fac pack dec13

The Actuarial Education Company © IFE: 2014 Examinations

FAC – P C – 14

Combined Materials Pack

ActEd Study Materials: 2014 Examinations

Foundation ActEd Course (FAC)

Contents

Study Guide for the 2014 exams Course Notes

Question and Answer Bank Summary Test

If you think that any pages are missing from this pack, please contact ActEd’s admin team by email at [email protected] or by phone on 01235 550005.

How to use the Combined Materials Pack

Guidance on how and when to use the Combined Materials Pack is set out in the Study Guide for the 2014 exams.

Important: Copyright Agreement

This study material is copyright and is sold for the exclusive use of the purchaser. You may not hire out, lend, give out, sell, store or transmit electronically or photocopy any

part of it. You must take care of your material to ensure that it is not used or copied by anybody else. By opening this pack you agree to these conditions.

© IFE: 2014 Examinations The Actuarial Education Company

All study material produced by ActEd is copyright and is sold for the exclusive use of the purchaser. The copyright is owned

by Institute and Faculty Education Limited, a subsidiary of the Institute and Faculty of Actuaries.

You may not hire out, lend, give out, sell, store or transmit electronically or photocopy any part of the study material.

You must take care of your study material to ensure that it is not used or copied by anybody else.

Legal action will be taken if these terms are infringed. In addition, we may seek to take disciplinary action through the

profession or through your employer.

These conditions remain in force after you have finished using the course.

FAC: Study Guide Page 1


Foundation ActEd Course (FAC)

Study Guide

Objectives of the Study Guide The purpose of this Study Guide is to give you the information that you should have before studying FAC.

0 Introduction

This document has the following sections: Section 1 Syllabus Page 2 Section 2 The FAC course Page 10 Section 3 Study skills Page 12 Section 4 Contacts Page 13 Section 5 Course index Page 15

FAC Online Classroom

Please note that by purchasing this FAC you receive complimentary access to the FAC online classroom. This is a series of pre-recorded tutorials covering the main points from the course with examples as well as a dedicated forum for queries staffed by tutors. To access the online classroom please visit:

https://learn.bpp.com

You should have received an email with your access details. If you have lost this then enter your username (which is your email address used by ActEd) and click the “Forgotten your password?” to have a new password emailed to you. Should you have any problems with accessing the online classroom then please do email our admin team at [email protected].

ACET Mock Exam

A practice exam containing questions of the same standard as the ACET exam can be found in the reference resources section of the FAC online classroom.

FAC: Study Guide


1 Syllabus

1.1 Syllabus

This syllabus has been written by ActEd to help in designing introductory tuition material for new students. The topics are all required within the Core Technical subjects. From past experience we know that some students can be a little rusty on mathematical techniques so we have designed a course to help students brush up on their knowledge. Unlike other actuarial subjects, there is no official syllabus or Core Reading written by the profession. (a) Mathematical Notation, Terminology and Methods (a)(i) Be familiar with standard mathematical notation and terminology, so as to be

able to understand statements such as the following:

1. , , , , 3 : n n na b c n n a b c

2. 2( ,0], {0}x x

3. { : 1,2,3, }x x

4. x x 163 5. “Zero is a non-negative integer; is a positive real number.”

6. “ f x( ) tends to 0 as x tends to , is not defined when x 0 , but

takes positive values for sufficiently large x .” (a)(ii) Know the representations and names of the letters of the Greek alphabet that are

commonly used in mathematical, statistical and actuarial work, including in particular, the following letters:

– lower-case: – upper-case:

(a)(iii) Understand the meaning of the following commonly used conventions:

– round brackets used to denote negative currency amounts, – K and m used as abbreviations for “thousand” and “million” – used to denote the change in a quantity – “iff” used as an abbreviation for “if and only if”.



(a)(iv) Understand the concept of a mathematical proof and the meaning of “necessary”, “sufficient” and “necessary and sufficient” as they are used in mathematical derivations.

(a)(v) Prove a result using the method of mathematical induction. (b) Numerical methods (b)(i) Evaluate numerical expressions using an electronic calculator with the

following features: arithmetic functions ( ), powers ( y x ) and roots

( yx ), exponential ( ex ) and natural log ( ln x ) functions. (The following

features are also useful but not essential: factorial function ( n!), combinations

( n rC ), hyperbolic tangent function and its inverse ( tanh x and tanh1 x ),

fraction mode, at least one memory and an “undo” facility. Statistical and financial functions are not required.) Students should be able to make efficient use of memories and/or brackets.

(b)(ii) Estimate the numerical value of expressions without using a calculator and

apply reasonableness tests to check the result of a calculation. (b)(iii) Quote answers to a specified or appropriate number of decimal places or

significant figures (using the British convention for representing numbers), and be able to assess the likely accuracy of the result of a calculation that is based on rounded or approximated data values.

(b)(iv) Be able to carry out consistent calculations using a convenient multiple of a

standard unit (eg working in terms of £000s). (b)(v) Express answers, where appropriate, in the form of a percentage (%) or as an

amount per mil (‰). (b)(vi) Calculate the absolute change, the proportionate change or the percentage

change in a quantity (using the correct denominator and sign, where appropriate) and understand why changes in quantities that are naturally expressed as percentages, such as interest rates, are often specified in terms of “basis points”.

(b)(vii) Calculate the absolute error, the proportionate error or the percentage error in

comparisons involving “actual” versus “expected” values or approximate versus accurate values (using the correct denominator and sign, where appropriate).

FAC: Study Guide


(b)(viii)Determine the units of measurement (dimensions) of a quantity and understand the advantages of using dimensionless quantities in certain situations.

(b)(ix) Use linear interpolation to find an approximate value for a function or the

argument of a function when the value of the function is known at two neighbouring points.

(b)(x) Apply simple iterative methods, such as the bisection method or the Newton-

Raphson method, to solve non-linear equations. (b)(xi) Carry out simple calculations involving vectors, including the use of row/column

vectors and unit vectors, addition and subtraction of vectors, multiplication of a vector by a scalar, scalar multiplication (“dot product”) of two vectors, determining the magnitude and direction of a vector, finding the angle between two vectors and understanding the concept of orthogonality.

(b)(xii) Carry out calculations involving matrices, including transposition of a matrix,

addition and subtraction of matrices, multiplication of a matrix by a scalar, multiplication of two appropriately sized matrices, calculating the determinant of a matrix, calculating and understanding the geometrical interpretation of eigenvectors and eigenvalues, finding the inverse of a 2 2 matrix and using matrices to solve systems of simultaneous linear equations.

(c) Mathematical Constants and Standard Functions (c)(i) Be familiar with the mathematical constants and e . (c)(ii) Understand and apply the definitions and basic properties of the functions

x n (where n may be negative or fractional), c x (where c is a positive constant),

exp( )x [ ex ], and ln x [ loge x or log x ].

(c)(iii) Sketch graphs of simple functions involving the basic functions in (c)(ii) by

identifying key points, identifying and classifying turning points, considering the sign and gradient, and analysing the behaviour near 0, 1, or other critical

values. (c)(iv) Simplify and evaluate expressions involving the functions x (absolute value),

[ ]x (integer part), max( ) and min( ) , and understand the concept of a

bounded function. [The notation ( )x 100 will also be used as an

abbreviation for max( , )x 100 0 .]



(c)(v) Simplify and evaluate expressions involving the factorial function n! for non-negative integer values of the argument and the gamma function ( )x for

positive integer and half-integer values of the argument. (c)(vi) Understand the concept of a complex number and be able to simplify

expressions involving i 1 , including calculating the complex conjugate. (c)(vii) Calculate the modulus and argument of a complex number, represent a complex

number on an Argand diagram or in polar form ( z rei ).

(c)(viii) Apply Euler’s formula e ii cos sin and use the basic properties of the sine and cosine functions to simplify expressions involving complex numbers, including determining the real and imaginary parts of an expression.

(c)(ix) Understand the correspondence between the factors of a polynomial expression

and the roots of a polynomial equation and appreciate that a polynomial equation of degree n with real coefficients will, in general, have n roots consisting of conjugate pairs and/or real values.

(d) Algebra (d)(i) Manipulate algebraic expressions involving powers, logs, polynomials and

fractions. (d)(ii) Solve simple equations, including simultaneous equations (not necessarily

linear) by rearrangement, substitution, cancellation, expansion and factorisation.

(d)(iii) Solve an equation that can be expressed as a quadratic equation (with real

roots) by factorisation, by “completing the square” or by applying the quadratic formula, and identify which of the roots is appropriate in a particular context.

(d)(iv) Solve inequalities (“inequations”) in simple cases and understand the concept of

a “strict” or “weak” inequality. (d)(v) State and apply the arithmetic-geometric mean inequality, and know the

conditions under which equality holds. (d)(vi) Understand and apply the and notation for sums and products, including

sums over sets (eg i0

) and repeated sums.

FAC: Study Guide


(d)(vii) Calculate the sum of a series involving finite arithmetic or geometric progressions or infinite geometric progressions using the formulae:

AP a n d a ln n 2 22 1( ( ) ) ( ) or ,

GPa r

r

n

( )1

1 and

GPa

r1,

and be able to determine when an infinite geometric series converges.

(d)(viii) Apply the formulae: k n nk

n

1

12

1( ) and k n n nk

n2

1

16

1 2 1 ( )( ) .

(d)(ix) Solve simple first or second order difference equations (recurrence relations),

including applying boundary conditions, by inspection or by means of an auxiliary equation.

(d)(x) Recognise and apply the binomial expansion of expressions of the form ( )a b n

where n is a positive integer, and ( )1 x p for any real value of p and, in the

latter case, determine when the series converges. (e) Calculus (e)(i) Understand the concept of a limit (including limits taken from one side) and

evaluate limits in simple cases using standard mathematical notation, including the use of “order” notation O x( ) and o x( ) , and the sup/ lub and inf / glb

functions (considered as generalisations of max and min ). (e)(ii) Understand the meaning of a derivative as the rate of change of a function when

its argument is varied (in particular, for functions dependent on t , the time measured from a specified reference point), including the interpretation of a derivative as the gradient of a graph.

(e)(iii) Differentiate the standard functions x n , c x , ex and ln x . (e)(iv) Evaluate derivatives of sums, products (using the product rule), quotients (using

the quotient rule) and “functions of a function” (using the chain rule). (e)(v) Understand the concept of a higher-order (repeated) derivative and be familiar

with the mathematical notation used to denote such quantities.



(e)(vi) Use differentiation to find the maximum or minimum value of a function over a specified range (including the application of a monotone function, such as the natural log function, to simplify the calculation) and determine the nature of stationary points.

(e)(vii) Understand the meaning of a partial derivative and how to express a partial

derivative in standard mathematical notation, and be able to evaluate partial derivatives in simple cases. Find extrema of functions of two variables.

(e)(viii)Use the method of Lagrangian multipliers. (e)(ix) Understand the meaning of an indefinite integral as the anti-derivative of a

function and the meaning of a definite integral as the limit of a sum of infinitesimal elements, including the interpretation of a definite integral as the area under a graph.

(e)(x) Integrate the standard functions x n , c x and ex . (e)(xi) Evaluate indefinite and definite integrals by inspection, by identifying and

applying an appropriate substitution, by integration by parts, by using simple partial fractions or by a combination of these methods.

(e)(xii) Determine when a definite integral converges. (e)(xiii)Understand the meaning of a multiple integral and how to express a multiple

integral in standard mathematical notation, and be able to evaluate a double integral as a repeated integral in simple cases, including determining the correct limits of integration. Swap the order of integration.

(e)(xiv) Apply the trapezium rule to find the approximate value of an integral. (e)(xv) State and apply Taylor series and Maclaurin series in their simplest form,

including using these to determine the approximate change in a function when the argument is varied by a small amount. (Knowledge of the error terms is not required.)

(e)(xvi) Recognise and apply the Taylor series expansions for ex and ln( )1 x and, in

the latter case, determine when the series converges. (e)(xvii)Solve simple ordinary first-order differential equations, including applying

boundary conditions, by direct integration (which may involve a function of the dependent variable), by separation of variables or by applying an integrating factor.

FAC: Study Guide


(e)(xvii)Differentiate expressions involving definite integrals with respect to a parameter, including cases where the limits of integration are functions of the parameter.

(g) General (g)(i) Be familiar with the currency systems of the United Kingdom (pounds and pence

sterling), the United States (dollars and cents), the European monetary system (Euro and cent) and other major economies, and be able to interpret and write down currency amounts using these systems.

(g)(ii) Be familiar with the Gregorian calendar, including determining when a

specified year is a leap year, the concepts of calendar years, quarters and tax years, and the abbreviations commonly used to represent dates in the United Kingdom, Europe and the United States.

(g)(iii) Understand the distinction between “expression”/“equation”/“formula” and

“term”/“factor”. (g)(iv) Understand the meaning of the words “gross” and “net”. (g)(v) Be able to spell the following words correctly: actuarial, appropriately,

basically, benefit, benefiting, bias(s)ed, calendar, cancelled, commission, consensus, correlation, cyclically, deferred, definitely, formatted, fulfil, gauge, hierarchy, immediately, independence, instalment (British spelling), lose, loose, millennium, necessary, occasion, occurred/occurring, offered, orthogonal, paid, particularly, pensioner, precede, proceed, receive, referred, relief, seize, separate, similarly, specifically, supersede, targeted, theorem, until, yield.

(g)(vi) Be able to determine the correct member of word pairs according to context: eg

affect/effect, principal/principle, dependant/dependent. (g)(vii) Be able to distinguish between the singular and plural forms of words of Latin

or Greek origin, including the following: criterion/criteria, formula/formulae, analysis/analyses. [The word “data” may be treated as singular or plural, according to the preferences of individual authors/speakers.]

(g)(viii)Be familiar with commonly used Latin expressions and abbreviations such as

“per annum”, “vice versa”, “status quo”, “pro rata”, “ie”, “eg”, “cf”, “sic” and “stet”.



1.2 The Profession’s Copyright

All of the course material is copyright. The copyright belongs to Institute and Faculty Education Ltd, a subsidiary of the Institute and Faculty of Actuaries. The material is sold to you for your own exclusive use. You may not hire out, lend, give, sell, transmit electronically, store electronically or photocopy any part of it. You must take care of your material to ensure it is not used or copied by anyone at any time. Legal action will be taken if these terms are infringed. In addition, we may seek to take disciplinary action through the profession or through your employer. These conditions remain in force after you have finished using the course.

FAC: Study Guide


2 The FAC course

2.1 Course Notes

The Foundation ActEd Course consists of a set of eight chapters of notes covering the following ideas: Chapter 1 Notation Chapter 2 Numerical Methods I Chapter 3 Mathematical constants and standard functions Chapter 4 Algebra Chapter 5 Numerical Methods II Chapter 6 Differentiation Chapter 7 Integration Chapter 8 Vectors and matrices We recommend that you work through the sections that you are unsure of, completing the questions that are given. If you need further practice, there is a Question and Answer Bank and a Summary Test. These both cover material from all the chapters. Section 5 of this study guide gives you an index of the topics that are covered in order to identify quickly which chapters you need to look at. When you are working through the Core Technical subjects you can continue to use this course as a reference document if you come across areas of mathematics that you are still unhappy about.



For further guidance, this grid shows which chapters from FAC are needed for Subjects CT1, CT3, CT4, CT5, CT6 and CT8: Chapter Section Course Notes

1 1-7 All This is just general information and notation. 2 1-4 All 3 1-3 All 4 1-9 All 5 1-6 All 5 7 CT6 Syllabus items (c)(vi), (c)(vii), (c)(ix) 5 8 CT6 Syllabus item (d)(ix) 6 1-8 All 6 9 CT8 Syllabus item e(viii) 7 1, 2 All 7 3.1-3.4 All 7 3.5 CT4, CT8 Syllabus item (e)(xviii) 7 4-7 All 7 8.1-8.3 CT4, CT8 Syllabus item (e)(xvii) 8 1, 2.1-2.4 CT4, CT6, CT8 Syllabus items (b)(xi), (b)(xii) 8 2.5 CT4 Syllabus item (b)(xii) (part)

2.2 Online Classroom

The Online Classroom is available to provide tuition on the material covered in FAC. It is a comprehensive, easily-searched collection of recorded tutorial units covering the same topics as the FAC Course Notes. These tutorial units are a mix of:

teaching, covering the relevant theory to help you get to grips with the course material, and

worked questions or examples, illustrating the various techniques you should be familiar with.

To find out more about the Online Classroom, and to watch example tutorial units, please visit the ActEd website at www.ActEd.co.uk.

FAC: Study Guide


3 Study skills

All the mathematical techniques covered in this course will be used in the context of one of the Core Technical subjects and will therefore not be directly tested on their own. For example, in the Core Technical exams you will not be set a question saying “Integrate this function by parts” but you will be asked to work out expectations in Subject CT3 which may involve integration by parts. It is therefore essential that you feel really comfortable with each method. You should study this course actively. In particular we recommend the following: 1. Annotate your Notes with your own ideas and questions. This will make your

study more active. 2. Attempt the questions in the Notes as you work through the course. Write down

your answer before you check against the solution. 3. Attempt the Question and Answer Bank and the Summary Test on a similar

basis, ie write down or work out your answer before looking at the solution provided.



4 Contacts on the course material

Queries From time to time, you may come across something in the study material that is not clear to you. If you cannot resolve the query through discussion with friends and colleagues, then you can use one of ActEd’s discussion forums: If you have access to the FAC Online Classroom, then you can post your query in

the forums within the Online Classroom itself.

Alternatively, you can post your query on our forum at www.ActEd.co.uk/forums (or use the link from our homepage at www.ActEd.co.uk). This forum includes a section for each actuarial exam – there’s one for “FAC and StatsPack”.

Our forums are dedicated to actuarial students so that you can get help from fellow students on any aspect of your studies from technical issues to general study advice. ActEd tutors monitor the forums to answer queries and ensure that you are not being led astray. If you are still stuck, then you can send queries by email to [email protected] (but we recommend you try a forum first). We will endeavour to contact you as soon as possible after receiving your query, but you should be aware that it may take some time to reply to queries, particularly when tutors are away from the office running tutorials. At the busiest teaching times of year, it may take us more than a week to get back to you. Corrections and feedback We are always happy to receive feedback from students, particularly concerning any errors, contradictions or unclear statements in the course. If you find an error, or have any comments on this course, please email them to [email protected].

FAC: Study Guide


Further reading If you feel that you would find it useful to obtain a different viewpoint on a particular topic, or to have access to more information and further examples, then the best place to look would be a mathematics textbook. The level of mathematics covered in FAC is broadly similar to that covered by those examinations taken immediately prior to going to university (A-Level or Higher exams in the UK). You may still have your old textbooks, or know which ones you used and be able to track them down. If not, one set of textbooks published to help students prepare for A-Level exams is: Edexcel AS and A Level Modular Mathematics – Core Mathematics 1, 2, 3 & 4 These are available from internet retailers, including www.amazon.co.uk. Each textbook covers different topics, so you can choose which would be most suitable for you.



5 Course index

Topic Chapter Page Absolute change 5 5 Arithmetic-geometric inequality 4 20 Arithmetic progressions 4 24 Binomial expansion 4 35 Calculator, use of 2 4 Complex numbers 5 17 Convergence 7 16 Curve sketching 6 25 Determinants 8 12 Difference equations 5 24 Differential equations 7 31 Differentiating an integral 7 14 Differentiation, products and quotients 6 12 Differentiation, standard functions 6 11 Dimensions 5 9 Double integrals 7 18 Eigenvectors and eigenvalues 8 21 Errors 5 7 Estimation 2 9 Extrema 6 29 Factorial notation 3 12 Fractions, algebraic 4 5 Functions and graphs 3 2 Gamma function 3 13 Geometric progressions 4 26 Greek symbols 1 5 Indices 4 2 Induction 1 10 Inequalities 4 16 Infimum 6 7 Integer part 3 9 Integration, by parts 7 12 Integration, partial fractions 7 9 Integration, standard functions 7 5

FAC: Study Guide


Integration, substitution 7 11 Interpolation 5 12 Iteration 5 15 Lagrangian multipliers 6 30 Leibniz’s formula 7 14 Limits 6 2 Logarithms 4 2 Maclaurin series 7 30 Mathematical notation 1 2 Matrices 8 8 Max and min notation 3 10 Modulus 3 8 Newton Raphson iteration 5 16 Order notation 6 3 Partial differentiation 6 27 Percentages 5 2 notation 4 21 Proof 1 7 Proportionate change 5 5 Quadratic equations 4 7 Rounding 2 2 Scalar product 8 5 Series 4 31 notation 4 21 Simultaneous equations 4 11 Stationary points 6 20 Supremum 6 7 Taylor series 7 27 Trapezium rule 7 25 Vectors 8 2

FAC-01: Notation Page 1


Chapter 1

Notation

You need to study this chapter to cover: ● standard mathematical notation and terminology

● the letters of the Greek alphabet

● conventions commonly used in financial and actuarial mathematics

● mathematical proof

● mathematical induction

● currencies, dates and ages.

0 Introduction

This chapter deals with the notation and terminology that you must be familiar with in order to study the actuarial exams. Much of this may be familiar to you already, in which case read the chapter quickly or use it as a reference guide.

FAC-01: Notation


1 Mathematical notation

Symbol Meaning Further explanation/examples Types of numbers: Integers (whole numbers) {...,–3, –2, –1, 0, 1, 2, 3,...} Natural numbers {1, 2, 3, ...} (counting numbers)

Rational numbers 32 , 6

51.2 = , 14990.141414... = etc (all can

(fractions) be written as pq )

Real numbers rational numbers plus irrational numbers

(such as 2 , p and e) ie no imaginary component

Complex numbers can be written in the form a ib+ , where

1i = - Logic:

" For all (values) 2 x x " Œ $ Œ

: Such that (see next example) $ There exists : 1 5x x$ Œ + =

/$ There doesn’t exist 2 : 4x x/$ Œ = -

fi Implies 22 4x x= - fi =

¤ Implies and is implied by 32 8x x= ¤ = iff If and only if equivalent to ¤ Set Theory:

{ }1,2,3,... A set

etc ( ,0]-• A set containing numbers from -• to 0, not

including -• (since the bracket is round next to -• ), but including 0 (since the bracket is square next to 0)

{ } or ∆ Empty set the set of odd numbers divisible by 2

Œ Is a member of 2½ Œ A B» Union of two sets A or B or both, ie the things that are in one

or other or both



A B« Intersection of two sets A and B, ie only the things that are in both

A Complement of a set not A, ie the things that are not in A Miscellaneous: p Pi 3.14159...(the ratio of the circumference of

a circle to its diameter) e base of the natural logarithm 2.7182818... • Infinity Æ Tends to approaches eg x Æ• means that x is

becoming very large

Note: ● A superscript “+” or “ - ” on symbols such as refers to the positive or

negative numbers within the group ie + means 1, 2, 3,… (excluding zero), ie the same as .

● When a superscript “+” or “- ” is used in situations such as 1x +Æ , it means that x is approaching 1 “from above”, in other words x is taking values slightly bigger than 1.

● is the set of complex numbers. These can be written in the form a ib+ ,

where ,a b Œ and 2 1i = - , a being called the real part and b being called the

imaginary part. You may have seen j used for 1- rather than i.

FAC-01: Notation


Example (i) Interpret the statement:

{ }: 4x x+Œ <

(ii) For the function 1( ) xf x = , describe what happens as x tends towards 0 and ±• .

Solution (i) This is the set of whole numbers which are less than 4 but are positive ie the set

{ }1,2,3=

(ii) ( )f x is not defined for 0x = , ( ) as 0f x x +Æ • Æ , ( ) as 0f x x -Æ -• Æ ,

( ) 0 as f x xÆ Æ ±• .

Question 1.1

(i) If { } { } and P prime numbers Q even numbers= = what is M P Q= « ?

(ii) What values are in the set 2{ : 10}x xŒ < ?



2 Greek symbols

You need to know the following Greek letters:

Letter Lower case

Used for Upper case Used for

alpha a parameter

beta b parameter B beta function

gamma g parameter G gamma function

delta d small change D difference

epsilon e small quantity

theta q parameter Q number of deaths

kappa k parameter

lambda l parameter

mu m mean and force of mortality

nu n force of mortality

when sick

pi p 3.14159= P product

rho r correlation coefficient and force of recovery

sigma s standard deviation and

force of sickness S sum

tau t parameter (pronounced

as in “torn”)

phi f probability density

function of standard normal distribution

F cumulative distribution

function of standard normal distribution

chi c

2c distribution

(pronounced as first syllable of “Cairo”)

psi y probability of ruin

omega w limiting age in a life

table

FAC-01: Notation


3 Conventions

There are many conventions and short-hand notations that are used in mathematical and financial work. The following are used in the course notes: ● Abbreviations are used for thousands and millions to save writing many zeros.

For example, £9K means £9,000 (the “K” comes from the “kilo” prefix seen in words such as kilometre, meaning 1,000 metres) and $6.2m or $6.2M means $6,200,000. This is used in preference to using “standard form”, where

$6,200,000 would be written as 6$6.2 10¥ .

● D is used to denote a change in a quantity, for example D profit = £534K means that the profit has risen by £534,000.

● If interest rates were 6% in January, 8% in February and 347 % in March, this

would often be described as an increase of 2 percentage points, followed by a reduction of 25 basis points (one basis point being one-hundredth of a percentage point). Basis points are sometimes abbreviated to bps.

● In accounting, negative amounts of money are represented by placing them in brackets eg 5 (5)- = .

Example Here is an example of a very simple income statement for a company (sometimes called a profit and loss account), showing the negative cashflows in brackets. Don’t worry if you don’t understand what the individual items represent. £ Pre-tax profit 9.6m Tax (2.4m)

Net profit 7.2m Dividends (1.7m)

Retained profit 5.5m

Similarly you will see things like “calculate the profit (loss) made last year”, which means calculating the income minus the outgo and writing it as a positive number for a profit, or in brackets if it is negative, ie a loss.



4 Proof

To prove that something is true in mathematics, it is not sufficient to show that something works for a particular case if you are trying to prove it in general. However you can prove that something is not true in general by showing that it doesn’t work for a particular case (this is called a counterexample). You need to be familiar with the terms sufficient, necessary and necessary and sufficient. If A is necessary for B, then B Afi (ie B implies A or B is true only if A is true). If A is sufficient for B, then A Bfi (ie A implies B or B is true if A is true). If A is necessary and sufficient for B, then A B¤ (ie A implies and is implied by B or A and B are equivalent statements or A is true if and only if B is true).

Example In a group of 50 people, there are 25 men, 11 people with beards (who are all male!) and 25 people who like football who are all men too. If we use the notation M for “is a man”, B for “has a beard” and F for “likes football”, then we have B Mfi and M F¤ . The first implication is true since if we know someone has a beard, they must be male. The second implication is true since if we know someone is male, they automatically like football and vice versa.

Example (i) If A is the statement “the integer x ends in a 5”, and B is the statement “the

number x is divisible by 5”, then A is sufficient for B, but not necessary (since a number ending in a 0 is also divisible by 5). So we can write A Bfi but not B Afi .

(ii) If x is a solution of the equation 2 0ax bx c+ + = , then if P is the statement

“ 2 4 0b ac- ≥ ” and Q is the statement “x is a real number”, then P is a necessary and sufficient condition for Q so that P Q¤ . (More about quadratic equations

later!)

FAC-01: Notation


5 Expressions, equations and formulae

While you’re studying for the Core Technical exams you’ll use a lot of expressions, equations and formulas.

5.1 Expressions

A mathematical expression is any combination of mathematical symbols that can be evaluated to give an “answer”. Usually the expression involves more than one symbol, often it includes some letters ( , ,x y z etc), and usually the answer is numerical (but not

necessarily). For example, the following are all mathematical expressions:

2 2+ , 251.09 , 1

nt

t

v=Â ,

5(1 ) 1i

d+ -

, (½)G

In Subjects CT1 and CT5 you’ll meet some of the special symbols used in actuarial calculations, and your answers to assignment questions and exam questions will involve

expressions containing actuarial symbols such as |20a and 1|[30]:25A .

If a question says “Find an expression for …”, you should simplify your final expression as much as possible.

5.2 Equations

An equation is just a statement that two expressions are equal. For example, the following are all equations:

2 2 4+ = , 251.09 8.6231= , |1

nt

nt

v a=

=Â , 5

|5(1 ) 1i

sd

+ - = , (½) pG =

The issue is slightly confused by the fact that a lot of word processing packages use the word “equation” to refer to anything that contains mathematical symbols. A word processing “equation” may really be an expression, a formula, an inequality(!) or just

nonsense (eg * p ∆"≈ ).



5.3 Formulas (or formulae)

A formula is just an expression that can be used specifically for calculating the answer to a particular type of problem. A formula may be stated in the form of an expression or as an equation, eg:

The quadratic formula is 2 4

2

b b ac

a

- ± -

The formula for calculating the present value of a unit annuity-certain payable annually

in arrears is |1 n

nv

ai

-= . You will meet this in Subject CT1.

A formula usually involves standard letters for the variables (eg you know that the

, ,a b c in the quadratic formula are the coefficients of 2x , x and the constant term).

Also the letters often stand for the quantities involved, as in F ma= (force = mass ¥ acceleration).

5.4 Terms and factors

A term is an element in an expression that is added or subtracted. A factor is an element in an expression that is multiplied or divided.

FAC-01: Notation


6 Induction

One method of proving a general mathematical result is the method of (mathematical) induction. To prove that a result is true for all positive integers, we prove that if the result is true for any particular integer k then it must also be true for the next integer

1k + . If we can also show that it is true when 1k = , then it must be true for all positive integers 1,2,k = .

Example

Prove by induction that 121 2 3 ( 1)n n n+ + + + = + .

Solution Assume the result is true for n k= , ie:

121 2 3 ( 1)k k k+ + +◊◊◊ + = +

Adding the next term on to both sides it follows that:

[ ]

12

12

12

1 2 3 ( 1) ( 1) ( 1)

( 1)( 2)

( 1) ( 1) 1

k k k k k

k k

k k

+ + + + + + = + + +

= + +

= + + +

Since this is what the original equation “predicts” when 1n k= + , we have shown that if the result is true for n k= then it is also true for 1n k= + . Consider whether the result is true when 1n = :

LHS = 1 RHS = 1 So the result is true for 1n = and by the above result it must also be true for

2,3,4,n = ie for all positive integer values of n.

Question 1.2

Prove by induction that 2 2 2 2 161 2 3 ( 1)(2 1)n n n n+ + + + = + + .



7 Some general knowledge

Currencies

Country Main currency unit Sub-division

United Kingdom Pound (£) Pence (£1 = 100p)

United States Dollar ($) Cent ($1 = 100¢) European Union Euro (€) Cent (€1 = 100¢)

Japan Yen (¥) –

Question 1.3

Today’s exchange rates are shown as: €/£ = 0.61 and £/$ = 1.44 How much would 1,000 Euro be worth in US dollars?

FAC-01: Notation


Dates Gregorian calendar The calendar system used in Western Europe and America is officially called the Gregorian calendar (named after Pope Gregory XIII who introduced it). This system is recognised and understood worldwide, although a number of countries in other parts of the world have alternative calendar systems that they use as well. Leap years Calendar years usually have 365 days but, in order to prevent the seasons gradually drifting, an extra “leap” day is added at the end of February in some years. These leap years have 366 days, the extra day being 29 February. The general rule for determining whether a particular calendar year is a leap year is as follows:

LEAP YEAR OR NOT? A calendar year IS NOT a leap year …

… unless it divides exactly by 4, in which case it IS a leap year … … unless it also divides exactly by 100, in which case it IS NOT a leap year … … unless it also divides exactly by 400, in which case it IS a leap year!

Question 1.4

In actuarial calculations involving weekly payments it is often assumed that there are 52.18 weeks in an “average” year. Where does this figure comes from?

In a lot of actuarial applications the exact number of days in each month makes very little difference to the numerical answers. In these cases you can assume that the

months are of equal length ie each month is exactly 112 of a year long. This simplifies

the calculations considerably.



Calendar years, quarters and tax years Many organisations divide each calendar year into four quarters for budgeting and accounting purposes. For example, the calendar year 2012 would be broken into the four quarters: 2012 Q1: 1 January 2012 – 31 March 2012 2012 Q2: 1 April 2012 – 30 June 2012 2012 Q3: 1 July 2012 – 30 September 2012 2012 Q4: 1 October 2012 – 31 December 2012 In actuarial calculations where payments are made quarterly it is normally sufficiently

accurate to assume that each quarter is exactly 14 of a year long.

In the UK the amount of tax payable by individuals is calculated based on the transactions during each tax year (sometimes also referred to as a “fiscal” year), which run from 6 April to 5 April. So, for example, the 2011/12 tax year is the period from 6 April 2011 to 5 April 2012 (both days inclusive). The actual dates will differ between countries, for example the New Zealand tax year runs from 1 April to 31 March. Fencepost errors

Question 1.5

If you need to erect a fence 10 metres long in an open field using 1 metre-long strips of wood, how many posts will you need to support it?

If you got the answer wrong, you’ll see that it’s very easy to make these fencepost errors. It’s particularly easy to make a mistake in calculations involving dates. Almost everyone gets one wrong at some point.

FAC-01: Notation


Question 1.6

(i) Five payments are made at 9-month intervals with the first payment on 1 January 2012. On what date will the last payment be made?

(ii) A man was born on 9 September 1960. In New Zealand, how many complete

tax years are there between 1 May 1998 and his 60th birthday? (iii) How long is the period from 1 March 2005 to 28 February 2015?

Conventions for writing dates To save time, dates are often written in numbers, rather than in words. So make sure you know the numbers of the months (eg October = 10, November = 11). Also, just to make life difficult, there are two different conventions in use. In the UK and Europe we use the DD/MM/YY order, whereas Americans use MM/DD/YY. This can cause a lot of confusion since 01/11/12 would mean 1 November 2012 in the UK, but 11 January 2012 in the US. (The reason for this discrepancy is that in the UK we tend to say “the first of November”, whereas in the US they tend to say “November one”.) To decide which convention is being used, look which position contains numbers greater than 12. This must be the days bit. In actuarial symbols a fixed period of time is represented by using a right-angle symbol,

so that “5 years”, for example, is usually represented by 5 . Some of your actuarial colleagues may use this as a shorthand notation. For example, they might write: “The

pension incorporates a |5 guarantee” or they might even use 312 as an abbreviation for

“3 months”.



Ages In life insurance work and pensions work, you’ll often have to work out people’s ages. This might sound easy, but there are actually three different ways commonly used to express ages: Age last birthday: This is the age one of your friends would tell you if you asked them how old they were. It’s just the number of candles they had on their last birthday cake. Age nearest birthday: This is the person’s age at their nearest birthday (which could be either the previous one or the following one). Pension fund calculations usually use this definition because, for a large group of people, age nearest birthday usually averages out at the true age. You’ll get some people who are slightly older and some who are slightly younger, and these will normally balance out. However, age last birthday will always understate the true age. Age next birthday: This is the person’s age at their next birthday. This definition is the one usually used by insurance companies. This will always overstate the age. These age definitions are often abbreviated to “age last”, “age nearest” and “age next”. You’ll use these ways of defining ages in Subject CT4. You may hear people in life offices referring to their policyholders as “a female aged 50 next” (say). Some people are born on one of the “leap days” eg 29 February 2012. For calculation purposes they are normally treated as if they were born on the following day ie 1 March 2012 in this example.

Question 1.7

A man was born on 6 May 1954. What will his age be on 1 January 2015 using each of these three age definitions?

FAC-01: Notation


Chapter 1 Solutions Solution 1.1

(i) {2}M = . (Don’t forget that a prime number is one that has no factors other than

1 and itself!)

(ii) { }10 10x- < <

Solution 1.2

Assume the result is true for n k= , ie:

2 2 2 2 161 2 3 ( 1)(2 1)k k k k+ + + + = + +

Adding the next term on to both sides:

( )

2 2 2 2 2 216

16

216

216

16

16

1 2 3 ( 1) ( 1)(2 1) ( 1)

( 1)[ (2 1) 6( 1)]

( 1)(2 6 6)

( 1)(2 7 6)

( 1)( 2) 2 3

( 1)[( 1) 1][2( 1) 1]

k k k k k k

k k k k

k k k k

k k k

k k k

k k k

+ + + + + + = + + + +

= + + + +

= + + + +

= + + +

= + + +

= + + + + +

So we have shown that if the result is true for n k= then it is also true for 1n k= + . Consider when 1n = : LHS = 1 RHS = 1 So the result is true for 1n = and by mathematical induction it is also true for n = 2, 3, 4, … ie for all positive integer values of n.



Solution 1.3

1,000 Euro must be equivalent to 1,000 0.61 £610¥ = .

£610 must be equivalent to 610 1.44 $878.40¥ = . The precise rules that banks have to use for converting currencies when Euro are involved are actually quite complicated eg you have to work to 6 decimal places. Solution 1.4

It’s 14365 7∏ (rounded to 2 DP).

Solution 1.5

11 The “obvious” answer was to divide 10 by 1 and say 10. But, because you need a post at each end of the fence, you actually need an extra one. If you said 10, you’ve made a “fencepost error”. To avoid these, you need to pay careful attention to which, if either, of the endpoints is included. This problem comes up when you are trying to work out the number of payments in a stream of payments. Solution 1.6

(i) 1 January 2015 (There are 4 gaps of 9 months between these 5 payments. This makes a total period of 36 months, which equals 3 years.)

(ii) 21 (The period from 1 April 1999 to 31 March 2020 consists of 21 complete tax

years ie 1999/2000, 2000/01, … , 2019/20.) (iii) 10 years (When you’re dealing with a period of time, it’s a straight subtraction.) Solution 1.7

Age last = 60 (because his last birthday would be 6 May 2014 and 2014 1954 60- = ). Age next = 61 (add 1 to his age last). Age nearest = 61 (because his nearest birthday would be 6 May 2015).




Unless prior authority is granted by ActEd, you may not hire out, lend, give out, sell, store or transmit electronically or

photocopy any part of the study material.





FAC-02: Numerical methods I Page 1


Chapter 2

Numerical methods I

You need to study this chapter to cover: ● accurately rounding numbers

● the use of an electronic calculator

● estimation

● abbreviations.

0 Introduction

This chapter reminds you of the conventions about rounding and accuracy. Since the Core Technical subjects rely heavily on numerical accuracy, rounding is more important in your actuarial studies than perhaps it has been during your university studies.

FAC-02: Numerical methods I


1 Rounding

Numerical answers should be rounded to a “suitable” number of decimal places (DP) or significant figures (SF). Marks can be lost in examinations if a final answer is not rounded to the accuracy that the question requires. Use common sense when deciding what “suitable” means. For example, when giving an answer in monetary terms, round to two decimal places. If a question is asking for an interest rate it is unlikely that you will need to work to more than three decimal places eg 6.125%. Do be aware though that we are talking about final answers here: if you are going on to use the interest rate later you will need an accurate figure to work with. More of that later!

Example Round the following numbers to two decimal places: 3.784, 15.239, 6.028, 6, 2002,

0.399- . Solution The answers are 3.78, 15.24, 6.03, 6.00, 2002.00, –0.40. Notice that all answers have two figures (digits) after the decimal point.

Example Round these numbers to two significant (ie non-zero) figures: 3.784, 15.239, 6.028, 6, 2002, 0.399- . Solution The answers are 3.8, 15, 6.0, 6.0, 2000, –0.40. Notice that the first “significant figure” is the first non-zero figure.



Question 2.1

Round these numbers to three decimal places:

0.0678, 15.3489, 9.9999

Question 2.2

Round these numbers to three significant figures:

14.3678, 5.9879, 0.08006

Notes: ● When a number is written as 6.00, it implies that the author believes it is correct

to 2DP ie the exact value lies somewhere in the range [5.995,6.005) , using the

convention that a digit greater than or equal to 5 causes a number to be rounded up.

● In some countries (continental Europe in particular) commas and full stops in numbers are used with the opposite meaning from in the UK and the US ie the decimal point is written as a comma and full stops (or spaces) are used to separate a large number. For example 3,142p = and the population of the UK

is approximately 55.000.000. You should use the UK notation in the exam.

● Do not use accuracy that is not valid, for example quoting the price of something as £2.78643 is not appropriate.



2 Use of a calculator

For a description of the functions you will need to have on your calculator, refer to the syllabus objectives for this course. In the exams you are not allowed graphical calculators. You can get guidance from the Profession as to which types are permissible. You must be familiar with the calculator that you are going to use in the exam. Since there are so many calculators on the market, this section is not going to be about how to use your particular calculator but instead it will enable you to practise getting the correct answer with your calculator. For technical help see your manual (if you can find it!). It is really important for you to try each example below for yourself to get used to working with your calculator. Most numerical answers have been rounded to three decimal places. Always ensure that you copy numbers correctly when you are half way through a calculation or reading numbers from actuarial tables.

Example

Calculate 1.4 3.66.23 5.8+ . Solution

Using the power key ( or y xx y ), we get 573.1474.

Example

Calculate ( )5.141 1.723+ .

Solution

Using the root key ( x or 1

yx ) and the power key, we get 49.091.

If you don’t have this key, you can treat the expression as 14 5.1(1 1.723 )+ . Since

1xx y y= you can use your or y xx y key.



Example

Calculate 2

21

3

.

Solution

The answer is 7

29

or 2.778. If your calculator has a fractions facility, be aware of the

order in which your calculator performs fractions!

Example

Calculate ( )247 ln 5.2+ .

Solution The answer is 4.929. Your calculator will have loge or ln for the natural logarithm

key. (We will deal with this function later in the course.) Most calculators have a

fraction key ( bca ) which can be helpful. Notice also that most calculators have a

squared key ( 2x ) to avoid you having to use the power key.

Example

Calculate 3.1243 e .

Solution

The answer is 1.961. On many calculators, xe and ln are on the same key. On recent

calculators, to get 3.1e , you would need to type 3.1xe (followed by = or Ans), whereas on older models you would need to type these in reverse. This is the same as for the square root key.



Example Calculate 4! This is read as “four factorial”. Solution There should be a factorial key ( !n ) on your calculator. The answer is 24. Notice that !n means ( 1) ( 2) 1n n n¥ - ¥ - ¥◊◊◊¥ . So here 4! 4 3 2 1= ¥ ¥ ¥ .

Example

Calculate tanh 2 and 1tanh 0.4- . Solution These are hyperbolic tangents and inverse hyperbolic tangents respectively, and are needed for the Fisher transformation in Subject CT3. If your calculator is a recent model then you will need to press hyp then tan before the number; otherwise it will be

the number then hyp then tan. The answers are 1tanh 2 0.964, tanh 0.4 0.424-= = .

Example

Calculate 215.2 3.74 4 1.68 2.49

2.7 19.86 3

- - ¥±+

.

Solution The answers are 2.493 and –1.477. Notice here that you can use brackets (or stack) for the first part of the expression and the memory function for the second, or alternatively the memory for both parts.



Question 2.3

Calculate 3.5 121 (1.04)

0.04 1.04

--.

Question 2.4

Calculate 1 1 0.1

ln2 1 0.1

.

Question 2.5

Calculate 2 4.5

3

2 (3.789 2.5)

5.5 2.1

+ +-

.

Question 2.6

Calculate 23 ( 3) 4 2 4

2 2

± - - ¥ ¥ -¥

.

Question 2.7

Calculate 10

10

117 1

100 1.035ln1.035(1.035)

.

Question 2.8

Calculate 10!

7!2!.



When substituting a value for a variable that occurs several times within an expression, it can be helpful to use a shortcut called “nested multiplication”. The following example shows what this shortcut involves.

Example

Calculate 2 33 7v v v+ + , for 1

0.92591.08

v

.

Solution We can do this as a nested multiplication:

2 3 23 7 (3 7 )

[1 (3 7 )]

v v v v v v

v v v

+ + = + +

= + +

Start by multiplying v by 7, then add 3, then multiply by v, then add 1, and finally multiply by v. Using this version we get an answer of 9.055.

Question 2.9

Calculate 2 3 42 5 6v v v v+ + + , when 1

0.90911.1

v

.

Question 2.10

For Question 2.9, work out the number of keystrokes you need to make on your calculator using nested multiplication and also for an alternative method that you can think of.



3 Estimation and accuracy

3.1 Estimation

We have looked at how to use calculators efficiently, but it is always possible to make a mistake. So it is important to have a rough idea of what the numerical answer to a calculation is likely to be. To do this, you can round the numbers involved to a convenient figure and then carry out the calculation without using a calculator (or you might make the same mistake again!).

Example

Estimate the value of 2 42.7 3.1

5.2 7.8

.

Solution

2 42.7 3.1

5.2 7.8

is roughly 2 43 3 9 81

305 8 3

.

The actual answer is –38.3.

Question 2.11

Find estimates for the answers to Question 2.5 and Question 2.6. Compare your estimates with the actual answers.

Note: You need to be very accurate with intermediate values in estimates and rounding when you are doing any of the following: ● subtracting two numbers of a similar size

● dividing by a small number

● raising a number to a high power.



3.2 Accuracy

When numbers have been rounded, and are then subsequently used in a calculation, the final answer is affected by the rounding. Therefore, when answering questions, it is essential to realise how accurately you need to quote your final answer. For example, if all figures in the question are given to three significant figures, do not give your final answer to five significant figures.

Example

In the formula 101 (1 )i

pi

, the accurate value of i is 0.034724. Compare the

values of p obtained using: (i) the accurate value of i (ii) i rounded to three significant figures (iii) i rounded to one significant figure. Solution (i) 8.32819 (ii) 8.32920 (iii) 8.53020 Rounding to three significant figures gives a value of p that is still accurate to two significant figures. However rounding to one significant figure results in losing all accuracy.

Question 2.12

Using the same formula as in the example above (which is one that you will meet in Subject CT1) but with the accurate value of i being 0.04562378, how many significant figures can you round i to, in order to maintain three significant figures of accuracy in your final answer?



4 Convenient abbreviations

Rather than writing several zeros when dealing with large numbers, it is often more convenient to work in, say thousands or millions. It is common to write £000s to mean thousands of pounds, and £m to mean millions of pounds.

Example

If 8

9 (1 )R vT Pv

i

, where

1£14,000, £700, , 0.05

1P R v i

i

, what is T?

(Work in £000s). Solution Working in £000s, 14, 0.7P R . Then 13.549T ie £13,549.

Example

If 4,000A , calculate 2 2,000 1,000,000A A

(i) directly (ii) working in units of 1,000 Solution

(i) 25,000,000 5,000

(ii) In units of 1,000:

1,000,000 becomes 21 1 , 2,000 becomes 2 and A becomes 4 The calculation is then done as follows:

24 2 4 1 25 5 ie the “real” answer is 5,000.



This page has been left blank so that you can keep the chapter summaries together for revision purposes.



Chapter 2 Summary You need to round off your final answers to a sensible degree of accuracy, but intermediate figures should not be rounded if that will mean an inaccurate final answer. You should be able to use your calculator efficiently and correctly. Remember to make a rough estimate of what the answer should be, in order to pick up careless numerical errors.




The rounded numbers are:

0.068, 15.349, 10.000 Solution 2.2

The rounded numbers are:

14.4, 5.99, 0.0801 Solution 2.3

The numerical answer is 0.295729. Solution 2.4


The numerical answer is –153.614. Solution 2.6

The numerical answer is 2.351 or –0.851. Solution 2.7


The numerical answer is 360.



Solution 2.9

Using the nested multiplication method, we get:

2 3 4 2 3

2

2 5 6 2 (5 6 )

[2 (5 6 )]

(1 [2 (5 6 )])

v v v v v v v v

v v v v

v v v v

Now using 1

1.1v , we get the answer to be 10.42.

Solution 2.10

On the calculator I’m using, nested multiplication took 30 key presses. Using another method (just working from left to right) I took 43 key presses, or using the memory I took 25 key presses. Nested multiplication can be more efficient. Solution 2.11

For Question 2.5 one estimate is 2 4 2

3

2 (4 3) 2 19 2 400201

2 26 2

.

For Question 2.6 one estimate is 3 9 32 3 6

2.25 or 0.754 4

.

In each case you can see that the estimate is fairly close to the accurate answer.



Solution 2.12

Using the accurate value of i we get p to be 7.888505805. Rounding i to 6SF we get p to be 7.888505. Rounding i to 5SF we get p to be 7.888497. Rounding i to 4SF we get p to be 7.888652. Rounding i to 3SF we get p to be 7.889427. Rounding i to 2SF we get p to be 7.873956. So i can be rounded off to 3SF to keep three significant figures of accuracy. Notice that your first guess might have been 4SF, in which case the preceding work would not be necessary.

FAC-03: Mathematical constants and standard functions Page 1


Chapter 3

Mathematical constants and standard functions

You need to study this chapter to cover:

● the definitions and basic properties of the functions nx , xc , exp( )x , and ln x

● the functions x , max( ) and min( )

● the factorial function !n and the gamma function.

0 Introduction

This chapter covers some standard functions and notation that will be needed in the following chapters.

FAC-03: Mathematical constants and standard functions


1 Standard functions and graphs

1.1 Exponential function

The exponential function, xe , can be defined by:

lim 1n

x

n

xe

n

or the series expansion 2 3

12! 3!

x x xe x .

It is the inverse of the natural logarithmic function (which we will look at in Chapter 4),

so that ln xe x . Do remember that e is just a number which you can find from your calculators:

1 2.718...e

If the power is a long expression then a convenient alternative notation is exp( )xe x ,

for example 21

22

12

expx x

e

. This makes things clearer to read.

However don’t mix up these two notations by writing something like expx , as this is

meaningless.

The graph of xy e looks like this:

y=exp(x)

0

2

4

6

8

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

x



Notice that xe can never be negative. This is because you can never get a negative answer if you raise a positive number to any power. Notice also that:

● 0e is 1

● as x gets large, so does xe

● and as x gets large but negative, 0xe .

Since xe is just “a number to the power of x”, then this graph is also the basic shape of

the graph of xy c , where c is a positive constant greater than 1. When c is a positive

constant less than 1, its graph is a reflection in the y-axis of the graph of xy c , with c

replaced by 1c . The diagram below shows the graphs of 0.5xy and 2xy :

0.5xy 2xy

Again notice that both graphs have an intercept of 1, ie they cross the y-axis at 1, but you must look at the value of c carefully to judge when the graph tends towards infinity or zero.

0

1

2

3

4

5

-3 -2 -1 0 1 2 3

x



1.2 Log function

A logarithm is the inverse of a power. The graph of the natural logarithm function lny x is:

-6

-5

-4

-3

-2

-1

0

1

2

3

0 1 2 3 4 5 6 7

We will meet the idea of a base in the next chapter, but it is worth noting that this is the shape of the logarithmic graph whatever base is used. We will look at logarithms and bases in a later chapter. Notice that log1 0 , and that you can’t take the log of a negative number (or zero).

The limits are log as x x , and log as 0x x . These will be important

in statistical work in later subjects. You will see the notation log x and ln x used for

the natural logarithm function.



1.3 Powers of x

The graph of ny x , where n is an even positive integer has the same general shape as 4y x which is shown below.

-202468

1012141618

-2 -1 0 1 2

x

y

Notice that the values of y for these functions can never be less than zero.

The graph of ny x , where n is an odd positive integer has the same general shape as 3y x which is shown below:

-10-8-6-4-202468

10

-3 -2 -1 0 1 2 3

x

y

You may notice that out of the last two graphs, one was symmetrical about the y-axis and the other one wasn’t. In general a function ( )f x which has the property

( ) ( )f x f x is called an even function. If ( ) ( )f x f x then ( )f x is called an odd

function.



Remembering that 1nn

xx

, the graph of ny x will have a discontinuity (where the

function is not defined) at 0x . For example, look at the graph of 3y x :

-300

-200

-100

0

100

200

300

-2 -1 0 1 2

x

y

The graph of y x is (remembering that x represents the positive square root):

0

0.5

1

1.5

2

2.5

0 1 2 3 4 5

x

y

This is the inverse function of 2y x , so the graph of y x is the reflection of

2y x in the line y x . The same relationship applies to the graphs of log x and xe

which we have already looked at. Notice that this graph does not exist for negative values of x since a real value of the square root of a negative number cannot be found.



When you see an expression involving a power, such as zt , you need to think carefully whether:

● it is like nx , ie t is a variable and z is a fixed number, or

● it is like xc , ie t is a fixed number and z is a variable.

1.4 Transformations

These standard graphs ( )y f x= can be transformed as follows:

( )y f x d= + causes a vertical translation (ie slide) of d

( )y f x c= + causes a horizontal translation left of c

( )y af x= causes a vertical stretch by a factor of a

( )y f bx= causes a horizontal stretch by a factor of 1 b (ie the graph is

squashed by a factor of b ).

Note how constants outside the function affect the function vertically, whereas constants inside the function affect it horizontally. The following question illustrates these points.

Question 3.1

Sketch the graphs of the following functions:

(i) 1y x

(ii) 53y x

(iii) 4 1xy e

(iv) ln(1 )y x

(v) 4( 3)y x

Question 3.2

Describe what the graphs of 2 ( )y f x and (2 )y f x look like in relation to the graph

of ( )y f x .



2 Other functions

2.1 Modulus function

x is the absolute value of x or the modulus of x. In practice, if x is a number, it means

ignore any negative sign, for example, 4 | 4 | 4 . Note that | |x c is the same as

saying c x c because x must either be a positive number in the range [0, )c or a

negative number in the range ( ,0)c .

Example

Write the expression 2 1 3x without the use of the modulus sign.

Solution

2 1 3 3 2 1 3x x

For those of you that can remember dealing with inequalities, we will further simplify this expression. For those that can’t, we will pick up on this again in the next chapter. Simplifying the inequality:

3 2 1 3 2 2 4

1 2

x x

x

Question 3.3

Write the expression 3 2x without the modulus sign.



2.2 Integer part

x means the integer part of x, for example 2.89 is 2. This could be used, amongst

other contexts, to give the complete number of years that someone has lived. It is also used a lot in computer calculations.

When you divide m by n, where both are positive integers, you get m

n

as the quotient

with a remainder of m

m nn

.

When dealing with negative numbers you need to check whether x really means the

integer part, so that [ ] 3 , or whether it means the greatest integer not exceeding x,

so that [ ] 4 . The second definition is more common.

Example A baby boy was born on 26 November 1979. If x is defined to be his exact age, what is

x on 11 February 2012?

Solution

He is 32 on 26 November 2011, so x is 32.



2.3 Max and min

The notation max( ) or min( ) is used to denote the largest or smallest of a set of

values. If the quantities involved are variables, the value of these functions may depend on the ranges of values that you are considering.

Example What is max( 2,10)x for the region 0 20x ?

Solution We could write max( 2,10)x as:

10 if 0 8

2 if 8 20

x

x x

So max( 2,10) 10 for 0 8x x , and max( 2,10) 2 for 8 20x x x .

The abbreviation ( 100)x is often used for max( 100,0)x .

Question 3.4

What is 2min( ,15)x for 0 6x ?



Question 3.5

In the UK the calculation of an employee’s National Insurance contributions is based on their Upper Band Earnings (UBE), which is calculated from the formula: min( , ) min( , )UBE S UEL S LEL

In this formula, S is annual salary, and UEL and LEL are two published figures called the Upper and Lower Earnings Limits (which might for example be £30K and £4K). Describe in words what UBE represents and give an alternative mathematical formula in the form:

if

if

if

UBE



3 Factorial and gamma functions

3.1 Factorial notation

We have already looked at how to evaluate !n on a calculator. The definition of !n is as follows: ! ( 1) ( 2) 1n n n n

where n is a non-negative integer. The factorial function satisfies the relationship ! ( 1)!n n n . If we put 1n , this

tells us that 1! 1 0! 0! 1 .

Question 3.6

Evaluate 5!

Example

Simplify the expression !(2 1)!

(2 )!( 2)!

n n

n n

.

Solution This expression can be written as:

! (2 1)!

( 2)! (2 )!

n n

n n

The first factor equals ( 1)n n , because !n contains an extra n and 1n in its

expansion, which are not contained in the expansion of ( 2)!n , and the second factor is

1

2n.

So we get 12

!(2 1)! ( 1)( 1)

(2 )!( 2)! 2

n n n nn

n n n

.



Question 3.7

Simplify the expression (3 )!(2 1)!

(3 2)!(2 1)!

n n

n n

.

3.2 Gamma function

( )x , where 0x , is defined by the integral 10

x tt e dt . This is the gamma

function. This function is used in statistics in connection with the gamma distribution, and there are several properties of the function that you are going to need to know. In this chapter we will quote the results without proof. We will look at their proofs later in the course after we have covered the necessary integration techniques. Result 1: ( ) ( 1) ( 1)x x x where 1x

Result 2: ( ) ( 1)!n n where 1, 2, 3,...n

Result 3: 12

( )

These results can be found in the Tables page 5. The graph of the gamma function looks like this:

1 2 3 4

1

2

3

4

5

6

7



Example Evaluate (6.5) .

Solution Using Result 1:

(6.5) 5.5 (5.5)

By repeating this process:

(6.5) 5.5 4.5 3.5 2.5 1.5 0.5 (0.5)

162.42

287.89

Question 3.8

Evaluate (5) .

Question 3.9

Evaluate (4.5) .

Question 3.10

Simplify ( 1)(2 1)!

(2 )( 1)!

n n

n n

where n is a natural number.

Question 3.11

Show that, if n is a nonnegative whole number, 2

(2 )!( ½)

2 !n

nn

n .



Stirling’s approximation Working out a factorial by multiplying all the numbers together would be very tedious for large values of n . The following approximations (called Stirling’s approximation) are sometimes useful in derivations involving large values of n :

½! 2n nn n e

and ½( ) 2n nn n e

(Don’t worry that these appear to be inconsistent with the relationship ( ) ( 1)!n n

Remember that for very large values of n , the ratio 1n

n

is very close to 1.)



Chapter 3 Summary Standard functions

You should know the properties and be able to sketch the graphs of

, , , and logn x xx c e x .

Transforming functions

( )y af bx c d= - + causes the function ( )y f x= to be:

Stretched vertically by a factor a

Squashed horizontally by a factor b

Translated horizontally to the right by c

Translated vertically up by d Modulus

The modulus of x, written as x , is the absolute value of x ie negative signs are ignored.

Integer part

The integer part of x is written as x .

Max and min

max( ) or min( ) is used to denote the largest or smallest of a set of values.

Factorial

The definition of n factorial is ! ( 1) ( 2) 1n n n n . Note that 0! 1 .

The gamma function

The gamma function is defined by 10

( ) x tx t e dt , when 0x .

It has the following properties:

( ) ( 1) ( 1)x x xG = - G - where 1x >

( ) ( 1)!n nG = - where 1,2,3,n =

(½) pG = .




(i) The graph of 1y x is as follows:

-10

-5

0

5

10

-1.5 -1 -0.5 0 0.5 1 1.5

x

y

(ii) The graph of 53y x is as follows:

-100

-50

0

50

100

-2 -1 0 1 2

x

y



(iii) The graph of 4xe is like that of xe , only steeper. In fact the whole graph is squeezed horizontally to a quarter of the original width. Adding 1 then shifts everything up by 1 unit.

0123456789

-1.5 -1 -0.5 0 0.5

x

y

(iv) This is like the graph of ln x shifted 1 unit to the left

-2.5

-2

-1.5

-1

-0.50

0.5

1

1.5

2

-1 -0.5 0 0.5 1 1.5

x

y



(v) This is like the graph of 4x , but shifted 3 units to the right.

024

68

10121416

0 1 2 3 4 5

x

y

Solution 3.2

The graph of 2 ( )f x will have all vertical distances doubled.

The graph of (2 )f x will have all horizontal distances halved.

Solution 3.3

3 2x or 3 2x

If you wish to simplify these expressions further, proceed as follows:

22 3

3x x

or 2

2 33

x x

Here there are two separate ranges of values of x where the inequality holds.



Solution 3.4

2 2min( ,15)x x for 0 3.87x

2min( ,15) 15x for 3.87 6x

Solution 3.5

UBE is the portion of the employee’s salary that falls in the “band” ( , )LEL UEL , which

would be (£4,000,£30,000) using the illustrative figures given.

The graphs in the diagram represent the two min functions. UBE is the difference between them, which corresponds to the height of the gap between the upper and lower lines.

LEL UEL

LEL

UEL

The alternative mathematical formula is:

0 if

if

if

S LEL

UBE S LEL LEL S UEL

UEL LEL S UEL

Solution 3.6

5! 5 4 3 2 1 120



Solution 3.7

(3 )!(2 1)! (3 )! (2 1)! 1 2 (2 1)2 (2 1)

(3 2)!(2 1)! (3 2)! (2 1)! (3 1)(3 2) (3 1)(3 2)

n n n n n nn n

n n n n n n n n

Solution 3.8

(5) 4! 24

Solution 3.9

(4.5) 3.5 (3.5) 3.5 2.5 1.5 0.5 (0.5) 6.5625 11.63

Solution 3.10

( 1)(2 1)! ( 2)!(2 1)! 1

(2 )( 1)! (2 1)!( 1)! ( 1)( 1)

n n n n

n n n n n n n

or

3

1

n n



Solution 3.11

We can prove this using mathematical induction.

When 0n , the equation says 0

0!(½)

2 0! , which is correct.

If we assume it’s true for a typical value of n , say n k , then we know that:

2

(2 )!( ½)

2 !k

kk

k

This would imply that:

2

(2 )!( 1½) ( ½) ( ½) ( ½)

2 !k

kk k k k

k

The last expression can be written as:

2 2 2 1

(2 )! 2 1 (2 )! (2 1)!( ½)

22 ! 2 ! 2 !k k k

k k k kk

k k k

This doesn’t quite match the formula we were hoping for. But, if we include an extra

factor of 2 2

2( 1)

k

k

(which won’t affect the answer), we get:

2 1 2 2

2 2 (2 1)! (2 2)!

2( 1) 2 ! 2 ( 1)!k k

k k k

k k k

This now matches the formula given when 1n k . So, if it is true for n k , it’s also true for 1n k , and by the principle of mathematical induction it must be true for all values of n .

FAC-04: Algebra Page 1


Chapter 4

Algebra

You need to study this chapter to cover: ● manipulating algebraic expressions involving powers, logs, polynomials and

fractions

● solving quadratic equations

● solving simultaneous equations

● solving inequalities

● the arithmetic-geometric mean inequality

● using the S and P notation for sums and products

● arithmetic and geometric progressions and other series

● the binomial expansion of expressions of the form ( )na b+ where n is a positive

integer, and (1 ) px+ for any real value of p .

0 Introduction

This chapter reminds you of the algebraic results that you need to be able to handle very easily in order to cope with the Core Technical subjects.

FAC-04: Algebra


1 Algebraic expressions

1.1 Indices

There are three laws governing indices (or powers):

( )

a b a b

a b a b

a b ab

x x x

x x x

x x

+

-

¥ =

=

=

Question 4.1

Simplify the following:

(i) 3 55 2x x¥

(ii) 2 716 6y y

(iii) 3 2(5 )b

1.2 Logarithms

A logarithm is the inverse of a power:

21010 100 log 100 2= ¤ = (read as log base 10 of 100)

So if we want to find a log, we need to answer the question: “To what power must we raise the base in order to get the number whose log we are trying to find?” They have bases (the 10 here), the most commonly used one being e. Logarithms to these bases can be calculated directly from a scientific calculator, eg log 15 2.708e = .

An alternative abbreviation for log to base e is ln. On your calculator, the ln button gives you loge and the log button gives you 10log .



There are three laws of logarithms that can be derived from the laws of indices:

log log log

log log log

log log

a a a

a a a

na a

x y xy

xx y

y

x n x

+ =

- =

=

In words, “the log of a product is the sum of the logs”, and “you can bring down the power”.

Question 4.2

Starting from the laws of indices, prove the second of these three laws.

Notice that the base of the logarithm has been missed out in the following examples. Here we are looking at the properties of logs in general. So the logarithm used can be to any base, but if you need to use a calculator then you will have to use base 10 or base e.

Example (i) Simplify log 2 log 4 log5x x x+ - .

(ii) Write 101 2log x in the form log ( )f x .

Solution

(i) This simplifies to 2 4 8

log log5 5

x x x

x

¥ = .

(ii) 2 210 10 10log log 10 log 10x x .

Question 4.3

Simplify ln e.

FAC-04: Algebra


Logarithms are also used to solve equations where the unknown forms part of a power.

Example

Solve the equation 2 11.1 2n+ = . Solution Take logs:

2 1log1.1 log 2

(2 1)log1.1 log 2

log 2(2 1)

log1.1

1 log 21 3.136

2 log1.1

n

n

n

n

Question 4.4

Solve the equation 2 2 12 1.05 1.1025t t- -¥ = .

Note:

The value of log

logb

b

x

y is independent of the base b used.



1.3 Fractions

It is possible to manipulate algebraic fractions using the same rules that are used for numerical fractions. If you have forgotten how to factorise a quadratic you might want to look at the next section now!

Example Simplify the following:

(i) 2 3 3 2

1 6 1

x x

x x

+ +-+ -

(ii) 2

2

3 4

2 7 4

x x

x x

- -- -

(iii) 2 2

2 2

2 3 2

2 5 12 2 5 3

x x x x

x x x x

- - + +∏- - + +

Solution (i) Putting these over a common denominator:

2 2

2

2 3 3 2 (2 3)(6 1) (3 2)( 1)

1 6 1 ( 1)(6 1)

(12 16 3) (3 5 2)

( 1)(6 1)

9 11 5

( 1)(6 1)

x x x x x x

x x x x

x x x x

x x

x x

x x

+ + + - - + +- =+ - + -

+ - - + +=+ -

+ -=+ -

(ii) Factorising the numerator and denominator:

2

2

3 4 ( 1)( 4)

(2 1)( 4)2 7 4

1

2 1

x x x x

x xx x

x

x

- - + -=+ -- -

+=+

FAC-04: Algebra


(iii) Remembering that when you divide by a fraction you “invert and multiply”:

2 2

2 2

2 3 2 ( 2)( 1) ( 2)( 1)

( 4)(2 3) (2 3)( 1)2 5 12 2 5 3

( 2)( 1) (2 3)( 1)

( 4)(2 3) ( 2)( 1)

( 2)( 1)

( 4)( 2)

x x x x x x x x

x x x xx x x x

x x x x

x x x x

x x

x x

- - + + - + + +∏ = ∏- + + +- - + +

- + + += ¥- + + +

- +=- +

Example

Simplify

1 1

a ba b

b a

+

+.

Solution Multiplying the numerator and denominator of a fraction by the same quantity leaves the value of the fraction unaffected. We can pick a suitable quantity to simplify the fraction:

2 2

1 1ab b aa b

a b ab a bb a

+ +¥ =++

Question 4.5

Simplify the expression 2 2

2 2

3 4 2 5 7

3 10 8 4 17 4

x x x x

x x x x

+ - + -∏- - - +

.



2 Quadratic equations

2.1 Solution by factorisation

This is where you try to write the quadratic 2 0ax bx c+ + = as ( )( ) 0dx e fx g+ + =

(the process being called factorisation) and then the solutions would be or ged fx = - - .

Notice that by comparing the coefficients of 2,x x and the constant term, a df= ,

c eg= , and b ef dg= + .

Example

Solve the equation 22 3 0x x+ - = by factorisation. Solution

22 3 0 (2 3)( 1) 0

1.5 or 1

x x x x

x

+ - = fi + - =

fi = -

Question 4.6

Solve the following equations by factorisation:

(i) 26 2 0x x- - =

(ii) 216 1 0x - =

Some useful results:

2 2 ( )( )a b a b a b- = + -

3 3 2 2( )( )a b a b a ab b± = ± +

Notice that in the last result to get the correct signs you need to read the top sign on ± and (or alternatively the bottom sign).

FAC-04: Algebra


2.2 Solution by completing the square

All quadratic equations can be written as 2( ) 0px q r+ - = , where , ,p q r Œ . If the

coefficient of 2x is negative, you would have to multiply through by –1 first. If 0r > ,

then this method enables the quadratic to be solved (in the example here, q r

xp

- ±= ).

Example Solve the following equations by completing the square:

(i) 4 8 1 02x x

(ii) 3 13 1 02x x Solution (i) There are several methods available to complete the square, for example you can

multiply out the expression 2( )px q r+ - , giving 2 2 22p x pqx q r+ + - , and

compare coefficients to solve the equations:

( ) ( )

2 2

2

1 12 2

4 8 1 0 (2 2) 3 0

(2 2) 3

2 2 3

2 3 or 2 3 1.866 or 0.134

x x x

x

x

x

- + = fi - - =

fi - =

fi - = ±

fi = + - =

(ii) You need to multiply through by –1 first:

( )( )

2 2 13 13 3

213 169 16 36 3

213 1576 36

13 1576 36

3 13 1 0 3 0

3 ( ) 0

( )

4.255 or 0.078

x x x x

x

x

x

- + = fi - + =

fi - - + =

fi - =

fi = ± =



Question 4.7

Solve the following equations by completing the square:

(i) 2 4 8 0x x+ - =

(ii) 216 17 2 0x x+ + =

Question 4.8

If 2 2 2

1 2

1 2

x x x

, for all values of x, express m and s , in

terms of 1 2 1 2, , , , , and a b m m s s .

2.3 Solution by formula

For the general quadratic equation, 2 0ax bx c+ + = the solutions are given by 2 4

2

b b acx

a

- ± -= . This can be proved by applying the method of completing the

square to the general quadratic equation.

FAC-04: Algebra


Example

Solve the equation 23 8 2 0x x+ + = , using the quadratic formula. Solution If the equation is compared with the general quadratic, then: 3, 8, 2a b c= = =

so:

( ) ( )1 16 6

8 64 24

6

8 40 or 8 40

0.279 or 2.387

x- ± -=

= - + - -

= - -

Question 4.9

Solve the equation 21 4 7 0x x+ - = .

Question 4.10

Solve the equation 2 13 4 3 4 0x x+ + ¥ - = .

Whenever solving equations, especially quadratics, you should check that the answer that you give is reasonable. For example, you might be working out the number of people or another value that has to be positive, so state that a negative answer is impossible. There may also be situations when the value you require lies between particular limits, for example a probability must lie between 0 and 1. Under these circumstances reject any values which lie outside the limits, stating why you have rejected them.



3 Simultaneous equations

When solving simultaneous equations it is important to use the most efficient way of solving them, in order to make best use of your time. Generally with two linear equations, rearrange them to a suitable form and then add or subtract to eliminate the variables. For non-linear equations, consider using substitution first, otherwise use cancelling or factorising.

Example (Linear equations solved by subtraction) Solve the simultaneous equations:

2 5 4 0

7 2 4

x y

y x

+ - =

= -

Solution Looking at the two equations in turn:

2 5 4 0 2 5 4x y x y+ - = fi + = (1)

7 2 4 4 7 2y x x y= - fi + = (2)

(2) 2 (1)- ¥ (or equation (2) take away 2 times equation (1)):

3 6 2y y- = - fi =

3x\ = -

FAC-04: Algebra


Example (Non-linear equations solved by substitution) Solve the simultaneous equations:

2 2 20x y+ =

2 3 14x y- =

Solution The second equation can be rearranged to give:

12 (14 3 )x y= +

Substituting this into the first equation gives:

2 214

2 2

2

(14 3 ) 20

196 84 9 4 80

13 84 116 0

y y

y y y

y y

+ + =

fi + + + =

fi + + =

Using the quadratic formula or otherwise:

5813

413

2 or

4 or

y

x

fi = - -

fi =



Example (Non-linear equations solved by division) Solve the simultaneous equations:

2

3 4

8

30

ar

ar ar

=

+ =

Solution Dividing the second equation by the first equation:

2 3.75r r+ = Solving this as a quadratic: 1.5 or 2.5r = -

32 329 25or a =

FAC-04: Algebra


Example (Non-linear equations solved by factorisation) Solve the simultaneous equations:

2 2

2 2

2 3 50

3 4 35

p pq q

p pq q

- - = -

+ + =

Solution Factorising both equations:

( )(2 3 ) 50

( )(3 ) 35

p q p q

p q p q

+ - = -

+ + =

Dividing:

3 7

2 3 10

10(3 ) 7(2 3 )

4

p q

p q

p q p q

p q

+ -=-

fi + = - -

fi =

Substituting:

2 2 23 16 16 35

1 or 1

4 or 4

p p p

p

q

+ + =

fi = -

fi = -



Question 4.11

Solve the simultaneous equations:

(i) 2 4

4 6

5

20

ab ab

ab ab

+ =

+ =

(ii) 2 2

2 2

2 5 3 5 0

2 5 0

x xy y

x xy y

+ - + =

+ - - =

Example

If 3

211.7

al

= , 2 9.3al

= and 4.1kal+ = , find , and ka l .

Solution Dividing the second equation by the first, we get:

9.3

1.59011.7 2

l l= fi =

Substituting into the second equation gives us:

29.3 1.590 23.50a = ¥ = Finally substituting both of these into the final equation gives us: 10.68k = -

Question 4.12

If 21

2 3em s+ = , and

2 22 ( 1) 11.2e em s s+ - = , what are m and s ?

FAC-04: Algebra


4 Inequalities

Inequalities are similar to equations, but rather than finding an exact value for the variable, we find a range in which the variable can lie. In some cases solving inequalities is like solving equations whereas other examples can be more complicated.

Example Solve the inequality 2 7 7 4x x- £ + . Solution Starting with 2 7 7 4x x- £ + , if we subtract 4 from both sides, and subtract 2x from both sides, we get: 11 5x- £ Then dividing by 5 we get:

115 x- £

Note: ● You can add a constant to both sides (or subtract a constant from both sides).

● You can divide or multiply both sides by a positive constant.

● You can divide or multiply both sides by a negative constant, providing you reverse the inequality sign.

● You can take logs of both sides (provided both sides are positive).



Example For what values of x is (2 11)( 7) 57x x+ - < - ?

Solution Manipulate the inequality to get zero on one side:

2

2

(2 11)( 7) 57

2 3 77 57

2 3 20 0

(2 5)( 4) 0

x x

x x

x x

x x

+ - < -

- - < -

- - <

+ - <

The inequality can be solved by drawing a graph of the function ( ) 22 3 20f x x x= - -

and working out when the graph lies below the x-axis. However this method relies upon the fact that the graph is easy to plot (or that you have a graphical calculator!). An alternative is as follows: The expression (2 5)( 4)x x+ - will change sign when 2.5 or 4x = - , so we can

consider whether the expression is positive or negative between these points. (See table below.) So the solution is 2.5 4x- < < , since we want the expression to be less than 0 ie negative.

2.5x < - 2.5 4x- < < 4x >

2 5x + – + +

4x - – – +

(2 5)( 4)x x+ - + – +

Example Solve the inequality | 1 | 3x - < .

Solution This is equivalent to 3 1 3x- < - < , which simplifies to 2 4x- < < .

FAC-04: Algebra


Example Solve the inequality:

2

2

8 5 72

3 5 2

x x

x x

+ - £+ -

Solution Getting zero on one side and using a common denominator, we get:

2 2

2

2

2

8 5 7 2(3 5 2)0

3 5 2

2 5 3 (2 1)( 3)0 0

(3 1)( 2)3 5 2

x x x x

x x

x x x x

x xx x

+ - - + - £+ -

- - + -£ fi £- ++ -

This will change sign at 1 12 32, , , 3x = - - .

By considering the sign between each of these values (see table below), the solution

includes 122 x- < < - or 1

3 3x< < . However, we need also to see what happens at

1 12 32, , , 3x = - - . Substituting these values in, we can see that our solution is

1 12 32 or 3x x- < £ - < £ .

2x < - 122 x- < < - 1 1

2 3x- < < 13 3x< < 3x >

2x + – + + + + 2 1x + – – + + + 3 1x - – – – + +

3x - – – – – + (2 1)( 3)

(3 1)( 2)

x x

x x

+ -- +

+ – + – +



Question 4.13

Solve the inequality 3 22 5 17 20x x x+ - < .

Note: ● For a double inequality, for example 10 2 1 20n< - < , you can treat it as two

separate inequalities and then combine the answers at the end.

● Be careful that your answer matches the question eg or < £ .

FAC-04: Algebra


5 Arithmetic-geometric mean inequality

The arithmetic mean of n numbers, 1, , na a , is defined to be 1 na a

n

+ +. The

geometric mean of n positive numbers, 1, , na a , is defined to be 1n

na a .

The Arithmetic-geometric mean inequality states that “the geometric mean of a set of positive numbers is less than or equal to the arithmetic mean of the same set of numbers” ie:

11

nnn

a aa a

n

+ +£

Example Show that the arithmetic-geometric mean inequality holds for the numbers 7 and 9. Solution

Arithmetic mean 7 9

82

+= =

Geometric mean 7 9 63 7.94= ¥ = = So it does hold for these two numbers.

Question 4.14

When does the arithmetic mean of two numbers equal the geometric mean?



6 Sums and products

Earlier in the course, when we considered proof by induction, sums were written out in full eg 1 2 3 n+ + + ◊◊ ◊ + . This section shows how to write sums and products of numbers using the S (sigma) or P (pi) notation as an abbreviation.

Example Write the following using the S notation: (i) 3 3 3 3 3n+ + + ◊◊ ◊ + =

(ii) 2 2 2 21 2 3 n+ + + ◊◊ ◊ +

(iii) 3 3 3 31 2 3 ( 1)n+ + + ◊◊ ◊ + -

(iv) 1 3 5 71+ + + ◊◊ ◊ +

(v) 2 2 2

1 1 1

2 4 8+ + + ◊◊ ◊

Solution

(i) 1

3 3n

i

n=

=Â (ii) 2

1

n

i

i=Â (iii)

13

1

n

i

i-

=Â

(iv) 36

1

(2 1)i

i=

-Â (v) 21

1

2 ii

•

=Â

There is always more than one way to write a series in this notation. For example, (iv)

could be written as 35

0

(2 1)j

j=

+Â . However, it will often be that there is a ‘clearest’

approach which you should use in order to make life easier for your reader.

FAC-04: Algebra


Question 4.15

Write the following using the S notation: (i) 2 5 10 226+ + + ◊◊ ◊ +

(ii) 22 4 8 2 n+ + + ◊◊ ◊ +

(iii) 1 1 1 13 5 7 2 1n++ + + ◊◊ ◊ +

(iv) 4 9 14 99- + - + ◊◊ ◊ +

Example Write the following using the P notation: (i) 3 5 7 (2 1)n¥ ¥ ¥ ◊◊ ◊ ¥ +

(ii) 2 3 2 3ne e e e +¥ ¥ ¥ ◊◊ ◊ ¥

(iii) 2 3 15

1 1 1 1

4 9 16 256a a a a¥ ¥ ¥ ◊◊ ◊ ¥

Solution

(i) 1

(2 1)n

i

i=

+’

(ii) 2 3

1

ni

i

e+

=’

(iii) 15

21

1

( 1) ii i a= +’



Question 4.16

Write this expression using the P notation:

1 11 1

1 2 31 1 1 1

n

t t t t

Example Simplify the expression:

2

1

(1 4 )n

i

i i=

+ +Â

Solution

2 2 2

1 1 1 1 1 1

(1 4 ) 1 4 4n n n n n n

i i i i i i

i i i i n i i= = = = = =

+ + = + + = + +Â Â Â Â Â Â

This can be further simplified, but we will deal with this later.

FAC-04: Algebra


7 Arithmetic and geometric progressions

7.1 Arithmetic progressions (AP)

An arithmetic progression (AP) is a set of numbers, such as 3,7,11,15, ,39 , in which

the difference between each two successive numbers is a constant. In this case it is 4. We call this the common difference and denote it by d . Since the numbers go up by the same amount each time we can use this to determine future terms. The first term is 3, the second term is 3 4 7+ = , the third term is 3 4 4 11+ + = and so on. So the 10th term will be 3 9 4 39+ ¥ = and in general using a to stand for the first term then the n th term is: ( 1)a n d+ -

Suppose, we want to sum the first 10 terms of this AP: 10 3 7 11 15 19 23 27 31 35 39S = + + + + + + + + +

One clever way of doing this is to note what happens when we sum pairs taken from each end: 3 39 42+ = 7 35 42+ = 11 31 42+ = 15 27 42+ = 19 23 42+ = So the total of these 10 terms is 5 42 210¥ = .

In general, if we sum n terms then we will have 12 n of these pairs. Using l to stand

for the last term then the sum of n terms will be 12 ( )n a l+ . However, the last term in

this sum of n terms is the n th term and so equals ( 1)a n d+ - . So our sum of n terms

is:

[ ]12 2 ( 1)nS n a n d= + -

Note that this formula is still correct, even if there is an odd number of terms in the AP.



Example A woman made a New Year’s resolution to save money. She decided to put £1 in a savings account in the first week of the year, £3 in the second, £5 in the third and so on. Assuming that there are 52 weeks in a year, what is her last payment, and how much will she have put into the account by the end of the year? Solution 1, 3, 5,… is an arithmetic progression with 1, 2a d= = .

Last payment ( 1) 1 51 2 £103a n d= + - = + ¥ = .

Total 522 22 ( 1) 2 51 2 £2,704n a n d .

Question 4.17

A mathematics student borrows some money from his parents (interest free) and agrees to pay it back as follows: the first monthly payment is £298, payments decrease by £28 per month, and the last payment is £18. How many payments were made? How much money did the student borrow?

FAC-04: Algebra


7.2 Geometric progressions (GP)

A geometric progression (GP) is a set of numbers, such as 3, 6, 12, ..., 192, in which the ratio of each term to the preceding term is a constant. In this case it is 2 (as 6 3 2= ,

12 6 2= , etc). We call this the common ratio and denote it by r .

Since the numbers are multiplied by the same each time we can use this to determine future terms. The first term is 3, the second term is 3 2 6¥ = , the third term is

3 2 2 12¥ ¥ = and so on. So the 7th term will be 63 2 192¥ = and in general using a to stand for the first term the n th term is:

1nar - Suppose we want to sum the first 7 terms in this GP: 7 3 6 12 24 48 96 192S = + + + + + + (1)

A clever trick is to look at what happens if we multiply this sum by the common ratio: 72 6 12 24 48 96 192 384S = + + + + + + (2)

Subtracting these two series we get: 7(2) (1) (2 1) 384 3 381S- fi - = - =

In general, when we subtract we will have ( 1) nnr S ar a- = - . So the sum is:

( 1)

1

n

na r

Sr

-=-

Or multiplying top and bottom by 1- we could write this as (1 )

1

n

na r

Sr

-=-

. This

formula is useful when the common ratio r lies between 1- and 1.



Example A boy gets his father (who is not a mathematician) to agree to a new system for pocket money. In the first week the father puts 1p in the first square on a chessboard. In the second week the father puts 2p in the second square. In the third week the father puts 4p in the third square, and so it continues doubling each time. How much money will the father have to put on the last square? Solution

1, 2, 4, 8, …is a geometric progression with 1, 2a r= = and general term of 1nar - .

A chessboard has 64 squares, so the last square will need 631 2¥ pence, or (to 1 SF) £90,000,000,000,000,000!

Question 4.18

Evaluate 10

1

5

1.06kk=Â .

FAC-04: Algebra


7.3 Convergence

An infinite geometric progression is a GP which does not end such as 1 1 12 4 8, , , . In the

case of infinite geometric progressions, the sums can either converge to a finite limit or diverge. For example consider the following two series:

1 2 4+ + + ◊◊ ◊ 1 1 12 4 81+ + + + ◊◊ ◊

The first GP has 2r = , so the terms get larger and larger and so the sum clearly does

not converge to a finite limit. In the second series we have 12r = , so the terms get

smaller and smaller and the sum does converge to a finite limit:

1

2

3

4

5

6

7

8

9

10

1

1.5

1.75

1.875

1.9375

1.96875

1.984375

1.9921875

1.99609375

1.998046875

S

S

S

S

S

S

S

S

S

S

==========

We can see that the sum gets closer and closer to 2 without ever quite getting there. So we say that the sum of the infinite GP is 2. In general the sum of an infinite geometric progression converges to a finite limit if

1 1r- < < , since 0nr Æ as n Æ• , so the terms 1nar - will tend to zero also. The sum is:

1

aS

r• =-

We get this by letting n Æ• in our GP sum, remembering that 0nr Æ .



Example Write 1.231231231… as a fraction. Solution

231 231 2311,000 1,000,000 1,000,000,0001.231231231... =1+ + + + ◊◊ ◊

which is 1 plus an infinite geometric progression, which converges since the common

ratio is 11,000 .

231

1,000 231 231 771 999 999 333

1,000

1.231231231... =1+ 1 1 11

= + = =-

s

Question 4.19

Calculate the value of the sum to infinity of: 4, 3.2, 2.56, 2.048, ...

FAC-04: Algebra


Note the techniques used in the following example.

Question 4.20

The sum of the first n terms of a sequence of numbers (which is not an AP or a GP) is given by the formula:

16 ( 4)( 5)nS n n n= - - ( 1,2,3,n = )

Find: (i) the sum of the first 10 terms (ii) the sum of the 11th to the 20th terms (inclusive) (iii) the 20th term (iv) a formula for nu , the n th term

(v) the smallest term in the sequence.



8 Further series work

8.1 Standard summations

There are two formulae for series that we proved as part of the work on proof by induction.

12

1

( 1)n

k

k n n=

= +Â and 2 16

1

( 1)(2 1)n

k

k n n n=

= + +Â

These can be used to simplify more complicated expressions.

Example

Simplify the expression 2

1

(1 4 )n

i

i i=

+ +Â .

Solution

( )

( )( )

2 2

1 1 1 1

2

1 1

4 12 6

16

216

216

(1 4 ) 1 4

4

( 1) ( 1)(2 1)

6 12( 1) ( 1)(2 1)

6 12 12 2 3 1

2 15 19

n n n n

i i i i

n n

i i

i i i i

n i i

n n n n n n

n n n n

n n n n

n n n

= = = =

= =

+ + = + +

= + +

= + + + + +

= + + + + +

= + + + + +

= + +

Â Â Â Â

Â Â

FAC-04: Algebra


Example

Sum the series 2 2 2 21 2 3 (2 1)n+ + + ◊◊ ◊ + + .

Solution

( )( )

2 12 2 2 2 2

1

16

16

13

1 2 3 (2 1)

(2 1) (2 1) 1 2(2 1) 1

(2 1)(2 2)(4 3)

(2 1)( 1)(4 3)

n

k

n k

n n n

n n n

n n n

+

=+ + + ◊◊ ◊ + + =

= + + + + +

= + + +

= + + +

Â

8.2 Swapping the order of summation

When we have summation over two variables, we are able to swap the order of the summation. This may make the actual summation easier. Suppose we have:

10

1 1

x

x y= =ÂÂ

Then x takes the values between 1 and 10 inclusive, ie 1 10x£ £ . However, y takes

the values between 1 and x . So inserting that in the correct position of the inequality, we would have 1 10y x£ £ £ .

By swapping the order, we mean that instead of x summing between the extreme numbers we will let y take these values. Then by the position of x in the inequality it

will be between y and 10. Hence:

10 10 10

1 1 1

x

x y y x y= = = ==ÂÂ Â Â

We have swapped the order of summation.



Example

(i) Calculate 5

0 0

2y

y x

y x= =Â Â , where 0 5x y£ £ £ .

(ii) Reverse the order of the summation and re-calculate the value.

You are given that 3 2 214

1

( 1)n

k

k n n=

= +Â .

Solution (i) Using the formula detailed in the last section:

5 5 5

3 2

0 0 0 0

12 2 ( 1) ( )

2

y

y x y y

y x y y y y y

This can be simplified using the results for the sum of squares and the sum of

cubes:

5 5 5

3 2 3 2 2 2

0 0 0

1 1( ) (5) (6) (5)(6)(11) 280

4 6y y y

y y y y= = =

+ = + = + =Â Â Â

(ii) To change the order of the summation, we consider the inequality 0 5x y£ £ £ .

Previously we summed from 0x = to x y= , then from 0y = to 5y = .

Reversing this, we sum from y x= to 5y = then from 0x = to 5x = . To get

the limits for y , see what it is bounded by in the inequality 0 5x y£ £ £ . So:

5 5 5

0 0 0

2 2y

y x x y x

y x x y= = = =

=Â Â Â Â

Using the formula detailed in the last section:

5 5 5 5 1 5

0 0 0 0 0

1 12 2 2 (5)(6) ( 1)( )

2 2

x

x y x x y y x

x y x y y x x x

FAC-04: Algebra


Simplifying this:

5 5 5 5 5 5

3 2

0 0 0 0 0

2 (30 ( 1)) 30x y x x x x x

x y x x x x x x= = = = = =

= - - = - +Â Â Â Â Â Â

Again, using the standard summations:

5 5

2 2

0

1 1 12 30 (5)(6) (5) (6) (5)(6)(11) 280

2 4 6x y x

x y

Note that it is easier in this case to use the first version of the summation.

Question 4.21

(i) If 0 10t s£ £ £ , calculate 10 10

0

1.04 1t

t s t

-

= =Â Â .

(ii) Swap the order of summation to re-calculate the answer.

You may use the result 1

0

1 1 1.04

0.041.04 1 1.04 1.04

bb

t bt

t b

.



9 Binomial expansions

9.1 Positive powers

The binomial theorem states that: If n is a positive integer, then, for all values of a and b:

1 2 2 1( )0 1 2 1

n n n n n r r n nn n n n n na b a a b a b a b ab b

r n n

Note that !

!( )!

n n

r r n r

, which appears as nrC on most calculators. The

n

r

expression comes in because there are many different ways of getting the term in n r ra b .

This can be used to expand expressions of the form ( )na b and the result is given on

page 2 of the Tables. If we rewrite the combinatorial function as follows:

! ( 1) ( 1)( )! ( 1) ( 1)

!( )! !( )! !

n n n n r n r n n n r

r n r r n r r

¥ - ¥ ¥ - + - ¥ - ¥ ¥ - += =- -

Then we can rewrite the binomial expansion as:

1 2 2 1( 1) ( 1)( 2) ( 1)

2! !n n n n r r n nn n n n n n r

a na b a b a b nab br

- - - -- - - ◊◊ ◊ - ++ + + ◊◊ ◊ + + ◊ ◊ ◊ + +

Notice that the sum of the powers of a and b is the power on the bracket being expanded, and that the coefficients of the terms can be obtained from Pascal’s triangle:

1 1 1 1 2 1 1 3 3 1

1 4 6 4 1 where the numbers are found by adding the two numbers in the row above it.

FAC-04: Algebra


For example, the row 1, 4, 6, 4, 1 gives the coefficients for the expansion of 4( )a b+ ,

ie:

1, 4, 4 3

2!

¥,

4 3 2

3!

¥ ¥,

4 3 2 1

4!

¥ ¥ ¥.

Example

Expand 7(2 3 )x- as far as the term in 3x .

Solution

7 7 6 1 5 2 4 3

2 3

7 6 7 6 5(2 3 ) 2 7 2 ( 3 ) 2 ( 3 ) 2 ( 3 )

2! 3!

128 1,344 6,048 15,120

x x x x

x x x

¥ ¥ ¥- = + ¥ ¥ - + ¥ ¥ - + ¥ ¥ - + ◊◊ ◊

= - + - +

Alternatively you could get the coefficients by looking at the row of Pascal’s triangle which begins 1, 7, … However you will still have to multiply in the appropriate powers of 2 and –3 to each term.

Example

Find the coefficient of the term in 4a in the expansion of 6(2 5 )a b+ .

Solution

The general term is 6 6(2 ) (5 )r rrC a b- , and we want 4a , so here 2r = .

The coefficient is 6 4 2 2 22 2 (5 ) 15 16 25 6000C b b b¥ ¥ = ¥ ¥ ¥ = .

Question 4.22

Expand 8(1 2 )x+ as far as the term in 4x .



9.2 Fractional or negative powers

Very often in practice, the power on the bracket will not be a positive integer, so alternatively we can use:

2 3( 1) ( 1)( 2)(1 ) 1

2! 3!n n n n n n

x nx x x- - -+ = + + + + ◊◊ ◊

where n is negative or fractional, and 11 x< £- . This series is derived using Maclaurin series which is covered in Chapter 7.

Example

Expand (1 2 )x- as far as the term in 3x . For what values of x is this expansion

valid? Solution

12

31 1 1 12 31 2 2 2 2 2

2

2 31 12 2

( ) ( )( )(1 2 ) (1 2 ) 1 ( 2 ) ( 2 ) ( 2 )

2! 3!

1

x x x x x

x x x

- - -- = - = + - + - + - + ◊◊ ◊

= - - - + ◊◊ ◊

This expansion is valid for 1 2 1 x , or 1 12 2

x .

Question 4.23

Expand 12(1 )x+ and hence approximate the following (without using a calculator!):

121.04 ,

141.10 ,

120.99

-,

12

12

1.09

1.07, 10200 ,

1 12 2a b

a b

--

where 1.11a = and 1.10b = .

FAC-04: Algebra


Example

Expand 1

2 3

x

x

++

as far as the term in 3x . For what values of x is this expansion valid?

Solution We start by manipulating the expression, since we want a binomial expansion which starts with a 1:

1 1 132

2 31 3 3 32 2 2

1 2 33 9 272 4 8

2 3 2 33 9 27 3 912 2 4 8 2 4

2 33 91 12 4 8 16

1(1 )(2 3 ) 2 (1 )(1 )

2 3

( 1)( 2) ( 1)( 2)( 3)2 (1 ) 1

2! 3!

2 (1 ) 1

1

xx x x x

x

x x x x

x x x x

x x x x x x

x x x

This is valid for 321 1 x- < £ , ie 2 2

3 3x- < £ .

Question 4.24

Expand 1

(1 4 )x- as far as the term in 3x . For what values of x is this expansion

valid? Use your expansion to estimate 120.98

- to 6DP.

Question 4.25

Simplify 10

1 k x

x

k xS p q

x

, and 2

0

nx n x

x

nS x p q

x

, given that 1p q+ = and

k is a positive integer.



Chapter 4 Summary Indices

The three laws of indices (powers) are:

a b a bx x x +¥ =

a b a bx x x -=

( )a b abx x=

Logarithms

Logarithms are the inverse of powers, eg logx b a= is equivalent to ax b= . The three

laws of logarithms are:

log log loga a ax y xy+ =

log log log ( )a a ax y x y- =

log logna ax n x=

Logarithms are used to solve equations where the unknown is a power, eg 2 50x = . Algebraic fractions

Operations are dealt with in the same way as ordinary fractions:

addition/subtraction – requires a common denominator:

eg a c ad bc ad bc

b d bd bd bd

±± = ± = .

multiplication – just multiply numerator and denominator, eg a c ac

b d bd¥ =

division – multiply by the inverse, eg a c a d ad

b d b c bc∏ = ¥ =

FAC-04: Algebra


Quadratic equations

2 0ax bx c+ + = can be solved by:

factorisation, eg ( )( ) 0 ,x x xa b a b- - = fi =

completing the square, eg 2( ) 0xa b g- + =

the quadratic formula: 2 4

2

b b acx

a

- ± -= .

Simultaneous equations

Linear simultaneous equations can be solved by elimination or substitution:

eg 3 5 29 (1)

2 11 (2)

x y

x y

+ =+ =

eg 3 (1) (2)¥ - will eliminate x , or we could rearrange (2) as 11 2x y= - and substitute

into (1). Non-linear simultaneous equations may also require division or factorisation to solve. Inequalities

Linear inequalities can be solved in the same way as equations, except that multiplying or dividing by a negative number reverses the sign of the inequality. Quadratic inequalities are best solved by considering the possibilities in a table. Arithmetic-geometric mean inequality

11

nnn

a aa a

n

+ +£

Notation

The P and S notations are used to write products and sums in shorthand. Arithmetic Progression (AP)

A series with first term a and common difference, d has:

general term: ( 1)a n d+ - sum: [ ]2 2 ( 1)nnS a n d= + -



Geometric Progression (GP)

A series with first term a and common ratio r has:

general term: 1nar - sum:(1 )

1

n

na r

Sr

-=-

1

aS

r• =-

Standard summations

12 ( 1)k n n= +Â

2 16 ( 1)(2 1)k n n n= + +Â

3 2 214 ( 1)k n n= +Â

Binomial expansion

The binomial expansion is given on page 2 of the Tables:

1 2 2( )1 2

n n n n nn na b a a b a b b n- - +Ê ˆ Ê ˆ+ = + + + + ŒÁ ˜ Á ˜Ë ¯ Ë ¯

2 3( 1) ( 1)( 2)(1 ) 1 1 1

2! 3!p p p p p p

x px x x x- - -+ = + + + + - < <

FAC-04: Algebra






(i) 810x

(ii) 5

8

3y

(ii) 625b Solution 4.2

Let log mam x a x= ¤ = and let log n

an y a y= ¤ = .

Now consider m na - , using the second law of indices, this can be simplified as follows:

m n m na a a x y- = ∏ = ∏

Using the definition of logs:

log log logm na a a

x xa m n x y

y y- = ¤ = - = -

and hence we have proved the result. Solution 4.3

ln 1e =

FAC-04: Algebra


Solution 4.4

2 2 12 1.05 1.1025t t- -¥ = Taking logs:

2 2 1log 2 1.05 log1.1025

log 2 ( 2) log1.05 (2 1)log1.1025

t t

t t

- -¥ =

+ - = -

Gathering up the terms:

log 2 2log1.05 log1.1025 (2log1.1025 log1.05)

log 2 2log1.05 log1.1025

2log1.1025 log1.05

4.736

t

t

- + = -

- +=-

=

Solution 4.5

2 2

2 2

3 4 2 5 7 ( 4)( 1) ( 4)(4 1)

(3 2)( 4) ( 1)(2 7)3 10 8 4 17 4

( 4)(4 1)

(3 2)(2 7)

x x x x x x x x

x x x xx x x x

x x

x x

+ - + - + - - -∏ = ¥+ - - +- - - +

+ -=+ +

Solution 4.6

(i) (2 1)(3 2) 0x x+ - = so 1 22 3or x = -

(ii) (4 1)(4 1) 0x x+ - = so 14x = ±



Solution 4.7

2(i) ( 2) 12 0

2 12

2 12

1.464 or 5.464

x

x

x

+ - =

+ = ±

= - ±

= -

( )

217 2898 64

17 1618 64

17 16114 8 64

(ii) (4 ) 2 0

(4 )

0.135 or 0.928

x

x

x

+ - + =

+ = ±

= - ±

= - -

FAC-04: Algebra


Solution 4.8

2 2 21 2

1 2

x x x

Comparing the coefficients of 2x and x:

2 2 21 2

1 1 as s s

+ = and 1 22 2 21 2

2 2 2m m mas s s

- - = -

Dividing the second equation by –2, and then dividing the equations:

1 22 21 2

2 21 2

1 1

m ms s m

s s

+=

+

So 2 2

1 2 2 12 22 1

m s m sms s

+=+

.

From the equation above for the coefficients of 2x , we get:

2 21 2

2 2 21 2

s s as s s+ =

Rearranging this equation gives us:

2 2 2 22 1 2 1 2

2 2 2 21 2 1 2

as s as ss ss s s s

= fi =+ +



Solution 4.9

Rearranging the equation so that the 2x term is positive, and then using the formula:

4 16 280.76 or 0.19

14x

± += = -

Solution 4.10

The equation can be written as 23(3 ) 4(3 ) 4 0x x+ - = , which is a quadratic in 3x . If

3xy = , then the equation can be written as 23 4 4 0y y+ - = , so that:

23

23

(3 2)( 2) 0

or 2

3 or 2x

y y

y

- + =

= -

fi = -

so:

23ln

0.369ln3

x = = -

Note that it is impossible to raise a positive number to any power and get a negative

number. So the alternative equation 3 2x = - has no solutions, and 0.369x = - is the only solution to the equation.

FAC-04: Algebra


Solution 4.11

(i) Dividing the second equation by the first 4 2

22 2

(1 )4

(1 )

ab b bab b

+ = =+

ie 2b = ± .

Substituting this back into the first equation:

14

4 16 5a a

a

+ =

=

So the two solutions are 14 and 2a b= = or 1

4 and 2a b= = - .

(ii) Factorising and dividing the second equation by the first:

(2 )( 3 ) 5

(2 )( ) 5

x y x y

x y x y

- + = -

- + =

13

3 2

x y

x y

x y x y x y

+ = -+

+ = - - fi = -

Substituting this into the first factorised equation:

( 5 )( ) 5

1

2

y y

y

x

- = -

fi = ±

fi =



Solution 4.12

Since 21

2em s+

squared is 22e m s+ , we can substitute the first equation into the second:

2

9( 1) 11.2es - =

Rearranging this, we get:

22.244es =

Which can be rearranged to give:

0.899s = ± Substituting this into the first equation gives 0.694m = .

Solution 4.13

3 22 5 17 20 0

(2 5)( 4)( 1) 0

x x x

x x x

+ - - <

fi - + + <

4x < - 4 1x- < < - 1 2.5x- < < 2.5x > 4x + – + + + 1x + – – + +

2 5x - – – – + (2 5)( 4)( 1)x x x- + + – + – +

ie the solution is 4x < - and 1 2.5x- < < . Solution 4.14

The arithmetic and geometric means are equal when the two numbers are equal. This can be proved as follows:

22( )

( ) 42 4

a b a bab ab a b ab

+ += fi = fi + =

This simplifies to 2( ) 0a b- = , ie a b= .

FAC-04: Algebra


Solution 4.15

(i) 15

2

1

( 1)k

k=

+Â

(ii) 2

1

2n

k

k=Â

(iii) 1

1

2 1

n

k k= +Â

(iv) 20

1

( 1) (5 1)k

k

k=

- -Â

Solution 4.16

1

1

1n

kk

t

Solution 4.17

If the number of payments is n, then, using the formula for the general term of an arithmetic progression:

18 298 ( 1) 28

11

n

n

= + - ¥ -

=

So, the total amount of the payments can be found by using the sum of an arithmetic progression, ie:

( )112 2 298 (11 1) 28 £1,738¥ + - ¥ - =



Solution 4.18

If we write out the terms of the series it will make things clearer:

10

2 3 101

5 5 5 5 5

1.061.06 1.06 1.06 1.06kk=

= + + + +Â

We see that we have a geometric progression with 5

1.06a = and

1

1.06r = . So the sum

of the first 10 terms of this GP would be:

10

10

5 11

1.06 1.0636.800

11

1.06

S

È ˘Ê ˆ-Í ˙Á ˜Ë ¯Í ˙Î ˚= =-

Solution 4.19

Dividing consecutive terms we can see that the common ratio is 3.2 4 0.8= . Hence,

the sum to infinity is:

4

201 0.8

S• = =-

FAC-04: Algebra


Solution 4.20

(i) The sum of the first 10 terms is found by putting 10n = into the formula for nS :

1

10 6 10 6 5 50S = ¥ ¥ ¥ =

(ii) We have just calculated the sum of the first 10 terms to be 50. We can calculate

the sum of the first 20 terms to be:

120 6 20 16 15 800S = ¥ ¥ ¥ =

The sum of the 11th to the 20th terms is just the difference between these two, ie:

20 10 800 50 750S S- = - =

(iii) The 20th term can be calculated as the difference between 19S and 20S , ie:

20 19 800 665 135S S- = - =

(iv) Similarly, the n th term is:

1 11 6 6

16

1 16 2

( 4)( 5) ( 1)( 5)( 6)

( 5)[ ( 4) ( 1)( 6)]

( 5)[3 6] ( 5)( 2)

n n nu S S n n n n n n

n n n n n

n n n n

-= - = - - - - - -

= - - - - -

= - - = - -



(v) To find the smallest term, we need to look at how two consecutive terms compare. The n th term will be at least as big as the one before it iff:

1 0n nu u -- ≥

ie 1 12 2( 5)( 2) ( 6)( 3) 0n n n n- - - - - ≥

This simplifies to: 4 0n - ≥ ie 4n ≥ So a given term will be at least as big as the previous term whenever 4n ≥ (with

equality only when 4n = ). This means that: 5 4u u≥ and 4 3u u= but 3 2u u≥/

Combining these, we get: 2 3 4 5u u u u> = <

So 3u and 4u are (jointly) the smallest terms. They each take the value 1- .

FAC-04: Algebra


Solution 4.21

(i) We have:

10

1 11s t

t=

= -Â

So the summation is:

10 10 10

0 0 0

1.04 (11 ) 11 1.04 1.04t t t

t t t

t t- - -

= = =- = - ¥Â Â Â

The first series is a geometric progression and the second series can be summed

using the formula given in the question:

11 1010

1 1 100

1 1.04 1 1 1.04 101.04 (11 ) 11 58.23

0.041 1.04 1 1.04 1.04t

t

t

(ii) Swapping the order of the summation:

10

0 0

1.04s

t

s t

-

= =ÂÂ

The sum over the variable t can be carried out using the formula for the sum of

a geometric progression:

10 10 10( 1)

10 0 0 0

1 1.04 1.04 1.041.04

0.041 1.04

s s st

s t s s

- + --

-= = = =

- -= =-ÂÂ Â Â

Simplifying this and using the formula for the sum of a geometric progression:

10 10 11

10 0

1.04 1 11 1.04 1 1 1.041 1.04 58.23

0.04 0.04 0.04 0.04 1 1.04s

s s

--

-= =

¥ -- = - =-Â Â



Solution 4.22

8 8 7 6 2 5 3 4 4

2 3 4

8 7 8 7 6 8 7 6 5(1 2 ) (1) 8(1) (2 ) (1) (2 ) (1) (2 ) (1) (2 )

2! 3! 4!

1 16 112 448 1,120

x x x x x

x x x x

¥ ¥ ¥ ¥ ¥ ¥+ = + + + + +

= + + + + +

Solution 4.23

The binomial expansion tells us that:

½ 12(1 ) 1x x+ ª +

In other words, if you want to approximate the square root of a number close to 1, you just halve the extra bit. So we get:

½ 121.04 1 0.04 1.02ª + ¥ =

¼1.10 is the 4th root of 1.10 , which is the same as square rooting twice. So we get:

¼ ½ ½ ½1.10 (1.10 ) 1.05 1.025= ª ª

The same method works with a number a bit less than one and/or a negative power:

½ ½ 120.99 (1 0.01) 1 ( 0.01) 1.005- -= - ª - ¥ - =

The next one gives:

½

½

1.09 1.045

1.0351.07ª

To approximate this ratio, we can subtract 0.035 from the top and bottom (which doesn’t affect the answer very much):

½

½

1.09 1.045 0.035 1.011.01

1.035 0.035 11.07

-ª = =-

Note that the 0.01 bit is just half the difference between the 1.09 and the 1.07.

FAC-04: Algebra


We must have numbers close to 1 for our approximation to work. So, for the next one we need to take out a factor first:

10200 10000 1.02 100 1.02 100 1.01 101= ¥ = ¥ ª ¥ = And finally:

½ ½ 1.055 1.05 0.0050.5

1.11 1.10 0.01

a b

a b

- -ª = =- -

Calculations involving rates of return earned on investments often involve expressions similar to these. It is useful to be able to approximate them to check that the calculations look sensible. Solution 4.24

12

3 3 51 12 32 2 2 2 2

2 3

(1 4 ) 1 2 ( 4 ) ( 4 )2! 3!

1 2 6 20

x x x x

x x x

- - ¥ - - ¥ - ¥ -- = + + - + - +

= + + + +

This is valid for 1 1

4 4x- £ < .

If we now substitute in 0.005x = , we get:

120.98 1 0.01 0.00015 0.0000025

1.0101525

- = + + +

=

So to 6DP 120.98 1.010153

- = .



Solution 4.25

If we write out the first sum without using sigma notation (and dividing through by kp ), it is:

2 31

2 3

1 1 2

0 1 2 3

( 1) ( 2)( 1)1

2 6

k

k k k kSq q q

p

k k k k kkq q q

It’s not obvious how to deal with this because the k factors are going up, rather than down, as in an ordinary binomial expansion. However, we can introduce some negatives to enable us to simplify the expression:

2 31 ( )( 1) ( )( 1)( 2)1 ( )( ) ( ) ( )

2 6

[1 ( )]

k

k k

S k k k k kk q q q

p

q p- -

- - - - - - - -= + - - + - + - +

= + - =

So: 1 1S =

This is actually a calculation of the sum of the probabilities for a negative binomial distribution. If we write out the second sum without using sigma notation, we have:

1 2 22

0

0 21 2

nx n x n n n

x

n n n nS x p q pq p q n p

x n

If we take a typical coefficient in this series, we get for example:

1( 1)( 2) ( 1)( 2)

3 33 23 2 1 2 1

n nn n n n nn n

So the series is:

1 2 22

1 1 1

0 1 1n n nn n n

S n pq n p q n pn

FAC-04: Algebra


Dividing through by the n and a p factor:

1 2 12 1 1 1

0 1 1n n nn n nS

q pq pnnp

The RHS is just the binomial expansion of 1( )np q , which equals 1 (since 1p q ).

So:

2 1S

np 2S np

This is actually the calculation of the mean value of a binomial distribution.

FAC-05: Numerical methods II Page 1


Chapter 5

Numerical methods II

You need to study this chapter to cover: ● expressing quantities as percentages or per mil

● absolute change, proportionate change and percentage change

● absolute error, proportionate error and percentage error

● dimensions

● linear interpolation

● iteration

● complex numbers

● difference equations.

0 Introduction

This chapter completes the numerical methods required for the Core Technical subjects.

FAC-05: Numerical methods II


1 Percentages

A percentage means “out of a hundred”. Percentages are used a lot in actuarial work so you must be familiar with how to use them efficiently and accurately. To find a percentage, p , of a quantity you use:

quantity100

p ¥

Example Find 7.5% of $2,500. Solution 7.5

$2,500 0.075 $2,500 $187.50100

¥ = ¥ =

Next we’ll look at how to work quickly with percentage increases and decreases:

Example The price of an item in a shop is £15.75. Find the new price if it is: (i) increased by 1.2% (ii) decreased by 5%. Solution (i) Increasing the price by 1.2% means that it will now be 100% 1.2% 101.2%+ =

of the original price:

101.2

£15.75 1.012 £15.75 £15.94100

¥ = ¥ =



(ii) Decreasing the price by 5% means that it will now be 100% 5% 95%- = of the original price:

95

£15.75 0.95 £15.75 £14.96100

¥ = ¥ =

Notice that in each case the final answer has to be rounded off to two decimal places because we are working in pounds and pence.

The next example looks at how we can calculate the original price given a percentage increase or decrease:

Example The new price of an item in a shop is £60. Find the original price if it was: (i) increased by 20% (ii) decreased by 20%. Solution (i) If the price was increased by 20% then the new price is 100% 20% 120%+ = of

the original price. Using x to stand for the original price, we have:

120 £60

1.20 £60 £50100 1.20

x x x¥ = ¥ = fi = =

(ii) If the price was decreased by 20% then the new price is 100% 20% 80%- = of

the original price. Using x to stand for the original price, we have:

80 £60

0.80 £60 £75100 0.8

x x x¥ = ¥ = fi = =



Question 5.1

A population increases by 5% each year. (i) How many years would it take for the population to at least double? (ii) If the population was 66,150,000 in 2012, what was it in 2010?

Finally to express a quantity A as a percentage of a quantity B we use:

100%A

B¥

Example At the start of year 1 a group consists of 51,890 people. By the start of year 2, there are 51,548 people. If the only reason for leaving the group is death, what percentage of the group died in year 1? Solution In the group, 51,890 51,548 342- = people have died. Expressing this as a percentage

of the original group:

342

100% 0.66%51,890

¥ =

Sometimes you may be asked to express something “per mil”, or out of a thousand, which is written as ‰. This notation is often used in connection with insurance premiums which are often small percentages so writing them per thousand rather than per hundred makes them easier to interpret.

Question 5.2

Express the number of people who have died in the previous example as a rate per mil.



2 Changes

The previous example is really asking about a percentage change in the population, and in this section we are going to look at different ways of expressing change.

2.1 Absolute change

An absolute change is asking for the difference between two figures, where the difference can be positive or negative. Absolute change should not be confused with absolute value which was covered in Chapter 4. absolute change new value original value= -

Example What is the absolute change in the population described in the last example? Solution The absolute change is –342, since the population decreased by 342 people.

2.2 Proportionate change

The disadvantage of looking at absolute changes is that it gives no indication of the size of the change compared to the original figure. For example, there is an absolute change of 1 when a population goes from 3 to 4 people, or from 340,025 to 340,026 people! Proportionate change looks at the change relative to the original number in the population, by dividing the absolute change by the original amount:

absolute change

proportionate changeoriginal value

=

Proportionate change is also sometimes called relative change.



Example What is the proportionate change in the price of a car that goes from £9,950 to £9,550? Solution

The absolute change is –£400, so the proportionate change is £400

0.04£9,950

- = - .

Notice that the answer has no units of measurement, since we have divided pounds by pounds. We will return to this idea in Section 4.

2.3 Percentage change

A percentage change expresses the proportionate change as a percentage.

absolute change

percentage change 100%original value

= ¥

Example What is the percentage change in the previous example? Solution The proportionate change is –0.04 so the percentage change is –4%.

Question 5.3

A man’s wage has risen from £14,567pa to £15,034pa. What are the absolute, proportionate and percentage changes in his wage?



3 Errors

Errors are similar to changes in that they look at differences, but errors refer specifically to differences between the “actual” (or “true” or “accurate”) value and the “expected” (or “approximate” or “experimental”) value: absolute error approx value true value= -

absolute error

proportionate errortrue value

=

absolute error

percentage error 100%true value

= ¥

Note that some texts define the formulae as the absolute values of each of the quantities.

Example

When calculating 1

1v

i=

+, the true value of i is 0.0372534. Calculate the absolute,

proportionate, and percentage error that will be introduced in the value of v by rounding i to two decimal places. Solution

The true value of v is 1

1.0372534. When rounded to 2 DP, i becomes 0.04, which

gives v to be 1

1.04.

Absolute error 1 1

0.00251.04 1.0372534

= - = - (4 DP).

Proportionate error (absolute error divided by true value of v ) 0.002641= - (6 DP). Percentage error (proportionate error multiplied by 100%) 0.2641%= - .



Question 5.4

In an actuarial calculation, a student uses 0.027i = having rounded the precise value of i to 2 significant figures. What is the largest percentage error that could have been

made in the calculated value of 1

1v

i=

+?



4 Dimensions

In the example in Section 2.2, we mentioned that the answer had no units, since we divided a number of pounds by a number of pounds. We are going to look further at that situation now by considering dimensions. The units in which we measure a value, (eg pounds, metres, kilograms), affect the numerical value given, eg lengths of 2.7m and 270cm represent the same quantity. This creates problems when we wish to compare values, since we need to ensure that they are measured in the same units, and this conversion can be complicated (for example converting a speed in metres per second into one in miles per hour). Dimensions are used to show what a value represents, in other words we would say that 2.7m or 270cm both have the dimension of length. In actuarial work the dimensions usually involve currency, time and people. For example, if it was stated that the average salary in the UK was £20,000, the units of this would be pounds per year per person. Numbers or coefficients (including constants such as p ) have a dimension of zero and they are referred to as dimensionless. If two values that have the same dimension are divided, then the resulting value is dimensionless. This is true of percentage or proportionate errors. Other values that you will meet in Subject CT3, which are dimensionless, are the correlation coefficient and the coefficient of skewness. Dimensions can give us a convenient way of telling if a formula is correct.

Example What is the dimension of the mean length of life of a light bulb? Solution The mean is defined to be the sum of a set of lifetimes, divided by the number of

lifetimes considered, ie 1

1 n

ii

x xn =

= Â . “Length of life” is measured in time, and n is

dimensionless, so the dimension of the mean is T, where T stands for time.



Question 5.5

If the variance is defined to be ( )21

1

1

n

ii

x xn =

-- Â , what is the dimension of variance for

the previous example?

Question 5.6

Given the definition of the variance in Question 5.5, by considering dimensions, decide which of the following could be other possible formulae for variance.

(i) 2 2

1

1

1

n

ii

x xn =

-- Â

(ii) 2

1

( 1)n

ii

x n x=

- -Â

(iii) 2

1 1

1 1

1 1

n n

i ii i

x x xn n= =

-- -Â Â

The argument of functions such as xe and log x must be dimensionless.

Question 5.7

Why must they be dimensionless?



The dimensions also tell you how the result of a formula will be affected if the values of the components are rescaled eg multiplied by 10.

Example A discussion is being held about the wages for the workers in a small firm. The director has said “If I double all wages then the mean and variance of the wages will also double”. Is she correct? Solution The dimensions of the mean of wages will be in the currency unit eg pounds. However the dimensions of the variance will be in the currency unit squared, eg pounds squared. This means that the variance will be multiplied by 4 and not 2, so the director is wrong.



5 Interpolation and extrapolation

Sometimes it is not possible to find an exact solution to an equation such as ( ) 0f x =

by algebraic methods. Under such circumstances an approximate value can be found if you know two values of the function, evaluated at values of x close to the true value, say 1 2 and x x . If the true value is in between these, it is called interpolation; otherwise

it is extrapolation. The method we are going to look at is called linear interpolation. As the name suggests, this method relies on the function being approximately linear between 1 2and x x .

The disadvantages of this method are: ● It can only give an approximate value (unless the function is truly linear)

● The answer obtained will only be close to the true value of x if 1 2 and x x are

relatively close to the true value

● The answer obtained will only be close to the true value of x if ( )f x is

approximately linear between 1 2and x x .

In practice, as long as you are aware that your answer may be slightly inaccurate, this is often an appropriate method to use and it is the method usually expected in exam solutions. To illustrate what we mean consider the following diagram:

0 x1 x

y

x2

error



This shows the graph of ( )y f x= , with the close values 1 2and x x , and the error that

this interpolation has introduced. Do be aware however that we have drawn a large-scale diagram and not chosen two values close to the true values just for illustration! To show how interpolation works in practice, consider the following diagram:

x1 x x2

f(x1) f(x2)f(x)

By considering the ratio of ‘lengths’ above and below the line, we can write:

1 2 1

1 2 1( ) ( ) ( ) ( )

x x x x

f x f x f x f x

- -=- -

or 1 1

2 1 2 1

( ) ( )

( ) ( )

x x f x f x

x x f x f x

- -=- -

Consider the first version. This can be rearranged to give:

11 2 1

2 1

( ) ( )( )

( ) ( )

f x f xx x x x

f x f x

-= + --

This gives us a formula for x . It is better for you to understand how this formula is derived rather than trying to learn it!

Question 5.8

If x does not lie between 1 2and x x , ie we are dealing with extrapolation, derive a

formula for finding x.

To see how this works in practice, look at the next example.



Example If (1.34) 0.523f = , and (1.65) 0.42f = - , what is an approximate value for x if

( ) 0f x = ?

Solution We know that x lies somewhere between 1.34 and 1.65, so:

1.65 1.65 1.34

0 ( 0.42) 0.42 0.523

x - -=- - - -

which gives 1.51x = .

Question 5.9

By finding two consecutive integer values which x lies between, and using linear

interpolation, find a value of x that satisfies the equation 3 25 3 7 0x x x- + + = .

Interpolation can be used to find the value of “x” or the value of the function.

Question 5.10

This is an extract from the tables of a function ( )F x :

x ( )F x

0.54 0.70540 0.55 0.70884 0.56 0.71226 Using linear interpolation, work out: (i) the value of ( )F x when 0.548x =

(ii) the value of x when ( ) 0.71F x = .



6 Iteration

There are some equations that cannot be solved using ‘standard’ methods to find an exact solution, but an approximate solution can usually be found using a numerical method based on iteration.

6.1 Bisection

The bisection method is based on finding two values between which the solution lies and then using the midpoint of the values for the next approximation.

Example

Find an approximate negative solution to the equation 3 12 7 0x x- - = . Give your answer to 1 DP. Solution

Let 3( ) 12 7f x x x= - - , so we are trying to solve ( ) 0f x = .

(0) 7 and ( 1) 4 the root lies between 0 and 1

( 0.5) 1.125 the root lies between 0.5 and 1

( 0.75)=1.578125 the root lies between 0.5 and 0.75

( 0.625) 0.255859 the root lies between 0

f f

f

f

f

= - - = fi -

- = - fi - -

- fi - -

- = fi - .5 and 0.625

( 0.5625) 0.42798 the root lies between 0.5625 and 0.625f

-

- = - fi - -

Therefore the answer is –0.6.

Question 5.11

Using the bisection method, find i to 3DP if i satisfies the equation:

( )2

320 1 (1 )

25(1 ) 61.5i

ii

--

- ++ + =



6.2 Newton-Raphson iteration

The Newton-Raphson iterative formula states that if nx is an approximate solution to the

equation ( ) 0f x = then a better approximation is 1( )

( )n

n nn

f xx x

f x+ = -¢

.

Example Using the Newton-Raphson formula, find an approximate positive solution to the

equation 5 24 7x x+ = , giving your answer to 2 DP. Solution

Let 5 2( ) 4 7f x x x= + - , so that:

4( ) 5 8

(1) 2

(2) 41

f x x x

f

f

Therefore there is a solution between 1 and 2. Try 1 1x = .

2

3

4

1.15385

1.13336

1.13290

x

x

x

=

=

=

Therefore the solution is 1.13 (to 2DP).

Question 5.12

Using the Newton-Raphson formula, find an approximate solution to the equation:

5

4 10(1 (1 ) )5(1 ) 41

ii

i

-- - ++ + =

Give your answer to 3 DP.



7 Complex numbers

7.1 Basic algebra

A complex number is represented in the form a ib+ , where 1i = - , and a and b are real numbers. a is called the “real part” and b is called the “imaginary part”. You may also see complex numbers written with j’s, rather than i’s. Complex numbers can be added, subtracted and multiplied as follows: (i) ( ) ( ) ( ) ( )a ib c id a c i b d+ + + = + + +

(ii) ( ) ( ) ( ) ( )a ib c id a c i b d+ - + = - + -

(iii) ( )( ) ( ) ( )a ib c id ac bd i ad bc+ + = - + +

The rules for addition and subtraction just require the real and imaginary parts to be added separately. The logic of the rule for multiplication can be seen by multiplying

out the brackets and using the fact that 2 1i = - .

Question 5.13

Find: (i) (2 3 ) (3 4 )i i+ + - (ii) (4 6 ) (3 7)i i- - -

(iii) (3 4 )(6 2 )i i- + (iv) (4 2 )(4 2 )i i+ -

The answer to last part of this question, where the product of two complex numbers is a real number, leads to an important property of complex numbers. a ib+ and a ib- are called “complex conjugates”. Their product is the real number

2 2a b+ . The complex conjugate of the number a ib+ is a ib- , and vice versa. This enables us to divide complex numbers.



Example

Simplify 2 3

1 2

i

i

-+

.

Solution If we multiply the numerator and denominator by the complex conjugate of the denominator, we get: 2 3 1 2 2 3 4 6 4 7 1

( 4 7 )1 2 1 2 1 4 5 5

i i i i ii

i i

- - - - - - -¥ = = = - -+ - +

This is in the form a ib+ , where 0.8a = - and 1.4b = - .

Question 5.14

Simplify 3 4

1 4

i

i

--

.

7.2 Argand diagrams

Complex numbers can be represented on an “Argand diagram”, where you plot the imaginary part against the real part. For example, the complex number 1 3z i= - can be shown as follows:

0

z

Re

Im

1

3

2

1 2 3-1-2-3

-1

-2

-3



Notice that it is the point that represents the complex number, not the line. There is a reason for including the line though, which we will describe shortly. By representing a complex number on an Argand diagram, it is easy to picture two further properties of complex numbers.

The modulus, r, of a complex number a ib+ is given by 2 2r a b= + , ie the square

root of the sum of the real part squared and the imaginary part squared. It is equivalent to the length of the line shown on the Argand diagram. The argument, q , of a complex number a ib+ is defined to be the angle between the line and the positive x-axis on the Argand diagram. In radians, p q p- < £ . This diagram illustrates this for two complex numbers 1z and 2z , with moduli 1r and

2r , and arguments 1q and 2q respectively.

Note that in the diagram, 2q is negative. By convention, angles measured

anticlockwise from the positive real axis are positive, and angles measured clockwise are negative.

Re

Im

z2

z1

r1

1

r2

2



Example Find the modulus and argument of the complex number 3 i- - . Solution

The modulus is 2 21 3 10 3.16+ = = .

The argument is 1 1tan 2.82

3

.

Question 5.15

Find the modulus and argument of the complex number 1 2

4 3

i

i

--

.

7.3 Euler’s formula

As we have seen, complex numbers can be represented as points on an Argand diagram.

The number ie q is represented by the point on the unit circle (the circle centred at the origin with radius 1) that forms an angle q with the real axis.

Re

Im

1

y

x

ei



We can work out the real and imaginary parts of this number (ie the horizontal and vertical co-ordinates) by drawing the triangle shown in the diagram. From basic

trigonometry, we know that adjacent

cosinehypotenuse

= , so that cos1

xq = ie cosx q= .

Similarly siny q= . So ie q has real part cosq and imaginary part sinq . In other

words:

cos sinie iq q q= + This is known as “Euler’s formula” and it holds for any value of q (including complex ones). Note, however, that q must be measured in radians not degrees.

Question 5.16

What are the co-ordinates of the numbers exp4

ip and 2exp

6

ip on the Argand diagram?

Note that any complex number can be expressed in the form ire q , which is known as polar form. q is the argument and r is the modulus.

Question 5.17

Express the complex number 7 5i+ in polar form.

It is useful to look at the complex conjugate of ie q . The complex conjugate can be

found by replacing i by i- in the number, so here it is ie q- . Since cos sinie iq q q= + ,

we can replace i by i- to give cos sinie iq q q- = - .

Re

Im

ei

e– i



Question 5.18

How can we deduce from this that cos( ) cosq q- = and sin( ) sinq q- = - ?

Question 5.19

Show that the product of ie q- and ie q does give a real number.

In many applied maths applications of complex numbers you will find that you need to

add ie q- and ie q . This gives:

cos sin cos sin 2cosi ie e i iq q q q q q q- + = - + + = Note that this also gives a purely real number.

Question 5.20

Show that 51 12 2 4(1 )(1 ) cosi ie ew w w-+ + = + .



7.4 Solution of polynomial equations

If a polynomial has real coefficients, then any complex roots must occur in complex conjugate pairs. Allowing complex roots means that any quadratic equation will have possible solutions.

Example

Find the roots of the equation 2 3 4 0z z- + = . Solution Using the quadratic formula, we get:

3 ( 1) 73 9 16 3 7 3 7

2 2 2 2

iz

± -± - ± - ±= = = =

Simplifying this we get:

7 73 32 2 2 2 or z i i= + -

Notice that these roots are a complex conjugate pair.

Question 5.21

You are given that 1 2z i= + is a root of the cubic equation 3 24 9 10 0z z z- + - = . Find the other two roots.



8 Difference equations

A difference equation is typically of the form 1 2 0t t tay by cy- -+ + = , where a , b and

c are constants. The equation has a solution of the form ( )ty f t= and it has an

associated auxiliary equation 2 0a b cl l+ + = . The general solution of the difference equation depends on the roots, 1l and 2l , of the

auxiliary equation.

2 4b ac- Solution of difference equation Note

0> 1 2t t

ty A Bl l= +

The auxiliary equation has two distinct real roots

1 2l lπ

0= ( ) tty A Bt l= +

There is only one (repeated) root

1 2l l l= =

0< ( cos sin )tty r A t B tq q= +

The complex roots of the auxiliary equation are

written in polar form ie

1 2,i ire req ql l -= =

These results are given on page 4 of the Tables. Once we have the general solution, we can then obtain a particular solution (ie we can obtain values of A , B and r ) if some extra information is given (called “boundary conditions”).

Example Solve the difference equations: (i) 1 25 6 0t t ty y y- -- + = given that 0 0y = and 1 1y =

(ii) 2 16 13 0t t ty y y+ +- + = .



Solution

(i) The auxiliary equation is 2 5 6 0l l- + = which has roots 2 and 3.

The general solution is of the form 2 3t tty A B= ¥ + ¥ .

However we know that 0 0y = and 1 1y = , so:

0 00

1 11

0 2 3 0

1 2 3 1 2 3

y A B A B

y A B A B

= = ¥ + ¥ fi = +

= = ¥ + ¥ fi = +

These can be solved to give 1B = and 1A = - , so our particular solution is:

( 1) 2 1 3 3 2t t t tty = - ¥ + ¥ = -

(ii) The auxiliary equation is 2 6 13 0l l- + = , which has roots

6 36 523 2

2i

± - = ± .

The general solution is of the form ( cos sin )tty r A t B tq q= + .

The modulus of these complex numbers is 2 23 2 13r = + = , and the

argument is 1 23tan 0.588q -= = , so the roots can be written in polar form as

0.58813 ie± . The general solution of the difference equation is then

( ) ( )13 cos0.588 13 sin 0.588t t

ty A t B t= + .

Question 5.22

Solve the difference equations: (i) 1 28 16 0t t ty y y- -- + = given that 0 1y = - and 1 3y =

(ii) 1 24 5 0t t ty y y- -- + = .



Chapter 5 Summary Percentages

To find p % of a quantity: quantity100

p ¥

To find a p % increase or decrease of a quantity:

100

quantity100

p+ ¥ 100

quantity100

p- ¥

To express A as a percentage of B : 100%A

B¥

Changes

absolute change new value original value= -

absolute change

proportionate changeoriginal value

=

absolute change

percentage change 100%original value

= ¥

Errors

absolute error approx value true value= -

absolute error

proportionate errortrue value

=

absolute error

percentage error 100%true value

= ¥

Dimensions

Dimensions tell us what a value represents and can be used to tell if formulae are correct.



Numerical methods

When an equation cannot be solved directly to get an exact solution, a numerical method such as interpolation or iteration needs to be used. Linear interpolation

1 1 11 2 1

2 1 2 1 2 1

( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( )

x x f x f x f x f xx x x x

x x f x f x f x f x

- - -= fi = + -- - -

Bisection If solution x lies between 1x and 2x then use 1 2½( )x x+ as next approximation.

Newton-Raphson If nx is approximate solution to ( ) 0f x = then a better approximation is obtained by:

1( )

( )n

n nn

f xx x

f x+ = -¢

.

Complex numbers

Any complex number z can be written in the form z a ib= + , where 1i = - . The complex conjugate of z is a ib- .

Complex numbers can be represented on an Argand diagram:

The modulus of z is given by 2 2r x y= + .

The argument of z is given by 1tan yxq -= .

The complex number z can be written in polar form iz re q= .

Euler’s formula states that cos sinie iq q q= + .

Complex roots of polynomial equations with real coefficients must occur in complex conjugate pairs. Second order difference equations

The general solution 1 2 0t t tay by cy- -+ + = depends on the roots 1 2( and )l l of the

auxiliary equation, 2 0a b cl l+ + = : See formulae on page 4 of the Tables.

y

x

z = x + yi

r




(i) In each year the population increases by 5%, so the new population is 100% 5% 105%+ = of the old population. So we will multiply by 1.05 each year. If the starting population is x then we need to find out how many years we multiply by 1.05 before we get 2x :

1.05 2 1.05 2n nx x¥ = fi =

Taking logs of both sides and rearranging:

ln 2

ln1.05 ln 2 14.2067ln1.05

n n= fi = =

So we will need at least 15 years for the population to double in size. (ii) Each year we will multiply by 1.05. So using x to stand for the population in

2010 then we have:

22

66,150,0001.05 66,150,000 60,000,000

1.05x x¥ = fi = =

Solution 5.2

0.66% is equivalent to saying 6.6‰. Solution 5.3

Absolute change is £467. Proportionate change is 0.0321. Percentage change is 3.21%. Solution 5.4

The true value could have been any number from 0.0265 to 0.0275. So we will use the extreme precise values to find the largest percentage error.



If 0.0265i = then the true calculation would have been 11.0265v -= compared to the

approximate calculation of 11.027v -= so this gives a percentage error of:

1 1

1

1.0265 1.027100% 0.048685%

1.0265

- -

-- ¥ =

Similarly, if 0.0275i = then the percentage error would be:

1 1

1

1.0275 1.027100% 0.048685%

1.0275

- -

-- ¥ = -

In this case they are both the same magnitude to 5 significant figures. So the largest percentage change is 0.048685% (whether up or down). Solution 5.5

Since the variance is calculated by squaring the x values, the dimension of the variance

here is 2T , where T is the dimension of time. Solution 5.6

The other possible formulae are (i) and (iii). The second one can’t be correct because

the dimension of the first term is 2T , while that of the second term is T, and terms of different dimensions cannot be subtracted. Solution 5.7

Consider the series for xe which is 2

12!

xx+ + + . If x had a dimension, then each

term of this series would have a different dimension and so this series would be meaningless. Therefore x must be dimensionless. A similar reasoning applies to log x .

Solution 5.8

You use exactly the same formula for extrapolation as you do for interpolation!



Solution 5.9

By trial and error, the consecutive integers can be found. They are 0 and –1, or 2 and 3, or 3 and 4.

Between 0 and –1, interpolating gives ( 1) 0 ( 2)

0.7780 ( 1) 7 ( 2)

xx

- - - -= fi = -- - - -

.

Between 2 and 3, interpolating gives 2 0 1

2.3333 2 2 1

xx

- -= fi =- - -

.

Between 3 and 4, interpolating gives 3 0 ( 2)

3.44 3 3 ( 2)

xx

- - -= fi =- - -

.

Solution 5.10

(i) Using the formula for interpolation:

(0.70884 0.70540)(0.548 0.54)(0.548) 0.70540

0.55 0.54

0.70815

F- -= +

-

=

(ii) Using the formula:

(0.56 0.55)(0.71 0.70884)0.55

0.71226 0.70884

0.553

x- -= +

-

=



Solution 5.11

When 0.03i = the LHS gives 61.15, and when 0.02i = the LHS gives 62.39. Using the bisection method, we get:

Value of i Value of equation

0.025 61.76 0.0275 61.45 0.02625 61.61 0.026875 61.53

So the value of i is 0.027 to 3DP. Solution 5.12

Let 5

4 10(1 (1 ) )( ) 5(1 ) 41

if i i

i

-- - += + + - , so that:

6 5

52

6 55

2

10 5(1 ) 10(1 (1 ) )( ) 20(1 )

50 (1 ) 10 10(1 )20(1 )

i i if i i

i

i i ii

i

- --

- --

¥ ¥ + - - += - + +¢

+ - + += - + +

The Newton-Raphson formula gives the formula for the next approximation to be:

54

6 55

2

10(1 (1 ) )5(1 ) 41

50 (1 ) 10 10(1 )20(1 )

ii

iii i i

ii

--

- --

- ++ + --

+ - + +- + +

(0.1) 0.323f = and (0.11) 0.747f = - , so the root lies between 0.1 and 0.11.

Try 1 0.1i = .

The formula gives 2 0.1029557i = , 3 0.1029739i = , ie the root is 0.103 to 3 DP.



Solution 5.13

(i) (2 3 ) (3 4 ) 5i i i+ + - = -

(ii) (4 6 ) (3 7) 11 9i i i- - - = -

(iii) 2(3 4 )(6 2 ) 18 24 6 8 26 18i i i i i i- + = - + - = -

(iv) 2(4 2 )(4 2 ) 16 8 8 4 20i i i i i+ - = - + - =

Solution 5.14

Multiplying the numerator and denominator by the complex conjugate of the denominator:

2

2

3 4 1 4 3 4 16 12 16 13

1 4 1 4 171 16

i i i i i i

i i i

- + - - + - -¥ = =- + -

Solution 5.15

We first need to simplify the complex number:

21 2 4 3 4 8 3 6 10 5 2

4 3 4 3 16 9 25 5

i i i i i i i

i i

- + - + - - -¥ = = =- + +

The modulus is then 2 2

2 10.4472

5 5

.

The argument is 1 1tan 0.46

2-- = - .



Solution 5.16

Using Euler’s formula 4 4

1 1exp cos sin

4 2 2

ii ip pp = + = + so the co-ordinates are:

1 1,

2 2

Similarly 6 6

3 12exp 2(cos sin ) 2

6 2 2

ii i

, so the co-ordinates are ( 3,1) .

Solution 5.17

Drawing a diagram, we have:

r

7

5

Using Pythagoras’ Theorem the modulus is:

2 27 5 74r = + =

Using trigonometry, the argument is:

15 5tan tan 0.62025

7 7q q - Ê ˆ= fi = =Á ˜Ë ¯

Don’t forget that your calculator should be in radians!



Solution 5.18

We could work out ie q- by thinking of it as ( )ie q- and applying the original form of Euler’s formula:

( ) cos( ) sin( )ie iq q q- = - + -

Comparing this with the other version for ie q- , we see that cos( )q- must equal cosq

and sin( )q- must equal sinq- .

These properties of cosines and sines can be described by saying that cosine is an “even” function and sine is an “odd” function. Solution 5.19

Consider their product:

2 2 2

2 2

(cos sin )(cos sin )

cos sin cos sin cos sin

cos sin 1

i ie e i i

i i i

q q q q q q

q q q q q q

q q

- = - +

= - + -

= + =

which is a real number. You could, of course, have used the basic properties of powers

to say that 0 1i i i ie e e eq q q q- - += = = .



Solution 5.20

Multiplying out gives:

1 1 1 1 1 12 2 2 2 2 2

1 12 4

5 14 2

54

(1 )(1 ) 1

1 ( )

2cos

cos

i i i i i i

i i

e e e e e e

e e

w w w w w w

w w

w

w

- - -

-

+ + = + + + ¥

= + + +

= + ¥

= +

Solution 5.21

We are told that 1 2i+ is a root, and since the polynomial has real coefficients, we know that 1 2i- must also be a root. The factors are therefore:

( (1 2 ))( (1 2 ))z i z i- + - -

and some other factor which is, as yet, unknown. Multiplying these factors out we get:

2 2 2( 1 2 )( 1 2 ) ( 1) 4 2 5z i z i z i z z- - - + = - - = - +

The cubic can then be written as:

3 2 24 9 10 ( 2 5)( 2)z z z z z z- + - = - + -

Therefore the other two roots are 2 and 1 2i- .



Solution 5.22

(i) The auxiliary equation is 2 8 16 0l l- + = , which has a repeated root of 4.

The general solution is given by ( ) 4tty A Bt= + ¥ .

However we know that 0 1y = - and 1 3y = , so we have the simultaneous

equations:

00

11

1 ( 0) 4 1

3 ( 1) 4 3 4( )

y A B A

y A B A B

= - = + ¥ ¥ fi - =

= = + ¥ ¥ fi = +

which are solved to give 1A = - and 1.75B = .

The particular solution is then ( 1 1.75 ) 4tty t= - + ¥ .

We can check, for example, that 2 8 3 16( 1) 40y = ¥ - - = , which agrees with the

formula just found.

(ii) The auxiliary equation is 2 4 5 0l l- + = , which has roots:

4 16 202

2i

± - = ± .

These complex numbers have modulus:

2 22 1 5r = + =

and argument:

11 12 2tan tan 0.464q q -= fi = =

So they can be written in polar form as 0.4645 ie± .

The general solution of the difference equation is then:

( ) ( )5 cos0.464 5 sin 0.464t t

ty A t B t= +

FAC-06: Differentiation Page 1


Chapter 6

Differentiation

You need to study this chapter to cover: ● limits and the order notation

● derivatives as rates of change of functions

● derivatives of the standard functions

● derivatives of sums, products, quotients and “functions of a function”

● higher-order derivatives

● maxima, minima and stationary points

● partial derivatives

● extrema of functions of two variables

● Lagrangian multipliers.

0 Introduction

This chapter covers all the methods of differentiation that you need in order to study the Core Technical subjects.

FAC-06: Differentiation


1 Limits

1.1 Limits

The value of 1

( )1

f xx

=-

cannot be directly calculated at the point 1x = . However,

we can say that when x is slightly bigger than 1, say 1x e= + , where e is a small positive number, then ( )f x is going to be very large and positive. This is obviously a

long-winded and cumbersome way of saying what happens, so we use the “limit notation”:

1

1lim

1x x+Æ= •

-

Remember that 1x +Æ means that x is approaching 1 from above.

Question 6.1

What is 1

1lim

1x x-Æ -?

Question 6.2

What is 2

2

3 2lim

3 4x

x x

x xÆ•

- +-

?

In order to find limits, you may want to consider the value of the function close to the limit. For example, in Question 6.2 you might wish to calculate the value of the function for larger and larger positive values of x. In Question 6.1 you could substitute in, say, 0.9, 0.99, 0.999 and so on. This will often give you some idea of whether the limit is finite or not (and if so, what its value is).



1.2 The order notation

If you studied series at school you will have seen series such as:

2 31 12 61xe x x x= + + + +

This series goes on for ever, and in some contexts it is important to know how the “remainder” terms, the bits contained in the “” behave. One way to be a little more accurate would be to write this series as:

2 3121 terms in xe x x x= + + + etc

What we are thinking here is that, if x is a fairly small number (say 0.1) then the term

in 3x will be much smaller than the earlier terms, and the terms in even higher powers will be smaller still. In algebraic calculations where it is important to be clear about the behaviour of the remainder term, we can use a special notation where we would write:

2 3121 ( )xe x x O x= + + + as 0x Æ

or:

2 2121 ( )xe x x o x= + + + as 0x Æ

Note that the first equation uses a capital “O” while the second uses a small “o”. O() Notation

The 3( )O x in the first equation is read as “Big-Oh of x cubed”. You can think of this as

representing some function that is no bigger than a fixed multiple of 3x .

Mathematically speaking, a function ( )f x is 3( )O x if you can find at least one

constant 0K > such that 3( )f x K x£ for all sufficiently small values of x .

Another way of writing this would be to say that 3

( )f xK

x£ , ie 3( )f x x is bounded by

some constant K for sufficiently small values of x .



o() Notation

The 3( )o x in the second equation is read as “Small-Oh of x cubed”. You can think of

this as representing some function that is smaller than any fixed multiple of 3x .

Mathematically speaking, a function ( )f x is 3( )o x if for all constants 0K > ,

3( )f x K x£ for all sufficiently small values of x .

Another way of writing this would be to say that 3

( )f xK

x£ , ie 3( )f x x is bounded by

all positive constants K for sufficiently small values of x , which means that

30

( )lim 0x

f x

xÆ= .

So small o is a stronger condition than big O . Note that if you’re consistently using big O ’s or small o ’s in a derivation you can just say “order of x cubed” etc.

Example

Which of the following functions are 3( )O x and/or 3( )o x for small values of x ?

(i) 25x

(ii) 30.00001x

(iii) 4999,999,999x

(iv) x xe e

x

--

Solution

(i) 2

3

5 5x

xx= so as x gets smaller 5 x gets larger so it will not be bounded by

some constant M , so it’s not 3( )O x . It’s also not heading to zero, so it’s not 3( )o x .



(ii) 3

3

0.000010.00001

x

x= so it’s clearly bounded by 0.00001M = (or any bigger

number!) and hence it is 3( )O x . But it doesn’t approach zero, so it’s not 3( )o x .

Notice that the 0.00001 factor doesn’t affect the orders here.

(iii) 4

3

999,999,999999,999,999

xx

x= so it will be bounded for small values of x

(for example it will be bounded by 999,999,999 for values of x smaller than 1)

and hence it is 3( )O x . It also heads to zero as 0x Æ so it is also 3( )o x .

Again the 999,999,999 factor doesn’t make any difference.

Note that being 3( )o x is a stronger condition than being 3( )O x . Any function

that is 3( )o x is automatically 3( )O x .

(iv) For x xe e

x

--, we have to do a “calculation” involving the order symbols. We

know that:

2 3121 ( )xe x x O x= + + +

and:

2 3121 ( )xe x x O x- = - + +

Notice that we’ve written 3( )O x+ rather than 3( )O x- in the second equation.

We can do this because these two symbols are equivalent. If we now evaluate the numerator, we get:

2 3 2 31 12 21 ( ) 1 ( )x xe e x x O x x x O x- È ˘ È ˘- = + + + - - + +Î ˚ Î ˚

Simplifying:

32 ( )x xe e x O x-- = +



Again notice that adding two 3( )O x terms just gives 3( )O x .

We now need to divide by x :

3( )

2x xe e O x

x x

-- = +

This last term is a function that is no bigger than a fixed multiple of 3x divided

by x . This will give a function that is no bigger than a fixed multiple of 2x ie 2( )O x .

So:

22 ( )x xe e

O xx

-- = +

Since 2( )O x is “bigger” than 3( )O x , we can conclude that x xe e

x

-- is neither

3( )O x nor 3( )o x .

You may also see the order notation used to describe the behaviour of a function for large values of x .



Example

Explain what the statement 211 ( )

1

xO x

x x-= + +

- as x Æ +• means.

Solution

This means that the behaviour of the function 1

x

x - is very similar to the behaviour of

the function 1

1x

+ for large values of x. The remainder (discrepancy) is a term that

approaches zero more quickly than any fixed multiple of 2

1

x.

You can see this if you expand 1

x

x - by writing it as:

( ) 111 2 3

1 1 1 11 1

1 1 xx

x

x x x x

-= = - = + + + +

- -

1.3 Supremums and infimums

For finite sets of numbers, you can always identify which element(s) is/are the largest/smallest. For infinite sets, this is not always possible, and we have to generalise the idea of maximum/minimum. The supremum (sup), or least upper bound of a set A is defined as the number a such that and a a A a a aa a≥ " Œ " < $ >¢ . In other words sup( )A is the number that is “never quite” exceeded by any member of

the set. It is literally the “least upper bound”.

For example, consider the sequence 31 22 3 4, , , . This sequence has no maximum, but

the supremum of the sequence is 1, since any number less than 1 will eventually be exceeded and any other upper bound would be greater than 1.



Similarly the idea of the minimum element is generalised to give the infimum or greatest lower bound. The infimum (inf), or greatest lower bound of a set A is defined as the number a such that and a a A a a aa a£ " Œ " > $ <¢ . Note that any other lower bound a ¢ is such that a a£¢ . Hence a is the greatest lower bound.

For example, consider the sequence 1 1 12 4 8, , , . This sequence has no minimum, but

the infimum of the sequence is 0. Note that for finite sets A, sup( ) max( )

a AA A

Œ= and inf( ) min( )

a AA A

Œ= .



2 Differentiation

Consider the function ( )y f x= , with two points lying on the graph of this function,

( )1 , ( )P x f x= and ( )2 , ( )P x h f x h= + + . Then the average rate of change of ( )y f x=

from 1P to 2P is given by ( ) ( )f x h f x

h

+ -.

We then define 0

( ) ( )limh

f x h f x

hÆ

+ - to be the rate of change of y with respect to x at

the point 1P , which is written as dy

dx or ( )

df x

dx or ( )f x¢ , and is known as the

derivative of y with respect to x.

Since ( ) ( )f x h f x

h

+ - is also the gradient of the straight line joining 1P to 2P ,

(technically known as a “chord”) then 0

( ) ( )limh

f x h f x

hÆ

+ - is the gradient of the curve

( )y f x= at the point 1P , since as 0h Æ , the chord gets closer and closer to the tangent

at 1P , and the gradient of the tangent is the same as the gradient of the curve at 1P .

P1 (x, f(x))

P2 (x+h, f(x+h))



In summary then:

0

( ) ( )( ) lim

h

f x h f xf x

hÆ

+ -=¢

which is the gradient of the graph of ( )y f x= at the point ( , )x y .

Example

By considering the gradient of the chord joining two points on the curve 2y x= , find

dy

dx.

Solution

Consider the points 21( , )P x x and 2

2(( ),( ) )P x h x h+ + . The gradient of the chord

joining them is given by 2 2( )x h x

h

+ -.

Simplifying this gives:

2 2 2 2 2 2( ) 2 22

x h x x xh h x xh hx h

h h h

+ - + + - += = = +

Since 2 2

0

( )lim 2h

x h xx

hÆ

+ - = , we have:

2dy

xdx

=

Note that the method used in the above example is called differentiation from first principles. For most work we will develop rules for finding derivatives that avoid the need for differentiation from first principles. However, it is important for you to understand the concept of differentiation first.

Question 6.3

Differentiate 22 3 4y x x= + + from first principles.



3 Differentiation of standard functions

Some standard functions and their derivatives are shown in the table below:

Function Derivative nx 1nnx - 0n π xc lnxc c xe xe

ln x 1

x

Example Differentiate the following with respect to x:

(i) 7y x= (ii) 9xy = (iii) ln(3 )y x=

Solution

(i) 121

2

77

2

dyx

dx x

-= ¥ = (Remember that 12x x= )

(ii) 9 ln9xdy

dx=

(iii) Rewriting ln(3 ) ln3 lny x x= = + hence we have 1dy

dx x=

Question 6.4

Differentiate the following functions with respect to x:

(i) 2

3

x (ii) 3 42 x (iii) 22ln(3 )x (iv) 3 1xe +



4 Products and quotients

Very often the functions that you need to differentiate are more complicated than those dealt with in Section 3. For example they may be a product or quotient of a pair of the standard functions given in the previous section.

4.1 Products

If u and v are functions of x, and ( )f x uv= , then ( )dv du

f x uv vu u vdx dx

= + = +¢ ¢ ¢ . This

is called the product rule.

Example If ( ) lnf x x x= , what is ( )f x¢ ?

Solution

1( ) 1 ln 1 lnf x x x x

x= ¥ + ¥ = +¢

To differentiate a product you differentiate each factor in turn (leaving the other factor the same) and add the results together. This rule also works when there are more than two factors involved, provided you do it properly. For example if y uvw= ,

( )dy d du

u vw vwdx dx dx

= ¥ + ¥ .

Question 6.5

A sum of money is paid continuously at a rate 2( ) 100 tt ter = , for 0 2t< < . At what

rate is the payment increasing at time 1.5t = ?



4.2 Quotients

If u and v are functions of x, and ( )u

f xv

= , then 2 2( )du dvdx dxv uvu uv

f xv v

--¢ ¢= =¢ . This

is called the quotient rule.

Example

If ( )3 1

xf x

x=

+, what is ( )f x¢ ?

Solution

2 2

(3 1) 1 3 1( )

(3 1) (3 1)

x xf x

x x

+ ¥ - ¥= =¢+ +

4.3 Chain rule

This is used to differentiate expressions involving nested functions. So instead of nx

we may have [ ( )]nf x . Similarly, instead of xe we may have ( )f xe and instead of ln x

we may have ln ( )f x .

Differentiating these nested functions will be too hard/messy using standard techniques.

For example, to differentiate 2 5(3 2 1)x x+ - we’d have to multiply out the bracket

which would be extremely time consuming! So what we do is call the nested function “u ” and then use the chain rule:

dy dy du

dx du dx= ¥



Example

If 2 5(3 2 1)y x x= + - , what is dy

dx?

Solution

So first of all we set 23 2 1u x x= + - . So we have 5y u= . Now we’ll differentiate our

two parts:

6 2du

xdx

= +

45dy

udu

=

Hence:

45 (6 2)dy dy du

u xdx du dx

= ¥ = ¥ +

Replacing u gives:

2 45(3 2 1) (6 2)dy

x x xdx

= + - ¥ +

In practice these stages are not written out in full as the answer can be written out

immediately. When differentiating [ ( )]nf x , think of differentiating the bracket as

“normal” ie bring the power down and decrease the power by one which gives 1[ ( )]nn f x - . Then multiply by the derivative of the bracket, ( )f x¢ .



Question 6.6

If 2 6(3 4)p r= - , what is dp

dr?

Let’s now have a look at an example involving the exponential function.

Example

If 35 1xy e -= , what is

dy

dx?

Solution

First of all we set 35 1u x= - . So we have uy e= . Now we’ll differentiate our two

parts:

215du

xdx

=

udye

du=

Hence:

215udy dy due x

dx du dx= ¥ = ¥

Replacing u gives:

35 1 215xdy

e xdx

-= ¥

When differentiating ( )f xe , think of differentiating the exponential as “normal” ie leave

it as it is ( )f xe . Then multiply by the derivative of the power ( )f x¢ .



Question 6.7

Differentiate 23xy e-= with respect to x.

Let’s now have a look at an example involving the log function.

Example Differentiate ln(3 2)y x= + .

Solution First of all we set 3 2u x= + . So we have lny u= . Now we’ll differentiate our two

parts:

3du

dx=

1dy

du u=

Hence:

1

3dy dy du

dx du dx u= ¥ = ¥

Replacing u gives:

1 3

33 2 3 2

dy

dx x x= ¥ =

+ +

When differentiating ln ( )f x , think of differentiating the log as “normal” ie 1 ( )f x .

Then multiply by the derivative of the function ( )f x¢ .

Question 6.8

Differentiate 32ln(4 5 1)y x x= - + with respect to x.



In summary our “shortcut” rules are:

Function Derivative

[ ( )]nf x 1[ ( )] ( )nn f x f x- ¥ ¢

( )f xe ( ) ( )f xe f x¥ ¢

ln ( )f x 1

( )( )

f xf x

¥ ¢

We’ll now look at an example which involves using the chain rule twice!

Example

Differentiate 2

log(1 2 )xeÈ ˘+Î ˚ with respect to x .

Solution

Using the shortcut chain rule to differentiate [ ( )]nf x we have:

2

log(1 2 ) 2 log(1 2 ) log(1 2 )x x xd de e e

dx dxÈ ˘ È ˘+ = + ¥ +Î ˚ Î ˚

But to find log(1 2 )xde

dx+ we need to use the chain rule for ln ( )f x :

1 1 2log(1 2 ) (1 2 ) 2

1 2 1 2 1 2

xx x x

x x x

d d ee e e

dx dxe e e+ = ¥ + = ¥ =

+ + +

Returning to our original expression, the required answer is:

2 2 4log(1 2 ) 2 log(1 2 ) log(1 2 )

(1 2 ) (1 2 )

x xx x x

x x

d e ee e e

dx e eÈ ˘ È ˘+ = + ¥ = +Î ˚ Î ˚ + +

Question 6.9

If { }22exp ( 1)xy el= - , find dy

dx.



Finally, combinations of the product, quotient and chain rules can be used in the same question:

Example

If 4

2 1

(3 4)

xy

x

+=-

, find dy

dx.

Solution Using the quotient rule and chain rule:

{ } { }

( )

4 3

8

3

8

5

(3 4) 2 (2 1) 4(3 4) 3

(3 4)

2(3 4) (3 4) 6(2 1)

(3 4)

2(9 10)

(3 4)

x x xdy

dx x

x x x

x

x

x

- ¥ - + ¥ - ¥=

-

- - - +=

-

+= --

Question 6.10

Differentiate the following with respect to x:

(i) 2( 1) xy x e= +

(ii) 2 2

ln

( 1)x

xy

e=

+

(iii) 4 2(2 7) ln(4 3)y x x= + +



5 Higher order derivatives

Expressions can be differentiated repeatedly. The notation for higher order derivatives is:

2 3 4

2 3 4, ,d y d y d y

dx dx dx etc

where 2

2

d y

dx is pronounced “dee-2-why-by-dee-x-squared” etc.

or, in function notation, ( ), ( )f x f x¢¢ ¢¢¢ etc.

The second derivative corresponds to the rate of change of the rate of change etc. The idea of a second or third derivative may not seem to have many practical applications. However, we shall see in the next section that the second derivative at least is often a very useful function.

Example

If ln(2 3)x t= + , what is 2

2

d x

dt?

Solution

21

2 2

2 42(2 3)

2 3 (2 3)

dx d xt

dt t dt t- -= = + =

+ +

Question 6.11

What is ( )f x¢¢ in each of the following cases?

(i) 2 2( ) (3 2 3)f x x x= + +

(ii) ( )1

xf x

x=

+

(iii) 1( ) ln xf x x +=



6 Stationary points

6.1 Maxima, minima and points of inflexion

A stationary point is one where the tangent to a graph is horizontal. Stationary points can be maxima, minima or points of inflexion, as illustrated in the diagram below. They can also be called turning points.

0.2 0.4 0.6 0.8 1

– 0.01

0.01

0.02

0.03

0.04

This is the graph of the function 3 2(1 )y x x= - , and it has 3 stationary points, a

minimum at 1x = , a maximum at 0.6x = , and a point of inflexion (neither a maximum or a minimum) at 0x = . A function ( )f x has a local maximum at the point x b= , if the values of ( )f x close to

x b= are less than ( )f b , or more formally, ( ) ( )f b f be± < , where e is a small

positive number. Similarly, ( )f x has a local minimum at the point x b= , if

( ) ( )f b f be± > .

Generally at these points, ( ) 0f x =¢ , and this is the method used to find maxima and

minima. However, the following graph illustrates that this is not always the case:

– 2 – 1 1 2

1

2

x

This is the graph of | |y x= , and it has a minimum at 0x = .



Once the points have been found, it is necessary to check whether they are maxima, minima or points of inflexion by finding the second derivative, and substituting in the value of x. If the second derivative is negative, the point is a maximum, if it is positive, the point is a minimum. If the second derivative is zero, further investigation is required.

Example

Find the maxima and minima of the function 3 2( ) 8 12f x x x x= - - + .

Solution

3 2

2

( ) 8 12

( ) 3 2 8 (3 4)( 2)

f x x x x

f x x x x x

= - - +

= - - = + -¢

The stationary points are at ( ) 0f x =¢ which occurs when 43 or 2x = - .

( ) 6 2f x x= -¢¢

If 43 , ( ) 10 0x f x= - = - <¢¢ , so there is a maximum point at ( )4 14

3 27,18- .

If 2, ( ) 10 0x f x= = >¢¢ , so there is a minimum point at (2,0) .

In order to understand why the second derivative is positive for a minimum, it is

necessary to consider the sign of the gradient of the graph (which we know is just dy

dx

or ( )f x¢ ). For a minimum point, the gradient of the graph changes from negative, to

zero, to positive, in other words the gradient is increasing. Since 2

2

d y

dx is the rate of

change of dy

dx,

2

2

d y

dxmust be positive for a minimum. By similar reasoning, it must be

negative for a maximum. If 2

20

d y

dx= , you need to check the sign of

dy

dx to see how the

graph is behaving. For example, 4y x= gives both dy

dx and

2

2

d y

dx to be zero at 0x = .

However dy

dx is negative just below zero, and positive just above zero, so 0x = gives a

minimum.



To find a point of inflexion, it is necessary to set ( ) 0f x =¢¢ .

You can have points of inflexion that are not horizontal. For example, the graph of the standard normal distribution has a maximum at 0x = and points of inflexion (at angles of about 13.6∞ ) at 1x = ± .



Example

Find the stationary points of 3 2( ) 2 3 72 3f x x x x= + - + , and determine their nature.

Solution To find stationary points, set ( )f x¢ equal to zero.

2

2

( ) 6 6 72

6( 12)

6( 3)( 4)

f x x x

x x

x x

= + -¢

= + -

= - +

The stationary points are at ( ) 0f x =¢ which is when 3 or 4x = - .

To find the points of inflexion, set ( )f x¢¢ equal to zero.

( ) 12 6f x x= +¢¢

The points of inflexion are at ( ) 0f x =¢¢ which is when 0.5x = - .

If 3, ( ) 42 0x f x= = >¢¢ , so there is a minimum point at (3, 132)- .

If 4, ( ) 42 0x f x= - = - <¢¢ , so there is a maximum point at ( 4,211)- .

There is a point of inflexion at ( )0.5, 39.5- .

Question 6.12

Find any turning points on the curve 3 24 3 90 6y x x x= - - + , and determine their

nature.

If you are trying to find the maxima and minima of a positive function ( )f x , then you

can find the maxima and minima of ln ( )f x . Since ln x is a steadily increasing

function, the x values at which the maxima/minima occur will be the same. This technique will be used frequently in the statistical subjects. It is useful when the functions to be differentiated are algebraically complex. Taking logs makes the algebra easier.



Example Find the maximum value of:

30 2 15 2 5( ) (1 ) ( ) ( )f e e e el l l ll - - - -= - -

( )f l is a positive-valued function. So we can take logs:

{ }{ }{ }

30 2 15 2 5

30 15 15 10

45 25

ln ( ) ln (1 ) ( ) ( )

ln (1 ) (1 )

ln (1 )

25 45ln(1 )

f e e e e

e e e e

e e

e

l l l l

l l l l

l l

l

l

l

- - - -

- - - -

- -

-

= - -

= - -

= -

= - + -

Now differentiating with respect to l , we get:

45ln ( ) 25

1

d ef

d e

l

lll

-

-= - +-

and this must be zero at a maximum, so:

2570

4525

1

25 25 45

25 70

ln 1.0296

e

e

e e

e

l

l

l l

l

l

-

-

- -

-

=-

- =

=

= - =



Checking that this is a maximum (and writing ln ( )d

fd

ll

as 45

251el

- +-

to simplify

calculations):

2

2

2

45ln ( ) 25

1

45

1

d df

dd e

e

e

Since this is always negative, the point we have found must be a maximum, and the

value here gives 20ln ( ) 45.6230 ( ) 1.535 10f fl l -= - fi = ¥ .

Question 6.13

Find the maximum value of:

( ) p nf e ll l -=

by first finding ln ( )f l .

You don’t always have to look at the second derivative when the nature of the turning points is obvious from the shape of the graph. For example, in the previous question

( )f l is always positive, (0) 0f = and ( ) 0f • = . So any stationary point we find must

be a “hump in the middle” ie a maximum, because the function is continuous in the range of values that we are interested in.



6.2 Curve sketching

It is often useful to be able to sketch the graphs of functions, in order to see what is happening. If you have a graphical calculator, this can be straightforward, although you are not allowed to take graphical calculators into the exams. If not you may be able to deduce the shape by transforming the graph of a standard function, as already seen. However, there are some useful techniques available if sketching by hand! Standard functions In Chapter 3 we looked at the standard functions and their graphs, including variations

of those functions, for example, 21 xy e= + .

Other curve sketching techniques The ideas in this chapter are also helpful when trying to decide how a particular function behaves. For example, in order to sketch the graph of a curve, the following techniques may be useful: 1. Find where the function crosses the x and y axes 2. Find any stationary points and their nature 3. Consider the sign and gradient of the function and the ranges of values for which

the function is positive or negative 4. Consider the behaviour of the function at extreme values ie when x or y tend to

zero or infinity, or at “impossible” values, such as where the denominator of a fraction would become zero.



Example

Sketch the graph of 21

2x

y e-= .

Solution

212

xdyxe

dx-= -

So 0dy

dx= when 0x = for max/min.

2 2 21 1 1

2 2 2

22 2

2 ( 1)x x xd y

x e e e xdx

- - -= - = -

When 0x = , 2

2 0d y

dx< , so there is a maximum at (0, 1) .

As , 0x yÆ ± • Æ .

Graph:

0

0.2

0.4

0.6

0.8

1

1.2

-2 -1 0 1 2

x

y

Question 6.14

Sketch the graph of 2

2

4

4 3

x xy

x x

-=- +

.



7 Partial differentiation

If ( , )f x y is a function of two independent variables x and y, then the partial derivatives

of ( , )f x y with respect to x and y are defined to be:

0

( , ) ( , )limh

f f x h y f x y

x h

∂∂ Æ

+ -= 0

( , ) ( , )limh

f f x y h f x y

y h

∂∂ Æ

+ -=

respectively. Effectively they can be calculated by differentiating f with respect to x

treating y as a constant for f

x

∂∂

, and differentiating f with respect to y treating x as a

constant for f

y

∂∂

.

f

x

∂∂

, the partial derivative of f with respect to x, tells you the rate of change of the

function f when x is varied but all other variables are kept constant.

Example

Find f

x

∂∂

and f

y

∂∂

for the function 2 3( , ) 2 ( 2 )f x y x y x y= + + .

Solution

24 3( 2 )f

xy x yx

∂∂

= + + 2 22 6( 2 )f

x x yy

∂∂

= + +

Higher derivatives can be found in a similar way. The notation used here is:

2 3

2 3,f f

x x

∂ ∂∂ ∂

etc 2 3

2 3,f f

y y

∂ ∂∂ ∂

etc 2 f

x y

∂∂ ∂

etc

2 f

x y

∂∂ ∂

means partially differentiate f

y

∂∂

with respect to x.



Example

If ( , ) ( ) xyf x y x y e= + , show that 2

2xyf f

y exx

.

Solution

( ) xy xyfx y ye e

x

∂∂

= + + 2

22 ( ) xy xy xyf

x y y e ye yex

∂∂

= + + +

2( )xy xy xy xyfy e x y y e ye ye

x

so the relationship is true.

Question 6.15

Find the 2 2 2

2 2, , , , and f f f f f

x y x yx y

∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂

for the function:

4 2 2 3( , ) (3 ) 2 (4 7 )f x y x y x y x= + - + -

Question 6.16

Show that 2 2f f

x y y x

∂ ∂∂ ∂ ∂ ∂

= for 1

( , )xye

f x yx

--= .

When x and y are both functions of some other variable t, the partial derivatives can be used to find the change in the function f when t is varied by calculating the total derivative using the relationship:

df f dx f dy

dt x dt y dt

∂ ∂∂ ∂

= +



8 Extrema of functions of two variables

Extrema in three-dimensional space are equivalent to turning points in two-dimensional space. Functions of two variables can have three types of extrema; maxima, minima and saddle points. Maxima and minima have their usual interpretations whereas a saddle point can be thought of as a horse’s saddle ie such a point is a minimum in one direction and a maximum in the other. To find these turning points we need to set the two partial derivatives:

,f f

x y

∂ ∂∂ ∂

equal to zero to find the turning point 0 0,x x y y= = .

To discover the nature of the turning point, the roots of the following equation for l need to be found:

0 0 00 0 0

2

2 2 2

2 20

x x x x x xy y y y y y

f f f

y xy x

The following conditions then apply. If the roots are: both positive there is a local minimum both negative there is a local maximum of differing signs there is a saddle point

Question 6.17

Find the turning points and their nature for the function 2 2( , ) 2f x y x y= + .



9 Lagrange multipliers

We are now going to extend the idea of maximisation to more than two variables. To

find the local extrema of a function ( )1, , nf x x we solve the equations:

1 20, 0, , 0

n

f f f

x x x

∂ ∂ ∂∂ ∂ ∂

= = =

However the variables may be constrained in some way. We can then use the method of Lagrangian multipliers. To explain this method we will refer to this simple problem.

Example A farmer wishes to enclose an area within a rectangular field. He wants the area enclosed to be as large as possible given the fact that he has 100m of fencing available. A straight river forms one of the boundaries of the field. What dimensions should he make the field in order to maximise the area?

In this example, we can set up the problem by letting the length and width of the rectangle be x and y respectively. Here x is the length of side perpendicular to the river and y is the length of the side parallel to the river. The problem is to maximise xy

subject to the condition that 2 100x y+ = .

In this very simple case it is fairly easy to see the solution: From the constraint 100 2y x= - , we are effectively maximising (100 2 )x x- .

Differentiating this we get 100 4x- , which gives 25x = when we set it equal to zero. By differentiating again we can see that 25x = does give a maximum, so the solution is to have a field which is 25 metres by 50 metres. However we can approach this in a different way, using Lagrangian functions. We will look at the specific case here and then generalise what you would need to do to tackle any problem. The Lagrangian function is defined to be (2 100)L xy x yl= - + - . (Note that the

constraint equation can be written as 2 100 0x y+ - = .)

We then solve the problem by solving 0, 0, 0L L L

x y

∂ ∂ ∂∂ ∂ ∂ l

= = = .



Question 6.18

Solve the problem using the Lagrangian function.

For more than two variables, any constraint equation can be written in the form

( )1 2, , , 0ng x x x = , in other words, we want to solve:

Find the (local) extrema of ( )1, , nf x x subject to ( )1 2, , , 0ng x x x =

Lagrangian theory tells us to construct the Lagrangian function:

( ) ( )1 1, , , ,n nL f x x g x xl= -

and then solve the equations:

1 20, 0, , 0, 0

n

L L L L

x x x

∂ ∂ ∂ ∂∂ ∂ ∂ ∂ l

= = = = .

However, there may be more than one constraint. Generally, if we want to solve the problem:

Find the (local) extrema of ( )1, , nf x x subject to the m constraints:

( ) ( )1 1 2 1 2, , , 0, , , , , 0n m ng x x x g x x x= =

Lagrangian theory tells us to set up the Lagrangian function:

( ) ( ) ( )1 1 1 1 1, , , , , ,n n m m nL f x x g x x g x xl l= - - -


1 2 10, 0, , 0, 0, , 0

n m

L L L L L

x x x

∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ l ∂ l

= = = = =

Question 6.19

Find the extrema of ( ) 2 2 2, ,f x y z x y z= + + subject to 2 3x y z- + = .



Chapter 6 Summary Order notation

( )f x is ( ( ))O g x if there exists at least one constant 0K > such that:

( )

( )

f xK

g x£ for all 0x x<

( )f x is ( ( ))o g x if the inequality is true for all constants 0K > , which means:

0

( )lim 0

( )x

f x

g xÆ=

Differentiation

Differentiating a function gives the gradient of the function at a general point. It is defined as:

0

( ) ( )( ) lim

h

f x h f xf x

hÆ

+ -=¢

Standard results include:

Function Derivative nx 1nnx - 0n π xc lnxc c xe xe

ln x 1

x

Product and Quotient rules

If u and v are functions of x , then:

( )d

uv u v uvdx

= +¢ ¢

2

d u u v uv

dx v v

-¢ ¢Ê ˆ =Á ˜Ë ¯



The chain rule:

To differentiate nested functions, we use:

dy dy du

dx du dx= ¥

This gives the following shortcut rules:

Function Derivative

[ ( )]nf x 1[ ( )] ( )nn f x f x- ¥ ¢

( )f xe ( ) ( )f xe f x¥ ¢

ln ( )f x 1

( )( )

f xf x

¥ ¢

Stationary (turning) points

These occur when 0dy

dx= . To determine their nature, we use:

2

20 max

d y

dx< fi

2

20 max

d y

dx> fi

2

2 0 point of inflexion (but needs further investigation)d y

dx= fi

In some situations to find the maximum or minimum of a function it is easier to find the maximum or minimum of the log of the function. Curve sketching

Find where the function crosses the x and y axes

Find any stationary points and their nature

Consider the sign and gradient of the function

Consider the behaviour of the function at extreme values or at impossible values.



Partial differentiation

Given a function ( , )f x y then f

x

∂∂

means differentiating ( , )f x y with respect to x

treating y as a constant. Extrema of ( , )f x y

These occur when 0f

x

∂ =∂

and 0f

y

∂ =∂

. To determine their nature we find the roots of:

0 0 00 0 0

2

2 2 2

2 20


f f f

y xy x

if both roots are positive there is a local minimum

if both roots are negative there is a local maximum

if roots are of differing signs there is a saddle point Lagrangian multipliers

To find the (local) extrema of ( )1, , nf x x subject to:

( ) ( )1 1 2 1 2, , , 0, , , , , 0n m ng x x x g x x x= =

Define the Lagrangian function:

( ) ( ) ( )1 1 1 1 1, , , , , ,n n m m nL f x x g x x g x xl l= - - -


1 2 10, 0, , 0, 0, , 0

n m

L L L L L

x x x

∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ l ∂ l

= = = = =




As x gets closer to 1 (from below), 1x - is going to get very small and negative.

Thus 1

1lim

1x x-Æ= -•

-.

Solution 6.2

We can find this limit if we divide the top and bottom through by 2x :

2 2

2

3 21

3 2 1lim lim

4 33 4 3x x

x x x xx x

xÆ• Æ•

- +- + = =- -

This is because all the terms with x ’s in their denominator tend to zero. Solution 6.3

Consider the points 1 2( , ( )) and ( , ( ))P x f x P x h f x h+ + , where 2( ) 2 3 4f x x x= + + .

The gradient of the chord joining them is ( ) ( )f x h f x

h

+ -, which when expanded and

simplified gives:

2 2 2 2 2

2

2( ) 3( ) 4 (2 3 4) 2 4 2 3 3 4 (2 3 4)

4 2 34 3 2

x h x h x x x xh h x h x x

h h

xh h hx h

h

+ + + + - + + + + + + + - + +=

+ += = + +

Now taking the limit as 0h Æ , we get that 4 3dy

xdx

= + .



Solution 6.4

(i) ( )2 32 3

3 63 6

d dx x

dx dxx x- -Ê ˆ = = - = -Á ˜Ë ¯

(ii) 4 13 3

33 4 8 8

2 23 3

d d xx x x

dx dx

(iii) First we use log rules to simplify this expression:

22ln(3 ) 2(ln3 2ln ) 2ln3 4lnx x x= + = +

Differentiating this gives 4

x.

(iv) ( )3 1 3x xde e

dx+ =

Solution 6.5

To find the rate of increase, we want the derivative ( )tr¢ . Using the product rule:

2 2

2

( ) 100 100 2

100 (1 2 )

t t

t

t e t e

e t

r = ¥ + ¥¢

= +

So at 1.5t = , 2 1.5(1.5) 100 (1 2 1.5) 8,034er ¥= + ¥ =¢ .

The payment is increasing at a rate of 8,034 at this point. We cannot specify the units since the units of the payment were not given in the question.



Solution 6.6

First of all we set 23 4u r= - , so we have 6p u= . Differentiating, we get:

56 6du dp

r udr du

= =

Hence:

56 6dp dp du

u rdr du dr

= ¥ = ¥

Replacing the u gives:

2 5 2 56(3 4) 6 36 (3 4)dp

r r r rdr

= - ¥ = -

Solution 6.7

First of all we set 23u x= - , so we have uy e= . Differentiating, we get:

6 udu dyx e

dx du= - =

Hence:

( 6 )udy dy due x

dx du dx= ¥ = ¥ -


2 23 3( 6 ) 6x xdy

e x xedx

- -= ¥ - = -



Solution 6.8

First of all we set 34 5 1u x x= - + , so we have 2lny u= . Differentiating, we get:

2 212 5

du dyx

dx du u= - =

Hence:

22(12 5)

dy dy dux

dx du dx u= ¥ = ¥ -


2

23 3

2 2(12 5)(12 5)

4 5 1 4 5 1

dy xx

dx x x x x

-= ¥ - =- + - +

Solution 6.9

Using the shortcut rule to differentiate ( )f xe , we have:

{ } { } { }2 2 22 2 2exp ( 1) exp ( 1) ( 1)x x xd de e e

dx dxl l l- = - ¥ -

But to find 22( 1)xd

edx

l - we need to use the chain rule for ( )f xe again:

2 2 22 2 2( 1) 4 4x x xd

e xe xedx

l l l- = ¥ =

But we then need to multiply by the original function, so our answer is:

{ }2 22 24 exp ( 1)x xdyxe e

dxl l= -



Solution 6.10

(i) Using the product rule we get:

{ }

2 2( 1) ( 1) 2( 1)

( 1) 1 2

( 1)( 3)

x x x

x

x

dx e x e e x

dx

x e x

x x e

È ˘+ = + + +Î ˚

= + + +

= + +

(ii) Using the quotient rule we get

2 2 2 2

2 2 2 4

2 2

2 3

2

2 3

1( 1) 2( 1)2 lnln

( 1) ( 1)

1( 1) 4 ln

( 1)

1 4 ln 1

( 1)

x x x

x x

x x

x

x

x

e e e xd x xdx e e

e e xx

e

e x x

x e

(iii) Using the product rule we get

( )

{ }

4 2 4 3 2

32

3

8(2 7) ln(4 3) (2 7) 4(2 7) 2 ln(4 3)

4 3

8(2 7)(2 7) (4 3)ln(4 3)

4 3

2 78(2 7) 2ln(4 3)

4 3

dx x x x x

dx x

xx x x

x

xx x

x

+ + = + ¥ + + ¥ ¥ ++

+= + + + ++

+Ï ¸= + + +Ì ˝+Ó ˛

We have used the fact that 2ln(4 3) 2ln(4 3)x x+ = + to make life easier.



Solution 6.11

(i) Differentiating gives:

2( ) 2(3 2 3)(6 2)f x x x x= + + +¢

Differentiating again, this time using the product rule, we get:

2

2 2

2

2

( ) 2(3 2 3) 6 (6 2) 2(6 2)

36 24 36 72 48 8

108 72 44

4(27 18 11)

f x x x x x

x x x x

x x

x x

= + + ¥ + + ¥ +¢¢

= + + + + +

= + +

= + +

(ii) Differentiating using the quotient rule we get:

{ }

12

12

32

12

12

( 1) ( 1)( )

1

( 1) 2( 1)

1

2

2( 1)

x x xf x

x

x x x

x

x

x

-

-

+ - ¥ +=¢

+

+ + -=

+

+=+

Differentiating again using the quotient rule gives:

{ }

3 12 2

12

52

3

3

2( 1) ( 2) 3( 1)( )

4( 1)

( 1) 2( 1) 3( 2)

4( 1)

4

4( 1)

x x xf x

x

x x x

x

x

x

+ - + ¥ +=¢¢+

+ + - +=

+

+= -+



(iii) The expression can be rewritten as ( ) ( 1) lnf x x x= + .

Differentiating using the product rule, we get:

1

( ) ( 1) ln

11 ln

f x x xx

xx

= + ¥ +¢

= + +

Differentiating again we get 2

1 1( )f x

xx= - +¢¢ .

Solution 6.12

To find the turning points we have to differentiate:

212 6 90dy

x xdx

= - -

Setting this equal to zero we get:

2 212 6 90 0 2 15 0 (2 5)( 3) 0

3 or 2.5

x x x x x x

x x

- - = fi - - = fi + - =

fi = = -

We need to differentiate again to find out what sort of turning points we have:

2

224 6

d yx

dx= -

Setting this equal to zero for points of inflexion, we get 0.25x = .

If 2

23, 0d y

xdx

= > , so there is a minimum point at (3, 183)- .

If 2

22.5, 0

d yx

dx= - < , so there is a maximum point at ( 2.5,149.75)- .

There is a point of inflexion at (0.25, 16.625)- .



Solution 6.13

( )ln ( ) ln ln ln

ln

p n p nf e e

p n

l ll l l

l l

- -= = +

= -

Differentiating with respect to l , we get:

ln ( )d p

f nd

ll l

= -

So for maxima or minima, setting this equal to zero, we get p

nl = .

Differentiating again:

2

2 2ln ( ) 0d p

fd

ll l

= - < (assuming that 0p > )

So p

n gives a maximum value of

ppp

en

or p

p

ne

.



Solution 6.14

2

2

4 ( 4)

( 3)( 1)4 3

x x x xy

x xx x

- -= =- -- +

, so when 0y = , 0 or 4x x= = . There is no need to

consider when 0x = , since that possibility has already been covered.

2

41

4 31

xy

x x

-=

- +, so as x Æ• , 1y -Æ .

The graph has a discontinuities when the denominator is zero, ie at 3 and 1x x= = . To find maxima and minima, we need to differentiate. Using the quotient rule we get:

2 2

2 2

( 4 3)(2 4) ( 4 )(2 4)

( 4 3)

dy x x x x x x

dx x x

- + - - - -=- +

Multiplying out the numerator:

2 2

6 12

( 4 3)

dy x

dx x x

-=- +

Setting this equal to zero gives 2x = . Differentiating again:

2 2 2 2

2 2 4

( 4 3) 6 (6 12) 2( 4 3)(2 4)

( 4 3)

d y x x x x x x

dx x x

- + ¥ - - ¥ - + -=- +

so when 2x = , 2

26 0

d y

dx= > , ie there is a minimum at (2, 4) .

There are no solutions to the equation 2

2 0d y

dx= , so there are no points of inflexion.



The graph is: You could also have deduced the general shape by writing the function in the form

31

( 1)( 3)y

x x= -

- -.

Solution 6.15

3 2 2

3 2 2

4(3 ) 3 4 3(4 7 ) 7

12(3 ) 4 21(4 7 )

fx y xy x

x

x y xy x

∂∂

= + ¥ - + - ¥ -

= + - - -

2

2 22

2 2

36(3 ) 3 4 42(4 7 ) 7

108(3 ) 4 294(4 7 )

fx y y x

x

x y y x

∂∂

= + ¥ - - - ¥ -

= + - + -

3 24(3 ) 4f

x y x yy

∂∂

= + - , therefore 2

2 22 12(3 ) 4f

x y xy

∂∂

= + -

x-4 -2 2 4 6

y

-4

-2

2

4

6

8



2 f

x y

∂∂ ∂

means partially differentiate f

y

∂∂

with respect to x, so:

2

236(3 ) 8f

x y xyx y

∂∂ ∂

= + -

Solution 6.16

To do this we need to find the first partial derivatives and then partially differentiate them again.

xyxyf xe

ey x

∂∂

--= = and 2 2

( ) (1 ) (1 ) 1xy xy xyf x ye e xy e

x x x

∂∂

- - -- - + -= =

Differentiating again, doing f

x y

for simplicity we get:

2

xyfye

x y

∂∂ ∂

-= -

Similarly:

2 2

2 2

(1 ) xy xy xy xy xyxyf xy xe e x xe xe x ye

yey x x x

∂∂ ∂

- - - - --+ ¥ - + ¥ - -= = = -

Therefore we can see that 2 2f f

x y y x

∂ ∂∂ ∂ ∂ ∂

= .

This result is generally true.



Solution 6.17

Partially differentiating:

4 2f f

x yx y

∂ ∂∂ ∂

= =

Setting these equal to zero, we get 0x = and 0y = , ie (0,0) is a turning point.

Finding the second derivatives:

2 2 2

2 24 2 0f f f

y xx y

∂ ∂ ∂∂ ∂∂ ∂

= = =

Substituting these into the required equation:

(4 )(2 ) 0 2,4l l l- - = fi =

Since both these values are positive, the turning point is a local minimum. Solution 6.18

2 0

0

2 100 0

Ly

x

Lx

y

Lx y

∂ l∂

∂ l∂

∂∂ l

= - =

= - =

= + - =

Substituting the first two equations into the third, we get:

2 2 100 0 25l l l+ - = ¤ =

This gives 25x = and 50y = as required.



Solution 6.19

We need to find the partial derivative equations:

0, 0, 0, 0L L L L

x y z

∂ ∂ ∂ ∂∂ ∂ ∂ ∂ l

= = = =

This gives us:

2 2 0, 2 0, 2 0, 2 3x y z x y zl l l- = + = - = - + =

Substituting the first three equations into the fourth:

1 12 3 1

2 2l l l l+ + = fi =

Hence:

1 11, ,

2 2x y z= = - =

1l = doesn’t mean a lot in this situation. Often this method is applied in mechanics to

real physical systems. In such problems the Lagrangian multiplier, l , does have physical meaning.

FAC-07: Integration Page 1


Chapter 7

Integration

You need to study this chapter to cover: ● indefinite and definite integrals, including the area under a graph

● integrating standard functions

● integrating by inspection, by substitution, and by parts

● convergence of definite integrals

● multiple integrals

● numerical methods for integration

● Taylor series and Maclaurin series

● simple ordinary differential equations.

0 Introduction

This chapter covers all the methods of integration that you need in order to study the Core Technical subjects.

FAC-07: Integration


1 Integrals

Suppose we have a function ( )f x , and that ( ) ( )F x f x=¢ . Then ( )F x is an anti-

derivative of ( )f x . The general form of the anti-derivative is the indefinite integral of

( )f x which is written as ( ) f x dxÚ . Integration and differentiation are connected in

that they are the “reverse” of each other.

Since ( )( ) ( ) ( )d

F x c F x f xdx

+ = =¢ , then, in general, ( ) ( )f x dx F x c= +Ú , where c is

known as the constant of integration. Integrals with limits are called definite integrals. They are evaluated between two

values. For example [ ]( ) ( ) ( ) ( )b

ba

a

f x dx F x F b F a= = -Ú . This is read as “the integral

from a to b of ( )f x ”. The numbers a and b are called the lower and upper limits of

integration respectively. For such integrals there is no “ c+ ”, since the constants cancel when the integral is evaluated. The practical application of integration is that it also gives the sum of infinitesimally small elements and, as illustrated below, this can be used to find areas under curves. Consider the area enclosed by the curve ( )y f x= , between x a= , and x b= , as shown

in the diagram below. We are going to look at the small section of area under the curve contained between x and x xd+ . 0

y

xb

y = f(x)

a x x + x



If ( )A x is the area under the curve from x a= to x, then the area bounded by x and

x xd+ is given by ( ) ( )A x x A xd+ - . Looking at the diagram and considering the

areas of the smaller and bigger rectangles in the diagram:

( ) ( ) ( ) ( )x f x x A x x A x x f xd d d d+ £ + - £

Dividing by xd , we get:

( ) ( )( ) ( )

A x x A xf x x f x

x

ddd

+ -+ £ £

As 0xd Æ , ( ) ( )A x x A x

x

dd

+ - lies between two values which get closer and closer

together, so:

0

( ) ( )lim ( )x

A x x A xf x

xd

ddÆ

+ - =

The LHS is just the definition of dA

dx, so:

( )dA

f xdx

= or ( ) ( ) A x f x dx= Ú

This is a general expression and we wanted the area enclosed between x a= and x b= ,

so the area is actually given by ( ) ( ) ( )b

a

f x dx A b A a= -Ú .

In the diagram given, the curve was sloping down between x and x xd+ . You get the same result if the function is sloping up. It is also important to note that if the area is below the x-axis then the integral, when evaluated, will give rise to a negative number. If the graph of ( )y f x= crosses the

x-axis between x a= and x b= , then ( ) b

a

f x dxÚ will not give the total area because

some of the area will be below the x-axis and some above. Under these circumstances it is necessary to split up the integral. For example to find the area enclosed between the

x-axis, 2x = - , 2x = and the curve 3( )f x x= , you would evaluate 2 0

3 3

0 2

x dx x dx-

-Ú Ú ,

using “–” since the second integral will be negative.

FAC-07: Integration


You also need to be careful if the function you are integrating tends to ±• in the range you are looking at. In this case the integral may be infinite or it may have a finite value.



2 Integration of standard functions

Some standard functions and their integrals are shown in the table below:

Function Integral k kx c+

nx 1

1

nxc

n

++

+, 1n π -

1

x ln x c+

( )

( )

f x

f x¢

ln ( )f x c+

xa ln

xac

a+

kxe kxe

ck

+

Example Integrate the following functions:

(i) x

(ii) 27 x

(iii) 43 xe

(iv) 2

1x +

Solution

(i) Writing x as 12x and using the formula above:

32

323 32

xc x c+ = +

FAC-07: Integration


(ii) This is the same as 2(7 )x . So using the formula with 27a = :

27

2ln 7

x

c+

(iii) 434

xe c+

(iv) 2ln 1x c+ +

If you need to make it clear which quantity in an expression is the variable that is changing, you can say “integrating with respect to x” etc.

Question 7.1

Integrate the following functions with respect to x:

(i) 3 22

24 x

x+

(ii) 212

xe

(iii) 126

x

(iv) 2

2 5

5 7

x

x x

++ +

When evaluating a definite integral, the integrated function is written in square brackets with the limits on the right. This notation means evaluate the function shown at the top and bottom limits and subtract. American textbooks use a line after the function with limits at the top and bottom, rather than the square brackets notation.



Example

Evaluate ln 2 5

02 xe dxÚ .

Solution

ln 2ln 2 5 5 5ln 2 02 2 25 5 50 0

2 x xe dx e e eÈ ˘= = -Î ˚Ú

Since 55ln 2 ln 2 52 32e e= = = , our expression simplifies to be:

64 6225 5 5- =

Constant factors can be pulled out of the square brackets if it helps to simplify the working.

Question 7.2

Evaluate 3 2

13x dxÚ . What does this integral represent?

FAC-07: Integration


3 Further integration

3.1 Integration by inspection

Unlike differentiation, where there are foolproof procedures for differentiating any function, integration requires you to be aware of the techniques available and think through each one until you find the correct method. As yet we have not covered the

methods needed to integrate functions such as 2xxe or 2 52 ( 6)x x + . These can be done

by inspection. This includes spotting an exact derivative as part of your integral. If you are not convinced by any of the answers given below don’t forget you can check your answers by differentiating.

Example Find the following:

(i) 2

3 xxe dxÚ (ii) 4( 2)x dx+Ú (iii) 2 5( 6)x x dx+Ú

Solution

(i) If the integral was 2

2 xxe , we could integrate it directly since the expression on the front would be the derivative of the power. So we pull out the 3, insert a 2 in the integral and divide by 2 outside:

2 2 23 3

3 22 2

x x xxe dx xe dx e c= = +Ú Ú

(ii) The derivative of the bracket is 1 so we don’t have to multiply or divide by

factors in this part:

4 51( 2) ( 2)

5x dx x c+ = + +Ú

(iii) The derivative of the bracket is 2x so we insert a 2 into the integral and divide

by 2 outside:

2 6 2 6

2 5 2 51 1 ( 6) ( 6)( 6) 2 ( 6)

2 2 6 12

x xx x dx x x dx c c

+ ++ = + = + = +Ú Ú



This technique fails if you try to “compensate” with anything other than numbers! You cannot pull outside the integral factors containing x or any function of x.

Question 7.3

Find the following:

(i) 2 5

4 6

( 3 4)

xdx

x x

++ +Ú

(ii) 23 47 xxe dx-Ú

(iii) 22xx e dxÚ

3.2 Integration using partial fractions

Partial fractions is the name of the technique used to split up a single fraction into

separate parts. For example 7

( 1)( 2)

x

x x

++ -

is equivalent to 3 2

2 1x x-

- +. Using this

technique, expressions that cannot be integrated directly can be written as expressions that can.

Example

Find 9 2

(3 2)(3 2)

xdx

x x

-+ -Ú .

FAC-07: Integration


Solution We need to split up the single fraction into two fractions that are added together. Using A and B as the numerators we get:

9 2 (3 2) (3 2)

(3 2)(3 2) 3 2 3 2 (3 2)(3 2)

x A B A x B x

x x x x x x

- - + += + =+ - + - + -

Comparing the numerators from the first and last terms we get 9 2 (3 3 ) 2 2x A B x A B- = + - + , and comparing coefficients we get 2, 1A B= = .

Substituting this into our integral we get:

1 2

3 2 3 2dx

x x+

- +Ú

We can now evaluate the integral:

2

1 2 1 2ln 3 2 ln 3 2

3 2 3 2 3 3

1ln (3 2)(3 2)

3

dx x x cx x

k x x

+ = - + + +- +

= - +

Ú

The two constants in the last two lines are not the same, which is why we have changed the letter.

Question 7.4

What is the relationship between c and k in the last example?

The denominator in the last example could have been given in the question as 29 4x - , so you will sometimes have to be careful to spot that partial fractions are required.

Question 7.5

Integrate 2

7

1

x

x

--

.



3.3 Integration by substitution (change of variable)

This method involves the replacement of one variable by another, enabling expressions to be simplified and hence integrated. The steps you need to follow are: ● Decide what substitution to use, making u your new variable. Generally you

will make u the “complicated” part of the expression.

● Find dx

du, and hence change dx into du .

● Express the function in terms of u.

● If you have limits, change them so that they correspond to u.

● Evaluate the simpler integral.

Example

Evaluate 1

4

0

(2 3)x x dx+Ú .

Solution

Let 2 3u x= + , then 2du

dx= which means that

1

2

dx

du= , and

3

2

ux

-= . We need to

change the limits, when 0x = , 3u = , and when 1x = , 5u = . The integral then becomes:

1 54 4

0 3

5 5 4

3

56 5

3

6 5 6 5

3(2 3)

2

3 1

2 2

1 3

4 6 5

1 5 3 5 1 3 3 3 11188

4 6 5 4 6 5 30

u dxx x dx u du

du

u udu

u u

FAC-07: Integration


This example could also have been carried out using integration by parts. For details of this method, see the next section.

Question 7.6

Find 2

( 1)

xdx

x -Ú .

Question 7.7

Evaluate 1

30

4

2 1x dxe +Ú , by using the substitution 32 1xu e= + .

3.4 Integration by parts

This method is used to integrate products. The product is broken down into one part that can be integrated and one part that can be differentiated. The formula for integration by parts is:

dv du

u dx uv v dxdx dx

= -Ú Ú

This formula is in the Tables page 3. You may see this written as u dv uv v du= -Ú Ú .

Note that the formula only replaces one integral with another integral. So you should aim to replace an integral that you cannot do by one that you can do, ie you should try to get a simpler integral than the one you started with.

Example

Find xxe dxÚ .



Solution Both parts of this integral can be differentiated and integrated, so we have to pick u and v so that the resulting integral is possible (and simpler than the original one!)

Let u x= , and xdve

dx= , so that 1

du

dx= and xv e= . So:

( 1)

x x x

x x

x

xe dx xe e dx

xe e c

x e c

= -

= - +

= - +

Ú Ú

Sometimes the choice of u and dv

dx can be less obvious.

Example

Find 23 xx e dxÚ

Solution

The 2xe term cannot be directly integrated, and putting 3dv

xdx

= will mean that the

second integral will be more complicated than the first, so we need to split up the 3x

term. The expression 2xxe can be integrated, so we can proceed as follows. Let

2u x= , and 2xdv

xedx

= , so that 2du

xdx

= and 21

2xv e= . Then:

2 2

2 2

2 2

2

3 2

212

21 12 2

21( 1)

2

x x

x x

x x

x

x e dx x xe dx

x e xe dx

x e e c

x e c

= ¥

= -

= - +

= - +

Ú Ú

Ú

FAC-07: Integration


Notice that this example used integration by inspection. Often a combination of several techniques will be required to evaluate an integral.

Question 7.8

Find:

(i) 6( 1)x x dx+Ú

(ii) ln x x dxÚ

(iii) 2

1

ln x dxÚ (Hint: think of it as 1 ln x¥ )

There is not always a unique way of integrating. Part (i) of this question could be done by substitution. Question 7.6 could have been carried out using integration by parts.

3.5 Differentiating an integral (Leibniz’s formula)

We have already mentioned the concept of integration being the reverse of differentiation. We will now look at differentiating an integral.

This section uses the result that ( , ) ( , ) b b

a a

d ff x t dt x t dt

dx x

∂∂

=Ú Ú , where a and b are

constants. This can be thought of as taking the d

dx inside the integral. This can be

generalised to:

( ) ( )

( ) ( )

( , ) ( ) ( , ( )) ( ) ( , ( )) ( , ) b x b x

a x a x

d ff x t dt b x f x b x a x f x a x x t dt

dx x

∂∂

= - +¢ ¢Ú Ú

This formula can be found on page 3 of the Tables. The proof of this result is beyond the scope of this course, but it is really just an application of the function of a function rule.



The formula is very useful in cases where the integral cannot easily be evaluated

directly, for example 2

0

xtde dt

dx

•-Ú .

Example

Evaluate 2

0

xd

x t dtdx

+Ú .

Solution

Here ( ) 0a x = , ( )b x x= , 2( , )f x t x t= + , so:

2 2

0 0

2 2

2

+ 1( ) 0 2

2

3

x xd

x t dt x x x dtdx

x x x

x x

= + - +

= + +

= +

Ú Ú

In this case we can show that this is the same as integrating directly:

2 2 2 3 2

00

1 1

2 2

x x

x t dt x t t x x

So 2 3 2 2

0

13

2

xd d

x t dt x x x xdx dx

.

Question 7.9

Evaluate 2 3

2

0

( 1) + x

dx tx dt

dx

+È ˘+Î ˚Ú .

FAC-07: Integration


4 Convergence

So far we have only dealt with integrals that have finite limits. We define:

( ) lim ( ) k

a akf x dx f x dx

•

Æ•=Ú Ú and ( ) lim ( )

b b

kk

f x dx f x dxÆ-•

-•

=Ú Ú

If these limits exist, the integrals are said to converge, otherwise the integrals do not converge. Convergence is also an issue when the function being integrated tends to ±• at a

certain point eg 1

x when 0x = .

Example

Does 21

1 dx

x

•Ú converge? If so, what is its value?

Solution

2 21 11

1 1 1 1 lim lim lim 1

kk

k k kdx dx

x kx x

Since this limit exists, the integral converges, and its value is 1.

You can usually tell whether an integral converges by trying to work it out and seeing whether you end up dividing by zero in the calculation. Note that the following integrals do not exist:

20

1 dx

x

•Ú

0

1

Ndx

xÚ 1

1dx

x

•Ú



Question 7.10

Explain why these integrals do not exist.

You will find in practice that any integral that has a real-life meaning will have a finite value, and that the integrated function will work out to be zero at the infinity end.

Question 7.11

Evaluate 2

0

xxe dx• -Ú .

Question 7.12

Use integration by parts to show that ( ) ( 1) ( 1)x x xG = - G - . Remember that the

definition of the gamma function is 1

0

( ) x tx t e dt•

- -G = Ú .

FAC-07: Integration


5 Double integrals

If you have a function of two variables, ( , )f x y , the graph of the function gives you a

surface, which might look like this:

-1

0

1x

-2

-1

0

1

2

y

0

1

z

-1

0

1x

The two horizontal axes are the x- and y-axes, and the z axis goes vertically upwards. We can extend the concept of an integral to calculate the volume of the solid region under a part of the surface. In order to calculate this volume we need to evaluate a double integral of the function

over the region R. This is written as ( , ) R

f x y dx dyÚÚ .

The mathematical definition of a double integral is just an extension of the definition for an ordinary integral. It involves taking the limit of the sum of the volumes of a large number of rectangular blocks. In practice you work out a double integral by first integrating with respect to one of the variables (x or y) then integrating with respect to the other variable. The limits of integration must be chosen so that they describe the boundaries of the region R. This is the footprint of the volume on the x-y plane.



To evaluate these integrals, we treat x as a constant when integrating with respect to y ,

and y as a constant when integrating with respect to x .

The case where the region R is a rectangle parallel to the axes is straightforward.

Question 7.13

Evaluate the double integral 15 5

10 0

2

3000x y

x ydy dx

= =

+Ú Ú .

However, in other cases the limits will be a function of x when first integrating with respect to y , and of y when first integrating with respect to x . If you need to work

out the limits, you will need to consider the required region, as illustrated in the following example.

FAC-07: Integration


Example

Integrate 2xy over the region R shown below:

(i) by first integrating with respect to y and then integrating with respect to x (ii) by first integrating with respect to x and then integrating with respect to y. Solution (i) In this part x is the variable in the “outside” integral and the range of x values is

0 2x£ £ . For any given value of x the range of values of y is 1 3 1y x£ £ + .

This gives us our limits of integration.

2 3 12

0 1

3 12 3

0 1

2 3

0

24 3 2

0

25 43

0

3

(3 1)

3 3

9 9 3

9 9

5 4

101.6

x

x

I xy dy dx

xydx

x x xdx

x x x dx

x xx

0

y

x

y = 3x + 1

2

1

7

R



(ii) In this part y is the variable in the “outside” integral and the range of y values is

1 7y£ £ . For any given value of y the range of values of x is 1

23

yx

- £ £ .

This gives us our limits of integration.

7 22

113

27 2 2

113

7 2 22

1

73 5 4 3

1

2

( 1)2

18

2 1 2

3 18 5 4 3

102.26 0.66 101.6

y

y

I xy dx dy

x ydy

y yy dy

y y y y

The following results can be used. The proofs are beyond the scope of this course.

( , ) ( , ) = ( , ) + ( , ) R R R

af x y bg x y dx dy a f x y dx dy b g x y dx dy+ÚÚ ÚÚ ÚÚ

1 2

( , ) = ( , ) + ( , ) R R R

f x y dx dy f x y dx dy f x y dx dyÚÚ ÚÚ ÚÚ

Here R is the combined region consisting of the separate regions 1R and 2R , ie

1 2R R R= » and 1 2R R« = ∆ .

Question 7.14

Evaluate the double integral 3 2

2 3

0

x y

y

xe dxdy+Ú Ú .

FAC-07: Integration


Sometimes double integrals with explicit limits are written with the dx and dy at the

front next to the integral sign each belongs to. In this case the last question would be

written as 3 2

2 3

0

x y

y

dy dx xe +Ú Ú .

The results for double integrals can be extended to triple integrals.

Question 7.15

Evaluate 3 3 2

1 0 1

z zx ye dx dy dz

++Ú Ú Ú .

If the integrand (the expression being integrated) factorises into functions of the three separate variables, and the limits are constants, you can multiply together the individual integrals. This is not obvious but it is true. This would mean:

2 2 2 2

1 1 1 1

( ) ( ) ( ) ( )x y x y

x y x y

dx dy f x g y f x dx g y dy=Ú Ú Ú Ú

Question 7.16

Evaluate 1 2 3

0 0 0

dx dy dz xyzÚ Ú Ú .



5.1 Swapping the order of integration

In a similar way to when we changed the order of summation in Chapter 4, we can change the order of integration.

Example

Given that 0 5t s< < < , calculate the value of 5

2

0 0

s

s t

s t dt ds= =Ú Ú , first by using this order

and then by swapping the order of the two variables. Solution Firstly, we will use the order given:

55 5 52 4 52 2

0 0 0 00 0

312.52 2 10

ss

s t s s

t s ss t dt ds s ds ds

Previously we integrated from 0t = to t s= , then from 0s = to 5s = . Reversing this, we integrate from s t= to 5s = then from 0t = to 5t = . To get the limits for s , see what it is bounded by in the inequality 0 5t s< < < . So:

5 5 5

2 2

0 0 0

s

s t t s t

s t dt ds t s ds dt= = = =

=Ú Ú Ú Ú

Carrying out this integration:

5 55 5 5 53 3 2 52

0 0 0 0

125 1 125312.5

3 3 3 2 5t s t t tt

s t t tt s ds dt t dt t dt

FAC-07: Integration


Question 7.17

(i) Carry out the integration 30

0 0

yy x x

y x

e e e dx dym n dm- - -

= =Ú Ú , where 0 30x y< < < .

(ii) Confirm that you get the same answer if you reverse the order of integration and

use the values 0.001, 0.06, 0.005n d m= = = .



6 Numerical methods for integration

Sometimes it is not possible to find the area under the curve ( )y f x= by using

integration techniques because you cannot integrate the function ( )f x . In such cases

we can find an approximation to the integral by using a numerical method. The method that we will look at here is called the trapezium rule. The technique involves approximating the area under the curve by finding the sum of the areas of several trapezia.

6.1 The trapezium rule

In the following diagram, we have split the area under the curve into trapeziums.

The area of a trapezium is given by the sum of the lengths of the two parallel sides, multiplied by the distance between them, divided by 2, ie it is the base multiplied by the average height. Therefore the total area of the trapezia shown in the diagram (ie the estimate of the area under the curve between 1x and 5x ) is:

1 2 2 3 3 4 4 51 1 1 1

( ) ( ) ( ) ( )2 2 2 2

y y d y y d y y d y y d+ + + + + + +

This can be rewritten as:

1 2 3 4 51

( 2 2 2 )2

y y y y y d+ + + +

y1 y2 y3 y4 y5

x

y

d d d d

x1 x5

FAC-07: Integration


This can also be expressed in words: “The sum of the first and last ordinate (or y value), plus twice the sum of the

other ordinates, multiplied by 12 d .”

We can choose the number of ordinates to suit the area we are trying to estimate, and obviously the smaller the value of d , the more accurate the approximation will be.

Example

Use the trapezium rule to estimate the area under the curve 2y x= between 1x = and

3x = . Use 9 ordinates. Also find the true value of the area and comment on your answers. Solution Using 9 ordinates means that we need to split the area into 8 sections, each of width 0.25. The trapezium rule will then give the approximate area to be:

( )2 2 2 2 2 2 2 2 211 3 2(1.25 1.5 1.75 2 2.25 2.5 2.75 ) 0.25

2

8.6875

+ + + + + + + + ¥

=

The true area is:

3 3

2 3 3 3

11

1 1(3 1 ) 8.67

3 3x dx x

È ˘= = - =Í ˙Î ˚Ú

You can see that here the trapezium rule has slightly over estimated the true area under

the curve. If you plotted an accurate graph of the function 2y x= and drew on the

trapeziums you should be able to see that the over-estimation has occurred due to the shape of the curve.

Question 7.18

Estimate the area under the curve 2

yx

= between 2x = and 3x = , using 5 ordinates.

Will this be an overestimate or an underestimate of the true value.



7 Taylor and Maclaurin series

You may need to expand a function as a power series. Taylor and Maclaurin series can be used to do this expansion.

7.1 Taylor series

If you want to approximate the value of ( )f x when x aª , then Taylor’s series states

that:

2 3

( ) ( ) ( ) ( ) ( )2! 3!

h hf a h f a hf a f a f a+ = + + + + ◊◊ ◊¢ ¢¢ ¢¢¢

Here we are thinking that x a h= + , where h is a small quantity.

Example

Expand ln x for x eª in ascending powers of x e- , up to the term in 3( )x e- .

Solution Applying the formula for Taylor’s series with a e= and h x e= - so that

( ) ln ( ( ))f x x f e x e= = + - , we get:

( ) lnf x x= ( ) 1f e =

1( )f x

x=¢

1( )f e

e=¢

2

1( )f x

x

-=¢¢ 2

1( )f e

e

-=¢¢

3

2( )f x

x=¢¢¢

3

2( )f e

e=¢¢¢ and so on

So:

2 32 3

1 1 1ln 1 ( ) ( ) ( )

2 3x x e x e x e

e e e= + - - - + - +

FAC-07: Integration


Question 7.19

Expand 2 5(1 ) (1 )i i- -+ + + as a series about 0.1i = .

What can you say about the value of this function if i increases from 0.1 to a slightly higher value (for example 0.11)?

We can also have a Taylor’s series expansion when there are two variables. If you want to approximate the value of ( , )f x y about ( , )a b then Taylor’s series states

that:

( , ) ( , )

2 2 22 2

2 2( , ) ( , ) ( , )

1( , ) ( , ) ( ) ( )

1!

1( ) 2 ( )( ) ( )

2!

a b a b

a b a b a b

f ff x y f a b x a y b

x y

f f fx a x a y b y b

x yx y

∂ ∂∂ ∂

∂ ∂ ∂∂ ∂∂ ∂

È ˘Í ˙= + - + -Í ˙Î ˚

È ˘Í ˙+ - + - - + -Í ˙Î ˚

+ As for the single variable case, a Taylor series is often used to look at the effect of a small change in the value of one of the arguments. So, for example, if x a h= + and y b k= + (where h and k are small) then we would have:

( , ) ( , )

2 2 22 2

2 2( , ) ( , ) ( , )

( , ) ( , )

12

2

a b a b

a b a b a b

f ff a h b k f a b h k

x y

f f fh hk k

x yx y

∂ ∂∂ ∂

∂ ∂ ∂∂ ∂∂ ∂

+ + = + +

È ˘Í ˙+ + +Í ˙Î ˚



Example

Expand 3xye about (2,1) as far as the terms in 2( 2)x - , 2( 1)y - etc.

Solution The required partial derivatives in their general form and then evaluated at 2x = and

1y = are:

3 6

3 6

22 3 6

2

22 3 6

2

23 3 6

3 3

3 6

9 9

9 36

3 3 3 21

xy

xy

xy

xy

xy xy

fye e

x

fxe e

y

fy e e

x

fx e e

y

fy xe e e

y x

∂∂

∂∂

∂∂

∂∂

∂∂ ∂

= =

= =

= =

= =

= ¥ + =

So the expansion is:

6 6 6 6 2 6 6 23 ( 2) 6 ( 1) 4.5 ( 2) 21 ( 2)( 1) 18 ( 1)e e x e y e x e x y e y+ - + - + - + - - + -

Question 7.20

Expand 1

x y- about (3,2) as far as the terms in 2( 3)x - , 2( 2)y - etc.

FAC-07: Integration


7.2 Maclaurin series

A Maclaurin series is just a Taylor series about 0x = ie change a to 0 and h to x. If ( )f x can be expanded as an infinite, convergent series of powers of x, then:

2 3(0) (0)( ) (0) (0)

2! 3!

f ff x f f x x x

¢¢ ¢¢¢= + + + + ◊◊ ◊¢

Maclaurin’s series are most useful for finding the series expansions for basic functions

such as xe , ln(1 )x+ etc.

Example

Obtain the expansion of xe as far as the term in 3x . Solution

Let ( ) xf x e= , then ( ) ( )n xf x e= , where ( ) ( )nf x means the n th derivative, and ( ) (0) 1nf = , so:

2 3

( ) 12! 3!

x xf x x= + + + +

Question 7.21

Obtain the expansion of ln(1 )x+ as far as the term in 3x .



8 Differential equations

Any equation which contains n

n

d y

dx in some form, for example, 2 3

dyx y

dx= + , is called

a differential equation. In this section we are going to consider some of the many techniques used to solve these equations. Solving differential equations means finding

( )y f x= .

8.1 Solution by direct integration

If we integrate dy

dx, then we get y and if we integrate

2

2

d y

dx, then we get

dy

dx, etc. This

enables us to solve differential equations of the form ( )n

n

d yf x

dx= .

Example

Solve the differential equation 2

22 2

d rt

dt= .

Solution Integrating once with respect to t:

323

drt c

dt= +

Integrating again:

416r t ct d= + +

As seen in this example, we can end up with several constants in the solution. This is called a “general solution”. You can find a “particular solution” if there are “boundary conditions” given in the question, giving you particular values of the variables involved.

FAC-07: Integration


Example

Solve the differential equation xdye

dx= , given that 1y = when ln 2x = .

Solution

The general solution of the differential equation is xy e c= + , but we know that when

ln 2, 1x y= = .

So:

1 2 1c c= + fi = - and the particular solution is:

1xy e= -

Question 7.22

Find the general solution of the differential equation 2

22

d xat

dt= . Given that when

0, 4t x= = , when 97121, t x= = , and when 1

121, t x= - = , find a and the particular

solution.

8.2 Solution by separation of variables

If ( , )dy

f x ydx

= , and ( , )f x y can be expressed as ( ) ( )p x q y , then the differential

equation can be solved by separating the variables, ie by writing it in the form:

1

( ) ( )

dy p x dxq y

=Ú Ú



Example

Solve the differential equation 2 1

dr r

dt t=

+, where 0r > and 0t > .

Solution

We can rearrange this in the form 2 1

dr dt

r t=

+, so:

( )12

1 1

2 1

ln ln(2 1) ln ln 2 1

dr dtr t

r t c r t c

=+

= + + fi = + +

Ú Ú

Note that we do not need to include modulus signs in the integrated expressions as both r and 2 1t + are positive. Note also that we need to include a constant on one side of the equation, not both. Taking exponentials of both sides:

( )ln 2 1ln 2 1 2 1t cr ce e r e t r k t+ +

= fi = + fi = +

where ck e= .

Question 7.23

Find the particular solution of the differential equation 2( 9) 2dy

x xy xdx

+ = + , where

0y > , given that when 4, 0.5x y= = .

Question 7.24

Find the general solution to the differential equation ( )dN

a bN Ndt

= - , where

0a

Nb

< < . This is known as the logistic equation or the Verhulst equation after the

mathematician who used it in his study of populations.

FAC-07: Integration


8.3 Solution by integrating factor

A differential equation of the form ( ) ( )dy

f x y g xdx

+ = can be solved by use of an

integrating factor. The integrating factor here is exp ( )f x dxÈ ˘Î ˚Ú , and the solution of the

differential equation is then the solution of:

exp ( ) ( )exp ( )y f x dx g x f x dx dxÈ ˘ È ˘=Î ˚ Î ˚Ú Ú Ú

Example

Find the general solution of 34 2 xdyy e

dx+ = .

Solution

The integrating factor is 4exp 4 xdx eÈ ˘ =Î ˚Ú . So we multiply through by 4xe which

gives 4 4 74 2x x xdye ye e

dx+ = .

We see that the LHS is the derivative of 4xye , derived from the product rule. So

4 7( ) 2x xdye e

dx= .

We then need to solve 4 72x xye e dx= Ú , which gives 4 72

7x xye e c= + . The solution of

our differential equation is 3 42

7x xy e ce-= + , where c is an arbitrary constant.

Question 7.25

Find the particular solution of the differential equation 23 2dy

xy y xydx

= + , where 0x > ,

given that 1y = when 27x = .



Chapter 7 Summary Integration

Integration is the reverse of differentiation:

if ( ) ( ) ( ) ( )F x f x f x dx F x c= fi = +¢ Ú

A definite integral has limits, in which case our answer does not require a constant:

( ) ( ) ( )b

a

f x dx F b F a= -Ú

This finds the area under the curve ( )y f x= between x a= and x b= .

Standard results include:

Function Integral k kx c+

nx 1

1

nxc

n

++

+, 1n π -

1

x ln x c+

xa ln

xac

a+

xe xe c+ To “undo” the chain rule we could use integration by inspection:

Function Integral

( )[ ( )]nkf x f x¢ 1[ ( )]

1

nk f xc

n

++

+, 1n π -

( )

( )

kf x

f x

¢ ln ( )k f x c+

( )( ) f xkf x e¢ ( )f xke c+

Alternatively, we could use integration by substitution by setting ( )u f x= in the above.

FAC-07: Integration


Partial fractions can be used to simplify algebraic fractions before integrating:

eg 1

( 1)( 3) 1 3

A Bdx dx

x x x x= +

+ + + +Ú Ú

If the above methods don’t work, then we could use integration by parts:

dv du

u dx uv v dxdx dx

= -Ú Ú

To evaluate integrals with infinite limits, you need to consider if they converge. Trapezium rule

The area under a curve (ie the integral) is approximately equal to:

11 2 12 ( 2 2 )n ny y y y d-+ + + +

Differentiating an integral

[ ] [ ]( ) ( )

( ) ( )

( , ) ( ) ( ), ( ) ( ), ( , )b x b x

a x a x

df x t dt b x f b x t a x f a x t f x t dt

dx x

∂= - +¢ ¢∂Ú Ú

Double integrals

These calculate the volume of a solid region under the surface ( , )f x y :

( , )x y

f x y dy dxÊ ˆÁ ˜Á ˜Ë ¯Ú Ú

For a domain a y x b£ £ £ the limits can be swapped as follows:

( , ) ( , )b x b b

x a y a y a x y

f x y dy dx f x y dx dy= = = =

Ê ˆ Ê ˆ=Á ˜ Á ˜

Á ˜ Á ˜Ë ¯ Ë ¯Ú Ú Ú Ú

Series

Taylor and Maclaurin series are used to expand functions as infinite series.

The formula for a Taylor series (in one variable) is:

2 3

( ) ( ) ( ) ( ) ( )2! 3!

h hf a h f a hf a f a f a+ = + + + + ◊◊ ◊¢ ¢¢ ¢¢¢



The formula for a Taylor series (in two variables) is:

( , ) ( , )

2 2 22 2

2 2( , ) ( , ) ( , )

1( , ) ( , ) ( ) ( )

1!

1( ) 2 ( )( ) ( )

2!

a b a b

a b a b a b


x y


x yx y

An alternative form for a Taylor series is:

( , ) ( , )

2 2 22 2

2 2( , ) ( , ) ( , )

( , ) ( , )

12

2

a b a b

a b a b a b

f ff a h b k f a b h k

x y

f f fh hk k

x yx y

The formula for a Maclaurin series is:

2 3(0) (0)( ) (0) (0)

2! 3!

f ff x f f x x x

¢¢ ¢¢¢= + + + + ◊◊ ◊¢

Differential equations

A differential equation is an equation which contains n

n

d y

dx.

( )n

n

d yf x

dx= can be solved by direct integration

( ) ( )dy

p x q ydx

= can be solved by separating the variables

( ) ( )dy

f x y g xdx

+ = can be solved by multiplying by the integrating factor ( )f x dx

eÚ

The general solution of a differential equation contains unknown constants. If you are given appropriate boundary conditions, you can find a particular solution.

FAC-07: Integration






(i) The function can be written as 2322 4x x- + , which can then be integrated to give:

5 53 31 12 2 12

25 5

x x c x cx

-- + + = - + +

(ii) 214

xe c+

(iii) This is equivalent to 12(6 )x , so using the formula given in the notes, we get:

126

2ln 6

x

c¥ +

(iv) We have the derivative of the denominator on the numerator, so this integrates to

give:

2ln 5 7x x c+ + +

Check this by differentiating if you are not sure.

Solution 7.2

332 3

11

3 26x dx xÈ ˘= =Î ˚Ú

This represents the area between the curve 23y x= , and the x-axis from 1x = to 3x = .

FAC-07: Integration


Solution 7.3

2 52 5

2 4

2 4

4 6(i) 2(2 3)( 3 4)

( 3 4)

2( 3 4)

4

1

2( 3 4)

xdx x x x dx

x x

x xc

cx x

-

-

+ = + + ++ +

+ += +-

= - ++ +

Ú Ú

(ii) 2 2 23 4 3 4 3 47 7

6 67 6 x x xxe dx xe dx e c- - -= = +Ú Ú

(iii) 2 2 2 22 1 1

2 2 2 x x x xx e dx xe dx xe dx e c= = = +Ú Ú Ú

Solution 7.4

Expanding the last line of the solution, we get:

21 1 1ln 3 2 ln(3 2) ln

3 3 3x x k- + + +

So it can be seen that 1

ln3

c k= .



Solution 7.5

2

7

1

x

x

--

can be written as 7

( 1)( 1)

x

x x

-+ -

.

Using partial fractions, we can split it as follows:

7 ( 1) ( 1)

( 1)( 1) 1 1 ( 1)( 1)

x A B A x B x

x x x x x x

- - + += + =+ - + - + -

Comparing coefficients, we get 4, 3A B= = - .

The integral then becomes:

4

3

4 3 ( 1)4ln 1 3ln 1 ln

1 1 ( 1)

xdx x x c c

x x x

+- = + - - + = ++ - -Ú

Solution 7.6

One possibility is to let 1u x= - , then the integral becomes:

31 1 12 2 2 2

12

43

4 43 3

2( 1) (2 2 ) 4

( 3) ( 2) 1

udu u u du u u c

u

u u c x x c

-+ = + = + +

= + + = + - +

Ú Ú

It is also possible to solve this by making the substitution 1u x= - .

FAC-07: Integration


Solution 7.7

If 32 1xu e= + , then 36 3( 1)xdue u

dx= = - . When 0x = , 3u = and when 1x = ,

32 1u e= + . Substituting gives:

3 32 1 2 1

3 3

4 1 4 1

3( 1) 3 ( 1)

e e

du duu u u u

+ +

¥ =- -Ú Ú

Splitting up the integral using partial fractions gives:

( ) ( )

3 3

3

2 1 2 1

3 3

2 1

3

3 3

3

3

4 1 4 1 1

3 ( 1) 3 ( 1)

4ln 1 ln

3

4 4ln 2 ln(2 1) ln 2 ln3

3 3

4 3ln

3 2 1

e e

e

du duu u u u

u u

e e

e

e

+ +

+

= -- -

È ˘= - -Î ˚

= - + - -

=+

Ú Ú



Solution 7.8

(i) Let u x= , and 6( 1)dv

xdx

= + , ie 1du

dx= , and 71

7 ( 1)v x= + , then we get:

6 7 71 1

7 7

7 81 17 56

7156

( 1) ( 1) ( 1)

( 1) ( 1)

( 1) (7 1)

x x dx x x x dx

x x x c

x x c

+ = + - +

= + - + +

= + - +

Ú Ú

It is also possible to use the substitution 1u x= + to do this integral.

(ii) Let lnu x= , and dv

xdx

= , ie 1du

dx x= , and 21

2v x= . So:

2 21 1 1

2 2

21 12 2

2 21 12 4

ln ln

ln

ln

xx x dx x x x dx

x x x dx

x x x c

= - ¥

= -

= - +

Ú Ú

Ú

This example is unusual since it is more common to set u equal to a power of x when integrating by parts.

(iii) Let lnu x= , and 1dv

dx= , ie

1du

dx x= , and v x= .

[ ] [ ]2 2

2 21 1

1 1

ln ln 1 lnx dx x x dx x x x= - = -Ú Ú

This can be evaluated to give:

(2ln 2 2) (1ln1 1) 2ln 2 1 ln 4 1- - - = - = -

FAC-07: Integration


Solution 7.9

We need to find 2 3

2

0

( 1) + x

dx tx dt

dx

+

+Ú , so comparing with the general equation:

2( ) 0, ( ) 2 3, and ( , ) ( 1)a x b x x f x t x tx= = + = + +

giving:

( ) 0, ( ) 2, and 2( 1)f

a x b x x tx

∂∂

= = = + +¢ ¢ .

So:

( )

2 3 2 32 2

0 0

2 32 212 0

2 212

2 12

( 1) + 2 ( 1) (2 3) 0 2( 1)

2(3 5 1) 2( 1)

2(3 5 1) 2( 1)(2 3) (2 3)

12 26 12

x x

x

dx tx dt x x x x t dt

dx

x x x t t

x x x x x

x x

+ +

+

È ˘+ = + + + - + + +Î ˚

È ˘= + + + + +Î ˚

= + + + + + + +

= + +

Ú Ú

Solution 7.10

The first one does not exist because 0

1lim

k

k xÆ•

È ˘-Í ˙Î ˚ is not defined for the lower limit.

The second one does not exist because 0

lim lnN

kkx

ÆÈ ˘Î ˚ is not defined for the lower limit.

The last one does not exist because 1

lim lnk

kx

Æ•È ˘Î ˚ does not converge.

One consequence of these integrals not existing is that the area under the graphs of these functions is infinite.



Solution 7.11

We will do this using integration by parts.

Let u x= , and 2xdve

dx-= , ie 1

du

dx= , and 21

2xv e-= - . Then:

2 2 21 12 20

0 0

2 21 12 40 0

2 21 1 12 4 4

14

lim

lim

lim

kkx x x

k

k kx x

k

k k

k

xe dx xe e dx

xe e

ke e

•- - -

Æ•

- -Æ•

- -Æ•

È ˘È ˘Í ˙= - - -Î ˚Í ˙Î ˚

È ˘È ˘ È ˘= - -Í ˙Î ˚ Î ˚Î ˚

È ˘= - - +Î ˚

=

Ú Ú

We have used the fact that exponential functions dominate polynomials for large x, ie

2lim 0k

kke-

Æ•= .

In practice when evaluating such integrals, you consider the limit, but don’t write it, as seen in the next solution. Solution 7.12

From the definition of the gamma function 1

0

( ) x tx t e dt•

- -G = Ú ( 0)x > .

So let 1xu t -= , ie 2( 1) xdux t

dt-= - , and tdv

edt

-= , ie tv e-= - . This gives:

1 1 2

00 0

2

0

( ) ( 1)

( 1) ( 1) ( 1)

x t x t x t

x t

x t e dt t e x t e dt

x t e dt x x

• ••- - - - - -

•- -

È ˘G = = - - - -Î ˚

= - = - G -

Ú Ú

Ú

where 1x > .

FAC-07: Integration


Solution 7.13

15 5 15521

2 010 0 10

15

10

152

10

2 1 2

3000 3000

110 12.5

3000

15 12.5

3000

11

48

x y

x ydy dx xy y dx

x dx

x x

= =

+ È ˘= +Î ˚

= +

È ˘= +Î ˚

=

Ú Ú Ú

Ú

Solution 7.14

We need to integrate by parts.

Use u x= , and 2 3x ydve

dx+= so that 1

du

dx= and 2 31

2x yv e += , to give:

3 2 3 22

2 3 2 3 2 3

0 0

3 23 4 5 2 3

0

33 4 3 4 5

0

33 4 5

0

1 1

2 2

1 1

2 4

1 1 1

4 4 2

3 1 1

4 4 2

x y x y x y

yy y

y y x y

y

y y y

y y

xe dxdy xe e dx dy

e ye e dy

e e y e dy

e y e dy



Again using integration by parts, this time with 1 1

4 2u y= - , we get:

3 33 5

3 4 5

0 00

33 4 5 5

0

13 15 4

1 1 1 1

4 4 2 5 10

1 1 1 1 1

4 5 4 2 50

1 23 1 7

4 100 4 100

641,284

yy y

y y y

ee y e dy

e y e e

e e e

Solution 7.15

3 3 2 3 32

11 0 1 1 0

3 32 1

1 0

332 1

01

32 4 1 3 2

1

32 4 1 3 21 14 3 1

14 10 6 5 4 31 1 1 14 3 4 3

2

293,103

z z zzx y x y

zz y y

zz y y

z z z

z z z

e dx dy dz e dy dz

e e dy dz

e e dz

e e e e dz

e e e ez

e e e e e e e

+ ++ +

+ + +

+ + +

+ + +

+ + +

È ˘= Î ˚

= -

È ˘= -Î ˚

= - - +

È ˘= - - +Î ˚

= - - - + + +

=

Ú Ú Ú Ú Ú

Ú Ú

Ú

Ú

Solution 7.16

1 2 3 1 2 3

0 0 0 0 0 0

1 9 12 4

2 2 2dx dy dz xyz x dx y dy z dz= ¥ ¥ = ¥ ¥ =Ú Ú Ú Ú Ú Ú

FAC-07: Integration


Solution 7.17

(i) Integrating first with respect to x :

30

0 0

30( )

00

30( )

00

30( )

0

1

1

1(1 )

yy x x

y x

yy x

y

yy x

y

y y

y

e e e dx dy

e e dy

e e dy

e e dy

Then integrating with respect to y , we get:

30( )

0

30( )

0

30 30( )

1 1

1 1 1 1

y y

y

y y

e e dy

e e

e e

(ii) Evaluating (i) numerically, we get 1.213. Changing the order of integration, consider 0 30x y< < < . So:

30 30 30

0 0 0

yy x x x x y

y x x y x

e e e dx dy e e e dy dxm n d n d mm m- - - - - -

= = = =

=Ú Ú Ú Ú



Then integrating with respect to y , we get:

30 30

0

3030

0

30( ) ( ) 30

0

30( ) ( ) 30

0

30( ) 30( ) 30

1 1

1 1(1 ) ( )

x x y

xx

x x x

x

x x

x

x x

e e e dx

e e e e dx

e e e dx

e e e

e e e

Substituting in the given values, we also get 1.213, confirming that the two

methods give the same numerical value. Solution 7.18

Using 5 ordinates means that we need to split the area into 4 sections, each of width 0.25. The trapezium rule will then give the approximate area to be:

1 2 4 4 4 2

0.25 0.8122 2 2.25 2.5 2.75 3

This will give an over estimate of the true value since the gradient of the curve is increasing over the range of values. We can check the exact value by integrating:

3

3

22

22ln 2ln3 2ln 2 0.811dx x

xÈ ˘= = - =Î ˚Ú

FAC-07: Integration


Solution 7.19

2 5( ) (1 ) (1 )f i i i- -= + + + , so (0.1) 1.447f = , and:

3 6

4 7

5 8

( ) 2(1 ) 5(1 )

( ) 6(1 ) 30(1 )

( ) 24(1 ) 210(1 )

f i i i

f i i i

f i i i

- -

- -

- -

= - + - +¢

= + + +¢¢

= - + - +¢¢¢

giving:

(0.1) 4.325

(0.1) 19.493

(0.1) 112.869

f

f

f

= -¢

=¢¢

= -¢¢¢

Using the Taylor series, this gives:

2 32 5 ( 0.1) ( 0.1)

(1 ) (1 ) 1.447 4.325( 0.1) 19.493 112.8692! 3!

i ii i i- - - -+ + + = - - + -

(as far as the term in 3( 0.1)i - ).

Because the term 4.325( 0.1)i- - has a negative coefficient, increasing i would reduce

the value of the function. In fact, changing i to 0.11 (ie 0.1 0.01i - = ) would increase the value of the function by approximately 4.325 0.01 0.04325- ¥ = - .



Solution 7.20

The required partial derivatives in their general form and then evaluated at 3x = and 2y = are:

2

2

23

2

23

2

23

( ) 1

( ) 1

2( ) 2

2( ) 2

2( ) 2

fx y

x

fx y

y

fx y

x

fx y

y

fx y

y x

∂∂

∂∂

∂∂

∂∂

∂∂ ∂

-

-

-

-

-

= - - = -

= - =

= - =

= - =

= - - = -


2 2

2 2

11 ( 3) ( 2) 2( 3) 2 2( 3)( 2) 2( 2)

2

2 ( 3) 2( 3)( 2) ( 2)

x y x x y y

x y x x y y

È ˘- - + - + - + ¥ - - - + -Î ˚

= - + + - - - - + -

Solution 7.21

( ) ln(1 )f x x= + , so (0) 0f = , and:

2 3

1 1 2( ) , ( ) , ( )

1 (1 ) (1 )f x f x f x

x x x= = - =¢ ¢¢ ¢¢¢

+ + +

giving:

(0) 1 (0) 1 (0) 2f f f= = - =¢ ¢¢ ¢¢¢

Using Maclaurin’s series, this gives:

2 31 12 3ln(1 )x x x x+ = - + (as far as the term in 3x )

FAC-07: Integration


Solution 7.22

Integrating twice gives:

3

3

dx atc

dt= +

4

12

atx ct d= + +

But using the boundary conditions:

0 4t d= fi =

971 4

12 12

at c= fi = + +

1

1 412 12

at c= - fi = - +

Solving these simultaneously, we get 1, 4a c= = , so the particular solution is:

4

4 412

tx t= + +



Solution 7.23

Separating the variables gives us:

2

1

2 9

xdy dx

y x=

+ +Ú Ú

Integrating both sides gives us:

212ln( 2) ln( 9)y c x+ = + +

Note that no modulus signs are required here as 2 9x + is always positive, and 2y + is

always positive since we are given that 0y > .

So:

2ln( 2) ln 9 2 22 9 9 2y c x ce e y e x y k x+ + += fi + = + fi = + -

where ck e= . But if 0.5y = when 4x = , then 0.5k = , so the particular solution is:

20.5 9 2y x= + -

Solution 7.24

The differential equation can be solved by separating the variables:

( )

1

1 1ln ln( )

dNdt

a bN N

a b adN dt

N a bN

N a bN c ta a

=-

+ =-

- - + =

Ú Ú

Ú Ú

Note that 0N > and 0a bN- > from the condition in the question, so no modulus signs are required here.

FAC-07: Integration


Setting 1

lnc ka

= , for some positive constant k , this can be simplified to give the

general solution:

1ln

atat

at

kN kN aet e N

a a bN a bN k be= fi = fi =

- - +

Solution 7.25

The differential equation can be rewritten in the form 1 2

3 3

dyy

dx x- = , providing that

0y π . 0y = is a trivial solution to the differential equation.

Using this version, we can see that the integrating factor is

13

1 1exp exp ln

3 3dx x x

x-È ˘ È ˘- = - =Í ˙ Í ˙Î ˚ Î ˚Ú .

Note that since 0x > , we don’t need to include modulus signs here. Multiplying through by the integrating factor and integrating both sides with respect to x gives the general solution to the differential equation:

1 1 23 3 3

2

3yx x dx x c

- -= = +Ú

But we know that 1y = when 27x = , so:

23

19 8

3c c= + fi = -

So the particular solution is 1 23 3 2

38yx x- = - or

132

38y x x= - .

FAC-08: Vectors and matrices Page 1


Chapter 8

Vectors and matrices

You need to study this chapter to cover: ● calculations involving vectors

● multiplication by a scalar

● scalar product of two vectors

● magnitude

● finding the angle between vectors

● orthogonality

● transposition of a matrix

● addition and subtraction of matrices

● multiplication of matrices

● determinants and inverses.

0 Introduction

You will meet the work covered in this chapter in Subjects CT4, CT6 and CT8. If you have not studied vectors and matrices before then you will need to look at this chapter very carefully.

FAC-08: Vectors and matrices


1 Vectors

1.1 Notation and arithmetic

Vectors are defined as quantities that have a magnitude (or length) and a direction. They can be thought of as representing the position of a point in two dimensions, three dimensions etc. We will deal with two and three-dimensional vectors in this chapter but the results can easily be extended to the n-dimensional case. All vectors in three-dimensional space can be written in terms of the base vectors i, j and k, which are the unit vectors (ie the vectors of length 1) in the x, y and z directions of cartesian space. There are different ways of writing vectors:

2

2 3 2 3 2 3 2

2

Taa i j k

(we’ll see what the T means shortly). In this chapter we will use the convention that vectors are in bold. In handwritten work a calculation involving the scalar l , the vector v and the matrices M and I (matrices will be introduced in Section 2) might look like this:

(M I )v 0l- =

where underlining is used to represent a vector or a matrix.

Note that the column vector

2

3

2

shows the coefficients of i, j and k, so that

2

3 2 3 2

2

i j k

where

1

0

0

i

,

0

1

0

j

, and

0

0

1

k

. Column vectors are usually

used in preference to row vectors because matrix equations then have a similar format to the corresponding set of simultaneous equations.



Vectors can be added, subtracted or multiplied by a scalar:

a d a d

b e b e

c f c f

a d a d

b e b e

c f c f

a ka

k b kb

c kc

ie you add, subtract or multiply the individual elements. Showing addition and multiplication pictorially:

Example

If

2

4

4

a

and

3

5

1

b

, what are:

(i) 2a (ii) a b (iii) 3 4a b ? Solution

(i)

4

2 8

8

a

(ii)

5

1

3

a b

(iii)

6

3 4 32

16

a b

ba

a + b

b

a2

2



Question 8.1

(i) If

3

2

8

a

and

1

6

2

b

what is 3 2b a ?

(ii) What are the values of p and q such that

5

50

46

p qa b

?

1.2 Magnitude

The magnitude of a vector is just its length, which can be calculated using an extended version of Pythagoras’ theorem. The magnitude of vector a is written as a or | |a .

In general, if

p

q

r

a

, then 2 2 2a p q r= + + .

Example

Find the magnitude of the vector

1

2

4

a

.

Solution

2 2 21 ( 2) 4 21a = + - + =

This gives us a way of finding a unit vector ie a vector with length 1. For example if a

is as defined in the last example, then the unit vector in the direction of a is

11

221

4

.



Question 8.2

Find the magnitude of the vector 2 5 3- +i j k .

Question 8.3

Find the unit vector in the direction of 3 4 2+ -i j k .

1.3 Scalar product

There are different types of product you can work out for vectors: the scalar (or dot) product and the vector (or cross) product. We will only look at the scalar product here, as the vector product is not needed for the actuarial exams. The scalar product of two vectors can be defined to be cosab q=a . b where q is the angle between them in the plane containing and a b , as shown below. It can also be calculated by multiplying the corresponding coefficients of the unit vectors, i, j and k, and then summing.

b

a



Example

Find a . b , where

1

4

5

a

and

2

3

6

b

.

Solution

(1)(2) ( 4)(3) (5)( 6) 2 12 30 40= + - + - = - - = -a . b

Note the answer here is a scalar not a vector, hence the name “scalar product”.

The scalar product can be used to find the angle between two vectors:

Example

Find the angle between the vectors

3

6

1

a

and

3

3

4

b

.

Solution

9 18 4 5= - + - =a . b , 9 36 1 46a = + + = , and 9 9 16 34b = + + = So:

cos 5 46 34 cosab q q= fi =a . b

5cos

46 34

83

q

q

\ =

=

Question 8.4

Find the angle between the vectors 2 3 5= + -a i j k and 3 2 8= - + +b i j k .



Since the scalar product is related to the angle between vectors, an important result

occurs when 0=a . b , since this means that the angle between the vectors is 90 , ie they are perpendicular. When this occurs, the vectors are called “orthogonal”. Note that in the following question, the vectors are in two-dimensional space.

Question 8.5

Find p such that a and b are perpendicular, where 1

7a

, 3

4

pb

.



2 Matrices

Matrices are arrays of numbers whose size is referred to as the number of rows by the number of columns. Notice that a vector is a special case of a matrix where the number of columns is one. A square matrix is one where the number of columns equals the number of rows.

Example

Describe the size of this matrix 1 1 3 5

3 4 6 4

.

Solution It is a 2 4¥ matrix.

Matrices have many uses, such as describing transformations, solving simultaneous equations, carrying out calculations involving multivariate statistical distributions, and showing state transition probabilities. We will look here at the techniques that you will need for the actuarial exams.

2.1 Basic arithmetic

The transpose of a matrix is found by swapping the rows and the columns. The

transpose of a matrix A is written as TA or ¢A .

Example

If

1 3

2 5

0 9

A

, what is TA ?

Solution

1 2 0

3 5 9TA



Transposing converts a column vector to a row vector and vice versa, and an m n¥ matrix to an n m¥ matrix. Column vectors are often written as the transpose of row

vectors in textbooks so that they can be written on one line eg (1 2 3)T .

Addition and subtraction are performed by adding or subtracting corresponding elements in the matrix.

So for 2 2¥ matrices a b e f a e b f

c d g h c g d h

.

Example

If

2 4 5

4 2 6

7 1 3

A

and

4 3 2

8 5 1

3 6 0

B

, what is A B ?

Solution

2 4 5 4 3 2 6 7 3

4 2 6 8 5 1 12 3 5

7 1 3 3 6 0 10 7 3

A B

The matrix with zero as all of its elements is called the zero matrix. For example the

2 2¥ zero matrix is 0 0

0 00

. A matrix such as

4 0 0

0 3 0

0 0 1

, where all the elements

not on the main diagonal are zero, is called a diagonal matrix.

Question 8.6

If a matrix is equal to its transpose, what can you say about the matrix?



2.2 Multiplication

Matrices can be multiplied by a scalar or by another matrix. When multiplying by a scalar, multiply each element in the matrix by that number.

Example

If 3 3

5 6A

, what is 2A ?

Solution

6 62

10 12A

Multiplying two matrices together gives another matrix. The elements in the rows of the first matrix are multiplied individually by the elements in the columns of the second and then summed.

So for 2 2¥ matrices a b e f ae bg af bh

c d g h ce dg cf dh

.

Example

If 3 1

5 4A

and 4 2

3 1B

what is AB ?

Solution

3 1 4 2 (3 4) (1 3) (3 2) (1 1) 15 7

5 4 3 1 ( 5 4) ( 4 3) ( 5 2) ( 4 1) 32 14AB

Note: Not all matrices can be multiplied together – the number of columns in the first matrix must be the same as the number of rows in the second. An m n¥ matrix multiplied by an n q¥ matrix gives an m q¥ matrix. You will soon realise if it is

impossible to multiply your matrices as you will run out of numbers.



Question 8.7

If

1 2 4

3 2 3

0 1 1

A

and

4 1 1

5 0 2

1 5 3

B

, what are AB and BA ?

Here AB and BA are not equal. Unlike in ordinary arithmetic, matrix multiplication is not commutative ie you get different answers if you multiply matrix B in front by matrix A (called pre-multiplying by A ) or if you multiply behind by matrix A (called post-multiplying by A ).

Question 8.8

What is

0 2 3 1

4 2 1 2

0 5 1 1

?

These could not have been multiplied the other way round.

Question 8.9

What is 1

2 1 3 3

2

?



Example

What do you get if you multiply the matrix a b

c dM

by 1 0

0 1I

?

Solution

a b

c dMI

and a b

c dIM

Note that multiplying by 1 leaves the original matrix unchanged. Because of this property the matrix with 1’s along the leading diagonal and 0’s elsewhere is called the identity matrix and is written as I .

2.3 Determinants and inverses

Determinants A determinant is a scalar quantity associated with a square matrix. The determinant of a 2 2¥ matrix is equal to the product of the numbers on the leading diagonal (top left corner to bottom right corner) minus the product of the numbers on the other diagonal. It is written as det , | | , or DA A when it is clear which matrix is

involved.

Example

What is det A if 2 6

4 3A

?

Solution det (2 3) (4 6) 30= ¥ - ¥ - =A



Question 8.10

What is the determinant of 3 5

2 7P

?

This is a specific definition, and we need to generalise. For any matrix M , if ijc is the

element in the matrix corresponding to row i and column j, and ijM is the determinant

of the matrix formed when we strike out row i and column j, then the determinant is

defined as 11 1

1

( 1)n

jj j

j

c M+

=-Â .

This looks quite daunting, but in practice it is quite simple to use.

Example

Calculate the determinant of

1 1 2

3 0 4

2 3 1

.

Solution The determinant is given by:

0 4 3 4 3 01 ( 1) ( 1) 2

3 1 2 1 2 3

1(0 12) 1(3 8) 2( 9 0)

12 11 18

19

- -D = ¥ + - ¥ - ¥ + ¥

- -

= - + + + - -

= - + -

= -

We have taken the elements in the top row and multiplied every other one by –1. These are then multiplied by the determinant of the 2 2¥ matrix that is left when the row and column containing the element are removed. Finally the results are summed.



Question 8.11

Find the determinant of

2 1 0

0 1 1

4 3 6

.

Inverses

The inverse of a matrix A , written as 1-A , is such that 1 1- -= =AA A A I . It is the matrix equivalent to a reciprocal. To find the inverse of a 2 2¥ matrix, swap the elements on the leading diagonal, change the sign of the elements on the other diagonal and divide by the determinant, ie

11a b d b

c d c aad bc

.

If the determinant is 0 then the matrix is said to be singular and the inverse does not exist. This is equivalent to trying to divide by zero with ordinary numbers.

Example

If 3 2

5 4A

, what is 1A ?

Solution

1 4 21

5 32A



Example

Using the matrix in the last example, show that 1 1- -= =AA A A I . Solution Note that the factor can be brought to the front.

1 3 2 4 2 2 0 1 01 1

5 4 5 3 0 2 0 12 2AA

1 4 2 3 2 2 0 1 01 1

5 3 5 4 0 2 0 12 2A A

Question 8.12

Find the inverse of 1 9

2 10B

, and check it by finding 1-BB .

There is no simple rule for finding the inverse of a 3 3¥ matrix. You need to use a more complicated method, such as solving a set of simultaneous equations.



2.4 Simultaneous equations

Matrices can be used to solve simultaneous linear equations, by first writing them in matrix form and then pre-multiplying by the inverse.

Example (Method 1) Solve the simultaneous equations (using matrices):

2 3 14

3 4 4

x y

x y

- =

+ =

Solution Note firstly that these simultaneous equations can be written as the following matrix equation. If you need convincing, multiply out the matrices.

2 3 14

3 4 4

x

y

Pre-multiplying by the inverse matrix on both sides, we get:

4 3 2 3 4 3 141 1

3 2 3 4 3 2 417 17

x

y

This gives the identity matrix on the left hand side, so we simplify to get:

1 0 681

0 1 3417

4

2

x

y

x

y

So the solution is 4x = and 2y = - .



Question 8.13

Solve the simultaneous equations (using matrices):

2 4 14

2 3 13

p q

q p

- = -

- =

Matrices can also be used in a different way to solve simultaneous equations. This method more closely resembles the method used to solve them algebraically that we covered earlier in the course. We will look at the last example again, this time looking at an elimination method.



Example (Method 2) Solve the simultaneous equations (using matrices):

2 3 14

3 4 4

x y

x y

- =

+ =

Solution We can write this in the abbreviated form:

2 3 14

3 4 4

x

y

This can in turn be written as:

2 3 14

3 4 4

We would like to get zero in the bottom left hand corner. We are allowed to replace any row with a linear combination of the rows. We will do this in two stages. Replacing the first row by second row minus first row ( (2) (1) ):

1 7 10

3 4 4

Replacing the second row by (2) 3 (1) :

1 7 10 1 7 10

0 17 34 0 1 2

This tells us that 2y = - , and so 10 7 4x y= - - = .



Example Solve the simultaneous equations (using the elimination method):

2 2 11

3 2 4 8

4 3 15

x y z

x y z

x y z

+ - =

- + = -

+ - =

Solution These can be written as:

2 1 2 11

3 2 4 8

1 4 3 15

This time we would like to get zeros in the bottom left hand corner. Replacing the first row by (1) (3) and the second row by (2) 3 (3) we get:

1 3 1 4

0 14 13 53

1 4 3 15

Replacing the third row by 2 ((3) (1)) (2) , we get:

1 3 1 4 1 3 1 4

0 14 13 53 0 14 13 53

0 0 5 15 0 0 1 3

Replacing the second row by (2) 13 (3) , we get:

1 3 1 4

0 14 0 14

0 0 1 3

so 3, 1, 2z y x= - = = .



When solving simultaneous equations, you do not always have as many different equations as it first appears. Any equation that can be written as a linear combination of the other equations (ie you can obtain this equation from the others by adding, subtracting or multiplying) is said to be linearly dependent on the others. When solving simultaneous equations you should ensure that the equations you are working with form a linearly independent set. You only need to solve linearly independent equations. For example, suppose you were given the simultaneous equations:

3 4 5

2 3 8

7 13

x y

x y

x y

- = -

+ =

- = -

Here you would only need to consider, say, the first two equations since the third is a linear combination of them (equation 3 is equation 1 minus equation 2). You could equally ignore the first or second equation. Careful choice of which equations to use can make things considerably easier for you, so do watch out!

Question 8.14


3 2 13

7 6 9

x y

x y

+ =

- =

using the method of the last example.

Question 8.15


2 4

2 3 2 1

3 3 17

x y z

x y z

x y z

+ - = -

- + =

+ - = -

using the method of the last example.



2.5 Eigenvectors and eigenvalues

There is an important aspect of matrix theory that arises from the fact that matrices can be used to represent transformations. If a vector v is transformed by a matrix A, then the resulting vector is Av. If the direction of the vector is unchanged once it has been transformed we can write: l=Av v where l is a constant. The vector v is called an “eigenvector” of the matrix, and the corresponding value of l is called an “eigenvalue”. To find the eigenvectors and eigenvalues, you have to work with the equation

l=Av v : ( )l l= fi - =Av v A I v 0

where 0 is the zero matrix. Notice that we had to insert the identity matrix I into the equation since we cannot subtract a scalar from a matrix. Thinking of our ways of solving “ordinary” equations, we see that this equation would be true if either ( )l- =A I 0 , or =v 0 , but these are not going to be very helpful since

this is just the trivial solution and is not the one we are interested in. To solve a general matrix equation =Bx C , we would normally find the inverse of B

and calculate 1-=x B C . Using this technique to try to solve the equation

( )l- =A I v 0 , then if ( )l-A I has an inverse we get 1( )l -= -v A I 0 , ie =v 0 .

Since we are looking for a non-trivial solution, we must prevent this method from working. The only thing that would stop us getting =v 0 would be if l-A I does not have an inverse ie it is singular. Remembering that singular matrices have a determinant of zero this gives us a way to find the eigenvalues and eigenvectors. In summary, to find the eigenvalues of a matrix A we must solve the equation det( ) 0l- =A I . The equation obtained from det( ) 0l- =A I is called the

“characteristic equation of the matrix”. Once you have the eigenvalues, you can return to the equation ( )l- =A I v 0 , to find

the eigenvectors. This method will work for a general n n¥ matrix, but due to the complication of calculating the determinant we will only look at 2 2 and 3 3¥ ¥ matrices here.



Example

Find the eigenvectors and corresponding eigenvalues of the matrix 2 1

2 5B

.

Solution First we use the equation det( ) 0B I to find the eigenvalues:

2

2 1det 0

2 5

(2 )(5 ) 2 0

7 12 0

( 3)( 4) 0

so 3 or 4 .

We now use the equation ( ) 0B I v to find the eigenvectors, letting x

yv

.

There is a separate eigenvector corresponding to each eigenvalue.

If 3 , then 2 1 0

2 5 0

x

y

, ie 1 1 0

2 2 0

x

y

, ie

0

2 2 0

x y

x y

.

Any value of x and y such that 0x y ie y x will make both rows work out.

So the eigenvector corresponding to 3 is 1

1k

, where k is a constant.

If 4 , then 2 1 0

2 1 0

x

y

, so the eigenvector corresponding to 4 is

1

2k

.

Notice that we include k in the eigenvector, since any value of the constant k will give an eigenvector.



Question 8.16

Find the eigenvectors and corresponding eigenvalues of the matrix 1 2

3 4B

.

Question 8.17

Show that 1l = is an eigenvalue of the matrix

2 2 3

1 1 1

1 3 1

, and hence find all the

eigenvalues and eigenvectors.



Chapter 8 Summary Vectors

Vectors have both a magnitude and a direction. For example:

( )2

2 3 3 2 3 1

1

Ta

Ê ˆÁ ˜= = + - = = -Á ˜-Ë ¯

a i j k

The magnitude is 2 2 22 3 ( 1)a = + + - .

Vectors can be added, subtracted and multiplied by a scalar:

eg

a d a d

b e b e

c f c f

+Ê ˆ Ê ˆ Ê ˆÁ ˜ Á ˜ Á ˜+ = +Á ˜ Á ˜ Á ˜

+Ë ¯ Ë ¯ Ë ¯

a d a d

b e b e

c f c f

-Ê ˆ Ê ˆ Ê ˆÁ ˜ Á ˜ Á ˜- = -Á ˜ Á ˜ Á ˜

-Ë ¯ Ë ¯ Ë ¯

a ka

k b kb

c kc

Ê ˆ Ê ˆÁ ˜ Á ˜=Á ˜ Á ˜Ë ¯ Ë ¯

Scalar product

cosab q=a.b

If the vectors are perpendicular (orthogonal) then . 0=a b .

Hence if x y z= + +a i j k and l m n= + +b i j k then . xl ym zn= + +a b .

Matrices

A matrix is a rectangular array of numbers. Matrices can be multiplied by a scalar:

eg a b ka kb

kc d kc kd

Ê ˆ Ê ˆ=Á ˜ Á ˜Ë ¯ Ë ¯

Matrices of the same dimensions can be added or subtracted:

eg a b c g h i a g b h c i

d e f j k l d j e k f l

+ + +Ê ˆ Ê ˆ Ê ˆ+ =Á ˜ Á ˜ Á ˜+ + +Ë ¯ Ë ¯ Ë ¯

The transpose of a matrix can be found by swapping the rows and the columns.

eg T a d

a b cb e

d e fc f

Ê ˆÊ ˆ Á ˜=Á ˜ Á ˜Ë ¯

Ë ¯



Matrices A and B can be multiplied as AB if they are compatible, ie if the dimensions of A are a n¥ then the dimensions of B must be n b¥ . To multiply matrices the elements in the rows of the first matrix are multiplied individually by the elements in the columns of the second and then summed.

eg a b e f ae bg af bh

c d g h ce dg cf dh

+ +Ê ˆ Ê ˆ Ê ˆ=Á ˜ Á ˜ Á ˜+ +Ë ¯ Ë ¯ Ë ¯

Determinants

If a b

c d

Ê ˆ= Á ˜Ë ¯

A then det ad bc= = -A A .

If

a b c

d e f

g h i

Ê ˆÁ ˜=Á ˜Ë ¯

A then dete f d f d e

a b ch i g i g h

= = - +A A

Inverses

1-A is the inverse of A if 1 1- -= =AA A A I where I is the identity matrix.

If a b

c d

Ê ˆ= Á ˜Ë ¯

A then 1 1 d b

c aad bc- -Ê ˆ= Á ˜-- Ë ¯

A .

A singular matrix has no inverse, ie the determinant is zero.

Matrices can be used to solve simultaneous equations. Eigenvalues and eigenvectors

The eigenvalues of a matrix A can be found by solving the equation: det( ) 0l- =A I

The corresponding eigenvectors v can then be found by solving the equation: ( )l- =A I v 0




(i)

3

3 2 22

22

b a

(ii) If

5

50

46

p qa b

then 3 5 and 2 6 50p q p q+ = - + = - . Solving these

equations simultaneously, we get 4 and 7p q= = - .

You should also check that the third equation, 8 2 46p q- = , also works out

with these values of p and q. Otherwise there are would be no solutions. Solution 8.2

The magnitude is 2 2 22 ( 5) 3 38+ - + = .

Solution 8.3

The magnitude of the vector is 2 2 23 4 ( 2) 29+ + - = , so the unit vector is

( )13 4 2

29+ -i j k .

Solution 8.4

To find the angle, we take the dot product.

cosab q=a . b ie 2 2 2 2 2 240 2 3 ( 5) ( 3) 2 8 cosq- = + + - - + +

Hence 40

cos or 13838 77

q q-= = .



Solution 8.5

If a and b are perpendicular, then their dot product is zero.

3 28 0p= - =a . b , hence 283p = .

Solution 8.6

The matrix must be a square matrix, otherwise the matrix and its transpose would be different sizes and could therefore not be equal. Also elements reflected about the leading diagonal must be equal. Such a matrix is called symmetric. Solution 8.7

(1 4) ( 2 5) (4 1) (1 1) ( 2 0) (4 5) (1 1) ( 2 2) (4 3)

(3 4) (2 5) ( 3 1) (3 1) (2 0) ( 3 5) (3 1) (2 2) ( 3 3)

(0 4) (1 5) ( 1 1) (0 1) (1 0) ( 1 5) (0 1) (1 2) ( 1 3)

10 21 9

25 12 2

6 5

AB

1

Similarly:

1 11 20

5 12 22

16 11 14

BA

Solution 8.8

0 2 3 1 7

4 2 1 2 7

0 5 1 1 11



Solution 8.9

1

2 1 3 3 1 1

2

A 1 1¥ matrix is just a scalar, so note that multiplying a row vector by a column vector in that order will give you a scalar. Solution 8.10

det 21 ( 10) 11= - - - = -P .

Solution 8.11

The determinant is 2(6 3) 1(0 ( 4)) 0 2- - - - + = .

Solution 8.12

1 10 91

2 18B

To check:

1 1 9 10 9 8 0 1 01 1

2 10 2 1 0 8 0 18 8BB



Solution 8.13

Noticing that the letters are in different orders, these equations can be written in the form:

2 4 14

3 2 13

p

q

Pre-multiplying both sides by the inverse of 2 4

3 2

, we get:

2 4 141

3 2 138

24 31

16 28

p

q

So 3p = - and 2q = .

Solution 8.14

These equations can be written as:

3 2 13

7 6 9

Replacing (1) by (2) 2 (1) :

1 10 17

7 6 9

Replacing (2) by (2) 7 (1) :

1 10 17

0 64 128

So 3x = and 2y = .



Solution 8.15

These equations can be written as:

1 2 1 4

2 3 2 1

3 1 3 17

Replacing (2) by (2) 2 (1) , and (3) by (3) (2) (1) , we get:

1 2 1 4

0 7 4 9

0 2 4 14

Replacing (2) by (2) (3) , we get:

1 2 1 4 1 2 1 4

0 5 0 5 0 1 0 1

0 2 4 14 0 2 4 14

Replacing (3) by (3) 2 (2) , we get:

1 2 1 4 1 2 1 4

0 1 0 1 0 1 0 1

0 0 4 16 0 0 1 4

This gives 4, 1, 2z y x= = = - .



Solution 8.16

For eigenvalues, 1 2

det 03 4

, so we have the characteristic equation:

2(1 )( 4 ) 6 0 3 10 0

This gives 2, 5 .

If 2 , then 1 2 0 1 2 0

3 4 0 3 6 0

x x

y y

.


1k

.

If 5 , then 1 2 0 6 2 0

3 4 0 3 1 0

x x

y y

.


3k

.



Solution 8.17

For the eigenvalues we must solve the equation:

2 2 3

det 1 1 1 0

1 3 1

So we have the equation:

{ } { } { }(2 ) (1 )( 1 ) 3 2 ( 1 ) 1 3 3 (1 ) 0l l l l l- - - - - + - - - + - - =

and this equation simplifies to 3 22 5 6 0l l l- - + = . Factorising the given cubic equation completely gives:

( 1)( 3)( 2) 0l l l- - + =

The eigenvalues are 1, 3, and –2.

When 1 ,

2 2 3 0 1 2 3 0

1 1 1 0 1 0 1 0

1 3 1 0 1 3 2 0

x x

y y

z z

, which gives

the eigenvector of

1

1

1

k

.

When 3 ,

2 2 3 0 1 2 3 0

1 1 1 0 1 2 1 0

1 3 1 0 1 3 4 0

x x

y y

z z

, which

gives the eigenvector of

1

1

1

k

.



When 2 ,

2 2 3 0 4 2 3 0

1 1 1 0 1 3 1 0

1 3 1 0 1 3 1 0

x x

y y

z z

, which

gives the eigenvector of

11

1

14

k

.

FAC: Glossary Page 1


Glossary

0 Introduction

This glossary contains useful information for your studies. As well as being mathematically correct your answers should demonstrate accurate use of language and grammar. This is especially important in subjects where you will have to produce essay style answers, such as Subject CT7 and the later subjects. The topics have been picked out from looking at assignment scripts and working out what reminders are necessary! The contents are: 1. Commonly misspelt words 2. Confusing word pairs 3. Latin and Greek 4. Words you will meet in actuarial work

FAC: Glossary


1 Commonly mis-spelt words

These are the correct spellings of words that have been spelt wrongly on assignments: actuarial ageing appropriately basically benefit benefiting biased calendar cancelled commission consensus correlation cyclically deferred definitely formatted fulfil gauge hierarchy immediately independence instalment interest millennium necessary occasion occurred occurring offer offered offering orthogonal paid particularly pensioner precede proceed receive referred referring relief seize separate similarly supersede targeted theorem until Notes: ● Single or double letters. With –ED and –ING words the rule is that you double

up the letter if the stress in the original word was on the last syllable (as in oc-CUR and re-FER) but not if the stress comes earlier in the word (as in OFF-er, TAR-get, BEN-e-fit). Words like “FOR-MAT”, which have equally stressed syllables, are considered to be stressed on the last syllable. With BIAS you can spell it either way – BIASED is more common, but you can use BIASSED if you prefer.

● British and American English treat L’s differently. In British spelling an L is always doubled-up before –ED and –ING, whereas the Americans don’t. So the British spelling is CANCELLED and the American spelling is CANCELED. Also in American English INSTALLMENT is spelt with two L’s. (In the exams it is best to use the British spelling conventions.)

● “I before E except after C” usually works provided the combination in question sounds like “EE”. However, the rule doesn’t work in plurals (eg POLICIES) or names (eg NEIL) or in the word SEIZE. (If the sound isn’t “EE”, you just have to remember the correct spelling eg HIERARCHY.)

● Adverbs derived from –IC words (eg CYCLICALLY) always have a silent AL in the ending, even if the corresponding –ICAL word doesn’t exist (eg BASICALLY, SPECIFICALLY). However, the word PUBLICLY is an exception for some reason.

● For some reason people (especially non-actuaries!) often miss the middle A out of ACTUARIAL.

● Make sure you put the right number of O’s in the words LO(O)SE. LOSE rhymes with “booze” and is the opposite of “find”. LOOSE rhymes with “moose” and is the opposite of “tight”.



2 Confusing word pairs

The following pairs of words sound similar but are spelt differently according to the context: Advice/advise (and practice/practise) ADVICE and PRACTICE are nouns. Example: He took the advice he was given about passing the exams. ADVISE and PRACTISE are verbs. Example: I advised him how to pass the exams. Affect/effect AFFECT is a verb meaning “to influence”. Example: Studying will affect your exam performance. EFFECT can be a noun meaning a “noticeable change”. Example: Taking this drug causes a strange effect. EFFECT can also be a verb meaning “to put into effect”. (This is rather a specialist meaning, but it comes up often in actuarial work.) Example: She effected her car insurance policy on 1 July. Dependant/dependent DEPENDANT is a noun meaning “someone who is supported by someone else”. Example: He has two dependants: his son and his sick mother. DEPENDENT is an adjective meaning “being supported by someone else” or “reliant on something else”. Example: This formula is dependent on the assumption of equal variances. In fact, there is some flexibility with this pair, but these are the conventions usually adopted.

FAC: Glossary


Precede/proceed PRECEDE is a verb meaning “to come before”. Example: Will payment precede or follow delivery? PROCEED is a verb meaning “to continue”. Example: Do you wish to proceed with reformatting your hard disk? Principal/principle PRINCIPAL is an adjective meaning “main” or “primary”. Example: The principal reason was financial. This spelling is also used for the head of a school or college. (This is really an abbreviation for “principal teacher” or “principal director”.) PRINCIPLE is a noun meaning a “rule” or “moral guidance”. Example: He was a man of principle. This spelling is also used in the phrases “on principle” and “in principle”, which are derived from this meaning.



3 Latin and Greek

Plurals A lot of words that come directly from Latin and Greek keep their original Latin and Greek plural endings when they’re used in English. The patterns you’ll see are:

Singular Plural Examples

LA

TIN

–US –I stimulus/stimuli

–UM –A maximum/maxima

medium/media

–A –AE formula/formulae

–IX or –EX –ICES index/indices

appendix/appendices

GR

EE

K –ON –A

criterion/criteria phenomenon/phenomena

–IS –ES basis/bases

analysis/analyses crisis/crises

Notes: ● A lot of Latin and Greek words use the regular English –(E)S ending. So you

should say: SURPLUSES, CENSUSES, SYLLABUSES, STATUSES, PREMIUMS, LEMMAS.

● With a lot of words you have the choice of a classical or an English plural. For example, with formula you can use either FORMULAE or FORMULAS. (Americans tend to use formulas.) You can also say TRAPEZIA or TRAPEZIUMS.

● The plural of INDEX is INDICES when you’re talking about an economic index, but INDEXES if you mean the pages at the back of a book. The plural of APPENDIX is APPENDICES when you’re talking about the section in a book, but APPENDIXES if you mean the thing that gives people appendicitis.

● The plural of SERIES is the same as the singular.

● The word DATA can be treated as plural (eg “The data are complete”). However, many people think of DATA as singular (eg “The data has arrived”). In the course notes you will probably come across both uses.

● The word DICE also causes problems. Strictly speaking, the thing with six sides you play games with is called a DIE and DICE is the plural form.

FAC: Glossary


Words and abbreviations PER ANNUM is a common phrase meaning a year eg “Some actuaries are paid more than £100,000 per annum.” The abbreviation pa is universally recognised. We also use pm (for a month), pcm (for a calendar month) and pq (for a quarter), but these are not so well known. PRO RATA means in proportion. For example, “Your pension will be calculated as 1/60th of your salary for each year of service, with months counting pro rata” means

that, if you worked for 5 years and 5 months (say), this would be counted as 5125 in the

calculation. It is also used as a verb eg “We can pro rata the payments to allow for holidays”. SIC. This is the Latin word for thus. People use it to mean “it really said this” when they’re quoting a passage from a document that has a mistake in. For example, “In his letter the client said that the company would be increasing its contribution rate from 6½% to 6¼% [sic].” I’ve put “sic” to show that I realised that the client’s secretary had probably typed these numbers the wrong way round, but this was what the letter actually said. STATUS QUO means the current position. So “trying to maintain the status quo” means trying to keep things as they are. STET. This means “Let it stand” in Latin. If you cross something out or change something, but then realise it was right after all and you wish you hadn’t changed it, you can write stet next to it. This tells the person reading it that you crossed it out by accident. This one can be useful in exams too. VICE VERSA means the other way round. eg. This means for example and is used where you want to give an example that could have been one of several. ie. This means that is and is used when you want to pin down exactly what you mean. cf. This just means compare. For example, you might write, “Using the approximation, I got £73.98 (cf £74.02 when calculated accurately).”



4 Words you will meet in actuarial work

The BASIS for an actuarial calculation is used to mean the set of assumptions (eg mortality rates, interest rates) used in the calculation. A STRONG basis is one with very pessimistic assumptions. A WEAK basis is one with optimistic assumptions. A LIFE just means a person. A FIRST-CLASS LIFE is a person in perfect health. Otherwise, they are IMPAIRED. IMMEDIATE is the opposite of “deferred”. It doesn’t necessarily mean “straight away”. A “deferred pension” would normally start making payments a number of years in the future. An “immediate pension” would make the first payment at some time during the coming year, but not necessarily at the start of that year. LEVEL means constant eg “level payments” are for the same amount each time. A NET payment is one where something has been deducted. Net monthly pay generally refers to the amount of your “take home” pay after your employer has deducted any amounts due in tax, pension contributions etc. Your GROSS monthly pay ignores these deductions. In actuarial contexts the words “net” and “gross” are used a lot and need not refer to tax. So you should always ask yourself “net of what?” ie what is it that has been deducted? An OFFICE (short for LIFE OFFICE) just means an insurance company. OUTGO is (very logically) the opposite of INCOME. The word PAYABLE means “must be paid” rather than “may be paid”. For example: “£1000 tax is payable on 31 January” doesn’t mean you have an option. The word SECULAR is used to mean, “in relation to time measured by reference to the calendar”. It clarifies the meaning when time could be measured relative to some other reference point eg the time since you were born or since you took out your life insurance policy (which is called the DURATION). So, for example, if I said “Mortality can be expected to improve over your lifetime because of secular effects”, I would mean that in the future they will find a cure for cancer etc, so you are likely to live longer than your parents’ generation. STOCHASTIC means allowing for random variation over time. It is the opposite of DETERMINISTIC.

FAC: Glossary


There are some words that put the S in a strange place when you make them plural eg SUMS ASSURED, ANNUITIES-CERTAIN, CLAIMS EXPERIENCE and NO

CLAIMS DISCOUNT.

FAC: Question & Answer Bank – Questions Page 1


Question & Answer Bank – Questions

Chapter 1 Question A1

Interpret the statement { }: 1 5x xŒ - £ < .

A { }1,0,1,2,3,4,5x Œ -

B { }1,0,1,2,3,4x Œ -

C { }0,1,2,3,4x Œ

D { }1,2,3,4x Œ [1]

FAC 1 1 Question A2

If {2,4,6,8}A = and {1,2,3,4}B = and {1,3,6,10}C = which are subsets of

{1,2,3,4,5,6,7,8,9,10} , which one of the following is ( )A B C« » ?

A {1,3,6}

B {6}

C {1,3,6,8,10}

D {1,3,6,10} [1]

FAC 1 1 Question A3

If {1,3,5,7,9}A = and {1,3,6,10}B = which are subsets of , which one of the

following would be in A B« ? A 3 B 5 C 6 D 8 [1] FAC 1 1

FAC: Question & Answer Bank – Questions


Question A4

What are the Greek letters used to represent a sum and a product? A ,S P

B ,s p

C ,rD

D ,pS [1]

FAC 1 2 Question A5

In June, interest rates were 2%. They rose by 40 basis points in July. What was the interest rate after this change? A 42% B 6% C 2.4% D 2.04% [1] FAC 1 3 Question A6

A is the statement “the integer x is odd”, B is the statement “the integer x is divisible by 3”. Which of the following statements is TRUE? A A is necessary for B B A is sufficient for B C A is necessary and sufficient for B D none of the above [2] FAC 1 4 Question A7

Which of the following was a leap year? A 1900 B 1910 C 1920 D 1950 [1] FAC 1 7



Question A8

Using the exchange rates below, how much would £100 be worth in US dollars?

£/€ = 1.227 and €/$ = 1.314 A $62.02 B $93.38 C $107.09 D $161.23 [1] FAC 1 7 Question A9

Ten payments of $100 are made at half-yearly intervals. The first payment is on 1 September 2011. On what date was the last payment made? A 1 March 2016 B 1 September 2015 C 31 August 2015 D 1 September 2016 [1] FAC 1 7 Question A10

For each of the following statements, either state that it is true or state that it is false and write down a correct version of the statement. (i) Ending in a 5 is a necessary condition for an integer to be a multiple of 5. [1]

(ii) 2{ : , 4, 2}x x x xŒ = π = ∆ [1]

(iii) 2 5 7 0,x x x- + > " Œ [2]

(iv) The natural logarithm function is defined on the set of values (0, ]• . [1]

[Total 5]



Question A11

A life office operates an investment fund that allows UK investors to buy investment units whose value is denominated in US dollars. Investors who wish to purchase units pay in sterling and an initial charge of 2½% is deducted before the payment is converted to US dollars and used to purchase the investment units. Last December, when the exchange rate between Sterling and the US dollar was £1 $1.67= , a woman invested £64,000 in this fund. Since then the quoted unit price has gone up by 5% and the exchange rate is now 1.60. What is the current sterling value of her investment? [3] FAC 1 7


What is 2.89951 rounded to 3 decimal places? A 2.899 B 2.900 C 2.90 D 2.810 [1] FAC 2 1 Question A13

What is 4,716 rounded to 3 significant figures? A 472 B 716 C 4,700 D 4,720 [1] FAC 2 1



Question A14

What is 0.020581 rounded to 2 significant figures? A 0.02 B 0.021 C 0.0206 D 0.02058 [1] FAC 2 1 Question A15

Calculate 1tanh 0.6- . A 0.0105 B 0.537 C 0.693 D 30.963 [1] FAC 2 2 Question A16

Calculate 1 (1 )

1

ni

i i

-- ++

where 0.062i = and 10n = .

A 14.130- B 12.528- C 6.865 D 7.743 [2] FAC 2 2



Question A17

Calculate 2 3 46 8 10 12v v v v where 1

1.08v = . Give your answer to 3 significant

figures. A 29.173 B 29.2 C 33.3 D 33.333 [2] FAC 2 2 Question A18

Calculate the value of 2

1 1 ln5 1.02exp

2 0.80.8 2

to 3 decimal places.

A 0.112 B 0.210 C 0.345 D 0.380 [2] FAC 2 2 Question A19

Calculate 5 12 46!4 20

3+ + .

A 5.897 B 243.9 C 259.7 D 327.4 [1] FAC 2 2



Question A20

Evaluate

1 nnv

nvd

i

- -, given that 0.10i = , 10n = , 1(1 )v i -= + and d iv= . [2]

FAC 2 2 Question A21

The interest rate i charged for a financial arrangement satisfies the following equation:

10 543,600,000(1 ) 12 10 366,000(1 ) 60,192,000i i+ = ¥ ¥ + +

By working in units of 1 million, or otherwise, calculate the value of i , expressing your answer as a percentage to three significant figures. [4] FAC 2 2



Question A22

An independent financial advisor (IFA) uses the form shown in the diagram to calculate the contribution rates for clients with personal pension policies. A colleague has carried out the calculation shown, which shows that, if Mr Smith wishes to retire at age 60 with an index-linked pension of 50% of his salary at retirement, he will have to make payments of 12.7% of his salary into his pension plan. This figure has been communicated to the client.

(i) Study the form and check that the calculations are correct. Then answer the

following additional questions that Mr Smith has raised:



(a) What contribution rate would you recommend if I made a one-off extra contribution now of £10,000 to my existing fund?

(b) If I decide to pay 15% of my salary instead, what percentage pension can

I expect to receive at retirement? (c) If I decide to retire at age 55 instead, what contribution rate would you

recommend if I still want a 50% pension? [6] (ii) Ms Jones, another client, has telephoned. She is aged 42 years and 6 months, has

a current salary of £35,000 and an existing fund of £175,000, and wishes to

retire at age 60 with a 23 rds pension. Calculate the recommended contribution

rate for her. [3] [Total 9] Question A23

The rules of a pension scheme state that employees who leave the company before retirement age are entitled to receive an annual pension payable monthly from retirement age. The amount of each payment is calculated using the formula:

Pension payment = Pension entitlement at date of leaving 1.05t¥ where t is the number of complete calendar years between the date of leaving and the date of payment. Pension payments are made on the first day of each month, starting in the month following the member’s 60th birthday. One member’s details are as follows: Sex = Male Date of birth = 14.04.1956 Date of leaving = 23.07.2003 Pension entitlement at date of leaving = £3,620 per annum (i) Calculate the amount of the first pension payment this member will receive and

write down the date on which this will be paid. (Assume that all months are of equal length.) [4]

(ii) Calculate the total amount of pension this member will receive during the first

ten years of retirement assuming that he is alive throughout that period. Apply a reasonableness check to your answer. [5]

[Total 9]




Which of these is the graph of xy e-= ?

A

x

y

B

x

y

C

x

y

D

x

y

[1] FAC 3 1



Question A25

Which of these is the graph of 1ny

x= where n is an odd number?

A

x

y

B

x

y

C

x

y

D

x

y

[1] FAC 3 1



Question A26

If the dotted line represents the graph of ( )y f x= , which of these is the graph of

(2 1)y f x= - ?

A

x

y

B

x

y

C

x

y

D

x

y

[2] FAC 3 1



Question A27

If a function ( )f x has the property ( ) ( )f x f x- = - , the function is known as:

A an odd function B a symmetrical function C an even function D an inverse function [1] FAC 3 1 Question A28

A person was born on 15 June 1955. If x is defined to be his exact age, what is [x] on 21 March 2011? A 55 B 55.75 C 55.83 D 56 [1] FAC 3 2 Question A29

Write the expression 4 1 5x + < without the use of the modulus sign.

A 1x < B 1.5x < - C 1.5 1x- < < D 1 1x- < < [1]


Calculate V, where 0.8x = and min( , 4)xV e p-= .

A 0.443 B 0.449 C 0.886 D 2.226 [1] FAC 3 2



Question A31

Calculate the value of 100!

97!3! to 2 decimal places.

A 161,700 B 323,400 C 15,684,900 D 31,369,800 [1] FAC 3 3 Question A32

Evaluate (5.5)G . You may use the fact that (½) pG = .


Simplify the expression( 1)(3 1)!

(3 )( 1)!

n n

n n

G + +G -

, where n is an integer.

A ( 1)(3 1)n n n+ +

B ( 1)(3 1)

3 ( 1)

n n

n n

+ +-

C (3 1)

(3 1)( 1)

n n

n n

+- -

D 23 (3 1)n n + [2]

FAC 3 3



Question A34

Evaluate each of the following, quoting your answers to 2 significant figures:

(i) e e [1] (ii) log 0.00001e [1]

(iii) 1tanh 0.9- [1] (iv) (12)G [1]

[Total 4]


Which of the following is the correct simplification of ( )2 32 2bb bx x x x¥ + ?

A 62 bx

B 2 32 2b b bx x+ ++

C 34 bx

D 3 32 2b bx x ++ [1] FAC 4 1 Question A36

Simplify 33 2x xe e .

A ½e

B 33 8x xe -

C 31

xe

D 33 8x xe [1]

FAC 4 1



Question A37

If loga x A= then:

A xa A=

B ax A=

C Aa x=

D aA x= [1] FAC 4 1 Question A38

Simplify 2log log .a ax

yy

A 2loga x

B 2

logax

y

C loga yx

D 2

logax

y

[1]

FAC 4 1



Question A39

Express 3 5 2

3 1 1

x x

x x

as a single fraction:

A 25 15 8

(3 1)( 1)

x x

x x

B 27 11 2

(3 1)( 1)

x x

x x

C 27 11 8

(3 1)( 1)

x x

x x

D 25 15 2

(3 1)( 1)

x x

x x

[2]


Simplify 2

2 2

3 6

2 3 2 2 3 1

x x x

x x x x

+ + -∏- - + +

.

A 3 2

4 3 2

2 9 10 3

2 17 16 12

x x x

x x x x

B 2

1

( 2)

x

x

C 2

2

4 3

4 4

x x

x x

+ +- +

D 2( 3)

(2 1)( 1)

x

x x

[5]

FAC 4 1



Question A41

Solve the quadratic equation 22 5 3 0x x . A 0.5 and 3x = - B 0.5 and 3x = - C 0.5 and 3x = D The equation has no real solutions [2] FAC 4 2 Question A42

Find the solution to 2 12 12 0x x in surd form.

A 6 2 6

B 6 4 3

C 12 6

D 12 2 3 [2] FAC 4 2 Question A43


95,432 64,717 57,963

92,404 64,522 80,068

x y

x y

+ =

+ = [2]



20.75 and 0.0225( ) ( 1)

a aba b a b a b

= =+ + + +

[2]

FAC 4 3



Question A45


91

la

=-

2

2 162( 1) ( 2)

a la a

=- -

[2]


If 2 3 7x - < then:

A 5 2x- < < B 2 5x- < < C 5 5x- < < D 5x < [1] FAC 4 4 Question A47

If 2 12 0x x- - < then: A 4x < B 3 and 4x x< - >

C 3x < - D 3 4x- < < [1] FAC 4 4 Question A48

Find the range(s) of values of x for which 2

( 2)(9 8)1

2 4

x x

x x

- - >+ -

. [5]

FAC 4 4



Question A49

12 2

0 1

n n

k k

k k

A 2

0

2n

k

k

B 1

2

0

n

k

k

C 2 1 2

1

2n

n

k

k k

D 2 2

0

1 2n

k

n k

[1]


A child receives pocket money of £5 in the first week, then £6 in the second week, increasing by £1 each week. How much does he (she) receive in the first year? A £1,456 B £1,586 C £1,612 D £3,172 [1] FAC 4 7 Question A51

Evaluate 1

210

4kk

•

=Â

A 70 B 210 C 280 D 840 [2] FAC 4 7



Question A52

Simplify 2

1

(1 2 )n

i

i i=

+ +Â . You may use the results 12

1

( 1)n

k

k n n=

= +Â and

2 16

1

( 1)(2 1)n

k

k n n n=

= + +Â .

A 3 22 4 3n n n+ +

B 3 22 9 7 6

6

n n n+ + +

C 28 15 1

6

n n+ +

D 3 22 9 13

6

n n n+ + [2]


20 202

1y x y

y x= =

=Â Â

A 20 20

2

1 1x y

x y= =Â Â

B 20 20

2

1x y x

x y= =Â Â

C 20 20

2

1x y y

x y= =Â Â

D 20

2

1 1

x

x y

x y= =Â Â [1]

FAC 4 8



Question A54

Calculate 6 6

1

2 ( 1)x y x

x y= =

-Â Â . You may use the results 12

1

( 1)n

k

k n n=

= +Â ,

2 16

1

( 1)(2 1)n

k

k n n n=

= + +Â and 3 2 214

1

( 1)n

k

k n n=

= +Â .

A 350 B 420 C 540 D 630 [5] FAC 4 8 Question A55

What is the coefficient of 5 3x y in the expansion of the expression 8(2 5 )x y+ ?

A 56 B 224,000

C 448,000

D 5,600,000 [2]


Use the result ( 1) ( 1)( 2)2 32! 3!

(1 ) 1 p p p p ppx px x x to simplify 2

(2 5 )

(3 2)

x

x

by expanding as far as the term in 3x .

A 2 31 11 57 243

2 4 8 16x x x

B 2 357 2432 11

2 4x x x

C 2 323 28 2442

3 3 27x x x

D 2 32 23 28 244

9 27 27 243x x x [5]

FAC 4 9



Question A57

The interest rate i for a financial arrangement satisfies the following equation:

½ ¼20,000(1 ) 5,000(1 ) 2,500(1 ) 30,000i i i+ + + + + =

(i) Find an approximate value for i by using the approximate relationship

(1 ) 1ti ti+ ª + , quoting your answer to the nearest 0.01%. [3]

(ii) By refining the approximate relationship to include an additional term, calculate

a more accurate value of i . [4] [Total 7] Question A58

You are given the following results involving series:

0 !

kx

k

xe

k

•

== Â

1

1

( 1)log (1 )

k k

ek

xx

k

• +

=

-+ = Â

1

( 1)...( 1)(1 ) 1

!p k

k

p p p kx x

k

•

=

- - ++ = + Â

Use these to simplify the following series expressions:

(i) 2 3 41 1 12 6 24x x x x- + - +

[2]

(ii) 2 41 12 241 x x+ + +

[2]

(iii) 2 3 41 1 12 3 4x x x x+ + + +

[2]

(iv) 2 34 5 6 7

0 1 2 3x x x

Ê ˆ Ê ˆ Ê ˆ Ê ˆ+ + + +Á ˜ Á ˜ Á ˜ Á ˜Ë ¯ Ë ¯ Ë ¯ Ë ¯

[2]

[Total 8]



Question A59

Let i be a positive quantity and let n be an integer greater than 1. (i)(a) By considering the arithmetic-geometric mean inequality as it would apply to the

quantities 1 and 1 i+ , show that ½1 (1 )2

ii+ > + .

(i)(b) Show that 1/1 (1 ) nii

n+ > + . [5]

(ii) Show that 1 (1 )nin i+ < + . [2]

[Total 7] Question A60

Let 2 3( ) 2 3 nnS x x x x nx= + + + + .

(i) By considering the expression ( )

( )nn

S xS x

x- , or otherwise, show that:

1

2

(1 )( )

(1 )(1 )

n n

nx x nx

S xxx

+-= ---

[3]

(ii) Write down an expression for lim ( )n

nS x

Æ•, stating the range of values of x for

which convergence occurs. [2] [Total 5]




The population of a country at the start of the year is 62,400,000. The population at the end of the year is 63,760,000. Express the change in population as a rate per mil. A 2.13‰ B 2.18‰ C 21.8‰ D 21.3‰ [1]


An item has increased in cost by 30% to $58.50. What was the original cost? A $38.50 B $40.95 C $45 D $76.05 [1] FAC 5 1 Question A63

What is the proportionate change in the price of a house that goes from £180,000 to £210,000?

A 14.29% B 16.67% C 83.33% D 116.7% [1] FAC 5 2



Question A64

When calculating 3y x= , a student uses 50x = to 2 significant figures. What is the

maximum absolute error possible here? A 0.0122 B 0.0123 C 0.119 D 0.127 [1] FAC 5 3 Question A65

You know that a equals 5,000 correct to one significant figure and b equals 0.20 correct to two decimal places. You calculate /a b to be 25,000. What is the greatest percentage by which you might be overestimating the true value? A 8.78% B 11.4% C 12.8% D 13.9% [2] FAC 5 3 Question A66

The following values from a standard normal distribution table were obtained: ( 1.20) 0.88493 ( 1.21) 0.88686P Z P Z< = < =

Use linear interpolation to calculate the value of ( 1.207)P Z < .

A 0.88507 B 0.88551 C 0.88590 D 0.88628 [1] FAC 5 5



Question A67

The Newton-Raphson method ( )( )1

nn n

n

f xx x

f x+ = -¢

is being used to find the solution

between 1 and 2 to the equation ( ) 2 3 5 0f x x x= + - = . What is 4x if 1 1x = ?


Expand (2 4)(3 )i i- - and express it in its simplest form, where 1i = - .

A 22 10 12i i- + - B 10 14i - C 10 10i - D 2 14i - [1] FAC 5 7 Question A69

What is the result if ( )2 3i- is multiplied by its complex conjugate?

A 5- B 5 12i- - C 13 D 13 12i- [1] FAC 5 7



Question A70

Simplify 2

2

i

i

-+

, where 1i = - .

A 3 45 5 i-

B 431 i-

C 451 i-

D 5 43 3 i- [2]


What is the modulus, r , and argument, q , in radians, of the complex number ( )5 3i- ?

A 4, 0.540r q= =

B 5.831, 0.540r q= = -

C 4, 30.96r q= =

D 5.831, 30.96r q= = - [2]


Find the roots of 2 3 2.5 0x x+ + = . A 0.5 1.5 or 0.5 1.5x i i= - + - - B 0.5 1.5 or 0.5 1.5x i i= + - C 1.5 0.5 or 1.5 0.5x i i= + - D 1.5 0.5 or 1.5 0.5x i i= - + - - [2]

FAC 5 7



Question A73

In the following equations, k is measured in £ and a is a dimensionless quantity:

1 kakl

= + , 2 2

akl

= , 3 3

2akl

=

(i) Determine the units of measurement of 1 2 3, ,k k k . [3]

(ii) Find the value of the constant c that would make the quantity 3

2c

kk

dimensionless. [1] [Total 4] Question A74

Let 1 2 4z i= + and 2 1 2z i= - .

(i) Write each of the following in the form a bi+ where ,a b RŒ :

1 2z z+ , 1 2z z- , 1 2z z , 1

2

z

z and 2 2

1 2z z+ [5]

(ii) Calculate 1 2z z and 1 2z z , and comment on your answers. [3]

(iii) Calculate 1 2z z and 1 2z z , and comment on your answers. [3]

Note that z denotes the complex conjugate of z .

(iv) Express 1 23 2z z+ in the form ire q . [2]

[Total 13]



Question A75

Consider the equation 3 22 9 14 5 0z z z- + - = . (i) Show that 2 i+ is a root of this equation. [3] (ii) Write down the other complex root. [1] (iii) Hence find all the roots of the equation. [2]

(iv) Which of these roots would lie outside the circle 1z = when represented on an

Argand diagram? [1] [Total 7] Question A76

Find the solution of the difference equation: 1 26 8 0t t ty y y- -+ + =

given that 0y and 1y are both equal to 5. [4]

FAC 5 8




Which of the options best describes the function 23 5x x ? A ( ) and ( )O x o x

B ( ) but not ( )O x o x

C ( ) but not ( )o x O x

D none of the above [1] FAC 6 1 Question A78

What is the supremum of the sequence 1 2 3

, , ,...4 7 10

?

A 1 3

B C 0.333 D 0 [1] FAC 6 1 Question A79

Which of the following best describes what dy dx represents?

A y xd d for small values of xd and yd

B the change in the value of the graph between 0 and x C the gradient of ( )y f x=

D the rate of change of a chord between x a= and x a h= + [1] FAC 6.2



Question A80

If ( ) 5f x x= , which of the following gives 0

( ) ( )lim

h

dy f x h f x

dx h+Æ

+ -= ?

A 0 B 1

C 5 D x [1] FAC 6.2 Question A81

Differentiate ( ) 5(4 )tf t with respect to t .

A 120t

B 5(4 )ln 4t

C 15 4tt D 20ln t [1] FAC 6 3 Question A82

Find the gradient of 3 22 3y x x= - + when 2x = .

A 15 B 20 C 23 D 44 [1] FAC 6 3



Question A83

Find the derivative of 3 5

4( )f x

x .

A 8 5

12

5x

B 2 5

12

5x

C 2 53

5

4

x

D 8 53

5

4

x [1]


Find ( )f u¢ where 2 9( ) (5 7)f u u .

A 2 109(5 7)u -

B 2 89(5 7)u -

C 2 845 (5 7)u u -

D 2 890 (5 7)u u - [1]


Find (0)M ¢ if 2 2½( ) t tM t e .

A

B em

C e

D 1 [2] FAC 6 4



Question A86

Find (0)M ¢ if ( ) ln 1t

M t

.

A a- B a l

C al D a l- [2]


Find ( )f x¢ where 2

2

4( ) ln 2 x

xf x e

e

.

A 2

2

14

2 xx

ee

B

2

2

2

2

42

42

xx

xx

xee

ee

C

2

2

2

2

84

42

xx

xx

xee

ee

D

2

2

2

2

42

2

xx

xx

xee

ee

[2]

FAC 6 4



Question A88

Find ( )f x¢ where 26(5 3)( ) ln 4 xf x e .

A 60(5 3)x

B 26(5 3)60(5 3)ln 4 xx e

C 26(5 3)

60(5 3)

4 x

x

e

D

2

2

6(5 3)

6(5 3)

60(5 3)

4

x

x

x e

e

[5]


Differentiate 2 5(4 3 )( 2 1)y x x x with respect to x .

A 2 4 2( 2 1) (3( 2 1) 5(4 3)(2 2))x x x x x x

B 2 4 2( 2 1) ( 33 76 43)x x x x

C 2 5 43( 2 1) 5(4 3 )(2 2)x x x x

D 2 415(2 2)( 2 1)x x x [5]

FAC 6 4



Question A90

Find ( )f x¢¢ where 2 4( ) 32 xf x x e .

A 2 464(1 8 ) xx e

B 2 464(1 8 8 ) xx x e

C 41,024 xe

D 2 4 464(1 8 8 ) xx x e [5]


Differentiate 23

x

xy

e twice with respect to x .

A 26 12 3

x

x x

e

- +

B 26 12 3

x

x x

e

- -

C 26 12 3

x

x x

e

+ +

D 26 12 3

x

x x

e

+ - [5]


Calculate the maximum value of the function 2( ) 5 2 3f x x x .

A 6 B 4

C 163

D 13- [2]

FAC 6 6



Question A93

Which of the following is a graph of ( 3)( 5)y x x x= + - ?

A

x-5 -4 -3 -2 -1 1 2 3 4 5

y

B

x-5 -4 -3 -2 -1 1 2 3 4 5

y

C

x-5 -4 -3 -2 -1 1 2 3 4 5

y

D

x-5 -4 -3 -2 -1 1 2 3 4 5

y

[2] FAC 6 6



Question A94

Find and distinguish between the turning points of the function: 3 2( ) 7f x x x x .

A maximum 1,6 and minimum 513 27

( ,7 )

B maximum 1,6 and minimum 1613 27

( ,6 )

C maximum 513 27

( ,7 ) and minimum 1,6

D maximum 1613 27

( ,6 ) and minimum 1,6 [5]


Using log differentiation or otherwise, find the value of x for which 4 3902( ) xf x x e-=

is a maximum.

A 43-

B 34-

C 34

D 43 [2]


What is ( ) ( )( )2 3axy b xy c xy

x

∂ + +∂

where a , b and c are constants?

A ( ) ( )22 3a b yx c xy+ +

B 2 2 3 3 4

2 3 4

ax y by x cy x+ +

C 22 3ay byx cyx+ +

D 2 3 22 3ay by x cy x+ + [1]

FAC 6 7



Question A97

For the function ( ) ( )2, ,f x y z xyz= , what is 3 f f

x y z x

∂ ∂¥∂ ∂ ∂ ∂

?

A ( )28 2xyz x yz-

B 8yz

C ( )328x yz

D ( )3216x yz [2]


Find the turning points and their nature for the function 3 2 2( , ) 3 2f x y x x xy y

by considering the roots of the equation:

0 0 00 0 0

2

2 2 2

2 20


f f f

x yy x

A Local maximum at 0, 0x y

B Saddle point at 4 43 3

,x y

C Local maximum at 0, 0x y and saddle point at 4 43 3

,x y

D Local minimum at 0, 0x y and maximum at 4 43 3

,x y [5]

FAC 6 8



Question A99

Determine the extrema of the function ( , ) 5 3f x y x y subject to the constraint 2 2 136x y .

You may make use of the Lagrangian function: 1 1( , , ) ( , , )n nL f x x g x x .

A maximum at (10, 6)f -

B minimum at ( 5,3)f -

C maximum at (10, 6)f - and minimum at ( 10,6)f -

D maximum at (5, 3)f - [5]


Evaluate:

(i) 3 3

0lim

x x

x xx

e e

e e

-

-Æ

--

[2]

(ii) 1

lim 12

n

n n

[2]


Differentiate the following function with respect to i :

10

10

100 1 (1 )5

(1 )

i

ii

-- +++

and evaluate the derivative at 0.05i = . [3] FAC 6 4



Question A102

Find a formula for the n th derivative of the function 1

2x

x+ . [4]


By considering log y , or otherwise, find an expression in terms of l for the maximum

value attained by the function 1 1( )y x xa g g aagl l- - -= + when 1a = and 4g = . [5]


A function is defined by:

( ) (1 )f x x xa b= + , [0,1]x Œ

where , Ra b Œ are parameters.

By considering the shape of the graphs, or otherwise, find the range of values attained by the function ( )f x for values of x in the range [0,1] in each of the following cases:

(i) 1, 3a b= = - [3]

(ii) 2a b= = [3]

(iii) ½, 1½a b= - = . [3]


Find the positions of the extrema of 3 2 2( , ) 2 2f x y x x y= - + and determine their

nature. [5] FAC 6 8



Question A106

Find the positions of the extrema of ( , , )f x y z x y z= + + , subject to the constraints

2 0x y+ = and 2 2 2 1x y z+ + = . [5]


According to a mortality table, the instantaneous rate of mortality at age x , which is denoted by ( )xm , is calculated from the formula:

0 1 0 170 70

( ) exp50 50

x xx a a b b

, 17x ≥

and 0 0.00338415a = - , 1 0.00386512a = - , 0 3.352236b = - and 1 4.656042b = .

(i) Show that (20) 0.000814m = and find the value of (40)m . [3]

(ii) Find the age in the range 20 40x£ £ at which the function m is stationary,

rounding your answer to the nearest month, and indicate the nature of this point. [4] (iii) Another mortality function ( )q x is related to ( )xm by the relationship:

1

0( ) 1 exp ( )q x x t dt

Find the value of (20)q . [5]

[Total 12]




Which of the following is NOT true?

A ( ) ( )b a

a b

f x dx f x dx= -Ú Ú

B ( ) ( ) ( )0 0

b b a

a

f x dx f x dx f x dx= -Ú Ú Ú

C ( ) ( ) ( )b

a

df x dx f b f a

dx= -Ú

D ( )b

a

df x dx

dxÚ is used to find the area under the ( )f x curve [1]


What is the integral of 26 6 6x x+ + ? A 12 6x c+ +

B 3 22 3 6x x x c+ + +

C 3 26 6 6x x x c+ + +

D 3 23 6 6x x x c+ + + [1] FAC 7 2

Question A110

What is the integral of 2

0.75

1

x dxÚ ?

A 1.351 B 0.636 C 0.477 D 0.119- [1]

FAC 7 2



Question A111

What is the integral of 2 2x x-+ ?

A 2 2

2

x x-+c+

B 3 3

3

x x-+c+

C 3 13

3

x x--c+

D 3 3

3

x x--c+ [1]


What is 4

2

5 x dxÚ ?

A 1,260

ln 6

B 600

ln5

C 13583

D 625 25

ln 4 ln 2- [1]

FAC 7 2



Question A113

What is xBc dxÚ ?

A ln

xBcconst

x+

B xBc const+

C 1

1

xBcconst

x

++

+

D ln

xBcconst

c+ [1]


Evaluate 3

0

xe dxll -Ú .

A 3( 1)e ll - -

B 31 e l--

C 3(1 )e ll --

D 4.51 e l-- [1] FAC 7 2



Question A115

What is the integral of 5

1x +?

A ( )5ln 1x + c+

B ( )2

10

1x

-+

c+

C ( )2

5

1x

-+

c+

D ( )5

ln 1x +c+ [1]


Evaluate 43 5

0

xx e dx• -Ú .

A 0 B 0.05 C 1 D the integral does not converge [2]


Evaluate 51

6 30 (10 )

xdx

x+Ú .

A 0.00000132 B 0.00000792 C 0.000145 D 0.000868 [2] FAC 7 2



Question A118

What is the integral of2

3

3 3

2 6 1

x

x x

++ +

, for 0x > ?

A 3

4 2

3

0.5 3

x xc

x x x

+ ++ +

B ( )2ln 3 3x c+ +

C ( )30.5ln 2 6 1x x c+ + +

D ( ) ( )3 33 3 ln 2 6 1x x x x c+ + + + [2]

FAC 7 2 & 3 Question A119

Use partial fractions to find the integral of ( )( )2 5

2 4 1

x

x x

+- +

.

A ( )( )

31

2

2 4ln

1

xk

x

Ï ¸-Ô ÔÌ ˝+Ô ÔÓ ˛

B 3 ½ln ( 1) (2 4)k x x+ -

C 123ln( 1) ln(2 4)x x c+ - - +

D 123ln(2 4) ln( 1)x x c- + + + [5]


What is 1

2

0

xxe dx ?

A 10.778- B 0.5 C 1 D 2.097 [2] FAC 7 3



Question A121

Leibniz’s formula states that:

( ) ( )

( ) ( )

( , ) ( ) ( , ( )) ( ) ( , ( )) ( , ) b x b x

a x a x

d ff x t dt b x f x b x a x f x a x x t dt

dx x

∂∂

= - +¢ ¢Ú Ú .

What is 2

0

3 2x

dx t dt

dx-Ú ?

A 23x

B 26x

C 26 8x x-

D 29 2x x- [2] FAC 7 3

Question A122

Find the value of 15 10

5 5

(5 )x y

x y dy dx= =

+Ú Ú .

A 2,875 B 3,250 C 2,812.5 D 1,406.25 [2]

FAC 7 5



Question A123

Which of the following does NOT converge?

A 2

1

xe dx•

-Ú

B 1

0

ln x dxÚ

C 2

1

x dx•

-Ú

D 31

6dx

x

•

Ú [1]


Using the trapezium rule and 7 ordinates, what is the approximate area under the curve 2(7 )y x= - between 1x = and 4x = ?


If xe is expanded to its fourth term, using the Maclaurin expansion

2(0)( ) (0) (0)

2!

ff x f f x x

¢¢= + + + ◊◊ ◊¢ , what is the value of 2e based on this?

A 5 B 6.333 C 7.389 D 7.667 [2]

FAC 7 7



Question A126

Which of the following is the correct first five terms of the Taylor expansion of

[ ]ln (3 )(3 )x y+ + about (0,0) ? You are given that the Taylor’s expansion of ( , )f x y

about ( , )a b is:

( , ) ( , )

2 2 22 2

2 2( , ) ( , ) ( , )

1( , ) ( , ) ( ) ( )

1!

1( ) 2 ( )( ) ( )

2!

a b a b

a b a b a b


x y


x yx y

∂ ∂∂ ∂

∂ ∂ ∂∂ ∂∂ ∂

È ˘Í ˙= + - + -Í ˙Î ˚

È ˘Í ˙+ - + - - + -Í ˙Î ˚

+

A 2 21 1 1 1ln9

3 3 18 18x y x y+ + - -

B 2 21 1 1 1ln9

3 3 9 9x y x y+ + - -

C 2 21 1 1 1ln9

3 3 9 9x y x y+ + + +

D 2 21 1 1 1ln9

3 3 18 18x y x y+ + + + [5]


Given that 3 1

dy y

dx x=

+, 2y e= when 0x = , what is the value of y when 2x = ?

A 51.723 B 7 C 14.135 D 6 [2] FAC 7 8



Question A128

The gamma function is defined by 1

0( ) x tx t e dt

• - -G = Ú .

(i) Show that ( ) ( 1) ( 1)x x xG = - G - for values of x for which both gamma functions

are defined. [3] (ii) Show that (1) 1G = . [2]

(iii) You are given that (½) pG = . By considering (½)G and applying the

substitution 212t z= , or otherwise, show that

212

11

2

ze dz

p• --•

=Ú . [4]


Find the area of the region enclosed by the x axis and the curve 26 2y x x= + - . [4]


Evaluate ( , )A

f x y dxdyÚÚ where ( , ) 2 6f x y x y= + and the region of integration is the

shaded area shown in the diagram. [6] FAC 7 5

1x

1y



Question A131

Show that 2

210.035103

(5 )

xI dx

x= =

+Ú using at least three different methods.

[3 marks for each correct method] Question A132

(i) Calculate the value of the following integral, where 0.5 1x y< < < :

2

12

0.5 0.5

(4 1)y

y x

y x

y e xe dx dy= =

+Ú Ú [4]

(ii) Explain why it may be considered easier to integrate with respect to x first

rather than integrating with respect to y first. [5]


(i) Write out the first four terms in the Taylor series for xe and 1(1 )y -+ , stating

the range of values of x and y for which these series are valid. [2]

(ii) Hence determine the coefficients ,a b in the series expansion:

2 31 ( )1x

xax bx O x

e= + + +

- [5]

[Total 7]



Question A134

(i) Find constants A and B such that 1

(2 5 ) 2 5

A B

P P P P∫ +

- -. [2]

(ii) Hence, or otherwise, solve the differential equation:

1

2 5dP

PP dt

= -

for the function ( )P t , where 250 ( )P t< < , subject to the boundary condition that

215(0)P = . [4]


Find the particular solution of the differential equation:

( 1) ( 1)dy

x x x ydx

+ - + =

where 0x ≥ , given that 0y = when 0x = . [5]

FAC 7 8


If

2

1

4

a

and

2

5

3

b

then the values x and y such that

10

7

18

x ya b

are:

A 1x and 6y

B 4x and 1y

C 3x and 2y

D 2x and 1y [2]

FAC 8 1



Question A137

The unit vector in the direction of 3 2 4i j k is given by:

A 29(3 2 4 )i j k

B

31

221

4

C

31

229

4

D 3 2 421 21 21

i j k [1]


The angle between

4

1

3

and

1

3

2

is:

A 65.4º B 74.8º C 105.2º D 114.6º [2] FAC 8 1 Question A139

Which one of the vectors is orthogonal to 3 2i j k ?

A 4 4 4i j k

B 4 4 4i j k

C 4 4 4i j k

D 4 4 4i j k [2]

FAC 8 1



Question A140

If 1 2

3 4A

and 2 4

3 5B

then AB is given by:

A 2 8

9 20

B 8 14

6 32

C 8 14

18 32

D 14 20

18 26

[2]


If

3 2 4

1 0 3

3 1 2

A

then det A is given by:

A 33 B 5

C 1- D 19- [2] FAC 8 2



Question A142

If 2 1

4 3A

then 1A is given by:

A 3 11

4 22

B 0.3 0.1

0.4 0.2

C 1 31

2 42

D 2 1

1.5 0.5

[2]


The eigenvalues, , of matrix A solve: A det( ) 0A I

B det( ) 0A I

C det( ) 0A I

D det( ) 0A [1]

FAC 8 2



Question A144

Which of the following is NOT an eigenvector of 3 2

6 1A

:

A 2

6

B 3

1

C 1

3

D 2

2

[2]


(i) By considering expressions involving scalar products, find a unit vector of the form a b c+ +i j k that is perpendicular to the displacement vectors 2 3- +i j and

10 +i k . [4] (ii) Write down the other unit vector perpendicular to 2 3- +i j and 10 +i k . [1]


By solving a set of simultaneous linear equations, or otherwise, find the inverse of the

matrix

0.2 0.2 0.2

0.2 0 0.2

0 0.4 0.2

P

. [5]

FAC 8 2



Question A147

The probability density function of the multivariate normal distribution is given by:

11 1( ) exp ( ) ( )

2(2 ) det( )

T

nf x x x

Evaluate ( )f x when 2n , 1

1

, 2 1

1 4

and 2

0x

. [5]

FAC 8 2

Glossary Question A148

The abbreviation eg means: A for example B compared with C and so on D that is to say [1] FAC Gloss Question A149

What is the plural of matrix? A matrices B matrixes C matrises D matrix [1] FAC Gloss



Question A150

A weak basis for an actuarial calculation: A makes many assumptions B makes optimistic assumptions C makes few assumptions D makes pessimistic assumptions [1] FAC Gloss Question A151

A deterministic model has: A optimistic assumptions B assumptions which allow for random variation C many assumptions D assumptions which remain fixed over time [1] FAC Gloss Question A152

Which of the following word pairs should be used to complete the blanks in this sentence? The ___________ from ActEd is to ___________ lots of past papers before the exam. A advise, practise B advise, practice C advice, practise D advice, practice [1] FAC Gloss



Question A153

A friend has a full-year salary equivalent to £18,000 per annum. The company she works for has a 6 day working week. She is paid pro rata as she only works 2½ days per week. How much does she earn each year? A £18,000 B £9,000 C £7,500 D £6,000 [1] FAC Gloss

FAC: Question & Answer Bank – Solutions Page 1


Question & Answer Bank – Solutions Solution A1

D The natural numbers are 1, 2, 3, 4, 5, ... Solution A2

C

( )A B« is everything in A and not in B which is {6, 8}. So ( )A B C« » is these two

values and all the values in C which is {1, 3, 6, 8 ,10}.

Solution A3

D

A B« means not in A and not in B.

Solution A4

A

Solution A5

C 40 basis points is 0.40%. Solution A6

D Both odd and even numbers are divisible by 3.

FAC: Question & Answer Bank – Solutions


Solution A7

C Solution A8

D £100 100 1.227 €122.7 122.7 1.314 $161.23= ¥ = = ¥ = Solution A9

A The first payment is 1 Sept 2011, the second payment is 1 Mar 2012, the third payment is 1 Sept 2012, the fourth payment is 1 Mar 2013, the fifth payment is 1 Sep 2013, the sixth payment is 1 March 2014, the seventh payment is 1 Sept 2014, the eighth payment is 1 Mar 2015, the ninth payment is 1 Sept 2015 and the tenth payment is 1 Mar 2016. Solution A10

(i) FALSE. Should say: “Ending in a 5 is a sufficient condition for an integer to be a multiple of 5.” [1]

(ii) FALSE. 2x = - is a member of this set. Various corrections are possible eg

changing to + or , adding the extra condition 2x π - or changing = ∆ to π ∆ (!). [1]

(iii) TRUE. If we “complete the square”, we can write 2 25 32 45 7 ( )x x x- + = - + ,

which always takes positive values. [2] (iv) FALSE. The natural logarithm function doesn’t have a finite value when the

argument is either 0 or • . So the set of values should be (0, )• . [1]



Solution A11

The investor invests £64,000. After the initial charge is deducted this would leave £64,000 0.975¥ .

When converted to dollars (at the exchange rate effective at that time) this would give $64,000 0.975 1.67¥ ¥ . [1]

If the unit price then was $P per unit, this would buy 1

64,000 0.975 1.67P

¥ ¥ ¥ units.

The unit price has now increased to $1.05P . So the dollar value of the units would be

164,000 0.975 1.67 1.05 64,000 0.975 1.67 1.05P

P¥ ¥ ¥ ¥ = ¥ ¥ ¥ . [1]

Applying the current exchange rate to find the current sterling value gives

1£64,000 0.975 1.67 1.05 £68,386.50

1.60¥ ¥ ¥ ¥ = . [1]

Solution A12

B Solution A13

D Solution A14

B Significant figures are counted from the first non-zero digit. Solution A15

C



Solution A16

D

101 1.0627.743

0.062 1.062

-- =

Solution A17

B

2 3 4

6 8 10 1229.2

1.08 1.08 1.08 1.08 to 3 SF

Solution A18

D

2

21 1 ln5 1.02 1 1exp exp 0.7367974

2 0.8 20.8 2 0.8 2

1exp( 0.2714352) 0.380

0.8 2

Solution A19

B

5 12 5 124 46! 7204 20 4 20 240 1.782 2.115 243.9

3 3+ + = + + = + + =



Solution A20

Using the formulas given:

1 1(1 ) 1.1 ( 0.90909)v i - -= + = = [1]

and: 0.1 / 1.1 ( 0.09091)d iv= = = [1]

So:

10101 1 1.1

10 1.10.1 / 1.1 29.04

0.10

nnv

nvd

i

--- -- - ¥

= = [1]

Comment We’ve kept the exact figures throughout the calculation here to ensure that the final answer is accurate. However, you could equally use the calculated values of v and d , provided you retain enough decimals. Usually 5 significant figures is about right for intermediate calculations. You’ll meet these symbols when you study compound interest in Subject CT1.

Solution A21

Expressing the coefficients in units of 1 million and bringing all the terms onto the LHS gives:

10 543.6(1 ) 43.92(1 ) 60.192 0i i+ - + - = [1]

This is a quadratic equation in 5(1 )i+ . Using the quadratic formula, we get:

2

5 43.92 (43.92) 4(43.6)( 60.192) 43.92 111.474(1 )

2 43.6 87.2i

± - - ±+ = =¥

[1]

So:

5(1 ) 1.782 or 0.775i+ = - [1]



Disregarding the second possibility (which corresponds to a negative interest rate), we find that:

1/5(1.782) 1 0.12248i = - = ie 12.2% [1]

Solution A22

While checking the calculation you should have noticed that factors needed for intermediate ages have been calculated using linear interpolation. For example, the

term of 1421 years is a quarter of the way from 20 to 25. So F1 is calculated as:

1416.5 (19.7 16.5) 17.3+ - =

(i)(a) Contribution rate This would bring the existing fund to £82,500 and the calculation would be: 4.125 17.3 50% 0.657 17.7k+ ¥ = ¥ ¥ 9.8%kfi = [2] (i)(b) Percentage pension We now have: 3.625 15% 17.3 0.657 17.7t+ ¥ = ¥ ¥ 53.5%tfi = [2] (i)(c) Contribution rate

We now have 31216n = and the calculation is:

3.625 13.875 50% 0.7255 19.9k+ ¥ = ¥ ¥ 25.9%kfi = [2]

Comment In any calculation you should ask yourself whether you believe the answer you’ve come up with. You can do this here by checking that the alterations have the effect you would expect. For example, the contribution rate in part (a) falls slightly, which makes sense. In other questions, you can apply a reasonableness check by simplifying the calculation using approximations and “ball park” figures.



(ii) Recommended contribution rate The calculation for Ms Jones (with 17½n = ) is:

235 14.75 0.708 19.9k+ ¥ = ¥ ¥ 29.8%kfi = [3]

Note that you need to use the female retirement factor here. The m and f in the columns for F3 stand for male and female.

Comment Although we chose factors that were fairly realistic, there is no guarantee that this calculation form would be appropriate for a real life calculation. There are also limits as to how much you’re allowed to contribute to a personal pension plan.

Solution A23

(i) Pension first payment and date Pension payments will start on 1 May 2016. The complete calendar years between leaving and retirement are the twelve years 2004, 2005, … , 2015. So 12t = when the payments start, but it will increase by 1 on 1 January each year. [2]

So the member’s first payment will be 1213,620 1.05 £541.75

12¥ ¥ = . [2]

(ii) Total pension during first 10 years The payments the member will receive during the first ten years of retirement will consist of: 8 payments of £541.75 during 2016 12 payments of £541.75 1.05¥ during 2017

12 payments of 2£541.75 1.05¥ during 2018 …

12 payments of 9£541.75 1.05¥ during 2025

4 payments of 10£541.75 1.05¥ during 2026 [2]



The total amount is:

9 108 541.75 12 541.75 (1.05 1.05 ) 4 541.75 1.05¥ + ¥ ¥ + + + ¥ ¥ [1]

The middle terms can be summed as a GP, which gives:

10

101.05 1.058 541.75 12 541.75 4 541.75 1.05 £83,132

1 1.05

-¥ + ¥ ¥ + ¥ ¥ =-

[1] (The exact amount will depend on how rounding is applied by the scheme’s administrators.) As a reasonableness check, we can say that the typical payment will be

5541.75 1.05 £691.43¥ = and there are 120 payments, which would make a total of 120 691.43 £82,971¥ = . This is close to our calculated answer. [1]

Solution A24

B

A is the graph of xy e=

C is the graph of lny x=

D is the graph of 2y x=

Solution A25

B

A is the graph of 1

yx

= -

C is the graph of 2

1y

x=

D is the graph of 2y x=



Solution A26

A

(2 1)y f x= - squashes the graph horizontally by a factor of 2 and shifts it 1 unit to the

right. B is (0.5 1)y f x= +

C is (2 1)y f x= +

D is (0.5 1)y f x= -

Solution A27

A Solution A28

A The person is 55 years and approximately 9 months. Solution A29

C

4 1 5

5 4 1 5

6 4 4

1.5 1

x

x

x

x

+ <

- < + <

- < <

- < <

Solution A30

A

0.8 0.449e- = and 4 0.443p = so min( , 4) 0.443xe p- =



Solution A31

A 100! 100 99 98

161,70097!3! 3!

¥ ¥= =

Solution A32

A

(5.5) 4.5 (4.5) 4.5(3.5) (3.5) 4.5(3.5)(2.5) (2.5)

4.5(3.5)(2.5)(1.5) (1.5) 4.5(3.5)(2.5)(1.5)(0.5) (0.5)

4.5(3.5)(2.5)(1.5)(0.5) 52.34p

G = G = G = G

= G = G

= =

Solution A33

D

( 1)(3 1)! !(3 1)!(3 1)(3 )

(3 )( 1)! (3 1)!( 1)!

n n n nn n n

n n n n

G + + += = +G - - -

Solution A34

To 2 significant figures, these work out as follows:

(i) 1.5 4.5e e e= = [1] (ii) log 0.00001 12e = - [1]

(iii) 1tanh 0.9 1.5- = [1] (iv) (12) 11! 40,000,000G = = [1]



Solution A35

D

( )2 3 2 3 3 32 2 2 2 2 2bb b b b b b bx x x x x x x x x x +¥ + = + = +

Solution A36

C

33 2 3 6 33

1x x x x xx

e e e e ee

Solution A37

C See the definition in Chapter 4 Section 1.2 of the notes. Solution A38

D

2 2

2log log log log loga a a a ax x x

y yy y y

Solution A39

C

2 2

2

3 5 2 ( 3)( 1) (5 2 )(3 1)

3 1 1 (3 1)( 1) ( 1)(3 1)

2 3 6 13 5

(3 1)( 1) ( 1)(3 1)

7 11 8

(3 1)( 1)

x x x x x x

x x x x x x

x x x x

x x x x

x x

x x



Solution A40

B

2

2 2

3 6 3 ( 3)( 2)

(2 1)( 2) (2 1)( 1)2 3 2 2 3 1

3

x x x x x x

x x x xx x x x

x

+ + - + + -∏ = ∏+ - + +- - + +

+=(2 1)x +

(2 1)

( 2)

x

x

+¥

-( 1)

( 3)

x

x

++

2

( 2)

( 1)

( 2)

x

x

x

-

+=-

Solution A41

A

Factorising gives (2 1)( 3) 0x x+ - = . Hence 0.5,3x = - . Solution A42

A

212 ( 12) 4(1)(12) 12 96 12 4 66 2 6

2(1) 2 2x

± - - ± ±= = = = ±

Solution A43

The coefficients in these equations are so horrible that it is better to represent them by letters, find a general solution, and then substitute the numbers at the end. If we write the equations as: ax by e cx dy f

we can then find x by multiplying them by d and b respectively and subtracting, to get: ( )ad bc x de bf



64522 57963 64717 80068

8.13095432 64522 64717 92404

de bfx

ad bc

[1]

Similarly (ie if we just swap /a b and /c d in this formula), we get:

92404 57963 95432 80068

12.88464717 92404 95432 64522

ce afy

bc ad

[1]

These calculations are much easier if you have a calculator that can store variables to memory A, B, C etc. (An alternative method would be to find the inverse of the 2 2 matrix.)

Comment We put this question in to illustrate the point that you’re allowed to introduce your own symbols and abbreviations if it makes the calculations more tractable.

Solution A44

From the first equation:

3

4

aa b

=+

4 3( )a a bfi = + 3a bfi = [½]

Changing all the a ’s to b ’s in the second equation then gives:

2

2

30.0225

(4 ) (4 1)

bb b

=+

3

0.022516(4 1)b

fi =+

[½]

Rearranging:

1 3

1 1.8334 16 0.0225

[½]

3 5.5a bfi = = [½]



Solution A45

The easiest way to solve these equations is to square the first equation and substitute it into the second equation:

2

2162 9 1621 2 2

l a aa a aÊ ˆ = fi =Á ˜Ë ¯- - -

[½]

This gives:

22

aa

=-

[½]

From which: 2( 2) 4a a a= - fi = [½]

This gives 27l = . [½] Solution A46

B

2 3 7

7 2 3 7

4 2 10

2 5

x

x

x

x

- <- < - <- < <- < <

Solution A47

D Factorising gives ( 4)( 3) 0x x- + < . So either 4 0x - < and 3 0x + > (ie 3 4x- < < )

or 4 0x - > and 3 0x + < (but it is not possible for both of these to be true at the same time).



Solution A48

Rearranging the inequality so that we have zero on the RHS:

2

( 2)(9 8)1 0

2 4

x x

x x

- - - >+ -

Expressing the LHS as a single fraction:

2

2

( 2)(9 8) ( 2 4)0

2 4

x x x x

x x

- - - + - >+ -

[½]

Simplifying:

2 2

2

(9 26 16) ( 2 4)0

2 4

x x x x

x x

- + - + - >+ -

2

2

8 28 200

2 4

x x

x x

- + >+ -

[1]

We can factorise the numerator:

2

4(2 5)( 1)0

2 4

x x

x x

- - >+ -

[½]

The denominator equals zero when 1 5x = - ± (ie 3.236- and 1.236). [1] If we call these ,a b (say), the graphs of the numerator and the denominator look like

this:

a b1 2.5

[1] So the ratio will be positive when both graphs are positive or both are negative, ie when

1 5 ( )x a< - - = or 1 1 5 ( )x b< < - + = or 2.5x > . [1]



Solution A49

D

12 2 2 2 2 2 2

0 1 0 0 0

( 1) 2 ( 1)n n n n n

k k k k k

k k k k n k n

Solution A50

B This is an arithmetic series with 5a = , 1d = and 52n = . Hence:

152 2 (52)(2 5 51 1) £1,586S = ¥ + ¥ =

Solution A51

A

2 31

210 210 210 210

44 4 4kk

•

== + + +Â

This is an infinite geometric series with 210

4a = and

1

4r = . Hence:

210 4

701 1 4

S• = =-



Solution A52

D

( )

2 2

1 1 1 1

1 12 6

2 3 216

3 2

(1 2 ) 1 2

2 ( 1) ( 1)(2 1)

(2 3 )

2 9 13

6

n n n n

i i i i

i i i i

n n n n n n

n n n n n n

n n n

= = = =+ + = + +

= + + + + +

= + + + + +

+ +=

Â Â Â Â

Solution A53

D Considering the order of x and y we have 1 20y x£ £ £ . So if x sums over the

numbers 1 to 20, then looking at the order y must sum from 1 to x .

Solution A54

C Swapping the order of summation gives:

6 6 6

1 1 1

612

1

63 2

1

6 63 2

1 1

2 21 14 6

2 ( 1) 2( 1)

2( 1) ( 1)

6 (6 1) 6(6 1)(2(6) 1)

350

y

x y x y x

y

y

y y

x y y x

y y y

y y

y y

= = = =

=

=

= =

- = -

= - +

= -

= -

= + - + +

=

Â Â Â Â

Â

Â

Â Â



Solution A55

B

The term will be 8 5 3 5 3 5 3 5 35(2 ) (5 ) 56 2 5 224,000C x y x y x y= ¥ ¥ =

Solution A56

A

2 2 232

2314 2

( 2)( 3) ( 2)( 3)( 4)2 33 3 314 2 2! 2 3! 2

2 327 2714 4 2

2 3 2 327 13514 2 4

571 112 4 8

(2 5 ) (2 5 )

(3 2) 2 ( 1)

(2 5 )(1 )

(2 5 ) 1 ( 2)( ) ( ) ( )

(2 5 )(1 3 )

(2 6 27 ) ( 5 15 )

x x

x x

x x

x x x x

x x x x

x x x x x x

x x

2 324316

x

Solution A57

(i) Approximate value for i Using the approximate relationship, the equation given becomes:

1 12 420,000(1 ) 5,000(1 ) 2,500(1 ) 30,000i i i+ + + + + = [1]

ie 27,500 23,125 30,000i+ = 10.81%ifi = [2]

We could have divided this through by 100 to make the figures easier. (ii) Improved approximate value for i If we included the quadratic term in the approximate relationship, it would become:

212(1 ) 1 ( 1)ti ti t t i+ ª + + - [1]



The equation given would then become:

( )( )( )( )

21 1 1 12 2 2 2

231 1 14 2 4 4

20,000(1 ) 5,000 1

2,500 1

30,000

i i i

i i

È ˘+ + + + -Î ˚

È ˘+ + + -Î ˚

= [1]

ie:

2

2

27,500 23,125 859.375 30,000

2,500 23,125 859.375 0

i i

i i

+ - =

fi - + - = [1]

Solving this quadratic equation, we get:

223,125 (23,125) 4( 859.375)( 2,500) 23,125 22,938.4

2( 859.375) 1,718.75i

- ± - - - - ±= =- -

To get the root consistent with part (i), we need to take the + sign, which gives

10.85%i ª . [1] Solution A58

(i) 2 3 41 1 12 6 24 1 xx x x x e-- + - + = - [2]

(ii) 2 41 1 12 24 21 ( )x xx x e e-+ + + = + [2]

Comment When you add up the two series the odd powers cancel. You might recognise this as the hyperbolic cosine function cosh x .

(iii) 2 3 41 1 12 3 4 log (1 )ex x x x x+ + + + = - - [2]

(iv) 2 3 5 24 5 6 7 ( 5)( 6)(1 ) 1 5 ( )

0 1 2 3 2!x x x x x x

[2]



Comment

This last one isn’t at all obvious, but if you write out the terms in the series 5(1 )x --

and cancel all the minus signs, you’ll see that they do match up. You’ll need this type of result when you study the properties of the negative binomial distribution in Subjects CT3 and CT6.

Solution A59

(i)(a) Arithmetic-geometric mean inequality 1 and 1 i+ are unequal positive quantities. So the AGM inequality applies, and tells us that:

1 (1 )

1(1 )2

ii

+ + > + ie ½1 (1 )2

ii+ > + [2]

(i)(b) Show that If we apply the AGM inequality to the n quantities consisting of 1n - 1’s and a 1 i+ , we get:

( 1) 1 (1 )

1 1 1 (1 )nn ii

n

- ¥ + + > ¥ ¥ ¥ ¥ + ie 1/1 (1 ) nii

n+ > + [3]

(ii) Show that If we expand the RHS of the expression in the question using a binomial expansion, we get:

2(1 ) 11 2

n nn ni i i i

[1]

Since 0i > , all the terms in this series are positive. So, if we just keep the first two, we know that:

(1 ) 1ni ni+ > + ie 1 (1 )nin i+ < + [1]



Solution A60

(i) Show that We have:

2 1

2 1

( )1 2 3

( ) 2 ( 1)

nn

n nn

S xx x nx

x

S x x x n x nx

-

-

= + + +

= + + + - +

Subtracting gives:

2 1( )( ) 1 n nn

nS x

S x x x x nxx

-- = + + + + - [1]

The LHS can be factorised and the terms except the last on the RHS can be summed as a GP:

1 1 1 1

1 ( ) ( )1 1

n nn n

n nx x x

S x nx S x nxx x x x

[1]

Multiplying both sides by 1

x

x

, then gives the required result:

1

2

(1 )( )

1(1 )

n n

nx x nx

S xxx

+-= ---

[1]

Comment This multiply-and-subtract “trick” can be a useful for summing certain types of series. This method is used for deriving formulae for annuity functions in Subject CT1.

(ii) Limit and convergence

Provided 1 1x- < < , nx will vanish as n Æ• and we will have:

2lim ( )(1 )

nn

xS x

xÆ•=

- [2]



Solution A61

C

The change is 1,360,000

0.0218 21.8‰62,400,000

= =

Solution A62

C $58.50 1.3 $45∏ = . Solution A63

B

The change is 30,000

16.67%180,000

= .

Solution A64

B 50 to 2 SF gives a range of 49.5 to 50.5.

3 350.5 50 3.6963 3.6840 0.0122- = - =

3 350 49.5 3.6840 3.6717 0.0123- = - = So the maximum absolute error is 0.0123.



Solution A65

D The true value of a could be anywhere in the range [4500,5500) .

The true value of b could be anywhere in the range [0.195,0.205) .

The least possible true value is therefore 4500

21,9510.205

= , which is less than our

estimate by 3,049 ie 3,049

13.9%21,951

= .

Solution A66

D

[ ]( 1.207) ( 1.20) 0.7 ( 1.21) ( 1.20)

0.88493 0.7(0.88686 0.88493)

0.88628

P Z P Z P Z P Z< = < + < - <= + -=

Solution A67

C

( )2

21

3 53 5 0 ( ) 2 3

2 3n n

n nn

x xf x x x f x x x x

x++ -= + - = fi = + fi = -¢

+

21 1

2 11

22 2

3 22

23 3

4 33

3 5 11 1.2

2 3 5

3 5 0.041.2 1.1926

2 3 5.4

3 5 0.000054861.1926 1.1926

2 3 5.3852

x xx x

x

x xx x

x

x xx x

x

+ - -= - = - =+

+ -= - = - =+

+ -= - = - =+



Solution A68

C

2(2 4)(3 ) 2 10 12 2 10 12 10 10i i i i i i- - = - + - = + - = -

Solution A69

C

( )( ) 22 3 2 3 4 9 4 9 13i i i- + = - = + = Solution A70

A

23 4

2 5 5

2 2 2 4 4 4 4 1 3 4

2 2 2 4 1 54

i i i i i i ii

i i i i

- - - - + - - -= ¥ = = = = -+ + - +-

Solution A71

B The Argand diagram is as follows:

r

5

3

2 25 ( 3) 34 5.831r = + - = =

13 35 5tan tan 0.540q q -- -= fi = = - radians



Solution A72

D

23 3 4(1)(2.5) 3 1 31.5 0.5

2(1) 2 2

ix i

- ± - - ± - - ±= = = = - ±

Solution A73

(i) In the calculation of 1k , k is added to al

. So these must have the same

dimensions. Since k is measured in £ and a is a dimensionless quantity, this

means that l must have dimensions of 1£- , and 1k has dimensions of £. [1]

The dimensions of 2k and 3k are then seen to be 22

1£

£-= and 3

3

1£

£-= ,

respectively. [2]

(ii) 3

2c

kk

has dimensions 3

3 22

££

£c

c-= . This will be dimensionless if 3 2 0c- = ie

32c = . [1]

Comment

1 2 3, ,k k k and 33 2

2

kk

are actually the mean, variance, skewness and coefficient of

skewness of the translated gamma distribution.



Solution A74

(i) Write in the form a + bi

1 2 (2 4 ) (1 2 ) 3 2z z i i i+ = + + - = + [1]

1 2 (2 4 ) (1 2 ) 1 6z z i i i- = + - - = + [1]

2

1 2 (2 4 )(1 2 ) 2 8 2 8 10z z i i i= + - = - = + = [1]

1

2

2 4 2 4 1 2 2 8 8 6 81.2 1.6

1 2 1 2 1 2 1 4 5

z i i i i ii

z i i i

+ + + + - - += = ¥ = = = - +- - + +

[1]

2 2 2 2

1 2 (2 4 ) (1 2 ) ( 12 16 ) ( 3 4 ) 15 12z z i i i i i+ = + + - = - + + - - = - + [1]

(ii) Calculate and comment

1 2 2 4 1 2 20 5 100 10z z i i= + - = ¥ = = 1 2 10 10z z = = [2]

These two quantities are always equal, whatever the values of 1z and 2z . [1]

(iii) Calculate and comment

1 22 4

1.2 1.61 2

iz z i

i

-= = - -+

1 2 1.2 1.6 1.2 1.6z z i i= - + = - - [2]

Again, these two quantities are always equal, whatever the values of 1z and 2z . [1]

(iv) Express in the form iθre

1 23 2 3(2 4 ) 2(1 2 ) 8 8 8(1 )z z i i i i+ = + + - = + = + [1]

This has magnitude 2 28 1 1 8 2+ = and an argument of 45∞ or 4

p radians.

So /41 23 2 8 2 iz z e p+ = . [1]



Solution A75

Consider the equation 3 22 9 14 5 0z z z- + - = . (i) Show 2 + i is a root When 2z i= + :

2 2(2 ) 3 4z i i= + = + 3 2 (2 )(3 4 ) 2 11z zz i i i= = + + = + [2]

So the LHS of the equation is: 2(2 11 ) 9(3 4 ) 14(2 ) 5 0i i i+ - + + + - = [1]

(ii) The other complex root Since this is a polynomial equation with real coefficients, the complex roots come in conjugate pairs. So the other complex root is 2z i= - . [1] (iii) Find the third root The third root (a , say) of this cubic equation must be real. So the equation must take the form: ( )[ (2 )][ (2 )] 0C z z i z ia- - + - - =

Multiplying out the two complex factors in square brackets, we must have the identity:

2 3 2( )( 4 5) 2 9 14 5C z z z z z za- - + = - + - [1]

For the 3z terms to match up, we must have 2C = . For the constant term to match up, we must have 5 5Ca- = - ½afi = . So the third factor is ½z - and the third root is ½z = . [1] (iv) Which roots lie outside the unit circle

Both the complex roots have magnitude 5 and so lie outside the unit circle. The real root has magnitude ½, and so lies inside. [1]



Solution A76

The auxiliary equation is 2 6 8 0v v+ + = which has roots 4- and 2- . [1]

The general solution is of the form ( 2) ( 4)t tty A B= ¥ - + ¥ - . [1]

However we know that 0 5y = and 1 5y = , so:

5

5 2 4

A B

A B

= +

= - - [1] These can be solved to give 7.5B = - and 12.5A = , so our particular solution is:

12.5( 2) 7.5( 4)t tty = - - - [1]

Check: 2y should equal 1 06 8 70y y- - = -

The formula gives 2 212.5 ( 2) 7.5 ( 4) 70¥ - - ¥ - = - .

Solution A77

B

23 53 5

x xx

x

- = -

3 5 3x- < so it is ( )O x but

0lim (3 5 ) 0x

xÆ

- π so it is not ( )o x .

Solution A78

A

The general term is 3 1

n

n + which will be bounded above by

1

3 3

n

n= .



Solution A79

C Solution A80

C

0 0 0 0

( ) ( ) 5( ) 5 5lim lim lim lim 5 5

h h h h

dy f x h f x x h x h

dx h h h+ + + +Æ Æ Æ Æ

+ - + -= = = = =

Solution A81

B Solution A82

B

26 2dy

x xdx

= - which is 20 when 2x = .

Solution A83

A

3 5 8 5 8 53 123 5 5 5 8 5

4 12( ) 4 ( ) 4( )

5f x x f x x x

x x

Solution A84

D

Using the chain rule 2 8 2 8( ) 9(5 7) 10 90 (5 7)f u u u u u .



Solution A85

A

Using the chain rule 2 22 ½( ) ( ) t tM t t e . Hence 0(0)M em m= =¢ .

Solution A86

B

Using the chain rule 1

( )1 1

M tt t

. Hence (0)Mal

=¢ .

Solution A87

D

Now 2 2 22

4( ) ln 2 ln 2 4x x x

xf x e e e

e

. Using the chain rule gives:

2

2

2 2

22

22

42

1( ) (4 8 )

22 4

xxx x

x x xx

xeef x xe e

e e ee

--

-= ¥ - =¢

+ +

Solution A88

D Using the chain rule gives:

22

2 2

6(5 3)1 6(5 3)

6(5 3) 6(5 3)

1 60(5 3)( ) 6(2)(5 3) 5

4 4

xx

x x

x ef x x e

e e

- -- -

- - - -

-= ¥ - - ¥ = -¢+ +



Solution A89

B Using the product and chain rules:

2 5 2 4

2 4 2

2 4 2 2

2 4 2

( 3)( 2 1) (4 3 )5( 2 1) (2 2)

( 2 1) 3( 2 1) 5(4 3 )(2 2)

( 2 1) ( 3 6 3) 5( 6 14 8)

( 2 1) ( 33 76 43)

dyx x x x x x

dx

x x x x x x

x x x x x x

x x x x

Solution A90

B Using the product rule:

4 2 4 4 2 4( ) 64 32 ( 4 ) 64 128x x x xf x xe x e xe x e

4 4 4 2 4

4 4 2 4

2 4

( ) 64 64 ( 4 ) 256 128 ( 4 )

64 512 512

64(1 8 8 )

x x x x

x x x

x

f x e x e xe x e

e xe x e

x x e

Solution A91

A

Now 2

233 x

x

xy x e

e . So using the product rule gives:

26 3x xdyxe x e

dx- -= -

2 2

2 22

6 12 36 6 6 3 (6 12 3 )x x x x x

x

d y x xe xe xe x e x x e

dx e- - - - - - += - - + = - + =



Solution A92

C

Now ( ) 2 6f x x . Setting this equal to zero gives a turning point at 13x = - .

Since ( ) 6f x this turning point is a maximum.

So the maximum value of this function is 2 161 1 13 3 3 3( ) 5 2( ) 3( )f - = - - - - = .

Solution A93

D Since ( 3)( 5)y x x x= + - we can see that this crosses the x -axis when 0, 3x = - and 5.

This means it is either C or D. Since we have a positive 3x term then y Æ• as

x Æ• which means it is D. Alternatively, we could calculate and determine the turning points using differentiation. Solution A94

C Differentiating and setting the derivative equal to zero gives:

2 13

( ) 3 2 1 (3 1)( 1) 0 ,1f x x x x x x

The second derivative is: ( ) 6 2f x x

When 1x = we have (1) 0f >¢¢ so we have a minimum. When 13x = - we have

13( ) 0f - <¢¢ so we have a maximum.



Solution A95

D

We have 902ln ( ) ln 4ln 3f x x x= + - , so:

4

ln ( ) 3d

f xdx x

= -

Setting this equal to zero and solving gives 43x = .

Solution A96

D

( ) ( )( ) ( )2 3 2 2 3 3

2 2 32 3

axy b xy c xy axy bx y cx yx x

ay bxy cx y

∂ ∂+ + = + +∂ ∂

= + +

Solution A97

D

( ) ( )2 2 2 2, ,f x y z xyz x y z= =

2 22f

xy zx

∂fi =∂

224

fxyz

x y

∂fi =∂ ∂

3

8f

xyzx y z

∂fi =∂ ∂ ∂

Hence:

32 2 2 3 38 2 16

f fxyz xy z x y z

x y z x

∂ ∂¥ = ¥ =∂ ∂ ∂ ∂



Solution A98

C Finding the first derivatives and setting them equal to zero and solving gives:

23 6 2 0f

x x yx

(1)

2 2 0f

x yy

(2)

Equation (2) tells us that y x= . Substituting this into equation (1) gives:

2 43

3 4 0 0,f

x x xx

So our two turning points are 0x = , 0y = and 43x = , 4

3y = .

Calculating the second derivatives gives:

2

2

2

2

2

6 6

2

2

fx

x

f

y

f

x y

So for 0x = , 0y = we have:

0 0 00 0 0

2

2 2 22

2 2

2

( 2 )( 6 ) 2

8 8 0


f f f

x yy x

Solving this gives roots of 1.172x = - and 6.828x = - . Since both of these roots are negative this means that we have a maximum at 0x = , 0y = .



For 43x = , 4

3y = we have:

0 0 00 0 0

2

2 2 22

2 2

2

( 2 )(2 ) 2

8 0


f f f

x yy x

Solving this gives roots of 2.828x = ± . Since these roots have different signs this

means that we have a saddle point at 43x = , 4

3y = .

Solution A99

C

The Lagrangian function is 2 2(5 3 ) ( 136)L x y x yl= - - + - . Finding the derivatives

and setting them equal to zero gives:

5

5 2 02

Lx x

xl

l∂ = - = fi =∂

(1)

3

3 2 02

Ly y

yl

l∂ = - - = fi = -∂

(2)

2 2 136 0L

x yl∂ = - - + =∂

(3)

Substituting (1) and (2) into (3) gives:

2 2

2

5 3 34 1136 0 136

2 2 44l

l l l-Ê ˆ Ê ˆ- - + = fi = fi = ±Á ˜ Á ˜Ë ¯ Ë ¯

When 14l = we have 10x = and 6y = - .

When 14l = - we have 10x = - and 6y = .

We could then check to see what kind of turning points we have, but only one answer has these two values – so it must be the correct one.



Solution A100

(i) Limit

We can’t evaluate 3 3

0lim

x x

x xx

e e

e e

-

-Æ

--

by substituting 0x = because this gives 0

0.

If we use the exponential series, we find that the denominator is:

( ) ( )2 2 31 12 21 1 2 ( )x xe e x x x x x O x-- = + + + - - + - = + [½]

The numerator is the same, but with x replaced by 3x :

3 3 36 ( )x xe e x O x-- = + [½]

So the limit is:

3 3 3 2

3 20 0 0

6 ( ) 6 ( )lim lim lim 3

2 ( ) 2 ( )

x x

x xx x x

e e x O x O x

e e x O x O x

-

-Æ Æ Æ

- + += = =- + +

[1]

(ii) Limit

This limit is the special case of the result lim 1n

x

n

xe

n

when ½x .

So:

½1lim 1 1.649

2

n

ne e

n

[2]



Solution A101

Differentiating, we get:

11 10

112

10(1 ) (1 (1 ) ) 110 100(1 ) 5

i i ii

i

- -- ¥ + - - + ¥- ¥ + + [2]

Evaluating this when 0.05i = gives:

11 10

112

0.5(1.05) 1 (1.05)1,000(1.05) 5 772.17

0.05

- -- - +- + = - [1]

Solution A102

Let 1

2xyx

= + . [1]

If we work out the first few derivatives, we get:

22 log 2xdyx

dx-= -

2

2 32 2 (log 2) 2xd y

xdx

-= +

3

3 43

2 (log 2) 6xd yx

dx-= - [2]

The pattern is then apparent:

12 (log 2) ( 1) !n

x n n nn

d yn x

dx- -= + - [1]

Note the trick of using ( 1)n- to obtain the alternating signs in the second term. We

could use mathematical induction to show that we have guessed the pattern correctly.



Solution A103

Substituting the values given for the parameters gives:

3 4 24 ( )y x xl l -= + [1]

Taking logs, as suggested:

4log log 4 log 3log 2log( )y x xl l= + + - + [1]

Since the log function is monotonic, this will have its maximum value for the same value of x as the original function. So, differentiating using the function-of-a-function rule, and equating to zero:

3

4

3 2(4 )log 0

d xy

dx x xl= - =

+ [1]

Rearranging:

4 43( ) 8x x 43 5x

1

43

5x

[1]

Substituting this back into the original function to find the maximum value of y :

3 32 2

4 4 3/4 5/4 1/43 3 3 8 14 4 3 5

5 5 5 5 16y

[1]



Solution A104

(i) α = 1,β = -3

Here:

3( ) (1 )f x x x -= +

so that:

(0) 0f = and 18(1)f = [1]

The derivative (found using the product rule) is:

3 4( ) (1 ) 3 (1 )f x x x x- -= + - +¢

This equals zero when:

3 4(1 ) 3 (1 ) 0x x x- -+ - + =

ie (1 ) 3 0x x+ - = 12xfi = and 4

27(½)f = [1]

So the graph has a maximum value at ½x = and looks like this:

½ 1

427

The range of values of the function here is:

4270 ( )f x£ £ [1]



(ii) α = β = 2

Here:

2 2( ) (1 )f x x x= +

So that: (0) 0f = and (1) 4f = [1]

The derivative is:

2 2( ) 2 (1 ) 2 (1 ) 0f x x x x x= + + + >¢ when [0,1]x Œ [1]

So the function increases steadily over the range and the graph looks like this:

1

4

The range of values of the function here is: 0 ( ) 4f x£ £ [1]



(iii) α = -½,β = 1½

Here:

½ 1½( ) (1 )f x x x-= +

So that:

0

lim ( )x

f xÆ +

= +• and (1) 2 2f = [1]

The derivative is:

1½ 1½ ½ ½312 2( ) (1 ) (1 )f x x x x x- -= - + + +¢

This equals zero when:

1½ 1½ ½ ½312 2(1 ) (1 ) 0x x x x- -- + + + =

ie (1 ) 3 0x x- + + = 12xfi = and

3 3(½)

2f = [1]

So the graph looks like this:

½ 1

4

The range of values of the function here is:

3 3

( )2

f x ≥ [1]



Solution A105

If 3 2 2( , ) 2 2f x y x x y= - + , then:

23 4

4

fx x

x

fy

y

∂∂

∂∂

= -

= [1]

Setting these equal to zero we get:

4

0 or 3

x = , 0y =

So the extrema occur at 0, 0x y= = and at 4

, 03

x y= = . [1]

To determine their nature, the second partial derivatives are needed:

2

2

2

2

2

6 4

4

0

fx

x

f

y

f

y x

∂∂

∂∂

∂∂ ∂

= -

=

= [1]

Setting up the required equations: For 0, 0x y= = the equation is (4 )( 4 ) 0l l- - - = , ie 4l = ± . This means that

0, 0x y= = is a saddle point. [1]

For 4

, 03

x y= = the equation (4 )(4 ) 0l l- - = , ie 4l = , twice. This means that

4, 0

3x y= = is a local minimum. [1]



Solution A106

The Lagrangian function is:

2 2 2( 2 ) ( 1)L x y z x y x y zl m= + + - + - + + - [1]

Finding the partial derivatives:

2 2 2

1 2 (1) 2 (4)

1 2 2 (2) 1 (5)

1 2 (3)

L Lx x y

x

L Ly x y z

y

Lz

z

∂ ∂l m∂ ∂l

∂ ∂l m∂ ∂m

∂ m∂

= - - = - -

= - - = - - - +

= - [1]

Setting these equal to zero, we get:

1 1 2(1) (2)

2 2

1(3)

2

1 2 4(4) 0 3 5 0

2 2

x y

z

l lm m

m

l l lm m

- -fi = fi =

fi =

- + - +fi + = fi - + = [1]

So 0.6l = . Finally substituting into (5) we get:

2 2 2

2

0.4 0.2 11

2 2 2

1.2 4

0.3

[1]

This gives the positions of the extrema as:

1 1 1

, ,5 0.3 10 0.3 2 0.3

x y z= - = = -

and: 1 1 1

, ,5 0.3 10 0.3 2 0.3

x y z= = - = [1]



Solution A107

(i) Show µ(20) and obtain µ(40)

20x 70

150

x

0 1 0 1(20) exp 0.000814a a b b [1]

40x 70

0.650

x

0 1 0 1(40) 0.6 exp 0.6 0.001077a a b b [2]

(ii) Find and determine the turning point The derivative of this function is:

11 0 1

1 70( ) exp

50 50 50

b xx a b b

[1]

Equating this to zero to find the stationary point gives:

1 1 0 170

exp 050

xa b b b

Rearranging:

10

1 1

50log 70 29.82e

ax b

b b

To the nearest month, this is 29 years and 10 months. [2] To determine the nature of this point (ie whether it is a minimum, a maximum or point of inflexion), we can look at the second derivative:

2

10 1

70( ) exp

50 50

b xx b bm È ˘-Ê ˆ Ê ˆ= +¢¢ Á ˜ Á ˜Í ˙Ë ¯ Ë ¯Î ˚



This must be a positive quantity since the squared factor and the exponential function must both be positive. So this point is a minimum. [1]

Comment Mortality reaches a low point in the late 20s before starting to increase into “old age”.

(iii) Find q(20) The integral in the formula for (20)q is:

1 10 1 0 10 0

10 1 0 10

20 70 20 70(20 ) exp

50 50

1 exp 150 50

t tt dt a a b b dt

t ta a b b dt

[2]

Although this looks complicated, we can integrate it directly:

121

0 1 0 101 0

0 1 0 1 0 11 1

50(20 ) exp 1

100 50

1 50 1 501 exp 1 exp

100 50

0.000791

t tt dt a t a t b b

b

a a b b b bb b

[2]

We then find that:

0.000791(20) 1 0.000791q e-= - = [1]

Comment You will meet the mortality functions featuring in this question in Subject CT4 and CT5. They are usually written as xq and xm .



Solution A108

D

( )b

a

f x dxÚ is used to find the area under the ( )f x curve.

Solution A109

B We simply use the rule “raise the power by 1 and divide by the new power”. Solution A110

A

22 1.75 1.750.75

1 1

2 11.351

1.75 1.75

xx dx

È ˘ -= = =Í ˙Í ˙Î ˚

Ú

Solution A111

C

3 12 2 3 11 3

3 3

x xx x dx x x c c

-- - -+ = - + = +Ú

Solution A112

B

44 4 2

2 2

5 5 5 6005

ln5 ln5 ln5

xx dx

È ˘ -= = =Í ˙Í ˙Î ˚

Ú



Solution A113

D This is a standard result, but can be proved as follows:

ln

ln

ln ln

x c xx x c Be Bc

Bc dx Be dx const constc c

¥¥= = + = +Ú Ú

Solution A114

B 3

3 3 0 3

00

( ) 1x xe dx e e e el l l ll - - - -È ˘= - = - - - = -Î ˚Ú

Solution A115

A Solution A116

B

Using the substitution 45u x= , we see that 320dudx x= , so 3

20du x dx= . Hence we get:

43 5

0 0 0

0 ( 1)0.05

20 20 20

u ux e e

x e dx du

•• • - -- È ˘ - -= = - = =Í ˙

Í ˙Î ˚Ú Ú



Solution A117

C

Using the substitution 610w x= + (so that 56dw

xdx

= ), we get:

115 21 11 36 30 10

10

2 2

1 1

6 6 2(10 )

1 1 10.000145

12 11 10

x wdx w dw

x

Solution A118

C

Using the substitution 32 6 1u x x= + + (so that 26 6dudx x= + ), we get:

2

33

3 3 10.5 0.5ln 0.5ln(2 6 1)

2 6 1

xdx du u c x x c

ux x

+ = = + = + + ++ +Ú Ú

Solution A119

A

Let ( )( )2 5

2 5 ( 1) (2 4)2 4 1 2 4 1

x A Bx A x B x

x x x x

+ ∫ + fi + = + + -- + - +

.

Substituting 1x = - gives 3 6 ½B B= - fi = - .

Substituting 2x = gives 9 3 3A A= fi = .

( )( )

3 12 2

312

2 5 3 ½ln(2 4) ln( 1)

2 4 1 2 4 1

(2 4)ln

( 1)

xdx dx x x c

x x x x

xk

x

+ = - = - - + +- + - +

Ï ¸-Ô Ô= Ì ˝+Ô ÔÓ ˛

Ú Ú



Solution A120

D Using integration by parts gives:

1 112 2 21 12 20

0 0

12 21 12 4 0

2 21 1 12 4 4

2.097

x x x

x

xe dx xe e dx

e e

e e

Solution A121

D Using Leibniz’s formula gives:

( ) ( )

[ ]

2 2 2

0 0

2

0

20

2 2

2

3 2 1 3 2 0 3 2

(3 2 ) 6

(3 2 ) 6

(3 2 ) 6

9 2

x x

t x

x

x

dx t dt x t x t dt

dx x

x x x dt

x x xt

x x x

x x

=

∂- = - - + -∂

= - +

= - +

= - +

= -

Ú Ú

Ú



Solution A122

A 15 10 15

102

55 5 5

15

5

152

5

(5 ) 5 0.5

25 37.5

12.5 37.5

2,875

yx y x

x

x

x y dy dx xy y dx

x dx

x x

== = =

=

=

È ˘+ = +Î ˚

= +

È ˘= +Î ˚

=

Ú Ú Ú

Ú

Solution A123

B Using integration by parts we get:

[ ]1

10

0

ln lnx dx x x x= -Ú

But ln x is undefined when 0x = . To be honest it is much easier to eliminate the alternatives:

2 21 1 12 2 21

1

0 ( )x xe dx e• •- -È ˘= - = - - =Î ˚Ú

2 1

11

0 ( 1) 1x dx x• •- -È ˘= - = - - =Î ˚Ú

3 2

11

6 30 ( 3) 3dx

x x

• •È ˘= - = - - =Í ˙Î ˚Ú



Solution A124

B Working with 7 ordinates, the area under the curve is approximately:

( )2 2 2 2 2 2 216 2 5.5 5 4.5 4 3.5 3 0.5 63.125

2È ˘+ + + + + + ¥ =Î ˚

Solution A125

B Using the Maclaurin expansion, we have:

2 3

12! 3!

x x xe x= + + + +

So:

2 3

2 2 21 2 6.333

2! 3!e + + + =



Solution A126

A First note that ln(3 )(3 ) ln(3 ) ln(3 )x y x y+ + = + + + . Hence:

0, 00, 0

0, 00, 0

2

2 20, 00, 0

2

2 20, 00, 0

2

0, 0

1 1

3 3

1 1

3 3

1 1

9(3 )

1 1

9(3 )

0

x yx y

x yx y

x yx y

x yx y

x y

f

x x

f

y y

f

x x

f

y y

f

y x

∂∂

∂∂

∂∂

∂∂

∂∂ ∂

= == =

= == =

= == =

= == =

= =

= =+

= =+

= - = -+

= - = -+

=


2 2 2 21 1 1 1 1 1 1 1 1ln9 ln9

3 3 2 9 9 3 3 18 18x y x y x y x y

È ˘+ + + - - = + + - -Í ˙Î ˚

Solution A127

C

13ln ln(3 1)

3 1 3 1

dy y dy dxy x c

dx x y x= fi = fi = + +

+ +Ú Ú

When 0x = , 2y e= which means:

2 13ln ln1 2 0 2e c c c= + fi = + fi =

Hence when 2x = , we have 13ln ln 7 2 14.135y y= + fi = .



Solution A128

(i) Show Γ( ) = ( - 1)Γ( - 1)x x x

Starting from the definition given, and integrating by parts with 1xu t -= and tdv e dt-=

(so that 2( 1) xdu x t dx-= - and tv e-= - ), we get:

1 1 2

0 00( ) ( ) ( 1)x t x t x tx t e dt t e x t e dt

•• •- - - - - -È ˘G = = - + -Î ˚Ú Ú [2]

The term in square brackets is zero. So we get:

2

0( ) ( 1) ( 1) ( 1)x tx x t e dt x x

• - -G = - = - G -Ú [1]

(This relationship will be valid provided the integral for ( 1)xG - converges, which

requires 1 0x - > ie 1x > .) (ii) Show Γ(1) = 1

Using the integral definition with 1x = , we find that:

0

0 0 0(1) 0 ( 1) 1t t tt e dt e dt e

•• •- - -È ˘G = = = - = - - =Î ˚Ú Ú [2]

(iii) Obtain integral Using the integral definition with ½x = , we know that:

½

0(½) tt e dt p

• - -G = =Ú [1]

If we apply the substitution 212t z= (so that

dtz

dz= ), we find that:

( ) 212

½2120

zz e zdz p

-• - =Ú [1]

Simplifying, and taking the constants to the RHS:

21

2

0 2

ze dz

p• - =Ú [1]



Since the function 21

2z

e-

is an even function (ie it takes the same values when the sign of z is reversed), the area under the graph for negative values of z is the same as for the positive values. So:

2 21 1

2 2

02 2

z ze dz e dz p

• •- --•

= =Ú Ú21

21

12

ze dz

p• --•

fi =Ú [1]

Comment This result is important in connection with the standard normal distribution.

Solution A129

The curve 26 2y x x crosses the x axis when:

2 32

6 2 0 (2 )(3 2 ) 0 or 2x x x x x [2]

So the required area is:

22 2 2 3 71 2

2 3 243 2 3 2

343(6 2 ) 6 14

24x x dx x x x

[2]

Solution A130

The shaded region can be specified as {( , ) : 0 1,0 }x y x y x£ £ £ £ . So the integral is:

1

0 0( , ) (2 6 )

x

A

f x y dxdy dx dy x y= +ÚÚ Ú Ú [2]

The inner integral is:

2 2

0 0(2 6 ) 2 3 5

xxx y dy xy y xÈ ˘+ = + =Î ˚Ú [2]

11 2 3530 0

5( , ) 5

3A

f x y dxdy x dx xÈ ˘fi = = =Î ˚ÚÚ Ú [2]

Note that you could also evaluate the integral in the opposite order as

1 1

0(2 6 )

ydy dx x y+Ú Ú .



Solution A131

(1) Substitution Using the substitution 5t x= + , we get:

7

7 7

2 26 66

5 1 5 5 7 5log log 0.035103

6 42

tI dt dt t

t tt t

[3]

(2) Integration by parts

Integration by parts, with u x= and 2

1

(5 )

dv

dx x=

+ (so that du dx= and

1

5v

x= -

+),

gives:

2

2 2

111

1 1 2 1 5 7log 5 log

5 5 7 6 42 6I x dx x

x xÈ ˘ È ˘= ¥ - - - = - + + + = - +Î ˚Í ˙+ +Î ˚ Ú [3]

(3) Partial fractions If we add and subtract 5 from the numerator of the integrand, it then takes a form that we can integrate directly:

2

2 2

2 21 11

(5 ) 5 1 5 5log 5

5 5(5 ) (5 )

xI dx dx x

x xx x

È ˘+ - È ˘= = - = + + =Í ˙ Í ˙+ ++ + Î ˚Î ˚Ú Ú [3]



Solution A132

(i) Calculate the integral Integrating with respect to x :

2

2

2

2

12

0.5 0.5

12

0.50.5

12 0.5

0.5

1 12 0.5

0.5 0.5

(4 1)

1(4 1)

4

1 1(4 1)

4 4

1 1(4 1) (4 1)

4 4

yy x

y x

yy x

y

y y

y

y y y

y y

y e xe dx dy

y e e dy

y e e e dy

y e dy e y e dy

= =

=

=

+

= =

+

È ˘= + Í ˙Î ˚

È ˘= + -Í ˙Î ˚

= + - +

Ú Ú

Ú

Ú

Ú Ú [2]

The first integral can be done by inspection, the second one needs to be done by parts:

{ }{ }

211 12 0.5

0.50.5 0.5

3 1 0.5 1 0.5 10.5

3 1 0.5 1 0.5 1 0.5

1 1(4 1) 4

4 4

1 1( ) 5 3 4[ ]

4 4

1 1( ) 5 3 4( ) 2.54

4 4

y y y y

y

y

e e y e e dy

e e e e e e

e e e e e e e

+

=

Ï ¸Ô ÔÈ ˘ È ˘- + -Ì ˝Î ˚Í ˙Î ˚ Ô ÔÓ ˛

= - - - -

= - - - - - =

Ú

[2]



(ii) Calculate the integral by reversing the order We can reverse the order of integration:

2 2

1 1 12 2

0.5 0.5 0.5

(4 1) (4 1)y

y x x y

y x x y x

y e xe dx dy xe y e dy dx= = = =

+ = +Ú Ú Ú Ú [1]

To integrate with respect to y , we integrate by parts. We get:

{ }

{ }

{ }

2

2

2

2

2 2 2

1 112

0.5

12 1 1

0.5

12 1 1

0.5

12 1

0.5

12 1 2 2 2

0.5

(4 1) 4

5 (4 1) [4 ]

5 (4 1) 4 4

4 3

( 4 3 )

x y y

y xx y x

x x yx

x

x x x

x

x x x

x

x x x x x

x

xe y e e dy dx

xe e x e e dx

xe e x e e e dx

xe e xe e dx

xe x e xe dx

== =

=

=

=

+ + +

=

Ï ¸Ô ÔÈ ˘+ -Ì ˝Î ˚Ô ÔÓ ˛

= - + -

= - + - +

= - +

= - +

Ú Ú

Ú

Ú

Ú

Ú [3]

This would be complicated to integrate, as we would need to integrate numerically. This confirms that integrating with respect to x first is the best option. [1]



Solution A133

(i) Taylor’s series The Taylor series for these functions are:

2 31 12 61xe x x x= + + + + ( x-• < < • ) [1]

and:

1 2 3(1 ) 1y y y y-+ = - + - + ( 1 1y- < < ) [1]

(ii) Determine the coefficients

Using the series for xe , we find that the series for the denominator is:

2 3 41 12 61 ( )xe x x x O x- = + + +

So the original function is:

2 3 41 12 61 ( )x

x x

e x x x O x=

- + + + [1]

Cancelling an x on the top and bottom:

{ } 12 31 1

2 3 2 61 12 6

11 ( )

1 1 ( )x

xx x O x

e x x O x

-È ˘= = + + +Î ˚- + + +

[1]

We can now use the series for 1(1 )y -+ with 2 31 12 6 ( )y x x O x= + + , which gives:

22 3 2 3 31 1 1 1

2 6 2 61 ( ) ( ) ( )1x

xx x O x x x O x O x

eÈ ˘ È ˘= - + + + + + +Î ˚ Î ˚-

[1]

Multiplying out the terms and ignoring any terms of power greater than 2, we get:

2 2 3 2 31 1 1 1 12 6 4 2 121 ( ) 1 ( )

1x

xx x x O x x x O x

eÈ ˘ È ˘= - + + + = - + +Î ˚ Î ˚-

[1]

So 12a = - and 1

12b = . [1]



Comment If you follow the calculations through keeping more terms you’ll find that the

coefficient of the 3x term is zero. So 21 12 121

1x

xx x

eª - +

- is actually a very good

approximation when x is small.

Solution A134

(i) Partial fractions The coefficients A and B satisfy the identity:

1

(2 5 ) 2 5

A B

P P P P∫ +

- -

Multiply through by (2 5 )P P- :

1 (2 5 )A P BP∫ - +

ie 1 2 ( 5 )A P B A∫ + - [1]

Equating the coefficients of the constant terms and the P terms:

12A =

and 5 0B A- = 525B Afi = =

So:

1 1 / 2 5 / 2

(2 5 ) 2 5P P P P∫ +

- - [1]



(ii) Solve the differential equation Rearranging the differential equation and integrating:

1

(2 5 )dP dt

P P=

-Ú Ú

Using the identity in part (i):

1 / 2 5 / 2

2 5dP dt

P P

ie 1 12 2ln ln(2 5 )P P t c- - = +

12 ln

2 5

Pt c

P= +

- [1]

Note that 0P > and 2 5 0P- > , so no modulus signs are required here.

We are told that, when 0t = , 215P = . So:

12

2 / 15ln

2 5(2 / 15)c=

- 1 1 1

2 10 2ln ln10cfi = = - [1]

So:

1 12 2ln ln10

2 5

Pt

P= -

- [1]

We can make P the subject of this as follows. Doubling and rearranging:

ln ln10 22 5

Pt

P+ =

-

10

ln 22 5

Pt

P=

-

210

2 5tP

eP=

-

210 (2 5 ) tP P e= -

2 2(10 5 ) 2t tP e e+ = [1]

2

2( )

5 10 tP te-

fi =+



Solution A135

The differential equation can be re-written in the form 1

1

dyy x

dx x- =

+. [1]

This can be solved using an integrating factor, where the integrating factor is:

[ ]1 1exp exp ln(1 )

1 1dx x

x xÈ ˘- = - + =Í ˙+ +Î ˚Ú [1]

Note that no modulus signs are required here since we know 0x ≥ , so 1 0x+ > . Multiplying through by the integrating factor and integrating both sides with respect to x , we obtain:

1 1

1 1y x dx

x x= ¥

+ +Ú [1]

Since 1

11 1

x

x x= -

+ +, we find that:

1 1

1 ln( 1)1 1

y dx x x cx x

= - = - + ++ +Ú [1]

where c is a constant. Since 0y = when 0x = , we find that 0c = , so the particular solution is

[ ]( 1) ln( 1)y x x x= + - + . [1]

Solution A136

C We need to solve the simultaneous equations:

2 2 10

5 7

4 3 18

x y

x y

x y

- + =+ =- = -

Rearranging the first equation gives 5y x= + . Substituting this into the second

equation gives 6 18 3x x= - fi = - . Hence 2y = .



Solution A137

C

The magnitude of 3 2 4i j k is 2 2 23 ( 2) 4 29+ - + = . So the unit vector in the

same direction is:

33 2 4 1

229 29

4

i j k

Solution A138

C The scalar product is: 4 1 ( 1) 3 3 ( 2) 5¥ + - ¥ + ¥ - = -

Hence:

2 2 2 2 2 25 4 ( 1) 3 1 3 ( 2) cos

26 14 cos

q

q

- = + - + + + -

=

So:

5

cos 105.2º26 14

q q= - fi =



Solution A139

B The easiest way to do this is to see which of these vectors has a scalar product of zero with 3 2i j k . (4 4 4 ) (3 2 ) 12 8 4 0i j k i j k ( 4 4 4 ) (3 2 ) 12 8 4 0i j k i j k ( 4 4 4 ) (3 2 ) 12 8 4 0i j k i j k (4 4 4 ) (3 2 ) 12 8 4 0i j k i j k Solution A140

C

1 2 2 4 2 6 4 10 8 14

3 4 3 5 6 12 12 20 18 32AB

Solution A141

C

0 3 1 3 1 0det 3 ( 2) 4

1 2 3 2 3 1

3 3 2 ( 7) 4 1

1

- -= - - +

- -= ¥ + ¥ - + ¥= -

A

Solution A142

B

If a b

c d

Ê ˆ= Á ˜Ë ¯

A then 1 3 11 14 26 ( 4)

d b

c aad bc- - -Ê ˆ Ê ˆ= =Á ˜ Á ˜-- - -Ë ¯ Ë ¯

A . So we have:

1 3 1 3 1 0.3 0.11 14 2 4 2 0.4 0.26 ( 4) 10

- - - -Ê ˆ Ê ˆ Ê ˆ= = =Á ˜ Á ˜ Á ˜- - Ë ¯ Ë ¯ Ë ¯

A



Solution A143

B Solution A144

B

The easiest way to check this is to find out which vectors satisfy 3 2

6 1

x xk

y y

Ê ˆ Ê ˆ Ê ˆ=Á ˜ Á ˜ Á ˜-Ë ¯ Ë ¯ Ë ¯

.

Those that do are eigenvectors.

3 2 2 6 23

6 1 6 18 6

-Ê ˆ Ê ˆ Ê ˆ Ê ˆ= = -Á ˜ Á ˜ Á ˜ Á ˜- - -Ë ¯ Ë ¯ Ë ¯ Ë ¯

3 2 3 7 3

6 1 1 19 1k

Ê ˆ Ê ˆ Ê ˆ Ê ˆ= πÁ ˜ Á ˜ Á ˜ Á ˜- - -Ë ¯ Ë ¯ Ë ¯ Ë ¯

3 2 1 3 13

6 1 3 9 3

-Ê ˆ Ê ˆ Ê ˆ Ê ˆ= = -Á ˜ Á ˜ Á ˜ Á ˜- - -Ë ¯ Ë ¯ Ë ¯ Ë ¯

3 2 2 10 25

6 1 2 10 2

Ê ˆ Ê ˆ Ê ˆ Ê ˆ= =Á ˜ Á ˜ Á ˜ Á ˜-Ë ¯ Ë ¯ Ë ¯ Ë ¯

Solution A145

(i) Unit perpendicular vector The scalar (dot) product of two perpendicular vectors is zero. So: ( ).( 2 3 ) 0a b c+ + - + =i j k i j and ( ).(10 ) 0a b c+ + + =i j k i k [1]

ie 2 3 0a b- + = and 10 0a c+ = So:

23b a= and 10c a= - [1]



For a unit vector we also need 2 2 2 1a b c+ + = , which gives:

( )22 223 ( 10 ) 1a a a+ + - =

3

913afi = [1]

and the required unit vector is:

3 2 30

913 913 913+ -i j k [1]

Comment If the question hadn’t directed you which method to use (and your knowledge of vectors is good) you could also have done this by working out the cross product ( 2 3 ) (10 )- + Ÿ +i j i k , then rescaled to get a unit vector. You will not need cross

products in the actuarial exams.

(ii) Other unit perpendicular vector The other unit vector perpendicular to these two vectors is the unit vector pointing in the opposite direction, which is just the negative of the one we’ve found ie:

3 2 30

913 913 913- - +i j k [1]

Solution A146

If we think of the matrix P as the matrix that transforms the column vector ( )Tx y z

into the column vector Ta b c , then we have the relationship:

0.2 0.2 0.2

0.2 0 0.2

0 0.4 0.2

a x

b y

c z

(*)



Premultiplying both sides by 1P , we see that:

1

a x

b y

c z

P

Thinking in terms of simultaneous equations, Equation (*) corresponds to:

0.2 0.2 0.2

0.2 0.2

0.4 0.2

x y z a

x z b

y z c

+ - =+ =+ =

[1]

To solve these, we need to turn them round and express , ,x y z in terms of , ,a b c . We

can simplify these equations by multiplying through by 5:

5

5

2 5

x y z a

x z b

y z c

+ - =+ =+ =

[1]

We can use the last two equations to express x and y in terms of z :

5x b z= - 2.5 0.5y c z= -

From the first equation, we then get: (5 ) (2.5 0.5 ) 5b z c z z a- + - - = 2 2z a b cfi = - + +

and so: 5 ( 2 2 ) 2 3x b a b c a b c= - - + + = + -

2.5 0.5( 2 2 ) 2y c a b c a b c= - - + + = - + [1]

Writing these in the form of simultaneous equations:

2 3

2

2 2

a b c x

a b c y

a b c z

+ - =- + =

- + + =



In matrix notation this is:

2 3 1

1 1 2

2 2 1

a x

b y

c z

[1]

So:

1

2 3 1

1 1 2

2 2 1

P

[1]

Solution A147

Working out the various parts of this expression first, we find that:

2 1

det( ) (2)(4) (1)(1) 71 4

[1]

1

1 2 1 4 111 4 1 27

[2]

and:

2 1 1

( )0 1 1

x

and ( ) 1 1Tx

Putting these together, we get:

1 4 1 11 1 1( ) ( ) 1 1

1 2 12 2 7

311 1

114

4

14

2

7

Tx x

[1]



and

2/7

2

1( ) 0.0452

(2 ) 7f e

p-= =

¥x [1]

Comment Note that a 1 1¥ matrix is really just a scalar. So when we calculate

11( ) ( )

2T -- - -x x we can treat the answer as a simple number.

Solution A148

A Solution A149

A Solution A150

B Solution A151

D Solution A152

C Solution A153

C

½£18,000 £7,500

6

2 ¥ =

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