[Spyros Kalomitsines] How to Solve Problems New M(BookFi.org)

HOW TO SOLVE PROBLEMS: NEW METHODS AND IDEAS

HOW TO SOLVE PROBLEMS: NEW METHODS AND IDEAS

SPYROS KALOMITSINES

Nova Science Publishers, Inc. New York

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NOTICE TO THE READER The Publisher has taken reasonable care in the preparation of this book, but makes no expressed or implied warranty of any kind and assumes no responsibility for any errors or omissions. No liability is assumed for incidental or consequential damages in connection with or arising out of information contained in this book. The Publisher shall not be liable for any special, consequential, or exemplary damages resulting, in whole or in part, from the readers’ use of, or reliance upon, this material. Independent verification should be sought for any data, advice or recommendations contained in this book. In addition, no responsibility is assumed by the publisher for any injury and/or damage to persons or property arising from any methods, products, instructions, ideas or otherwise contained in this publication. This publication is designed to provide accurate and authoritative information with regard to the subject matter covered herein. It is sold with the clear understanding that the Publisher is not engaged in rendering legal or any other professional services. If legal or any other expert assistance is required, the services of a competent person should be sought. FROM A DECLARATION OF PARTICIPANTS JOINTLY ADOPTED BY A COMMITTEE OF THE AMERICAN BAR ASSOCIATION AND A COMMITTEE OF PUBLISHERS. LIBRARY OF CONGRESS CATALOGING-IN-PUBLICATION DATA Kalomitsines, Spyros. How to solve problems : new methods and ideas / Spyros Kalomitsines (author). p. cm. ISBN 978-1-60456-429-7 (hardcover) 1. Problem solving. I. Title. QA63.K36 2008 510.76--dc22 2008008039

Published by Nova Science Publishers, Inc. New York

CONTENTS

Preface vii Acknowledgements ix Chapter 1 The Description Method 1 Chapter 2 The Method of Getting Out of Loops 21 Chapter 3 The Spiral Method for Solving Problems 35 Chapter 4 Other Methods 43 Chapter 5 Two Models for Teaching Mathematics and Problem Solving 81 Chapter 6 A Formal Computer Model of Our Methods 125 Solutions of Unsolved Problems 133 References 171 About the Author 173 Index 175

PREFACE

To my wife Matina who has been a continuous source

of encouragement and support Teaching mathematics to high-school students, for more than thirty years, I was troubled

by such questions as: − Can we get rid of the fear of problems, and in general of mathematics? − Is there a way to enhance our aptitude in problem solving and become more creative? − What causes more difficulties when we face problems? After about thirty years of effort, I developed some new problem solving methods. These

methods have been tested many times in class, while teaching high-school students, and were found to be very effective. They were also rigorously tested using the concepts and methods of cognitive psychology. At the suggestion of professor J. Greeno of the University of Pittsburgh, where I was a Fulbright scholar, I wrote a computer program based on my problem solving ideas, which solved a substantial number of problems automatically.

In this book you will find suggestions and methods that will help you overcome many difficulties when you solve problems. These methods are given in simple and easy-to-follow ways, always illustrated by examples. If one example requires mathematical knowledge beyond your grasp, you can skip it. There is a sufficient variety of examples for you to be able to comprehend the methods described.

After reading this book, your problem solving skills will have improved appreciably. Even if you still cannot solve a problem, you will have approached it in refreshing and exciting ways.

In Chapter 1, the “description method” is introduced. This method shows how to generate ideas towards solutions, based on existing knowledge.

This is perhaps the most important method; it encompasses a large number of problems and is easily combined with most other methods.

In Chapter 2, a method is given that helps to avoid loops when solving a problem. In Chapter 3, difficult problems are solved by a certain combination of the description

method, the means-end analysis and the method of getting out of loops.

Spyros Kalomitsines viii

In Chapter 4, additional methods or alternative techniques for certain kinds of problems are discussed, as: contradiction, working backward, defining a procedure, setting subgoals, techniques for word problems and problems in matrix algebra, etc.

In Chapter 5, you will find a teaching model for problems with long or difficult solutions. You will also find a model for helping students when they solve hard problems. The model shows how to group problems according to their common characteristics and to search for useful patterns.

A number of such groups from calculus are presented. In Chapter 6, you will find an answer to the question: Can the previous methods be

programmed on a computer? This can be done through a formalization of the methods. Such a procedure sets the stage for a fruitful production of new problem solving techniques.

This book is intended to help teachers but also to help students who want to improve their

problem solving skill on their own.

ACKNOWLEDGEMENTS A number of people have assisted and supported me throughout the course of writing this

book. I express my gratitude to all of them. Professor W. Wickelgren of MIT encouraged me to plunge into the field of problem

solving after he read my ideas. His book "How to Solve problems" provided me with new hints and convinced me that my work was on the right track.

My former high school student and now professor at the Technical University of Crete, Yannis Phillis and his wife Nili edited most parts of the book carefully and made many valuable remarks. Professor Phillis encouraged me to contact American researchers in problem solving, so my work reached its present level of maturity. Without his encouragement my ideas would probably have filled only a few scraps of paper.

Professor J. Greeno formerly of the University of Pittsburgh, currently Stanford, gave me several suggestions thus assisting me in formulating my methods rigorously. I am grateful to him.

At Carnegie Mellon I attended several lectures of the late Herbert Simon on cognitive psychology. These lectures provided me with perspective vis-à-vis my problem solving directions. I also recall gratefully the discussions I held with the wise old man about my subject.

Spyros Kalomitsines

Summer 2007, Athens – Greece

Chapter 1

THE DESCRIPTION METHOD

1.1. THE SUBTLE POINTS OF PROBLEMS Imagine that you are in a classroom. Your teacher of mathematics suddenly poses an

original problem to be solved. “Original problem” means that it has no similarity to other problems known to you. What are you going to do?

Many such occasions have led me to the observation that, if the students face a problem for the first time, they either do nothing, or make some unsuccessful trials and then just stare at it without making any effort. Exceptions occur, especially among talented students, but here my concern is with the majority of students. Even good students would not solve an original problem unless given a crucial hint. The process of finding such useful hints is just our question, here. Can we devise a method, which would teach students how to generate the crucial idea to solve a given problem?

I have investigated this question for many years. Here are my conclusions in brief: The “blank” minutes people spend staring at a problem are often useless. If we could

eliminate them or reduce them to a minimum, then we would shorten the time of arriving at the crucial idea. To succeed in that, we have to change the way we usually approach original problems, as it happened in physics in the days of Galileo: “Not only thinking, but experimenting and thinking.” An analogy for problem solving would be: “Not only thinking (which makes solvers spend blank minutes) but thinking and acting.” Let me make this clearer.

The crucial idea can very often be produced, together with a lot of additional

information, which comes from the problem itself. Every problem contains information, often hidden in subtle ways. This is a kind of very clever coding of information. How can we break this code in such a way that the crucial idea would emerge?

Here is a technique, which I call “The description method.”

Spyros Kalomitsines 2

1.2. THE DESCRIPTION METHOD IN TWO STEPS

STEP 1 (Preliminary Step)

Preparing Our Mind Before starting to apply our method we read the problem many times until we have fully

grasped the facts and the goal. We try for a while to solve it. Now suppose that the problem is not a routine one; none of the strategies we know seem to be applicable and no other idea comes to mind about how to solve it. The problem appears to be new and difficult. In this case start applying the method of description. Our method should be applied after having made such an investigation, or after having made attempts to solve the problem and hit a dead end.

STEP 2

Generating Ideas To use the method of description, we take the various parts or details of the problem and

try to recall all relevant knowledge. Because there is danger that some constraints may intervene and prevent the crucial idea from emerging, we should do this: While looking at a part of the problem, we temporarily try hard to ignore the rest of it. We focus our attention on this part exclusively and then we start recalling anything relevant. We write down in short notes everything we recall.

Two examples follow

Example 1.2.1. If your problem contains the phrase «odd numbers», you may proceed as follows: They have the form 2n+1 odd+odd = even odd+even = odd odd×even = even odd+even ≠ 0 the difference of two successive odd numbers is always 2 so the sequence 1,3,5,.. is an arithmetic sequence properties of arithmetic sequences? and so on. Once in a while you stop and take a quick look at what you have written. But now you

should have in mind the whole problem. Is any piece of the extracted information useful for your problem? If yes, use it, if not, go on retrieving more information. Try your best.

The Description Method 3

Your mind must be free of constraints Be a young child again who writes, or sings a naive poem, or song with no set patterns

and no fear of critical reactions. The difficulties in generating new ideas are often due to psychological reasons.

Example 1.2.2.

What information do you derive from the phrase “isosceles triangle?” It has two equal sides. It has two equal angles. It is symmetrical about an axis. The bisectors of the two equal angles are equal in length. There are two equal medians. There are two equal altitudes. The equal altitudes are also sides of two equal triangles. The same is true of the medians as well as the bisectors. What are the properties of equal triangles? Etc. After proper training, one could extend the above process almost indefinitely. Ultimately

one should be able to do it mentally, writing down only a few items, if any.

1.3. EXPLANATION OF THE METHOD The central idea of the method is to fully activate the mind. It is a kind of bottom up

approach where we start by describing the individual parts of the problem until we hit the crucial idea.

− Not only thinking but thinking and acting − Acting means: use language and describe in short notes all your thoughts − Write your thoughts down − Draw a diagram if necessary − Use simple symbols − In other words, if you are stuck with a problem, change your approach. Don’t forget that the most difficult thing in problem solving is to see another aspect of the same detail − Keep writing thoughts coming from other thoughts − Write short lines, as if you write a simple poem or song. Be a young child again. Set your mind free − Keep a rhythm while writing. Do this in a nice and creative environment (neat paper, convenient chair and lighting).

You need persistence, good temper and self-confidence. Try your best to surpass

yourself.


In Chapter 6 you will see a computer simulation of the description method. It is a rigorous test of the method, based on concepts of cognitive psychology.

This way of writing down short notes tunes your mind to a productive rhythm, something like music that liberates and helps the mind to recall more information, and the eye to recognize the required idea quickly while scanning the notes later. Fears and hesitations prevent the mind from generating ideas. Writing, i.e. moving your hand together with thinking helps generate ideas. While writing we generate ideas. It has been proven time and again that many ideas are generated while one is not only thinking but is also engaging in using his language in a discussion and in writing his thoughts down.

Finally, to generate good ideas one has to be in a good psychological state. Illustration of

the method will be given through a number of examples.

1.4. SOLUTIONS OF PROBLEMS USING THE DESCRIPTION METHOD

The following problems are presented as they were actually solved by my students in

class, following the description method. I reminded them of the method and I urged them: Write down as much as you can in short notes, do not hesitate.

Problem 1.4.1. You are given three equations: ⎫⋅ = ⎪⋅ = ⋅ ⎬⎪+ + = ⎭2

2

3 3 5

a b cb c 2 aa 2b 3c 0

a, b, c are real numbers

Prove that a = b = c = 0. In a while one of the students provided the following solution:

2a b c⋅ = c2 is positive or zero So a b⋅ is also positive or zero 2a2 is positive or zero So b c⋅ is positive or zero b c 0⋅ ≥ , a b 0⋅ ≥ A positive product implies that both factors b and c or a and b have the same sign Hence, a, b and c are either all positive, all negative, or all equal to zero What else can I do?


There are several things

2a b c⋅ = hence 2a b c 0⋅ − = or bca

2= etc.

Let me examine now 3 3 5a 2b 3c 0+ + = There are three powers All three exponents are odd a, b and c are either positive, negative, or zero. How can this expression equal zero? It is either positive, negative, or zero, since a3, b3 and c5 are either all positive, all

negative, or all zero So, necessarily a = b = c = 0 Now, it is easy to write down a rigorous proof using some of the above ideas. By setting

his mind free, the student was able to generate many ideas, some of which were the crucial ones.

Note:

If you have prepared your mind and if you have written down the crucial idea, it is rather easy to recognize it. Psychologists say that it is easier to recognize, than to recall.

The crucial information exists often implicitly and by describing we make it explicit. Einstein said God is sophisticated, but not evil, meaning that information exists; it is not hidden.

Problem 1.4.2.

Consider the puzzle shown in Figure 1.1. You are given a checkerboard and 32

dominoes. Each domino covers exactly two adjacent squares on the board. Thus, the 32 dominoes can cover all 64 squares of the checkerboard. Now suppose one square is cut off from each of two diagonally opposite corners of the board. Is it possible to place 31 dominoes on the board so that all of the remaining 62 squares are covered? If so, show how it can be done; if not, prove it impossible. You could try to solve it for a while before reading on.

Figure 1.1. Checkerboard puzzle.


Solution by the method of description. Many people find it very hard to solve this problem. Let us see how one student used the method of description.

What do I see in this problem?

I have 62 squares I have 31 dominoes Each domino is rectangular What are the properties of rectangles? Parallel sides, congruent angles, congruent diagonals, etc. What else do I see?

Lets focus on another detail

Each domino covers two squares The two squares have a common side The two squares have different colors So one domino covers one black and one white square Here I see a simple correspondence 1 domino covers 1 black square and 1 white square

So, 31 dominoes cover 31 black squares and 31 white squares

Is that possible? How may squares do I have? What are their colors? Is there any subtle point? Let me be careful here I have 62 squares Two white squares have been removed

So there are 30 white and 32 black squares Let me stop and take a quick look at everything I have written so far. Is it possible that

the crucial idea is contained in what I have noticed? Yes, obviously! I have proved that 31 dominoes can cover 31 black and 31 white squares, but our checkerboard has only 30 white and 32 black squares, so it is impossible for the 31 dominoes to cover them all.

Important advantages of short notes:

(a) They are produced easily (b) One can easily recognize the crucial idea among them Researchers in anthropology say that before using language, humans had invented a small

number of simple tools, and used them, unchanged, for about one million years. Just after starting to use language, they invented a large number of tools of higher complexity in a very short time. It was a real revolution. This demonstrates that language enhances thought and increases the human ability to solve problems.


If you cannot come up with any good idea, leave the problem for a while. Come back to it later. Start describing again.

I urge all mathematics teachers to try the method of description in class, as well as the methods of chapters 2, 3. They will very soon realize that these methods make a difference. The teacher can give a difficult problem to the students and then encourage them to generate ideas from every part of it. Sometimes unexpected solutions will emerge in class. It has happened to me many times. It makes the class exciting!

1.5. DISCUSSION What most people will have overlooked in the case of the checkerboard problem are the

facts that both of the absent squares are white, and that each domino covers one white and one black square. These two observations make up the crucial idea, which, unless you go about the problem in the proposed way by describing in short notes, will probably escape you, as experiments have proved. The color of the squares is the information that most people fail to use. This information is given in an implicit form and by description we make it explicit.

Using the method of description means following this advice: If you cannot solve your problem, don’t go straight at it. Go around it. Describe everything fully. Include all the features that you notice. In a very short time the crucial idea will be written on the paper in front of you, and then all you have to do is recognize it. It is rather easy to ignore the irrelevant data. This can be done by re-examining all the information you have recorded. If you prepare your mind, as Pasteur said, it will be easier to recognize the crucial idea, and you will not be lost in the larger problem space. This is one of the reasons we use short notes. Another reason for writing short notes, as has already been mentioned, is that this technique helps keep a pleasant rhythm that produces more ideas.

Some readers may argue that the danger of getting lost in the large problem space reduces the value of the method. Here is a twofold response to this concern:

(a) Before starting to describe, we urge the student to go over the problem and perform

preliminary work (see STEP 1 of section 1.2). This is necessary, because then the mind will be better prepared to look for a solution. Furthermore, searching around the problem to see whether any similarity exists between this problem and another with a known solution, or thinking whether any known strategy can be applied is part of the description. After such an investigation we come to the detailed description, if no solution has yet been found.

(b) It is preferable to search in a somewhat large space that most probably contains the crucial idea than to abandon the problem or to search in a narrow but sterile space.

If you cannot find the crucial idea immediately, continue to draw more inferences by

feeding back, describing, and scanning your data repeatedly. You must have patience, perseverance, a good understanding of the method, and the will to solve the problem. With practice, the usual hesitation to bring out the right idea will disappear. Inhibitions or lack of daring prevent us from making information explicit that is presented in subtle ways. Many


great philosophers emphasize that a virtue of great researchers is to be audacious when searching for ideas.

It is rather like a battle. If we know only a few things about the enemy, it is unwise to go straight at him. We had better go around the enemy first, gather intelligence and pinpoint his weak points. When dealing with problems we have to find their subtle points.

The Role of Training Is Important I would like to emphasize that a certain kind of training in the method of description

proved to be useful. Before giving students problems to solve, they should be given training in which they read little phrases and are asked to say as many things as they can.

If we train ourselves properly, we can extend this descriptive process almost indefinitely. When this method is applied in class, we shall undoubtedly observe individual differences in the inferences they draw. What does this situation lead to? First, there is the possibility that different ways of solving a problem will emerge. Second, if a problem is very hard, the variety of ideas generated in class when people work together is large and thus the chance of hitting on the crucial idea is greatly enhanced. This is why it is often easier to arrive at a solution if a number of people work together.

1.6. PROBLEMS FROM MATHEMATICAL OLYMPIADS

Problem 1.6.1. Prove that 55552222 + 22225555 is divisible by 7. Stop reading and try to solve it by yourself. I once gave this problem to my students, who

had been trained in the method of description. One of them gave the following description: What do I see here? I have the numbers 5555, 2222 What inferences can I draw? 5555 5 1111= × , 2222 2 1111= × It reminds me of the properties of powers: ( )n n na b a b⋅ = ⋅ , ( )knk na a= ( ) ( ) ( ) ( )1111 1111 1111 11112 2 5 55 1111 2 1111× + ×

or ( ) ( )1111 11112 2 5 55 1111 2 1111× + ×


What do I see now? The sum of two powers with the same odd exponent, 1111. It reminds me of the identity: ( )( )n n n 1 n 2 n 1a b a b a a b ... b− − −+ = + − + +

So I obtain ( ) ( ) ( )( )2 1110 5 11102 2 5 55 1111 2 1111 5555 ... 2222× + × − +

Subtle point here? Let me now try to see what is the remainder of the division. ( )2 2 5 55 1111 2 1111 : 7× + ×

It is easy now to find this remainder, either by using number theory or by dividing

2 2 5 55 1111 2 1111× + × by 7. So I see that the remainder is zero and thus the problem is solved.

Another student gave a different description: What do I see here? I see the sum of two powers I see 5s and 2s only What is the goal? To prove that the given sum is divisible by 7. Let me repeat 5s, 2s, 7…5, 2, 7 Any subtle point here? Is there anything to connect these numbers? 5 + 2 = 7 or 2 = 7 – 5 What about our problem? 2222 7777 5555= − Let me use this:

According to the binomial expansion formula ( )n n n 1 na b a na b ... b−− = − + −

So ( )555522225555 7777 5555+ − = 2222 5555 55555555 7777 ... 5555+ − −

55557777 ...− is obviously divisible by 7

Hence the rest of the expression must be divisible by 7 ( )2222 5555 2222 33335555 5555 5555 1 5555− = −


Since 22225555 is not divisible by 7, 33331 5555− must be divisible by 7, or the remainder of 33335555 : 7 must be 1. But since the remainder of 5555 : 7 is 1, the remainder of 11115555 : 7 is also 1 as is well known from number theory.

Finally 33331 5555− is divisible by 7 and the problem is solved. Note 1: At first sight this problem seems difficult. The numbers are extremely large. However,

never lose courage, start describing carefully and continue until the crucial idea appears. Note 2: To draw inferences from equations or algebraic expressions in general, perform as many

transformations as possible: factor, simplify, express one variable in terms of the others, etc.

Problem 1.6.2. Another example involves a rather difficult problem of Euclidian geometry. I gave it once

to a class of students, untrained in the method of description, as homework. The next day, they all came in complaining that it was too difficult to solve. I also gave it to a class of students trained in the method of description. At first I reminded them of the method and I urged them to use it intensively. A number of students solved it. Here is the problem (see Figure 1.2) and the solution given by one student:

The problem. You are given a sharp triangle ABC and its altitude AD. The goal is to

construct an isosceles triangle EDF, ( )=ED EF with angle DEF having a given measure θ, and the vertices Ε, F lying on the sides AC and AB correspondingly.

A

E

CDB

F θ

Figure 1.2. A triangle.

Solution. Here is the solution that the student gave, using the method of description:


A

CB

a)A

CDB

b)

E

DF

)c

A

E

CDB

F φ

φ

d)

Figure 1.3.

I can describe the main parts, but let me first separate them (see Figure 1.3). 1. Seeing ABC, what relations can I recall?

Let me start writing

Angles oA B C 180+ + = (A is the abbreviation for BAC ) BC AC AB BC AC− < < + etc.

2. AD is an altitude AD is perpendicular to BC So we have right angles We also have right triangles Let me write any relations I recall: ( ) ( ) ( )2 2 2AB AD BD= + ( ) ( ) ( )2 2 2AC AD DC= + ( ) ( ) ( ) ( )( )2 2 2AB AC BC 2 BC DC= + − ( ) ( ) ( ) ( )( )2 2 2AC AB BC 2 BC BD= + −

triangle oABD D 90⇒ = , oA B 90+ = etc. 3. Triangle EDF


It is isosceles, ED EF=

E θ= , D F=

But oD F E 180+ + = Solution of the equation ( ) ( )o o2 D E 180 D 180 θ / 2+ = ⇒ = −

Hence the measure of D is known: ( )o180 θ / 2−

That is, the angles of triangle EDF are constant Triangle with constant angles The ratio of two of its sides is a constant: ( )DE / DF λ= 1

Also, ( )EF / FD λ= , ( )DF / ED 1/ λ=

4. (See Figure 1.3d) What do I see now?

FDE is known, let us call it φ

DFE is also φ ratio ( )DE / DF λ= , etc

Can I find the position of E? Can I exploit the above relations? A known angle, a known ratio Yes, it reminds me of a known algorithm, the combination of a rotation and a dilation2

1.7. DIFFERENT WAYS OF APPLYING THE METHOD If despite your efforts, the information you have collected does not lead you to a solution,

then check your textbooks. Work carefully to find something that fits your problem: definition, postulate, theorem, algorithm or heuristic strategy.

Take notes as you did in previous examples. Additionally you may go back to problems you have solved in the past that might be related to the present one. It is possible that a technique you have used in the past can be applied again. Here is a good criterion that helps you choose that technique.

Use the technique that requires most of the data of your problem. You’ve already transformed these data into pieces of information, written down in the form of short notes. Match these notes and the technique to the utmost.

1 Let T be the set of all the isosceles triangles with angles θ and ( )180 θ / 2− , and let KLM be a certain triangle

belonging to T. If K L M′ ′ ′ is any other triangle of T, then KL/ LM K L / L M′ ′ ′ ′= ; that is, for any triangle, this ratio equals the constant KL / LM .

2 In the students’ geometry textbook, there is a chapter where various geometrical transformations are studied. One such transformation is rotation about a point followed by dilation. In our case, the rotation of AB about point


until we decide to apply it, that is, to put one angle into the other, then it becomes explicit.

Note: Do it with any definition related to a part of your problem. Apply definitions and

make information explicit. Also apply formulas or theorems which have some relation to your problem. Do not hesitate at all. It is best to take implicit information and make it explicit.

1.8. PROBLEMS FROM PHYSICS

1.8.1. A Tank Containing Water Balances a Weight as Shown in the Figure. Will the Equilibrium Be Affected If You Put Your Finger into the Water without Touching the Tank?

Try to find the right answer by description What happens when you put your finger into the water? The level of the water rises somewhat Your finger feels lighter The principle of Archimedes applies The water exerts an upward force on your finger Water molecules move against your finger Your finger exerts a force on the water in accordance with Newton’s third law Pressure on the bottom will increase Since the finger loses weight, where has the weight gone? What do the laws of physics tell us? Now the answer is obvious. Equilibrium will be destroyed. The tank becomes heavier

because of the force exerted by the finger (Newton’s third law).


1.8.2. The Sun Attracts Objects on the Surface of the Earth with Greater Force Than the Moon. Yet the Phenomenon of Tides Is Caused Chiefly by the Moon, Not by the Sun. Why?

Stop reading for a while and try to answer the question. It looks strange. Draw a figure. Strangely enough the sea moves under the pull of the moon. How can it happen? Does

Newton‘s law cease to be valid? Of course not. So there must be something subtle here; implicit information which we have to make explicit. Let us try the method of description.

Water is attracted by the moon Water is attracted by the sun Water prefers the lesser force An apparent contradiction occurs, but there are no contradictions in nature.

Earth

Earth

tide

Moon

Moon

Sun

Sun

attractive force

attractive force

What conclusions can you draw from this? Tide is independent of the magnitude of forces Then what does it depend on? Continue describing; examine a part of the problem Look at the earth and the moon only What do you observe? The earth is attracted by the moon


What is the attractive force?

You can calculate it by Newton’s law: 2dmMKF = etc.

Water is attracted by the moon Water moves in the direction of the moon and tends to leave the earth So the moon causes different accelerations to the earth as a whole and to the water alone Now examine the earth and the sun Does the sun cause different accelerations? Of course it does However, the difference here must be smaller (crucial idea) Now, the problem becomes straightforward. According to the theory of gravitation, the

acceleration of the earth due to the moon is 2a KM / d= , where M = mass of the moon, K = gravitational constant and d = distance from the center of the moon to the center of the earth (the acceleration of the earth equals that of a body with the same mass located at its center).

Earth

Moon

R

d

Therefore, the difference between the acceleration of the surface water and the

acceleration of the earth is:

2

2 2 2 2

KM KM KM(2dR R )(d R) d d (d R)

−− =− −

But R is very small compared to d, so d R d− ≈ and the above result becomes

3d2RKM

.

If M1 is the mass of the sun, then the difference of accelerations will be 31

1

d2RKM

,

where d1 is the distance between sun and earth But d is approximately 60R d1 is approximately 25000R M1 is approximately 627 10 M⋅ .


The ratio of the differences becomes 3d2RKM

: 31

1

d2RKM

= 3MdMd

13

31 ≈

Therefore the effect of the moon on tides is almost three times that of the sun.

1.9. FURTHER DISCUSSION People usually tend to attack whole problems although this process is often at odds with

the workings of the short-term memory (STM). The capacity of STM is limited. To attack then the whole problem is like demanding from the brain to do more than it can. The method of description, together with the rest of the methods in this book, provide the tools for (a) retrieving information (b) transforming information into convenient forms.

When we look for an object we have lost, say a key, we do not just look around. We look underneath other objects, turn them over, empty drawers and so on- but not randomly. We start with the most likely places. Prospectors do not expect to find gold by chance, but make chance work for them by creating suitable conditions. “Fortune favors the brave” as the saying goes. If fortune had a hand in great inventions and discoveries, it was because luck was challenged in the most audacious of ways. The genius of even the most brilliant inventors is no more than this: they worked cleverly and hard to enable the crucial idea to emerge.

It is incontrovertibly true that no invention or discovery has ever been made at random. No progress has ever taken place overnight. No problem has ever been solved at first sight. The human brain is not capable of making great leaps. It combines earlier knowledge with new observations, thus creating suitable conditions for new knowledge.

What is the difference between great researchers and laymen? The difference is that great researchers are gifted with an extraordinary ability to scrutinize concepts in many ways and synthesize quickly and courageously. They are able to relate seemingly unrelated concepts and obtain new, powerful insights.

Einstein discovered the theory of special relativity after having set up the basic conditions. First of all, time was ripe. This tells us never to start solving a problem if we are unaware of related facts, similar problems, or pertinent theory. For example, do not work on a probability problem if you ignore probability theory. For Einstein the ground had been prepared, all necessary knowledge was available, several important experiments had been carried out. Also, he worked hard, with enthusiasm, having understood his problem deeply.

The ancient Greek mathematicians were unable to solve the problem of squaring the circle because crucial knowledge was missing. They were unable to make the necessary deductive leaps. A long process of evolution in mathematics led to advanced algebra and calculus, thus creating the necessary tools to find the right answer: It is impossible to square a circle by means of straightedge and compass.

So be attentive: don't try to solve a problem before you have grasped what is given and what is required. Do not hope for a successful solution unless you are determined and enthusiastic. Follow the instructions given here.

However, in order for instructions to be fully understood and used in practice, you should examine a variety of different examples to see the method in action. Then you may begin to apply these instructions in problems of your own. It is not enough to learn to drive; you should practice if you want to become a skillful driver.


Note: The brainstorming method for individuals or groups of people consists of generating as many ideas as possible uncritically. It welcomes unusual ideas and progresses by combining and improving these ideas. The description method has some similarities with the brainstorming method, but also an important difference: The description method shows HOW someone can generate ideas related to a given problem. The use of language, as well as writing short notes help to generate good ideas.

1.10. CREATIVE WRITING Science is just one field where problems appear. In reality life is full of problems and one

might benefit from the ideas of this book when asked to tackle a problem. Creative writing and the generation of ideas became an intriguing problem for me while I was director of the Experimental School of the University of Athens in the 1980’s.

I had heard several teachers complain that students in their Greek composition writing classes had enormous difficulties to generate ideas. The school provided the ideal laboratory for me and I set out to apply the description method to composition or creative writing. I was fortunate enough to be assisted by several teachers of Greek, including my wife, Matina.

The idea was simple. The students posses knowledge stored in their long term memories (LTM). This knowledge is called declarative knowledge. My goal was to encourage them to generate ideas from their LTM, by using the description method. Schematically the procedure is presented in the table below. The table has three columns A, B and C.

Declarative knowledge Procedural knowledge (A) time

(B) environment, etc. (C)

today yesterday last week last month last summer a year ago two years ago when I was young …

Proceed slowly, try to remember as many-many things relative to your topic as possible Write in short notes anything you recall.

your home your neighborhood your country your parents your friends your school your TV your sports your books your magazines, etc movies or plays you have seen etc.

Record positive or negative aspects. Write in short notes.

health education culture happiness entertainment food living economy family freedom justice art science …

Chapter 2

THE METHOD OF GETTING OUT OF LOOPS

2.1. HOW TO GET OUT OF LOOPS People usually try to solve problems via known methodologies. If they did not operate

thus, they would often be lost in a large number of trials, and progress would be limited. Available ways, if any, are followed. However, it has been observed that sometimes exactly this leads to failure. We become trapped in loops because of our previous knowledge and make circles around the problem but the solution lies elsewhere. The second method to be presented here addresses this difficulty.

This will be done using examples.

Example 2.1.1. Problem. You are given six matches and you are asked to make four equilateral triangles

with them. Try for a while before reading on. Now, here is the method. Describe in short notes the attempts you have made. Write them down on the left half of a

page. On the right half of the page, do your best to express all possible opposite statements and actions, which we call negations. Write as many negations as you can. One of them often gives you a good idea of how to get out of a loop.

Description of attempts Negations First I make one triangle I don't make only one triangle, but

several

Then I make another with a common side I don't make another, but two more

This leaves only one match

Is there anything I have not mentioned?


Description of attempts Negations I am trying to construct it on a table, etc.

Let me not try constructing it on a table.

Let me try constructing it not in a plane

Let me try constructing it in space (crucial idea)

The solution is to make a tetrahedron in three dimensions.

Note: Using this method you might succeed in getting rid of a common hesitation to change

procedures. Previously learned associations prevent our minds from grasping the correct procedure to a solution and unnecessary constraints are often introduced which keep us on the wrong track. Perhaps the word "triangle," a planar figure, constrains our brain to the plane.

After some practice you shall be able to go through this process without repeating all the steps. You will find the right idea or the right procedure quickly by writing down only a few pieces of information. This method can be summarized as follows: Analyze fully what you have already done, then express negations in all possible ways. This increases flexibility and helps you change procedures after you have gotten rid of all unnecessary constraints.

2.2. ANOTHER FORM OF THE METHOD Take one sentence from the left side, e.g., "I' m trying to construct it on a table." Now repeat the sentence inserting the word "not" between words. Some of the resulting

sentences may have no meaning. However, others may reveal an important idea. Let us do it. I am not trying to construct it on a table I am trying not to construct it on a table I am trying to not construct it on a table

The Method of Getting Out of Loops 23

I am trying to construct it not on a table I am trying to construct it on not a table Obviously, some of these sentences contain the crucial idea. The probability of solving

the problem is now very good.

Example 2.2.1. Problem. You are given four separate pieces of chain, each having three links (see Figure

2.1). It costs 2 cents to open and 3 cents to close a link. All links are closed in the beginning. Your goal is to join all 12 links into a single circle (see Figure 2.2) at a cost of no more than 15 cents.

Figure 2.1. Open Chains. Figure 2.2. Closed chains.

Try for a while to solve it before reading on. Here is our scheme. Description of attempts Negations First, I take one of the pieces

I don't take one piece

I take two pieces

I open its two links

I don't open its two links

I open one link or I open its three links

I join them with the end links of two other pieces

I don't join them with the end links


Description of attempts Negations Then I open the end links of the connected piece

I don't open the end links. I open other links

I join them with the end links of the last piece

I don't join them with the end links, etc.

I failed

Then I start with another piece I obviously enter a loop

Don't hesitate. Write in the right-hand column as many negations as you can. Look now at the negations. Do you recognize a good idea towards the solution? Here it is. Open three links of one piece It costs 2 cents x 3 = 6 cents. Then we connect two end links with each opened link and pay 3 cents x 3 = 9 cents. The total cost is 6 cents + 9 cents = 15 cents.

Alternative Form of the Method Take sentences from the left column and insert the word "not" or "don't" between words.

Such, sentences as the following will be created I do not open its two links I open not its two links I join them with not the end links of two other pieces, etc. The crucial idea is once more written down. All you have to do is to recognize it. It is easier to recognize than recall an idea. To solve

a problem you go through an often long process of generating and evaluating ideas. The present method takes you almost halfway this process. It almost solves the problem. If your mind is prepared, negations are followed by solutions.

Example 2.2.2.

The problem. Draw four straight lines through the three-by-three array of nine dots below without lifting your pencil. No segment may be drawn twice, i.e., never go backwards on the same line.


Stop reading and try to solve the problem. If you have not yet solved it, you may have produced a number of attempted solutions such as these

Go through the usual process to get out of the loop.

Attempts Negations I started from one dot and drew lines, but reached a dead end I started from another dot and moved in different directions, but again reached a dead end When I started from one dot, I always ended at another one etc.

Starting from one dot, I should continue in a different manner I should not start from one dot I should not continue through only one dot I should not start from a dot I should start or end a line outside the framework of dots

Do you now see the crucial idea?

Note: The statement of the problem did not oblige you to move from one dot to another. What

might have prevented your mind from reaching the correct procedure? Ask yourself while re-examining the problem and its solution.

Example 2.2.3.

The problem. Eliminate α from the equations

3 3 3x sinα y cosα c sinα cosα+ =


3 3 3x cosα y sinα c cos2α+ = Try for a while before reading on. A hint: If you try to eliminate α by expressing α or sinα or cosα in terms of x and

y, you will come to a dead end. Go on to apply the method by yourself.

Attempts Negations I try to express sinα in terms of x, y Let me not try to express sinα in terms

of x, y

I try to express cosα in terms of x, y, but the same difficulty arises

Let me not try with cosα either

I try combinations by trial and error, but again I come to a dead end, etc.

Let me try another transformation. Let me try to express not sinα Let me express x and y in terms of sinα , cosα

The previous equations are of the form

3 3ax by c+ = 3 3a x b y c′ ′ ′+ =

At this point we do not know where the new setting of the problem will lead, nor if it

produces anything useful. On the other hand, in the absence of other propositions, it would be unproductive to question this setting. We hope that the new form of the problem will lead us to the solution, which often happens. After solving this system, we simplify and finally we find

3 3 3

3 3 3

x c cos αy c sin α

⎫= ⎬= ⎭ which yield x c cosαy c sinα= ⋅= ⋅

Thus we arrive at two simple equations and the solution becomes obvious. Substitute

cosα = x/c into 2y c sinα c 1 cos α= ⋅ = ± − . Then ( )2 2 2y c 1 cos α= − , therefore, ( )( )22 2

2xy c 1 c= −

Alternatively use the identity 2 2sin α cos α 1+ = and substitute xcosαc

=


ysin αc

=

Discussion

In this problem, many people cannot free themselves of the unnecessary constraint of

expressing the variables to be eliminated in terms of the others, because they have done so in other problems. Free your mind from unnecessary constraints.

2.3. A LITTLE EXPERIMENT IN CLASS Chapter 5 of a Greek high school textbook of Euclidean geometry discusses the definition

and properties of parallel lines and gives theorems on the sum of the angles of a triangle and a convex polygon.

Chapter 6 gives definitions and theorems about parallelograms. After teaching chapter 5 to two groups of 20 students each, I gave one of them the

following problem.

2.3.1. In the Figure below, Angle A = Angle C, and Angle B = Angle D

Prove that AB is parallel to DC and AD is parallel to BC. A few students quickly arrived at the following simple proof

angle A = angle C angle B = angle D

A + B + C + D = 360o These equalities yield 2C + 2D = 360o or C + D = 180o By a well-known theorem about parallel lines we deduce that AD is parallel to BC. The same equalities yield o2C 2B 360+ = or oC B 180+ = . Hence AB is parallel to

DC.


Two weeks later, I gave the same problem to the other group, but I changed the phrasing a little, as they had entered the sixth chapter about parallelograms:

If angle A = angle C and angle B = angle D, prove that ABCD is a parallelogram. The students tried and tried, but they got nowhere although they were the better group in

mathematics. Then I asked for their notebooks, wishing to see what they were doing. They had all been

trapped in the same loop. That day we had proven several properties of parallelograms, e.g., if ABCD is a

parallelogram, then AB = DC, AD = BC and vice versa, the diagonals bisect each other, and so on.

We had used the same method to prove all the theorems. We drew diagonals, compared

triangles etc. Then all the students tried to apply the same method persistently although they had hit a dead end. They had been trapped into a loop because of their recent experience.

Then I suggested going back to chapter 5. In a few minutes a number of students had solved the problem.

If You Feel That You Are in a Loop

Make a list of opposite actions Do other possible ways exist? Negate the method you have been using Negate the form of the algebraic statement of the problem, if there is one, and then

continue, that is, make all possible transformations Refuse to look at a form from one particular angle Look at it from others, too Draw auxiliary lines – then more auxiliary lines Write down more details What other descriptions could have been made? Where else could you have started from? Provide an alternative interpretation Reinterpret each detail See what other people have to say about your problem If you are tired, leave the problem for a while


Leave it longer Do something else If it is late at night, forget it and go to bed Perhaps in the morning a fresh idea will occur to you If you are taking an examination, move on to another problem Come back to the initial problem later Perhaps you will be able to use a new approach now We can think of an almost infinite number of examples drawn from daily life. Take for

example the insistence of people or groups of people on particular ideas. The Ptolemaic system in astronomy, whereby the sun moves around the earth is such an example. The principles of the Ptolemaic system were rooted in people's minds for thousands of years because they were convenient. This was the case until Copernicus rejected the Ptolemaic system and replaced it with the one we know today. As a matter of fact, the Greek Aristarchus had proposed a similar system, but humanity did not immediately accept his ideas, as conditions had not matured sufficiently*. Another example is the change in the course of geometry brought about by the rejection of Euclid's axiom of parallels by Gauss, Riemann and Lobatchevsky, perhaps the most revolutionary negation in mathematics.

Physicists insisted on classical Newtonian physics despite its inability to explain certain phenomena. Physics had to await Einstein's revolutionary ideas for a change of course. He negated the absoluteness of space and time, and he also changed the study of gravity by using the principle of equivalence. This was one of the most daring negations ever made by man and its result was a complete overturn of certain scientific and philosophical theories.

Ironically Einstein himself was resistant to the new and revolutionary ideas of quantum theory and war broke out between him and supporters of the new theory. We see how human the weakness of using unnecessary constraints is. Patterns become easily ingrained in the human brain. Even giants of thought were unable to avoid it.

For this reason: If you get stuck with a problem of whatever kind If your research project bogs down If you feel unable to press forward despite the description you have made, make a careful analysis of what you have already done Classify your actions into similar groups, if any

Historians have noted that Copernicus knew the system of Aristarchus (see Will and Ariel Durant, “The Story of

Civilization, the Renaissance”, Vol. 6, Simon and Schuster, New York, 1957).


Finally, make all possible negations One of them will show you the correct way to the solution If it does not, don't give up Carry on and find more negations Don’t forget! There are always more patterns to be recognized than those you already

have found

2.4. THE METHOD OF GETTING OUT OF LOOPS MORE RIGOROUSLY Here is an example.

2.4.1. If a, b and c are rational numbers, and if + + =33a 4 b 2 c 0 , then prove a = b = c = 0

We write down possible attempts and negations. Attempts Negations

3 3a 4 b 2 c= − − hence ( ) ( )3 33 3a 4 b 2 c= − − , which leads to

a similar form

3 3b 2 a 4 c= − − yields ( ) ( )3 33 3b 2 a 4 c= − − , which also

leads to a similar form Finally, transfer terms from one side to the other and raise to a power There is a loop because all attempts are equivalent

Do not raise to powers after transferring terms Raise to powers without transferring terms Raise to powers after making a different transformation Follow a different procedure Since no more negations can be found, go back to descriptions, in the hope of discovering subtle points

I have 3a 4 and 3 2 , but ( )23 34 2= so I have the number 3 2 and its square ( )2

3 32 4=


The initial expression has the form 2ax bx c 0+ + = hence 2

3 b b 4ac22a

− ± −=

2b 4ac− cannot be a perfect square, as it would lead to 3 2 = rational

Now raising to the third power, we finally find 2b ac 0− = , 3 316a 4b 12abc 0+ − =

These equalities yield 3 316a 8b= , thus 3 2a b= or a b 0= = . Therefore c 0= too. Instead of this procedure, however, let us summarize the main thrust of the left column.

We Want to Eliminate an Irrational The above procedure has led us into a loop. Assume that you possess a computer program containing all your knowledge, theorems

and techniques. Then you give the machine the instruction: "Find and print all techniques about how to eliminate an irrational."

The following will be printed (1) Raise to powers. (2) Raise to powers after transformations. (3) Use two simultaneous equations. (This technique is commonly used to eliminate

variables.) Of these methods, 1 and 2 led us into a loop. We then go to 3 and look for two

simultaneous equations. We have already enriched our problem space with the previous trials and can immediately spot two such equations, e.g.

3 3a 4 b 2 c 0+ + =

3 3 3 2 23 32b c a 4 3c a 4 3ca 2 2− = + ⋅ + +

by eliminating 3 2 we obtain ( )4 3 3 2 2 3 232b c b 4ba 6c a 4 6ca 3c ab− − − + = − + ,

which yields

4 3 3 2 22b c b 4ba 6c a 0− − − + = (1)

3 26ca 3c ab 0− + = (2)


By combining (1) and (2) (simplifying, substituting etc.) we obtain

6 3 3 6c 2a c 8a 0− − = , from which 3 3c 4a= or 3 3c 2a= − .

These equalities yield 3c 4 a= ⋅ or 3c 2 a= − ⋅ hence c a 0= = . Consequently b 0= , too. Note: By looking for two simultaneous equations, we can find alternative ways to solve the

problem, e.g., multiply

3 3a 4 b 2 c 0+ + = (3)

by 3 3 332 : a 8 b 4 c 2 0+ + =

or 3 3b 4 c 2 2a 0+ + = (4) Applying Crammer's rule to (3), (4) we get

23

2

c b2a c c 2ab4a b ac bb c

−− − += = − , 3 4 = rational which is a contradiction.

Thus 2c 2ab 0− + = and 2ac b 0− = .

Eliminating a we get 3 3c 2b= or 3c b 2= which yields c = b = 0, hence a 0= , too.

How did multiplication by 3 2 occur?

Describing rationals and irrationals, we see that 3 34 2 2⋅ = , which gives the crucial idea.

If knowledge or techniques are missing, then the chance to solve the problem is small.

The method here helps to select the right existing technique. To create a new technique by means of this method would be a lot more difficult.

Chapter 3

THE SPIRAL METHOD FOR SOLVING PROBLEMS

3.1. THE THREE BASIC METHODS The spiral method is a flexible technique for solving difficult problems. It is a

combination of the description method, means-end analysis, and the method of getting out of loops. The description method (DSB) has already been presented in Chapter 1. Another important method called Means-end analysis (MEA), encourages you to find the difference between what is given and the goal, and then cover that distance. The method of getting out of loops (GOL) (Chapter 2) is employed whenever you go round in circles. Simple techniques illustrate the procedure and aid you in your quest for alternative ways.

When you get stuck with a problem, the spiral method recommends the DSB. If this does not lead to a solution, then try MEA. If the latter leads to a loop, move on with the GOL method. A flowchart below illustrates the idea (see Figure 3.1).

END

WRI TE SOL UTI ON

SOLV ED

GOL

L OOP

SOLV ED

MEA

SOLV ED

DSB

YES

YES

YESNO

NO

START

Figure 3.1.


Attention should be paid when using the DSB or MEA to avoid repetition of description or subgoals. Otherwise the spiral method will itself become a loop. Alternating DSB and MEA forms a spire, hence the method’s name.

The fine details of the procedure depend on your knowledge. All the above methods have been tested in class as well as via computer and their effectiveness has been proven. From personal experience I know that the spiral method led most students to a solution. Without it the students had difficulties of varying degrees or simply gave up.

Now let us see a few examples.

3.2. APPLICATIONS OF THE SPIRAL METHOD If an example is difficult, skip it. Go to the next one.

Example 3.2.1.

If α, b, c are the sides of triangle ABC, prove that + + < + +2 2 2α b c 2αb 2αc 2bc .

A

αB C

bc

Solution

Given Differences? Goal Α, b, c are sides Description? α b c< + b α c< + (1) c α b< +

2 2 2α b c 2bc< + + 2 2 2b α c 2αc< + + (3) 2 2 2c α b 2αb< + +

2α αb αc< + 2b αβ βc< + (4) 2c αc bc< +

What is the difference? − three inequalities on the left side, one on the right − different powers

Is there a way to eliminate differences? − Raise to the second power. Nothing useful − Is there another way to make

squares on the left side? Multiply by a, b, c respectively. Is there a difference now? By adding, we reach the goal

2 2 2α b c2αb 2αc 2bc+ +< + + (2)

Description?


Let us now present another method by giving an alternative description.

Use the equalities OA OB BA− = , OB OC CB− = , OC OA AC− = . Then

Given Goal

OA 4OB 2OC 0+ + = (1) 2 2 2BA CB AC= = (2)

OA OC 0⋅ =

Is there a difference now?

There is no B in the goal How can you eliminate it? Use geometry Draw two axes as in the figure below

The coordinates of A, are α α 3A ,2 2

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

Attention! If you use two axes, place them so that you are left with the smallest number of variables.

B (0,0) C (a,0)

O (x,y)

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠a a 3A ,2 2

Then,

α xα α 3 2OA x i y j2 2 α 3 y

2

⎡ ⎤−⎢ ⎥⎛ ⎞⎛ ⎞ ⎢ ⎥= − + − =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎢ ⎥⎝ ⎠ −⎢ ⎥⎣ ⎦

Now, equality (1) becomes The goal becomes α x

0 x α x2 4 2 00 y yα 3 y

2

⎡ ⎤−⎢ ⎥ − −⎡ ⎤ ⎡ ⎤⎢ ⎥ + + =⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎣ ⎦ ⎣ ⎦−⎢ ⎥⎣ ⎦ (3)

( )α α 3x α x y y 02 2

⎛ ⎞⎛ ⎞− − − − =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ (4)

Is there a difference now?

Equation (3) yields: 5αx14

= , α 3y14

=

Substitute into (4) and find 0 0= , therefore the goal is true. Conclusion: A fast and efficient solution requires a minimum number of variables.

The Spiral Method for Solving Problems 39

Example 3.2.3. Consider a polynomial ( )f x with real coefficients. The remainder of the division

( ) −f x : x α is r and the remainder of ( ) −f x : x b ( ≠α b ) is also r. Prove that the

remainder of ( ) ( )( )− −f x : x α x b is r, too.

Proof

Since the divisions ( )f x r : x α− − and ( )f x r : x b− − leave the same remainder 0,

and since x α− , x b− have no other polynomial common factor, we conclude that

( ) ( )( )f x r : x α x b− − − leaves also a remainder of 0.

Hence ( ) ( )( ) ( )f x x α x b q x r= − − + , which proves that r is the remainder of

( ) ( )( )f x : x α x b− − .

Experiments have shown that students do not follow this line of short reasoning. They usually find long solutions or just give up. Let us now apply the spiral method.

Given Goal Remainder of ( ) ( )f x : x α− is r

Remainder of ( ) ( )f x : x b− is r

α b≠ Using DSB we obtain

( ) ( ) ( )f x x α q x r= − +

( ) ( ) ( )1f x x b q x r= − +

remainder of ( ) ( )( )f x : x α x b− − is r

( ) ( )( ) ( )2f x x α x b q x r= − − +

All students readily transform the problem to the set of equations above. Using MEA we obtain Difference? The product ( )( )x α x b− − is on the right hand column.

Subgoal: Eliminate this difference. Multiply the equalities of the first column. This is a dead end. Alternative? Multiply after

transforming. A dead end again. Obviously trying to form the product ( )( )x α x b− − traps

you in a loop. Getting out of loops (GOL) Look for alternatives. Use a textbook and look for a theorem that might connect most

elements of the left column with those of the right one. Here is one such theorem:


If two polynomials ( x α− , x b− in our case) divide another polynomial f(x) (the remainder is 0) and if the two polynomials have no other polynomial common factor, then the division of f(x) by their product leaves also a remainder of 0.

MEA Here we should have 0 instead of r. To remove r we write: ( ) ( ) ( )f x x α q x r= − + ,

hence, ( ) ( ) ( )f x r x α q x− = − , and the way to the solution opens up.

Note: We often manage to solve problems which we consider easy. Sometimes problems, when

first posed, may appear painfully difficult. In fact no original problem can be solved in a straightforward manner. To find solutions to new problems requires hard work and persistence. Inspiration plays a role but without hard work it can do next to nothing.

Inspiration is not an idea that comes suddenly out of nowhere, rather it might be something similar to what we call pattern recognition. So, the recognition of the best and most fruitful idea out of many trials and combinations of ideas could be called inspiration.

The scientist employs a number of strategies to attack a problem but inspiration may show him the best one. Inspiration favors only a prepared mind.

Inspiration often comes after a gestation period. A problem may appear to be forgotten when suddenly a crucial idea comes to mind. This probably suggests that trials are also made by our mind unconsciously. Our mind seems to work in the background, even if we appear to have consciously forgotten about the problem.

Example 3.2.4.

If ( )4 4 4 2 2 2 2 2 2α b c α b b c c α 2αbc α b c+ + + + + = + + , prove that α b c= = .

If a teacher gives this problem to students and they cannot solve it, the following dialogue

may ensue.

Teacher. “Go through your books and try to find a similar problem.” Student. “I see many problems of the form “if A, then prove B,” but I am not sure what to

do.” Teacher. “Can you classify them according to special characteristics?”

Students, assisted by the teacher, make the following classification.

Class A. Problems of the form: If 1 2E E= , then prove 1 2G G= , where E1, E2, G1, G2 are algebraic expressions

Class B. If 1 2E E= , then prove that 1 2G G= or 3 4G G=

Class C. If 1 2E E= and 3 4E E= , then prove that 1 2G G=

Class D. If 1 2E E= , then prove that 1 2G G= and 3 4G G=

The Spiral Method for Solving Problems 41

Class E. If 1 2E E= and 3 4E E= , then prove that 1 2G G= or 3 4G G= Teacher. “Now, do you recognize anything? Which class does your problem belong to?” Student. “I think class D.” Teacher. “OK. Do you know a strategy for this class of problems? Give an example.” Student. “If 2 2 2α b c αb αc bc 0+ + − − − = , prove that α b c= = .” Teacher. “How is it proved?” Student. “ 2 2 2α b c αb αc bc 0+ + − − − = yields

( ) ( ) ( )2 2 2α b b c c α 0− + + + − = , so α b 0− = , b c 0− = , etc.”

Teacher. “Can you apply a similar strategy? Locate differences between this problem and

the problem you have to solve.” Student. “I should transform the given expression into a sum of squares.” Teacher. “What is the difference?” Student. “The equality

4 4 4 2 2 2 2 2 2α b c α b b c c α+ + + + + = ( )2αbc α b c= + + must take the form ( ) ( ) ( )2 2 2. . . 0+ + =

What are the differences? There is a number of them but I can easily reduce them by

consecutive transformations. I finally reach the form

( ) ( ) ( )2 2 22 2 2α bc b cα c αb 0− + − + − =

This yields 2α bc 0− = , 2b cα 0− = , 2c αb 0− = It is now easy to prove thatα b c= = . There are many ways to prove it, e.g., add the

three equations above to obtain

2 2 2α b c αb αc bc 0+ + − − − = , hence, α b c= = .” Note: Grouping problems together enhances the power of the description method. Now

you will be in a position to attack whole classes of problems; not just individual ones.

Chapter 4

OTHER METHODS In this chapter additional methods are introduced, which come in handy when solving

problems. They are presented with illustrative examples where needed. Some of these methods have been applied over the ages and are now classics. Others are often applied in class by experienced teachers, while some have been tried in my classes.

4.1. THE IMPORTANCE OF GOOD NOTATION Use appropriate symbols. Present the information of the problem in simple form or in the

form of a precise diagram, if necessary. The symbols should be simple and kept to a minimum. For instance, if you use a to

denote the section AB of a straight line, and x to describe AC, where C is a point between A and B, then do not use y to denote CB but rather a – x.

The use of simple symbols, such as using a instead of AB, and keeping their number

small makes for greater clarity. This method is in fact a good description, which improves your chances of drawing conclusions and solving the problem.

Example 4.1.1. ABCD is a rectangle, BD a diagonal, AA΄ perpendicular to BD, CC΄ perpendicular to

BD, E the middle of AB and F the middle of BC. Prove that lines A΄E and C΄F are perpendicular.


A E

D

A΄

C΄

F

B

C I once gave this problem to my students. They tried for a while without success. Then I

urged them to use simple symbols for the angles and keep them to a minimum. Here is what some students did: They started from any angle, say DAA΄ which was called x.

Then angle πA AE x2

′ = −

angle π πEA C x x2 2

⎛ ⎞′ ′ = − − =⎜ ⎟⎝ ⎠

because EA EA′= ,

so angle EAA angle EAΑ′ ′= angle EBA x′ = because EA EB′ = . Similarly

angle πC BF x2

′ = −

angle C CF x′ = angle FC C x′ =

angle πFC B x2

′ = −

A E

D

A´

C´F

B

C

G

x

x

xπ πx x2 2- -

x

x

Other Methods 45

Therefore, two angles of the triangle GA΄C΄ sum up to π πx x2 2

+ − = , hence G is a right angle.

This and other solutions were found in a few minutes by several students. Note: Using good notation takes you half way to a solution. If you have to use more than one symbol (but not too many), use a, b, c, rather than a1, a2,

a3, as the former are easier for the brain to manipulate than the latter. Pay attention to your goal. Often it is easier to start from the goal describing and drawing

conclusions, than it is to start from the data. Check your solution step by step. Make sure that each step you take is correct. People

make mistakes because they do not check on their progress at each stage. It is absolutely essential to do this when solving a problem. Even great men make mistakes. “To err is human” is a well-known dictum verified time and again in the history of science.

4.2. LOOKING FOR SIMILAR PROBLEMS When confronted with a problem try to recall whether you have ever solved similar

problems – it should be the first step in your description. It has been said that the most difficult problems are those which do not remind us of any

that we have ever seen before. We realize how true this is if we recall that the mind cannot make leaps. But in what ways could we relate one problem to another? Let us see some examples.

Example 4.2.1. Prove that if a, b and c are real numbers, then ( )2 2 2 2 2 2a b b c b c abc a b c+ + ≥ + + .

If you happen to know the solution to the following problem "If x, y and z are real numbers, prove that 2 2 2x y z xy xz yz+ + ≥ + + ," then by

substituting ab x= , ac y= , bc z= , you realize that both problems are equivalent. In the

end, 2 2 2x y z xy xz yz+ + ≥ + + can be expressed as

( ) ( ) ( )2 2 2x y y z x z 0− + − + − ≥ .

Note: If you recall a problem bearing some similarity to yours, then use the spiral

method: Locate differences, etc.


Example 4.2.2. If you are given an equality, as a b c 0+ + = , and you are asked to prove another

equality, for example 3 3 3a b c 3abc+ + = , then start from a b c 0+ + = . Solve for a: = − −a b c . Substitute this value for a in + + = ⇒3 3 3a b c 3abc

( ) ( )⇒ − − + + = − −3 3 3b c b c 3 b c bc

⇒− − − − + + = − −3 2 2 3 3 3 2 2b 3b c 3bc c b c 3b c 3bc

⇒ − − = − −2 2 2 23b c 3bc 3b c 3bc

This technique can be used in all similar problems. Now solve the following

If a b c 0+ + = , prove that 2 2 2

2 2 2

a b c 12a bc 2b ca 2c ab

+ + =+ + +

.

This problem is similar to the one above.

( )a b c= − + then substitute:

( )( ) ( ) ( )

2 2 2

2 2 2

b c b c 12b c b c 2c b c b2 b c bc

++ + =

− + − ++ +.

add ( )

( )( ) ( )( ) ( )( )

2 2 2b c b c2b c b 2c b c 2b c b c b 2c

++ + =

+ + − + − − +

( )( )( ) ( )( )( )

2 3 2 2 3b c b 2b c 2c b c2b c b 2c b c b 2c 2b c

+ + − −= + =

+ + − + +

( )( )( )

( )( )( )( )( )

2 2 2b c b bc c 2bcb c2b c b 2c b c b 2c 2b c

− + + ++= + =

+ + − + +

( )( )( )( )( )( )

2 2 2b c b 2c2b 2c bc 4bc 1b 2c 2b c b 2c 2b c

+ ++ + += = =

+ + + + .

Other Methods 47

Example 4.2.3. If x, y and z are positive real numbers such that + + =x y z 1 (1), prove that

≤2 1x yz64

(2).

Solution

From x y z 1+ + = we have y 1 x z= − − . We substitute this value for y into (2):

( )2 2 3 2 21 1x 1 x z z x z x z x z64 64

− − ≤ ⇒ − − ≤

( )2 2 2 3 1x z x x z 064

⇒ − + − − ≤

Now, what does it remind us of? The inequality 2αz bz c 0+ + ≤ is always true if α 0<

and 2D b yαc 0= − ≤ .

Here we have 2α x 0= − < and ( )22 3 21D x x 4 x 064

= − − ⋅ ≤

( )24 21x x 1 x 016

⇔ − + − ≤

( )22 1x 1 x 016

⇔ − − ≤

( )22 1x 1 x16

⇔ − ≤

( ) 1x 1 x4

⇔ − ≤ because ( )x 1 x 0− >

2 1x x 04

⇔ − + − ≤

21x 02

⎛ ⎞⇔ − − ≤⎜ ⎟⎝ ⎠

which is always true.

Note: Ask yourself if the problem you have to solve is a special case of another familiar

problem. Naturally, everyone knows such cases, as for example, theorems and general for-mulas.


Example 4.2.4 You are given the two simultaneous equations:

a b c d e 8+ + + + =

2 2 2 2 2a b c d e 16+ + + + = Find the maximum value of e.

Solution Using the method of description we notice that the left sides of the equations resemble an

inequality familiar from elementary algebra:

( )( ) ( )22 2 2 2 2 21 2 n 1 2 n 1 1 n na a ... a b b ... b a b ... a b+ + + + + + ≥ + + (1)

Note: Watch for hidden equalities or inequalities. For example in 2 2a b 2a 2b 2 0+ + + + = ,

two identities ( ) ( )2 2a 1 b 1 0+ + + = are hidden.

In the above problem, inequality (1) is hidden. Now we have to remove the differences between our problem and (1). To eliminate a, b, c, d, from the given equations, we write

a b c d 8 e+ + + = − , 2 2 2 2 2a b c d 16 e+ + + = − . As a special case we set 1 2 3 4a a a a 1= = = = and 1b a= , 2b b= , 3b c= , 4b d= .

Hence, by substituting we have ( ) ( )2 2 2 2 2a b c d a b c d 4+ + + ≤ + + + ⋅

thus ( ) ( )2 28 e 16 e 4− ≤ − ⋅ . We solve this inequality 25e 16e 0− ≤ , hence 160 e5

≤ ≤ .

The maximum value of e is 165

, which can be attained by choosing a b c d= = = .

Note: You will often find it easier to solve a problem if you first examine special cases.

Example 4.2.5. Let f(x) be a polynomial and ( ) ( )f x 1 f x+ = for every real number x. Prove that f(x)

is constant.

Other Methods 49

Solution Examine special cases. First, replace x with the values 0, 1, 2, …

( ) ( )( ) ( )( ) ( )

( ) ( )

f 0 f 1f 1 f 2f 2 f 3..................f n f n 1

===

= +

Therefore, ( ) ( ) ( ) ( )f 0 f 1 f 2 ... f n ....= = = =

What conclusions can you draw now? Here is one useful piece of information. The

equation ( ) ( )f x f 0= has an infinite number of roots 1, 2, 3, … Consequently, the

polynomial ( ) ( )f x f 0− is identically equal to zero. Therefore ( ) ( )f x f 0= is a constant.

Indeed it is known from polynomial theory that if a polynomial has one more root than its degree, then the polynomial is identically equal to zero.

Example 4.2.6. If A, B, C are sets, prove that ( ) ( ) ( )A B C A B A C∩ ∪ = ∩ ∪ ∩ .

Solution

We shall prove that, if an element x belongs to ( )A B C∩ ∪ , it also belongs to

( ) ( )A B A C∩ ∪ ∩ and vice versa.

If ( )x A B C∈ ∩ ∪ , then x A∈ and x B C∈ ∪ .

From x B C∈ ∪ , we derive the following special cases. If x B∈ , then x A B∈ ∩ , hence ( ) ( )x A B A C∈ ∩ ∪ ∩

If x C∈ , then x A C∈ ∩ , hence ( ) ( )x A B A C∈ ∩ ∪ ∩

If x B∈ , and x C∈ then ( ) ( )x A B A C∈ ∩ ∪ ∩

The first part of the solution has been established. Conversely, if

( ) ( )x A B A C∈ ∩ ∪ ∩ we examine three cases.

(i) If x A B∈ ∩ , then x A∈ and x B∈ , hence ( )x A B C∈ ∩ ∪

(ii) If x A C∈ ∩ , then x A∈ and x C∈ , hence ( )x A B C∈ ∩ ∪

(iii) If x A B∈ ∩ and x A C∈ ∩ , then ( )x A B C∈ ∩ ∪


Thus we have proved that every element of one side belongs to the other side too and, consequently, both sides are equal.

Many problems in set theory can be solved by using the method of special cases.

Generalizing the Problem An attempt to generalize our problem helps sometimes because it forces us to make a

description of critical properties and thus some implicit information is made explicit. In mathematics, generalizing is a very powerful procedure and it often leads to new discoveries. Many branches of mathematics were born by generalizing previous ones. At first sight it seems to be painful to learn a new branch of mathematics, but if we manage to do it, we will be rewarded. Difficult old problems, are solved now quite easily.

Example 4.2.7. Find a plane that passes through a given line (e) and bisects the volume of a given regular

octahedron, ABCDEF. (BCDE is a square and A, F are symmetrical about it. K is the center of symmetry. This is a regular octahedron).

K

(e)A

D

FK

B J

H

L

G E

I

C

Solution Generalize the problem. You are to construct a plane through a given straight line which

divides the volume of a solid object with a center of symmetry into two equal parts. The solution is rather clear. The plane passes through the center of symmetry K and (e)

and is described by GHIJKL.

Other Methods 51

4.3. INTENSIVE USE OF THE DATA Perhaps, it would not be out of place here to point out a familiar instruction. Check to see

whether you have used all the information given. This instruction alone can often lead to the solution of a problem. Here is an example.

Example 4.3.1. Express the volume of the pyramid of the figure below in terms of a and b, if it is given

that A΄ΟΒ΄ is a right angle, AA΄ and ΒΒ΄ are perpendicular to the plane Α΄ΟΒ΄ and ΟΑΒ is an equilateral triangle.

A

B

E

O

l k

a

b

l-k

yh

x

a

a

B΄ Α΄b

Solution

The formula for the volume of a pyramid is 1V3

= . ( )ABB A h′ ′ = ( )b k h1

3 2+

,

where h is the altitude of the pyramid (the vertical distance of O from the plane ABB΄A΄) and (ΑΒΒ΄Α΄) is the area of ABB΄A΄.

Let us summarize the procedure to the solution. We write down the formula for the volume and then, using all the data, we find all the relevant equations. Finally we eliminate the auxiliary variables.

Let us look at the problem analytically. How can we use the given fact that BB΄O is a

right triangle? From the theorem of Pythagoras,

2 2 2y a+ = . (1)

k−


AOA΄ is also right-angled. Pythagoras again: 2 2 2x k a+ = . (2) Similarly from Β΄ΟΑ΄ 2 2 2x y b+ = . (3)

h is the altitude of Β΄ΟΑ.΄ Do we know any relevant theorems? Yes y x b h⋅ = ⋅ . (4)

How can we use the given fact that AA΄ and ΒΒ΄ are perpendicular to the same plane?

Remember that Α΄Β΄ΒΒ is a trapezoid with two right angles, and altitude A΄Β΄= b (this is so because by considering the right triangles ΒΟΒ΄ and ΑΟΑ΄ we have already used in full the information that ΒΒ΄ and ΑΑ΄ are perpendicular to the same plane).

What relationships can we extract from a trapezoid with two right angles? We draw a line AE parallel to Α΄Β΄. The right triangle AEB yields

( )2 2 2k b a− + =

(5)

Have we used the fact that OAB is equilateral? Yes, by denoting each of its sides by a.

AE parallel to Α΄Β΄ is a common auxiliary line in trapezoids. But if we had not known this, how could we ever have thought of it? Which piece of the given information have we not used yet? At first glance we see that we have not yet used AB = a. Can we find some relationship for Α΄Β΄? Can it be the side of a triangle?

We now have five independent equations with five unknowns. We do not need the figure any longer. Algebra takes over from this point. The unknowns are k, , h, x, y. There are several methods to solve those equations. Eliminate x, y so that the resulting equations contain only k, , h.

From (1), (2), (3) we have

2 2 2 2k 2a b+ = − (6)

(1), (2), (4) yield b h⋅ = ( ) ( )2 2 2 2yx a a k= − − =

( )4 2 2 2 2 2a k a k= − + + (7)

(5) yields 2 2 2 2k 2 k b a+ − + = , because of (6)

then 2 2 2 22a b 2k b a− − + = thus 2ak

2= (8)

(6), (7), (8) yield ( )4

4 2 2 2 abh a 2a b a4

= − − + =

4 4 2 2 4 4 2 24a 8a 4a b a 3a 4a b4 4

− + + − += =

Other Methods 53

(6) yields ( )2 2 2k 2k 2a b+ − = − hence ( )2

2 2 2 ak 2a b 22

+ = − + =

2 23a b= − thus 2 2k 3a b+ = − Substituting we obtain

4 2 22 21 3a 4a bV 3a b

6 4− +

= ⋅ − .

4.4. USE ONLY A PART OF YOUR DATA If you cannot find the solution to a problem, work only with part of your data and see

how you may transform the problem. This method is described in Polya's book "How to solve it". Here are two examples.

Example 4.4.1. In a triangle ABC find a point P such that the angles BAP, CBP and ACP are equal.

Solution

B C

A

x

x

xP

Try without reading further for a while, working with part of what is requested. Suppose

that only ACP and CBP are equal. Then there is clearly an infinite number of solutions. What conclusions can you draw about the various positions of P? Where does a description lead you? Let angle PBC = x, and PCB = c – x . Use the symbol c for angle ACB.

Then o oBPC 180 x c x 180 c= − − + = − which means that the locus of P is an arc, which can easily be drawn. Also angle oAPC 180 A= − , so we can draw another arc. Then the point at which the two arcs intersect inside the triangle is the point P you seek.


Example 4.4.2. You are given a triangle ABC and three straight lines d1, d2, d3 . Find three points A1, B1,

C1 on BC, AC, AB respectively, so that the sides of the triangle A1 B1 C1 will be parallel to d1, d2 , d3 , i.e., A1 B1 parallel to d2 , A1 C1 parallel to d3 and B1 C1 parallel to d1.

Solution

A

B

C

B ΄

B ΄́

B1

C΄

C ΄́

C1

A1

A΄

A ΄́

d3

d2

d1

Focus on part of what is asked. Assume that only the vertices C΄ and B΄ will be on the

sides of the triangle ABC. Then Α΄ may not be on BC. It is then easy to construct several triangles Α΄Β΄C΄, with sides parallel to d1, d2 and d3. Now you observe how the vertices A΄, Α΄΄, … evolve.

At first glance you see that Α, Α΄, Α΄΄ … lie on a straight line. So you may guess that the intersection of this straight line with BC gives the required point A1. This is the crucial idea. Now the solution follows. Take a point C΄ at random on the side AB (see figure) and draw a parallel from there to d1. This way, you obtain Β΄. From C΄ draw a parallel to d3 and from B΄ another to d2, thus constructing the triangle A΄Β΄C΄. Now draw a straight line passing through A, A΄. BC and AA΄ intersect at point A1. From this point draw parallels to d2 and d3 and obtain B1 and C1. The proof that B1C1 is parallel to d1 is easily reached from the similar triangles 1 1AA B , AA B′ ′ . By similarity

Other Methods 55

1 1

AA ABAA AB

′ ′= (1)

Also from the similar triangles AA1C1 and

AA΄C΄, AA ACAA AC1 1

′ ′= (2)

Hence (1), (2) yield 1 1

AB ACAB AC

′ ′= , thus B C′ ′ is parallel to B1C1 .

Note: It is sometimes important to restate the problem. Then state it again, differently.

This helps to obtain a better description.

4.5. WORKING BACKWARDS Sometimes it is better to start with the goal and look for a preceding step.

Example 4.5.1.

If >x 0 , >y 0 , + =x y 1 (1) prove that ⎛ ⎞⎛ ⎞+ + + ≥⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

221 1 25x yx y 2

(2).

Solution

The first side of (2) reminds us of the inequality ( )2

2 2 a ba b

2+

+ ≥ .

So, a preceding step of (2) might be:

“Prove that

21 1x yx y 25

2 2

⎛ ⎞+ + +⎜ ⎟

⎝ ⎠ ≥ ”

Now, we substitute 2

1 1x y 1 1 25x y

⎛ ⎞+ = ⇒ + + ≥⎜ ⎟

⎝ ⎠

1 1 x y 11 5 4 4x y xy xy

+⇒ + + ≥ ⇒ ≥ ⇒ ≥

1xy4

⇒ ≤

This problem is similar to problem 4.2.3.


From (1) y 1 x⇒ = − .

So ( ) 21 1 1xy x 1 x x x 04 4 4

≤ ⇔ − ≤ ⇔ − + − ≤

21x 02

⎛ ⎞⇔ − − ≤⎜ ⎟⎝ ⎠

, which is true.

Note: You will find more examples in chapter 5.

4.6. THE METHOD OF CONTRADICTION This is a well known method which can be applied to an immense number of problems

and theorems.

Example 4.6.1. Three people A, B, C, some of whom always tell the truth and some always lie, make the

following statements: A said that B and C are liars; B denied being a liar; C said that B is a liar. How many of them are lying and how many are telling the truth?

Solution Examine all the possibilities. First introduce a suitable notation.

A B C 1. all liars L L L 2. all truth-tellers T T T 3. two truth-tellers T T L 4. two truth-tellers T L T 5. two truth-tellers L T T 6. two liars L L T 7. two liars T L L 8. two liars L T L

T here are then eight different possibilities. Assume (1) is true. A lies when he says that B

and C are liars, which means that at least one of them is a truth-teller. Therefore, (1) is false. Similarly we reject possibilities 2, 3, 4, 5 and 7, and the only ones we are left with are 6

and 8, that is, that there are two liars, any two, and one truth-teller.

Other Methods 57

Example 4.6.2. Prove that the polynomial 5 2x x 2x 1− + − has no negative roots.

Solution Of course you cannot try all the negative values, since they are infinite. Suppose 1r 0< ,

is a root, hence 5 21 1 1r r 2r 1 0− + − = .

How can you obtain a contradiction? By description the expression can be transformed to

( )251 1r r 1 0− − = , thus ( )25

1 1r r 1= − .

But the left hand side is negative and the right positive, which is a contradiction.

Another Solution 5

1r 0< , 21r 0− < , 12r 0< , 1 0− < , hence 5 2

1 1 1r 2r 2r 1 0− + − < which is a contradiction.

Note: Try the contradiction method in such cases as: (a) You need to determine whether the goal is inconsistent with the given information. (b) Exactly one of several alternative goals can be derived from the data. The method

also applies when you see statements of the kind “no”, “never”, “unique”, “ a b≠ ”.

Example 4.6.3. Given a polynomial ( ) n

n n 1 n 1f x a x a x ... a− −= + + + with integer coefficients and if f(0)

and f(1) are odd integers, prove that the equation f(x) = 0 has no integer roots.

Solution Assume that an integer a is a root of f(x), which implies that f(x) is divisible by

( ) ( ) ( )x a thus, f x x a Q x− = −

where Q(x) is the quotient of the division. Hence

( ) ( )f 0 aQ 0= − and ( ) ( ) ( )f 1 1 a Q 1= − .

If you now make a description of what you have obtained, you run into a contradiction.

f(0) is an odd number. f(0) must be divisible by –a and by Q(0). So -a as well as Q(0) must be odd. f(1) is odd.


It must be divisible by 1– a and Q(1). So 1 - a and Q(1) are odd. Finally, -a , 1 - a , Q(0) , Q(1) are odd. However, -a , 1 - a are successive integers. They cannot both be odd. If -a is odd, then -a + 1 is even. The contradiction is obvious -a , -a + 1 are both odd. -a , -a + 1 are successive integers and cannot both be odd.

Example 4.6.4. Prove that the prime numbers are infinite (prime numbers are those divisible only by 1

and themselves such as 2, 3, 5, 7, 11, 13, …).

Solution Assume they are finite. Call the last prime P. Consider now the number

( )K 2 3 5 P 1= ⋅ ⋅ ⋅⋅⋅ + . But K is greater than P, so K is not a prime. Consequently K must be

divisible by a prime e. But if you divide, you obtain (remember that all primes are included in the product 2.3.5…P). K 2 3 5 e P 1e e e

⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅ ⋅= +

integer = integer + fraction

This is a contradiction. You have then proved that the series of prime numbers is infinite. Note: The above is a famous proof due to Euclid.

4.7. DEFINING A PROCEDURE Often a problem solver may decide from the start or after several trials about the

correctness of a procedure. If a procedure appears to be promising, continue by eliminating the difference between what is given and the goal. If, on the other hand, a procedure leads you nowhere, then change it and follow a new one patiently. This variation of the spiral method may be applied to any problem in a variety of fields, even everyday life.

Note: A procedure is not an algorithm but rather a systematic trial-and-error process.

Other Methods 59

Example 4.7.1.

ABC is a triangle, D a point on AB, and E a point on AC, such that AD ECDB EA

= .

If K, L and M are the middle points of AB, AC, and DE respectively, prove that K, L and

M lie on the same straight line.

Solution You need to prove that KM rKL= , for r real.

A

B C

E

LK

D

M

To find an expression of the type KM rKL= , try a more specific procedure. Express

KM and KL in terms of vectors, e.g., BA and AC . Follow this procedure carefully. Start reducing differences.

DE DA AEKM KD DM KD KD2 2

+= + = + = + (1)

Try to express all vectors in terms of AB and AC .

If AD aDB

= , then AD aAB a 1

=+

, hence ( )a 1 AD aAB+ =

thus aAD AB

a 1=

+ or

aDA BAa 1

=+

(2)

Similarly EC aEA

= yields AC a 1EA

= +

1EA AC

a 1=

+ hence

1AE ACa 1

=+

(3)


( )BA a 1 aKD KA DA BA BA2 a 1 2 a 1

−= − = − =

+ + (4)

(1), (2), (3) and (4) yield

( ) ( ) ( )1 a a 1KM BA BA AC

2 a 1 2 a 1 2 a 1−

= + + =+ + +

( ) ( )1 BA aBA aBA AC2 a 1

= − + + =+

( ) ( )1 1 BA AC 1BA AC KL2 a 1 a 1 2 a 1

+= + = ⋅ =

+ + +.

Thus, 1KM KL

a 1=

+.

4.8. SETTING SUBGOALS If your problem is difficult, or if a large number of trials is necessary to find the solution,

divide it into simpler problems. In other words, set subgoals. If you cannot find another method to reach the solution, and if you have time, then subgoals should be attempted, to increase your chances of arriving at the solution. Examples follow.

Example 4.8.1. There are 20 boys and 15 girls in a class. If we have to send a delegation of 6 boys and 5

girls to a reception, in how many ways can that delegation be made up? Try to solve the problem yourself before reading on. Hint: First reach the following subgoals: find out how many ways there are to select the 6

boys, and then the 5 girls.

Solution We can select the 6 boys in as many ways as there are combinations of six out of twenty;

that is:

( )n 20r 6

n! 20! 15 16 17 20C Cr! n r ! 6! 14! 1 2 3 6

⋅ ⋅ ⋅ ⋅ ⋅= ⇒ = = =

− ⋅ ⋅ ⋅ ⋅ ⋅

17 120 19 38,760= ⋅ ⋅ =

Other Methods 61

Similarly for the girls we obtain

155

15!C 3,0035! 10!

= =

But any given combination of boys can be attached to all the combinations of girls. Thus,

the total number of combinations is equal to the product 38,760 3,003 116,396,280× = .

Example 4.8.2. Seven men and two children want to cross a river in a small boat which is capable of

carrying only one man or two children. How many times must the boat cross the river to get them all across?

Let a subgoal be “one man crosses the river and the boat returns.” Four trips are needed for a man to go across and for the boat to return; first the two

children cross and one of them brings the boat back. Then a man takes the boat and goes across and the second child returns the boat. After four trips the first man has crossed and the children are back where they started. Since there are seven men, the boat needs to cross the river 7 4 28× = times, plus one final trip to bring across the children, in all 29 trips.

Setting a subgoal (subproblem), which in this case is repeated seven times, leads to a quick route to the solution of the original problem.

4.9. WORD PROBLEMS In order to solve a word problem, one should translate it into mathematical terms. This is

common practice in science. I have observed that many students are confused by complex word problems. They often

do not know how to start and proceed from there. The answer to these questions emerges through the following examples.

Example 4.9.1. Peter is 24 years old. His age is twice the age that John was when Peter was the age

John is today. What is John’s age?

Solution This problem puzzles students a lot. To tackle it let us follow an alternative use of short

notes of the description method. We start by writing short lines once more. Under each line we try to put symbols, so that

an equation may emerge little by little.


Peter is 24 years old. His age is twice

( )24 : 2 12=

the age that John was

( )12↓

when Peter was

( )x↓

the age John is today

( )x↓

Look carefully at these notes. Write them in a brief form Peter → today → 24 John → then → 12 Peter → then → x John → today → x Here we have ages: 24 and x (now); x and 12 (then). Does an equation exist? Is there any relation of ages then and now? Is there anything invariant? After some thought, the difference of ages emerges as invariant. Therefore, 24 x x 12 2x 36 x 18− = − ⇒ = ⇒ = . Now let us see another method: Reduction to a unit.

Other Methods 63

Example 4.9.2. A farmer hires workers to pick 12,500 kg of tomatoes in one day. Every worker picks 125

kg per hour and he is paid $8 per hour. Also the farmer hires a foreman who is paid $12 per hour. Additionally, the farmer pays $12 per employee to cover their social security.

How many workers must he hire to minimize his cost?

Solution A very useful method is to find first what happens to one worker (reduction to a unit). Suppose that x workers are hired.

They will work for a total of12500 100

125= hours.

One worker will work for 100

x hours. (Reduction to a unit)

Cost = (daily wor ker's wages) (foreman 's wages) (social security)+ + =

100100 8 12 (x 1) 12x

⋅ + ⋅ + + ⋅

Now to compute x, for which the cost is minimized we differentiate the cost function

with respect to x and set the result equal to zero

2

12000 12 0 x 10x

+ − = ⇒ =

The second derivative of the cost function with respect to x is 32400x− , which is positive for x=10, therefore, the cost is minimized at x=10.

Example 4.9.3. The daily production of 40 oil wells is 16,000 barrels. This production decreases by 8

barrels for every new well. Find how many new wells must be opened to maximize oil production.

Solution

Reduction to a unit. 40 wells produce 16,000 barrels Therefore, one well produces 400 barrels Suppose that there are x new wells

Now every well will produce 400 8x− barrels


So, total production is ( )( ) 240 x 400 8x 8x 80x 16,000+ − = − + + Maximization now is a simple exercise. We differentiate with respect to x and set the

result equal to zero.

16 80 0x− + = or

80 516

x = =

The second derivative of the production function with respect to x is 16 0− < , therefore, 5x = maximizes production.

Example 4.9.4. Why do the wheels of a car moving forward in movies often appear to be turning

backwards on the screen?

Solution Some information here is implicit. Let us describe what turning forward mean How does the eye get this impression? Draw a figure taking a point A on the wheel (reduction to a unit) Draw the wheel in successive positions while moving forward

I Α

Α

Α

Α

successive frames moving forward The wheel appears to turn forward Where should A be if we are given the impression of the wheel turning backward?

II Α

Α

Α

Α

successive frames moving backward

Other Methods 65

The wheel appears to turn backward Is there any more implicit information? How does the projector operate? Frames succeed each other at a given speed If successive frames are as in I, we get the impression that the wheel turns forward But if they are as in II, we have the impression that the wheel is turning backward How can this happen? The answer now becomes rather transparent. If the speed of the frames in the film is

greater than the speed of revolution of the wheel, then we observe a backward turning. Indeed in the time one frame succeeds another, the wheels perform less than one revolution, say 3/4 of a revolution, as illustrated in II.

Note: Drawing a figure and introducing suitable notation often is the best approach to description. Figures help us draw inferences. Another important point is that if your problem has many facets and you are about to become confused, choose only one item and draw inferences from that, e.g., you choose only one point A of the wheel and start describing.

Example 4.9.5.

The Mathematicians’ Forum Nine mathematicians meet at a conference and it becomes apparent that at least two out

of every three of them speak a common language. If each mathematician can speak at most three languages, show that there are at least three of them who speak the same language.

Hint

Symbolize the mathematicians by the numbers 1, 2, …, 9, and examine the case of mathematician 1. Assume that 1 and 9 speak no common language. Also assume that 1 speaks three languages. Take all possible combinations of the numbers 1 and 9 taken 3 at a time.

1,9,2 1,9,3 1,9,4 1,9,5 1,9,6 1,9,7 1,9,8

Solution A common language may exist between 1 and 2, 1 and 3 etc., or between 9 and 2, 9 and 3

etc., but never between 1 and 9. Since there are seven cases, either 1 or 9 will be repeated at least four times. Let us assume that 1 is repeated four times, say (1, 2), (1, 3), (1, 4), (1, 5) speak a common language. If 1 spoke the same language as all of them, the problem would be solved. If 1 spoke a different language than everyone else, then he would speak four


languages which is a contradiction. By necessity he speaks the same language as at least two others and the problem is solved (see figure).

Thus 1, 4, and 5 speak the same language. All possible cases are examined in a similar way.

Note: The crucial idea is that you cannot map three items onto four other items. One of

the items will pair with two others. This is the so called pigeonhole principle.

Example 4.9.6. A tank with capacity 100 m3 contains 20 kg pure water. We start pouring into the tank a

liquid solution, which contains 1 kg/m3 salt at a rate of 3 m3/min. The contents of the tank are stirred thoroughly so that the content is homogeneous at all times. Liquid is removed from the tank through an outlet at the bottom at a rate of 2 m3/min. How much salt will the tank contain when it is filled up?

Solution

We first compute the time it takes for the tank to fill up

( ) ( )100 20 : 3 2 80 :1 80− − = = min

Then we proceed with reduction to a unit. Suppose that the initial time is 0 and at some time 0t ≠ there are Q kg of salt in the tank.

Other Methods 67

How does Q change with time?

dQ amount added amount going outdt 1 min

−= =

( )Q3 1 2

20 3t 2t= ⋅ − ⋅

+ −, because at time t the tank contains

( ) 320 3t 2t 20 t m+ − = + of liquid and Q kg of salt.

If the tank contains 1 3m of liquid (reduction to a unit), then it contains Q

20 t+ kg of salt.

At 2 m3 it contains Q 2

20 t⋅

+ kg of salt, or

dQ Q3 2dt 20 t

= − ⋅+

This is a differential equation of the form

( ) ( ) ( ) ( )Q t P t Q t K t′ + = (1)

whose solution is

( ) ( )( ) ( ( ) )

P t dt P t dtQ t e c e K t dt

−∫ ∫= + ∫ (2)

where 2( )

20P t

t=

+, ( ) 3K t = and

22( ) ln(20 )20

P t dt dt tt

= = ++∫ ∫ , therefore,

2( ) ln(20 )

2

1(20 )

P t dt te et

− − +∫ = =+

Finally ( )2ln 202

1( ) [ ](20 )

tQ t c e dtt

+= + =+ ∫

2 32 2

1 1[ 3(20 ) ] [ (20 ) ](20 ) (20 )

c t dt c tt t

= + + = + ++ +∫


We find c

32

10 [ 20 ]20

c= + or 320c = −

Hence, 3 32

1( ) [ 20 (20 ) ](20 )

Q t tt

= − + ++

For t=80, 3 33

1(80) [ 20 100 ] 99.2100

Q = − + = kg of salt.

The Counting Rule

Let ( )n s be the number of all the different ways that a certain selection can be made.

If we can break the selection process into a finite sequence of selections 1 2 KS ,S , ..., S

so that the number of making any selection rS does not depend upon the results of the

previous selections, then ( ) ( ) ( ) ( )1 2 Kn S n S n S ...n S=

DERIVATION OF BASIC FORMULAS OF COMBINATORIAL ANALYSIS

Example 4.9.7. Let S be a set of n distinct objects. In how many ways can we select a subset of r objects, if we are interested in the order

they are arranged?

Solution STEP 1: Select from n objects one object for the first position. This can be done in n ways STEP 2: Select from the n 1− remaining objects one object for the second position. This can be done in n 1− ways

We continue this process until finally we select the rth object from the

( )n r 1 n r 1− − = − + objects.

This last selection can be done in n r 1− + ways So, the number of arrangements is ( )( ) ( )n n 1 n 2 ... n r 1− − − +

Note: We call each subset a permutation of r objects written as

( )( ) ( ) ( )nr

n!P n n 1 n 2 ... n r 1n r !

= − − − + =−

.

Other Methods 69

Example 4.9.8. In how many ways can we select r objects from a set of n distinct objects, without regard

for the order? We call such sets unordered combinations.

Solution Let x be the number of unordered combinations. Select one combination of r objects. The number of permutations of these r objects is r

rP r!= . Hence the number of all permutations of the x combinations is r!x . But this must be

equal to ( ) ( )nrP n n 1 ... n r 1= − − + .

Therefore, ( )( ) ( )r!x n n 1 n 2 ... n r 1= − − − + ⇒

( )( ) ( )( )

n n 1 n 2 ... n r 1 n!xr! r n r !

− − − +⇒ = =

−

4.10. PROBLEM SOLVING BY REDUCTION TO VERY SMALL OR

INFINITESIMAL ITEMS

The Method of Exhaustion This is an ingenious idea due to Eudoxus, who was a student at Plato’s Academy. This

idea was later used by Archimedes to compute areas and volumes of spherical and conical objects. Newton and Leibniz, among others, perfected it in the context of calculus.

Examples (Here you must have basic knowledge of calculus, otherwise skip it) How to find the volume of a solid by rotation (quick answer)


Case 1

O x

y

a b

If a planar region is rotated about the x-axis, then a solid is generated. Its volume is

( )b

2

a

π f x dx⎡ ⎤⎣ ⎦∫ .

Why?

O x

y

a bΔx

f(x)

Consider the rectangle of width Δx and height f(x) revolving about the x-axis. It produces

a cylinder whose volume is ( )( )2π f x Δx . Therefore ( )( )2

dV π f x dx= , for dx very

small.

By integrating we obtain ( )( )b

2

a

V π f x dx= ∫ .

Example 4.10.1.

Case 1 y cos x=

Other Methods 71

O x

y

π6 4

π

π π4 4

2

π π6 6

1 cos2xV πcos xdx π dx2

+= = =∫ ∫

π π4 4

π π6 6

1 sin 2x π π 3x 12 4 24 4 2

⎛ ⎞⎡ ⎤ ⎡ ⎤= + = + −⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎜ ⎟⎣ ⎦ ⎣ ⎦ ⎝ ⎠

Case 2

x

y

O

πx dy2

f(x)

If the planar region is rotated about the y-axis, then the volume is a

2

b

V πx dy= ∫ ,

( )y f x= .

Example 4.10.2.

( ) 2f x x= , a = 2 22 2 2

2

0 0 0

yV πx dy πydy π 2π2⎤

= = = =⎥⎦

∫ ∫


Case 3 If the planar region is rotated about a line not on the boundary:

x

y

y=x2

O

x=-2

( )2 2dV π x 2 dy π 2 dy= + − ⋅

( )a

2 2

0

V π x 2 π 2 dy⎡ ⎤= + − ⋅⎣ ⎦∫

Case 4

Volume of a torus

A solid torus is generated by rotating the disc 2 2x y 4+ = about the line x 9= . Let us take a thin rectangle of width dy

Other Methods 73

x

y

O

9+√4-y2

9–√4-y2

dy

The length of the rectangle is 22 4 y−

Then ( ) ( )2 22 2 2dV π 9 4 y dy π 9 4 y dy π 4 9 4 y dy= + − − − − = ⋅ ⋅ −

By using symmetry, 2

2

0

V 2 36π 4 y dy= −∫

Substitute y 2sin u= y 0 u 0= ⇒ =

πy 2 u2

= ⇒ =

dy 2cosudu=

π π2 2

2 2

0 0

V 72π 4 4sin u 2cos udu 72π 4 cos udu⇒ = ⋅ − = ⋅ =∫ ∫

ππ2 2 2

00

1 cos 2u y sin 2y288π du 288π 72π2 2 4

+ ⎡ ⎤= = + =⎢ ⎥⎣ ⎦∫

Area of a Surface by Revolution Find the area of the surface generated by rotating the arc 3y x= from (0, 0) to (1, 1)

about the x-axis.


Solution

General formula: b

a

Area 2π radius of revolution ds= ⋅∫

The proof of this formula is obtained as previously. We take a small segment on the

tangent line of length ( )( )2ds 1 f x dx′= + .

This segment generates approximately a cylinder of radius ( )f x .

The area of the cylinder is ( )2π f x ds , etc.

x

y

O

In our case 1

3 4

0

s 2πx 1 9x dx= +∫ .

Substitute 4u 1 9x= + or 4 21 9x u+ =

3 3 2u36x dx 2udu x dx du36

= ⇒ =

for x 0 u 1= ⇒ =

for x 1 u 10= ⇒ =

Therefore ( )1010 3 3

1 1

2u 1 u 1 π πs 2π du π π 10 10 10 136 9 3 27 27 27

⎤= ⋅ = ⋅ = − = −⎥

⎦∫

4.11. MATRIX ALGEBRA PROBLEMS

Some Problem Solving Techniques We remind you of the following:


This is a “system” If we substract, then 3AB 3BA 0 AB BA− = ⇒ =

4.11.2 If 2 23B A= (1) 4BA 3AB I 0− + = (2) , then prove that 3A 3B= − .

Solution Premultiply (2) by B

24B A 3BAB B 0− + = Using (1) we obtain 34A 9BAB 3B 0− + = (3) Postmultiply (2) by B

24BAB 3AB B 0− + = Because of (1)

34BAB A B 0− + = (4) We eliminate BAB from (3) and (4) and obtain 3A 3B= −

4.11.3. If 4 4A B= , 3 3BA AB= , and if 3 3A B+ is invertible, then prove that A B= .

Solution Since 3 3A B+ has an inverse, there exists and inverse matrix X such

that ( )3 3A B X I+ = .

Multiply by A, then ( )4 3A AB X A+ =

Multiply by B, then ( )3 4BA B X B+ =

Hence A B= 2. How to prove AB BA= (if this is the goal of a problem)

Method 1 Use the previous method of making a system.

Method 2 Factorize the given expression to obtain the form ( )( ) I⋅ ⋅ =

Other Methods 77

4.11.4. If kA mB nI+ = , *k , m, n∈ , show that AB BA= .

Solution Premultiply and postmultiply the equation above by A. Two expressions result

2kA mAB nA+ = 2kA mBΑ nA+ = Substract the last two equations AB BA=

4.11.5. If AB A B 0+ + = (1), prove that AB BA= .

Solution Start with AB A B 0+ + = and add the unit matrix on both sides. AB A B I I+ + + = Factorizing yields ( ) ( )A B I B I I+ + + =

Hence, ( )( )A I B I I+ + =

This means that ( )( )B I A I I+ + =

Which yields BA B A I I+ + + = And finally, BA B A 0+ + = (2) From (1) and (2) AB BA=

4.11.6. If 2 22B A AB 2BA 3I+ + + = (1), prove that AB BA= .

Solution Factorizing (1) yields

( ) ( )2 22B A AB 2BA 2B B A A A B+ + + = + + + =

( )( )2B A B A 3I= + + =

Hence, ( )( )B A 2B A 3I+ + = (2)

from (1) and (2) we deduce that AB BA= .


4.11.7.If ( ) ( )n2 A I 3 B I+ = − (1) prove that AB BA= .

Solution

Premultiply and postmultiply by (A I)+ The resulting equations are

( ) ( )( )n 12 A I 3 A I B I++ = + − (2)

( ) ( )( )n 12 A I 3 B I A I++ = − + (3)

Hence, ( )( ) ( )( )3 A I B I 3 B I A I AB BA+ − = − + ⇒ =

3. Use of the spiral method

4.11.8. If 2A A I 0+ + = , prove that the determinant nA I 3 A− = ±

(A is a n n× matrix)

Solution We are given the equation 2A A I 0+ + = Given the identity

( )( )3 2A I A A I A I− = + + −

it can be proven that 3A I 0− =

The goal is to prove nA I 3 A− = ±

Taking a backward step from the goal yields2 nA I 3 A− =

Manipulating 2A A I 0+ + = gives ( ) ( )2 2A I 3A 0 A I 3A− + = ⇒ − = −

Hence, ( )2 nA I 3 A− = −

Using the equation 3A I 0− = gives

33A I A 1 A 1= ⇒ = ⇒ =

Hence ( )2 nA I 3− = − . It is sufficient to prove than n is even. This comes from

( ) 2n3 A I 0− = − ≥ .

Other Methods 79

4.11.9. If 7A I= , 2AB BA= , then show that 3B A BA=

Solution To eliminate differences, start with 3A BA and gradually move A to the right, using the

data

3 2 2 2 2 3 3A BA A ABA A BA A A BA AABA= = = = =

2 3 5 2 5 7ABA A ABA BA A BA B= = = = . 4. A method for diagonal matrices In order to prove that a matrix A is diagonal, it is often sufficient to prove that A κI= .

4.11.10. If Α Β 2Ι+ = , 2 2Α Β 3Ι+ = , show that 3 3Α Β+ is diagonal

Solution The equality Α Β 2Ι+ = , from technique (1) yields ΑΒ ΒΑ= . The factors 2 2Α Β+ and 3 3Α Β+ remind us of the identities

( )22 2Α Β Α Β 2ΑΒ 4Ι 2ΑΒ+ = + − = − so 14Ι 2ΑΒ 3Ι ΑΒ Ι2

− = ⇔ =

and ( )( ) ( )3 3 2 2Α Β Α Β Α ΑΒ Β 2Ι 3Ι ΑΒ+ = + − + = − =12Ι 3Ι Ι 5Ι2

⎛ ⎞− =⎜ ⎟⎝ ⎠

4.11.11. If 3 2Α Α 2Ι 0+ − = and Α Ι− is invertible, prove that A4 is diagonal

Solution

Factorizing 3 2Α Α 2Ι 0+ − = gives

( )( )3 2 2Α Α 2Ι A I A A 2I 0− − = − + + =

Α Ι− is invertible

Multiplying by ( ) 1Α Ι −− yields 2Α 2Α 2Ι 0+ + =

Therefore, ( ) ( )22 4 2A 2A 2I A 4 A I 4 A 2A I= − − ⇒ = + = + + =

( ) ( )24 A 2A 2I I 4 0 I 4I= + + − = − = −

Chapter 5

TWO MODELS FOR TEACHING MATHEMATICS AND PROBLEM SOLVING

The following two problems tormented me for years.

Problem 1 Every year I observed that some of my students have great difficulties to understand and

retain in their memory long proofs or short proofs with difficult steps.

Problem 2 A similar problem is that of students who take the difficult countrywide university

entrance examination in Greece. Beyond problem solving techniques, they have to memorize the solutions to over 700 problems. How could I help them do it?

Cognitive psychology provided me with excellent ideas to solve these two problems. It has been found that people learn faster if at the beginning of a chapter the teacher provides a summary using small and informative sentences which are numbered using the numbers 1, 2, etc. Sentences should be spaced out.

5.1. A MODEL FOR TEACHING THEOREMS, PROBLEMS WITH LONG PROOFS OR SOLUTIONS

I started applying a very simple technique to teach long proofs. The effectiveness of this

technique is great. It makes the class pleasant, provides the opportunity to exchange ideas, and enhances the quality of teaching. The method goes as follows:

(1) Study the proof carefully and find its main points or steps. These should be at least

two and no more than five. (2) The main points amount to a summary of the whole procedure and should be

recorded in a simple, brief and pleasant form.


(3) Each point should be numbered and there should be blank spaces between them. (4) After this brief procedure is understood, the remaining details should be filled in

slowly by the teacher, preferably with the participation of the students. The method is illustrated with two examples. The first is a theorem from number theory.

Ιts proof often daunts students because it is long and difficult. The second is a proof from geometry, which also often daunts the average student

although it is very brief.

Example 5.1.1. If a and b are integers, b 0≠ , prove that there are two unique integers q and r such that

a qb r= + , 0 r b≤ < .

Solution

The following proof is usually given in textbooks.

Case I α 0≥ and b 0> Let A be the set of all integers of the form a bx− , where x is an integer such that

0a bx− ≥ Since 1b ≥ , then ab a≥ , hence 0a ab a a+ ≥ + ≥ If x a= − , the inequality 0a ab+ ≥ implies that a ab+ is an element of A, so A is

not the null set According to the axiom "Every nonempty set A of nonnegative integers contains a unique

element which is less than any other element in A," there is a least element r in A. Since r A∈ , there is an integer q such that α bq r− = , from which α bq r= + , 0 r≤ We shall now prove that r b< Assume that r b≥ , then r b 0− ≥ Since ( ) ( 1)r b a qb b a q b− = − − = − + , r – b belongs to A This is a contradiction since r – b and r belong to A, r is the least element of A and

r b r− < Therefore r b< , and we have proved that a bq r= + , 0 r b≤ < Let us now prove that q and r are unique Assume that there are two numbers q΄, r΄ such that a bq r′ ′= + , 0 r b′≤ < Without loss

of generality we assume that q q′ ≤

Since a bq r= + , bq r bq r′ ′+ = + , which yields ( )b q q r r′ ′− = − (1)

The inequalities 0 r≤ and r b′ < yield r b r′ < + or r r b′ − < , hence ( )b q q b′− <

We deduce that q q 1′− < and q q 0′− ≥ , hence q q 0′− = or q q′ =

Two Models for Teaching Mathematics and Problem Solving 83

From q q′= and relation (1) we obtain r r′ = Case I of the theorem has been proved.

Case II α < 0 , b > 0 A similar procedure can be followed after proving that a ba− belongs to A.

Case III α is any integer, b < 0 There are two unique integers q and r such that α b q r= + , 0 r b≤ <

But b b= − , so ( )α b q r b q r= − ⋅ + = − + .

Now let us apply the method of summaries for case I. We write the following summary on the blackboard: (1) Prove that there are two integers q and r such that α bq r− = . Use the axiom "Every

non-empty…" (2) Prove that r b< . Use the method of contradiction.

(3) Prove that q and r are unique. Assume that two numbers q΄ and r΄ exist such that a bq r′ ′= + , r b′ < . Subtract ( )b q q r r′ ′− = − (contradiction method).

(4) Prove that ( )b q q b′− < using the inequalities r b< , r b′ < .

The student is then urged to study these summaries carefully. A number of questions

arise about how to complete the proof. It is a great teaching method to make the students ask questions. The above summary is not necessarily the best but it illustrates the method.

Example 5.1.2.

Let p be a plane and xOy a right angle whose side Oy is parallel to p. Let OO΄, yy΄, and xx΄ be perpendicular straight lines from O, y, and x to p (see the figure below). Prove that x΄O΄y΄ is a right angle.

Solution

Since Oy is parallel to p, Oy is parallel to O΄y΄. ΟΟ΄ is perpendicular to p, therefore, it is perpendicular to O΄y΄. Since Oy is parallel to O΄y΄, Oy is perpendicular to OO΄. Oy is perpendicular to both Ox and OO΄, therefore it is perpendicular to their plane OO΄x΄x, which suggests that Oy is orthogonal to O΄x΄. O΄x΄ is perpendicular to the plane Oyy΄O΄ because it is perpendicular to OO΄ and orthogonal to Oy. Finally O΄x΄ is perpendicular to O΄y΄, since O΄y΄ is a line of Oyy΄Ο΄.


p

O΄

O

x

y

x΄

y΄

Here is a summary of this proof: 1. Oyy΄O΄ is a rectangle, hence Oy is perpendicular to OO΄. 2. Oy is perpendicular to plane OO΄X΄X, hence it is orthogonal to O΄x΄. 3. O΄x΄ is perpendicular to OO΄ and orthogonal to Oy, hence O΄x΄ is perpendicular to

plane OO΄y΄y.

5.2. A MODEL FOR MEMORIZING A LARGE NUMBER OF DIFFICULT PROBLEMS

How can a student remember the solutions of about 700 difficult problems? Here is the answer. Problems are not discrete entities but they share similarities. Solution

strategies and techniques may usually apply to 10 or 15 problems. It is then advisable to group problems together according to solution procedure. Thus, students learn about 70 techniques instead of 700 solutions.

If techniques are recorded in the style of summaries the time spent studying is minimized, the methods of chapters 1-4 are applied with greater effect, and mathematics is enjoyed to a fuller extent.

Grouping is one step towards tackling problems. One then proceeds by applying some of the methods already developed in previous chapters to obtain a detailed solution.

The teacher now has to work harder but the reward is greater. To illustrate the idea, problems of calculus are grouped together according to certain

common features. Teachers of course, can do their own grouping. This is just an example.


Example 5.3.2. If ( )( )fof x x= − , prove that:

(i) f is one-to-one (ii) f is neither increasing nor decreasing (iii) ( )f 0 0=

Solution

(i) The description method suggests that the definition of a function be used

If ( ) ( )1 2f x f x= , then ( )( ) ( )( )1 2f f x f f x= , hence 1 2x x− = − or 1 2x x=

(i) Use the contradiction method of chapter 4. If there is “not”, “neither”, “unique”, try contradiction. Suppose that fof is increasing Then 1 2x x< yields ( )( ) ( )( )1 2f f x f f x< hence 1 2x x− < − or 1 2x x> , which

is a contradiction Similarly it can be proven that fof is not decreasing.

(iii) Use the description method. Make intensive use of data. If x = 0, then ( )( )f f 0 0=

If x = f(0), then ( )( )( ) ( )f f f 0 f 0= −

By replacing ( )( )f f 0 in the second equality by 0 we find ( ) ( )f 0 f 0= − , or

( )2f 0 0=

therefore, ( )f 0 0=

Example 5.3.3. If ( ) ( )2003f x f x x 0+ + = (1), for every real number x, show that 1f − exists and find

its expression.

Solution Use the definition of a function STEP 1 Let ( ) ( )1 2f x f x=

Then ( ) ( )2003 20031 2f x f x=

From (1) we have ( ) ( )20031 1 1f x f x x 0+ + = and


5.4. GROUP 2. LIMITS

Case 1

How to evaluate limits of the form 00

=→ )x(g

)x(flimax

Solution

Eliminate the indeterminacy 00

using one of the following

(i) If f(x) and g(x) are polynomials, we factor them out and simplify )x(g)x(f

(ii) If there is a square root, multiply numerator and denominator by the conjugate as follows

Multiply: ba + by ba −

ba − by ba +

ba + by ba − iii) Try L’ Hopital’s rule

Example 5.4.1. Find the limits:

(i) 2

2x 0

x 1 1limx 16 4→

+ −

+ − (ii)

x 0

sin7xlimx 4 2→ + −

Hint: Follow instruction (i) or (ii).

Case 2 How to find )x(flim

ax→, if f(x) is included in an expression of the type

x alim(.) L→

= .

Solution

STEP 1 Let the given expression be (.) g(x)=

STEP 2 Solve the above equation for f(x) and find )x(flimax→


Example 5.4.2.

If x 1

2 f ( x ) 3lim 1x 1→

−= −

− find

x 1lim f ( x )→

.

Solution

STEP 1 Let 1

32−−

=x

)x(f)x(g , then 11

−=→

)x(glimx

STEP 2 Solve for f(x)

g(x)(x 1) 2f (x) 3− = − , or g(x)(x 1) 3 f (x)

2− +

= .

Therefore, 23

23111

231

11=

+−−=

+−=

→→

))(()x)(x(glim)x(flimxx

Example 5.4.3.

If x 1

f ( x ) 1lim 3x 1→

−=

−, find

2

2x 1

x f ( x ) 1limx 1→

−−

.

Solution

STEP 1 Denote )x(gx

)x(f=

−−1

1.

STEP 2 Solve for f(x), 11 +−= )x(g)x()x(f

STEP 3 Substitute, 1

111

12

22

2

2

−−+−

=−−

xx)x(g)x(x

x)x(fx

STEP 4 Expand the fraction into a sum of fractions

11

111

11

2

2

2

2

2

2

+−

−=

−−

+−

−x

)x(g)x(xxx

x)x(g)x(x

Finally, 251

231

11311

111 2

12

2

1=+=+

+⋅

=++

=−−

→→ x)x(gxlim

x)x(fxlim

xx

Note: The method above could be called the method of the four verbs

Denote Solve Substitute Expand


Example 5.4.4.

If x

f ( x )lim 8x→+∞

= and [ ]xlim f ( x ) 5x 2→+∞

− = , find 2x

f ( x ) 2xlimxf ( x ) 5x 4→+∞

+− +

Solution

The previous methods do not apply here. We may, on the other hand, incorporate the given limits into the limit we want to find.

xx)x(f

x)x(f

)x

x)x(f(x

)x

)x(f(x

x)x(xfx)x(f

45

2

45

2

452

2+−

+=

+−

+=

+−+

Finally, 50228

45

2=

++

=+−

+

+∞→

xx)x(f

x)x(f

limx

Case 3

How to find )x(flimax→

given that )x(g)x(f)x(h ≤≤

Solution

If L)x(glim)x(hlimaxax

==→→

, then L)x(flimax

=→

.

Example 5.4.5.

1. Find x 2lim f ( x )→

if 2x 4x 4 f ( x ) 2

x 2−

+ ≤ ≤ +−

.

2. Find x 0lim f ( x )→

if 3 4sin5x x xf ( x ) sin5x x− ≤ ≤ + , for all x R∈

Hint: Divide by x for x>0, and find limits. Then divide by x for x<0, and also find limits.

Case 4

How to find ( )( )x

f xlim

g x→±∞ or ( )

xlim f x→±∞

.


Solution Apply Bolzano's theorem (b). STEP 1 Clear fractions, if any. STEP 2 Bring everything to the left hand side. STEP 3 Denote the left hand side f(x). STEP 4 Prove that ( ) ( )f α f b 0⋅ < .

Note: If the interval (α, b) is not given, try your own numbers based on experience and

intuition.

Example 5.5.1.

Show that 20 8x 1 x 3 0x 2 x 3+ +

+ =− −

has at least one root in (2, 3).

Solution

STEP 1 Clear fractions: ( )( ) ( )( )20 8x 3 x 1 x 2 x 3 0− + + − + =

STEPS 2, 3 ( ) ( )( ) ( )( )20 8f x x 3 x 1 x 2 x 3= − + + − +

STEP 4 Try Bolzano's theorem

( ) ( ) ( ) ( ) ( ) ( )20 8 20f 2 2 3 2 1 2 2 2 3 2 1 0= − ⋅ + + − ⋅ + = − + <

( ) ( ) ( ) ( ) ( )20 8 8f 3 3 3 3 1 3 2 3 3 3 3 0= − ⋅ + + − ⋅ + = + >

from which ( ) ( )f 2 f 3 0⋅ < , hence there is at least one root in (2, 3).

Example 5.5.2. Let f and g be continuous functions in [α, b] such that ( ) ( )f α g α≤ and

( ) ( )f b g b≥ .

Show that there exists at least one number [ ]0x a,b∈ , such that ( ) ( )0 0f x g x= .

Solution

STEPS 2, 3 Let ( ) ( ) ( )h x f x g x= −

STEP 4 From ( ) ( ) ( )h α f α g α 0= − ≤ and ( ) ( ) ( )h b f b g b 0= − ≥ , we obtain

a) If ( ) ( ) ( )h α f α g α 0= − = , then one root is α, that is 0x α= .

b) If ( ) ( ) ( )h b f b g b 0= − = , then one root is b, that is 0x b= .


c) If ( ) ( )f α g α≠ and ( ) ( )f b g b≠ , then ( ) ( )h α h b 0⋅ < , hence,

by Bolzano's theorem there exists ( )0x α,b∈ , such that ( )0h x 0= , that is

( ) ( )0 0f x g x= .

Example 5.5.3.

If [ ] [ ][ ] [ ]

⎫→ ⎪⎬→ ⎪⎭

f : a, b a, bg : a, b a, b

, show that there exists a number

[ ] ( )( )∈ =0 0 0x a, b : f g x x .

There is also a number ( )( ) ( )( )+ =1 1 1 1x : f g x g f x 2x .

Solution

Backward step: Apply Bolzano’s theorem to

( ) ( )( )h x f g x x= − or ( ) ( )( ) ( )( )h x f g x g f x 2x= + − .

Note: We try to find hidden inequalities.

[ ] [ ]f : a, b a, b→ implies

( )a f x b≤ ≤ , hence, ( )f x a 0− ≥ and ( )f x b 0− ≤ .

Also [ ] [ ]g : a, b a, b→ implies

( )a g x b≤ ≤ , hence, ( )g x a 0− ≥ and ( )g x b 0− ≤

Apply now Bolzano’s theorem, etc.

Example 5.5.4.

If ( ) ( )[ ]= ⎫⎪

⎬→ ⎪⎭

f 0 f 2πf : 0, 2π

show that there exists a number

[ ] ( ) ( )∈ + =1 1 1x 0, π : f x π f x .

Solution Backward step: Apply Bolzano’s theorem to

( ) ( ) ( ) ( ) ( ) ( )h x f x π f x h 0 f π f 0= + − ⇒ = −

Also ( ) ( ) ( ) ( ) ( )h π f 2π f π f 0 f π= − = − , therefore ( ) ( )h 0 h π 0⋅ ≤ , etc.


In General If [ ]f : a, b → , ( ) ( )f a f b= , show that there exists a number x1, such that

( )1 1 1a b b ax 0, : f x f x

2 2+ −⎡ ⎤ ⎛ ⎞∈ + =⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠

.

Solution

We form the difference: ( ) ( )b ah x f x f x2−⎛ ⎞= + −⎜ ⎟

⎝ ⎠. Now

( ) ( )

( ) ( )

b ah a f f a2

a b a b a bh f b f f a f2 2 2

+ ⎫⎛ ⎞= −⎜ ⎟ ⎪⎝ ⎠ ⎪⇒⎬+ + +⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎪= − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎪⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎭

( ) ( )1a bh a h 0 h x 0

2+⎛ ⎞⇒ ⋅ ≤ ⇒ =⎜ ⎟

⎝ ⎠

Example 5.5.5. If < < <0 a 1 b , [ ] [ ]→f : a, b a, 1 , [ ] [ ]→g a, b 1, b , ( ) =f a a , ( ) =f b b ,

show that there exists a number ( ) ( ) =1 1 1 1x : f x g x x .

Solution

Backward step: Consider the function ( ) ( ) ( )h x f x g x x= − ⇒

( ) ( ) ( ) ( ) ( )( )h a f a g a a ag a a a g a 1 0= − = − = − ≥

(because of the hidden inequality ( )1 g a b≤ ≤ ).

Also ( ) ( ) ( ) ( ) ( )( )h b f b g b b bf b b b f b 1 0= − = − = − ≤

Problems on Continuity 1. Let a function →f : be continuous at 0, and a number ∈m such that

( ) ( ) ( )≥ + + 2xf x sin 7 x sin 11x m for ∈x . Find m and ( )f 0 .

2. Let a function →f : such that ( )− ≤ ≤ +2 21 x f x 1 x for ∈x . Show

that f is continuous at 0, and ( )

→

− += −

x 0

f x x 1 1limx 2

.


5.6. GROUP 4. DIFFERENTIATION

Problems Using the Definition

( )h 0

f (α h) f (α)f α limh→

+ −′ =

There are problems where you cannot apply the formulas of differentiation. In such cases

use the definition of a derivative.

Example 5.6.1.

Show that the function ( )3 1x sin , x 0f x x

0, x 0

⎧ ≠⎪= ⎨⎪ =⎩

is differentiable at 0. Find f΄(x).

Solution

Applying the definition of the derivative

h 0

f (h) f (0)limh→

−=

3

h 0

1h sinhlim

h→= 0

h1sinhlim 2

0h=

→

Also =′)x1sin(x3 ⎟

⎠⎞

⎜⎝⎛−+ 2

32

x1

x1cosx

x1sin3x .

Therefore, ( )2 1 13x sin x cos , x 0

f x 2 x0, x 0

⎧ − ≠⎪′ = ⎨⎪ =⎩

Example 5.6.2. If f(x) = P(x) x 2− is differentiable at 2 and P(x) is a polynomial, prove that P(2) = 0.

Solution

Apply the definition of the derivative

-h 0

P(2 h) 2 h 2 0lim

h→

+ + − −=

x 0

P(2 h) 2 h 2 0lim

h+→

+ + − − (1)


If -0h → , then h < 0, and h h= −

If +→ 0h , then h > 0, and hh =

Therefore, (1) yields P(2)(–1) = P(2).1 or –2P(2) = 0, thus P(2) = 0.

Example 5.6.3. If x3 – 3x + 2 ≤ f(x) ≤ x5 + 2x3 – 3x + 2 (1), show that f is differentiable at 0.

Solution

We have to find h 0

f (h) f (0)limh→

−

If x = 0, then 2f(0)2 ≤≤ , or f(0) = 2. The expression f(h) – f(0), or f(h) – 2 can be computed from (1)

3 5 3h 3h f (h) 2 h 2h 3h− ≤ − ≤ + −

If h > 0, then 4 2f (h) 2h 3 h 2h 3h−

− ≤ ≤ + − , hence h 0

f (h) 2lim 3h+→

−= −

If h < 0, then 4 2f (h) 2h 3 h 2h 3h−

− ≥ ≥ + − , hence h 0

f (h) 2lim 3h−→

−= −

Finally f΄(0) = –3.

Example 5.6.4. If f and g are differentiable functions such that f(0) = g(0) and f(x)g(x)=x2 show that

f΄(0)g΄(0) = 1.

Solution Describe the goal f΄(0)g΄(0) = 1.

Obviously h 0

f (h) f (0)f (0) limh→

−′ = and h 0

g(h) g(0)g (0) limh→

−′ =

What are f(0) and g(0) equal to? From f(x)g(x) = x2 for x = 0, we obtain f(0) . g(0) = 0. Since f(0) = g(0), we conclude that f(0) = g(0) = 0.


Therefore, it is sufficient to prove that h

f(h)lim0h→

1h

g(h)lim0h

=⋅→

This would be true if h

f(h) 1h

g(h)=⋅ .

But this last equality is true, since f(x)g(x) = x2 for all x.

Example 5.6.5. If f 2(x)+g2(x)=x2 for all x, show that: (i) f and g are continuous at 0 (ii) f and g are differentiable at 0 and [f΄(0)]2 + [g΄(0)]2 = 1

Solution (i) We have to prove that f(0)f(x)lim

0x=

→ and g(0)g(x)lim

0x=

→

From f

2(x) + g2(x) = x2 we obtain 22 x(x)f ≤ and 22 x(x)g ≤ or

xf(x) ≤ , xg(x) ≤ , hence 0f(x)lim0x

=→

and 0g(x)lim0x

=→

.

In order to find f(0), g(0) substitute x=0 f

2(0) + g2(0) = 0, which implies f(0) = 0, g(0) = 0 or f(0)f(x)lim

0x=

→, g(0)g(x)lim

0x=

→

(ii) A backward step is +⎥⎦⎤

⎢⎣⎡ −

→

2

0h hf(0)f(h)lim 1

hg(0)g(h)lim

2

0h=⎥⎦

⎤⎢⎣⎡ −

→

or +⎥⎦⎤

⎢⎣⎡

→

2

0h hf(h)lim 1

hg(h)lim

2

0h=⎥⎦

⎤⎢⎣⎡

→

Another backward step is +⎥⎦⎤

⎢⎣⎡

2

hf(h) 1

hg(h) 2

=⎥⎦⎤

⎢⎣⎡ , x ≠ 0

but the last equality is true because f2(x) + g2(x) = x2.

Example 5.6.6.

If f is continuous at 0 and 2x 0

f(x) xlim 1x→

−= , show that f is differentiable at 0.


Solution

We have to prove that h 0

f(h) f(0)limh→

− exists.

Let 2

f (x) x g(x)x−

= .

Solve for f(x): f(x) = x2g(x) + x, hence 0f(x)lim

0x=

→ or f(0) = 0, because

f is continuous at 0.

Make appropriate substitutions in h 0

f (h) f (0)limh→

−

2

h 0

h g(h) h 0limh→

+ −= [ ]1hg(h)lim

0h+

→ = 0 . 1 + 1 = 1

5.7. GROUP 5. PROBLEMS ON TANGENTS If m is the gradient of a line at point (x1,y1), then its equation is y – y1 = m(x – x1). If the line is tangent to a differentiable function f at x1, then m = f΄(x1) and the equation of the tangent line is y – f(x1) = f΄(x1)(x – x1). If the tangent is parallel to the y-axis, then the equation of the tangent is x = x1 (gradient does not exist).

In this case 1 1

h 0

f (x h) f (x )limh→

+ −= +∞ ∞−or

Τhe normal line at x1, has gradient m1

−)f(x

1

1

−= , f(x1) ≠ 0 .

Note: In problems on tangents, start as follows: «Let (a,b) be the common point of the tangent line and the curve...» Then make full use of the data of the problem.


Example 5.7.1.

Let 1f(x)x

= .

(i) find the equation of the tangent lines through point ( )1,1 2 .

(ii) if the tangent line of case (i) intersects the x-axis at A and the y-axis at B, show that the area of the triangle OAB is independent of α, where (a, b) is the common point of the tangent and f.

Solution

(i) Follow the instruction. Let (α,b) be the common point of the tangent and f, then the equation of the tangent is

( )2

1y b x αα

− = − − .

Use the data now. Since the tangent lines pass through ( )1,1 2 you can substitute x = 1,

21y = , also

1bα

= (from x1f(x) = )

Therefore, ( )2

1 1 1 1 α2 α α− = − − . Solve the equation

α2 –2α = -2 + 2α or α2 – 4α + 2 = 0, hence 4 16 8α 2 2

2± −

= = ± .

Thus there are two tangent lines for α 2 2= + and α 2 2= − .

(ii) For x=0 we find 2yα

= , hence 2OBα

= .

For y=0 we find x=2α, hence ΟΑ=2α.

Therefore the area of triangle OAB is 1 2 2α 22 α

⋅ = , which is independent of α.


Example 5.7.2. Find k so that the line y = 9x – 14 be a tangent to f(x) = x2 – 3kx+2

Solution Follow the instruction. Let (α, b) be the common point. Using all the data we obtain

b = 9α – 14 (the point (α, b) belongs to the line) (1) b = α2 – 3kα + 2 (the point (α, b) belongs to f) (2) f΄(α) = 9 or 2α – 3k = 9 (3)

Solve equations (1) - (3).

From (3) 2α 9k

3−

=

from (1) and (2) 9α – 14 = α2 – 3kα + 2 Substitute the value of k

9α – 14 = α2 – 3 2α 9

3−

α + 2, or

9α – 14 = α2 – 2α2 + 9α + 2, then

α2 = 16, thus α 4= ± , k =31

− , or k= 3

17−

5.8. GROUP 6. RATE OF CHANGE PROBLEMS

The rate of change of y with respect to x equals the derivative dxdy

.

Assume that you are given a problem where dxdy

is required. The following steps could

be helpful.

STEP 1 To find dxdy

, take a backward step by writing down an equation involving y

and x. Make full use of everything given. Use formulas from geometry, trigonometry, physics or anything related to your problem. Draw a figure, if necessary, and describe all details.


STEP 2 Differentiate the equation or equations found in STEP 1 with respect to x or any variable given in your problem or you may first eliminate unnecessary variables and then differentiate.

STEP 3 Solve for dxdy

or any other derivative asked in the problem.

Note: If a quantity decreases, the rate of change is negative. For example, if a balloon

looses air at a rate 4cm3/s, then its volume V(t) satisfies ( )V t 4′ = − .

Example 5.8.1. The semicircle in the figure below has diameter AB=20m. A point C is moving from A to

B with velocity v(t)=2m/s. DC is perpendicular to AB at C. Find the rate of change of the angle θ(t) at the time t when BD=AB/2.

D

A B

b

θ(t)

α

C

2t

Solution STEP 1 Draw conclusions from the data

Assume that point C started moving from A at time t = 0. Then AC = 2t and CB = 20 – 2t. θ(t) has two aspects: it is an angle of both triangles ACD and ABD.

Describe fully Triangle ACD Triangle ABD

sinθ(t)=b

DC sinθ(t)=

20α

cosθ(t)= b2t

cosθ(t)= 20b

tanθ(t)= 2t

DC tanθ(t)=

bα


Now take a look at all the above equations. Do not forget your goal. You wish to find an equation for θ(t). It seems best to multiply the equalities

( )

( )

2tcosθ tb

bcosθ t20

⎫= ⎪⎪⎬⎪

= ⎪⎭

or cos2θ(t) =202t

STEP 2 Differentiate now with respect to t

-2cosθ(t) · sinθ(t) · θ΄(t) =101

STEP 3 Now it is time to use the equality BD=AB/2. This means that θ(t) =π6

or 30°

Therefore, -2 ⋅⋅21

23

θ΄(t)1

10= , or θ΄(t) =

351

−

Example 5.8.2. Water is poured into a conic tank at a rate of 2m3/min. The radius of the base is 2m and

the height is 4m. Find the rate of change of h, where h is the water level when h = 2m.

B

h

4

AO 2

r

D C


2.Theorem 2 (Rolle' S Theorem) If f is continuous in [α, b] and differentiable in (α, b) and if ( ) ( )f α f b= , then there

exists at least one number x1 in (α, b) such that ( )1f x 0′ = .

3. Theorem 3 (Mean Value Theorem) If ( )f x is continuous in [α, b] and differentiable in (α, b), then there exists at least one

number x1 in (α, b) such that ( ) ( ) ( )1

f b f αf x

b α−

′ =−

, or ( ) ( ) ( ) ( )1f b f α b α f x′− = − .

O

y

xa bx1 x2

f(a)

f(b)

( ) ( ) ( )1

f b f αf x

b α−

′ =−

( ) ( ) ( )2

f b f αf x

b α−

′ =−

Geometrically this theorem says that there exists a tangent line that is parallel to the line

passing through (α, f(α)) and (b, f(b)) .

4. Theorem 4 If ( )f x 0′ > in an interval, then ( )f x is increasing.

If ( )f x 0′ < in an interval, then ( )f x is decreasing.

If ( ) ( )f x f b≤ , f(x) has a maximum at b.


Example 5.9.2.

If ( ) 2 x2g x x e 2ln x 1− − ≤ + for x > 0, ( ) eg 1 12

= + for g differentiable at 1, show

that ( ) eg 1 22

′ = + .

Solution

STEP 1 Clearly ( ) 2 x2g x x e 2ln x 1 0− − − − ≤ .

Let ( ) ( ) 2 xh x 2g x x e 2ln x 1= − − − − .

STEP 2 The number 1 is a root of ( )h x since

( ) ( ) eh 1 2g 1 1 e 0 1 2 2 1 e 1 02

= − − − − = + − − − =

therefore, ( ) ( )h x h 1≥ .

STEP 3 By Fermat's theorem

( )h 1 0′ = or ( ) 22g 1 2 e 0 01

′ − − − − = , hence ( ) eg 1 22

′ = + .

GROUP 2 How to prove inequalities of the type A < B < C .

Answer

STEP 1 Give the inequality the form ( ) ( )f b f α

k λb α−

< <−

Identify f(x), and apply the mean-value theorem ( ) ( ) ( )1

f b f αf x

b α−

′ =−

,

for some ( )1x α,b∈

STEP 2 Find ( )f x′ and try to prove that ( )1k f x λ′< <

STEP 3 Substitute )( 1xf ′ from Step 1 into the expression of Step 2.

Example 5.9.3.

If π0 α b2

< < < , show that 2 2

b α b αtanb tanαcos α cos b− −

< − < .


Solution

STEP 1 2 2

1 tan b tanα 1cos α b α cos b

−< <

− (1)

A suitable function is ( )f x tan x= .

By the mean-value theorem ( )1 21

1 tan b tanαf xcos x b α

−′ = =−

for some ),(1 bax ∈ STEP 2 Since cos2x is a decreasing function, we have:

If 1α x b< < then 2 2 21cos b cos x cos α< < or

2 2 21

1 1 1cos α cos x cos b

< <

STEP 3 Substitute 21

tan b tanα 1b α cos x−

=−

and obtain the goal (1).

GROUP 3 How to prove inequalities

Answer STEP 1 If the inequality has the form ( ) ( )f x g x≥ , then take the difference

( ) ( ) ( )h x f x g x= − .

STEP 2 Study the sign of h΄(x). Follow the solved problems below.

Example 5.9.4.

If πx 0,2

⎡ ⎞∈ ⎟⎢⎣ ⎠, prove that sin x tan x 2x+ ≥ .

Solution

STEP 1 Let ( )h x sin x tan x 2x= + − .

( ) 2

1h x cos x 2cos x

′ = + − .

STEP 2 Differentiate once more

( )3

4 3

2sin x cos x 2 cos xh x sin x sin xcos x cos x

−′′ = − + =


Since sin x 0≥ , 32 cos x 0− > and 3cos x 0> , then ( )h x 0′′ ≥ .

Thus, h΄ is increasing for: x 0≥ , ( ) ( )h x h 0 0′ ′≥ = .

Since ( )h x 0′ ≥ , h is increasing and x 0≥ yields

( ) ( )h x h 0 0≥ = .

GROUP 4

Characterization of Functions In problems involving functions which are not necessarily given explicitly one may use

the Bolzano, Rolle, or mean-value theorems to prove certain relations. To do that use the methods of Chapter 4, especially the backward procedure.

Example 5.9.5. If [ ] +→ *f : a,b R , [ ]→g : a,b R , are differentiable functions, and if

( ) ( ) ( ) ( )− = −ln f a ln f b g b g a , show that there is a number θ, ( )∈θ a,b , such that

( ) ( ) ( )′ ′+ =f θ f θ g θ 0 .

Solution

Use Rolle’s theorem. A possible backward step? Which function h(θ) was differentiated to obtain

( ) ( ) ( )f θ f θ g θ 0′ ′+ = ?

This expression becomes ( )( ) ( )f θ

g θ 0f θ′

′+ = .

Therefore, the required function is ( ) ( ) ( )h x ln f x g x= + .

Here ( ) ( )h a h b= , because ( ) ( ) ( ) ( )ln f a ln f b g b g a− = − .

Then Rolle’s theorem applies.

Example 5.9.6. Proof of the mean value theorem: Let f(x) be continuous in [a, b] and differentiable in (a, b). Then there exists at least one

( )∈c a,b such that ( ) ( ) ( ) ( )′− = −f b f a b a f c .


Solution Backward procedure: Find a function g, to which you apply Rolle’s theorem to obtain

( ) ( ) ( ) ( )f b f a b a f c′− = − .

Backward step: ( ) ( ) ( ) ( )f b f a b a f c 0′− − − = .

By inspection, assume that g has the form

( ) ( ) ( ) ( ) ( )g x f b f a x b a f x= − − −⎡ ⎤⎣ ⎦

Does Rolle’s condition ( ) ( )g a g b= hold?

( ) ( ) ( ) ( ) ( )g a f b f a a b a f a= − − − =⎡ ⎤⎣ ⎦

( ) ( ) ( ) ( ) ( ) ( )af b af a bf a af a af b bf a= − − + = −

( ) ( ) ( ) ( ) ( )g b f b f a b b a f b= − − − =⎡ ⎤⎣ ⎦

( ) ( ) ( ) ( ) ( ) ( )bf b bf a bf b af b af b bf a= − − + = −

Hence, ( ) ( )g a g b= and

g(x) is continuous in [a, b] and differentiable in (a, b).

Example 5.9.7. If [ ] +→ *f : a,b R is a differentiable function, where:

( ) ( ) ( )( )ln f a ln f b a b a b− = − + ,

then show that there exists a number θ for ( )∈θ a,b , such that ( )( )′ =ln θ 2θ .

Solution

Which is a backward step for ( )( )ln θ 2θ′ = ?

It is easy to see that Rolle’s theorem applies for ( ) ( ) 2h x ln f x x= − .


Example 5.9.8. f is a continuous function in [a, b], differentiable in (a, b) and ( ) ( )≠f a f b .

Show that: i) ( ) ( ) ( )= +2 f x f a f b has at least one root x1 in (a, b).

ii) There exist at least two numbers x2 and x3 in (a, b) such that

( ) ( )( )

( ) ( )−

+ =′ ′ −2 3

2 b a1 1f x f x f b f a

.

Solution

i) A backward step is to apply Bolzano’s theorem on ( ) ( ) ( ) ( )h x 2f x f a f b= − − .

ii) Backward steps? We need two intervals to apply the mean-value theorem. Question (ii) is related to (i) and the intervals [ ]1a, x , [ ]1x ,b are suitable.

Note: To answer question (ii), use question (i) fully.

Example 5.9.9. If f is twice differentiable and ( ) ( ) ( )≥ +7 f x 4 f 1 3 f 2 for all ∈x R , show that there

exists an ( )∈1x 1,2 such that ( )′′ =1f x 0 .

Solution

A backward step is to prove ( ) ( )f 1 f 2′ ′= .

Try the values 1 and 2: ( ) ( ) ( )7f 1 4f 1 3f 2≥ + or ( ) ( )f 1 f 2≥ .

Similarly ( ) ( ) ( )7f 2 4f 1 3f 2≥ + or ( ) ( )f 2 f 1≥ .

Therefore, ( ) ( )f 1 f 2= .

The given inequality becomes ( ) ( )7f x 7f 1≥ = ( )7f 2 .

By describing, we conclude that there are two externals, at 1 and 2, therefore, ( ) ( )f 1 f 2 0′ ′= = .

Example 5.9.10. Show that the solutions of equation + = +x x x x7 4 6 5 are 0 and 1.


Solution The contradiction method is suitable. Assume that a root ρ 0≠ , ρ 1≠ exists. By describing observe that 7 4 6 5 7 6 5 4+ = + ⇒ − = − Change the expression, so

that some known theorem could be applied. Hence ρ ρ ρ ρ7 6 5 4− = − . The function ( ) ρf x x= seems to be reasonable.

Apply the mean-value theorem

( )ρ ρ

17 6f x7 6−′ =−

or ρ ρ ρ 117 6 ρx −− =

( )ρ ρ

25 4f x5 4−′ =−

or ρ ρ ρ 125 4 ρx −− =

Therefore ρ 1 ρ 1

1 2x x− −= , but this is a contradiction since 2 1x x< . Thus the roots are 0 and 1.

Example 5.9.11. If [ ]→f : 0, 1 R is differentiable and ( ) =f 0 0 , prove that there exists an

( )∈1x 0, 1 such that ( ) ( )′ =−

11

1

f xf x

1 x.

Solution

A backward step: Apply Rolle’s theorem to ( ) ( ) ( )φ x 1 x f x= − , [ ]φ : 0,1 R→ .

Problems

1. If [ ]→f : a, b is a twice differentiable function and a bf

2+⎛ ⎞ =⎜ ⎟

⎝ ⎠,

( ) ( )f a f b2+

= show that there exists a number ( )∈0x a, b such that

( )′′ =0f x 0 .


5.10. L' HOPITAL'S RULE

Group 5

1. If a limit has the form 00

, ±∞±∞

, use l' Hôpital's rule

( )( )

( )( )x α x α

f x f xlim lim

g x g x→ →

′=

′, given that

( )( )x α

f xlim

g x→

′′

exists.

This also holds for one-sided limits. 2. If you have the form 0 ⋅∞ , reduce it to:

0 00 1 0⋅∞ = =

∞

or 0 10

∞ ∞⋅∞ = =

∞

For example x 0 x 0

ln xlim x ln x lim 1x

+ +→ →= =

−∞+∞

.

3. If you have one of the forms 00, 1∞, 1-∞, ∞0, use the formula g g ln ff e= . Then work with glnf which is of the form 0 ⋅∞ or 0∞⋅ .

4. If you find the form ∞−∞ , then either factor out

( ) ( ) ( ) ( )( )

f xf x g x g x 1

g x⎡ ⎤

− = − = ∞⋅⎢ ⎥⎣ ⎦

or use conjugate expressions such as

( ) ( )( )2 2

2

2x x

x 1 x x 1 xlim x 1 x lim

x 1 x→+∞ →+∞

+ − + ++ − = =

+ +

2

x

xlim→+∞

=21 x+ −

2...

x 1 x=

+ +

Note: The rule holds in (i) α is a real number, or +∞ , or –∞ .

(ii) in all cases +∞−∞

, −∞−∞

, −∞+∞

.

(iii) in cases of side limits: x α+→ , x α−→ .


Solved Problems

Example 5.10.1. Evaluate the following limits

i) 2

1x

x 0

1lim ex+

−

→ ii) cot x

x 0

xlim sin e2+→

iii) 1

cot xsin x

x 0lim e e

+→

⎛ ⎞−⎜ ⎟

⎝ ⎠

Solution

i)

22 2

11 10 xx x0 3

3x 0 x 0 x 0

2 ee 2exlim lim limx 1 x+ + +

−⎛ ⎞− −⎜ ⎟⎝ ⎠

→ → →= = (you are getting away from a solution)

Try to get out of the loop

2

2 2

1x 2

1 1x 0 x 0 x 0x x

3

1 1e x xlim = lim lim

x 2e ex

+ + +

∞⎛ ⎞− ⎜ ⎟∞⎝ ⎠

→ → →

−= =

−

2 21 1x 0 x 0x x

x 1lim lim x 0 0 02e 2e

+ +→ →= = = ⋅ =

ii) cot xx 0 x 0 cot x2

x 1 xsin cos2 2 2lim lim 1e e

sin x

+ +−→ → −=

⋅

But x 0

1 x 1lim cos2 2 2+→

= and the denominator is 00

, so you are trapped in a loop.

Exit the loop using

cot x cot x 2cot x 2

x 0 x 0 x 0 2

2

1 xe e 2sine sin x 2lim = lim lim1 x xcos cos sin xx 2 2sin x2 2sin2

+ + +

∞∞

→ → →

−

= =− ⋅

cot x 2cot x

x 0 x 02 2 3

xe 2sin e2lim limx x x xcos 4sin cos 2cos2 2 2 2

+ +→ →= = = +∞

⋅ ⋅


5.11. GRAPHING A FUNCTION – CONCAVITY – INFLECTION POINTS

Group 6 1. To draw the graph of a function, first find its critical points by setting ( )f x 0′ = .

Example

Suppose that ( )f x′ has three roots r1, r2 and r3 and its sign is as follows:

_+ + +

r1 r2 r3

0 0

functionincreasing

localmaximum

decreasingincreasing

localminimum

increasing

2. Find the asymptotes, if any. There are three possibilities.

Case 1 If ( )

x alimf x→

= +∞ or −∞ , then x a= is a vertical asymptote.

Case 2

If ( )xlim f x a→+∞

= or ( )xlim f x a→−∞

= , then y a= is a horizontal asymptote.

Case 3

If the following two limits exist

(i) ( )

x

f xlim a

x→+∞=

(ii) ( )xlim f x ax b→+∞

− =⎡ ⎤⎣ ⎦ , then y ax b= + is an asymptote at +∞ .

(Similarly work for −∞ ) Note: Following these steps you may obtain a rough graph of f. For a more accurate

graph, you have to take an extra step. Let f be twice differentiable. 3. Find ( )f x′′ , its roots and sign.


Example

_+

1

Here is another case.

(37) No root of f΄΄ exists at x1, only a vertical tangent line exists, that is

( ) ( )1

1

x x1

f x f xlim

x x→

−= +∞

− or −∞ .

Finally (i) An inflection point exists where the concavity changes and a tangent line exists. Note: For the function of the graph below the concavity changes but no tangent line

exists. Hence there is no inflection point.

x 1 A function is (ii) -concave up if ( )f x 0′′ > or ( )f x′ is increasing.

- concave down if ( )f x 0′′ < or ( )f x′ is decreasing.


Example 5.11.1. Find the asymptotes = +y ax b of the graph of

( )⎧ ≠⎪= ⎨⎪ =⎩

2 1x sin , x 0f x x

0 , x 0

Solution

( )x x x u 0

1sinf x 1 sin uxlim lim x sin lim lim 1 a1x x ux

→+∞ →+∞ →+∞ →= = = = =

( )( ) 2

x x x

1 1lim f x x lim x sin x lim x xsin 1x x→+∞ →+∞ →+∞

⎛ ⎞ ⎛ ⎞− = − = − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

x u 0

11 sin u 1x sin 1uxlim lim1 u

x→+∞ →

−−= =

(we simplify the expression)

2u 0

sin u u 0limu 0→

− ⎛ ⎞= =⎜ ⎟⎝ ⎠

(l’ Hôpital’s rule)

u 0 u 0

cos u 1 sin ulim lim 0 b2u 2→ →

− −= = = =

Therefore, y x= is an asymptote.

Problems

1. ( )+∞ →f : 0, is a differentiable concave up function.

If ( )

→+∞

′=

x

f xlim

x, show that

( ) ( )→+∞

+ −=

x

f x 1 f xlim

x.

2. [ ]→f : 0, a is a differentiable concave down function with ( ) =f 0 0. Show

that ( ) ( )=

f xg x

x, ( ]∈x 0, a is decreasing.


3. If f is a differentiable function in and concave up, show that ( ) ( )++⎛ ⎞ <⎜ ⎟

⎝ ⎠

f a f ba bf2 2

, ≠a b .

(Solutions on page 177)

5.12. THE FUNDAMENTAL THEOREM OF CALCULUS

How to use the first fundamental theorem ( )( ) ( )x

af t dt f x

′=∫ ,

( )( )( ) ( )( ) ( )φ x

af t dt f φ x φ x

′′=∫ .

Example 5.12.1.

If ( ) ( )+

= ∫2 x 1

xf x g t dt , show that ( ) ( ) ( )′ = + −f x 2g 2x 1 g x .

Solution

Differentiate

( ) ( ) ( )( )0 2x 1

x 0f x g t dt g t dt

+ ′′ = + =∫ ∫

( ) ( ) ( ) ( ) ( )g x g 2x 1 2x 1 2g 2x 1 g x′= − + + ⋅ + = + −

Example 5.12.2. Evaluate the limits:

i) +→

⎛ ⎞⎜ ⎟⎝ ⎠∫

x

2 0x 0

1lim t sin t dtx

ii) →

⎛ ⎞⎜ ⎟−⎝ ⎠∫

x

3x 3

1 sintlim dtx 3 t

Solution

i)

xx 0

2 20

t sin t dt1 0t sin t dtx x 0

⎛ ⎞= ⎜ ⎟⎝ ⎠

∫∫ (l’ Hôpital’s rule)

x 0 x 0

x sin x 1 sin x 1lim lim2x 2 2x+ +→ →

= =

Chapter 6

A FORMAL COMPUTER MODEL OF OUR METHODS

Our methods can be modelled and programmed on a digital computer. Such a model has

two parts.

1. ACTIVATION OF THE THEORY To solve a problem a certain theory should be used consisting of definitions, axioms,

theorems, algorithms and possibly heuristics. A theory is represented by reflexive statements such as x=x and IF-THEN statements which are called productions. The IF part of a production is called condition and the THEN part action. The general form of productions is IF (condition) THEN (action).

2. CLASSES OF PRODUCTIONS There are four classes of productions. (i) Class 1 contains algorithms that lead to a solution of a given problem. (ii) Class 2 contains heuristics such as the backward procedure. (iii) Class 3 contains axioms, definitions and theorems. This class does not introduce new

elements but relates given quantities. For example, if ABC is a right triangle, then 2 2 2( ) ( ) ( )AB AC BC+ = .

(iv) Class 4 contains productions that introduce new elements. For example, “if α belongs to a group ( , )G ∗ , then there is an inverse element α′ such that

eα α α α′ ′∗ = ∗ = ” or from geometry “if A and B are two points, then one and only one straight line joins them.”

Let us see the solutions of the classical unsolved problems 3, 4 and 5 of chapter 1. These

solutions are usually given in textbooks.


Solutions of Problems 3, 4 and 5 3. If b A∈ then b 0 b+ = (0 is the identity element) (1)

( )α b 0 α b⇒ ⋅ + = ⋅ (2)

α b α 0 α b⇒ ⋅ + ⋅ = ⋅ (distributive law) α 0 0⇒ ⋅ = because the identity element is unique, that is x q x q 0+ = ⇒ = .

Note: Human problem solvers have a difficulty to use steps (1) and (2). 4. Since α′ is an element, there is a left inverse α′′ such that α α e′′ ′⋅ = (1) This is the crucial idea. Hence,

( ) ( ) ( ) ( )( )α α e α α α α α α α α α α′ ′ ′′ ′ ′ ′′ ′ ′⋅ = ⋅ = ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅

( )( ) ( )α α α α α e α α α e′′ ′ ′ ′′ ′ ′′ ′= ⋅ ⋅ ⋅ = ⋅ ⋅ = ⋅ =

Note: Even the brightest students can rarely think of step (1), and they give up. 5. Draw line BE.

From triangles BCE and BED we have: BE common side, CE BD= and

DBE BEC> , because DBE is an external angle of triangle ABE. Therefore, BC DE< . B

D E

C

A

Note: Students find it difficult to solve this problem. Let us now see how these three problems were solved by the computer.

Problem 3 We want to prove that α 0 0⋅ = in a ring. First of all relevant theory is identified and activated. From any standard textbook one

finds four laws of equalities, seven properties for rings and three theorems (cancellation law, uniqueness of identity element, and uniqueness of inverse). No algorithm or heuristics are found. Thus we obtain class 3 and class 4 productions. However, one could augment these classes with classes 1 and 2 in the following general form.

Work backwards. Use contradiction. Use a method for equalities: if you have to prove that A B= , then (a) transform A till you reach B: 1 2A A A ... B= = = = , or (b) transform

A Formal Computer Model of Our Methods 127

to B till you reach A. (c) Apply productions to the data till you reach the goal. (d) Use the spiral method.

List of productions CLASS 3 Condition Actions 1. α element α α= 2. α b= b α= 3. α b= , b c= α c=

4. α b= , ( )f α s= ( )f b s=

5. α b= , c d= α c b d+ = + , α c b d⋅ = ⋅ 6. ( )α b c d+ + = ( )α b c d+ + =

7. ( )α b c d+ + = ( )α b c d+ + =

8. α b c+ = b α c+ = 9. α, 0 = element, 0 identity α 0 α+ = , 0 0 0+ =

10. ( )α b c d⋅ ⋅ = ( )α b c d⋅ ⋅ =

11. ( )α b c d⋅ ⋅ = ( )α b c d⋅ ⋅ =

12. ( )α b c d⋅ + = ,

( )b c α d+ ⋅ = α b α c d⋅ + ⋅ = , b α c α d⋅ + ⋅ =

13. α q α+ = q 0= (the identity element is unique)

14. α q 0+ = q α′= (the inverse element is unique) CLASS 4 15. α element There is α′ such that: α α 0′+ = Now we start applying these productions to the data of our problem, i.e., to α and 0 quite

mechanically. The following is the actual output of a computer program containing the above

productions. 1. Since α and 0 are elements, we write (or the computer builds) α α= , 0 0= . Now

these two equalities belong to the enriched space of the data 2. The law is applied to the equalities α α= , 0 0= but no new equalities are

generated 3. It fails (this condition does not match any part of the data space 4. Fails 5. Since there are two equalities α 0 α 0⇒ ⋅ = ⋅ , α 0 α 0+ = + 6. Fails 7. Fails 8. Fails


9. α 0 α+ = , 0 0 0+ = 10. Fails 11. Fails 12. Fails 13. Fails 14. Fails End of first cycle. Let us examine what has been recorded up to now. α α= , 0 0= , α 0 α 0⋅ = ⋅ , α 0 α 0+ = + , α 0 α+ = , 0 0 0+ = We do not proceed with 15, because a new element “α′ ” is introduced. We might use 15

after several trials with productions 1 – 14. Now we consider all these equalities as data of the problem and we begin again with the application of productions 1 – 14.

1. Fails 2. α α 0= + , etc. 3. Nothing new 4. Nothing new 5. All possible combinations will occur here: ( )α α 0 α α⋅ + = ⋅ ,

( )0 α 0 0 α⋅ + = ⋅ , ( ) ( )α α 0 α α 0⋅ ⋅ = ⋅ ⋅ , ( ) ( ) ( )α 0 α 0 α 0 α+ ⋅ + = + ⋅ ,

( )α 0 0 α 0⋅ + = ⋅

6. Fails 7. Fails 8. Fails 9. Fails 10. Fails 11. ( ) ( )α α 0 α α 0⋅ ⋅ = ⋅ ⋅

12. α α α 0 α α⋅ + ⋅ = ⋅ , 0 α 0 0 0 α⋅ + ⋅ = ⋅ , ( ) ( ) ( )α 0 α α 0 0 α 0 α+ ⋅ + + ⋅ = + ⋅ ,

α 0 α 0 α 0⋅ + ⋅ = ⋅ 13. α 0 0⋅ = , 0 0 0⋅ = , α 0 0⋅ = End of second cycle. We find that the required relation α 0 0⋅ = has appeared in the

enriched space. Following backwards the equalities involved in the course towards the goal we obtain

two complete proofs.

( )α 0 α

α α 0 α α α α α 0 α α α 0 0α α

+ = ⎫⇒ ⋅ + = ⋅ ⇒ ⋅ + ⋅ = ⋅ ⇒ ⋅ =⎬= ⎭

as well as an unexpected one


( )0 0 0 α 0 0 α 0 α 0 α 0 α 0 α 0 0+ = ⇒ + = ⋅ ⇒ ⋅ + ⋅ = ⋅ ⇒ ⋅ =

Note” The solution was the result of a deterministic procedure. This means that anyone

can solve this problem. However, there are other things we must discuss here before proceeding to the second problem.

(a) The goal of the problem was not taken into consideration in the above procedure. We

just started from the given data and we made trials. We proceeded by enriching the problem space. Is this what the method means? Partly, yes, but we should not forget that solving problems is not a simple job; it is a very complicated matter. Although no recipes can be given, I can say this here: Whenever you get stuck with a problem, try this method for a while. It will either give you the solution or some good ideas about it.

(b) When using this method, after a few steps check whether you have reduced the difference between data and goal. This may shorten the way to the goal.

A more sophisticated method of extracting information follows. This method reduces the

number of trials by eliminating the distance between the data and the goal (spiral method, chapter 3).

The enriched problem the goal is space after applying productions to a, 0 becomes: α 0 0⋅ = α α= , 0 0= α 0 α 0⋅ = ⋅ α 0 α 0+ = + α 0 α+ = 0 0 0+ = The solver is encouraged to go through productions and spot those which are likely to

eliminate or reduce the distance from the left to the right side α 0 0⋅ = . The solver should also try to make backward steps from the goal, if possible. Production 13. “If α q α+ = , then q 0= ” is needed. But here q must be equal to α 0⋅ . Which of the productions 1 – 15 may lead to production 12 which contains α 0⋅ ? α α= , α 0 α+ = by productions 5, 12 and 13 yield

( )α α 0 α α⋅ + = ⋅ , hence α α α 0 α α⋅ + ⋅ = ⋅ , which yields α 0 0⋅ = .

An alternative proof can be derived similarly: 0 0 0+ = , ( )α 0 0 α 0 α 0 α 0 α 0 α 0 0⋅ + = ⋅ ⇒ ⋅ + ⋅ = ⋅ ⇒ ⋅ =


Problem 4 We now proceed with problem 4 which is harder than problem 3 for human solvers,

although the number of productions is only nine.

CLASS 3 Condition Actions 1. α element α α= 2. α b= b α= 3. α b= , b c= α c=

4. α b= , ( )f α s= ( )f b s=

5. α b= , c d= α c b d⋅ = ⋅

6. ( )α b c d⋅ ⋅ = ( )α b c d⋅ ⋅ =

7. ( )α b c d⋅ ⋅ = ( )α b c d⋅ ⋅ =

8. α, e e α α⋅ = CLASS 4 9. α α′ , α α e′ ⋅ = Class 3 productions fail to start the problem; consequently we proceed with Class 4. The crucial idea then is immediately generated but there it is difficult to detect it.

The new data after using 9 are α, α′ , e, α α′⋅ , α′′ , α α e′′ ′⋅ = , α α′ , α α e′ = Well, after a number of cycles, the program reported the following: From 1 we obtain α α= , α α′ ′= , e e= , αα αα′ ′= , α α′′ ′′= , α α e′′ ′⋅ = ,

α α α α′ ′⋅ = ⋅ and from 5 ( ) ( ) ( )α α α α e α α′′ ′ ′ ′⋅ ⋅ ⋅ = ⋅ ⋅ and

( ) ( ) ( )α α α α e α α′′ ′ ′ ′⋅ ⋅ ⋅ = ⋅ ⋅

( )( )α α α α α α′′ ′ ′ ′⇒ ⋅ ⋅ ⋅ = ⋅

( )( )α α α α α α′′ ′ ′ ′⇒ ⋅ ⋅ ⋅ = ⋅

( )α e α α α′′ ′ ′⇒ ⋅ ⋅ = ⋅

α α α α′′ ′ ′⇒ ⋅ = ⋅ e α α′⇒ = ⋅

Here the program reported “Finish”, which means proof is found. Some alternatives now:


The associative law can be applied as follows: Ignore the parentheses and concentrate on substitutions of two adjacent elements: Left hand side: α α α α α e α′′ ′ ′ ′′ ′⋅ ⋅ ⋅ = ⋅ ⋅ = α α e′′ ′⋅ = .

Right hand side: ( )e αα α α′ ′⋅ = ⋅ . Therefore α α e′⋅ =

One could detect the crucial idea by means of a host of trials by giving them a direction,

e.g., by combining the two equalities α α e′ ⋅ = , α α e′′ ′⋅ =

( ) ( )( )α α e α e α e α α α α e′′ ′ ′′ ′ ′′ ′ ′⋅ = ⇒ ⋅ ⋅ = ⇒ ⋅ ⋅ ⋅ =

( )( ) ( ) ( )α α α α e α α α α e′′ ′ ′ ′′ ′ ′⇒ ⋅ ⋅ ⋅ = ⇒ ⋅ ⋅ ⋅ =

( )e α α e α α e′ ′⇒ ⋅ ⋅ = ⇒ ⋅ = ,

thus a second solution is obtained. One could transform the left hand side of the goal:

( ) ( ) ( ) ( )( )α α e α α α α α α α α α α′ ′ ′′ ′ ′ ′′ ′ ′⋅ = ⋅ ⋅ = ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅

( )( ) ( )α α α α α e α α α e′′ ′ ′ ′′ ′ ′′ ′= ⋅ ⋅ ⋅ = ⋅ ⋅ = ⋅ = .

Problem 5

B

D E

C

A

1. The program uses backwards production, and tries to find theorems with the same

goal. Three such theorems are found.

(a) If ABC, A B C′ ′ ′ are triangles and AB A B′ ′= and AC A C′ ′= , A A′> , then BC B C′ ′> .

(b) If ABC is a triangle and B C> , then AC AB> . (c) If AB DC> and DC EF> , then AB EF> .


2. Then the program comes to the next list of productions, and tries to see if the data of 1a, 1b or 1c are included in the data of the problem. This fails and so description starts.

3. Next, the program was run after the introduction of an auxiliary line segment, BE. The program applied the definition: If P, Q are points, then PQ is a line segment, as well as all applicable theorems of geometry. Thus:

4. It noted the existence of the triangles ABC, ABE, BCE, BDE and ADE. 5. It noted their sides and their angles. 6. It noted the external angles of triangles ABC, ABE and BCE. 7. It found relations between external and internal angles

DBC BAC>

DBC ACB>

DBE BAE>

DBE AEB> , etc.

8. The program noted equalities between the angles such as BEC BEA= ,

EDB EDA= , etc.

9. The program applied substitution in (6) and (7), and yielded DBE CEB> . 10. After this, the program came back to the backward procedure. It found that a

matching occurs with case 1a, and reported “Finish,” that is, proof is found.

SOLUTIONS OF UNSOLVED PROBLEMS

Chapter 1

B

D C

F

E

A

Let EF be such a straight line. The area of EFCD is three times the area of ABFE. Describe in detail. Area of EFCD How can we find it? What is its shape? It is a trapezoid

Then area FC DE DC

2+

= ⋅

What does FC DE

2+

remind me of?

FC DE

2+

median of trapezoid, FC DEMN

2+

=


M

ND C

F

E

M medium of EF N medium of DC ΜΝ parallel to FC

M

N

L B

D C

F

E

A

Continue describing L, M and N are collinear LN is parallel to AD Area of ( ) ( )EFCD MN DC= ×

Area of ( )ABFE LM (AB)= ×

( ) ( ) ( ) ( )MN DC 3 LM AB⇒ × = ×

AB DC BC AD LN= = = = Hence ( )MN 3 LM=

Therefore, M is constant.

If I want the upper part to be 13

of the lower one,

then all EF pass through the fixed point M

M

N

L

M3

M1

M2

B

D C

A

Solutions of Unsolved Problems 135

How many of these can exist? Four: M, M1, M2, M3 (see figure) Consequently 9 straight lines cross these four points M, M1, M2, M3 This means that at least 3 lines will cross the same point, otherwise a contradiction will

result: if only 2 lines crossed each point, there would be no more than 8 lines. This is an application of the Pigeonhole Principle: How can 9 pigeons stay in 4 nests? At least 3 of them will be in the same nest. Note: Good knowledge of the Pigeonhole Principle has helped you make a correct

statement. The description method led you to the crucial idea. 2. Describe ( )( )2 2b c b c b c− = − + .

Therefore ( )( )3α b c b c= − +

Further description. Describe the detail 3α

( )( )3 2 2α αα α α b c b c= = = − +

Hence it suffices that:2

b c αb c α− = ⎫

⎬+ = ⎭

(i)

or 2b c α

b c α⎫− =⎬

+ = ⎭ (ii)

More description. There are two equations. By adding

(i) 2

2 α α2b α α b2+

⇒ = + ⇒ = . Is it an integer?

Yes, because ( )α α 1

b2+

= is an integer.

Also, ( ) ( )2 α α 12c α α α α 1 c

2−

= − = − ⇒ =


CHAPTER 2

1. The Problem of The Stars Description of trials Possible negations I try all available places I try to draw it in all possible places I try the blank spaces I try each and every space

I try not only in all blank spaces I try on the stars, around the stars, around one star

2. The Horseshoe Problem

Description of trials Negations First I cut once I cut in one direction I cut it into two pieces Then I deal with one of them I continue cutting either of the two pieces

I should cut more than once I should cut in more than one direction I shouldn’t deal with one of them alone I should deal with one of them together with the one before I shouldn’t continue cutting either of them I should continue cutting both of them


First cut along AB. Then position the pieces so that the straight line EF lies on top of the straight line CD. Now one cut will be enough to cut the remaining pieces.

Note: What unnecessary constraints have you observed here? Simply that you should cut

the pieces one by one. What other experiences prevented you from finding the correct procedure?

CHAPTER 3 1. Goal

r rational means αrb

= , where α, b are integers

and b 0≠ .

Step 1 A soft description

Given Since f is periodic there is a number T 0≠ such that

( )( ) ( )( ) ( ) ( )sin 2 x T cos r x T sin 2x cos rx+ + + = +

Step 2

What is the difference between what is given and the goal? Difference: sine , cosine, x To eliminate the difference make a more dynamic description. Periodic means ( ) ( )f x T f x+ = , for each x

Τhus x can take any value x must be eliminated sine, cosine must also be eliminated How can I eliminate the sine or cosine? Can I make the sine or cosine equal zero? From trigonometry it is known that sin x sinα= yields x 2kπ α= + or x 2kπ π α= + −

sin 0 0= , πcos 02=

cos x cosα= yields x 2κπ α= ± , …


Following the above, try to eliminate x, sine, and cosine by giving x various values. What is a good value for x? A good value is one that gives sin0.

If x 0= , then

( ) ( )sin 2T cos rT sin 0 cos 0+ = + , or

( ) ( )sin 2T cos rT 1+ = (1)

If x T= − , then

( ) ( )sin 0 cos0 sin 2T cos rT+ = − + − or ( ) ( )sin 2T cos rT 1− + = (2)

I thus obtain two equations with two unknowns. From (1) and (2) I obtain ( )2cos rT 2= , hence, ( )cos rT 1 cos0= = ,

therefore, rT 2κπ= (3)

from (3) and (2) ( )sin 2T 0= I obtain 2T k π′= , thus, k πT2′

= (4)

and from (3) and (4) 4k αrk b

= =′

, where k, k΄ are integers.

2. The Following Proof Is Commonly Found in Textbooks

From the information given we see that nlimα 0≥ . If nlimα 0> then

( )n n 1lim n α α ++ = ∞ , which contradicts ( )n n 1lim n α α 1++ = . Hence, nlimα 0= . Since

αn is a decreasing sequence n 1 n n 1α α α+ −< < , and this yields n n 1 n n n 1α α 2α α α+ −+ < < + or

( ) ( )n n 1 n n n 1n α α 2nα n α α+ −+ < < + (1)

Clearly

( ) ( ) ( )( )n n 1 n n 1 n n 1n 1 nn α α n α α n 1 α αn 1 n 1− − −−

+ = + = − +− −

(2)

but

nlim 1n 1

=−

and ( )( )n n 1lim n 1 α α 1−− + =

because the last expression has the form ( )n n 1n α α ++ .


Hence, from (1), nlim 2nα 1= . Students have great difficulty arriving at (1) and (2). So then let us see if the spiral method can fare better.

First write the given information and the goal in brief. Given Goal

nα 0> nlim 2nα 1= αn decreasing

( )n n 1lim n α α 1++ =

Start DSB

1α 0> , 2α 0> , …

1 2 3 n 1 n n 1α α α ...α α α ...− +> > > > MEA What is the difference? How can you reach the goal using the given information? On the

left you have inequalities and ( )n n 1lim n α α 1++ = . On the right you have nlim 2nα 1= . Is

there a theorem relating these two? Yes, it is. If n nlim x lim y L= = and n n nx α y≤ ≤ ,

then nlimα L= . A subgoal then is to create the double inequality n n nx 2nα y≤ ≤ , where

n nlim x lim y 1= = . Given Transformed goal

n 1 n n 1α α α+ −< < n n nx 2nα y≤ ≤ I observe one difference. The quantity αn is on the left hand side but 2nαn is on the right

hand side. How can I rectify it? The factor 2n makes the difference. I eliminate it. From n 1 n n 1α α α+ −< < I obtain n 1 n n 12nα 2nα 2nα+ −< <

Is n 1lim 2nα 1+ = or n 1lim 2nα 1− = ? I do not know. On the other hand I have to use

the data ( )n n 1lim n α α 1++ = . Let me look for alternatives to (1).

Note. If you insist on continuing with (1) you will be caught in a loop, so the GOL

method is activated immediately. Alternatives of 2, n and nα give n2nα or (1): Do not multiply by 2n Multiply only by n Add to obtain n 1 nα α+ + Multiply and add Do things successively

n 1 n n 1α α α+ −< < yields n n 1 n n 1 nα α α α α+ −+ < < +


Hence ( ) ( )n n 1 n n 1 nn α α 2nα 2n α α+ −+ < < +

What is the difference now? A new subgoal is set: Prove that ( )n 1 nlim n α α 1− + = .

Start the spiral method again. Given Goal

( )n n 1lim n α α 1++ = ( )n 1 nlim n α α 1− + =

MEA

n 1 nα α− + is the term preceding n n 1α α ++ . One difference here is that we have

( )n 1 nn α α− + instead of ( )( )n 1 nn 1 α α−− + . If we eliminate the difference, we create

another difference, so there may be a new subgoal.

( ) ( )( )n 1 n n 1 nnn α α n 1 α α

n 1− −+ = − +−

DSB

( )n n 1limn α α 1++ = yields ( )( )n 1 nlim n 1 α α 1−− + =

MEA Difference? Now we have ( )n / n 1− . To eliminate this difference, we have to prove that

( )lim n / n 1 1− = , but this is straightforward.

Several variations to the above procedure may be followed. (a) One may look for alternative ways to manipulate n 1 n n 12nα 2nα 2nα+ −< < . For example

n 1 n n 12nα 2nα 2nα+ −< < yields ( ) ( )n 1 n 1 n n 1 n 1n α α 2nα n α α+ + − −+ < < +

Now working backwards

( ) ( )n 1 n n n 1 nn α α 2nα n α α+ −+ < < +

because

( ) ( )n 1 n 1 n 1 nn α α n α α+ + ++ < +

and ( ) ( )n 1 n n 1 n 1n α α n α α− − −+ < + therefore, it suffices to prove

( )n 1 nlimn α α 1− + = ..


(b) To prove ( )n n 1lim n α α 1−+ = use DSB.

Given Goal

( )( )n n 1lim n 1 α α 1−− + = ( )n n 1lim n α α 1−+ =

DSB

( )n n 1 n n 1lim n α α α α 1− −+ − − =⎡ ⎤⎣ ⎦

MEA Difference? Eliminate n n 1α α −− −

Since nlimα 0= and

n 1limα 0− = , then ( )n n 1lim n α α 1−+ =

3. Description

Given Goal x 0> , y 0> , z 0> x 1> or y 1>

xyz 1= backwards

1 1 1x y zx y z

+ + > + + x 1 0− > or y 1 0− >

xyz 1= , therefore, some of them are greater than 1 while others are smaller Let z 1< without loss of generality Then x 1> or y 1> Difference? Therefore, I must work with x, y. On the left side I have to find So let me eliminate z

1zxy

⇒ = ⇒ x 1− , y 1−

1 1 1x y xyxy x y

+ + > + + Then I have to factor

2 2 2 2x y xy 1 y x x y⇒ + + > + +


2 2 2 2x y xy 1 y x x y 0⇒ + + − − − >

( )( )( )1 y x 1 xy 1 0⇒ − − − >

Clearly y 1> or x 1> If y 1> and x 1> , then

( )( )( )1 y x 1 xy 1 0− − − < , which is wrong

Therefore, only y 1> , or only x 1> .

4. An Easy Method to Prove the First Fundamental Theorem of Calculus Is the Following

First, let us recall a few things about the definition of ( )b

af x dx∫ using Riemann’s Sum

Riemann’s Sum

f is a continuous function

We divide [ ]a, b into n subintervals of equal length b a

n−

each.

Then the sum of the area of the resulting rectangles is

( ) ( ) ( ) ( )n

n 1 2 n ii 1

b a b aS f x f x ... f x f xn n =

− −= + + + =⎡ ⎤⎣ ⎦ ∑ (Riemann’s Sum)

The area of the region enclosed by ( )y f x= , x a= , x b= and the x-axis is

( )n

in i 1

b alim f xn→+∞

=

− ∑ , which is denoted ( )b

af x dx∫ .

Using this definition of ( )b

af x dx∫ , we can very easily prove the following theorems:


1. ( )b

acdx c b a= −∫

2. If ( )f x 0≥ , [ ]x a, b∈ , then ( )b

af x dx 0≥∫ .

3. If ( )f x 0> , then ( )b

af x dx 0>∫ .

4. If ( ) ( )f x g x≤ , [ ]x a, b∈ , then ( ) ( )b b

a af x dx g x dx≤∫ ∫

5. ( ) ( )b b

a af x dx f x dx≤∫ ∫

6. ( ) ( ) ( )x

kF x F k f t dt− = ∫ , F is any antiderivative of f.

7. ( ) ( )( ) ( ) ( )b b b

a a aλf x μg x dx λ f x dx μ g x dx+ = +∫ ∫ ∫

Let us now come to the first fundamental theorem of Calculus. Let f(x) be a continuous function for all x [ a,b ]∈

If ( ) ( )=∫x

af t dt F x , then ( ) ( )′ =F x f x

Try to prove it, using the methods of this book

Given Goal

( ) ( )x

af t dt F x=∫

f continuous Describe: f continuous ⇒ for ε>0 ↓

( ) ( )f t f k ε− <

↓ Describe again theorems 1, 2, 3 ↓

( ) ( )x

k

f t f k dt x k− < −∫

Set x=k, that is

( ) ( )F k f k′ = .

Write in short notes and eliminate differences. Apply the most appropriate theorems and definitions. The difference is completely eliminated

( ) ( )F x f x′ =

Backward step?

( ) ( ) ( )x k

F x F klim f k

x k→

−=

−

↓ backward step again?

( ) ( ) ( )F x F kf k ε

x k−

− <−

, ε>0

↓ backward step again?

( ) ( ) ( ) ( )F x F k x k f kε

x k− − −

<−

↓

backward step?

( ) ( )x x

k k

f t dt f k dt x k ε− < −∫ ∫

Now you can write a complete proof.


CHAPTER 4 1. From (1) ( )x y z= − + . We substitute this value of x into (2) and obtain

( )2 2 2 2 2 2 2α y z 2yz b y c z+ + + + −

( ) ( )2bcyz 2αc y z z 2αb y z y 0− + + + + >

or ( ) ( ) ( )2 22 2 2α b y 2z α bc αc αb y α c z 0+ + − + + + + >

Now we take the discriminant

( ) ( ) ( )2 2 22 2 2D 4z α bc αc αb 4 α b α c z= − + + − + + =

( )2 28z bc 2α 2αc 2αb 0= − + + <

2. We examine special cases first.

If we have 1 object, we have 1 permutation. If we have 2 objects, we have 2 permutations. If we have 3 objects, we have 6 permutations.

For three objects a1, a2, a3, we obtain the permutations a1a2a3 , a1a3a2 , a2a1a3 , a2a3a1 ,

a3a1a2 , a3a2a1 . We find that each time the number of objects increases by one, the number of

permutations is multiplied by the new number of objects. Let Pn , be the permutations of n objects. We now obtain

1P 1= , 2P 1 2= ⋅ , 3P 1 2 3= ⋅ ⋅ , 4P 1 2 3 4= ⋅ ⋅ ⋅ Hence nP 1 2 3 ... n= ⋅ ⋅ ⋅ ⋅ which is written n!. By examining special cases, we discovered

the required formula nP n!= , but we have not obtained a rigorous mathematical proof. Using

mathematical induction, we could prove that ( )n 1P n 1 !+ = + given that nP n!= .

Take the permutation a1a2…an . A new object n 1a + can be placed before a1 , between a1 ,

a2 or between a2 , a3 , etc., or finally after an . Therefore, there are n 1+ different places for

n 1a + , hence a1a2 … an will give n 1+ new permutations. As this can happen with every one

of the Pn permutations, finally we obtain ( ) ( ) ( )nn 1 P n 1 n! n 1 !+ = + = + different

permutations, therefore, ( )n 1P n 1 !+ = +


3. Complete the table working backwards

A B C end of game 3 80 80 80 end of game 2 160 40 40 end of game 1 80 140 20 start 40 70 130

4. The only way to put one liter of water in the 4-liter bucket is to transfer one liter from

the large bucket. Fill it and then remove 4 liters twice. 5. To reject all other numbers, as roots, you have to make an infinite number of trials. So

you divide the numbers into two classes. There are those less than 2, (x<2) and those greater than 2 (x>2).

Examine first the numbers less than 2, x < 2.

Transform x x x5 3 4= + into x x3 41

5 5⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

This transformation is often very useful. If you have an expression with terms raised to

the same power, divide it to obtain ratios. This method facilitates manipulations. Here you have one unknown less with exponent x.

Since x 23 3x 2 then

5 5⎛ ⎞ ⎛ ⎞< >⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

x 24 45 5

⎛ ⎞ ⎛ ⎞>⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

. Add x x 2 23 4 3 2 1

5 5 5 5⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ > + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

hence 1 > 1, a contradiction.

Similarly, if x 2> then x 23 3

5 5⎛ ⎞ ⎛ ⎞<⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

x 24 45 5

⎛ ⎞ ⎛ ⎞<⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

, then add: x x 2 23 4 3 2 or 1 1

5 5 5 5⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ < + <⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

, a contradiction.

Consequently the only root is 2. 6. Since we deal with ≠ , the contradiction method seems appropriate. Suppose that ( ) ( )f x g x x= .

Describe


Differentiate both sides ( ) ( ) ( ) ( )f x g x f x g x 1′ ′+ =

Continue describing Substitute ( ) ( )x 0 : f 0 0 0 g 0 1′ ′= + = , or 0 1= , which is impossible.

7. ( )2 2 2A B I A B AB BA I AB BA I+ = ⇒ + + + = ⇒ + = − 2A B ABA A⇒ + = − (1) and 2AB BAB B+ = − (2)

Subtracting (1) – (2) we find 2 2A B ABA AB BAB A B+ − − = − +

B ABA A BAB A B ABA BAB⇒ + − − = − + ⇒ =

CHAPTER 5

Chapter 5.3. 1. Describe Use the definition If ( ) ( )1 2g x g x= , then ( )( ) ( )( )1 2f g x f g x= . But since f g is a one-to-one

function, then 1 2x x= . 2. Describe To prove ( )f 0 0= , let x be 0 in

( ) ( ) ( ) ( ) ( ) ( )f x f x f 0 f 0 2f 0 0 f 0 0− = − ⇒ = − ⇒ = ⇒ =

Now ( ) 3x f x x x cos x≥ + , for x 0≠ ,

yields ( )3x x cos xf xx

+≥ (1)

How to prove that ( )3x x cos xf xx

+=

Continue describing and use the data The equality ( ) ( )f x f x− = − , suggests substituting x by –x:


( ) ( ) ( )3 3x x cos x x x cos xf x f x

x x− − − − −

− ≥ ⇒ − ≥ ⇒−

( )3x x cos xf xx

+⇒ ≤ (2)

From (1) and (2) ( )3x x cos xf xx

+= .

3. Since there are only two alternatives for the goal, that is ( )f x x> or ( )f x x< , you

should try the contradiction method. Let ( )f x x> be true. Describe. f is increasing, thus

( )( ) ( ) ( )f f x f x x f x> ⇒ >

The last inequality contradicts ( )f x x> . The same happens if we assume ( )f x x< .

Therefore, ( )f x x= .

4. Describe. Make full use of the data. The equality holds for all x 0≠ . What does it

mean?

Replace x by any other expression, for example 1x

. Why? Never ask this question. It will

prevent you from generating good ideas. Use many possibilities without any specific purpose. It is like looking for something in

total darkness. You feel random objects hoping to bump into something useful, which in the context of problem solving is the crucial idea.

Substituting 1x

for x, obtain ( )1 1f 2f x 1x x

⎛ ⎞ + = +⎜ ⎟⎝ ⎠

(2)

Go a step back and check whether you have hit on the crucial idea.

Eliminating 1fx

⎛ ⎞⎜ ⎟⎝ ⎠

in (1) and (2) find ( )f x .

5. ( ) ( )( ) ( ) ( )1 1f x x f f x f x x f x− −< ⇒ < ⇒ < contradiction.

6. f is increasing, because if 1 2x x< then ( ) ( )1 2f x f x< .

What can you do about the inequality ( )1f 3x 2 1− − > ?

Describe. Use any possibility.


f increasing ( )( ) ( )1 3f f 3x 2 f 1 3x 2 1 1 1−⇒ − > ⇒ − > + − ⇒

3x 2 1 x 1⇔ − > ⇒ > 7. Describe

( )0 0f x x= means ( )0 0f x x 0− = or ( )0h x 0= .

As h is one-to-one, that x0 is unique. Now f g g f= yields ( )( ) ( )( ) ( )0 0 0f g x g f x g x= = .

Thus, ( )( ) ( )0 0f g x g x= and the uniqueness of x0 at ( )f x x= mean that

( )0 0g x x= .

8. If x 0= , then ( )( )f f 0 0= .

Describe Try any possibility Replace x by ( )f 0 . Why? Don’t ask why;

Just create new expressions, substitute, hope. What else could you do? Use substitutions, as you have nothing else to do.

( )( )( ) ( ) ( )( )f f f 0 f 0 f f 0=

Every now and then pause to evaluate what you have found. Notice ( )( )f f 0 0= ( ) ( )f 0 f 0 0 0⇒ = ⋅ =

9. Describe If x y 0= = , then ( ) ( )2f 0 f 0= . This yields ( )f 0 0= , or ( )f 0 1= .

Now, if y 0= , then ( ) ( ) ( )f x f x f 0= ⋅ . If ( )f 0 0= , then ( )f x 0= ,

which is constant If ( )f 0 1= , then what? ( ) ( )f x f x= means nothing.

Continue describing. If x y= , then ( ) ( ) ( ) ( )2 2f 0 f x f x 1 f x 1= ⇒ = ⇒ = ± .

If ( )f x 1= , the result makes sense; if ( )f x 1= − , then ( )( )1 1 1− = − − ,

which is impossible.


Chapter 5.4. (Limits) 1. Describe

Square roots of the form α b± suggest multiplication by a conjugate.

You would like to obtain α b± , but not α b c± − . Can I derive expressions like those in group 2? Set a goal

To create smaller fractions of the type 00

.

Then ( )( )

( )( )

f x 2 2 f x 7 3f x 2 f x 2

+ − + −+

− −.

Now take each fraction separately. 2. Describe There is a similarity to group 2 solved problem 1.

Let ( ) ( ) ( ) ( )f x x

h x f x xh x xx−

= ⇒ = +

Let ( ) ( ) ( ) ( )g x 1

κ x g x xκ x 1x−

= ⇒ = +

Then, ( ) ( ) ( )( ) ( )xh x x xκ x 1 xf x g x x

x x

+ + −⎡ ⎤⋅ − ⎣ ⎦= , etc.

3. Describe

Transform the expression ( )

( ) ( )( )

( ) ( )

3 2f x g xf x g x f x g x

++ +

Recall related theorems

Use one fraction, e.g. , ( )

( ) ( )

3f xf x g x+

What is it you are looking for? You would like to obtain two functions ( )h x and ( )κ x such that,

( ) ( )( ) ( ) ( )

3f xh x κ x

f x g x≤ ≤

+ with ( )lim h x = ( )lim κ x 0= .

One choice could be ( )h x 0= and ( )κ x =( )( )

3f xf x

,


because ( )

( ) ( )( )( ) ( )

3 32f x f x

f xf x g x f x

< =+

, etc.

Similarly for ( )

( ) ( )( )

( ) ( )( )( ) ( )

2 2 2g x g x g x0 g x

f x g x f x g x fg x⇒ < < =

+ +, etc.

4. The numerator reminds of the well-known identity

( ) ( )33 3f g f g 3fg f g+ = + − +

So the numerator becomes

( ) ( ) ( )3f g 3fg f g f g 3fg+ − + − + + =

( ) ( ) [ ]2f g f g 1 3fg f g 1⎡ ⎤= + + − − + − =⎣ ⎦

( )( )( ) ( )f g f g 1 f g 1 3fg f g 1= + + + + − − + − =

( ) ( )( )f g 1 f g f g 1 3fg= + − + + + −⎡ ⎤⎣ ⎦

Hence, the fraction is simplified, etc.

5. It is known that ( )

( )x α

sin f xlim 1

f x→= , if ( )

x αlim f x 0→

=

You try to create such expressions as the above

If ( ) ( )sin 3x sin 5xsin xx 0 κ 3 5c

x 3x 5x> ⇒ + ≤

Taking limits you find κ 3 5c+ ≤ Similarly for x 0< you find κ 3 5c+ ≥ . Hence, κ 3 5c+ = .

6. This is similar to problem (3) because: 1lim 0f= ,

1lim 0g=

It suffices to show that ( ) ( )( ) ( )2 2

f x g xlim 0

f x g x+

=+

However ( ) ( )( ) ( )2 2

f x g x0

f x g x+

< =+

( )( ) ( )

( )( ) ( )

( )( )

( )( )2 2 2 2 2 2

f x g x f x g xf x g x f x g x f x g x

= + < + =+ +

( ) ( )1 1 0

f x g x= + → , etc.


7. See problem (5) and follow a similar procedure

( ) ( )2 22 sin κx sin 3x

κ 9 6κκx 3x

⎛ ⎞ ⎛ ⎞⇒ + ≤⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Taking limits ( )22κ 9 6κ κ 3 0+ ≤ ⇒ − ≤ , hence, κ 3= .

8. Describe

sin 4x reminds us of the x 0

sin 4xlim 14x→

=

Divide by x

( )2 22x x f x 2x x− ≤ ≤ +

If x 0> , then ( )f x

2 x 2 xx

− ≤ ≤ +

If x 0< , then ( )f x

2 x 2 xx

− ≥ > +

Therefore, ( )

x 0

f xlim 2

x→=

Now ( )( )

( )

( )

f x sin 4x3 43f x sin 4x x 4xf x5f x 9x 5 9

x

++=

++

, etc.

9. ( ) ( ) ( )2 2 40 f x f x g x≤ ≤ +

10. i) ( ) ( ) ( ) ( )2 2

x xxf x f xf x 2 f x 2 2

= ⇒ = <+ +

, x 0

xlim 0

2→= , etc.

ii) ( ) ( ) ( )2 f x f xf x 2 1 0

x x⎡ ⎤

+ − =⎢ ⎥⎣ ⎦

Hence, ( ) ( )f x f x 1lim 2 1 0 limx x 2

⎡ ⎤− = ⇒ =⎢ ⎥

⎣ ⎦.

11. ( ) ( )2 22 2f g 2f 4g 5 f 1 g 2+ − + + = − + +

( ) ( ) ( )2 2 20 f 1 f 1 g 2≤ − ≤ − + + , etc.

12. Use the equality ( ) ( ) ( )f x y f x f y xy+ = + −


Set x 1 u x u 1− = ⇒ = + Hence, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )f x f 1 f u 1 f 1 f u f 1 u 1 f 1 f u

1x 1 u u u− + − + − ⋅ −

= = = −−

Chapter 5.5. (Continuity)

1. Describe Use the definition of continuity ( ) ( )

x 0lim f x f 0→

= .

If x 0> , then ( ) ( ) ( ) 2sin 7x sin 11x mf xx x x

≥ + +

If x 0< , then ( ) ( ) ( ) 2sin 7x sin 11x mf xx x x

≤ + +

Hence, ( ) ( ) ( ) 2

x 0 x 0

sin 7x sin 11x mlim f x lim 7 117x 11x x→ →

⎛ ⎞= + +⎜ ⎟

⎝ ⎠

If m 0≠ , then 2

x 0

mlimx→

= ±∞ , therefore, m 0= and ( )f 0 18= .

2. For x 0= , ( )f 0 1=

Also ( )

x 0lim f x 1→

= , hence, f is continuous at 0

Now to evaluate ( )

x 0

f x x 1lim

x→

− +, use what is given

( ) ( )x 0lim f x f 0 1→

= =

Thus, ( ) ( )

x 0 x 0

f x x 1 f x 1 x 1 1lim lim

x x→ →

− + − − + += =

( )x 0 x 0

f x 1 x 1 1lim limx x→ →

− + −= − .

To evaluate these two limits notice

( ) ( )2 2 2 21 x f x 1 x x f x 1 x− ≤ ≤ + ⇒ − ≤ − ≤

If x 0> then ( )f x 1

x xx−

− ≤ ≤ , and if ( )f x 1

x 0 x xx−

< ⇒ − ≥ ≥


Therefore, ( )

x 0

f x 1lim 0

x→

−= etc.

3. Describe

( ) ( ) ( ) ( )( ) ( )( )2f x α b f x αb f x α f x b 0− + + = − − <

The phrase “at least one root in ( )α, b ” leads one to use Bolzano’s theorem for

( ) ( )h x xf x αb= −

( ) ( ) ( ) ( ) ( ) ( )h α h b αf α αb bf b αb αb f α α f b b 0= − − = − − <⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

4. A root of the equation ( ) ( )f x g x= is a root of ( ) ( ) ( )h x f x g x= − .

This leads to the use of Bolzano’s theorem

( ) ( ) ( ) ( ) ( ) ( )h α h b f α g α f b g b= − −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

Start from the data and describe

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )f α f b g α g b f α g α f b g b 0+ = + ⇒ − + − =⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

or ( ) ( ) ( ) ( )f α g α f b g b 0− − ≤⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ .

5. Describe

( ) ( ) ( ) ( ) ( )f x g x h x f x g x 0≠ ⇒ = − ≠

( )h x 0⇒ > for all x, or ( )h x 0< for all x,

otherwise by Bolzano’s theorem ( )h x 0= for some x.

Describe the goal now

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )f α f b g α g b f α g b f b g α+ − − =

( ) ( ) ( ) ( ) ( ) ( )f α f b g b g α f b g b= − − − =⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

( ) ( ) ( ) ( ) ( ) ( )f b g b f α g α h b h α 0= − − = >⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

6. Describe

for ( ) ( )( ) ( ) ( ) 1x 100 f 100 f f 100 1 99 f 99 1 f 9999

= ⇒ = ⇒ ⋅ = ⇒ =

Hence, f takes all values from 199

to 99.


Therefore, there is a number x0 such that ( )0f x 50= .

Hence ( ) ( )( ) ( ) ( )0 01f x f f x 1 50 f 50 1 f 5050

= ⇒ ⋅ = ⇒ = .

7. Let ( ) ( ) ( )h x f x g x x= − . It suffices to show that

( ) ( ) ( ) ( ) ( ) ( )h 1 h 1 0 f 1 g 1 1 f 1 g 1 1 0− < ⇒ − − − + ≤⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

( ) ( ) ( )( ) ( ) ( )

( )

2 22 2

2 22 2

2 f g f g 2fg 2 fg 1 f g 0full useof data 2 f g f g 2fg 2 1 fg f g 0

2 fg 1 0

= + = + − ⇒ + = + ≥⎫⇒⎬

= + = − + ⇒ − = − ≥ ⇒⎭

⇒ − ≤

8. From the given data we conclude ( )0 f x 1≤ ≤ , ( )0 g x 1≤ ≤

Let ( ) ( ) ( ) ( ) ( ) ( )( )h x f x x h 0 h 1 f 0 f 1 1 0= − ⇒ = − ≤

Hence, there is [ ]m 0, 1∈ such that ( )f m m= . Since f is decreasing, m is unique.

Now ( )( ) ( )( ) ( )f g m g f m g m= =

But ( )( ) ( )f g m g m= means that ( )g m m= , because the equation ( )f x x= has a

unique solution. 9. Describe

( )( ) ( )( )2 2f α 3 f b 3 0− + + ≤

( ) ( )f α 3, f b 3⇒ = = − ⇒

( ) ( )f α f b 9 0⇒ = − < Use Bolzano’s theorem

10. ( )22 2 2 2 2f 4f 4cos x f 4f 4 4sin x 0 f 2 4sin x+ + = + + − ≤ ⇒ + ≤

for ( ) ( )2x 0 f 2 0 f 0 2= ⇒ + ≤ ⇒ = −

Also ( )f 2 2 sin x lim f 2 0+ ≤ ⇒ + = ⇒

( ) ( )x 0lim f x 2 f 0→

⇒ = − =

11. ( )( ) ( )( )f x κ f x 0− − ≤

But ( )m f x M≤ ≤


m = minimum of f M = maximum of f Therefore, m κ= , M =

12. ( )( )( )

( )( )( )

m f 0 M m f 0 Mm f 5 M 2m 2f 5 2Mm f 10 M 3m 3f 10 3M

≤ ≤ ≤ ≤⎫⎪≤ ≤ ⇒ ≤ ≤⎬⎪≤ ≤ ≤ ≤⎭

Add ( ) ( ) ( ) 6m f 0 2f 5 3f 10 6M⇒ ≤ + + ≤

( ) ( ) ( )f 0 2f 5 3f 10m M

6+ +

≤ ≤

Hence there exists a number x0: ( ) ( ) ( ) ( )0

f 0 2f 5 3f 10f x

6+ +

= .

Chapter 5.6. – 5.8. 1. Describe

( ) ( ) ( )x 0

g x g 0g 0 lim

x 0→

−′ =

− , ( ) ( ) ( )

x 0

f x f 0f 0 lim

x 0→

−′ =

−

What is the difference between the data ( ) ( )g x f x x− ≤ , ( ) ( )f 0 g 0= and the goal?

In order to eliminate the difference, you must form something like ( ) ( )g x g 0

x−

,

( ) ( )f x f 0x−

starting from the data.

Proceed as follows:

( ) ( ) ( ) ( ) ( ) ( )g x f x x g x g 0 f x f 0 x− ≤ ⇒ − − − ≤⎡ ⎤⎣ ⎦

If ( ) ( ) ( ) ( )g x g 0 f x f 0

x 0 1x x− −

> ⇒ − ≤ ,

Hence, ( ) ( ) ( ) ( ) ( ) ( )

x 0

g x g 0 f x f 0lim 1 g 0 f 0 1

x x→

− −⎡ ⎤′ ′− ≤ ⇒ − ≤⎢ ⎥

⎣ ⎦ (1)

Similarly, if x 0< ,

( ) ( ) ( ) ( ) ( ) ( )g x g 0 f x f 01 g 0 f 0 1

x x− −

′ ′− ≥ ⇒ − ≥ (2)


From (1) and (2) the result ( ) ( )g 0 f 0 1′ ′− = is obtained.

2. When the word “not” appears in problems the contradiction method seems to be

appropriate.

Assume that ( ) ( )

0

0

x x0

f x f xlim

x x→

−=

−, ∈

Using the given inequality ( ) ( )

( ) ( )0

0 0

f x f x 1x x f x f x−

≥− −

, we find

( ) ( )( ) ( )0

0

x x0 0

f x f x 1lim limx x f x f x→

−≥ = +∞

− −, that is > +∞ .

Therefore, f is not differentiable. In order to show that f is one-to-one, we shall start from the definition

Let ( ) ( ) ( ) ( )( )21 2 1 2 1 2f x f x f x f x x x= ⇒ − ≥ − ⇒

1 2 1 20 x x x x⇒ ≥ − ⇒ =

3. Since the derivative ( )f κ′ appears in the problem, use the definition

( ) ( ) ( )x κ

f x f κf κ lim

x κ→

−′ =

−

How to eliminate the difference between the two parts of the equality? Change the first part

( ) ( )xf κ κf xx κ−−

so that ( ) ( )f κ f x

x κ−−

will emerge

Therefore, ( ) ( ) ( ) ( ) ( ) ( )xf κ κf x xf κ κf x κf κ κf κ

x κ x κ− − − +

= =− −

( ) ( ) ( ) ( ) ( ) ( ) ( )xf κ κf κ f x f κ f x f κκ f κ κ

x κ x κ x κ− − −

= − = −− − −

, etc.

4. Describe From the data: ( ) ( ) ( )f xy f x f y= and ( )f 1 2′ =


( ) ( ) ( )x 1

f x f 1f 1 lim

x 1→

−′ =

−

Evaluate ( )f 1

( ) ( ) ( ) ( ) ( ) ( ) ( )f xy f x f y f 1 1 f 1 f 1 f 1 0= ⋅ ⇒ ⋅ = ⋅ ⇒ = or ( )f 1 1=

If ( )f 1 0= , then ( ) ( ) ( ) ( )f x 1 f x f 1 0 f x 0⋅ = ⋅ = ⇒ =

This contradicts ( )f 1 2′ = . Therefore ( )f 1 1=

( ) ( ) ( ) ( ) ( ) ( )( )h 1 h 1

f xh f x f x f h f xf x lim lim

xh x x h 1→ →

− −′ = = =

− −

( ) ( ) ( ) ( ) ( )h 1

f x f h 1 f x f xlim 2 f x 2

x h 1 x x→

−′= = ⋅ ⇒ =

−

5. Work backwards. A backward step is to prove ( )f x 1′ =

Use the definitions

If y 0> , then ( ) ( ) ( ) ( ) ( )f x y f x f y f y f 0

y y y 0+ − −

≤ =−

Hence, ( ) ( ) ( ) ( ) ( ) ( )

y 0

f x y f x f y f 0lim lim f x f 0 1

y y 0+→

+ − −′ ′≤ ⇒ ≤ =

−

If y 0< , then ( )f x 1′ ≥ , therefore, ( )f x 1′ = , etc.

6. Intensive use of data If x 0= , then ( )0 f 0 1 0 β β 1⋅ = + + ⇒ = −

Also ( ) ( )2x

2x 0 x 0

e αx 1 0f 0 lim f x lim x 0→ →

+ − ⎛ ⎞= = ⎜ ⎟⎝ ⎠

From L’ Hopital’s rule ( )( )

( )( )x α x α

f x f xlim lim

g x g x→ →

′=

′

( )2x

0

x 0

2e αf 0 lim 2e α 0 α 22x→

+⇒ = ⇒ + = ⇒ = −

Hence, ( )2x 2x

x 0 x 0

2e 2 0 4ef 0 lim lim 22x 0 2→ →

− ⎛ ⎞= = =⎜ ⎟⎝ ⎠


Therefore, ( )2x

2

e 2x 1, x 0f x x2 , x 0

⎧ − −≠⎪= ⎨

⎪ =⎩

Then ( )2x 2

3x 0

e 2x 1 2x 0 4f 0 lim ...x 0 3→

− − − ⎛ ⎞′ = = =⎜ ⎟⎝ ⎠

7. It is sufficient to prove ( ) ( )f 0 g 0= and ( ) ( )f 0 g 0′ ′=

If x 0= , then ( ) ( ) ( ) ( )f 0 g 0 0 f 0 g 0− ≤ ⇒ =

Eliminate the difference between the data and the goal

( ) ( ) ( ) ( ) ( ) ( )2 f x f 0 g x g 0f x g x x x

x 0 x 0− −

− ≤ ⇒ − ≤ ⇒− −

( ) ( ) ( ) ( ) ( ) ( )g x g 0 f x f 0 g x g 0x x

x 0 x 0 x 0− − −

⇒ − ≤ ≤ +− − −

Therefore, if x 0→ then ( ) ( )g 0 f 0′ ′= .

8. Since the problem contains a negation, the contradiction method should be applied Assume that there are two points ( )( )1 1A x , p x , ( )( )2 2B x , p x , and the tangent lines

at A and B are perpendicular. Then ( ) ( )1 2p x p x 1′ ′⋅ = −

Now, using Bolzano’s theorem, there is some x0 such that ( )0p x 0′ = . This contradicts

( )p x 0′ ≠ .

9. ( ) ( ) ( ) ( ) ( ) ( )h 0 h 0

f x h f x f x f h f xf x lim lim

h h→ →

+ − ⋅ −′ = = =

( ) ( ) ( ) ( ) ( ) ( )h 0 h 0

f h 1f x lim f x lim g h f x 1 f x

h→ →

−= = = ⋅ =

10. Describe f differentiable at 0 ⇒


( ) ( ) ( ) ( )h 0 h 0

f 0 h f 0 f h 1f 0 lim lim

h h→ →

+ − −′ = =

[ ( ) ( ) ( )2x y 0 f 0 f 0 f 0 1= = ⇒ = ⇒ = ]

Now, ( ) ( ) ( ) ( ) ( ) ( )h 0 h 0

f x h f x f x f h 2xh f xf x lim lim

h h→ →

+ − + −′ = = =

( ) ( ) ( ) ( )h 0

f h 1lim f x 2x f x f 0 2x

h→

−⎡ ⎤′= + = ⋅ +⎢ ⎥

⎣ ⎦

If it is differentiable at α 0≠ , it suffices to prove that ( )f 0′ exists. We find (see the

procedure above and set x α= )

( ) ( ) ( ) ( )h 0 h 0

f h 1 f h 1f α f α lim 2α lim

h h→ →

− −′ = + ⇒ exists, that is ( )f 0′ exists.

11.

x y

AB

O

60°

speed 12

speed 10

STEP 1 Use the cosine rule in triangle OAB

AB2 = OA2 + OB2 –2OA · OB · cos60°

or, S2(t) = x2(t) + y2(t) – 2x(t)y(t)21

= x2(t) + y2(t) – x(t) · y(t)

STEP 2 Differentiate 2S(t)S΄(t)=2x(t)x΄(t)+2y(t)y΄(t)-x΄(t)y(t)-x(t)y΄(t) STEP 3 Substitute x΄(t)=10, y΄(t)=-12 (decreasing), x(t)=10, y(t)=8

For S(t) use the cosine rule S2(t) = 102 + 82 – 2 · 10 · 8 · cos60° = 84


12.

O

A

B

STEP 1 By the cosine law we obtain:

AB2 = OA2 + OB2 – 2OA · OB · cosθ, ΟΑ = ΑΒ = R Let AB = S(t), arcAB = α(t), given α΄(t) = 2, therefore, S2(t) = 2R2 – 2R2 cosθ(t) ⇔ S2(t) =2R2(1 – cosθ(t)).

STEP 2 Differentiate 2S(t)S΄(t) = 2R2sinθ(t)θ΄(t) (1)

But θ(t)R = α(t), so θ΄(t) =α (t)

R′

= R2

.

STEP 3 ( )S t OA R= = (when ( ) oθ t 60′ = ). Hence, (1) yields

( ) ( ) ( ) ( )2S t S t R sinθ t θ t′ ′⋅ = or

( ) 2 o 2R S t R sin 60R

′⋅ = ⋅ , thus ( )S t 3′ =

Chapter 5.9. 1. Describe

The numbers a, a b

2+

and b suggest the use of the mean value theorem in the intervals

a ba, 2+⎡ ⎤

⎢ ⎥⎣ ⎦, and

a b , b2+⎡ ⎤

⎢ ⎥⎣ ⎦

⇒ There is 1a bx a,

2+⎛ ⎞∈⎜ ⎟

⎝ ⎠ such that

( )( ) ( ) ( )

1

a bf f a f b f a2f x a b b aa2

+⎛ ⎞ −⎜ ⎟ −⎝ ⎠′ = =+ −−


where the equality ( ) ( )f a f ba bf

2 2++⎛ ⎞ =⎜ ⎟

⎝ ⎠ was used

For the interval a b , b

2+⎡ ⎤

⎢ ⎥⎣ ⎦ the following is found

( ) ( ) ( )2

f b f af x

b a−

′ =−

Therefore, ( ) ( )1 2f x f x′ ′= and Rolle’s theorem gives ( )0f x 0′′ = .

2. Describe

2ax sin x x+ ≥ suggests the use of derivatives of

( ) 2f x ax sin x x= + − . It suffices to prove ( )f x 0≥

( )f x 2ax cos x 1′⇒ = + −

( ) ( )1f x 2a sin x 2 1 0 f x 02

′′ ′′= − > ⋅ − = ⇒ >

Hence f ′ is an increasing function. Also see ( )f 0 0′ =

Thus, ( ) ( )x 0 f x f 0 0′ ′< ⇒ < =

( ) ( )x 0 f x f 0 0′ ′> ⇒ > =

Therefore, there is a minimum at ( ) ( )0 f x f 0 0⇒ ≥ =

3. Follow the method for inequalities

( )3 5x xf x x sin x

6 120= − + −

( )2 4x xf x 1 cos x

2 24′ = − + −

( )3xf x x sin x

6′′ = − + +

( )2xf x 1 cos x

2′′′ = − + +

( ) ( )4f x x sin x= − ( ) ( )5f x 1 cos x= −

Start from the last equality:


( ) ( )41 cos x 0 f x− ≥ ⇒ increasing ( ) ( ) ( ) ( )4 4f x f 0 0⇒ ≥ = ( ) ( ) ( ) ( )4 3f x 0 f x⇒ ≥ ⇒ increasing ( ) ( ) ( ) ( )3 3f x f 0 0⇒ ≥ =

( )f x′′⇒ increasing

( ) ( )f x f 0 0′′ ′′⇒ ≥ =

( )f x′⇒ increasing

( ) ( )f x f 0 0′ ′⇒ ≥ =

f⇒ increasing

( ) ( )f x f 0 0⇒ ≥ =

4. 2 2 2 22

ln x ln y x y 2y ln x 2y ln y x y 0x y 2y− +

< ⇔ − − + <−

Let y be constant and ( ) 2 2 2 2f x 2y ln x 2y ln y x y= − − + ⇒

( ) ( )( )2 y x y x1f x 2y 2x 2 0x x

− +′⇒ = − = <

f⇒ is decreasing in [ )y, +∞

therefore, for ( ) ( )x y f x f y 0> ⇒ < =

5. ( ) ( ) ( ) ( ) ( )f x f xf x e x f x f x e 1′ ′+ = ⇒ + =

( ) ( ) ( )f x

1f x 1 0 f x 11 e

′ ′⇒ = < ⇒ < <+

If x y≠ , the goal becomes ( ) ( )f x f y

1x y−

<−

Describe and apply the mean value theorem

( )0f x 1′ < which is true.

If x y= , the goal becomes 0 0≤ . 6. ( ) ( )f x 2x 1 f x 2x 1 0′ ′> − ⇒ − + >

( )( )2f x x x 0′⇒ − + >


( ) ( ) 2h x f x x x⇒ = − + is increasing

Thus ( ) ( ) ( )0 x 1 h 0 h x h 1< < ⇒ < < ⇒

( ) ( ) ( ) ( ) ( )2f 0 f x x x f 1 f 0 f 1⇒ < − + < ⇒ <

7. i) Work backwards

A backward step is to prove ( )( ) ( )( )2 2f x g x 0

′⎡ ⎤+ =⎣ ⎦

, which is easy, then

( ) ( )2 2f x g x c+ =⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ for 2 2x 0 0 1 c c 1= ⇒ + = ⇒ =

ii) How can I prove that two equalities are true simultaneously? Take a backward step to prove

( )( ) ( )( )2 2f x sin x g x cos x 0− + − = , which is also easy.

8. i) If ( )f x 0< for some x, then by Bolzano’s theorem in the interval [ ]0, 1 ,

( )0f x 0= , ( )0x 0, 1∈ .

Hence 02x0 e= which is not true. Therefore, ( )f x 0> for x∈ .

ii) ( ) ( ) ( ) 22x 2xf x f x e f x e′ ′⎡ ⎤′ ⎡ ⎤= ⇒ = ⇒⎡ ⎤⎣ ⎦ ⎣ ⎦⎣ ⎦

( ) 2 2xf x e c= +⎡ ⎤⎣ ⎦

For x 0 1 1 c c 0= ⇒ = + ⇒ =

Hence ( ) xf x e= ±

But since ( ) ( ) xf x 0 f x e> ⇒ = .

9. ( )x 2e x 1 ln x 1 1⎡ ⎤+ + + =⎣ ⎦ . Simplify

( ) ( ) ( )2 x 2 xx 1 ln x 1 e f x x 1 ln x 1 e− −⇒ + + + = ⇒ = + + + −

Differentiate ( )2

x x2 2

2x x 1 2xf x 1 e ex 1 x 1

− −+ +′⇒ = + + = + =+ +

( )2x

2

x 1e 0

x 1−+

= + >+

Hence ( )f x is increasing, therefore it has only one root 0x 0= .


10. Apply Rolle’s theorem to the antiderivative of ( ) k 2 mf x ax bx cx= + + .

( ) k 1 1 m 1a b bF x x x xk 1 1 m 1

+ + += + ++ + +

(antiderivative, ( ) ( )F x f x′ = )

Now, ( )F 0 0= , ( ) a b cF 1 0k 1 1 m 1

= + + =+ + +

.

Rolle’s theorem can be applied to ( )F x which means that

there is a number ( )0x 0, 1∈ such that ( ) ( )0 0F x f x 0′ = = .

11. Use Fermat’s theorem a) Since ( )f 1 0= , ( ) ( )f x f 1≥

then ( ) af 1 0 1 0 a 11

′ = ⇒ + = ⇒ = −

hence, ( ) 1f x ln x 1 0x

= + − > . Now, ( ) 2

1 1 x 1f xx x x

−′ = − =

The sign of f ′ is Therefore ( ) ( )f x f 1 0≥ = .

b) Rearrange the inequality, so that the form of ( )f x will be apparent

( ) ( )2 22 2

1 1ln 2x 2 ln x 3x 3 2x 2

+ − > + − ⇒+ +

( ) ( )2 22 2

1 1ln 2x 2 1 ln x 3 12x 2 x 3

+ + − > + + −+ +

( ) ( )2 2f 2x 2 f x 3⇒ + > + (1)

Since 22x 2 1+ > , 2x 3 1+ > , and since in the interval [ )1, +∞

f is increasing, inequality (1) implies 2 2 22x 2 x 3 x 1 x 1+ > + ⇒ > ⇒ >

12) Since f is differentiable, f is continuous too Hence there is a number ( )1x 0, 1∈ such that ( )1f x 3=

Apply Rolle’s theorem in the interval ( )1x , 2

( )0f x 0′⇒ = , ( )0x 0, 2∈ .

0 1 –


Chapter 5.10. 1. Try to simplify

1 yx

23

2 3

1 1 1x sin 1 sin1 sin y yx x xx x sin 1 1 1x yx x

⎛ ⎞=⎜ ⎟⎝ ⎠− − −⎛ ⎞− = = = ⇒⎜ ⎟

⎝ ⎠

00

3 2y 0 y 0 y 0 y 0

sin y y cos y 1 sin y cos x 1lim lim lim limy 3y 6y 6 6+ + + +→ → → →

− − − −⇒ = = = = −

2. Simplify the expression

2 2 2 2

22 2 2 2 4

1 1 sin x x sin x x 1x sin x x sin x x sin x

x

− −− = = ⋅

⎛ ⎞⎜ ⎟⎝ ⎠

It is well-known that x 0

sin xlim 1x→

=

Apply L’ Hopital’s Rule 0

2 2 0

4 3 3x 0 x 0 x 0

sin x x 2sin x cos x 2x sin 2x 2xlim lim limx 4x 4x→ → →

− − −= = =

00

2x 0 x 0

2cos 2x 2 4sin 2x 1 sin 2x 1lim lim lim12x 24x 3 2x 3→ →

− − −⎛ ⎞= = = = −⎜ ⎟⎝ ⎠

.

3. Manipulate

1 yx x21 1 1 1x ln 1 1 x x ln 1 1 x ln 1

x x x x

⎛ ⎞=⎜ ⎟⎝ ⎠⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − = + − = + − =⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

( )2

ln 1 y yy+ −

=

Then ( )

( )

00

2y 0 y 0 y 0

1 1ln 1 y y 1 11 ylim lim limy 2y 2 1 y 2+ + +→ → →

−+ − −+= = = −

+ .

4. ( ) ( ) ( )

00

2h 0

2f x 3h 5f x 3f x 2hlim

h→

− − + +=


( ) ( )00

h 0

6f x 3h 6f x 2hlim

2h→

′ ′− − + += =

( ) ( ) ( )h 0

18f x 3h 12f x 2hlim 15f x

2→

′′ ′′− + +′′= =

Hence ( ) ( ) ( ) ( ) ( ) x15f x 15f x f x f x f x ce′′ ′ ′′ ′ ′= ⇒ = ⇒ =

( ) x1f x ce c⇒ = + .

Now using ( )f 0 1= , ( )f 0 2′ = we find 1c c 1+ = , c 2 c 2= ⇒ = , 1c 1= − .

Therefore ( ) xf x 2e 1= − .

5. The expression ( ) ( )2 2

2 2

ln x x 2 ln x 3x 6x x 1 x 3x 5

+ + + +=

+ + + +

has the form ln y lnωy 1 ω 1

=− −

Examine ln yy 1−

Differentiate ( ) ( )2 2

1 1(y 1) ln y 1 ln yln y y yy 1 y 1 y 1

− − − −′⎛ ⎞= =⎜ ⎟− − −⎝ ⎠

Differentiate the numerator 2 2

1 1 1 1 y1 ln yy y y y

′⎛ ⎞ −− − = − =⎜ ⎟

⎝ ⎠

Therefore if y 1> , then ln yy 1−

is decreasing and if 0 y 1< < , it is increasing.

The quantities 2y x x 2= + + and 2ω x 3x 6= + + are greater than 1 (simple proof).

Therefore, ln yy 1−

is decreasing, hence the function

( ) ln yf yy 1

=−

is one-to-one.

Hence, y ω= , or 2 2x x 2 x 3x 6 2x 4 0 x 2+ + = + + ⇒ + = ⇒ = − .


Chapter 5.11. 1. The description method here suggests to use the mean value theorem

( ) ( ) ( ) ( )1f x 1 f x x 1 x f x′⇒ + − = + − , 1x x x 1< < +

As f is concave up, f ′ is increasing, so ( ) ( ) ( )1f x f x f x 1′ ′ ′< < +

( ) ( ) ( ) ( )f x f x 1 f x f x 1′ ′⇒ < + − < +

( ) ( ) ( ) ( )f x f x 1 f x f x 1 x 1x x x 1 x′ ′+ − + +

⇒ < < ⋅+

But ( ) ( )

x x

f x f x 1 x 1lim limx x 1 x→+∞ →+∞

′ ′ +⎡ ⎤+= = ⋅⎢ ⎥+⎣ ⎦

Hence ( ) ( )

x

f x 1 f xlim

x→+∞

+ −= .

2. It suffices to prove ( )g x 0′ < . Now work backwards

( ) ( ) ( ) ( )2

f x xf x f xg x 0 0 0

x x

′ ′ −⎡ ⎤′ < ⇔ < ⇔ < ⇔⎢ ⎥

⎣ ⎦

( ) ( ) ( ) ( ) ( ) ( ) ( )1

f x f x f 0xf x f x 0 f x f x

x x 0−

′ ′ ′⇔ − < ⇔ < = =−

[mean value

theorem] The last result is obtained by applying the mean value theorem in the interval

[ ]0, x , 10 x x< < .

The last inequality ( ) ( )1f x f x′ ′< , 1x x< is true as f ′ is decreasing (concave down

function). 3. Assume a b< . Work backwards

( ) ( ) ( ) ( )f a f ba b a b a bf f f a f b f2 2 2 2

++ + +⎛ ⎞ ⎛ ⎞ ⎛ ⎞< ⇔ − < −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Using the mean value theorem

( ) ( )1 2a b a bf x a f x b

2 2+ +⎡ ⎤ ⎡ ⎤′ ′⇔ − < −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦


( 1 2a ba x x b

2+

< < < < )

( ) ( )1 2b a b af x f x

2 2− −′ ′⇔ <

( ) ( )1 2f x f x′ ′⇔ < which is true, because f ′ is increasing.

Chapter 5.12.

1. Method 1 Show that ( )F x 0′ = , therefore, ( )F x c= . But since ( )F 0 0= , then c = 0 .

Method 2

Substitute 1 ux= , then the second integral becomes

x

21

1 du1 u

−+∫ .

2. Method 1

For any number λ, ( ) ( )( )b 2

aλf x g x dx 0+ ≥∫ , hence,

( ) ( ) ( ) ( )b b b2 2 2

a a aλ f x dx 2λ f x g x dx g x dx 0+ + ≥∫ ∫ ∫

This means that the discriminant D is nonpositive. Therefore, D 0≤ yields

( ) ( )( ) ( ) ( )2b b b2 2

a a a4 f x g x dx 4 f x dx g x dx 0− ⋅ ≤∫ ∫ ∫

Note: It is very difficult to derive this proof. To proceed alternatively follow the methods

of the book

Method 2 Follow the method for inequalities. Let a be a constant and b = x a variable. Then prove

( ) ( ) ( )( ) ( )( ) ( )( )2x x x2 2

a a aF x f t g t dt f t dt g t dt 0= − ≤∫ ∫ ∫ .

Differentiate


( ) ( ) ( )( ) ( ) ( ) ( )

( )( ) ( ) ( )( )

x 2

a

x x22 2

a a

F x 2 f t g t dt f x g x f x

g t dt g x f t dt

′ = − ⎡ ⎤⎣ ⎦

− =⎡ ⎤⎣ ⎦

∫

∫ ∫

( ) ( ) ( ) ( )( ) ( )( ) ( )( )

( )( ) ( )( )

x x 2 2

a a

x 2 2

a

2 f t g t f x g x dt g t f x dx

f t g x dt

⎡= − − + +⎢⎣⎤+ =⎥⎦

∫ ∫

∫

( ) ( ) ( ) ( )( )x 2

ag t f x f t g x dt 0= − − ≤∫

Therefore, ( )F x 0′ ≤ , or F is decreasing.

Hence, a x< yields ( ) ( )F x F a 0< = .

Now for x = b, ( )F b 0≤ .

3. Method 1

Since F(0) = 0 , from the mean-value theorem

( ) ( ) ( ) ( )1

F x F 01 F x F xx x 0

−′= =

−

Take a step back 10 x x< < yields ( ) ( )1F x F x′ ′< (1)

but ( ) ( )F x f x′ = , thus (1) becomes ( ) ( )1f x f x< .

This is true since f is increasing.

Method 2 (Method for Inequalities) Backward step: ( ) ( )xF x F x 0′ − > or ( ) ( )xf x F x 0− > .

Differentiate ( ) ( )( ) ( ) ( ) ( ) ( )xf x F x f x xf x F x xf x 0′ ′ ′ ′− = + − = ≥ ,

therefore, ( ) ( )xf x F x− is increasing, which means that if 0 x< , then

( ) ( )xf x F x 0− >

REFERENCES

[1] Anderson J. R., Cognitive psychology and its implications, Freeman, San Francisco, 1980.

[2] H. Dörrie, 100 great problems of elementary mathematics, their history and solution, Dover, New York, 1965.

[3] Gardner, M., ed, Mathematical Puzzles of Sam Loyd, Dover, New York, 1959. [4] Greeno J., A theory of knowledge for problem solving. Problem solving and Education:

Issues in teaching and research, Laurence Erlbaum Associates, Hillsdale, N. Jersey, 1980.

[5] Greeno J. G., Nature of problem-solving abilities, W.K. Estes (Ed.) Handbook of learning and Cognitive process (Bol. 5). Laurence Erlbaum Associates, Hillsdale, N. Jersey, 1978.

[6] Kalomitsines S., How to solve problems (in Greek), Athens, 1979. [7] Kalomitsines S., Attack your problem (published in Greece), 1980. [8] Kalomitsines S., Methods for mathematical problem solving (in Greek), Athens, 1998. [9] Kalomitsines S., Two methods for solving problems, Educational studies in

Mathematics, Vol. 14, 3 pp. 251-274, 1983. [10] Kalomitsines S., The spiral method for solving problems, International Journal of

Mathematical Education in Science and Technology, Vol. 16, 4 pp.1-6, 1985. [11] Kalomitsines S., Some new ways of proceeding in problem solving, Report Pittsburgh

University, 1985. [12] Libet, B., Mind time, Harvard University, Cambridge, 2005. [13] Newell, A. and Simon H. A., Human problem solving, Prentice Hall Englewood Cliffs,

N. H., 1972. [14] Newell, A., Production systems-models of control structures in visual information

processing, Ed. W.C. Chases, Academic Press, New York, 1973. [15] Polya G., How to solve it, Garden City, N. J. Doubleday, 1957. [16] Polya G., Mathematical discovery vol. 1, Wiley, New York, 1962. [17] Polya G., Mathematical discovery vol. 2, Wiley, New York, 1964. [18] Rubinstein, M. F., Patterns of problem solving, Prentice Hall, New Jersey, 1975. [19] Schoenfeld, A. H., Mathematical problem solving, Academic Press, New York, 1985 [20] Wickelgren, W. A., How to solve problems, Freeman, New York 1974.

ABOUT THE AUTHOR Spyros Kalomitsines has been teaching High School Mathematics at the Experimental

School of the University of Athens and other schools in Greece for about forty years. In the class he was always giving emphasis on problem solving methods and he invented some new ones, which are included in this book. The author, as a Fulbright scholar, at the suggestion of Professor W. Wickelgren went to the Learning Research and Development Center of Pittsburgh University, where top researchers on problem solving were working. There, encouraged by Professor J. Greeno, he formulated the heuristic principles of one of his methods explicitly and implemented them as a computational model in problem solving. He also had long discussions about his methods with H. Simon (Nobel 1978) at the Carnegie Mellon University.

INDEX

A

access, 13 activation, 19 actual output, 127 age, 19, 61, 62 aid, 35 air, 103 algorithm, 12, 13, 58, 126 alternative, viii, 28, 32, 35, 38, 57, 61, 129, 140 alternatives, 13, 39, 130, 139, 147 anthropology, 6 application, 128, 135 aptitude, vii Archimedes, 14, 69 Ariel, 29 arithmetic, 2 associations, 22 astronomy, 29 Athens, ix, 18, 171, 173 attention, 2, 45

B

black, 6, 7 bogs, 29 boys, 60, 61 brain, 17, 22, 45 brainstorming, 18

C

calculus, viii, 17, 42, 69, 84 capacity, 17, 66 children, 61 classes, 18, 41, 43, 125, 126, 145 classical, 29, 125 classification, 40

classroom, 1 coding, 1 cognitive, vii, ix, 4 cognitive psychology, vii, ix, 4 colors, 6 complexity, 6 composition, 18, 19 computer, vii, viii, 4, 13, 20, 31, 36, 125, 126, 127 concave, 120, 121, 122, 167 concrete, 19 constraints, 2, 3, 22, 27, 29, 137 continuing, 139 continuity, 92, 152 control, 171 convex, 27 cosine, 137, 138, 159, 160 costs, 23, 24 Crete, ix critical points, 119 culture, 18, 19 cycles, 130

D

danger, 2, 7 declarative knowledge, 18 definition, 12, 13, 14, 27, 86, 97, 132, 142, 146, 152,

156 degree, 49 derivatives, 161 deterministic, 129 differentiation, 97 dilation, 12 division, 9, 39, 40, 57 DSB(description method), 35, 36, 39, 139, 140, 141

E

earth, 15, 16, 29

Index 176

economy, 18 education, 18, 19, 171 Einstein, 5, 17, 29 encouragement, vii, ix entertainment, 18 enthusiasm, 17 environment, 3, 18, 19 equality, 38, 41, 46, 79, 85, 86, 99, 104, 146, 147,

151, 156, 161 Euclidean, 13, 27 Euclidian geometry, 10 evil, 5 evolution, 17 exercise, 64 eye, 4, 64

F

failure, 21 false, 56, 75 family, 18 fear, vii, 3 feeding, 7 film, 65 flexibility, 22 food, 18 freedom, 18

G

Galileo, 1 games, 80 generation, 18 gestation, 40 gifted, 17 girls, 60, 61 goals, 57 God, 5 gold, 17 graph, 106, 119, 120, 121 gravitation, 16 gravitational constant, 16 gravity, 29 Greece, ix, 81, 171, 173 grouping, 84 groups, viii, 18, 27, 29

H

happiness, 18 Harvard, 171 health, 18, 19

height, 70, 104 heuristic, 12, 173 high school, ix, 27 homework, 10 homogeneous, 66 host, 131 human, 6, 17, 29, 45, 130 human brain, 17, 29 humanity, 29 humans, 6

I

identity, 9, 19, 26, 75, 78, 126, 127, 150 indeterminacy, 88 individual differences, 8 induction, 144 inequality, 47, 48, 55, 82, 95, 108, 109, 110, 113,

139, 147, 156, 164, 167 inferences, 7, 8, 10, 65 infinite, 29, 49, 53, 57, 58, 145 information processing, 171 inspection, 112 inspiration, 40 instruction, 31, 51, 88, 101, 102 intelligence, 8 interpretation, 28 interval, 93, 107, 161, 163, 164, 167 intuition, 93 inventions, 17 inventors, 17

J

justice, 18

L

language, 3, 4, 6, 18, 65, 66 law, 14, 15, 20, 75, 126, 127, 131, 160 laws, 14, 20, 126 lead, 8, 12, 13, 26, 31, 35, 51, 53, 125, 129 learning, 171 Leibniz, 69 links, 23, 24 locus, 53 lying, 10, 56

M

magazines, 18

Index 177

mathematical, vii, 13, 61, 144, 171 mathematical knowledge, vii mathematicians, 17, 65 mathematics, vii, 1, 7, 17, 28, 29, 50, 84, 171 matrix, viii, 75, 76, 77, 78, 79 matrix algebra, viii means-end analysis, vii, 35 mechanical, 13 median, 133 memory, 81 men, 45, 61 MIT, ix models, 171 molecules, 14 money, 80 Moon, 15 morning, 29 multiplication, 32, 149 music, 4

N

natural, 42, 115 New Jersey, 171 New York, 29, 171 Newton, 15, 69 Newton’s law, 16 Newtonian physics, 29 normal, 100

O

observations, 7, 17 oil, 63 oil production, 63

P

paper, ix, 3, 7 parents, 18, 19 pattern recognition, 40 periodic, 42, 137 perseverance, 7 personal, 36 philosophers, 8 philosophical, 29 physics, 1, 14, 102 planar, 22, 70, 71, 72 play, 19, 80 polynomial, 13, 39, 40, 48, 49, 57, 97, 106 polynomials, 40, 88, 91, 92 power, 30, 31, 36, 41, 145

powers, 5, 8, 9, 30, 31, 36 probability, 17, 23 probability theory, 17 problem solving, vii, viii, ix, 1, 3, 13, 81, 147, 171,

173 problem space, 7, 31, 129 procedures, 22 production, viii, 63, 64, 125, 129, 131 production function, 64 program, vii, 13, 31, 127, 130, 131, 132 projector, 65 psychological, 3, 4 psychology, 81, 171 pure water, 66

Q

quantum theory, 29

R

radius, 74, 104 rain, 19 random, 13, 17, 54, 147 reading, vii, 5, 8, 15, 21, 23, 25, 26, 53, 60 real numbers, 4, 42, 45, 47, 80, 115 reality, 18 reasoning, 39 recall, ix, 2, 4, 5, 11, 18, 24, 45, 142 recalling, 2, 19 reception, 60 recognition, 40 reduction, 63, 64, 66, 67 regular, 50 rejection, 29 relationships, 52 Renaissance, 29 research, 29, 171 Research and Development, 173 researchers, ix, 8, 17, 173 returns, 61 revolutionary, 29 rhythm, 3, 4, 7 rings, 126

S

salt, 66, 67, 68 scalar, 75 school, vii, 18, 173 science, 18, 45, 61 scientific, 29

Index 178

search, viii, 7 searching, 7, 8, 13 self-confidence, 3 semicircle, 103 semigroup, 19, 20 sentences, 22, 23, 24, 81 series, 58 set theory, 50 shape, 133 short-term memory, 17 sign, 4, 110, 119, 164 similarity, 1, 7, 45, 54, 149 simulation, 4 sine, 137, 138 skills, vii social, 63 social security, 63 solutions, vii, viii, 7, 20, 24, 25, 39, 40, 45, 53, 81,

84, 113, 125 special relativity, 17 spectrum, 19 speed, 65, 106 sports, 18 stars, 33, 136 sterile, 7 storms, 19 strategies, 2, 40, 84 students, vii, viii, 1, 4, 7, 8, 10, 18, 19, 27, 28, 36,

39, 40, 44, 45, 61, 81, 82, 83, 84, 126 substitution, 132 summaries, 83, 84 summer, 18 Sun, 15 surface water, 16 symbols, 3, 43, 44, 61 symmetry, 50, 73 systematic, 58 systems, 171

T

teachers, viii, 7, 18, 43 teaching, vii, viii, 27, 81, 83, 171, 173 tellers, 56

textbooks, 12, 82, 125 theory, 9, 10, 16, 17, 29, 49, 82, 125, 126, 171 thinking, 1, 3, 4, 7 tides, 17 time, 1, 4, 6, 7, 17, 18, 29, 45, 60, 65, 66, 67, 84,

103, 104, 106, 144, 171 tomato, 63 torus, 72 total product, 64 training, 3, 8, 19 transfer, 30, 145 transformation, 12, 26, 30, 145 transformations, 10, 12, 28, 31, 41 transparent, 65 traps, 39 trial, 26, 58 trial and error, 26

V

values, 49, 57, 113, 138, 153 variable, 10, 103, 168 variables, 27, 31, 38, 51, 85, 87, 103 variation, 58 velocity, 103 visual, 171

W

walking, 106 war, 29 water, 14, 16, 80, 104, 145 weakness, 29 wells, 63 wet, 19 workers, 63 writing, ix, 3, 4, 7, 11, 18, 19, 22, 61, 102

Y

yield, 26, 27, 31, 32, 52, 55, 60, 82, 129

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