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Life and Travel

in 4D

Life and Travel

in 4DLife and Travel

in 4DTic-Tetris-Toe

Tic-Tetris-Toe

Tic-Tetris-Toe Shark

A

ttacks

Shark

A

ttack

s

Shark

A

ttack

s

December 200

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in the Sky is a semi-annual publication of

PIMS is supported by the Natural Sciences and Engineer-

ing Research Council of Canada, the British Columbia

Information, Science and Technology Agency, the Al-

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University, the University of Alberta, the University of

British Columbia, the University of Calgary, the Univer-

sity of Victoria, the University of Washington, the Uni-

versity of Northern British Columbia, and the University

of Lethbridge.

This journal is devoted to cultivating mathematical rea-

soning and problem-solving skills and preparing students

to face the challenges of the high-technology era.

Editors in ChiefNassif Ghoussoub (University of British Columbia)Tel: (604) 8223922, E-mail: director@pims.math.caWieslaw Krawcewicz (University of Alberta)Tel: (780) 4927165, E-mail: wieslawk@v-wave.com

Associate Editors

John Bowman (University of Alberta)Tel: (780) 4920532 E-mail: bowman@math.ualberta.caDragos Hrimiuc (University of Alberta)Tel: (780) 4923532 E-mail: hrimiuc@math.ualberta.caVolker Runde (University of Alberta)Tel: (780) 4923526 E-mail: runde@math.ualberta.ca

Editorial Board

Peter Borwein (Simon Fraser University)Tel: (640) 2914376, E-mail: pborwein@cecm.sfu.caFlorin Diacu (University of Victoria)Tel: (250) 7216330, E-mail: diacu@math.uvic.caKlaus Hoechsmann (University of British Columbia)Tel: (604) 8225458, E-mail: hoek@math.ubc.caMichael Lamoureux (University of Calgary)Tel: (403) 2203951, E-mail: mikel@math.ucalgary.ca

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in the Sky in the SkyPacific Institute for Pacific Institute forthe Mathematical Sciences the Mathematical Sciences449 Central Academic Blg 1933 West MallUniversity of Alberta University of British ColumbiaEdmonton , Al berta Vancouver, B.C.T6G 2G1, Canada V6T 1Z2, Canada

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Contributions Welcome

in the Sky accepts materials on any subject related tomathematics or its applications, including articles, problems,cartoons, statements, jokes, etc. Copyright of material submit-ted to the publisher and accepted for publication remains withthe author, with the understanding that the publisher may re-produce it without royalty in print, electronic and other forms.Submissions are subject to editorial revision.

We also welcome Letters to the Editor from teachers,

students, parents or anybody interested in math education (besure to include your full name and phone number).

Cover Page: In October 2001, in the Sky was invited bythe principal of Edmontons Tempo School, Dr. Kapoor, tomeet with students in grades 10 and 11. The picture on thecover page was taken by Henry Van Roessel at Tempo Schoolduring our visit. More photos from that visit are published onpage 28.If you would like to see your school on the cover page of in the

Sky, please invite us for a short visit to meet your students and staff.

CONTENTS:The Language of MathematicsTimothy Taylor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

Tic-Tetris-ToeAndy Liu, University of Alberta . . . . . . . . . . . . . . . . . . . . . . . 4

WeierstraVolker Runde, University of Alberta . . . . . . . . . . . . . . . . . . . . 7

Life and Travel in 4DTomasz Kaczynski, Universite de Sherbrooke . . . . . . . . . . . . . 1 0

Shark Attacks and the Poisson ApproximationByron Schmuland, University of Alberta . . . . . . . . . . . . . . . . 1 2

The Rose and the NautilusA Geometric Mystery Story

Klaus Hoechsmann, University of British Columbia . . . . . . . . 15

Three Easy TricksTed Lewis, University of Alberta . . . . . . . . . . . . . . . . . . . . . 18

Inequalities for Convex Functions (Part I)Dragos Hrimiuc, University of Alberta . . . . . . . . . . . . . . . . . 20Math Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

Math Links . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Our School VisitDon Stanley, University of Alberta . . . . . . . . . . . . . . . . . . . . . 2 8

cCopyright 2001Sidney Harris

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This column is an open forum. We welcome opin-ions on all mathematical issues: research; educa-tion; and communication. Please feel free to write.Opinions expressed in this forum do not necessarily reflectthose of the editorial board, PIMS, or its sponsors.

The Language of Mathematics

by Timothy Taylor

Im a writer. I write storiesnovels, short stories, novel-

las. Im aware that sometimes the creative arts (like myfiction) and the hard sciences (like mathematics) are con-sidered uncomfortable companions. People tend to imag-ine themselves as being attracted to one or the other, butnot to both. In my case, to be truthful, I found mathe-matics difficult in school and for a long time I thought Ididnt like the subject. But I was wrong. Not only have Isurprised myself in recent years by discovering an interestin mathematics and its applications, I have surprised my-self more by discovering this interest through my creativewriting.

Somebody once asked me, Whats the most importantskill required to write fiction? I told them that you hadto be able to sit alone in a room and type on a computer

for long periods of time. This is partly true. When youhave your story idea and you know your characters, therecomes a point when you simply have to sit down and writefor as long as it takes to finish. But thats not the onlyrequirement, of course. There is also a lot of researchrequired to prepare yourself. In my writing, some of thisresearch might be considered incidental. If my charactervisits Rome, I make it my business to learn about the city.I dont just write, He went to Rome. I write somethinglike, He stayed at the Albergo Pomezia in Campo diFiorinot far from the old Jewish ghetto. This detail might nothelp us understand the character better, but it serves tocreate a sense of reality.

Ultimately, you do want the reader to understand thecharacters however, and heres where a more integralkind

of research comes into play. Characters in my stories tendto have a fairly clear set of desires and objectives in life.These can range from grand to mundane. But in any case,there will be a particular set of issues that concern a char-acter, some variety of problem that he or she must solve.For example, a character arriving at a new school mightwish to meet new friends. My character in Rome (hesan art critic) is consumed with the work of a particularpainter. The crucial thing is that the set of issues, orthe problems that confront a character, determine whichlanguage they use.

When I say language, I dont mean tongue, whereChinese, Russian, English, or French might be examples.Instead, I mean the set of words and concepts that a char-acter is inclined to use within a given tongue. These arisedirectly from the issues that concern that character. Andso, there is a unique language of art criticism (colour, com-position, theme, culture, aesthetic etc.), just as there is alanguage of business, of church, of personal relationships,

anddrum roll pleaseof mathematics!

I hadnt really considered this until I wrote a story calledSilent Cruise a few years back. In that story, I introduceDett and Sheedy. Sheedy is a businessman and thinks onlyin those terms. Dett is a young man who is consumedby his own way of calculating probabilities (he does this,in part, because he likes betting at the racetrack). Inorder to put words into Detts mouth that make sensegiven his very peculiar obsessions, I had to re-acquaintmyself with a language I hadnt thought about in sometime: mathematics. I should emphasize that Detts wayof making calculations is not rigourous. A math studentreading the story would see this right away. But the pointis that he thinks not in a literate language (like Sheedy),

but in a numerate one. And Detts way of expressinghimself, to a large extent, defines who he is, how well hecommunicates with Sheedy, and what kinds of problemshe is able or unable to solve. As I wrote the story, Ienjoyed trying to think through Dett in his numerateway, even though I had a hard time doing so at school.In the process, I came to think that of all the languages Ihad researched for characters over the years, mathematicsis very special. I would even go so far as to say that it isa precious language. Its difficult to learn and, as a result,it is rare and valuable. But it is also very powerful, andperhaps this interests me more. As I wrote Detts storyeven though he applied his numeracy in an unconventionalwayhe had the tools to solve many, many problems thatSheedy did not. And that fact, boiled right down, was the

essence of my story.

An editor once commented on Silent Cruise in a news-paper article. He said that I had drawn a picture of acharacter who thought primarily with numbers and howthat was very unusual in Canadian fiction. I take thesewords as a compliment beyond any other that I have re-ceived in connection with my writing. They mean that Ihad not only done my research well enough to convincethis editor, but that I had communicated some of whatis preciousrare and valuablein the language of math-ematics. I only wish I spoke it better.

Timothy Taylor is the author of the national bestsellerStanley Park, a novel. His book, Silent Cruise and OtherStories, will be published next year. The short story,Silent Cruise, was short-listed for the Journey Prize 2000.Taylor won the prize for a different story, where the lan-guage spoken was concerned mostly with cheese.

We should also mention that an article written by Timo-thy Taylor for Saturday Night Magazine won a Gold Medalat the National Magazine Awards.

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Tic-Tetris-Toe

by Andy Liu

Part I: Introduction

Tic-Tetris-Toe is very much like Tic-Tac-Toe. Theclassic game is played on a 3 by 3 board, taking a squarein each turn. Whoever is first to get 3 squares in a rowor 3 on a diagonal wins. However, in this new game, wemake two changes.

First, while we still play on a square board, it does nothave to be 3 by 3. Tic-Tetris-Toe is actually five differentgames, each with a board of a different size. Second, wetry to get different shapes. We use those from the popular

video game Tetris, as shown in Figure 1.

Figure 1

There are actually seven different pieces, but since theyare allowed to turn over, we have only five Tic-Tetris-Toe games. We call these pieces N4, L4, T4, I4 and O4,because they each have four squares and look like the let-ters N, L, T, I and O, respectively.

For N4, we play on a 3 by 3 board. For L4, we play ona 4 by 4 board. For T4, we play on a 5 by 5 board. ForI4, we play on a 7 by 7 board. For O4, we play on a 9 by9 board. Of course, the advantage is with the first player.Can you figure out a way for a sure win if I let you gofirst? Try these games with your friends, and then checkbelow. Dont peekthat will spoil the fun!

Part II: The N4 Game

Let us label the rows of the board 1, 2, and 3, and thecolumns a, b, and c. That way, each square will have aname. For example, the square at the bottom left cornerwill be called a1.

You will need at least four moves to win, and you willhave an extra fifth move that may come in handy in somescenarios. Mapping out a winning strategy requires thatyou look quite far ahead. On the other hand, I (the op-posing player) may be able to stop you from winning withone or two moves. Perhaps you should first consider whatmy strategy will be.

ab

c

1

2

3

O

ab

c

1

2

3

O O

ab

c

1

2

3

O

O

Figure 2

Figure 2 shows three ways in which I can stop you fromwinning. In each case, even if I give you all of the remain-ing squares, you still cannot complete N4. This tells youthat you must take b2 in your first move, and make surethat you take at least one of b1 and b3, and at least oneof a2 and c2.

Note that once you have taken b2, you do not have toworry about me sneaking up on you for a surprise win.You will do no worse than a draw. It would be too em-barrassing to lose as the first player. Can I still stop youfrom winning? After you have taken b2, I really have twodifferent choices: taking an edge square or a corner one.

Suppose I take a2. You already know that you musttake c2. Now I give up. In your third move, you can takeb1 and create a double-threat at a3 and c1. If I preventyou from doing this by taking one of these three squares,you will take b3 and create a double-threat at a1 and c3.Figure 3 shows that the key to your success is the W5shape.

a b c

1

2

3

O1 X1 X2

a b c

1

2

3

O1 X1 X2

Figure 3

Am I better off if I start with a corner square, say a1?Suppose you still take c2. After all, it has worked once.Now I know that I must take one of a3, b3, and c1. Youcan force me to take c3 on my next move by taking b1yourself. Then you can create a double-threat by takingone of c3, a2, and a3, depending on my move.

a b c

1

2

3

O1

O2 O3

X1 X2

X3 X4

a b c

1

2

3

O1

O2 O3

X1 X2

X3

X4

a b c

1

2

3

O1 O2

O3

X1 X2

X3

X4

Figure 4

Can you remember all of this? You do not have todo that. Just understand that you must have b2, one ofa2 and c2, and one of b1 and b3. Then look for doublethreats. With a little bit of practice, you will always winif you move first.

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Part III: The L4, T4 and I4 Games

Label the extra rows 4, 5 and so on, and the extracolumns d, e and so on. You can have an easy win inthe L4 game. Start by taking b2. You are guaranteed toget b3 or c2 in your next move. If both are still there andboth b1 and b4 are still empty, take b3. Otherwise, a2and d2 will be empty, so take c2. On your third move,

complete 3 squares in a row, and I cannot stop you fromcompleting L4 on your fourth move. Since I have onlymade three moves so far, I cannot beat you to it.

You can win the T4 game by starting at the obviousplace, c3. I can make one of five essentially different re-sponses, at a1, a2, a3, b2, or b3. On your second move,you take d4. On your third move, you take either c4 tocreate a double-threat at b4 and c5, or d3 to create adouble-threat at d2 and e3. I can neither stop you norbeat you to it.

1

2

3

4

5

6

7

a b c d e f g

O1

O2

O3

O4O5

X1

X2

X3

X4X5

X6

Figure 5

The I4 game is the only one of the five that can beplayed competitively. While you have a sure win, it cannotbe forced until your eighth move. In trying for the win,it is possible that you may set up a double-threat for me.

Figure 5 shows a sample game in which I put up a goodfight. On the seventh move, you either take c6 for thedouble-threat at a6 and e6, or take b4 for the double-threat at b3 and b7. I cannot stop both.

Part IV: The O4 Game

This game holds a big surprise. Even though the boardlooks more than large enough, you will not be able to forcea win. I have a very simple but effective counter strategythat will prevent you from winning, even if we play on aninfinite board. It is an elegant idea which demonstratesthe beauty of mathematics.

Figure 6

I will combine pairs of adjacent squares into dominoesin the pattern of a brick wall, into which you will bashyour head in vain. Whenever you take a square, I willtake the other square of the same domino. As shown inFigure 6, no matter how you fit in O4, it must contain acomplete domino. Since you can only have half of it, youcannot win!

Part V: Further ProjectsProblem 1.Find four connected shapes of three squares or less, joinededge-to-edge.

Remark: These shapes are called the monomino O1, thedomino I2, and the trominoes I3 and V3. None of themprovides much challenge as a gamethe first player has aneasy win if the board is big enough. This is because eachof these pieces form parts of other pieces for which the firstplayer can win. Our Tetris pieces are the tetrominoes. Ifwe go to the pentominoes, you will find these games muchmore challenging. There are twelve such pieces, called F5,I5, L5, N5, P5, T5, U5, V5, W5, X5, Y5, and Z5, asshown in Figure 7. Pentomino is a registered trademark

ofSolomon Golomb, who has written a wonderful bookcalled Polyominoes. This word means, shaped or formedof many squares. After the pentominoes come the 35hexominoes, 108 heptominoes, 369 octominoes, and so on.

Problem 2.Since P5 contains O4, and the domino strategy of Figure6 works for the O4 game, the second player can also forcea draw in the P5 game, even if it is played on an infiniteboard. On the other hand, there are four pentominoes thatdo not contain O4, but for which the domino strategy ofFigure 6 also works. Which pentominoes are they?

Problem 3.Show how the first player can force a win for the N5 gameon a 6 by 6 board, and for the L5 and Y5 games on a 7by 7 board.

Figure 7

Problem 4.Match each of the other four pentominoes with one of thepatterns in Figure 8 for a domino strategy.

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Figure 8

Problem 5.How many of the hexominoes contain a pentomino forwhich the second player has a domino strategy?

Figure 9

Problem 6.

Find domino strategies for the second player in games us-ing the hexominoes in Figure 9. For one of them, you willhave to find a new pattern.

Figure 10

Problem 7.With the possible exception of the hexomino in Figure 10,no polyominoes formed of six or more squares offer thefirst player a sure win, even on an infinite board. Can a

win be forced in a game using this hexomino?

Problem 8.Returning to Tic-Tetris-Toe, it is easy to see that there isno win for the first player in the N4 game if it is playedon a 2 by 2 board, because it is not even big enough tohold the piece. The L3 game on a 3 by 3 board is also adraw, if played properly. The classic Tic-Tack-Toe is stilla draw even with additional winning configurations. Canthe first player still force a win in the T4 games on a 4 by4 board, or the I4 game on a 6 by 6 board?

Part VI: Acknowledgement

This article is based on Martin Gardners Mathemati-cal Games column in Scientific Americanmagazine, April,1979. It has since been collected into the anthology Frac-tal Music, Hypercards and More, as Chapter 13, under thetitle Generalized Tic-tac-toe. This book was published byW. H. Freeman and Company, New York, in 1992. Theoriginal work was done by the noted graph theorist FrankHarary.

A mother of three is pregnant with her fourth child. Oneevening, her eldest daughter says to her dad, Do you know,daddy, what Ive found out?

No.The new baby will be Chinese!What?!Yes. Ive read in the paper that statistics show that every

fourth child born nowadays is Chinese. . . .

A father who is very much concerned about his sons poorgrades in math decides to register him at a religious school.After his first term there, the son brings home his report card;he gets As in math.

The father is, of course, pleased, but wants to know, Whyare your math grades suddenly so good?

You know, the son explains, when I walked into the class-room the first day and saw that guy nailed to a plus sign onthe wall, I knew one thingthis place means business!

What happened to your girlfriend, that really smart mathstudent?

She is no longer my girlfriend. I caught her cheating onme.

I dont believe that she cheated on you!Well, a couple of nights ago I called her on the phone,

and she told me that she was in bed wrestling with three un-

knowns....

Q: Why do mathematicians often confuse Christmas andHalloween?

A: Because Oct 31 = Dec 25.

Q: How do you make one burn?A: Differentiate a log fire.

Write to us if you get this joke!

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Weierstra

Volker Runde

Each university and each department develops a pecu-liar kind of folkloreanecdotes about those of its gradu-ates (or dropouts) that somehow managed to become fa-mous (or notorious). Very often, there is an element ofglee to these stories: Well, he may now be a governmentminister, but I flunked him in calculus! And also veryoften, it is impossible to tell the truth from the legend.

Karl Weierstra

When I was a math stu-dent at Munster, Germanyin the 1980s, such anecdotescentered mainly around twopeople: Gerd Faltings, thefirst and only German to winthe Fields medal1, mathe-matics equivalent of the No-bel prize; and Karl Weier-stra, the man who (be-sides many other mathemat-ical accomplishments) intro-duced and into calculus.Weierstra had been a stu-dent at Munster in the 1830sand 1840s. There was no onearound anymore who knew

anybody who had known anybody who had known any-body who had known Weierstra, but this didnt preventthe folklore from blooming. According to his legend,Weierstra flunked out of law school because he spent

most of his time there drinking beer and doing math-ematics. Then he worked for more than ten years asa school teacher in remote parts of Prussia, teachingnot only mathematics, but also subjects like botany,calligraphy, and physical education. Finally, when almost40 years old, he became a famous mathematician, andwas eventually appointed a professor at Berlinwithoutever having received a PhD. This story may sound wild,and in some ways it simplifies the facts, but it is not farfrom the truth.

Karl Theodor Wilhelm Weierstra was born on October31, 1815, in the village of Ostenfelde, which is located inwhat was then the Prussian province of Westphalia. Astreet and an elementary school in Ostenfelde are named

after him, and his birth housestill occupied todayis The last letter is not a , but an , a letter unique to the

German alphabet, which is pronounced like an s. Books writtenin English usually spell the name Weierstrass.

Volker Runde is a professor in the Department of Math-ematical Sciences at the University of Alberta. His web site ishttp://www.math.ualberta.ca/runde/runde.html and his E-mailaddress is vrunde@ualberta.ca .

1 If you want to know more about the Fields medal: there is anarticle on itThe Top Mathematics Award by Florin Diacuin theJune 2001 issue of in the Sky.

listed in a local tourist guide. His father, who worked forPrussias customs and taxation authorities, was sent fromone post to the next within short periods of time. For the

Karl Weierstraas a young man

first 14 years of Karl Weier-stra life, his family wasmore or less constantly onthe move. In 1829, Karls fa-ther obtained an assistants

position at the tax office inthe city of Paderborn (alsoin Westphalia), and the fam-ily could finally settle down.Young Karl enrolled at thelocal Catholic Gymnasium inPaderborn, where he excellednot only in mathematics, butalso in German, Latin, andGreek. Not only was he astrong student, he was alsoquite capable of putting hisbrains to work on much more

practical matters. At age 15, Karl contributed to hisfamilys income by doing bookkeeping for a wealthymerchants widow.

Throughout his life, Weierstra Senior suffered from theknowledge that he did not have the right education to riseto a rank in the Prussian civil service that would havebetter suited his abilities. Instead, he had to content him-self with relatively low-level positions, not very challengingand not very well paid. Like many a father in this situ-ation, he was determined to prevent such a fate befallinghis bright eldest son. When Karl graduated in 1834, hisfather decided to send him to Bonn to study Kameralis-tik (a combination of law, finance, and administration).Being a dutiful son, Karl went. . .

. . . and did all he could to sabotage the life his fatherhad planned for him. He joined a schlagende Verbindung,

a kind of student fraternity typical of German universitiesin the 19th century. Besides keeping the brewing indus-try busy, fraternity members engaged in a peculiar rit-ual: the Mensur, a swordfight with a peculiar twist. Un-like in todays athletic competitions, the students foughtwith sharp sabers. They wore protective gear that cov-ered most of their bodiesexcept the cheeks. During aMensur, the opponents tried to inflict gashes on one an-others cheeks. The scars were borne with pride as signsof honour and manhood.2 Almost two metres tall, quickon his feet, and with strong arms, Karl Weierstra was afearsome swordsman. His face remained unscarred, andafter a while nobody was keen on challenging him any-more. Having escaped from under his fathers tutelage,he spent his years at Bonn drinking beer and wielding thesaberand seriously studying mathematics. Although hewas not enrolled in mathematics, he read some of the mostadvanced math books of his time. In 1838, when it wastime for him to take his exams, he simply dropped out.

His family was desperate. They had made considerablefinancial sacrifices to secure a better future for Karl, whohad let them down. Having wasted four years of his life,he needed a bread-winning degree, and fast. So, in 1839,

2 If you find such ideas of honour and manhood absolutely re-volting, youre absolutely right.

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he enrolled at the Akademie in Munster, the forerunner totodays university, to become a secondary school teacher.Although this was not really a university, but rather ateacher training college, they had one good mathemati-cian teaching thereChristoph Gudermann. He is saidto have been an abysmal teacher: very often, he had justone student sitting in his classKarl Weierstra. In 1840,Weierstra graduated. His thesis was so good that Gud-

ermann believed it to be strong enough for a doctoral de-gree. However, the Akademie was not really a university;it did not have the right to grant doctorates. So, insteadof receiving a doctorate and starting an academic career,Weierstra left the Akademie as a mere school teacher.

His first job (probationary) was in Munster. One yearlater, he was sent to Deutsch Krona3 in the province ofWest Prussia as an auxiliary teacher, then, in 1848, toBraunsberg4 in East Prussia. Of course, he taught math-ematics, but also physics, geography, history, German,andbelieve it or notcalligraphy and physical educa-tion. Besides the demands of working full time as a teacherand having a social life (remember, he liked beer), he foundtime to do research in mathematics. During his time in

Braunsberg, he published a few papers in his schools year-book. High school year books are not exactly where peo-ple look for cutting edge research in mathematics, and sonobody noticed them. Then, in 1854, he published a pa-per entitled, Zur Theorie der Abelschen Functionen in awidely respected journal. I wont even make an attempt toexplain what it was about. But unlike his previous work,this one was noticed.

It dawned on mathematicians all over Europe that theman who was probably the leading analyst of his daywas rotting in a small East Prussian town, spending mostof his time teaching youngsters calligraphy and physicaleducation. On March 31, 1854, Weierstra finally re-ceived a doctorate, an honorary one from the university ofKonigsberg.5 In 1856, he accepted a position at the Gewer-beinstitut in Berlin, an engineering school, and a year laterhe joined the faculty of the University of Berlin as an ad-junct professor. As a teacher, he attracted large audiences.Often, he taught in front of more than 200 students. In1869, when he was almost 50 years old, Weierstra was ap-pointed full professor at the university of Berlin. In 1873and 1874, he was Rektor magnificus of the university; in1875, he became a knight of the order Pour le Meritein the category of Arts and Sciences, the highest honournewly unified Germany could bestow upon one of its citi-zens; and, in 1885, on the occasion of his 70th birthday, acommemorative coin was issued in his honour.

The years of leading a double life as a secondary schoolteacher and a mathematical researcher took their toll on

Weierstra health. A less vigorous man would probablyhave collapsed under the double burden much earlier. In1850, Weierstra began to suffer from attacks of dizziness,which culminated in a collapse in 1861. He had to pausefor a year before he could teach again, and he never re-covered fully. In 1890, at age 75, Weierstra retired fromteaching because of his failing health. The last years of

3 Now Walcz in Poland.4 Now Braniewo in Poland.5 Now Kaliningrad in Russia.

his life were spent in a wheelchair. In 1897, he died.

Weierstra published few papershe was very criticaltoward his own work. But although he was a brilliant re-searcher, the greatest impact he had on mathematics wasas a teacher. At Berlin, he repeatedly taught a two-yearcourse on analysis, the predecessor of all modern introduc-tions to calculus and analysis. Although he never wrote

a textbook, notes taken in class by his students have sur-vived and convey an impression of his lectures. Perhapsthe longest lasting legacy of those lectures is their em-phasis on rigour. When calculus was created in the 17thcentury, mathematicians did not worry about rigourouslyproving their results. For example, the first derivativedy/dx of a function y = f(x) was thought of as a quotientof two infinitesimals (i.e., infinitely small quantities dyand dx). Nobody could really tell what infinitely smallquantities were supposed to be, but mathematicians thendidnt really care. The new mathematics enabled themto solve problems in physics and engineering that hadbeen beyond the reach of the human mind before. So whybother with rigour? In the 18th century, mathematicianswent so far as to proclaim that rigour was for philosophers

and theologians, not for mathematicians. But with thelack of rigour, contradictory results cropped up with dis-turbing frequencypeople often arrived at formulae thatwere obviously wrong. And, if a particular formula deter-mines whether or not a bridge collapses, you dont wantit to be wrong. Weierstra realized that if calculus wasto rest on solid foundations, its central notion, that of thelimit, had to be made rigourous. He introduced the defi-nition that (essentially) is still used today in classrooms:

A number y0 is the limit of a function f(x) as xtends to x0 if, for each > 0, there is > 0 suchthat |f(x) y0| < for each x with |x x0| < .

Students may curse it, but it will not go away.

Weierstra was not only an influential lecturer, but alsoone of the most prolific advisors of PhD theses of all time.There is a database on the Internet6 that lists 31 PhD stu-dents of Weierstra and 1,346 descendants (i.e., PhDs ofPhDs of PhDs etc.) of Weierstra. Interestingly, the twoformer students who generated the most folklore werenthis students in a technical sense.

Sofya Kovalevskaya was a young Russian noblewomanwho had come to Germany to study mathematics. Thisalone was no small feat at a time when the very idea of a

woman receiving a university education was revolutionary.For two years, she studied at Heidelberg, where authoritieswould not let her enroll officially, but eventually allowedher to attend lectures unofficially (provided the instructordid not object). Then she moved to Berlin to work withWeierstra, only to find that she was not even allowed toaudit lectures.

6 The Mathematics Genealogy Project athttp://hcoonce.math.mankato.msus.edu/ .

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Sofya Kovalevskaya

This prompted Weier-stra, by all we know a po-litically conservative man,to tutor her privately.Since Kovalevskaya couldnot receive a doctoratefrom Berlin, Weierstraused his influence to per-suade the University of

Gottingen to award herthe degree in 1874. Shespent the following nineyears jobhunting. Be-ing a woman didnt help.The best job she couldfind was teaching arith-metic at an elementaryschool. Finally, in 1883,she was offered a pro-fessorship at Stockholm,where she worked until her

death in 1891 at the age of 41. Weierstra and Ko-valevskaya stayed in touch throughout her mathematicalcareer. After her death, Weierstra destroyed their

correspondence. This fact, along with Kovalevskayasstriking beauty, gave rise to innuendos that she may havebeen more to Weierstra (who never married) than just astudent. Perhapsbut we dont know.

Karl Weierstrain old age

Gosta Mittag-Leffler,another of the great math-ematicians protegees, wasalso not Weierstra stu-dent strictly speaking.Already enrolled at theUniversity of Uppsala,Sweden, he came to Berlinin 1875 to attend Weier-stras lectures, which hadan enormous impact onhis mathematical develop-ment. He then returnedto his native Sweden,where he received his doc-torate. Over the years,Mittag-Leffler became in-disputably the mostinfluential mathematicianof his time in Sweden. Hemade use of his clout to

overcome the obstacles faced by Sofya Kovalevskayaregarding her appointment at Stockholm. What Mittag-Leffler is most famous for, however, is not a mathematical

accomplishment, but a piece of mathematical folklore.To this day, mathematicians suffer quietly from thelack of a Nobel prize, and, some say, Mittag-Leffleris to blameaccording to legend, the first version ofNobels will mentioned a prize in mathematics. Then,Nobel found out that his wife had had an affair withMittag-Leffler. Infuriated that his wifes lover could wellbe the first prize winner, Nobel changed his will andremoved the math prize. Thats a fine piece of juicyfolklore, but nothing more; like Weierstra, Nobel was alifelong bachelor.

At the end of his course on mathematical methods in opti-mization, the professor sternly looked at his students and said,There is one final piece of advice Im going to give you nowwhatever you have learned in my course, never, ever try toapply it to your personal lives!

Why? the students asked.

Well, some years ago, I observed my wife preparing break-fast, and I noticed that she wasted a lot of time walking backand forth in the kitchen. So, I went to work, optimized thewhole procedure, and told my wife about it.

And what happened?

Before I applied my expert knowledge, my wife needed

about half an hour to prepare breakfast for the two of us. Andnow, it takes me less than fifteen minutes. . . .

Q: What is an extroverted mathematician?

A: One who, in conversation, looks at the other personsshoes instead of at his own.

In a dark, narrow alley, a function and a differential operatormeet, Get out of my way or Ill differentiate you til yourezero!

Try itIm ex. . . .

cCopyright 2001Sidney Harris

Same alley, same function, but a different operator: Getout of my way or Ill differentiate you til youre zero!

Try itIm ex. . . .

Too bad. . . Im d/dy.

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Life and Travel in 4D

by Tomasz KaczynskiIt is the spirit of the age to believethat any fact, no matter how suspect,is superior to any imaginative exercise,no matter how true.Gore Vidal

In higher-level mathematics courses, the study of vectorspaces of arbitrary dimension n or even infinite dimensionis often required. A natural question that arises is:

How can we visualize a space of a dimensionhigher than 3?

We can draw pictures in 2D, we can make models in 3Dbut 4D? That seems very strange! Most teachers have nochoice but to introduce an n-dimensional space Rn as apurely algebraic object consisting of sequences

x := (x1, x2, x3, . . . , xn)

of n real numbers. But this does not satisfy people whodabble more in geometry than in algebra. Some physi-cists and mathematicians claim to be able to see in 4DEinstein supposedly could even see in 5D! I do not quitebelieve those stories, but rather think that all we can do isaccept 4D or 5D, learn to cope with it, and maybe one daywe will believe that we really see it. When I was in college,I once attended a series of talks given by university stu-

dents with the aim of popularizing mathematics. There Ilearned a beautiful way of coping with a multi-dimensionalspace. The answer is in the concept of empathy, the ca-pacity for participation in anothers feelings or ideas. Whois that other person? He is called Flatman.

Figure 1

Tomasz Kaczynski is a professor in the Departement demathematiques et dinformatique at Universite de Sherbrooke. Hisweb site is http://www.dmi.usherb.ca/kaczyn/index.html andhis E-mail address is kaczyn@dmi.usherb.ca .

We recommend the book by Edwin A. Abbott, Flatland: aromance of many dimensions, London, Seeley, 1884.

Let us put ourselves in the position of Flatman, a 2Dman who lives in a 2D world and is unable to imagine a3D space. What does his 2D world look like? His worldis a straight line. His skin is not a surface, it is a closedcurve. This is shown in Figure 1, which is not just another2D projection of our World, but an image showing how hisFlat World would actually appear.This presents Flatmanwith some technical problems. For example, if Flatman

wants to pass to the right of Flatwoman, he has to jumpover her. The anatomy of Flatman or Flatwoman is alsonot evident.

Now, lets look at Figure 2. Flatwomans digestivesystem cannot be a tube like the one we have insideus, because it would split her into at least two separateparts (called by mathematicians connected components),as shown on Figure 2(a). We may guess that Flatman orFlatwoman must digest food similarly to the bacteria inFigure 2(b).

Note that the Flatman in Figure 1 cannot see the in-terior of the Flatwoman. But we, 3D-beings, can see herinterior from our outer dimension. Maybe some superior4D-being can see what we have inside without using X-rays, and could remove a tumor from a patients body

without any surgery. It is thus clear that concepts such asinterior, exterior, and boundary are relative to the outerspace in which our world is embedded.

Part APart B

Food

(a) (b)

Figure 2

Lets give this discussion some mathematical meaning.Suppose a prisoner is kept at a point P = (0, 0) of a 2D-world R2 whose points are denoted by the Cartesian coor-dinates (x, y). His cell is limited by the circle S1 given bythe equation x2 + y2 = 1. If there is no hole in the circle,there is no way he could get outside of the circle, e.g., tothe point Q = (2, 0). This fact is intuitively acceptable,but very hard to proveit is the famous Jordan ClosedCurve Theorem. But everything changes if we embed the

plane into the 3D-space R3

, whose points are denoted byCartesian coordinates (x, y, z). Our planar world is givenby the equation z = 0. The prisoners position is nowP = (0, 0, 0), the boundary circle is given by the pair ofequations

x2 + y2 = 1,z = 0,

and the destination point is Q = (2, 0, 0). The prisonercan easily get out by jumping over the circle. A possible

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trajectory in time t can be given by

x = t,y = 0,z = t(2 t),

which starts at t = 0 and reaches the point Q at t = 2.

This is shown in Figure 3(a).

01 2

Q x

y

z

P

S1 S2

P Q

(a) (b)

Figure 3

We may now raise the dimension. Consider the pointP = (0, 0, 0) inside a balloon-shaped cell called sphere S2

and given by the equation x2 + y2 + z2 = 1. Consider thepoint Q = (2, 0, 0) on the outside of the sphere. Again,there is no way a prisoner staying at P could move out toQ without cutting through the sphere. Lets add an extradimension: the points ofR4 are (x, y, z, u), P = (0, 0, 0, 0),and Q = (2, 0, 0, 0). Our 3D-space is given by the equationu = 0, and the limiting 2D-sphere S2 is now given by

x2 + y2 + z2 = 1,u = 0.

With the extra fourth dimension, the prisoner can jumpoutside the sphere without cutting it. The trajectory isnow

x = t,y = 0,z = 0,u = t(2 t).

This scenario is shown in Figure 3(b). Now, lets tacklea more serious topic. Once upon a time, a scientist namedFlatilei discovered that the Earth is not a line, but a circlearound a disc. Many flat sailors rushed to attempt a cruise

around the world, the first one being Flatellan. Severalcenturies pass before some flastronomers claim that theuniverse is not a 2D space; it is actually limited. It mightbe, for example, a huge balloon in a 3D space, i.e., thesphere S2R given by the equation

x2 + y2 + z2 = R2,

where R is the radius of the sphere. This is shown inFigure 4.

Figure 4

The flat circular Earth is a small round patch in thatspherical universe. Science-fiction writers imagine storiesof space missions where a spacecraft is sent along a straightline and returns to the Earth. How can that be? Becausewhat was believed to be a straight line is actually a greatcircle on the sphere. We may attempt the same mentalexercise by adding one more dimension, and view our uni-verse as a 3D-sphere S3 given by the equation

x2 + y2 + z2 + t2 = R2.

One may ask: If the universe is limited, why must it bea 3D-sphere? Actually, it does not have to be. In the nextissue, we will investigate other possibilities; look forwardto the article Travelling on the Surface of a Giant Donut.

Q: Do you already know the latest stats joke?A: Probably....

Q: What is the fundamental principle of engineering mathe-matics?A: Every function has a Taylor series that converges to thefunction and breaks off after the linear term.

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Shark Attacks andthe Poisson

Approximation

by Byron Schmuland

A story with the dramatic title Shark attacks at-tributed to random Poisson burst appeared on September7, 2001 in the National Post. In the story, Professor DavidKelton of Penn State University used a statistical modelto try to explain the surprising number of shark attacks

that occurred in Florida last summer. According to thearticle, Kelton suggests the spate of attacks may have hadnothing to do with changing currents, dwindling food sup-plies, the recent rise in shark-feeding tourist operations, orany other external cause.

Just because you see events happening in a rash like thisdoes not imply that theres some physical driver causingthem to happen. It is characteristic of random processesthat they exhibit this bursty behaviour, he said.

What was the professor trying to say? Can mathematicsreally explain the increase in shark attacks? And what arethe mysterious Poisson bursts?

The main point of the Professor Keltons comments was

that unpredictable events, like shark attacks, do not occurat regular intervals as in Figure 1(a), but tend to occurin clusters, as in Figure 1(b). The unpredictable natureof these events means that there are bound to be periodswith a higher than average number of events, as well asperiods with a lower than average number of events, oreven no events at all.

Find more about the author at the following web site:http://www.stat.ualberta.ca/people/schmu/dept page.html

You can also send your comments directly to the author atschmu@stat.ualberta.ca

Regular Events

Time

(a)

Random Events

Time

(b)

Figure 1

The statistical model used to study the sequences ofrandom events gets its name from French mathematicianSimeon Denis Poisson (1781-1840), who first wrote aboutthe Poisson distribution in a book on law. The Poissondistribution can be used to calculate the chance that aparticular time period will exhibit an abnormally largenumber of events (Poisson burst), or that it will exhibitno events at all. Since Poissons time, this distribution hasbeen applied to many different kinds of problems, such asthe decay of radioactive particles, ecological studies onwildlife populations, traffic flow on the Internet, etc. Hereis the marvellous formula that helps to predict the proba-

bility of random events:

The chance of exactly k events occurring is

k

k! e,

for k = 0, 1, 2, . . .

The funny looking symbol is the Greek letter lambda,and it means the average number of events. The symbolk! = k(k1) . . .21 is the factorial ofk, and e isthe exponential function ex with the value x = pluggedin. Lets take this new formula out for a spin.

Shark Attack!If, for example, we average two shark attacks per sum-

mer, then the chance of having six shark attacks next sum-mer is obtained by plugging = 2 and k = 6 into theformula above. This gives

probability of six attacks (26/6!) e2 = 0.01203,which is a little more than a 1% chance. This means thatsix shark attacks are quite unlikely in any one year, al-though it is likely to happen about once every 85 years.The chance that the whole summer passes without anyshark attacks can also be calculated by plugging = 2and k = 0 into the formula. This gives

probability of no attacks (20/0!) e2 = 0.13533,which is a 13% chance. Thus, we can expect a sharklesssummer every seven or eight years.

In this hypothetical shark problem, the number of at-tacks followed the Poisson distribution exactly. The Pois-son distribution is most often used to find approximateprobabilities in problems with n repeated trials and prob-ability p of success. Let me show you what I mean.

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Lotto 649One of my favourite games to study is Lotto 649. Six

numbers are randomly chosen from 1 to 49, and if youmatch all six numbers you win the jackpot. Since the num-ber of possible ticket combinations is

496

= 13,983,816,

your chance of winning the jackpot with one ticket is onein 13,983,816, which is p = 7.15 108. Lets say you area regular Lotto 649 player and that you buy one ticket

twice a week for 100 years. The total number of ticketsyou buy is n = 100 52 2 = 10,400. What is the chancethat you will win a jackpot sometime during this 100 yearrun?

This is a pretty complex problem, but the Poisson for-mula makes it simple. First of all, the average num-ber of jackpots during this time period is = np =10,400/13,983,816 = 0.0007437. Plugging this into theformula with k = 0 shows that the chance of a jackpot-less 100 years is

probability of no jackpots e0.0007437 = 0.99926.Wow! Even if you play Lotto 649 religiously for 100 years,there is a better than 99.9% chance that you will never,ever win the jackpot.

CoincidencesTake two decks of cards and shuffle both of them thor-

oughly. Give one deck to a friend and place both decksface down. Now, at the same time, you and your friendturn over the top cards. Are they the same? No? Thentry again with the second card, the third card, etc. If yougo through the whole deck, what is the chance that, atsome point, you and your friend will turn over the samecard?

In this problem, there are n = 52 trials and the chanceof a success (coincidence) on each trial is p = 1/52. Theaverage number of coincidences is = np = 52/52 = 1,

and so putting k = 0 in the Poisson formula givesprobability of no coincidences e1 = 0.36788.

The chance that you will see a coincidence is 1 0.36788 = 0.63212. You will get a coincidence about 63%of the time you play this game. Try it and see!

Birthday ProblemSuppose there are N people in your class. What are the

odds that at least two people share a birthday? Imaginemoving around the class checking every pair of people tosee if they share a birthday. The number of trials is equalto the number of pairs of people, i.e., n =

N2

= N(N

1)/2. The probability of success in a given trial is the

chance that two randomly chosen people share a birthday,i.e., p = 1/365. This gives the average number of sharedbirthdays as = N(N 1)/(2 365), so the probabilityof no shared birthdays is

probability of no shared birthdays eN(N1)/(2365).Therefore, the probability of at least one shared birth-day is approximately 1 eN(N1)/(2365). Heres whathappens when you try using different values of N in thisformula.

Probability of a shared birthday

N Prob N Prob

10 0.115991 60 0.99216620 0.405805 70 0.99866230 0.696320 80 0.99982640 0.881990 90 0.99998350 0.965131 100 0.999999

With N = 10 people, there is only about an 11.5%chance of a shared birthday, but with N = 30 people thereis a 69.6% chance. In a large class (like at a university!)with N = 100 students, a shared birthday is 99.9999%certain.

In a large class, perhaps it is possible to have a triplebirthday. Following the same pattern, lets work out thechance that there is at least one triple shared birthday ina class of N people. This time, as each triple of people ischecked, there are N3 = N(N 1)(N 2)/6 trials, andthe chance of success on each trial is p = 1/3652.

This gives = N(N 1)(N 2)/(6 3652), so theprobability of no triple shared birthdays is

probability of notriple shared birthdays

eN(N1)(N2)/(63652).

Therefore, the probability of at least one triple shared

birthday is 1 eN(N1)(N2)/(63652). Now, lets lookat different values of N in this formula.

Probability of a triple shared birthday

N Prob N Prob

10 0.002699 60 0.53725420 0.025344 70 0.70848130 0.087370 80 0.84277940 0.199470 90 0.92902750 0.356838 100 0.973779

My large first year statistics courses usually have about100 students, and I always check their birthdays. Accord-ing to the table, there should be a triple shared birthdaymore than 97% of the time. It really is true; there hasalways been a triple shared birthday in my classes.

The Great OneDuring Wayne Gretzkys days as an Edmonton Oiler, he

scored a remarkable 1669 points in 696 games, for a rate of = 1669/696 = 2.39 points per game. Using the Poissonformula, with k = 0, we estimate that the probability ofGretzky having a pointless game is

probability of no points (2.39)0

0!e2.39 = 0.0909.

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Over 696 games, this ought to translate into about 696 0.0909 = 63.27 pointless games. In fact, during that pe-riod, he had exactly 69 pointless games.

For one-point games, we find an approximate probabil-ity of

probability of one point (2.39)1

1!e2.39 = 0.2180,

for a predicted value of 696 0.2180 = 151.71 one-pointgames. Lets try the same calculation for other values ofk,and compare the Poisson formula prediction to the actualstatistics.

Points Actual # Games # Predicted by Poisson

0 69 63.271 155 151.712 171 181.903 143 145.404 79 87.175 57 41.816 14 16.717 6 5.728 2 1.729 0 0.46

As you can see, there is remarkable agreement betweenthe predictions of the Poisson formula, and the actualnumber of games with different point totals. This showsthat Gretzky was not only a high scoring player, but aconsistent one as well. The occasional pointless game, orPoisson burst in seven- or eight-point games, was notdue to inconsistent play, but was exactly what would beexpected in any random sequence of events. Another rea-

son why he really was the Great One!

A physicist, a statistician, and a pure mathematician go tothe races and place bets on horses.

The physicists horse comes in last. I dont understand it.I have determined each horses strength through a series ofcareful measurements.

The statisticians horse does a little bit better, but still failsmiserably. How is this possible? I have statistically evaluatedthe results of all races for the past month.

They both look at the mathematician, whose horse came infirst. How did you do it?

Well, he explains. First, I assumed that all horses wereidentical and spherical. . . .

Two men are having a good time in a bar. Outside, theresa terrible thunderstorm. Finally, one of the men thinks thatits time to leave. Since he has been drinking, he decides towalk home.

But arent you afraid of being struck by lightning? hisfriend asks.

Not at all. Statistics shows that, in this part of the country,one person per year gets struck by lightningand that oneperson died in the hospital three weeks ago.

Isnt statistics wonderful?How so?Well, according to statistics, there are 42 million alligator

eggs laid every year. Of those, only about half get hatched. Ofthose that hatch, three-fourths of them get eaten by predatorsin the first 36 days. And of the rest, only 5 percent get to be ayear old for one reason or another. Isnt statistics wonderful?

Whats so wonderful about all that?If it werent for statistics, wed be up to our arses in alliga-

tors!

cCopyright 2001Sidney Harris

Do you know that 87.166253% of all statistics claim a preci-sion that is not justified by the method employed?

A mathematician has been invited to speak at a conference.His talk is announced as, Proof of the Riemann Hypothesis .

When the conference takes place, he speaks about somethingcompletely different. After his talk, a colleague asks him, Didyou find an error in your proof?

He replies: No, I never had one.But why did you make this announcement?Thats my standard precautionin case I die on my way

to the conference. . . .

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The Rose and

the Nautilus

A Geometric Mystery Story

by Klaus Hoechsmann

The image belowor some variation thereofappearsin many books and web sites. The story behind it dealswith such diverse concepts as

golden rectangles,Fibonacci numbers,

regular pentagons,and logarithmic spirals.

Figure 1

Its main appeal lies in uncovering the invisible threadsconnecting these items. Its origins are lost in the mists ofantiquity, with some parts as old as the Pyramids, and oth-ers surfacing in Euclids Elements (250 BC) and PtolemysAlmagest(150 AD). The artwork by Leonardo da Vinci inPaciolis Divina Proportione (1509 AD) helped spread itsfame beyond the mathematical crowd.

Older accounts are entirely geometric, but with theirfussy monochrome diagrams and awkward notation, theyare difficult to follow. More recent versions, on the other

hand, tend to achieve brevity by abundant use of algebraicformulas, which most people find incomprehensible. Thisarticle will try to provide a simple but complete accountin the geometric vein and in full colour.

Figure 2 is a simplification of Figure 1. To recreate thelatter, you need only cut a quarter circle out of the yellowsquare, another one out of the light green square, thenthe dark green square, and so onand draw the brown

Find more about the author and other interesting articles athttp://www.math.ubc.ca/hoek/Teaching/teaching.html .

diagonals in again.

Figure 2

Here you have a rectangle made up of smaller andsmaller squares arranged in spiral fashion. This is knownas a golden rectangle.

A rectangle is golden if it has the following special prop-erty: if you cut a square (shown in yellow) away from it,you are left with a rectangle of the same shapeso youcan continue cutting off squares indefinitely.

A rose is a rose is a roseand shares its pentagonal sym-metry with many other flow-ers, from buttercups to petu-nias. The underlying penta-gram (shown here in blue) ismade up of golden triangles,each of which is obtained from

a golden rectangle by collapsingone of the shorter sides.

This is the connection between golden rectangles andregular pentagons.

The nautilus, a tropical sea-shell, provides a beautiful ex-ample of a logarithmic spiral,described by a point revolving

around a center while, at thesame time, moving outward ex-ponentially (like the tip of a

clock-hand that grows as compound interest). Hence, itcould just as well be called an exponential spiral. Butpeople love the L-wordit sounds so impressiveandsince it merely refers to the reverse way of looking at thesame pattern, it is quite as legitimate as the E-word.

The scaffolding for such a curve is given by a kind ofrectangular spiral (shown here in green) wrapping around

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a rectangular cross (in red).

Figure 3

Figure 3 reveals an analogous structure within thegolden rectangle. In fact, if you start with anynon-squarerectangle, draw a diagonal and (from another corner) aperpendicular to it, and then wrap a rectangular spiralaround the resulting cross, you always get the frameworkfor some logarithmic spiral. Doing this with a goldenrect-angle, gives you a bonus: your spiral can now be approxi-

mated by a bunch of (easily drawn) quarter circles. Alas,only approximated! The golden spiral shown in Figure1 is not truly logarithmicit is not even smooth: if youwere driving along it with constant speed, you would feela sudden jerk in the steering wheel as you changed circlesin passing from the yellow square to the green one.

So much for the geometric inhabitants of this zoobutwhat about Fibonacci and his numbers

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, . . . ?

This is the hub of the mystery, although it looks in-nocent and even playful at first sight. Through this se-quence, Fibonacci (about 1200 AD) was attempting tomodel the growth of a population in which the newbornhave to sit out one mating season before getting involvedin the game. He actually thought of rabbits! What doesthis rabbit sequence have to do with the divine propor-tion?

The truth is that the squares shown in Figure 2 are notin an ideal plane, but are actually made up of coloureddots (called pixels), and have side lengths of 144, 89,55, 34, 21 and 13 pixels, respectivelyrabbit numbers!

The red rectangle stuck in themiddle of Figure 2 measures 13-by-8 pixels, and when enlarged10-fold looks like what you seeon the left.

If you continue to cut off squares, you willafter 5stepswind up with the white square. And then? Ina truly golden rectangle, you would be able to continuecutting off squares indefinitely. So, this is a fakebut aclose imitation! This is no coincidence, as we shall nowsee.

Inflating Away the Imbalance

Inflating a rectangle means adding a square to its longerside, as shown on the right. The original rectangle is pur-

ple, while the added square is yellow. The enlarged rectan-gle will be referred to as the inflationof the old one. Thisis not standard terminology; it is an ad hoc term coinedfor the sole purpose of making this text easier to read.(Mathematics allows that kind ofpoetic license, as long as it is con-sistent.) We want to study theway inflation affects the shape ofa rectangle. If it remains un-altered, we say the rectangle isgolden.

We shall eventually see that continued inflation leadsto a more and more golden shape. But what does thatmean? How can we compare the shapes of rectangles andcheck that one is more golden than another?

Of course, we all know thatrectangles come in variousshapes, from squat squares tothe thinnest of strips. Hereis a way to compare them:when two rectangles have the

same shapelike the blueand the yellow ones depicted on the lefttheir diagonalsline up as shown, splitting the whole diagram exactly intwo. The two grey rectangles must therefore be equal inarea.

If the blue and yellow shapes are not the same, their di-agonals do not line up, and the complementary grey areasare not equal. This is the situation shown below on theright.

To compare the shape ofa rectangle with the shapeof its inflation, we thereforehave to look at the diagrambelow, where the blue and

purple rectangles are congru-ent (i.e., their correspondingsides are equal in length).The difference between theareas of the upper grey squareand the lower grey strip willbe called the imbalance of thepurple rectangle.

This is another ad hoc term intended to simplify thelanguage. In some sense, the imbalance of a rectanglemeasures how far it is from being golden. The imbalanceis zero in a golden rectangle.

The grey square and stripin the diagram above, showup againcoloured yellowand blue, respectivelyin thediagram at the right. (Makesure you can argue this in de-tail!).

This suggests another way of defining the imbalance ofthe purple rectangle: It is the difference between the areasof (a) the square that must be added to inflate it and (b)the strip that must be added to the inflation to completethe larger square.

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Now let us look at theimbalance of the inflation,shown all in purple on theleft. The square used to in-flate the inflation is yellow,the strip required to com-plete the larger square is blue.What is the difference be-

tween the areas of these twoshapes? The answer is shown

on the right: since the blueand yellow rectangles are con-gruent, they contribute noth-ing to the new imbalance. Itis still based on the differencebetween the two grey areasjust like the old one!

Conclusion: Inflation doesnot change the size of theimbalance. In particular,inflation keeps golden rectangles golden. Even better

since continued inflation keeps pumping up the area with-out changing the imbalance, the latter fades into insignif-icance compared with the size of the rectangles. Moreconcretely, in the last four diagrams above, the imbalancehappens to be five (square) pixelsa puny amount if wekeep inflating the rectangles to wall size. In the end, ourrectangles willfor all practical purposesbe golden.

Back to the Rabbits

So, where is the promised insight into the Fibonaccinumbers? Well, if you attach a square to the longer side ofa one-by-two rectangle, you get a two-by-three rectangle;further inflation yields a three-by-five rectangle, and so on,generating the sequence

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, . . . ,

already seen on the previous page. Comparing the squareof any of these numbers with the product of its two neigh-bours, you will always get a difference of 1. That is the im-balance. If, however, you begin with a one-by-three rect-angle, continued inflation will spawn the Lucas sequence,

1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, . . . ,

named after the French number wizard Edouard Lucas,who lived in the 19th century. Its imbalance is 5, and, ofcourse, it too marches toward the divine proportion.

Algebraically, the preservation of the imbalance is veryeasy to see. After all, we are dealing here with number

sequences

a, b, c, d, ...with c = a + b and d = b + c = a + 2b.

The imbalance for a, b, c is b2 ac = b2 a(a + b). Forb, c, d, it is c2 bd = (a + b)2 b(a + 2b), which comesto ac b2, as you can (and should) easily check. In otherwords, its size remains constant but it changes from plus tominus and vice versa. Can you see this switching of signsin the pictorial proof above? How would you comparethis algebraic proof with the geometric one? Would the

conclusions of this comparison hold in every case whereboth algebra and geometry can be used? Try to come upwith examples. . . .

What about the perfectly golden case, with zero imbal-ance: b2 = a(a + b)? Is this possible with integers a andb? Remember: an a by b rectangle would be golden,and could be deflated (as in Figure 2 on the previous

page) ad infinitum, always keeping integer values for itssides. Could that be?. . . TO BE CONTINUED. . . .

A mathematician gives a talk intended for a general audi-ence. The talk is announced in the local newspaper, but heexpects few people to show up because anyone who is not amathematician will be unable to make any sense of the title:Convex Sets and Inequalities.

To his surprise, the auditorium is full when his talk begins.After he has finished, someone in the audience raises a hand.

But you said nothing about the actual topic of your talk!What topic do you mean?Well, the one that was announced in the paper: Convicts,

Sex, and Inequality.

cCopyright 2001Sidney Harris

Q: What does the little mermaid wear?

A: An algae-bra.

One day, Jesus said to his disciples, The Kingdom of Heavenis like 3x squared plus 8x minus 9.

A man who had just joined the disciples looked very con-fused, and asked Peter, What on Earth, does he mean bythat?

Peter replied, Dont worryits just another one of hisparabolas.

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Three Easy Tricks

by Ted Lewis

These three math tricks will baffle most people.In each case, I will describe how you will presentthe trick to Jessica and Jonathan, two of yourspectators. The challenge for you is to figure outhow it is done before peeking at the explanationsat the end of the article. Each trick uses a differentbut simple mathematical principle.

Finding a CardYou hand a deck of cards to Jessica. Ask her to think

of a number no smaller than 11 and no larger than 19,and to deal that many cards face down in a pile on thetable. Turn your back so you cannot see how many cardsshe deals, and tell her to do it silently so that you cannotcount the cards.

When she has finished, take the rest of the deck fromher and place it alongside the small pile she has dealt.Ask her to pick up the small pile of cards that she dealt,add the two digits of her chosen number, then deal thatmany cards face down onto the top of the deck. So, ifthe number she chose was 17, she would now deal eightcards onto the top of the deck. To make the cards dealtcompletely random, ask her to shuffle the cards she is leftholding, and to remember the card that is on the bottomwithout showing it to you.

Then have her put this little packet of cards face downon top the deck. The card she looked at is now sandwichedsomewhere in the middle the deck.

Now, have her cut the deck twice. Although there seemsto be no way for you to do so, you look through the deckand immediately display her card.

A Prediction

Ted Lewis is a professor in the Department of Mathemat-ical Sciences at the University of Alberta. His web site ishttp://www.math.ualberta.ca/tlewis.

A sealed envelope is placed on the table.

You fan the cards and let Jonathan freely choose oneand place it face up on the table. You may even let himlook at the faces of the cards as you fan them. Let us sup-pose that card is the 8. You hand the deck to Jonathan,face down, and ask him to deal some cards face down nextto the 8, counting from the value of the card up to theKing. In this case, he would he would count, Nine, 10,Jack, Queen, King as he deals five face down cards. Askhim to turn the next card face up and repeat the process.Say it is the J: Jonathan deals two face down cards ashe counts, Queen, King. He turns up the next card, saythe 4, counting, Five, six, . . ., Jack, Queen, King ashe deals the face down cards. Now, he totals the face upcards: 8 + 11 + 4 = 23, and he counts 23 more cards facedown on the table.

He places the next card face up the table; it is, say,the ace of hearts. He opens the envelope and finds a pa-per inside with the message, You will choose the ace ofhearts!

In this trick, the cards that Jonathan turns face up do

not have to be an eight, a Jack and a fourthe trick worksno matter what they are.

Two SpectatorsBoth Jessica and Jonathan participate in this trick.

Three cards are placed face down on the table. Jessicais going to do some calculations, and using a calculatorwould be a good idea. Jonathan will not need to do anycalculations.

Ask Jessica to write down any three-digit number. Youturn your back so that you cannot see the number, and

you ask Jessica to create a six-digit number by writing thenumber alongside itself. For example, you explain, Ifthe original number was 123, the six-digit number wouldbe 123123.

Ask Jonathan to choose one of the three cards, and turnit face up. Then ask Jessica to divide the six-digit numberby the value of the face up card, ignoring any remainder,and to write down the result and circle it. A Jack countsas 11, a Queen as 12, and a King as 13. You may mentionthat, Although I dont have any idea what your result is,I have a very strong feeling that there was no remainder.

Ask Jonathan to choose one of the two remaining facedown cards and turn it face up. Ask Jessica to write downthe answer when the circled result is divided by the valueof the card. Surprisingly, again there is no remainder.

Ask Jessica to divide the latest result by the originalthree-digit number, and to write down the result of thiscalculation. This is the final result, and incredibly, thereis still no remainder.

Now review the situation for Jessica and JonathanJessica was free to choose any three-digit number whatso-ever. Jonathan was free to choose any two of the threeface down cards. Would it not be quite a miracle if the

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final result happened to be the value of the remaining facedown card? But that is exactly what happens.

There is some chicanery involved herethe three cardsare not arbitrary. Can you figure out what they are? Ifyou are good at arithmetic, you should be able to do so.

How the Tricks Work

Finding a Card

Before handing the cards to Jessica, memorize the cardthat is tenth from the top. This is the locator card, andher chosen card will always be the one directly above it.

There is an interesting mathematical reason why thisworks. A positive integer and the sum of its digits havethe same remainder when you divide by nine (see the boxat the end of this article). This means that if you havemulti-digit number and from it you subtract the sum ofthose digits, the result must be a number that is divisibleby nine, and it has to be exactly nine if the multi-digit

number is between 10 and 19. Now recall what Jessicadoes: she deals 1n cards on the table, where 1n is one ofthe numbers 11, 12, 13, . . . , 19. This reverses the order ofthe 1n cards. Then she picks up these cards and deals(1 + n) of them onto the top of the deck. Since the cardsare in reverse order, she is left holding the original topnine cards of the deck, and the tenth cardthe locatorcardis now back in position on top of the deck. Jessicashuffles the nine cards that she is holding, remembers thebottom card, and puts everything on top of the deck. So,the chosen card ends up directly on top of the locator card.Now I think thats rather neat.

A Prediction

Put the force card (the A) as the 43rd card fromthe top of the deck. The counting method ensures thatJonathan will put 42 cards onto the table. The chart be-low shows why:

Face up cardNumber of

face down cards

8J4

13 813 1113

4

The total number of cards on the table at this point is3+(138)+(1311)+(134). Jonathan then proceedsto deal (8 + 11 + 4) more cards, for a total of 3 + 3(13)cards. Incidentally, when doing this trick, make sure thatJonathan chooses his first card from the part of the deckabove the 43rd card. Do this by slowly fanning out onlythe top half of the deck for him to choose from.

Two Spectators

The three cards that are placed face down must be aseven, a Jack, and a King. The product of the valuesof the three cards is 1001, and the product of the three-digit number xyz with 1001 is xyzxyz. In other words,xyzxyz = xyz 7 11 13. By following your instructions,Jessica is dividing the six-digit number by three of its fac-

tors. Whats left is the fourth factorthe value of theremaining face down card.

If you would like to learn more about magic that is basedon simple mathematics, here are two excellent books:

Mathematical Magic by William Simon, preface byMartin Gardner, Dover Publications, New York, 1964.

Self-Working Card Tricks by Karl Fulves, Dover Pub-lications, New York, 1976.

Simons book explains the mathematics behind thetricks quite thoroughly. Fulves book includes many tricksthat are not mathematically based, and you will have tothink a bit to understand the mathematics behind those

that are.

The three tricks in this article are not from these books.Nevertheless, I would like to quote the advice given byBill Simon in the forward to his book: Be sure to runthrough the items you plan to use so that you can demon-strate them with certainty, with complete understanding,and in an entertaining manner. You will then be sure ofperforming real mathematical magic!

Dividing by 9

Suppose that xyz is a positive 3-digit number. Thismeans that the number is

100x + 10y + z .

Now rearrange it:

(99x + 9y) + (x + y + z) .

So, when we divide the number xyz by 9, we get thesame remainder as we do when we divide x + y + z by9. The same reasoning works regardless of the numberof digits.

A logician at Safeway:Paper or plastic?Not not paper and not plastic!

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Inequalities forConvex Functions

(Part I)

by Dragos Hrimiuc

1. Convex functions.Convex functions are powerful tools for proving a largeclass of inequalities. They provide an elegant and unifiedtreatment of the most important classical inequalities.

A real-valued function on an interval I is called convexif

f(x + (1 )y) f(x) + (1 )f(y) 1for every x, y I and [0, 1]; it is called strictly convexif

f(x + (1 )y) < f(x) + (1 )f(y) 2for every x, y I, x = y and (0, 1).

Notice: f is called concave (strictly concave) on I if fis convex (strictly convex) on I.

The geometrical meaning of convexity is clear: f isstrictly convex if and only if for every two points P =(x, f(x)) and Q = (y, f(y)) on the graph of f, the pointR = (z, f(z)) lies below the segment P Q for every z be-tween x and y.

x x

y y

0

P

Q P

Q

R

R

The graph of aconvex function

The graph of aconcave function

How to recognize a convex function without the graph?We can use 1 directly, but the following criterion is oftenvery useful:

Test for Convexity: Let f be a twice differentiablefunction on I. Then

f is convex on I if f(x) 0 for every x I. f is strictly convex on I if f(x) > 0 for every x

in the interior of I.

Remark: If f is a continuous function on I, then it canbe proved that f is convex if and only if for all x1, x2 I

f

x1 + x2

2

f(x1) + f(x2)

2;

Dragos Hrimiuc is a faculty member in the Department ofMathematical Sciences at the University of Alberta.

and it is strictly convex if and only if

f

x1 + x2

2

1;(iii) f(x) = 1(x+a)p , x > a, p > 0;(iv) f(x) = tan x, x 0, 2 ;(v) f(x) = ex, x R.

The following are examples of strictly concave functions:(i) f(x) = sin x, x [0, ];(ii) f(x) = cos x, x 2 , 2 ;(iii) f(x) = ln x, x (0, );(iv) f(x) = xp, x 0, p (0, 1).

Notice:

1. The linear function f(x) = ax + b, x

R is convex

and also concave.2. The sum of two convex (concave) functions is a convex

(concave) function.

2. Jensens Inequality.Jensens inequality is an extension of 1 . It was named

after the Danish mathematician who proved it in 1905.

Jensens Inequality: Let f : I R be a convexfunction. Let x1, . . . , xn I and 1, . . . , n 0 suchthat 1 + 2 + . . . + n = 1. Then

f(1x1 +2x2 . . .+nxn) 1f(x1) + . . .+nf(xn). 3

Proof: Lets use mathematical induction. The inequalityis true for n = 1. Now assume that it is true for n = k,and lets show that it remains true for n = k + 1.

Let x1, . . . , xk, xk+1 I and let 1, . . . , k, k+1 0with 1 + 2 + . . . + k + k+1 = 1. At least one of1, 2, . . . , k+1 must be less then 1 (otherwise the inequal-ity is trivial). Without loss of generality, let k+1 < 1 andu = 11k+1x1 + . . . +

k1k+1xk. We have

11 k+1 + . . . +

k1 k+1 = 1,

and also

1x1 + . . . + kxk + k+1xk+1 = (1 k+1)u + k+1xk+1.Now, since f is convex,

f((1k+1)u+k+1xk+1) (1k+1)f(u)+k+1f(xk+1)and, by our induction hypothesis,

f(u) 11 k+1 f(x1) + . . . +

k1 k+1 f(xk).

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Hence, combining the above two inequalities, we get:

f(1x1 + . . . + k+1xk+1) 1f(x1) + . . . + k+1f(xk+1).

Thus, the inequality is established for n = k + 1, andtherefore, by mathematical induction, it holds for any pos-itive integer n.

Remarks:

1. For strictly convex functions, the inequality in 3holds if and only if x1 = x2 = . . . = xn. Use mathe-matical induction to prove it.

2. If1 = 2 = . . . = n =1n , then 3 becomes

f

x1 + . . . + xn

n

f(x1) + . . . + f(xn)

n. 4

3. Iff is a concave function, then 3 and 4 read as

f(1x1 + . . . + nxn) 1f(x1) + . . . + nf(xn), 3and

f

x1 + . . . + xn

n

f(x1) + . . . + f(xn)

n. 4

Jensens Inequality has variety of applications. It canbe used to prove many of the most important classicalinequalities.

Weighted AMGM Inequality:Let x1, . . . , xn 0, 1, . . . , n > 0 such that1 + . . . + n = 1. Then

1x1 + . . . + nxn x1

1 x

2

2 . . . x

n

n 5The equality holds if and only ifx1 = x2 = . . . = xn.

Proof: We may assume that x1, . . . , xn > 0. Let f(x) =ln x, x (0, ). Since f is strictly concave on (0, ), byusing 3 we get:

ln(1x1 + . . . + nxn) 1 ln x1 + . . . + n ln xn,

or, equivalently, ln(1x1 +. . .+nxn) ln x11 . . . xnn , andhence,

1x1 + . . . + nxn x11 . . . xnn(since f(x) = ln x is a strictly increasing function).

By taking 1 = 2 = . . . = n =1n in 5 , we obtain:

AMGM Inequality:If x1, . . . , xn 0, then

x1 + x2 + . . . + xnn

nx1x2 . . . xn, 6

with equality if and only if x1 = x2 = . . . = xn.

Let x1, x2, . . . , xn, 1, 2, . . . , n > 0 be such that1 + . . . + n = 1. For each t R, t = 0, the weightedmean Mt of order t is defined as

Mt =

1x

t1 + 2x

t2 + . . . + nx

tn

n

1t

.

Some particular situations are significant:

M1 =1x1 + 2x2 + . . . + nxn

n

is called the weighted arithmetic mean (WAM); and

M1 =n

1x1

+ 2x2 + . . . +nxn

is called the weighted harmonic mean (WHM),

M2 =

1x21 + 2x

22 + . . . + nx

2n

n

is called the weighted root mean square (WRMS).It can be shown by using lHopitals Rule that

limt0

Mt = x11 x

22 . . . x

nn .

So, if we denote M0 = limt0

Mt, we see that

M0 = x11 x

22 . . . x

nn ,

which is called the weighted geometric mean (WGM).Also, if we set

M = limt

Mt and M = limt

Mt,

we obtain

M = max{x1, . . . , xn}, M = min{x1, . . . , xn}.

Power Mean Inequality:Let x1, x2, . . . , xn, 1, . . . , n > 0 be such that1 +. . .+n = 1. Ift and s are non-zero real numberssuch that s < t, then

1xs1 + . . . + nx

sn

n

1s

1xt1 + . . . + nxtn

n

1t

. 7

Proof: If 0 < s < t or s < 0 < t, the inequality 7 isobtained by applying Jensens Inequality 3 to the strictlyconvex function f(x) = x

ts . Indeed, if a1, a2, . . . , an, 1,

. . . , n > 0 and 1 + . . . + n = 1, then

1a1 + 2a2 + . . . + nan

n

ts 1a

ts1

+ 2ats + . . . + na

tsn

n.

By choosing a1 = xs1, . . . , an = x

sn, we immediately ob-

tain 3 .

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If s < t < 0, then 0 < t < s, and by applying 7 for1x1

, 1x2 , . . . ,1xn

, we get

1(

1x1

)t + . . . + n( 1xn )t

n

1t

1(

1x1

)s + . . . + n( 1xn )s

n

1s

,

which can be rewritten as 7 .

Remark: If t < 0 < s, then Mt M0 Ms. Also, wehave the following classical inequality:

M M1 M0 M1 M2 M.

Holders Inequality: If p, q > 1 are real numberssuch that 1p +

1q = 1 and a1, . . . , an, b1, . . . , bn are

real (complex) numbers, then

n

k=1 |

ak||

bk|

n

k=1 |

ak|p

1p

n

k=1 |

bk|q

1q

. 8

Proof: We may assume that |ak| > 0, k = 1, . . . , n.(Why?) The function f(x) = xq is strictly convex on(0, ), hence, by Jensens Inequality,

nk=1

kxk

q

nk=1

kxqk,

where x1, . . . , xn, 1, . . . , n > 0 and 1 + . . . + n = 1.Let A =

nk=1 |ak|p. By choosing k = 1A |ak|p and xk =

1

k |ak

||bk

|in the above inequality, we obtain 8 .

Remarks:

1. In 8 , the equality holds if and only if x1 = x2 =. . . = xn. That is,

|a1|p|b1|q =

|a2|p|b2|q = . . . =

|an|p|bn|q .

Notice that this chain of equalities is taught in thefollowing way: if a certain bk = 0, then we shouldhave ak = 0.

2. If p = q = 2, Holders Inequality is just Cauchys

Inequality:n

k=1

|ak||bk|2

nk=1

|ak|2

nk=1

|bk|2

.

The equality occurs when

|a1||b1| =

|a2||b2| = . . . =

|an||bn| .

Minkowskis Triangle Inequality: If p > 1 anda1, a2, . . . , an, b1, b2, . . . , bn 0, then

nk=1

(ak + bk)p

1p

nk=1

apk

1p

+

n

k=1

bpk

1p

. 9

Proof: We may assume ak > 0, k = 1, . . . , n. (Why?)

The function f(x) =

1 + x1p

p, x (0, ), is strictly

concave since f(x) = 1pp

1 + x

1p

p2 x 1p2 < 0. By

Jensens Inequality,1 +

n

k=1

kxk

1p

p

n

k=1

k

1 + x

1p

k

p,

where x1, . . . , xn, 1, . . . , n > 0 and 1 + . . . + n = 1.

Let A =

nk=1 a

pk. By taking k =

apk

A and xk =bpk

apk

for

k = 1, . . . , n in the above inequality, we obtain 9 .

Remarks.

1. The equality in 9 occurs if and only if

b1a1

=b2a2

= . . . =bnan

.

2. Ifp = 2 we get the so-called Triangle Inequality: nk=1

(ak + bk)2 n

k=1

a2k +

nk=1

b2k.

Example 1. If a, b 0 and a + b = 2, then1 + 5

a5

+ (1 +5

b)5 26.

Solution: Since f(x) = (1 + 5

x)5 is strictly concave on[0, ), by using Jensens Inequality 4 we get

2

1 +

5

a + b

2

5 (1 + 5a)5 + (1 + 5

b)5.

By substituting a + b = 2, we get the required inequality.The equality occurs when a = b = 1.

Example 2. If a, b, c > 0, then

aa bb cc

a + b + c

3

a+b+c.

Solution: The above inequality is equivalent to

ln(aa bb cc) ln

a + b + c

3

a+b+c,

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or

a ln a + b ln b + c ln c (a + b + c) ln

a + b + c

3

.

Let f(x) = x ln x, x (0, ). Since f(x) = 1x > 0, thefunction f is strictly convex on (0, ). Now, the aboveinequality follows from 4 .

Example 3. If a, b, c > 0 then

a

a + 3b + 3c+

b

3a + b + 3c+

c

3a + 3b + c 3

7.

Solution: Let s be a positive number and f(x) = xsx =s

sx 1, x (0, s). The function f is strictly convex sincef(x) = 2s(sx)3 > 0. We get:

2a

s

2a+

2b

s

2b+

2c

s

2c 3

13 (2a + 2b + 2c)

s

1

3(2a + 2b + 2c)

,

or

a

s 2a +b

s 2b +c

s 2c 3(a + b + c)

3s 2(a + b + c) .

If we take s = 3(a + b + c), the required inequality follows.

Example 4. If a1, a2, . . . , an 1, thenn

k=11

1 + ak n

1 + n

a1a2 . . . an.

Solution: Let f(x) = 11+ex , x [0, ). The function fis strictly convex since f(x) = e

x(ex1)(ex+1)3 > 0 on (0, ).

Using 4 , we get

nk=1

1

1 + exk n

1 + e1n

nk=1

xk.

By taking xk = ln ak, k = 1, . . . , n, we obtain the requiredinequality.

Example 5. For a triangle with angles , and , the

following inequalities hold:

sin + sin + sin 3

32 ;

sin + sin + sin 3 4

34 ;

sin sin sin 3

38 ;

cos cos cos 18 ; sec 2 + sec 2 + sec 2 2

3.

Solution: Use the Jensen Inequality for the strictly con-cave functions sin x,

sin x, lnsin x, lncos x, and for the

strictly convex function sec x2 , x (0, ).

Example 6. Let a1, . . . , an, 1, . . . , n > 0 and 1 +. . . + n = 1. If a

11 . . . a

nn = 1, then

a1 + a2 + . . . + an 111 . . .

nn

.

The equality occurs if and only if ak =k

11

...nnfor k =

1, . . . , n.

Solution: By using the Weighted AMGM Inequality, wehave

a1 + . . . + an = 1

a11

+ . . . + n

ann

a11

1. . .

ann

n= 1

11

...nn.

The equality occurs if and only if

a11

=a22

= . . . =ann

.

in which case the constraint a11 . . . ann = 1 leads to ak =

k11

...nnfor k = 1, . . . , n.

Example 7. If a1, . . . , an > 0 and a1a2 . . . an = 1, then

a1 +

a2 + . . . + n

an n + 12

.

Hint: Use the Weighted AMGM Inequality.

Example 8.(i) If a, b, c > 0, then

a10 + b10 + c10

a5 + b5 + c5

a + b + c

3

5.

(ii) If a1, a2, . . . , an > 0 and k > p 0, then

ak1 + . . . + akn

ap1 + . . . + apn

a1 + . . . + ann

kp.

Solution: (i) A particular case of (ii).

(ii) Let Mt =at1

+at1

+...+atnn

1t

. Then, by using the

Power Mean Inequality 7 , we get

ak1 + ak2 + . . . + a

kn = nM

kk = nM

pkM

kpk

nMppMkp1 by (7)= (ap1 + . . . + a

pn)

a1+...+an

n

kp.

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Example 9. If a1, a2, . . . , an > 0, then

(i) an+11 + . . . + an+1n a1 . . . an(a1 + . . . + an),

(ii) an11 + . . . + an1n a1 . . . an

1a1

+ . . . + 1an

.

Solution: (i)

an+11 + . . . + an+1n = nMn+1n+1 = nMnn+1M1n+1 nMn0 M11 = a1 . . . an(a1 + . . . + an).

(ii) See (i).

Example 10. If a, b, c, x, y, z, n > 0, and

(an + bn + cn)n+1 = xn + yn + zn,

thenan+1

x+

bn+1

y+

cn+1

z 1.

Solution:

an + bn + cn =an+1

x

nn+1 x nn+1 +

bn+1

y

nn+1 y nn+1

+cn+1

z

nn+1 z nn+1 .

By using Holders Inequality with p = n+1n and q =p

p1 =n + 1, we obtain:

an+bn+cn

an+1

x+

bn+1

y+

cn+1

z

nn+1

(xn+yn+zn)1

n+1 ,

and the required inequality follows.

Example 11. If a1, . . . , an, b1, . . . , bn > 0, then(a1 + . . . + an)n

+1

(b1 + . . . + bn)n a

n+11

bn1+ . . . +

an+1nbnn

.

Solution: By using Holders Inequality with p = n + 1,q = n+1n , we get

a1 + . . . + an =

an+11

bn1

1n+1

bn

n+1

1 + . . . +an+1nbnn

1n+1

bn

n+1n

an+11

bn1

+ . . . +an+1nbnn

1n+1

(b1 + . . . + bn)n

n+1 ,

so

(a1 +. . .+an)n+1

an+11

bn1+ . . . +

an+1nbnn

(b1 +. . .+bn)

n,

from which we get the required inequality.