Fac pack dec13 by vedant

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 ActEd@bpp.com 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.

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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.

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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.

Introduction This document has the following sections: Section 1 Section 2 Section 3 Section 4 Section 5

Syllabus The FAC course Study skills Contacts Course index

Page 2 Page 10 Page 12 Page 13 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 ActEd@bpp.com.

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.

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FAC: Study Guide

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.

 a, b, c, n  , n  3 : a n  b n  c n

2. 3. 4. 5. 6.

x  (, 0], x 2     {0}   {x : x  1, 2,3,} x   x  163 “Zero is a non-negative integer;  is a positive real number.” “ 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”.

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FAC: Study Guide

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(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 ( x y ), exponential ( e x ) and natural log ( ln x ) functions.

(The following

features are also useful but not essential: factorial function ( n! ), combinations ( n Cr ), hyperbolic tangent function and its inverse ( tanh x and tanh 1 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).

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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 NewtonRaphson 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 ) [  e x ], and ln x [  log e 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) .]

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FAC: Study Guide

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(c)(v) Simplify and evaluate expressions involving the factorial function n! for nonnegative 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  re i ). (c)(viii) Apply Euler’s formula e i  cos  i 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  ) and repeated sums. i0

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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  n2 (2a  (n  1)d ) or

n (a 2

 l) ,

a (1  r n ) and 1 r a   GP  , 1 r GP 

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

(d)(viii) Apply the formulae:

 k  12 n( n  1) and

k 1

 k 2  16 n( n  1)( 2n  1) .

k 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 , e x 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.

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FAC: Study Guide

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(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 e x . (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 e x and ln(1 ď&#x20AC;Ť 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.

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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”.

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FAC: Study Guide

1.2

Page 9

The Professionâ&#x20AC;&#x2122;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.

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FAC: Study Guide

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 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8

Notation Numerical Methods I Mathematical constants and standard functions Algebra Numerical Methods II Differentiation Integration 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.

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FAC: Study Guide

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For further guidance, this grid shows which chapters from FAC are needed for Subjects CT1, CT3, CT4, CT5, CT6 and CT8: Chapter Section Course 1 1-7 All 2 1-4 All 3 1-3 All 4 1-9 All 5 1-6 All 5 7 CT6 5 8 CT6 6 1-8 All 6 9 CT8 7 1, 2 All 7 3.1-3.4 All 7 3.5 CT4, CT8 7 4-7 All 7 8.1-8.3 CT4, CT8 8 1, 2.1-2.4 CT4, CT6, CT8 8 2.5 CT4

2.2

Notes This is just general information and notation.

Syllabus items (c)(vi), (c)(vii), (c)(ix) Syllabus item (d)(ix) Syllabus item e(viii)

Syllabus item (e)(xviii) Syllabus item (e)(xvii) Syllabus items (b)(xi), (b)(xii) Syllabus item (b)(xii) (part)

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.

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FAC: Study Guide

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.

Attempt the questions in the Notes as you work through the course. Write down your answer before you check against the solution.

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.

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FAC: Study Guide

Page 13

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 FAC@bpp.com (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 FAC@bpp.com.

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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.

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FAC: Study Guide

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Course index Topic Absolute change Arithmetic-geometric inequality Arithmetic progressions

Chapter 5 4 4

Page 5 20 24

Binomial expansion

Calculator, use of Complex numbers Convergence Curve sketching

2 5 7 6

4 17 16 25

Determinants Difference equations Differential equations Differentiating an integral Differentiation, products and quotients Differentiation, standard functions Dimensions Double integrals

8 5 7 7 6 6 5 7

12 24 31 14 12 11 9 18

Eigenvectors and eigenvalues Errors Estimation Extrema

8 5 2 6

21 7 9 29

Factorial notation Fractions, algebraic Functions and graphs

3 4 3

12 5 2

Gamma function Geometric progressions Greek symbols

3 4 1

13 26 5

Indices Induction Inequalities Infimum Integer part Integration, by parts Integration, partial fractions Integration, standard functions

4 1 4 6 3 7 7 7

2 10 16 7 9 12 9 5

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FAC: Study Guide

Integration, substitution Interpolation Iteration

7 5 5

11 12 15

Lagrangian multipliers Leibniz’s formula Limits Logarithms

6 7 6 4

30 14 2 2

Maclaurin series Mathematical notation Matrices Max and min notation Modulus

7 1 8 3 3

30 2 8 10 8

Newton Raphson iteration

Order notation

Partial differentiation Percentages  notation Proof Proportionate change

6 5 4 1 5

27 2 21 7 5

Quadratic equations

Rounding

Scalar product Series  notation Simultaneous equations Stationary points Supremum

8 4 4 4 6 6

5 31 21 11 20 7

Taylor series Trapezium rule

7 7

27 25

Vectors

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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.

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.

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FAC-01: Notation

Mathematical notation Symbol

Meaning

Further explanation/examples

Types of numbers:

 

Integers (whole numbers) Natural numbers (counting numbers)



Rational numbers

{...,–3, –2, –1, 0, 1, 2, 3,...} {1, 2, 3, ...} 3, 2

1.2 = 65 , 0.141414... = 14 etc (all can 99 p q

(fractions)

be written as

)



Real numbers



Complex numbers

rational numbers plus irrational numbers (such as 2 , p and e) ie no imaginary component can be written in the form a + ib , where i = -1

Logic: " : $

For all (values) Such that There exists

" x Œ  $ x2 Œ  (see next example) $ x Œ : x +1 = 5

There doesn’t exist

$/ x Œ  : x 2 = -4

ﬁ

Implies

x = -2 ﬁ x 2 = 4

¤ iff

Implies and is implied by If and only if

x = 2 ¤ x3 = 8 equivalent to ¤

Set Theory:

{1,2,3,...}

A set

 ( -•, 0]

etc A set

∆ or

{}

Œ A» B

Empty set Is a member of Union of two sets

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) the set of odd numbers divisible by 2 2½ Œ  A or B or both, ie the things that are in one or other or both

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FAC-01: Notation

A« B A

Intersection of two sets Complement of a set

Page 3

A and B, ie only the things that are in both not A, ie the things that are not in A

Miscellaneous:

p e • Æ

3.14159...(the ratio of the circumference of a circle to its diameter) base of the natural logarithm 2.7182818... Infinity 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 x Æ 1+ , 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 i 2 = -1 , a being called the real part and b being called the

imaginary part. You may have seen j used for

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-1 rather than i.

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FAC-01: Notation

Example

(i)

Interpret the statement:

{x Œ (ii)

}

:x<4

For the function f ( x) = 1x , 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 x = 0 , f ( x) Æ • as x Æ 0 + , f ( x) Æ -• as x Æ 0 - , f ( x) Æ 0 as x Æ ±• .

Question 1.1

(i)

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

(ii)

What values are in the set {x Œ  : x 2 < 10} ?

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FAC-01: Notation

Page 5

Greek symbols You need to know the following Greek letters: Letter

Lower case

Used for

alpha

parameter

beta

gamma

Upper case

Used for

parameter

beta function

parameter

gamma function

delta

small change

difference

epsilon

small quantity

theta

parameter

number of deaths

kappa

parameter

lambda

parameter

mean and force of mortality

force of mortality when sick

= 3.14159

product

rho

correlation coefficient and force of recovery

sigma

sum

tau

standard deviation and force of sickness parameter (pronounced as in “torn”) probability density function of standard normal distribution

cumulative distribution function of standard normal distribution

phi

chi

c 2 distribution (pronounced as first syllable of “Cairo”)

psi

probability of ruin

omega

limiting age in a life table

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FAC-01: Notation

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 $6.2 ¥ 106 .

●

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 7 34 % 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 Tax

£ 9.6m (2.4m)

Net profit Dividends

7.2m (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.

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FAC-01: Notation

Page 7

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 ﬁ A (ie B implies A or B is true only if A is true). If A is sufficient for B, then A ﬁ B (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 ﬁ M 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 ﬁ B but not Bﬁ A.

(ii)

If x is a solution of the equation ax 2 + bx + c = 0 , then if P is the statement “ b 2 - 4ac ≥ 0 ” 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!)

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FAC-01: Notation

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: 25

2 + 2 , 1.09 ,

Âv

(1 + i )5 - 1

t =1

, G (½)

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

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 , 1.0925 = 8.6231 ,

Â vt = an| , t =1

(1 + i )5 - 1

= s5| , G (½) = p

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 ∆ ).

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FAC-01: Notation

5.3

Page 9

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

-b ± b 2 - 4ac 2a

The formula for calculating the present value of a unit annuity-certain payable annually in arrears is an| =

1 - vn . You will meet this in Subject CT1. i

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 x 2 , 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.

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FAC-01: Notation

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

Prove by induction that 1 + 2 + 3 +  + n = 12 n(n + 1) . Solution

Assume the result is true for n = k , ie: 1 + 2 + 3 + ◊◊◊ + k = 12 k (k + 1) Adding the next term on to both sides it follows that: 1 + 2 + 3 +  + k + (k + 1) = 12 k (k + 1) + (k + 1) = 12 (k + 1)(k + 2) = 12 (k + 1) [(k + 1) + 1]

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

RHS = 1

So the result is true for n = 1 and by the above result it must also be true for n = 2,3, 4, ie for all positive integer values of n.

Question 1.2

Prove by induction that 12 + 22 + 32 +  + n 2 = 16 n(n + 1)(2n + 1) .

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FAC-01: Notation

Page 11

Some general knowledge Currencies Country United Kingdom United States European Union Japan

Main currency unit Pound (£) Dollar ($) Euro (€) Yen (¥)

Sub-division Pence (£1 = 100p) Cent ($1 = 100¢) Cent (€1 = 100¢) –

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?

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Page 12

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

1 12

of a year long. This simplifies

the calculations considerably.

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FAC-01: Notation

Page 13

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

1 4

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.

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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”.

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FAC-01: Notation

Page 15

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?

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FAC-01: Notation

Chapter 1 Solutions Solution 1.1

(i)

(ii)

M = {2} . (Don’t forget that a prime number is one that has no factors other than 1 and itself!)

10 < x < 10

}

Solution 1.2

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

12 + 22 + 32 +  + k 2 = 16 k (k + 1)(2k + 1) Adding the next term on to both sides: 12 + 22 + 32 +  + k 2 + (k + 1) 2 = 16 k (k + 1)(2k + 1) + (k + 1) 2 = 16 (k + 1)[k (2k + 1) + 6(k + 1)] = 16 (k + 1)(2k 2 + k + 6k + 6) = 16 (k + 1)(2k 2 + 7k + 6) = 16 (k + 1)(k + 2) ( 2k + 3) = 16 (k + 1)[(k + 1) + 1][2(k + 1) + 1] So we have shown that if the result is true for n = k then it is also true for n = k + 1 . Consider when n = 1 : LHS = 1

RHS = 1

So the result is true for n = 1 and by mathematical induction it is also true for n = 2, 3, 4, … ie for all positive integer values of n.

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FAC-01: Notation

Page 17

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 365 14 ∏ 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).

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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.

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.

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Page 2

FAC-02: Numerical methods I

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.

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FAC-02: Numerical methods I

Page 3

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 p = 3,142 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.

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Page 4

FAC-02: Numerical methods I

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 6.231.4 + 5.83.6 . Solution

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

Example

(

Calculate 1 + 4 1.723

)

5.1

Solution

Using the root key ( x

or x

) and the power key, we get 49.091. 1

If you donâ&#x20AC;&#x2122;t have this key, you can treat the expression as (1 + 1.723 4 )5.1 . Since x

y = y x you can use your x y or y x key.

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FAC-02: Numerical methods I

Page 5

Example 2

 2 Calculate 1   .  3

Solution

7 or 2.778. If your calculator has a fractions facility, be aware of the 9 order in which your calculator performs fractions!

The answer is 2

Example

Calculate

( 74 + ln 5.2)

Solution

The answer is 4.929. Your calculator will have log e or ln for the natural logarithm key. (We will deal with this function later in the course.) Most calculators have a fraction key ( a b c ) which can be helpful. Notice also that most calculators have a squared key ( x 2 ) to avoid you having to use the power key.

Example

Calculate

4 2 e3.1 3

Solution

The answer is 1.961. On many calculators, e x and ln are on the same key. On recent calculators, to get e3.1 , you would need to type e x 3.1 (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.

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FAC-02: Numerical methods I

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 n ¥ (n - 1) ¥ (n - 2) ¥ ◊◊◊ ¥ 1 . So here 4! = 4 ¥ 3 ¥ 2 ¥ 1 .

Example

Calculate tanh 2 and tanh -1 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 tanh 2 = 0.964, tanh -1 0.4 = 0.424 .

Example

Calculate

15.2 - 3.74 42 - 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.

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FAC-02: Numerical methods I

Page 7

Question 2.3

Calculate

1 - (1.04) -3.5 12 . 0.04 1.04

Question 2.4

Calculate

1  1  0.1  ln  . 2  1  0.1 

Question 2.5

Calculate

2 + (3.7892 + 2.5) 4.5 5.5 - 2.13

Question 2.6 3 ± ( -3) 2 - 4 ¥ 2 ¥ -4 Calculate . 2¥2

Question 2.7

Calculate

100 (1.035)10

1   17 1  10    1.035  . ln1.035

Question 2.8

Calculate

10! . 7!2!

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FAC-02: Numerical methods I

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

1   Calculate v + 3v 2 + 7v3 , for v  0.9259   .  1.08  Solution

We can do this as a nested multiplication: v + 3v 2 + 7v3 = v + v 2 (3 + 7v) = v[1 + v(3 + 7v)]

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

 1  Calculate v + 2v 2 + 5v3 + 6v 4 , when v  0.9091  .  1.1 

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.

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FAC-02: Numerical methods I

Estimation and accuracy

3.1

Estimation

Page 9

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

2.7 2  3.14 . Estimate the value of 5.2  7.8 Solution

2.7 2  3.14 32  34 9  81 is roughly   30 . 5.2  7.8 58 3 The actual answer is –38.3.

Question 2.11

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

Compare your

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.

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Page 10

3.2

FAC-02: Numerical methods I

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

1  (1  i ) 10 , the accurate value of i is 0.034724. Compare the i values of p obtained using: In the formula p 

(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?

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FAC-02: Numerical methods I

Page 11

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

R(1  v8 ) 1 , where P  £14, 000, R  £700, v  , i  0.05 , what is T? i 1 i (Work in £000s). If T  Pv9 

Solution

Working in £000s, P  14, R  0.7 . Then T  13.549 ie £13,549.

Example

If A  4, 000 , calculate

A2  2, 000 A  1, 000, 000

(i)

directly

(ii)

working in units of 1,000

Solution

(i) (ii)

25, 000, 000  5, 000

In units of 1,000: 1,000,000 becomes 12  1 , 2,000 becomes 2 and A becomes 4 The calculation is then done as follows: 42  2  4  1  25  5 ie the “real” answer is 5,000.

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FAC-02: Numerical methods I

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

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FAC-02: Numerical methods I

Page 13

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.

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FAC-02: Numerical methods I

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

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FAC-02: Numerical methods I

Page 15

Chapter 2 Solutions Solution 2.1

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 0.100.

Solution 2.5

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 214.734.

Solution 2.8

The numerical answer is 360.

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FAC-02: Numerical methods I

Solution 2.9

Using the nested multiplication method, we get: v  2v 2  5v3  6v 4  v  2v 2  v3 (5  6v)  v  v 2 [2  v(5  6v)]  v(1  v[2  v(5  6v)])

Now using v 

1 , we get the answer to be 10.42. 1.1

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

For Question 2.6 one estimate is

2  (42  3) 4 6  23



2  192 2  400   201 . 2 2

3  9  32 3  6   2.25 or  0.75 . 4 4

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

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FAC-02: Numerical methods I

Page 17

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.

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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 x n , c x , exp( x) , and ln x

●

the functions x , max() and min()

●

the factorial function n ! and the gamma function.

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

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FAC-03: Mathematical constants and standard functions

Standard functions and graphs

1.1

Exponential function The exponential function, e x , can be defined by:  x e  lim 1   n  n x

or the series expansion e x  1  x 

x 2 x3  . 2! 3!

It is the inverse of the natural logarithmic function (which we will look at in Chapter 4), so that eln x  x . Do remember that e is just a number which you can find from your calculators: e1  2.718... If the power is a long expression then a convenient alternative notation is e x  exp( x) ,

 

x 2

for example e

 12 

  x   2   exp   12   .     

This makes things clearer to read.

However don’t mix up these two notations by writing something like exp x , as this is meaningless. The graph of y  e x looks like this: y=exp(x) 8 6 4 2 0 -2

-1.5

-1

-0.5

0.5

1.5

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Page 3

Notice that e x 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: ●

e0 is 1

●

as x gets large, so does e x

●

and as x gets large but negative, e x  0 .

Since e x is just “a number to the power of x”, then this graph is also the basic shape of the graph of y  c x , 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 y  c x , with c replaced by 1 . The diagram below shows the graphs of y  0.5 x and y  2 x : c y  0.5 x

y  2x

5 4 3 2 1 0 -3

-2

-1

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.

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1.2

FAC-03: Mathematical constants and standard functions

Log function A logarithm is the inverse of a power. The graph of the natural logarithm function y  ln x is:

3 2 1 0 -1 0

-2 -3 -4 -5 -6

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 x   as x   , and log x   as x  0 . 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.

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FAC-03: Mathematical constants and standard functions

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

18 16 14 12 10 8 6 4 2 0 -2 0

-2

-1

Notice that the values of y for these functions can never be less than zero. The graph of y  x n , where n is an odd positive integer has the same general shape as y  x3 which is shown below:

1.3

Page 5

-3

-2

-1

10 8 6 4 2 0 -2 0 -4 -6 -8 -10

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.

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FAC-03: Mathematical constants and standard functions

Remembering that x  n 

1 x

, the graph of y  x  n will have a discontinuity (where the

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

300 200 100

0 -2

-1

-100

-200 -300

The graph of y  x is (remembering that

x represents the positive square root):

2.5 2

1.5 1 0.5 0 0

This is the inverse function of y  x 2 , so the graph of y  x is the reflection of y  x 2 in the line y  x . The same relationship applies to the graphs of log x and e x 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.

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Page 7

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

1.4

●

it is like x n , ie t is a variable and z is a fixed number, or

●

it is like c x , ie t is a fixed number and z is a variable.

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)

y  x 1

(ii)

y  3 x5

(iii) (iv)

y  e4 x  1 y  ln(1  x)

(v)

y  ( x  3) 4

Question 3.2

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

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FAC-03: Mathematical constants and standard functions

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 (c, 0) . Example

Write the expression 2 x  1  3 without the use of the modulus sign. Solution 2 x  1  3  3  2 x  1  3

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 x  1  3  2  2 x  4  1  x  2

Question 3.3

Write the expression   3 x  2 without the modulus sign.

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FAC-03: Mathematical constants and standard functions

2.2

Page 9

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. m When you divide m by n, where both are positive integers, you get   as the quotient n m with a remainder of m  n    . n 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.

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2.3

FAC-03: Mathematical constants and standard functions

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( x  2,10) for the region 0  x  20 ? Solution

We could write max( x  2,10) as:  10 if 0  x  8   x  2 if 8  x  20 So max( x  2,10)  10 for 0  x  8 , and max( x  2,10)  x  2 for 8  x  20 . The abbreviation ( x  100)  is often used for max( x  100, 0) . Question 3.4

What is min( x 2 ,15) for 0  x  6 ?

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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: UBE  min( S ,UEL)  min( 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:   UBE    

if  if  if 

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FAC-03: Mathematical constants and standard functions

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: n !  n  (n  1)  (n  2) 1 where n is a non-negative integer. The factorial function satisfies the relationship n !  n  (n  1)! . If we put n  1 , this tells us that 1!  1 0!  0!  1 . Question 3.6

Evaluate 5!

Example

Simplify the expression

n !(2n  1)! . (2n)!(n  2)!

Solution

This expression can be written as: n! (2n  1)!  (n  2)! (2n)! The first factor equals n(n  1) , because n ! contains an extra n and n  1 in its expansion, which are not contained in the expansion of (n  2)! , and the second factor is 1 . 2n So we get

n !(2n  1)! n(n  1) 1   2 (n  1) . (2n)!(n  2)! 2n

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Question 3.7

Simplify the expression

3.2

(3n)!(2n  1)! . (3n  2)!(2n  1)!

Gamma function ( x) , where x  0 , is defined by the integral

 x 1 t

0 t

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:

( x)  ( x  1)( x  1) where x  1

Result 2:

(n)  (n  1)! where n  1, 2, 3,...

Result 3:

( 12 )  

These results can be found in the Tables page 5. The graph of the gamma function looks like this: 7 6 5 4 3 2 1 1

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FAC-03: Mathematical constants and standard functions

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

(n  1)(2n  1)! where n is a natural number. (2n)(n  1)!

Question 3.11

Show that, if n is a nonnegative whole number, (n  ½) 

(2n)! 22n n !

 .

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Page 15

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 :

n !  n n½ e  n 2

and

(n)  n n½ e n 2

(Don’t worry that these appear to be inconsistent with the relationship (n)  (n  1)! n 1 Remember that for very large values of n , the ratio is very close to 1.) n

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FAC-03: Mathematical constants and standard functions

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Page 17

Chapter 3 Summary Standard functions

You should know the properties and be able to sketch the graphs of x n , c x , e x , and log 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 n !  n  (n  1)  (n  2) 1 . Note that 0!  1 . The gamma function 

The gamma function is defined by ( x)   t x 1et dt , when x  0 . 0

It has the following properties: 

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



G (n) = (n - 1)! where n = 1, 2,3,



G (½) = p .

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FAC-03: Mathematical constants and standard functions

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Chapter 3 Solutions Solution 3.1

(i)

The graph of y  x 1 is as follows:

10 5

0 -1.5

-1

-0.5

0.5

1.5

-5 -10

(ii)

The graph of y  3 x5 is as follows:

100 50 y

0 -2

-1

-50 -100

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The graph of e 4x is like that of e x , 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.

9 8 7 6 5 4 3 2 1 0

(iii)

FAC-03: Mathematical constants and standard functions

-1.5

-1

-0.5

0.5

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

(iv)

-1

-0.5

2 1.5 1 0.5 0 -0.5 0 -1 -1.5 -2 -2.5

0.5

1.5

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FAC-03: Mathematical constants and standard functions

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

(v)

Page 21

16 14 12 10 8 6 4 2 0 0

Solution 3.2

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

Solution 3.3

  3x  2 or   3x  2 If you wish to simplify these expressions further, proceed as follows:

  2  3x  x 

  2  3x  x 

 2 3

2 3

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

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FAC-03: Mathematical constants and standard functions

Solution 3.4 min( x 2 ,15)  x 2 for 0  x  3.87 min( x 2 ,15)  15 for 3.87  x  6

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.

UEL

LEL LEL

UEL

The alternative mathematical formula is: 0  UBE   S  LEL  UEL  LEL

if S  LEL if LEL  S  UEL if S  UEL

Solution 3.6 5!  5  4  3  2 1  120

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Page 23

Solution 3.7

(3n)!(2n  1)! (3n)! (2n  1)! 1 2n(2n  1)     2n(2n  1)  (3n  2)!(2n  1)! (3n  2)! (2n  1)! (3n  1)(3n  2) (3n  1)(3n  2)

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

(n  1)(2n  1)! (n  2)!(2n  1)! 1 1 or 3   (2n)(n  1)! (2n  1)!(n  1)! n(n  1)(n  1) n n

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FAC-03: Mathematical constants and standard functions

Solution 3.11

We can prove this using mathematical induction. When n  0 , the equation says (½) 

0! 20 0!

   , which is correct.

If we assume it’s true for a typical value of n , say n  k , then we know that: (k  ½) 

(2k )! 22 k k !



This would imply that: (k  1½)  (k  ½)(k  ½)  (k  ½) 

(2k )! 22 k k !



The last expression can be written as: (k  ½) 

(2k )! 22 k k !

 

2k  1 (2k )! (2k  1)!   2k 1   2k 2 2 k! 2 k!

This doesn’t quite match the formula we were hoping for. But, if we include an extra 2k  2 factor of (which won’t affect the answer), we get: 2(k  1) 2k  2 (2k  1)! (2k  2)!   2k 2   2 k 1 2(k  1) 2 k! 2 (k  1)!

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

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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 (a + b) n where n is a positive integer, and (1 + x) p for any real value of p .

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.

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FAC-04: Algebra

Algebraic expressions

1.1

Indices There are three laws governing indices (or powers): x a ¥ xb = x a + b x a xb = x a - b ( x a )b = x ab

Question 4.1 Simplify the following:

1.2

(i)

5 x3 ¥ 2 x5

(ii)

16 y 2 6 y 7

(iii)

(5b3 ) 2

Logarithms A logarithm is the inverse of a power: 102 = 100 ¤ log10 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 e 15 = 2.708 . An alternative abbreviation for log to base e is ln. On your calculator, the ln button gives you log e and the log button gives you log10 .

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Page 3

There are three laws of logarithms that can be derived from the laws of indices: log a x + log a y = log a xy log a x - log a y = log a

x y

log a x n = n log a 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 x + log 4 x - log 5 x .

(ii)

Write 1  2log10 x in the form log f ( x) .

Solution

2x ¥ 4x 8x = log . 5x 5

(i)

This simplifies to log

(ii)

log10 x 2  log10 10  log10 10 x 2 .

Question 4.3

Simplify ln e.

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FAC-04: Algebra

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

Solve the equation 1.12 n+1 = 2 . Solution

Take logs: log1.12 n 1  log 2  (2n  1)log1.1  log 2  (2n  1) 

log 2 log1.1

 1  log 2 n   1  3.136 2  log1.1 

Question 4.4

Solve the equation 2 ¥ 1.05t - 2 = 1.10252t -1 .

Note: The value of

logb x is independent of the base b used. logb y

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FAC-04: Algebra

1.3

Page 5

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 x + 3 3x + 2 x + 1 6x - 1

(ii)

x 2 - 3x - 4 2 x2 - 7 x - 4

(iii)

x2 - x - 2 x2 + 3x + 2 â&#x2C6;? 2 x 2 - 5 x - 12 2 x 2 + 5 x + 3

Solution

(i)

Putting these over a common denominator: 2 x + 3 3x + 2 (2 x + 3)(6 x - 1) - (3 x + 2)( x + 1) = x + 1 6x - 1 ( x + 1)(6 x - 1)

(ii)

(12 x 2 + 16 x - 3) - (3 x 2 + 5 x + 2) ( x + 1)(6 x - 1)

9 x 2 + 11x - 5 ( x + 1)(6 x - 1)

Factorising the numerator and denominator: x2 - 3x - 4 ( x + 1)( x - 4) = 2 2 x - 7 x - 4 (2 x + 1)( x - 4) =

x +1 2x + 1

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(iii)

FAC-04: Algebra

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

x2 - x - 2 x 2 + 3x + 2 ( x - 2)( x + 1) ( x + 2)( x + 1) ∏ 2 = ∏ 2 2 x - 5 x - 12 2 x + 5 x + 3 ( x - 4)(2 x + 3) (2 x + 3)( x + 1) =

( x - 2)( x + 1) (2 x + 3)( x + 1) ¥ ( x - 4)(2 x + 3) ( x + 2)( x + 1)

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

Example

1 1 + Simplify a b . a 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: 1 1 + a b ¥ ab = b + a a b ab a 2 + b 2 + b a

Question 4.5

Simplify the expression

x 2 + 3x - 4 2 x2 + 5x - 7 . ∏ 3 x 2 - 10 x - 8 4 x 2 - 17 x + 4

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Page 7

Quadratic equations

2.1

Solution by factorisation This is where you try to write the quadratic ax 2 + bx + c = 0 as (dx + e)( fx + g ) = 0 (the process being called factorisation) and then the solutions would be x = - de or -

g f

Notice that by comparing the coefficients of x 2 , x and the constant term, a = df , c = eg , and b = ef + dg . Example

Solve the equation 2 x 2 + x - 3 = 0 by factorisation. Solution 2 x 2 + x - 3 = 0 ﬁ (2 x + 3)( x - 1) = 0 ﬁ x = -1.5 or 1

Question 4.6

Solve the following equations by factorisation: (i)

6 x2 - x - 2 = 0

(ii)

16 x 2 - 1 = 0

Some useful results: a 2 - b 2 = (a + b)(a - b) a 3 ± b3 = (a ± b)(a 2  ab + b 2 )

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).

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2.2

FAC-04: Algebra

Solution by completing the square All quadratic equations can be written as ( px + q ) 2 - r = 0 , where p, q, r Œ . If the coefficient of x 2 is negative, you would have to multiply through by –1 first. If r > 0 , then this method enables the quadratic to be solved (in the example here, x =

-q ± r ). p

Example

Solve the following equations by completing the square: (i)

4 x 2  8x  1  0

(ii)

3x 2  13x  1  0

Solution

(i)

There are several methods available to complete the square, for example you can multiply out the expression ( px + q ) 2 - r , giving p 2 x 2 + 2 pqx + q 2 - r , and compare coefficients to solve the equations: 4 x 2 - 8 x + 1 = 0 ﬁ (2 x - 2) 2 - 3 = 0

fi (2 x - 2) 2 = 3 fi 2x - 2 = ± 3 fix= (ii)

1 2

(2 + 3 ) or 12 (2 - 3 ) = 1.866 or 0.134

You need to multiply through by –1 first:

( ﬁ 3 (( x -

)

3 x 2 - 13 x + 1 = 0 ﬁ 3 x 2 - 13 x + 13 = 0 3 13 2 ) 6

)

- 169 + 13 = 0 36

) 2 = 157 ﬁ ( x - 13 6 36 ﬁ x = 13 ± 6

157 36

= 4.255 or 0.078

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Question 4.7

Solve the following equations by completing the square: (i)

x2 + 4 x - 8 = 0

(ii)

16 x 2 + 17 x + 2 = 0

Question 4.8 2

2  x  1   x  2  x   If          , for all values of x, express m and s , in     1    2  terms of a , b , m1, m2 ,s 1, and s 2 .

2.3

Solution by formula For the general quadratic equation, ax 2 + bx + c = 0 the solutions are given by -b ± b 2 - 4ac x= . This can be proved by applying the method of completing the 2a square to the general quadratic equation.

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FAC-04: Algebra

Example

Solve the equation 3x 2 + 8 x + 2 = 0 , using the quadratic formula. Solution

If the equation is compared with the general quadratic, then: a = 3, b = 8, c = 2 so: x= =

-8 ± 64 - 24 6 1 6

(-8 +

)

40 or

1 6

( -8 -

)

= -0.279 or - 2.387

Question 4.9

Solve the equation 1 + 4 x - 7 x 2 = 0 .

Question 4.10

Solve the equation 32 x +1 + 4 ¥ 3x - 4 = 0 . 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.

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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: 2x + 5 y - 4 = 0 7 y = 2 - 4x Solution

Looking at the two equations in turn:

2x + 5 y - 4 = 0 ﬁ 2x + 5 y = 4

(1)

7 y = 2 - 4x ﬁ 4x + 7 y = 2

(2)

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

-3 y = -6 ﬁ y = 2 \ x = -3

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FAC-04: Algebra

Example (Non-linear equations solved by substitution)

Solve the simultaneous equations: x 2 + y 2 = 20 2 x - 3 y = 14

Solution

The second equation can be rearranged to give: x = 12 (14 + 3 y ) Substituting this into the first equation gives: 1 (14 + 3 y ) 2 4

+ y 2 = 20

fi 196 + 84 y + 9 y 2 + 4 y 2 = 80 fi 13 y 2 + 84 y + 116 = 0 Using the quadratic formula or otherwise: 58 fi y = -2 or - 13

ﬁ x = 4 or

4 13

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Example (Non-linear equations solved by division)

Solve the simultaneous equations: ar 2 = 8 ar 3 + ar 4 = 30

Solution

Dividing the second equation by the first equation: r + r 2 = 3.75 Solving this as a quadratic: r = 1.5 or - 2.5

32 9

32 25

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FAC-04: Algebra

Example (Non-linear equations solved by factorisation)

Solve the simultaneous equations: 2 p 2 - pq - 3q 2 = -50 3 p 2 + 4 pq + q 2 = 35

Solution

Factorising both equations:

( p + q)(2 p - 3q) = -50 ( p + q)(3 p + q) = 35 Dividing: -7 3p + q = 2 p - 3q 10 fi 10(3 p + q) = -7(2 p - 3q) fi 4p = q Substituting: 3 p 2 + 16 p 2 + 16 p 2 = 35 fi p = 1 or - 1 fi q = 4 or - 4

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Question 4.11

Solve the simultaneous equations:

ab 2 + ab 4 = 5

(i)

ab 4 + ab 6 = 20 2 x 2 + 5 xy - 3 y 2 + 5 = 0

(ii)

2 x 2 + xy - y 2 - 5 = 0

Example

= 11.7 ,

a a = 9.3 and + k = 4.1 , find a , l and k . 2 l l

Solution

Dividing the second equation by the first, we get: 9.3 l = ﬁ l = 1.590 11.7 2 Substituting into the second equation gives us:

a = 9.3 ¥ 1.5902 = 23.50 Finally substituting both of these into the final equation gives us: k = -10.68

Question 4.12

If e

m + 12 s 2

= 3 , and e2 m +s (es - 1) = 11.2 , what are m and s ?

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FAC-04: Algebra

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 x - 7 £ 7 x + 4 . Solution

Starting with 2 x - 7 £ 7 x + 4 , if we subtract 4 from both sides, and subtract 2 x from both sides, we get: -11 £ 5x

Then dividing by 5 we get: - 11 £x 5 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).

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Example

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

Manipulate the inequality to get zero on one side: (2 x + 11)( x - 7) < -57 2 x 2 - 3 x - 77 < -57 2 x 2 - 3 x - 20 < 0 (2 x + 5)( x - 4) < 0 The inequality can be solved by drawing a graph of the function f ( x ) = 2 x 2 - 3 x - 20 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 x + 5)( x - 4) will change sign when x = -2.5 or 4 , so we can consider whether the expression is positive or negative between these points. (See table below.) So the solution is -2.5 < x < 4 , since we want the expression to be less than 0 ie negative.

2x + 5

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

x < -2.5

-2.5 < x < 4

x>4

– – +

+ – –

+ + +

Example

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

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

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FAC-04: Algebra

Example

Solve the inequality: 8x2 + 5x - 7 £2 3x 2 + 5 x - 2

Solution

Getting zero on one side and using a common denominator, we get: 8 x 2 + 5 x - 7 - 2(3 x 2 + 5 x - 2) £0 3x 2 + 5 x - 2 2x2 - 5x - 3 (2 x + 1)( x - 3) £0ﬁ £0 2 (3 x - 1)( x + 2) 3x + 5 x - 2

This will change sign at x = -2, - 12 ,

1, 3

By considering the sign between each of these values (see table below), the solution includes -2 < x < - 12 or x = -2, - 12 ,

1, 3

-2 < x £ - 12 or

3. 1 3

1 3

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

Substituting these values in, we can see that our solution is

< x £ 3. x < -2

-2 < x < - 12

- 12 < x < 13

– – – –

+ – – –

+ + – –

+ + + –

+ + + +

–

x+2 2x + 1 3x - 1 x-3 (2 x + 1)( x - 3) (3x - 1)( x + 2)

1 3

< x<3

x>3

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Question 4.13

Solve the inequality 2 x3 + 5 x 2 - 17 x < 20 . Note: ●

For a double inequality, for example 10 < 2n - 1 < 20 , 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 £ .

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FAC-04: Algebra

Arithmetic-geometric mean inequality The arithmetic mean of n numbers, a1,, an , is defined to be geometric mean of n positive numbers, a1,, an , is defined to be

a1 + + an . n

The

a1 an .

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: n

a1 an £

a1 +  + an n

Example

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

Arithmetic mean =

7+9 =8 2

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?

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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)

12 + 22 + 32 + ◊◊◊ + n 2

(iii)

13 + 23 + 33 + ◊◊◊ + (n - 1)3

(iv)

1 + 3 + 5 + ◊◊◊ + 71

(v)

1 1 1 + 2 + 2 + ◊◊◊ 2 2 4 8

Solution n

(i)

Â 3 = 3n

(ii)

i =1 36

(iv)

Â (2i - 1)

(v)

i =1

n -1

i =1

Â i 2 (iii) •

Â i3

Â 2 2i i =1

There is always more than one way to write a series in this notation. For example, (iv) 35

could be written as

Â (2 j + 1) .

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

j =0

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

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FAC-04: Algebra

Question 4.15

Write the following using the S notation: (i)

2 + 5 + 10 + ◊◊◊ + 226

(ii)

2 + 4 + 8 + ◊◊◊ + 22n

(iii)

1 3

(iv)

-4 + 9 - 14 + ◊◊◊ + 99

+ 15 + 71 + ◊◊◊ +

1 2 n +1

Example

Write the following using the P notation: (i)

3 ¥ 5 ¥ 7 ¥ ◊◊◊ ¥ (2n + 1)

(ii)

e ¥ e 2 ¥ e3 ¥ ◊◊◊ ¥ e2 n + 3

(iii)

1 1 1 1 ¥ ¥ ¥ ◊◊◊ ¥ 2 3 4a 9a 16a 256a 15

Solution n

(i)

’ (2i + 1) i =1

(ii)

2n +3

’ ei i =1

(iii)

’ (i + 1)2a i i =1

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Question 4.16

Write this expression using the P notation: 1

1

    t  t  t  t        1   1    1    1   1   2   3   n 

1

Example

Simplify the expression: n

Â (1 + 4i + i 2 ) i =1

Solution n

i =1

Â (1 + 4i + i 2 ) = Â1 + Â 4i + Â i 2 = n + 4Â i + Â i 2 This can be further simplified, but we will deal with this later.

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FAC-04: Algebra

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: a + (n - 1)d

Suppose, we want to sum the first 10 terms of this AP: S10 = 3 + 7 + 11 + 15 + 19 + 23 + 27 + 31 + 35 + 39

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 for the last term then the sum of n terms will be

1 n of these pairs. Using l to stand 2 1 n( a + l ) . However, the last term in 2

this sum of n terms is the n th term and so equals a + (n - 1)d . So our sum of n terms is: Sn = 12 n [2a + (n - 1)d ]

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

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

n 2

 2a  (n  1)d   522  2  51  2   £2,704 .

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?

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7.2

FAC-04: Algebra

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 3 ¥ 26 = 192 and in general using a to stand for the first term the n th term is:

ar n -1

Suppose we want to sum the first 7 terms in this GP: S7 = 3 + 6 + 12 + 24 + 48 + 96 + 192

(1)

A clever trick is to look at what happens if we multiply this sum by the common ratio: 2S7 = 6 + 12 + 24 + 48 + 96 + 192 + 384

(2)

Subtracting these two series we get: (2) - (1) ﬁ (2 - 1) S7 = 384 - 3 = 381 In general, when we subtract we will have (r - 1) S n = ar n - a . So the sum is: Sn =

a(r n - 1) r -1

a(1 - r n ) . This Or multiplying top and bottom by -1 we could write this as Sn = 1- r formula is useful when the common ratio r lies between -1 and 1.

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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 a = 1, r = 2 and general term of ar n -1 . A chessboard has 64 squares, so the last square will need 1 ¥ 263 pence, or (to 1 SF) £90,000,000,000,000,000!

Question 4.18 10

Evaluate

Â 1.06k .

k =1

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7.3

FAC-04: Algebra

Convergence An infinite geometric progression is a GP which does not end such as

1 , 1 , 1 , . 2 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 + 12 + 14 + 18 + ◊◊◊

The first GP has r = 2 , 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 r = 12 , so the terms get smaller and smaller and the sum does converge to a finite limit: S1 = 1 S2 = 1.5 S3 = 1.75 S4 = 1.875 S5 = 1.9375 S6 = 1.96875 S7 = 1.984375 S8 = 1.9921875 S9 = 1.99609375 S10 = 1.998046875 

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

S• =

a 1- r

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

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Example

Write 1.231231231… as a fraction. Solution 231 231 231 1.231231231... =1+ 1,000 + 1,000,000 + 1,000,000,000 + ◊◊◊

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

1 1,000

1.231231231...

231 1,000 =1+ 1 1 - 1,000

231 231 77 = 1 + 999 = 1 999 = 1 333 s

Question 4.19

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

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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: Sn = 16 n(n - 4)(n - 5) ( n = 1, 2,3,ď &#x2039; ) 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 un , the n th term

(v)

the smallest term in the sequence.

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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. n

Â k = 12 n(n + 1)

Â k 2 = 16 n(n + 1)(2n + 1)

and

k =1

These can be used to simplify more complicated expressions. Example n

Â (1 + 4i + i 2 ) .

Simplify the expression

i =1

Solution n

i =1

Â (1 + 4i + i 2 ) = Â1 + Â 4i + Â i 2 n

i =1

= n + 4Â i + Â i 2 = n + 42 n(n + 1) + 16 n(n + 1)(2n + 1) = 16 n (6 + 12(n + 1) + (n + 1)(2n + 1) )

( n ( 2n

)

= 16 n 6 + 12n + 12 + 2n 2 + 3n + 1 =

1 6

+ 15n + 19

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FAC-04: Algebra

Example

Sum the series 12 + 22 + 32 + ◊◊◊ + (2n + 1) 2 . Solution

12 + 22 + 32 + ◊◊◊ + (2n + 1) 2 =

2 n +1

Â k2

k =1

= 16 (2n + 1) ((2n + 1) + 1)( 2(2n + 1) + 1) = 16 (2n + 1)(2n + 2)(4n + 3) = 13 (2n + 1)(n + 1)(4n + 3)

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

Â Â x =1 y =1

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

Â Â = x =1 y =1

10 10

Â Â y =1 x = y

We have swapped the order of summation.

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Page 33

Example

(i)

(ii)

Calculate

y =0

x =0

Â 2 y Â x , where 0 £ x £ y £ 5 .

Reverse the order of the summation and re-calculate the value. n

You are given that

Â k 3 = 14 n2 (n + 1)2 .

k =1

Solution

(i)

Using the formula detailed in the last section: 5

y 0

x 0

 2y x 

1   2 y  2 y( y  1)   y 0

 ( y3  y 2 )

y 0

This can be simplified using the results for the sum of squares and the sum of cubes: 5

Â (y

+y )=

y =0

(ii)

Ây

y =0

Â y 2 = 4 (5)2 (6)2 + 6 (5)(6)(11) = 280

y =0

To change the order of the summation, we consider the inequality 0 £ x £ y £ 5 . Previously we summed from x = 0 to x = y , then from y = 0 to y = 5 . Reversing this, we sum from y = x to y = 5 then from x = 0 to x = 5 . To get the limits for y , see what it is bounded by in the inequality 0 £ x £ y £ 5 . So: 5

y =0

x =0

Â 2yÂ x = Â x Â 2y x =0 y = x

Using the formula detailed in the last section: 5

 x  2y 

x 0 y  x

x 1   1   5  5  1    2 x y y          x 2  2 (5)(6)  2 ( x  1)( x)  x 0  y 0    y 0  x 0 5

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FAC-04: Algebra

Simplifying this: 5

x=0

x =0

Â x Â 2 y = Â x(30 - x( x - 1)) = 30 Â x - Â x

x =0 y = x

Â x2

x=0

Again, using the standard summations: 5

5 1 1 1 x   2 y  30  2  (5)(6)  4 (5)2 (6)2  6 (5)(6)(11)  280 x 0 y  x

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

Question 4.21

(i)

If 0 £ t £ s £ 10 , calculate

(ii)

Â1.04 Â1.

t =0

-t

s =t

Swap the order of summation to re-calculate the answer. t 1  1  1.04b b    1.04t 0.04  1  1.041  1.04b  . t 0   b

You may use the result

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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:

n n n n  n  n 1  n  n (a  b) n    a n    a n 1b    a n  2b 2        a n  r b r        ab    b 0 1  2 r  n  1 n n n n! , which appears as nCr on most calculators. The   Note that     r  r !(n  r )! r expression comes in because there are many different ways of getting the term in a nrbr . This can be used to expand expressions of the form (a  b) n and the result is given on page 2 of the Tables. If we rewrite the combinatorial function as follows: n! n ¥ (n - 1) ¥  ¥ (n - r + 1)(n - r )! n ¥ (n - 1) ¥  ¥ (n - r + 1) = = r !(n - r )! r !(n - r )! r! Then we can rewrite the binomial expansion as: a n + na n -1b +

n(n - 1) n - 2 2 n(n - 1)(n - 2) ◊◊◊ (n - r + 1) n - r r a b + ◊◊◊ + a b + ◊◊◊ + nab n -1 + b n 2! r!

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 1

1 2

3 4

1 3

1 4

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

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FAC-04: Algebra

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

4 ¥ 3 4 ¥ 3 ¥ 2 4 ¥ 3 ¥ 2 ¥1 , , . 2! 3! 4!

Example

Expand (2 - 3 x)7 as far as the term in x3 . Solution

(2 - 3 x)7 = 27 + 7 ¥ 26 ¥ ( -3x)1 +

7¥6 7¥6¥5 ¥ 25 ¥ ( -3x) 2 + ¥ 24 ¥ ( -3x)3 + ◊◊◊ 2! 3!

= 128 - 1,344 x + 6,048 x 2 - 15,120 x3 + 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 a 4 in the expansion of (2a + 5b)6 . Solution

The general term is 6Cr (2a)6 - r (5b) r , and we want a 4 , so here r = 2 . The coefficient is 6C2 ¥ 24 ¥ (5b) 2 = 15 ¥ 16 ¥ 25 ¥ b 2 = 6000b 2 .

Question 4.22

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

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9.2

Page 37

Fractional or negative powers Very often in practice, the power on the bracket will not be a positive integer, so alternatively we can use: (1 + x) n = 1 + nx +

n(n - 1) 2 n(n - 1)(n - 2) 3 x + x + ◊◊◊ 2! 3!

where n is negative or fractional, and -1 < x £ 1 . This series is derived using Maclaurin series which is covered in Chapter 7. Example

Expand

(1 - 2 x) as far as the term in x3 . For what values of x is this expansion

valid? Solution 1 2

(1 - 2 x) = (1 - 2 x) = 1 +

1 (- 1 ) 1 ( -2 x ) + 2 2 ( -2 x ) 2 2 2!

1 ( - 1 )( - 3 ) 2 2 2 ( -2 x )3

+ ◊◊◊

= 1 - x - 12 x 2 - 12 x3 + ◊◊◊

This expansion is valid for 1  2 x  1 , or  12  x  12 .

Question 4.23 1

Expand (1 + x) 2 and hence approximate the following (without using a calculator!): 1 2

1 4

1.04 , 1.10 , 0.99

- 12

a2 - b2 , , 10200 , 1 a-b 1.07 2 1.09 2

where a = 1.11 and b = 1.10 .

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FAC-04: Algebra

Example

Expand

1+ x as far as the term in x3 . For what values of x is this expansion valid? 2 + 3x

Solution

We start by manipulating the expression, since we want a binomial expansion which starts with a 1: 1 x  (1  x)(2  3 x) 1  21 (1  x)(1  32 x) 1 2  3x (1)(2)   21 (1  x) 1  32 x  2! 

 32 x 





x3      21 (1  x) 1  32 x  94 x 2  27 8 

1 2

1 

3 2

(1)(2)(3) 3!

 32 x 

     



x  94 x 2  27 x3  x  32 x 2  94 x3     8



9 x3   12  14 x  83 x 2  16

This is valid for -1 < 32 x £ 1 , ie - 23 < x £ 23 .

Question 4.24

Expand

1 as far as the term in x3 . For what values of x is this expansion (1 - 4 x)

valid? Use your expansion to estimate 0.98

- 12

to 6DP.

Question 4.25 

n  k  x  1 k x n , and  p q S  x   x  x  p x q n  x , given that p + q = 1 and 2  x 0  x 0   k is a positive integer.

Simplify S1 

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FAC-04: Algebra

Page 39

Chapter 4 Summary Indices

The three laws of indices (powers) are:



x a ¥ xb = x a + b



x a xb = x a - b



( x a )b = x ab

Logarithms

Logarithms are the inverse of powers, eg log x b = a is equivalent to x a = b . The three laws of logarithms are:



log a x + log a y = log a xy



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



log a x n = n log a x

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



division – multiply by the inverse, eg

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a c ac ¥ = b d bd

a c a d ad ∏ = ¥ = b d b c bc

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FAC-04: Algebra

Quadratic equations

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



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



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



the quadratic formula: x =

-b ± b 2 - 4ac . 2a

Simultaneous equations

Linear simultaneous equations can be solved by elimination or substitution: eg

3 x + 5 y = 29 x + 2 y = 11

(1) (2)

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

a1 an £

a1 +  + an 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: a + (n - 1)d

sum: Sn =

n 2

[2a + (n - 1)d ]

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Page 41

Geometric Progression (GP)

A series with first term a and common ratio r has: general term: ar n -1

sum: Sn =

a(1 - r n ) 1- r

S• =

a 1- r

Standard summations

Â k = 12 n(n + 1) Â k 2 = 16 n(n + 1)(2n + 1) Â k 3 = 14 n2 (n + 1)2 Binomial expansion

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

Ê nˆ Ê nˆ (a + b) n = a n + Á ˜ a n -1b + Á ˜ a n - 2b 2 +  + b n Ë 1¯ Ë 2¯ (1 + x) p = 1 + px +

n Œ+

p( p - 1) 2 p( p - 1)( p - 2) 3 x + x + 2! 3!

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-1 < x <1

Page 42

FAC-04: Algebra

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

ÂŠ IFE: 2014 Examinations

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FAC-04: Algebra

Page 43

Chapter 4 Solutions Solution 4.1

(i)

10x8

(ii)

8 3y 5

(ii)

25b6

Solution 4.2

Let m = log a x ¤ a m = x and let n = log a y ¤ a n = y . Now consider a m - n , using the second law of indices, this can be simplified as follows: am-n = am ∏ an = x ∏ y

Using the definition of logs: am-n =

x x ¤ log a = m - n = log a x - log a y y y

and hence we have proved the result.

Solution 4.3 ln e = 1

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FAC-04: Algebra

Solution 4.4

2 ¥ 1.05t - 2 = 1.10252t -1 Taking logs: log 2 ¥ 1.05t - 2 = log1.10252t -1 log 2 + (t - 2) log1.05 = (2t - 1)log1.1025

Gathering up the terms: log 2 - 2log1.05 + log1.1025 = t (2log1.1025 - log1.05) t=

log 2 - 2log1.05 + log1.1025 2log1.1025 - log1.05

= 4.736

Solution 4.5 x 2 + 3x - 4 2 x2 + 5x - 7 ( x + 4)( x - 1) ( x - 4)(4 x - 1) ∏ = ¥ 2 2 3 x - 10 x - 8 4 x - 17 x + 4 (3 x + 2)( x - 4) ( x - 1)(2 x + 7) =

( x + 4)(4 x - 1) (3 x + 2)(2 x + 7)

Solution 4.6

(i)

(2 x + 1)(3x - 2) = 0 so x = - 12 or

(ii)

(4 x + 1)(4 x - 1) = 0 so x = ± 14

2 3

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FAC-04: Algebra

Page 45

Solution 4.7 (i)

( x + 2) 2 - 12 = 0 x + 2 = ± 12 x = -2 ± 12 = 1.464 or - 5.464

(ii)

(4 x + 17 )2 8

289 64

+2=0

(4 x + 17 )=± 8 x=

1 4

161 64

17 8

161 64

)

= -0.135 or - 0.928

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FAC-04: Algebra

Solution 4.8 2

2  x  1   x   2  x               1    2 

Comparing the coefficients of x 2 and x: 1

s 12

s 22

a s2

and

2 m1

s 12

2 m2

s 22

2 ma

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

m1 m2 + s 12 s 22 1

s 12 So m =

s 22

m1s 22 + m2s 12 . s 22 + s 12

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

s 12 + s 22 a = 2 s 12s 22 s Rearranging this equation gives us:

s2 =

as 12s 22 as 12s 22 ﬁ = s s 12 + s 22 s 12 + s 22

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FAC-04: Algebra

Page 47

Solution 4.9

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

4 ± 16 + 28 = 0.76 or - 0.19 14

Solution 4.10

The equation can be written as 3(3x ) 2 + 4(3x ) - 4 = 0 , which is a quadratic in 3x . If y = 3x , then the equation can be written as 3 y 2 + 4 y - 4 = 0 , so that:

(3 y - 2)( y + 2) = 0 y=

2 3

or - 2

ﬁ 3x =

2 3

or - 2

so: x=

ln 23 ln 3

= -0.369

Note that it is impossible to raise a positive number to any power and get a negative number. So the alternative equation 3x = -2 has no solutions, and x = -0.369 is the only solution to the equation.

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FAC-04: Algebra

Solution 4.11

(i)

ab 4 (1 + b 2 ) = b 2 = 4 ie b = ±2 . Dividing the second equation by the first 2 2 ab (1 + b ) Substituting this back into the first equation: 4a + 16a = 5

1 4

So the two solutions are a = (ii)

1 4

and b = 2 or a =

1 4

and b = -2 .

Factorising and dividing the second equation by the first: (2 x - y )( x + 3 y ) = -5 (2 x - y )( x + y ) = 5 x+ y = -1 x + 3y x + y = - x - 3 y fi x = -2 y Substituting this into the first factorised equation: ( -5 y )( y ) = -5 fi y = ±1 fi x = 2

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Solution 4.12

Since e

m + 12 s 2

squared is e 2 m +s , we can substitute the first equation into the second:

9(es - 1) = 11.2 Rearranging this, we get: 2

es = 2.244

Which can be rearranged to give:

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

Solution 4.13 2 x3 + 5 x 2 - 17 x - 20 < 0 ﬁ (2 x - 5)( x + 4)( x + 1) < 0

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

x < -4 – – – –

-4 < x < -1 + – – +

-1 < x < 2.5 + + – –

x > 2.5 + + + +

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

Solution 4.14

The arithmetic and geometric means are equal when the two numbers are equal. This can be proved as follows: a+b ( a + b) 2 = ab ﬁ = ab ﬁ (a + b) 2 = 4ab 2 4 This simplifies to (a - b) 2 = 0 , ie a = b .

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FAC-04: Algebra

Solution 4.15 15

(i)

Â (k 2 + 1)

k =1 2n

(ii)

Â 2k

k =1 n

(iii)

Â 2k + 1

k =1 20

(iv)

Â (-1)k (5k - 1)

k =1

Solution 4.16 n

 t   1    k  k 1 

1

Solution 4.17

If the number of payments is n, then, using the formula for the general term of an arithmetic progression: 18 = 298 + (n - 1) ¥ -28 n = 11

So, the total amount of the payments can be found by using the sum of an arithmetic progression, ie: 11 2

(2 ¥ 298 + (11 - 1) ¥ -28) = £1,738

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Page 51

Solution 4.18

If we write out the terms of the series it will make things clearer: 10

Â 1.06k

k =1

5 5 5 5 + + +  + 1.06 1.062 1.063 1.0610

We see that we have a geometric progression with a =

5 1 and r = . So the sum 1.06 1.06

of the first 10 terms of this GP would be:

S10 =

5 È Ê 1 ˆ Í1 - Á ˜ 1.06 ÍÎ Ë 1.06 ¯ 1-

10 ˘

1 1.06

˙ ˙˚ = 36.800

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: S• =

4 = 20 1 - 0.8

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Solution 4.20

(i)

The sum of the first 10 terms is found by putting n = 10 into the formula for Sn : S10 = 16 ¥ 10 ¥ 6 ¥ 5 = 50

(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: S20 = 16 ¥ 20 ¥ 16 ¥ 15 = 800 The sum of the 11th to the 20th terms is just the difference between these two, ie: S20 - S10 = 800 - 50 = 750

(iii)

The 20th term can be calculated as the difference between S19 and S20 , ie: S20 - S19 = 800 - 665 = 135

(iv)

Similarly, the n th term is: un = Sn - Sn -1 = 16 n(n - 4)(n - 5) - 16 (n - 1)(n - 5)(n - 6) = 16 (n - 5)[n(n - 4) - (n - 1)(n - 6)] = 16 (n - 5)[3n - 6] = 12 (n - 5)(n - 2)

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(v)

Page 53

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: un - un -1 ≥ 0 ie

1 ( n - 5)( n - 2) - 1 ( n - 6)( n - 3) 2 2

≥0

This simplifies to: n - 4 ≥ 0 ie n ≥ 4

So a given term will be at least as big as the previous term whenever n ≥ 4 (with equality only when n = 4 ). This means that: u5 ≥ u4 and u4 = u3 but u3 ≥/ u2 Combining these, we get: u2 > u3 = u4 < u5 So u3 and u4 are (jointly) the smallest terms. They each take the value -1 .

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Solution 4.21

(i)

We have: 10

Â1 = 11 - t s =t

So the summation is: 10

t =0

Â1.04-t (11 - t ) = 11Â1.04-t - Â t ¥ 1.04-t

The first series is a geometric progression and the second series can be summed using the formula given in the question:

 1  1.0411  1  1  1.0410 10  t t      58.23 1.04 (11 ) 11     1 1 10     1  1.04  0.04  1  1.04  1.04 t 0     10

(ii)

Swapping the order of the summation: 10

Â Â1.04-t

s =0 t =0

The sum over the variable t can be carried out using the formula for the sum of a geometric progression: 10

Â Â1.04-t =

s =0 t =0

1 - 1.04- ( s +1) 10 1.04 - 1.04- s Â 1 - 1.04-1 = Â 0.04 s =0 s =0 10

Simplifying this and using the formula for the sum of a geometric progression: 1.04 10 1 10 11 ¥ 1.04 1 1 - 1.04-11 -s Â1 Â 1.04 = 0.04 - 0.04 1 - 1.04-1 = 58.23 0.04 s = 0 0.04 s = 0

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Solution 4.22

(1 + 2 x)8 = (1)8 + 8(1)7 (2 x) +

8¥7 6 8¥7¥6 5 8¥7¥6¥5 4 (1) (2 x) 2 + (1) (2 x)3 + (1) (2 x) 4 + 2! 3! 4!

= 1 + 16 x + 112 x 2 + 448 x3 + 1,120 x 4 +

Solution 4.23

The binomial expansion tells us that: (1 + x)½ ª 1 + 12 x 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: 1.04½ ª 1 + 12 ¥ 0.04 = 1.02 1.10¼ is the 4th root of 1.10 , which is the same as square rooting twice. So we get: 1.10¼ = (1.10½ )½ ª 1.05½ ª 1.025

The same method works with a number a bit less than one and/or a negative power: 0.99 -½ = (1 - 0.01) -½ ª 1 - 12 ¥ ( -0.01) = 1.005 The next one gives: 1.09½ 1.045 ª 1.07½ 1.035

To approximate this ratio, we can subtract 0.035 from the top and bottom (which doesn’t affect the answer very much): 1.09½ 1.045 - 0.035 1.01 ª = = 1.01 1 1.07½ 1.035 - 0.035

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

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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 ª 100 ¥ 1.01 = 101 And finally: a½ - b½ 1.055 - 1.05 0.005 ª = = 0.5 a-b 1.11 - 1.10 0.01 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 (1 - 4 x)

- 12

= 1 + 2x +

- 12 ¥ - 32 2!

( -4 x ) 2 +

- 12 ¥ - 32 ¥ - 52 3!

( -4 x)3 +

= 1 + 2 x + 6 x 2 + 20 x3 +

This is valid for -

1 1 £x< . 4 4

If we now substitute in x = 0.005 , we get: 0.98

- 12

= 1 + 0.01 + 0.00015 + 0.0000025 = 1.0101525

So to 6DP 0.98

- 12

= 1.010153 .

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Solution 4.25

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

 k  1 2  k  2  3 S1  k  1  k     q      q    q  pk  0   1   2   3   1  kq 

(k  1)k 2 (k  2)(k  1)k 3 q  q  2 6

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: S1 ( - k )( - k - 1) ( - k )( - k - 1)( - k - 2) = 1 + ( - k )( - q ) + ( - q)2 + ( - q )3 +  k 2 6 p = [1 + ( - q )]- k = p - k

So: S1 = 1 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: S2 

n

x 0

 

 x  x  p x q n  x  0   1  pq n 1  2  2  p 2q n  2   n  n  p n

If we take a typical coefficient in this series, we get for example: n  n  1 n(n  1)(n  2) (n  1)(n  2)  n  n 3   3   3  2 1 2 1  3  2 

So the series is:  n  1 n 1  n  1 2 n  2  n  1 n   n  S2  n   pq  n  p q p  0   1   n  1

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Dividing through by the n and a p factor: S2 np

 n  1 n 1  n  1 n  2  n  1 n 1     q    pq p  0   1   n  1

The RHS is just the binomial expansion of ( p  q ) n 1 , which equals 1 (since p  q  1 ). So: S2  1  S2  np np This is actually the calculation of the mean value of a binomial distribution.

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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.

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

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FAC-05: Numerical methods II

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:

p ¥ quantity 100 Example

Find 7.5% of $2,500. Solution

7.5 ¥ $2,500 = 0.075 ¥ $2,500 = $187.50 100 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 = £15.94 100

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FAC-05: Numerical methods II

(ii)

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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 = £14.96 100

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 ¥ x = 1.20 ¥ x = £60 ﬁ x = = £50 100 1.20

(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 ¥ x = 0.80 ¥ x = £60 ﬁ x = = £75 100 0.8

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FAC-05: Numerical methods II

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: A ¥ 100% 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.

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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 â&#x20AC;&#x201C;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: proportionate change =

absolute change original value

Proportionate change is also sometimes called relative change.

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FAC-05: Numerical methods II

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. percentage change =

absolute 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. proportionate and percentage changes in his wage?

What are the absolute,

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Page 7

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 proportionate error =

percentage error =

absolute error true value

absolute error ¥ 100% true value

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

1 , the true value of i is 0.0372534. Calculate the absolute, 1+ i proportionate, and percentage error that will be introduced in the value of v by rounding i to two decimal places. When calculating v =

Solution

The true value of v is gives v to be

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

1 . 1.04

Absolute error =

1 1 = -0.0025 (4 DP). 1.04 1.0372534

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

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FAC-05: Numerical methods II

Question 5.4

In an actuarial calculation, a student uses i = 0.027 having rounded the precise value of i to 2 significant figures. What is the largest percentage error that could have been 1 made in the calculated value of v = ? 1+ i

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Page 9

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 x =

1 n Â x . “Length of life” is measured in time, and n is n i =1 i

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

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FAC-05: Numerical methods II

Question 5.5

If the variance is defined to be

1 n 2 xi - x ) , what is the dimension of variance for ( Â n - 1 i =1

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)

1 n 2 Âx - x2 n - 1 i =1 i n

(ii)

Â xi2 - (n - 1) x i =1

(iii)

n 1 n 2 1 x x Â Âx n - 1 i =1 i n - 1 i =1 i

The argument of functions such as e x and log x must be dimensionless. Question 5.7

Why must they be dimensionless?

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Page 11

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.

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FAC-05: Numerical methods II

Interpolation and extrapolation Sometimes it is not possible to find an exact solution to an equation such as f ( x) = 0 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 x1 and x2 . 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 x1 and x2 . 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 x1 and x2 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 x1 and x2 .

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: y

x2 0

error

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This shows the graph of y = f ( x) , with the close values x1 and x2 , and the error that this interpolation has introduced. Do be aware however that we have drawn a largescale 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: f(x1)

f(x)

f(x2)

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

x - x1 f ( x) - f ( x1 ) x - x1 x2 - x1 = = or f ( x) - f ( x1 ) f ( x2 ) - f ( x1 ) x2 - x1 f ( x2 ) - f ( x1 ) Consider the first version. This can be rearranged to give: x = x1 +

f ( x) - f ( x1 ) (x - x ) f ( x2 ) - f ( x1 ) 2 1

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 x1 and x2 , ie we are dealing with extrapolation, derive a formula for finding x. To see how this works in practice, look at the next example.

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FAC-05: Numerical methods II

Example

If f (1.34) = 0.523 , and f (1.65) = -0.42 , what is an approximate value for x if f ( x) = 0 ? Solution

We know that x lies somewhere between 1.34 and 1.65, so: x - 1.65 1.65 - 1.34 = 0 - ( -0.42) -0.42 - 0.523 which gives x = 1.51 .

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 x3 - 5 x 2 + 3x + 7 = 0 . 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 0.54 0.55 0.56

F ( x) 0.70540 0.70884 0.71226

Using linear interpolation, work out: (i)

the value of F ( x) when x = 0.548

(ii)

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

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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 x3 - 12 x - 7 = 0 . Give your answer to 1 DP. Solution

Let f ( x) = x3 - 12 x - 7 , so we are trying to solve f ( x) = 0 . f (0) = -7 and f ( -1) = 4 fi the root lies between 0 and - 1 f ( -0.5) = -1.125 fi the root lies between - 0.5 and - 1 f ( - 0.75)=1.578125 fi the root lies between - 0.5 and - 0.75 f ( -0.625) = 0.255859 fi the root lies between - 0.5 and - 0.625 f ( -0.5625) = -0.42798 fi the root lies between - 0.5625 and - 0.625 Therefore the answer is –0.6.

Question 5.11

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

25(1 + i )

-3

(

20 1 - (1 + i ) -2 i

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) = 61.5

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6.2

FAC-05: Numerical methods II

Newton-Raphson iteration The Newton-Raphson iterative formula states that if xn is an approximate solution to the equation f ( x) = 0 then a better approximation is xn +1 = xn -

f ( xn ) . f ¢( xn )

Example

Using the Newton-Raphson formula, find an approximate positive solution to the equation x5 + 4 x 2 = 7 , giving your answer to 2 DP. Solution

Let f ( x) = x5 + 4 x 2 - 7 , so that: f ( x)  5 x 4  8 x f (1)  2 f (2)  41

Therefore there is a solution between 1 and 2. Try x1 = 1 . x2 = 1.15385 x3 = 1.13336 x4 = 1.13290 Therefore the solution is 1.13 (to 2DP).

Question 5.12

Using the Newton-Raphson formula, find an approximate solution to the equation: 5(1 + i )

-4

10(1 - (1 + i ) -5 ) + = 41 i

Give your answer to 3 DP.

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Complex numbers

7.1

Basic algebra

Page 17

A complex number is represented in the form a + ib , where i = -1 , 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 i 2 = -1 . Question 5.13

Find: (i)

(2 + 3i ) + (3 - 4i )

(ii)

(4 - 6i ) - (3i - 7)

(iii)

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

(iv)

(4 + 2i )(4 - 2i )

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

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

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FAC-05: Numerical methods II

Example

Simplify

2 - 3i . 1 + 2i

Solution

If we multiply the numerator and denominator by the complex conjugate of the denominator, we get: 2 - 3i 1 - 2i 2 - 3i - 4i - 6 -4 - 7i 1 ¥ = = = ( -4 - 7i ) 1 + 2i 1 - 2i 1+ 4 5 5 This is in the form a + ib , where a = -0.8 and b = -1.4 .

Question 5.14

Simplify

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 z = 1 - 3i can be shown as follows: Im 3 2 1

-3

-2

-1

-1 -2 -3

7.2

3i - 4 . 1 - 4i

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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 r = a 2 + b 2 , 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 z1 and z2 , with moduli r1 and r2 , and arguments q1 and q 2 respectively. Im

z1 r1

1

Re 2 r2 z2

Note that in the diagram, q 2 is negative.

By convention, angles measured

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

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FAC-05: Numerical methods II

Example

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

The modulus is 12 + 32 = 10 = 3.16 .   1  The argument is     tan 1     2.82 .  3  

Question 5.15

Find the modulus and argument of the complex number

7.3

1 - 2i . 4 - 3i

Euler’s formula As we have seen, complex numbers can be represented as points on an Argand diagram. The number eiq 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. Im

i

e 1

 x

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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 adjacent x ie x = cosq . , so that cosq = trigonometry, we know that cosine = hypotenuse 1 Similarly y = sin q . So eiq has real part cosq and imaginary part sin q . In other words: eiq = cosq + i sin 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 exp

ip ip and 2exp on the Argand diagram? 4 6

Note that any complex number can be expressed in the form reiq , 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 eiq . The complex conjugate can be found by replacing i by -i in the number, so here it is e -iq . Since eiq = cosq + i sin q , we can replace i by -i to give e - iq = cosq - i sin q . Im

i

e 



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– i

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FAC-05: Numerical methods II

Question 5.18

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

Question 5.19

Show that the product of e -iq and eiq does give a real number. In many applied maths applications of complex numbers you will find that you need to add e -iq and eiq . This gives: e - iq + eiq = cosq - i sin q + cosq + i sin q = 2cosq Note that this also gives a purely real number. Question 5.20

Show that (1 + 12 eiw )(1 + 12 e - iw ) =

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5 4

+ cos w .

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7.4

Page 23

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 z 2 - 3 z + 4 = 0 . Solution

Using the quadratic formula, we get: z=

3 ± 9 - 16 3 ± -7 3 ± ( -1) 7 3 ± i 7 = = = 2 2 2 2

Simplifying this we get: z=

3 2

7 2

i or

3 2

7 2

Notice that these roots are a complex conjugate pair.

Question 5.21

You are given that z = 1 + 2i is a root of the cubic equation z 3 - 4 z 2 + 9 z - 10 = 0 . Find the other two roots.

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FAC-05: Numerical methods II

Difference equations A difference equation is typically of the form ayt + byt -1 + cyt - 2 = 0 , where a , b and c are constants. The equation has a solution of the form yt = f (t ) and it has an

associated auxiliary equation al 2 + bl + c = 0 . The general solution of the difference equation depends on the roots, l1 and l2 , of the auxiliary equation. b 2 - 4ac

Solution of difference equation

yt = Al1t + Bl2t

yt = ( A + Bt )l

yt = r t ( A cosq t + B sin q t )

Note The auxiliary equation has two distinct real roots l1 π l2 There is only one (repeated) root l1 = l2 = l The complex roots of the auxiliary equation are written in polar form ie

l1 = re iq , l2 = re -iq 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)

yt - 5 yt -1 + 6 yt - 2 = 0 given that y0 = 0 and y1 = 1

(ii)

yt + 2 - 6 yt +1 + 13 yt = 0 .

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Solution

(i)

The auxiliary equation is l 2 - 5l + 6 = 0 which has roots 2 and 3. The general solution is of the form yt = A ¥ 2t + B ¥ 3t . However we know that y0 = 0 and y1 = 1 , so: y0 = 0 = A ¥ 20 + B ¥ 30 ﬁ 0 = A + B y1 = 1 = A ¥ 21 + B ¥ 31 ﬁ 1 = 2 A + 3B These can be solved to give B = 1 and A = -1 , so our particular solution is: yt = ( -1) ¥ 2t + 1 ¥ 3t = 3t - 2t

(ii)

The

auxiliary

equation

l 2 - 6l + 13 = 0 ,

which

has

roots

6 ± 36 - 52 = 3 ± 2i . 2 The general solution is of the form yt = r t ( A cosq t + B sin q t ) . The modulus of these complex numbers is r = 32 + 22 = 13 , and the argument is q = tan -1 23 = 0.588 , so the roots can be written in polar form as 13 e ±0.588i .

The general solution of the difference equation is then

( )

yt = A 13 cos 0.588t + B

( 13 ) sin 0.588t . t

Question 5.22

Solve the difference equations: (i)

yt - 8 yt -1 + 16 yt - 2 = 0 given that y0 = -1 and y1 = 3

(ii)

yt - 4 yt -1 + 5 yt - 2 = 0 .

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

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Page 27

Chapter 5 Summary Percentages

To find p % of a quantity:

p ¥ quantity 100

To find a p % increase or decrease of a quantity: 100 + p ¥ quantity 100 To express A as a percentage of B :

100 - p ¥ quantity 100 A ¥ 100% B

Changes

absolute change = new value - original value absolute change proportionate change = original value absolute change percentage change = ¥ 100% original value Errors

absolute error = approx value - true value absolute error proportionate error = true 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.

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FAC-05: Numerical methods II

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 x - x1 f ( x) - f ( x1 ) = x2 - x1 f ( x2 ) - f ( x1 )

ﬁ x = x1 +

f ( x) - f ( x1 ) (x - x ) f ( x2 ) - f ( x1 ) 2 1

Bisection If solution x lies between x1 and x2 then use ½( x1 + x2 ) as next approximation. Newton-Raphson If xn is approximate solution to f ( x) = 0 then a better approximation is obtained by: xn +1 = xn -

f ( xn ) . f ¢( xn )

Complex numbers

Any complex number z can be written in the form z = a + ib , where i = -1 . The complex conjugate of z is a - ib . z = x + yi Complex numbers can be represented on an Argand diagram:

The modulus of z is given by r = x 2 + y 2 . The argument of z is given by q = tan -1 x . y



The complex number z can be written in polar form z = reiq . Euler’s formula states that eiq = cosq + i sin q . Complex roots of polynomial equations with real coefficients must occur in complex conjugate pairs. Second order difference equations

The general solution ayt + byt -1 + cyt - 2 = 0 depends on the roots (l1 and l2 ) of the auxiliary equation, al 2 + bl + c = 0 : See formulae on page 4 of the Tables.

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Chapter 5 Solutions Solution 5.1

(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 : x ¥ 1.05n = 2 x ﬁ 1.05n = 2

Taking logs of both sides and rearranging:

n ln1.05 = ln 2 ﬁ n =

ln 2 = 14.2067 ln1.05

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: x ¥ 1.052 = 66,150,000 ﬁ

66,150,000 = 60,000,000 1.052

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.

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If i = 0.0265 then the true calculation would have been v = 1.0265-1 compared to the approximate calculation of v = 1.027 -1 so this gives a percentage error of: 1.0265-1 - 1.027 -1 ¥ 100% = 0.048685% 1.0265-1

Similarly, if i = 0.0275 then the percentage error would be: 1.0275-1 - 1.027 -1 ¥ 100% = -0.048685% 1.0275-1

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 T 2 , 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 T 2 , while that of the second term is T, and terms of different dimensions cannot be subtracted.

Solution 5.7

x2 +  . If x had a dimension, then each 2! term of this series would have a different dimension and so this series would be meaningless. Therefore x must be dimensionless. Consider the series for e x which is 1 + x +

A similar reasoning applies to log x .

Solution 5.8

You use exactly the same formula for extrapolation as you do for interpolation!

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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

x - ( -1) 0 - ( -2) = ﬁ x = -0.778 . 0 - ( -1) 7 - ( -2)

Between 2 and 3, interpolating gives

x - 2 0 -1 = ﬁ x = 2.333 . 3 - 2 -2 - 1

Between 3 and 4, interpolating gives

x - 3 0 - ( -2) = ﬁ x = 3.4 . 4 - 3 3 - ( -2)

Solution 5.10

(i)

Using the formula for interpolation:

F (0.548) =

(0.70884 - 0.70540)(0.548 - 0.54) + 0.70540 0.55 - 0.54

= 0.70815 (ii)

Using the formula: x=

(0.56 - 0.55)(0.71 - 0.70884) + 0.55 0.71226 - 0.70884

= 0.553

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FAC-05: Numerical methods II

Solution 5.11

When i = 0.03 the LHS gives 61.15, and when i = 0.02 the LHS gives 62.39. Using the bisection method, we get: Value of equation 61.76 61.45 61.61 61.53

Value of i 0.025 0.0275 0.02625 0.026875

So the value of i is 0.027 to 3DP.

Solution 5.12

Let f (i ) = 5(1 + i ) -4 +

10(1 - (1 + i ) -5 ) - 41 , so that: i

f ¢(i ) = -20(1 + i )

-5

i ¥ 10 ¥ 5(1 + i ) -6 - 10(1 - (1 + i ) -5 ) + i2

= -20(1 + i ) -5 +

50i (1 + i ) -6 - 10 + 10(1 + i ) -5 i2

The Newton-Raphson formula gives the formula for the next approximation to be: 10(1 - (1 + i ) -5 ) - 41 i i50i (1 + i ) -6 - 10 + 10(1 + i ) -5 -20(1 + i ) -5 + i2 5(1 + i ) -4 +

f (0.1) = 0.323 and f (0.11) = -0.747 , so the root lies between 0.1 and 0.11. Try i1 = 0.1 . The formula gives i2 = 0.1029557 , i3 = 0.1029739 , ie the root is 0.103 to 3 DP.

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FAC-05: Numerical methods II

Page 33

Solution 5.13

(i)

(2 + 3i ) + (3 - 4i ) = 5 - i

(ii)

(4 - 6i ) - (3i - 7) = 11 - 9i

(iii)

(3 - 4i )(6 + 2i ) = 18 - 24i + 6i - 8i 2 = 26 - 18i

(iv)

(4 + 2i )(4 - 2i ) = 16 - 8i + 8i - 4i 2 = 20

Solution 5.14

Multiplying the numerator and denominator by the complex conjugate of the denominator: 3i - 4 1 + 4i 3i - 4 - 16i + 12i 2 -16 - 13i ¥ = = 1 - 4i 1 + 4i 17 1 - 16i 2

Solution 5.15

We first need to simplify the complex number: 1 - 2i 4 + 3i 4 - 8i + 3i - 6i 2 10 - 5i 2 - i ¥ = = = 4 - 3i 4 + 3i 16 + 9 25 5 2

The modulus is then

2 1       0.4472 .  5 5

The argument is - tan -1

1 = -0.46 . 2

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FAC-05: Numerical methods II

Solution 5.16

Using Euler’s formula exp

ip 1 1 so the co-ordinates are: = cos p4 + i sin p4 = +i 4 2 2

 1 1  ,    2 2

 3 i 1  2(cos 6  i sin 6 )  2   i  , so the co-ordinates are ( 3,1) . 6 2  2

Similarly 2exp

Solution 5.17

Drawing a diagram, we have:

r 5  7

Using Pythagoras’ Theorem the modulus is: r = 7 2 + 52 = 74

Using trigonometry, the argument is: tan q =

5 Ê 5ˆ ﬁ q = tan -1 Á ˜ = 0.62025 Ë 7¯ 7

Don’t forget that your calculator should be in radians!

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FAC-05: Numerical methods II

Page 35

Solution 5.18

We could work out e -iq by thinking of it as ei ( -q ) and applying the original form of Euler’s formula: ei ( -q ) = cos( -q ) + i sin( -q )

Comparing this with the other version for e -iq , we see that cos( -q ) must equal cosq and sin( -q ) must equal - sin q . 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: e - iq eiq = (cosq - i sin q )(cosq + i sin q ) = cos 2 q - i sin q cosq + i sin q cosq - i 2 sin 2 q = cos 2 q + sin 2 q = 1

which is a real number. You could, of course, have used the basic properties of powers to say that e - iq eiq = e -iq + iq = e0 = 1 .

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FAC-05: Numerical methods II

Solution 5.20

Multiplying out gives: (1 + 12 eiw )(1 + 12 e - iw ) = 1 + 12 eiw + 12 e -iw + 12 eiw ¥ 12 e -iw = 1 + 12 (eiw + e - iw ) + 14 =

5 4

+ 12 ¥ 2cos w

5 4

+ cos 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:

( z - (1 + 2i ))( z - (1 - 2i)) and some other factor which is, as yet, unknown. Multiplying these factors out we get: ( z - 1 - 2i )( z - 1 + 2i ) = ( z - 1) 2 - 4i 2 = z 2 - 2 z + 5

The cubic can then be written as: z 3 - 4 z 2 + 9 z - 10 = ( z 2 - 2 z + 5)( z - 2)

Therefore the other two roots are 2 and 1 - 2i .

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Page 37

Solution 5.22

(i)

The auxiliary equation is l 2 - 8l + 16 = 0 , which has a repeated root of 4. The general solution is given by yt = ( A + Bt ) ¥ 4t . However we know that y0 = -1 and y1 = 3 , so we have the simultaneous equations: y0 = -1 = ( A + B ¥ 0) ¥ 40 ﬁ - 1 = A y1 = 3 = ( A + B ¥ 1) ¥ 41

ﬁ 3 = 4( A + B)

which are solved to give A = -1 and B = 1.75 . The particular solution is then yt = ( -1 + 1.75t ) ¥ 4t . We can check, for example, that y2 = 8 ¥ 3 - 16( -1) = 40 , which agrees with the formula just found.

(ii)

The auxiliary equation is l 2 - 4l + 5 = 0 , which has roots: 4 ± 16 - 20 = 2 ±i. 2 These complex numbers have modulus: r = 22 + 12 = 5

and argument: tan q =

1 2

ﬁ q = tan -1 12 = 0.464 5e ± 0.464i .

So they can be written in polar form as

The general solution of the difference equation is then:

yt = A

( 5 ) cos 0.464t + B ( 5 ) sin 0.464t t

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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.

Introduction This chapter covers all the methods of differentiation that you need in order to study the Core Technical subjects.

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FAC-06: Differentiation

Limits

1.1

Limits 1 cannot be directly calculated at the point x = 1 . However, x -1 we can say that when x is slightly bigger than 1, say x = 1 + 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”: The value of f ( x) =

1 =• x -1

lim

x Æ1

Remember that x Æ 1+ means that x is approaching 1 from above. Question 6.1

What is lim

x Æ1

1 ? x -1

Question 6.2 x 2 - 3x + 2 ? x Æ• 3 x 2 - 4 x

What is lim

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).

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FAC-06: Differentiation

1.2

Page 3

The order notation If you studied series at school you will have seen series such as: e x = 1 + x + 12 x 2 + 16 x3 +

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:

e x = 1 + x + 12 x 2 + terms in x3 etc What we are thinking here is that, if x is a fairly small number (say 0.1) then the term in x3 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: e x = 1 + x + 12 x 2 + O( x3 )

as x Æ 0

e x = 1 + x + 12 x 2 + o( x 2 )

as x Æ 0

or:

Note that the first equation uses a capital “O” while the second uses a small “o”.

O() Notation The O ( x3 ) 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 x3 . Mathematically speaking, a function f ( x) is O ( x3 ) if you can find at least one constant K > 0 such that f ( x) £ K x3 for all sufficiently small values of x .

f ( x) £ K , ie f ( x) x3 is bounded by 3 x some constant K for sufficiently small values of x .

Another way of writing this would be to say that

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FAC-06: Differentiation

o() Notation The o( x3 ) 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 x3 . Mathematically speaking, a function f ( x) is o( x3 ) if for all constants K > 0 , f ( x) £ K x3 for all sufficiently small values of x .

f ( x) £ K , ie f ( x) x3 is bounded by 3 x all positive constants K for sufficiently small values of x , which means that f ( x) lim =0. x Æ 0 x3 Another way of writing this would be to say that

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 O( x3 ) and/or o( x3 ) for small values of x ? (i)

5x 2

(ii)

0.00001x3

(iii)

999,999,999x 4

(iv)

e x - e- x x

Solution

(i)

5x2 5 = so as x gets smaller 5 x gets larger so it will not be bounded by 3 x x some constant M , so it’s not O( x3 ) . It’s also not heading to zero, so it’s not o( x 3 ) .

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FAC-06: Differentiation

(ii)

Page 5

0.00001x3 = 0.00001 so it’s clearly bounded by M = 0.00001 (or any bigger x3 number!) and hence it is O( x3 ) . But it doesn’t approach zero, so it’s not o( x3 ) . Notice that the 0.00001 factor doesn’t affect the orders here.

(iii)

999,999,999 x 4 = 999,999,999 x so it will be bounded for small values of x x3 (for example it will be bounded by 999,999,999 for values of x smaller than 1) and hence it is O( x3 ) . It also heads to zero as x Æ 0 so it is also o( x3 ) . Again the 999,999,999 factor doesn’t make any difference. Note that being o( x3 ) is a stronger condition than being O( x3 ) . Any function that is o( x3 ) is automatically O( x3 ) .

(iv)

e x - e- x , we have to do a “calculation” involving the order symbols. We For x know that: e x = 1 + x + 12 x 2 + O( x3 ) and: e - x = 1 - x + 12 x 2 + O( x3 ) Notice that we’ve written +O( x3 ) rather than -O( x3 ) in the second equation. We can do this because these two symbols are equivalent. If we now evaluate the numerator, we get: e x - e - x = ÈÎ1 + x + 12 x 2 + O( x3 ) ˘˚ - ÈÎ1 - x + 12 x 2 + O( x3 ) ˘˚

Simplifying: e x - e - x = 2 x + O ( x3 )

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FAC-06: Differentiation

Again notice that adding two O( x3 ) terms just gives O( x3 ) . We now need to divide by x : e x - e- x O ( x3 ) =2+ x x This last term is a function that is no bigger than a fixed multiple of x3 divided by x . This will give a function that is no bigger than a fixed multiple of x 2 ie O( x 2 ) .

So: e x - e- x = 2 + O( x 2 ) x e x - e- x is neither Since O( x ) is “bigger” than O( x ) , we can conclude that x 2

O( x3 ) nor o( x3 ) .

You may also see the order notation used to describe the behaviour of a function for large values of x .

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FAC-06: Differentiation

Page 7

Example

Explain what the statement

x 1 = 1 + + O ( x -2 ) as x Æ +• means. x -1 x

Solution

This means that the behaviour of the function

x is very similar to the behaviour of x -1

1 for large values of x. The remainder (discrepancy) is a term that x 1 approaches zero more quickly than any fixed multiple of 2 . x the function 1 +

You can see this if you expand

(

x 1 = = 1 - 1x x - 1 1 - 1x

1.3

)

-1

x by writing it as: x -1 =1+

1 1 1 + 2 + 3 + x x x

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 a ≥ a " a Œ A and " a < a $ a ¢ > a . 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

1, 2, 3, 2 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.

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FAC-06: Differentiation

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 a £ a " a Œ A and " a > a $ a ¢ < a . 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, 1, 2 4 8

 . This sequence has no minimum, but

the infimum of the sequence is 0. Note that for finite sets A, sup( A) = max( A) and inf( A) = min( A) . a ŒA

a ŒA

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FAC-06: Differentiation

Page 9

Differentiation Consider the function y = f ( x) , with two points lying on the graph of this function,

P1 = ( x, f ( x) ) and P2 = ( x + h, f ( x + h) ) . Then the average rate of change of y = f ( x) from P1 to P2 is given by

f ( x + h) - f ( x ) . h

P2 (x+h, f(x+h)) P1 (x, f(x))

f ( x + h) - f ( x ) to be the rate of change of y with respect to x at hÆ 0 h dy d the point P1 , which is written as f ( x) or f ¢( x) , and is known as the or dx dx derivative of y with respect to x. We then define lim

f ( x + h) - f ( x ) is also the gradient of the straight line joining P1 to P2 , h f ( x + h) - f ( x ) is the gradient of the curve (technically known as a “chord”) then lim hÆ 0 h y = f ( x) at the point P1 , since as h Æ 0 , the chord gets closer and closer to the tangent

Since

at P1 , and the gradient of the tangent is the same as the gradient of the curve at P1 .

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FAC-06: Differentiation

In summary then: f ¢( x) = lim

hÆ 0

f ( x + h) - f ( x ) h

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 y = x 2 , find dy . dx Solution

Consider the points P1 ( x, x 2 ) and P2 (( x + h),( x + h) 2 ) . The gradient of the chord joining them is given by

( x + h) 2 - x 2 . h

Simplifying this gives: ( x + h) 2 - x 2 x 2 + 2 xh + h 2 - x 2 2 xh + h 2 = = = 2x + h h h h ( x + h) 2 - x 2 = 2 x , we have: hÆ 0 h

Since lim

dy = 2x dx 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 y = 2 x 2 + 3 x + 4 from first principles.

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FAC-06: Differentiation

Page 11

Differentiation of standard functions Some standard functions and their derivatives are shown in the table below: Function

Derivative

nx n -1 n π 0

c x ln c

ex 1 x

ln x

Example

Differentiate the following with respect to x: y=7 x

(i)

(ii)

y = 9 x (iii)

y = ln(3x)

Solution

(i)

dy 7 -1 = 7 ¥ 12 x 2 = (Remember that dx 2 x

(ii)

dy = 9 x ln 9 dx

(iii)

Rewriting y = ln(3 x) = ln 3 + ln x hence we have

x = x2 )

dy 1 = dx x

Question 6.4

Differentiate the following functions with respect to x: (i)

3 x2

(ii) 2 x 4

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(iii) 2ln(3 x 2 )

(iv) 3e x + 1

Page 12

FAC-06: Differentiation

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 f ¢( x) = uv ¢ + vu ¢ = u

dv du +v . This dx dx

is called the product rule. Example

If f ( x) = x ln x , what is f ¢( x) ? Solution

f ¢( x ) = x ¥

1 + 1 ¥ ln x = 1 + ln 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) + ¥ vw . dx dx dx Question 6.5

A sum of money is paid continuously at a rate r (t ) = 100te2t , for 0 < t < 2 . At what rate is the payment increasing at time t = 1.5 ?

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FAC-06: Differentiation

4.2

Page 13

Quotients If u and v are functions of x, and f ( x) =

du dv vu ¢ - uv ¢ v dx - u dx u = . This , then f ¢( x) = v v2 v2

is called the quotient rule. Example

If f ( x) =

x , what is f ¢( x) ? 3x + 1

Solution

f ¢( x ) =

4.3

(3 x + 1) ¥ 1 - x ¥ 3 1 = (3x + 1) 2 (3x + 1) 2

Chain rule This is used to differentiate expressions involving nested functions. So instead of x n we may have [ f ( x)]n . Similarly, instead of e x we may have e f ( x ) 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 (3 x 2 + 2 x - 1)5 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

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FAC-06: Differentiation

Example

If y = (3 x 2 + 2 x - 1)5 , what is

dy ? dx

Solution

So first of all we set u = 3 x 2 + 2 x - 1 . So we have y = u 5 . Now we’ll differentiate our two parts: du = 6x + 2 dx dy = 5u 4 du Hence: dy dy du = ¥ = 5u 4 ¥ (6 x + 2) dx du dx Replacing u gives: dy = 5(3x 2 + 2 x - 1) 4 ¥ (6 x + 2) dx In practice these stages are not written out in full as the answer can be written out immediately. When differentiating [ f ( x)]n , think of differentiating the bracket as “normal” ie bring the power down and decrease the power by one which gives n [ f ( x)]n -1 . Then multiply by the derivative of the bracket, f ¢( x) .

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FAC-06: Differentiation

Page 15

Question 6.6

If p = (3r 2 - 4)6 , what is

dp ? dr

Let’s now have a look at an example involving the exponential function. Example

If y = e5 x

-1

, what is

dy ? dx

Solution

First of all we set u = 5 x3 - 1 . So we have y = eu . Now we’ll differentiate our two parts: du = 15 x 2 dx dy = eu du Hence: dy dy du = ¥ = eu ¥ 15 x 2 dx du dx Replacing u gives: 3 dy = e5 x -1 ¥ 15 x 2 dx

When differentiating e f ( x ) , think of differentiating the exponential as “normal” ie leave it as it is e f ( x ) . Then multiply by the derivative of the power f ¢( x) .

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FAC-06: Differentiation

Question 6.7 2

Differentiate y = e -3x with respect to x. Let’s now have a look at an example involving the log function. Example

Differentiate y = ln(3 x + 2) . Solution

First of all we set u = 3 x + 2 . So we have y = ln u . Now we’ll differentiate our two parts: du =3 dx dy 1 = du u Hence: dy dy du 1 = ¥ = ¥3 dx du dx u Replacing u gives: dy 1 3 = ¥3= dx 3x + 2 3x + 2 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 y = 2ln(4 x3 - 5 x + 1) with respect to x.

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FAC-06: Differentiation

Page 17

In summary our “shortcut” rules are: Function

Derivative

[ f ( x)]n

n [ f ( x)]n -1 ¥ f ¢( x)

e f ( x)

e f ( x ) ¥ f ¢( x )

ln f ( x)

1 ¥ f ¢( x ) f ( x)

We’ll now look at an example which involves using the chain rule twice! Example 2

Differentiate ÈÎ log(1 + 2e x ) ˘˚ with respect to x . Solution

Using the shortcut chain rule to differentiate [ f ( x)]n we have: d È d x ˘2 log(1 2 ) ˚ = 2 ÈÎ log(1 + 2e x ) ˘˚ ¥ log(1 + 2e x ) e + Î dx dx But to find

d log(1 + 2e x ) we need to use the chain rule for ln f ( x) : dx

d 1 d 1 2e x x x log(1 + 2e x ) = (1 2 e ) 2 e ¥ + = ¥ = dx 1 + 2e x dx 1 + 2e x 1 + 2e x

Returning to our original expression, the required answer is: x d È 4e x x ˘2 x È log(1 + 2e x ) ˘ ¥ 2e + = = log(1 2 e ) 2 ˚ Î ˚ (1 + 2e x ) (1 + 2e x ) log(1 + 2e ) dx Î

Question 6.9

{

}

If y = exp l (e2 x - 1) , find

dy . dx

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Page 18

FAC-06: Differentiation

Finally, combinations of the product, quotient and chain rules can be used in the same question: Example

If y =

2x + 1 dy , find . dx (3 x - 4) 4

Solution

Using the quotient rule and chain rule:

{

} {

}

(3x - 4) 4 ¥ 2 - (2 x + 1) ¥ 4(3x - 4)3 ¥ 3 dy = dx (3x - 4)8 =

2(3x - 4)3 ((3x - 4) - 6(2 x + 1) ) (3 x - 4)8

2(9 x + 10) (3x - 4)5

Question 6.10

Differentiate the following with respect to x: (i)

y = ( x + 1) 2 e x

(ii)

(iii)

y = (2 x + 7) 4 ln(4 x + 3) 2

ln x (e + 1)2 2x

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FAC-06: Differentiation

Page 19

Higher order derivatives Expressions can be differentiated repeatedly. The notation for higher order derivatives is: d2y d3y d4y , , etc dx 2 dx3 dx 4

where

d2y is pronounced “dee-2-why-by-dee-x-squared” etc. dx 2

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 x = ln(2t + 3) , what is

d 2x ? dt 2

Solution

dx 2 = = 2(2t + 3) -1 dt 2t + 3

d 2x -4 = 2 dt (2t + 3) 2

Question 6.11

What is f ¢¢( x) in each of the following cases? (i)

f ( x) = (3 x 2 + 2 x + 3) 2

(ii)

f ( x) =

(iii)

f ( x) = ln x x +1

x x +1

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FAC-06: Differentiation

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. 0.04 0.03 0.02 0.01 0.2

0.4

0.6

0.8

– 0.01

This is the graph of the function y = x3 (1 - x) 2 , and it has 3 stationary points, a minimum at x = 1 , a maximum at x = 0.6 , and a point of inflexion (neither a maximum or a minimum) at x = 0 . 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 ± e ) < f (b) , where e is a small positive number. Similarly, f ( x) has a local minimum at the point x = b , if f (b ± e ) > f (b) . Generally at these points, f ¢( x) = 0 , and this is the method used to find maxima and minima. However, the following graph illustrates that this is not always the case: 2

–2

–1

This is the graph of y =| x | , and it has a minimum at x = 0 .

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FAC-06: Differentiation

Page 21

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 f ( x) = x3 - x 2 - 8 x + 12 . Solution f ( x) = x3 - x 2 - 8 x + 12 f ¢( x) = 3 x 2 - 2 x - 8 = (3 x + 4)( x - 2)

The stationary points are at f ¢( x) = 0 which occurs when x = - 43 or 2 .

f ¢¢( x) = 6 x - 2

(

)

If x = - 43 , f ¢¢( x) = -10 < 0 , so there is a maximum point at - 43 ,18 14 . 27 If x = 2, f ¢¢( x) = 10 > 0 , so there is a minimum point at (2,0) . In order to understand why the second derivative is positive for a minimum, it is dy necessary to consider the sign of the gradient of the graph (which we know is just dx or f ¢( x) ). For a minimum point, the gradient of the graph changes from negative, to d2y is the rate of zero, to positive, in other words the gradient is increasing. Since dx 2

change of

dy d 2 y must be positive for a minimum. By similar reasoning, it must be , dx dx 2

negative for a maximum. If

d2y dy = 0 , you need to check the sign of to see how the 2 dx dx

graph is behaving. For example, y = x 4 gives both However

dy d2y to be zero at x = 0 . and dx dx 2

dy is negative just below zero, and positive just above zero, so x = 0 gives a dx

minimum.

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FAC-06: Differentiation

To find a point of inflexion, it is necessary to set f ¢¢( x) = 0 . You can have points of inflexion that are not horizontal. For example, the graph of the standard normal distribution has a maximum at x = 0 and points of inflexion (at angles of about 13.6∞ ) at x = ±1 .

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FAC-06: Differentiation

Page 23

Example

Find the stationary points of f ( x) = 2 x3 + 3 x 2 - 72 x + 3 , and determine their nature. Solution

To find stationary points, set f ¢( x) equal to zero. f ¢( x) = 6 x 2 + 6 x - 72 = 6( x 2 + x - 12) = 6( x - 3)( x + 4)

The stationary points are at f ¢( x) = 0 which is when x = 3 or - 4 . To find the points of inflexion, set f ¢¢( x) equal to zero. f ¢¢( x) = 12 x + 6 The points of inflexion are at f ¢¢( x) = 0 which is when x = -0.5 . If x = 3, f ¢¢( x) = 42 > 0 , so there is a minimum point at (3, -132) . If x = -4, f ¢¢( x) = -42 < 0 , 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 y = 4 x3 - 3 x 2 - 90 x + 6 , 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.

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FAC-06: Differentiation

Example

Find the maximum value of: f (l ) = (1 - e - l )30 (e - l - e -2 l )15 (e -2 l )5

f (l ) is a positive-valued function. So we can take logs:

{ = ln {(1 - e l ) = ln {(1 - e l )

} l }

ln f (l ) = ln (1 - e - l )30 (e - l - e -2 l )15 (e -2 l )5 -

30 -15l

(1 - e - l )15 e -10

45 -25l

}

= -25l + 45ln(1 - e - l )

Now differentiating with respect to l , we get: d 45e - l ln f (l ) = -25 + dl 1 - e-l

and this must be zero at a maximum, so: 25 =

45e - l 1 - e- l

25 - 25e - l = 45e - l 25 = 70e - l 25 l = - ln 70 = 1.0296

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Page 25

Checking that this is a maximum (and writing

d 45 to simplify ln f (l ) as -25 + l dl e -1

calculations): 45  d2 d  ln f ( )  25     2 d  d e 1 

45e

 e  1

Since this is always negative, the point we have found must be a maximum, and the value here gives ln f (l ) = -45.6230 ﬁ f (l ) = 1.535 ¥ 10-20 .

Question 6.13

Find the maximum value of: f ( l ) = l p e - nl

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, f (0) = 0 and f (•) = 0 . 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.

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6.2

FAC-06: Differentiation

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, y = 1 + e 2 x .

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

Find any stationary points and their nature

Consider the sign and gradient of the function and the ranges of values for which the function is positive or negative

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.

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FAC-06: Differentiation

Page 27

Example

Sketch the graph of y = e

- 12 x 2

Solution

dy - 1 x2 = - xe 2 dx So

dy = 0 when x = 0 for max/min. dx 1 2 d2y - 12 x 2 - 12 x 2 2 2 -2x x e e e = = ( x - 1) dx 2

When x = 0 ,

d2y < 0 , so there is a maximum at (0, 1) . dx 2

As x Æ ± •, y Æ 0 . Graph: 1.2 1 0.8 y

0.6 0.4 0.2 0 -2

-1

Question 6.14

Sketch the graph of y =

x2 - 4 x . x2 - 4 x + 3

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FAC-06: Differentiation

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:

∂f f ( x + h, y ) - f ( x, y ) = lim ∂ x hÆ 0 h

∂f f ( x , y + h ) - f ( x, y ) = lim ∂ y hÆ 0 h

respectively. Effectively they can be calculated by differentiating f with respect to x ∂f , and differentiating f with respect to y treating x as a treating y as a constant for ∂x ∂f . constant for ∂y

∂f , the partial derivative of f with respect to x, tells you the rate of change of the ∂x function f when x is varied but all other variables are kept constant. Example

Find

∂f ∂f and for the function f ( x, y ) = 2 x 2 y + ( x + 2 y )3 . ∂x ∂y

Solution

∂f = 4 xy + 3( x + 2 y ) 2 ∂x

∂f = 2 x 2 + 6( x + 2 y ) 2 ∂y

Higher derivatives can be found in a similar way. The notation used here is:

∂2f ∂3f etc , ∂ x 2 ∂ x3

∂2f ∂3f etc , ∂ y 2 ∂ y3

∂2f etc ∂ x∂ y

∂2f ∂f means partially differentiate with respect to x. ∂ x∂ y ∂y

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FAC-06: Differentiation

Page 29

Example  f  2f If f ( x, y ) = ( x + y )e , show that  y  e xy  . 2 x x  xy

Solution

∂f = ( x + y ) ye xy + e xy ∂x

∂2f = ( x + y ) y 2e xy + ye xy + ye xy ∂ x2

 f  y  e xy   ( x  y ) y 2e xy  ye xy  ye xy x 

so the relationship is true.

Question 6.15

Find the

∂ f ∂ f ∂2 f ∂2 f ∂2 f , , , , and for the function: ∂ x ∂ y ∂ x2 ∂ y 2 ∂ x∂ y

f ( x, y ) = (3 x + y ) 4 - 2 x 2 y 2 + (4 - 7 x)3

Question 6.16

∂2f ∂2f 1 - e - xy Show that = for f ( x, y ) = . ∂ x∂ y ∂ y∂ x x 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

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FAC-06: Differentiation

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 x = x0 , y = y0 . To discover the nature of the turning point, the roots of the following equation for l need to be found:  2  f  2  y 

x  x0 y  y0

 2   f    2   x 

x  x0 y  y0

  2    f     y x  

   0 x  x0  y  y0 

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 f ( x, y ) = 2 x 2 + y 2 .

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Page 31

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 f ( x1,, xn ) we solve the equations:

∂f ∂f ∂f = 0, = 0, , =0 ∂ x1 ∂ x2 ∂ xn 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 x + y = 100 . In this very simple case it is fairly easy to see the solution: From the constraint y = 100 - 2 x , we are effectively maximising x(100 - 2 x) . Differentiating this we get 100 - 4x , which gives x = 25 when we set it equal to zero. By differentiating again we can see that x = 25 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 L = xy - l (2 x + y - 100) . (Note that the constraint equation can be written as 2 x + y - 100 = 0 .) We then solve the problem by solving

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∂L ∂L ∂L = 0, = 0, = 0. ∂x ∂y ∂l

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FAC-06: Differentiation

Question 6.18

Solve the problem using the Lagrangian function. For more than two variables, any constraint equation can be written in the form g ( x1, x2 ,, xn ) = 0 , in other words, we want to solve: Find the (local) extrema of f ( x1,, xn ) subject to g ( x1, x2 ,, xn ) = 0 Lagrangian theory tells us to construct the Lagrangian function: L = f ( x1, , xn ) - l g ( x1, , xn )

and then solve the equations:

∂L ∂L ∂L ∂L = 0, = 0,  , = 0, = 0. ∂ x1 ∂ x2 ∂ xn ∂l However, there may be more than one constraint. Generally, if we want to solve the problem: Find the (local) extrema of f ( x1,, xn ) subject to the m constraints: g1 ( x1, x2 ,, xn ) = 0, , g m ( x1, x2 ,, xn ) = 0

Lagrangian theory tells us to set up the Lagrangian function: L = f ( x1, , xn ) - l1g1 ( x1, , xn ) - - lm g m ( x1,, xn )

and then solve the equations:

∂L ∂L ∂L ∂L ∂L = 0, = 0, , = 0, = 0, , =0 ∂ x1 ∂ x2 ∂ xn ∂ l1 ∂ lm Question 6.19

Find the extrema of f ( x, y, z ) = x 2 + y 2 + z 2 subject to 2 x - y + z = 3 .

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Page 33

Chapter 6 Summary Order notation

f ( x) is O( g ( x)) if there exists at least one constant K > 0 such that: f ( x) £ K for all x < x0 g ( x)

f ( x) is o( g ( x)) if the inequality is true for all constants K > 0 , which means: lim

xÆ0

f ( x) =0 g ( x)

Differentiation

Differentiating a function gives the gradient of the function at a general point. It is defined as: f ¢( x) = lim

hÆ 0

f ( x + h) - f ( x ) h

Standard results include: Function

Derivative

nx n -1 n π 0

c x ln c

ex 1 x

ln x

Product and Quotient rules

If u and v are functions of x , then: d (uv) = u ¢v + uv ¢ dx d Ê u ˆ u ¢v - uv ¢ Á ˜= dx Ë v ¯ v2

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FAC-06: Differentiation

The chain rule:

To differentiate nested functions, we use: dy dy du = ¥ dx du dx This gives the following shortcut rules: Function

Derivative

[ f ( x)]

n [ f ( x)]n -1 ¥ f ¢( x)

e f ( x)

e f ( x ) ¥ f ¢( x )

ln f ( x)

1 ¥ f ¢( x ) f ( x)

Stationary (turning) points

These occur when

dy = 0 . To determine their nature, we use: dx

d2y < 0 ﬁ max dx 2

d2y > 0 ﬁ max dx 2

d2y = 0 ﬁ point of inflexion (but needs further investigation) dx 2

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.

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FAC-06: Differentiation

Page 35

Partial differentiation

Given a function f ( x, y ) then

∂f means differentiating f ( x, y ) with respect to x ∂x

treating y as a constant. Extrema of f ( x , y )

These occur when  2  f  2  y 

x  x0 y  y0

∂f ∂f = 0 . To determine their nature we find the roots of: = 0 and ∂y ∂x  2   f    2   x 

x  x0 y  y0

  2    f   y x    

   0 x  x0  y  y0 



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 f ( x1,, xn ) subject to: g1 ( x1, x2 ,, xn ) = 0, , g m ( x1, x2 ,, xn ) = 0

Define the Lagrangian function: L = f ( x1, , xn ) - l1g1 ( x1, , xn ) - - lm g m ( x1,, xn )

and then solve the equations:

∂L ∂L ∂L ∂L ∂L = 0, = 0, , = 0, = 0, , =0 ∂ x1 ∂ x2 ∂ xn ∂ l1 ∂ lm

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FAC-06: Differentiation

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

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Page 37

Chapter 6 Solutions Solution 6.1

As x gets closer to 1 (from below), x - 1 is going to get very small and negative. Thus lim

x Æ1-

1 = -• . x -1

Solution 6.2

We can find this limit if we divide the top and bottom through by x 2 :

x 2 - 3x + 2 = lim lim x Æ• 3 x 2 - 4 x x Æ•

3 2 + x x2 = 1 4 3 3x

This is because all the terms with x ’s in their denominator tend to zero.

Solution 6.3

Consider the points P1 ( x, f ( x)) and P2 ( x + h, f ( x + h)) , where f ( x) = 2 x 2 + 3 x + 4 . The gradient of the chord joining them is

f ( x + h) - f ( x ) , which when expanded and h

simplified gives: 2( x + h) 2 + 3( x + h) + 4 - (2 x 2 + 3 x + 4) 2 x 2 + 4 xh + 2h 2 + 3 x + 3h + 4 - (2 x 2 + 3 x + 4) = h h =

Now taking the limit as h Æ 0 , we get that

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4 xh + 2h 2 + 3h = 4 x + 3 + 2h h

dy = 4x + 3 . dx

Page 38

FAC-06: Differentiation

Solution 6.4

( )

(i)

d Ê 3ˆ d 6 3x -2 = -6 x -3 = - 3 ÁË 2 ˜¯ = dx x dx x

(ii)

d 3 4 d  4  8 1 83 x 2 x   2x 3   x 3  dx dx  3  3

(iii)

First we use log rules to simplify this expression:





2ln(3 x 2 ) = 2(ln 3 + 2ln x) = 2ln 3 + 4ln x

Differentiating this gives

(iv)

(

4 . x

)

d 3e x + 1 = 3e x dx

Solution 6.5

To find the rate of increase, we want the derivative r ¢(t ) . Using the product rule:

r ¢(t ) = 100 ¥ e2t + 100t ¥ 2e2t = 100e 2t (1 + 2t )

So at t = 1.5 , r ¢(1.5) = 100e 2 ¥1.5 (1 + 2 ¥ 1.5) = 8,034 . 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.

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Solution 6.6

First of all we set u = 3r 2 - 4 , so we have p = u 6 . Differentiating, we get: du = 6r dr

dp = 6u 5 du

Hence: dp dp du = ¥ = 6u 5 ¥ 6r dr du dr Replacing the u gives: dp = 6(3r 2 - 4)5 ¥ 6r = 36r (3r 2 - 4)5 dr

Solution 6.7

First of all we set u = -3 x 2 , so we have y = eu . Differentiating, we get: du = -6 x dx

dy = eu du

Hence: dy dy du = ¥ = e u ¥ ( -6 x ) dx du dx Replacing the u gives: 2 2 dy = e -3 x ¥ ( -6 x) = -6 xe -3 x dx

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FAC-06: Differentiation

Solution 6.8

First of all we set u = 4 x3 - 5 x + 1 , so we have y = 2ln u . Differentiating, we get: du = 12 x 2 - 5 dx

dy 2 = du u

Hence: dy dy du 2 = ¥ = ¥ (12 x 2 - 5) dx du dx u Replacing the u gives: dy 2 2(12 x 2 - 5) = 3 ¥ (12 x 2 - 5) = 3 dx 4 x - 5 x + 1 4 x - 5x + 1

Solution 6.9

Using the shortcut rule to differentiate e f ( x ) , we have:

{

}

{

} {

}

2 2 2 d d exp l (e 2 x - 1) = exp l (e 2 x - 1) ¥ l (e2 x - 1) dx dx

But to find

2 d l (e2 x - 1) we need to use the chain rule for e f ( x ) again: dx

2 2 2 d l (e2 x - 1) = l ¥ 4 x e2 x = 4l xe2 x dx

But we then need to multiply by the original function, so our answer is:

{

}

2 2 dy = 4l xe 2 x exp l (e 2 x - 1) dx

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Page 41

Solution 6.10

(i)

Using the product rule we get: d È ( x + 1) 2 e x ˘˚ = ( x + 1) 2 e x + e x 2( x + 1) Î dx = ( x + 1)e x { x + 1 + 2} = ( x + 1)( x + 3)e x

(ii)

Using the quotient rule we get 1 (e2 x  1) 2   2(e 2 x  1)2e 2 x ln x d  ln x  x  2x  2 dx  (e  1)  (e 2 x  1) 4 



(iii)

1  4e2 x ln x x (e2 x  1)3

(e2 x  1) 

e 2 x 1  4 x ln x   1 x(e 2 x  1)3

Using the product rule we get

(

)

d 8 (2 x + 7)4 ln(4 x + 3) 2 = (2 x + 7) 4 ¥ + 4(2 x + 7)3 ¥ 2 ¥ ln(4 x + 3) 2 dx 4x + 3 =

{

8(2 x + 7)3 (2 x + 7) + (4 x + 3)ln(4 x + 3)2 4x + 3

}

Ï 2x + 7 ¸ = 8(2 x + 7)3 Ì + 2ln(4 x + 3) ˝ Ó 4x + 3 ˛

We have used the fact that ln(4 x + 3) 2 = 2ln(4 x + 3) to make life easier.

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FAC-06: Differentiation

Solution 6.11

(i)

Differentiating gives: f ¢( x) = 2(3x 2 + 2 x + 3)(6 x + 2)

Differentiating again, this time using the product rule, we get: f ¢¢( x) = 2(3 x 2 + 2 x + 3) ¥ 6 + (6 x + 2) ¥ 2(6 x + 2) = 36 x 2 + 24 x + 36 + 72 x 2 + 48 x + 8 = 108 x 2 + 72 x + 44 = 4(27 x 2 + 18 x + 11)

(ii)

Differentiating using the quotient rule we get: ( x + 1) - x ¥ 12 ( x + 1)

f ¢( x ) =

= =

- 12

x +1

1 (x 2

+ 1)

- 12

{2( x + 1) - x}

x +1

x+2 3

2( x + 1) 2

Differentiating again using the quotient rule gives: 3

2( x + 1) 2 - ( x + 2) ¥ 3( x + 1) 2 f ¢¢( x) = 4( x + 1)3 1

( x + 1) 2 {2( x + 1) - 3( x + 2)}

4( x + 1)3

x+4 5

4( x + 1) 2

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(iii)

Page 43

The expression can be rewritten as f ( x) = ( x + 1)ln x . Differentiating using the product rule, we get: f ¢( x) = ( x + 1) ¥ =1+

1 + ln x x

Differentiating again we get f ¢¢( x) = -

1 1 + . x2 x

Solution 6.12

To find the turning points we have to differentiate: dy = 12 x 2 - 6 x - 90 dx Setting this equal to zero we get: 12 x 2 - 6 x - 90 = 0 fi 2 x 2 - x - 15 = 0 fi (2 x + 5)( x - 3) = 0 fi x = 3 or x = -2.5

We need to differentiate again to find out what sort of turning points we have: d2y = 24 x - 6 dx 2

Setting this equal to zero for points of inflexion, we get x = 0.25 . If x = 3,

d2y > 0 , so there is a minimum point at (3, -183) . dx 2

If x = -2.5,

d2y < 0 , so there is a maximum point at ( -2.5,149.75) . dx 2

There is a point of inflexion at (0.25, -16.625) .

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FAC-06: Differentiation

Solution 6.13

(

)

ln f (l ) = ln l p e - nl = ln l p + ln e - nl = p ln l - nl Differentiating with respect to l , we get: d p ln f (l ) = - n dl l So for maxima or minima, setting this equal to zero, we get l =

p . n

Differentiating again: d2 p ln f ( l ) = < 0 (assuming that p > 0 ) dl2 l2 p

p  p  p So   gives a maximum value of   e  p or   . n n  ne 

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Page 45

Solution 6.14

x2 - 4 x x( x - 4) y= 2 = , so when y = 0 , x = 0 or x = 4 . There is no need to x - 4 x + 3 ( x - 3)( x - 1) consider when x = 0 , since that possibility has already been covered.

4 x

4 3 1- + 2 x x

, so as x Æ • , y Æ 1- .

The graph has a discontinuities when the denominator is zero, ie at x = 3 and x = 1 . To find maxima and minima, we need to differentiate. Using the quotient rule we get: dy ( x 2 - 4 x + 3)(2 x - 4) - ( x 2 - 4 x)(2 x - 4) = dx ( x 2 - 4 x + 3) 2 Multiplying out the numerator: dy 6 x - 12 = 2 dx ( x - 4 x + 3) 2 Setting this equal to zero gives x = 2 . Differentiating again: d 2 y ( x 2 - 4 x + 3) 2 ¥ 6 - (6 x - 12) ¥ 2( x 2 - 4 x + 3)(2 x - 4) = dx 2 ( x 2 - 4 x + 3)4 so when x = 2 ,

d2y = 6 > 0 , ie there is a minimum at (2, 4) . dx 2

There are no solutions to the equation

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d2y = 0 , so there are no points of inflexion. dx 2

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FAC-06: Differentiation

The graph is:

y 8 6 4 2 -4

-2

-2 -4 You could also have deduced the general shape by writing the function in the form 3 y =1. ( x - 1)( x - 3)

Solution 6.15

∂f = 4(3 x + y )3 ¥ 3 - 4 xy 2 + 3(4 - 7 x) 2 ¥ -7 ∂x = 12(3 x + y )3 - 4 xy 2 - 21(4 - 7 x) 2

∂2f = 36(3 x + y ) 2 ¥ 3 - 4 y 2 - 42(4 - 7 x) ¥ -7 2 ∂x = 108(3x + y ) 2 - 4 y 2 + 294(4 - 7 x)

∂f ∂2f = 4(3 x + y )3 - 4 x 2 y , therefore = 12(3 x + y ) 2 - 4 x 2 2 ∂y ∂y

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FAC-06: Differentiation

Page 47

∂2f ∂f means partially differentiate with respect to x, so: ∂ x∂ y ∂y ∂2f = 36(3x + y ) 2 - 8 xy ∂ x∂ y

Solution 6.16

To do this we need to find the first partial derivatives and then partially differentiate them again.

∂ f xe - xy = = e - xy and ∂y x Differentiating again, doing

∂ f x( ye - xy ) - (1 - e - xy ) (1 + xy )e - xy - 1 = = ∂x x2 x2

  f  for simplicity we get:  x   y 

∂2f = - ye - xy ∂ x∂ y Similarly:

∂2f (1 + xy ) ¥ - xe - xy + e - xy ¥ x xe - xy - xe - xy - x 2 ye - xy = = = - ye - xy 2 2 ∂ y∂ x x x Therefore we can see that

∂2f ∂2f = . ∂ x∂ y ∂ y∂ x

This result is generally true.

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FAC-06: Differentiation

Solution 6.17

Partially differentiating:

∂ f ∂ f = 4x = 2y ∂x ∂y Setting these equal to zero, we get x = 0 and y = 0 , ie (0,0) is a turning point. Finding the second derivatives:

∂2 f ∂2 f ∂2 f =4 =2 =0 ∂ y∂ x ∂ x2 ∂ y2 Substituting these into the required equation: (4 - l )(2 - l ) = 0 ﬁ l = 2, 4 Since both these values are positive, the turning point is a local minimum.

Solution 6.18

∂L = y - 2l = 0 ∂x ∂L = x-l =0 ∂y ∂L = 2 x + y - 100 = 0 ∂l Substituting the first two equations into the third, we get: 2l + 2l - 100 = 0 ¤ l = 25 This gives x = 25 and y = 50 as required.

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FAC-06: Differentiation

Page 49

Solution 6.19

We need to find the partial derivative equations:

∂L ∂L ∂L ∂L = 0, = 0, = 0, =0 ∂x ∂y ∂z ∂l This gives us: 2 x - 2l = 0, 2 y + l = 0, 2 z - l = 0, 2 x - y + z = 3 Substituting the first three equations into the fourth: 1 1 2l + l + l = 3 ﬁ l = 1 2 2 Hence: 1 1 x = 1, y = - , z = 2 2

l = 1 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.

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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.

Introduction This chapter covers all the methods of integration that you need in order to study the Core Technical subjects.

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FAC-07: Integration

Integrals Suppose we have a function f ( x) , and that F ¢( x) = f ( x) . Then F ( x) is an antiderivative of f ( x) . The general form of the anti-derivative is the indefinite integral of

f ( x) which is written as

Ú f ( x) dx .

Integration and differentiation are connected in

that they are the “reverse” of each other.

d ( F ( x) + c) = F ¢( x) = f ( x) , then, in general, dx known as the constant of integration.

Since

Ú f ( x) dx = F ( x) + c , where c is

Integrals with limits are called definite integrals. They are evaluated between two b

values. For example

Ú f ( x) dx = [ F ( x)]a = F (b) - F (a) . b

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 + d x . y

y = f(x)

x+x

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FAC-07: Integration

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If A( x) is the area under the curve from x = a to x, then the area bounded by x and x + d x is given by A( x + d x) - A( x) . Looking at the diagram and considering the areas of the smaller and bigger rectangles in the diagram:

d x f ( x + d x) £ A( x + d x) - A( x) £ d x f ( x) Dividing by d x , we get: f ( x + d x) £

As d x Æ 0 ,

A( x + d x) - A( x) £ f ( x) dx

A( x + d x) - A( x) lies between two values which get closer and closer dx

together, so: A( x + d x) - A( x) = f ( x) d xÆ0 dx lim

The LHS is just the definition of

dA , so: dx

dA = f ( x) or A( x) = Ú f ( x) dx dx This is a general expression and we wanted the area enclosed between x = a and x = b , b

so the area is actually given by

Ú f ( x) dx = A(b) - A(a) . a

In the diagram given, the curve was sloping down between x and x + d x . 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 b

x-axis between x = a and x = b , then

Ú f ( x) dx

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 3

x-axis, x = -2 , x = 2 and the curve f ( x) = x , you would evaluate

Úx 0

dx -

Úx

dx ,

-2

using “–” since the second integral will be negative.

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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.

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FAC-07: Integration

Page 5

Integration of standard functions Some standard functions and their integrals are shown in the table below: Function k

1 x f ¢( x ) f ( x)

x n +1 + c , n π -1 n +1 ln x + c ln f ( x) + c

ax +c ln a

ekx +c k

a e

Integral kx + c

Example

Integrate the following functions: (i)

(ii)

72 x

(iii)

3e 4 x

(iv)

2 x +1

Solution

(i)

x as x 2 and using the formula above:

Writing 3

x2 3 2

+c =

2 3

x3 + c

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(ii)

FAC-07: Integration

This is the same as (7 2 ) x . So using the formula with a = 7 2 : 72 x +c 2ln 7

(iii)

3 4x e 4

(iv)

2ln x + 1 + 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)

2 3 + 4 x2 2 x

(ii)

1 e2 x 2

(iii)

(iv)

2x + 5 x2 + 5x + 7

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.

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FAC-07: Integration

Page 7

Example

Evaluate

ln 2

Ú0

2e5 x dx .

Solution ln 2

Ú0

2e5 x dx = ÈÎ 52 e5 x ˘˚

ln 2 0

= 52 e5ln 2 - 52 e0

Since e5ln 2 = eln 2 = 25 = 32 , our expression simplifies to be: 64 5

2 5

62 5

Constant factors can be pulled out of the square brackets if it helps to simplify the working. Question 7.2

Evaluate

Ú1 3x

dx . What does this integral represent?

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FAC-07: Integration

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 2

methods needed to integrate functions such as xe x or 2 x( x 2 + 6)5 . 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)

x Ú 3xe dx

Ú ( x + 2) dx 4

(ii)

(iii)

Ú x( x

+ 6)5dx

Solution

(i)

If the integral was 2 xe x , 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:

Ú 3xe (ii)

dx =

The derivative of the bracket is 1 so we don’t have to multiply or divide by factors in this part:

Ú ( x + 2) (iii)

2 3 3 2 2 xe x dx = e x + c Ú 2 2

1 dx = ( x + 2)5 + c 5

The derivative of the bracket is 2x so we insert a 2 into the integral and divide by 2 outside: 1 1 ( x 2 + 6)6 ( x 2 + 6)6 2 5 Ú x( x + 6) dx = 2 Ú 2 x( x + 6) dx = 2 6 + c = 12 + c 2

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FAC-07: Integration

Page 9

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:

3.2

4x + 6

(i)

Ú ( x2 + 3x + 4)5

(ii)

3x Ú 7 xe

(iii)

Úx

-4

e2x dx

Integration using partial fractions Partial fractions is the name of the technique used to split up a single fraction into x+7 3 2 separate parts. For example is equivalent to . Using this ( x + 1)( x - 2) x - 2 x +1 technique, expressions that cannot be integrated directly can be written as expressions that can. Example

Find

9x - 2

Ú (3x + 2)(3x - 2) dx .

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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: 9x - 2 A B A(3 x - 2) + B(3x + 2) = + = (3 x + 2)(3 x - 2) 3 x + 2 3 x - 2 (3 x + 2)(3 x - 2) Comparing the numerators from the first and last terms we 9 x - 2 = (3 A + 3B ) x - 2 A + 2 B , and comparing coefficients we get A = 2, B = 1 .

get

Substituting this into our integral we get: 1

Ú 3x - 2 + 3x + 2 dx We can now evaluate the integral: 1

Ú 3x - 2 + 3x + 2 dx = 3 ln 3x - 2 + 3 ln 3x + 2 + c 1 = ln k (3x - 2)(3x + 2) 2 3 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 9 x 2 - 4 , so you will sometimes have to be careful to spot that partial fractions are required. Question 7.5

Integrate

x-7 . x2 - 1

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FAC-07: Integration

3.3

Page 11

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

●

Express the function in terms of u.

●

If you have limits, change them so that they correspond to u.

●

Evaluate the simpler integral.

dx , and hence change dx into du . du

Example 1

Evaluate

Ú x(2 x + 3) dx . 4

Solution

du dx 1 u -3 = 2 which means that = , and x = . We need to dx du 2 2 change the limits, when x = 0 , u = 3 , and when x = 1 , u = 5 . The integral then becomes: Let u = 2 x + 3 , then

u  3 4 dx  u  du 2 du 3

 x(2 x  3) dx   0

u 5  3u 4 1 du 2 2 3



1  u 6 3u 5      4  6 5  3

1  56 3  55  1  36 3  35  11          188  4 6 5  4 6 5  30

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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

2x dx . ( x - 1)

Question 7.7 1

Evaluate

Ú 2e3x + 1 dx , by using the substitution u = 2e

+1.

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

Ú u dx

dx = uv - Ú v

du dx dx

This formula is in the Tables page 3. You may see this written as Ú u dv = uv - Ú v du . 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

Ú xe dx . x

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FAC-07: Integration

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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

du dv = e x , so that = 1 and v = e x . So: dx dx

Ú xe dx = xe - Ú e dx x

= xe x - e x + c = ( x - 1)e x + c

Sometimes the choice of u and

dv can be less obvious. dx

Example

Find

3 x Ú x e dx

Solution

The e x

term cannot be directly integrated, and putting

dv = x3 will mean that the dx

second integral will be more complicated than the first, so we need to split up the x3 2

term. The expression xe x can be integrated, so we can proceed as follows. Let 2 2 du dv u = x 2 , and = xe x , so that = 2 x and v = 12 e x . Then: dx dx 3 x2

Úx e

dx = Ú x 2 ¥ xe x dx 2

= 12 x 2e x - Ú xe x dx 2

= 12 x 2e x - 12 e x + c =

2 1 2 ( x - 1)e x + c 2

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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)

Ú x( x + 1) dx

(ii)

Ú x ln x dx

(iii)

Ú ln x dx

(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. b

d ∂f f ( x, t ) dt = Ú ( x, t ) dt , where a and b are This section uses the result that Ú ∂x dx constants. This can be thought of as taking the

d inside the integral. This can be dx

generalised to: b( x)

b( x)

d ∂f f ( x, t ) dt = b ¢( x) f ( x, b( x)) - a ¢( x) f ( x, a( x)) + Ú ( x, t ) dt Ú ∂ dx a ( x ) x a( x) 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.

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FAC-07: Integration

Page 15

The formula is very useful in cases where the integral cannot easily be evaluated •

d - xt 2 e dt . directly, for example dx Ú 0

Example x

Evaluate

d x 2 + t dt . Ú dx 0

Solution

Here a( x) = 0 , b( x) = x , f ( x, t ) = x 2 + t , so: x

d x 2 +t dt = 1( x 2 + x) - 0 + Ú 2 x dt dx Ú0 0

= x2 + x + 2 x2 = 3x 2 + x In this case we can show that this is the same as integrating directly: x

 2 1 2 2 3 1 2  x  t dt   x t  2 t  0  x  2 x 0 x

d d  1  x 2  t dt   x3  x 2   3 x 2  x .  dx 0 dx  2 

Question 7.9

d Evaluate dx

2 x +3

Ú 0

È ( x + 1) 2 +tx ˘ dt . Î ˚

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FAC-07: Integration

Convergence So far we have only dealt with integrals that have finite limits. We define: •

Úa

f ( x) dx = lim Ú f ( x) dx and k Æ• a

f ( x) dx = lim

-•

k Æ-•

Ú f ( x) dx k

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 1 certain point eg when x = 0 . x Example

Does

•

Ú1

1 dx converge? If so, what is its value? x2

Solution 

1

k 1 1  1  1  dx  dx  lim     lim    1 lim  2 2 1 k  k   x 1 k   k x 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: •

Ú0

1 dx x2

Ú0

1 dx x

•1

Ú1

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FAC-07: Integration

Page 17

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

•

Ú0

xe -2 x dx .

Question 7.12

Use integration by parts to show that G ( x) = ( x - 1)G ( x - 1) .

Remember that the

•

definition of the gamma function is G ( x) = Ú t x -1e -t dt . 0

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FAC-07: Integration

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 2

z 1 0 0 -1

-1

0 x

-2

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

ÚÚ f ( x, y) dx dy . R

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.

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FAC-07: Integration

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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 15

Evaluate the double integral

x =10 y = 0

2x + y dy dx . 3000

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.

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FAC-07: Integration

Example

Integrate xy 2 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. y

7 y = 3x + 1 R 1 0

Solution

(i)

In this part x is the variable in the “outside” integral and the range of x values is 0 £ x £ 2 . For any given value of x the range of values of y is 1 £ y £ 3 x + 1 . This gives us our limits of integration. 2 3 x 1   I     xy 2 dy  dx  1  0 2

3 x 1

 xy 3      3 1 0

x(3 x  1)3 x  dx 3 3 0

 2

  9 x 4  9 x3  3 x 2 dx 0

 9 x5 9 x 4     x3  4  5  0  101.6

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FAC-07: Integration

(ii)

Page 21

In this part y is the variable in the “outside” integral and the range of y values is y -1 £ x £ 2. 1 £ y £ 7 . For any given value of y the range of values of x is 3 This gives us our limits of integration.     I     xy 2 dx  dy  1  y 1  3  7 2

 x2 y 2     dy 2    y 1 1 3

  2 y2  1

( y  1)2 y 2 dy 18 7

 2 y3 1  y5 2 y 4 y3        18  5 4 3    3 1  102.26  0.66  101.6

The following results can be used. The proofs are beyond the scope of this course.

ÚÚ af ( x, y) + bg ( x, y) dx dy =a ÚÚ f ( x, y) dx dy +bÚÚ g ( x, y) dx dy R

ÚÚ f ( x, y) dx dy = ÚÚ f ( x, y) dx dy + ÚÚ f ( x, y) dx dy R

Here R is the combined region consisting of the separate regions R1 and R2 , ie R = R1 » R2 and R1 « R2 = ∆ . Question 7.14 3 2

Evaluate the double integral

Ú Ú xe

2 x +3 y

dxdy .

0 y

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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 3

written as Ú dy Ú dx xe 2 x + 3 y . The results for double integrals can be extended to triple integrals. Question 7.15 3 3z 2+ z

Evaluate

Ú Ú Ú 1

e x + y dx dy dz .

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: x2

Ú dx Ú dy f ( x) g ( y) = Ú

f ( x) dx Ú g ( y ) dy y1

Question 7.16 1

Evaluate Ú dx Ú dy Ú dz xyz .

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FAC-07: Integration

5.1

Page 23

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 5

Given that 0 < t < s < 5 , calculate the value of

s =0

t dt ds , first by using this order

t =0

and then by swapping the order of the two variables. Solution

Firstly, we will use the order given: 5

2 t

2 s

 s5  s4  s  t dt ds   s  2  ds   2 ds  10   312.5  0  0 s 0 t 0 s 0 s 0 2

Previously we integrated from t = 0 to t = s , then from s = 0 to s = 5 . Reversing this, we integrate from s = t to s = 5 then from t = 0 to t = 5 . To get the limits for s , see what it is bounded by in the inequality 0 < t < s < 5 . So: 5

s =0

t dt ds =

t =0

ÚtÚ

s 2 ds dt

t =0 s =t

Carrying out this integration: 5

5  s3  125  t 3 1 125t 2 t 5  t s ds dt t dt t dt       312.5       3  3 3  2 5  t t  0 s t t 0  t 0 0 2

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FAC-07: Integration

Question 7.17 30

(i)

Carry out the integration

y =0

(ii)

-m y

e -n x e -d x dx dy , where 0 < x < y < 30 .

x =0

Confirm that you get the same answer if you reverse the order of integration and use the values n = 0.001, d = 0.06, m = 0.005 .

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FAC-07: Integration

Page 25

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. y

y2 d

y3 d

y4 d

d x5

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 x1 and x5 ) is: 1 1 1 1 ( y1 + y2 )d + ( y2 + y3 )d + ( y3 + y4 )d + ( y4 + y5 )d 2 2 2 2 This can be rewritten as: 1 ( y + 2 y2 + 2 y3 + 2 y4 + y5 )d 2 1

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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

1d 2

.”

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 y = x 2 between x = 1 and x = 3 . 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:

(

)

1 2 1 + 32 + 2(1.252 + 1.52 + 1.752 + 22 + 2.252 + 2.52 + 2.752 ) ¥ 0.25 2 = 8.6875 The true area is: 3

1 3 3 È1 3˘ 2 Ú x dx = ÍÎ 3 x ˙˚1 = 3 (3 - 1 ) = 8.67 1 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 y = x 2 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

2 between x = 2 and x = 3 , using 5 ordinates. x Will this be an overestimate or an underestimate of the true value. Estimate the area under the curve y =

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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 ª a , then Taylor’s series states that: f (a + h) = f (a) + hf ¢(a) +

h2 h3 f ¢¢(a) + f ¢¢¢(a) + ◊◊◊ 2! 3!

Here we are thinking that x = a + h , where h is a small quantity. Example

Expand ln x for x ª e in ascending powers of x - e , up to the term in ( x - e)3 . Solution

Applying the formula for Taylor’s series with a = e and h = x - e so that f ( x) = ln x = f (e + ( x - e)) , we get:

f ( x) = ln x 1 f ¢( x ) = x -1 f ¢¢( x) = 2 x 2 f ¢¢¢( x) = 3 x

f ( e) = 1 1 f ¢ (e) = e -1 f ¢¢(e) = 2 e 2 f ¢¢¢(e) = 3 e

and so on

So: 1 1 1 ln x = 1 + ( x - e) - 2 ( x - e) 2 + 3 ( x - e)3 + e 2e 3e

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Question 7.19

Expand (1 + i ) -2 + (1 + i ) -5 as a series about i = 0.1 . 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:

f ( x, y ) = f ( a , b ) +

È 1 Í∂ 2 f 2! Í ∂ x 2 Î

1 È∂ f Í 1! Í ∂ x Î

( x - a) + ( a ,b )

( x - a)2 + 2 ( a ,b )

∂ f ∂y

∂2 f ∂ x∂ y

˘ ( y - b) ˙ ˙˚ ( a ,b ) ( x - a )( y - b) +

( a ,b )

∂2 f ∂ y2

˘ ( y - b) 2 ˙ ˙ ( a ,b ) ˚

+

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: f (a + h, b + k ) = f (a, b) + h

∂ f ∂x

È 1Í 2∂2 f + h 2 Í ∂ x2 Î

+k ( a ,b )

( a ,b )

∂ f ∂y

( a ,b )

∂2 f + 2hk ∂ x∂ y

˘ ˙ +k ∂ y 2 ( a ,b ) ˙ ˚ 2∂

( a ,b )

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Example

Expand e3 xy about (2,1) as far as the terms in ( x - 2) 2 , ( y - 1) 2 etc. Solution

The required partial derivatives in their general form and then evaluated at x = 2 and y = 1 are:

∂ f = 3 ye3 xy = 3e6 ∂x ∂ f = 3xe3 xy = 6e6 ∂y ∂2 f = 9 y 2e3 xy = 9e6 2 ∂x ∂2 f = 9 x 2e3 xy = 36e6 2 ∂y ∂2 f = 3 y ¥ 3xe3 xy + 3e3 xy = 21e6 ∂ y∂ x So the expansion is: e6 + 3e6 ( x - 2) + 6e6 ( y - 1) + 4.5e6 ( x - 2) 2 + 21e6 ( x - 2)( y - 1) + 18e6 ( y - 1) 2

Question 7.20

Expand

1 about (3, 2) as far as the terms in ( x - 3) 2 , ( y - 2) 2 etc. x- y

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7.2

FAC-07: Integration

Maclaurin series A Maclaurin series is just a Taylor series about x = 0 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: f ( x) = f (0) + f ¢(0) x +

f ¢¢(0) 2 f ¢¢¢(0) 3 x + x + ◊◊◊ 2! 3!

Maclaurin’s series are most useful for finding the series expansions for basic functions such as e x , ln(1 + x) etc. Example

Obtain the expansion of e x as far as the term in x3 . Solution

Let f ( x) = e x , then f ( n ) ( x) = e x , where f ( n ) ( x) means the n th derivative, and f ( n ) (0) = 1 , so:

f ( x) = 1 + x +

x 2 x3 + + 2! 3!

Question 7.21

Obtain the expansion of ln(1 + x) as far as the term in x3 .

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Differential equations dy dny in some form, for example, = 2 x + 3 y , is called n dx dx 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) .

Any equation which contains

8.1

Solution by direct integration If we integrate

dy dy d2y , then we get , then we get y and if we integrate , etc. This dx dx dx 2

dny = f ( x) . enables us to solve differential equations of the form dx n

Example d 2r Solve the differential equation 2 = 2t 2 . dt

Solution

Integrating once with respect to t: dr 2 3 = t +c dt 3 Integrating again: r = 16 t 4 + 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.

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Example

Solve the differential equation

dy = e x , given that y = 1 when x = ln 2 . dx

Solution

The general solution of the differential equation is y = e x + c , but we know that when x = ln 2, y = 1 . So: 1 = 2 + c ﬁ c = -1

and the particular solution is: y = ex - 1

Question 7.22

Find the general solution of the differential equation t = 0, x = 4 , when t = 1, x =

97 12

d 2x = at 2 . Given that when 2 dt

1 , find a and the particular , and when t = -1, x = 12

solution.

8.2

Solution by separation of variables dy = f ( x, y ) , and f ( x, y ) can be expressed as p ( x)q ( y ) , then the differential dx equation can be solved by separating the variables, ie by writing it in the form:

Ú q( y) dy = Ú p( x) dx

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Example

Solve the differential equation

dr r , where r > 0 and t > 0 . = dt 2t + 1

Solution

We can rearrange this in the form 1

dr dt = , so: r 2t + 1

Ú r dr = Ú 2t + 1 dt ln r = 12 ln(2t + 1) + c ﬁ ln r = ln

(

)

2t + 1 + c

Note that we do not need to include modulus signs in the integrated expressions as both r and 2t + 1 are positive. Note also that we need to include a constant on one side of the equation, not both. Taking exponentials of both sides: eln r = e

(

)

2t +1 + c

ﬁ r = ec 2t + 1 ﬁ r = k 2t + 1

where k = ec .

Question 7.23

Find the particular solution of the differential equation ( x 2 + 9)

dy = xy + 2 x , where dx

y > 0 , given that when x = 4, y = 0.5 .

Question 7.24

Find the general solution to the differential equation 0<N <

dN = (a - bN ) N , where dt

a . This is known as the logistic equation or the Verhulst equation after the b

mathematician who used it in his study of populations.

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8.3

FAC-07: Integration

Solution by integrating factor dy + f ( x) y = g ( x) can be solved by use of an dx integrating factor. The integrating factor here is exp È Ú f ( x) dx ˘ , and the solution of the Î ˚ differential equation is then the solution of:

A differential equation of the form

y exp È Ú f ( x) dx ˘ = Ú g ( x) exp È Ú f ( x) dx ˘ dx Î ˚ Î ˚

Example

Find the general solution of

dy + 4 y = 2e3 x . dx

Solution

The integrating factor is exp È Ú 4 dx ˘ = e 4 x . So we multiply through by e 4x which Î ˚ dy gives e 4 x + 4 ye 4 x = 2e7 x . dx We see that the LHS is the derivative of ye 4x , derived from the product rule. So d ( ye4 x ) = 2e7 x . dx We then need to solve ye 4 x = Ú 2e7 x dx , which gives ye 4 x = our differential equation is y =

2 7x e + c . The solution of 7

2 3x e + ce -4 x , where c is an arbitrary constant. 7

Question 7.25

Find the particular solution of the differential equation 3xy

dy = y 2 + 2 xy , where x > 0 , dx

given that y = 1 when x = 27 .

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Chapter 7 Summary Integration

Integration is the reverse of differentiation:

Ú f ( x) dx = F ( x) + c

if F ¢( x) = f ( x) ﬁ

A definite integral has limits, in which case our answer does not require a constant: b

Ú f ( x) dx = F (b) - F (a) a

This finds the area under the curve y = f ( x ) between x = a and x = b . Standard results include: Function k

Integral kx + c

x n +1 + c , n π -1 n +1

1 x

ln x + c

ax +c ln a

ex + c

To “undo” the chain rule we could use integration by inspection: Function

Integral

kf ¢( x)[ f ( x)]

k[ f ( x)]n +1 + c , n π -1 n +1

kf ¢( x) f ( x)

k ln f ( x) + c

kf ¢( x)e f ( x )

ke f ( x ) + c

Alternatively, we could use integration by substitution by setting u = f ( x) in the above.

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FAC-07: Integration

Partial fractions can be used to simplify algebraic fractions before integrating: eg

Ú ( x + 1)( x + 3) dx = Ú x + 1 + x + 3 dx

If the above methods don’t work, then we could use integration by parts: dv

Ú u dx

dx = uv - Ú v

du dx dx

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: 1 (y 2 1

+ 2 y2 +  + 2 yn -1 + yn )d

Differentiating an integral b( x)

b( x)

∂ d f ( x, t ) dt = b ¢( x) f [b( x), t ] - a ¢( x) f [ a( x), t ] + Ú f ( x, t ) dt Ú dx a ( x ) ∂ x a( x) Double integrals

These calculate the volume of a solid region under the surface f ( x, y ) : Ê ˆ f ( x , y ) dy Á ˜ dx ÚÁÚ ˜¯ xË y For a domain a £ y £ x £ b the limits can be swapped as follows: Ê x ˆ f ( x , y ) dy ˜ dx = Ú ÁÁ Ú ˜¯ x=a Ë y =a b

Ê b ˆ f ( x , y ) dx ˜ dy Ú ÁÁ Ú ˜¯ y =a Ë x= y b

Series

Taylor and Maclaurin series are used to expand functions as infinite series. The formula for a Taylor series (in one variable) is: f (a + h) = f (a) + hf ¢(a) +

h2 h3 f ¢¢(a) + f ¢¢¢(a) + ◊◊◊ 2! 3!

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The formula for a Taylor series (in two variables) is: f ( x, y )  f ( a , b ) 



 1  2 f 2!   x 2 

1  f  1!   x 

( x  a)  ( a ,b )

( x  a)2  2 ( a ,b )

 f y

2 f  x y

 ( y  b)   ( a ,b ) ( x  a)( y  b) 

( a ,b )

2 f  y2

 ( y  b) 2   ( a ,b ) 

 An alternative form for a Taylor series is: f (a  h, b  k )  f (a, b)  h

 f x

 1 22 f  h 2   x2 

k ( a ,b )

( a ,b )

 f y

( a ,b )

2 f  2hk  x y

2 f k  y2 ( a ,b ) 2

   ( a ,b ) 



The formula for a Maclaurin series is: f ( x) = f (0) + f ¢(0) x +

f ¢¢(0) 2 f ¢¢¢(0) 3 x + x + ◊◊◊ 2! 3!

Differential equations dny A differential equation is an equation which contains . dx n dny = f ( x) can be solved by direct integration dx n dy = p( x)q( y ) can be solved by separating the variables dx dy f ( x ) dx + f ( x) y = g ( x) can be solved by multiplying by the integrating factor e Ú dx

The general solution of a differential equation contains unknown constants. If you are given appropriate boundary conditions, you can find a particular solution.

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FAC-07: Integration

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

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Chapter 7 Solutions Solution 7.1 2

The function can be written as 2 x -2 + 4 x 3 , which can then be integrated to give:

(i)

-2 x -1 +

12 53 2 12 5 x + c = - + x3 + c 5 x 5

(ii)

1 e2 x 4

(iii)

This is equivalent to (6 2 ) x , so using the formula given in the notes, we get:

+c 1

62 2¥ +c ln 6

(iv)

We have the derivative of the denominator on the numerator, so this integrates to give: ln x 2 + 5 x + 7 + c

Check this by differentiating if you are not sure.

Solution 7.2 3

Ú 3x 1

dx = ÈÎ x3 ˘˚ = 26 1

This represents the area between the curve y = 3 x 2 , and the x-axis from x = 1 to x = 3 .

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FAC-07: Integration

Solution 7.3

(i)

4x + 6

Ú ( x 2 + 3x + 4)5

dx = Ú 2(2 x + 3)( x 2 + 3 x + 4) -5 dx

2( x 2 + 3x + 4) -4 +c -4

-4

e 2 x dx = Ú xe x dx =

1 2

(ii)

3x Ú 7 xe

(iii)

Úx

1 +c 2( x + 3x + 4)4 2

-4

dx =

7 6

3x Ú 6 xe 2

dx = 76 e3 x

Ú 2 xe

-4

+c 2

dx = 12 e x + c

Solution 7.4

Expanding the last line of the solution, we get: 1 1 1 ln 3x - 2 + ln(3x + 2) 2 + ln k 3 3 3 1 So it can be seen that c = ln k . 3

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Solution 7.5 x-7 x-7 can be written as . 2 ( x + 1)( x - 1) x -1

Using partial fractions, we can split it as follows: x-7 A B A( x - 1) + B( x + 1) = + = ( x + 1)( x - 1) x + 1 x - 1 ( x + 1)( x - 1)

Comparing coefficients, we get A = 4, B = -3 . The integral then becomes: 4 3 ( x + 1) 4 = + + = +c dx x x c 4ln 1 3ln 1 ln Ú x +1 x -1 ( x - 1)3

Solution 7.6

One possibility is to let u = x - 1 , then the integral becomes: 3 1 1 2(u + 1) -1 2 + 2u 2 ) du = 4 u 2 + 4u 2 + c du = (2 u Ú u Ú 3 1

= 43 (u + 3)u 2 + c = 43 ( x + 2) x - 1 + c It is also possible to solve this by making the substitution u = x - 1 .

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FAC-07: Integration

Solution 7.7

If u = 2e3 x + 1 , then

du = 6e3 x = 3(u - 1) . dx

When x = 0 , u = 3 and when x = 1 ,

u = 2e3 + 1 . Substituting gives: 2e3 +1

Ú 3

4 1 4 ¥ du = u 3(u - 1) 3

2e3 +1

1 du u (u - 1)

Splitting up the integral using partial fractions gives: 4 3

2e3 +1

Ú 3

1 4 du = u (u - 1) 3

2e3 +1

Ú 3

1 1 - du (u - 1) u

2e3 +1 4 ÈÎ ln u - 1 - ln u ˘˚ 3 3

4 4 ln 2e3 - ln(2e3 + 1) - (ln 2 - ln 3) 3 3

4 3e3 ln 3 3 2e + 1

(

)

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Solution 7.8

(i)

Let u = x , and

dv du = ( x + 1)6 , ie = 1 , and v = 17 ( x + 1)7 , then we get: dx dx

Ú x( x + 1)

dx = 17 x( x + 1)7 - Ú 17 ( x + 1)7 dx 1 ( x + 1)8 + c = 17 x( x + 1)7 - 56

1 (x 56

+ 1)7 (7 x - 1) + c

It is also possible to use the substitution u = x + 1 to do this integral. (ii)

Let u = ln x , and

dv du 1 = x , ie = , and v = 12 x 2 . So: dx dx x

Ú x ln x dx = 12 x

ln x - Ú 12 x 2 ¥

1 x

= 12 x 2 ln x - Ú 12 x dx = 12 x 2 ln x - 14 x 2 + c This example is unusual since it is more common to set u equal to a power of x when integrating by parts. (iii)

Let u = ln x , and

du 1 dv = , and v = x . = 1 , ie dx x dx

Ú ln x dx = [ x ln x] - Ú 1dx = [ x ln x - x]1 1

2 1

This can be evaluated to give: (2ln 2 - 2) - (1ln1 - 1) = 2ln 2 - 1 = ln 4 - 1

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FAC-07: Integration

Solution 7.9

d We need to find dx

2 x +3

( x + 1) 2 +tx dt , so comparing with the general equation:

a ( x) = 0, b( x) = 2 x + 3, and f ( x, t ) = ( x + 1) 2 + tx

giving: a ¢( x) = 0, b ¢( x) = 2, and

∂f = 2( x + 1) + t . ∂x

So: d dx

2 x +3

Ú 0

( x + 1) +tx dt = 2 ÈÎ ( x + 1) 2 + x(2 x + 3) ˘˚ - 0 + 2

2 x +3

2( x + 1) + t dt

= 2(3x 2 + 5 x + 1) + ÈÎ 2( x + 1)t + 12 t 2 ˘˚

2 x +3 0

(

= 2(3x 2 + 5 x + 1) + 2( x + 1)(2 x + 3) + 12 (2 x + 3) 2

)

= 12 x 2 + 26 x + 12 12

Solution 7.10 k

È 1˘ The first one does not exist because lim Í - ˙ is not defined for the lower limit. k Æ• Î x ˚ 0 N

The second one does not exist because lim ÈÎ ln x ˘˚ k is not defined for the lower limit. k Æ0 k

The last one does not exist because lim ÈÎ ln x ˘˚1 does not converge. k Æ• One consequence of these integrals not existing is that the area under the graphs of these functions is infinite.

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Solution 7.11

We will do this using integration by parts. dv du = e -2 x , ie = 1 , and v = - 12 e -2 x . Then: dx dx

Let u = x , and •

Ú xe

-2 x

k È ˘ -2 x ˘ k 1 È dx = lim Í Î - 2 xe ˚ - Ú - 12 e -2 x dx ˙ 0 k Æ• Í ˙˚ 0 Î k k˘ È = lim Í ÈÎ - 12 xe -2 x ˘˚ - ÈÎ 14 e -2 x ˘˚ ˙ 0 0˚ k Æ• Î

= lim ÈÎ - 12 ke -2k - 14 e -2 k + 14 ˘˚ k Æ• =

1 4

We have used the fact that exponential functions dominate polynomials for large x, ie lim ke -2k = 0 .

k 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 G ( x) = Ú t x -1e -t dt ( x > 0) . 0

So let u = t x -1 , ie •

G ( x) = Ú t 0

du dv = ( x - 1)t x - 2 , and = e - t , ie v = - e - t . This gives: dt dt •

x -1 - t

•

e dt = ÈÎ -t x -1e -t ˘˚ - Ú -( x - 1)t x - 2e -t dt 0 0

•

= ( x - 1) Ú t x - 2e - t dt = ( x - 1)G ( x - 1) 0

where x > 1 .

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FAC-07: Integration

Solution 7.13 15

x =10

2x + y 1 2 5 Ú 3000 dy dx = 3000 Ú ÈÎ 2 xy + 12 y ˘˚0 dx y =0 10 15

1 = 10 x + 12.5 dx 3000 Ú 10

15 1 È 2 ˘ x x 5 + 12.5 ˚10 3000 Î

11 48

Solution 7.14

We need to integrate by parts. Use u = x , and

du dv 1 = e 2 x + 3 y so that = 1 and v = e 2 x + 3 y , to give: dx dx 2

3 2 2 1 2 x  3 y   1 2 x 3 y  2 x 3 y   xe dxdy xe e dx  dy    2   2 y  y y 0 

3 2

 0

  e3 y  4  0

1 5 y  1 2 x 3 y  ye   e  dy 2 4 y

1 1 1    e3 y  4  e3 y  4    y  e5 y dy 4 4 2  0 3

3 1 1    e3 y  4    y  e5 y dy 4 4 2  0

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Again using integration by parts, this time with u = 3

1 1 - y , we get: 4 2

5y 1 5y  1 3 y  4   1 1  e  e y         e dy  4  4 2 5 10    0   0 0 3

1  1 1 1  1   e3 y  4    y  e5 y  e5 y  5 4 2  50 4 0 

1 13 23 15 1 4 7 e  e  e  4 100 4 100

 641, 284

Solution 7.15 3 3z 2+ z

Ú Ú Ú 1

3 3z

e x + y dx dy dz = Ú 1

Ú 0

3 3z

=Ú 1

2+ z

Èe x + y ˘ Î ˚1

dy dz

e 2 + z + y - e1+ y dy dz

= Ú ÈÎ e2 + z + y - e1+ y ˘˚ dz 0 1

= Ú e 2 + 4 z - e1+ 3 z - e2 + z + e dz 1

= ÈÎ 14 e 2 + 4 z - 13 e1+ 3 z - e 2 + z + ez ˘˚

= 14 e14 - 13 e10 - 14 e6 - e5 + 13 e 4 + e3 + 2e = 293,103

Solution 7.16 1

Ú dx Ú dy Ú dz xyz = Ú x dx ¥ Ú y dy ¥ Ú z dz 0

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FAC-07: Integration

Solution 7.17

(i)

Integrating first with respect to x : 30



 y



y 0



e  x e x dx dy

x 0

1 (  ) x   y   e      e  dy 0 y 0







 y

y 0

1 (  ) x   e    dy    0

 1  e  y   (1  e(  ) y )  dy     y 0





Then integrating with respect to y , we get: 30

  



e  y  e (    ) y dy

y 0 30

   1  y 1   e  e (    ) y           0  (ii)

   1 1 30  1 1  e  e30(    )                

Evaluating (i) numerically, we get 1.213. Changing the order of integration, consider 0 < x < y < 30 . So: 30

e- m y m

y =0

x =0

e -n x e -d x dx dy =

x =0

e -n x e -d x

e - m y m dy dx

y=x

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Then integrating with respect to y , we get: 30



x 0

e x e  x  e   y  dx  x





x 0 30





e  x e  x  e  x  e 30   dx   e (    ) x  e(  ) x e 30  dx

x 0 30

 1 1 (  ) x 30     e (    ) x  e e         0 

   

(1  e 30(    ) ) 

1 (e 30(    )  e 30  )  

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  0.812   2  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

Ú x dx = ÈÎ 2ln x ˘˚2 = 2ln 3 - 2ln 2 = 0.811 2

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FAC-07: Integration

Solution 7.19 f (i ) = (1 + i ) -2 + (1 + i ) -5 , so f (0.1) = 1.447 , and: f ¢(i ) = -2(1 + i ) -3 - 5(1 + i ) -6 f ¢¢(i ) = 6(1 + i ) -4 + 30(1 + i ) -7 f ¢¢¢(i ) = -24(1 + i ) -5 - 210(1 + i ) -8

giving: f ¢(0.1) = -4.325 f ¢¢(0.1) = 19.493 f ¢¢¢(0.1) = -112.869 Using the Taylor series, this gives: (1 + i )

-2

+ (1 + i )

-5

(i - 0.1) 2 (i - 0.1)3 = 1.447 - 4.325(i - 0.1) + 19.493 - 112.869 2! 3!

(as far as the term in (i - 0.1)3 ). Because the term -4.325(i - 0.1) has a negative coefficient, increasing i would reduce the value of the function. In fact, changing i to 0.11 (ie i - 0.1 = 0.01 ) would increase the value of the function by approximately -4.325 ¥ 0.01 = -0.04325 .

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Solution 7.20

The required partial derivatives in their general form and then evaluated at x = 3 and y = 2 are:

∂ f = - ( x - y ) -2 = -1 ∂x ∂ f = ( x - y ) -2 = 1 ∂y ∂2 f = 2( x - y ) -3 = 2 2 ∂x ∂2 f = 2( x - y ) -3 = 2 2 ∂y ∂2 f = -2( x - y ) -3 = -2 ∂ y∂ x So the expansion is:

1 1 - ( x - 3) + ( y - 2) + ÈÎ 2( x - 3) 2 + 2 ¥ -2( x - 3)( y - 2) + 2( y - 2) 2 ˘˚ 2 = 2 - x + y + ( x - 3)2 - 2( x - 3)( y - 2) + ( y - 2) 2

Solution 7.21

f ( x) = ln(1 + x) , so f (0) = 0 , and: f ¢( x ) =

1 1 2 , f ¢¢( x) = , f ¢¢¢( x) = 2 1+ x (1 + x) (1 + x)3

giving: f ¢(0) = 1

f ¢¢(0) = -1

f ¢¢¢(0) = 2

Using Maclaurin’s series, this gives: ln(1 + x) = x - 12 x 2 + 13 x3 (as far as the term in x3 )

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FAC-07: Integration

Solution 7.22

Integrating twice gives: dx at 3 = +c dt 3 x=

at 4 + ct + d 12

But using the boundary conditions: t =0ﬁd =4

t =1ﬁ

97 a = +c+4 12 12

t = -1 ﬁ

1 a = -c+4 12 12

Solving these simultaneously, we get a = 1, c = 4 , so the particular solution is: x=

t4 + 4t + 4 12

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Solution 7.23

Separating the variables gives us: 1

Ú y+2

dy = Ú

x dx x2 + 9

Integrating both sides gives us: ln( y + 2) = c + 12 ln( x 2 + 9) Note that no modulus signs are required here as x 2 + 9 is always positive, and y + 2 is always positive since we are given that y > 0 . So: eln( y + 2) = ec + ln

x2 +9

ﬁ y + 2 = ec x 2 + 9 ﬁ y = k x 2 + 9 - 2

where k = ec . But if y = 0.5 when x = 4 , then k = 0.5 , so the particular solution is: y = 0.5 x 2 + 9 - 2

Solution 7.24

The differential equation can be solved by separating the variables: dN

Ú (a - bN ) N = Ú dt Ú

ba 1a + dN = Ú dt N a - bN

1 1 ln N - ln(a - bN ) + c = t a a

Note that N > 0 and a - bN > 0 from the condition in the question, so no modulus signs are required here.

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FAC-07: Integration

1 a

Setting c = ln k , for some positive constant k , this can be simplified to give the general solution: 1 kN =t ﬁ ln a a - bN

kN = eat a - bN

ﬁ N=

ae at k + be at

Solution 7.25

dy 1 2 y = , providing that dx 3x 3 y π 0 . y = 0 is a trivial solution to the differential equation.

The differential equation can be rewritten in the form

Using this version, we can 1 ˘ -1 È È 1 ˘ exp Í Ú - dx ˙ = exp Í - ln x ˙ = x 3 . Î 3x ˚ Î 3 ˚

see

that

the

integrating

factor

Note that since x > 0 , 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: yx

- 13

2 2 -1 = Ú x 3 dx = x 3 + c 3

But we know that y = 1 when x = 27 , so: 1 = 9 + c ﬁ c = -8 23 3 So the particular solution is yx

- 13

= x 3 - 8 23 or y = x - 8 23 x 3 .

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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.

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.

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FAC-08: Vectors and matrices

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 T   a  a  2i  3 j  2k   3    2 3 2   2   

(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 - l I )v = 0 where underlining is used to represent a vector or a matrix.  2   Note that the column vector  3  shows the coefficients of i, j and k, so that  2     2 1 0 0          3   2i  3 j  2k where i   0  , j   1  , and k   0  . Column vectors are usually  2  0 0 1         used in preference to row vectors because matrix equations then have a similar format to the corresponding set of simultaneous equations.

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Page 3

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:

b 2a 2a + b

Example  2  3     If a   4  and b   5  , what are:  4  1    

(i)

(ii)

ab

(iii)

3a  4b ?

Solution  4   (i) 2a   8   8   

5   (ii) a  b   1   3   

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 6    (iii) 3a  4b   32   16   

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FAC-08: Vectors and matrices

Question 8.1

1.2

(i)

 3 1     If a   2  and b   6  what is 3b  2a ? 8  2     

(ii)

 5    What are the values of p and q such that pa  qb   50  ?  46   

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 | .  p   In general, if a   q  , then a = r  

p2 + q2 + r 2 .

Example 1   Find the magnitude of the vector a   2  .  4  

Solution a = 12 + ( -2) 2 + 42 = 21

This gives us a way of finding a unit vector ie a vector with length 1. For example if a 1 1   2 . is as defined in the last example, then the unit vector in the direction of a is 21    4

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Question 8.2 Find the magnitude of the vector 2i - 5 j + 3k .

Question 8.3

Find the unit vector in the direction of 3i + 4 j - 2k .

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 a . b = ab cosq where q is the angle between them in the plane containing a and 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 ď ą

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FAC-08: Vectors and matrices

Example 1  2     Find a . b , where a   4  and b   3  .  5  6     

Solution

a . b = (1)(2) + ( -4)(3) + (5)( -6) = 2 - 12 - 30 = -40 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 3  3      Find the angle between the vectors a   6  and b   3  .  1  4    

Solution a . b = -9 + 18 - 4 = 5 , a = 9 + 36 + 1 = 46 , and b = 9 + 9 + 16 = 34

So: a . b = ab cosq ﬁ 5 = 46 34 cosq \ cosq =

5 46 34

q = 83

Question 8.4

Find the angle between the vectors a = 2i + 3 j - 5k and b = -3i + 2 j + 8k .

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Since the scalar product is related to the angle between vectors, an important result occurs when a . b = 0 , 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  1 3p Find p such that a and b are perpendicular, where a    , b    .  7   4

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FAC-08: Vectors and matrices

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  1 1 3 5  Describe the size of this matrix  .  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 AT or A ¢ . Example  1 3    If A   2 5  , what is AT ? 0 9   

Solution  1 2 0 AT     3 5 9 

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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. a b  e So for 2 ¥ 2 matrices   c d g

f  ae b f   . h  c  g d  h

Example  2 4 5  4 3 2      If A   4 2 6  and B   8 5 1  , what is A  B ? 7 1 3  3 6 0    

Solution 7 3  2 4 5   4 3 2   6       A  B   4 2 6    8 5 1    12 3 5   7 1 3   3 6 0   10 7 3      

The matrix with zero as all of its elements is called the 4 0 0  2 ¥ 2 zero matrix is 0    . A matrix such as  0 0 0 0 

zero matrix. For example the 0 0  3 0  , where all the elements 0 1 

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?

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2.2

FAC-08: Vectors and matrices

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  3 3 If A    , what is 2 A ?  5 6  Solution

6  6 2A     10 12  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.  a b  e So for 2 ¥ 2 matrices    c d  g

f   ae  bg  h   ce  dg

af  bh  . cf  dh 

Example  4 2   3 1 If A    and B    what is AB ?  5 4   3 1  Solution

(3  2)  (1  1)   15 7   3 1  4 2   (3  4)  (1  3) AB        5 4  3 1   (5  4)  (4  3) (5  2)  (4  1)   32 14  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.

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Question 8.7  1 2 4   4 1 1      If A   3 2 3  and B   5 0 2  , what are AB and BA ?  0 1 1   1 5 3    

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  0 2 3  1     What is  4 2 1  2  ?  0 5 1  1   

These could not have been multiplied the other way round. Question 8.9 1   What is  2 1 3  3  ?  2   

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FAC-08: Vectors and matrices

Example a b 1 0 What do you get if you multiply the matrix M    by I   ? c d  0 1 Solution a b  a b  MI    and IM    c d  c d 

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 A , | A | , or D when it is clear which matrix is involved. Example  2 6  What is det A if A   ? 4 3  Solution

det A = (2 ¥ 3) - (4 ¥ -6) = 30

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Question 8.10  3 5 What is the determinant of P   ?  2 7 

This is a specific definition, and we need to generalise. For any matrix M , if cij is the element in the matrix corresponding to row i and column j, and M ij is the determinant of the matrix formed when we strike out row i and column j, then the determinant is n

defined as

Â (-1) j +1c1 j M1 j . j =1

This looks quite daunting, but in practice it is quite simple to use. Example  1 1 2    Calculate the determinant of  3 0 4  .  2 3 1   

Solution

The determinant is given by: D = 1¥

0 -4 3 -4 3 0 + ( -1) ¥ ( -1) ¥ +2¥ 2 1 2 -3 -3 1

= 1(0 - 12) + 1(3 + 8) + 2( -9 - 0) = -12 + 11 - 18 = -19 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.

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FAC-08: Vectors and matrices

Question 8.11  2 1 0   Find the determinant of  0 1 1  .  4 3 6   

Inverses The inverse of a matrix A , written as A -1 , is such that AA -1 = A -1A = 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 a b    c d 

1



1  d b   . ad  bc  c a 

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

3 2 1 If A    , what is A ? 5 4 Solution

1  4 2  A 1    2  5 3 

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Example

Using the matrix in the last example, show that AA -1 = A -1A = I . Solution

Note that the factor can be brought to the front. 1 3 AA 1   2 5 1 4 A 1A   2  5

2  4 2  1  2    4  5 3  2  0 2  3 2  1  2    3  5 4  2  0

0 1 0   2 0 1 0 1 0   2 0 1

Question 8.12

1 9  -1 Find the inverse of B    , and check it by finding BB .  2 10  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.

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2.4

FAC-08: Vectors and matrices

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 x - 3 y = 14 3x + 4 y = 4 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  x  14         3 4  y   4  Pre-multiplying by the inverse matrix on both sides, we get: 1  4 3  2 3  x  1  4 3 14          17  3 2  3 4  y  17  3 2  4  This gives the identity matrix on the left hand side, so we simplify to get:  1 0  x  1  68         0 1  y  17  34   x  4      y   2 

So the solution is x = 4 and y = -2 .

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Question 8.13

Solve the simultaneous equations (using matrices):

2 p - 4q = -14 2q - 3 p = 13 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.

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Example (Method 2)

Solve the simultaneous equations (using matrices): 2 x - 3 y = 14 3x + 4 y = 4 Solution

We can write this in the abbreviated form:  2 3  x  14         3 4  y   4 

This can in turn be written as:  2 3  3 4

14   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  3 4

10   4 

Replacing the second row by (2)  3  (1) : 1 7   0 17

10   1 7  34   0 1

10   2 

This tells us that y = -2 , and so x = -10 - 7 y = 4 .

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Example

Solve the simultaneous equations (using the elimination method): 2 x + y - 2 z = 11 3x - 2 y + 4 z = -8 x + 4 y - 3z = 15 Solution

These can be written as:  2 1 2   3 2 4  1 4 3 

11   8  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   0 14 13  1 4 3 

4   53  15 

Replacing the third row by 2  ((3)  (1))  (2) , we get:  1 3 1   0 14 13 0 0 5 

4   1 3 1   53    0 14 13 15   0 0 1

4   53  3 

Replacing the second row by (2)  13  (3) , we get:  1 3 1   0 14 0 0 0 1 

4   14  3 

so z = -3, y = 1, x = 2 .

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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: 3x - 4 y = -5 2x + 3y = 8 x - 7 y = -13 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

Solve the simultaneous equations: 3x + 2 y = 13 7x - 6 y = 9 using the method of the last example.

Question 8.15

Solve the simultaneous equations: x + 2 y - z = -4 2x - 3y + 2z = 1 3x + y - 3 z = -17 using the method of the last example.

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2.5

Page 21

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: Av = l 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 Av = l v :

Av = l v ﬁ ( A - l 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 ( A - l 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 x = B -1C .

Using this technique to try to solve the equation

( A - l I ) v = 0 , then if ( A - l I ) has an inverse we get v = ( A - l I ) -1 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 A - l 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( A - l I ) = 0 . The equation obtained from det( A - l I ) = 0 is called the “characteristic equation of the matrix”. Once you have the eigenvalues, you can return to the equation ( A - l 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.

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Example

 2 1 Find the eigenvectors and corresponding eigenvalues of the matrix B   . 2 5  Solution

First we use the equation det(B   I )  0 to find the eigenvalues: 1  2   det  0 5  2 (2   )(5   )  2  0

 2  7  12  0 (  3)(  4)  0 so   3 or   4 .  x We now use the equation (B   I ) v  0 to find the eigenvectors, letting v    .  y There is a separate eigenvector corresponding to each eigenvalue. 1  x   0  2    1 1 x   0   x  y   0  If   3 , then       , ie       , ie     . 5    y   0   2  2 2  y   0   2x  2 y   0  Any value of x and y such that x  y  0 ie y   x will make both rows work out.

1 So the eigenvector corresponding to   3 is k   , where k is a constant.  1  2 1 x   0  If   4 , then       , so the eigenvector corresponding to   4 is  2 1  y   0  1 k .  2  Notice that we include k in the eigenvector, since any value of the constant k will give an eigenvector.

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Question 8.16

1 2  Find the eigenvectors and corresponding eigenvalues of the matrix B   .  3 4 

Question 8.17  2 2 3    Show that l = 1 is an eigenvalue of the matrix  1 1 1  , and hence find all the    1 3 1  eigenvalues and eigenvectors.

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FAC-08: Vectors and matrices

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

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Chapter 8 Summary Vectors

Vectors have both a magnitude and a direction. For example:

Ê 2ˆ T a = a = 2i + 3j - k = Á 3 ˜ = (2 3 -1) Á ˜ Ë -1¯ The magnitude is a = 22 + 32 + ( -1) 2 . Vectors can be added, subtracted and multiplied by a scalar: eg

Ê aˆ Ê d ˆ Ê a + d ˆ Á b˜ + Á e ˜ = Á b + e ˜ Á ˜ Á ˜ Á ˜ Ë c¯ Ë f ¯ Ë c + f ¯

Ê aˆ Ê d ˆ Ê a - d ˆ Á b˜ - Á e ˜ = Á b - e ˜ Á ˜ Á ˜ Á ˜ Ë c¯ Ë f ¯ Ë c - f ¯

Ê a ˆ Ê ka ˆ k Á b ˜ = Á kb˜ Á ˜ Á ˜ Ë c ¯ Ë kc ¯

Scalar product

a .b = ab cosq If the vectors are perpendicular (orthogonal) then a . b = 0 . Hence if a = xi + yj + zk and b = li + mj + nk then a .b = xl + ym + zn . Matrices

A matrix is a rectangular array of numbers. Matrices can be multiplied by a scalar: eg

Ê a b ˆ Ê ka kb ˆ = kÁ Ë c d ˜¯ ÁË kc kd ˜¯

Matrices of the same dimensions can be added or subtracted: eg

Êa b ÁË d e

cˆ Êg + f ˜¯ ÁË j

h iˆ Ê a + g b + h = k l ˜¯ ÁË d + j e + k

c+iˆ f + l ˜¯

The transpose of a matrix can be found by swapping the rows and the columns.

Êa b ÁË d e

Êa T cˆ = Áb Á f ˜¯ Ëc

dˆ e˜ ˜ f¯

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FAC-08: Vectors and matrices

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.

Ê a bˆ Ê e ËÁ c d ˜¯ ÁË g

f ˆ Ê ae + bg = h ˜¯ ÁË ce + dg

af + bhˆ cf + dh ˜¯

Determinants

Ê a bˆ If A = Á then det A = A = ad - bc . Ë c d ˜¯

Êa b If A = Á d e Á Ëg h

cˆ e f ˜ then det A = A = a ˜ h i¯

f d -b i g

f d +c i g

e h

Inverses A -1 is the inverse of A if AA -1 = A -1A = I where I is the identity matrix.

Ê a bˆ 1 Ê d -b ˆ If A = Á then A -1 = . ˜ ad - bc ÁË -c a ˜¯ Ëc d¯ 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( A - l I ) = 0 The corresponding eigenvectors v can then be found by solving the equation:

( A - l I) v = 0

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Chapter 8 Solutions Solution 8.1

(i)

(ii)

 3    3b  2a   22   22     5    If pa  qb   50  then 3 p + q = 5 and - 2 p + 6q = -50 .  46    equations simultaneously, we get p = 4 and q = -7 .

Solving these

You should also check that the third equation, 8 p - 2q = 46 , also works out with these values of p and q. Otherwise there are would be no solutions.

Solution 8.2

The magnitude is

22 + ( -5) 2 + 32 = 38 .

Solution 8.3

The magnitude of the vector is

32 + 42 + ( -2) 2 = 29 , so the unit vector is

1 (3i + 4 j - 2k ) . 29

Solution 8.4

To find the angle, we take the dot product. a . b = ab cosq ie -40 = 22 + 32 + ( -5)2 ( -3) 2 + 22 + 82 cosq

Hence cosq =

-40 or q = 138 . 38 77

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Solution 8.5

If a and b are perpendicular, then their dot product is zero. a . b = 3 p - 28 = 0 , hence p =

28 . 3

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)    AB   (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)  21 9   10     25 12 2   6 5 1   

Similarly:  1 11 20    BA   5 12 22   16 11 14   

Solution 8.8  0 2 3  1   7        4 2 1  2    7   0 5 1  1 11     

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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 P = -21 - ( -10) = -11 .

Solution 8.11

The determinant is 2(6 - 3) - 1(0 - ( -4)) + 0 = 2 .

Solution 8.12

1  10 9  B 1     8  2 1  To check: 1  1 9  10 9  1  8 0   1 0  BB 1         8  2 10  2 1  8  0 8   0 1 

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Solution 8.13

Noticing that the letters are in different orders, these equations can be written in the form:  2 4  p   14         3 2  q   13   2 4  Pre-multiplying both sides by the inverse of   , we get:  3 2   p 1  2 4  14       8  3 2  13  q 1  24   3      8  16   2 

So p = -3 and q = 2 .

Solution 8.14

These equations can be written as: 3 2   7 6

13   9

Replacing (1) by (2)  2  (1) :  1 10   7 6

17   9 

Replacing (2) by (2)  7  (1) :  1 10   0 64

17   128 

So x = 3 and y = 2 .

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Solution 8.15

These equations can be written as:  1 2 1   2 3 2  3 1 3 

4   1  17 

Replacing (2) by (2)  2  (1) , and (3) by (3)  (2)  (1) , we get:  1 2 1   0 7 4  0 2 4 

4   9  14 

Replacing (2) by (2)  (3) , we get:  1 2 1   0 5 0  0 2 4 

4   1 2 1   5    0 1 0 14   0 2 4

4   1  14 

Replacing (3) by (3)  2  (2) , we get:  1 2 1  0 1 0  0 0 4 

4   1 2 1   1   0 1 0 16   0 0 1

4   1 4 

This gives z = 4, y = 1, x = -2 .

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Solution 8.16

1   For eigenvalues, det   3

2    0 , so we have the characteristic equation: 4   

(1   )(4   )  6  0   2  3  10  0

This gives   2,   5 . 1   If   2 , then   3

2  x   0   1 2  x   0              . 4    y   0   3 6  y   0 

 2 So the eigenvector corresponding to   2 is k   . 1 1   If   5 , then   3

2  x   0   6 2  x   0              . 4    y   0   3 1  y   0 

1 So the eigenvector corresponding to   5 is k   .  3 

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Solution 8.17

For the eigenvalues we must solve the equation:  2   2  det  1 1   1 3 

  1 0 1   

So we have the equation:

(2 - l ) {(1 - l )( -1 - l ) - 3} + 2 {( -1 - l ) - 1} + 3{3 - (1 - l )} = 0 and this equation simplifies to l 3 - 2l 2 - 5l + 6 = 0 . Factorising the given cubic equation completely gives: (l - 1)(l - 3)(l + 2) = 0 The eigenvalues are 1, 3, and –2.  2   2  When   1 ,  1 1   1 3 

 x   0   1 2 3  x   0           1  y    0    1 0 1  y    0  , which gives         1     z   0   1 3 2  z   0 

 1   the eigenvector of k  1  . 1    2   2  1  When   3 ,  1  1 3 

 x   0   1 2 3  x   0           1  y    0    1 2 1  y    0  , which         1     z   0   1 3 4  z   0 

1   gives the eigenvector of k 1 . 1  

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Page 34

When

FAC-08: Vectors and matrices

  2 ,

 2   2  1   1  1 3 

 x   0   4 2 3  x   0           1  y    0    1 3 1  y    0  , which         1     z   0   1 3 1  z   0 

 11    gives the eigenvector of k  1  .  14   

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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

Confusing word pairs

Latin and Greek

Words you will meet in actuarial work

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Page 2

FAC: Glossary

Commonly mis-spelt words These are the correct spellings of words that have been spelt wrongly on assignments: actuarial biased cyclically hierarchy necessary offering proceed separate

ageing calendar deferred immediately occasion orthogonal receive similarly

appropriately cancelled definitely independence occurred paid referred supersede

basically commission formatted instalment occurring particularly referring targeted

benefit consensus fulfil interest offer pensioner relief theorem

benefiting correlation gauge millennium offered precede seize 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 ocCUR 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”.

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FAC: Glossary

Page 3

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.

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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.

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FAC: Glossary

Latin and Greek Plurals

LATIN

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:

GREEK

Page 5

Singular –US

Plural –I

Examples stimulus/stimuli

–UM

–A

maximum/maxima medium/media

–A

–AE

formula/formulae

–IX or –EX

–ICES

index/indices appendix/appendices

–ON

–A

–IS

–ES

criterion/criteria phenomenon/phenomena 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.

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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 5 that, if you worked for 5 years and 5 months (say), this would be counted as 5 12 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).”

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FAC: Glossary

Page 7

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.

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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.

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FAC: Question & Answer Bank – Questions

Page 1

Question & Answer Bank – Questions Chapter 1 Question A1 Interpret the statement { x Œ  : -1 £ x < 5} .

B C D

x Œ{-1,0,1, 2,3, 4,5} x Œ{-1,0,1, 2,3, 4} x Œ{0,1, 2,3, 4} x Œ{1, 2,3, 4}

[1] FAC 1 1

Question A2 If

A = {2,4,6,8} and B = {1, 2,3, 4} and C = {1,3,6,10} which are subsets of

{1,2,3, 4,5,6,7,8,9,10} , which one of the following is ( A « B ) » C ? A B C D

{1,3,6} {6} {1,3,6,8,10} {1,3,6,10}

[1] FAC 1 1

Question A3 If A = {1,3,5,7,9} and B = {1,3,6,10} which are subsets of  , which one of the following would be in A « B ? A B C D

3 5 6 8

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[1] FAC 1 1

Page 2

FAC: Question & Answer Bank – Questions

Question A4 What are the Greek letters used to represent a sum and a product? A B C D

S, P s ,p D, r S, p

[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 B C D

42% 6% 2.4% 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 B C D

A is necessary for B A is sufficient for B A is necessary and sufficient for B none of the above

[2] FAC 1 4

Question A7 Which of the following was a leap year? A B C D

1900 1910 1920 1950

[1] FAC 1 7

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FAC: Question & Answer Bank – Questions

Page 3

Question A8 Using the exchange rates below, how much would £100 be worth in US dollars? £/€ = 1.227 A B C D

and

€/$ = 1.314

$62.02 $93.38 $107.09 $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 B C D

1 March 2016 1 September 2015 31 August 2015 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)

{x : x Œ , x 2 = 4, x π 2} = ∆

[1]

(iii)

x 2 - 5 x + 7 > 0, "x Œ

[2]

(iv)

The natural logarithm function is defined on the set of values (0, •] .

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[1] [Total 5]

Page 4

FAC: Question & Answer Bank – Questions

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

Chapter 2 Question A12 What is 2.89951 rounded to 3 decimal places? A B C D

2.899 2.900 2.90 2.810

[1] FAC 2 1

Question A13 What is 4,716 rounded to 3 significant figures? A B C D

472 716 4,700 4,720

[1] FAC 2 1

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FAC: Question & Answer Bank – Questions

Page 5

Question A14 What is 0.020581 rounded to 2 significant figures? A B C D

0.02 0.021 0.0206 0.02058

[1] FAC 2 1

Question A15 Calculate tanh -1 0.6 . A B C D

0.0105 0.537 0.693 30.963

[1] FAC 2 2

Question A16 Calculate A B C D

1 - (1 + i ) - n where i = 0.062 and n = 10 . i 1+ i

-14.130 -12.528 6.865 7.743

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[2] FAC 2 2

Page 6

FAC: Question & Answer Bank – Questions

Question A17 Calculate 6v  8v 2  10v3  12v 4 where v =

1 . 1.08

Give your answer to 3 significant

figures. A B C D

29.173 29.2 33.3 33.333

[2] FAC 2 2

Question A18  1  ln 5  1.02  2  Calculate the value of exp      to 3 decimal places. 0.8 0.8 2    2 

A B C D

0.112 0.210 0.345 0.380

[2] FAC 2 2

Question A19 Calculate A B C D

6! 5 12 4 +4 + 20 . 3

5.897 243.9 259.7 327.4

[1] FAC 2 2

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FAC: Question & Answer Bank – Questions

Page 7

Question A20 1 - vn - nv n Evaluate d , given that i = 0.10 , n = 10 , v = (1 + i ) -1 and d = iv . i

[2] FAC 2 2

Question A21 The interest rate i charged for a financial arrangement satisfies the following equation: 43,600,000(1 + i )10 = 12 ¥ 10 ¥ 366,000(1 + i )5 + 60,192,000

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

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FAC: Question & Answer Bank â&#x20AC;&#x201C; Questions

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:

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FAC: Question & Answer Bank – Questions

(ii)

Page 9

(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]

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

2 rds 3

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]

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FAC: Question & Answer Bank – Questions

Chapter 3 Question A24 Which of these is the graph of y = e - x ? A y

B y

C y

D y

[1] FAC 3 1

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FAC: Question & Answer Bank – Questions

Page 11

Question A25 Which of these is the graph of y =

1 where n is an odd number? xn

A y

B y

C y

D y

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[1] FAC 3 1

Page 12

FAC: Question & Answer Bank – Questions

Question A26 If the dotted line represents the graph of y = f ( x) , which of these is the graph of y = f (2 x - 1) ? A y

B y

C y

D y

[2] FAC 3 1

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FAC: Question & Answer Bank – Questions

Page 13

Question A27 If a function f ( x) has the property f ( - x) = - f ( x) , the function is known as: A B C D

an odd function a symmetrical function an even function 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 B C D

55 55.75 55.83 56

[1] FAC 3 2

Question A29 Write the expression 4 x + 1 < 5 without the use of the modulus sign. A B C D

x <1 x < -1.5 -1.5 < x < 1 -1 < x < 1

[1] FAC 3 2

Question A30 Calculate V, where x = 0.8 and V = min(e - x , p 4) . A B C D

0.443 0.449 0.886 2.226

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[1] FAC 3 2

Page 14

FAC: Question & Answer Bank – Questions

Question A31 Calculate the value of A B C D

100! to 2 decimal places. 97!3!

161,700 323,400 15,684,900 31,369,800

[1] FAC 3 3

Question A32 Evaluate G (5.5) . You may use the fact that G (½) = p . A B C D

52.34 104.69 230.31 287.89

[2] FAC 3 3

Question A33 Simplify the expression A B C D

G (n + 1)(3n + 1)! , where n is an integer. G (3n)(n - 1)!

n(n + 1)(3n + 1) (n + 1)(3n + 1) 3n(n - 1) n(3n + 1) (3n - 1)(n - 1) 3n 2 (3n + 1)

[2] FAC 3 3

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FAC: Question & Answer Bank – Questions

Page 15

Question A34 Evaluate each of the following, quoting your answers to 2 significant figures: (i)

e e

[1]

(ii)

log e 0.00001

[1]

(iii)

tanh -1 0.9

[1]

(iv)

G (12)

[1] [Total 4]

Chapter 4 Question A35

( )

Which of the following is the correct simplification of 2 xb ¥ x 2

2 x 6b

2 xb + b + 2 x3+ b

4 x3b

2 x3b + 2 x3+ b

+ 2 xb x3 ?

[1] FAC 4 1

Question A36

    e2 x  .

Simplify e3 A

e½

1 e3 x

- 8 x3

8 x3

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[1] FAC 4 1

Page 16

FAC: Question & Answer Bank – Questions

Question A37 If log a x = A then: A

ax = A

xa = A

aA = x

Aa = x

[1] FAC 4 1

Question A38 x Simplify 2log a    log a y.  y

A B C D

log a x 2 x log a    y log a yx

 x2  log a    y   

[1] FAC 4 1

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FAC: Question & Answer Bank – Questions

Page 17

Question A39 x  3 5  2x as a single fraction:  3x  1 x  1

Express

5 x 2  15 x  8 (3x  1)( x  1)

7 x 2  11x  2 (3 x  1)( x  1)

7 x 2  11x  8 (3 x  1)( x  1)

5 x 2  15 x  2 (3x  1)( x  1)

[2] FAC 4 1

Question A40 Simplify

A B

x+3 x2 + x - 6 . ∏ 2 x 2 - 3x - 2 2 x 2 + 3x + 1 2 x3  9 x 2  10 x  3

2 x 4  x3  17 x 2  16 x  12 x 1 ( x  2) 2

x2 + 4 x + 3 x2 - 4 x + 4

( x  3) 2 (2 x  1)( x  1)

[5] FAC 4 1

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FAC: Question & Answer Bank – Questions

Question A41 Solve the quadratic equation 2 x 2  5 x  3  0 . A B C D

x = -0.5 and 3 x = 0.5 and - 3 x = 0.5 and 3 The equation has no real solutions

[2] FAC 4 2

Question A42 Find the solution to x 2  12 x  12  0 in surd form. A

62 6

64 3

12  6

12  2 3

[2] FAC 4 2

Question A43 Solve the simultaneous equations: 95, 432 x + 64,717 y = 57,963

[2]

92, 404 x + 64,522 y = 80,068

FAC 4 3

Question A44 Solve the simultaneous equations:

a a +b

= 0.75

and

ab = 0.0225 (a + b ) (a + b + 1) 2

[2] FAC 4 3

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FAC: Question & Answer Bank – Questions

Page 19

Question A45 Solve the simultaneous equations:

a -1 a l2 = 162 (a - 1) 2 (a - 2)

[2] FAC 4 3

Question A46 If 2 x - 3 < 7 then: A B C D

-5 < x < 2 -2 < x < 5 -5 < x < 5 x<5

[1] FAC 4 4

Question A47 If x 2 - x - 12 < 0 then: A B C D

x<4 x < -3 and x < -3 -3 < x < 4

x>4

[1] FAC 4 4

Question A48 Find the range(s) of values of x for which

( x - 2)(9 x - 8) > 1. x2 + 2 x - 4

[5] FAC 4 4

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Page 20

FAC: Question & Answer Bank – Questions

Question A49 n

k

k 0

n 1

  k2  k 1 n

2 k2 k 0 n 1

 k2

k 0

2 n 1 k    2 k 2 k 1 n

 n  12  2  k 2

[1]

k 0

FAC 4 6

Question A50 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 B C D

£1,456 £1,586 £1,612 £3,172

[1] FAC 4 7

Question A51 Evaluate

A B C D

•

210 k k =1 4

70 210 280 840

[2] FAC 4 7

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Question A52 n

Simplify

Â (1 + 2i + i 2 ) . i =1

You may use the results

Â k = 12 n(n + 1)

and

k =1

Â k 2 = 16 n(n + 1)(2n + 1) .

k =1

2n3 + 4n 2 + 3n

2n3 + 9n 2 + 7 n + 6 6

8n 2 + 15n + 1 6

2n3 + 9n 2 + 13n 6

[2] FAC 4 8

Question A53 20

Â y Â x2 = y =1 x = y

x =1

y =1

Â x2 Â y 20

x =1 20

y=x 20

x2 Â y

x =1

y =1

x= y

Âx Â y 2

y =1

Â x2 Â y

[1] FAC 4 8

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FAC: Question & Answer Bank – Questions

Question A54 6

Calculate

Â Â 2 x( y - 1) .

You

may

use

the

x =1 y = x

Â k 2 = 16 n(n + 1)(2n + 1) and

k =1

A B C D

results

Â k = 12 n(n + 1) ,

k =1 n

Â k 3 = 14 n2 (n + 1)2 .

k =1

350 420 540 630

[5] FAC 4 8

Question A55 What is the coefficient of x5 y 3 in the expansion of the expression (2 x + 5 y )8 ? A B C D

56 224,000 448,000 5,600,000

[2] FAC 4 9

Question A56 Use the result (1  x) p  1  px 

p ( p 1) 2 x 2!



p ( p 1)( p  2) 3 x 3!

  to simplify

(2  5 x) (3 x  2) 2

by expanding as far as the term in x3 . A B C

1 11 57 2 243 3  x x  x 2 4 8 16 57 2 243 3 2  11x  x  x 2 4 23 28 2 244 3 2 x x  x 3 3 27 2 23 28 2 244 3  x x  x 9 27 27 243

[5] FAC 4 9

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Question A57 The interest rate i for a financial arrangement satisfies the following equation: 20,000(1 + i ) + 5,000(1 + i )½ + 2,500(1 + i )¼ = 30,000

(i)

Find an approximate value for i by using the approximate relationship (1 + i )t ª 1 + ti , quoting your answer to the nearest 0.01%.

(ii)

[3]

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: •

xk e =Â k =0 k ! x

•

( -1) k +1 x k log e (1 + x) = Â k k =1 •

p ( p - 1)...( p - k + 1) k x k! k =1

(1 + x) p = 1 + Â

Use these to simplify the following series expressions: (i)

x - 12 x 2 + 16 x3 -

(ii)

1 + 12 x 2 +

(iii)

x + 12 x 2 + 13 x3 + 14 x 4 +

[2]

(iv)

Ê 4ˆ Ê 5ˆ Ê 6ˆ 2 Ê 7 ˆ 3 + x + ÁË 0˜¯ ÁË 1˜¯ ÁË 2˜¯ x + ÁË 3˜¯ x +

[2]

1 x4 24

+

[2] [2]

[Total 8]

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FAC: Question & Answer Bank – Questions

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 i quantities 1 and 1 + i , show that 1 + > (1 + i )½ . 2 (i)(b) Show that 1 + (ii)

i > (1 + i )1/ n . n

[5]

Show that 1 + in < (1 + i ) n .

[2] [Total 7]

Question A60 Let Sn ( x) = x + 2 x 2 + 3x3 +  + nx n . (i)

By considering the expression

S n ( x) = (ii)

Sn ( x) - Sn ( x) , or otherwise, show that: x

x(1 - x n ) nx n +1 (1 - x) 2 (1 - x)

[3]

Write down an expression for lim Sn ( x) , stating the range of values of x for n Æ•

which convergence occurs.

[2] [Total 5]

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Chapter 5 Question A61 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 B C D

2.13‰ 2.18‰ 21.8‰ 21.3‰

[1] FAC 5 1

Question A62 An item has increased in cost by 30% to $58.50. What was the original cost? A B C D

$38.50 $40.95 $45 $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 B C D

14.29% 16.67% 83.33% 116.7%

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[1] FAC 5 2

Page 26

FAC: Question & Answer Bank â&#x20AC;&#x201C; Questions

Question A64 When calculating y = 3 x , a student uses x = 50 to 2 significant figures. What is the maximum absolute error possible here? A B C D

0.0122 0.0123 0.119 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 B C D

8.78% 11.4% 12.8% 13.9%

[2] FAC 5 3

Question A66 The following values from a standard normal distribution table were obtained: P ( Z < 1.20) = 0.88493

P( Z < 1.21) = 0.88686

Use linear interpolation to calculate the value of P( Z < 1.207) . A B C D

0.88507 0.88551 0.88590 0.88628

ÂŠ IFE: 2014 Examinations

[1] FAC 5 5

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Question A67 The Newton-Raphson method xn +1 = xn -

f ( xn ) is being used to find the solution f ¢ ( xn )

between 1 and 2 to the equation f ( x ) = x 2 + 3 x - 5 = 0 . What is x4 if x1 = 1 ? A B C D

1.2 1.1925 1.1926 1.2080

[2] FAC 5 6

Question A68 Expand (2i - 4)(3 - i ) and express it in its simplest form, where i = -1 . A B C D

-2i 2 + 10i - 12 10i - 14 10i - 10 2i - 14

[1] FAC 5 7

Question A69 What is the result if ( 2 - 3i ) is multiplied by its complex conjugate? A B C D

-5 -5 - 12i 13 13 - 12i

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[1] FAC 5 7

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FAC: Question & Answer Bank – Questions

Question A70 Simplify

2-i , where i = -1 . 2+i

3 5

1 - 43 i

1 - 54 i

5 3

- 54 i

- 43 i

[2] FAC 5 7

Question A71 What is the modulus, r , and argument, q , in radians, of the complex number (5 - 3i ) ? A B C D

r r r r

= 4, q = 0.540 = 5.831, q = -0.540 = 4, q = 30.96 = 5.831, q = -30.96

[2] FAC 5 7

Question A72 Find the roots of x 2 + 3x + 2.5 = 0 . A B C D

x = -0.5 + 1.5i or - 0.5 - 1.5i x = 0.5 + 1.5i or 0.5 - 1.5i x = 1.5 + 0.5i or 1.5 - 0.5i x = -1.5 + 0.5i or - 1.5 - 0.5i

[2] FAC 5 7

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Question A73 In the following equations, k is measured in £ and a is a dimensionless quantity:

k1 =

a a 2a + k , k2 = 2 , k3 = 3 l l l

(i)

Determine the units of measurement of k 1,k 2 ,k 3 .

(ii)

Find the value of the constant c that would make the quantity

[3]

dimensionless.

k3 k 2c

[1] [Total 4]

Question A74 Let z1 = 2 + 4i and z2 = 1 - 2i . (i)

Write each of the following in the form a + bi where a, b Œ R : z1 + z2 , z1 - z2 , z1z2 ,

z1 and z12 + z22 z2

[5]

(ii)

Calculate z1 z2 and z1z2 , and comment on your answers.

[3]

(iii)

Calculate z1 z2 and z1 z2 , and comment on your answers.

[3]

Note that z denotes the complex conjugate of z . (iv)

Express 3z1 + 2 z2 in the form reiq .

[2] [Total 13]

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FAC: Question & Answer Bank â&#x20AC;&#x201C; Questions

Question A75 Consider the equation 2 z 3 - 9 z 2 + 14 z - 5 = 0 . (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 z = 1 when represented on an Argand diagram?

[1] [Total 7]

Question A76 Find the solution of the difference equation: yt + 6 yt -1 + 8 yt - 2 = 0

given that y0 and y1 are both equal to 5.

[4] FAC 5 8

ÂŠ IFE: 2014 Examinations

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Chapter 6 Question A77 Which of the options best describes the function 3x  5 x 2 ? A B C D

O( x) and o( x) O( x) but not o( x) o( x) but not O( x) none of the above

[1] FAC 6 1

Question A78 What is the supremum of the sequence A B C D

1 2 3 , , ,... ? 4 7 10

13  0.333 0

[1] FAC 6 1

Question A79 Which of the following best describes what dy dx represents? A B C D

d y d x for small values of d x and d y the change in the value of the graph between 0 and x the gradient of y = f ( x) the rate of change of a chord between x = a and x = a + h

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[1] FAC 6.2

Page 32

FAC: Question & Answer Bank – Questions

Question A80 If f ( x) = 5 x , which of the following gives A B C D

dy f ( x + h) - f ( x) ? = lim + dx hÆ 0 h

0 1 5 x

[1] FAC 6.2

Question A81 Differentiate f (t )  5(4t ) with respect to t . A

20t 1

5(4t )ln 4

C D

5t 4t 1 20ln t

[1] FAC 6 3

Question A82 Find the gradient of y = 2 x3 - x 2 + 3 when x = 2 . A B C D

15 20 23 44

[1] FAC 6 3

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Question A83 Find the derivative of f ( x) 



C D

4 x

12 5x8 5 12 5x 2 5 4

3 x2 5 5

[1]

3 x8 5 5

FAC 6 3

Question A84 Find f ¢(u ) where f (u )  (5u 2  7)9 . A

9(5u 2 - 7)10

9(5u 2 - 7)8

45u (5u 2 - 7)8

90u (5u 2 - 7)8

[1] FAC 6 4

Question A85 Find M ¢(0) if M (t )  e t ½



C D

e 1

2 2

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[2] FAC 6 4

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FAC: Question & Answer Bank – Questions

Question A86 t  Find M ¢(0) if M (t )   ln 1   .   A B C D

-a

a l al -a l

[2] FAC 6 4

Question A87 2 4   Find f ¢( x) where f ( x)  ln  2e x  2 x  . e  

1 4 e2 x 2 4 2 xe x  2 x e 2 4 2e x  2 x e 2 8 4 xe x  2 x e 2 4 2e x  2 x e 2 4 2 xe x  2 x e 2 2 ex  2x e 2

2e x 

[2] FAC 6 4

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Question A88



Find f ¢( x) where f ( x)  ln 4  e6(5 x 3) A

60(5 x  3)

60(5 x  3) ln 4  e6(5 x 3)





.



60(5 x  3) 4  e6(5 x 3)

60(5 x  3)e 6(5 x 3) 4e

6(5 x  3)2

[5] FAC 6 4

Question A89 Differentiate y  (4  3 x)( x 2  2 x  1)5 with respect to x . A

( x 2  2 x  1) 4 (3( x 2  2 x  1)  5(4 x  3)(2 x  2))

( x 2  2 x  1) 4 (33 x 2  76 x  43)

3( x 2  2 x  1)5  5(4  3 x)(2 x  2) 4

15(2 x  2)( x 2  2 x  1) 4

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[5] FAC 6 4

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FAC: Question & Answer Bank – Questions

Question A90 Find f ¢¢( x) where f ( x)  32 x 2e 4 x . A

64(1  8 x 2 )e 4 x

64(1  8 x  8 x 2 )e 4 x

1,024e4 x

64(1  8 x 2  8 x 4 )e 4 x

[5] FAC 6 5

Question A91 Differentiate y 

3x 2

6 - 12 x + 3 x 2 ex

6 - 12 x - 3 x 2 ex

6 + 12 x + 3 x 2 ex

6 + 12 x - 3 x 2 ex

twice with respect to x .

[5] FAC 6 5

Question A92 Calculate the maximum value of the function f ( x)  5  2 x  3 x 2 . A B

6 4

16 3

- 13

[2] FAC 6 6

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Question A93 Which of the following is a graph of y = x( x + 3)( x - 5) ? A y

1 2 3 4 5 x

-5 -4 -3 -2 -1

B y

1 2 3 4 5 x

-5 -4 -3 -2 -1

C y

1 2 3 4 5 x

-5 -4 -3 -2 -1

D y

-5 -4 -3 -2 -1

1 2 3 4 5 x

[2] FAC 6 6

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Question A94 Find and distinguish between the turning points of the function: f ( x)  x3  x 2  x  7 . A

5 ) maximum 1,6  and minimum ( 13 ,7 27

) maximum  1,6  and minimum ( 13 ,6 16 27

5 maximum ( 13 ,7 27 ) and minimum 1,6 

maximum ( 13 ,6 16 ) and minimum  1,6  27

[5] FAC 6 6

Question A95 Using log differentiation or otherwise, find the value of x for which f ( x) =

90 4 -3 x x e 2

is a maximum. A

- 43

- 34

3 4 4 3

[2] FAC 6 6

Question A96 What is

(

)

∂ 2 3 axy + b ( xy ) + c ( xy ) where a , b and c are constants? ∂x

a + 2b ( yx ) + 3c ( xy )

ax 2 y by 2 x3 cy 3 x 4 + + 2 3 4

ay + 2byx + 3cyx 2

ay + 2by 2 x + 3cy 3 x 2

[1] FAC 6 7

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Question A97 ∂3 f ∂f ¥ ? For the function f ( x, y, z ) = ( xyz ) , what is ∂x∂y∂z ∂x 2

8 xyz - 2 x ( yz )

8 yz

8x 2 ( yz )

16x 2 ( yz )

3 3

[2] FAC 6 7

Question A98 Find the turning points and their nature for the function f ( x, y )  x3  3 x 2  2 xy  y 2 by considering the roots of the equation:  2  f  2  y 

x  x0 y  y0

 2   f    2  x 

x  x0 y  y0

  2   f    xy  

   0 x  x0  y  y0 

Local maximum at x  0, y  0

Saddle point at x  43 , y 

Local maximum at x  0, y  0 and saddle point at x  43 , y 

Local minimum at x  0, y  0 and maximum at x  43 , y 

4 3

[5] FAC 6 8

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Question A99 Determine the extrema of the function f ( x, y )  5 x  3 y subject to the constraint x 2  y 2  136 .

You may make use of the Lagrangian function: L  f ( x1,, xn )   g ( x1,, xn ) . A B C D

maximum at minimum at maximum at maximum at

f (10, -6) f ( -5,3) f (10, -6) and minimum at f ( -10,6) f (5, -3)

[5] FAC 6 9

Question A100 Evaluate: (i)

e3 x - e -3 x lim xÆ0 e x - e - x

(ii)

1   lim  1   n   2n 

[2]

[2] [Total 4]

Question A101 Differentiate the following function with respect to i : 100 1 - (1 + i ) -10 +5 i (1 + i )10 and evaluate the derivative at i = 0.05 .

[3] FAC 6 4

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Question A102 Find a formula for the n th derivative of the function 2 x +

1 . x

[4] FAC 6 5

Question A103 By considering log y , or otherwise, find an expression in terms of l for the maximum value attained by the function y = agl a xg -1 (l + xg ) -a -1 when a = 1 and g = 4 . [5] FAC 6 6

Question A104 A function is defined by: f ( x) = xa (1 + x) b , x Œ[0,1]

where a , b Œ R 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)

a = 1, b = -3

[3]

(ii)

a =b =2

[3]

(iii)

a = -½, b = 1½ .

[3] [Total 9]

Question A105 Find the positions of the extrema of f ( x, y ) = x3 - 2 x 2 + 2 y 2 and determine their nature. [5] FAC 6 8

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Question A106 Find the positions of the extrema of f ( x, y, z ) = x + y + z , subject to the constraints x + 2 y = 0 and x 2 + y 2 + z 2 = 1 .

[5] FAC 6 9

Question A107 According to a mortality table, the instantaneous rate of mortality at age x , which is denoted by m ( x) , is calculated from the formula:   x  70   x  70     exp b0  b1    , x ≥ 17  50   50   

 ( x)  a0  a1 

and a0 = -0.00338415 , a1 = -0.00386512 , b0 = -3.352236 and b1 = 4.656042 . (i)

Show that m (20) = 0.000814 and find the value of m (40) .

(ii)

Find the age in the range 20 £ x £ 40 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 m ( x) by the relationship:

[3]

1 q ( x)  1  exp     ( x  t ) dt   0 

Find the value of q (20) .

[5] [Total 12]

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Chapter 7 Question A108 Which of the following is NOT true? b

Ú f ( x) dx = - Ú f ( x) dx a

Ú f ( x) dx = Ú f ( x) dx - Ú f ( x) dx b

Ú dx f ( x) dx = f (b) - f (a ) a

Ú dx f ( x) dx is used to find the area under the

f ( x) curve

[1]

FAC 7 1

Question A109 What is the integral of 6 x 2 + 6 x + 6 ? A

12 x + 6 + c

2 x3 + 3 x 2 + 6 x + c

6 x3 + 6 x 2 + 6 x + c

3x3 + 6 x 2 + 6 x + c

[1] FAC 7 2

Question A110 2

What is the integral of

Úx

0.75

dx ?

A B C D

1.351 0.636 0.477 -0.119

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[1] FAC 7 2

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FAC: Question & Answer Bank – Questions

Question A111 What is the integral of x 2 + x -2 ? A

x 2 + x -2 +c 2

x3 + x -3 +c 3

x3 - 3x -1 +c 3

x3 - x -3 +c 3

[1] FAC 7 1

Question A112 4

What is Ú 5 x dx ? 2

A B

1, 260 ln 6 600 ln 5

583 13

625 25 ln 4 ln 2

[1] FAC 7 2

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Question A113 What is

Ú Bc

Bc x + const ln x

Bc x + const

Bc x +1 + const x +1

Bc x + const ln c

dx ?

[1] FAC 7 2

Question A114 3

Evaluate Ú l e - l x dx . 0

l (e -3l - 1)

1 - e -3l

l (1 - e -3l )

1 - e -4.5l

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[1] FAC 7 2

Page 46

FAC: Question & Answer Bank – Questions

Question A115 What is the integral of A B C D

5 ? x +1

5ln ( x + 1) +c -10

( x + 1)2 -5

( x + 1)

+c +c

5 +c ln ( x + 1)

[1] FAC 7 2

Question A116 Evaluate A B C D

• 3 -5 x 4

Ú0

dx .

0 0.05 1 the integral does not converge

[2] FAC 7 2

Question A117 Evaluate A B C D

x5 Ú0 (10 + x6 )3 dx . 1

0.00000132 0.00000792 0.000145 0.000868

[2] FAC 7 2

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Question A118 3x 2 + 3 What is the integral of 3 , for x > 0 ? 2x + 6x + 1

x3 + 3 x +c 0.5 x 4 + 3 x 2 + x

ln 3 x 2 + 3 + c

0.5ln 2 x3 + 6 x + 1 + c

(3x

(

)

(

) (

)

+ 3 x ln 2 x3 + 6 x + 1 + c

[2] FAC 7 2 & 3

Question A119 Use partial fractions to find the integral of

(2 x - 4)3 ¸Ô ( x + 1) ˝˛Ô

1 Ô ln k 2 Ì

ln k ( x + 1)3 (2 x - 4)½

3ln( x + 1) - 12 ln(2 x - 4) + c

3ln(2 x - 4) + 12 ln( x + 1) + c

ÓÔ

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

[5] FAC 7 3

Question A120 1

What is

 xe

dx ?

A B C D

-10.778 0.5 1 2.097

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[2] FAC 7 3

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FAC: Question & Answer Bank – Questions

Question A121 Leibniz’s formula states that: b( x)

b( x)

d ∂f f ( x, t ) dt = b ¢( x) f ( x, b( x)) - a ¢( x) f ( x, a( x)) + Ú ( x, t ) dt . Ú dx a ( x ) x ∂ a( x) x

d 3x 2 - 2t dt ? What is dx Ú 0

3x 2

6x 2

6 x2 - 8x

9 x2 - 2 x

[2] FAC 7 3

Question A122 15 10

Find the value of

Ú Ú

(5 x + y ) dy dx .

x =5 y =5

A B C D

2,875 3,250 2,812.5 1,406.25

[2] FAC 7 5

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Question A123 Which of the following does NOT converge? •

Úe

-2 x

Ú ln x dx

•

Úx

-2

•

Ú x3 dx

[1]

FAC 7 4

Question A124 Using the trapezium rule and 7 ordinates, what is the approximate area under the curve y = (7 - x) 2 between x = 1 and x = 4 ?

A B C D

40.1875 63.125 80.375 138.25

[2] FAC 7 6

Question A125 If

to its fourth term, using the Maclaurin expansion f ¢¢(0) 2 f ( x) = f (0) + f ¢(0) x + x + ◊◊◊ , what is the value of e2 based on this? 2!

A B C D

expanded

5 6.333 7.389 7.667

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[2] FAC 7 7

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FAC: Question & Answer Bank – Questions

Question A126 Which of the following is the correct first five terms of the Taylor expansion of ln [(3 + x)(3 + y )] about (0,0) ? You are given that the Taylor’s expansion of f ( x, y ) about (a, b) is: f ( x, y ) = f ( a , b ) +

È 1 Í∂ 2 f 2! Í ∂ x 2 Î

1 È∂ f Í 1! Í ∂ x Î

( x - a) + ( a ,b )

( x - a)2 + 2 ( a ,b )

∂2 f ∂ x∂ y

∂ f ∂y

˘ ( y - b) ˙ ˙˚ ( a ,b ) ( x - a )( y - b) +

( a ,b )

∂2 f ∂ y2

˘ ( y - b) 2 ˙ ˙ ( a ,b ) ˚

+

A B C D

1 1 1 1 ln 9 + x + y - x 2 - y 2 3 3 18 18 1 1 1 1 ln 9 + x + y - x 2 - y 2 3 3 9 9 1 1 1 1 ln 9 + x + y + x 2 + y 2 3 3 9 9 1 1 1 1 ln 9 + x + y + x 2 + y 2 3 3 18 18

[5] FAC 7 7

Question A127 Given that A B C D

dy y = , y = e 2 when x = 0 , what is the value of y when x = 2 ? dx 3x + 1

51.723 7 14.135 6

[2] FAC 7 8

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Question A128 •

The gamma function is defined by G ( x) = Ú t x -1e -t dt . 0

(i)

Show that G ( x) = ( x - 1)G ( x - 1) for values of x for which both gamma functions are defined. [3]

(ii)

Show that G (1) = 1 .

(iii)

You are given that G (½) = p .

[2]

By considering G (½) and applying the • 1 - 12 z 2 e dz = 1 . [4] substitution t = 12 z 2 , or otherwise, show that Ú -• 2p [Total 9]

Question A129 Find the area of the region enclosed by the x axis and the curve y = 6 + x - 2 x 2 . [4] FAC 7 3

Question A130 y 1

Evaluate

ÚÚ f ( x, y) dxdy

where f ( x, y ) = 2 x + 6 y and the region of integration is the

shaded area shown in the diagram.

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[6] FAC 7 5

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FAC: Question & Answer Bank – Questions

Question A131 Show that I = Ú

x dx = 0.035103 using at least three different methods. (5 + x) 2 [3 marks for each correct method]

Question A132 (i)

Calculate the value of the following integral, where 0.5 < x < y < 1 : y

(4 y + 1)e

y = 0.5

(ii)

xe 2 x dx dy

[4]

x = 0.5

Explain why it may be considered easier to integrate with respect to x first [5] rather than integrating with respect to y first. [Total 9]

Question A133 (i)

Write out the first four terms in the Taylor series for e x and (1 + y ) -1 , stating [2] the range of values of x and y for which these series are valid.

(ii)

Hence determine the coefficients a, b in the series expansion: x = 1 + ax + bx 2 + O( x3 ) e -1 x

[5] [Total 7]

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Question A134 1 A B ∫ + . P(2 - 5P) P 2 - 5P

(i)

Find constants A and B such that

(ii)

Hence, or otherwise, solve the differential equation:

[2]

1 dP = 2 - 5P P dt for the function P (t ) , where 0 < P (t ) < 52 , subject to the boundary condition that 2 . P(0) = 15

[4] [Total 6]

Question A135 Find the particular solution of the differential equation: ( x + 1)

dy - x( x + 1) = y dx

where x ≥ 0 , given that y = 0 when x = 0 .

[5] FAC 7 8

Chapter 8 Question A136  2  2  10      If a   1  and b   5  then the values x and y such that xa  yb   7  are:  4  3   18       

A B C D

x  1 and y  6 x  4 and y  1 x  3 and y  2 x  2 and y  1

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FAC: Question & Answer Bank – Questions

Question A137 The unit vector in the direction of 3i  2 j  4k is given by: A

29(3i  2 j  4k )

 3 1   2 21    4

 3 1   2 29    4

3 i 2 21 21

j

4 k 21

[1] FAC 8 1

Question A138 4 1   The angle between  1 and  3  is: 3  2     

A B C D

65.4º 74.8º 105.2º 114.6º

[2] FAC 8 1

Question A139 Which one of the vectors is orthogonal to 3i  2 j  k ? A B C D

4i  4 j  4k 4i  4 j  4k 4i  4 j  4k 4i  4 j  4k

[2] FAC 8 1

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Question A140  2 4  1 2 If A    then AB is given by:  and B    3 5   3 4  A B C D

 2 8     9 20   8 14     6 32   8 14     18 32   14 20     18 26 

[2] FAC 8 2

Question A141  3 2 4  If A   1 0 3  then det A is given by:  3 1 2   

A B C D

33 5 -1 -19

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[2] FAC 8 2

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FAC: Question & Answer Bank – Questions

Question A142  2 1 1 If A    then A is given by:  4 3 

A B C D

1  3 1     2  4 2   0.3 0.1    0.4 0.2  1  1 3     2  2 4  1   2   1.5 0.5 

[2] FAC 8 2

Question A143 The eigenvalues,  , of matrix A solve: A B C D

det( A   I )  0 det( A   I)  0 det( A  I )  0 det( A   )  0

[1] FAC 8 2

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Question A144 3 2  Which of the following is NOT an eigenvector of A   :  6 1

A B C D

 2    6  3    1  1    3   2    2

[2] FAC 8 2

Question A145 (i)

By considering expressions involving scalar products, find a unit vector of the form ai + bj + ck that is perpendicular to the displacement vectors -2i + 3j and 10i + k . [4]

(ii)

Write down the other unit vector perpendicular to -2i + 3 j and 10i + k .

[1] [Total 5]

Question A146 By solving a set of simultaneous linear equations, or otherwise, find the inverse of the  0.2 0.2 0.2    matrix P   0.2 0 [5] 0.2  .  0 0.4 0.2    FAC 8 2

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Question A147 The probability density function of the multivariate normal distribution is given by: f ( x) 

 1  exp   (x   )T  1 (x   )   2  (2 ) n det( )

1 2 1  2 Evaluate f (x) when n  2 ,     ,     and x    . 1 4 0  1 

[5] FAC 8 2

Glossary Question A148 The abbreviation eg means: A B C D

for example compared with and so on that is to say

[1] FAC Gloss

Question A149 What is the plural of matrix? A B C D

matrices matrixes matrises matrix

[1] FAC Gloss

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Question A150 A weak basis for an actuarial calculation: A B C D

makes many assumptions makes optimistic assumptions makes few assumptions makes pessimistic assumptions

[1] FAC Gloss

Question A151 A deterministic model has: A B C D

optimistic assumptions assumptions which allow for random variation many assumptions 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 B C D

advise, practise advise, practice advice, practise advice, practice

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[1] FAC Gloss

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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 B C D

£18,000 £9,000 £7,500 £6,000

[1] FAC Gloss

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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.

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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. x = -2 is a member of this set. Various corrections are possible eg changing  to  + or  , adding the extra condition x π -2 or changing = ∆ to π ∆ (!). [1]

(iii)

TRUE. If we “complete the square”, we can write x 2 - 5 x + 7 = ( x - 52 ) 2 + 43 ,

which always takes positive values. (iv)

[2]

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]

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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 64,000 ¥ 0.975 ¥ 1.67 ¥

1 units. P

The unit price has now increased to $1.05P . So the dollar value of the units would be 1 [1] 64,000 ¥ 0.975 ¥ 1.67 ¥ ¥ 1.05P = 64,000 ¥ 0.975 ¥ 1.67 ¥ 1.05 . P

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] 1.60

Solution A12 B

Solution A13 D

Solution A14 B Significant figures are counted from the first non-zero digit.

Solution A15 C

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Solution A16 D 1 - 1.062-10 = 7.743 0.062 1.062

Solution A17 B 6 8 10 12     29.2 to 3 SF 2 3 1.08 1.08 1.08 1.084

Solution A18 D  1  ln 5  1.02 2  1 2  1 exp    exp    0.7367974     0.8 0.8 2   0.8 2  2   2  1



1 0.8 2

exp(0.2714352)  0.380

Solution A19 B 6! 5 12 4 720 +4 + 20 = + 45 12 + 4 20 = 240 + 1.782 + 2.115 = 243.9 3 3

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Solution A20 Using the formulas given:

and:

v = (1 + i ) -1 = 1.1-1 ( = 0.90909)

[1]

d = iv = 0.1 / 1.1 ( = 0.09091)

[1]

So: 1 - vn 1 - 1.1-10 - nv n - 10 ¥ 1.1-10 0.1 / 1.1 d = = 29.04 0.10 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: 43.6(1 + i )10 - 43.92(1 + i )5 - 60.192 = 0

[1]

This is a quadratic equation in (1 + i )5 . Using the quadratic formula, we get:

(1 + i )5 =

43.92 ± (43.92) 2 - 4(43.6)( -60.192) 43.92 ± 111.474 = 2 ¥ 43.6 87.2

[1]

So: (1 + i )5 = 1.782 or - 0.775

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[1]

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FAC: Question & Answer Bank – Solutions

Disregarding the second possibility (which corresponds to a negative interest rate), we find that: i = (1.782)1/5 - 1 = 0.12248 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 21 14 years is a quarter of the way from 20 to 25. So F1 is calculated as: 16.5 + 14 (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 + k ¥ 17.3 = 50% ¥ 0.657 ¥ 17.7 ﬁ k = 9.8%

[2]

(i)(b) Percentage pension We now have: 3.625 + 15% ¥ 17.3 = t ¥ 0.657 ¥ 17.7 ﬁ t = 53.5%

[2]

(i)(c) Contribution rate 3 We now have n = 16 12 and the calculation is:

3.625 + k ¥ 13.875 = 50% ¥ 0.7255 ¥ 19.9 ﬁ k = 25.9%

[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.

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(ii)

Page 7

Recommended contribution rate

The calculation for Ms Jones (with n = 17½ ) is:

5 + k ¥ 14.75 =

2 3

¥ 0.708 ¥ 19.9 ﬁ k = 29.8%

[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 t = 12 when the payments start, but it will increase by 1 on 1 January each year. [2] So the member’s first payment will be (ii)

1 ¥ 3,620 ¥ 1.0512 = £541.75 . 12

[2]

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 £541.75 ¥ 1.052 during 2018 … 12 payments of £541.75 ¥ 1.059 during 2025 4 payments of £541.75 ¥ 1.0510 during 2026

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[2]

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FAC: Question & Answer Bank – Solutions

The total amount is: 8 ¥ 541.75 + 12 ¥ 541.75 ¥ (1.05 + + 1.059 ) + 4 ¥ 541.75 ¥ 1.0510

[1]

The middle terms can be summed as a GP, which gives: 1.05 - 1.0510 8 ¥ 541.75 + 12 ¥ 541.75 ¥ + 4 ¥ 541.75 ¥ 1.0510 = £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 541.75 ¥ 1.055 = £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 y = e x C is the graph of y = ln x D is the graph of y = x 2

Solution A25 B A is the graph of y = C is the graph of y =

1 x

1 x2

D is the graph of y = x 2

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Solution A26 A

y = f (2 x - 1) squashes the graph horizontally by a factor of 2 and shifts it 1 unit to the right. B is y = f (0.5 x + 1) C is y = f (2 x + 1) D is y = f (0.5 x - 1)

Solution A27 A

Solution A28 A The person is 55 years and approximately 9 months.

Solution A29 C 4x + 1 < 5 -5 < 4 x + 1 < 5 -6 < 4 x < 4 -1.5 < x < 1

Solution A30 A

e -0.8 = 0.449 and

p 4 = 0.443 so min(e - x , p 4) = 0.443

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FAC: Question & Answer Bank – Solutions

Solution A31 A 100! 100 ¥ 99 ¥ 98 = = 161,700 97!3! 3!

Solution A32 A G (5.5) = 4.5G (4.5) = 4.5(3.5)G (3.5) = 4.5(3.5)(2.5)G (2.5) = 4.5(3.5)(2.5)(1.5)G (1.5) = 4.5(3.5)(2.5)(1.5)(0.5)G (0.5) = 4.5(3.5)(2.5)(1.5)(0.5) p = 52.34

Solution A33 D G (n + 1)(3n + 1)! n!(3n + 1)! = = n(3n + 1)(3n) G (3n)(n - 1)! (3n - 1)!(n - 1)!

Solution A34 To 2 significant figures, these work out as follows: (i)

e e = e1.5 = 4.5

[1]

(ii)

log e 0.00001 = -12

[1]

(iii)

tanh -1 0.9 = 1.5

[1]

(iv)

G (12) = 11! = 40,000,000

[1]

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Solution A35 D

( )

2 xb ¥ x 2

+ 2 xb x3 = 2 xb x 2b + 2 xb x3 = 2 x3b + 2 xb + 3

Solution A36 C

 e3    e 2 x  x

 e3 x  e6 x  e 3 x 

1 e3 x

Solution A37 C See the definition in Chapter 4 Section 1.2 of the notes.

Solution A38 D 2  x2  x x 2log a    log a y  log a    log a y  log a    y   y  y  

Solution A39 C x  3 5  2 x ( x  3)( x  1) (5  2 x)(3 x  1)    3 x  1 x  1 (3x  1)( x  1) ( x  1)(3 x  1) x2  2 x  3 6 x 2  13 x  5   (3 x  1)( x  1) ( x  1)(3 x  1) 

7 x 2  11x  8 (3 x  1)( x  1)

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Solution A40 B ( x + 3)( x - 2) x+3 x2 + x - 6 x+3 ∏ = ∏ 2 2 2 x - 3x - 2 2 x + 3 x + 1 (2 x + 1)( x - 2) (2 x + 1)( x + 1) =

(2 x + 1) ( x + 1) x+3 ¥ (2 x + 1) ( x - 2) ( x + 3) ( x - 2)

( x + 1) ( x - 2) 2

Solution A41 A Factorising gives (2 x + 1)( x - 3) = 0 . Hence x = -0.5,3 .

Solution A42 A x=

12 ± ( -12) 2 - 4(1)(12) 12 ± 96 12 ± 4 6 = = =6±2 6 2(1) 2 2

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

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x

Page 13

de  bf 64522  57963  64717  80068   8.130 ad  bc 95432  64522  64717  92404

[1]

Similarly (ie if we just swap a / b and c / d in this formula), we get: ce  af 92404  57963  95432  80068   12.884 bc  ad 64717  92404  95432  64522

y

[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:

a a +b

3 ﬁ 4a = 3(a + b ) ﬁ a = 3b 4

[½]

Changing all the a ’s to b ’s in the second equation then gives: 3b 2 3 = 0.0225 ﬁ = 0.0225 2 16(4 b + 1) (4 b ) (4 b + 1)

[½]

Rearranging:

1



 1  1.833   4  16  0.0225 

[½]

ﬁ a = 3b = 5.5

[½]

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Solution A45 The easiest way to solve these equations is to square the first equation and substitute it into the second equation: 2

a Ê l ˆ a = 162 ﬁ 92 = 162 ÁË ˜¯ a -1 a - 2 a -2

[½]

This gives:

a a -2

[½]

From which:

a = 2(a - 2) ﬁ a = 4 This gives l = 27 .

[½] [½]

Solution A46 B 2x - 3 < 7 -7 < 2 x - 3 < 7 -4 < 2 x < 10 -2 < x < 5

Solution A47 D Factorising gives ( x - 4)( x + 3) < 0 . So either x - 4 < 0 and x + 3 > 0 (ie -3 < x < 4 ) or x - 4 > 0 and x + 3 < 0 (but it is not possible for both of these to be true at the same time).

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Solution A48 Rearranging the inequality so that we have zero on the RHS: ( x - 2)(9 x - 8) -1 > 0 x2 + 2 x - 4

Expressing the LHS as a single fraction: ( x - 2)(9 x - 8) - ( x 2 + 2 x - 4) >0 x2 + 2 x - 4

[½]

Simplifying: (9 x 2 - 26 x + 16) - ( x 2 + 2 x - 4) >0 x2 + 2 x - 4

8 x 2 - 28 x + 20 >0 x2 + 2 x - 4

[1]

We can factorise the numerator: 4(2 x - 5)( x - 1) >0 x2 + 2 x - 4

[½]

The denominator equals zero when x = -1 ± 5 (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:

1 b

2.5

[1] So the ratio will be positive when both graphs are positive or both are negative, ie when x < -1 - 5 ( = a) or 1 < x < -1 + 5 ( = b) or x > 2.5 .

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Page 16

FAC: Question & Answer Bank – Solutions

Solution A49 D n

n 1

k 0

k 1

 k2   k2 

 k2 

k 0

 k 2  (n  1)2  2  k 2  (n  1)2

Solution A50 B This is an arithmetic series with a = 5 , d = 1 and n = 52 . Hence: S52 = 12 (52)(2 ¥ 5 + 51 ¥ 1) = £1,586

Solution A51 A •

210 210 210 210 = + 2 + 3 + k 4 4 4 4 k =1

This is an infinite geometric series with a =

S• =

210 1 and r = . Hence: 4 4

210 4 = 70 1-1 4

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Solution A52 D n

i =1

Â (1 + 2i + i 2 ) = Â1 + 2Â i + Â i 2 =n+2

(

1 n( n + 1) 2

) + 16 n(n + 1)(2n + 1)

= n + n 2 + n + 16 (2n3 + 3n 2 + n) =

2n3 + 9n 2 + 13n 6

Solution A53 D Considering the order of x and y we have 1 £ y £ x £ 20 . 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

Â Â 2 x( y - 1) = x =1 y = x

y =1

x =1

Â 2( y - 1) Â x 6

Â 2( y - 1) 12 y( y + 1) y =1

Â y3 - y 2 y =1

Â y - Â y2 3

y =1

= 14 62 (6 + 1) 2 - 16 6(6 + 1)(2(6) + 1) = 350

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FAC: Question & Answer Bank – Solutions

Solution A55 B The term will be 8C5 (2 x)5 (5 y )3 = 56 ¥ 25 ¥ 53 x5 y 3 = 224,000 x5 y 3

Solution A56 A (2  5 x) (3 x  2)



(2  5 x) 22 ( 32

x  1) 2

 14 (2  5 x)(1  32 x) 2  14 (2  5 x) 1  (2)( 32 x)    14 (2  5 x)(1  3 x   14 (2  6 x  

27 2 x 2

27 2 x 4

( 2)( 3) 3 (2 2!

x)2 

( 2)( 3)( 4) 3 (2 3!

x )3    

 27 x3   ) 2

 27 x3   )  (5 x  15 x 2  135 x3   )  4 

x3    12  11 x  57 x 2  243 4 8 16

Solution A57 (i)

Approximate value for i

Using the approximate relationship, the equation given becomes:

20,000(1 + i ) + 5,000(1 + 12 i ) + 2,500(1 + 14 i ) = 30,000

[1]

27,500 + 23,125i = 30,000 ﬁ i = 10.81%

[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: (1 + i )t ª 1 + ti + 12 t (t - 1)i 2

[1]

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The equation given would then become:

20,000(1 + i ) + 5,000 È1 + 12 i + 12 Î

( 12 )(- 12 ) i 2 ˘˚

+ 2,500 È1 + 14 i + 12 Î

( 14 ) (- 34 ) i2 ˘˚

= 30,000

[1]

ie: 27,500 + 23,125i - 859.375i 2 = 30,000 ﬁ - 2,500 + 23,125i - 859.375i 2 = 0

[1]

Solving this quadratic equation, we get: i=

-23,125 ± (23,125) 2 - 4( -859.375)( -2,500) -23,125 ± 22,938.4 = 2( -859.375) -1,718.75

To get the root consistent with part (i), we need to take the + sign, which gives i ª 10.85% . [1]

Solution A58 (i)

x - 12 x 2 + 16 x3 -

(ii)

1 + 12 x 2 +

1 x4 24

+ = 1 - e - x

[2]

+ = 12 (e x + e - x )

[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)

x + 12 x 2 + 13 x3 + 14 x 4 + = - log e (1 - x)

[2]

(iv)

 4 5 6 2 7 3 (5)(6) 5 ( x) 2       x    x    x   (1  x)  1  5 x  2!  0 1  2  3

[2]

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FAC: Question & Answer Bank – Solutions

Comment

This last one isn’t at all obvious, but if you write out the terms in the series (1 - x) -5 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: i 1 + (1 + i ) > 1(1 + i ) ie 1 + > (1 + i )½ 2 2

[2]

(i)(b) Show that If we apply the AGM inequality to the n quantities consisting of n - 1 1’s and a 1 + i , we get: (n - 1) ¥ 1 + (1 + i ) n i > 1 ¥ 1 ¥ ¥ 1 ¥ (1 + i ) ie 1 + > (1 + i )1/ n n n (ii)

[3]

Show that

If we expand the RHS of the expression in the question using a binomial expansion, we get:

n n (1  i ) n  1    i    i 2   i n 1 2

[1]

Since i > 0 , all the terms in this series are positive. So, if we just keep the first two, we know that: (1 + i ) n > 1 + ni ie 1 + in < (1 + i ) n

[1]

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Solution A60 (i)

Show that

We have: Sn ( x) = 1 + 2 x + 3x 2 +  nx n -1 x Sn ( x) = x + 2 x 2 +  + (n - 1) x n -1 + nx n Subtracting gives: Sn ( x) - Sn ( x) = 1 + x + x 2 +  + x n -1 - nx n x

[1]

The LHS can be factorised and the terms except the last on the RHS can be summed as a GP: n

1 x 1   nx n   1 Sn ( x)  1 x x 

1 x 1 x     nx n  Sn ( x)  1 x  x 

[1]

 x  Multiplying both sides by   , then gives the required result: 1 x  S n ( x) =

x(1 - x n ) nx n +1 1- x (1 - x) 2

[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 < x < 1 , x n will vanish as n Æ • and we will have: lim Sn ( x) =

n Æ•

x (1 - x) 2

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FAC: Question & Answer Bank – Solutions

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

50.5 - 3 50 = 3.6963 - 3.6840 = 0.0122

50 - 3 49.5 = 3.6840 - 3.6717 = 0.0123

So the maximum absolute error is 0.0123.

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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 estimate by 3,049 ie

4500 = 21,951 , which is less than our 0.205

3,049 = 13.9% . 21,951

Solution A66 D

P ( Z < 1.207) = P ( Z < 1.20) + 0.7 [ P ( Z < 1.21) - P( Z < 1.20)] = 0.88493 + 0.7(0.88686 - 0.88493) = 0.88628

Solution A67 C f ( x ) = x 2 + 3x - 5 = 0 ﬁ

f ¢( x) = 2 x + 3 ﬁ xn +1 = xn -

xn2 + 3 xn - 5 2 xn + 3

-1 x12 + 3 x1 - 5 =1= 1.2 x2 = x1 2 x1 + 3 5

0.04 x22 + 3 x2 - 5 = 1.2 = 1.1926 x3 = x2 2 x2 + 3 5.4 x32 + 3 x3 - 5 0.00005486 = 1.1926 = 1.1926 x4 = x3 2 x3 + 3 5.3852

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FAC: Question & Answer Bank – Solutions

Solution A68 C (2i - 4)(3 - i ) = -2i 2 + 10i - 12 = 2 + 10i - 12 = 10i - 10

Solution A69 C

(2 - 3i )(2 + 3i ) = 4 - 9i 2 = 4 + 9 = 13 Solution A70 A 2 - i 2 - i 2 - i 4 - 4i + i 2 4 - 4i - 1 3 - 4i 3 4 = ¥ = = = = 5 - 5i 2+i 2+i 2-i 4 +1 5 4 - i2

Solution A71 B The Argand diagram is as follows:

5 

r = 52 + ( -3) 2 = 34 = 5.831 tan q =

-3 5

ﬁ q = tan -1 -53 = -0.540 radians

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Solution A72 D x=

-3 ± 32 - 4(1)(2.5) -3 ± -1 -3 ± i = = = -1.5 ± 0.5i 2(1) 2 2

Solution A73 (i)

a . So these must have the same l dimensions. Since k is measured in £ and a is a dimensionless quantity, this

In the calculation of k 1 , k is added to

means that l must have dimensions of £ -1 , and k 1 has dimensions of £. The dimensions of k 2 and k 3 are then seen to be respectively. (ii)

[1]

1 1 = £ 2 and -3 = £ 3 , -2 £ £ [2]

k3 £3 has dimensions = £3- 2c . This will be dimensionless if 3 - 2c = 0 ie 2c c k2 £ c = 32 .

Comment

k 1,k 2 ,k 3 and

[1]

k3 are actually the mean, variance, skewness and coefficient of k 23 2

skewness of the translated gamma distribution.

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FAC: Question & Answer Bank – Solutions

Solution A74 (i)

Write in the form a + bi

z1 + z2 = (2 + 4i ) + (1 - 2i ) = 3 + 2i

[1]

z1 - z2 = (2 + 4i ) - (1 - 2i ) = 1 + 6i

[1]

z1z2 = (2 + 4i )(1 - 2i ) = 2 - 8i 2 = 2 + 8 = 10

[1]

z1 2 + 4i 2 + 4i 1 + 2i 2 + 8i - 8 -6 + 8i = = ¥ = = = -1.2 + 1.6i z2 1 - 2i 1 - 2i 1 + 2i 1+ 4 5

[1]

z12 + z22 = (2 + 4i ) 2 + (1 - 2i )2 = ( -12 + 16i ) + ( -3 - 4i ) = -15 + 12i

[1]

(ii)

Calculate and comment

z1 z2 = 2 + 4i 1 - 2i = 20 ¥ 5 = 100 = 10

z1z2 = 10 = 10

[2]

These two quantities are always equal, whatever the values of z1 and z2 . (iii)

Calculate and comment

z1 z2 =

2 - 4i = -1.2 - 1.6i 1 + 2i

z1 z2 = -1.2 + 1.6i = -1.2 - 1.6i

[2]

Again, these two quantities are always equal, whatever the values of z1 and z2 . (iv)

[1]

Express in the form re iθ

3z1 + 2 z2 = 3(2 + 4i ) + 2(1 - 2i ) = 8 + 8i = 8(1 + i )

[1]

This has magnitude 8 12 + 12 = 8 2 and an argument of 45 ∞ or So 3z1 + 2 z2 = 8 2eip / 4 .

p 4

radians.

[1]

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Solution A75 Consider the equation 2 z 3 - 9 z 2 + 14 z - 5 = 0 . (i)

Show 2 + i is a root

When z = 2 + i : z 2 = (2 + i ) 2 = 3 + 4i

z 3 = zz 2 = (2 + i )(3 + 4i ) = 2 + 11i

[2]

So the LHS of the equation is:

2(2 + 11i ) - 9(3 + 4i ) + 14(2 + i ) - 5 = 0 (ii)

[1]

The other complex root

Since this is a polynomial equation with real coefficients, the complex roots come in [1] conjugate pairs. So the other complex root is z = 2 - i . (iii)

Find the third root

The third root ( a , say) of this cubic equation must be real. So the equation must take the form:

C ( z - a )[ z - (2 + i )][ z - (2 - i )] = 0 Multiplying out the two complex factors in square brackets, we must have the identity: C ( z - a )( z 2 - 4 z + 5) = 2 z 3 - 9 z 2 + 14 z - 5

[1]

For the z 3 terms to match up, we must have C = 2 . For the constant term to match up, we must have -5a C = -5 ﬁ a = ½ . So the third factor is z - ½ and the third root is z = ½ . (iv)

[1]

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]

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FAC: Question & Answer Bank – Solutions

Solution A76 The auxiliary equation is v 2 + 6v + 8 = 0 which has roots -4 and -2 .

[1]

The general solution is of the form yt = A ¥ ( -2)t + B ¥ ( -4)t .

[1]

However we know that y0 = 5 and y1 = 5 , so: 5= A+ B 5 = -2 A - 4 B

[1]

These can be solved to give B = -7.5 and A = 12.5 , so our particular solution is: yt = 12.5( -2)t - 7.5( -4)t

[1]

Check: y2 should equal -6 y1 - 8 y0 = -70 The formula gives 12.5 ¥ ( -2) 2 - 7.5 ¥ ( -4) 2 = -70 .

Solution A77 B 3x - 5 x 2 = 3 - 5x x 3 - 5 x < 3 so it is O( x) but lim (3 - 5 x) π 0 so it is not o( x) . xÆ0

Solution A78 A The general term is

n 1 n which will be bounded above by = . 3n 3 3n + 1

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Solution A79 C

Solution A80 C dy f ( x + h) - f ( x ) 5( x + h) - 5 x 5h = lim = lim = lim = lim 5 = 5 dx hÆ 0+ h h hÆ 0+ hÆ0+ h hÆ 0+

Solution A81 B

Solution A82 B dy = 6 x 2 - 2 x which is 20 when x = 2 . dx

Solution A83 A f ( x) 

4 x

 4 x 3 5 

f ( x)  4( 53 ) x 8 5   12 x 8 5   5

12 5 x8 5

Solution A84 D Using the chain rule f (u )  9(5u 2  7)8  10u  90u (5u 2  7)8 .

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FAC: Question & Answer Bank – Solutions

Solution A85 A Using the chain rule M (t )  (    2t )e t ½

2 2

. Hence M ¢(0) = m e0 = m .

Solution A86 B Using the chain rule M (t )  

 1









 

 1 

t  

. Hence M ¢(0) =

a . l

Solution A87 D





2 2 4   Now f ( x)  ln  2e x  2 x   ln 2e x  4e2 x . Using the chain rule gives: e  

4 e2 x 2 + 2x e

f ¢( x ) =

1 2e

+ 4e

-2 x

¥ (4 xe x - 8e -2 x ) =

2 xe x ex

Solution A88 D Using the chain rule gives: f ¢( x ) =

1 4 + e -6(5 x - 3)

¥ -6(2)(5 x - 3) ¥ 5e

-6(5 x - 3)2

60(5 x - 3)e -6(5 x - 3) 4 + e -6(5 x - 3)

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Solution A89 B Using the product and chain rules: dy  (3)( x 2  2 x  1)5  (4  3 x)5( x 2  2 x  1)4  (2 x  2) dx  ( x 2  2 x  1) 4  3( x 2  2 x  1)  5(4  3 x)(2 x  2)     ( x 2  2 x  1) 4 (3 x 2  6 x  3)  5(6 x 2  14 x  8)     ( x 2  2 x  1) 4 (33 x 2  76 x  43)

Solution A90 B Using the product rule: f ( x)  64 xe 4 x  32 x 2  (4e 4 x )  64 xe 4 x  128 x 2e 4 x f ( x)  64e 4 x  64 x(4e 4 x )  256 xe 4 x  128 x 2 (4e 4 x )  64e 4 x  512 xe4 x  512 x 2e 4 x  64(1  8 x  8 x 2 )e 4 x

Solution A91 A Now y 

3x 2 e

 3 x 2e  x . So using the product rule gives:

dy = 6 xe - x - 3 x 2e - x dx d2y 6 - 12 x + 3 x 2 -x -x -x 2 -x 2 -x = 6 e 6 xe 6 xe + 3 x e = (6 12 x + 3 x ) e = dx 2 ex

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FAC: Question & Answer Bank – Solutions

Solution A92 C Now f ( x)  2  6 x . Setting this equal to zero gives a turning point at x = - 13 . Since f ( x)  6 this turning point is a maximum. So the maximum value of this function is f ( - 13 ) = 5 - 2( - 13 ) - 3( - 13 ) 2 = 16 . 3

Solution A93 D Since y = x( x + 3)( x - 5) we can see that this crosses the x -axis when x = 0, -3 and 5. This means it is either C or D. Since we have a positive x3 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: f ( x)  3 x 2  2 x  1  (3 x  1)( x  1)  0  x   13 ,1

The second derivative is:

f ( x)  6 x  2 When x = 1 we have f ¢¢(1) > 0 so we have a minimum. When x = - 13 we have f ¢¢( - 13 ) < 0 so we have a maximum.

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Solution A95 D + 4ln x - 3 x , so: We have ln f ( x) = ln 90 2 d 4 ln f ( x) = - 3 dx x Setting this equal to zero and solving gives x = 43 .

Solution A96 D

(

)

(

∂ ∂ 2 3 axy + b ( xy ) + c ( xy ) = axy + bx 2 y 2 + cx3 y 3 ∂x ∂x

)

= ay + 2bxy 2 + 3cx 2 y 3

Solution A97 D f ( x, y, z ) = ( xyz ) = x 2 y 2 z 2 2

ﬁ

∂f = 2 xy 2 z 2 ∂x

ﬁ

∂2 f = 4 xyz 2 ∂x∂y

ﬁ

∂3 f = 8 xyz ∂x∂y∂z

Hence: ∂3 f ∂f ¥ = 8 xyz ¥ 2 xy 2 z 2 = 16 x 2 y 3 z 3 ∂x∂y∂z ∂x

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FAC: Question & Answer Bank – Solutions

Solution A98 C Finding the first derivatives and setting them equal to zero and solving gives: f  3x 2  6 x  2 y  0 x f  2x  2 y  0 y

(1) (2)

Equation (2) tells us that y = x . Substituting this into equation (1) gives: f  3x 2  4 x  0  x  0, 43 x

So our two turning points are x = 0 , y = 0 and x = 43 , y = 43 . Calculating the second derivatives gives: 2 f x 2 2 f y 2

 6x  6  2

2 f 2 xy

So for x = 0 , y = 0 we have:  2  f  2  y 

x  x0 y  y0

 2   f    2  x 

x  x0 y  y0

  2   f    xy  

  2   (2   )(6   )  2 x  x0  y  y0    2  8  8  0

Solving this gives roots of x = -1.172 and x = -6.828 . Since both of these roots are negative this means that we have a maximum at x = 0 , y = 0 .

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For x = 43 , y =  2  f  2  y 

4 3

x  x0 y  y0

Page 35

we have:  2   f    2  x 

x  x0 y  y0

  2   f    xy  

  2   (2   )(2   )  2 x  x0  y  y0   2  8  0

Solving this gives roots of x = ±2.828 . Since these roots have different signs this means that we have a saddle point at x = 43 , y = 43 .

Solution A99 C The Lagrangian function is L = (5 x - 3 y ) - l ( x 2 + y 2 - 136) . Finding the derivatives and setting them equal to zero gives: ∂L 5 = 5 - 2l x = 0 ﬁ x = ∂x 2l ∂L 3 = -3 - 2l y = 0 ﬁ y = ∂y 2l ∂L = - x 2 - y 2 + 136 = 0 ∂l

(1) (2) (3)

Substituting (1) and (2) into (3) gives: 2

Ê 5 ˆ Ê -3 ˆ - Á ˜ - Á ˜ + 136 = 0 ﬁ Ë 2l ¯ Ë 2l ¯

When l =

1 4

34 1 = 136 ﬁ l = ± 2 4 4l

we have x = 10 and y = -6 .

When l = - 14 we have x = -10 and y = 6 . 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.

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FAC: Question & Answer Bank – Solutions

Solution A100 (i)

Limit e3 x - e -3 x 0 by substituting x = 0 because this gives . x x xÆ0 e - e 0

We can’t evaluate lim

If we use the exponential series, we find that the denominator is:

(

) (

)

e x - e - x = 1 + x + 12 x 2 + - 1 - x + 12 x 2 - = 2 x + O( x3 )

[½]

The numerator is the same, but with x replaced by 3x : e3 x - e -3 x = 6 x + O( x3 )

[½]

So the limit is: e3 x - e -3 x 6 x + O ( x3 ) 6 + O( x 2 ) = lim = lim =3 xÆ0 e x - e - x x Æ 0 2 x + O ( x3 ) xÆ0 2 + O( x 2 ) lim

(ii)

[1]

Limit n

x  This limit is the special case of the result lim  1    e x when x  ½ . n   n

So: n

1   ½ lim  1    e  e  1.649 n   2n 

[2]

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Solution A101 Differentiating, we get: -10 ¥ 100(1 + i ) -11 + 5

i ¥ 10(1 + i ) -11 - (1 - (1 + i ) -10 ) ¥ 1 i2

[2]

Evaluating this when i = 0.05 gives: -1,000(1.05)

-11

0.5(1.05) -11 - 1 + (1.05) -10 +5 = -772.17 0.052

[1]

Solution A102 Let y = 2 x +

1 . x

[1]

If we work out the first few derivatives, we get: dy = 2 x log 2 - x -2 dx d2y = 2 x (log 2) 2 + 2 x -3 2 dx d3y = 2 x (log 2)3 - 6 x -4 3 dx

[2]

The pattern is then apparent: dny = 2 x (log 2) n + ( -1) n n ! x - n -1 dx n

[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.

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FAC: Question & Answer Bank – Solutions

Solution A103 Substituting the values given for the parameters gives: y = 4l x3 (l + x 4 ) -2

[1]

Taking logs, as suggested: log y = log 4 + log l + 3log x - 2log(l + x 4 )

[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: d 3 2(4 x3 ) log y = =0 dx x l + x4

[1]

Rearranging: 1

3(  x )  8 x

 3  5x

 3  4  x   5 

[1]

Substituting this back into the original function to find the maximum value of y : 3

3   3  4  y  4       5   5  

2

 3  4  8   4      5   5 

2



1  33/ 455/ 4  1/ 4 16

[1]

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Solution A104 (i)

α = 1, β = -3

Here: f ( x) = x (1 + x) -3

so that: f (0) = 0 and f (1) =

1 8

[1]

The derivative (found using the product rule) is: f ¢( x) = (1 + x) -3 - 3 x (1 + x) -4

This equals zero when: (1 + x) -3 - 3 x (1 + x) -4 = 0

(1 + x) - 3 x = 0 ﬁ x =

1 2

and f (½) =

4 27

[1]

So the graph has a maximum value at x = ½ and looks like this: 4 27

The range of values of the function here is: 0 £ f ( x) £

4 27

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(ii)

FAC: Question & Answer Bank – Solutions

α= β=2

Here: f ( x) = x 2 (1 + x) 2

So that: f (0) = 0 and f (1) = 4

[1]

The derivative is: f ¢( x) = 2 x(1 + x)2 + 2 x 2 (1 + x) > 0 when x Œ[0,1]

[1]

So the function increases steadily over the range and the graph looks like this:

The range of values of the function here is: 0 £ f ( x ) £ 4

[1]

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α = -½, β = 1½

Here: f ( x) = x -½ (1 + x)1½

So that: lim f ( x) = +• and f (1) = 2 2

[1]

xÆ0 +

The derivative is: f ¢( x) = - 12 x -1½ (1 + x)1½ + 32 x -½ (1 + x)½ This equals zero when: - 12 x -1½ (1 + x)1½ + 32 x -½ (1 + x)½ = 0 ie

- (1 + x) + 3x = 0 ﬁ x =

1 2

and f (½) =

3 3 2

[1]

So the graph looks like this:

The range of values of the function here is: f ( x) ≥

3 3 2

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FAC: Question & Answer Bank – Solutions

Solution A105 If f ( x, y ) = x3 - 2 x 2 + 2 y 2 , then:

∂f = 3x 2 - 4 x ∂x ∂f = 4y ∂y

[1]

Setting these equal to zero we get: x = 0 or

4 , y=0 3

4 So the extrema occur at x = 0, y = 0 and at x = , y = 0 . 3

[1]

To determine their nature, the second partial derivatives are needed:

∂2f = 6x - 4 ∂ x2 ∂2f =4 ∂ y2 ∂2f =0 ∂ y∂ x

[1]

Setting up the required equations: For x = 0, y = 0 the equation is (4 - l )( -4 - l ) = 0 , ie l = ±4 . x = 0, y = 0 is a saddle point.

This means that [1]

4 For x = , y = 0 the equation (4 - l )(4 - l ) = 0 , ie l = 4 , twice. This means that 3 4 [1] x = , y = 0 is a local minimum. 3

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Solution A106 The Lagrangian function is: L = x + y + z - l ( x + 2 y ) - m ( x 2 + y 2 + z 2 - 1)

[1]

Finding the partial derivatives:

∂L = 1 - l - 2m x ∂x ∂L = 1 - 2l - 2 m y ∂y ∂L = 1 - 2m z ∂z

∂L = -x - 2y ∂l

(1)

(4)

∂L = - x2 - y 2 - z 2 + 1 ∂m

(2)

(5)

[1]

(3)

Setting these equal to zero, we get: (1) ﬁ x =

1- l 2m

(3) ﬁ z =

1 2m

(2) ﬁ

1 - 2l 2m

-1 + l -2 + 4l + = 0 ﬁ - 3 + 5l = 0 2m 2m

(4) ﬁ

[1]

So l = 0.6 . Finally substituting into (5) we get: 2

 0.4   0.2   1   2    2    2   1        1.2  4  2     0.3

[1]

This gives the positions of the extrema as: x=-

and:

1 1 1 ,y = ,z = 5 0.3 10 0.3 2 0.3

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FAC: Question & Answer Bank – Solutions

Solution A107 (i)

Show µ(20) and obtain µ(40)

 x  70  x  20     1  50 

 (20)  a0  a1  exp b0  b1   0.000814

[1]

 x  70  x  40     0.6  50 

 (40)  a0  0.6a1  exp b0  0.6b1   0.001077 (ii)

[2]

Find and determine the turning point

The derivative of this function is:

 ( x) 

 1 b  x  70   a1  1 exp b0  b1   50 50  50   

[1]

Equating this to zero to find the stationary point gives:

  x  70   a1  b1 exp b0  b1    0  50    Rearranging: x

  a1  50  log e     b0   70  29.82 b1   b1  

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

È Êb ˆ Ê x - 70 ˆ ˘ m ¢¢( x) = Á 1 ˜ exp Íb0 + b1 Á Ë 50 ¯ Ë 50 ˜¯ ˙˚ Î

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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 q (20) is: 1

0

  20  t  70   20  t  70     exp b0  b1    dt 50 50     

 (20  t )dt   a0  a1  0

1   t   t    a0  a1   1  exp b0  b1   1  dt 0  50   50   

[2]

Although this looks complicated, we can integrate it directly: 1

  t2  50   t   (20 ) exp 1 t dt a t a t b b            0 1  0 1  50    0        100  b1  0 1

  1  50  1   50  a0  a1   1  exp b0  b1   1   exp b0  b1   100  b1  50   b1   0.000791

[2]

We then find that: q (20) = 1 - e -0.000791 = 0.000791

[1]

Comment

You will meet the mortality functions featuring in this question in Subject CT4 and CT5. They are usually written as qx and m x .

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FAC: Question & Answer Bank – Solutions

Solution A108 D b

Ú f ( x) dx 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 2

Úx

0.75

È x1.75 ˘ 21.75 - 1 dx = Í = 1.351 ˙ = 1.75 ÎÍ 1.75 ˚˙1

Solution A111 C -2 2 Ú x + x dx =

1 3 x3 - 3 x -1 x - x -1 + c = +c 3 3

Solution A112 B 4

È 5x ˘ 54 - 52 600 dx = = = 5 Í ˙ Ú ln 5 ln 5 ln 5 Í ˙ Î ˚2 2 x

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Solution A113 D This is a standard result, but can be proved as follows:

Ú Bc

dx = Ú Be

x ¥ ln c

Be x ¥ ln c Bc x dx = + const = + const ln c ln c

Solution A114 B 3

0 -l x -l x -3l -3l Ú l e dx = ÈÎ -e ˘˚ = -e - (-e ) = 1 - e 0

Solution A115 A

Solution A116 B Using the substitution u = 5 x 4 , we see that •

3 -5 x 4

Úx e 0

• -u

•

du dx

= 20 x3 , so

du 20

= x3dx . Hence we get:

È e-u ˘ e 0 - ( -1) dx = Ú du = Í = 0.05 ˙ = 20 20 ÍÎ 20 ˙˚

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FAC: Question & Answer Bank – Solutions

Solution A117 C Using the substitution w = 10 + x 6 (so that

dw = 6 x5 ), we get: dx 11

x5 1 11 3 1  w2  dx w dw     0 (10  x6 )3 6 10 6  2  10 1



1 1 1   2  2   0.000145 12  11 10 

Solution A118 C Using the substitution u = 2 x3 + 6 x + 1 (so that

du dx

= 6 x 2 + 6 ), we get:

3x 2 + 3 1 3 Ú 2 x3 + 6 x + 1 dx = 0.5Ú u du = 0.5ln u + c = 0.5ln(2 x + 6 x + 1) + c

Solution A119 A Let

2x + 5 A B ∫ + ﬁ 2 x + 5 = A( x + 1) + B(2 x - 4) . (2 x - 4)( x + 1) 2 x - 4 x + 1

Substituting x = -1 gives 3 = -6 B ﬁ B = -½ . Substituting x = 2 gives 9 = 3 A ﬁ 2x + 5

A = 3. ½

Ú (2 x - 4)( x + 1) dx = Ú 2 x - 4 - x + 1 dx = 32 ln(2 x - 4) - 12 ln( x + 1) + c ÏÔ (2 x - 4)3 ¸Ô = 12 ln Ì k ˝ ÓÔ ( x + 1) ˛Ô

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Solution A120 D Using integration by parts gives: 1

2x 2x 2x  xe dx   12 xe  0   12 e dx 0

 12 e 2   14 e 2 x   0  12 e 2  14 e 2  14  2.097

Solution A121 D Using Leibniz’s formula gives: x

(

d 3 x 2 - 2t dt = 1 3 x 2 - 2t Ú dx 0

)

t=x

-0+Ú 0

(

)

∂ 3 x 2 - 2t dt ∂x

= (3 x 2 - 2 x) + Ú 6 x dt 0

= (3 x 2 - 2 x) + [6 xt ]0

= (3 x 2 - 2 x) + 6 x 2 = 9 x2 - 2 x

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FAC: Question & Answer Bank – Solutions

Solution A122 A 15 10

Ú Ú

(5 x + y ) dy dx =

x =5 y =5

x =5

È5 xy + 0.5 y 2 ˘ Î ˚ y = 5 dx

25 x + 37.5 dx

x =5 15

= ÈÎ12.5 x 2 + 37.5 x ˘˚

x =5

= 2,875

Solution A123 B Using integration by parts we get: 1

Ú ln x dx = [ x ln x - x]0 1

But ln x is undefined when x = 0 . To be honest it is much easier to eliminate the alternatives: •

Úe 1

-2 x

•

dx = ÈÎ - 12 e -2 x ˘˚ = 0 - ( - 12 ) =

•

1 2

•

-2 -1 Ú x dx = ÈÎ - x ˘˚ = 0 - (-1) = 1 1

•

Ú 1

•

6 È 3˘ dx = Í - 2 ˙ = 0 - ( -3) = 3 3 x Î x ˚1

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Solution A124 B Working with 7 ordinates, the area under the curve is approximately:

(

)

1 2 6 + 2 ÈÎ 5.52 + 52 + 4.52 + 42 + 3.52 ˘˚ + 32 ¥ 0.5 = 63.125 2

Solution A125 B Using the Maclaurin expansion, we have: ex = 1 + x +

x 2 x3 + + 2! 3!

e2  1 + 2 +

22 23 + = 6.333 2! 3!

So:

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FAC: Question & Answer Bank – Solutions

Solution A126 A First note that ln(3 + x)(3 + y ) = ln(3 + x) + ln(3 + y ) . Hence:

∂ f ∂x ∂ f ∂y

1 3+ x

1 3+ y

x = 0, y = 0

∂2 f ∂ x2 ∂2 f ∂ y2

=x = 0, y = 0

∂2 f ∂ y∂ x

x = 0, y = 0

1 (3 + x) 2 1 (3 + y ) 2

1 3

1 3 =-

1 9

x = 0, y = 0

=0 x = 0, y = 0

So the expansion is: 1 1 1È 1 1 ˘ 1 1 1 1 ln 9 + x + y + Í - x 2 - y 2 ˙ = ln 9 + x + y - x 2 - y 2 3 3 2Î 9 9 ˚ 3 3 18 18

Solution A127 C dy y = ﬁ dx 3 x + 1

dy dx =Ú ﬁ ln y = 13 ln(3x + 1) + c y 3x + 1

When x = 0 , y = e 2 which means: ln e 2 = 13 ln1 + c ﬁ 2 = 0 + c ﬁ c = 2

Hence when x = 2 , we have ln y = 13 ln 7 + 2 ﬁ

y = 14.135 .

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Solution A128 (i)

Show Γ( x ) = ( x - 1)Γ( x - 1)

Starting from the definition given, and integrating by parts with u = t x -1 and dv = e -t dt (so that du = ( x - 1) t x - 2 dx and v = -e - t ), we get: •

•

G ( x) = Ú t x -1e -t dt = ÈÎt x -1 ( - e -t ) ˘˚ + Ú ( x - 1) t x - 2e - t dt 0 0 0

[2]

The term in square brackets is zero. So we get: •

G ( x) = ( x - 1) Ú t x - 2e -t dt =( x - 1)G ( x - 1)

[1]

(This relationship will be valid provided the integral for G ( x - 1) converges, which requires x - 1 > 0 ie x > 1 .) (ii)

Show Γ(1) = 1

Using the integral definition with x = 1 , we find that: •

•

G (1) = Ú t 0e - t dt = Ú e -t dt = ÈÎ - e - t ˘˚ = 0 - ( -1) = 1 0 0 0 (iii)

[2]

Obtain integral

Using the integral definition with x = ½ , we know that: •

G (½) = Ú t -½ e - t dt = p

[1]

If we apply the substitution t = 12 z 2 (so that

Ú0 ( •

)

-½ - 1 z 2 1 z2 e 2 zdz 2

dt = z ), we find that: dz

= p

[1]

Simplifying, and taking the constants to the RHS: • - 1 z2 2

Ú0 e

dz =

p 2

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FAC: Question & Answer Bank – Solutions

- 1 z2

Since the function e 2 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: •

Ú-•

- 12 z 2

• - 1 z2 2

dz = 2 Ú e 0

dz = 2p ﬁ

1 2p

•

Ú-•

- 12 z 2

dz = 1

[1]

Comment

This result is important in connection with the standard normal distribution.

Solution A129 The curve y  6  x  2 x 2 crosses the x axis when: 6  x  2 x 2  0  (2  x)(3  2 x)  0  x   32 or 2

[2]

So the required area is: 2

3 2 (6  x  2 x

2 343 7 ) dx  6 x  12 x 2  23 x3    14 24   3 2 24

[2]

Solution A130 The shaded region can be specified as {( x, y ) : 0 £ x £ 1,0 £ y £ x} . So the integral is:

ÚÚ A

f ( x, y ) dxdy = Ú dx Ú dy (2 x + 6 y )

[2]

The inner integral is: x

Ú0

(2 x + 6 y )dy = ÈÎ 2 xy + 3 y 2 ˘˚ = 5 x 2 1

ÚÚ f ( x, y) dxdy = Ú0 5 x

ﬁ

[2]

dx = ÈÎ 53 x3 ˘˚ = 0

5 3

[2]

Note that you could also evaluate the integral in the opposite order as 1

Ú0 dy Úy dx (2 x + 6 y) .

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Page 55

Solution A131 (1)

Substitution

Using the substitution t = 5 + x , we get: I 

(2)

7 t 5 dt   6 t2

5 7 5 1 5    t  2  dt   log t  t   log 6  42  0.035103 t   6 

[3]

Integration by parts

Integration by parts, with u = x and

dv 1 1 = (so that du = dx and v = ), 2 dx (5 + x) 5+ x

gives: 2

2 1 ˘ 1 2 1 5 7 2 È I = Íx ¥ dx = - + + ÈÎ log 5 + x ˘˚1 = -Ú + log ˙ 5 + x ˚1 1 5 + x 7 6 42 6 Î

(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: I =Ú

2 2È 1 x) - 5 5 ˘ 5 ˘ È dx dx log 5 x = = + + = Ú1 ÎÍ 5 + x (5 + x)2 ˚˙ ÍÎ 5 + x ˙˚1 (5 + x) 2

2 (5 +

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FAC: Question & Answer Bank – Solutions

Solution A132 (i)

Calculate the integral

Integrating with respect to x : y

(4 y + 1)e

y = 0.5

xe 2 x dx dy

x = 0.5

y È 1 2 x2 ˘ Ú (4 y + 1)e ÍÎ 4 e ˙˚0.5 dy y = 0.5 1

2 1 È1 ˘ (4 y + 1)e y Í e2 y - e0.5 ˙ dy 4 Î4 ˚ y = 0.5

1 = 4

(4 y + 1)e

2 y2 + y

y = 0.5

1 dy - e0.5 4

(4 y + 1)e y dy

[2]

y = 0.5

The first integral can be done by inspection, the second one needs to be done by parts: 1 ¸ 1 È 2 y 2 + y ˘1 1 0.5 ÏÔ È Ô y ˘1 e - e Ì Î (4 y + 1)e ˚ - Ú 4e y dy ˝ 0.5 ˚˙0.5 4 4 ÎÍ y = 0.5 ÓÔ ˛Ô

{

}

1 3 1 1 0.5 1 (e - e ) - e 5e - 3e0.5 - 4[e y ]10.5 4 4

1 3 1 1 0.5 1 (e - e ) - e 5e - 3e0.5 - 4(e1 - e0.5 ) = 2.54 4 4

{

}

[2]

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(ii)

Page 57

Calculate the integral by reversing the order

We can reverse the order of integration: y

(4 y + 1)e

y = 0.5

2 x2

dx dy =

x = 0.5

2 x2

x = 0.5

(4 y + 1)e y dy dx

[1]

y=x

To integrate with respect to y , we integrate by parts. We get: 1 Ï ¸ Ô 2 x2 ÔÈ y ˘1 y + xe (4 y 1) e 4 e dy ÌÎ ˝ dx Ú Ú ˚ y=x ÔÓ Ô˛ x = 0.5 y=x 1

xe 2 x

xe2 x

x = 0.5 1

( xe2 x

{5e - (4 x + 1)e

- [4e y ]1x dx

{5e - (4 x + 1)e

- 4e1 + 4e x dx

{e - 4 xe 1

}

+ 3e x dx

- 4 x 2e 2 x

+ 3 xe 2 x

) dx

[3]

x = 0.5

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]

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FAC: Question & Answer Bank – Solutions

Solution A133 (i)

Taylor’s series

The Taylor series for these functions are: e x = 1 + x + 12 x 2 + 16 x3 + ( -• < x < • )

[1]

(1 + y ) -1 = 1 - y + y 2 - y 3 + ( -1 < y < 1 )

[1]

and:

(ii)

Determine the coefficients

Using the series for e x , we find that the series for the denominator is: e x - 1 = x + 12 x 2 + 16 x3 + O( x 4 )

So the original function is:

x x = e x - 1 x + 12 x 2 + 16 x3 + O( x 4 )

[1]

Cancelling an x on the top and bottom:

{

}

x 1 = = 1 + ÈÎ 12 x + 16 x 2 + O( x3 ) ˘˚ 2 3 x 1 1 e - 1 1 + 2 x + 6 x + O( x )

-1

[1]

We can now use the series for (1 + y ) -1 with y = 12 x + 16 x 2 + O( x3 ) , which gives: 2 x È 1 x + 1 x 2 + O ( x3 ) ˘ + È 1 x + 1 x 2 + O ( x3 ) ˘ + O ( x3 ) = 1 6 6 Î2 ˚ Î2 ˚ ex - 1

[1]

Multiplying out the terms and ignoring any terms of power greater than 2, we get: x 1 x 2 + O ( x3 ) = 1 - ÈÎ 12 x + 16 x 2 ˘˚ + ÈÎ 14 x 2 ˘˚ + O( x3 ) = 1 - 12 x + 12 e -1 x

1 . So a = - 12 and b = 12

[1]

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Comment

If you follow the calculations through keeping more terms you’ll find that the x 1 x 2 is actually a very good coefficient of the x3 term is zero. So x ª 1 - 12 x + 12 e -1 approximation when x is small.

Solution A134 (i)

Partial fractions

The coefficients A and B satisfy the identity: 1 A B ∫ + P(2 - 5P) P 2 - 5P Multiply through by P(2 - 5 P) :

1 ∫ A(2 - 5P) + BP 1 ∫ 2 A + P( B - 5 A)

[1]

Equating the coefficients of the constant terms and the P terms: A= and

1 2

B - 5A = 0 ﬁ B = 5A =

5 2

So: 1 1/ 2 5/2 ∫ + P(2 - 5P ) P 2 - 5P

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[1]

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(ii)

FAC: Question & Answer Bank – Solutions

Solve the differential equation

Rearranging the differential equation and integrating: 1

Ú P(2 - 5P) dP = Ú dt Using the identity in part (i): 5/2  1/ 2   dP   dt P 2  5P 

 

1 ln P 2

1 ln 2

- 12 ln(2 - 5P) = t + c

P =t +c 2 - 5P

[1]

Note that P > 0 and 2 - 5P > 0 , so no modulus signs are required here. 2 . So: We are told that, when t = 0 , P = 15

1 ln 2

2 / 15 1 = - 1 ln10 = c ﬁ c = 12 ln 10 2 2 - 5(2 / 15)

[1]

1 ln 2

P = t - 12 ln10 2 - 5P

[1]

So:

We can make P the subject of this as follows. Doubling and rearranging: P + ln10 = 2t 2 - 5P 10 P ln = 2t 2 - 5P 10 P = e 2t 2 - 5P ln

10 P = (2 - 5 P)e 2t P(10 + 5e 2t ) = 2e2t 2 ﬁ P(t ) = 5 + 10e -2t

[1]

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Solution A135 The differential equation can be re-written in the form

dy 1 y = x. dx x + 1

[1]

This can be solved using an integrating factor, where the integrating factor is: 1 1 È ˘ exp Í Ú dx ˙ = exp [ - ln(1 + x)] = x +1 Î x +1 ˚

[1]

Note that no modulus signs are required here since we know x ≥ 0 , so 1 + x > 0 . Multiplying through by the integrating factor and integrating both sides with respect to x , we obtain: 1 1 y = Úx¥ dx x +1 x +1 Since

[1]

x 1 =1, we find that: x +1 x +1 1 1 y = Ú1 dx = x - ln( x + 1) + c x +1 x +1

[1]

where c is a constant. Since

y = 0 when x = 0 , we find that c = 0 , so the particular solution is

y = ( x + 1) [ x - ln( x + 1)] .

[1]

Solution A136 C We need to solve the simultaneous equations: -2 x + 2 y = 10 x + 5y = 7 4 x - 3 y = -18 Rearranging the first equation gives y = 5 + x . Substituting this into the second equation gives 6 x = -18 ﬁ x = -3 . Hence y = 2 .

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FAC: Question & Answer Bank – Solutions

Solution A137 C The magnitude of 3i  2 j  4k is

32 + ( -2) 2 + 42 = 29 . So the unit vector in the

same direction is: 3i  2 j  4k 29

 3 1    2 29    4

Solution A138 C The scalar product is:

4 ¥ 1 + ( -1) ¥ 3 + 3 ¥ ( -2) = -5 Hence: -5 = 42 + ( -1) 2 + 32 12 + 32 + ( -2) 2 cosq = 26 14 cosq So: cosq = -

5 26 14

ﬁ q = 105.2º

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Solution A139 B The easiest way to do this is to see which of these vectors has a scalar product of zero with 3i  2 j  k . (4i  4 j  4k )(3i  2 j  k )  12  8  4  0 (4i  4 j  4k )(3i  2 j  k )  12  8  4  0 (4i  4 j  4k )(3i  2 j  k )  12  8  4  0 (4i  4 j  4k )(3i  2 j  k )  12  8  4  0

Solution A140 C 4  10   8 14   1 2  2 4   2  6 AB        3 4  3 5   6  12 12  20   18 32 

Solution A141 C

det A = 3

0 -3 1 -3 1 0 - ( -2) +4 -3 2 -3 1 1 2

= 3 ¥ 3 + 2 ¥ ( -7) + 4 ¥ 1 = -1

Solution A142 B Ê 3 -1ˆ Ê a bˆ 1 Ê d -bˆ 1 = If A = Á then A -1 = . So we have: Á ˜ ˜ ad - bc Ë - c a ¯ 6 - ( -4) ÁË 4 2 ˜¯ Ëc d¯ A -1 =

Ê 3 -1ˆ 1 Ê 3 -1ˆ Ê 0.3 -0.1ˆ 1 = = 6 - ( -4) ÁË 4 2 ˜¯ 10 ÁË 4 2 ˜¯ ÁË 0.4 0.2 ˜¯

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FAC: Question & Answer Bank – Solutions

Solution A143 B

Solution A144 B Ê 3 2 ˆ Ê xˆ Ê xˆ = kÁ ˜ . The easiest way to check this is to find out which vectors satisfy Á ˜ Á ˜ Ë 6 -1¯ Ë y ¯ Ë y¯ Those that do are eigenvectors. Ê 3 2 ˆ Ê 2 ˆ Ê -6 ˆ Ê 2ˆ ÁË 6 -1˜¯ ÁË -6˜¯ = ÁË 18 ˜¯ = -3 ÁË -6˜¯ Ê3 2 ˆ Ê 3 ˆ Ê 7 ˆ Ê 3ˆ = π k ÁË 6 -1˜¯ ÁË -1˜¯ ÁË19˜¯ ÁË -1˜¯ Ê 3 2 ˆ Ê 1 ˆ Ê -3ˆ Ê1ˆ ÁË 6 -1˜¯ ÁË -3˜¯ = ÁË 9 ˜¯ = -3 ÁË -3˜¯ Ê 3 2 ˆ Ê 2ˆ Ê10ˆ Ê 2ˆ ÁË 6 -1˜¯ ÁË 2˜¯ = ÁË10˜¯ = 5 ÁË 2˜¯

Solution A145 (i)

Unit perpendicular vector

The scalar (dot) product of two perpendicular vectors is zero. So:

(ai + bj + ck ).( -2i + 3 j) = 0 and (ai + bj + ck ).(10i + k ) = 0 ie

[1]

-2a + 3b = 0 and 10a + c = 0

So: b = 23 a and c = -10a

[1]

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For a unit vector we also need a2 +

( 23 a )

Page 65

a 2 + b 2 + c 2 = 1 , which gives:

+ ( -10a) 2 = 1 ﬁ a =

3 913

[1]

and the required unit vector is: 3 2 30 i+ jk 913 913 913

[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 ( -2i + 3j) Ÿ (10i + 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 ij+ k 913 913 913

[1]

Solution A146 If we think of the matrix P as the matrix that transforms the column vector ( x

into the column vector  a b c  , then we have the relationship: T

 a   0.2 0.2 0.2  x       0.2  y   b    0.2 0  c   0 0.4 0.2  z      

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(*)

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FAC: Question & Answer Bank – Solutions

Premultiplying both sides by P 1 , we see that: a  x    P b   y c  z     1 

Thinking in terms of simultaneous equations, Equation (*) corresponds to: 0.2 x +0.2 y -0.2 z = a +0.2 z = b

0.2 x

[1]

+0.2 z = c

0.4 y

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: x + y - z = 5a + z = 5b

x 2y

[1]

+ z = 5c

We can use the last two equations to express x and y in terms of z : x = 5b - z y = 2.5c - 0.5 z

From the first equation, we then get: (5b - z ) + (2.5c - 0.5 z ) - z = 5a ﬁ z = -2a + 2b + c and so:

x = 5b - ( -2a + 2b + c) = 2a + 3b - c y = 2.5c - 0.5( -2a + 2b + c) = a - b + 2c

[1]

Writing these in the form of simultaneous equations: 2a

+3b

-c = x

-b

+2c = y

-2a +2b

+c = z

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In matrix notation this is:  2 3 1 a   x        1 1 2  b    y   2 2 1  c   z      

[1]

 2 3 1     1 1 2   2 2 1   

[1]

So:

1

Solution A147 Working out the various parts of this expression first, we find that: det( ) 



1

2 1  (2)(4)  (1)(1)  7 1 4

2 1   1 4

1

1  4 1     7  1 2 

[1]

[2]

and:  2   1  1 (x   )          and (x   )T  1 1  0   1   1 

Putting these together, we get: 1 1 1  4 11  (x  )T  1 (x   )   1 1    2 2 7  1 2 1



 3 1 1 1   14 1



4 14



2 7

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FAC: Question & Answer Bank – Solutions

and f ( x) =

1 2

(2p ) ¥ 7

e -2/7 = 0.0452

[1]

Comment

Note that a 1 ¥ 1 matrix is really just a scalar. So when we calculate 1 - (x -  )T  -1 (x -  ) we can treat the answer as a simple number. 2

Solution A148 A

Solution A149 A

Solution A150 B

Solution A151 D

Solution A152 C

Solution A153 C 2½ ¥ £18,000 = £7,500 6

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