Mathematics

the Practical,

the Logical, and

the Beautiful,

by

Benjamin Baumslag and Frank Levin


PREFACE

This book is intended for:

1.     Readers who liked mathematics at school but never studied it further.

2.     Young people with mathematical talent.

3.     Teachers who are looking for inspirational material for school. 

Each reader will study different parts of the book in different ways. Not many are likely to read every page, but then, not every visitor to an art museum looks at every painting. You do what you find enjoyable and find the time for; you may even think of coming several times.  Chapters can be read in almost any order.  Nor is it necessary to read the whole of a chapter.  Many sections indicate optional material.  In particular, the solved problems are optional. The more difficult calculations are often put into the solved problems, and even if one does not read these, one still gets an understandable account. 

The aim of the book is expressed by its title, namely to give examples of mathematics which is practical and useful in everyday life, examples of beautiful mathematics, and to illustrate the logical arguments used in mathematics, i.e. proof.  Practical topics include  approximation and the use of the powers of 10 notation. Then there is percentages, and estimating various quantities with simple calculations (Chapter 4), some knowledge of graphs, some probability and statistics. There is also a chapter on units like Watts and horsepower.

Proof is demonstrated for instance by proving Pythagoras’s theorem, and using numbers to derive Euclidean Geometry. 

There are some beautiful classical results[1] such as examples of a finite Geometry and a Projective Geometry, the existence of an infinite number of primes, and the irrationality of the square root of 2.  Fermat’s little and last theorems and the estimation of the number of primes less than a given number N, are discussed. Finally, we return to counting but this time counting infinite sets, and have the striking results of Cantor such as there are as many points on a line of length 1 as on a line of length 2. 

Much could be added, but we have chosen to be brief in order to present a more easily comprehendible book.  There is much of value and interest anyway.   

We have tried as far as possible to provide mathematics, which the reader can verify for himself or herself, and not have to rely on our authority.

 

The book begins with topics that would normally be discussed in school, and ends with topics, which would normally appear in a university course on mathematics.  The careful choice of material and presentation provides an account which is understandable by those who have studied secondary school mathematics.  Because we give a self-contained account, the reader who has forgotten the school mathematics will be reminded of some of the details.  What is required of the reader is a flexible mind, curiosity, and the sort of patience and determination that is required to play bridge or chess or solve cross- word or sudoku puzzles. 

The examples are from different countries, England or Sweden or the U.S.A. But these are only examples which are meant to illustrate various methods, and the reader will with their help apply these techniques to their own interests and  needs.

The material in this book is not original to us.

There are many brilliant ideas in mathematics. If we can introduce you to some of them it will be an honor and a privilege. 

 Acknowledgements

We gratefully acknowledge help from the following: David Baumslag, Pia Baumslag, Jeff Bourne, János Hegedüs, Geoffrey Howson, Lars-Göran Larsson, Marko Marin, Anatoliy Malyarenko, Julia Medin, Nils Mårtenson, Gunilla Sandberg, Abe Shenitzer, Naomi Yodaiken, Ralph Yodaiken.

 

Västerås and Swansea 2007

Benjamin Baumslag and Frank Levin

 

 

ISBN

© xxxxxxxxx

CONTENTS

PREFACE   iii

Acknowledgements. iv

CONTENTS  iv

1.  PRELIMINARIES AND A LITTLE FUN   9

§1 How to read this book. 9

§2 Ask questions. 10

§3 Indispensable tools for reading this book. 10

§4 Sets, numbers and infinity. 11

§5 A litle fun. 12

2.  SQUARING, CHECKING AND APPROXIMATION   14

§1 Introduction. 14

§2 Finding a square. 14

§3 Checking. 15

§4 Powers of 10. 16

§5 Accuracy. 18

§6 Solved Problems. 18

§7 Optional items. 20

3.  MATHEMATICS AND COMPUTING   22

§1 Bytes and bits. 22

§2 Turing Machines. 22

§3 Speed of operation. 23

§4 Will the computer replace the mathematician?. 24

4.  FERMI PROBLEMS  25

§1 Weight of a baby. 26

§2 Number of people in the world. 27

§3 Circumference of the earth. 27

§4 Maximum number of inhabitants on the earth. 27

§5 Viability of running a shop. 27

§6 General Comments on Fermi calculations. 28

§7 Solved problems. 28

5.  PERCENTAGES  33

§1 Definition and examples. 33

§2 A fictitious company’s accounts. 35

§3 Calculating percentages. 35

§4 Examples of judging figures in the news. 36

§5 Keep an eye on the total figures. 37

§6 Abortions. 37

§7 Compound increases. 37

§8 Worked examples. 37

6.  MEASUREMENT SENSE OR DIMENSIONAL ANALYSIS  40

§1 Length. 40

§2 Mass. 41

§3 Time. 42

§4 Speed: 42

§5 Acceleration. 44

§6 Force. 45

§7 Work and Power, Joules and Watts. 45

§8 Dimensional Analysis. 46

§9 Solved problems. 48

§10 Additional Topics. 49

7.  MEASURING HEIGHTS OR TRIGONOMETRY   55

§1 Shadow heights. 55

§2 The artist’s method. 56

§4 Dropping a stone over a cliff 59

§5 Solved problems and other optional topics. 59

8.  LOGARITHMS AND NATURAL LOGARITHMS  63

§1 Multiplying by adding. Logarithms. 63

§2 Doubling your money. 65

§3 Euler’s e. 65

§4 The natural logarithm is roughly 2.3 times the ordinary logarithm.. 66

§5 Theory of logarithms. 67

§6 Optional worked examples. 68

§7 Logarittihms and Planeetary Motion. 68

§8 Earth quakes measured on the Richter scale. 69

§9 Some Historical remarks. 70

9.  COORDINATE GEOMETRY   72

§1 Coordinates. 73

§2 Graphs as a concise source of information. 74

§3 Plotting a graph. 75

§4 Equations and Geometry. 78

§6 The Greeks and their curves. 79

§7 Equations and conic sections. 81

§8 Applications of coordinate geometry. 81

§9 Solved Problems. 82

§10 Solving problems in Geometry with algebra and vice - versa. 82

10.  SOLVING EQUATIONS AND GAUSS’S METHOD   84

§1 Solving an equation. 84

§2 Method 1: Take a guess. 84

§3 Method 2: Draw a graph. 85

§4 Method 3: Do what your mathematics teacher told you. 85

§5 Two unknowns. 86

§6 Some general results. 87

§7 Solved problems and other optional material 88

11.  FUNCTIONS  91

§1 What do we mean by a set?. 91

§2 Functions defined geometrically. 92

§3 Awkward functions. 93

§4 Functions described algebraically. 94

§5 Contrast with the definition in the calculus books. 95

§6 What comes after 1, 2, 3? Could it be 34?. 95

§7 Solved Problems and other Optional items. 96

12.  GEOMETRY   98

§1 Reducing the assumptions. 99

§2 Euclidean Geometry assuming only the real numbers. 100

§3 A finite Geometry. 100

§4 A finite projective geometry. 101

§5 Hyperbolic geometry. 102

§6 Sketch of the arguments for completing §2. 103

13.  PROBABILITY AND STATISTICS  105

§1 Probability. 105

§2 De Méré ’s Bet 106

§3 The same birthday. 107

§4 Two boys?. 107

§3 Three wise men. 108

§5 Monte Carlo method of calculating p approximately. 108

§6 It is impossible to send a rocket to the moon. 109

§7 Lotteries. 110

§8 Nuclear power plants exploding. 110

§9 Sampling. 111

§10 General remarks about sampling. 112

§11 Our own beliefs. 113

§12 Average. 114

§13 Solved Problems. 114

14.  PYTHAGORAS’S THEOREM    116

§1 Proof of Pythagoras’s Theorem.. 117

§2 The converse of Pythagoras’s theorem.. 119

§3 Pythagorean triples. 119

§4 Estimating the square root of a number 120

§5 The square root of 2 is irrational 121

§6 Distance to the Horizon. 121

§7 Solved Problems. 123

§8 Fermat’s Last Theorem.. 126

15.  MODULAR ARITHMETIC. PRIME NUMBERS AND LOGARITHMS  128

§1 Modular arithmetic. 128

§2 Congruence modulo n. 129

§3 Prime numbers. 131

§4 Gauss’s estimate of number of primes. 131

§6 Fermat’s little theorem.. 132

§7 Proof of the casting out nines check. 132

§8 ISBN check digit. 133

§9 Make a cipher code for your credit cards. 134

§10 Application to codes. 134

§11 Solved Problems and other optional items. 135

16.  COUNTING   138

§1 What do we mean by a set?. 138

§2 The same cardinality. 139

§3 Lines of lengths 1 and 2 have the same cardinality. 140

§4 A list 141

§5 The rational numbers have the same cardinality as the whole numbers. 141

§6 An infinite set which does not have the same cardinality as the whole numbers. 142

§7 Is there a set of cardinality less than the reals but greater than the natural numbers?. 142

§8 Solved problems. 143

§9 Hilbert’s Infinite Hotel 144

§10 Russell’s Paradox. 145

17.  THE TRACHTENBERG METHOD OF MULTIPLICATION   146

§1 The Method. 146

§2 Solved Problems and some optional ideas. 149

18.  BIBLIOGRAPHY   151

 


Chapter 1           

PRELIMINARIES AND A LITTLE FUN

Mathematics begins with a guess, just as naturally as love begins with a kiss.

 

§1 How to read this book

This is a book for people who choose to read for fun and enlightenment. Doing mathematics voluntarily means you can pick and choose what you want to do.

Reading mathematics is slower than reading other subjects. You may expect to read a page in a few minutes if you read a novel: with mathematics you can be lucky to read a sentence at that speed. The subject is concentrated. So do not try to study too much at one sitting, it being better to learn a little well rather than a lot badly.

We have laid out this book in the best way for our minds. Since your mind is different, you may prefer to change the order. You may skip sections you find boring. Do so. But be prepared to return to them later, when maybe they make more sense.

The chapters are on the whole independent and so can be read in the order you prefer.  .  There are three main exceptions;  Chapter 10 §1 to §4 inclusive on linear equations is needed for Chapter 12 §2 on Geometry.  Chapter 9 §1 on coordinates is needed for Chapter 11 on functions.

Chapter 8 §1 to §4 inclusive on logarithms is needed for Chapter 15 §4 on the number of primes less than a given number.

 

Solved problems can be skipped at without loss of intelligibility. Among these problems will be more detailed or technical arguments and laving them out will make following the text easier.  One can always return to them later if one feels like it.  Many problems are of interest in themselves.  Also often a solved problem can explain a difficulty-

Solved problems are also good for practice.   It is more fun and instructive to do them oneself before reading the solution.  Solving problems on one’s own is not so easy.  You may find it useful to read the book “How to solve it” by Polya if you are interested.  But in any case, the problems are optional.

 

 

figure 1. George Pólya (1887-1985), a talented mathematician of the 20th century. Towards end of his life he was much interested in the teaching of mathematics, see his book “How to solve it.”

 

§2 Ask questions

One is sometimes admonished not to ask stupid questions. There are two ways of doing this, either ask no questions, i.e. give up learning, or else know the subject so well that you can avoid asking the stupid questions. In other words, ignore the advice to avoid asking stupid questions.

And don’t be afraid to estimate or to guess. Even a wild guess can help you. I would hesitate to give this advice in say Chemistry. You may run the danger of an explosion if you guess which chemicals to combine, but in mathematics nothing so drastic can occur.

§3 Indispensable tools for reading this book

Pencil and paper are essential for reading this mathematics book and any other mathematics book. This is because the best and easiest way to follow an argument is to do the calculations yourself. Not only must you have pencil and paper to hand, you must use them all the time. The text in the book you use will give you a clue as to what you should be doing. Often it helps to write the definitions or the assertions in your own hand to absorb and understand them. We have assumed that enough of what you studied at school either remains or else you will be reminded of it as you read. If that is not the case, the book may be very hard to read, since it does not begin from the beginning. However we feel that most people will be able to cope.

In particular, we hope you remember the use of symbols in mathematics. To write a product like 3´4 we use ´ in between, but if we have represented an unknown number by a symbol x say, and we take twice this quantity, we leave out the multiplication sign and simply write 2x. Thus if one is searching for an unknown quantity x and twice this quantity plus 4 is 10, then this is written briefly as 2x + 4 = 10.

Example of how symbols help understanding:

Here is a party trick. You ask somebody to do the following:

·       Think of a whole number between 1 and 9

·       Multiply by 5. Add 3. Multiply by 2.

·       Then think of another whole number between 1 and 9 and add it to your total.

·       Give me your answer.

You can then at once tell the person the two numbers. All you have to do is to take away 6 from the total he gives you. Then the tens will be the first number thought of and the units the next number. For instance, suppose he thinks of the numbers 4 and 7. The first step he has to do is to multiply by 5 thus getting 20. To that he adds 3 getting 23. He then multiplies by two to get 46. Adding 7 gives a total of 53. So in telling him the numbers he thought of you subtract 6 to get 47. Thus 4 was the first number he thought of and 7 the second.

How it works is easy to see with the use of symbols

Thus let x and y denote the numbers thought of. Multiplying x by 5 gives 5x. Adding 3 gives 5x + 3. When we multiply by 2 we get 10x + 6. We then add y to get 10x + y + 6. When we subtract 6 we get 10x + y. Since x and y are numbers between 1 and 9, x becomes the tens and y becomes the units.

Another example of how symbols help understanding:

This is the hand method of learning the product of two numbers lying between 6 and 10

Baumslag’s father used this method for teaching young children who had mastered multiplying numbers lying between 1 and 5 how to deal with larger numbers. Place both hands in front of you with palms facing.  The thumb in each hand represents 6, the next finger 7 and so on, till the little finger which represents 10.  To find the product of two numbers, say 7 and 8, place the finger representing 7 on the left hand on the finger representing 8 on the right hand.   Count the fingers touching and those up to and including the thumbs.  In this case 5, and count 5 tens, i.e. 50.  There are three fingers on the left hand, and 2 on the right hand which have not been counted.  Multiply the two and three to get 6, and add to the 50 to get the product 56. This method helps children to multiply two numbers each lying between 6 and 10.  

We can explain how this method works as follows:  Let s be the finger on the left hand and t the finger on the right hand.  Then this represents the product of  5 + s and 5 + t.   This is 25 + 5s + 5t + st. 

The calculation we do is to count the number of fingers from the touching fingers to the thumbs and this is (s + t), and count them as 10s, i.e. we get 10s + 10t.  To this we add the product of the remaining fingers, i.e.  5 - s and 5 - t, getting  25 – 5s - 5t + st.

Adding this to  the 10s + 10t gives us 25 +5s + 5t + st, i.e. the same as we got before.  .

 

 

§4 Sets, numbers and infinity

The numbers 1,2,3,  etc., are called whole numbers. A set is a synonym for a collection: for instance, the collection of all whole numbers. This is denoted by {1, 2, 3, …}, where the squiggly brackets are a convention for denoting a set. The objects or elements of the set are the whole numbers, 1, 2, 3, … and the dots are understood to mean that the list continues forever.

Modern Mathematics explains much in terms of sets, and we will do so as well in this book.

The set of whole numbers is an infinite set. If you tried to list them in a finite number of steps you could not. An example of a finite set would be {1, 2, 3}.

The set of all atoms on the Earth, although huge, is not an infinite set. In theory one could list all the atoms, and after a while the list would come to a stop.

For centuries mathematics consisted of the study of numbers and geometry. This has long ago ceased to be the case, but we will stick to these parts of mathematics.

In addition to the whole numbers, we also have the fractions, numbers which are one whole number divided by another, like ¾. The fractions are also called rational numbers. The rational numbers also include the negative fractions.

The rational numbers can also be expressed as decimal expansions.   These can be finite, or else infinite, like 1/3 = ,33333…  But if a rational number’s decimal expansion is infinite, then it has a repeat pattern after a while.  If we allow these, and all other possible  decimal expansions, both positive and negative, then we obtain the set of all numbers These are called Real Numbers to distinguish them from the Imaginary Numbers, which involve the square root of – 1. 

 

A useful word in mathematics is “theorem,” which means “important deduction or result.”

§5 A little fun.

1. Multiplying any two numbers using only the two times table.

 

We can multiply any two numbers by multiplying solely by 2 and dividing solely by 2.  We illustrate the method by working out  46 ´33.

We write the numbers in two columns.  The next row is produced by multiplying the first number by 2 and dividing the second number by 2.  We ignore any halves that appear. We continue in this way till we get to 1 in the right-hand column. 

46

33

92

16

184

8

368

4

736

2

1472

1

We then add all the numbers in the left-hand column, which are opposite an odd number in the right-hand column.  The result is the product of the  two numbers.  Thus in the example we have chosen, we get

1472 + 46 = 1518 which is the answer.

 

2. The game of Nim 

The game of Nim is played with matches (or tooth-picks for non-smokers.) 

There are two players.  One arranges the matches in as many piles as desired, with as many matches as desired in each pile.  Each player at each move chooses a pile and takes as many matches as he or she wants.   But each player must take at least one match on every move.  At each move one must restrict oneself to one pile only.  The LOSER is the person who takes the last match.

Now there is a strategy for winning.  We take two special cases rather than the general case. 

Two piles Nim

If there are only two piles and it is your turn, then if one pile has only one match, take the other pile away.  If each pile has more than one match, then take matches away from the larger pile to leave two piles with the same number of matches.  If on your turn both piles have the same number of matches, then you will lose if your opponent knows the strategy.  Otherwise you can take away only one match from one pile, hoping that your opponent does not know the method. 

One pile with one match and two other piles.

When there are three piles each with one match, the person to makes the first move loses. So if there are two piles with one match in each and a third pile with two or more matches, leave only one match in the larger pile.

If two piles have more than one match, check the number of matches in each.

Case a) The two piles have the same number of matches in them and this is more than one  match. Take away the pile with one match to get the two pile situation described above. 

Case b) The smaller of the two piles with more than one match has an odd number of matches.  Take matches away from the larger pile so that it has one match less than the smaller.

Case c) The smaller of the two piles with more than one match has an even number of matches.  

Take matches away from the larger pile so that it has one match more than the smaller.

 

 

 

 

CHAPTER 2

SQUARING, CHECKING AND APPROXIMATION

 Mathematicians like the rest of us make mistakes. But, being accustomed to checking, they usually detect their mistakes before it is too late.

§1 Introduction

All of us have learnt the basic skills we need to do many arithmetic calculations.

We are suggesting that instead of letting our skills degenerate, we try to use them every day to think and make interesting conclusions. It is so seldom that we use these basic skills that we may now no longer be adept in adding, multiplying and dividing. Many think that does not matter, we can always use an electronic calculator, and so we can. But there is still a place for these basic skills, and most important, personal satisfaction in being able to do these calculations oneself. So in this chapter we will begin by practicing. In these initial sections, we will also find methods of checking the calculation.

§2 Finding a square

When we write (45)2 we mean 45 times 45. More generally, if x is any number, x2 means x times itself. We say “x-squared”, because it represents the area of a square of side x.

There is a quick way of working out the square of a number ending in 5. The general rule is: Remove the last digit, 5, multiply the remaining number, call it r, by r+1 and attach 25. For instance, to calculate (45)2, remove the 5, leaving 4, multiply 4 by 4 + 1 = 5 to get 20, attach 25 to get the answer 2025. (Why this method works is explained in problem 11 §7.)

 

Example: (995)2. (Some advice: Before reading the solution to an example, work through the problem yourself and read the solution only if you get stuck or wish to verify that you are correct.)

Solution: Take away the 5 to get 99 (r in this case), add 1 to get 100 (i.e. r + 1), multiply 99 and 100 to get 9900, and attach 25 to get the answer of 990025.

 

Of course you can calculate this result by multiplying 995 by 995, using the usual method of multiplying numbers, but this method is simpler and quicker.

 

You may say, why should I bother with such a calculation. Is that not the reason for calculators? Yes, you are right. But using a calculator is tantamount to letting some authority tell you the result. Which is a pity, since mathematics is the one subject where you can rely on yourself, and not on authority.

§3 Checking

Casting out nines

In §2 we calculated the square of 995. Can we check the result?

It may seem over the top to bother about checking in this particular case, since it is not a very complicated calculation, but by discussing what happens in such a calculation we are preparing the ground for how to handle much more difficult problems. Mathematicians are keen to check their calculations, because often other things depend on them, and one wants to be absolutely certain. It is also surprisingly easy to make mistakes, even using electronic devices. Repeating a calculation is also a good method of checking, but often one tends to repeat the same mistakes when doing the second calculation.

If the following check fails, we will know that the result is false. If the check succeeds, the result may nevertheless be wrong. But this check is useful for all sorts of arithmetic calculations. It is called the method of casting out nines. But first, we must define the checksum of a number. We can illustrate by 867. We begin by adding the digits of 867: 8+6+7 = 21. Since the sum is greater than 8, we repeat the process with 21, that is, we add the digits of 21 to get 3. Since 3 is less than 8, the checksum of 867 is 3. One further point, if the number 9 occurs anywhere in our calculations, we replace it by 0. For example, the checksum of 9 is 0. Also, the checksum of 291 is 2+0+1, or 3,which we obtained by replacing the middle digit, 9, by 0.

The method is based on the following fact: If the result of a product is correct, then the product of the checksums of the factors must be the same as the checksum of the answer. As an example of the method, suppose we are to check that the product, 35´23, is 805. We begin by replacing the numbers 35, 23 and 805 by their checksums, 8, 5 and 4, obtained by adding their digits. (For instance, 805 is replaced by 8+0+5 = 13, but since 13 is larger than 8, we further replace 13 by the sum of its digits, 1+3=4.) Multiply 8 and 5, the checksums of 35 and 23, to get 40. Add the digits of 40 to get 4. Since this matches the checksum of 805, the check is positive, and we have increased our confidence in the result, though the accuracy is not guaranteed.

The reason the method is called casting out nines is the rule that to obtain the checksums all 9s in the calculation are replaced by 0. For instance, applying the check to 9´9=81 gives 0´0 for the product, while the answer has checksum 8 + 1 = 9 which we replace by 0. Thus the check works.

We check the example in §2: 995´995 is supposed to be 990025. We replace the 9’s with 0s to get for the sum of the digits 0 + 0 +2 + 5 = 7. The sum of the digits in 995 with 9s replaced by 0 is 5, and 5x5 is 25 which has checksum 7, the checksum of the answer. This increases our confidence in the result.

(Chapter 15 §7 gives an explanation of why casting out nines works.)

Example

Use the method of casting out nines to check whether 74´38 =2,712.

Solution: Sum of digits of 74 is 11, and the sum of digits in 11 is 2. Sum of digits in 38 is also 11, which becomes 2 when we sum its digits. The product of the two checksums is 4, and this should agree with the check on the given answer. The sum of the digits in the given answer is 2, namely 2 + 7 + 1 + 2 = 12, but the sum of the digits of 12 is 3, not 4. Thus, 74´38 is not 2,712. Indeed, a more careful calculation gives the correct answer of 2,812.

Casting out 9s can also be used to check a sum of numbers against the total. For instance, to check that 75 + 236 = 305, replace 75 by its checksum, 3, and 236 by its checksum, 2. The sum of these two checksums is 5, which should be the checksum of 305. However, the checksum of 305 is 8, so there is a mistake in the addition.

Remark

The Welsh mathematician Jim Wiegold used to emphasize the importance of checking by his code of practice: Whenever he used a result he felt bound to check the proof of the result so as to ensure correctness of the previous result as well as his own. This is despite the fact that a referee has checked all articles printed. However, even Wiegold has not been able to carry out his code of practice always. The amount of checking is just too much in some cases.

Two Quick Checks

A very crude check of a product of two numbers is obtained by counting the number of digits in each factor and adding. Suppose the sum of these digits is S. Then the product should have either S or S - 1 digits. In §2 we claimed that 45´45 was 2025. Thus the product we have calculated has 4 digits as it should. Although this is a very crude check, it does bring to light errors we might otherwise miss.

A similar but more accurate method is to use only single digits. We again take the example of the square of 45, which we calculated in §2.

 (45)2 = 45´45, and this is approximately 50´40 =2000. We chose this approximation by increasing 45 to 50, i.e. a whole number of tens and then reducing the other factor 45 to 40, arguing that as we had increased one factor, we should compensate by reducing the other. Of course multiplying 50 by 40 is easy. The result 2000 is strikingly close to the result obtained in §2, that is, 2025. These two checks, casting out 9s and approximating, give further evidence that the method in §2 both works and probably there is no serious mistake in the calculation.

§4 Powers of 10

We have already explained in the meaning of 102 as the product of 10 and itself, i.e. 100. Similarly 103 is the product of three 10’s, and so on. Thus

102 = 10 ×10= 100

103= 10×10×10 = 1,000

104= 10×10×10 ×10 = 10,000

105= 10×10 ×10×10×10 = 100,000

106= 10×10×10×10×10 ×10 = 1,000,000

107= 10×10×10×10×0×10×10 = 10,000,000

and so on. These are called the powers of 10. A useful word is exponent: the exponent of 105 is 5; that of 107 is 7. Note that 105 is 1 followed by five 0s, 107 is 1 followed by seven 0s and so on. It seems reasonable to define 101 to be 1 followed by one 0, i.e. 10, and 100 as 1 followed by no 0s, i.e. 1. Thus101 = 10 and 100 = 1.

 

Multiplying these powers of 10 is easy: one simply adds the exponents. Thus 107×105

= 107+5 = 1012.

 

You will notice that an advantage of this way of writing is that it is much easier to comprehend, for instance, 109 rather than 1,000,000,000.  Even more important, this notation gives one a way of expressing very large numbers.  Problem 8 in §6 illustrates this.

 

figure 1. Archimedes (287BC - 212BC). One of the greatest mathematicians of all times. He had an alternative of the power of tens notation to denote large numbers which enabled him to calculate the grains of sand in the entire universe (as known then).

A very useful method is to express a number as a product of a number lying strictly between 10 and 1 and powers of 10. Thus we can express 887 as 8.87´102. The number 64789 is expressible as 6.4789´104. The exponents make it very easy to compare these numbers. Obviously the one with exponent 4 is very much bigger than the one with exponent 2. Also, if the numbers were written out in detail, it could be rather awkward to perform calculations with them. For example, to square 65,000, if we rewrite this as 65×103, the answer, using our previous formula, is easily seen to be 4225×106, which can also be expressed as 4.225×109or, without exponents, 4,225,000,000.

The advantage of this calculation is that it is so simple to do; it also gives us a very good idea of the powers of 10 that are in the answer. As such it is a useful test. With it we will certainly discover large errors.

Example

(995)2.  995 is approximately 1´103, and so the square is approximately

1´103´1´103 = 1´106. If we look at the example in §2 we calculated (995)2 to be 990025, which is equal to 9.90025´105. This is very close to 106.

Negative powers of 10

Positive powers of 10 are useful for large numbers. Negative powers are used for small numbers. For instance, 10-5 means 1 divided by 105, i.e. 1/100,000. In general the same rule applies that multiplying by powers of 10, whether positive or negative or mixed, we simply add the exponents. Thus

10-5 ´10-15´10-5= 10-25.

Rounding

The number 3.467 can be rounded to 2 decimal places by changing it to 3.47, i.e. we drop the last digit and if it is 5 or larger add 1 to the second decimal. If the last digit is less than 5, we simply drop it and leave the other digits unchanged.

Examples

To three decimal places we replace 18.8244 by 18.824, to 4 decimal places replace 5.67185 by 5.6719, to one decimal place replace 299.95 by 300.0. In this last example, we drop the 5 and add .1 to 299.9 thus getting 300.0.

In Science and Engineering most numbers are not exact, being the result of measurement, which is always subject to some inaccuracy. In the scientific notation that we have discussed above, the convention is that all digits given are correct, with the possible exception of the last digit, which could be 1 larger if the number has been rounded up. Thus 5.678´108 means that the number lies between 5.6775´108 and 5.6784´108.

§5 Accuracy

Taking 10 instead of 14 is an approximation. The error is 4/14, i.e. approximately 28%. So when we consider a product and approximate by taking the nearest single digit numbers we can incur large errors. When we take the product of two such numbers the errors compound. For instance, to calculate 14´23, if we approximate by 10´20 = 200, instead of the correct product of 322, the error is 122/322 i.e., an error of about 40%. The method of approximating by taking a single digit is subject to considerable errors. Often it is still useful to do so, but we recommend using two digit numbers to approximate, thus getting a much closer approximation. We recommend this because multiplying two digit numbers is relatively easy. In fact, if one uses the method advocated by Trachtenberg (described in Chapter 17) one can write down the answer in a few seconds.

In working out an approximation multiplying two numbers, note that the error is approximately the sum of the percentage errors in each of the factors as explained in 2 of §6. In the example above, approximate 14´23 by 10´20, the percentage errors of the factors are 28% and 13%, so the error in the product is approximately 41%. In the actual calculation we found an error of 40%.

§6 Solved Problems

1. Practicing squaring a number that ends in 5. Find the squares of 85, 75, 125.

Solution: (85)2 : Drop the 5 to get 8, add 1 to what remains to get 9. Multiply 8 and 9 to get 72. Attach 25 to get the result of 7,225.

(75)2 : Drop the 5 to get 7, add 1 to what remains to get 8. Multiply 7 and 8 to get 56. Attach 25 to get the result of 5,625.

(125)2 : Drop the 5 to get 12, add 1 to what remains to get 13. Multiply 12 and 13 to get 156. Attach 25 to get the result of 15,625.

2. Checking by casting out nines – Find 82×36. Check by the method of casting out nines.

Solution: 82´36 = 2952 by direct calculation.

To check we sum the digits of the answer replacing 9 with a 0, getting 2 + 0 + 5 + 2 = 9 which we replace by 0.

The check for 82 is 8 + 2 = 10, 1 + 0 = 1 on adding the digits of 10. The check for 36 is 3 + 6 = 9, which we replace by 0.

We multiply the check digits for 82 and 36 to get 1´0 = 0, the same as the check digit for the answer.   So our check does not indicate an error.

3. A quick check. Check problem 2 by checking the number of digits.

Solution: We add the number of digits in each factor: 82, has two digits and 36 has two digits. The answer should have 4 or maybe 4 – 1 = 3 digits. In fact the answer 2,952 has 4 digits.

5. Single digit check. Check problem 2 by replacing the product by one with single digit numbers.

Solution: To calculate 82´36 we replace 82 by 80 and 36 by 40, the product is 3,200, which agrees reasonably well with 2,952.

6. The scientific notation. In the scientific notation, what is the meaning of 7.5´104?

Solution: This means 7.5 times 104, which is 1 followed by four 0s, i.e. 10,000. Furthermore the result lies between 7.45´104 and 7.54´104.

 

7. Powers of 10. Use powers of 10 to multiply 2.5´104 by 3.1´103.

2.5´3.1 = 7.75.

104 ´103 = 107 and so the product is

7.75´107

 

8. Powers of 10 continued. Estimate the volume of a sphere with center the earth and extending to the moon.  (This example illustrates how easily the powers of ten notation can handle very large numbers. Indeed, without the powers of ten notation we could not even give an answer.)

Solution: The volume of a sphere of radius r is 4p r3/3. The moon is approximately 1.6´106 km. So the volume is 4p(1.6)3 ´1018/3 = 17.157 ´ 1018 = 1.7157´ 1019 km3.

9. Negative powers of 10. What is the meaning of 10-3?

103 means 1 followed by 3 zeros. 10-3 means 1/103 = 1/1000 = 0.001.

10. Errors. Calculate the percentage error in the calculation in example 5 above. Does it agree with the assertion that the error is approximately the sum of the percentage errors?

Solution: Replacing 82 by 80 gives a percentage error of (2/82)´100, i.e. approximately 2%. Replacing 36 by 40 gives a percentage error of (4/36)´100. i.e. approximately 11%, the sum gives an error of approximately 13%. The actual error in the answer is less than (3/32)´100, i.e. approximately 10%. This agrees well with the calculated error.

§7 Optional items

The method of checking modulo 11

1. There is another useful method of checking which we can explain by means of problem 2 of §6. This is 82´36.

For this check, we add the first digit to the third digit and then add the result to the fifth digit and so on. We then add the second digit to the fourth digit and then add the result to the sixth digit and so on. We then subtract the second sum from the first to get our check number.

For instance, 93,546 gets check number (6 + 5 + 9) – (4 + 3) = 20 – 7 = 13, which is then replaced by 3 – 1 = 2. Sometimes the check number will be negative, for instance the check number of 82 is 2 – 8 = - 6. In such a case, we add 11 to get a positive number. So 82 which had check number – 6 has check number 11 – 6 = 5.

Our problem is 82´36. The first factor 82 has check number 5 as we have just explained. The other factor is 36, which has check number 6 –3 = 3. We then multiply the check number of the first factor by the check number of the second factor to get 15, which is further replaced by 5 – 1 = 4. In problem 2 the answer was 2,952, whose check number is (9 + 2) - (2+5) = 4, which agrees with our previous check digit. (See Chapter 15 §6 problem 2 for an explanation of why this method of checking works.)

2. Demonstration of summation of percentage errors in a product.

The percentage error in a product is approximately the sum of the percentage errors in each approximation.

Solution; Before giving the explanation we note that the product of two small numbers is very much smaller than each of the individual numbers. For instance, if we multiply .01 by .02, the result is .0002, which is considerably smaller than both .01 and .02. So the product of two small numbers can be neglected if we are looking for an approximate result.

Suppose now that we are multiplying two numbers, n and m, by approximating to n by

n + a and to m by m + b. Our approximate answer will then be (n + a) ´ (m + b).

[As an example, say we are multiplying 2.9 by 4.8. Suppose we use n = 2.9 and a = .1, and m = 4.8 and b = .2. Thus instead of 2.9´4.8 we take 3´5.]

The difference between our approximate answer and the correct answer, n´m, will be

(n + a)´ (m + b) - n´m = nb + ma + ab, which is approximately nb + ma, if we assume that a and b are relatively small, and so ab can be neglected by the remark at the beginning of this solution. The percentages of error in the approximations are (a/n) ´100 and (b/m) ´100 and, for the product, (nb + ma)/(nm) ´100. Finally, (nb + ma)/(nm) ´100 = b/m´100 + a/n´100. That is, the sum of the percentage errors of each of the factors, which is what we were to prove.

[If we repeat the above argument using the example we have chosen of 2.9´4.8, we have the difference between our approximate answer of 3´4 = 12 and the correct answer 2.9´4.8 will be (2.9 + .1) ´(4.8 + .2) – 2.9´4.8 = 2.9´.2 + 4.8´.1 + .1´.2, which is approximately 2.9´.2 + 4.8´.1 since .1´.2 can be neglected. The percentages of error in the approximations of the factors are (.1/2.9)´100 and (.2/4.8)´100 and, for the product, {2.9´.2+4.8´.1 2.9´4.8})´100 = (.2/4.8)´100 + (.1/2.9)´100, that is the sum of the percentage errors in each of the factors]

3. Prove the quick method of squaring a number ending in 5 described in §2.

Note that 95 = 9´10 + 5. In general a number ending in 5 can be written 10n + 5. Thus

(10n + 5)2 = (10n + 5) (10n + 5) = 100n2 + 50n + 50n + 25 = 100(n2 + n) + 25.

Since (n2 + n) = n(n + 1), the result follows.

 

Chapter 3

MATHEMATICS AND COMPUTING

 

 

This chapter is quite diffent from the others.  Whereas most of what appears in the rest of the book will be correct in thousands of years, much of what appears here, and especially the specifications, will be out of date probably before the book is published.

§1 Bytes and bits

Computers’ memories are specified in bytes.  A byte corresponds to a character, such as a number or a letter of the alphabet or a punctuation mark.  The computer has a very limited vocabulary.  It understands only 0 and 1.  This is called a bit.  A byte consists of 8 bits.   

 

The computer has two types of memory.  The first is called RAM memory, and stands for random access memory.  It is the memory that the computer has for calculating and thinking, and corresponds to what we would normally use a sheet of paper for our calculations, which can then be thrown away.  This, for instance, we would normally use for recording a telephone number when somebody phones.  Later on we would transfer this to a telephone list, which is kept.  This corresponds to the memory on the hard disc of the computer.

 

The byte is a small unit, and we have a number of other units to describe the memory capacity of a computer.    A kilobyte is 1000 bytes, a megabyte is a million bytes and a gigabyte is one thousand million bytes.  One thousand million is a billion and so a gigabyte is a billion bytes.    It is not unusual for a computer to have  200 gigabytes of memory, i.e. it has more characters that it can remember than there are people in  the world (6 billion.)  With such a memory, it can remember 5 words describing each person in the world.

 

§2 Turing Machines

The Turing Machine is a theoretical model of the computer.  It is a very simple device, but then the computer is also a very simple device.  A Turing Machine has a infinite strip of paper.  It can make a mark÷on the paper or delete a mark and then move one position to the left or one position to the right.  It can also be in a number of states, and depending on its state, it will do other things. 

 

Perhaps an example of how the Turing Machine adds will help.  A Turing Machine to add has two states, State A and State B. 

 

Initially it is given two numbers to add.  The two numbers are indicated by a number of marks.  There is a space in between them.  For instance, 5 + 3 will appear on the infinite strip as follows:

 

÷÷÷÷÷ ÷÷÷. 

 

The tape is read at the beginning by the machine in state A.  If in this state the machine sees a mark, it moves one space to the right.  It then reads the next item and if it is a mark it moves one step to the right again.  If it sees a blank it  moves one step to the right. It now changes to state B. If it sees a mark it moves one step to the left and makes a mark and then moves one step to the right and erases the mark.  It then moves one step to the right and continues.  If it sees a blank it simply stops.  The result is of course that all the marks are now all together and there are now eight of them, and so the Turing Machine has added the two numbers to get the total of eight.

 

§3 Speed of operation

From this description of the Turing Machine you get the impression that the computer is not very smart. Of course the Turing machine is only a theoretical model of the computr, a nd the computer works completely differently. 

 

Perhaps a more usful way to think of the computr, is that it is a device  that carries out th4e instructions of algorithms.  An algorithm is a step by step procedure.  At each step the algorithm tells one exactly what one has to do.  As an example we will consider the algorithm of finding the highest common factor of two numbers.  This is the largest number which divides both numbers exactly.  For instance, the highest common factor of 54 and 30 is 6.  This is easy to see by dividing the numbers mentally.  A simple algorithm for doing this is as follows:

 

We prepare two columns.  In the first row we write 54 in the first column and in the second we write 30. If the two numbers are the same the highest common factor is that common number.   In this case of course 30 is the least number, and we write it in the same column in the new row.  We subtract it from other number and place that in the same row under itself.  We continue in this way till we get two equal numbers in a row.  This the highest common factor.

 

Thus we have in this example

54         30  (Begin with the two initial numbers.)

24          30  (30 unchanged and subtracted from 54.)

24            6  (24 unchanged and subtracted from 30.)

18            6  (6  unchanged and subtracted from 24.) 

12            6  (6 unchanged and subtracted from 18.)

6              6  (6 unchanged and subtracted from 12.)

As the two numbers are now the same, that is the highest common factor.

As the computer is limited to carrying out such laborious processes this agaam suggests that it is rather stupid.  How come then it is so effective.? 

The answer lies in its ability to do each step incredibly fast.  Typically performance is measured in giga cycles per second, i.e. a billion times per second.  The word hertz after a famous physicist means cycles per second.  Thus 2 gigahertz is a typical speed for computers. 

The electricity in your house has a frequency of 50 or 60 hertz, radio waves are measured in kilocycles or at most megacycles. Per second.  The more hertz the faster the computer can do calculations.  So the computer is outstandingly clever because it does everything incredibly fast, even though it uses quite laborious methods.  …..

§4 Will the computer replace the mathematician?

This has already occurred in some respects. 

Many people do not calculate the sum of or the product of two numbers, they use a pocket calculatior.  At most shops nowadays the assistanct taking your money does not need to calculate how much change to give you.  This is done automatically by the cash register.  Nor does an accountant nowadays need to be quick and accurate at adding numbres, he or she simply uses a computing program which does all the adding automatically.  Similarly the payment clerk, does not need to calculate your tax, it is all done automatically by the computer program. 

At University the first two parts of mathematics that are usually studied are Calculus and Linaear Algebra.   There are many computer programs that can do all that a clever student can do and more quickly and more accurately.  Although students still study these two subjects, it is surely only a matter of time before the subjects will be modified and at the very least, much of the techniques and methods will prove to be redundant and will not be studied.  How far this will go is hard to say.  Att the moment, most mathematics lecturers have a built in tendency to teach the subject very much in the old way.  But time will certainly change this and we can expect the computer to be used much more

 

Just how much we can leave to the computer is hard to say.  We must avoid the danger that after a while there will be nobody who really understands the principles and we simply rely on the computer as an oracle.  And of course there are going to be times when the computer is going to be wrong.  Either because there are bugs in the program (problems and conflicts that arise which nobody thought of at the time) and also because conditions may have changed and so are no longer applicable. 

 

In this book we stick mainly to our own understanding, and do not rely on authority, whether it be computers or famous professors. 

 

 

 

 

 

 

 

 

Chapter 4

 

FERMI PROBLEMS

To scribble a few figures on the back of an envelope and yet get a reasonable approximation to something whose value you have no idea about, sounds a bit like cheating. It isn’t. It’s a Fermi calculation.

The Nobel Prize winner Enrico Fermi was a physicist who had great skill in estimating with little information. For instance, as a standard question he would ask his students, ”How many piano tuners are there in Chicago?” With ingenuity one could find an estimate. How good the estimate is depends on the skill of the estimator. At the first ever explosion of a nuclear bomb Fermi noted how a piece of paper had been blown away by the blast, which was many miles away, and produced very quickly an estimate of the yield. His result was remarkably accurate.

As another example of his methods, knowing the distance between Los Angeles and New York and the time difference, we will be able to estimate the circumference of the earth.

In this chapter we will solve some Fermi-type problems. The idea is to get numerical values with very little information, and, of course, with results that are only rough approximations, For instance, one of our calculations will be an estimate for the weight of the earth. After our calculation we will be able to replace a vague remark that “the earth weighs a great deal” with the remark that the earth weighs approximately so many kilograms. This is relatively easy to check against published figures, and it will turn out that our rough calculation is out by a factor of 2. But this is much more precise than the original estimate that “the earth weighs a great deal”.

Enrico Fermi Photo

figure 1 Enrico Fermi (1901-1954). Nobel Prize in Physics 1938

Most of the examples below can also be checked in reference books, which enables us to see the advantages and the limitations of our calculations. But there are many cases where there is no reliable answer, and then the Fermi calculations give us guidance.

To simplify the calculations, it will be useful to express numbers using powers of 10. We remind the reader of these powers, which were discussed in Chapter 2. By 103 we mean 10 multiplied by itself three times, or 1000. Generally, for any whole number x, 10x means ten multiplied by itself x times. For example, 107 means 1 followed by seven 0s, i.e. 10,000,000, and 3.8´ 107 means 3.8 multiplied by 10,000,000, i.e., 38,000,000.

The x in 10x is called an exponent. As we explained previously when we multiply powers of 10 we simply add the exponents, and if we divide by a power of 10, we subtract the exponents, e.g. 104´105 = 109, and 106 divided by 102 is 10(6 - 2) = 104.

§1 Weight of a baby

Assuming that a man of 2 meters height weighs 100 kilograms, how much should a baby 50 cm high weigh? Since the baby is a fourth of the height of the man, as a first guess one might divide 100 kilos by 4 to get 25 kilos. This is clearly too naïve, since weight depends on volume, and the baby is not only shorter, it is also not as wide. Moreover, among many other considerations, skin and bones are less dense. To get a better approximation think of two solid rectangular boxes, one of which is one-fourth the length, width and breadth of the other. The volume of the smaller one will be (1/4)3 = 1/64th of the larger. If the same reasoning applies to the baby, its weight should be 100/64, or approximately 1½ kilos.

This answer shows both the weaknesses and strengths of this type of calculation. Everybody knows that a 50 cm baby is likely to weigh about 3 kilograms: twice as much as our estimate of 1½ kilos. However, with very little effort we have obtained a rough value, which is somewhere near the right answer. Clearly we should not expect this simple approach to give us an accurate result. Nevertheless, it has provided us with a rough idea of the result.

There is another possible approximation, which is based on the body mass index. This is a way of checking on being underweight or overweight. The body mass index is obtained by dividing the weight in kilograms by the square of the height.

For instance, consider the man who weighs 100 kilos and has a height of 2 meters. The square of his height is 4 and so the body mass index is his weight divided by 4, that is 25. And indeed, 25 is given empirically as the maximum body mass index for a person not to be overweight.

If we use this method for the baby, with its weight denoted by w, and its length 0.5 m, then, since (0.5)2 = 0.25, the body mass index for the baby is w/0.25. Assuming that w/0.25 = 25, the maximum for the ideal body mass index for a man, the baby’s weight must be 6 kilograms. Again, we are some distance from a reasonable result.

One can regard these rough calculations as, at least, giving us some quantitative information. This might be sufficient for our needs, but it will often be just the first step in trying to get a useful result.

§2 Number of people in the world

There are something like 200 countries in the world, and Great Britain has some 60 million people, i.e. 6´107 people. There are other countries much larger, like the USA, Russia, China, India, but many much smaller Assuming that all on average, 200 countries have half the population of Great Britain, this would make the total population of the world about

200´3´107 = 6´109. This very crude calculation has given the correct result.

§3 Circumference of the earth

Enrico Fermi came up with this clever way to deduce the circumference of the earth.

The distance from New York to Los Angeles is approximately 5,000 kilometers, and the difference in time is 3 hours. Since the earth is divided into 24 time zones, the distance from New York to Los Angeles corresponds to 1/8 the circumference of the earth. Hence, his estimate for the circumference of the earth is 8´5000 = 40,000 km. The equatorial circumference is in fact 40,074 km. Note that the circumference varies as the earth is not a perfect sphere.

Fermi’s argument is very much the same as that used by the Greek astronomer, Eratosthenes, about 240 B.C., who chose Aswan and Alexandria in Egypt instead of New York and Los Angeles, and came to a result of a little over 40,000 km. His method was based on the relative position of the sun at noon in the two places.

§4 Maximum number of inhabitants on the earth

The radius of the earth, r, is about 6´103 km so the surface area of the earth is given approximately by the formula 4πr2 (the surface area of a sphere of radius ) i.e.

4p ´(36 ´106) = 452´106 km = 4.52´108 km2. However, since 70% of the earth’s surface is water, this leaves 30% of the area on dry land, that is, approximately 1.5´108´106 m2.

Then if we allow 100 m2 of space for each person, this comes to a maximum of 1.5´1012 people. It is interesting to compare this figure with the actual number of people inhabiting the earth today, something like 6´109.

§5 Viability of running a shop

Suppose you wish to earn a net income of £20,000 per year. Working a whole year with 40 hours a week, and 50 weeks gives 2,000 hours. This means that you need to make at least £10 per hour. However, it would be foolish to regard that as sufficient, because there are always extra expenses. So we assume that what is needed is £20 per hour, just asking double since we do not have any better idea. Assuming that the profit margin is one third, we will need to sell some £60 per hour, or £5 per five minutes. This seems rather optimistic so we conclude that the profit margin must be higher, say 50%. With this profit margin you can reduce your average hourly sales to £40, or £3.50 per five minutes.

Of course, there’s a trade-off here, since increasing your prices may result in fewer customers.

§6 General Comments on Fermi calculations

It would be foolish to use these rough calculations as being correct conclusions. But it is remarkable how often they give one quite a good idea.

It is a sobering thought that often the figures that are quoted by the authorities have been made by similar calculations. Your own calculations can give you some reason to agree or disagree.

Indeed, in general when you are given some official number, you should always add in your mind plus or minus 20%. We suggest this margin of error because we know that usually it is not possible to give a very exact number. There are always errors. It is also not uncommon for people to err in a direction that makes them look better.

It is important to regard a Fermi calculation as the first stage in a more detailed investigation. Also, of course, the more you know about a subject the more accurate you can make your Fermi calculation. 

§7 Solved problems

1. Our town has 120,000 inhabitants. What are the numbers of births and deaths?

Solution: Since the town does not seem to be growing or declining, to a first approximation the numbers of births and deaths should be about the same. If the average life span is 70, for our rough calculation we may assume that, on average, an inhabitant will die at age 70. Thus the number of deaths should be 120,000 divided by 70, i.e. about 1,700 per year.

2. How many hairdressers in our town of 120,000 people?

Solution: Most men need a haircut once a month. Most of the hairdressers take about quarter of an hour per haircut, presumably more for women, but then women have a haircut less frequently.

On average, in a town of 120,000 people each month there will be about 100,000 people needing a haircut. Divide by 20 to get the number per day, 5,000. Each hairdresser can do 20 haircuts a day. Divide by 20 to get 250 hairdressers. This can be checked by using the yellow pages to count the number of hairdressers. In our phone book the total is 120. So the calculation is wrong by a factor of 2. This is not bad for a rough calculation, but in any case, one thing we did not take into account is that many hairdresser salons employ more than one hairdresser. If we assume the average is 2, then the rough calculation should have been improved by dividing by 2: 250 divided by 2 is 125 hairdresser salons.

3. Weight of a page of paper.

Solution: Paper is measured at weight per square meter. Suppose our paper weighs 80g per square meter. If the size of a page is 210´297 mm, i.e. approximately

0.2m´0.3m, then the area will be 0.06 square meters. Hence, the weight must be 80´0.06 or approximately 5 g. As a check we use a scale and found the weight to be 4 g.

4. Weight of the earth

Solution: The volume of a sphere of radius r is (4pr3)/3. Since the radius of the earth is approximately 6,000km, and since 4p/3 is approximately 4, the volume is roughly 4´ (6´103)3 = 4´216´109, or 864´109 km3. The weight of one cubic centimeter of water is 1 gram, so the weight of one cubic meter of water is 106 grams, or 103 kg, which means that one cubic kilometer weighs 1012 kilograms.

If we guess that earth is four time as heavy as water, the weight of the earth must be approximately 864´109´4´1012 kilograms, or 3456´1021 kg or 3.456´1024 kg. Checking on the internet we found a value of 5.9763´10 24 kg. for the weight of the earth. Our guess is out by a factor under 2, not too bad for a first approximation.

5. Quick calculation of the tax burden.

Solution: In a country with a tax rate of 20% and a value added or sales tax of 20%, for every income of 100 one has to pay 20 in tax. This leaves one with 80. A total purchase price of 80 breaks down to 67 plus 13 value added tax. So when one purchases an item for 80 one has to pay value added tax of about 13. The person who receives the 67 has to pay 20% tax on that, which gives a further 13 tax to the government. Thus, the total tax so far paid on the original 100 is

20 + 13 + 13 = 45.

6. Conservatives claim that reducing taxes will encourage sales, which, in turn, will result in more tax being collected despite the lower tax rate. How sound is the economics?

Solution: To verify this imagine that there is a 5% tax reduction, say, from 20% to 15%. Suppose this results in everybody earning 15% more, a rather extravagant estimate. Previously, if somebody earned 100 he paid tax of 20. Now the same person earns 115 and pays tax at 15%, i.e. 17.2. This is lower than the original 20, and results in a net loss for the government.

7. A person is selling a product at £100 with a profit of £50. He decides to hold a sale giving a 10% reduction. How many more of his product must he sell in order make the same profit as before?

Solution: Since the sale price will be £90, his new profit will be £40, instead of £50. Thus, one needs to sell 25% more to get the same result as before. Of course selling 25% more is a lot more work, so unless many sales result from this maneuver, in the long run it will not be a good idea

8 Three for the price of two.

 

Some stores offer three items for the price of two.  For instance, one is offered three books for the price of two..  In the following table, we calculate the profits. Take the example when the books are sold at a price of £10 each.  The profit will of course depend on what the books cost the  store.  We will consider two possibilities, assuming first that each book costs £5.  Then we consider the profit if each books costs £4.

Thus if each book costs the store  £5, selling one at £10 gives a profit of £5.  Selling three for the price of two costs the store £15, and they sell them for £20.  This gives them a profit of £5.  The rest of the following table has been calculated in a similar way.

 

Cost of book to the store

Profit selling one book

Profit 3 books for the price of 2

£5

£5

£5

£4

£6

£8

  From the table we see that selling three for the price of two gives the same profit as selling just one book..  On the other hand, with a lower cost, of £4 for the store, three for the price of two gives a larger profit than selling simply one.  Thus for this method of selling to be profitable for the store, there must be a substantial mark-up.

 

9. Pyramid selling.

There is a type of selling which sounds good for all participants. The first person sells franchises to 10 subagents, and gets a percentage of their profits. Each agent then sells franchises to 10 subagents and also gets a percentage of their sales. And so on. This system is untenable. Why?

Solution: Suppose we are dealing with a town of 100,000 people. Suppose there are 5 stages of agents and subagents. The very first in the chain has sold franchises to 10 subagents and each of these sells to10 more, making a total of 100 agents. Each of these 100 sells to 10 more, making a total of 1,000. Each of these again sells to 10 making a total of 10,000 participants. Each of these sells to 10 more, making a total of 100,000 at the fourth stage. At the fifth stage there would be 1,000,000 participants, and these are only the ones who have been appointed at the fifth stage, and do not even include all the others. In other words, after 5 steps the whole system collapses because there are not sufficiently many people to participate.

10. Estimate the lower of the two blood pressure readings for a person.

Solution: The blood pressure is measured by two readings in mm (millimeters) of mercury. When you are standing upright the lower pressure must be sufficient to keep the blood in the brain, otherwise you would faint. The heart is roughly 50 cm from the top of the head, and so the pressure must be sufficient to support a 50 cm column of blood. If we make a rough guess that blood and water weigh the same, we would need to support a 50 cm column of water. However, since we are not always at the minimum requirement, let us add 50% to get 75 cm of water.

Blood pressure is measured not in the lengths of water columns, but in the lengths of columns of mercury. To change from a column of water of 75 cm to a column of mercury divide by 13.9 to get about 5.4 cm, which is 54 mm. Doctors would say that this is too low, and a value of 70 or 80 would be more reasonable.

11. Pension. How much must one save for a pension of T pounds per year?

Solution: Suppose you want to have a pension of T pounds per year. Once one reaches the retirement age of 65, one has probably not much more than 15 years left of life. So one will need 15´T pounds in savings. Of course one will invest this sum, but the usual idea is to invest in very safe funds, which means that you can not expect a very high return, but hopefully enough to cover inflation and perhaps give you enough money for a few extra years if you live longer than 80.

As a check we note that the annuity rate is about 6% to 7% which is also about 1/15. This means that insurance companies that provide annuities want a payment of 15´T to provide a payment of T pounds per year.

12. Pension. What percentage of his salary should a person save for a satisfactory pension?

Solution: If one receives a salary of S, in practice one hopes for a pension of S/2, which is regarded as satisfactory in England. Thus if we use the calculation in the preceding example, one should save a minimum of 15´S/2. Here one can afford to take greater risks and thus get a greater return on the money one saves. One can at a guess expect to get double or even three times the money one has saved because of a reasonable return. (See the following problem for an explanation of this.)  Let’s be cautious and say double.  Thus one needs to save 15´S/4. Assume that one works for 40 years. Thus each year one needs to save 15´S/160, i.e., about 9% of salary.

13. If one saves one pound per year for 40 years at 5% interest, what is the final sum?

Solution: Note that 5% is a good average return, allowing for inflation and taxes.

After the first year one has (1 + .05) pounds. One then puts in another 1 in savings, making a total of (1 + (1 + .05)) pounds. Since this accumulates interest at 5%, after another year we will have (1 + .05) + (1 + .05)2 pounds. One then adds an extra saving of 1 to get a total, after the second year, of

1 + (1 + .05) + (1 + .05) 2 pounds.

Continuing in the same way, after 40 years we will accumulate a total of

1 + (1 + .05) + (1 + .05) 2 + …+ (1 + .05)40 pounds.

In order to calculate this sum, we replace (1 + .05) by the symbol r and thus the sum S we wish to calculate can be written as

S = 1 + r + r2 + … + r40.

Next, we multiply this by r we get

rS = r + r2 + … + r40 + r41.

The differences between rS and S are the 1 in the expression for S and r41 in the expression for rS. Hence, if we subtract S from rS, the common terms cancel out and we see that

rS – S = (r – 1)S = r41 – 1.

Hence, S = (r41 – 1)/(r – 1), and since r = 1.05, we see that

S = (7.392-1)/(1.05-1) = 6.392/.05 = 127.84.

Without interest, we would of course have saved £1 each year for 40 years, i.e. £40. The interest has meant that we have something like 3 times the amount.

14 A twenty year old receives a prize of one million pounds which he puts in a bank. Should he give up his job and “live happily ever after”?

Solution: Assume his present job pays £25,000 per year. With a million pounds one would expect to have a more luxurious life, say £50,000 per year. This would last 20 years. So he would be 40 when he had used up all his winnings. Clearly our 20 year old must think this over more carefully. He should certainly consider interests. Suppose he invests in an interest bearing account at 6% and that inflation is 3% and tax is 20% per annum. So he is receiving 3% after allowing for inflation, and after tax, he is getting 2.4%.  Thus he has an income of £24,000 per annum.  It does not look as if he can afford to use £50,000 per annum. These calculations suggest that he seeks a financial adviser.      

15. In September 2005 a hurricane threatened to destroy Houston Texas. It was essential to leave Houston. If you have a car, how urgent is it to leave?

Solution: To assess how urgent this was one could do the following rough calculation: Suppose that 1 million cars need to leave Houston. Not knowing Houston we have to guess. 

 It would be nice to have some basis for this guess, but as usual Fermi calculations are made on insufficient knowledge.  Inhabitants of Houston would do better. 

Suppose there are 5 ways of leaving the city and each road has 4 lanes.  Suppose that with the emergency, traffic moves slowly, say 20 km per hour.  Suppose we allow 20 m per car.  Then in an hour each lane will take 20000/20 = 1000 cars, so five four lane highways will take 20000 cars per hour.,  A million cars will require about 50 hours, about 2 days.  With only two day’s notice it would be sensible to leave as soon as possible.   . 

16. How long do you need in order to learn a foreign language?

Solution: To manage in a foreign language, one needs say 3,000 words. One can learn say 6 words in an hour. Hence one needs 500 hours. At ten hours a week, this is about 50 weeks, i.e. a year to get a useable knowledge of a language.

17. Estimate the proportion of teachers in the population.

Solution:  Assuming that ages in the population vary from 0 to 80 and that they are equally distributed.  In England one goes to school from the age of 5 to 17.  That means  the proportion of school children is 12/80 = .15. or 15%.    Assuming that each pupil is in a class of 30 others, and each such class needs a teacher, then we have that the percentage of teachers must be ½%.

 

 

Chapter 5

PERCENTAGES

 ½ could be a small or a large number. It depends what you compare it with. But ½% is small. That is the advantage of percentages. You know the importance of each item.

§1 Definition and examples

A simple but extremely valuable method of understanding numbers is to interpret them as percentages. This is especially so for figures one is not intimately concerned with. So, for instance, financial reports can be understood better by converting them to percentages. Similarly the budget for a country is more readily understood when expressed in terms of percentages.

As an example: In a prison population of 6,000, 12 prisoners escaped one year. The opposition called for the resignation of the Justice Minister who is also responsible for prisons. In percentage terms this means that 0.2% of the prison population have escaped, and even if this occurs every year, it is quite a small percentage, and calling for the resignation of the Minister of Justice seems a bit overboard.

Another example: The Rector of our university explained we were some ten thousands of pounds in the red. It was a tremendous figure. We were all shocked. So we asked the Rector what the deficit was as a percentage of the income. He had not thought of this but in the end said about 5%. This did not seem so serious a problem as it did at first. This is the value of percentages. They enable you to make sense of the figures. Nowadays with pocket calculators or spreadsheets they are easily calculated.

Examples

On holiday with a budget of £500 for a week a couple has estimated the following expenses.

 

Bus, tube and train travel

£70

Food including restaurants

£210

Museum and theatre charges

£150

Miscellaneous

£70

figure 1. Holiday Money

In the next table we have expressed the expenses in terms of percentages of the budget.

 

Bus, tube and train travel

14%

Food including restaurants

42%

Museum and theatre charges

30%

Miscellaneous

14%

figure 2. Holiday money in percentages

The percentages are calculated by dividing each expense by 500, the total cost in pounds for the week, and multiplying the result by 100. Thus bus, tube and train travel percentage are calculated by (70/500) ×100 = 14. The sum of all percentages must be 100%. It is easy to understand the significance of the figures, for instance, 75% is three quarters, etc.

Percentages give a clearer way of seeing how the money is spent. For instance, these figures seem to indicate that it might be worthwhile spending less on food, by reducing the number of restaurant meals and eating more sandwiches prepared at home. This would leave more of the budget to spend on museums and theatre visits.

Also useful to keep in mind is the connection between expenditures and percentages of the weekly and daily budgets in the following table.

 

£

% per week

% per day

10

2

14

50

10

70

100

20

140

500

100

700

figure 3. Holiday expenditure as percentages of the total available to spend

Here again we have assumed a total weekly budget of £500. Spending £10 amounts to a percentage of the week’s budget of 10/500×100, i.e. 2%. But the daily budget is £500/7=£71.43. So £10 as a percentage of the daily budget is about 13%. This makes it easier to judge whether it is worth spending that £10 on a particular day.

In calculating a percentage one starts by deciding on the reference value. In this example we have chosen the total amount available for the week to be the reference. Choosing a different reference will result in different percentages and give different impressions, so it is important to consider carefully what one should use as a reference. In this case, we might have used the cost of the total holiday as a reference, i.e. the cost of travel, the cost of the hotel, and the cost of the food, museums, bus etc. In this case the idea was to help decide how £500 was to be used to enjoy the holiday, and so it seemed the right quantity to choose. In general, if the reference is denoted by R, the percentage for each item is calculated by the formula

(Item/R)´100.

Caution: Do not try to average percentages.  For instance, if an investment goes up 100% one year, and then down by 50% the next year, the net result is not 25%  (100 – 50)/2) the average of the two percentages, but 0%.  For if you have £100, a 100% increase gives you £200, and a 50% decrease gives you £100, i.e. you are back where you started from.  . 

§2 A fictitious company’s accounts

In the following table we have listed a company’s accounts.

 

 

This year

Last year

Sales

£386,234

£320,234

Advertising

£50,000

£25.000

Postage & telephone

£35,874

£30,874

Wages

£140,000

£130,000

Directors fees

£70,000

£70,000

Consultant’s fees

£568

£368

Accountant’s fees

£1,865

£1,865

Profit

£87,927

£62,127

Working capital

£25,000

£25,000

figure 4. Accounts of a fictitious company

We can re-express this in terms of percentages of the previous year’s results as in the following table:

 

This year/last year %

Sales

120

Advertising

200

Postage &telephone

116

Wages

108

Directors fees

100

Consultant’s fees

154

Accountant’s fees

100

Profit

141

Working capital

100

figure 5. Accounts re-expressed in percentages of previous year y

Thus we see at a glance that sales are up by 20% and profits are up by 40% and advertising has doubled. The other items have shown relatively no change.

 

§3 Calculating percentages

We can calculate the percentages quickly with the aid of a pocket calculator. Even quicker is to use a spreadsheet. If you are familiar with say Excel, then the following description may be sufficient, though usually handling computer programs is more easily done with help. To illustrate how this method works we will take the holiday budget described in Fig. 1.

Begin by opening Excel. In cell D2 type “Percentage with respect to R =”. In cell H2 put in 500, which is the reference figure we chose for the holiday budget. Type in the table beginning in cell C4 where you type in “Bus, tube and train travel.” In cell H4 type 70, and continue to type in the rest of Fig.1. The table for the budget is now in cells C4 to C7 with the numbers in cells H4 to H7,

 

Bus, tube and train travel

70

Food including restaurants

210

Museum and theatre charges

150

Miscellaneous

70

figure 6. Example used to illustrate use of Excel for calculating percentage

In cell I3 type “%”. In cell I4 type = H4/$H$2*100 and press return. This calculates the percentage that 70 is of the reference you have placed in cell H2 (which in this case is 500). Go back to cell I4 and press ctrl and C at the same time. Then go down to the next cell I5 and press ctrl and v at the same time. You then go to the next row down to cell I6 and press ctrl and v at the same time. You go on doing this till you have covered the whole column. What Excel does is to copy your instruction of how to work out a percentage to each of the cells. It changes the H4 successively to H5, H6 and H7 as you go down the column, but the H2 which is the reference remains unchanged because Excel interprets the $ sign to mean leave this unchanged.

Thus, you will get the percentages in this way. If you should decide to change the reference R, go back to cell H2 and change it accordingly.

§4 Examples of judging figures in the news

It is very difficult to comprehend large numbers, so it is particularly useful to use percentages to describe them. The problem is: what percentage of what? The following examples give ways of understanding the significance of the numbers.

Example: In the Swedish news two items were mentioned. The first was that there would be an extra amount of one million crowns to assist further employment. The other was that 2,300 million crowns was the estimate of how much money was spent on illegal drugs per year.

Since a reasonable salary in Sweden is a quarter of a million crowns per year, the million crowns correspond to the wages of 4 people in a year. We can see immediately that the million crowns is not worthwhile bothering about. It can’t make much difference to the overall job situation.

The second figure, of 2,300 million crowns should be considered in relationship to the population. Since the population of Sweden is 9 million or roughly 10,000,000, or 107, this means that on average 2.3×102 crowns are spent on illegal drugs by each person. Of course, only a small part of the total population can be taking drugs, so as a rough guess we can start by excluding half the people, those between the ages of 0 and 20 and those from 60 to 80. Of those remaining, at a rough guess, suppose a quarter take drugs. This means that these people are spending 2,000 crowns each per year. This is a considerable sum of money and so the drug consumption is significant.

§5 Keep an eye on the total figures

Occasionally percentages may deceive. Suppose for argument that we have a study of 2,000 people, half of whom drank water with meals, and half of whom did not. Suppose in the first group there were 3 cases of cancer, and in the other group there were two cases of cancer. Then we could claim:

“In a study of 2,000 people, those who drank water with meals had 50% more cancers than those who did not drink water with their meals.”

Strictly speaking the statement is correct, but totally misleading, in that the number of cases is not sufficient to make a sensible conclusion.

§6 Abortions

The problem of allowing legal abortions is one of considerable importance. There are a number of different views, the most usual being

  1. Abortion is killing and killing is not allowed and so abortion should be illegal.
  2. While the foetus is inside the woman, it is really her right to decide what is to be done, and so abortion is acceptable.
  3. Contraceptive and abortion advice encourages sexual intercourse and so should be banned.
  4. In view of the need to avoid abortion; sexual and contraceptive advice should be freely available.

We do not wish to take sides but simply wish to point out the urgency of the problem. In Sweden the number of abortions per year is 30,000. Assuming that the ages of most people in the risk zone for needing an abortion are between 15 to 25, i.e. a ten-year span. If one argues that the population goes up to 80 that means that is roughly 1/8 of the population. Since Sweden has a population of 9 million, there are approximately one million people in that range and half of them are women. That means that one twentieth or about 5% of the women population per year are affected, a very large percentage.

§7 Compound increases

If you look back at prices and wages over say the last twenty years it is striking how much they have risen. This may be a consequence of the fact that we always think of increases in percentages per year. We expect our salaries to increase by a certain percentage each year. Similarly we accept with resignation but as being at least reasonable, an increase of say 5% each year in costs. However these costs are compounded, and occurring year after year they become large. For instance an increase of 5% per annum becomes a doubling in 14 or 15 years. Is it possible that the very concept of a percentage increase per year is the reason for the huge increases in costs and salaries?

§8 Worked examples

1. Your salary increases from £30,000 to £35,000. What is the percentage increase?

Solution:

The percentage increase is (5,000/30,000)×100 = 16.6%.

2. Your salary decreases from £30,000 to £25,000. What is the percentage decrease?

Solution

The percentage decrease is (5,000/30,000)×100 = 16.6%.

3. Your salary of £30,000 increases by 5%. What is your new salary?

Solution:

The increase is ((5/100)×30,000 = £1,500.

4. Your salary increases by 10% one year, only to fall by 10% the following year. Are you back to where you started?

Solution: For each £100 you received before the increase, you now receive £110. The decrease of 10% means that you now receive £99, so you are not back to where you started.

5. Your salary goes up 10% for two consecutive years. Is this the same as a 20% increase?

Solution: £100 increases to £110, which in turn increases to £121. Thus your salary after two increases of 10% is greater than after a single increase of 20%.

6. Using Fig. 4 of §2 calculate the percentages of expenditures with respect to sales for this year.

Solution:

 

Percentages with respect to Sales

Sales

100

Advertising

13

Postage & telephone

9

Wages

36

Director´s fees

18

Consultant’s fees

0.1

Accountant’s fees

0.48

Profit

23

Working capital

6

figure 7. Calculating the percentages in Figure 4.

 

7. Estimating a percentage helps thinking. 

In the Swedish election held in September 2006 an alliance of conservative parties went to the polls with the promise of reducing unemployment.  They had a number of measures, which included making it cheaper for employers to employ a long time unemployed, but they were also going to reduce unemployment benefits.  Was it smart for the unemployed to vote for this alliance?

Solution:  Probably not.  For one must always ask by how much could the new government reduce unemployment.  It won’t be 100% and a reasonable guess would be 10% because on the whole it is difficult to make a big change in a society.  This means that 10% of the unemployed would get a job but that the other 90% would have lower benefits. 

 

Chapter 6

MEASUREMENT SENSE OR DIMENSIONAL ANALYSIS

If the units are right, the formula is often right.

This chapter is about length, time, force, work and power, fundamental physical concepts, which we will discuss informally to provide an intuitive feel for these concepts. Kilometres, kilograms, watts, horsepower, seconds: these are the substance of our discussion.

In the early 1900s, the advent of mass production required precise measurement. In the beginning of the motor industry each part was handcrafted to each car, but with the introduction of mass production methods it became important to build parts which were so standard that they would fit any unit on the production line. Precision measurements and, consequently, accurate units, became imperative.

The international system of units, normally denoted by SI (short for Système International), is most common, although the Imperial system is popular in America and Great Britain. We will concentrate almost exclusively on the SI system.

§1 Length

Since some measurements are very small and others very large, it is convenient to express these in terms of powers of 10, which we will now define. The convention is that 10, stands for 10 multiplied by itself n times if n is a positive number while 10-n is 1/10n. For example, 5X103 = 5000, while 5X10-3 = 0.005. This has the added advantage that calculations involving products become easier, since 10m×10n = 10m+n, for both positive or negative values of m and n. [These powers of 10 have also been discussed in Chapter 2, Section 4.]

The standard unit of length is the meter, abbreviated by m. Originally it was defined as one ten thousandth of the distance from the equator to the North Pole. Later a special rod was kept in Paris to be the standard for the meter. The present method involves using light to define the standard, but the technical details need not concern us here. All we need to know is that there is a system that ensures an accurate and uniform definition. Several prefixes are standard for SI units. We use kilo-, for “thousand”; centi-, for “hundredth”; milli- for “thousandth”. For example, a kilometer is a thousand meters.

A centimeter (cm) is one hundredth of a meter, that is, 1 cm = 10-2 m and 1 m = 102 cm, To have a reasonable understanding of the units of length, the width of a hand is about 10 cm, a meter is the measurement from the ground to your belly button, and the length of a small European car is about 4 meters. A three-story block of flats is about 10 meters high, which is also tree height. Also, while 100 meters is a short stroll, 100 meters vertically is extremely high, the height of a 30-story skyscraper.

Turning to measuring the smallest of distances, such as the atom, we need another unit called the angstrom. An angstrom is 10-8 cm., which is about the width of an atom. The smallest distance that can be seen in an electron microscope is three-fifths of an angstrom.

The following table lists a few other approximations.

 

Finger width

2 cm

Width of thumb

2.5 cm

Height of step in a flight of stairs

16 cm

Length of a foot, or height of a head

25 cm

From the foot till the knee

50 cm

Width of kitchen units

60 cm

Foot to navel

1 m

Tall man or height of door

2 m

Length of small car

4 m

One story of a block of flats

3 m

Height of airplane flight

5 – 12 km

Height of TV satellite

36,000 km

§2 Mass

The SI system has a standard unit of mass, the kilogram, abbreviated kg, which is a fixed body kept under careful conditions

The standard mass is kept well protected and used only to make a very few secondary standards, which are themselves used to make further standards, and in this way the standard of mass, the kilogram, is uniform throughout the world.

The gram – written g - is the mass of a body one thousandth of the kilogram. To give some idea of masses we have the following examples:

 

 

Mass

A4 sheet of paper

4 g

Standard letter

20 g

Slice of bread

40g

An egg

60g

Banana or apple

120 g

Meal of two eggs, two slices of bread, and a potato

350 g

A litre of milk

1 kg

A packed suitcase

20 kg

A man

70 kg

A medium sized European car

800 kg

Substituting for the unit as shown in the following example is a useful technique for changing from one unit to another.

Example: Changing units, milligrams to kilograms. Express 34 mg as kilograms.

The first step is to convert mg to g:

1000 mg = 1 g, so 1mg = 1/1000g = 10-3g.

We simply take the mg in 34mg and replace it by ×10-3g. Thus

34mg = 34 ×10-3g.

The next step is to convert g to kg:

1000g = 1 kg, so 1g = 1/1000 kg = 10-3 kg, or g = 10-3 kg.

To continue, we replace the g by ×10-3 kg. Hence,

34mg = 34 ×10-3g = 34 ×10-3 ×10-3kg = 34 ×10-6kg.

Note that we have simplified 10-3 ×10-3 to 10-6 .

§3 Time

The SI unit of time is the second, and this is denoted by s. Originally this was defined to be the time for a pendulum of length one meter to swing from one extreme to the other. The modern measurement is based on the Cesium-133 atom, but we will not bother about the details in this book.

“One thousand and one”, “one thousand and two”, etc. spoken deliberately measures time in seconds. The number of heart beats is about 70 per minute, so that one heart beat is about one second. Counting fast from one to ten lasts about 2½ seconds.

We denote hours by the symbol h. The following table lists some time markers.

 

Hours worked in a year (40 hours/week)

2,000

Hours in a year

8,760

Time spanned by great grandfather, grandfather and father

100 years

§4 Speed:

Speed is defined as distance traveled divided by the time traveled.

Speed = distance/time

In symbols, the speed of an object traveling a distance d in time t is

d/t.

Example: Calculating speed

A horse travels 56 kilometers in 3 hours. What is the speed of the horse?

Solution

Speed = distance/time = 56km/3 h = 18.67 km/h. Here is a table of speeds.

Travel at 11 km/h

3 m/s

Travel at 108 km/h

30 m/s

Man running 100 meters in 10 seconds

36 km/h

Speed of sound

340 m/s

Speed of light

3 ×105 km/s

For instance, if a man shouts out to you at a distance of 100 meters, the sound will take one third of a second to reach you. On the other hand, the light reflected from the man will take 102 times 1/3 ×10-8 seconds or 1/3 ×10-6 seconds to reach you. The sun is so far away that it takes 8 minutes for the light from the sun to reach the earth.

Estimating the speed of sound

Two people stand about 150 meters from one another. The first one blows a whistle, and starts his stopwatch. The other whistles back as soon as he hears the first one. When the first one hears the other whistle, he stops his stopwatch. Thus the sound has travelled 300 m, and if the time is about 1 second, the speed can be calculated as 300m/sec.

Another method of estimating the speed of sound is in the spirit of Chapter 4, using Fermi calculations, in which we guess and estimate a rough value.

We know that commercial aeroplanes fly at a speed of about 900km/h. This is less than the speed of sound, so this is a lower estimate for the speed of sound. Now 900km/h = 15km/ min = 250 m/s. So knowing the speed of commercial aircraft, we can estimate the speed of sound to be about 250m/s, which is roughly in agreement with the known value.

Estimating the speed of light by satellite

This is also a calculation in the spirit of Chapter 4. If you listen to a TV sending a report from a correspondent who is a considerable distance away so that the message comes via satellite, you will notice a pause between the question asked in the studio and the reply. This is partially due to the distance that the radio signal must travel. The radio signal travels at the speed of light. The satellite is normally in what is called a geostationary orbit, which is some 40,000 km above the earth. The signal must therefore travel from the sender in your country to the satellite, and then travel from the satellite to the correspondent. His reply must also travel up to the satellite and from the satellite back to the studio. That is a total journey of 4 ×40,000 km or 160,000 km. By dividing this distance by the delay in replying one can get a rough estimate of the speed of light. In one program we viewed, there was a ½ seconds delay, which gives an estimate for the speed of light to be 320,000 km/s.

Rohmer estimate of the speed of light

In 1676 the Danish astronomer Rohmer gave an estimate for the speed of light. He had measured when the Jovian moon would cross the face of Jupiter, and found that this time varied depending on how far away Jupiter was from the earth. He suggested that the time difference was due to the time it took for the light to travel to the earth and in this way gave the estimate, 227,000 km/s.

Some wind speeds

60 km/h is a strong breeze, large branches in motion, umbrellas handled with difficulty, telegraph wires can be heard whistling. Wind at 70 km/h is called a severe gale, and will cause structural damage, such as chimney pots being displaced and slates removed. A wind speed of 100 km/h is a violent storm; when it reaches 118 km/h it is classified as a hurricane, thankfully not very often experienced.

Examples of speed conversion

To convert m/s to km/h we note that 1km = 1000m, so that 1 m = 10-3 km and 1 s = 1/3600 h. Substituting these values for m and s, we get
1 m/s = 10-3 km/(1/3600) h = 10-3 X 3600 km/h = 3.6 km/h.
Converting from km/h to m/sec is a bit easier. Since 1km = 1000m and 1hr=3600s, we conclude that 1 km/h = 1000m/3600s = 5/18 m/s.

§5 Acceleration

Acceleration is defined as rate of change of speed. That is, if the speed at time t = t0 is s0, the acceleration necessary to achieve a speed of s1 at time t1 is defined to be

(s1 - s0)/(t1 - t0).

That is, the difference in speeds divided by the time.

Example 1: Acceleration.

A car running at 40km/h speeds up to 50 km/h in 2 minutes, i.e., 1/30 h.. The acceleration is given by (50 – 40) km/h/ (1/30) h = 300 km/h/h. This is usually written 300 km /h2.

Example 2: Acceleration.

Calculate the acceleration of a sports car if it moves from 0 to 100km/h in 4 seconds. The change in speed from 0 to 100 is 100km/h = 1/36 km/s. Since this occurs in 4 seconds, the acceleration is 1/144 km/s2. In this example we chose to work in seconds, whereas in the previous example we worked in hours. Either way is acceptable, but since the SI units include the second as a basic unit, it is probably better to use seconds.

figure 1 Galileo Galilei (1564-1642). Introduced the modern scientific approach based on experiment or theory supported by experiment. Father of Mechanics the study of moving bodies, forces and gravitation. Also made magnificent discoveries in Astronomy.

Acceleration due to gravity

It is a remarkable fact that all bodies acted on by gravity fall to the earth at the same speed if air resistance is insignificant. Galileo (1564-1642) demonstrated this by dropping two grossly different sized cannon balls from the leaning tower of Pisa and observing when they struck the ground. The expert knowledge at that time was that the larger the mass the shorter the time of impact. When Galileo tried the experiment he proved this was wrong, and showed that the difference perceived earlier was due to air resistance.

One way in which to argue this logically is to consider a mass of 10 kg and regard it as being split into 10 masses of mass 1 kg each. Drop all ten one kg masses simultaneously and, of course, they will all fall to the ground in the same time. Two adjacent masses will not affect one another so if we glue all the masses together to form one 10 kg mass, they will still fall to the ground in the same time.

You may prefer to argue this case using Newton’s laws of motion, as given in §10 problem 3.

Experiment reveals that the acceleration of any mass when wind resistance is disregarded is 9.80 meters per second per second.

§6 Force

Force depends on two factors. You notice that a force is acting when a mass accelerates. The mass and the acceleration are both needed to define the force. It is defined by multiplying the mass by the acceleration. The SI unit of force is the Newton, which is the force needed to cause a mass of one kilogram to accelerate one meter per second per second. If you hold a mass of 100 grams in your hand, it is pulled to the earth with a force of approximately one Newton. This is because the acceleration, as we remarked above, is 9.80 meters per second per second, and so the force is 0.1 kg × 9.80 m/s2; that is 0.98 Newtons, or approximately one Newton. The symbol to denote a Newton is N.

§7 Work and Power, Joules and Watts

In this section we discuss the material in an intuitive way, and provide a more precise and detailed discussion in §10.

Work depends on two factors. There is a force you are struggling against and a distance that you move through, and indeed, work is defined to be the product of the force and the distance. For instance, if you lift a suitcase a meter high from the ground the work is less than if you lift two suitcases a meter high, or lift one suitcase 2 meters high. How long you take to do this does not change the work done; a second for the job entails the same amount of work as taking an hour, just as if you travel from one point to another, you still have travelled the same distance, whether it takes a day or an hour. It is the speed which changes if you take more or less time, not the work.

However, if you take the time into account, then instead of work you measure power. Power is the work divided by the time; so the shorter the time, the more the power.

Power is measured in watts – indicated by W. A kilowatt is a 1,000 W, and is denoted by kW.

Example: Lifting a suitcase weighing 25 kilogram one meter high (i.e. up to your waist) in one second is a power of roughly 250 W. If you do the same work in one half a second, the power is 500 W.

How exactly these ideas are defined and the values calculated appear in the optional material below in §10. Here the aim is to give an intuitive idea of these concepts.

Thus, a person at rest is working at a rate of something like 30 W. A normal size electric light bulb works at a rate of 60 W, and an electric kettle works at a rate of about 1,000 W, i.e. a kilowatt. The power of a human being, i.e. the rate at which he can work, is approximately 90 W.

During the steam age it was the custom to visualize the power of the new steam engines in comparison with horses. This led to the unit of a horsepower. One horsepower is the equivalent of 746 W. Actually this corresponds to an idealized horse, and the value is a little optimistic, in that working at the rate of one horsepower is more the rate of work of one and a half horses.

The concept of horsepower gives one a graphic way of viewing power. When you boil a kettle of water in an electric kettle, the electricity is working at a rate that is more than that of a horse. A car driving fast at constant speed along a level road works at the rate of about 20 horsepower.

Cars can typically develop 100 horsepower. Super sports cars boast 350 horsepower and more.

The typical electricity consumption in a modern flat means that we have roughly the equivalent of a horse working for us for ten hours every day or 80 man-hours of labour. In other words, we have the equivalent of 8 slaves working for us every day for ten hours. No wonder modern man is like a king of only a century ago. [See §9 problem 3 for the calculation.]

Kilowatt hours: A kilowatt hour is a measurement of work. It is the work done at a rate of one kilowatt for one hour. For example, a typical electric kettle boiling water non-stop for an hour performs one kilowatt hour of work. It is denoted by kWh,

§8 Dimensional Analysis

A useful technique for checking a physics formula, or even getting a suggestion as to what the formula should look like, is obtained by using the fundamental dimensions of length [L], time [T], and mass [M]. The square brackets are used to indicate that we are talking about dimensions. It is a fundamental law in Physics that in any equation the dimensions on the right-hand side of the equation must be equal to the dimensions on the left-hand side of the equation.

Example 1

Find the dimensions of area and volume. What are the dimensions of speed and acceleration?

Solution: Area is calculated by multiplying length by breadth. Thus the dimension is [L]2. Volume is obtained by multiplying length by breadth by height with dimension [L]3. Speed is distance divided by time, giving us the dimension [L]/[T] or [L][T]-1. Acceleration is speed divided by time. Since the dimensions of speed, as we have just calculated, are [L][T]-1, for acceleration we must divide by time again, and the dimension must be [L][T]-2

Example 2

Find the units of force and work.

Solution: Force is defined to be the product of mass and acceleration. Acceleration has the units [L][T]-2, thus the units of force are [M][L][T]-2. Work is force, [M][L][T]-2, times distance, [L], so the units for work are [M][L]2[T]-2.

Example 3

A pendulum of length l is subject to gravity, which has acceleration g. Find the form of the formula for the period of the pendulum, that is, the time required for the pendulum to swing from its initial starting position back to its starting position.

Solution: It seems reasonable to assume that the period will be a constant times a power say a of the length, l, of the pendulum and the b-th power of the acceleration due to gravity, g. That is, T should have the form

T = constant× lagb ,

where the powers a and b are to be determined. The dimensions of g are the dimensions of acceleration, i.e. [L][T]-2, and the dimensions of l are [L]. Hence, the dimensions of the right-hand side of the formula are [L]a[L]b[T]-2b = [L]a+b[T]-2b . These must match with the dimensions of the left-hand side, and since T has the units of [T], this means that [T] = [L]a+b[T]-2b. This will be possible only if -2b = 1, and a+b = 0, that is, b = - ½, and a = ½. Hence, the formula for the period must look like:

T = constant × l½g-½.

(Note that an exponent of ½ means the square root, and an exponent of - ½ means 1 divided by the square root.) In fact, the correct formula is

T= 2pl½g-½ = 2pÖ(l/g).

Of course, this analysis does not give an exact result for the constant, but it is remarkable how frequently dimensional analysis points towards the right formula.

Example 4

A body in free fall (i.e., subject only to gravity and with no wind resistance) has an acceleration g due to gravity. Find the units in a formula relating distance fallen, x, and time of fall, t.

Solution: Assuming a formula of the form x = constant×gatb, where g is the acceleration due to gravity, in terms of units this becomes

[L] = [L]a[T]-2a[T]b.

Since the left-hand side of the equation has [L] to the power 1, a must also be 1. On the left-hand side [T] does not appear, but on the right-hand side we see that with a having the value 1, we have [T]b - 2 so this entails that b = 2. The actual formula is

x = ½ gt2.

We now come to what is at once the most wonderful and the most terrible of all formulae in Physics, namely the formula from Einstein’s paper on Relativity relating energy, E, mass, m, and the speed of light, c, in one formula:

E = mc2,

the product of the mass m and c2, which itself is the product of the speed of light by itself. Note that in Physics energy means the ability to do work, and the equation says that if the mass m were converted entirely into work, the amount of work would be given by the formula. Thus the units of energy are the same as the units of work.

figure 2. Albert Einstein (1879-1955), whose theoretical discoveries
contributed to the development of nuclear power and the nuclear bomb.

It is the most wonderful equation because it explains so much. For instance, the sun, which is the source of all energy and light on the earth, gives out an enormous amount of energy. For a long time it was thought that this was the result of burning: the same sort of burning we have on Earth, where matter combines with oxygen. That meant that it was possible to estimate how much energy the sun had, but when the calculation was done, it was obvious that the sun should have burnt out long ago. Ordinary burning could not explain the energy, which the sun emitted and had emitted for so many years. Einstein’s equation could. The energy of the sun came from the conversion of mass to energy.

Einstein’s equation also meant that we ourselves could convert matter into energy in our nuclear reactors, which has turned out to be a mixed blessing, with plenty of energy but with  awful dangers. Einstein’s theory has also led to nuclear weapons with the ability to destroy our civilisation completely.

Example 5: Check that the dimensions on both sides of Einstein’s equation match.

Solution: The dimensions of E the left-hand side of the equation are those of work, which we found out in Example 2 immediately above, were [M][L]2[T]-2. The dimensions of the RHS are those of [M] multiplied by the dimensions of velocity squared, i.e.[L]2[T]-2, Hence the dimensions on both sides of the equation are the same.

§9 Solved problems

1. What is the acceleration in going from 0 to 100 km/h in 10 seconds?

Solution. The acceleration is given by difference in speed divided by time. We need to use the same unit of time. 10 seconds is 1/360 of an hour.  The difference in speed is 100. So the acceleration is

100/(1/360) = 3.6X104 km h-2.

2. In a recent Winter Olympics the difference in time between the first and second places in the women’s skeleton slide was 0.65 seconds. Assuming that the final speed attained in the event was 120 km/h, what was the distance in meters between the two at the finish?

Solution. 120km/h = 120 ×1000 /3600 m /s = 33.33 m/sec. Hence, 0.65 seconds corresponds to

0.65 × 33.33 m = 21.66 m.

3. Assuming that your household uses 2,500 kWh per year, what is the average daily consumption? What is the equivalent in horses per day?

Solution. There are 365 days in a year, so the average daily consumption is 2500/365 = 6.8 kWh. One horsepower is 746 W. So it would require roughly 9 to 10 horses to work at the rate of 6.8 kW for an hour. So 9 or 10 horses working for an hour, or one horse working for 9 or 10 hours, will give about 6.8 kWh. In terms of manpower, or slave power, this means we would need 8 slaves working each day for each household for 10 hours, since a man can work at only about 90 W.

4. If a man drives at the rate of 50 km/h for the first 300 km of a trip, how fast must he drive over the next 300 km to have an average speed of 60 km/h for the 600 km journey?

Solution. At an average speed of 60 km/h for 600 km, it would take 10 hours to complete the journey. At 50 km/h, the first 300 km drive would have taken 6 hours, which leaves him 4 hours to complete the final 300 km. Hence, to make up the time he would have to drive at 75 km/h over the final 300 km.

 

5. One man can dig a hole in 2 days, another man can dig the hole in 3 days. How long will they take to dig the hole together?

Solution: There is an ingenious way of solving this problem. One calculates the rate of digging the hole. The first man has a rate of ½ a hole per day, the second man a rate of 1/3 a hole per day. Thus the combined rate is ½ + ⅓ = 5/6 of a hole per day. Thus it takes 6/5 of day to dig the hole.

§10 Additional Topics

1. Definition of force

The force acting on a mass is defined by multiplying the mass M in kg by the acceleration a in m/s2, i.e.

Force = Ma.

The unit of force is the Newton. Thus a Newton is the force that accelerates a mass of 1 kg by 1 m/s2. For example, a mass of 15 kg which when acted on by a force accelerates 40 m/s2 has a force of 15 ×40 = 600 N acting on it.

2. Newton’s laws of motion (published in 1687)

figure 3. Newton (1643–1727) was co-discoverer of the Calculus and developed
a far-reaching theory of moving bodies such as the motion of the planets.

Newton formulated three laws of motion. With these and the mathematical theory that Newton developed, which is called Calculus, Newton was able to explain all the laws of planetary motion. Newton’s methods are still fundamental in calculating, for instance, the movement of satellites as well as the laws of physics which regulate our everyday lives.

·       The first law states that a body continues moving in a straight line at the same speed unless acted on by a force. One result of this law was to dispel the commonly held belief that planets moved because angels or spirits were pushing them.

·       The second law defines force in terms of mass and acceleration as described in item 1 immediately above.

·       The third law is that to every action there is an equal and opposite reaction. That law governs the motion of rockets as well as collisions of billiard balls.

Newton also assumed that there is a force of attraction between two bodies with masses m1, m2, which is a constant G times the product of their masses divided by the square of the distance d between them, that is,

F = G m1m2/d2 .

The value of G has been found experimentally to be 6.67 ×10-11. The units of G are Nm2/kg2.where N stands for Newton, m for meter and kg for kilogram.

3. Explain why all bodies fall to the ground at the same time if air resistance is not a factor, using Newton’s Laws of Motion.

Solution; If M is the mass of the earth and r is the distance to the centre of the earth, then a mass m will experience a force of

GmM/ r2

according to point 2 above. Since the force equals ma where a is the acceleration, we have the equation

ma = GmM/r2.

If we cancel m from both sides, the result is

a = GM/r2

in which m doesn’t play a part.  This means that all bodies accelerate at the same rate independent of their masses and, hence, will reach the ground at the same time if released simultaneously.

 

4. Definition of work

Work is defined to be the product of the distance a force is moved through times the force. The unit of work is the joule, abbreviated to J.

The Joule is the work done in moving a force of 1 Newton a distance of 1 meter. Since the force due to gravity on a 100 gram mass is about 1 Newton, if you raise a 100 gram weight from the floor to you waist, you will have done one joule of work.

5. Definition of Watts

The rate of work in Joules per second is called the power. A Joule per second is also known as a Watt.

6. A man weighing 70 kilos walks up three flights of stairs in one minute. What is the work done and what is the rate of work, i.e. the power?

Solution: Every step is 16 cm high and there are 16 steps per level. That means 768 cm, i.e. roughly 8 m. He is lifting 70 kg. The force acting on the man is calculated by taking the mass and multiplying by the acceleration due to gravity, which is 9.81 m/s. Thus the force in Newtons is

70´9.81 = 686.7 N.

To calculate the work done we need to multiply the force by the distance, to get

8´ 686.7 = 5,493.6 J.

To find the rate or work we divide by the time in seconds, which is 60. Thus the result is

5,493.6 /60 = 91.56 J/s = 91.56 W.

7. Approximate the power in raising a shoe from the floor to a height of 2m. in three seconds.

Solution: A shoe of mass say 400 g =0 .4 kg has a force of 0.4× 9.81 N . i.e. the mass 0.4 kg multiplied by the acceleration due to gravity 9.81 m/s2. This is approximately 4 N. So the work done on lifting the shoe two meters is approximately 8J. Thus the power required is 8/3 J/s. or 2.67 W.

8. A car journey of 100 km takes an hour and uses 7 litres of fuel. What is the work done and what is the rate of work.

Solution: This is another of those Fermi type calculations. According to the label on a bottle of cooking oil it has an energy value of 3,700 kJ per decilitre. It would be better to know the value for a decilitre of petrol but this figure is not available, so we use the cooking oil as a rough estimate. Hence a litre of petrol has about 40,000 kJ and 7 litres about 300,000 kJ. Per second we have therefore been using 100 kJ per second, or. 100kw. Since a horsepower is roughly 740w or, even more roughly, a kilowatt , this corresponds to 100 horsepower. We know however that there are inefficiencies in engines, and so if we assume only one quarter of the energy is available at the wheels, the car is developing roughly 25 horsepower.

9. Calories and conversions

Another unit of energy is the calorie. This is defined to be the amount of energy required to raise the temperature of one gram of water by one degree Centigrade. Since calories and Joules are measurements of energy one can convert the one to the other, and a calorie is the larger by approximately 4 times. More accurately,

1 calorie = 4.184 J.

Just as there is a kJ, there is also the concept of kilocalorie, a thousand times larger. This is also the unit used to measure human nutrition. It is often written as kcal or Calories, with a large C. A daily intake is roughly 2,000 kcal.

10. Calories of some foods per 100 gram.

·       Bread 230 kcal

·       Olive oil 884 kcal

·       Butter 700 kcal

·       Fried hamburger 280 kcal

·       Cheese 30% 400 kcal

·       Cottage cheese 4% 100 kcal

·       Sugar 400 kal

·       Baked soya beans 100 kcal

Thus, each gram of food gives between 1 and 9 kilocalories.

11. Estimate the weight of food eaten each day.

This is an exercise in the spirit of Chapter 4 Fermi Problems. Since one eats about 2,000 kcal per day, and food gives about 2 kcal per gram (rough guess based on the list of foods above), we need 2,000 divided by 2, i.e. 1,000 g or 1 kg of food per day

12 Estimate the rate of energy used by a person.

Solution: Since 2,000 kilocalories is consumed in 24 hours (and this corresponds to the minimum), the rate of energy is 2000/24 ×3600 kilo calories per second = about 6 calories per second. From point 10 immediately above, 1 calorie = 4.184 J. Multiply by 4.18 to get roughly 24 J per second or 24 W. This seems to corroborate the roughly 30 W given in the text above,

13. Given that 1W = 1J/s, that 1 calorie = 4.18 J, and that 1 horse-power = 745 W, …

a)     Express J in terms of calories

b)     Find the relationship between kJ and kW hours

c)     Find kW in terms of horse-power.

Solution:

a)     Since 1 calorie = 4.18 J, dividing by 4.18, 1J = 0.239 calories.

b)     To express kW hours in terms of kJ. 1J/s = 1W, so 1 W hr = 3600J.

So 1kW hour = 3600 kJ. Hence 1 kJ = 1/3600 kW hours = 2.778 X 10-5 kW hours.

c)     1 horsepower = 745 W. Thus 1000 horsepower = 745 kW. Thus dividing by 745, 1 kW = 1000/745 horsepower = 1.342 horsepower.

14. Temperature

The most commonly used temperature scales are the Centigrade (also called Celsius) and the Fahrenheit.

There is a third scale, called Kelvin. Zero degrees Celsius is approximately 273 degrees Kelvin, that is, 0 degrees Kelvin is 273 degrees below 0 Celsius. The Kelvin is of scientific interest because at a temperature of 0 degrees Kelvin molecules have no energy and have stopped vibrating.

The Celsius scale, which dates from 1743, is based on a temperature of 0 degrees for water freezing and 100 degrees for water boiling. On the Fahrenheit scale, which dates from 1724, water freezes at 32 degrees and boils at 212 degrees. 

The founder of the scale, Fahrenheit, used 32 to avoid negative temperatures in winter. Also, Fahrenheit’s idea was to have a temperature of 100 degrees for the human body, which is close to the actual figure of 98.6 degrees Fahrenheit. Unfortunately, when he did his experiment, his assistant had a slight fever.

To convert from Centigrade to Fahrenheit multiply by 9/5 and add 32. To convert from Fahrenheit to Centigrade, subtract 32 and multiply by 5/9.

Example: Convert 20 degrees Celsius to Fahrenheit. We multiply 20 by 9 and divide by 5 to get 36. We then add 32 to get the Fahrenheit equivalent of 68 degrees Fahrenheit.

Convert 80 degrees Fahrenheit to Centigrade. We subtract 32 from 80 to get 48, multiply by 5 and divide by 9 to get approximately 27 degrees Celsius.

15. When does the Celsius reading equal the Fahrenheit reading?

Solution Let t denote the temperature on the Celsius scale which is to be the same reading on the Fahrenheit scale. Since t degrees Celsius is the same as (9/5)t + 32 degrees Fahrenheit, the problem is to find a t such that (9/5)t + 32 = t. Subtracting t from both sides gives (4/5)t = -32, so t = - 40. That is, when it’s -40 degrees Celsius it is also -40 degrees Fahrenheit.

16. Check the following verse which accentuates the bewildering variety of old English units:

A dozen, a gross and a score,

Plus three times the square root of four,

Divided by seven, plus five times eleven,

Is nine squared

And nothing more.

Solution: This is indeed an exercise in English units. A dozen is 12, a gross is 144 and a score is 20, and so their sum is 176, Three times the square root of 4 is 6, giving a total of 182. If we divide 182 by 7, we get 26. Add 5 times 11, i.e. 55, we get 81. 81 is 9 squared, as claimed.

17. Derive the units of G.

Solution: We have the formula: force = GmM/r2.  Force has the units [M][L][T]-2, and r has the units [L].

From the formula

[M][L][T]-2 = [G][M]2/[L]2, it follows that [G] = [M]-1[L]3[T]-2 .

CHAPTER 7

MEASURING HEIGHTS OR TRIGONOMETRY

 A knowledge of right-angled triangles with largest side 1

enables us to calculate the lengths of any other right-angled  triangle

§1 Shadow heights

It is easy to measure heights by means of shadows. If you want to decide on the height of a tree then compare the length of its shadow with the shadow of a vertical pole. If the shadow of the tree is four times longer than the shadow of the pole, then its height is 4 times the height of the pole.

Often it is easy to use your own shadow to make the comparison, since you know your own height. By calculating how many lengths of your shadows fit inside the shadow of the tree you can estimate the height of the tree. You can do the same for a building or any vertical structure.

This method for measuring heights is based on comparing similar triangles. Similar triangles are the same shape, but of different size; like two shirts of different sizes. More formally, two triangles ABC and A¢B¢C¢ are similar if their angles are equal in pairs, as in Fig. 1, where angle A = angle A´, angle B = B´ and angle C = C´. If two triangles are similar, the one with sides of length x, y and z, then there is some number k so that the lengths of the other triangle’s sides are k×x, k×y, k×z..

Fig. 1 Similar triangles

How does this concern measuring heights from shadows? The reason comes about as follows. First of all, since the sun is so far away we may think of light from the sun as consisting of parallel rays.

Suppose that the top of the tree we are concerned with is at the point T (see Fig. 2) and its bottom at B. The sun casts a shadow that extends from B to the point E, the end of the shadow. Thus we have a triangle BTE with a right angle at the point B since the tree is at right angles to the ground. By measuring the length BE of the shadow of the tree, we can determine the tree’s height BT if we have a reference.

The reference we obtain by measuring the shadow of a person standing upright, i.e. also at right angles to the ground. Suppose the top of the person is represented by the point P, the person is standing at A, and the person’s shadow is from A to Q as in Fig. 2.

Now the two triangles BTE and APQ are similar, because each has a right-angle and the angles TEB and PQA are equal, since the sun’s rays are parallel. This means that the third angle of each triangle is the same. We can therefore find the lengths of the triangle BTE by multiplying the lengths of the triangle APQ by a constant k. In particular, if we know that any side of BTE is k times the length of the corresponding side of APQ, this will mean that all sides of BTE are k times the sides of APQ.

For example, suppose the shadow cast by the person is 3 meters and that the tree has a shadow of 30 m, then the constant k must be 10. If the person has a height of 2 m, then the height of the tree must be 10 times the height of the man, i.e. 20 m.

We can write this in the form of an equation in general. To find the constant k we simply divide the length of the shadow of the tree by the length of the shadow of the man, i.e.

length of shadow of tree/length of shadow of man.

We then multiply by the height of the man, so that

Height of tree = (shadow of tree/shadow of man) × (height of man),

figure 2 Shadows

We checked the height of a building this way: By comparison, we climbed from floor one to floor two, measuring the number of steps. There were 16. Each step was 16 centimetres high, so the total height for one floor is 16×16 = 256 cm, i.e., roughly 2.60 meters. The three-storey flat was therefore 7.80 meters. But since the ground floor was about one meter from the ground, we made the height to be 8.80 meters, i.e. about 9 meters. This agreed with the height calculated from the shadows.

These are rough calculations, but the surveyor uses the same principle of similar triangles and measuring very accurately gets precise results.

§2 The artist’s method

A standard method used by artists is shown in Fig. 3. Here he holds his arm straight in front of him and compares the various heights on his pencil. For instance, if he is drawing a head, he gets the tip of the pencil at the top of the head he is drawing, and then places his thumb to be in line with the eyebrows. He then uses the pencil to mark this distance on his sketch. Then in a similar way, he measures the distance of the eyebrows and the tip of the nose, and then to the mouth and finally to the chin. In this way he gets accurate measurements. This is also based on similar triangles. The explanation is given in §5 problem 8.

 

 

 

 

 

 

 

figure 3 The artist’s method

Right-angled triangles appear often in practice, for instance in surveying. Given a right-angled triangle ABC and knowing the angle at A and the length of the largest side h, we can with the aid of a table calculate the lengths of the other two sides. This is the idea of sine and cosine. The largest side of a right-angled triangle is also called the hypotenuse.

We would have to produce an infinite table if we listed all right-angled triangles. We reduce our list by considering right-angled triangles with hypotenuse 1. The triangle is then determined by one of its angles, since the sum of the angles of a triangle must be 180 degrees. The sine and cosine are simply the lengths of the two sides in a triangle with hypotenuse 1.

Thus in the right-angled triangle ABC in Fig. 4, AB has length 1.

figure 4 Definition of sine and cosine

In Fig.4, the side BC, the side straight in front of the angle A, is called sine of the angle at A, and written sin(A). The side AC is called the cosine of the angle at A and written cos(A). (Note that it is customary to write the sine or cosine of a given angle without the final e.)

To make sure you remember which is which, you can think of sin as S in, i.e. short for straight in front. Also you can think of cos and that the C stands for the closer side.

The sine and cosine are useful for surveying and navigating.  It is interesting to note that the gps system, which from satellites determines the latitude and longitude, uses them as well. 

 

Extensive tables of sine and cosine exist. Some of the values are listed below in Fig. 5.

Angle in degrees

Sine

Cosine

0

1

15º

0.2588

0.9659

30º

0.5

0.8660

45º

0.7071

0.7071

60º

0.8660

0.5

75º

0.9659

0.2588

90º

1

0

figure 5. Table of sines and cosines to 4 decimal places

There is also a useful approximation if we are dealing with very small angles. If d is the value of an angle in degrees, and d is small, then the value of sin(d) is approximately dp/180. For instance, sin(0.180°) is 0.001p, or 0.00314, correct to 5 decimal places.

Now suppose we are given a right angled-triangle XYZ with Z a right angle, see Fig. 6. Then consider the right-angled triangle ABC with right angle at C and with angle A equal to angle X. Then triangle ABC is similar to triangle XYZ. But the sides of triangle ABC are already tabulated in the tables of sine and cosine. Since the hypotenuse of the triangle XYZ is h, which is h times the length of the hypotenuse of ABC, this tells us that the multiplication factor connecting the sides of XYZ to ABC is h. So to find the sides of triangle XYZ it is only necessary to multiply the sides of ABC by the length of the hypotenuse h of XYZ, that is the sides of XYZ are hsin(A) and hcos(A).

figure 6. A right-angled triangle XYZ with hypotenuse
h, and a similar triangle ABC with hypotenuse 1

Example 1: A right-angled triangle has angle 45 degrees. Calculate the length of the sides if the largest side, the hypotenuse, has length 14m.

Solution. The sides will be given by 14×sin(45°) and 14×cos(45°). From the above table the sine of 45 degrees is 0.7071, as is the cosine, so the length of either of the shorter sides is 14 ×0.7071 = 9.9.

Example 2: A right-angled triangle has angle 30º. Calculate the length of the sides if the largest side has length 29m.

Solution. The sine of 30º is 0.5. Hence, the length of the side opposite the 30 degree angle is 29×sin(30°) = 29 ×0.5 = 14.5 m. The length of the side adjacent to the 30 degree angle is 29 ×cos(30°) = 29 ×0.8660 = 25.114 m.

Example 3: Determine the lengths of all the sides in the right-angled triangle in Fig. 7.

Solution: The sides are AC = 10×sin(15°) = 2.588 and AB = 10×cos(15°) = 9.658, from the table in Fig. 5.

figure 7. A right-angled triangle with 15º angle at B and hypotenuse 10 m.

§4 Dropping a stone over a cliff

One way of estimating the height of a cliff, which is over a deserted stretch of water, is to drop a stone and with a stopwatch measure the time for the stone to reach the water. If this time is t, then the height is given by the formula

Height of cliff = ½gt2,

where g is the acceleration due to gravity, i.e. about 9.8 ms-2. (This formula was mentioned in Example 4 §8 of Chapter 6.) In our case we got a time of 1.5 s, and so the height of the cliff was

½ ×9.8 ×1.5 ×1.5 = 11.025 m.

Example: Using a ruler to calculate reaction time.

Use a centimetre ruler. One person holds the ruler at the top and the other holds his hand near the bottom, with forefinger and thumb almost clasping the ruler, but loosely, so that if the ruler is released at the top it will fall.

The first person suddenly says now, and releases the ruler. The second person must clasp his finger and thumb together so stopping the falling ruler. One then notes the distance the ruler has fallen before the second person reacts and grabs the ruler. The reaction time is then calculated from the formula

s = ½gt2

From this formula we see on multiplying by 2 and dividing by g that

2s/g = t2

We then calculate the time in seconds by measuring the distance s in cm. Then calculate 2s/g and take the square root. That is the reaction time. For g take 980 cm s-2.

In an actual test one of us stopped the ruler in 10 cm. Thus the reaction time was 0.14 s..

§5 Solved problems and other optional topics

1. Suppose you are measuring the height of a building with a shadow 20 meters long, and your shadow is 5 m. long. If you are 2 m tall, how high is the building?

Solution. We use the formula we found in §1 (with the slight modification that it is a building and not a tree that we are concerned with).

Height of building = (shadow of building/shadow of man) × (height of man)

If h is the height of the building, then h = 20/5 ×2 = 8 m.

2. Explain why sine of 0º is 0 and sine of 90º is 1.

Solution. For an angle of 0º it is not exactly clear how to define the sine. But if we take a right-angled triangle with hypotenuse 1 in which the angle is very small, we can see that the side, which defines the sine, becomes very small indeed, and the smaller the angle the smaller the side becomes. It seems reasonable to define the sine of as 0.

Similarly we see what happens when the angle approaches 90º, where it is clear that the side opposite the angle becomes larger and larger and seems to be getting closer and closer to the hypotenuse 1. It is thus natural to define the sine of 90º as being 1.

 

3. Determine the sine of 30º and 60º without the use of tables.

Solution. The triangle in Fig. 8 has all three sides of length 1.

                                                         B

                                                                           

 

 

                                  A                                              C

                                                           D

figure 8. Equilateral triangle with sides of length 1.

The angles are also equal, so each measures 60º. A perpendicular dropped from the vertex B to the base bisects the base, that is, cuts the base in half, so AD = DC. and AD = 1/2. Also, BAD is a right-angled triangle with hypotenuse BA of length 1.

Since the angle BAC is 60º, the angle BAD is 30º, so sin(30º) = AD = 1/2. Also, since ADB is a right triangle, by Pythagoras’s theorem, AD2 + BD2 = AB2. Hence, (1/2)2 +BD2 = 1, so BD, the side opposite the 60º angle at A, is the square root of

1 - (1/2) 2 = 3/4,

which means cos(30º) = 0.866.

 

4. A surveyor measures the angle of a top of a building at 15 degrees at a distance of 20 meters. His instrument is at a height of 1.5m. What is the height of the building?

Solution: Refer to Fig. 9.

We know that the triangle ABC is similar to a triangle XYZ with hypotenuse 1 and angle 15º, which has sides YZ = sin(15°)= 0.2588 and XZ = cos(15°) = 0.9659 as listed in Fig. 5. Thus, the sides of ABC are simply a multiple k of the sides of XYZ, so, in particular, AC = kXZ. Thus the factor k is calculated by taking

AC/XZ = 20/cos (15º) = 20/0.9659 = 20.706.

Thus, BC = 20.706× sin(15º)= 20.706 × 0.2588 = 5.359. To get the height of the building we must then add 1.5 meters, thus getting a final height of 6.859 meters.

figure 9. A surveyor calculates the height of a building.

5. I stood on the top of a cliff and dropped a stone. It took 3 seconds before I heard the splash when the stone hit the water below. What is the height of the cliff?

Solution: The distance is calculated from the formula

distance fallen = ½gt2.

In this case, t = 3 and g, as we know, is 9.8 m/s2. Thus, the distance is ½×9.8×9 = 44.1 m.

6. Remark.

For the next two problems we will accept the following fact from physics: A force F at an angle a can be regarded as the result of two forces F1 and F2 acting at right angles to each other, as shown in Figure 10. In particular, the two forces are

F1 = F×sin(a) and F2 = F×cos(a).

figure 10. The force F can be regarded as the sum of two forces at right-angles to one another

7. Explain why if the sides of a military battle tank are oblique rather than vertical, a shell is less able to penetrate it. For example, consider the shell hitting the tank at an angle of 90° and then, if the tank’s sides are slanted, at an angle of 30°.

Solution; We use he remark in 6 above. If the shell hits a vertical side of a tank standing at an angle of 90 degrees with a force F, then the force acting at that point is going to be F. However, if the angle at which it is directed is 30 degrees, then the force can be regarded as the result of the two forces, F×cos(30°) and F×sin (30°). The force of F×cos(30°) is parallel to the tanks armour, and so does not help to penetrate it. The other force of F×sin(30°) is ½×F and this is the effective penetrating force. This explains why angling the tank’s armour helps reduce the shell’s impact.

 

8. Sunbathing at an angle rather than directly. Suppose you are lying in front of the sun with the rays coming in at 90 or at 30 degrees. The effect is considerably reduced according to sine of 30 degrees.

Solution: If the intensity of the sun’s radiation is I, then a similar result to 6 above holds for it, that is the intensity can be regarded as two intensities, at right-angles, the one of

I×sin(30°) and I×cos(30°), The intensity parallel to the skin does not cause any sunburn, and the only part that gives sunburn is I×sin(30°) = ½×I.

The same type of argument helps to explain why, when the sun’s rays come in at lower angles, winter is colder than summer.

9.     Explain how the artist’s method works,

 

 

 

 

 

 

 

 

 


figure 10. The artist’s method explained

In Fig. 10 the bottom of the vertical object we are drawing is at the point D and the pencil is held vertically at the point A. The artist’s eye is at O.

The point C represents a point on the object which is at a height x above the bottom point. Seen by the artist this point is at a height of x¢ on the pencil which is fixed at the point A. Since he triangles OAB and ODC are similar, we have that

x¢/OA = x/OC, or that x¨= (OA/OD) x.

DDddddddddddd

 

B

 

A

 

 

 

D

 

x

 

O

 
Thus is proportional to x and with this method we get a direct scaling of the object to be drawn.

Chapter 8

LOGARITHMS AND NATURAL LOGARITHMS

“The invention of logarithms saved astronomers a lot of trouble and doubled their lives” - Laplace

 

§1 Multiplying by adding. Logarithms

figure 11. John Napier (1550-1617)

Multiplication is more time consuming than addition. To multiply two 8-digit numbers, we need to perform 64 multiplications and several additions. On the other hand, to add two 8-digit numbers we need only add 8 times. The difference is large and gets much larger with every increase in the number of digits, so after the invention of the telescope ushered in a revolution in astronomy, which, in turn, necessitated large and precise calculations, it became a matter of time before a method of simplifying calculations would be found. The answer was: logarithms.

John Napier in Edinburgh published the first table of logarithms in 1614. It is said that Henry Briggs, professor of geometry in Gresham College, London, was so impressed by Napier’s system of logarithms that he was speechless for fifteen minutes when they first met, gazing at Napier with admiration. Then he explained that he had travelled especially to see Napier and enquired, “… by what engine of wit or ingenuity you came first to think of this most excellent help in astronomy... “.

Ten years later, in partial collaboration with Napier, Briggs published a new table of logarithms, which he called “common logarithms”, based on an improved system still in use today.

In Fig. 2 we have a short table of common logarithms. In each column the number on the right is the logarithm of the number on the left. Briggs produced a table of logarithms to 17 places, which meant it could be used to find products of up to 17-digit numbers, correct to 16 decimal places. This was a major effort and took almost the whole ten years. Since then, much larger tables to over 200 places have been constructed.

 

1.0 .000

2.0 .301

3.0 .477

4.0 . 602

5.0 .699

6.0 .778

7.0 .845

8.0. .903

9.0 .954

1.1 .041

2.1 .322

3.1 .491

4.1 .613

5.1 .708

6.1 .785

7.1 .851

8.1 .908

9.1 .959

1.2 .079

2.2 .342

3.2 .505

4.2 .623

5.2 .716

6.2 .792

7.2 .857

8.2 .914

9.2 .964

1.3 .114

2.3 .362

3.3 .519

4.3 .633

5.3 .724

6.3 .799

7.3 .863

8.3 .919

9.3 .968

1.4 .146

2.4 .380

3.4 .531

4.4 .643

5.4 .732

6.4 .806

7.4 .869

8.4 .925

9.4 .973

1.5 .176

2.5 .398

3.5 .544

4.5 .653

5.5 .740

6.5 .813

7.5 .875

8.5 .929

9.5 .978

1.6 .204

2.6 .415

3.6 .556

4.6 .663

5.6 .748

6.6 .820

7.6 .881

8.6 .934

9.6 .982

1.7 .230

2.7 .431

3.7 .568

4.7 .672

5.7 .756

6.7 .826

7.7 .886

8.7 .840

9.7 .987

1.8 .255

2.8 .497

3.8 .580

4.8 .681

5.8 .763

6.8 .833

7.8 .892

8.8 .944

9.8 .901

1.9 .279

2.9 .462

3.9 .591

 4.9 .690

5.9 .771

6.9 .839

7.9 .898

8.9 .949

9.9 .996

 

 

 

 

 

 

 

 

10.0 1.000

figure 2. Table of common logarithms

This is the way logarithms are used to multiply. If we want to find the product of two numbers, x and y, we look up their logarithms in a table of logarithms. The sum of these logarithms will be the logarithm of their product. The answer is the number with this logarithm, which then can be found from the table of logarithms, such as the one in the above table. For example, from the table,

logarithm of 2 = 0.301 and logarithm of 3 = 0.477.

The sum of these two logarithms, 0.778, will be the logarithm of the product of 2 and 3, which we find from the table to be the logarithm of 6. This is a trivial example, but, in general, we can use tables of logarithms to calculate more difficult products easily.

If we denote the common logarithms of x and y by Log(x) and Log(y), then what we said in the previous paragraph translates to the equation:

Log(x) + Log(y) = Log(xy).

Logarithms look like magic, but this is actually nothing but the law of exponents as explained in Chapter 2 §4. For the theory see the explanation in §5 of this chapter.

The initial impetus for logarithms as we said was the need to simplify multiplication. However, we can now do this extremely quickly with the aid of computers, so what is the point of studying logarithms? The answer is that logarithms are still extremely useful in theory. For example, as we shall see later (in Chapter 15) the idea of logarithms gives a formula for approximating the number of prime numbers less than a given number n. (A prime number is a number like 5 or 7 or 17 which is not itself the product of smaller numbers. An example of a non-prime number is 6, the product of 2 and 3.) It is difficult to see what the logarithm could possibly have to do with prime numbers, and the result is indeed a surprise.

Other uses of logarithms are in chemistry to define acidity of a substance and in acoustics, to define decibels, the unit of sound intensity. We will not discuss these but point out that they are important applications of the logarithm.

§2 Doubling your money

As an application of logarithms, suppose you invest one pound at x% interest per annum. A good approximation to the number of years required to double your money at this rate of interest is 69/x. For example, if the rate of interest is 7%, it will take just under 10 years to double your money. This rough approximation is more accurate with small values of x, but is still useful. We will explain how this works in §6, problem 1.

Example.

If you receive interest at 3%, how long will it be before you double your money?

Solution: Divide 69 by 3 to get 23 years.

As this is a long time, maybe in this case one would want to know how long it takes for your money to increase by 40%. The rough rule in this case is to divide 34 by the interest rate. Thus with a 3% interest rate, to increase your money by 40% will take approximately 34 divided by the interest rate, i.e. 34 divided by 3 or between 11 and 12 years. See §6 problem 2 for an explanation of why this works.

§3 Euler’s e

Common logarithms are not the only ones in use. There are other logarithms, called natural logarithms, which have many uses, mainly scientific. In Chapter 15 we will use this logarithm to explain a formula giving an estimate for the number of prime numbers less than a particular number. To understand the difference between the natural logarithms and common logarithms, we change direction. We shall define a new constant e, as famous in mathematics as the constant p.

To see how e comes about imagine that you invest a unit of money, say, a pound or a dollar, at a rate of x per cent per annum. That means that your return at the end of the year will be

(1 + x/100).

If we replace x/100 by y, the return can be expressed as (1 + y). Next, suppose that the rate is x percent per year, but paid every six months. That means that every pound or dollar invested will return (1 + y/2) pounds or dollars after the first six months, and this money will itself get an interest of x/2 percent for the next six months. Hence, the total you will get for one year is will be

(1+y/2)×(1+y/2) = (1+y/2)2..

Similarly, if the rate is x% payable every 3 months, that is, 4 times a year, the return after 1 year will be

(1 + y/4) 4.

In general, if interest is paid at the end of n equal periods per year, the total return for one year will be

(1 + y/n) n.

The table below shows the values of (1 + y/n) n for y = 1 and various values of n.

 

Value of n

Value of (1 + 1/n) n

10

2.593742

100

2.704814

1000

2.716924

10000

2.718146

1,000,000

2.71828


figure 3. Approximating (1+1/n)n

The question is, what happens as n increases further? Here we come face to face with fundamental and subtle concepts, which are the subjects in more advanced courses in mathematics. The best we can do is to calculate (1 + 1/n) n, with n as large as we can manage. We then hope that this is close to the correct result and that nothing unexpected will happen with larger n. Sometimes this method works reasonably well, sometimes it goes disastrously wrong. To be sure of what we are talking about requires a course after the first course in calculus. Neither Newton nor Leibniz, the discoverers of the calculus, had a complete understanding of these concepts, and indeed it has taken some 300 years after them to develop the proper ideas.

In this case, however, large n produces no surprises, and we will get a number approximately 2.718. This is the number the Swiss mathematician, Leonard Euler (1707-1783), called e. It has retained this name to this day.

§4 The natural logarithm is roughly 2.3 times the ordinary logarithm

The logarithms discovered by Briggs are called common logarithms. They are also called logarithms to the base 10, they are defined in terms of exponents of 10 and satisfy

Log(10) = 1.

Natural logarithms, usually denoted by ln(x), are defined in terms of exponents of e and have the same property as common logarithms, in that multiplication can be replaced by addition. They are also called logarithms to the base e because they are based on exponents of e rather than of 10. In particular, ln(e) = 1, and, as we already have said, the natural logarithm retains the important property that

ln(xy) = ln(x) + ln(y).

For ordinary calculations logarithms base 10 are more convenient, but for theoretical questions logarithms base e have important advantages. One disadvantage is that a table is much more difficult to construct than for common logarithms. Fortunately, there is a conversion factor: To obtain ln(x), multiply Log(x) by ln(10), which is approximately 2.3026.

An interesting approximation for ln(1 + x) when x is small is x. For instance, ln(1 + .001) = .000999950033, which is very close to .001 The smaller x is the better the approximation.

If x is a very large number with n digits, a good approximation to Log(x) is (n-1). For instance, Log(5,613,678) = 6.749, to 3 decimal places. The error in approximating this as one less than the number of digits, i.e. by 6, is less than 1, so this is accurate to 20%. If the number has 100 digits, the error is at most 1, and so the approximation is correct to 1%. In general, the larger the number, the better the accuracy of this rough approximation.

 

§5 Theory of logarithms

Theory of logarithms. To explain Briggs’ common logarithms we need the concept of exponents, introduced in Chapter 2. For any number, a, and each whole number n, an is defined as the product of n copies of a and n is called the exponent or power of a. For example, 23 = 2×2×2 = 8. We define a1 = a, and a 0 = 1.

There are two laws of exponents we need:

a m×a n = a m + n, and (a m)n = a mn.

For example, 102×103 = 105, that is, the product of two 10s times the product of three 10s is the product of five 10s. Also, (102)3 = 106.

In Briggs’ system the logarithm of a number x, written as Log(x), is defined as the exponent of 10 which gives x, that is, Log(x) = y if

10y = x.

For example, Log(100) = 2, since 102 = 10×10 =100, Log(10) =1, since 101=10.

To define logarithms we will need to define 10y for exponents y other than whole numbers.

Fractions as exponents. Let n be a positive integer and let x = 1/n. Then ax is defined to be the nth root of a. Thus 4½ is that number whose square is 4, and, hence, 4½ = 2. If x = m/n, i.e., one positive integer divided by another, we define ax to be the m-th power of a(1/n). For instance,

23/2 = (2½)3 = (1.4142)3 = 2.828.

So far we have defined ax for all exponents which can be written in the form m/n, where m and n are whole numbers. Such numbers are called rational numbers. But there are many numbers which are not rational numbers (for example, the square root of 2, as explained in Chapter 14, §5. Nevertheless it is possible to define ax for any positive number x, but a precise definition uses the concept of limit, which we do not discuss here. Instead we will accept that this can be done, and we will also assume that the closer a rational number y is to x, the better ar is an approximation to ax. For example, the square root of 2 is approximately 1.414, which, is 1414/1000, so a√2 is approximately the 1414-th power of the 1000-th root of a.

Negative exponents. If the above law of exponents is to hold for both positive and negative numbers, then, ax × a-x should be ax-x = a0 = 1. Therefore, we define a-x = 1/ax. For example,

2-3 = 1/23 = 1/8.

Definition of logarithms to the base a. If ax = y, then x is defined to be the logarithm of y to the base a, and is written as loga(y). In particular, common logarithms can be defined as logarithms to the base 10.

Definition of the natural logarithm (also called logarithm to the base e). If ex = y then x is the natural logarithm of y, written as x = ln(y).

Laws of logarithms

  1. Log(x) + Log(y) = Log(xy)

To see this let Log(x) = x´and Log(y) = y´, Then, xy = 1010= 10x´+y´ so, by definition,

x´+ y´= Log(xy), i.e., Log(x) + Log(y) = Log(xy),

  1. xLog(y) = Log(yx).

This follows from raising both sides of the defining equation, y = 10Log(y), to the power x. Thus, yx = (10Log(y))x = 10xLog(y), i.e., xLog(y) is the exponent of 10 which gives yx, so, by definition, Log(yx) = xLog(y).

§6 Optional worked examples

1. Why does the rough rule for doubling your money work?

Solution: If the interest rate is x%, the return on M invested for n years is M(1 + x/100)n. If this is to double, (1 + x/100)n = 2. Taking the natural logarithm on both sides of the equation, the left hand side equals ln(1 + x/100)n = n ln(1 + x/100), and, since x/100 is small,

ln(1 + x/100) is approximately, x/100 – see §4.

On the other hand, the right-hand side must be ln(2) = .6931471.

Hence, we have the equation nx/100 = .69 to solve. Multiplying the right side of the equation by 100 and dividing by x we see that n is approximately 69 divided by x.

 

2. Why does the rough rule for the money increasing by 40% work?

Solution. An increase of 40%, means a multiplication factor of 1.4 instead of 2 in the above example, and since ln(1.4) is approximately .34, for 40% the above solution becomes nx/100 = .34, so n = 34/x.

3. Why is ln(x), the natural logarithm of x, equal to ln(10) times Log(x)?

Solution: By definition, x = eln(x). Also, x = 10Log(x), and since 10 = eln(10), x = (eln(10))Log(x). Hence, x = eln(10)Log(x), so ln(x) = ln(10)Log(x).

4. Given that Log(2) is approximately 0.3, find Log(5).

Solution. Since 2×5 = 10, Log(2×5) = Log(10) = 1. Hence, Log(2) + Log(5) = 1, so Log(5) is approximately 0.7.

5. Use 210 to find an approximation to Log(2).

Solution. Since 210 = 1024, 210 is approximately 103. Taking the 10th root of both sides gives, 100.3 = 2, so Log(2) = 0.3, approximately. Actually, to 4 decimal places, Log(2) = 0.3010.^

§7 Logarittihms and Planeetary Motion

Kepler’s Laws

Astronomers by the 1600s had observed the planets, established their distance to the sun, and their period, i.e. the time it took for the planet to return to its orginal position.  These figures are summarized in the table bleow.  The period is measured in days, and the distance is measured with the distance of the earth to the sun as unit.   Thus from the table, the distance of Jupiter from the sun is a little more than 5 times the distance of the earth to the sun.

 

Plantet

Mean distance D to sun

Period  T of rotation in days

Mercury

0.387

0.241

Venus

0.723

0.615

Earth

1.000

1.000

Mars

1.524

1.881

Jupiter

 5.203

11.862

Saturn

 9.555

29.458

 

The relationship between T and D is not easy to see, but by  condiering  ln(T) and ln(D, it  seems as if  ln(T) is one and a half times ln(D). 

We have listed ln(D  and ln(T) in the table below.   We have calculated a fourth column by multiplying ln(D) by 3/2.  This last column agrees well with the third column ln(T), with an error of at most 1 in a 1000..

 

Plantet

 ln(D)

 ln(T)

3/2´ln(D)

Mercury

0.949

1.423

1.424

Venus

0.324

0.486

0.486

Earth

0.000

0.000

0.000

Mars

0.421

0.632

0.632

Jupiter

1.649

2.473

2.474

Saturn

2.257

3.383

3.386

 

 

 

 

Thus we have Kepler’s law

 

3/2´ln(D)  = ln(T)

 

This is the third of Kepler’s famous laws.

 We can simplify the left-hand side to ln(D3/2) and so if we calculate e  to this power we get

 

D3/2 = T..

 

This is indeed a remarkable formula.

 

 

§8 Earth quakes measured on the Richter scale

 

Earthquakes are usually measured on the Richter scale.

The details are involved but we will give a simplified version.  The severity of an earthquake 100 kn away can be calculated as follows.

To set up the scale, Richter decided on a certain reading S on the seismograph whould be taken as a standard.  Then if a seismograph 100 km away recorded I on the seismograph, the Richter reading was defined to be

 

Log(I/S).

 

That is, we divide the intensity I by the standard S and then take the logarithm,  So if an earthquake gave a reading 10 times as large as the standard S, its Richter number would 1, if it was 100 times larger, its reading would be 2 and if it was a 1000 times larger, its reading would be 3. 

 

The seismograph of course has to be specified and there has to be a way for dealing with earthquakes which are at other distances than 100 km.  The purpose of this example is to illustrate the use of the logarithm for this scale.  There is also a logarithmic scale used for sound and also one for the apparent magnitude of stars.

 

A Richter value of 4 corresponds to light eathrquakes, usually without significant damage, but easily noticed shaking a nd ratling.

 

 

 

 

 

A Richter value of 8 corresponds to a severe earthquake causing serious damage over an area of several hundreds o kilometres.,

 

 A Richter ,value of 9 would be devastating over an area of many thousand kilometeres. A reading of 10 has fortunately not yet been recorded. ,

 

 

 

 

§9 Some Historical remarks

Around 1600, Lippershay invented the telescope in Holland. At first this invention was classified as a military secret, but when Galileo got wind of it he fashioned one on his own. His observations led to great strides in astronomy, which, in turn, led to efforts by astronomers to simplify the many precise numerical calculations involved. The main problems came from spherical trigonometry with the need to multiply numbers with a large number of digits precisely.

In 1524, Stifel described the basic principles of logarithms, but did not carry his ideas through to constructing a table. Almost a hundred years later the Scot, John Napier, after working twenty years, published the first table. It was he who called his exponents logarithms. His table was an immediate success and made an impact similar to that made by the introduction of computers in our time.

Napier’s original system was not based on powers of 10. This had the big disadvantage that his table of logarithms was very long. It had to include logarithms of all numbers, not just from 1 to 10. Briggs realized the convenience of using powers of 10, and in 1624 published his table, which greatly simplified the use of logarithms.

The advantage of using base 10 is in the construction of tables. A table of common logarithms of numbers from 1 to 10 is readily extendable to any number. The rule is to express the number as a number, say x, between 1 and 10, times 10 to some exponent. The common logarithm is then the exponent of 10 plus Log(x). For example, 5,613,678 = 5.613678 × 106, so

Log(5,613,678) = 6 + Log(5.613678).

Chapter 9

COORDINATE GEOMETRY

Algebra was one thing, and Geometry another. Descartes made them one.

figure 1. René Descartes (1596-1650)

The idea of describing the position of a point in a plane by giving its distances from each of two lines that are perpendicular to one another is deceptively simple. But it leads to two important consequences: The ability to visualise how quantities depend on one another, and also a link between algebra and geometry. A very difficult problem in algebra may in its geometrical translation prove to be solvable, and, vice-versa, a difficult geometrical problem may turn out to be easier to handle in its algebraic translation. We owe this ingenious idea to the French mathematician and philosopher, René Descartes (1596-1650). His methods have been expanded and improved by many other mathematicians, to make the impressive subject we have today.

§1 Coordinates

figure 2. Coordinates

In Fig. 2 we have drawn two straight lines, OX and OY, which intersect in a point O, and are at right-angles. The position of any point can be described by a pair of numbers, the first gives the horizontal distance of the point P to the vertical line OY, while the second gives the vertical distance of P from the horizontal line OX. Distances above OX are taken to be positive, while distances below OX are taken to be negative. Distances to the right of OY are taken to be positive, while distances to the left of OY are taken to be negative. The position of a point is indicated by bracketing these two numbers together. And this bracketed pair of numbers are called the coordinates of the point.. For instance, in Fig. 3, the coordinates of the point P are (2,1), those of the point Q are (1,2), those of R are (-2,-1.5), and the coordinates of the point S are (-3, 0.75). We call the line OX the x-axis and OY the y-axis, while the point O is called the origin. Any collection of points on the plane is called a graph.

 S(-3,0.75)

 

 R(-2,-1.5)

 

 P(2,1)

 

 Q(1,2)

 

figure 3. Examples of coordinates

 

 

§2 Graphs as a concise source of information

The first important application is the listing of information. Figure 4 illustrates a graph of temperature scale conversions. For example, to find the equivalent on the Fahrenheit scale of 20 degrees Celsius we carry out the following:

figure 4. Graph of degrees Celsius vs. degrees Fahrenheit

We locate 20 on the OC axis and draw a vertical line till it reaches the curve, which in this case is a straight line. The corresponding F value is 68. So 68ºF corresponds to 20º C. By using the same procedure in reverse, we can convert from Fahrenheit to Centigrade.

Another example is illustrated in Fig. 5. This enables us to convert from pounds, £, to euros, €. Using this graph the equivalent of 5 euros in pounds is at the intersection of the vertical line drawn from 5 on the €-axis to meet the curve. The corresponding value on the £-axis is 3.50, which is the corresponding value in pounds.

figure 5. Graph of pound versus euro

A graph like this is very useful when travelling abroad. It is easy to make one that is accurate at the time of travel. Simply check the number of euros corresponding to £10, €14, for example, and plot the point (10,14). Draw the straight line joining this point and the point O. This then gives you the required conversions. Of course, this scheme can be used for converting from other currencies, using the relevant correct rates.

As a third example of the conveying of information, consider Fig. 6, the 2005 U.K. postal rates for a 1st class letter, in which we can relate the weight of a letter to the postage charge to send it.

figure 6. Cost versus weight of the package

§3 Plotting a graph

Example 1 The number of cells in time t.

A biologist checks under a microscope the number of cells that he is growing in a culture. At a time t = 0 there is only one cell. The table below summarises his findings.

 

 

 

 

Time t

0

1

2

3

Number of cells

1

2

4

8

We plot the points with coordinates (t, number of cells at time t). For instance, at time t = 2 we have 4 cells so we plot the point with coordinates (2,4).

After plotting all these points, we then draw a smooth curve joining them as shown in Fig. 7.

figure 7. Graph of number of cells at a time t

Example 2

This is an example where a quantity y depends on another quantity x and we have the following table of values.

x

0

1

2

3

4

y

1

3

4

4.5

3

We again plot the points (x, y) and then draw a smooth curve joining these points as in Fig. 8.

figure 8. Graph of Example 2

One important advantage of this visual representation is that although the data was given for 5 points only, this rough graph allows us to approximate a value of y for intermediate values of x. For instance, a reasonable guess for y for x = 3.5 would seem to be 4.6.

Example 3: Stretched rubber band.

An elastic band has various weights attached to it and its length is measured. The graph is then sketched as in Fig. 9. We get a straight line. But we know if the weight is too great the elastic band will snap. So it is not possible to assume that the graph will continue as it seems to do from the beginning.

figure 9. Graph of a stretched rubber band’s length as the load is increased.

Example 4: Carbon dating.

The graph in Fig.10 gives a way of finding out how old objects are by carbon-14 content.

All plants, animals and people absorb carbon-12, normal carbon, and to a much less extent, carbon-14, which is radioactive. While alive, both carbons exist in the same ratios in plants, animals and people. After death the level of normal carbon remains constant, but carbon-14 decays. In fact, carbon-14 has a half life of 5,700 years, that is, one-half will decay in 5,700 years, half of the rest will decay in another 5,700 years, etc. The amount of normal carbon in the specimen determines how much carbon-14 there was originally, so the remaining carbon-14 determines its age. The following graph illustrates the relationship between the percentage of the carbon-14 of the original remaining to the age. For instance, if just 5 percent of the original carbon-14 remains the specimen is 24,640 years old,

figure 10. Carbon-14 dating.

§4 Equations and Geometry

Descartes then had this brilliant idea: He set up a way of associating equations with curves.

Consider the equation y +  x2 = 0. This is satisfied by a variety of values of x and y. Descartes idea was to consider all such values (x, y) that satisfied this equation and plot the corresponding points. For instance, (2,- 4) and (3,- 9). The collection of all such points form a curve as shown in Fig. 11. This also called the graph of the equation

 

figure 11. Graph of y + x2 = 0.

In this way Descartes set up a relationship between equations and curves in the plane. Depending on the problem, it allows us to use geometrical techniques to assist in solving algebraic problems and algebraic techniques for solving geometrical problems. This study initiated by Descartes is known as analytic geometry or as coordinate geometry.

But first, let us consider some typical equations and the corresponding graphs.

Example

2x + 3y = 2. We consider the set of all points with coordinates (x, y), which satisfy this equation. To do this we draw up a table of values. We chose various values of x and then find the corresponding value of y to satisfy the equation. For instance, for x = 0, the equation
2x + 3y = 2 simplifies to 3y = 2 and hence y = 2/3. We do the same for various values of x and in this way we obtain the table below:

x

y

0

2/3

1

0

2

-2/3

3

-4/3

-1

4/3

-2

2

-3

8/3

Next, we plot these points and try to join these points with a smooth curve. The result is illustrated in Fig.12.

figure 12. The graph corresponding to 2x + 3y = 2.

In fact, it turns out that the graph of any equation of the form ax + by + c = 0, with a, b, c real numbers, is always a straight line (excluding the trivial case a = b = 0). Thus, the graph could have been more quickly drawn by joining any two points (x,y) which satisfy the equation.

The next example is the graph of x2 + y2 = 1. The corresponding curve is a circle, radius 1, centre at (0,0), as shown in Fig. 13. As in the above example, this can be verified directly by constructing a table of values and plotting the points.

figure 13. The graph of the equation x2 + y2 = 1.

 

§6 The Greeks and their curves

Two thousand years ago the Greeks studied the straight line, the circle and a group of curves, which are related to the circle, namely the conic sections. They studied these curves for their own sake and their work, certainly for about 2000 years, could only be described as entirely intellectual, without any application - useless knowledge. All this was to change. These curves became of fundamental importance in understanding our universe, and in Satellite TV.

What is a conic section? A conic section is the intersection of a double cone and a plane, see Fig. 14. When we talk about the cone we mean a hollow cone, something like a double dunce’s cap made out of paper. So the intersection with a plane produces a curve,


figure 14. The intersection of a plane and a cone
gives a curved line, which is called a conic section.

These curves were given the names parabola, ellipse, and hyperbola, depending on how they intersect the cone, but we shall not elaborate. We simply have drawn examples to illustrate the different types of curves in Fig. 14. In Fig.15 and Fig. 16 we have drawn the curves without showing how they arise from the intersection of a plane and the cone.

If you have a torch that has a well-formed cone of light coming from it, you can shine it in a darkened room on the ceiling, and by positioning it in different ways, you will get the circle, the ellipse and part of the hyperbola.

One type of these conic sections is easily drawn as follows. To draw a circle we can take a piece of string and tie its ends together, thus forming a loop. Put a nail in a board and holding a pencil in the loop at maximum extension, you can then draw a circle. If you take two nails, positioned at a and b, separated of course, and put the loop round the nails, again you can draw a closed figure, which is called an ellipse. Thus in this case there are two key points, the nails, and each of these is called a focus of the ellipse. (The plural of focus is foci).

figure 15. Ellipse with foci at a and b.

A property of the ellipse is that if you shine a light from one focus to the circumference on the ellipse (for instance, if the circumference of the ellipse is in the form of a mirror), the light will be reflected in the other focus, as illustrated in Fig, 15. Here the straight line we have drawn from a is shown reflected in the circumference and then it passes through the other focus b:

An example of a parabola is the curve that a cannon ball traces out when shot out. See Fig. 16.

figure 16. A parabola, the path of a cannon ball.

§7 Equations and conic sections

As mentioned above, there is an association between equations and curves. Here is a short table with some examples.

 

Equation

Curve

2x + 3y = 4

Straight line

x2 + y2 = 1

Circle with centre at O and radius 1

x2/4 + y2/9 = 1

Ellipse

xy = 1

Hyperbola

The table gives examples of what are general results. We state these results without proof.

The first is that any equation which involves only x and y and numbers, such as 2x + 3y = 4 (such equations are called linear) is a straight line.

Even more remarkable is the following theorem:

It concerns equations which involve only x, x2, y, y2 and xy and numbers, the so called quadratic equations, for instance, 7x + 4x2 +8y - 9y2 + 17 = 0.  These differ from linear equations in that they must involve at lest one term which is a square or else the product of x and y. 

Then the theorem states that the graph of such an equation will correspond to a conic section.

This is a very striking result. That an equation with x and y appearing only to the powers of 1 and 2 should have anything to do with taking a section of a cone is remarkable.

§8 Applications of coordinate geometry

The first major application of the conic sections occurred between 1609 and 1616 when the astronomer Kepler discovered three important rules of motion. The first was that the planets moved in ellipses with the sun at a focus. Moreover, the paths of comets like Halley’s comet also have the form of an ellipse.

Using his three rules of motion and his law of gravity, Newton was able in 1687 to show that Kepler's laws were a consequence. He used the connection between equations and curves that we have described above. The same method is used to determine the orbits of satellites. The reason why we can watch TV in so many different places is a consequence of these calculations.

 

The reflecting property described in §6 is important for the antennas that are used to send and receive the TV signals. Newton had the idea of using the reflecting property of the parabola to design a telescope. The parabola has a focus, similar to the ellipse; but only one, not two. And the reflecting property means that parallel light will be reflected into the focus. The same property holds for radio waves. This is the basis of the reflecting antenna, the so-called parabolic antenna. The TV programs that are sent all round the earth use the same principle. A more down to earth application is in the reflectors for headlights in cars, in torches and in spotlights.

 Another important application of this reflecting principle is medical. Lithotripsy eliminates the need for surgery to remove kidney stones. To pulverize the stones the lithotripter uses shock waves, which pass harmlessly through soft tissue. The patient is placed in an elliptical tank of water with the kidney stone at one focus. The shock waves are generated at the other focus. The procedure lasts about an hour during which time about 8,000 shock waves are administrated.

The property of reflection from one focus to the other also explains the workings of so-called whispering galleries, such as in St. Paul’s Cathedral and in the rotunda of the Capitol in Washington, where, if a person stands at one focus his voice is “reflected” to the other focus.

Finally to emphasise the importance of conic sections, even the paths of electrons rotating around the nucleus of an atom are ellipses.

 

§9 Solved Problems

1. Draw the graph of y = x + 3.

Solution. The graph of any equation of the form y = ax + b is a straight line. Hence, the graph is determined by any two points on the graph. For example, the points (0,3) and (-3,0) are on the graph, so it we plot these points the graph is the straight line joining them

2. Draw the graph of y = 5x.

Solution. As in the previous example this is the graph of a straight line. Two points which satisfy y = 5x, are (0,0) and (1,5), so the graph is the line joining these points.

3. Draw the graph of y = 5x + 3.

Solution. A straight line joining, for example, the points (0,3) and (1,8).

§10 Solving problems in Geometry with algebra and vice - versa

To find a common solution to two equations, we draw the curves corresponding to each of the equations and observe intersections. In Figure 14 we have done this for the two equations,

y + x = 4 and y - x = 2.

From Fig. 17 we see that these two lines intersect in the point (1,3). Hence, x = 1, y = 3 is a common solution to the two equations.


figure 17. Solving two equations with a graph.

Algebraically, we can solve the two equations as follows: If we add the two equations, we have (y + x) + (y – x )= 2 + 4. i.e., 2y = 6 or y = 3. We then substitute y = 3 in the first equation to get 3 + x = 4, so x = 1, which agrees with our first solution.

A more difficult problem algebraically is to find a common solution of the two equations

x2 + y2 = 4 and y = x3 – x.

In Fig. 18, the graphs of the circle x2 + y2 = 4 and the curve y = x3-x have been plotted.

figure 18. Common solutions to x2 + y2 = 4 and y = x3–x.

We can see from the graph that there are exactly two intersections, which are, approximately, (1.5,1.5) and (-1.5, -1.5).

The solutions suggest a bit more. Since for both points the x-coordinate and the y-coordinate are equal, we might investigate what happens if we set x = y in both equations.

If we set y = x in the first equation, the result is x2 + x2 = 4, that is, 2x2 = 4, and dividing both sides by 2 gives x2 = 2. Substituting y = x in the second equation gives x = x3x, or, 2x = x3, and dividing both sides by x gives 2 = x2, which agrees with what we found for the first equation. Thus, it is true that at the common solution, x = y, and x2 = 2. Therefore, more precisely, the common solutions are (+√2,+ √2) and (-√2, -√2 ). Note that √2 is approximately 1.414, not too far off our estimate from the graph.

Here we can see the use of geometry and algebra, each contributing to the solution.

Chapter 10

SOLVING EQUATIONS AND GAUSS’S METHOD

 Back substitution is the essence of this ingenious method.

§1 Solving an equation

Consider the following problem: We are designing a budget hotel. An area, 25 meters long, is to contain 11 rooms; and each room is to be the same size, say, x-by-3.5 meters. The wall in each room is to be ¼ meter wide. What value of x will give the best possible size of each room? See Fig. 1.

figure 1. Designing a hotel.

The number of walls is 10, and these add 10 ×¼ = 2.5 meters to the total. In addition there are eleven rooms each of width x, making a total of 11x. Thus the equation we need to solve for x is 11x + 2.5 = 25.

If you remember your school mathematics you will be able to solve this problem straight away. Otherwise, try guessing! Our mathematics teacher would have been horrified at this suggestion. “Don’t guess, boy,” he would say. We now know that he was wrong. Why not guess?

§2 Method 1: Take a guess

An easy first guess is: x = 1. When we substitute x = 1 in 11x + 2.5 = 25, the left hand side (abbreviated LHS) becomes 13.5, while the right hand side (RHS) is 25. So the first guess was a bit off, but it was not too bad. The next guess, so as to make the LHS bigger, is x = 2, which makes LHS = 24.5.  This is still not quite correct since RHS = 25, but very good for a guess. 


The next guess, x = 2.5 makes the LHS = 30, which overshoots the mark of 25 by 5, and indicates that the actual answer is between 2 and 2.5, but very close to 2, since x =2 undershot the mark of 25 by only 0.5, while x = 2.5 overshot the mark by 5. We could continue in this way, trying say x = 2.1.

§3 Method 2: Draw a graph

We re-write the equation 11x + 2.5 = 25 by subtracting 25 from both sides of the equation, thus getting 11x –22.5 = 0. The problem can be formulated as finding the value of x to make y = 0 in the equation y = 11x - 22.5.

figure 2. Solving the equation 11x + 2.5 = 25 by drawing the line y = 11x – 22.5

Fig. 2 is the graph of y = 11x - 22.5. The graphical solution to our problem can be obtained as follows:

·       First, we know from Chapter 9 that this graph is a straight line. The graph can be drawn by finding two distinct points on the line and joining them. For the first point we choose x = 1.5. Substituting x = 1.5 in the equation y = 11x – 22.5 we find y = -6. Thus the point (1.5,- 6) lies on this straight line. For the second point we chose x = 2. Substituting x = 2 in the equation y = 11x – 22.5, shows that (2,-0.5) is also a point on the line. We can then draw the graph by joining these two points with a straight line. (There is nothing special about the choices x = 1.5 and x = 2, any other choices would have done just as well.)

·       At the point where the straight line crosses the x-axis the value of y is 0, which means that the corresponding value of x is the value which solves the equation 0 = 11x – 22.5. From the graph we observe that y = 0 when x is approximately 2. This is a good approximation since when x = 2, y = 22-22.5, that is, y = -0.5.

§4 Method 3: Do what your mathematics teacher told you

The method your mathematics teacher might have taught you for solving the above problem is based on the principle that equal mathematical operations done to both sides of an equality produce a new equality. Using this principle the solution to the problem proceeds in the following steps.

·       Subtract 2.5 from both sides of 11x + 2.5 = 25 to get 11x = 25 - 2.5 = 22.5.

·       Finally, divide both sides by 11 to get x = 22.5/11, which is approximately 2.05. As a check we see that 11× 2.05 + 2.5 = 25.05, with an error of 0.05 in 25 or 0.22%.

§5 Two unknowns

Mathematicians are ambitious. Having solved a problem they immediately seek to generalise the problem. The problem we solved in the previous section involved one unknown, x. Why not solve problems with two unknowns? With two unknowns and one equation there can be an infinite number of solutions. However, this situation changes if we have to find a numbers xand y to satisfy two equations.

A typical example is to find values for x and y satisfying both equations

x + y = 2          (1)

2x+3y = 5        (2)

The problem here is to find values for x and y which will satisfy both equations.

figure 3. C.F. Gauss (1777 – 1855), one of the greatest mathematicians.

The great German mathematician, C.F.Gauss, devised a systematic method for solving any number of equations in any number of unknowns. In this chapter we will apply it to two equations in two unknowns and to three equations in three unknowns in Section 7 Problem 5, but the method is easily adaptable to any number of equations and unknowns.

For two equations in two unknowns, x and y, Gauss’s method involves changing to an equivalent but simpler pair of equations having the same solution, the first equation of which still involves both unknowns but the second of which no longer involves one of the two unknowns.

The two operations we can use to do this are: (1) multiply any equation by a non-zero constant and (2) add or subtract one equation from another. Since both operations are reversible we can be sure that the method has neither eliminated a possible solution nor added a new solution to the system of equations. For our example, if equation (1) is multiplied by 2 this results in

2x + 2y =4

and subtracting this from equation (2), gives

2x2x + 3y – 2y = y = 5 – 4,

from which we see that y = 1 is the only possible value for y in the solution of the system. The value of x can now be found by substituting y = 1 for y in equation (1), so x + 1 = 2, or x = 1. Hence, the common solution is (1,1).

As with any method, it is advisable to check that x = 1, y = 1, solves both equations, which we do by substituting these values for x and y in the original equations and checking that the equations hold.

If we have three equations in three unknowns x, y and z, we can use the first equation to eliminate x from the second and third equations, giving a system in which the first equation is unchanged, but the second and third equations involve just the two unknowns y and z. We now use the above method to solve the system consisting of the new second and third equations for y and z. As above, the first of these two can be used to eliminate y from the third, leaving one equation in one unknown z. This can be easily solved for z, this value substituted back into the second equation to solve for y, and finally substitute both values obtained for y and z back in the first equation to solve for x.

That then is the general method. It means that the very last equation we obtain by this systematic elimination process has only one unknown. We then solve this equation, and with the value obtained for the last unknown, we substitute in the previous equations. This means that one of them has only one unknown. And so we can proceed. This method is called back-substitution, and the whole procedure is called Gaussian elimination.

Examples 5 and 6 in §7, below, give further examples of this method.

§6 Some general results

These general results will be useful in Chapter 12 on Geometry, so it is worthwhile noting them now.  Observe that the way we derive them depemds only on the properties of the real numbers.  This will be important in Chapter 12.

If we have two equations in two unknowns, then, the following possibilities arise:

1. There is no solution

As an example consider the system:,

2x + 3y = 1            (3)

4x + 6y = 6            (4)

We use our method of changing the equations by transforming the system into one in which x has been eliminated from one equation. In this case we can do this by subtracting twice the first equation from the second, which eliminates x from the second equation. Thus, our new system becomes

2x + 3y = 1            (3)

4x +6y – (4x + 6y) = 0x + 0y = 6-2 = 4 (4´)

Our subtraction has successfully removed x, but at the same time removed y. The net result has been the equation 0 = 4, which is clearly impossible and means that our system can have no solution.

 

2. There is a unique solution

For example, consider the following system:

2x + 3y = 1  (5)

x + y = 1  (6)

If we subtract twice the second equation from the first, x will be eliminated from the first and our new system will be:

0x + y = -1  …(5¢)

x + y = 1   …..(6)

Solving the first equation for y we find y = - 1.

We now use the method of back substitution putting in this value of y in equation (6) to get x – 1 = 1, so x = 2, y = - 1 is the unique solution.

3. There are an infinite number of solutions

For instance

x - y = 2.    ….(7)

                                                         2x – 2y = 4     ….(8)

Again we change to a system with two unknowns in the first equation and one in the second. We do this by subtracting twice the first equation from the second. The result is the system

xy = 2   …..(7)

0x + 0y = 0   …(8’)

In this case both x and y have disappeared in the second equation, but in contrast with the example in (1), the second equation (8) is true for all values of x and y. Thus the system has actually reduced to just the one equation, xy = 2. In this case, we can choose any value, say, t for y, and then we find x = 2 + t is a solution to the system of equations, and this works for any value of t. Thus, we have an infinite number of solutions.  What we have also proved is that the two equations are equivalent, so that any value of x and y which satisfies the one equation, will satisfy the other and vice-versa.

§7 Solved problems and other optional material

1. Find a solution to the equation 2.5x - 6 = 5 by guessing.

Solution: We guess 10 because that way we get rid of the decimal. Then the left-hand side becomes 25 – 6 = 19, which is some way from 5. Guessing x = 4 gives a left-hand side which is 4. So perhaps x = 5, and this we see gives a left-hand side of 6.5. Next we try x = 4.5 and this gives a left-hand side of 5.25 which is quite close to the right-hand side.

2. Find a solution to the equation 2.5x - 6 = 5 by the school teacher’s method

Solution: Adding 6 to both sides gives 2.5x = 11. Divide by 2.5 to get x = 11/2.5 = 4.4.

 

3. Find a solution to the equation 2.5x - 6 = 5 with a graph.

Solution. By subtracting 5 from both sides of the equation we get the equivalent equation of 2.5x - 11 = 0.  So we draw the graph of  y = 2.5x – 11 and find out when it crosses the x-axis. We choose two points satisfying the equation and then we can draw the graph which we know must be a straight line. We choose convenient values for x, for instance x = 2 and x = 4. If x = 2 then y = 2.5x – 11 = 5 – 11 = -6 and so the point (2, -6) lies on the line. Similarly choosing x = 4 we see that the point (4, -1) lies on the line. Thus, we join these two points and get our graph . (Fig. 4)

figure 4. C.F. Graph of y = 2.5x – 11.

We need to find where the curve touches the X-axis, and this is approximately at x = 4.4.

4. Why can we not divide by 0?

Solution: 6 divided by 3 is 2 is another way of saying that 2´3 = 6. In symbols a divides b means that there is a number c such that ac = b. Now 0 times any number is 0. For instance, 3 × 0 = 0 and 5 × 0 = 0. Thus 0 ¸ 0 could be 3 or 5 or indeed any number. So dividing 0 by 0 we would not get a unique answer. If we try to divide a non-zero number by 0 we are in worse trouble. For instance, we cannot divide 1 by 0 since there is no number c such that 0 ´ c = 1

5. Show that the following system of equations has no solution.

x + y + z = 4    (9)

                                                 2x + 3y + 2z = 5   (10)

                                                                 x + 2y + z = 3   (11)

Solution: First we arrange that x is removed from equations (10) and (11) by subtracting twice equation (9) and then just equation (9) from (10) and (11) respectively, thus getting the three equations

x + y + z = 4          (9)

0x + y + 0z = -3    (10´)

0x + y + 0z = -1    (11´)

Next we subtract (10´) from (11´) to remove y from the last equation, getting the three equations

x + y + z = 4    (9)

   0x + y + 0z = -3       (10´´)

   0x + 0y + 0z = 2       (11´´)

But this last equation is inconsistent, since the left-hand side is 0 and the right-hand side is 2. Thus these equations do not have a solution.

6. Solve the following system of equations for x, y and z:

x + y + 2z = 2              (12)

3x + 2y + 2z = 3          (13)

2x + 3y + 2z = 1          (14)

Solution: Subtract 3 times equation (12) from equation (13) to give the new equation (13’) and then subtract 2 times equation (12) from equation (14) to give the new equation (14’). This leaves equation (12) unchanged but replaces equations (13) and (14) by (13’) and (14’):

x + y + 2z = 2…… …..(12)

0×x - y - 4z = - 3…    …..(13´)

0×x + y - 2z = - 3……  ..(14´)

Next solve the system consisting of the two equations (13’) and (14’), by adding equation (13´) to equation (14’) to get rid of the y. This replaces (14’) by the new (14”), which no longer involves y.

- y - 4z = - 3    (13´)

0×y - 6z = - 6  (14´´).

The solution to (14´) is z = 1. Substitute z = 1 in equation (13´) to get y = -1. Finally substitute z = 1 and y = -1 in equation (12) to get x = 1.

 

7. Why is –1 × -1 = +1?  (Assume that 1 times any number leaves the number unchanged, that 0 times any number is 0 and that the distributive law a(b + c) = ab + ac holds for any choice of numbers a, b, and c, positive or negative.)

Solution.

(i)        Since 1 times any number does not change the number, 1 × -1 = -1.

(ii)       1 + (- 1) = 0. Multiply both sides by –1.  The right-hand side becomes 0.  The left-hand side becomes using the distributive law –1 × 1 + (-1) × (-1) and this is equal to the right-hand side which is 0.  Thus ,

 –1 × 1 + (-1) × (-1) = 0,and since
            -1 × 1= -1, we see that -1 + (-1) × (-1) = 0 and so (-1) ×(-1) must be +1.

Chapter 11

FUNCTIONS

The concept of function took centuries to evolve.
Which is very surprising, for although it is very general,
it is remarkably simple. It is fundamental to mathematics.

The concept of function is so central to mathematics, that we will devote this whole chapter to discussing it. Historically, it has not been easy to define this concept properly. Now that we have settled on a definition, it seems relatively straightforward, although abstract.

Knowledge of Chapter 9, Coordinate Geometry, is needed for this chapter. Indeed, we will use coordinates to define the concept of function. This is quite natural, as we shall see. We also need the concept of set.

From Chapter 9 we need to recall that we choose an origin O and draw two lines at right angles to one another, the x-axis OX and the y-axis OY. We specify the position of a point by means of its coordinates, i.e. the distances from the x-axis and the y-axis.

§1 What do we mean by a set?

In the 1960s the New Mathematics was introduced into the schools. It caused much confusion since most parents did not know what it was about and so could not help their children. Also, it did not seem to be of much practical importance.

The New Mathematics was based on set theory. The idea of set theory is to consider collections of objects, for instance, the collection of all schoolteachers. Such a collection is called a set, and the objects in the set, in this case, schoolteachers, are called the elements of the set.

Almost any collection can be a set. For instance, the set consisting of the number 2 is a set consisting of a single element, 2. On the other hand, the set of all numbers is an example of an infinite set. The idea of set is indeed quite basic, and nobody has any difficulty with it. The difficulty comes with trying to express as much as possible of mathematics in terms of set theory.

Just as it is difficult for a foreign tourist to express himself in English with a limited vocabulary of only one thousand words, so is it a struggle to express complicated mathematical ideas in terms of set theory alone.

§2 Functions defined geometrically

The idea of a function is a crucial concept in mathematics, and so it is worthwhile defining it precisely. We begin with a geometrical definition and then provide an algebraic one. Obviously one can have only one definition, so we will have to decide on which one we will take as our primary definition. In fact, this will turn out to be the algebraic one, but the geometric one is a good start. The method was suggested by the Polish-Swedish mathematician, Richard Bonner.

We begin by choosing the usual coordinate system in the plane with x-axis OX and y-axis OY. We define a relation to be any set, R, of points in the plane. If these points have an extra property, namely, that any vertical line parallel to the y-axis OY meets R in at most one point, we call R a function. That is, if (x, y) is a point in the function R there cannot be a point (x, y’) in the function R with y’≠ y. The graphs in Fig. 1(a) and (b) illustrate sets of points which define functions, while in Fig.2 the set S consisting of the points on either the drawn circle or the drawn curve is not a function, since there are lines parallel to the y-axis which meet the points of the set in more than one point. For instance, the y-axis itself meets the set S in three points.

figure 1.a. The points on the curve are a function.

figure 1.b. The points on the curve are a function.

figure 2. The set consisting of the points on the circle
together with the points on the curve is not a function.

You may object that you have studied functions before, and this definition is not at all in accordance with the usual definition. That may be true, but be patient, and the connection will become apparent to you soon.

If f is a function, such as the one described by the points of the curve drawn in Fig. 1(a), we define f(x) to be the number y if the point (x, y) is an element in the relation, that is, (x, y) are the coordinates of the intersection of the line parallel to the y-axis which passes through x on the x-axis. According to the definition of function there is at most one such number y, and so f(x) is unique. However, it is also possible that there is no such point, in which case f(x) is not defined. The collection of all x for which f(x) is defined is called the domain of f.

§3 Awkward functions

The advantage of the geometrical definition is that it is very easy to illustrate various kinds of functions. In Fig. 3 we have one awkward function. Its domain is the set of all real numbers between –3.5 and 7.2 with the exclusion of 0, because there is no point (0,y) in the function. We say, intuitively, that the function has a gap at x = 0.

figure 3. A function with a gap.

The points of the function in Fig. 4 are (x,1), for all x less than or equal to 0, but for all x greater than 0 the points are (x,2). It does not have a gap at x = 0, but we say it has a jump at x = 0.

figure 4. A function with a jump.

In Fig. 5 we have a curve that abruptly changes direction at (0.2)

figure 5. A function that changes direction abruptly.

Many of these functions, which we can draw so easily, are difficult to analyse algebraically. Curves with no gaps or jumps are said to be continuous. (This is not a proper definition but an intuitive one.)  If, in addition, the curve does not change direction abruptly, it is called a smooth curve. (Again this is not an exact definition.) .

§4 Functions described algebraically

Having defined function geometrically, we will abandon this definition and instead give the algebraic definition. The geometrical definition of a function is as a collection or set of points with an extra condition. For the algebraic definition we take the coordinates of these points as our definition, so a relation R is now a set of coordinates. For instance, the set of all coordinates of the form (x, 2x) is a relation.

In the geometrical definition we had the additional requirement that every line parallel to the y-axis meets our function in at most one point. For our algebraic definition that is the same thing as saying that for any x, there is at most one y for which (x, y) belongs to the function.

We can now state our definition:

Definition. A function f is a set of coordinates (x, y), in which for any number x, if (x, y) and (x, y’) belong to f, then y = y’. Also, if (x, y) belongs to f, we write f(x) = y.

Note that the possibility that two or more different x’s can occur with the same y is not ruled out. But the essential requirement is that we cannot have the same element x occurring twice with two different second values y and y`. For example, the set {(1,2), (1,3)} is not a function.

Example. The following are examples of functions.:

Function A: The set of all coordinates of the form (x, x+2) for all real numbers x.

·       Function B: The set of all coordinates (x, Log(x)) for all numbers x greater than 0.

·       Function C: The set of coordinates (x, 1) for all numbers x less than 5; (x, 3)
for x greater than 5 but less than 6; and (x, 2) for all numbers x greater than 6.

·       Function D: The set {(1,1), (2,1)}

§5 Contrast with the definition in the calculus books

A function from a set A to a set B is defined to be a rule so that to each number in A is associated precisely one number in B.

This is a common definition. The disadvantages are that we do not know what a “rule” is, and what “associated to” means. Given something, which is supposed to be a function, we do not have a way of checking that it is. We would first have to check that we are given a “rule”. But what is a “rule”? What do we mean by “associated to”? Our algebraic definition is simpler. We check that we have a set, and we check that each element a occurs as the first entry in an element belonging to the set with only one b.

§6 What comes after 1, 2, 3? Could it be 34?

In an IQ test if asked what comes after 1, 2, 3, one would be tempted to answer 4. But this is not the only logical answer. For instance, for the formula

f(x) = x + 5(x - 1)(x - 2)(x - 3),

f(1) = 1 + 5×0×(-1) ×(-1) = 1,
f(2) = 2+ 5×1×0×(
-1) = 2,
f(3) = 3 + 5×2×1×0 = 3.

However, f(4) = 34. That is, for this function, the number that comes after 1, 2, 3, is 34.

Note that to check what happens when x takes on the values 1, 2 and 3 in the above formula, the calculations are easy to carry out, since each (x - 1)(x - 2)(x - 3) is 0 for the values of 1, 2 and 3.

This approach was applied more generally by the French-Italian mathematician Lagrange. As an example of Lagrange’s method, suppose we have a function g(x) and we know that g(3) = 11, g(4) = 3 and g(6) = 10. What would be a reasonable estimate for g(5)?

figure 6. Lagrange (1736–1813).

Lagrange’s solution wa