adj − bdj 2 (ac + bd) + (bc − ad)j = 2 2 2 c + cdj − cdj − d j c2 + d 2 This doesn’t exactly look more friendly, but looking closely you will see that we now only have a real number on the bottom line. The division is now routine ac + bd bc − ad + 2 j = 2 c + d2 c + d2 It is particularly more pleasant to see this in real examples. =

5.4.5

Examples

Here are some concrete examples involving specific numbers. 1. 3 2 + 3j =1+ j 2 2 No trickery is required here, it’s easy to divide by real numbers. 2. 2 + 3j (2 + 3j)(−4 − j) = −4 + j (−4 + j)(−4 − j) Performing the multiplications fully gives rise to =

−5 − 14j 5 14 =− − j 17 17 17

Note that dividing the 15 in finally is not required, it is done here to stress that the answer really is just a complex number in x + yj form. 3. (2 + 3j)(j) 2j + 3j 2 2 + 3j = = j (j)(j) j2 Following the method above religiously, we should have multiplied by −j top and bottom not j, remember, we reverse the sign on the imaginary

74

5.5 Definitions

75

section. However, the aim of this is to create a real number on the bottom line, and in this case we can obtain that simply by multiplying by j. It’s quite ok to multiply by −j though, and some people would prefer this. =

5.5 5.5.1

−3 + 2j = 3 − 2j −1

Definitions Modulus

The modulus of a complex number z = x + yj, written |z| is given by |z| =

p x2 + y 2 .

Note that we met the concept of modulus before (see 3.8.1) and so a new definition seems odd. Close examination will show that this definition is exactly the same as the previous √ modulus on the real numbers (note that in this case y = 0 and so |z| = x2 ).

5.5.2

Conjugate

The conjugate of a complex number z = x + yj, written z ∗ or z is given by z ∗ = x − yj. On an Argand diagram this represents the reflection of the number in the real axis. Note that (z ∗ )∗ = z. The conjugate is particularly important in the operation of division, because z × z ∗ = |z|2 which is a real number. Observe that when a quadratic equation has two complex solutions, they are conjugates of each other.

5.6 Representation

5.5.3

76

Real part

The real part of a complex number z = x + yj, denoted Re(z) or 0 then f (a + x) − f (a) > 0 ⇒ f (a + x) > f (a) for small x; • If f 00 (a) < 0 then f (a + x) − f (a) > 0 ⇒ f (a + x) < f (a) for small x. These statements are precisely that there is a local minimum, or maximum at x = a respectively.

Chapter 12 Differential Equations A very important application of integration is that of differential equations. These are equations in terms of the derivatives of a variable.

12.1

Concept

We shall restrict ourselves to Differntial Equationss involving only two variables x and y. A Differential Equation or D.E. for short is an equation involving x, y and the derivatives of y with respect to x. The order of a differential equation is the number of the highest derivative present in the equation. In general we wish to find the function y = f (x) which satisfies the D.E., and in general, unless we have information to help us calculate them, we will have a constant for each order of the D.E. when we solve it.

12.2

Exact D.E.s

Differential equations of the form dy = f (x) dx are called exact differential equations, as the right hand side is exactly a derivative and all we need do is integrate on both sides with respect to x.

12.2 Exact D.E.s

201

This is a first order exact equation, and we could have second order or third order equations, but we will restrict ourselves to first order equations of this type here. Equations of the form d (f (x, y)) = g(x) dx are also exact, as all that has to be done is to integrate on both sides with respect to x to remove all derivatives.

12.2.1

Example

Solve the differential equation dy = 3x2 dx Solution Integrate on both sides with respect to x, remembering that integration is the reverse of differentiation and “cancels it”. Z Z dy dx = 3x2 dx dx The next step is usually ommitted when we feel comfortable. Z Z ⇒ dy = 3x2 dx ⇒ y = x3 + c is the final solution. R R Note that dy is a shorthand for 1dy which is y.

12.2.2

Example

Solve the differential equation x

dy √ = x+1 dx

12.2 Exact D.E.s

202

Solution This doesn’t look exact, but all we have to do is divide by x on both sides to obtain √ 1 dy x+1 dy x2 + 1 ⇒ = ⇒ = dx x dx x so that we obtain 1 dy 1 = x− 2 + ⇒ dx x Now integrate on both sides, with respect to x. Z Z 1 dy 1 dx = x− 2 + dx dx x so that we obtain 1

y = x2 ÷ which is

1 + ln x + c 2

1

y = 2x 2 + ln x + c

12.2.3

Example

Solve the following differential equation. x2

dy + 2xy = 4e2x dx

Solution This is an exact equation, but it doesn’t look like it. The left hand side can be written as the derivative of a product, as so ⇒


d x2 y = 4e2x dx

and so integrating both sides with respect to x removes the derivative (cancels it). Z ⇒ x2 y =

4e2x dx = 2e2x + c

and so dividing by x2 we obtain ⇒y=

2e2x + c x2

12.3 Variables separable D.E.s

12.3

203

Variables separable D.E.s

Differential equations of the form f (y)

dy = g(x) dx

are called variables separable differential equations, because it is possible to separate the function of y on the left hand side from that of the function of x on the right hand side. If we integrate with respect to x on both sides we obtain Z Z dy f (y) dx = g(x)dx dx which, by a variation of the chain rule can be shown to be Z Z f (y)dy = g(x)dx. In fact most people thing of rearranging the original D.E. to get terms involving x on one side, together with dx and those of y and dy on the other, like so: dy = g(x) ⇒ f (y)dy = g(x)dx f (y) dx and then placing integral signs in front of both sides. Although this does not formally reflect what is happening mathematically it is a useful way to think of these problems and makes solving problems much easier.

12.3.1

Example

Solve the following differential equation y

dy =x dx

Solution There are two ways of doing this, the formal way and the informal way. For this first example, we shall do both. Formally, we integrate with respect to x on both sides. Z Z dy y dx = xdx dx

12.3 Variables separable D.E.s

204

Z

Z

⇒

ydy =

xdx

y2 x2 = +c 2 2 Note that each integration technically gives rise to a constant, but we can absorb them into a single one. We’re finished with the calculus now, it’s only algebra to polish up. Multiply by 2 on both sides ⇒

y 2 = x2 + 2c but 2c is just a constant, say d so y 2 = x2 + d We could leave it here. The other way of thinking is to imagine we split the dy from the dx. So that here, we “multiply up” by dx on both sides ⇒ ydy = xdx and then we place integral signs in front of both sides Z Z ⇒ ydy = xdx this is not really what happens, but it works out the same, and is faster. From here we continue as above.

12.3.2

Example

Solve the differential equation1 dy = −λy dx where λ is a constant. 1

This is quite an important D.E. in science, if we let y = A and x = t it is the equation

which defines the activity of a nuclear sample over time (radioactive decay), in which case lambda is a poaitive constant known as the decay constant (which is related to the half life). If we let y = I and x = t and λ = in a discharging capacitor.

1 CR

we get the D.E. which describes the current

12.3 Variables separable D.E.s Solution This is variables separable, even though it is not of the classic form described above, we can put it into that form. Divide by y on both sides to obtain 1 dy = −λ y dx there are no x terms, which just makes things easier for us. Do our “trick” or rearranging the derivative (placing all the y and dy pieces on one side and the x and dx pieces on the other). 1 ⇒ dy = −λdx y and place integral signs in front of both sides (we could have done this formally as above of course). Z Z 1 ⇒ dy = −λdx y and we can take the constant out on the RHS to obtain Z Z 1 ⇒ − λ dx y which gives us ln y = −λx + c R

R

please remember dx = 1dx = x. We can remove the logs on both sides by taking e to the power of both sides (please see 3.9.5). eln y = e−λx+c ⇒ y = e−λx+c we can now use the laws of indices (see 3.5) (the first one, in reverse) to write ⇒ y = e−λx ec and note that c is a constant, so that ec is a constant, let’s say A. We finally obtain ⇒ y = Ae−λx Solution Solve the following differential equation dy 2 = dx (x + 3)(y − 2)

205

12.4 First order linear D.E.s Solution Again, this is variables separable, but the variables have not yet been seperated. Move all the x and dx terms to one side and the y and dy to the other. 2 dx ⇒ y − 2dy = x+3 Now place integral signs in front of both sides Z Z 2 ⇒ y − 2dy = dx x+3 and integrate on both sides. Now Z y2 y − 2dy = − 2y 2 don’t worry about the constant, we will do one for the whole equation as usual. The other integral Z Z x+3 2 dx = 2 x x+3 d which can be done in several ways, the numerator is the derivative of the denominator, so we can say Z x+3 2 x = 2 ln(x + 3) d but we could also have used substitution u = x+3, or the “quick” substitution method for ax + b instead of x. Therefore we have ⇒

y2 − 2y = 2 ln(x + 3) + c 2

which can’t be cleaned up with algebra too much more.

12.4

First order linear D.E.s

Differential equations of the form dy + f (x)y = g(x) dx are known as first order linear differential equations. To solve these we normally attempt to convert them to an exact differential equation. We

206

12.4 First order linear D.E.s

207

do this my multiplying through by an integrating factor. Let us call this function i(x). We attempt to find a formula for i(x). Please note this is not examinable, it is the final result here that is important. dy + i(x)f (x)y = i(x)g(x) dx Now if we consider that the left hand side is not exact (hopefully) it seems likely it may be a product i(x)

d (i(x)y) = i(x)g(x) dx in which case, from the product rule we have d di i(x) = i(x)f (x) ⇒ = i(x)f (x) dx dx which is first order variables separable Z Z Z di = f (x)dx ⇒ ln(i(x)) = f (x)dx ⇒ i(x) which finally gives us i(x) = e

12.4.1

R

f (x)dx

Z = exp( f (x)dx)

Example

Solve the following differential equation. dy + 2y = ex dx Solution This is a first order linear differential equation in classic form. In this case, we first find the integrating factor i(x) i(x) = e

R

f (x)dx

R

=e

2dx

= e2x

so we multiply through by this on both sides. ⇒ e2x

dy + 2ye2x = ex × e2x = e3x dx

12.4 First order linear D.E.s The left hand side is now exact, althought this might not be obvious. d 2x
⇒ e y = e3x dx and if we now integrate on both sides we obtain 1 ⇒ e2x y = e3x + c 3 2x and thus we can now divide by e on both sides 1 ⇒ y = ex + ce−2x 3

12.4.2

Example

Solve the following differential equation. dy x3 + x2 y = x − 1 dx Solution This is not yet in classic form - note that in the form we stated for this type dy term, so let’s divide by x3 of equation there is nothing in front of the dx throughout to try and achieve this. dy 1 x−1 ⇒ + y= dx x x2 Now this is in classic first order linear form, and we try to find the integrating factor i(x). R 1 R i(x) = e f (x)dx = e x dx = eln x = x So we now multiply throughout by this factor, which is simply x, to obtain x−1 dy ⇒x +y = dx x which is now an exact D.E., looking at the LHS we see that d x−1 ⇒ (xy) = dx x and so integrating both sides with respect to x yields Z Z x−1 1 ⇒ xy = dx = 1 − dx = x − ln x + c x x and finally, dividing by x both sides we obtain ln x c + y =1− x x

208

12.5 Second order D.E.s

12.5

Second order D.E.s

Second order differential equations, are in general more difficult to solve than their first order counterparts, but we shall only deal very simple cases of this class of D.E.s, which many consider simpler to solve than first order equations. We shall consider only second order differential equations with constant coefficients, which have the form a

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

When f (x) = 0 this is called a homogenous differential equation, otherwise it is called inhomogeneous. We shall only consider solutions to the homogeneous case. The solution to the homogenous equation is known as the Complementary function, or C.F. for short. There is a step-by-step procedure for solving these D.E.s.

12.5.1

Homogenous D.E. with constant coefficients

Consider

dy d2 y + b + cy = 0. 2 dx dx We begin by forming a quadratic equation known as the auxilliary equation. This equation is am2 + bm + c = 0. a

Now we know from 3.7 that this equation can have two, one or no real (two complex) solutions. Two real solutions Suppose that we have two real solutions m = α and m = β, then the C.F. is given by y = Aeαx + Beβx where A and B are constants2 which can often be determined from data in the question. 2

These arise out of the two integrations required for the second order derivative in the

equation in fact, but this process does not require us to do any integration at all.

209

12.5 Second order D.E.s

210

One real solution Suppose that we have one real solution m = α, then the C.F. is given by y = (A + Bx)eαx where once again A and B are constants. Two complex solutions Suppose that we have two complex solutions3 m = α + βj and m = α − βj. Then the C.F. is given by y = Ae(α+βj)x + Be(α−βj)x but this is a very clumsy representation and infrequently used. We can use the laws of indices (see 3.5) to improve it. Please note that this “proof ” is not examinable, it the final result we want here. ⇒ y = Aeαx eβjx + Beαx e−βjx Now if we use the identity from 5.6.3 we see that we obtain ⇒ y = Aeαx (cos βx + j sin βx) + Beαx (cos(−βx) + j sin(−βx)) which by th e nature of sin and cos may be shown to be ⇒ y = Aeαx (cos βx + j sin βx) + Beαx (cos βx − j sin βx) and finally with some basic manipulation ⇒ y = eαx ((A + B) cos βx + (Aj − Bj) sin βx)) . Recall that A and B are constants, and so (A + B) and (A − B)j are also constants. Let us call them C and D respectively. Thus ⇒ y = eαx (C cos βx + D sin βx)) . This is the form that should be used with two complex roots, and this result should be known. 3

Recall at this point that when we solve a quadratic equation to give two complex

solutions these solutions form a conjugate pair (see 5.5.2).

12.5 Second order D.E.s

12.5.2

211

Example

Solve the following differential equation d2 y dy − 5 + 6y = 0 2 dx dx Solution This is a second order homogeneous D.E. with constant coefficients, and moreover a = 1, b = −5 and c = 6. The auxilliary equation is thus m2 − 5m + 6 = 0 so solve this, plain vanilla, quadratic equation (see 3.7) ⇒ (m − 2)(m − 3) = 0 ⇒ m = 2, m = 3 this is two real roots, so the solution is y = Ae2x + Be3x

12.5.3

Example

Solve the following differential equation d2 y dy + 2 +y =0 dx2 dx Solution This is a second order homogeneous D.E. with constant coefficients, and moreover a = 1, b = 2 and c = 1. The auxilliary equation is thus m2 + 2m + 1 = 0 so solve this quadratic. ⇒ (m + 1)(m + 1) = 0 ⇒ m = −1 this is one real root, so the solution is y = (A + Bx)e−1x which is y = (A + Bx)e−x

12.5 Second order D.E.s

12.5.4

212

Example

Solve the following differential equation d2 y dy 2 2+ − 6y = 0 dx dx Solution This is a second order homogeneous D.E. with constant coefficients, and moreover a = 2, b = 1 and c = −6. The auxilliary equation is thus 2m2 + m − 6 = 0 so solve this quadratic. 3 ⇒ (2m − 3)(m + 2) = 0 ⇒ m = , m = −2 2 this is two real roots, so the solution is 3

y = Ae 2 x + Be−2x

12.5.5

Example

Solve the following differential equation d2 y dy + 2 + 5y = 0 2 dx dx Solution This is a second order homogeneous D.E. with constant coefficients, and moreover a = 1, b = 2 and c = 5. The auxilliary equation is thus m2 + 2m + 5 = 0 so solve this quadratic.

p √ 22 − 4(1)(5) −2pm −16 m= = 2 2 = −1pm2j ⇒ m = −1 + 2j, m = −1 − 2j −2 ±

So we have two complex roots, with α = −1 and β = 2 in the form used above, so that y = e−1x (C cos 2x + D sin 2x) which is y = e−x (C cos 2x + D sin 2x) which, incidentially reflects a simple harmonic motion under damping.

12.5 Second order D.E.s

12.5.6

213

Example

Solve the following differential equation4 d2 y = −ω 2 y 2 dx where ω is a constant. Solution First place it the more usual form d2 y + ω2y = 0 dx2 which is the usual second order homogeneous D.E. with constant coefficients. The auxilliary equation is m2 + ω 2 = 0 Remember, ω is just a constant. Solving this quadratic, we could use the formula of course, but it’s easier to say √ ⇒ m2 = −ω 2 ⇒ m = ± −ω = ±ωj and we have ⇒ m = ωj, m = −ωj and we have two complex solutions, this time with α = 0 and β = ω in the above form. Thus y = e0x (C cos ωx + D sin ωx) and finally we have y = C cos ωx + D sin ωx which is the general form of a wave with angular frequency ω, the phase angle and amplitude can be determined using a procedure demonstrated in our trigonometry section (see 4.9.3).

4

This is another very important D.E., as it is the D.E. which describes all simple

harmonic motion (behind all sin and cos waves and similar phenomonena).

Chapter 13 Differentiation in several variables The function z = f (x, y) may be represented in three dimensions by a surface, where the value z represents the height of the surface above the x, y plane.

13.1

Partial Differentiation

If z = f (x, y) is a function of two independent variables x and y, then we define the two partial derivatives of z with respect to x and y as ∂z f (x + δx, y) − f (x, y) = lim ∂x δx→0 δx and ∂z f (x, y + δy) − f (x, y) = lim ∂y δy→0 δy

13.1.1

Procedure

The actual practice of partial differentation rather than “ordinary” is simply that we differentiate with respect to one variable, assuming all the others are constant for this differentiation. There are some warnings associated with this that come later, but for now we look at some examples.

13.1 Partial Differentiation

13.1.2

215

Examples

For each of the following functions z = f (x, y), find

∂z ∂x

and

∂z . ∂y

1. z = xy 2 + 1 (see figure 13.1)

Figure 13.1: The graph of f (x, y) = xy 2 + 1 2. z = x3 (sin xy + 3x + y + 4) (see figure 13.2) Solutions 1.

2.

∂z ∂z = y2; = 2xy ∂x ∂y ∂z = 3x2 (sin xy + 3x + y + 4) + x3 (y cos xy + 3) ∂x ∂z = x3 (x cos xy + 1) ∂y

13.1 Partial Differentiation

Figure 13.2: The graph of f (x, y) = x3 (sin xy + 3x + y + 3)

13.1.3

Notation

Due to the ambiguity of the f 0 notation (we do not know which variable we differentiated with respect to) a different notation is adopted for partial derivatives. ∂f ∂ f= ∂x ∂x ∂f ∂ f= fy = ∂y ∂y

fx =

13.1.4

Higher Derivatives

If the partial derivatives exist, there is nothing to stop us differentiating repeatedly. Note however that we can differentiate either with respect to the same variable or a different one, giving some interesting notation. We show the short hand notations at the same time. Please note the order of differentiation and notation carefully.   ∂ ∂f ∂2 ∂ 2f fxx = = f= ∂x ∂x ∂x2 ∂x2

216

13.1 Partial Differentiation

217

  ∂ ∂f ∂2 ∂ 2f fyy = = 2f = ∂y ∂y ∂y ∂y 2   ∂2 ∂ 2f ∂ ∂f = f= fxy = ∂y ∂x ∂y∂x ∂y∂x   ∂ ∂f ∂2 ∂ 2f fyx = = f= ∂x ∂y ∂x∂y ∂x∂y while it is usually the case that fxy = fyx it is not always the case.

13.1.5

Example

For the functions in Example 13.1.2 find the higher order derivatives fx y, fy x, fy y and fx x. Solution 1.

∂ 2 y = 0; ∂x ∂ 2 fx y = y = 2y; ∂y ∂ fy x = 2xy = 2y; ∂y ∂ 2xy = 2x. fy y = ∂y fx x =

2.


∂ 3x2 sin xy + 9x3 + 3x2 y + 12x2 + x3 y cos xy + 3x3 ∂x 2 = 3x y cos xy + 6x sin xy + 27x2 + 6xy + 24x − x3 y 2 sin xy + 3x2 y cos xy + 9x2 ;
∂ 3x2 sin xy + 9x3 + 3x2 y + 12x2 + x3 y cos xy + 3x3 fx y = ∂y fx x =

= 3x3 cos xy + 3x2 − x4 y sin xy + x3 cos xy = 4x3 cos xy + 3x2 − x4 y sin xy; ∂ 4 fy x = x cos xy + x3 ∂y = −x4 y sin xy + 4x3 cos xy + 3x2 ; ∂ 4 fy y = x cos xy + x3 ∂y = −x5 sin xy.

13.2 Taylor’s Theorem

13.2

Taylor’s Theorem

We now attempt to extend Taylor’s theorem to two variables (see 11.3). Suppose that f (x, y) and its partial derivatives (including those of higher order) exist and are continuous in a neighbourhood of (a, b). Then    2 ∂ ∂ 1 ∂ ∂ f (a + h, b + k) = f (a, b) + h +k f (a, b) + h +k f (a, b) ∂x ∂y 2! ∂x ∂y  n−1  n 1 ∂ ∂ 1 ∂ ∂ + · · ·+ h +k f (a, b)+ h +k f (a+θh, b+θk) (n − 1)! ∂x ∂y (n)! ∂x ∂y where 0 < θ < 1 and the notation used indicates  r   r−1 ∂ ∂ ∂ ∂ ∂ ∂ h f (x, y) = h h f (x, y) +k +k +k ∂x ∂y ∂x ∂y ∂x ∂y where   ∂ ∂ h +k f (x, y) = hfx (x, y) + kfy (x, y) ∂x ∂y

13.3

Stationary Points

In simple two variable calculus, one of the most important problems to be solved is finding and classifying the turning points of curves. That is, the points where the curve becomes horizontal for however short a time. We wish to accomplish the same feat in three dimensions, and this corresponds to finding places where the ”landscape” is flat at a specific point.

13.3.1

Types of points

The most important types of points to find are local maxima and local minima. A local maximum (a, b) is a point so that for small h and k f (a + h, b + k) ≤ f (a, b) ⇒ f (a + h, b + k) − f (a, b) ≤ 0. In other words, even slight movements “off centre” result in a smaller value (and they must be small, for larger maxima may exist elsewhere). Conversely, a local minimum (a, b) is a point so that for small h and k f (a + h, b + k) ≥ f (a, b) ⇒ f (a + h, b + k) − f (a, b) ≥ 0. The conditions on the left are more understandable, but the form on the right is useful later.

218

13.3 Stationary Points

13.3.2

219

Finding points

At any such point the rate of change with respect to any variable must be zero (otherwise the surface would be “steep” rather than flat approaching from the direction of that variable’s axis). Therefore, in three variables, where z = f (x, y) ∂z ∂z = 0, = 0. ∂x ∂y These conditions are necessary for locating local maxima and minima but they are not sufficient. Points that satisfy these equations are known as critical points or stationary points. We need to classify points we find in that way.

13.3.3

Classifying points

If we use Taylor’s theorem (see 13.2) to expand at the critical point1 , we obtain  2 ∂ ∂ 1 h +k f (a, b) + · · · f (a + h, b + k) − f (a, b) = 2 ∂x ∂y and therefore 1 f (a + h, b + k) − f (a, b) = 2

  2 2 ∂2 2 ∂ 2 ∂ h + 2hk +k f (a, b) + · · · ∂x2 ∂x∂y ∂y 2

Therefore, if h and k are sufficiently small then the sign of f (a + h, b + k) − f (a, b) will be determined by   2 2 ∂2 1 2 ∂ 2 ∂ h + 2hk +k 2 ∂x2 ∂x∂y ∂y 2 For simplicity we shall make the substitutions that A = fxx , B = fxy and C = fyy evaluated at (a, b). So we wish to examine the sign of Ah2 + 2Bhk + Ck 2 1

Remember that at the critical point the first partial derivatives of z with respect to x

and y are zero, which will cause the first term in the Taylor expansion with derivatives to vanish.

13.3 Stationary Points and we start by completing the square. Provided that A 6= 0, then.   2B C 2 2 2 2 Ah + 2Bhk + Ck = A h + hk + k A A ( 2   ) B AC − B 2 =A h+ k + k2 A A2 As of course the squared term will always be positive, the sign of the whole expression (which remember is close to f (a + h, b + k) − f (a, b)) will be dictated by AC − B 2 . If AC − B2 < 0 In this case, then depending on the values of h and k the expression can have both positive and negative values however close to (a, b) we approach. Therefore this type of turning point is neither a maximum nor a minimum, but is in fact a saddle point. If AC − B2 > 0 In this case the expression will have the same sign as A itself, and we shall have either a maximum or minimum.

13.3.4

Summary

To find and classify turning points in three variables where z = f (x, y) the whole procedure is. Solve fx = fy = 0, and find all (a, b); 2 • If fxx fyy − fxy < 0 at (a, b) then we have a saddle point. 2 • If fxx fyy − fxy >0

– If fxx > 0 then (a, b) is a local minimum (and fyy will also be positive). – If fxx < 0 then (a, b) is a local maximum (and fyy will also be negative). 2 • If fxx fyy − fxy = 0 further investigation is required.

220

13.3 Stationary Points

13.3.5

221

Example

Find the critical points of z = f (x, y) = x3 + y 3 − 3x − 12y + 20 and determine their nature. Solution First of all we need to find the two partial derivative and set them equal to zero. fx = 3x2 − 3; fy = 3y 2 − 12 and now we place them equal to zero fx = 0 ⇒ x = ±1 fy = 0 ⇒ y = ±2 and both these equations must be satisfied at once. Therefore we have four critical points (−1, −2), (−1, 2), (1, −2), (1, 2) Now we calculate fxx , fyy and fxy . fxx = 6x; fyy = 6y; fxy = 0 so to determine the type of turning point in the first instance we use 2 fxx fyy − fxy = 36xy

Now look at each point in turn • (−1, −2) – 36xy is positive, but fx x < 0 we have a local maximum – Calculating z we find the point at (−1, −2, 38). • (−1, 2) – 36xy is negative, we have a saddle point – Calculating z we find the point at (−1, 2, 6). • (1, −2)

13.4 Implicit functions

222

– 36xy is negative, we have a saddle point – Calculating z we find the point at (1, −2, 34). • (1, 2) – 36xy is positive, but fx x > 0 we have a local minimum – Calculating z we find the point at (1, 2, 2). as shown in figure 13.3.

Figure 13.3: A graph with four turning points

13.4

Implicit functions

If x, y, . . . are all functions of one variable t then dz ∂z dx ∂z dy = + + ··· dt ∂x dt ∂y dt Thus if z = f (x, y) = 0

13.5 Lagrange Multipliers

223

then dz ∂z dx ∂z dy = + =0 dx ∂x dx ∂y dx Thus

13.5

− ∂z dy = ∂z∂x dx ∂y

Lagrange Multipliers

Lagrange2 problems where there is some constraint on one or more of the variables that must be satisfied. Normally the problem is expressed in findind the maxima and minima of u = f (x, y, z) provided that φ(x, y, z) = 0. We can regard φ(x, y, z) as expressing z in terms of x and y. That is z is a function z(x, y). This means we are in fact trying to find the maxima and minima of f (x, y, z(x, y)) which is only a function of two independent variables. If f (x, y, z(x, y)) has a critical point at (a, b). 0=

∂z ∂ f (x, y, z(x, y)) = fx + fz ∂x ∂x

0=

∂ ∂z f (x, y, z(x, y)) = fy + fz ∂y ∂y

Recall that φ(x, y, z) = 0 and this time we can regard z as a function of x and y in exactly the same way to give

We can eliminate only if

∂z ∂z

and

φx + φz

∂z =0 ∂x

φy + φz

∂z =0 ∂y

∂z ∂y

we see that (a, b, c) is a critical point if and

• φ(a, b, c) = 0 2

Comte Joeseph Lewis Lagrange (1736-1813) was a French Mathematician who did a

great deal of work on what is now called Classical Mechanics.

13.5 Lagrange Multipliers

224

• φx fz − φz fx = 0 • φy fz − φz fy = 0 when all functions are evaluated at (a, b, c). If we now define λ = −fz /φz , these conditions become • φ(a, b, c) = 0 • fx + λφx = 0 • fy + λφy = 0 • fz + λφz = 0 The value λ is called the Lagrange multiplier and we find turning points by working out the values of λ which satisfy the above four equations. Another way of putting this is that we define F = f + λφ and solve for the equations Fx = Fy = Fz = φ = 0

13.5.1

Example

Find the closest distance from the surface z = x2 + y 2 to the point (3, −3, 4). Solution The distance from (x, y, z) to the point (3, −3, 4), which we shall call d satisfies the equation d2 = f (x, y, z) = (x − 3)2 + (y + 3)2 + (z − 4)2 The surface specifies the constraint, given that (x, y, z) lies on the surface z = x2 + y 2 we have that φ(x, y, z) = x2 + y 2 − z 2 = 0 We now consider F = f + λφ. F = (x − 3)2 + (y + 3)+ (z − 4)2 + λ(x2 + y 2 − z)

13.6 Jacobians

225

and thus Fx = 2(x − 3) + 2λx = 0 Fy = 2(y + 3) + 2λy = 0 Fz = 2(z − 4) − λ = 0 φ = x2 + y 2 − z = 0 If we rearrange each of these equations for each variable in turn, and insert them into the fourth we obtain 9 9 λ+8 =0 + − 2 2 (1 + λ) (1 + λ) 2 ⇒ 36 − (λ + 8)(λ2 + 2λ + 1) = 0 ⇒ (λ − 1)(λ + 7)(λ + 4) = 0 We examine the distance at each value for λ. 3 3 9 19 λ = 1 ⇒ x = , y = − , z = , d2 = 2 2 2 4 1 1 1 147 λ = −7 ⇒ x = − , y = , z = , d2 = 2 2 2 4 2 λ = −4 ⇒ x = −1, y = 1, z = 2, d = 36 so the shortest distance was √

19 2

13.6

Jacobians

The determinant

∂x ∂u ∂y ∂u

∂y ∂v ∂x ∂v

13.7 Parametric functions is called the Jacobian 3 of x and y with respect to u and v. This is often denoted by ∂x ∂x   x, y ∂(x, y) ∂u ∂v J = = u, v ∂(u, v) ∂y ∂y ∂u ∂v This can be expanded to higher dimensions, so that for example the Jacobian of x, y and z with respect to u, v and w is ∂x ∂x ∂x ∂u ∂v ∂w ∂y ∂y ∂y ∂(x, y, z) = ∂(u, v, w) ∂u ∂v ∂w ∂z ∂z ∂z ∂u ∂v ∂w

13.6.1

Differential

If z = f (x, y), then if f (x, y) is differentiable δz = f (x + δx, y + δy) − f (x, y) = δxfx + δyfy + 1 deltax + 2 deltay where δx, δy → 0 ⇒ 1 , 2 → 0. The expression dz = fx dx + fy dy is called the total differential at z.

13.7

Parametric functions

Suppose z = f (x, y) is differentiable and suppose x = x(t) and y = y(t) are differentiable. Then dx dy dz = fx + fy dt dt dt 3

Karl Gustav Jacob Jacobi (1804-1851) was a German mathematician who worked on

various fields such as analysis, number theory and who helped to found the area known as elliptic functions, which were used to produce the recent proof of Fermat’s last Theorem.

226

13.8 Chain Rule

227

in other words

dz ∂z dx ∂z dy = + dt ∂x dt ∂y dt

The proof is ommitted.

13.7.1

Example

If z = x2 y − 4xy 20 + 1 where x = t3 − t and y = cos t find

dz dt

when t = 0.

Solution We need to evaluate fx and fy . ∂z = y(2x) − 4y 20 + 0 = 2xy − 4y 20 ∂x ∂z = x2 − 4x × 20y 19 + 0 = x2 − 80xy 19 ∂y We also require the ordinary derivatives of the x and y functions with respect to t. dx dy = 3t2 − 1, = − sin t dt dt Therefore dz = (2xy − 4y 20 )(3t2 − 1) + (x2 − 80xy 19 )(− sin t) dt Now when t = 0, we have that x = 03 − 0 = 0, y = cos 0 = 1 plugging this all in yields dz = (0 − 4(1))(0 − 1) + (0 − 0)(−0) = −4 × −1 = 4 dt

13.8

Chain Rule

Suppose z = f (x, y) and x = x(u, v) and y = y(u, v) are functions such that xu , xv , yu and yv all exist. Then zu = fx xu + fy yu

13.8 Chain Rule i.e.

228

∂z ∂f ∂x ∂f ∂y = + ∂u ∂x ∂u ∂y ∂u and similarly zv = fx xv + fy yv

i.e.

∂z ∂f ∂x ∂f ∂y = + ∂v ∂x ∂v ∂y ∂v

Chapter 14 Integration in several variables 14.1

Double integrals

Consider the volume between a function over a region R on the x, y plane. Provided f (x, y) is positive over the region R then X V = lim f (x, y)δxδy δx,δy→0

which we denote as Z Z f (x, y)dxdy

V = R

which, if R is considered to be the simple box shown in figure 14.1 (that is those points for which a ≤ x ≤ b and c ≤ y ≤ d) may be evaluated by Z Z Z dZ b V = f (x, y)dxdy = f (x, y)dxdy R

14.1.1

c

a

Example

Calculate the volume under the graph of f (x, y) = xy + 1 over the interval 0 ≤ x ≤ 2 and 0 ≤ y ≤ 4.

14.2 Change of order

230

Figure 14.1: Double integration over the simple region R. Solution This is very straight-forward, because the region of integration is simply a rectangle. Z

4

2

Z

Z xy+1dxdy =

V = 0

14.2

0

0

4



x2 y +x 2

2

Z dy =

0

0

4

 4 2y+2dy = y 2 + 2y 0 = 24

Change of order

It is possible, with care, to change the order in which the integration is performed (that is, whether we sum over x or y first). This is sometimes required because the order we may try first results in a very difficult integration. Usually it helps to sketch over the region of integration first, we shall consider a simple situation in which R is such that any line parallel to the x or y axis meets the boundary or R at most twice. If this is not the case then it is possible to subdivide R and treat each section in this way. One possibility is that we split R into two curves y = φ1 (x) and y = φ2 (x) such that φ(x) ≤ φ(x) for a < x < b. We divide R into vertical strips of

14.3 Examples

231

width δx, which is then divided into sections of height δy. See figure 14.2 for details.

Figure 14.2: Double integration over x then y. Therefore the integral in this manner is Z Z f (x, y)dydx R

Z x=b (Z

)

y=φ2 (x)

f (x, y)dy

=

dx

y=φ1 (x)

x=a

where normally we shall not emphasise that x = or y = in the limits. However, we can also turn the region into two functions x = ψ1 (y) and x = ψ2 (y) where φ1 (y) ≤ φ2 (y) for all c < y < d. This is shown in figure 14.3. In this case we can write the integral as Z Z f (x, y)dxdy R

Z

y=d

(Z

=

f (x, y)dx dy. y=c

14.3

)

x=ψ2 (y)

x=ψ1 (y)

Examples

Here are some examples of double integrations.

14.3 Examples

232

Figure 14.3: Double integration over y then x.

14.3.1

Example

Reverse the order of integration to evaluate the integral in example 14.1.1. Solution 4

Z

2

Z

V =

xy + 1dxdy 0

0

was the original integral, which, because the region of integration is so simple (just a rectangle) we obtain Z 2Z 4 V = xy + 1dydx 0

Z = 0

2



xy 2 +y 2

0

4

Z dx =

0

2

8x + 4dx 0

= [4x2 + 4x]20 = 24 which was the same result we achived before.

14.3.2

Example

Find the volume enclosed under the curve z = cos xy bounded in the xy plane by y = x2 , x > 0, y < 1.

14.4 Triple integrals

233

Solution This volume is given by Z

1

x2

Z

cos xydydx. 0

0

So evaluate the middle integral first Z 1 2 [−x sin xy]x0 dx = 0

Z =

1



−x sin x3 − {−x sin 0} dx

0

Z =−

1

x sin x3 dx.

0

This leaves us with a single integral in one variable.

14.4

Triple integrals

Double integrals sum a function over some region R, and if the function translates as a height then the result is a volume that is calculated. A volume could be obtained directly by a triple integral over some three dimension region V (over the function 1). Actually this is usually pointless, but sometimes we wish to sum a function over a volume and not just an area. For example, suppose that the density of some region of space is given by the function ρ = ρ(x, y, z), then the mass of the region could be found by multiplying the density by the volume elements over the whole volume. In other words Z Z Z m= ρdxdydz. V

14.4.1

Example

The density of a gas in a cubic box which extends in each axis from 0 to 5 is given by ρ = x2 y + z find the mass enclosed in the box.

14.5 Change of variable

234

Solution The mass will be given by Z Z Z

x2 y + zdxdydz

m= V

which is Z

5

Z

5

Z

= 0

0

5

x2 y + zdxdydz

0

and as before we do the integrals from the inside out 5 Z 5Z 5 3 Z 5Z 5 xy 125y = + xz dydz = + 5zdydz 3 3 0 0 0 0 0 5 Z 5 Z 5 125y 2 3125 = + 5yz dz = + 25zdz 6 6 0 0 0  5 3125 25z 2 5 1 4 3125 625 = + + = 625( + ) = 625( ) = 6 2 0 6 2 6 2 3 2500 = 3

14.5

Change of variable

Sometimes is is easier to calculate an integral by changing the variables it is integrated with repect to. Suppose that an integral in terms of x and y is changed to an integral in terms of u and v where x = φ(u, v), y = ψ(u, v) then the integral over the original region R in the xy plane changes to one over region R0 in the uv plane thus Z Z Z Z ∂(x, y) dudv f (x, y)dxdy = f (φ(u, v), ψ(u, v)) ∂u, v 0 R R where

∂x ∂x ∂(x, y) ∂u ∂v = =J ∂y ∂y ∂u, v ∂u ∂v is a Jacobian (see 13.6) related to the transformation.

14.5 Change of variable

14.5.1

235

Polar coordinates

If we wish to transform from x, y to r, θ we can use the transformation x = r cos θ; y = r sin θ; r ≥ 0; 0 ≤ θ < 2π and in this case the Jacobian is ∂x ∂x ∂r ∂θ cos θ −r sin θ = ∂y ∂y sin θ r cos θ ∂r ∂θ

= r(cos2 θ + sin2 theta) = r

This idea can be extended to triple integrals.

14.5.2

Example

Evaluate

Z Z (1 −

I=

p x2 + y 2 dxdy

R

where R is the region bounded by the circle x2 + y 2 = 1. Solution We note that the function is more or less 1 − r if r is the distance of the point from the origin, and the region is polar in nature. It follows that a transformation is useful here. Transforming to polars Z 2π Z 1 I= (1 − r)rdrdθ 0

0

where the region of the circle has been represented by letting r run from 0 to 1 and letting θ run from 0 to 2π, we have changed the integral in terms of the new variable and added an r from the Jacobian for this transformation. We now proceed Z = 0

2π



r2 r3 − 2 3

1

Z dθ =

0

0

2π

dθ = f racπ3 6

14.5 Change of variable

14.5.3

236

Cylindrical Polar Coordinates

In three dimensions two commonly used systems of coordinated are used. Cylindrical polar coordinates are simply two dimensions coordinates where the height is left in cartesian form. We use the transformation x = ρ cos φ, y = ρ sin φ, z = z where ρ ≥ 0, 0 ≤ φ < 2π and therefore

cos φ −ρ sin φ 0 ∂(x, y, z) J= = sin φ ρ cos φ 0 ∂(ρ, φ, z) 0 0 1 = ρ(cos2 φ + sin2 φ) = ρ

so note that in this coordination system, rho is the distance from the z axis, not the origin as such.

14.5.4

Spherical Polar Coordinates

To use three dimensional coordinates based on the distance from the origin and not the z-axis we need to define the three coordinates r, the distance from the origin, φ the angle from the projection of the point on the xy plane from the x axis, and θ the angle between the z-axis and the line joing the point to the origin. This means the transformation equations are x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ r ≥ 0, 0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π and thus sin θ cos φ ∂(x, y, z) = sin θ sin φ J= ∂(r, θ, φ) cos θ

r cos θ cos φ −r sin θ sin φ r cos θ sin φ r sin θ cos φ −r sin θ 0

= r2 {cos2 θ sin θ(cos2 φ + sin2 φ) + sin2 θ(cos2 φ + sin2 φ)} = r2 sin θ{cos2 θ + sin2 θ} = r2 sin θ



Chapter 15 Fourier Series Fourier1 series are a powerful tool. On one hand they simply allow a periodic function to be expressed as an infinite series of simple functions, usually trigonometric or exponential, but this also allows great insight into a function be splitting into component “frequencies”.

15.1

Periodic functions

A periodic function is one that repeats its values at regular intervals. So that if the repeat occurs ever T units on the x-axis. f (x) = f (x + T ) = f (x + 2T ) + · · · + f (x + nT ) The contant value T is known as the period of oscillation.

15.1.1

Example

Of course the classic examples of periodic functions are the graphs of sin x and cos x. Consider y = A sin(ωt + ) 1

Baron Jean Baptiste Fourier (1768-1830) was a French mathematician who narrowly

avoided the guillotine during the French Revolution. Fourier contributed greatly to the use of differential equations to solve problems in physics.

15.1 Periodic functions

238

where t is the time in seconds. This is the classic representation of a wave with amplitude A. The constant ω is known as the angular frequency and the constant  is known as the phase angle. Note that ω = 2πf where f is the traditional frequency measured in Hertz. As for all oscillations with a frequency f the time period of oscillation is given by T =

1 . f

The frequency f is known as the fundamental frequency of the signal since this governs the overall timing of the repetitive nature of the function.

15.1.2

Example

Find the period T and the fundamental frequency f for (a) f (t) = sin t (b) g(t) = 3.4 cos(4.5t + 1.2) where t is the time in seconds. Solution (a) In this case we know that sin t repeats every 2π radians. Therefore T = 2πs ≈ 6.283s ⇒ f =

1 = 0.159Hz 2π

(b) Everything here is a distraction apart from the constant 4.5 which precedes the t. This is the angular frequency ω and so ω = 2πf ⇒ f = and therefore T =

ω 4.5 = ≈ 0.716Hz 2π 2π

1 ≈ 1.396s. f

15.2 Sets of functions

15.2

239

Sets of functions

Fourier series require expansions in terms of certain collections, or sets, of functions.

15.2.1

Orthogonal functions

A sequence (φk (x)) of functions is said to be orthogonal on [a, b] if  Z b = 0 if m 6= n φm (x)φn (x)dx 6= 0 if m = n a

15.2.2

Orthonormal functions

A sequence (φk (x)) of functions is said to be orthonormal on [a, b] if  Z b 0 if m 6= n φm (x)φn (x)dx = 1 if m = n a

15.2.3

Norm of a function

The quantity s Z

b

(φk (x))2 dx

||φk || = a

is known as the norm of the function φk (x). If we divide each function in the collection of an orthogonal system by its norm we obtain an orthonormal system.

15.3

Fourier concepts

We now examine the most important concepts at the heart of the Fourier Series.

15.3.1

Fourier coefficents

If f is a function defined for a ≤ x ≤ b such that the integral Z b 1 f (x)φk (x)dx ck = kφk k2 a

15.4 Important functions

240

exists. Then the sequence (ck ) is called the Fourier co-efficients of f with respect to the orthogonal system (φk (x)).

15.3.2

Fourier series

Using the Fourier coefficients for f , the infinite series ∞ X

ck φk (x)dx

k=0

is called the Fourier series of f with respect to the orthogonal system.

15.3.3

Convergence

We already know that not all infinite series converge to a value. If we suppose that • f (x) is defined for a ≤ x ≤ b; • f (x + nT ) = f (x) for all integers n; • f (x) is integrable on −π ≤ x ≤ π. Then the Fourier series with respect to the trigonometric system exists and if f (x) • has a finite number of maxima and mimima; • has a finite number of discontinuities for a ≤ x ≤ b then the Fourier series converges to f (x) where f (x) is continuous, and to the midpoint of the two pieces on eitherside of x otherwise. These are known as the Dirichlet conditions.

15.4

Important functions

The most important sets of functions for Fourier Series are the trigonemetric and exponential functions.

15.4 Important functions

15.4.1

241

Trigonometric system

The system of trigonometric functions 1, cos x, sin x, cos 2x, sin 2x, cos 3x, sin 3x, . . . is orthogonal in the range −π ≤ x ≤ π. Furthermore √ √ k1k = 2π; kcos nxk = ksin nxk = π if n is 1, 2, 3, . . . . Proof 2

Z

π

k1k =

12 dx = 2π;

−π

Z

π

cos2 nxdx −π  π Z π 1 1 sin 2nx = (1 + cos 2nx)dx = x+ =π 2 −π 2 2n −π Z π 2 2 sin2 nxdx k sin nxk = π  π Z π 1 1 sin 2nx = (1 − cos 2nx)dx = x− =π 2 −π 2 2n −π Now we must show that the inner product of any function with any different function evaluates to zero. It is left as a trivial exercise to show that the products of 1 with sin nx and 1 with cos nx produce a zero integral. 2

2

k cos nxk =

π

Z 1 −π π cos(m − n)x + cos(m + n)xdx cos mx cos nxdx = 2 −π  π 1 sin(m − n)x sin(m + n)x + = = 0 for (m 6= n). 2 m−n m+n −π Z π Z 1 π sin mx sin nxdx = cos(m − n)x − cos(m + n)xdx 2 −π −π  π 1 sin(m − n)x sin(m + n)x = − = 0 for (m 6= n). 2 m−n m+n −π Z π Z 1 π sin mx cos nxdx = sin(m + n)x + sin(m − n)xdx 2 −π −π  π 1 cos(m + n)x cos(m − n)x =− + = 0. 2 m+n m−n π

Z

15.5 Trigonometric expansions

15.4.2

242

Exponential system

The system of exponential functions . . . e−3jx , e−2jx , e−jx , 1, ejx , e2jx , e3jx , . . . is orthogonal in the range −π ≤ x ≤ π.

15.5

Trigonometric expansions

Using the trigonometric system of functions in the range −π ≤ x ≤ π we can see that the Fourier co-efficients of f , if they exist, (ck ) can be more easily written by splitting them into the coefficeints for the function 1, a0 , the coefficents for cos nx, an and for sin nx, bn whence a0 2 where Z 1 a0 = πf (x)dx; π −π and for values of n = 1, 2, 3, . . . Z 1 π f (x) cos nxdx an = π −π Z 1 π f (x) sin nxdx bn = π −π and the Fourier series of f is given by ∞ a0 X (an cos nx + bn sin nx) + 2 n=1

15.5.1

Even functions

If f (x) is even on the range −π ≤ x ≤ π then it is easy to show Z 2 π an = f (x) cos nxdx, n ≥ 0, π 0

2

that

bn = 0, n > 0. Note these are not different expansions, but what the previously stated expansions collapse to in this very special case. 2

This follows from the discussion in 7.4 before, noting the nature of each product being

integrated.

15.6 Harmonics

15.5.2

243

Odd functions

If f (x) is odd on the range −π ≤ x ≤ π then it is easy to show that an = 0, n ≥ 0 2 bn = π

15.5.3

π

Z

f (x) sin nxdx, n > 0 0

Other Ranges

Note that in much of the theory above, we have not assumed that all expansions run between −π and π. In the trigonometric system it is relatively easy to adapt the process for periodic functions that repeat from −L to L. It can be shown quite easily that ∞

nπx nπx  a0 X  an cos + + bn sin f (x) = 2 L L n=1 where 1 a0 = L

15.6

Z

L

f (x)dx −L

L

1 an = L

Z

1 bn = L

Z

f (x) cos

nπx dx for n > 0 L

f (x) sin

nπx dx for n > 0 L

−L L

−L

Harmonics

When the Fourier series of a function f (x) is produced using the trigonometric system, it is clear that the function is made up of signals with specific frequencies. The functions an cos nx + bn sin nx could be combined together into one signal of the form cn sin(nx + n ) using the technique shown in 4.9.3. Therefore this whole term represents the component of the function or signal f (x) with angular frequency n.

15.6 Harmonics

244

This is called the nth harmonic. The 1st harmonic is therefore given by a1 cos x + b1 sin x = c1 sin(x + 1 ) is known as the first harmonic or fundamental harmonic. The term a0 2 can be looked at as a form of static background noise that does not rely on frequency at all.

15.6.1

Odd and Even Harmonics

Sometimes symmetry in the original function will tell us that some harmonics will not be present in the final result. Even Harmonics Only If f (x) = f (x + π) then there will be only even harmonics. Odd Harmonics Only If f (x) = −f (x + π) there there will be only odd harmonics. We can now surmise a great deal about the terms we expect to find in the series before expansion; there are shown in table 15.1. f (x) = f (−x) (Cosine Only) f (x) = −f (x) (Sine only) f (x) = f (x + π)

even cosine only

even sine only

f (x) = −f (x + π) odd cosine only

odd sine only

Table 15.1: Symmetry in Fourier Series

15.6.2

Trigonometric system

For the trigonometric system, as described above, it is often useful to combine the signals as described in 4.9.3. In general however this can be done to yield the formulae p cn = a2n + b2n

15.7 Examples

245 −1

n = tan



an bn



for integers n > 0.

15.6.3

Exponential system

The exponential system is less intuitive than the trigonometric system, but the constants cn are often simpler to determine.

15.6.4

Percentage harmonic

With the constants cn defined as above we define the percentage of the nth harmonic to be cn × 100. c1 That is, the percentage that the amplitude of the nth harmonic is of the amplitude of the fundamental harmonic.

15.7

Examples

15.7.1

Example

If f (x) = x for x such that −π < x ≤ π and f (x + 2π) = f (x) for all real x, evaluate the Fourier series of f (x). Solution Since f (x) is an odd function we have that an = 0 and  π Z Z π −2x cos nx 2 2 π x sin nxdx = + cos nxdx bn = π 0 nπ nπ 0 0 −2 cos nπ 2(−1)n+1 = . n n Thus the Fourier series of f (x) is ⇒ bn =

2(sin x −

sin 2x sin 3x sin 4x + − + · · · ). 2 3 4

15.8 Exponential Series

15.8

246

Exponential Series

We have noted above the exponential set of functions forms an orthogonal set of functions and hence it will be possible to produce Fourier series in this set of functions. We can approach this idea from another direction. We know from Euler’s identity (5.6.3) that ejθ = cosθ + jsinθ and therefore ejnx = cos nx + j sin nx while e−jnx = cos(−nx) + j sin(−nx) ⇒ e−jnx = cos nx − j sin nx due to the even and odd nature of the cosine and sine functions respectively. We can combine these two equations to find that cos nx =

ejnx − e−jnx 2

sin nx =

ejnx − e−jnx 2j

Now consider the formula for the Fourier series of f (x) which is ∞

a0 X (an cos nx + bn sin nx) f (x) = + 2 n=1 and we shall insert these terms  ∞ 
bn jnx
a0 X an jnx −jnx −jnx = + e +e e −e + 2 2 2j n=1 to simply things, we multiply top and bottom of the right most term by j which gives us j on the top line and means we are dividing by j 2 = −1. We absorb that into the brackets, inverting the subtraction to give  ∞ 
bn j −jnx
a0 X an jnx −jnx jnx + e +e + e −e = 2 2 2 n=1 ∞

a0 X + = 2 n=1



an − b n j 2



jnx

e

 +

an + b n j 2

 e

−jnx



15.8 Exponential Series

247

and this can be written more simply as =

∞ X

cn ejnx .

−∞

Note carefully here that n now runs from −∞ to ∞. The relationship of the contants cn are  1  2 (an − jbn ) n > 0 1 a0 n=0 cn =  21 (a + jbn ) n < 0 2 n

Chapter 16 Laplace transforms Laplace1 transforms are (among other things) a way of transforming a differential equation into an algebraic equation. This equation is then rearranged and we attempt to reverse the transform. This last part is usually the hardest unfortunately.

16.1

Definition

Suppose that f (t) is some function of t, then the Laplace transform of f (t), denoted L{f (t)} is given by Z ∞ F (s) = L {f (t)} = e−st f (t)dt 0

16.1.1

Example

Find L {eat }. Solution  L eat =

Z

at −st

e e 0

1

∞

Z

∞

dt =

e(a−s)t dt

0

These are named after Pierre Simon de Laplace (1749-1827), a brilliant French math-

ematician sometimes known as the “Newton of France”.

16.1 Definition

249 e(a−s)t = a−s 

∞ 0

which will be defined (convergent) when s > a. In which case 1 1 = a−s s−a

 L eat = 0 −

16.1.2

Example

Find L {1}. Solution ∞

Z

−st

L {a} =

1×e 0

Thus L {1} = 0 −

16.1.3

e−st dt = −s 

∞ 0

1 1 = −s s

Example

Find L {tn } where n is a positive integer. Solution n

Z

L {t } =

∞

tn e−st dt

0

Now, recall the integration by parts formula Z Z du dv u dx = uv − v dx dx dx and allow u = tn ⇒

du dv e−st = ntn−1 ; ⇒v=− dx dx s

Thus we obtain   Z ∞ −st ∞ est ne L {t } = −t − − ntn−1 dtdt s 0 s 0 n

The already integrated part of the expression evaluates to zero, when ∞ is used as a limit the exponential term “crushes” the polynomial term to zero

16.1 Definition

250

and when zero is used as a limit the polynomial term is also zero. Thus, tidying up the integral that remains, gives Z n ∞ n−1 −st n  n L {t } = t e dt = L tn−1 s 0 s so that we obtain a reduction formula. Now, we know L {t0 } = L {1} = 1s . Therefore, 1 11 = 2 L {t} = ss s  2 2 1 2 L t = = 3 2 ss s and in general n! L {tn } = n+1 s

16.1.4

Inverse Transform

We define the inverse Laplace transform, denoted L−1 {F (s)} in the “obvious” way. L {f (t)} = F (s) ⇔ L−1 {F (s)} = f (t)

16.1.5

Elementary properties

As the Laplace transform is an integral, we can employ automatically those laws we take for granted in integration. In particular L {f (t) ± g(t)} = L {f (t)} ± L {g(t)} and L {kf (t)} = kL {f (t)} where k is a constant and f (t) and g(t) are functions of t as indicated. In other words, we may split the transform (and for that matter the inverse transform) over addition and or subtraction, and we can take constants in and out of the transform.

16.1.6

Example

Find L {sin ωt} and L {cos ωt}.

16.2 Important Transforms

251

Solution We use Euler’s identity ejωt = cos ωt + j sin ωt Now  L ejωt =

1 s + jω s ω × = 2 +j 2 2 s − jω s + jω s +ω s + ω2

So  L ejωt = L {cos ωt + j sin ωt} and equating real and imaginary components we obtain L {cos ωt} = and L {sin ωt} =

16.2

s2

s + ω2

ω s2 + ω 2

Important Transforms

We present a table (see table 16.1) of important transforms, without proof, as this would require the precise definition of indefinite integrals. In most cases a condition must be satisfied for the integral to be convergent (and thus for the transform to exist).

16.2.1

First shifting property

Let f (t) be a function of t with Laplace transform F (s), which exists for s > b, then if a is a real number  L eat f (t) = F (s − a) for s > a + b. Proof Clearly, if Z 0

∞

e−st f (t)dt = F (s)

16.2 Important Transforms

252

f (t)

L {f (t)}

Condition

eat

1 s−a

s>a

k

k s

s>0

tn

sin at

cos at

sinh at

cosh at

n! sn+1

s>0

a + a2

s>0

s s 2 + a2

s>0

a − a2

s > |a|

s s 2 − a2

s > |a|

s2

s2

Table 16.1: Common Laplace transforms

16.2 Important Transforms

253

f (t)

L {f (t)}

Condition

eat tn

n! (s − a)n+1

s>a

eat sin bt

b (s − a)2 + b2

s>a

eat cos bt

s−a (s − a)2 + b2

s>a

eat sinh at

b (s − a)2 − b2

s > a + |b|

eat cosh at

s−a (s − a)2 − b2

s > a + |b|

Table 16.2: Further Laplace transforms then

Z

∞

F (s − a) =

e−(s−a)t f (t)dt

0

We note that 

at



Z

L e f (t) =

∞ at

e f (t)e 0

−st

Z

∞

dt =

e−(s−a)t f (t)dt

0

where the right most term is clearly F (s − a) as shown above.

16.2.2

Further Laplace transforms

The first shifting property of Laplace transforms gives rise to the following other transforms, shown in table 16.2.

16.3 Transforming derivatives

16.3

Transforming derivatives

The most important property of the Laplace transform is how derivates of f (t) transform.

16.3.1

First derivative

Let |f (t)| ≤ M eat when t ≥ T for non-negative constants M , a and T . Then L {f 0 (t)} = sL {f (t)} − f (0)

16.3.2

Second derivative

For the second derivative, we obtain L {f 00 (t)} = s2 L {f (t)} − sf (0) − f 0 (0)

16.3.3

Higher derivatives

We can continue (using proof by induction) to obtain  L f (n) (t) = sn L {f (t)} − sn−1 f (0) − sn−2 f 0 (0) − · · · − f (n−1) (0) The most important thing to note is that all these derivates transform to simple algebraic expressions in terms of the Laplace transform of f (t) itself. In other words, the differentiation is undone. In general, transforming an equation is relatively simple. It must then be rearranged for the transform of f (t) and then, and this is the hard part, we struggle to recognise forms for inverse transforms.

16.4

Transforming integrals

If L {f (t)} = F (s) then Z L 0

t

 1 f (t)dt = F (s). s

254

16.5 Differential Equations

255

Proof Let

t

Z

f (t)dt = g(t) 0

or in other words

d g(t) dt which under Laplace transformation yields f (t) =

L {f (t)} = sg − g(0) Note that Z g(0) =

0

f (x)dx = 0 0

where x has been used to stress the dummy variable. 1 1 g = L {f (t)} ⇒ L {f (t)} = F (s) s s

16.5

Differential Equations

We are now in a position to solve some differential equations using Laplace transforms.

16.5.1

Example

In nuclear decay, the rate of decays is directly proportional to the number of nucleii to decay. Therefore the number of nucleii remaining N (t) follows the following differential equation. dN = −λN dt Assuming that N (0) = N0 when t = 0 solve the differential equation. Solution There are much easier ways of solving this equation than resorting to Laplace transforms as the reader is no doubt aware, but we use it as a very simple example of the principles involved.

16.5 Differential Equations

256

First we transform the whole equation sL {N } − N (0) = −λL {N } Now we have an algebraic equation, into which we plug in our initial known conditions. ⇒ sL {N } − N0 = −λL {N } We now rearrange this to make L {N } the subject of the equation. ⇒ sL {N } + λL {N } = N0 ⇒ (s + λ)L {N } = N0 N0 s+λ In theory now we need only find the expression that transforms to the RHS and we have solved our differential equation. In practice this can be quite tricky and we often have to manipulate the expression a great deal first. In this very simple case, very little has to be done. ⇒ L {N } =

⇒ L {N } = N0 × Now the expression

1 s+λ

1 s+λ

is L {eat } where we take a = −λ and so.

  ⇒ L {N } = N0 × L e−λt ⇒ L {N } = L N0 e−λt Therefore, removing the transform on both sides we have N = N0 e−λt

16.5.2

Example

Solve the differential equation 2 given that x = 0 and

dx dt

dx d2 x + 5 − 3x = t − 4 2 dt dt

= 2 when t = 0.

16.5 Differential Equations

257

Solution First of all we transform the equation on both sides 2(s2 x − sx(0) − x0 (0)) + 5(sx − x(0)) − 3x =

1 4 − . 2 s s

Now we insert the initial conditions now, which are x(0) = 0 and x0 (0) = 2 to obtain 1 4 2s2 x − 4 + 5sx − 3x = 2 − . s s Next, we rearrange to make the Laplace transform of x(t), denoted by x for short, to be the subject of the equation. (2s2 + 5s − 3)x = ⇒x=

1 4 1 − 4s + 4s2 − + 4 = s2 s s2 4s2 − 4s + 1 . s2 (2s2 + 5s − 3)

Finally, we have the most difficult part, we have already transformed the whole differential equation into an algebraic equation and solved it, now we have to invert the transform. Factorize first, to simplify as much as possible. x=

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

and now use partial fractions 2s − 1 A B C = + 2+ + 3) s s s+3

s2 (s

and multiply both sides by the denominator of the L.H.S. 2s − 1 = As(s + 3) + B(s + 3) + Cs2 . This is an identity, or in other words it is true for all values of s, so certainly it is true for specific values. s = −3 ⇒ −7 = 9C ⇒ C = − s = 0 ⇒ −1 = 3B ⇒ B = −

7 9

1 3

16.5 Differential Equations

258

and comparing coefficients of s2 we obtain 7 0=A+C ⇒A= . 9 Thus x=

71 1 1 7 1 − 2− 9s 3s 9s+3

and therefore x=

16.5.3

7 1 7 − t − e−3t . 9 3 9

Example

Solve the differential equation d4 y − 81y = 0 dt4 given that y = 1,

dy d2 y d3 y = 2 = 3 =0 dt dt dt

when t = 0. Solution Transforming the equation yields s4 y − s3 y(0) − s2 y 0 (0) − sy 00 (0) − y 000 (0) − 81y = 0 and inserting initial conditions simplifies this to s4 y − s3 − 81y = 0 which rearranges to (s4 − 81)y = s3 ⇒y=

s3 s3 = . s4 − 81 (s2 − 9)(s2 + 9)

Another simplification is possible here. y=

s3 A B Cs + D = + + 2 2 (s − 3)(s + 3)(s + 9) s−3 s+3 s +9

16.5 Differential Equations

259

⇒ s3 = A(s + 3)(s2 + 9) + B(s − 3)(s2 + 9) + (Cs + D)(s − 3)(s + 3). We begin inserting key values of s once again. 1 s = −3 ⇒ −27 = B(−6)(18) ⇒ B = ; 4 1 s = 3 ⇒ 27 = A(6)(18) ⇒ A = ; 4 s = 0 ⇒ 0 = 27A − 27B + D(−9) ⇒ D = 0; 1 s = 1 ⇒ 1 = 40A − 20B + C(−8) ⇒ C = . 2 Therefore

1 1 1 1 1 s + + 2 4s−3 4s+3 2s +9 and we can easily now employ the inverse transform to obtain y=

1 1 1 y = e3t + e−3t + cos 9t 4 4 2 Alternative Solution Suppose that we missed the second factorization of the bottom line, then we would have proceeded as follows. s3 As + B Cs + D = 2 + 2 2 2 (s − 9)(s + 9) s −9 s +9 and multiply by denominator of the L.H.S. s3 = (As + B)(s2 + 9) + (Cs + D)(s2 − 9), s = 3 ⇒ 27 = 54A + 18B ⇒ 3 = 6A + 2B s = −3 ⇒ −27 = −54A + 18B ⇒ −3 = −6A + 2B which solved together yield A =

1 2

and B = 0.

s = 3j ⇒ −27j = (3jC + D)(−18) which gives D = 0 and C = 21 . Now we have 1 1 1 1 y= + 2 s2 − 9 2 s2 + 9

16.5 Differential Equations

260

and using our table of transforms we see that 1 y = (cosh 3t + cos 3t) 2 This looks like a different solution, but in fact, using the definition of the cosh function we see that   1 1 1 3t 1 −3t + cos 3t y= e + e 2 2 2 2 which can be seen to be equivalent.

16.5.4

Example

Solve the differential equation d2 x = −ω 2 x dt2 using Laplace transforms2 assuming that x(0) = A and x0 (0) = 0. Solution Taking the transforms both sides gives s2 x − sx(0) − x0 (0) = −ω 2 x and fill in initial conditions to give ⇒ s2 x − sA = −ω 2 x and rearrange for x ⇒ (s2 + ω 2 )x = As ⇒ x = A

s2

s + ω2

from which it follows simply that x = A cos ωt 2

This is the equation for Simple Harmonic Motion and is a very important differential

equation. It is not necessary to use Laplace transforms to solve it and this method is used as an example.

16.5 Differential Equations

16.5.5

261

Exercise

Using the previous differential equation d2 x = −ω 2 x 2 dt but the initial conditions x(0) = 0, and x0 (0) = v, show that x=

16.5.6

ω sin ωt v

Example

Consider the network shown in figure 16.1.

Figure 16.1: An L and R curcuit. Find i given that i = 0 when t = 0. Solution By Kirchoff’s laws di =E dt and taking Laplace transforms on both sides gives Ri + L

Ri + L(si − i(0)) = inserting initial conditions gives (R + Ls)i =

E s

E s

16.5 Differential Equations and so i=

262

E A B = + . s(Ls + R) s Ls + R

Therefore E = A(Ls + R) + Bs which is true for all values of s, so in particular s = 0 ⇒ E = AR ⇒ A =

E R

while comparing coefficients of s gives. 0 = AL + B ⇒ B = −

EL . R

Inserting these values for A and B gives i=

E 1 EL 1 E1 E 1 − = − Rs R Ls + R R s R s + R/L

So finally we have to invert the transform to obtain i= or i=

16.5.7

E E −Rt − e L R R

R E (1 − exp(− t)) R L

Example

Consider the circuit shown in figure 16.2.

Figure 16.2: An L, C, and R curcuit. Find i given that R = 250Ω, C = 10−4 F , E = 10V and L = 1H and that the inital current is zero.

16.6 Other theorems

263

Solution From Kirchoff’s laws we have di 1 Ri + L + dt C which when transformed gives

Z

t

idt = 10 0

Ri + L(si − i(0)) +

1 10 i= Cs s

then we fill in our conditions to give i(s + 250 +

104 10 )= s s

which with some work yields i=

10 (s + 50)(s + 200)

i=


1 −50t e − e−200t 15

finally leaving us with

16.6

Other theorems

Here are some more theorems about Laplace transforms for which no proof is given. In each case we assume that f (t) is a function and that F (s) = L {f (t)}

16.6.1

Change of Scale

Given a constant a, then 1 s L {f (at)} = F ( ). a a

16.6.2

Derivative of the transform

Given a constant n, then L {tf (t)} = −

d {F (s)} ds

or more generally L {tn f (t)} = (−1)n

dn {F (s)} dsn

16.6 Other theorems

16.6.3

264

Convolution Theorem

Given that g(t) is such that G(t) = L {g(t)} then L

−1

Z

t

f (r)g(t − r)dr.

{F (s)G(s)} = 0

This integral on the right is often denoted by g ∗ f or f ∗ g.

16.6.4

Example

Given that L {cos t} =

s2

s +1

find L {cos 3t} . Solution We already can work this out straight away from our table 16.1 but we do this as an exercise. From 16.6.1 we see that L {cos 3t} = =

16.6.5

s/3 1 s 1 = 2 2 3 (s/3) + 1 9 s /9 + 1

s 1 9s = 2 2 9s +9 s +9

Example

Given that  L e4t =

1 s−4

find  L te−4t

16.7 Heaviside unit step function

265

Solution Again, we can work this out directly from our table 16.2 but this is an exercise. We can see from 16.6.2 that we can write    −4t 1 1 d = (−1)(−1)(s − 4)2 = L te =− ds s − 4 (s − 4)2

16.6.6

Example

Find L

−1



1 (s − 2)(s − 3)



Solution Once more, this problem would normally be tackled using partial fractions, but we use it as a very simple application of the convolution theorem (see 16.6.3). Let F (s) =

1 1 ⇒ f (t) = e2t ; G(s) = ⇒ g(t) = e3t s−2 s−3

Then L

−1



1 (s − 2)(s − 3)



Z =

t 2r 3(t−r)

e e

Z dr =

0

t

e2r e3t e−3r dr

0

Note that t is a constant with respect to this integration, and so we may bring a term outside. Z t  t 3t =e e−r dr = e3t −e−r 0 0

giving a final answer of = −e2t + e3t .

16.7

Heaviside unit step function

The Heaviside unit step function 3 is defined by 3

Oliver Heaviside (1850-1925) was a British physicist with no university education.

He produced much of the theoretical underpinning of cable telegraphy.

16.7 Heaviside unit step function

 u(t) =

0 1

266

for t < 0 for t ≥ 0

and is shown in figure 16.3.

Figure 16.3: The unit step function u(t). At times we may wish to cause the unit step to occur earlier or later than t = 0, and we can do this quite easily by examining the function u(t − c) where c is some constant. This function takes the form  0 for t < c u(t − c) = 1 for t ≥ c and is shown in figure 16.4.

Figure 16.4: The displaced unit step function u(t − c). The effect of this function is to allow us to switch on and off other functions and thus help us make very complicated waveforms. We do this by multiplying the unit function against the function we wish to control, say f (t). Thus

16.7 Heaviside unit step function

 f (t) × u(t − c) =

267

0 for t < c f (t) for t ≥ c

and we have successfully “switched off” the function f (t) for times before t = c. Frequently we shall want much more fine control than this however, requiring that we can again switch off the function f (t) after some time interval. This can easily be done by combining variations of u(t − c).

Figure 16.5: Building functions that are on and off when we please. For example, consider figure 16.5 on page 267. Here we have shown the graphs of u(t − 2) and u(t − 4) and clearly, by subtracting the second graph

16.7 Heaviside unit step function

268

from the first, we obtain the third graph, which shows that we can “switch on” between t = 2 and t = 4 or any other values we please.

16.7.1

Laplace transform of u(t − c) e−cs s

L {u(t − c)} = Proof By definition Z

∞

e−st u(t − c)dt

L {u(t − c)} = 0

but we know that e

−st

 × u(t − c) =

0 for t < c e−st for t ≥ c

and thus

Z

∞

e−st dt.

L {u(t − c)} = c

Note in particular the change of limits of this integral. Therefore  −sc   −st ∞ −e −e = {0} − L {u(t − c)} = s s c and so esc s and allowing c to be zero, note in particular that L {u(t − c)} =

L {u(t)} =

16.7.2

1 s

Example

Find the function f (t), described by step functions, and the transform F (s) for the waveform shown in figure 16.6.

16.7 Heaviside unit step function

Figure 16.6: A positive waveform built from steps. Solution First of all we begin to compose the waveform. f (t) = 4{u(t) − u(t − 2)} + 2{u(t − 4) − u(t − 6)} + {u(t − 7) − u(t − 8)} In each case we multiply a combination of step functions which switch the function on and off in the appropriate places by the function we want in that region, here that is just a constant each time. It is easier to expand this before the transform is applied f (t) = 4u(t) − 4u(t − 2) + 2u(t − 4) − 2u(t − 6) + u(t − 7) − u(t − 8). This may look a little intimidating at first, but in fact it is easy to transform this when we recall that • We can split the transform over addition and subtraction; • We can take constants outside the transform; • Transforming u(t − c) is straightforward. So we obtain 1 e2s e4s e6s e7s e8s L {f (t)} = F (s) = 4 − 4 +2 −2 + − s s s s s s =

1 4 − 4e2s + 2e4s − 2e6s + e7s − e8s s

269

16.7 Heaviside unit step function

16.7.3

Example

Find the function f (t), described by step functions, and the transform F (s) for the waveform shown in figure 16.7.

Figure 16.7: A waveform built from steps.

Solution From the graph we see that f (t) = −2{u(t − 1) − u(t − 3)} + 1{u(t − 3) − u(t − 5)} which when expanded yields = −2u(t−1)+2u(t−3)+u(t−3)−u(t−5) = −2u(t−1)+3u(t−3)−u(t−5). Now we apply the transform and obtain e3s e5s es +3 − s s s 1 = −2es + 3e3s − e5s s

L {f (t)} = F (s) = −2

16.7.4

Delayed functions

If L {f (t)} = F (s), then, the Laplace transform of the delayed function f (t − a)u(t − a) is given by L {f (t − a)u(t − a)} = e−as F (s)

270

16.7 Heaviside unit step function

271

Proof Clearly Z L {f (t − a)u(t − a)} =

∞

e−st f (t − a)u(t − a)dt

0

but note that no area can occur before t = a (due to the switching off of the function with u(t − a). Therefore this integral becomes Z ∞ e−st f (t − a)dt. = a

We now make a substitution, let T = t − a. Then clearly dT = 1 ⇒ dT = dt dt under this substitution the integral becomes Z ∞ e−s(T +a) f (T )dT. 0

Note carefully the change of limits, and observe that s is a constant with respect to this integral. Z ∞ −as =e e−sT f (T )dT = e−as F (s) 0

which proves the result.

16.7.5

Example

Find the function f (t), described by step functions, and the transform F (s) for the waveform shown in figure 16.8. Solution Some simple examination shows that this function could be defined as  when 0 ≤ t < 2  t 2 when 2 ≤ t < 5 f (t) =  12 − 2t when 5 ≤ t < 6 and which is zero at all other times. Converting this into step function form, we obtain f (t) = t{u(t)−u(t−2)}+2{u(t−2)−u(t−5)}+(12−2t){u(t−5)−u(t−6)}.

16.8 The Dirac Delta

272

Figure 16.8: A waveform built from delayed linear functions. Actually there is more than one way of tackling the problem from this point, this is one method.4 We expand f (t) thinking ahead to what we can tackle.

f (t) = tu(t)−tu(t−2)+2u(t−2)−2u(t−5)+12u(t−5)−2tu(t−5)+2(t−6)u(t−6) where the last section has been expanded to make it a delayed function. If we proceed to do this in the other sections where possible we obtain. f (t) = tu(t) − (t − 2)u(t − 2) − 2(t − 5)u(t − 5) + (t − 6)u(t − 6) So now we appeal to the result for delayed functions, and noting that the function being delayed is f (t) = t in each case, to obtain F (s) = =

16.8

e−2s e−5s e−6s 1 − − 2 + 2 s2 s2 s2 s

1  1 − e−2s − 2e−5s + e−6s 2 s

The Dirac Delta

The impulse function or Dirac delta is defined by 4

Another way in this case would be to expand with no forethought wherupon we would

have a list of step functions, and step functions multiplied by t. We could then use the derivative of the transform (see 16.6.2) to crack the problem.

16.8 The Dirac Delta

273

δ(t) =

  0

for t < 0 for 0 ≤ t <   0 for t ≥  1 

where  → 0 and Z

∞

δ(t)dt = 1 −∞

The Laplace transform of δ(t) is given by L {δ(t)} = 1 Proof  L {δ(t)} = L

 1 1 [u(t) − u(t − )] = (1 − e−se )  s

Now, recall that e−x = 1 − x +

x2 x3 − + ··· 2! 3!

so that 1 s2 2 s3 3 (1 − 1 + s − + − ···) s 2! 3! s2 2 s3 3 = 1 − s + − + ··· 2! 3! and now, allowing  → 0 we obtain L {δ(t)} =

L {δ(t)} = 1

16.8.1

Delayed impulse

The Laplace transform of the delayed impulse funtion, δ(t − a), is given by L {δ(t − a)} = e−as Proof This is a direct consequence of the first shifting property ( 16.2.1) and the transform of the normal dirac delta ( 16.8).

16.9 Transfer Functions

16.8.2

274

Example

Find the Laplace transform of the wave train shown in figure 16.9.

Figure 16.9: An impulse train built from Dirac deltas.

Solution We could describe the train as a function as follows f (t) = 3δ(t − 1) − 2δ(t − 3) + δ(t − 4) − 3δ(t − 5) +2δ(t − 6) − δ(t − 8) and thus the transform will be L {f (t)} = F (s) = 3e−s − 2e−3s + e−4s − 3e−5s + 2e−6s − e−8s

16.9

Transfer Functions

Consider once again the circuit in figure 16.2, with a potentially varying EMF e. If we take Laplace transforms on both sides and adopt the shorthands i = L {i(t)} , e = L {e(t)} then we obtain Lsi + Ri +

1 i = e. Cs

16.9 Transfer Functions

275

Thus

LCs2 + RCs + 1 =e Cs Cs i . ⇒ = 2 e LCs + RCs + 1 This is called the transfer function for the circuit. In general the transfer function of a system, is given by. i

FT (s) =

L {input} . L {output}

The transfer function is determined by the system, and once found is unchanged by varying inputs and their corresponding outputs. The analysis of the transfer function can reveal information about the stability of the system. The system can be considered stable if the output remains bounded for all values of t, even as t → ∞. So terms in the output of the form et , t, t2 cos 3t are all unbounded. On the other hand, terms of the form e−t and e−2t cost show stability. We can determine the presence of such terms by analysing the poles 5 of the transfer function. This really comes down to examining the demonimator and finding what values of s cause it to become zero. We can then analyse the stability of the system by plotting the poles on an Argand diagram, and using the following simple rules. • If all the poles occur to the left of the imaginary axis then the system is stable; • If any pole occurs to the right of the imaginary axis then the system is unstable; • If a pole occurs on the imaginary axis the system is marginally stable if the pole is of order 6 1. The system is unstable if the pole is of higher order. So for example, consider the factors in the denominator of the transfer function shown in table 16.3 and what they indicate. In each case we can see an expression that would arise from the inverse and see the stability of the end system. 5

A pole of a function f (z) is a value which when inserted into z, causes an infinite

value of f (z). Note that z may be complex. 6 The order of a pole is essentially how often if occurs. If a pole is listed once, it has order 1, and if listed twice it has order 2, etc.

16.9 Transfer Functions Factor

276 Inverse Pole Order Stable?

(s − 3)

e3t

3

1

no

(s + 2)

e−2t

−2

1

yes

(s + 2)2

te−2t

−2

2

yes

(s2 + 4)

sin 2t

±2j

1

yes

s

1

0

1

yes

s2

t

0

2

no

Table 16.3: Examples of transfer function denominators

16.9.1

Impulse Response

If the input to the system is the impulse function δ(t), and the response is the function h(t), then the transfer function is given by FT (s) =

L {h(t)} = L {h(t)} . L {δ(t)}

since L {δ(t)} = 1. So that one way to determine the transfer function for a system is simply to take the Laplace transform of the impoulse response for that system. Supposing that now we change the input function to f (t) and that the corresponding output function is g(t), we obtain FT (s) =

L {g(t)} L {g(t)} ⇒ L {h(t)} = ⇒ L {g(t)} = L {f (t)}×L {h(t)} . L {f (t)} L {f (t)}

So we can obtain the transform of the general output, provided we know the Laplace transform of the input and impulse response for the system. To obtain the output from the system g(t) we now take the inverse transform on both sides. g(t) = L−1 {L {f (t)} × L {h(t)}} . From the convolution theorem (see 16.6.3), we see that g(t) = f ∗ h.

16.9 Transfer Functions

16.9.2

277

Initial value theorem

If f (t) is a function of t with Laplace transform F (s) then lim f (t) = lim sF (s). t→0

s→0

Proof We know that  L

dx dt



∞

Z

e−st f 0 (t)dt = sF (s) − f (0).

= 0

Now, let s → ∞, to obtain 0 = lim sF (s) − lim f (0) s→∞

s→0

as the integral will tend to zero, because as s → ∞ the term e−st tends rapidly to zero. ⇒ 0 = lim sF (s) − lim f (t) s→∞

t→0

and rearranging gives us the theorem.

16.9.3

Final value theorem

If f (t) is a function of t with Laplace transform F (s), and if limt→∞ f (t) exists, then lim f (t) = lim sF (s). t→∞

s→0

Proof Once again, we begin with the observation   Z ∞ dx L = e−st f 0 (t)dt = sF (s) − f (0). dt 0 This time we allow s → 0. On the left hand side (the integral) we obtain Z ∞ Z t 0 f (t)dt = lim f 0 (t)dt = lim f (t) − f (0). o

t→∞

0

t→∞

and on the right hand side we obtain simply lim F (s) − f (0).

s→0

16.9 Transfer Functions

278

So equating these again we obtain lim f (t) − f (0) = lim F (s) − f (0)

t→∞

s→0

and adding f (0) on both sides we obtain the final result.

Chapter 17 Z-transform 17.1

Concept

Consider a signal of a continuous function f (t) and how it appears when the signal is examined at discrete time intervals, say when t = kT, k = 0, 1, 2, . . . where T is some fixed time period.

Figure 17.1: A continuous (analog) function Then instead of seeing a continous curve we instead obtain a series of spikes indicating the value of the signal at each sampling period, these can be represented by Dirac deltas.

17.1 Concept

280

Figure 17.2: Sampling the function

Figure 17.3: The digital view

17.2 Important Z-transforms

281

In this way, the discrete “version” of the sample appears as fD (t) =

∞ X

f (kT )δ(t − kT )

k=0

where f (kT ) is the value of the signal at the sampling frequencies and δ(t − kT ) is the delayed Dirac delta. Note that k is a dummy variable in this summation and won’t appear in the final expansion, although T will. If we consider the Laplace transform of this sum, we obtain F (s) = L {fD (t)} = f (0) + f (T )e−T s + f (2T )e−2T s + · · · ⇒ F (s) =

∞ X

f (kT )e−kT s

k=0

in this sort of expansion the exponential term occurs very frequently. If we make the simplifying substitution z = eT s then the transformed equation becomes ∞ X

f (kT )z −k

k=0

This expression is the Z-transform of the discrete function, and we write. F (z) = Z {fD (t)} =

∞ X

f (kT )z −k

k=0

17.2

Important Z-transforms

17.2.1

Unit step function Z {u(t)} =

Proof From the definition of the Z-transform.

z z−1

17.2 Important Z-transforms

282

inf ty

U (z) = Z {u(t)} =

X

u(kT )z −k

k=0

Now u(kT ) means the value of the step function at each of the specified sampling periods, but this is always simply one. ⇒ U (z) =

∞ X

1z −k = 1 + z −1 + z −2 + z −3 + · · ·

k=0

which is a geometric progression. Using the formula for the sum to infinity we obtain that (if |z| > 1) U (z) =

1 1 − z −1

which although a completely transformed equation, is simpler in appearance if we multiply by the top and bottom by z which obtains the desired result. Note that T does not appear in the formula.

17.2.2

Linear function Z {t} =

Tz (z − 1)2

Proof From the definition F (z) = Z {t} =

∞ X

f (kT )z −k .

k=0

Now f (t) = t, and t = kT so f (kT ) = kT and therefore F (z) =

∞ X

kT z −k = T (z −1 + 2z −2 + 3z −3 + · · · ).

k=0

Which is another standard summation F (z) = T

z −1 (1 − z −1 )2

which upon multiplying through by z 2 top and bottom produces the desired result.

17.2 Important Z-transforms

17.2.3

283

Exponential function Z {e−at} =

z z − e−aT

Proof From the definition ∞ X  F (z) = Z e−at = f (kT )z −k k=0

=

∞ X

e−akT z −k

k=0

z + e−2aT z −2 + e−3aT z −3 + · · ·
−1
−2
−3 = 1 + eaT z + eaT z + eaT z + cdots =1+e

−aT −1

which is another geometric progression. Using the sum to infinity formula we obtain that 1 F (z) = −aT 1 − e z −1 which multiplying through top and bottom by z gives the required result.

17.2.4

Elementary properties

Just as for the Laplace transform, the Z-transform obeys the following basic rules. If f (t) and g(t) are functions of t and c is a constant. Z {f (t) ± g(t)} = Z {f (t)} ± Z {g(t)} and Z {cf (t)} = cZ {f (t)}

17.2.5

Real translation theorem

In Laplace transforms the most important property was that of the transform of the derivative which allowed differential equations to be solved easily. If f (t) is such that F (z) = Z {f (t)} then Z {f (t − nT )} = z −n F (z)

17.2 Important Z-transforms

284

Proof From the definition Z {f (t − nT )} =

∞ X k=0

f (kT − nT )z −k =

∞ X

f ((k − n)T )z −k

k=0

= f (−nT ) + f (1 − nT )z −1 + f (2 − nT )z −3 + · · ·

Chapter 18 Statistics We are frequently required to describe the properties of large numbers of objects in a simplified way. For example, political opinion polls seek to distill the complex variety of political opinions in a region or country into just a few figures. Similarly, when we talk about the mean time to failure of a component, we use a single number to reflect the behaviour of a large population of components. We shall review some definitions.

18.1

Sigma Notation

Throughout statistics and mathematics in general, we often use sigma notation as a shorthand for a sum of objects. A capital sigma is used, and the range of the summation is given above and below, unless this is obvious. b X

f (i) = f (a) + f (a + 1) + · · · + f (b)

i=a

Usually an index value (i in this example) assumes a start value and increases by one until it reaches a final value. The expression is evaluated in each case and the results are summed.

18.1.1

Example

Here are some examples of sigma notation.

18.1 Sigma Notation

286

n X

i = 1 + 2 + 3 + ··· + n

i=1 n X 1 1 1 1 = + + ··· + 2 2 i 1 4 n i=1

We use the symbol ∞ to represent the absence of an endpoint. ∞ X 1 1 1 = + + ··· = 1 n 2 2 4 i=1

In statistics we are usually concerned with summations where the i is just an index. For example n X

xi = x1 + x2 + · · · + xn

i=1

is a common abbreviation for adding up n items of data labelled appropriately. There are two specific summations which we examine carefully. If k is a constant, and f (i) is a function of i we can show easily: n X

k = nk.

i=1

This is because n X i=1

k=k | + k + k + k{z+ k + · · · + k} = nk. nof these

Another important result is n X i=1

kf (i) = k

n X

f (i).

i=1

This is because n X

kf (i) = kf (1) + kf (2) + kf (3) + · · · + kf (n)

i=1

= k {f (1) + f (2) + f (3) + · · · + f (n)} which gives us the result.

18.2 Populations and Samples

18.2

Populations and Samples

When we embark upon research, we usually have a specific population in mind. The population is the entire collection of objects, people, etc. which we wish to answer questions about. Populations are often vast, and problems in health care can have a population consisting of all the people on Earth, now over six billion. It is clear that in most cases it will be impossible to consider every element in the population. For this reason, we normally work with a sample, or several samples. A sample is a collection drawn from the population in some sensible way. By sensible, we mean that carelessly selecting a sample may give a distorted collection from the population. For example, if we conduct our opinion poll my choosing names from a phone book, we may exclude the other members of that house who are not listed, and those without phones. As the latter group may have different political opinions than the average, we can run into trouble here.

18.2.1

Sampling

There is a lot to consider in choosing a sample from a population with care, and we discuss here the most elementary sampling methods. Random sampling This term is applied to any method of sampling which ensures that each member of the population has an equal chance of being selected in the sample. We could do this by drawing from a hat, or using a table of random numbers. Stratified random sampling In this method, we attempt to contrive the sample so that the proportions of subsets in the parent population are preserved. For example, in a population of people that may be (typically) 55% female, we would attempt to ensure that 55% of the sample were female, but otherwise randomly selected.

287

18.3 Parameters and Statistics Systematic sampling In this method, we take some starting point in the population, and select every kth entry. Cluster sampling In this method, we divide the population are into several sections (clusters) and randomly select a few of the sections, choosing all the members from them. Convenience sampling In this method, we use the data that is readily available (cf. exit polls). This method has many potential pit-falls.

18.3

Parameters and Statistics

Technically, a number describing some property of the population is called parameter and a number describing some property of the sample is called statistic.

18.4

Frequency

In many situations, a specific value will occur many times within a sample. The frequency of a value is the number of times that value occurs within the sample. The relative frequency of a value is determined by dividing the frequency for that value by the sum of all frequencies (which is the same as the number of all values). Thus, relative frequency is always scaled between 0 and 1 inclusive and is closely related to the concept of probability.

18.5

Measures of Location

Measures of location attempt to generalise all the values with a single, central value. These are often called averages, and that quantity that is normally

288

18.5 Measures of Location

289

called the average in everyday speech is just one example of this class of measures. There are three main averages.

18.5.1

Arithmetic Mean

The mean, or more precisely arithmetic mean, of a set of n data elements x1 , x2 , . . . , xn is given by n

1X xi n i=1 It is customary to use the symbol x to denote the mean of a sample, and the symbol µ to denote the mean of the population. Clearly the mean provides a rough measure of the centre point of the data, note that it may be somewhat unreliable for many uses if the the data is very skewed (see 18.9). There are two other commonly used averages which may be used.

18.5.2

Mode

The mode of a sample is the value that occurs the most within the sample, that is, the value with the highest frequency (see 18.4). Then two values are tied with the highest frequency the sample is called bimodal and both values are modes. If more than two values are tied, we call the sample multimodal.

18.5.3

Median

The median of a sample is the middle value when the values of the sample are ordered in ascending or descending order. Note that when there are an even number of values, we usually use the mean of the middle two elements.

18.5.4

Example

Consider a sample formed of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9. State the mean, mode and median.

18.6 Measures of Dispersion Solution • The sample mean will be given by x=

1+2+3+4+5+6+7+8+9 = 5. 9

• The mode is the element with the highest frequency, but all the numbers 1 to 9 have the same frequency (1). Consequently there is no unique mode or one might even suggest that every number is the mode. • The median is the middle number when the sample is arranged in order (and it already is), so here the median is 5. This kind of a distribution is called the Uniform Distribution because (at least within a specific range) every item has an equal chance of being picked.

18.5.5

Example

State the mean, mode and median of the sample formed of the numbers 5, 5, 5, 5, 5, 5, 5, 5, 5. Solution • The sample mean will be given by x=

5+5+5+5+5+5+5+5+5 = 5. 9

• This time the mode is clear cut, there is only one element, and it has a frequency of 9, so the mode is 5. • The median is 5 once again.

18.6

Measures of Dispersion

The previous examples demonstrates that averages on their own do not tell us enough about the data, we also want to know how spread out or dispersed the data is.

290

18.6 Measures of Dispersion

18.6.1

Range

The simplest form of this measure is the range, which is simply the smallest value subtracted from the largest value. This measure is often unreliable, as the largest and smallest values can often be freakish and error-prone (called outliers). Note the whole measure depends on two values only.

18.6.2

Standard deviation

The standard deviation of a set of data elements x1 , x2 , . . . , xn is given by v u n u1 X (x − xi )2 σ=t n i=1 We use the symbol σ to refer to the standard deviation of the population, and the symbol s to refer to the standard deviation of the sample (which has a very slightly altered formula). This form of the standard deviation formula is quite intuitive, showing the way the measure is formed, but it is computationally ackward. A simple bit of sigma manipulation provides a computationally simpler expression. It can be shown that v v u n u n u1 X u1 X (x − xi )2 = t x2 − x2 σ=t n i=1 n i=1 A correction is made when we are finding the standard deviation of a sample as opposed to a population. Therefore v v u u n n u 1 X u 1 X s=t (x − xi )2 = t x2 − x2 n − 1 i=1 n − 1 i=1 In the former case we alter the n found on the bottom line to n−1. (Note that P in the frequency versions of the formulae below it can be easily seen that n = f ). The advantage of the standard deviation is that it uses all the values in the sample to calculate the dispersion. The standard deviation is strongly focused on the concept of the mean. Examining its formulation reveals that we are summing the squared differences from the mean. It follows that in situations were the mean is unsuitable, the standard deviation is usually unsuitable also.

291

18.7 Frequency Distributions

18.6.3

Inter-quartile range

Just as we did to calculate the median of a sample, we arrange the values in the sample in ascending order. The middle values is of course the median, the value found one quarter of the way through the values is the lower quartile and the value found three quarters through the values is the upper quartile. The value found when we subtract the lower quartile from the upper is called the inter-quartile range. This measure is superior to the range, and often harder to evaluate than the standard deviation. However, it is very useful with skewed samples (18.9).

18.7

Frequency Distributions

In practice, we often use frequencies in large samples to reduce the work. Note that many assumptions may rely on data with little or no skew. Suppose that in a sample, the only distinct values that occur are given by x1 , x2 , . . . , xn ; further suppose that each value has a frequency given by f1 , f2 , . . . , fn . Then, for example, the mean and standard deviation are given by Pn f i xi µ = Pi=1 n i=1 fi and sP n f (x − x)2 i=1 Pn σ= i=1 fi

18.7.1

Class intervals

We often consider, either for simplicity of some other reason, that data falls into certain ranges of values, known as class intervals. For example, suppose we table the ages of all people admitted to casualty in a specified time. Rather than consider every age value, we might consider ages tabulated as shown in table 18.1. There is no need to insist on equal widths of interval, we can change them as required. The fundamental trick to dealing with this situation is to imagine that all the items in each interval are concentrated on the central value. For example, we consider the 7 items in the first interval all to be concentrated at 4.5 years, and so on. The calculation can then go on unimpeded.

292

18.8 Cumulative frequency

293

Age

Frequency

0-9

7

10 - 19

4

20 - 29

6

30 - 39

8

40 - 49

5

50 - 59

13

60 - 69

6

70 - 79

1

80 +

1

Table 18.1: An example of class intervals Note that if we have a value which falls between the intervals, we use rounding to determine which interval it belongs to. Thus, a value of 9.2 years belongs to the 0-9 interval, while 9.8 years belongs to the 10-19 interval. Therefore the catchment areas are slightly wider than the intervals and these are called class boundaries.

18.8

Cumulative frequency

Frequently, it is useful to construct a cumulative frequency graph of our data. This is useful for finding the median, and other key points in the data range. The cumulative frequency is the sum of all the frequencies above or below a specified point.

18.8.1

Calculating the median

Once we have plotted our cumulative frequency graph, we find the value that if half of the largest cumulative frequency. If we draw a horizontal line from the cumulative frequency to the graph and then vertically down, the figure we land on is the median.

18.9 Skew

294

This is very useful for calculating (or estimating) the mean in data arranged by frequency tables.

18.8.2

Calculating quartiles

Finding the quartiles is very similar. To find the lower quartile take one quarter of the highest cumulative frequency and draw a horizontal line to the graph and vertically down to the axis we obtain the quartile. We start with three quarters of the highest cumulative frequency to find the upper quartile.

18.8.3

Calculating other ranges

We are not restricted to the 50%, or 25% and 75% marks in this procedure, we can equally find the 10%, 5% or any figure we care by working out this percentage of the total cumulative frequency and tracing to the right and down.

18.9

Skew

In addition to this, it is possible to measure the skew of a sample. Skew is a measure of how much the data tends to one side or the other of the mean. Calculation of skew is beyond the scope of this course.

18.10

Correlation

When we have two sets of interlinked numbers of equal size, we are often interested in whether the numbers show a correlation. That is to say, whether or not there is a link between the two sets. Suppose that we have two sets of numbers x1 , x2 , x3 , . . . , xn ; y1 , y2 , y3 , . . . , yn then clearly we can plot the values (x1 , y1 ), (x2 , y2 ), (x3 , y3 ), . . . , (xn , yn )

18.10 Correlation

295

on a graph. If the points are relatively scattered, this usually suggests no link, but if the points lie along a line we can make some guesses about a link. For example, if the xs represented the heights in metres of a certain population, and the ys represented the weights in kilograms, we would expect a certain correlation. Although it is possible to deal with curves, we shall consider the situation when the points lie approximately in a straight line. Of course, we prefer to have quantitive measures.

18.10.1

Linear regression

If we assume that the graph of y = mx + c approximately passes through the points, we can show that the best estimates for m and c are given by the formula P P P n( xy) − ( x)( y) P P m= n( x2 ) − ( x)2 and P P P P ( y)( x2 ) − ( x)( xy) P P c= n( x2 ) − ( x)2 after which we can use the formula of the line to estimate the value of y given a value for x. It is possible to reverse the variables when we wish to estimate x given y. Essentially this can be proven by measuring how far each point is from the proposed line, totalling these distances and then trying to make them as small as possible (calculus). All summations here are ranging from i = 1 to n as usual.

18.10.2

Correlation coefficient

The Pearson product moment correlation coefficient or simply linear correlation coefficient r is given by P P P x xy − ( x)( y) r=p P P p P P n( x2 ) − ( x)2 n( y 2 ) − ( y)2 is a measure of how well the scattered points fit the straight line above.

18.10 Correlation

296

Interpretation The value of r is bounded by -1 and 1. That is −1 ≤ r ≤ 1. A value of 1 or −1 signifies a perfect line up, with 1 representing a positive gradient (larger x produces larger y), and with −1 representing a negative gradient (larger x produces smaller y). So values close to −1 or 1 suggest a correlation, while values close to 0 in the middle suggest no correlation. Scaling The value of r is unaffected by scaling - that is the units used do not change it, nor does swapping the x and y values. Warnings Remember, this is designed for linear relationships. If you suspect a curved relationship you should transform to a straight line first. Note also that correlation does not indicate causality.

Chapter 19 Probability Probability is concerned with the likelyhood of certain events. In many ways then, probability can be thought of as an attempt to predict the likely outcomes of experiments, while statistics provides the means of analysing data after an event has occured.

19.1

Events

An event is a specific well defined occurrence. That is, it must be totally unabiguous whether or not the event has occured. It is common-place to use capital letters to denote events.

19.1.1

Probability of an Event

The probability of an event is a real number between 0 and 1 inclusive. For our purposes a probability of 0 denotes that the event cannot occur, and a probability of 1 denotes that the event must occur. We shall use the shorthand notation P (A) to denote the probability of an event A.

19.1.2

Exhaustive lists

We say that a set of events A1 , A2 , . . . , An form an exhaustive list of events if and only if they cover all possibilities. That is, at least one of the events

19.2 Multiple Events must occur.

19.2

Multiple Events

Commonly, we have several events, and we are interested in the probability of some combination of events. For example, we may wish to know the probability that A and B both occur, or the probability that A occurs or B does not.

19.2.1

Notation

Certain notation is used throughout probability theory to denote these logical combinations, and we now review this. The following notation is used. • The probability that both A and B occur is denoted by P (A ∧ B) or P (AB); • The probability that either A or B occurs is denoted by P (A ∨ B) or P (A + B); • The probability that A does not occur is denoted by P (A) or P (¬A); • The probability that A will occur given that B has already occurred is denoted P (A|B). It is vital that you note that P (A + B) is a notation for the probability of (A or B), and not (A and B) as you might have guessed.

19.2.2

Relations between events

Before we introduce methods for obtaining the probabilities of these combinations, we need to look at two important concepts. Mutually exclusive events Two events A and B are said to be mutually exclusive if and only if there is no way that both A and B can occur.

298

19.3 Probability Laws

299

Independent events Two events A and B are said to be independent if and only if the occurrence of A does not effect the probability of B, and the occurrence of B does not effect the probability of A. More than two events We can easily extend these ideas to more than two events. For example, we say that events A, B and C are mutually exclusive if and only if • A and B are mutually exclusive; • B and C are mutually exclusive; • A and C are mutually exclusive. We perform a similar extension to the notion of independence, and if we are dealing with more than three events.

19.3

Probability Laws

Now we are ready to examine simple probability laws.

19.3.1

A or B (mutually exclusive events)

Given two mutually exclusive events A and B, the probability of A or B occurring is given by P (A ∨ B) = P (A) + P (B). A helpful consequence of this result follows

19.3.2

not A

Given any event A P (A) = 1 − P (A). This is a simple application of the previous law. Clearly A and A cannot occur at once and so they are mutually exclusive. Furthermore, either A

19.3 Probability Laws happens or it does not (in which case A happens) and so P (A ∨ A) = P (A) + P (A) = 1. Rearranging this equation yields our result.

19.3.3

1 event of N

If a set of N events, A1 , A2 , . . . , AN are mutually exclusive, an exhaustive list and equally probable then the probability of each event is N1 .

19.3.4

n events of N

If an experiment has N equally likely exhaustive outcomes, and the event A corresponds to n of those outcomes then the P (A) = Nn . We use these result all the time, without even giving it any thought.

19.3.5

Examples

Given a fair coin, the probability of getting a head on a single toss of the coin is 12 . Given a fair die, the probability of getting a 6 on a single roll of the die is 16 . We now know how to find the probability of A or B in certain circumstances, so we turn to the problem of A and B.

19.3.6

A and B (independent events)

Given two independent events A and B, the probability of both A and B occurring is given by P (A ∧ B) = P (A) × P (B). Warning The result above does not work when the events are dependent; be especially careful here, it is usually simple to see if events are mutually exclusive, but more difficult to determine their independence.

300

19.3 Probability Laws

19.3.7

Example

A certain device consists of two components A and B, which are wired in series. If either component is not functioning the circuit will break. Each component is unaffected by the state of the other. The probability of A being functional is 0.9, and the probability of B being functional is 0.8. What is the probability of the device being functional? Solution We shall label A as the event that component A is functioning, and similarly label B as the event that component B is functioning. We are told that A and B are independent events, the components work independently of each other. Therefore we can use our simple law described in 19.3.6. P (A ∧ B) = P (A) × P (B) = 0.9 = 0.8 = 0.72 It is worth noting that the probability that the device is working is lower than the probability of each component. We shouldn’t be surprised that placing several vulnerable components in series has this effect. Of course many experiments concern many more than three events, but this does not cause us any concern as we can easily extend these two probability laws.

19.3.8

A or B or C or ...

Given n mutually exclusive events A1 , A2 , . . . , An , then the probability that one of them occurs is given by P (A1 ∨ A2 ∨ · · · ∨ An ) = P (A1 ) + P (A2 ) + · · · + P (An ). Clearly this is an extension of 19.3.1.

19.3.9

A and B and C and ...

Given n independent events A1 , A2 , . . . , An , then the probability that all of them occurs is given by P (A1 ∧ A2 ∧ · · · ∧ An ) = P (A1 ) × P (A2 ) × · · · × P (An ). Clearly this is an extension of 19.3.6.

301

19.3 Probability Laws

19.3.10

Example

A fair die is rolled once, show that the probability of rolling a number less than 6 is 56 . Solution There are at least three ways of tackling this problem. Let A1 be the event of rolling a 1, down to A6 being the probability of rolling a 6. 1. We could simply observe that there are only six equally likely outcomes, and our event pertains to five of them. That, with a simple application of 19.3.4, gives us a probability of 56 . 2. Clearly A1 , A2 , . . . , A5 are mutually exclusive (it is impossible to roll two numbers at the same time on one roll). Therefore (by 19.3.1) we can simply add the probabilities, and as each probability is 61 we arrive at an answer of 56 . 3. We could observe that the question is equivalent to the probability of not rolling a 6. We know the probability of rolling a 6 is simply 61 , so we subtract this from 1 to find 56 , (see 19.3.2). These results are useful, but we will frequently encounter situations where two events are not mutually exclusive, or not independent.

19.3.11

A or B revisited

Given two events A and B, the probability of A or B occurring is given by P (A ∨ B) = P (A) + P (B) − P (A ∧ B). Compare this result to 19.3.1. Let us use this result to examine the reliability of two objects in parallel.

19.3.12

Example

A certain device consists of two components A and B, which are wired in parallel. Each component has enough capacity individually to work the device. Each component is unaffected by the state of the other. The probability of A being functional is 0.9, and the probability of B being functional is 0.8. What is the probability of the device being functional?

302

19.3 Probability Laws

303

Solution We shall label A as the event that component A is functioning, and similarly label B as the event that component B is functioning. Unlike the example above, this device will work if A or B is functional. We note that A and B are certainly not mutually exclusive; it is quite possible for both components to be working at the same time. Therefore we have to use the general form of the probability law (19.3.11), and not the basic form (19.3.1). P (A ∨ B) = P (A) + (P ) − P (A ∧ B) = 0.9 + 0.8 − 0.72 = 0.98 Note that A and B were independent, so we were able to simply use the P (A∧B) = P (A)×P (B) relation (see 19.3.6) for that part of the calculation. Note also that the reliability of the parallel system is much greater than that of the individual components. Compare this to the previous example of the series system. So, we no longer require events to be mutually exclusive to find the probability of one of them occurring. So what about the notion of independence?

19.3.13

A and B revisited

Given any two events A and B, the probability that A and B both occur is given by P (A ∧ B) = P (A)P (B|A) = P (B)P (A|B). Compare this result to 19.3.6. This relationship also provides a neat way of working out conditional probability.

19.3.14

Conditional probability

Given any two events A and B, the probability that A occurs given B already has is given by P (A|B) = provided that P (B) 6= 0.

P (A ∧ B) P (B)

19.4 Discrete Random Variables

19.3.15

Example

The probability that two components A and B in a device are both working is 0.63. The probability that B is working is 0.7. Calculate the probability that A is working given that B is. Solution Clearly P (A ∧ B) = 0.63, from the above P (A|B) = 0.63/0.7 = 0.9. Question. Do we have enough information in this example to calculate P (A)? We can also rearrange our probability law 19.3.13 to produce the following important result.

19.3.16

Bayes Theorem P (B|A) =

P (B)P (A|B) . P (A)

Once again, this assumes that P (A) 6= 0. More generally we can write P (A|F ) =

P (A)P (F |A) . Σi P (F |Bi )P (Bi )

The letter F has been picked here to represent failure.

19.4

Discrete Random Variables

A discrete random variable is any variable quantity that can only take exact or separate numerical values. For example, while the length of feet measured in people may assume any (therefore continuous) values, the shoe sizes they have can have only discrete values.

19.4.1

Notation

It is usual to use capital letters to denote a variable, and lower case letters to denote possible outcomes. These outcomes should all be arranged to be mututally exclusive.

304

19.4 Discrete Random Variables

305

Therefore suppose that X can be one of the following outcomes x1 , x2 , x3 , . . . , xn , then

n X

P (X = xi ) = 1

i=1

and indeed, this is one definition of a discrete random variable.

19.4.2

Expected Value

The expected value of a discrete random variable is analogous to the mean value that the variable will have. We denote that expected value of the variable X by E(X). Let us examine the probability distribution for a simple random variable. If we roll two fair dice and add the scores, we obtain the following distribution. xi

f (xi )

2

1 36

3

2 36

4

3 36

5

4 36

6

5 36

7

6 36

8

5 36

9

4 36

10

3 36

11

2 36

12

1 36

Table 19.1: Probabilities for total of two rolled dice It is clear in this case that the probability distribution is symmetrical and that we would expect 7 to be the most common score produced by this experiment.

19.4 Discrete Random Variables

306

How could this be quantitively calculated however? Well suppose that we calculate the total of totals if we repeat the experiment 36 times. So we would expect to have 1occurence of 2, 2 occurences of 3 and so on. Therefore in total we would have T = 2 × 1 + 3 × 2 + 3 × 3 + · · · + 11 × 2 + 12 × 1 = 252 and therefore if 252 is the combined total of 36 experiments we could divide by 36 to work out the expected score from one such experiment. Thus we obtain E(X) = 252/36 = 7. Another way this could be written would be to perform the division by 36 earlier. E(X) =

1 2 3 2 1 T =2× +3× +3× + · · · + 11 × + 12 × 36 36 36 36 36 36

252 =7 36 The division has been moved to the second part of the multiplication to help highlight that each term is merely the product of the outcome with the probability of that outcome. We can generalise this as follows. =

Definition If a discrete random variable X has the possible outcomes x1 , x2 , . . . xn , with a probability density function f (x) such that f (xi ) = P (X = xi ), then E(X) =

n X

xi f (xi ).

i=1

19.4.3

Variance

Given that we have some measure of the mean outcome for a discrete random variable we turn our attention to how dispersed the outcomes may be around this mean. We do this by examining the variance which is the name for the square of standard deviation (see 18.6.2). Now the standard deviation is essentially the root, of the mean of the squared deviations from the mean, so the variance will be the same calculation without this final root. We shall denote the variance of the variable X with the notation var (X) .

19.4 Discrete Random Variables Definition If a discrete random variable X has the possible outcomes x1 , x2 , . . . xn , with a probability density function f (x) such that f (xi ) = P (X = xi ), and we adopt the notation that µ = E(x) then var (x) = E[(X − µ)2 ]. Now although this definition is intuitive, showing as it does that we are averaging the squares of the deviations of the variable from the mean, it is tedious to calculate. One can see that the mean has to be subtracted from all the figures, then these must be squared and totalled. We can use some algebra to find an alternative expression. var (x) = E[(X − µ)2 ] = E[x2 − 2µX + µ2 ] = E(X 2 ) − 2µE(X) + µ2 = E(X 2 ) − µ2 and so in summary var (X) = E(X 2 ) − µ2 which is easier to calculate even if it is less obviously true.

19.4.4

Example

We revisit a previous problem. Let X be the score obtained by totalling the rolls of two fair dice. Find E(X) and var (X). Solution We solve the whole problem as we would from the start, without the assumptions we made above. By definition E(X) is the sum of the products of the outcomes and their probabilities. This is the sum of the third column which is 7. We now let µ = 7. Now var (X) = E(X 2 ) − µ2 and E(X 2 ) is simply found by squaring the xi and multiplying them by their repspective probabilities (and summing these). This is the sum of the fourth column is 54 65 . Thus 5 5 35 var (X) = 54 − 49 = 5 = 6 6 6

307

19.5 Continuous Random Variables xi

308

f (xi ) xi f (xi ) x2i f (xi )

2

1 36

2 36

4 36

3

2 36

6 36

18 36

4

3 36

12 36

48 36

5

4 36

20 36

100 36

6

5 36

30 36

180 36

7

6 36

42 36

294 36

8

5 36

40 36

320 36

9

4 36

36 36

324 36

10

3 36

30 36

300 36

11

2 36

22 36

242 36

12

1 36

12 36

144 36

total

1

7

54 56

Table 19.2: Calculating E(X) and var (X) for two rolled dice.

19.5

Continuous Random Variables

A continuous random variable is not limited to discrete values, but instead can take any value in a specified range. As an analogy consider the real number line. The real numbers make up the entire line, and every point is a real number. No matter what two real numbers one may pick, there are infinitely more inbetween. By contrast there are huge gaps between the integers as they appear on the number line, and in fact it is easy to pick two such numbers, such as 2 and 3 so that there is no integer between them. So the Integers are a discrete collection of numbers while the real numbers are continuous. 1 1

In fact this nature of the real numbers is noted in the fact that they are often described

as the continuum. Furthermore when people discuss the sizes of the integers versus that of the real numbers the integers are of the size of the smallest infinity ℵ0 while the real numbers are of the size of the continuum, written c.

19.5 Continuous Random Variables

19.5.1

309

Definition

If X is a continuous variable such that there are real numbers a and b such that P (a ≤ X < b) = 1; P (X < a) = 0; P (X ≥ b) then X is a continuous random variable.

19.5.2

Probability Density Function

For a continuous random variable X the probability density function or p.d.f. will in general take the form of a function that varies with X over all the values in the range. Instead of asking for the probability that X has a certain value, we usually need to calculate the probability that X lies in a certain range. 2 We do this by calculating the area under the curve in that range. Thus Z b P (a ≤ X < b) = f (x)dx a

where f (x) is the p.d.f. for the continuous random variable X.

2

For example, if we want to find P (X = 2) for a continuous variable, we are normally

looking for all values of X that round to the number 2 and so we find P (1.5 ≤ x < 2) instead.

Chapter 20 The Normal Distribution 20.1

Definition

In many situations, data follows a very characteristic distribution known as the normal distribution, or Gaussian distribution.

Figure 20.1: The Normal Distribution This distribution is totally symmetrical (has no skew), and most data is to be found in the middle values, while large or small values are much rarer. The distribution is centered about the mean and about 99.7% of the data lies within three standard deviations on either side of the mean.

20.2 Standard normal distribution

Figure 20.2: Close up of the Normal Distribution

20.2

Standard normal distribution

The normal distribution with a mean of 0 and a standard deviation of 1 is called the standard normal distribution. It is important because questions about all normal distributions can be reduced to a problem on the standard distribution, for which a set of tables can be produced (see table A.1 below). Areas under the graph of the distribution represent the proportion of the population residing there.

20.2.1

Transforming variables

Essentially to use the standard distribution, we must be able to transform more difficult problems. Let us look at a specific problem. Suppose that the heights of 1000 people are normally distributed with mean 170 cm, and standard deviation 5 cm. How many people will have heights below 180 cm. We deal with this problem by calculating how far we are away from the mean, measured in standard deviations. In this example the distance between 180 and the mean 170 is 10 cm, but we want to find the number of standard deviations away, so dividing by the deviation, 5, we obtain 2. In general, if the boundary value we are interested in is x, the appropriate standard normal boundary value z is given by x−µ σ To complete our problem, we examine our tables which show the area under the graph (proportion) up to and including our value (which is 2). z=

311

20.2 Standard normal distribution From the tables we get a value of 0.977, and this fraction of our population represents 977 people.

20.2.2

Calculation of areas

Note that the tables provided, like most tables show areas calculated only in a certain way. To work out other areas we must use ingenuity combined with the following facts 1. The total area under the graph is 1; 2. The graph is perfectly symmetrical about 0; 3. Therefore the area under each side is 21 .

20.2.3

Example

Let us return to our example, and make it slightly more complex. Suppose that the heights of 1000 people are normally distributed with mean 170 cm, and standard deviation 5 cm. How many people will have heights between 167 cm and 180 cm. A sketch graph of each situation is highly advisable. Next we calculate z for each of our values. 167 − 170 180 − 170 = 2; z2 = = −0.6 5 5 We already know from the tables that the area enclosed under the graph up to z = 2 is 0.977. We are only interested in the region above 167 as well however. We simplify the problem by calculating the area from z = 0 to z = 2, this is the same area, without the region below z = 0 which we know has area 0.5. Therefore the contribution from z = 0 to z = 2 is 0.477. Now we work out the area from z = −0.6 to z = 0. Our table does not consider negative values of z and so we must do something different. By symmetry, the area from z = 0 to z = +0.6 will be the same. The area up to z = 0.6 is 0.726 from our tables, so once more we subtract 0.5 to find our desired area, with a value of 0.226. Adding these regions together gives an area of 0.477 + 0.226 = 0.703, so multiplying this area (proportion) by our population, we come up with an answer of 703 people. z1 =

312

20.2 Standard normal distribution

20.2.4

Confidence limits

Consider the “central” 95% of our population, that is a proportion of 0.950. As that section is spread symmetrically, half of this lies on either side of the mean, so 0.475 lies above the mean for example. If we wish to find the z boundary point for this, we need to get an area like one in our tables, that is all the area up to that point. The area under the mean is 0.5, and the area above to the boundary is 0.475, creating a total of 0.975. This corresponds to a z value of 1.96. The values µ − 1.96σ and µ + 1.96σ are called the 95% confidence limits for our data. Picking a member of the population at random we can be 95% sure they will fall into this region.

20.2.5

Sampling distribution

Suppose that we have a large population, out of which we select many samples of size n. If we calculate the sample mean for each sample, then we could consider the sample formed by these averages. This is a sample taken from the population of all possible averages of samples of size n from the original population. This is confusing - you must realise that there are two populations, the original one (P1 say), and the population (P2 say) of samples of size n taken from P1 . Example Take the following, extremely small example. Suppose our population P1 consists of the numbers 1, 2, 3, 4, 5 and that we are taking samples of size 3, then the population P2 of all possible samples looks like this.

313

20.3 The central limit theorem 1 1 1 2 1 1 2 1 2 3

20.3

314 2 2 3 3 2 3 3 4 4 4

3 4 4 4 5 5 5 5 5 5

The central limit theorem

This is an enormously important result in statistics. Many populations we deal with are not normally distributed, but this theorem gives us hope here. The central limit theorem suggests that as the sample size n increases, the sampling distribution approaches a normal distribution. 1 In other words, our population may not be very “normal” (see our example above), but the means of the samples drawn from it are roughly normal, and become more so as n increases. More specifically, if we have a distribution (which may or may not be normal) with mean µ and standard deviation σ, and we randomly select samples of size n from this population, then: The distribution of sample means x approaches a normal distribution as n becomes larger; The mean of the sample means will be the population mean µ; The standard deviation of the sample means will be

√σ . n

A value of n = 30 is quite safe to use this for most original parent populations, but any sample size is safe for a normally distributed parent. We write µx = µ and σ σx = √ n 1

There are exceptions, the Cauchy distribution does not obey the Central Limit The-

orem for example.

20.4 Finding the Population mean

20.4

Finding the Population mean

Imagine a population data is thought to have a mean µ, then if a sample of size n is taken at random we can be 95% confident that the sample mean x will lie in the following range σ σ µ − 1.96 √ ≤ x ≤ µ + 1.96 √ n n where the figure of 1.96 is the corresponding z score for a two-tailed 95% confidence interval, and we are taking samples from the sample mean population, and so it’s standard deviation as above. If n = 1 then clearly this just becomes the normal 95% confidence limit, but as n becomes larger this interval becomes smaller and smaller, giving us a more detailed and precise picture of the possible locations of x. Of course, normally the whole point is that we don’t have the population mean, but instead have simply the sample mean and wish the deduce the population mean from it. The same logic can be employed. We can interpret the interval above like so: σ |µ − x| ≤ 1.96 √ n so that the distance between µ and x has an upper limit (within our confidence level). Or to put it yet another way σ σ x − 1.96 √ ≤ µ ≤ x + 1.96 √ . c c so that given x and an estimate for σ we can produce a range of possible values for µ which will narrow as n increases given the relevant confidence level.

20.5

Hypothesis Testing

Another use for this result is to determine if a mean has moved significantly. This is generally only useful if we really were sure we knew it before. Suppose that a mean of a population was reliably found to be µ. Some time later a sample of size n was taken from the new population, and the mean found to be x, assume the standard deviation σ has not changed.

315

20.5 Hypothesis Testing

20.5.1

Two tailed tests

Suppose that we have these hypotheses • H1 : The means are significantly different µ 6= µ; • H0 : The means are in fact the same (the “null” hypothesis). Once again suppose that we wish to use a confidence level that corresponds to a score of Z for a two-tailed test. This is indeed a two tailed test since hypothesis H1 is simply whether the two means are different, in either direction. We first of all obtain a test statistic by working out how far away our new mean is from the old |µ − x| but we are really interested in how far this is in standard deviations. So we √ divide by the standard deviation for this situation which is σ/ n since we are looking at a sample of size n taken from the population. So our test statistic is √ |µ − x| n z= σ where once again, due to the two-tailed nature of the problem we are only interested in the magnitude of the difference, not the sign. If this test statistics is larger than the value Z obtained above then either 1. the mean has not changed but this sample lay outside the confidence interval even so; this is called a Type I error, and the probability of this occurring is usually denoted as α and called the significance level. Clearly α = 0.05 for a 95% confidence. 2. the hypothesis H1 is actually true. There is no way to determine for sure which is the case, which is why a high level of confidence is useful in order to make the error unlikely. We are therefore forced to accept H1 and reject H0 . Conversely, if the value of z < Z then either 1. the mean has changed significantly, but our random sample simply did not reflect this; this is called a Type II error, and probability of this occurring is usually denoted as β. 2. the hypothesis H0 is actually true. Again we reject H1 and accept the null hypothesis that the change is not statistically significant.

316

20.6 Difference of two normal distributions

20.6

Difference of two normal distributions

Another useful sampling distribution is that of the difference of two normal distributions. Suppose that we have two normal distributions, with means µ1 and µ2 and standard deviations σ1 and σ2 . Then if we take a sample of size n1 from the first distribution and n2 from the second then the difference in the means of the samples will have a mean µd of µd = µ1 − µ2 and standard deviation of s σd =

σ12 σ22 + . n21 n22

317

Appendix A Statistical Tables For the standard normal distribution, that is the normal distribution with mean 0 and standard deviation 1, the probability density function is given by: 1 2 1 1 1 φ(x) = √ e− 2 x = √ exp(− x2 ). 2 2π 2π Table A.1 shows the cumulative normal distribution calculated as Z x Φ(x) = φ(z)dz.

−∞ 2

The χ distribution, for a given confidence P and degrees of freedom ν is defined by P I(χ2 , ∞) = 100 I(0, ∞) where Z I(a, b) =

b

1

1

x 2 ν−1 e− 2 x dx.

a

is shown tabulated in tables A.2 and A.3.

319

x

0

1

2

3

4

5

6

7

8

9

0.0

0.500

0.504

0.508

0.512

0.516

0.520

0.524

0.528

0.532

0.536

0.1

0.540

0.544

0.548

0.552

0.556

0.560

0.564

0.567

0.571

0.575

0.2

0.579

0.583

0.587

0.591

0.595

0.599

0.603

0.606

0.610

0.614

0.3

0.618

0.622

0.626

0.629

0.633

0.637

0.641

0.644

0.648

0.652

0.4

0.655

0.659

0.663

0.666

0.670

0.674

0.677

0.681

0.684

0.688

0.5

0.691

0.695

0.698

0.702

0.705

0.709

0.712

0.716

0.719

0.722

0.6

0.726

0.729

0.732

0.736

0.739

0.742

0.745

0.749

0.752

0.755

0.7

0.758

0.761

0.764

0.767

0.770

0.773

0.776

0.779

0.782

0.785

0.8

0.788

0.791

0.794

0.797

0.800

0.802

0.805

0.808

0.811

0.813

0.9

0.816

0.819

0.821

0.824

0.826

0.829

0.831

0.834

0.836

0.839

1.0

0.841

0.844

0.846

0.848

0.851

0.853

0.855

0.858

0.860

0.862

1.1

0.864

0.867

0.869

0.871

0.873

0.875

0.877

0.879

0.881

0.883

1.2

0.885

0.887

0.889

0.891

0.893

0.894

0.896

0.898

0.900

0.901

1.3

0.903

0.905

0.907

0.908

0.910

0.911

0.913

0.915

0.916

0.918

1.4

0.919

0.921

0.922

0.924

0.925

0.926

0.928

0.929

0.931

0.932

1.5

0.933

0.934

0.936

0.937

0.938

0.939

0.941

0.942

0.943

0.944

1.6

0.945

0.946

0.947

0.948

0.949

0.951

0.952

0.953

0.954

0.954

1.7

0.955

0.956

0.957

0.958

0.959

0.960

0.961

0.962

0.962

0.963

1.8

0.964

0.965

0.966

0.966

0.967

0.968

0.969

0.969

0.970

0.971

1.9

0.971

0.972

0.973

0.973

0.974

0.974

0.975

0.976

0.976

0.977

2.0

0.977

0.978

0.978

0.979

0.979

0.980

0.980

0.981

0.981

0.982

2.1

0.982

0.983

0.983

0.983

0.984

0.984

0.985

0.985

0.985

0.986

2.2

0.986

0.986

0.987

0.987

0.987

0.988

0.988

0.988

0.989

0.989

2.3

0.989

0.990

0.990

0.990

0.990

0.991

0.991

0.991

0.991

0.992

2.4

0.992

0.992

0.992

0.992

0.993

0.993

0.993

0.993

0.993

0.994

2.5

0.994

0.994

0.994

0.994

0.994

0.995

0.995

0.995

0.995

0.995

2.6

0.995

0.995

0.996

0.996

0.996

0.996

0.996

0.996

0.996

0.996

2.7

0.997

0.997

0.997

0.997

0.997

0.997

0.997

0.997

0.997

0.997

2.8

0.997

0.998

0.998

0.998

0.998

0.998

0.998

0.998

0.998

0.998

2.9

0.998

0.998

0.998

0.998

0.998

0.998

0.998

0.999

0.999

0.999

3.0

0.999

0.999

0.999

0.999

0.999

0.999

0.999

0.999

0.999

0.999

Table A.1: Table of Φ(x) (Normal Distribution)

320

P

99

95

90

85

80

60

50

ν =1

0.000157

0.00393

0.0158

0.0358

0.064

0.275

0.455

2

0.0201

0.103

0.211

0.325

0.446

1.02

1.39

3

0.115

0.352

0.584

0.798

1.01

1.87

2.37

4

0.297

0.711

1.06

1.37

1.65

2.75

3.36

5

0.554

1.15

1.61

1.99

2.34

3.66

4.35

6

0.872

1.64

2.20

2.66

3.07

4.57

5.35

7

1.24

2.17

2.83

3.36

3.82

5.49

6.35

8

1.65

2.73

3.49

4.08

4.59

6.42

7.34

9

2.09

3.33

4.17

4.82

5.38

7.36

8.34

10

2.56

3.94

4.87

5.57

6.18

8.30

9.34

11

3.05

4.57

5.58

6.34

6.99

9.24

10.34

12

3.57

5.23

6.30

7.11

7.81

10.18

11.34

13

4.11

5.89

7.04

7.90

8.63

11.13

12.34

14

4.66

6.57

7.79

8.70

9.47

12.08

13.34

15

5.23

7.26

8.55

9.50

10.31

13.03

14.34

20

8.26

10.85

12.44

13.60

14.58

17.81

19.34

30

14.95

18.49

20.60

22.11

23.36

27.44

29.34

40

22.16

26.51

29.05

30.86

32.34

37.13

39.34

50

29.71

34.76

37.69

39.75

41.45

46.86

49.33

60

37.48

43.19

46.46

48.76

50.64

56.62

59.33

Table A.2: Table of χ2 distribution (Part I)

P

40

30

20

10

5

2

1

0.1

ν =1

0.708

1.07

1.64

2.71

3.84

5.41

6.63

10.83

2

1.83

2.41

3.22

4.61

5.99

7.82

9.21

13.82

3

2.95

3.66

4.64

6.25

7.81

9.84

11.34

16.27

4

4.04

4.88

5.99

7.78

9.49

11.67

13.28

18.47

5

5.13

6.06

7.29

9.24

11.07

13.39

15.09

20.51

6

6.21

7.23

8.56

10.64

12.59

15.03

16.81

22.46

7

7.28

8.38

9.80

12.02

14.07

16.62

18.48

24.32

8

8.35

9.52

11.03

13.36

15.51

18.17

20.09

26.12

9

9.41

10.66

12.24

14.68

16.92

19.68

21.67

27.88

10

10.47

11.78

13.44

15.99

18.31

21.16

23.21

29.59

11

11.53

12.90

14.63

17.28

19.68

22.62

24.73

31.26

12

12.58

14.01

15.81

18.55

21.03

24.05

26.22

32.91

13

13.64

15.12

16.98

19.81

22.36

25.47

27.69

34.53

14

14.69

16.22

18.15

21.06

23.68

26.87

29.14

36.12

15

15.73

17.32

19.31

22.31

25.00

28.26

30.58

37.70

20

20.95

22.77

25.04

28.41

31.41

35.02

37.57

45.31

30

31.32

33.53

36.25

40.26

43.77

47.96

50.89

59.70

40

41.62

44.16

47.27

51.81

55.76

60.44

63.69

73.40

50

51.89

54.72

58.16

63.17

67.50

72.61

76.15

86.66

60

62.13

65.23

68.97

74.40

79.08

84.58

88.38

99.61

Table A.3: Table of χ2 distribution (Part II)

Appendix B Greek Alphabet The greek alphabet has been included for completeness here. Of course, not all the greek letters are used in this course, but a full reference may prove useful.

322

Name

Lower Case Upper Case

Alpha

α

A

Beta

β

B

Gamma

γ

Γ

Delta

δ

∆

Epsilon



E

Zeta

ζ

Z

Eta

η

H

Theta

θ

Θ

Iota

ι

I

Kappa

κ

K

Lambda

λ

Λ

Mu

µ

M

Nu

ν

N

Xi

ξ

Ξ

Omicron

o

O

Pi

π

Π

Rho

ρ

P

Sigma

σ

Σ

Tau

τ

T

Upsilon

υ

Υ

Phi

φ

Φ

Chi

χ

X

Psi

ψ

Ψ

Omega

ω

Ω

Index !, 32 , 2 Im(z), 76 Re(z), 76 j, 68 =(z), 76 Π, 2