The Maths Workbook - St Peter's College, Oxford
The Maths WorkbookOxford University Department of EconomicsThe Maths Workbook has been developed for use by students preparing for the PreliminaryExaminations in PPE, Economics and Management, and History and Economics. It coversthe mathematical techniques that students will need for Introductory Economics, and part ofwhat is required for the Mathematics and Statistics course for Economics and Managementstudents.The Maths Workbook was written by Margaret Stevens, with the help of: Alan Beggs, DavidFoster, Mary Gregory, Ben Irons, Godfrey Keller, Sujoy Mukerji, Mathan Satchi, PatrickWallace, Tania Wilson.c Margaret Stevens 2003Complementary TextbooksAs far as possible the Workbook is self-contained, but it should be used in conjunction withstandard textbooks for a fuller coverage: Ian Jacques Mathematics for Economics and Business 3rd Edition 1999 or 4th Edition 2003. (The most elementary.) Martin Anthony and Norman Biggs Mathematics for Economics and Finance, 1996.(Useful and concise, but less suitable for students who have not previously studiedmathematics to A-level.) Carl P. Simon and Lawrence Blume Mathematics for Economists, 1994. (A goodbut more advanced textbook, that goes well beyond the Workbook.) Hal R. Varian Intermediate Microeconomics: A Modern Approach covers many ofthe economic applications, particularly in the Appendices to individual chapters,where calculus is used. In addition, students who have not studied A-level maths, or feel that their mathsis weak, may find it helpful to use one of the many excellent textbooks available forA-level Pure Mathematics (particularly the first three modules).How to Use the WorkbookThere are ten chapters, each of which can be used as the basis for a class. It is intendedthat students should be able to work through each chapter alone, doing the exercises andchecking their own answers. References to the textbooks listed above are given at the end ofeach section.At the end of each chapter is a worksheet, the answers for which are available for the use oftutors only.The first two chapters are intended mainly for students who have not done A-level maths:they assume GCSE maths only. In subsequent chapters, students who have done A-level willfind both familiar and new material.
Contents(1) Review of AlgebraSimplifying and factorising algebraic expressions; indices and logarithms; solvingequations (linear equations, equations involving parameters, changing the subject ofa formula, quadratic equations, equations involving indices and logs); simultaneousequations; inequalities and absolute value.(2) Lines and GraphsThe gradient of a line, drawing and sketching graphs, linear graphs (y mx c), quadratic graphs, solving equations and inequalities using graphs, budget constraints.(3) Sequences, Series and Limits; the Economics of FinanceArithmetic and geometric sequences and series; interest rates, savings and loans;present value; limit of a sequence, perpetuities; the number e, continuous compounding of interest.(4) FunctionsCommon functions, limits of functions; composite and inverse functions; supply anddemand functions; exponential and log functions with economic applications; functions of several variables, isoquants; homogeneous functions, returns to scale.(5) DifferentiationDerivative as gradient; differentiating y xn ; notation and interpretation of derivatives; basic rules and differentiation of polynomials; economic applications: MC,MPL, MPC; stationary points; the second derivative, concavity and convexity.(6) More Differentiation, and OptimisationSketching graphs; cost functions; profit maximisation; product, quotient and chainrule; elasticities; differentiating exponential and log functions; growth; the optimumtime to sell an asset.(7) Partial DifferentiationFirst- and second-order partial derivatives; marginal products, Euler’s theorem; differentials; the gradient of an isoquant; indifference curves, MRS and MRTS; thechain rule and implicit differentiation; comparative statics.(8) Unconstrained Optimisation Problems with One or More VariablesFirst- and second-order conditions for optimisation, Perfect competition and monopoly; strategic optimisation problems: oligopoly, externalities; optimising functionsof two or more variables.(9) Constrained OptimisationMethods for solving consumer choice problems: tangency condition and Lagrangian;cost minimisation; the method of Lagrange multipliers; other economic applications;demand functions.(10) IntegrationIntegration as the reverse of differentiation; rules for integration; areas and definiteintegrals; producer and consumer surplus; integration by substitution and by parts;integrals and sums; the present value of an income flow.
CHAPTER 1Review of AlgebraMuch of the material in this chapter is revision from GCSE maths (although some of the exercises are harder). Some of it – particularly thework on logarithms – may be new if you have not done A-level maths.If you have done A-level, and are confident, you can skip most of the exercises and just do the worksheet, using the chapter for reference wherenecessary.—./—1. Algebraic Expressions 1.1. Evaluating Algebraic ExpressionsExamples 1.1:(i) A firm that manufactures widgets has m machines and employs n workers. The numberof widgets it produces each day is given by the expression m2 (n 3). How many widgetsdoes it produce when m 5 and n 6?Number of widgets 52 (6 3) 25 3 75(ii) In another firm, the cost of producing x widgets is given by 3x2 5x 4. What is thecost of producing (a) 10 widgets (b) 1 widget?When x 10, cost (3 102 ) (5 10) 4 300 50 4 354When x 1, cost 3 12 5 1 4 3 5 4 12It might be clearer to use brackets here, but they are not essential:the rule is that and are evaluated before and .12(iii) Evaluate the expression 8y 4 6 ywhen y 2.4(Remember that y means y y y y.)8y 4 121212 8 ( 2)4 8 16 128 1.5 126.56 y6 ( 2)8(If you are uncertain about using negative numbers, work through Jacques pp.7–9.)Exercises 1.1: Evaluate the following expressions when x 1, y 3, z 2 and t 0:x 3(a) 3y 2 z (b) xt z 3 (c) (x 3z)y (d) yz x2 (e) (x y)3 (f) 5 2t z1
21. REVIEW OF ALGEBRA1.2. Manipulating and Simplifying Algebraic ExpressionsExamples 1.2:(i) Simplify 1 3x 4y 3xy 5y 2 y y 2 4xy 8.This is done by collecting like terms, and adding them together:1 3x 4y 3xy 5y 2 y y 2 4xy 8 5y 2 y 2 3xy 4xy 3x 4y y 1 8 4y 2 7xy 3x 3y 7The order of the terms in the answer doesn’t matter, but we often put a positive termfirst, and/or write “higher-order” terms such as y 2 before “lower-order” ones such as yor a number.(ii) Simplify 5(x 3) 2x(x y 1).Here we need to multiply out the brackets first, and then collect terms:5(x 3) 2x(x y 1) 5x 15 2x2 2xy 2x 7x 2x2 2xy 15(iii) Multiply x3 by x2 .x3 x2 x x x x x x5(iv) Divide x3 by x2 .We can write this as a fraction, and cancel:x x xxx3 x2 xx x1(v) Multiply 5x2 y 4 by 4yx6 .5x2 y 4 4yx6 5 x2 y 4 4 y x6 20 x8 y 5 20x8 y 5Note that you can always change the order of multiplication.(vi) Divide 6x2 y 3 by 2yx5 .6x2 y 3 2yx5 3x2 y 33y 36x2 y 3 2yx5yx5yx33y 2x3y(vii) Add 3xy and 2 .The rules for algebraic fractions are just the same as for numbers, so here we find acommon denominator :3x y y2 6x y 2 2y 2y6x y 22y
1. REVIEW OF ALGEBRA(viii) Divide3x2yby3xy 32 .3x2 xy 3 y2 3x2 26x23x22 3 yxyy xy 3xy 46xy4Exercises 1.2: Simplify the following as much as possible:(1) (a) 3x 17 x3 10x 8 (b) 2(x 3y) 2(x 7y x2 )(2) (a) z 2 x (z 1) z(2xz 3)(3) (a)3x2 y6x(b)(4) (a) 2x2 8xy(5) (a)2x y 2 y2x(6) (a)2x 1 x 4312xy 32x2 y 2(b) (x 2)(x 4) (3 x)(x 2)(b) 4xy 5x2 y 3(b)2x y 2 y2x(b)11 x 1 x 1(giving the answers as a single fraction)1.3. FactorisingA number can be written as the product of its factors. For example: 30 5 6 5 3 2.Similarly “factorise” an algebraic expression means “write the expression as the product oftwo (or more) expressions.” Of course, some numbers (primes) don’t have any proper factors,and similarly, some algebraic expressions can’t be factorised.Examples 1.3:(i) Factorise 6x2 15x.Here, 3x is a common factor of each term in the expression so:6x2 15x 3x(2x 5)The factors are 3x and (2x 5). You can check the answer by multiplying out thebrackets.(ii) Factorise x2 2xy 3x 6y.There is no common factor of all the terms but the first pair have a common factor,and so do the second pair, and this leads us to the factors of the whole expression:x2 2xy 3x 6y x(x 2y) 3(x 2y) (x 3)(x 2y)Again, check by multiplying out the brackets.(iii) Factorise x2 2xy 3x 3y.We can try the method of the previous example, but it doesn’t work. The expressioncan’t be factorised.
41. REVIEW OF ALGEBRA(iv) Simplify 5(x2 6x 3) 3(x2 4x 5).Here we can first multiply out the brackets, then collect like terms, then factorise:5(x2 6x 3) 3(x2 4x 5) 5x2 30x 15 3x2 12x 15 2x2 18x 2x(x 9)Exercises 1.3: Factorising(1) Factorise: (a) 3x 6xy(b) 2y 2 7y(2) Simplify and factorise: (a)x(x2(3) Factorise: xy 2y 2xz 4z(c) 6a 3b 9c 8) 2x2 (x 5) 8x(b) a(b c) b(a c)(4) Simplify and factorise: 3x(x x4 ) 4(x2 3) 2x1.4. PolynomialsExpressions such as5x2 9x4 20x 7 and 2y 5 y 3 100y 2 1are called polynomials. A polynomial in x is a sum of terms, and each term is either a powerof x (multiplied by a number called a coefficient), or just a number known as a constant. Allthe powers must be positive integers. (Remember: an integer is a positive or negative wholenumber.) The degree of the polynomial is the highest power. A polynomial of degree 2 iscalled a quadratic polynomial.Examples 1.4: Polynomials(i) 5x2 9x4 20x 7 is a polynomial of degree 4. In this polynomial, the coefficient ofx2 is 5 and the coefficient of x is 20. The constant term is 7.(ii) x2 5x 6 is a quadratic polynomial. Here the coefficient of x2 is 1.1.5. Factorising QuadraticsIn section 1.3 we factorised a quadratic polynomial by finding a common factor of each term:6x2 15x 3x(2x 5). But this only works because there is no constant term. Otherwise,we can try a different method:Examples 1.5: Factorising Quadratics(i) x2 5x 6 Look for two numbers that multiply to give 6, and add to give 5:2 3 6 and 2 3 5 Split the “x”-term into two:x2 2x 3x 6 Factorise the first pair of terms, and the second pair:x(x 2) 3(x 2)
1. REVIEW OF ALGEBRA5 (x 2) is a factor of both terms so we can rewrite this as:(x 3)(x 2) So we have:x2 5x 6 (x 3)(x 2)(ii) y 2 y 12In this example the two numbers we need are 3 and 4, because 3 ( 4) 12 and3 ( 4) 1. Hence:y 2 y 12 y 2 3y 4y 12 y(y 3) 4(y 3) (y 4)(y 3)(iii) 2x2 5x 12This example is slightly different because the coefficient of x2 is not 1. Start by multiplying together the coefficient of x2 and the constant:2 ( 12) 24 Find two numbers that multiply to give 24, and add to give 5.3 ( 8) 24 and 3 ( 8) 5 Proceed as before:2x2 5x 12 2x2 3x 8x 12 x(2x 3) 4(2x 3) (x 4)(2x 3)(iv) x2 x 1The method doesn’t work for this example, because we can’t see any numbers thatmultiply to give 1, but add to give 1. (In fact there is a pair of numbers that doesso, but they are not integers so we are unlikely to find them.)(v) x2 49The two numbers must multiply to give 49 and add to give zero. So they are 7 and 7:x2 49 x2 7x 7x 49 x(x 7) 7(x 7) (x 7)(x 7)The last example is a special case of the result known as “the difference of two squares”. Ifa and b are any two numbers:a2 b2 (a b)(a b)Exercises 1.4: Use the method above (if possible) to factorise the following quadratics:(1) x2 4x 3(2) y 2 10 7y(3) 2x2 7x 3(4) z 2 2z 15(5) 4x2 9(6) y 2 10y 25(7) x2 3x 1
61. REVIEW OF ALGEBRA1.6. Rational Numbers, Irrational Numbers, and Square RootsA rational number is a number that can be written in the form pq where p and q are integers.An irrational number is a number that is not rational. It can be shown that if a numbercan be written as a terminating decimal (such as 1.32) or a recurring decimal (such as3.7425252525.) then it is rational. Any decimal that does not terminate or recur is irrational.Examples 1.6: Rational and Irrational Numbers(i) 3.25 is rational because 3.25 3 41 (ii) 8 is rational because 8 81 .134 .Obviously, all integers are rational.(iii) To show that 0.12121212. is rational check on a calculator that it is equal to (iv) 2 1.41421356237. is irrational.433 .Most, but not all, square roots are irrational:Examples 1.7: Square Roots (i) (Using a calculator) 5 2.2360679774. and 12 3.4641016151. (ii) 52 25, so 25 5q(iii) 23 23 94 , so 49 23 Rules for Square Roots:r aaab a band bbExamples 1.8: Using the rules to manipulate expressions involving square roots (i) 2 50 2 50 100 10 (ii) 48 16 3 4 3qq 9849 49 7(iii) 98 84284 (iv) 2 2 20 1 220 1 52 4 1 5 2 8 2 4 2(v) 82 8 22 2q (vi) 27y 27y9 33y 3yppp p (vii) x3 y 4xy x3 y 4xy 4x4 y 2 4 x4 y 2 2x2 yExercises 1.5: Square Roots 15(1) Show that: (a) 2 18 6 (b) 245 7 5 (c) 5 33 p (2) Simplify: (a) 345 (b) 2x3 8x (c) 2x3 8x (d ) 13 18y 2Further reading and exercises Jacques §1.4 has lots more practice of algebra. If you have had any difficulty withthe work so far, you should work through it before proceeding.
1. REVIEW OF ALGEBRA2. Indices and Logarithms 72.1. IndicesWe know that x3 means x x x. More generally, if n is a positive integer, xn means “xmultiplied by itself n times”. We say that x is raised to the power n. Alternatively, n maybe described as the index of x in the expression xn .Examples 2.1:(i) 54 53 5 5 5 5 5 5 5 57 .x5x x x x x x x x x3 .2xx x 2(iii) y 3 y 3 y 3 y 6 .(ii)Each of the above examples is a special case of the general rules: am an am nma n am na (am )n am nNow, an also has a meaning when n is zero, or negative, or a fraction. Think about thesecond rule above. If m n, this rule says:ana0 n 1aIf m 0 the rule says:1a n naThen think about the third rule. If, for example, m 12 and n 2, this rule says: 1 2a2 awhich means that1a2 13 a1Similarly a is the cube root of a, and more generally a n is the nth root of a: 1an n aApplying the third rule above, we find for more general fractions: mma n n a n amWe can summarize the rules for zero, negative, and fractional powers: a0 1 (if a 6 0)1 a n n a1 an n a mm a n n a n am
81. REVIEW OF ALGEBRAThere are two other useful rules, which may be obvious to you. If not, check them usingsome particular examples: an bn (ab)nand a n a n bnbExamples 2.2: Using the Rules for Indices(i) 32 33 35 243 11(ii) 52 2 52 2 5 1 33(iii) 4 2 4 2 23 8 1 3311(iv) 36 2 36 2 6 3 3 6216 1 2 2227 3 227 332927 33 3(v) 3 8 2 2 2 182483832.2. LogarithmsYou can think of logarithm as another word for index or power. To define a logarithm wefirst choose a particular base. Your calculator probably uses base 10, but we can take anypositive integer, a. Now take any positive number, x.The logarithm of x to the base a is:the power to which the base must be raised to obtain x.If x an then loga x nIn fact the statement: loga x n is simply another way of saying: x an . Note that, sincean is positive for all values of n, there is no such thing as the log of zero or a negative number.Examples 2.3:(i) Since we know 25 32, we can say that the log of 32 to the base 2 is 5: log2 32 5(ii) From 34 81 we can say log3 81 4(iii) From 10 2 0.01 we can say log10 0.01 21(iv) From 9 2 3 we can say log9 3 0.5(v) From a0 1, we can say that the log of 1 to any base is zero: loga 1 0(vi) From a1 a, we can say that for any base a, the log of a is 1: loga a 1Except for easy examples like these, you cannot calculate logarithms of particular numbersin your head. For example, if you wanted to know the logarithm to base 10 of 3.4, you wouldneed to find out what power of 10 is equal to 3.4, which is not easy. So instead, you can useyour calculator. Check the following examples of logs to base 10:Examples 2.4: Using a calculator we find that (correct to 5 decimal places):(i) log10 3.4 0.53148(ii) log10 125 2.09691(iii) log10 0.07 1.15490
1. REVIEW OF ALGEBRA9There is a way of calculating logs to other bases, using logs to base 10. But the only otherbase that you really need is the special base e, which we will meet later.2.3. Rules for LogarithmsSince logarithms are powers, or indices, there are rules for logarithms which are derived fromthe rules for indices in section 2.1: loga xy loga x loga yx loga loga x loga yy loga xb b loga x loga a 1 loga 1 0To see where the first rule comes from, suppose:m loga x and n loga yThis is equivalent to:x am and y anUsing the first rule for indices:xy am an am nBut this means that:loga xy m n loga x loga ywhich is the first rule for logs.You could try proving the other rules similarly.Before electronic calculators were available, printed tables of logs were used calculate, forexample, 14.58 0.3456. You could find the log of each number in the tables, then (applyingthe second rule) subtract them, and use the tables to find the “anti-log” of the answer.Examples 2.5: Using the Rules for Logarithms(i) Express 2 loga 5 13 loga 8 as a single logarithm.12 loga 5 31 loga 8 loga 52 loga 8 3 loga 25 loga 2 loga 50 2 (ii) Express loga xy3 in terms of log x and log y. 2 loga x2 loga y 3loga xy3 2 loga x 3 loga yExercises 1.6: Indices and Logarithms(1) Evaluate (without a calculator):2(a) 64 3 (b) log2 64 (c) log10 1000(2) Simplify: (a) 2x5 x6 6(3) Simplify: (a) 3 ab(b)(xy)2x3 y 2(d) 4130 4131(c) log10 (xy) log10 x(d) log10 (x3 ) log10 x(b) log10 a2 31 log10 b 2 log10 abFurther reading and exercises Jacques §2.3 covers all the material in section 2, and provides more exercises.
101. REVIEW OF ALGEBRA 3. Solving Equations3.1. Linear EquationsSuppose we have an equation:5(x 6) x 2Solving this equation means finding the value of x that makes the equation true. (Someequations have several, or many, solutions; this one has only one.)To solve this sort of equation, we manipulate it by “doing the same thing to both sides.” Theaim is to get the variable x on one side, and everything else on the other.Examples 3.1: Solve the following equations:(i) 5(x 6) x 2Remove brackets:5x 30 x from both sides: 5x x 30Collect terms:4x 30 30 to both sides:4x both sides by 4:x(ii)(iii) x 2x x 223285 x 1 2x 43Here it is a good idea to remove the fraction first: all terms by 3:5 x 3 6x 12Collect terms:8 x 6x 12 6x from both sides:8 7x 12 8 from both sides: 7x 4 both sides by 7:x 475x 12x 9Again, remove the fraction by (2x 9):5x 2x from both sides: 3x both sides by 3:xfirst: 2x 9 9 3All of these are linear equations: once we have removed the brackets and fractions, each termis either an x-term or a constant.Exercises 1.7: Solve the following equations:(1) 5x 4 19(2) 2(4 y) y 17(3)2x 15 x 3 0(4) 2 (5)4 zz14 (3a 7 5) 32 (a 1)
1. REVIEW OF ALGEBRA113.2. Equations involving ParametersSuppose x satisfies the equation:5(x a) 3x 1Here a is a parameter : a letter representing an unspecified number. The solution of theequation will depend on the value of a. For example, you can check that if a 1, the solutionis x 3, and if a 2 the solution is x 5
The Maths Workbook Oxford University Department of Economics The Maths Workbook has been developed for use by students preparing for the Preliminary . each of which can be used as the basis for a class. It is intended that students should be able to work through each chapter alone, doing the exercises and .
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