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TheMathematicalGazette

Sir Wifred H. Cockcroft1923-1999

Volume 84: Number 499 March 2000THE MATHEMATICAL ASSOCIATION £16.00

THE MATHEMATICAL ASSOCIATIONAN ASSOCIATION OF TEACHERS AND STUDENTS

OF ELEMENTARY MATHEMATICS

'I hold every man a debtor to his profession, from the which asmen of course do seek to receive countenance and profit, so oughtthey of duty to endeavour themselves by way of amends to be ahelp and an ornament thereunto.' BACON

THE COUNCIL

PRESIDENT

Professor John Berry

IMMEDIATE PAST PRESIDENT Professor Chris RobsonPRESIDENT DESIGNATE Mr Steve AbbottCHAIR OF COUNCIL Dr Sue SandersSECRETARY Mr Roy AshleyTREASURER Mr Paul Metcalf

REPRESENTATIVESOF COMMITTEES

BRANCHES Mr Bob Francis (Chair)

CONFERENCES Mr Martin Bailey (Chair)

PROFESSIONAL DEVELOPMENT Ms Sue Jennings (Chair)

PUBLICATIONS Mr Peter Bailey (Chair)

PUBLICITY & MEMBERSHIP Mr David Hodgson (Chair)

TEACHING COMMITTEE Mr Doug French (Chair)

EDITOR IN CHIEF Mr Bill Richardson

MEMBERS

WITHOUT

OFFICE

Mr Robert Barbour Ms Trish MorganMr Neil Bibby Mr Michael MudgeMr David Carter Ms Robyn PicklesMiss Susie Jameson Mr Tony RobinDr Jim Message

OFFICE MANAGER Mrs Marcia Murray

EDITORIAL COMMITTEE OF THE MATHEMATICAL GAZETTEEditor Mr Steve Abbott

Production Editor Mr Bill RichardsonReviews Editors Mr Bud Winteridge

Mrs Rosalie McCrossanProblems Editors Mr Graham Hoare Mr Tim Cross

Assistant Editor Mr Gerry Leversha

CONTENTS (continued)

Notes 84.01 to 84.28 (continued)

Unexpected symmetry in a derived Alexander J. Gray 87Fibonacci sequence

A recurrence relation among Fibonacci sumsAlexander J. Gray 89

Some unusual iterations Mark Thornber 90

When the sum equals the product Leo Kurlandchik and 91Andrzej Nowicki

Never say never: some mistaken identities Mark J. Cooker 94

A curious property of the integer 24 M. H. Eggar 96

What do cycles of a given length generate?Mowaffaq Hajja 97

A game with positive and negative numbersM. H. Eggar 98

An inductive proof of the arithmetic mean Zbigniew Urmanin 101− geometric mean inequality

Weighted mean in a trapezium Larry Hoehn 102

A formula for integrating inverse functions S. Schnell and C. Mendoza103

Mathematician versus machine P. Glaister 105

On a conjecture of Paul Thompson Tim Jameson 107

Maximal volume of curved folding boxes Kenzi Odani 110

More on a sine product formula Walther Janous and 113Jeremy King

On a limit for prime numbers J. A. Scott 115

SHM and projections P. Glaister 116

Another cautionary chi-square calculation Nick Lord 119

More on dual Van Aubel generalisations Michael de Villiers 121

ObituarySir Wilfred Cockcroft 1923-1999 Peter Reynolds 123

Correspondence 125

Notices 127

Problem corner G. T. Q. Hoare 128

Student problems Tim Cross 135

Other Journals Anne C. Baker 139

Book Reviews 140

© The Mathematical Association 2000

TheMathematicalGazette

Sir Wifred H. Cockcroft1923-1999

Volume 84: Number 499 March 2000THE MATHEMATICAL ASSOCIATION £16.00

CONTENTS

Editorial 1

Articles

One hundred years on Graham T. Q. Hoare 2

Lewis Carroll − mathematician Canon D. B. Eperson 9and teacher of children

Snubbing with and without eta H. Martyn Cundy 14

The Fermat-Torricelli points of lines Roy Barbara 24n

Continued fractions Robert Macmillan 30

A construction of magic cubes Marián Trenkler 36

The factorial function: Stirling's formula David Fowler 42

A simple energy-conserving model Richard Bridges 51

The Hale-Bopp comet explored H. R. Corbishley 58with A level mathematics

Notes 84.01 to 84.28

A portrayal of right-angled triangles which I. Grattan-Guinness 66generate rectangles with sides in integral ratio

Circumradius of a cyclic quadrilateral Larry Hoehn 69

A neglected Pythagorean-like formula Larry Hoehn 71

An unexpected reduced cubic equation J. A. Scott 74

Touching hyperspheres D. F. Lawden 75

Comments on note 82.53—a generalised Andrejs Dunkels 79test for divisibility

A matrix method for a system of A. J. B. Ward 81linear Diophantine equations

On the application of Whittaker's theorem J. A. Scott 84

Digital roots and reciprocals of primes Alexander J. Gray 86

(The contents are continued inside the back cover.)Printed in Great Britain by J. W. Arrowsmith Ltd

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Mathematical GazetteA JOURNAL OF THE MATHEMATICAL ASSOCIATION

Vol. 84 March 2000 No. 499

Editorial: It's voting time again!The time has come to vote for theFifth Annual Mathematical Gazette

Writing Awards. Please use the address carrier from this issue of theGazetteto identify the articles and notes of 1999 that impressed you most. TheIndexfor 1999 will remind you of the many good submissions.

There will again be a prize draw among those who respond.Theprize, worth about £30, will be a copy of the bookMathematics: frontiersand perspectives, edited by Vladimir Arnold, Michael Atiyah, Peter Lax andBarry Mazur (AMS, 2000).

Previous Annual Mathematical Gazette Writing Awards

Year Best Article Best Note

Colin Fletcher David Fowler1996 Two prime centenaries A simple approach to

the factorial function

Ann Hirst and Keith Lloyd Colin Dixon

1997 Cassini, his ovals and aspace probe to Saturn

Geometry and thecosine rule

Robert M. Young Robert J. Clarke

1998 Probability, pi, and the primes

The quadratic equation formula

Please indicate, in the spaces provided on the voting form, the titles ofyour 3 favouriteArticles and your 3 favouriteNotes. Note thatMatters forDebate count asArticles. Alternatively, you can just write your choices in aletter or on a postcard. Each vote will be given equal weighting. The resultswill be announced in the July 2000 issue.

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STEVE ABBOTT

2 THE MATHEMATICAL GAZETTE

One hundred years onGRAHAM T. Q. HOARE

David Hilbert, one of the giants of mathematics, delivered a lecture atthe International Congress of Mathematics at Paris in 1900. The first part ofthe lecture, a preamble to his announcement of the now-famous 23problems, began with the words:

‘Who of us would not be glad to lift the veil behind which the futurelies hidden; to cast a glance at the next advances of our science andat the secrets of its development during future centuries? Whatparticular goals will there be toward which the leading mathematicalspirits of coming generations will thrive? What new methods andnew facts in the wide and rich fields of mathematical thought will thenew centuries disclose?History teaches the continuity of the development of science. Weknow that every age has its own problems, which the following ageeither solves or casts aside as profitless and replaces by new ones. Ifwe would obtain an idea of the probable development ofmathematical knowledge in the immediate future, we must let theunsettled questions pass before our minds and look over theproblems which the science of today sets and whose solution weexpect from the future. To such a review of problems the presentday, lying at the meeting of the centuries, seems to me well adapted.For the close of a great epoch not only invites us to look back intothe past but also directs our thoughts to the unknown future.The deep significance of certain problems for the advance ofmathematical science in general and the important role which theyplay in the work of the individual investigator are not to be denied.As long as a branch of science offers an abundance of problems, solong is it alive; a lack of problems foreshadows extinction or thecessation of independent development. Just as every humanundertaking pursues certain objects, so also mathematical researchrequires its problems. It is by the solution of problems that theinvestigator tests the temper of his steel; he finds new methods andnew outlooks, and gains a wider and freer horizon.’

Later we find the oft-quoted passage:‘This conviction of the solvability of every mathematical problem is apowerful incentive to the worker. We hear within us the perpetualcall: There is the problem. Seek its solution. You can find it bypure reason, for in mathematics there is no ignorabimus.’

Hilbert considered that the 23 problems he had chosen were those mostlikely to stimulate important new advances in mathematics. It redounds tohis perspicacity that much fruitful mathematical activity resulted inaddressing these problems in the twentieth century. As we shall see fromthe list below, which we give together with short commentaries and notes,

ONE HUNDRED YEARS ON 3

the so-called problems vary from specific mathematical questions toprogrammes of research. Some have been reformulated or extended withoutlosing their identity. We note the importance Hilbert attached to algebraicnumber theory, since the 8th problem, partly, and the 9th, 11th and 12th,entirely, are devoted to it. Problems 1, 2 and 10 belong to mathematicallogic, whereas 6, 19, 20 and 23 fall within the provinces of applications.Observe too that topology, then at an early stage of its development, featuresstrongly. Readers will appreciate that we cannot do justice to Hilbert'svision in a short article such as this.

Hilbert's Problems1. Cantor's continuum hypothesis (CH) and well-ordering1(a). Is ? Undecidable. Assuming the consistency of theZermelo-Fraenkel axioms for set theory (ZF), the work of K. Gödel (1938)and P. Cohen (1963) established that both the statement of the hypothesis(that ) and its negation are consistent with ZF. Thus thehypothesis is completely independent of the axioms of set theory.

2¼0 = ¼1

2¼0 = ¼1

1(b). Hilbert also asked whether the continuum of numbers can be well-ordered. This problem is related to the Axiom of Choice (AC), but in 1963P. Cohen proved the independence of AC from the other axioms of settheory, so the problem remains unresolved.

Note: Gödel believed that the AC and the CH were either true or falseand that ZF did not encapsulate what was ‘obviously’ true about set theory.The task was to think of some new axiom which would determine AC andCH. He did not succeed in devising such an axiom (the existence ofmeasurable cardinals was proposed as such, but was not in any sense‘obvious’) so this remains an unresolved consequence of the Hilbertchallenge.

2. To establish the consistency of the axioms of arithmeticGödel's two theorems shattered the Hilbert programme. The second of

these proves that the consistency of a theory at least as strong as arithmeticcannot be proved within the theory.

3. To show, using only the congruence axioms, whether two tetrahedrahaving the same altitude and base area have the same volume

Proved false by M. Dehn (1900).

4. To investigate geometries (metrics) in which the line segment betweenany pair of points gives the shortest path between the pair (geodesic)

Considered too vague.

5. Can the assumption of differentiability for the functions defining acontinuous transformation group be avoided?

Reformulated to encompass a larger domain of topological groups, theproblem was solved in the form that a locally Euclidean topological group is


a Lie group by A. Gleason (1952) and by D. Montgomery and L. Zippin(1955).

Note: If each point of a topological group has a neighbourhoodhomeomorphic to an open set of a given Euclidean space, then is called alocally Euclidean group. If the underlying topological space of atopological group has the structure of a real analytic manifold, where thegroup operations , are real analytic mappings, thenis a Lie group.

GG

(x, y) → xy x → x−1 G

S. Lie envisaged an approach to solving partial differential equationsanalogous to Galois' group-theoretic resolution of algebraic equations.

6. The mathematical axiomatisation of physicsHilbert considered that physics was too difficult to be left to physicists.

Progress has been minimal, not least because the meaning of 6 is unclear.Again, Hilbert could not have foreseen the many developments in 20thcentury physics. We can record, however, that the axiomatisation ofprobability theory was accomplished by A. Kolmogorov and that ofquantum physics by A. Wightman.

7. To establish the transcendence of certain numbersThe following generalisation of Lindemann's theorem was conjectured

by A. O. Gelfond (1929) and proved by A. Baker (1966).If are non-zero algebraic numbers such

that are linearly independent over the rationals then.

α1, α2, … , αr, β1, β2, … βr

ln α1, ln α2, … , ln αr

β1 ln α1 + β2 ln α2 + … +βr ln αr ≠ 0A special case of this, found independently by Gelfond and T.

Schneider (1934), which answers Hilbert's enquiry about the nature of ,states that ifα is an algebraic number and is an irrational number,then is a transcendental number.

2 2

≠ 0, 1 βαβ

8. To investigate problems concerning the distribution of prime numbers; inparticular, to show the correctness of the Riemann hypothesis

Tantalisingly, the Riemann hypothesis evades resolution. Note: The Riemann zeta function is defined by for

and . This converges when , and can becontinued to all of by a formula giving in terms of . TheRiemann hypothesis states that the non-trivial roots of the Riemann zetafunction all lie on the line . Riemann had already noted that, if

then . He believed, for example, that a proof ofthe hypothesis might establish the existence of an infinity of twin primes.

ζ (s) = ∑∞1 n−s

s = σ + iτ ∈ c σ > 1 σ > 1c ζ (1 − s) ζ (s)

σ = 12

ζ (s) = 0, 0 ≤ Re(s) ≤ 1

9. To find the most general law of reciprocity in an algebraic number fieldHilbert contributed to this, but it was E. Artin (1927) who established it

for Abelian extensions of ; the non-Abelian case is still open.q


Note: The quadratic reciprocity law state that if are different oddprimes then

p, q

(p

q) . (q

p) = (−1)(p − 1)(q − 1)/4 ,

where , Legendre's symbol, is defined for any integer and any odd

prime as

(a

p) a

p

(a

p) =

1 if x2 ≡ a (modp) is solvable forx

−1 if x2 ≡ a (modp) is not solvable forx

0 if a ≡ 0 (modp) .

Gauss was the first to solve the quadratic and cubic reciprocity laws.

10. To find an algorithm for deciding whether any given Diophantineequation has a solution

Following pioneering work by M. Davis, H. Putnam and J. Robinson,the problem was finally solved, negatively, by Y. Matijaseviè (1970).

11. To investigate the theory of quadratic forms over an arbitrary algebraicnumber field of finite degree

H. Hasse (1929) and C. L. Siegel (1936, 1951) obtained importantresults. A. Weil and T. Ono (1964-1965) demonstrated a connexionbetween the problem and algebraic groups. Generally, still incomplete.

12. Extension of Kronecker's theorem on Abelian fields to an arbitraryalgebraic field

Poorly posed by Hilbert, the problem was corrected and solved by T.Takagi. In 1922 he proved the following fundamental theorem: everyAbelian extension of an algebraic number field is a class field for the field(corresponding to a congruence class group in ) and, conversely, everyclass field of is an Abelian extension of .

FF

E F FNote: Given a group of automorphisms of a given field , and a

subfield of , the group consisting of all automorphisms of leaving everyelement of invariant is denoted by . A Galois extension is calledan Abelian extension when is Abelian. Kronecker's theorem statesthat cyclotomic fields are Abelian extensions of and, conversely, everyAbelian extension of is a subfield of a cyclotomic field.

G L KL LK G (L / K)

G (L / K)q

q

The problem is related to finding functions which, for an arbitrary field,play the same role as the exponential function for the rational field andelliptic modular functions for imaginary quadratic fields.

13. To show the impossibility of the solution of the general algebraicequation of the 7th degree by compositions of continuous functions of twovariables


Solved by V. I Arnol'd (1957) for continuous functions; still unsolved ifanalyticity is required.

Note: We mention, in passing, the beautiful results of Kolmogorov andArnol'd that arbitrary real-valued continuous functions of any number ofvariables can be representedexactly as compositions of a finite number ofsuch functions of only two variables.

14. To consider invariants which arise when only the transformations of asubgroup of the totality of linear transformations, the projective lineargroup, are permitted

By producing a counter-example, M. Nagata (1958) showed that theinvariants need not be finitely generated.

Note: An invariant is a mathematical object which remains unchangedunder certain kinds of transformation.

Recently there has been renewed activity in invariant theory; it haswidened its scope and has entered the realm of abstract algebra. IndeedProblem 14, in algebraic language, can be rendered as:

Given fields , , and , where ,the problem is to determine whether the ring , is finitelygenerated over . Here is the field of rational functions in

with coefficients in , and is the ring ofpolynomials with coefficients in .

k k(x1, … , xn) K k ⊆ K ⊆ k (x1, … , xn)K ∩ k [x1, … , xn]

K k (x1, … , xn)(x1, … , xn) k k[x1, … , xn]

k

15. To establish the foundations of algebraic geometry, in particular,H. Schubert's enumerative calculus

Solved by B. L. van der Waerden (1938-1940), A. Weil (1950) andothers. In the late 1950s and 1960s, A. Grothendieck rewrote thefoundations of algebraic geometry after Weil.

Note: Algebraic geometry is the study of algebraic curves, algebraicvarieties and their generalisations to dimensions. Suppose is an -dimensional vector space with scalars in some field . If is a subset ofcomposed of all points which satisfy each of a set of polynomialequations , , with coefficients in , then isan algebraic variety.

n V nF W V

(x1, … , xn)pi (x1, … , xn) = 0 i ∈ z

+ F W

Originally, enumerative calculus was developed for counting thenumber of curves touching a given set of curves, and enumerative geometryrefers to Schubert's application of the conservation of number principle [1].

16. To study the topology of real algebraic curves and surfacesSporadic results.

17. Suppose is a rational function with real coefficients thattakes a positive value for any -tuple . The problem is todetermine whether the function can be written as a sum of squares ofrational functions

f (x1, … , xn)n (x1, … , xn)f


Solved, affirmatively, by Artin (1926-1927) for real closed fields. In1967 DuBois gave a negative solution to the general case. In the same yearPfister gave the number of squares required.

18. To investigate the existence of non-regular space-filling polyhedraK. Reinhardt (1928), a student of Hilbert, showed that such a ‘tiling’

exists. In 1910, L. Bieberbach proved that, up to equivalence, there are onlyfinitely many -dimensional crystallographic groups.n

19. To determine whether the solution of regular problems in the calculusof variations are necessarily analytic

Solved by S. Bernstein, I. G. Petrovskii, and others.

20. To investigate the existence of solutions of partial differential equationswith prescribed boundary conditions

Hilbert contributed here by resurrecting Dirichlet's problem; a vastamount of work has been done in this area pre- and post-Hilbert.

Note: This ‘problem’ is closely linked to the 19th.

A typical boundary problem takes the form in some

region , with and on the boundary of .

∂ 2u

∂ t2− ∇2u = f

R u(0, t) = u1∂ u

∂ t(0, t) = u2 R

An elliptic partial differential equation, for example, is a real 2nd orderpartial differential equation of the form:

∑n

i ,j = 1

aij∂ 2u

∂ xi∂ xj+ F (x1, … , xn, u,

∂ u

∂ x1, … ,

∂ u

∂ xn) = 0

such that the quadratic form is non-singular and positive definite.

Typical examples are the Laplace (Dirichlet's problem) and Poissonequations. We might, in passing, mention the link with potential theory.

∑n

i , j = 1

aijxixj

21. To show that there always exists a linear differential equation of theFuchsian class with given singular points and monodromy group

Several special cases have been solved, for example by H. Röhrl (1957)and P. Deligne (1970), but a negative solution was found by B. Bolibruch(1989).

Note: The first indication of a deep relationship between groups anddifferential equations emerged in Riemann's investigation of thehypergeometric differential equation, which belongs to class of equations ofFuschian type. As it is linear, and of second order, its solutions areexpressible as a sum of basic solutions, the analytic continuation of whicharound each singular point gives rise to more branches of the solution that


depend linearly on those first chosen. The matrix of constants whichcharacterises this dependence is called a monodromy matrix and the groupgenerated by these matrices is called the monodromy group of the equation[2].

22. Uniformisation of complex analytic functions by means of automorphicfunctions

Parametrising all algebraic curves (representing simultaneously theirand values by functions of a single parameter) became known as theuniformisation problem. Poincaré conjectured that all but the simplest ofalgebraic curves arise from decompositions of the upper half-plane into atessellation by polygons. The problem was resolved by H. Poincaré and P.Koebe (1907). This result underpins much of modern complex analysis(and complex dynamical systems).

xy

23. To extend the methods of the calculus of variationsHilbert, and many others, have made contributions to this area, which

has grown apace, especially since the Second World War, and has beensubsumed under optimisation theory (operations research) which includessuch disciplines as control theory, decision theory, linear programming,Markov chains and queuing theory.

AcknowledgementsThe author wishes especially to thank the referees who read the first

draft of this paper and the editor, Stephen Abbott, for their invaluable help.

References1. E. T. Bell, The development of mathematics, Dover (1992) p. 340.2. I. Grattan-Guinness (ed.),Companion encyclopaedia of the history of

mathematical sciences, Routledge (1994) pp. 470-471.

Comprehensive sourcesE. J. Borowski and J. H. Borwein,Dictionary of mathematics, Collins(1989).Eric W. Weisstein,CRC concise encyclopaedia of mathematics, Chapmanand Hall (1999).Kiyosi Itô (ed.), Encyclopaedic dictionary of mathematics, MIT Press.

GRAHAM T. Q. HOARE3 Russett Hill, Chalfont St Peter, Bucks SL9 8JY

LEWIS CARROLL − MATHEMATICIAN AND TEACHER OF CHILDREN 9

Lewis Carroll − mathematician and teacher ofchildren

CANON D. B. EPERSON

It is well known that Lewis Carroll enjoyed the company of childrenand entertained them with fantastic stories, whilst on boating expeditions oron beaches, but his diaries* also reveal that he enjoyed teaching childrenmathematics and other school subjects. On April 16th 1855, he recorded hisconcern with the education of his younger sister Louisa, who had an aptitudefor mathematics: ‘Went into Darlington − bought Swale'sChamber's Euclidfor Louisa. I had to scratch out a good deal he had interpolated, (e.g.definitions of words of his own) and put in some he had left out. An authorhas no right to mangle the original writer whom he employs: all additionalmatter should be carefully distinguished from the genuine text. N. B. Pott'sEuclid is the only edition worth getting − both Capell and Chamber's aremangled editions.’ Three days later he recorded: ‘Advanced Louisa'smathematics to simple Equations (third day of Algebra), and the first 12propositions of Euclid.’ On the next day, he left his home at the rectory ofCroft, and returned to Oxford for the Easter Term.

At this time his own mathematical education was providing him withproblems. He was studying the monumental works of George Salmon onconic sections, in which he found deficiencies and inconsistencies thathindered the compilation of his ownNotes on Salmon. New subjects alsoworried him; ‘I talked over the calculus of variations with Price (his tutor)today, but without any effect. I see no prospect of understanding the subjectat all.’ Four days later he wrote: ‘I have spent a good deal of the daypuzzling over a difficulty in Salmon’, and again consultation with Price didnot help.

He was happier a few days later when Price lent him a little book onfinite differences by Knuff, ‘by which all kinds of series can be summed: Ihave not yet made it out, but it looks very neat.’

Mr Charles Lutwidge Dodgson was then a 23-year-old undergraduate atChrist Church, Oxford, who had achieved First Class Honours inMathematical Moderations and had been appointed a Student of ChristChurch (equivalent to being elected a Fellow of the college), a post thatinitiated him into the teaching profession, as his duties included givingprivate tuition to younger men. During the summer vacation he had his firstexperience of teaching a class of children at the new National School at

* I am indebted to my friend Edward Wakeling for his permission to quoteextensively from volumes I and II of his annotated edition ofLewis Carroll'sDiaries (the private journals of Charles Lutwidge Dodgson), published by theLewis Carroll Society. In places, the punctuation has been altered and the theinterested reader may wish to consult the originals.


Croft-on-Tees: ‘I went to the Boys' School in the morning to hear my father(Archdeacon Dodgson) teach, as I want to begin trying myself soon. Someof the boys were much more intelligent than I expected.’ On the followingSunday ‘I took the first and second class of the Boys' School in the morning− we did part of the life of St. John, one of the “lessons” on Scripture Lives.I liked my first attempt in teaching very much.’

On the next day he ‘Took the first class alone in Old Testament (part ofJudges). Mr Hobson (the headmaster) wants some assistance in Latin, andhe also proposes that I should teach Coates (who is trying for a Pupil-teachership) some Algebra—we made no definite arrangements’. On thenext day ‘As there was nothing for me to do in the Boys' School, I took thesecond class in the Girls', and liked the experiment very much. Theintelligence of the children seemed to vary inversely as their size. Theywere a little shy this first time, but answered well nevertheless.’

During the next week ‘I took the first class of boys: besides this I teachJames Coates Euclid and Algebra on Tuesday, Thursday and Fridayevenings, and read Latin with Mr Hobson on Wednesday evening andSaturday morning − so that I have tolerable practice in teaching.’ Later herecorded ‘My scheme of teaching now is

Boys' School Sunday morning.Girls' school, second class, all other mornings except Saturdayand Friday.Mr Hobson (Latin) Wednesday evening and Saturday morning.James Coates (Euclid and Algebra) Tuesday, Thursday andFriday evening.

besides these, I give the first class in the Girls' School a lesson in sums,every Friday morning, making about 9 hours teaching per week.’

Presumably this continued until the school closed for the summerholiday, as the last reference to Croft School in the diary was on SundayJuly 29th: ‘Took the Boy's School in texts for the first time − my regularwork now is with the second class girls.’

It is worth noting that the school was built in 1845 by the NationalSociety with generous help from the Dodgson family; it consisted of twolarge rooms, one for boys and one for girls, catering for 120 pupils. Therewere only two teachers, Mr Henry Hobson and his wife Sarah, but it isthought that Archdeacon Dodgson and other members of his family assistedregularly with the teaching in an honorary capacity.

On January 29th 1856, he breakfasted with the Revd Henry Swabey inorder to arrange about teaching in his school. ‘We settled that I am to comeat ten on Sunday, and at two on Tuesdays and Fridays to teach sums. I gavethe first lesson there today, to a class of 8 boys, and found it much morepleasant than I expected. The contrast is very striking between town andcountry boys: here they are sharp, boisterous, and in the highest spirits, thedifficulty of teaching being not to get an answer, but to prevent allanswering at once. They seem tractable and in good order: I stayed a short


time afterwards to watch: for want of teachers, the master (CharlesMayhew) had to conduct two lessons at once, while a third (a writinglesson) went on by itself.’

On Feb. 1st ‘The Master at St Aldate's School asked if I would join thefirst class of girls with the boys. I tried it for today, but I do not think theycan be kept together, as the boys are much the sharpest. This made a classof 15. I went on with “practice” as before.’ Later ‘Gordon (the SeniorCensor) suggested a question in ancient mathematics; viz. how did theRomans work multiplication? He, Lloyd, and I, tried it, but could not makemuch of it.’ On Feb. 5th he ‘varied the lesson at school with a story,introducing a number of sums to be worked out. I also worked for them thepuzzle of writing the answer to an addition sum, when only one of the fiverows have been written: this, and the trick of counting alternately up to 100,either putting on no more than 10 to the number last named, astonished themnot a little.’

This shows that Dodgson was well aware of the value of ‘recreationalmathematics’ in the classroom, and he may have been a pioneer in usingpuzzles and tricks in order to retain the interest of children in a subject thatthey found difficult, although he may have been using methods similar tothose he had experienced when at school. It is thought that he inventedseveral puzzles and pastimes, and that he planned after his retirement topublish a book containing some of them, in his series ofCuriosaMathematica.

On Feb. 8th he found ‘the school class noisy and inattentive, the noveltyof the thing is wearing off, and I find them rather unmanageable. Showedthem the “9” trick of striking out a figure, after subtracting a number fromits reverse’. (This depends upon the fact that the difference between anynumber and its reverse is always a multiple of 9, and so has a digit sum thatis 9 or a multiple of 9.)

On the next day he wrote to Swabey, asking ‘what he considered thebest way for my going on at the school: my idea is to form a new class,consisting only of the bright and attentive boys and girls: the system oftaking the whole of the two first classes does not answer well’ but when hemet him at the school, he ‘agreed to try a little longer taking the whole ofboth classes, and set them sums all round, so as to give each something todo. I taught them a little about fractions, and explained the trick of theaddition sum.’

The struggle to maintain discipline continued for a few more days; oneday he found the ‘school class again noisy and troublesome. I have not yetacquired the art of keeping order’, though four days later, ‘School classbetter, as I threatened to banish those who did not attend from the lesson’.A week later he recorded: ‘Class again noisy and inattentive, it is mostdisheartening, and I almost think I had better give up teaching there for thepresent.’ Three days later he ‘Left word at the school that I shall not be ableto come again for the present. I doubt if I shall try again next term: the gooddone does not seem worth the time and trouble.’


Fortunately his association with the children at Croft School ended on ahappier note. When he returned home during the next vacation, he ‘Heardthe singing lesson in the school, about 50 are learning, and there are manygood voices among them: one piece they sang in full harmony. They arealso learning a choral service, which Mr Baker hopes to introduce in churchon weekdays.’

When Dr Henry George Liddell came to Christ Church to reside in thedeanery with his wife and their four young children, Dodgson was afrequent visitor and soon made friends with Harry their only boy, whom hemet for the first time whilst watching the torpids* from the Christ Churchbarge. On March 6th 1856 he recorded: ‘he is certainly the handsomest boyI ever saw’. In the following summer term he taught Harry how to row, andon June 3rd he ‘spent the morning at the deanery, photographing thechildren’. Later ‘Frank and I, with Harry Liddell, went down to Sandford ina gig. We rowed with sculls down with Harry as stroke, and he steeredback.’ On June 5th ‘from half-past four to seven, Frank and I made aboating excursion with Harry and Ina: the latter (the Dean's eldest daughter),much to my surprise, having got permission from the Dean to come. Wewent down to the island, and made a kind of picnic there, taking biscuitswith us, and buying ginger beer and lemonade there. Harry, as before,rowed stroke most of the way, and (fortunately, considering the wild spiritsof the children) we got home without accident, having attracted by ourremarkable crew a good deal of attention from almost everyone we met.’Dodgson regarded this as one of the happiest days of his life.

In the Michaelmas term he soon ‘Fell in with Harry and Ina Liddell, andtook them up to see my book of photographs’, and a few days later, ‘MetMiss Prickett, the governess at the Deanery, walking with Ina, and settledthat I would come over on Wednesday morning, if fine,’ in order to takephotographs. ‘The morning was fair, and I took my camera over to thedeanery, just in time to see the whole party (except Edith) set off with thecarriage and ponies, a disappointment for me, as it is the last vacant morningI shall have in the term.’ Five days later, as there was a clear sun, he ‘wentto the deanery to take portraits at two, but the light failed, and I only got oneof Harry. I spent an hour or so afterwards with the children and thegoverness, up in the schoolroom, making them paper boats, etc.’

When Mrs Liddell informed Dodgson that she intended to send Harry toTwyford School after Easter, she ‘took me into the schoolroom to seespecimens of his sums and Latin: in the former he is well on.’ A few dayslater he ‘Called at the deanery. The Dean and Mrs L. are going abroad forfour months, for his health. The children are to remain in Oxford: Lloyd hasundertaken to teach Harry his Latin and Greek. I offered to teach him sumsetc. but Mrs L. seemed to think it would take up too much of my time. Twodays before the parents sailed for Madeira, he ‘called at the deanery, andtook Harry a Christmas box, a mechanical tortoise: (I gave Ina one the otherday, Mrs Rutherford's children).’ No mention was made of gifts to Alice

* A torpid was the second boat of a college.


and Edith, but one can be sure that they were not forgotten. Five days laterhe ‘met Harry and Ina in the Quadrangle, coming home from riding, andwent into the deanery with them, and stayed for luncheon (or rather theirdinner).’

Early in February 1857 whilst ‘walking in the afternoon, I fell in withIna Liddell and the governess, and returned with them to the deanery, whereI spent about an hour with the young party in the schoolroom. Miss Prickettshowed me a letter the other day from Mrs Reeve (Mrs Liddell's mother) inwhich she expressed great alarm at Harry's learning “mathematics” with me!She fears the effect of overwork on the brain. As far as I can judge, there isnothing to fear at present on that score, and I sent a message to that effect.’

On February 8th Dodgson ‘went to chapel in surplice for the first timesince the 14th of October 1855. I read the second lesson in the afternoon.Harry ran up to me afterwards to tell me “you've got your white gown on,and you read in the church!” Two days later he recorded ‘my pupil HarryLiddell is beginning to tire of the arithmetic lesson. I talked the subject overwith the governess, and settled that he had better give up coming to meunless we succeed better in future.’ But on the next day he ‘spent an hour atthe deanery in the afternoon by Harry's invitation,’ and on the followingmorning ‘Harry did well today, it is doubtful how long the change will last.’

In the autumn of the following year Dodgson visited Harry at TwyfordSchool, and, of course, saw him in Oxford during the school holidays, andenjoyed his company rowing on the Isis. Undoubtedly he had made aninvaluable contribution to the education of the ‘fine young man’ who cameto call on him in November 1862.

On May 5th 1857 Dodgson recorded in his diary ‘I went to the deaneryin the afternoon, partly to give little Alice a birthday present, and stayed fortea.’ She was then 5 years old, and had already been photographed morethan once. His friendship with Ina, Alice and Edith naturally developedafter Harry had gone to boarding school, and his visits to the deanery wereso frequent that some suspected that the attraction was Miss Prickett! Thereis no evidence that Dodgson ever offered to participate in thr education ofthe three young maidens, but his imaginative stories and the word puzzlesthat he invented must have developed their appreciation of the Englishlanguage and increased their vocabularies. This influence upon children andadults still continues, and, for this reason, Lewis Carroll is still honoured100 years after his death.

CANON D. B. EPERSONHillrise, 12 Tennyson Road, Worthing, Sussex BN11 4BY

An imperfect issueNo feature on perfect numbers forMath Gaz 496? Then there's always

8128!A comment from J. H. Evans


Snubbing with and without eta

H. MARTYN CUNDY

In the recentGazette [1], John Sharp has given us a most entertainingaccount of the number (eta) which is the only real root of the equation

, and its connection with the snub cube. I have beeninvestigating this a bit further, with some results that may be of interest.

ηx3 − x2 − x − 1 = 0

More about ηJohn Sharp suggests that may be a pervasive number like ore, but

these are transcendental numbers, while is an ordinary algebraic number,although its derivation is interesting. but, usingTartaglia's method − putting , we obtain its value explicitlyas

η πηη = 1.8392867552… ,

x = 13 + u + v

η = 13 [1 + (19 + 3 33)1/3 + (19 − 3 33)1/3] .

does have some useful properties: John Sharp gives some, but here is ashort list:η

(η − 1) (η2 − 1) = (η + 1) (η − 1)2 = 2,

(η − 1) (η2 + 1) = 2η,

(η + 1)2 (η − 1) = (η2 − 1) (η + 1) = 2η2,

(η + 1)2 = η (η2 + 1) ,

η + 1 / η = 2 / (η − 1) = (1 + 1 /η)2 ,

η4 + 1 = 2η3 η + 1 / η3 = 2.and thus

With a start near 2, on iteration converges rapidly to . Thegeneral equation of this type,

y = 2 − 1 /x3 η

xn = xn − 1 + xn − 2 + … + x + 1 = (xn − 1) / (x − 1)gives , which approaches 2 steadily with increasing .x = 2 − 1 /xn n

If we set , we obtain the equation , and weshall find this more meaningful and useful as an aid to the process ofsnubbing, to which I now turn.

η − 1 = t t2 (t + 2) = 2

A more general approachMeasurements of significant features of the snub cube are not too

difficult to make, but most people find themselves baffled by the snubdodecahedron and wonder how to get started on it. Let me tell two storieswhich may help.(i) Can you discover fromlocal measurements how big the earth is?

SNUBBING WITH AND WITHOUT ETA 15

Eratosthenes had to do some travelling about Egypt to make hisestimate, but I was brought up in a generation (before metrication) thatwas taught the following myth. If you go to East Anglia (where thereare lots of zero contours, a spot height of 2 indicates a ‘hill’, and thereis even a trig. point labelled −1) there are long stretches of level water.Put three posts in a straight stretch, one mile apart. All the posts stickequally far out of the water, but, due to the curvature of the earth, themiddle one is 8 inches above the line of sight of the other two. Fromthis and the geometrical theorem in Figure 1, which states that

we have at once that 8ins. × earth's diameter =(1 mile) . The diameter is therefore given by miles= 7920 miles, which is surprisingly accurate!

KA.KB = KC.KD2 d = 5280 × 12 / 8

CK

D

B

A

FIGURE 1

(ii) A friend recently was making a model with wooden balls of the carbonmolecule with 60 atoms, called a buckyball. (They are arranged likethe points on a football where hexagons and pentagons meet.) Hecould not get the angles right, nor find how big the model would be.We need the same theorem.

Every uniform polyhedron has equal edges and every vertex alike.Since the vertices are all alike, they are on a sphere whose centre is theircentroid. So the neighbours of a vertex , say ,, , , , all lie on twospheres: the circumsphere of the polyhedron, and a sphere with centre andradius equal to the edge-length. Two spheres meet in a plane circle. So the

form a plane cyclic polygon called thevertex figureof . If we know itsshape and how much the vertex lies above it, we can deduce all about thepolyhedron.

A B1 B2 B3 B4 B5

A

Bi AA

Snub polyhedraWe will confine our attention to the snub polyhedra that have 4

equilateral triangular faces and a (possibly different) regular polygonal facesurrounding each vertex. There are 6 of these (described in detail in [2]) if


we include the icosahedron (which is a snub tetrahedron). The other 5 (in anotation that is described in the next paragraph) are:

| 2 3 4 snub cuboctahedron

| 2 3 5 snub icosidodecahedron

| 2 3 5/2 great inverted snub icosidodecahedron

| 2 3 5/3 great snub icosidodecahedronand finally one with its triangles reflexed

| 2 3/2 5/3 great inverted retrosnub icosidodecahedron (!).These can be seen illustrated in all their glory in [2], nos. 17, 18, 113,

116, 117. Figure 2 shows the snub cube (or, better, snub cuboctahedron).All of these occur in enantiomorphic pairs, that is, with left-handed or right-handed twists, making mirror-images of one another. As to the names, thesnub cube is related to the octahedron in the same way as it is to the cube; itcontains faces in the planes of both solids. To be fair, then, it is preferred tocall it the snub cuboctahedron, and so on. Some people use other names, butthe numerical code is definitive. All these polyhedra can be generated byreflections in the sides of spherical triangles, and the numbers give theangles of these triangles. Thus| 2 3 5/2 indicates the use of a triangle withangles , , , see [2, 3].π / 2 π / 3 2π / 5

The side of a convex regular polygon with sides (a -gon) subtends anangle of at its centre . The pentagram, or regular star-pentagon, hassides subtending , so it is convenient to label it as a 5/2-gon; an -gon will have vertices equally spaced on a circle, joined by edges toneighbours points away.

p p2π / p K

4π / 5 m/ nm

n

FIGURE 2

If every edge is of length 2 units, the vertex figures of all thesepolyhedra have the form shown here in Figure 3: for to 4 the angle

will be , say, where is slightly greater than . will be adiagonal of a -gon so that and .Let the radius . Then from we have , andfrom and we have .

i = 1BiKBi + 1 2θ θ π / 6 B1B5

p ∠B1AB5 = π(1 − 2/p) ∠B1KB5 = 2(π − 4θ)KBi = ρ B1KB2 ρ sinθ = 1

B1KB5 B1AB5 2c = B1B5 = 2ρ sin 4θ = 4 cosπ / p


Thereforesin 4θsinθ

= c = 2 cos(π / p) . (1)

Let , so that . These equations combine togive

t = 2 cos 2θ t + 2 = 4 cos2 θ

t2 (t + 2) = c2 = 4 cos2 (π / p) . (2)For the six polyhedra listed above, = 3, 4, 5, 5/2, 5/3 (5/2 reversed), 5/3;but in the last two the vertex figure is crossed; the theory holds, but it is noteasy to see what. is happening! See [2].

p

So in equation (2) is equal to 1, 2, , , where isthe golden ratio. John Sharp used the letter for this number − atransatlantic name, but the English use of is commoner.

c2 τ2 τ−2 τ = ( 5 + 1) / 2φ

τ

1 1

B1

B1

B2

B2

B3

B4

B5

A

K

K

ρ ρθ θ

B1

B5

A

Kρρ

πp

π θ− 4

FIGURE 3

Resulting measurementsWell, after all that, what use is ? This is where Figure 1 comes into its

own. Let be the height of the vertex above the plane of the , and letbe the diameter of the snub polyhedron (Figure 4). Thenand . At once we have (a result we could also getfrom the similarity of triangles , ), and

th A Bi d

h (d − h) = ρ2

h2 + ρ2 = 4 d = 4 /hABD AKB

h2 = 4 − ρ2 = 4 − cosec2 θ = 4 (1 −1

2 − t ) =4(1 − t)

2 − t.

Therefore , the radius of the circumsphere of the snub polyhedron is givenby

R

R =2 − t

1 − t.


K

D

h

d − h

ρ ρB, Bi

A

FIGURE 4

To find the circumscribing polyhedronHere, as John Sharp showed, we need the radius of the sphere touching

the faces of the polyhedron we are seeking, the in-radius, which, followingCoxeter [4], we denote by . We are interested in two possible polyhedra:that with the -gon as face, and that with the triangle as face belonging tothe dual polyhedron. Which triangle depends on which enantiomorph wehave, but the choice does not affect the measurements. It is immediate thatfor a -gon , and for atriangle . If the edge of the circumscribing polyhedron is

, we need to find this from our knowledge of the radius of its shared in-sphere. The polyhedron has -gonal faces and -gonal vertex figures.If is its centre, a vertex, the midpoint of an edge, and the centre of aface, again following [3], we name the radii , ,

. , sincet is the length of an edge. We need, where is the dihedral angle. First we need. In the diagram (Figure 5), since are coplanar,, so is the radius of the vertex figure, which is

a -gon with edge-length , so its radius is. Therefore . The

rest is straightforward, but tiresome.

2Rp

p 2R2 = R2 − cosec2 (π / p) = 1 / (1 − t) − cot2 (π / p)2R2 = 1/(1 − t) − 1

32e 2R

p, q p qO V E F

OV = 0R OE = 1ROF = 2R VE = e 2e∠EOF = ψ π − 2ψ∠VOE = φ V, K, O, V′, E∠VV′K = φ V′K = 2e cosφ

q 4e cos(π / p)2e cos(π / p) cosec(π / q) cosφ = cos(π / p) cosec(π / q)

sin2 φ = 1 − cos2 (π / p) cosec2 (π / q) = k2 cosec2 (π / q) ,where .k2 = sin2 (π / q) − cos2 (π / p) = sin2 (π / p) − cos2 (π / q)Then , and 0R = e cosφ = (e sin (π / q)) / k

1R2 = OE2 = e2 (sin2 (π / q) − k2) / k2 = (e2 cos2 (π / p)) / k2.

Finally


2R2 = 0R

2 − VF2 = e2 sin2 (π / q) / k2 − e2 cosec2 (π / p)= (e2 / k2) cos2 (π / q) cot2 (π / p) .

cot2ψ = (cos2(π /q))/k2 = OF2/FE2 =[1/(1 − t) − cot2(π/p)]

e2 cot2π /p,So

leading to , or, finallye2 cos2 (π / p) = k2 [tan2 (π / p) / (1 − t) − 1]

e2 = sin2 (π / p)cos2 (π / q)

− 1

tan2 (π / p)

1 − t− 1

.

O

K

V

E

F

e

e

φ

φ

ψ

V′

FIGURE 5

Angle of twistWe now know the sizes of the polyhedra which enclose the snub

polyhedron, with faces containing its non-snub faces of a . Thesefaces lie on similar faces of the enclosing polyhedron − thecase − but aretwisted in relation to them. We now find the angle of twist. This looksformidable until we spot the secret. The polygons are forced to twistbecause one triangle edge links adjacent -gons, with its midpoint on ,where is the midpoint of the separating edge of the case. (See Figure 6).The central plane through this edge bisects , which lies in a planeperpendicular to it. (If you have made a model of a snub cube, this is seenvery clearly.) If as shown in the figure are coordinates of in its face,with origin , we have ; ;

, where is the inclination of to the -axis. Theseequations lead to the quadratic in :

p, q

p OMM

PP′

(x, y) PM x2 cos2ψ + y2 = 1 x = ecot(π/p) − cosec(π/p) cosα

y = cosec(π/p) sinα α 2OP xcosα

k2 cos2 α + 2e cosα cos(π / p) cos2 (π / q) −(e2 cos2 (π / q) + sin2 (π / p)) cos2 (π / p) = 0.


M(P′) L (P)

2O

O

ψ ψ

x cosψ

e

e

M

P

x

yα

π pL

2O

P′

FIGURE 6

Inserting the value of and a little manipulation produces for the positiveroot of this quadratic

e2

k cosec(π / p) cosα =sin (π / p)

1 − t− cos(π / q)

11 − t

− cos2 (π / p),

i.e. , where isnow the mid-radius of the snub polyhedron. This result is so unexpectedlysimple that one wonders if there is a simpler way of getting it!

cosα = [sin(π /p)/k] [1R′ sin(π /p)−2 Rcos(π /q)] 1R′ = R2 − 1

The angle of twist is , and is given in the accompanying table.π / p − α


Snubber's kit: [ only. For others, see [2] and be amazed].(34, p)Equations: Snub ; edge length 2 units.p,3

. . .t2 (t + 2) = 4 cos2 (π / p) 0R2 = 1 + 1 / (1 − t) 1R = 1 / 1 − tIf a with edges of length contains the -gonal faces of the snub

( , are now interchangeable), thenp, q 2e p

p q

2R2 =

11 − t

− cot2πp

. k2 = sin2 πp

− cos2πq

.

e = 2R k tanπp

secπq

.

Twist of snub on a -gonal face is , wherep β = π / p − α

k cosα = sinπp

1R sinπp

− 2R cosπq

.

TABLES: numerical results for snub edge = 2

TABLE 1 p = 3, c2 = 1

t Polyhedron R p, q e 2R k twist angle

−1 Same tet. 3 / 2 3, 3 1 1/ 6 12 2 0

1 /τ Icosa. 2 + τ 3, 3 τ2 2 τ2 / 3 22·23°−τ Great icosa. 3 − τ 3, 3 2 /τ2 3 / τ2 22·23°

TABLE 2 p = 4, c2 = 2


0·839287 Snub cube 2·687427 Cube 2·285227 2·28522712 16·468°

Octahedron 2·972103 2·426712 20·315°

TABLE 3 p = 5, c2 = τ2


0·943151 Snub 4·311675 Dodecahedron 1·778973 3·96183212τ−1 13·106°

dodeca. Icosahedron 2·748341 4·154179 19·518°

TABLE 4 p = 5 / 2, c2 = τ−2


0·399021 (W.113) 1·632161 Gt Stellateddodecahedron

6·216529 1·24835012τ 27·108°

Greaticosahedron

5·230732 1·153524 24·515°

−0·505561 (W.116) 1·290040 Gt Stellateddodecahedron

3·721987 0·747415 10·155°

Great icosa. 2·608352 0·575214 4·420°−1·893460 (W.117) 1·160003 Gt Stellated

dodecahedron2·439766 0·489933 3·672°

Great icosa. 0·551694 0·110786 0·544°


Notes on the tablesTetrahedron. 3 real roots for . means so that willcoincide with and with . So triangle will just be triangle

repeatedturned over. So at vertex we have 2 ordinary equilateraltriangles and one covered three times; finally we arrive at the originaltetrahedron covered 5 times—a rather peculiar icosahedron! givesus what we expected, but reminds us of what we have probablyforgotten. There are in fact two ways of dividing the edges of the coreoctahedron of the stella octangula in the ratioτ : 1. It may be divided

t t = −1 2θ = 2π / 3 B4

B1 B5 B2 KB5B1

KB1B2 A

t = τ−1

t = −τ

icosa-hedronface1

τ line of intersection

with second tetrahedron

of stella octangula

Great icosahedron face

Tetrahedron face

τ4

τ2

τ2 2

τ3 22.2°

FIGURE 7

internally, giving the familiar icosahedron, or externally, giving the greaticosahedron, a beautiful polyhedron with star-pointed vertices. Its edges areparallel to those of the internal icosahedron. See Figure 7 and the fullerdiscussion below.

The snub cube. The inverse ratio giving the edge of the snub cube as afraction of the encasing cube is 0·4375933…. John's eagle eye found asimilar number in a remote place—Figure 174 in [5]. It would indeed bemiraculous if a number which is the root of a cubic equation should turn upin the entirely Euclidean process of locating a tetrahedron in adodecahedron. Such has not happened; the ratio here is1 / (τ 2) = 0·437016…


The starry snubs. The negative roots for here again correspond to vertexfigures which cross over, i.e., which have intersecting sides. The resultingpolyhedra have intricate intersecting faces, triangles and pentagrams. Thereare 8 other uniform snub polyhedra involving more complicated vertexfigures.

t

Central inversionUnder this transformation in which each point of a solid is mapped to its

reflection in the centre, every snub polyhedron is transformed into itsenantiomorph, and the encasing polyhedron, usually symmetrical, remainsunchanged. Thus both enantiomorphic snubs are inscribable in the samecase. But there is one exception where the opposite situation occurs. Aregular tetrahedron is not centrally symmetrical; its central inverse is asecond tetrahedron with edges perpendicular to those of the first. The twotogether form the beautiful 8-pointed star which Kepler named the stellaoctangula. The core common to both tetrahedra is an octahedron with edgeshalf as long. But here the snub polyhedron is symmetrical—an icosahedron—which must therefore be inscribable in both tetrahedra, i.e. in the coreoctahedron of the stella octangula. Indeed, as John Sharp has so delightfullyreminded us, its 12 vertices lie one on each edge of this core octahedron,dividing it in the ratio . But there are 12 other points dividing thesesides externally in the same ratio (i.e. ) and they will be thevertices of another symmetrical icosahedron, the great icosahedron whosefaces are 20 pentagrams. This is a regular polyhedron, pictured in [2] no. 4l.

τ : 1τ : 1 τ : −1

References1. John Sharp, Have you seen this number?,Math. Gaz. 82 (July 1998) pp.

203-214.2. Magnus J. Wenninger,Polyhedron models, Cambridge University Press

(1971).3. H. S. M. Coxeter, Uniform polyhedra,Phil. Trans. 246 (1954) pp. 40l-

450.4. H. S. M. Coxeter, Introduction to geometry, Wiley (1961) §10.4.5. H. Martyn Cundy and A. P. Rollett,Mathematical models, Oxford

University Press (1961).H. MARTYN CUNDY

2 Applerigg, Kendal, Cumbria LA9 6EA

Use it or lose it!!His mind may have been trapped in an aged body, but it was active, and he

retained his sense of humour − usually at his own expense. Two years ago he wrote:‘I am settling down into an untidy compromise between the venerable and thedecrepit’. 18 months ago theMathematical Gazette printed an article by him headed‘An algorithm for square roots: an episode in the campaign against dotage’.

From an appreciation of Rev. L. M. Brown in The Scotsman, April 1999.


The Fermat-Torricelli points of n lines

ROY BARBARA

For convenience, if , are points in the Euclidean plane , willdenote the closed segment .Excluding the endpoints, we get the open segment . Further, if is ahalf line, will stand for .

P P′ r2 [PP′]

(1 − λ) P + λP′ ∈ r2 : 0 ≤ λ ≤ 1

] PP′ [ Px] Px Px − P

The problem denotes a set of distinct lines in .

Further, we associate with each a positive real number called itsweight. will denote the function , , where

is the distance from to . A question of interest is to minimise ,i.e. to find and to determine the set of , denoted by

, such that . In the extremal case where the are all parallel,by projection in the common direction of the , the problemreduces tofinding the Fermat points of collinear points; one easily finds that iseither aline or a closedband formed by two adjacent lines . Fromnow on, we assume the (most general) hypothesis, which is that

S = Li, i = 1, … , n n (≥ 2) r2

Li mi

f r2 → r f (X) = ∑n

i =1mid(X, Li)d (X, Li) X Li f

α = minimumf (X) XFT (S) f (X) = α Li

Li

n FT (S)Li Li, Li + 1

the are not all parallelLi

Some terminology

See Figure 1, is theboundary. is theinterior.

A vertex is the intersection of at least two distinct lines . If two vertices lie on some line such that contains no vertex, are

adjacent vertices. In such a case, is aside and is aside-

b = ∪n

i = 1Li Λ = r

2 − b

Li

V, V′ Li ] VV′ [ V, V′[VV′] P ∈ ] VV′ [

RAY-POINT

RAY

VERTEXVERTEX

SIDE-POINT

ZONE

CHORD SID

ESI

DE

PERIMETER

VERTEX

L1

L2

L3

L4

L5 L6

∞−CHORD

FIGURE 1

THE FERMAT-TORRICELLI POINTS OF N LINES 25

point. A half line , containing exactly one vertex , is aray,and is aray-point. A segment where and

is achord. A half line where and is an∞-chord.

Vx ⊆ someLi VP ∈ ] Vx [PQ] P, Q ∈ b

] PQ[ ⊆ Λ Px P ∈ b ] Px ⊆ Λ

Let , where denotes one of the two closed half planes

defined by ; such a closed convex set , if non-empty, is azone; aperimeter is the boundary of a bounded zone, hence a convex polygon.

Z = ∩n

i = 1Zi Zi

Li Z

The main result and examplesTheoremThe function always attains its minimum value at a vertex.f αExactly three cases are possible:1) is reached at precisely one vertex and .α V0 FT (S) = V02) is reached at precisely two adjacent vertices and

.α V1, V2

FT (S) = [V1V2]3) is reached at precisely the vertices of aperimeter and

, where denotes the compact convex zone delimited bythe perimeter.

α r (≥ 3)FT (S) = Z0 Z0

Note: The theorem provides an effective procedure to determine and. Indeed, the vertices beingfinite in number, denote them by

. Compute and find . Compare

to each and deduce , finding that, is equal to precisely one,precisely two, or at least three, of the .

αFT (S)V1, … , Vs f (V1) , … , f (Vs) α = min

s

j = 1f (Vj)

α f (Vj) FT (S) αf (Vj)

For the first three examples, denotes a triangle where, as usual,the side lengths are denoted by ( opposite to , etc.), and where

denote the lengths of the altitudes . Wetake , where is the line , is , and is .

T ABCa, b, c a A

ha, hb, hc (ha : fromA toBC, )etc.S = L1, L2, L3 L1 BC L2 CA L3 AB

Example 1 We choose . Then ; ;.

m1 = m2 = m3 = 1 f (A) = ha f (B) = hb

f (C) = hc

(i) Suppose, say, . Then is the unique point in the plane ofwhich the sum of the distances from the sides of is a minimum.

ha < hb, hc AT

(ii) Suppose is isosceles with apex angle less than . Then,. The theorem provides: . In particular,

is constant for all .

T A π / 3ha > hb = hc FT (S) = [BC]d (X, L2) + d (X, L3) X ∈ [BC]

(iii) Suppose equilateral. Then, . Here is theconvex hull of (interior and boundary of ). In other words,every point inside minimises . We get the familiar property that thesum of the distances from a point inside an equilateral triangle, to thesides, is a constant.

T f (A) = f (B) = f (C) FT (S)A, B, C T

T f


Example 2 We take ; ; . Then,. We conclude that is constant inside , which

expresses that , for all inside .

m1 = a m2 = b m3 = c f (A)(= aha) = f (B)= f (C) = 2 × areaT f T

f (X) = 2 × areaT X T

Example 3 We take now ; . Then,. According to the theorem, has constant (minimum)

value 1 inside . Let denote the incentre of and the inradius of . Wemust have . Since , we getthe familiar formula:

m1 = 1 /ha; m2 = 1 /hb m3 = 1 /hc f (A)= f (B) = f (C) = 1 f

T I T r Tf (I) = 1 f (I) = (1 /ha) r + (1 /hb) r + (1 /hc) r

1r

=1ha

+1hb

+1hc

.

In examples 4 and 5, the weights are all 1.

Example 4 (Corollary) Let denote a convexregular polygon. Wetake as the set of lines Clearly, by an argument ofsymmetry, . We conclude that isconstant(minimum) inside the polygon . (This seems not to be trivial if isodd: case of a pentagon, heptagon, etc.)

A1… An

S A1A2, A2A3, … , AnA1.f (A1) = f (A2) = … = f (An) f

A1… An n

Example 5 See Figue 2. More generally, let be arbitrary regularpolygons. We take as the union of all the sides of the . We assume thatthe corresponding polygonal regions have a non-empty common intersection

. Then, .

T1, … , Tr

S Ti

I0 FT (S) = I0

T1

T2

T3

I0

T1T2

I0

FIGURE 2

Proof of the TheoremNote first the following: the zones cover the set : is either avertex, a side-point, a ray-point, or interior to a (bounded or unbounded)zone ; given two distinct points , then, lie in a same zone if andonly if is not traversed by any line . (A set is said totraverse a set

(at ) if and are both non-empty.)

P ∈ r2 P

P, Q P, Q] PQ[ Li A

B A ∩ B A ∩ B B − A

Let be distinct points. Consider as an -axis, with arbitraryorigin . For , denotes the abscissa of . Let be aP0 P, R PR

→x

O ∈ linePR X ∈ [PR] x X L


line.(i) If does not traverse (possibly ), then, for some

, (for all ). Furthermore,does not depend on .

L ] PR[ L = linePRa, b ∈ r d (X, L) = a . x + b X ∈ [PR] a

O(ii) If traverses at , let such thatL ] PR[ Q a, b, a′, b′ ∈ r

if d (X, L) = ax + b X ∈ [PQ] if ,d (X, L) = a′x + b′ X ∈ [QR]

then, and .a < 0 a′ = −a > 0Proof If is parallel to , (i) is obvious as is constant. We

assume now that intersects the line at angle , for . Then,L PR d(X, L)

L PR θ x = x0

if d (X, L) = | x − x0 | sinθ = (− sinθ) (x − x0) x ≤ x0

if .= sinθ (x − x0) x ≥ x0

(i) and (ii) immediately follow.

Let (e.g. side; chord). Then, is alinear transformationon . In particular, is either constant or strictly monotonic on .P1 [PQ] ⊆ zoneZ f

[PQ] f [PQ]

Proof Consider as an -axis, with an arbitrary origin . Let denote the abscissa of . By 0, as does not traverse ,

for some . Hence, , with; .

PQ→

x O ∈ linePQx X ∈ [PQ] P Li ] PQ[d (X, Li) = aix + bi ai, bi ∈ r f (X) = Ax + BA = ∑n

i = 1 miai B = ∑ni = 1 mibi

Let . If is minimum at , then is minimum on .

P1′ (Corollary) [PR] ⊆ zoneZ f Q ∈ ] PR[f [PR]

Let half line (e.g. ray;∞-chord). Then strictly increaseson . In particular, is never minimum on .P2 Px ⊆ zoneZ f

Px→

f ] Px

No line traverses . Hence, as describes , isconstant or increasing. Since the are not all parallel, at least for one ,

, and hence , strictly increases.

Proof Li ] Px X Px→

d (X, Li)Li i

d (X, Li) f (X)

is not minimum, at any ray-point, or at any interior toan unbounded zone. (Such points lie in some , where lies in somezone .)

P2′ (Corollary) f X] Px Px

Z

Let be distinct points. Consider as an -axis, with some origin. denotes the abscissa of . Suppose is

such that: are in the same zone; are in the same zone; and arein different zones (at least one traverses at ). Set

and if . Then .

P3 P, R PR→

xO ∈ linePR x X ∈ [PR] Q ∈ ] PR[

P, Q Q, R PRLi ] PR[ Q f (X) = Ax + B

if X ∈ [PQ] f (X) = A′x + B′ X ∈ [Q, R] A < A′

Proof For convenience, we assume labelling such that fortraverses at , and, for , does not traverse . For ,

j ≤ r , Li

] PR[ Q j > r Lj ] PR[ j > r


write ( ). For , using , write if and if

. Clearly, ; .Hence, .

d(X, Lj) = αj x + βj

X ∈ [PR] 1 ≤ i ≤ r P0d (X, Li)= −aix + bi (ai > 0) X ∈ [PQ] d (X, Li) = ai x + bi′X ∈ [QR] A = ∑r

i =1mi (−ai) + ∑j > r mjαj A′ = ∑ri =1miai + ∑j > r mjαj

A < A′

Let be points not lying in the same zone. Assume minimum.Then, .P4 P, Q f (P)

f (P) < f (Q)

Proof Consider as an -axis, with some origin .denotes the abscissa of . The set of points , where istraversed by (at least) one line isnonempty and finite. Let

be the elements of , in this order, from to . Set ;. For , . By , ,

if . As minimum, . We deduce .By , . Hence, for , , so strictlyincreases on . Now, so

.

(P4) PQ→

x O ∈ linePQ xX ∈ [PQ] I R ] PQ[

Li R1, … , Rr

(r ≥ 1) I P Q R0 = PRr + 1 = Q i = 0, … , r Ri, Ri + 1 ∈ same zone P1 f (X) = Aix + Bi

X ∈ [Ri, Ri + 1] f (R0) f (R0) ≤ f (R1) A0 ≥ 0P3 A0 < A1 <… < Ar 1 ≤ i ≤ r Ai > 0 f

[Ri, Ri + 1] f (R0) ≤ f (R1) < f (R2) <… < f (Rr + 1) ,f (P) < f (Q)

Corollary If , then, lie in a same zone*.P4′ P, Q ∈ FT (S) P, Q

If is minimum at three different vertices , , , then is minimumat an interior point .P5 f V1 V2 V3 f

P ∈ ΛProof . By , is constant (minimum) on .Clearly, is either a side or a chord. We consider two cases:

Vi, Vj ∈ same zone P1 f [ViVj][ViVj]

Case 1: Some is a chord: is minimum at any .[ViVj] f P ∈ ] ViVj [ ⊆ ΛCase 2: , , areall sides: We note here thatare not collinear (otherwise, if, say, is between and , and wouldnot lie in a same zone). In such a situation, one can see that mustbe aperimeter. Let ; is minimum. By , is constant(minimum) on . In particular, is minimum at any point

.

[V1V2] [V2V3] [V3V1] V1, V2, V3

Vj Vi Vk Vi Vk

V1V2V3

U ∈ ] V2, V3 [ f (U) P1 f[UV1] f

P ∈ ] UV1 [ ⊆ Λ

Now, we conclude the proof of the theorem:Set , . By elementary topology, we know thatis reached.

α = minimumf (X) X ∈ r2 α

Case 1: is reached at .α O ∈ ΛClearly, lies in aunique zone ( is interior to ). By , is

bounded. Claim . Indeed, let . and must lie ina same zone. Hence, . Conversely, let . lies on some chord

, containing . By , is minimum on , hence at .

O Z O Z P2′ ZFT (S) = Z P ∈ FT (S) O P

P ∈ Z P ∈ Z PC ⊆ Z C O P1′ f C P

For the rest, we assume . By , is only reached at sidepoints, and, or, vertices.

FT (S) ⊆ b P2′ α

* The reason for which I write ‘a zone’ instead of ‘the zone’ is that a give point (namely, aboundary point) may lie in different zones.


Case 2: is reached at some side point :α ∈ ] V1V2 [By , is minimum on . By , is not reached at a third

vertex. Claim . Otherwise, would be reached at a sidepoint such that , that is, orand . By , would be minimum on and hence at 3 differentvertices, contradicting .

P1′ f [V1, V2] P5 aFT (S) = [V1V2] α

P ∈ ] VV′ [ side [VV′] ≠ [V1V2] V V′ ≠ V1

V2 P1′ f [V, V′]P5

Case 3: is only reached at vertices:αα is reached, say, at . Claim . Otherwise, let

, . and must lie in a same zone. By , must beconstant (minimum) on , hence minimum at . Sinceeither is a side point or , we get a contradiction.

V FT (S) = VV′ ∈ FT (S) V′ ≠ V V V′ P1 f

[VV′] P ∈ ] VV′ [P P ∈ Λ

RemarkThe results in this note can be generalised to other functions ,

providing is assumed to beconvex andzonewise linear. A theorem of theconvexity of the minimum set can be found in [1].

ff

AcknowledgementI am indebted to the referee for all his suggestions, as well as to Dr

Faruk Abi-Khuzam for introducing me to the problem and finding variousapplications.

Reference1. Luenberger:Introduction to Linear and Nonlinear Programming (p.

119).ROY BARBARA

Department of Mathematics, University of Beirut, Lebanon

A mistake we could all makeCOMMON MISTAKES IN TESTS

At seven: calculate the money left over from 54p after buying five pencils at10p (52 per cent wrong, most of those answering 44p).

From an article in The Times, 18 December 1998.

SCHOOL TEST RESULTSSir, Apparently, “the tests cannot be bettered as a way of giving ... information

on children's progress” according to the Qualifications and Curriculum Authority(report, today). However, one of the ‘common mistakes’ criticised in your report ismade by seven-year-olds, who ‘calculate the money left over from 54p after buyingfive pencils at 10p’ as 44p.

What would their answer have been after buying five pencils at 10p each?I remain Sir, yours, etc.Letter to The Times, 19 December 1998, both sent in by Frank Tapson.


Continued fractions

ROBERT MACMILLAN

Consider the vulgar fraction 257/47. We can convert it to an integerplus proper fraction. If we now invert the fractional part and repeat theprocess on it, we get

257 / 47 = 5 + 22 / 47 = 5 + 1 / (47 / 22)

47 / 22 = 2 + 3 / 22 = 2 + 1 / (22 / 3)

22 / 3 = 7 + 1 / 3.and

The original fraction can thus be expressed in the form

5 + 1 / (2 + 1 / (7 + 1 / 3)) ,which is called a continued fraction (CF). Since it is uniquely defined bythe numbers 5,2,7,3, such a continued fraction list is conventionally written[5;2,7,3], the first number being the integer part. Since , wesee that this CF could also be expressed as [5;2,7,2,1]: similarly, every finiteCF has two forms, ending either with d (greater than 1) or with .*

3 = 2 + 1 / 1

d − 1, 1Continued fractions are of interest not only because they can be used to

express any fraction in a form independent of number base, but also becausetheir iterative derivation and evaluation make them useful for computation.Every fraction can be converted into a CF of finite length, whereas mostfractions, such as 5/3, require infinite decimals for their exact representation.For example, 101/97 has for its CF [1;24,4], whereas the decimal has arecurring part with 96 digits.

Continued fractions used to be regarded as little more than interestingmathematical curiosities but, more recently, the advent of computers and thediscovery of methods of prime factorisation involving the CF algorithmhave shown that they can be of practical value. This article shows how toderive and evaluate continued fractions and indicates some of theirproperties, but no attempt is made to give all the proofs, since these arereadily available in texts such as Chapter 10 of the excellent book byKenneth Rosen [1].

Finite continued fractionsUsing symbols to express the form of the computation above, we have

* In general, a continued fraction is an expression of the form

a1 +b1

a2 +b2

a3 +…where the numbers may be real or complex numbers. If all the are equal to1 and if

is an integer and is a positive integer for , then it is a simple continued fraction;this article is restricted to simple continued fractions.

ai , bi bi

a1 aj j > 1

CONTINUED FRACTIONS 31

x(0)/y(0) = a(0) +1

a(1) + 1a(2)+ 1

…+ 1

a(n−1)+ 1a(n),

= a(0) + 1/(x(1)/y(1))

with in that case.n = 3Putting as the first terms

of the continued fraction list, we can write, as a general term,C (k) = [a(0) ; a (1) , a(2) , … , a (k)] k + 1

x (k)y(k)

= a (k) +y(k + 1)x (k + 1)

=a (k) x (k + 1) + y(k + 1)

x (k + 1)with .a (k) = INT (x (k) / y(k))

Since the fractions are in their lowest terms, we can equate thedenominators to find

x (k + 1) = y(k) , (1)and equate numerators to find

x (k) = a (k) x (k + 1) + y(k + 1) ,

y(k + 1) = x (k) − a (k) x (k + 1) = x (k) − a(k) y(k) (2)so

a (k + 1) = INT (x (k + 1) / y(k + 1)) . (3)with

Also from an earlier fromula. These recurrences express and in termsof their previously known values, starting with , , and and canthus be used to find the CF of any fraction. The equations are readilyprogrammed into a loop which repeats until , and each value ofis either printed as it is found, or stored in an array for .

x, y ax (0) y(0) a (0)

k = n a(k)C (n)

To find the CF of a decimal number, it is only necessary to express it asa fraction with an integer numerator and the appropriate power of 10 as thedenominator, then reduce this to its lowest terms. For example, 0.3125 =3125/10000 = 5/16 and the CF is found to be [0;3,5].

To evaluate a finite CF, the reverse process can be used, the recurrencerelations required being

y(k − 1) = x (k) (1a)

x (k − 1) = y(k) + a(k − 1) x (k) . (2a)and

Initially and , and the equationsabove can be programmed into a loop which is repeated until , giving

as the required value of the CF. For example, evaluating [0;3,5]gives 5/16 = 0.3125. Evaluating a CF will always give a fraction expressedin its lowest terms. The CF [3;7,16] = [3;7,15,1] has the value 355/113, thewell known approximation to , and [0;3,7,16] = 113/355, its reciprocal.

x (k) = x (n) = a (n) y(k) = y(n) = 1k = 0

x (0) / y(0)

π


ConvergentsEvery rational number can be expressed as a fraction with integral

numerator and denominator and can therefore be represented by a finite CF.For an irrational number, however, this is not so, and the CF has an infinitenumber of terms. Increasing the number of terms used gives progressivelybetter approximations to its value.

As an example, consider the CF for e, which is[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10, …].

The values of are called theconvergents to , and they can becomputed as below:

C (k) e

C(0) = 2/1 = 2.0 C(5) = 87/32= 2.71875 C(10) = 2721/1001C(1) = 3/1 = 3.0 C(6) = 106/39= 2.71795 = 2.7182817C(2) = 8/3 = 2.67 C(7) = 193/71= 2.7182796 C(11) = 23225/8544C(3) = 11/4= 2.75 C(8) = 1264/465= 2.7182796 = 2.718281835C(4) = 19/7= 2.714C(9) = 1457/536= 2.7182836

These approximations to the true value of the CF are alternately greaterand less than it. Each convergent is the best approximation having adenominator no greater than its own: any better approximation requires alarger denominator.

Similarly, the CF for , of which we previously quoted only the firstthree terms, actually extends to infinity, the first eleven terms being

π

[3;7,15,1,292,1,1,1,2,1,3, … ],

from which we can find = 4272943/1360120 = 3.14159265359 …C (10)Instead of computing the convergents, as above, by finding for

progressively larger values of , it is more convenient and elegant toevaluate successive convergents by using recurrence relations whichcompute from and . These relations are as follows andtheir validity is best proved by induction (see [1]).

C (k)k

C (k) C (k − 1) a (k)

C (k) = r (k) / s(k)If

r (k) = a (k) r (k − 1) + r (k − 2)then

s(k) = a(k) s(k − 1) + s(k − 2) .and

From and we can derive , and , as follows:a (0) a (1) r (0) r (1) s(0) s(1)

C (0) = r (0) / s(0) = a (0) , sor (0) = a (0) ands(0) = 1

C (1) = r (1) / s(1) = a (0) + 1 /a (1) , r (1) = a (0) a (1) + 1 ands(1) = a (1) .so

These equations can be programmed into a repeating loop which prints and continues until (or until , when ).r (k) / s(k) k = n a(k) = 0 k = n + 1

When the of the CF are large it is clear that the CF will convergemore rapidly, and it is apparent that the best approximations are when ,

a (k)a (k)


the first term to be cut off, is a large number. That is why [3;7,16] givessuch a good approximation to , since the CF has been cut off at the largenumber 292. Similarly, Feigenbaum's number, central to chaos theory andequal to 4.6692016..., is approximated well by the CF [4;1,2] = 4.667, sincethe next term is 43 and, taking two more terms, we find [4;1,2,43,2] =4.66920, the next term being 165.

π

Periodic continued fractionsSince the golden ratio, , is defined byφ

φ = 1 +1φ

,

we can see immediately that it is represented by the CF[1;1,1,1,1,1, … ],

the most slowly convergent of all CFs, the convergents being, as is wellknown, the ratio of successive Fibonacci numbers: 1/1, 2/1, 3/2, 5/3, 8/5,13/8, 21/13, 34/21, 55/34, 89/55…. The CF for is really a periodic one soit can be written [1;'1'], where the single quotes enclose the repeatingportion, having in this case the period 1. It can be shown that everyquadratic irrational has a periodic CF.

φ

The general quadratic irrational has the form (whereare integers, , are non-square), which is analogous to the form

for the general rational number; this suggests a possible form ofrecurrence to produce the CF, , equal to ,namely:

(x + z) / y x, y, zy ≠ 0 z > 0

x / yA(k) [a(0) ; a (1) , a(2) , … a (k)]

with A(k) = (x (k) + z) / y(k)

and a (k) = INT (A(k))

put x (k + 1) = a(k) y(k) − x (k)

and y(k + 1) = (z − (x (k + 1))2) / y(k)

then a (k + 1) = INT (A(k + 1)) .

Again, it can be proved by induction that , so derived, is the thpartial quotient in the CF of .

a (k) k(x + z) / y

In the particular case and , we have , sothese recurrences enable us to find the CF of the square root of any integerwhich, like every purely quadratic irrational, is periodic and of the form

x (0) = 0 y(0) = 1 A(0) = z

[a(0) ; ' a (1) , a(2) , a (3) , … , a (k) ']where the period is . Furthermore, it can be shown that andthat the sequence of terms from to is palindromic. Thus theCF of a square root always has the form

k a(k) = 2a(0)a (1) a (k − 1)

[a(0) ; ' a (1) , a(2) , a (3) , … , a (3) , a(2) , a (1) ,2a (0) '] .


To compute the CF of a square root, it is only necessary to program thelast recurrences above in a loop which starts with and , andrepeats until . We can immediately find that ,which is, of course, Hero's approximation; and ,

and . It can also be shown, withoutdifficulty, that, more generally,

x(0) = 0 y(0) = 1a (k) = 2a(0) 2 = [1 : '2']

3 = [1; ' 1,2 ']5 = [2; ' 4 '] 7 = [2; '1,1,1,4 ']

(n2 + 1) = [n; '2n '] (n2 − 1) = [n − 1; '1, 2(n − 1) '](n2 + n) = [n; '2, 2n '] (n2 − n) = [n − 1; '2, 2(n − 1) ']

(n2 + 2) = [n; ' n,2n '] (n2 − 2) = [n − 1; '1, n − 2,1,2(n − 1) ']For example, , ,

but .11 = [3; '3,6 '] 20 = [4; '2,8 '] 47= [6; '1,5,1,12 ']

46 = [6; '1,3,1,1,2,6,2,1,1,3,1,12 ']To find the convergents of a square root, it is probably more convenient

not to stop the loop repetition when , but to continue findingsuccessive values of and, as each one is found, to use the recurrencesalready applied above to compute the next value of . In this way wecan find the convergents of as , , ,

and and those of are ,, , , , and

. These two converge relatively slowly. , on the otherhand, converges rapidly: by the fourth convergent, 33294/5401, we alreadyhave the root accurate to 7 significant figures.

a (k) = 2a(0)a (k)

C (k)2 3/2= 1.5 7/5= 1.4 17/12= 1.4167

41/29= 1.4138 99/70= 1.4143 3 2/1= 2.05/3 = 1.67 7/4= 1.75 19/11= 1.727 26/15= 1.7333 71/41= 1.731797/56= 1.7321 38

ConclusionWe have seen how a CF can be used to convert a decimal to a fraction

by finding the CF of the decimal and evaluating it. The process ofcomputation is in fact closely similar to that of finding the highest commonfactor of numerator and denominator, using the Euclidean algorithm, anddividing out. In the eighteenth century CFs were used to find solutions toPell's equation

p2 − hq2 = 1.If is a convergent of , then it can be shown that all solutionsare given by and for suitable . The full details are in [1].

r (k) / s(k) hp = r (k) q = s(k) k

In more recent years, methods using the CF algorithm have beendevised to accomplish the factorisation of large numbers, an importantmatter for public key systems of security coding. Consider the equation

x2 ≡ y2 (modn)with and . Under these conditions must divide

and, since it does not divide either or, the highest common factor of and and that of and must be divisors of that do not equal 1 or . We can now make

equal to the number to be factorised and use the CF algorithm to find and, as described in [1]. Other, more powerful, techniques based on CF

0 < y < x < n x+ y < n nx2 − y2 = (x + y) (x − y) (x + y)(x − y) n (x + y) n(x − y) n n n

xy


expansions are to be found in [2].

AcknowledgmentI am indebted to the referee for trapping some errors and making several

helpful suggestions which I have used.

References1. Kenneth H. Rosen,Elementary number theory and its applications, (3rd

edn.), Addison-Wesley (1992).2. H. Riesel, Prime numbers and computer methods for factorisation,

Birkhaüser, Boston (1985).

ROBERT MACMILLAN43 Church Road, Woburn Sands, Bucks MK17 8TG

Within a few orders of magnitude . . .“Proton Armageddon” [News and Analysis, “In Brief”, January] contains an

error. The lower limit for the lifetime of a proton is described as being 100 billiontrillion years longer than the age of the universe. In fact, the lifetime of a proton is atleast 100 billion trilliontimes longer than the age of the universe. We apologize forthe confusion.

From Scientific American, April 1999. Spotted by Nick Lord who commentsthat if they are going to make a mistake, make a big one!

Not before time?A postmark spotted by Chris Pease, who speculateson the ‘true’ meaning of the date. Seven daysbefore the start of April 1999? Or perhaps, analternative to 7.4.99BC, in which case, should theGuinness Book of Records be informed of acandidate for the slowest delivery?

Excremental growth?There are 6.8 million dogs in Britain and each day 10 000 tons of dog mess are

deposited in public places.From The Times 3 June 1999 and gleaned by Brian Midgley who has averaged

this as 3.3 lbs per pooch (or 1.5kg) which seems somewhat out of proportion.

1 in 4 = 4 in 1? Perhaps schools have to offer reciprocal funding!. . . and one in four classrooms would have a computer as part of a technologicalrevolution.. . . there will be a minimum of four of them in every primary and secondaryclassroom.

From The Times 6 March 1999, sent in by Frank Tapson.


A construction of magic cubes

MARIÁN TRENKLER

In this paper amagic cube of order is a 3-dimensional matrix containing natural numbers such

that the sums of the numbers along each row ( -tuple of elements having thesame coordinates on two places) and also along each of its four greatdiagonals are the same. This contrasts with a previous article [1] whichconsidered magic cubes with no requirement for the diagonal sums.

nQn = qn(i , j, k); 1 ≤ i , j, k ≤ n 1,2, … n3

n

In Figure 1 a magic cube is depicted. The sum of the three numbersin every row is 42. The sums of the numbers on the diagonals (the triplets8, 14, 20, 19, 14, 9, 10, 14, 18 and 6, 14, 22) are 42 too.

Q3

A

CB8

24

10

15

1

26

19

17

6

B12

7

23

25

14

3

5

21

16

22

11

9

2

27

13

18

4

20

C

FIGURE 1 − Magic cube Q3

Using a pair of orthogonal Latin squares, it was proved that suchmatrices of order exist for each . Probably the first mention of amagic cube (of order 4) appeared in a letter from Fermat on April 1, 1640(see [2, p. 365].) More information on magic squares and cubes can befound in books [2] and [3]. In this paper we describe, in three steps, aconstruction of a magic cube for every integer . (In a similar waywe can construct a magic square for every integer .)

n n ≠ 2

Qn n ≠ 2Mn n ≠ 2

Step 1.If is an odd integer, then a magic cube can be constructed using the

following formulan Qn

qn(i , j, k) = [(i − j + k − 1) − ni − j +k−1n ] n2 +

[(i − j − k) − ni − j −kn ] n + [(i + j + k − 2) − ni + j +k−2

n ] + 1.The symbol denotes the integer part of .x x

The formula was derived using two mutually orthogonal Latin squaresof odd order , and

where and are two constantsand the formula is taken from [1, p. 57]. This formula can be rewritten as

n Rn = r (i , j) = (i + j + a) − i + j + an

Sn = s(i , j) = (i − j + b) − i − j + bn a b

q∗n(i , j, k) = s(i , s(j, k))n2 + s(i , r (j, k))n + r (i , r (j, k)) + 1

The constants and were chosen so that for a b m = (n + 1) / 2

s(m, s(m, m)) = s(m, r (m, m)) = r (m, r (m, m)) = n − 12 .

The proof of the correctness of our formula is similar to [1, p. 58]. The sum

A CONSTRUCTION OF MAGIC CUBES 37

on every diagonal is the same because for each triple from thedefinition of it follows ( denotes the number )

(i , j, k)Qn x¯ n + 1 − x

qn (i , j, k) + qn (i¯ , j¯ , k¯ ) = ∑2

k = 0

(n − 1) nk + 2 = n3 + 1

qn (n + 12 , n + 1

2 , n + 12 ) = n3 + 1

2 .and

The sum of numbers on each diagonal is (n − 1)2 (n3 + 1) + (n3 + 1)

2 = n(n3 + 1)2 .

Step 2.If , , then a magic cube can be constructed

by the following formulasn = 4k k = 1, 2, 3, … Qn

qn(i , j, k) = (k − 1)n2 + (j − 1)n + i if I(i , j, k) is odd,qn(i , j, k) = (n − k)n2 + (n − j)n + (n − i) + 1 if I(i , j, k) is even,

where .I(i , j, k) = (i + 2(i − 1)n + j + 2(j − 1)

n + k + 2(k−1)n )

In Figure 2 a magic cube is depicted. The sums of the four numbersin each row are 130. The sums of the numbers on the diagonals (thequadruples 1, 43, 22, 64, 4, 42, 23, 61, 13, 39, 26, 52 and 16, 38, 27,49) are 130 too.

Q4

AB

CD

BC

D1

60

56

13

63

6

10

51

62

7

11

50

4

57

53

16

48

21

25

36

18

43

39

30

19

42

38

31

45

24

28

33

32

37

41

20

34

27

23

46

35

26

22

47

29

40

44

17

49

12

8

61

15

54

58

3

14

55

59

2

52

9

5

64

FIGURE 2 − Magic cube Q4

The proof of the correctness of our formulas follows from the followingthree facts:

(i) No two elements with different coordinates are the same because is odd, if and only, if is odd.I(i¯ , j¯ , k¯ ) I(i , j, k)

(ii) The sums of numbers in the rows are the same, because, for every oddcoordinate (or , or ):i j k

qn(i , j, k) + qn(i + 1, j, k) = n3 or n3 + 2

qn(i , j, k) + qn(i , j + 1, k) = n3 − n + 1 or n3 + n + 1

qn(i , j, k) + qn(i , j, k + 1) = n3 − n2 + 1 or n3 + n2 + 1.In every row there are pairs of elements whose sum is (or

or ) and the same number of pairs whose sumis (or or ).

n / 4 n3

n3 − n + 1 n3 − n2 + 1n3 + 2 n3 + n + 1 n3 + n2 + 1

(iii) The sums on the diagonals are the same, because, for every triple:(i , j, k)

qn (i , j, k) + qn (i¯ , j¯ , k¯ ) = n3 + 1.


Step 3.If , , then a construction of a magic cube

starts from a magic cube and an auxiliary cube oforder . First we describe the construction of . In Figure 3 six layers of

are shown.

n = 4k + 2 k = 1, 2, 3, …Qn Qn/2 Vn = vn(i , j, k)

n Vn

V6

0 5 5 0 6 5

3 6 3 6 3 0

3 5 5 0 3 5

6 0 3 6 6 0

6 0 0 6 3 6

3 5 5 3 0 5

7 2 2 7 1 2

4 1 4 1 4 7

4 2 2 7 4 2

1 7 4 1 1 7

1 7 7 1 4 1

4 2 2 4 7 2

1st Layer 2nd Layer

0 5 5 6 0 5

3 6 0 3 3 6

6 0 3 6 6 0

3 5 5 0 3 5

6 0 3 6 6 0

3 5 5 0 3 5

7 2 2 1 7 2

4 1 7 4 4 1

1 7 4 1 1 7

4 2 2 7 4 2

1 7 4 1 1 7

4 2 2 7 4 2

3rd Layer 4th Layer

5 0 5 0 6 3

3 6 3 6 3 0

0 3 3 6 3 6

5 6 5 0 0 5

5 0 5 3 3 5

3 6 0 6 6 0

2 7 2 7 1 2

4 1 4 1 4 7

7 4 4 1 4 7

2 1 2 7 7 2

2 7 2 4 4 2

4 1 7 1 1 7

5th Layer 6th LayerFIGURE 3

This cube consists of 27 cubelets of order 2, each containing the numbers 0,1, … ,7. The arrangements of the numbers in cubelets is such that the sumsof numbers in each row and each diagonal are the same, i.e. 21.

If for then is obtained by thecomposition of copies of such that

n = 6 (2k + 1) k = 1, 2, 3, … , Vn

(2k + 1)3 V6

vn(i , j, k) = v6(i − 6i −16 , j − 6j −1

6 , k − 6k−16 ) for all1 ≤ i , j, k ≤ n.

If for , then our constructionstarts from , where and six matrices (in pairs of orders

n = 6 (2k + 1) + 4 k = 0, 1, 2, …Vm m = n − 4


and and ) called blocks. Blocksm × m × m × 2 m × 2 × 2 2 × 2 × 2

A = a(i , j, k) and A = a¯ (i , j, k), 1 ≤ i , j ≤ m, 1 ≤ k ≤ 2,B = b(i , j, k) and B

= b(i , j, k) , 1 ≤ i ≤ m, 1 ≤ j, k ≤ 2,

C = c(i , j, k) and C = c¯ (i , j, k) , 1 ≤ i , j, k ≤ 2

are defined by relations

a (i , j, k) = b (i , j, k) = c (i , j, k) = vm (i , j, k) ,

a (i , j, k) = b (i , j, k) = c¯ (i , j, k) = 7 − vm (i , j, k) ,for all define elements with coordinates .(i , j, k)

The auxiliary cube consists of a cube , three pairs of blocks and, six pairs of and and four pairs of and . The blocks and are

situated by the vertices of , blocks and are situated by the edges ofand blocks and at the opposite faces of . In Figure 4a there are thefirst two layers of the auxiliary cube , in Figure 4b there are the -th, for

, layers and in Figure 4c there are the last two (the-th and the -th) layers.

Vn Vm AA B B

C C

C C

Vn B B

Vn

A A Vm

Vn xx = 3, 4, 5, … , n − 2(n − 1) n

Every diagonal of is formed from diagonals of , and . Itfollows from this construction that the sum of numbers in every row andevery diagonal of is greater by 14 than the sum in any row of .

Vn C Vm C

Vn Vm

c (1,1, x) c (1,2, x) b (1,1, x) … b (m,1, x) c¯ (1,1, x) c¯ (1,2, x)

c (2,1, x) c (2,2, x) b (1,2, x) … b (m,2, x) c¯ (2,1, x) c¯ (2,2, x)

b (1,1, x) b (1,2, x) a (1,1, x) … a (1, m, x) b (1,1, x) b (1,2, x)

… … … … … … …

b (m,1, x) b (m,2, x) a (m,1, x) … a (m, m, x) b (2m,1x) b (m,2, x)

c¯ (1,1, x) c¯ (1,2, x) b (1,1, x) … b (m,1, x) c (1,1, x) c (1,2, x)

c¯ (2,1, x) c¯ (2,2, x) b (1,2, x) … b (m,2, x) c (2,1, x) c (2,2, x)

FIRST AND SECOND LAYER OF , .Vn x = 1, 2

b (x,1,1) b (x,1,2) a (x,1,1) … a (x, m,1) b (x,1,1) b (x,1,2)

b (x,2,1) b (x,2,2) a (x,1,2) … a (x, m,2) b (x,2,1) b (x,2,2)

a (x,1,1) a (x,1,2) vm(1,1, x) … vm(1, m, x) a (x,1,1) a (x,1,2)

… … … … … … …

a (x, m,1) a (x, m,2) vm(m,1, x) … vm(m, m, x) a (x, m,1) a (x, m,2)

b (x,1,1) b (x,1,2) a (x,1,1) … a (x, m,1) b (x,1,1) b (x,1,2)

b (x,2,1) b (x,2,2) a (x,1,2) … a (x, m,2) b (x,2,1) b (x,2,2)

-th LAYER OF , (x + 2) Vn x = 1, 2, … , m


c¯ (1,1, x) c¯ (1,2, x) b (1,1, x) … b (m,1, x) c (1,1, x) c (1,2, x)

c¯ (2,1, x) c¯ (2,2, x) b (1,2, x) … b (m,2, x) c (2,1, x) c (2,2, x)

b (1,1, x) b (1,2, x) a (1,1, x) … a (1, m, x) b (1,1, x) b (1,2, x)

… … … … … … …

b (m,1, x) b (m,2, x) a (m,1, x) … a (m, m, x) b (m,1, x) b (m,2, x)

c (1,1, x) c (1,2, x) b (1,1, x) … b (m,1, x) c¯ (1,1, x) c¯ (1,2, x)

c (2,1, x) c (2,2, x) b (1,2, x) … b (m,2, x) c¯ (2,1, x) c¯ (2,2, x)

-th LAYER OF , (n − x + 2) Vn x = 1, 2FIGURE 4

If , then we repeat the previous construction.n = 6 (2k + 1) + 8We define a magic cube by the following relationQn

qn(i , j, k) = vn(i , j, k)n3

8 + qn/2(i +12 , j +1

2 , k+12 ) .

A construction of magic squaresSimilarly we find that a magic square of

odd order can be constructed using the following formulaMn = mn(i , j); 1 ≤ i , j ≤ n

n

mn(i , j) = [(i − j − n−12 ) − ni − j + n− 1

2n ]n + [(i + j + n−3

2 ) − ni + j + n− 32

n ] + 1,and, for , by the formulasn = 4k

mn(i , j) = (i − 1)n + j if I(i , j) is odd,

mn(i , j) = (n − i)n + (n − j) + 1 if I(i , j) is even,

where I(i , j) = (i + 2(i − 1)n + j + 2(j − 1)

n ) .

RemarkAnalogous formulas can be derived for magic -dimensional

hypercubes of order , for every integer , and every .(See [1].) For example, if and is odd, then we start from theformula

dn ≠ 4k + 2 k ≥ 1 d ≥ 4

d = 4 n

q∗n (i , j, k, l) = s(i , s(j, s(k, l))) n3 + s(i , s(j, r (k, l))) n2 +

s(i , r (j, r (k, l))) n + r (i , r (j, r (k, l))) + 1.

Let for . Analogously to a magic cube, we can construct a magic square of order . It is

constructed using and an auxiliary square . Theconstruction of starts from (situated in the middle of Figure 5). Forexample, in Figure 5, is depicted which was obtained using . If

or , then (to preserve the sum of numbers

n = 4k + 2 k = 1, 2, 3, …Qn Mn n = 4k + 2

Mn/2 Wn = wn (i , j)Wn W6

W10 W6

n = 6 (2k + 1) + 4 6(2k + 1)


on its diagonals) the last two rows are changed. The elements of areMn

mn (i , j) = wn (i , j) n2

4 + mn/2 (i + 12 , j + 1

2 ) .

1 2 1 2 0 3 0 3 2 1

0 3 0 3 2 1 2 1 3 0

1 0 1 2 0 3 0 3 2 3

2 3 0 3 2 1 2 1 1 0

0 2 3 0 3 0 3 0 3 1

3 1 2 1 2 1 2 1 0 2

0 2 1 0 2 3 0 3 3 1

3 1 2 3 0 1 2 1 0 2

3 0 2 0 1 2 1 2 0 3

2 1 3 1 3 0 3 0 1 2

FIGURE 5 − AUXILIARY SQUARE V10

References1. M.Trenkler, Magic cubes, Math. Gaz. 82 (March 1998) pp. 56-61.2. W. S. Andrews, Magic squares and cubes, Dover, New York (1960).3. W. H. Benson, O. Jacoby,Magic cubes, new recreations, Dover, New

York (1981).MARIÁN TRENKLER

©afárik University, Jesenná 5, 041 54 Ko¹ice, Slovakia

Baby boomThe most recent figures are deeply alarming. Nearly one in 100 girls aged

between 13 and 15 between 1995 and 1996 got pregnant − up by 11%.From The Times, 13 December 1998 and spotted by Frank Tapson, who asks

‘11% of what?’

Formula 1 devalues the £He values the man as much as machine, one reason why he recruited the world's

best driver on the highest salary − $20 million (around £1.2 million) a year.From The Times, 4 May 1999, sent in by Frank Tapson.

Negative growth. . . These figures coincide to establish that the yew grows about a foot in girth

every 30 years. Rich or poor soil can add or subtract about five feet to this figure.From The Times 6 October 1998, sent in by Frank Tapson.

A fishy taleThen on another trip we took a whole ton, with several other hauls of 200 stone.From The Times 2 January 1999 spotted by Frank Tapson who wonders if it

implies that 1 ton is bigger than 200 stone? For younger readers, 1 ton is 160 stone.


The factorial function: Stirling's formula

DAVID FOWLER

How big is ? For example, to pluck a number out of thin air, what isthe order of magnitude of 272!?

n!

This is the third note of a series on the factorial. The two previous notes[1] and [2] dealt with the factorial function while here we only consider

when is an integer, but its main results also apply to non-integerarguments. This note is longer and more elaborate than the previous ones,but the principal result is in the opening section. Onlyaficionados need goany further.

x!n! n

The plan is very straightforward: integrating by parts gives. On the other hand, approximating the same integral

by the trapezium rule gives something very much like . Thisgives an identity which we then exponentiate.

∫n1 log t dt =

n logn − (n − 1)∫n1 log t dt logn!

So we start with

∫k + 1k log t dt = 1

2 (logk + log(k + 1)) + ek

by a one-step application of the trapezium rule, where we must find out justhow big the error term can be. (Proper analysis is all about error terms!)There is a general estimate for one step of the process applied to a function:

ek

f

|∫a + h

af (t) dt −

h

2(f (a) + f (a + h))| ≤

h3

12sup

x ∈ [a, a + h]|f ″ (x)| ,

and a proof of this will be given later. In the case of our integral, and, so that . Hence will converge (by

comparison with ), and so will tend to a limit astends to infinity. Also is ‘convex downwards’ — its graph lies aboveall of its chords — so all of the are positive, and will increase to .

h = 1d2

dt2 log t = −1 / t2 |ek| ≤ 1 / (12k2) ∑ ek

∑ 1 /k2 sn = ∑n − 11 ek s n

logxek sn s

Putting all of this together,

∫n1 logt dt = 1

2 log1+ log2+ … + log(n − 1) + 12 logn + sn

= logn! −12

logn + sn

and

∫n

1log t dt = n logn − (n − 1)

so

logn! = (n + 12) logn − n + 1 − sn

THE FACTORIAL FUNCTION: STIRLING'S FORMULA 43

and therefore, exponentiating and using the result that ,ea log b = ba

n! = n(n + 12)e−ne(1 − sn) = λnn

(n + 12)e−n,

where will decrease to some limit as tends to infinity.In fact, we shall show below that so that, for large , we arrive atStirling's formula, either in the logarithmic form given earlier, or in themultiplicative form

λn = e(1 − sn) λ > 0 nλ = 2π n

n! ∼ 2nπ (n

e)n

,

which gives an underestimate for . (Here as means that as . Their absolute difference may be large, but their

relative error , becomes small.)

n! f ∼ g n → ∞f (n)g(n) → 1 n → ∞

f (n) − g(n)g(n)

Back now to our ‘randomly’ chosen example of 272!; it was in factchosen specially for convenience, since , soe = 2.7183…

272! ∼ 2π 272( 2722.718…)272

and we can estimate this easily. (A lot of science in general, andmathematics in particular, is more meaningful if we have some idea of howbig things are, how close together things are, etc.)

First, by mental arithmetic,, which is smaller than it should be, and2π ≈ 61

4 = 254 = 2.5

, again smaller, and272 ≈ 256 = 16

, also smaller,( 2722.718…)272 ≈ ( 272

2.72)272 = 100272

and our formula gives an underestimate. So we should expect that

2.5 × 16 × 100272 = 4 × 10545

should be an underestimate for 272!.Next and boringly, evaluating it using a pocket calculator (though a bit

of special manipulation may be necessary to get the answer out) gives

272! ≈ 4.90927… ×10545,also an underestimate.

Finally, my personal computer can produce the exact value:

4 91077 75488 33688 95232 71661 18754 61955 68214 1619548789 25656 54460 51695 24519 12348 09306 31406 19315 4660897339 55555 21886 63186 49961 20305 14101 29305 12020 3893133375 31545 34412 98132 63073 29390 10373 33664 88089 7354975755 99796 32836 30443 52849 27493 28267 91661 18697 4181242313 86429 31079 89404 54836 03454 58203 61221 64128 2229829519 11515 27297 28254 15061 05307 70249 78972 01682 9649131942 63614 21090 85550 56425 17848 51393 20167 76733 0418695064 19795 77258 62474 98611 55053 49324 93984 49320 24411


88988 27907 73061 48222 04505 83307 87840 00000 00000 0000000000 00000 00000 00000 00000 00000 00000 00000 00000 00000.

(This number prompts the question: how many zeros does ! finish with?)nHaving got this first expression for the approximation so simply, we

should now go on to estimate the error more and more precisely, and sorefine this estimate. In fact

n! ∼ 2nπ (n

e)n (1 +1

12n+

1288n2

−139

51840n3−

5712488320n4

+ … ) ,

where this infinite series is not convergent, but ‘asymptotic’, and these kindsof series are important for describing and evaluating functions that tend toinfinity. For large , the absolute values of the terms of such series start bydecreasing rapidly but, further on, they eventually start increasing. Howeverthe partial sums give remarkably good and efficient approximations to thefunction. For example, the first digits, rounded, arising from using thisseries for 272! are

n

272! = 4 91077 75488 33689… ,

one term 4 90927 32525 25454…two terms: 4 91077 73190 85664…three terms: 4 91077 75494 88017…four terms: 4 91077 75488 33892…five terms 4 91077 75488 33686…

and the accuracy is improved for larger values of .nThis note will end with a first step in the refinement of the formula.

Interlude: A bit of historyJames Stirling gave a more general version of the formula for as

Proposition 28 of hisMethodus Differentialis (1730).* Translated,simplified, and with the notations made to look a bit more familiar, thisreads:

logn!

To find the sum of any number of logarithms, whose numbersare in arithmetic progression.Let , , , , …, , denote anynumbers in arithmetic progression, whose common differenceis . Then the sum of the logarithms set forth will be equal tothe difference of the following two series

x + d x + 3d x + 5d x + 7d z − d

2d

zlogz

2d−

z

2d−

d

12z+

7d3

360z3−

31d5

1260z5+

127d7

1680z7−

511d9

1188z9+ etc

and

* This section is entirely based on [5], which also contains details of correspondence betweenStirling, De Moivre, and Euler. I am grateful to Dr Ian Tweddle for help and explanations.


x logx

2d−

x

2d−

d

12x+

7d3

360x3−

31d5

1260x5+

127d7

1680x7−

511d9

1188x9+ etc.

As an illustration, he specialised this: Put and ,which gives the integers from 1 to ; and the second series, in , will nowsum to a number which, without explanation, Stirling set equal to(‘half of the logarithm of the circumference of the circle whose radius isunity’).* Then, ignoring the terms in powers of since these will be smallfor large , we get

x = d = 12 z = n + 1

2n x

−12 log 2π

1 /zz

log1+ log2+ … + logn ∼ (n + 12) log(n + 1

2) − (n + 12) + 1

2 log2πand, exponentiating,

n! ∼ 2π (n + 12

e )n + 12

.

This variant is slightly more accurate than the usual version of the formulathat we have been considering here.

At the same time that Stirling was deriving these formulae, De Moivrewas exploring a problem in gambling, which led him to estimating the sizeof the middle coefficient of the binomial expansion of , namely

, and some of his results were included in aSupplement to his

Miscellanea Analytica, published in 1730, a few months after Stirling'sbook. He noticed the occurrence of the same numbers 12, 360, 1260, and1680 in his own results, and adapted his methods to the sums of logarithms.Among other results, he found that

(1 + x)2n

( ) =(2n)!n! n!

2nn

log(n − 1)! = (n − 12) logn + 1

2 log2π − n + 112n − 1

360n3 + 11260n5 − 1

1680n7 + etc

which, after adding the term to both sides, yields the logarithmicversion of the formula we have been considering here. His methods wererather less rigorous than Stirling's, but he did evaluate the multiplierusing another of Stirling's results. The first known explicit occurrence of theactual multiplicative version that we have been dealing with is in a letterfrom Euler to Goldbach dated 23 June 1744. These manipulations ofStirling, De Moivre, and Euler are quite remarkable, considering themethods available to them at their time.

logn

2π

Mopping upTwo details have been left over. (These involve some elementary but

elaborate manipulations and should be omitted by anyone who is satisfiedby the derivation so far, though the first is very pretty.)

• The evaluation of the multiplier .λ

* This series is not convergent (its terms increase rapidly from 7/360 onwards) butasymptotic. When summed up to and including its smallest term, it gives

, a tolerably good approximation to.

12 log 1

2 − 12 − 1

12 + 7360 = −0.9189…

−12 log 2π = −0.9046…


There are many ways of doing this. Here is one using Wallis' product:

π2

= limn → ∞

21

.23

.43

.45

.65

.67

. … .2n

2n − 1.

2n

2n + 1

= limn → ∞

21

. (22

.22) .

23

.43

. (44

.44) .

45

.65

. (66

.66) .

67

. … .2n

2n − 1. (2n

2n.2n

2n) 2n

2n + 1

= limn→ ∞

24n(n!)4

(2n!)2.

12n + 1

= limn→ ∞

(( n!n(n

e)n)4

. ( 2n(2ne )2n

2n! )2

. ( 24n

2n + 1.

n2(ne)4n

2n(2ne )4n))

= limn→ ∞( n

2(2n + 1)) = λ4.λ−2 14,

so .λ = 2π(There is a nice derivation of Wallis' product using the reduction formula

, , but enough is enough; one must stop somewhere!)In = ∫π/20 sinnt dt n ≥ 2

• The estimate of the error in the trapezium rule.*

We start by deriving a form of Taylor's theorem using integration byparts. (This, as for so many proofs of different varieties of Taylor's theorem,will be rather like a rabbit pulled out of a hat, but there is no space here formore exploration and explanation.)

∫ a + x

a

(a + x − t)n − 1

(n − 1)! f (n) (t) dt

= (a + x − t)n − 1

(n − 1)! f (n − 1) (t)

a + x

a+ ∫

a + x

a

(a + x − t)n − 2

(n − 2)! f (n − 1) (t) dt

= −xn − 1

(n − 1)! f (n − 1) (a) +

(a + x − t)n − 2

(n − 2)! f (n − 2) (t)

a + x

a

+ ∫ a + x

a

(a + x − t)n − 3

(n − 3)! f (n − 2) (t) dt

= …

= −xn−1

(n − 1)! f (n−1)(a) −

xn−2

(n − 2)! f (n−2)(a) − … −xf ′ (a) − f (a) + f (a + x)

so

f (a + x) = f (a) + xf ′ (a) +… +xn−1

(n − 1)!f (n−1)(a) + ∫

a+x

a

(a + x − t)n−1

(n − 1)! f (n)(t)dt

where this final integral is the remainder term. (Proper analysis is all about

* This is a standard result, but I have included it here for completeness and because I wouldhave difficulty in giving an impromptu proof of it.


remainder terms, another name for error terms!)

Apply this to the function , so ,, etc.:

F (x) = ∫ x a f (t) dt F′ (x) = f (x)

F″ (x) = f ′ (x)

F(a + h) = F(a) + hf (a) +h2

2f ′ (a) +

12 ∫

a+h

a(a + h − t)2f ″(t)dt.

Now eliminate the term in by using the Taylor series for :f ′ (a) f (x)

f (a + h) = f (a) + h f ′(a) + ∫a + h

a(a + h − t) f ″(t) dt

so that

F(a + h) − F(a) =h

2 f (a) +

h

2 f (a + h) +

12 ∫

a+h

a(a + h − t)2f ″(t)dt

−h

2 ∫a+h

a(a + h − t)f ″(t)dt

=h

2(f (a) + f (a + h)) +

12 ∫

a+h

a(a + h − t)(a − t)f ″(t)dt,

and this last integral gives the error term for the approximation of theintegral using a one-step trapezium rule. But if and ,then while and hence

h ≥ 0 t ∈ [a, a + h](a + h − t) ≥ 0 (a − t) ≤ 0

|12 ∫a+h

a(a + h − t)(a − t)f ″(t)dt| ≤

12 ∫

a+h

a(a + h − t)(t − a)|f ″(t)|dt

≤12 ∫

a+h

a(a + h − t)(t − a)dt × sup

x∈[a, a+h]|f ″(x)| =

h3

12sup

x∈[a, a+h]|f ″(x)| ,

the estimate that we wanted.

A refined approximationLet us start again, and estimate the error a bit more delicately.* We

haveek

logn! = (n + 12) logn − n + 1 − sn

where and . Settingsn = ∑n − 11 ek ek = ∫

k+ 1k logt dt − 1

2 (logk + log(k + 1))

∫ k + 1

klog t dt = [t log t − t]k + 1

k

and simplifying, we find

ek =2k + 1

2log(k + 1

k ) − 1.

Now

log (1 + x

1 − x) = 2 (x +x3

3+

x5

5+ … ) |x| < 1,for

in which we substitute

* This section is based on [3].


x =1

2k + 1, so

1 + x

1 − x=

k + 1k

and

ek =2k + 1

2log(k + 1

k ) − 1

= (2k + 1)( 12k + 1

+1

3(2k + 1)3+

15(2k + 1)5

+ … ) − 1

=1

3(2k + 1)2+

15(2k + 1)4

+1

7(2k + 1)6+ … .

Manipulate this series to find upper and lower bounds for . First increaseeach term to give

ek

ek <1

3(2k + 1)2+

13(2k + 1)4

+1

3(2k + 1)6+ …

=1

3(2k + 1)2 (1 +1

(2k + 1)2+

1(2k + 1)4

+ … ) =112(1

k−

1k + 1)

after a bit of reorganisation. Now go the other way round:

ek > ( 13(2k + 1)2

+1

32 (2k + 1)4+

133 (2k + 1)6

+ … )=

13(2k + 1)2 − 1

=1

12k2 + 12k + 2>

112( 1

k + 112

−1

k + 1 + 112

).

(This last step is a bit more involved. Working backwards

( 1k + 1

12

−1

k + 1 + 112

) =1

(k + 112)(k + 1 + 1

12)=

1k2 + (1 + 1

6)k + 112(1 + 1

12)

=12

12k2 + 14k + (1 + 112)

>12

12k2 + 12k + 2,

since .)2k − 1112 > 0

Thus we have established that

112 ( 1

k + 112

−1

k + 1 + 112

) < ek <112 (1

k−

1k + 1) .

Adding these for and using again the notations and we get

k = 1, 2, 3, …sn = ∑n − 1

1 ek s = ∑∞1 ek

113

< s <112

,

while adding for , …, and writing , , we getk = n, n + 1, n + 2 tn = ∑∞ n ek

112n + 1

< tn <1

12n.


So, finally,

logn! ≈ (n +12) logn − n + 1 − s + tn.

Exponentiating, inserting the value and combining thesetwo results, we get the refined estimate

e1 − s = λ = 2π

2nπ (n

e)n

e1/(12n + 1) < n! < 2nπ (n

e)n

e1/(12n).

We can add this to our collection of estimates for 272!:4 91077 70887 06842 ... < 272! < 4 91077 75495 11548 ...,

matching the accuracy of two and three terms of the asymptotic expansionquoted earlier.

Coda

An asymptotic expansion for can be found using refinements ofthe procedure here, in which we find some expression for the error of onestep of the trapezium approximation. The most efficient way is to express itin terms of sums of multiples of the derivatives of , in a procedure creditedto Euler and MacLaurin, and two extracts from letters between Stirling andEuler about this form a nice conclusion to our story. First, from Stirling toEuler, 16 April 1738 [5, pp. 144-5], in reply to an earlier letter in whichEuler had described some of his results:

∑ f (n)ek

f

Most pleasing to me was your theorem for summing series bymeans of the area of a curve and derivatives or fluxions ofterms for it is general and well-suited for application. Iimmediately perceived of extending it to very many types ofseries, and what is extraordinary, it approximates very rapidlyin most cases. Perhaps you have not noticed that my theoremfor summing logarithms is nothing more than a special case ofyour general theorem.* But this discovery was all the morepleasing to me because I had also thought about the samematter some time ago; but I did not proceed beyond the firstterm ...At this point you should be advised that in due course MrMacLaurin, Professor of Mathematics at Edinburgh, will bepublishing a book on fluxions. He has communicated to mesome of its pages which have already been printed. In these hehas two theorems for summing series by means of derivativesof the terms, one of which is the self-same result that you sentme; I have informed him of this. Although he had willinglypromised that he would acknowledge this in his preface, Inevertheless submit to your judgement whether you wish topublish your letter in our Philosophical Transactions.

* Tweddle remarks about this, [5, p. 158 n. 15], that in fact it is De Moivre's version whichcomes from Euler's formula.


and then, in reply, from Euler to Stirling, 27 July 1738 [5, p. 146]:… in this matter I have very little desire for anything to bedetracted from the fame of the celebrated Mr MacLaurin sincehe probably came upon the same theorem for summing seriesbefore me, and consequently deserves to be named as its firstdiscoverer. For I found the theorem about four years ago, atwhich time I also described its proof and application in greaterdetail to our [Saint Petersburg] Academy; my dissertation onthis ... will shortly see public light in ourCommentarii whichcome out each year.

What a nice friendly bit of jostling over an issue of precedence. If onlyNewton, Leibniz, and their associates could have behaved like that!

PostscriptA simplified version of the contents of the first section of this note, with thefussy but important details omitted, should be a standard exercise in anyintroduction to analysis, and some version of Stirling's formula should beproved in any serious course, perhaps as an extended exercise. In fact thisderivation is in the book [4] by Stirling (no relation!), as Exercise 10 onp. 134, but I have replaced the evaluation of the multiplier, , hisExample 11, by the use of Wallis' product.

2π

References1. D. H. Fowler, A simple approach to the factorial function,Math. Gaz.

80 (1996) pp. 378-381.2 D. H. Fowler, The factorial function: the next steps,Math. Gaz.83

(1999) pp. 53-58.3. H. Robbins, A remark on Stirling's formula,American Mathematical

Monthly 62 (1955) pp. 26-29.4. D. S. G. Stirling, Mathematical analysis, a fundamental and

straightforward approach, Ellis Harwood & John Wiley, Chichester,Brisbane, & Toronto (1987).

5. I. Tweddle,James Stirling: ‘This about series and such things’, ScottishAcademic Press, Edinburgh (1988).

DAVID FOWLERMathematics Institute, University of Warwick, Coventry CV4 7AL

QED Requiescat In PaceThe long-running BBC science programmeQED has decided to change its

name to Living Proof after it was discovered that many viewers hadn't the faintestidea of what the title meant. A BBC spokesman said: “I think that in the Sixties andSeventies − in the days of grammar schools − a lot of people would have known, butnot now.”

From the Sunday Telegraph 26 September 1999, by Nick Lord.

A SIMPLE ENERGY-CONSERVING MODEL 51

A simple energy-conserving modelexhibiting elastic restitution

RICHARD BRIDGES

From the date of my own A level studies (many moons ago!) I havebeen dissatisfied with the presentation of the law of elastic restitution. Thetitle Newton's ‘Experimental’ Law implies that it is a purely empiricalphenomenon, yet if there is a wide range of conditions in which it isapproximately valid (and it isn't worth using at all if this isn't true), thenthere must be some reason for it. Equally unsatisfactorily, the law impliesloss of energy in most circumstances. Even in my school days I could seethat the usual argument about conversion into heat, sound etc didn't reallyhold water, as this process couldn't conceivably be instantaneous. A recentarticle by O'Connor [1] stimulated me to crystallise my thoughts on thesubject [2], and I was later able to develop a simple and plausiblemechanical model exhibiting restitution, which also conserved energy [3].My presentation in [3] was mainly focussed on the physical consequences ofthe model, and solution of the differential equations involved wasnumerical. Yet, in the simplest case, the model is amenable to fairly elegantexact analytical techniques, which it is the aim of this paper to present.Apart from the interest of the model, the methods used are a nice illustrationof the role of normal modes in the solution of simultaneous lineardifferential equations—a topic approachable by A level students of FurtherMaths, or by first-year undergraduates.

The basic modelMy contention in [2] was that the energy lost in inelastic rebound goes

initially into internal vibrational modes of the rebounding system. In [3] Ideveloped a one-dimensional model of a ‘ball’ in which two point-masses

are connected by a massless ideal spring of stiffness (I: internal), andinteract with a wall or other ‘balls’ through outwardly directed springs ofstiffness (S: surface)—see Figure 1.

m/ 2 kI

kS

While moving freely such a system is able to vibrate about its centre ofmass, the springs playing no part. Interaction with a rigid wall ismodelled by fixing the end of one spring to the wall: the system then hastwo degrees of freedom, and therefore two vibrational modes of differentfrequencies. When the ‘ball’ is first incident on the wall these modes areexcited to different amplitudes and phases, and when the interaction ends(when the compression-only spring goes slack) the motion smoothlytransfers into a centre of mass velocity plus some amplitude of the freeoscillation mode.

kS

kS

kS


−0.5 0.0 0.5 1.0 1.5 2.0Dimensionless time

m2

m2

kS

kI

1

2

3

4

5

6

7

kS

Dim

ensi

onle

ss d

ispl

acem

ent

FIGURE 1: Rebound with , kR = 2 e = 0.882

Analysis of the modelIf a body of mass compresses a spring of stiffness against a rigid

wall, then its extension (in the horizontal plane, or in the absence ofgravity) obeys the differential equation

m kx

mx¨ + kx = 0 (1)A solution of the form yields:x = αe ± iωt

(−ω2m + k) α = 0 (2)whence . Such a model exhibits perfect elasticity: if it isincident on the wall with an initial velocity , a half cycle of SHM takesplace in time and it rebounds with velocity so that thecoefficient of restitution , assuming of course the mass does not hitthe wall.

ω = k / mv

τ = π / ω ve = 1

Two bodies of mass connected to each other by a spring ofstiffness and repelled from a rigid wall by a spring of stiffness havedisplacements and obeying differential equations:

m/ 2kI kS

x1 x2

12mx¨ 1 + kSx1 − kI (x2 − x1) = 0

12mx¨ 2 + kI (x2 − x1) = 0. ( )3a

These may be rewritten as the matrix equation:

Mx + K Cx = 0 ( )3bwhere the column vector and matrices and (C: Constrained)are:

x 2 × 2 M K C


x = ( ) M = ( ) K C = ( ) .x1

x2

m/ 2 00 m/ 2

kS + kI −kI

−kI kI

When the spring is not compressed, the system vibrates freely and theterm is dropped from giving .

kS kS

K C K F

Generalisation to n massesSystems with masses (possibly different) joined in a line by springs

obey an equation of very similar form to (3b). The vectors and matrices are-dimensional, being diagonal and being tri-diagonal and symmetric

(retaining the symmetry of is useful, and is the reason for notautomatically dividing (3a) by . If the masses differ, this breaks thesymmetry of ).

n

n M KK

mK

If we substitute a solution of the form into (3b) we obtain:x = αe±iωt

(−ω2M + K C) α = 0 (4)Unlike (2), this is a matrix equation, for which must obey thecharacteristic equation:

ω

det(−ω2M + K C) = 0.This is an th degree equation in whose solutions, the eigenvalues,

are the natural vibrational frequencies of the system. Each eigenvaluehas a corresponding eigenvector direction representing the mode ofvibration. Replacing with in (4) and (5) leads to the vibrationalfrequencies and modes of the free system. Since one of thesolutions for is zero, corresponding to the translational motion (whichhas all components equal), leaving only free vibrational modes offrequency and eigenvector directions .

n ω2

n ωi

αi

K C K F

det(K F) = 0ω α0

n − 1ωFi

αFi

A standard piece of matrix manipulation shows that, sinceM is diagonaland (or ) is symmetric, the eigenvector directions corresponding todistinct eigenvalues are orthogonal and may be chosen to be normalised, inthe sense that (ie equal to 0 unless , when equal to ).

is arbitrary, but should be of the dimensions of a mass—the total mass ofthe system, , is a suitable value. The eigenvectors are thendimensionless.

K C K F αi

αTi Mαj = mδij i = j m

mTrace(M )

Modelling the collisionThe three stages of the collision, (I) before (during which there is no

vibration), (II) during and (III) after the impact, may be described by:(I) x (t) = −vtα0 (6a)

(II) x (t) = ∑n

i = 1

bi sin (ωi t) αi (6b)

(III) x(t) = (+ evt+ c0)α0 + ∑n−1

i =1

(ci sin(ωFit) + di cos(ωFi

t))aFi( )6c


The , and all have the dimensions of length. The impact stage (II) isassumed to start at when becomes 0. It ends after time , thecollision duration, determined by first becoming 0 again. Component 1of (6b) gives:

bi ci di

t = 0 x1 τx1

x1 (τ) = ∑n

i = 1

bi sin (ωiτ) αi1 = 0. (7)

The stages are required to join on to each other smoothly in the sense thatand maintain the same value across , also ensuring that energy isconserved. The coefficients are thus determined by:

xx˙ t = 0

b

(I) x˙ (0) = −vα0 = ∑n

i = 1

biωiαi = x˙ (0) (II )

The coefficients may be projected out one by one by taking the scalarproduct of the above equation with :

bαT

j M

vαTj Mα0 = −bjωjαT

j Mαj = −mbjωj ⇒ bj = −vαTj Mα0/mωj. (8)

Note that the are proportional to the initial velocity . The andcoefficients in (6c) of the system after rebound may be found by taking thescalar product with of the equations:

bi v c d

αTFjM

x (τ) = ∑n

i = 1

bi sin (ωiτ) αi(II)

= (evτ + c0)α0 + ∑n−1

i =1

(ci sin(ωFiτ) + di cos(ωFi

τ))αFi= x(τ) (III)

x˙ (τ) = ∑n

i = 1

ωibi cos(ωiτ) αi(II)

= evα0 + ∑n−1

i =1

ωFi (ci cos(ωFiτ) − di sin(ωFi

τ))αFi= x˙ (τ) (III)

and then solving simultaneously. The details are messy, but it is clear thatthe and , like the , are proportional to , which leads to the linearity ofthe restitution law. Finding the restitution coefficient is easy, however; thescalar product with gives:

ci di bi v

αT0M

e =1

mv ∑n

i = 1

ωibi cos(ωiτ) αT0Mαi = − ∑

n

i = 1

cos(ωiτ) (αT0Mαi / m)2

. (9)

We therefore have a compact exact expression for the coefficient ofrestitution in systems of this type. Evaluating it is not as straightforward asmight appear, as the eigenvalues and vectors of the -dimensional equation(4) have to be found, and the awkward trigonometrical equation (7) has tobe solved for the collision duration .

n

τ


Results for massesn = 2Returning to the simplest case where there are 2 masses, the

characteristic equation (5) is only quadratic in and yields for the modesof the free motion:

ω2

ω0 = 0, α0 = ( ) , ω2F = 4ki / m, αF = ( )1

1−11

and for the constrained motion, defining the surface to interior stiffness ratio:kR = kS/ kI

ωi =14

ω2F (2 + kR + (−1)i 4 + k2

R) , αi = ( ) .1 − 2 (ωi / ωF)2

1

Note that the eigenvectors have not been normalised, for simplicity. Since,for this case, (where is the identity) each needs to bemultiplied by and divided by its magnitude before using it to evaluatefrom (9). Note also that will depend solely on , although the collisionduration will also be proportional to .

M = (m/ 2) I I2 e

e kR

1 / ωF

Equation (7) for the collision duration does not appear to be analyticallysolvable even for . As a starting point, it is useful to consider the casewhere is small, so that the interior of the ‘ball’ is stiff in comparison withthe surface. One would expect the situation to be very similar to a singlemass on a spring, and indeed the lower frequency mode has

. Assuming that little energy goes into internalvibration (for which ) in this case, one would expectand . This also suggests that might be a reasonable firstapproximation to start Newton-Raphson iteration of (7) to find , and so itproves up to around .

n = 2kR

m kS

ω21 ≈ ω2

FkR / 4 = kS/ mω2

2 ≈ ω2F τ ≈ π / ω1

e ≈ 1 τ0 = π / ω1

τkR = 3.8

All the preceding algebra was put into a spreadsheet (Microsoft Excel 4)and evaluated for various values of . Figure 1 shows the result of a typicalcollision of a ‘ball’ with a rigid wall with , plotting the coordinatesof each mass against time (in time units such that ). The internaloscillation of the ‘ball’ after the rebound is obvious (though it is shown at alarger scale in relation to the length of the spring than is likely in practice).This internal oscillation represents the energy diverted from the reducedcentre of mass velocity, which in Figure 1 is 0.882 of the approach velocity.So for this collision. Figure 2 graphs against for a suitablerange of . The least value of , around 0.877, occurs for . Theseresults agree to better than 0.01% with those obtained by numericalintegration of the differential equations in [3].

kR

kR = 21 = 2π / ω1

e = 0.882 e kR

kR e kR ≈ 2.3

For the collision is complicated by the inner mass losingcontact with the wall at before striking it a second time. Asbecomes very large, in effect the inner mass rebounds from the wall throughthe spring with , reversing its velocity. Then the masses reboundoff each other through the much weaker spring, again with

kR > 3.8t ≈ 0.5τ kR

kS e = 1kI


−1 0 10.86

0.88

0.90

0.92

0.94

0.96

0.98

1.00

1.02e

log10 (kR)FIGURE 2: against for massese log10 (kR) n = 2

and reversing each of their velocities relative to the centre of gravityof the colliding pair. Finally, the inner mass rebounds from the wall asecond time through the spring, so that the masses both end up travellingaway from the wall with their initial velocities reversed, and overall .The second impact complicates the algebra, but the numerical results from[3] show that nothing surprising happens, rising steadily to 1 asincreases.

e = 1

kS

e = 1

e kR

Energy conservation Energy conservation in our model is demonstrated by pre-multiplying

(3b) by , which leads to:x˙ T

d

dt (12

x˙ TMx +12

xTK Cx) = 0 ⇒ E =12

x˙ TMx +12

xTK Cx = const. (10)

The first term is the KE, the second the elastic PE, and the symmetry ofhas been used. The constancy of through the changing stages of themotion is guaranteed by the continuity of and , and the fact that thechange of stage occurs when the spring is slack.

K C

Ex x˙

kS

The reader may verify that substitution of the expressions for from(6a, b, c) into (10) leads to

x (t)

E =12

mv2 = ∑n

i = 1

12

mω2i b

2i =

12

e2mv2 + ∑n − 1

i = 1

12

mω2Fi

(c2i + d2

i )

This demonstrates clearly how the energy has been divided between theinternal vibrational modes, leaving a reduced kinetic energy of translationalmotion. Total energy is conserved, however.


Concluding remarksWe have therefore arrived at a simple model, amenable to exact

analysis, which exhibits the phenomenon of elastic restitution, and in whichall the energy is accounted for. It is not meant to be particularly realistic,and indeed values of much less than our minimum are common inpractice. It is really meant to clarify thinking about the subject.

e

A cursory investigation of higher values of (by numerical integration)yielded values usually nearer to 1 than for two masses. A Fourier analysisof the modes of a one-dimensional uniform elastic bar rebounding from arigid wall seems to give identically equal to 1, though this appears to be aspecial case, which would probably not apply to a non-uniform bar. Severalother extensions suggest themselves, such as introducing damping into theequations of motion, or considering collisions between two ‘balls’ of thetype modelled. Some of these were explored or discussed in [3], but theresults obtained here are about as far as it is reasonable to go with our exactnormal mode analysis. For further investigation, numerical integration of thedifferential equations of motion as adopted in [3], though less elegant, seemsto be the most effective way to proceed.

ne

e

AcknowledgementsThanks are due to an anonymous referee, who suggested some

clarifications of the notation. He also drew my attention to the papers byRoper and Hartley [4] and Bridge [5]. Roper and Hartley note theunphysical implications of the normal treatment of impulsive jerks instrings, and suggest some alternatives. Although they do not present it, ananalysis along the lines of this paper would be possible. Bridge (norelation!) presents some very nice detailed experimental work on bouncingballs, with some relevant comments concerning theory.

References1. W. J. O'Connor, The famous ‘lost’ energy when two capacitors are

joined: a new law?, Phys. Educ. 32 (1997) pp. 88-91.2. R. T. Bridges, Joining capacitors, Phys. Educ. 32 (1997) p. 217.3. R. T. Bridges, Energy conservation and restitution in inelastic

collisions: a simple model, Phys. Educ. 33 (1998) pp. 311-315.4. Tom Roper and Ron Hartley, “… Assume the string is inextensible and

elastic …”, Math. Gaz. 75 (1991) pp. 15-23.

5. N. J. Bridge, The way balls bounce,Phys. Educ.33 (1998) pp. 174-181.

RICHARD BRIDGESKing Edward's School, Birmingham B15 2UA


The Hale-Bopp comet explored with A levelmathematics

H. R. CORBISHLEY

On 15 March 1997, when boarding an aeroplane bound for Spain, I washanded a copy of theIndependent newspaper. One page of the paper wasdevoted to the Hale-Bopp comet; from two pieces of information about thecomet's orbit I set out to find out as much as I could about the orbit using Alevel mathematics, and in some cases further mathematics. I was interestedin finding the fastest and slowest speeds of the comet, and the size of itsorbit in relation to the planetary orbits. I also found speeds and times atother points on the orbit, where distances from the sun could be calculated.These findings, along with an approximate calendar for the next orbit of thecomet, are presented in a diagram at the end of this article. During thecourse of my working, I made two discoveries of a more general nature.The first of these is that the radius of curvature of a conic at the pointnearest to a focus is equal to the semi-latus rectum. The second was one ofKepler's laws, that for an elliptical orbit , where is the time of acomplete orbit, and is the semi-major axis.

T2 ∝ a3 Ta

From an article by Charles Arthur, the science editor of theIndependent, I learnt that

(i) nearest approach to the sun is 85 million miles,(ii) the time of a complete orbit is 4000 years.

This information led to an elliptical orbit with eccentricity. I was later informed that the latest estimate of , published in

[1] is . This corresponds to an orbital time of approximately 2550years, to be found later. The nearness of these two values of , and the largedifference in orbital times, shows how sensitive the following calculationsare to small changes in data. The calculation of depends upon observingthe position of the comet at a number of related times.

e ≈ 0.99636 e0.99512

e

e

Newton's law of gravity applied to the comet tells us that, when it is adistance from the sun, it accelerates towards the sun with an acceleration

, where , being the gravitational constant, and the mass ofthe sun. A book of physical constants gave ,

. Hence, to 3 s.f. Since thisleads to a conic orbit with the sun at the nearest focus, the initial calculationswill be done with a general conic in polar form. Later this will become anellipse in Cartesian form.

rk/ r2 k = GM G M

G = 6.673× 10−11Nm2kg−2

M = 1.99× 1030kg k = 1.33× 1020Nm2kg−1

THE HALE-BOPP COMET EXPLORED WITH A LEVEL MATHEMATICS 59

L

PV

v

lr

q AS

= l

L'

q

FIGURE 1

Taking the sun, , as origin and , where is at the shortest distance ofthe comet from the sun, as the axis , let , , the semi-latus rectum of the conic. Let be the fastest speed of the comet (at ), and the speed at a general point . Then the equation of the orbit is

S SA Aθ = 0 SA = q SL = l

V Av P(r , θ)

r =l

1 + e cosθ( )i

where is the eccentricity of the conic.eThe equations for radial and transverse accelerations are

r¨ − rθ2 = −k

r2(ii)

r θ + 2r˙ θ = 0. ( )iii

(iii) × r ⇒ r2θ + 2rr˙ θ =d

dt(r2θ) = 0

⇒ r2θ = h, whereh is constant.

This is one of Kepler's laws of planetary motion.Now at , A r2θ = r (r θ) = qV

⇒ r2θ = qV ( )iv

Eliminating between (ii), (iv) leads toθ

r¨ =q2V2

r3−

k

r2.

Multiplying by and integrating w.r.t. leads to2r˙ t

r˙ 2 = −q2V2

r2+

2k

r+ c

At , , A r˙ = 0 r = q ⇒ c = V2 −2k

q


⇒ r˙ 2 = V2 (1 −q2

r2) − 2k (1q

−1r ) . ( )v

The speed at is given byP

v2 = r˙ 2 + (r θ)2= r˙ 2 +

(r2θ)2

r2.

Substituting for , from (iv), (v) and simplifying leads to r2θ r˙ 2

v2 = V2 − 2k (1q

−1r ) ( )vi

giving at in terms of , and . The next problem is finding .v P(r , θ) V, k, q r VMy first idea was to use the equation

V2

ρ=

k

q2( )vii

where is the radius of curvature of the conic at . This led to someinteresting mathematics involving , , but I later discovered this wasunnecessary. However equation (vii) will prove of use later on.

ρ As ψ

Returning to the general conic (i) with initial conditions

v = V, r˙ = 0, r = q =l

1 + e θ = 0. (when viii)

Differentiating (i) w.r.t t

⇒ r˙ =el sinθ θ

(1 + e cosθ)2=

e sinθl

× r2θ by (i)

⇒ r˙ =e sinθ

l× qV by (iv).

Then, using (viii) to eliminate ,q

l

r˙ =eV sinθ1 + e

.

Differentiating again . Then using (i), (iv), (viii) to

eliminate , , respectively we arrive at

⇒ r¨ =eV cosθ θ

1 + ecosθ θ q

r¨ =V2l (l − r)r3 (1 + e)2

( )ix

Now we are able to eliminate , , from (ii) using (ix), (iv), (viii)respectively to give

r¨ θ q

V2l (l − r)r3 (1 + e)2

−V2l2

r3 (1 + e)2= − k

r2.

From this we obtain


V = (1 + e)k

l. ( )x

Or, in terms of , by (viii)e, k, q

V =k (1 + e)

q. ( )xi

Given million miles .q = 85 = 8.5 × 107 × 1600 m to 3 s.f.,⇒ q = 1.36 × 1011 m

to 5 d.p. (from [1])e = 0.99512 to 3 s.f.,⇒ l = q (1 + e) = 2.71 × 1011 m

V =k (1 + e)

q= 44200 ms−1 to 3 s.f.

= 44200 ×94

mph = 99400 mph to 3 s.f.

i.e.V ≈ 100 000 mph.

Now, from (vii), . Then by (xi), (viii)ρ =V2q2

k

ρ =k (1 + e)

q×

q2

k= q (1 + e) = l

i.e. for any conic, the radius of curvature at the nearest point to a focus is ,the semi-latus rectum.

l

For the area swept out by the comet in a given time, by (iv),

A = ∫ 12r

2dθ = ∫ 12r

2θ dt = ∫ 12qV dt

, where is the area swept out in time . Then replacing, using (xi),

⇒ A = 12qVt A t

V

A = 12t kq(1 + e). ( )xii

We must now consider the full elliptical orbit.

y B

b

ae

lL

S q

V

A x

L'

O

b

B'

aA'

vmin

a

FIGURE 2 (Not to scale!)


Let the ellipse have equation , where .

Then

x2

a2+

y2

b2= 1 b2 = a2 (1 − e2)

SA = q = a − ae = a (1 − e) ⇒ a = q1 − e

to 3 s.f.⇒ a = 2.79 × 1013 m = 1.74 × 1010 milesThe area enclosed by the ellipse is . Let be the

time taken to complete the orbit, then by (xii)πab = πa2 1 − e2 T

πa2 1 − e2 =T

2kq(1 + e) =

T

2ka(1 − e2)

which simplifies to

T =2πa3/2

k1/2( )xiii

i.e. , where is the semi-major axis.T ∝ a3/2 aFrom (xiii), years to 3 s.f.T = 8.03× 1010s = 2550Let the slowest speed of the comet be , when the comet is at .

Then from (vii) with , replaced by , and bywe have

vmin A′ρ = l V vmin q SA′ = a (1 + e)

v2min

l=

k

a2 (1 + e)2(xiv)

⇒ vmin =kl

a (1 + e)= 108 ms−1 = 243 mph to 3 s.f.

Alternatively, dividing (xiv) by (vii) leads to

vmin =V (1 − e)

1 + e,

from which the same values as above are obtained.Earlier, I said that the speed at any point might be found in

terms of , , using (vi) and is now known. Unfortunately (vi) is notvery accurate when are only known approximately and is smallcompared with , since the equation involves finding a smaller quantity bysubtracting two much larger quantities. To give the worst possible exampleof this, using (vi) to find with values of as given above leads to

to 3 s.f. which is wildly out. Fortunately, this difficultymay be overcome by replacing in (vi) by using (xi). Then

v P(r , θ)V k q V

V, k, q vV

vmin V, k, qvmin = 1590 ms−1

V

v2 =k (1 + e)

q− 2k (1

q−

1r )

which simplifies to

v2 = k (2r

−1 − e

q ) . ( )xv

This equation does not have the failings of (vi) and may be used to find ata number of points on the orbit where is known, e.g. at , (and also ).

vr L, B A′


At (and ), by (viii)L L′ v2 = k (2l

−1 − e

q ) =k

l(1 + e2)

⇒ v = 31 300 ms−1 = 70 300 mph to 3 s.f.

i.e. v ≈ 70 000 mph L.at

Thus, at , , which would be an exact solution for a parabolicorbit.

L v ≈ V / 2

At (and ), B B′ v2 = k (2a

−1 − e

q ) =k

a

⇒ v = 2180 ms−1 = 4910 mph to 3 s.f.

i.e. v ≈ 4900 mph B.at

Again, at , A′ v2min = k ( 2

a (1 + e)−

1 − e

q ) =k (1 − e)a (1 + e)

⇒ vmin = 108 ms−1 = 243 mph to 3 s.f.

This agreement with the previous values found for confirms theaccuracy of (xv) at all points on the orbit.

vmin

I will conclude with a little astronomy. Astronomers call and , thepoints of the orbit nearest to and furthest from the sun, the perihelion and theaphelion, respectively.

A A′

Working within the solar system, lengths are measured in astronomicalunits (A.U.), and times in years, where 1 A.U. is the mean distance of theearth from the sun.

1 A.U. = 1.50 × 1011 m to 3 s.f.

Suppose now that is measured in A.U. and in years. Then, and .

Substituting these values in (xiii) leads to to 4 s.f. And to 3s.f. . Astronomers use this equation in the form

a Ta A.U. = a × 1.5 × 1011 m T yrs = T × 365 × 24 × 3600 s

T = 1.004a3/2

T = a3/2

T2 = a3 ( )xvi

with measured in years, and in A.U.T aChanging our measurements to A.U.

q =1.36 × 1011

1.5 × 1011= 0.907 A.U. to 3 s.f.

l = 1.81 A.U. to 3 s.f.Similarly,

a = 186 A.U. to 3 s.f.

Then, A.U. to 3 s.f. andA.U. to 3 s.f..

b = a 1 − e2 = 18.4 A′S = a (1 + e) = 371

By (xvi), years to 3 s.f. We will use this value of from here on.

T = 1863/2 = 2540T


Since the line to the comet from the sun sweeps out areas proportionalto the time taken, to find the time taken to reach any point from , we

multiply the area swept out by .

AT

πab= 0.237 yr(A.U.)−2

To find the time taken to reach from we need to find , the areaenclosed by and the arc . A rather difficult integration which youmight like to try gives

L A ASA, SL AL

A =l2 (arccose − e 1 − e2)

2 (1 − e2)3/2 = 1.095 (A.U.)2 to 4 s.f.

Since the orbit of the comet is so eccentric with , we mightapproximate the segment to the right of by the segment of a parabola,with the same width , and height . Then to4 s.f. Then the time taken from to is

e ≈ 1LL′

SA LL′ A = 23ql = 1.094 (A.U.)2

A L 1.094 × 0.237 yr = 0.259 yr

= 0.259 × 365 days= 95 days ≈ 3 months.In moving from to , the area swept out isA B

quadrantALBO − OSB = 14πab − 1

2aeb

= 983 (A.U.)2 to 3 s.f.

⇒ time taken from to And the timetaken from to will be yr.

A B = 983 × 0.237 = 233 yr to 3 s.f.A A′ 1

2 × 2540 yr = 1270The above-mentioned article also gave the date when the comet was at

as 1 April 1997. Taking this as our starting point, it is now possible to chartthe progress of the Hale-Bopp comet on its next circuit around the sun. Theerror limits in the diagram were obtained using some error analysis which isnot shown in this article.

A

B

L

A

L'

B'

A'

243 mph

3267 ± 19 AD

2231 ± 3 AD4900 mph

100 000

70 000

mph

mph

b = 18.4 ± 0.1 A.U.

a =186 ± 1A.U.

l = 1.81± 0.41 A.U.

q ≈ 0.91 A.U.

SA′ = 371± 2.5 A.U.4537± 38ADsun

4−6/7/97

1/4/97 returning

(26−28/12/96)

4303± 35AD

FIGURE 3

If we compare the Hale-Bopp comet's orbit with that of Pluto, theremotest planet in our solar system, for Pluto, , ,

⇒ I found this information in [2], a veryreadable book written for sixth formers.

T = 249 yr a = 39.5 A.U.e = 0.25 A′S = 49.5 ≈ 50 A.U.


An article by Professor Chandra Wickramasinghe on the same page ofthe Independent of 15 March 1997 states that the comet is a ‘mountain ofice’ approximately 40 km (25 miles) across. The bright tail we see when thecomet is nearest to the sun is due to a cloud of dust particles being blownaway by solar radiation. It is also posited that bacteria in some form may beliving on the comet, and that life on earth may have begun when a cometcrashed into the earth long ago.

I find it interesting that the comet we saw at its perihelion rushing alongat 100 000 mph and brighter than the stars, will, in the yearAD 3267(approximately), be a very cold ‘mountain of ice’ at its aphelion in outerdarkness, moving at 243 mph, a speed most of us will have exceededrelative to the earth when flying away on our holidays abroad. Also, thecomet which has a bright shining six months on one side of the sungradually disappears on the other side to spend two and a half millennia inorbit before returning for another six months of glory. We were lucky to seeit!

References1. The British Astronomical Circular (1 May 1997).2. Brian Milner, Cosmology, Cambridge Modular Sciences series,

Cambridge University Press (1995).H. R. CORBISHLEY

5 Bradley Grove, Silsden, Keighley BD20 9LX

Looks like it could be all of us!You are 750 times more likely to die in an asteroid impact than you have a

chance of winning the National Lottery tonight.. . .What we do know is that we are going to be struck by an asteroid. There are

1,500 objects out there that could hit the earth.From The Western Mail, Saturday 10 July 1999, sent in by Michael Mudge.

Europe v America?A mix-up between metric and imperial measurements caused a spacecraft to

crash on Mars. The makers of the booster rocket used pounds per square inch butNASA read them as metric newtons.

From the Daily Telegraph, 2 October 1999, by Nick Lord.

And another oneThe largest known Mersenne prime has been discovered by

Nayan Hajratwala, The USA-based technology consultant used the Great InternetMersenne Prime Search (GIMPS) software and distributed computing technologyprovided by the USA-based company Entropia.com.

(26972593 − 1)

From Scientific Computing World, August/September 1999, and sent in byMichael R. Mudge.


Notes84.01 A portrayal of right-angled triangles which generate

rectangles with sides in integral ratio

The basic theoremA few years ago I presented an elementary but beautiful and little-

known theorem, which states that, if the sides of the right-angled trianglewith sides 3, 4 and 5 are extended, then the three classes of rectangles set onthese pairs of lines as diagonals have their own sides in the respective ratios1 : 1, 2 : 1 and 3 : 1 (see [1] pp. 138-139). The accompanying diagram (in[1]) showed these rectangles necessarily going beyond the triangle; but nowI have found a new diagram which shows these properties entirely inside it(Figure 1). It has an attractive uniformity of layout, and also draws onsimple but nice divisions of the sides: , and

. The ratios are marked inside each sub-triangle from which therectangles can be generated; each one is that of the altitude to half the base.This version of the theorem reads:

3 = 1 + 2 4 = 1 + 35 = 3 + 2

AP : PY = 1 : 1, BQ : QZ = 2 : 1 CR: RX= 3 : 1. (1)and

The rectangles mentioned above have as respective centres and sidesA andYZ , B and ZX, and C and XY.

A

ZY

CX

P

QR

B

3

3

11

2

2

4

5

3

1:1

2:1

3:1

FIGURE 1

The case 1 : 1 is, of course, obvious, asYZ sides a square; the other twocases are not difficult to prove, and can be given to pupils as an exercise. Atrigonometric proof follows from the formulae relating functions of angleto those of . For , let , then the ratio isgiven by

2γγ YCX ∠RCY = γ p = CR: RY

p = cotγ =2 cos2γsin2γ

=1 + cos2γ

sin2γ=

1 + 45

35

=93

=31

(2)

NOTES 67

as required. A purely geometrical proof can be found by, for example,dropping the perpendicularYE into CB and drawingYF perpendicular toCAto meetCB at F (Figure 2), and working out from the similarity ofwith —again, a pleasant educational exercise. and 2 : 1 can behandled similarly.

YEXRCX ZBX

Y

CX

R

E F

FIGURE 2

A corollary of the two results states that

∠YCR+ ∠XBQ = ∠ZAP(= 45°) , (3)

Tan−1 13

+ Tan−1 12

= Tan−1 11

. (4)that is,

The extended theoremI also showed that the rectangles with sides in the ratio forany

positive integer could be produced, but by only one triangle with integralsides (discarding common factors, of course). The new correspondingdiagram is given in Figure 3; to avoid boring ½s, I have set the sides as ,

and , where integers , since any resulting common factorsare harmless. As expected, there is nice symmetry in the use of the threeletters. The desired ratio will always be found in the sub-triangleXCY,since only its angleC can be sufficiently small to generate the high ratiossought; Figure 3, while accurate, is drawn for clarity of layout and is not arealistic example.

n : 1n

2a2b 2c a > b > c

n : 1

The proofs follow those above; if the sides of the triangle are taken to be

2a = (r2 + s2) , 2b = (r2 − s2) 2c = 2rs, (5)and

with and integers, then, for the triangle atC, the ratiop corresponding to(2) is given by

r s

p = (a + b) : c = ((r2 + s2) + (r2 − s2)) : 2rs = r : s. (6)


A

ZY

CX

B

1:1

2b

2a

2c

b + a− c

b + a− c

b+c−

ab + c − a

c + a − b

c+a−

b

n:1

FIGURE 3

The integral cases stated above are given by and ;, say, comes from the triangle 122 : 120 : 22 (or 61 : 60 : 11).

r = n s = 1n = 11

However, unlike the basic theorem, the sides of the sub-triangle aroundB will give a rational but non-integral ratio :q

q = (c + a) : b = (2rs + (r2 + s2)) : (r2 − s2) = (r + s) : (r − s). (7)For example, the above case produces .n = 11 q = 6 : 5

The corollary corresponding to (4) is

Tan−1 ( c

a + b) + Tan−1 ( b

a + c) = Tan−1 11

. (8)

Their history?In my note, I report a sturdy effort to record the history of these

theorems, which however had ended in total failure. Since its appearance, Ihave received many delighted responses from correspondents and atconferences; butnobody had even seen either theorem before, never mindknow their origins! I suspect that the extended version is new; but somefurther historical speculations concerning the basic theorem have come tomind.

I found it in a nice book on sacred geometry, illustrated by a separatefigure for each ratio (see [2], pp. 82-84). No historical information wasprovided, but thehistorical context of this geometry is very suggestive; forthe ratios 2 : 1 and 3 : 1 were of major importance in the secret lore ofmedieval cathedral builders, who called them respectively ‘ad quadratum’and ‘ad triangul(at)um’ (see, for example, [3]). Further, in Figure 1,

and , two numerologically important numbers.Maybe this theorem was part of the secret knowledge which they are knownto have guarded (hence causing the historical obscurity). All editions ofmasons' workbooks which I have seen present the ratios separately: the ratio2 : 1 comes from the right-angled isosceles triangle ( in Figure 1);

∠ACB ≈ 18° ∠ABC ≈ 27°

AYZ

NOTES 69

3 : 1 arises from the equilateral triangle when a pair of these are arranged toform a hexagram, for all their sides are divided into thirds.

Another historical clue may lie in Freemasonry. The ‘Mason's square’was originally the right-angled isosceles triangle and hence half a normalsquare (for an important example see [4], pl. 21); but around the late 17thcentury it was changed to an ornament composed of the two shorter arms ofthe 3 : 4 : 5 triangle (see, for example, Mozart and his brothers so dressed in[5], pl. 26). I have not found the reasons for the change, but perhaps thistheorem played a role; for, thanks to it, the new triangle expressed both 2 : 1and 3 : 1 whereas the older one captured only 2 : 1.

It is strange that such a lovely theorem, a perfect example of simplebeauty in mathematics, should be so fugitive. Maybe the story will bedisclosed one day; in the meantime, an excellent educational and culturalgobbet is available.

References1. I. Grattan-Guinness, ‘Ad quadratum’ and beyond: right-angled triangles

generate all rectangles with sides in integral ratio,Zentralblatt fürDidaktik der Mathematik 29 (1995) pp. 138-139.

2. N. Pennick,Sacred geometry. Symbolism and purpose in religiousstructures, Turnstone (1980).

3. J. L. M. Lund,Ad quadratum. A study of the geometric bases of classicand medieval religious architecture, vol. 1, Batsford (1921).

4. C. Knight and R. Lomas,The Hiram key. Pharaohs, Freemasons andthe discovery of the secret scroll of Jesus, Arrow (1997).

5. H. C. Robbins Landon,Mozart and the Masons, (2nd edition), Thamesand Hudson (1991).

I. GRATTAN-GUINNESS Middlesex University, Enfield, Middlesex EN3 4SF

e-mail: [email protected]

84.02 Circumradius of a cyclic quadrilateralIt is reasonably well known [1, pp. 64-65] that the circumradius of an

arbitrary triangle with sides , and is given by the formula ,where is the area of the triangle. Sinceby Heron's formula, we have a formula for in terms of the three sides ofthe triangle.

a b c R= abc/4K = s(s − a) (s − b) (s − c)

r

It is not known, or certainly not so well known, that the circumradius ofa cyclic quadrilateral can also be found in terms of the sides of thequadrilateral. To see this, let be any cyclic quadrilateral with

, , , , , and . Also letand be the areas of and , respectively. For , from theformula above, we have so that . In a similarmanner for .

ABCDa = AB b = BC c = CD d = DA e = AC f = BD a

c ABD CBD ABDR = adf / 4a a = adf / 4R

c = bcf / 4R CBD


A

D

C

B

a

b

c

e

f

d

FIGURE 1

If is the area of quadrilateral , then. Similarly, by using and , .

Therefore

ABCD = a + c

= f (ad + bc)/4R BAC DAC = e(ab+ cd)/4R

2 =ef

16R 2(ad + bc) (ab + cd) .

Ptolemy's theorem [1, p.128] states thatin a cyclic quadrilateral, theproduct of the diagonals is equal to the sum of the products of the two pairsof opposite sides. Hence for quadrilateral . Therefore

, or equivalently,ef = ac + bd ABCD

2 = (ac+ bd)(ad + bc)(ab+ cd)/16R 2

R =(ac + bd) (ad + bc) (ab + cd)

4.

Since by Brahmagupta's formula [1,p. 135] for the area of a cyclic quadrilateral with ,we have

= (s− a)(s− b)(s− c)(s− d)s = 1

2 (a + b + c + d)

R =14

(ab + cd) (ac + bd) (ad + bc)(s − a) (s − b) (s − c) (s − d)

Reference1. Nathan A. Court,College Geometry, Barnes and Noble, New York

(1952).LARRY HOEHN

Department of Mathematics and Computer Science,Austin Peay State University, P.O. Box 4626, Clarksville, TN 37044 USA

NOTES 71

84.03 A neglected Pythagorean-like formulaGiven an isosceles triangle as shown in Figure 1, we have the

Pythagorean-like formula . If the segment is an altitude,then . This formula has surely been discovered many times,but yet it doesn’t seem to appear in the mathematical literature.

c2 = a2 + bd ac2 = a2 + b2

A

B

CD

a

b

c c

d

FIGURE 1

My discovery of this formula came from looking at one of the manyproofs of Pythagoras' theorem. That is, given , with a right angle atC,construct a circle with centre and radius as shown in Figure 2.By using the intersecting chords theorem, we have , so

.

ABCB BC = a

(c + a)(c − a) = b2

c2 = a2 + b2

A

B

C

a

a

b

c

FIGURE 2

Suppose on the other hand that is not a right angle but rather anobtuse angle for . By again constructing a circle with centre andradius , the circle will intersect extended side at some point asshown in Figure 3. By the intersecting chords theorem, we have

, so .

∠CABC B

BC = a AC D

(c + a) (c − a) = ((d − b) + b) b c2 = a2 + bd


AE

a

a

b

B

CD

a

b

c c

d − b

FIGURE 3

Since there are so many proofs of Pythagoras' theorem (see [1] and [2]),it is natural to suspect that this Pythagorean-like theorem can also be provedin many ways.

A

B

C

a

b

c c

d

q

FIGURE 4

For example, by the law of cosines in Figure 4, we have and . Hencec2 = a2 + b2 − 2abcosθ c2 = a2 + d2 − 2adcos(180− θ)

cosθ =a2 + b2 − c2

2aband cos(180 − θ) =

a2 + d2 − c2

2ad.

Therefore, . By simplifyingthis result we obtain .

(a2 + b2 − c2) / 2ab = − (a2 + d2 − c2) / 2adc2 = a2 + bd

This Pythagorean-like theorem can also be derived by two applicationsof Pythagoras' theorem. If is the altitude to the base of the isoscelestriangle in Figure 5, then and . Hence

hc2 = h2 + y2 a2 = h2 + x2

c2 − a2 = y2 − x2 = (y − x) (y + x) = bd, or c2 = a2 + bd.

NOTES 73

A

B

CD

a

b

c c

d

y

h

x

FIGURE 5

Finally, we note that this formula is a special case of Stewart’s theorem[3, p. 58]. ‘If and are any three collinear points, and is any otherpoint, then both in magnitude and sign

K, L M P

PK2 × LM + PL2 × MK + PM2 × KL + KL × LM × MK = 0.’Making the substitutions , we have

, , , , and , so. This simplifies to

, so .

P = B, K = D, M = A, L = CPK = PM = c PL = a KL = d LM = b MK = −(d + b)c2b + a2(−(b + d)) + c2d + db(−(b + d)) = 0c2(b + d) − a2(b + d) − bd(b + d) = 0 c2 = a2 + bd

Given the simplicity of this formula, and its close kinship to Pythagoras'theorem, it is quite curious that it is not prevalent in the mathematicalliterature.

References1. Elisha Scott Loomis,The Pythagorean proposition, National Council of

Teachers of Mathematics, Washington, D.C. (1968).2. Larry Hoehn, The Pythagorean theorem: an infinite number of proofs?

Mathematics Teacher90 (September 1997) pp. 438-441. [Correction toFigure 2 91 (January 1998), p. 73.]

3. Howard Eves,A survey of geometry (revised edition), Allyn and Bacon,Inc., Boston (1972).

LARRY HOEHNDepartment of Mathematics and Computer Science,

Austin Peay State University, PO Box 4626, Clarksville, TN 37044 USA


84.04 An unexpected reduced cubic equationy

Sb

PQ

C

−1 1

O x

R −b

f3

f23

f

FIGURE 1

Figure 1 shows the two circles

(x ± 1)2 + y2 = a2, (a > 1)intersecting at two points and , where .

and (where ) are two points on the circles, so thattriangle is isosceles. Furthermore, if where is the centre

, then

S (0, b) R(0, −b) b = a2 − 1P(x, y) Q (−x, y) x > 0

RPQ CP⊥ QR C(−1, 0)

x2 + x − y2 − by = 0, (b = a2 − 1) .Since lies on the circle with centre , the elimination of and removal ofa factor leads to the cubic equation

P C yx

4x3 + 12x2 + (9 − 3b2) x − 4b2 = 0,or .4(d / a)3 − 3(d / a) − 1 /a = 0 (d = x + 1)It may be shown graphically that this reduced cubic equation hasthree real roots, exactly one of these being positive.

f (d / a)

The roots may be found by means of the trigonometrical formula

4 cos3A − 3 cosA = cos 3A, (3A = φ + 2nπ, n = 0,1,2) ,

d = a cos(φ / 3) (a = secφ)so that

is the solution required.Note the interesting geometrical significance of the angle . The

isosceles triangle yields , and the areaφ

CPR PR= 2a sin (2φ / 3)

PQR = 12PR2 sin(2φ / 3) = 2a2 sin3 (2φ / 3) .

NOTES 75

Finally, the reader may wish to speculate as to whether triangle has thegreatest area of all possible triangles lying within the intersection of the twocircles.

PQR

J. A. SCOTT1 Shiptons Lane, Great Somerford, Chippenham, Wiltshire SN15 5EJ

84.05 Touching hyperspheresIf each of four circles lying in a plane touches the other three externally,

their radii are related by the equationr i (i = 1,2,3,4)

( 1r1

+1r2

+1r3

+1r4

)2

= 2 ( 1r2

1+

1r2

2+

1r2

3+

1r2

4) .

This can be proved by applying well-known trigonometrical formulas to thefigure (as for example in [1], pp. 13-15).

This result generalises to hyperspheres in an -dimensionalEuclidean space , the factor 2 being then replaced by . The caseis stated in [2] and the general case is established in [3].

n + 2 nEn n n = 3

In this note, we prove the general result using an elementary argument.Let us take the centre of one of the hyperspheres as origin and let

be unit vectors directed from towards thecentres of the other hyperspheres. Then, if denotes the radius of thehypersphere with centre and the radius of the hypersphere with centre

, we shall have

Oui (i = 1, 2, … , n + 1) O

Ci r0

O ri

Ci

CiCj

→= (r0 + r j) uj − (r0 + r i) ui, i ≠ j.

Since , by squaring the last equation we find that| CiCj→

| = r i + r j

(r i + r j)2 = (r0 + r j)2 + (r0 + r i)2 − 2 (r0 + r i) (r0 + r j) ui.uj,the dot denoting a scalar product. By expanding the three squares and thenputting

si =r i

r0 + r i, (1)

we can show thatui.uj = 1 − 2sisj, i ≠ j. (2)

Since is a unit vector, .ui ui.ui = 1In , the vectors must be linearly dependent, i.e. multipliers

can be found, not all zero, such thatEn n + 1 ui ki

k1u1 + k2u2 + … + kn + 1un + 1 = 0.Taking the scalar product of this equation with and making use ofequation (2), we obtain the linear equations

uj

n + 1

(1 − 2s1sj)k1 + (1 − 2s2sj)k2 + … + kj + … + (1 − 2sn+1sj)kn+1 = 0, (3)for .j = 1, 2, … , n + 1


The necessary and sufficient condition for the system of equations (3) tohave a non-null solution is that the determinant of the coefficients mustvanish; thus

| | = 0. (4)

1 1− 2s1s2 1 − 2s1s3 … 1 − 2s1sn+1

1 − 2s2s1 1 1− 2s2s3 … 1 − 2s2sn+1

… … … … …1 − 2sn+1s1 1 − 2sn+1s2 1 − 2sn+1s3 … 1

This is a relationship between the radii . It remains toreduce it to the required form.

r0, r1, … , rn + 1

The determinant in (4) can be expressed as a sum of determinants,all but of which vanish by virtue of repeated columns of 1s. Of thesesummands, there is the determinant

2n + 1

n + 2

A = | | .1 −2s1s2 −2s1s3 … −2s1sn + 1

1 0 −2s2s3 … −2s2sn + 1

1 −2s3s2 0 … −2s3sn + 1

… … … … …1 −2sn + 1s2 −2sn + 1s3 … 0

together with similar determinants in which the 1s occupy the othercolumns and finally there is the determinant

n

B = | | .0 −2s1s2 −2s1s3 … −2s1sn + 1

−2s2s1 0 −2s2s3 … −2s2sn + 1

−2s3s1 −2s3s2 0 … −2s3sn + 1

… … … … …−2sn + 1s1 −2sn + 1s2 −2sn + 1s3 … 0

By removing factors from rows and columns, we find

A = (−2)n s1s22s

23… s2

n + 1D,where

D = | | ,1s1

1 1 … 1

1s2

0 1 … 1

1s3

1 0 … 1

… … … … …1

sn + 11 1 … 0

and

B = (−2)n + 1 s21s

22… s2

n + 1E,

NOTES 77

where

E = | | .0 1 1 … 11 0 1 … 11 1 0 … 1… … … … …1 1 1 … 0

We now define th-order determinantsn

Fn = | | , Gn = | | .0 1 1 … 11 0 1 … 11 1 0 … 1… … … … …1 1 1 … 0

1 1 1 … 11 0 1 … 11 1 0 … 1… … … … …1 1 1 … 0

Subtracting the first column of from each of the other columns andthen expanding it from its first row, we obtain . Expandingfrom its first column into cofactors with coefficients , thencycling the leftmost columns of the th cofactor, we find that

.

Gn

Gn = (−1)n−1 Fn

(n − 1) ±1k k

Fn = − (n − 1) Gn − 1= (n − 1) (−1)n − 1

Next, expanding as we would , we haveD Fn + 1

D =1s1

Fn − ( 1s2

+1s3

+ … +1

sn + 1) Gn

=(−1)n − 1

s1(n − 1) + (−1)n ( 1

s2+

1s3

+ … +1

sn + 1) .

Further, .E = Fn + 1 = (−1)n nWe have expanded the relationship (4) to give

2ns1s22s

23… s2

n + 1 ( 1s2

+1s3

+ … +1

sn + 1−

n − 1s1

) + 2ns2

1s2s23… s2

n + 1 ( 1s1

+1s3

+ … +1

sn + 1−

n − 1s2

) + . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

+ 2ns21s

22s

23… sn + 1 ( 1

s1+

1s2

+ … +1sn

−n − 1sn + 1

) − 2n + 1ns2

1s22s

23… s2

n + 1 = 0.After division by , this can be written in the form2ns2

1s22… s2

n + 1

( 1s1

+1s2

+ … +1

sn + 1)2

= n ( 1s2

1+

1s2

2+ … +

1s2

n + 1) + 2n.


From (1), we have and substitution leads finally to the result1si

=r0

r i+ 1

( 1r0

+1r1

+ … +1

rn + 1)2

= n ( 1r2

0+

1r2

1+ … +

1r2

n + 1) .

Given , and , this relationship provides a quadraticequation for . It can be rearranged into the form

r1, r2, … rn + 1

1 / r0

1r0

−1

n − 1 ( 1r1

+1r2

+ … +1

rn + 1)

2

=n

(n − 1)2

( 1

r1+

1r2

+ … +1

rn+1)2

− (n − 1)( 1r2

1+

1r2

2

+ … +1

r2n+1

),

showing that the roots are real if, and only, if

( 1r1

+1r2

+ … +1

rn+1)2

≥ (n − 1)( 1r2

1+

1r2

2

+ … +1

r2n+1

) (5)

For (the case mentioned at the beginning of this note), strictinequality always occurs in (5), and the two values of are given by

n = 2r0

1r0

=1r1

+1r2

+1r3

± 2( 1

r1+

1r2

+1r3

)2

− ( 1r2

1+

1r2

2+

1r2

3)

=1r1

+1r2

+1r3

± 21

r2r3+

1r3 r1

+1

r1r2

=1r1

+1r2

+1r3

±2r

,

where is the radius of the inscribed circle of the triangle .r C1C2C3

The extreme case of equality in (5) is precisely the condition thathyperspheres in an should touch one another externally. If (5) is notsatisfied, no hyperspheres of radii in can beconstructed to touch one another externally. For example, in , fourspheres having given radii cannot always be constructed totouch one another externally. In the extreme case when

n + 1En − 1

n + 1 r1, r2, … , rn + 1 En

E3

r1, r2, r3, r4

( 1r1

+1r2

+1r3

+1r4

)2

= 2 ( 1r2

1+

1r2

2

+1r2

3+

1r2

4) ,

their centres must lie in a plane (an ), which intersects them in four greatcircles whose radii satisfy the condition (above) for them to touch oneanother externally. The figure is then symmetrical about this plane and twospheres having the same radius can be constructed, one on each side of theplane, to touch the other four; this is the case where the above quadraticequation for has equal roots.

E2

1 / r0

If the condition (5) is satisfied, it is easy to see that, because the sum of

NOTES 79

the roots of the quadratic equation determining is positive, it either hastwo positive roots, or one positive and one negative root. In the former case,two hyperspheres can be constructed to touch the other externally. Inthe latter case, the magnitude of the negative root equals the radius of ahypersphere which can be constructed to enclose the otherhyperspheres and to touch each of them internally; this interpretationfollows immediately by noting that our earlier argument is easily amendedto correspond to such a figure, by replacing by .

1 / r0

n + 1

n + 1

r0 −r0

The reader may like to confirm that the general relationship proved for holds also in the trivial case , where in (a line) a

hypersphere is a pair of points, its radius is half the distance between them,and ‘touching’ means ‘having a common point’. The relationship reduces to

and two hyperspheres touch externally, whilst the thirdencloses them both.

En (n ≥ 2) n = 1 E1

r0 + r1 + r2 = 0

The author is indebted to Graham Hoare and the referee for providinginformation relating to the history of this problem.

References1. H. S. M. Coxeter, Introduction to geometry, Wiley (1961).2. F. Soddy, The kiss precise,Nature137 (1936) p. 1021 [reproduced in

Math. Gaz. 79 (July 1995) p. 274 ]; Nature 139 (1937) p. 77.3. H. S. M. Coxeter, Aequationes Math. 1 (1968) pp. 104-121.

D. F. LAWDENNewhall, Temple Grafton, Alcester B49 6NU

84.06 Comments on note 82.53—a generalised test fordivisibility

Murray Humphreys and Nicholas Macharia present a test for divisibilityby 19 [1]. There are other ways of formulating that rule. I would like toshare one such formulation with the readers of theGazette. [1] starts byrestricting attention to three-digit numbers. In my formulation I do not needsuch a restriction. In my justification the 19 is revealed in a quiteconspicuous way.

Furthermore, I will show how to generalise the rule.GIVEN: A starting number, which is a non-negative integer.STEP 1. Delete the units digit of the starting number.STEP 2: To what is left add 2 times the deleted number. The result is

the new number—or, if you wish−the next number.CLAIM: The starting number is divisible by 19 if and only if the new

number is divisible by 19.The examples of this rule will, of course, be exactly the same as in [1].

Nothing but another formulation of the rule has been done:Example 1. 3672 → 367 + 4 = 371 → 37 + 2 = 39, not divisible by 19.


Example 2. 342 → 34 + 4 = 38 = 2 × 19.

Justification of the claim. Let us introduce the following notation: = the starting number,s = the number of tens in ,t s = the units digit of ,u s = the new number.n

Then according to STEP 1 and STEP 2, respectively, we have

⇔

s = 10t + un = t + 2u

2s = 20t + 2un = t + 2u

and so subtraction gives

n − 2s = −19t. (1)The numbers 2 and 19 are relatively prime, and so (1) shows that if one ofand has the factor 19 then the other must too.

nsNote that this proof does the job no matter what the number of digits of

is.nNow one might ask what would happen if we changed the multiplier 2

to some other positive integer, i.e., if we altered STEP 2 toSTEP 2(m): To what is left add times the deleted number. The result is

the next number.m

Then we would get . Repeating the reasoning we usedwhen we had we get the formula corresponding to (1) to be

n = t + mum = 2

n − ms = − (10m − 1) t. (2)We note that the numbers and in (2) have no common

factors greater than one. If they did, then we could write, for some integers, , and , where ,

m 10m − 1

c m1 p c ≥ 2

m = cm1 and 10m − 1 = cp,and so

10cm1 − 1 = cp, giving c (10m1 − p) = 1,which is a contradiction, for the number in brackets is an integer and is atleast 2, so their product cannot be equal to 1.

c

Returning to (2) we conclude that either both and are divisible by or else neither of them is.

n s10m − 1

Let us look at the particularly rewarding STEP 2(5).

Example 3. 686 → 98 → 49 49.→→

The conclusion is that 686 is divisible by 49.

NOTES 81

Example 4.119 → 56 → 35 → 28 → 42

↑ ↓7 ← 21 ← 14

We conclude that 119 is not divisible by 49. The loop that appears containsonly numbers divisible by 7, and so the idea emerges that we can formulatea divisibility rule for any factor of 49, or more generally, any factor of

.10m − 1Using (2) again it is easily seen that if is divisible by a proper factor of

the number then so is . I leave the details of this reasoning to thereader.

n10m − 1 s

I use this rule with various multipliers, starting with 5, in primary schooland teacher education, when treating the review of multiplication. I do notat the outset say anything about divisibility. Usually I let the pupils orstudent teachers use the starting number 18, so that we can agree on what todo when the number is a one-digit number. The pupils or students are askedto carry on till they discover something.

One gets a long loop very much worth exploring. See [2].Later on, when we have discussed multiplier 5, everybody is

encouraged to try other values of the multiplier, and there is much exploringand very much multiplying going on. Many get excited and happy whenthey realise how is involved (without, of course, using the symbol

).10m − 1

mSo here is a beautiful result in pure mathematics, and it can be used for

an interesting and rewarding review of multiplication. Isn't mathematicswonderful?References1. M. Humphreys and N. Macharia, Tests for divisibility by 19,Math.

Gaz. 82 (Nov. 1998) pp. 475-477.2. Andrejs Dunkels, Much more than multiplying by 5,Mathematics in

School 20 (3) (May 1991) pp. 9-11.ANDREJS DUNKELS

Dept. of Mathematics, Luleå University of Technology, SE-971 87LULEÅ, Sweden

Formerly:Head of Dept. of Mathematics, Kenya Science Teachers College,

Nairobi, Kenya

84.07 A matrix method for a system of linear Diophantineequations

In [1], Koshy applies a matrix method to the solution of asingle linearDiophantine equation. In [2] Cook solvesa special case of the equationsolved in [1]. (It is a special case because the right-hand side is given to bethe highest common factor of the coefficients on the left.) We now wish to


show how to apply a matrix method to thegeneral system ofsimultaneous linear Diophantine equations in unknowns. Such a systemmay be written in the form

mn

Ax = b (1)whereA is an integer matrix and are respectively ,integer vectors. Two integer matrices areequivalentif thereare square integer matrices of orders respectively, each withdeterminant , such that . A necessary and sufficient conditionfor (1) to have a solution is that the matrices and areequivalent. The necessity of the condition is easily proved, for if (1) has a

solution, , then . The proof of sufficiency is

more difficult, requiring the use ofSmith normal form [3]. Alternatively,sufficiency follows as an extreme special case of Roth's lemma [4].

m × n x, b n × 1 m × 1m × n C, D

P, Q m, n±1 PCQ = D

(A b)… (A O)…

x (A b) ( ) = (A 0)… In −x

01 × n 1

…

In order to obtain the general solution of (1) we use nothing morecomplicated than row and column reduction of the coefficient matrix todiagonal form. (The algorithm used is a simplified version of that used toobtain Smith normal form [3]). We use the following row and columnreductions:

A

(i) Multiplication of a row (column) by ,±1(ii) Interchange of two rows (columns),(iii) Addition to one row (column) of an integer multiple of another row

(column).In effect, such a sequence of row (column) operations is pre- (post-)

multiplication by a square matrix with determinant .±1We now describe the algorithm used to diagonalise the coefficient

matrix of (1). To simplify the exposition, we shall always refer to thecurrent occupant of the -position as . First use row and columninterchange to bring a non-zero entry inA of smallest absolute value into the

-position. Then add integer multiples of the first row to the other rows soas to make the entries in the first column as small as possible in absolutevalue. Do the same to columns, so as to make entries in the first row assmall as possible in absolute value. If the and ( ),

are not all zero at this stage, repeat the process. Each time theprocess is repeated, the entries in the first row and first column decrease inabsolute value so, by continued repetition, we end up with all entries (except

) in the first row and column being zero. Repeat the whole operationwith , and so on, ending with a matrix whose only non-zero entries lie onthe leading diagonal.

Ai , j aij

1,1

ai1 a1j 2 ≤ i ≤ m(2 ≤ j ≤ n)

a11

a22

For example, consider the system :Ax = b

NOTES 83

( ) ( ) = ( ) (2)−4 −2 2−2 2 −2

x1

x2

x3

4−4

We use the following sequence of row and column operations:Interchange rows 1 and 2, add −2(row 1) to row 2, add column 1 to

column 2, add −(column 1) to column 3, add column 2 to column 3,multiply columns 1 and 2 by −1. This sequence of operations reduces the

coefficient matrix of (2) to and has the same effect as pre-

multiplying by and post-multiplying by

; If we pre-multiply (2) by , then

we have . Putting we have

so , and is

arbitrary. Thegeneral solution is then . Then or

where is an arbitrary integer. This

is the general solution of the system (2).

( )2 0 00 6 0

A P = ( )0 11 −2

A

Q = ( )−1 −1 00 −1 10 0 1

PAQ= ( ).2 0 00 6 0

P

PAQQ−1( ) = P( )x1

x2

x3

4−4

u = ( ) = Q−1( )u1

u2

u3

x1

x2

x3

( ) ( ) = ( ) ( ) = ( )2 0 00 6 0

u1

u2

u3

0 11 −2

4−4

−412

2u1 = −4 6u2 = 12 u3

( ) = ( )u1

u2

u3

−22u3

x = Qu

( ) = ( )( ) = ( )x1

x2

x3

−1 −1 00 −1 10 0 1

−22u3

0−2 + u3

u3

u3

The example given in [1] is and using our

method we find that . Thus the

method of [1] is a variant of our method applied to thetransposed system. Likewise, the coefficient matrix of the example given in [2] is

and by our method we find that

. Again, the author of [2] has

worked with the transposed system rather than the given system. Also, bothauthors have confined their work to the extreme special case in which

( ) ( ) = (2072)1976 1776x1

x2

( ) ( ) = ( )1976 1776−71 22279 −247

8 0

xTAT = bT

( )15579 342

( ) ( ) = ( )15579 342−9 38410 −1731

9 0


(and in [2] the work is confined to a special case of this specialcase).m = 1

Finally, we note that the author of [1] mentions that a linear congruenceis virtually the same thing as a linear Diophantine equation. (Once again, asingle linear congruence is considered.) The reader may like to use ourpresent method to solve the simultaneous linear congruences (mod5), (mod 7); (Answer: ).

x ≡ 2x ≡ 3 x = 17 + 35n

References.1. T. Koshy, Linear Diophantine equations, linear congruences and

matrices, Math. Gaz. 82 (July 1998) pp. 274-277.2. I. Cook, Diophantine equations. A tableau, or spreadsheet for solving

, Math. Gaz. 82 (November 1998) pp. 463-468.xa + yb = h3. J. H. Wilkinson,The algebraic eigenvalue problem, Clarendon Press,

Oxford (1965) p. 18.4. A. J. B. Ward, A straightforward proof of Roth's lemma in matrix

equations, Int. J. Math. Educ. Sci. Technol. 30 (1) (1999) pp. 33-38.A. J. B. WARD

19 Woodside Close, Surbiton, Surrey KT5 9JU

84.08 On the application of Whittaker's theoremWhittaker's theorem (stated below) belongs to the theory of equations;

it was extended by Aitken in 1924. It can also be applied to someconvergent infinite power series.

Denote the determinant of the square matrix of real elements | A |

A =

a1 a2 a3 a4 … … an an + 1

1 a1 a2 a3 … … an − 1 an

0 1 a1 a2 … … an − 2 an − 1

0 0 1 a1 … … an − 3 an − 2

… … … … … … … …… … … … … … … …0 0 0 0 … … a1 a2

0 0 0 0 … … 1 a1

by , and let denote the minor for element . The theorem[1] states that, if the equation has a real root of least absolutevalue, where

An + 1 Bn (n + 1, 1)f (x) = 0

f (x) = 1 − a1x + a2x2 − a3x

3 + a4x4 − … ,

then it is given by the Whittaker series

B0

A0A1+

B1

A1A2+

B2

A2A3+

B3

A3A4+ … (A0 = B0 = 1)

NOTES 85

if this series converges. (The theorem admits a repeated root, but not rootsof equal modulus and opposite sign.) This series expression is notparticularly user-friendly, whereas the th partial sum has a simple formwhich can be obtained after the appropriate use of Jacobi's theorem on theadjugate [2] to express in terms of and . Jacobi's theoremstates that any minor of order here) in is equal totimes its signed complementary minor in . Applying this to the minorformed of the four corners of ,

n

Bn An − 1, An An + 1

m(< n + 1 | adjA | | A |m− 1

| A |A

| | = A2 − 1n + 1An − 1,

An (−1)n Bn

(−1)n An

Bn = A2n − An − 1An + 1 (n ≥ 1) ,from which

whence the th partial sum is given by . Thus computation isfacilitated with the help of the further relation

n Sn An − 1 / An

An = a1An − 1 − a2An − 2 + a3An − 3 − … + (−1)n − 1 anA0 (n ≥ 1) ,obtained by expanding from its first row, then expanding the cofactor ofeach by its first column times.

An

ai (i − 1)Several examples were tested using a BASIC program. Convergence

seemed to be consistent with that of first-order iterative processes.The example is exceptional and

leads to the surprising result , and for. When the odd positive integers here are replaced by their squares,

however, the limit of the sequence is , which also arises in the caseof . The reader may like to trace the relation between

and the quadratic equations in such cases.

f (x) = 1 − 3x + 5x2 − 7x3 + …A1 = 3 An = 4 (n > 1) Sn = 1

n > 23 − 2 2

f (x) = 1 − 6x + x2

f (x)Further, the series leads to

the same result as for the (Fibonacci) quadratic .f (x) = 1 + 1 × 5x + 2 × 6x2 + 3 × 7x3 + …

f (x) = 1 + x − x2

Certain standard series may be modified to provide estimates for someof the household constants (typically , , ). For example, although thelack of odd powers in the series for results in oscillatory behaviour, thechoice furnishes an approximation for . The first zeroof the Bessel function can be similarly investigated.

π e log 2cosx

f (x) = cos x π2 / 4J0 (x)

Interesting applications of Whittaker's theorem abound, but finally notethat , having no real roots, leads to the divergent scenario

, , .f (x) = exp(−x)

An = 1 /n! Bn = 1 / [n! (n + 1)!] Sn = n

References1. H. W. Turnbull, Theory of equations, Oliver and Boyd (1952).2. J. W. Archbold, Algebra, Pitman (1964).

J. A. SCOTT1 Shiptons Lane, Great Somerford, Chippenham, Wiltshire SN15 5EJ


84.09 Digital roots and reciprocals of primesThe digital root of a number is found by adding its digits, then adding

the digits of the result, and so on, until a single digit is obtained. Forexample: 142857→27→9. Most readers will be aware that the digital rootis equivalent to the starting number modulo 9.

Recently I computed the digital roots of the recurring parts of thedecimal representations of prime reciprocals (Table 1).

Prime Reciprocal as a decimal Digital root

2 0.50⋅

0

3 0.3⋅

3

5 0.20⋅

0

7 0.1⋅42857

⋅9

11 0.0⋅9⋅

9

13 0.0⋅76923

⋅9

17 0.0⋅588235294117647

⋅9

19 0.0⋅52631578947368421

⋅9

TABLE 1

For primes greater than 5, all the digital roots appear to have the samevalue, 9. We can confirm this if we note that recurring decimals can beturned into fractions with denominators of the form where

and are chosen to be as small as possible. For example, if then and so that

and . If p is a prime with and

10n (10m − 1) n ≥ 0 m ≥ 1x = 0.1352

⋅7⋅

1000x = 135.2⋅7⋅

100000x = 13527.2⋅7⋅

999000x = 13392 x = 13392 / 999000 p > 51p

=a

10n (10m − 1),

then . Since the firstpossibility is ruled out, so for some . The value ofmdetermines the period of the recurrence. For example, in the case , thenumbers 9, 99, 999, 9999 and 99999 are not divisible by 7, but .We can also determine that in the case of prime reciprocals since, for

, , and for thismeans that and , so that we can cancel the fraction to

where . This contradicts the choice ofnto be as small as possible.

ap= 10n(10m − 1) ⇒ p | 10n p | (10m − 1) or p ≠ 2,5 p | (10m − 1) m ≥ 1

p = 77 | 999999

n = 0n > 0 ap = 10n (10m − 1) = 2n5n (10m − 1) p ≠ 2,5

2n | a 5n | a1 /p = a′ / (10m − 1) a′ = a / 10n

We therefore have and , where (possiblypreceded by one or more zeroes) is the recurring part of the decimalrepresentation of . Hence , so that is a multiple of 9. If

thena must be a multiple of 9 and will therefore have a digital rootof 9.

n = 0 1 /p = a / (10m − 1) a

1 /p ap = 10m − 1 app ≠ 3

ALEXANDER J. GRAYFlat 2, 54 Bloomsbury Street, London WC1B 3QT

NOTES 87

84.10 Unexpected symmetry in a derived Fibonaccisequence

This note concerns a sequence related to the extendedFibonacci sequence, which is indexed by positive and negative terms. TheFibonacci sequence, which is given by , and

, may be extended to negatively indexed terms by rearranging therecurrence relation to for , giving,

gn : n ∈ z

f 0 = 0 f 1 = 1 f n+1 = f n + f n−1

(n ≥ 1)f n − 1 = f n + 1 − f n n < 0

f −1 = 1, f −2 = −1, f −3 = 2, f −4 = −3, …The derived Fibonacci sequence is then defined for all by

, so, for example,n ∈ z

gn = nfn − 1

… , g−3 = 9, g−2 = −4, g−1 = 1, g0 = 0, g1 = 0, g2 = 2, g3 = 3, g4 = 8, …I discovered the property described in the proposition below when I was

investigating the differences for the derived Fibonacci sequence. It is wellknown that the sequence of differences for the Fibonacci sequence isanother copy of the Fibonacci numbers. The situation for the derivedsequence is not so simple, but yields a surprising symmetry.

There is a standard notation for differences [1; p. 374] which uses thesymbol. If is a sequence, then the (first) differences are defined by

, the second differences by ,and, for , the th differences by . The tablebelow shows the derived Fibonacci sequence (for ) in the first rowand the differences in subsequent rows. Note that we maydefine to be and write for .

unn ∈zun = 1un = un+1 − un 2un = un+1 − un

k > 2 k kun = k− 1un+1 − k− 1un

gn n > 0kgn (k ≥ 1)

0gn gn 1gn gn

n 0 1 2 3 4 5 6 7 8 9 10 11

gn 0 0 2 3 8 15 30 56 104 189 340 605gn 0 2 1 5 7 15 26 48 85 151 2652gn 2 −1 4 2 8 11 22 37 66 1143gn −3 5 −2 6 3 11 15 29 484gn 8 −7 8 −3 8 4 14 19

The table shows an interesting pattern. The first diagonal columncontains the same numbers (with some changes of sign) as the first row.The second diagonal corresponds to the second row and so on. Inattempting to prove this relationship I was also led to a formula for ,which is given in Proposition 1 below. The proof makes use of a propertyof the Fibonacci numbers proved in the lemma.

kgn

LemmaThe negatively indexed and positively indexed Fibonacci numbers are

related:

f −n = (−1)n + 1 f n.


ProofWe use induction, noting that the result is true for and .

Suppose that the result is true for and and consider thecase . Using the defining recurrence relation we have

n = 0 n = 1n = k − 1 n = k

n = k + 1

f −(k + 1) = f −k − 1 = f −k + 1 − f −k = f −(k − 1) − f −k

so by the induction hypothesis

f −(k + 1) = (−1)(k − 1) + 1 f k − 1 − (−1)k + 1 f k

= (−1)k (f k − 1 + f k)= (−1)k + 2 f k + 1.

Since the result for is true whenever the results forand are true, the result is true for all by the principle ofinduction.

n = k + 1 n = k − 1n = k n ∈ n

Proposition 1If is the derived Fibonacci sequence given by

then, for and ,gnn ∈z gn = n f n − 1

k ≥ 0 n ∈ z

kgn = kfn − k + 1 + nfn − k − 1. (1)Proof

The result follows by induction on . We show how to deduce the resultin the case , given that the result for is valid for any .From the definition of and the induction hypothesis we have, for any

,

kk = r + 1 k = r n

n ∈ z

r + 1gn = rgn + 1 − rgn

= (rf n + 1 − r + 1 + (n + 1) f n + 1 − r − 1) − (rf n − r + 1 + nfn − r − 1)= r (f n − r + 2 − f n − r + 1) + n (f n − r − f n − r − 1) + f n − r.

Using the recurrence relation for Fibonacci numbers, we obtain

r + 1gn = rf n − r + nfn − r − 2 + f n − r

= (r + 1) f n − (r + 1) + 1 + nfn − (r + 1) − 1, .as required

To complete the proof, we note that the case follows directly forall from the definition of , since .

k = 0n gn 0gn = gn = nfn − 1

The symmetry, that was noted above, between the rows and columns inthe table, is made more precise in Proposition 2. In fact, the symmetry isnot perfect, as there is sometimes a factor of −1 involved.

NOTES 89

Proposition 2With defined as in Proposition 1, for we havegn n, k ≥ 0

ngk = (−1)n + k kgn.Proof

To prove this result, we note that (1) is valid for all and all. Therefore, if and we have, from the lemma

k ≥ 0n ∈ z n ≥ 0 k ≥ 0

ngk = nfk − n + 1 + kfk − n − 1

= (−1)k − n + 2 nf−(k − n + 1) + (−1)k − n kf−(k − n − 1)

= (−1)k − n (nfn − k − 1 + kfn − k + 1)= (−1)k + n kgn.

Note that because . This completes theproof.

(−1)k − n = (−1)k + n (−1)2n = 1

AcknowledgementI first noticed the symmetry among the differences in 1997. I would

like to thank the Editor for suggesting suitable notation and for hisassistance with the proofs.

Reference1. L. Råde, B. Westergren,Mathematics handbook for science and

engineering, (3rd edn.), Chartwell-Bratt/Studentliteratur, (1995).ALEXANDER J. GRAY

Flat 2, 54 Bloomsbury Street, London WC1B 3QT

84.11 A recurrence relation among Fibonacci sumsConsider the sums of Fibonacci numbers,

τn = ∑n

k = 0

f k

where etc. On tabulating the first few terms(Table 1)

f 0 = 0, f 1 = 1, f 2 = 1,

n 0 1 2 3 4 5 6

τn 0 1 2 4 7 12 20

TABLE 1

we find that for small values ofn. The proof of thisdepends only on the recurrence relation for Fibonacci numbers.

1 + τn + τn+1 = τn+2

We have

1 + τn + τn+1 = 1 + ∑n

k = 0

f k + ∑n+1

k = 0

f k


= f 1 + ∑n

k = 0

(f k + f k+1) + f 0 ( f 1 = 1)since

= f 0 + f 1 + ∑n

k = 0

f k+2 )(by the recurrence relation

= ∑n+2

k = 0

f k = τn+2 as required.

ALEXANDER J. GRAYFlat 2, 54 Bloomsbury Street, London WC1B 3QT

84.12 Some unusual iterationsA common A level question is to ask for the first three or four iterates of

a particular system and then say something about the value to which thesequence converges. With the advent of graphical calculators it is now easyto produce many iterates of a given formula with only a few keystrokes.The purpose of this note is to exploit this feature to illustrate how the firstfew iterates may give a false picture of the whole sequence. Good examplesof such behaviour can be quite tricky to find. The calculations were done ona TI85 and take a matter of minutes for an average student to do forthemselves. The results were identical on a TI86, and (with fewer digitsdisplayed) on a TI82. It would be interesting to see if the same sequencesare generated by other makes of calculator with different storage orcalculation methods.

Example 1

The system , , converges to 2. If we

modify it to , the sequence

becomes

xn + 1 = 1 + xn −x2

n

4x0 = 1.75

xn + 1 = 1 + xn −x2

n

4+

1100000

e10000(xn − 2)

x0 = 1.75

x1 = 1.984375

x2 = 1.99993896484

x3 = 2.00000002142

x4 = 2.00001002144

x5 = 2.00002724114

x6 = 2.00015242885

x7 = 43.6774066446

x8 = overflow!

NOTES 91

Example 2

The system , , converges to 2. If we modify

it to , we get

xn + 1 =xn

2+

2xn

x0 = 1.75

xn + 1 =xn

2+

2xn

−4

1 + 1012 (xn − 2)2

x0 = 1.75

x1 = 2.01785714279

x2 = 2.00007900136

The sequence on a TI85 continues in this fashion, close to , untilx = 2

x24 = 2.00000059441

x25 = −0.95568081953

x26 = −2.57058932682

x27 = −2.06332636966

x28 = −2.00097178739

x29 = −2.00000023598

x30 = −2

x31 = −2.The sequence has converged to −2 !!

It is a good exercise for more able students to explain what causes theunexpected results.

MARK THORNBERDurham Johnston Comprehensive School, Crossgate Moor, Durham DHI 4SU

84.13 When the sum equals the productFor , let denote the number of integer solutions ,

with , of the equationn ≥ 2 a (n) (x1, x2, … , xn)

1 ≤ x1 ≤ x2 ≤… ≤ xn

x1 + x2 + … + xn = x1x2… xn.For example, when , we have the solutions (2, 2), (1, 2, 3),(1, 1, 2, 4), because

n = 2, 3, 4

2 + 2 = 2 × 2, 1 + 2 + 3 = 1 × 2 × 3, 1 + 1 + 2 + 4 = 1 × 1 × 2 × 4.In general, we find that is always a solution; we shall callthis the trivial solution associated with , so that . It is easy tocheck that there are no non-trivial solutions when , so that

. When we have the two non-trivialsolutions (1, 1, 1, 3, 3) and (1, 1, 2, 2, 2), and it is not difficult to check that

(1, 1, … , 2, n)n a(n) ≥ 1

n = 2, 3, 4a (2) = a (3) = a (4) = 1 n = 5


these are the only solutions, so that . In general, such checks canbe carried out by finding an explicit bound for in a solution. Thus, from

, we deduce that , and isuniquely specified by when there is a solution, so that all thesolutions can be found by computation. The argument shows that isfinite; in fact , for example.

a (5) = 3xi

x1 + x2 + … + xn ≤ nxn x1x2… xn − 1 ≤ n xn

x1, x2, … xn − 1

a (n)a (n) ≤ nn − 1

We prove that

x1 + x2 + … + xn ≤ 2n

with equality only for the trivial solution. (This problem appeared in thePolish Mathematical Olympiad in 1990.) Let denote the number of unitelements in a solution , so that there arenon-unit elements having the form , with .It follows that

bn

(x1, x2, … , xn) k = n − bn ≥ 2yi + 1 1 ≤ y1 ≤ y2 ≤ … ≤ yk

(y1 + 1) (y2 + 1) … (yk + 1) = y1 + y2 + … + k + bn.When this becomes , and is maximum when

. Thus , and hence .When , we have

k = 2 y1y2 = n − 1 y1 + y2

y1 = 1 y1 + y2 ≤ n x1 + x2 + … + xn = n + y1 + y2 ≤ 2nk ≥ 3

x1 + x2 + … + xn = n + (y1 + y2 + … + yk)≤ n + (y1y2 + y2y3 + … + yky1)< n + (y1 + 1)(y2 + 1)… (yk + 1) − (y1 + y2 + … + yk)= 2n.

As a corollary, we find that , sothat , and hence

2k ≤ x1… xn = x1 + … + xn ≤ 2nk ≤ log2 n + 1

bn ≥ n − 1 − [log2 n] .This estimate is also sharp because there is equality when , with

; the relevant solution is (1, 1, …, 1, 2, 2, …, 2) with .n = 2s − s

s ≥ 2 bn = 2s − 2s

We do not know whether as or not, but we canshow that can be arbitrarily large. If , and

a (n) → ∞ n → ∞a (n) n = 22s + 1

xn − 1 = 2j + 1, xn = 22s − j + 1, j = 0, 1, 2, … , s,with

then each such is a solution, so that for such .

(1, 1, … ,1, xn−1, xn)a(n) ≥ s+ 1 > 1

2 log2n nIf each in a solution is odd, then the product is also odd, and hence the

sum is odd, which implies that has to be odd. Therefore, ifis even, at least one is even. Moreover, the product is now even, and sothat the sum is even, which in turn implies the existence ofanother even . The argument shows that when is even.

xi

x1 +… +xn n nxi

x1 +… +xn

xi 4 | x1 +… +xn n

NOTES 93

n a(n) n a(n) n a(n) n a(n) n a(n) n a(n) n a(n) n a(n) n a(n) n a(n)1 − 11 3 21 4 31 4 41 7 51 4 61 9 71 6 81 7 91 62 1 12 2 22 2 32 3 42 2 52 3 62 3 72 3 82 4 92 33 1 13 4 23 4 33 5 43 5 53 7 63 4 73 9 83 5 93 64 1 14 2 24 1 34 2 44 2 54 2 64 4 74 4 84 2 94 35 3 15 2 25 5 35 3 45 4 55 5 65 7 75 3 85 10 95 66 1 16 2 26 4 36 2 46 4 56 4 66 2 76 3 86 5 96 57 2 17 4 27 3 37 6 47 5 57 5 67 5 77 6 87 4 97 68 2 18 2 28 3 38 3 48 2 58 4 68 5 78 3 88 5 98 59 2 19 4 29 5 39 3 49 5 59 4 69 4 79 5 89 8 99 410 2 20 2 30 2 40 4 50 4 60 2 70 3 80 2 90 2 100 5

The above table gives the values for for , obtainedby exhaustive search for solutions using a computer. For example, when

, we find that for , and the remaining threevalues for are given by each of the following four possibilities

, , , . Similarly, when ,we need to have for , and the remaining five values

are given by each of the five possibilities 1, 1, 1, 2, 100,1, 1, 1, 4, 34, 1, 1, 1, 10, 12, 1, 1, 4, 4, 7, 2, 2, 3, 3, 3. We alsofound that , , , .

a (n) 1 < n ≤ 100

n = 50 xi = 1 1 ≤ i ≤ 47x48, x49, x50

1, 2, 50 1, 8, 8 2, 2, 17 2, 5, 6 n = 100xi = 1 1 ≤ i ≤ 95

x96, … , x100

a (1997) = 20 a (1998) = 8 a (1999) = 16 a (2000) = 10We do not know whether can take the value 1 infinitely often or

not. The table shows that the largest with is .We also found three values for with in the range

, namely , 174 and 444, but we have not foundany more solutions of .

a (n)n < 100 a (n) = 1 n = 24n a(n) = 1

100 ≤ n < 1000 n = 114a (n) = 1

If is composite, so that , with , then(1, 1, ... , 1, , ) is a non-trivial solution. Therefore, ifand , then is a prime. Next, if then (1, 1,...,1, 2, 2, ) is a non-trivial solution. We already know that, ifand , then is an even number not of the form , so that wemust have . We extend the argument to give the following necessarycondition for when .

n − 1 n = ab + 1 2 ≤ a ≤ ba + 1 b + 1 a (n) = 1

n ≥ 5 n − 1 n = 3m + 2m + 1 a (n) = 1

n ≥ 5 n 3m + 16 | n

a (n) = 1 n ≥ 100

Theorem. Let satisfy . Then150 or 174 (mod 210).

n ≥ 100 a(n) = 1 n ≡ 0, 24, 30, 84, 90, 114,

Proof. First, if , then we need to have or 6, becausen = 7m+ r r = 0, 2, 3

(1,1, … ,1,8, m+ 1), (1,1, … ,1,2,4, m+ 1), (1,1, … ,1,2,2,2, m+ 1)are non-trivial solutions corresponding to , respectively.r = 1, 4, 5

Since , if , then we need to have , 6, 12, 18 or24, and we can also rule out because has the factor 5. When

and 18, we have the non-trivial solutions

6 | n n = 30m + r r = 0r = 6 n − 1

r = 12

(1, 1, … ,1, 2, 2, 2, 2, 2m+ 1) (1, 1, … , 1, 2, 3, 6m+ 4)and


respectively. Therefore, we may only have . The required resultin the theorem follows.

r = 0, 24

We conclude with two conjectures.

Conjecture 1. If and , then n ≥ 5 a (n) = 1 n ≡ 24 (mod 30) .

Conjecture 2. If then .n > 444 a (n) > 1

Some additional information on this topic may be found in [1, 2].

AcknowledgementWe are grateful to the referee for his valuable remarks.

References1. W. Sierpinski, Number theory, PWN Warsaw (1959).2. (author needed), (Title needed),Amer. Math. Monthly78 (October

1971) p. 1021.LEO KURLANDCHIK and ANDRZEJ NOWICKI

Department of Mathematics and Informatics, Nicholaus CopernicusUniversity, 87-100 Toruñ, Poland


84.14 Never say never: some mistaken identitiesAfter an exasperating batch of marking, I told a lecture audience of

undergraduates:‘One over A plus B is NEVER equal to one over A plus one over B.’A doubt passed over my mind while I was saying this, but I had to get

on with the lecture. For a start, I would have been correct had I said that ifA and B are REAL then .1 / (A + B) ≠ 1 /A + 1 /B

Later, a discussion in the bar led a colleague and me to wonder if thefollowing is ever true of complex numbers:

1A + B

=1A

+1B

. (1)

The gist of the following argument fitted onto a beer mat. We supposedthe fractions to be finite, so that , and . Equation(1) leads to this quadratic equation

A ≠ 0 B ≠ 0 A + B ≠ 0

(B

A)2

+B

A+ 1 = 0, (2)

which has the pair of solutions and , whereB+ / A B− / A

B±

A=

−12

± i3

2. (3)

NOTES 95

This can also be written in two other ways:

B± = A cos(2π / 3) ± iA sin(2π / 3) = A exp(±i 2π / 3) . (4)We concluded that, given a non-zero complex number , there are two

complex values for which satisfy equation (1). Interestingly so these three points in the complex plane are

equidistant from the origin. Also, from (4), the angle subtended at the originbetween any two numbers chosen from the set is .Therefore the three points corresponding to are the vertices ofan equilateral triangle, centred at the origin. Put in a more symmetric way,take any equilateral triangle , centre , in the complex plane; then any twovertices of are complex numbers which satisfy equation (1).

AB

| B+ | = | B− | = | A |

A, B+, B− 2π / 3A, B+, B−

T OT

One beer mat was not enough to explore the possibility of threecomplex numbers satisfying

1A + B + C

=1A

+1B

+1C

. (5)

You may like to show that (5) implies that .From this we deduce that either or or .Consequently equation (5) always reduces to a trivial identity. For example,the case gives

(B + A) (A + C) (C + B) = 0B = −A A = −C C = −B

C = −B

1A + B − B

=1A

+1B

−1B

. (6)

which reduces to the trivial identity

1A

=1A

.

Despite this disappointment, I invite you to consider the possibility that

1A + B + C + D

=1A

+1B

+1C

+1D

. (7)

There are plenty of non-trivial examples of this, even when are constrained to the real numbers. For example,A, B, C, D

11 + 2 + 3 + (± 63/11− 3) =

11

+12

+13

+1

(± 63/11− 3).

(Hint: take as given real numbers and determine values ofDwhich satisfy equation (7); this is always possible unless

and . The same treatmentgeneralises to any finite number of constants.)

A, B, C

1 /A + 1 /B + 1 /C = 0 A + B + C ≠ 0

I was unable to find an example of (7) for which aredistinct, non-zero integers. But later Jim Buddenhagen* sent me severalclasses of solutions, each a family of quartics in one integer parameter. Ipay tribute to Jim's remarkable work with one of his examples:

A, B, C, D

* (e-mail: [email protected])


A = −2n (2n + 3) (n2 + 3) ;

C = 3(2n + 3) (n2 + 3n − 3) ;

B = 3n (n − 2) (n2 + 3n − 3) ;

D = (n2 + 3) (n2 + 3n − 3) .

wheren is any integer except 0 and 2. For this solution, both sides of (7)equal . If we put we have, for example, thesolution , , and : the reciprocal of theirsum is , which is the sum of their reciprocals.

1 / [6 (n − 2) (n2 + 3)] n = 1A = −40 B = −3 C = 15 D = 4

1 / (−24)Another maddening student howler, often remarked upon in common

rooms and mathematically inclined bars, is

log (a + b) = log(a) + log (b) [sic] . (8)If and are positive real numbers, then (whatever the base of the log) (8)implies that . Hence . In order that bepositive we must have . Consequently . This gives an interestingrelation if we put and consequently .Then, from equation (8), for any real angle (not an integer multiple of

), we have

a b1 /a + 1 /b = 1 b = a / (a − 1) b

a > 1 b > 1a = sec2 θ b = a/ (a − 1) = cosec2θ

θπ / 2

log (sec2 θ + cosec2 θ) = log(sec2 θ) + log (cosec2 θ) . (9)The reader is invited to prove (9) directly.

A different choice of and gives the following result for any non-zeroreal

a bφ

log (cosh2 φ + coth2 φ) = log (cosh2 φ) + log (coth2 φ) . (10)Lastly, in the bar, we wondered when and decided

that we had better stop there; we heeded the wise warning ‘never drink andderive’.

e(a + b) = ea + eb

Acknowledgments: Thanks to Paul Hammerton, Tom Ward, and Graham Everest at the

School of Mathematics, UEA, and Jim Buddenhagen, of San Antonio,Texas, for the mathematical refreshments.

MARK J. COOKERSchool of Mathematics, University of East Anglia, Norwich NR4 7TJ

84.15 A curious property of the integer 24

Proposition 1 The only positive integers n with the property

m2 ≡ 1 (modn) m n (1)for all integers coprime to

are , i.e. the divisors of 24 greater than 1.n = 2, 3, 4, 6, 8, 12, 24

Proof It is easily checked that the stated values of satisfy (1). Forexample, when , the condition that is coprime to implies that

(mod 24). In each case (mod24).

nn = 24 m n

m ≡ 1, 5, 7, 11, 13, 17, 19, 23 m2 ≡ 1

NOTES 97

It remains to show that no other integer has property (1). We firstobserve that, if satisfies (1) then the lowest common multiple of and24 also satisfies (1). To see this, consider a prime that divides and let

denote the highest power of that divides . Then divides at leastone of and 24. An integer is coprime to if and only if is coprime toboth and 24. Thus (mod ) by property (1) for , if divides, or by property (1) for 24, if divides 24. Since the congruence holds

for all primes that divide we have (mod .

nn N n

p Npr(p) p N pr(p)

n m N mn m2 ≡ 1 pr(p) n pr(p)

n pr(p)

p N m2 ≡ 1 N)Suppose now that satisfies (1) and .

Then, by the previous paragraph, satisfies (1) and . The inequalityshows that mod and we can conclude from (1) (for ) that isdivisible by 5. Hence . Thus (mod ) and so, by (1), isdivisible by 7. Hence . Let denote the th prime (so, ,

, , etc.). We show by induction on that forevery . The case has been established. Assume the result for . Inthe next paragraph it is shown that for . It thenfollows that (mod ). By (1), is divisible by , and hence

. The proposition follows, since no integer satisfies thisinequality for all .

n n ≠ 2, 3, 4, 6, 8, 12, 24N N > 24

52 ≡ ⁄ 1 N N NN ≥ 120 72 ≡ ⁄ 1 N NN ≥ 840 pi i p1 = 2

p2 = 3 p3 = 5 i N ≥ 4p1p2… pi

i i = 4 ip2

i + 1 ≤ p1… pi i ≥ 4p2

i + 1 ≡ ⁄ 1 N N pi + 1

N ≥ 4p1… pi + 1 Ni

To complete the proof, we show by induction on thatfor . The result is easily verified for . Assume the result for .Recall Bertrand's postulate that, for every positive integer , there is a prime satisfying (a simple proof by P.Erdõs may be found on page

231 of [1]). In particular, taking , we have that .Hence , as . Thiscompletes the inductive step.

i p2i + 1 ≤ p1… pi

i ≥ 4 i = 4 ic

p c < p ≤ 2cc = pi + 1 pi + 2 ≤ 2pi + 1

p1… pipi + 1 > p2i + 1pi + 1 ≥ 1

4p2i + 2pi + 1 > p2

i + 2 pi + 1 > 4

Reference1. H. E. Rose,A course in number theory,(2nd edn.). Clarendon Press

(1994).M. H. EGGAR

Department of Mathematics and Statistics, University of Edinburgh, EH9 3JZ

84.16 What do cycles of a given length generate?Let denote the symmetric group of degree , i.e. the group of all

permutations of symbols. Let denote the alternating group, i.e. thesubgroup of consisting of even permutations.

Sn nn An

Sn

It is a fundamental theorem in a first course in abstract algebra thattranspositions (i.e. cycles of length 2) generate all of . It is alsofundamental that cycles of length 3 generate all of . If one denotes thesubgroup of generated by all cycles of length by , thenthe above can be restated as

Sn

An

Sn r (2 ≤ r ≤ n) S(r)n

S(2)n = Sn, S(3)

n = An.

One expects this to create a hope that there are (beside and ) otherSn An


distinguished subgroups of that are yet to be discovered, namely thesubgroups

Sn

S(4)n , S(5)

n , … , S(n)n .

At least it raises the question of what these subgroups are. I find it amazingthat, during the many years of my career as an algebra instructor, I havenever been asked this question by any student.

Our question can be easily answered (and our hope, if any, dashed) byobserving that

( )( ) = ( ) ∀r ≥ 3.1 2 … r − 1 r r r − 1 … 3 1 2 1 3 2

This identity shows that contains all 3-cycles and thus .Hence is either or . Since a cycle of length is even if, and only if, is odd, it follows that if is odd and if is even.

S(r)n S(r)

n ⊇ An∀n ≥ 3S(r)

n Sn An rr S(r)

n = An r S(r)n = Sn r

MOWAFFAQ HAJJADepartment of Maths. and Comp. Sci., American University of Sharjah,

PO Box 26666, Sharjah, United Arab Emirates

84.17 A game with positive and negative numbers

The gameIn many games, simple rules generate a complex multitude of positions.

In our game a position consists of a finite sequence of real numbers. Weimpose the rule that a sequence of length may be reduced to ashorter sequence by omitting any zero or by replacing two adjacent entriesof opposite sign by their sum. The game ends when no further reduction ispossible.

(a1, … , an) n

Different choices of steps may reduce a given sequence to differentirreducible sequences. For example,

(4, −2, −1, 1) → (4, −2, 0) → (4, −2) → (2)

(4, −2, −1, 1) → (2, −1, 1) → (1, 1) .Various questions immediately spring to mind. Which sequences can bereduced to a single element, i.e. to a sequence of length 1? Is there analgorithm (apart from enumeration of all possible reductions) that achievesthis reduction for such sequences? Which numbers can be the lengths ofirreducible sequences obtained by reducing a given sequence? A challengefor the reader is to prove, before looking ahead, that the sequence

can only be reduced to irreduciblesequences of lengths 2 and 3. Two people can play a game by alternatelychoosing a step in the reduction of a sequence. One person aims tominimise the length of the final irreducible sequence and the other personaims to maximise it. They then play again, starting with the same initialsequence but with reversed roles.

(−2, −3, 8, −5, 7, −3, −1, 7, −4, 10, −7)

The game, for a single player or a pair of players, could be used in a

NOTES 99

secondary school to make more enjoyable routine practice of adding apositive number to a negative number, while simultaneously providingscope for serious mathematical thought.

Some analysisThe first obvious observation is that the sum of the entries in the

sequence remains unchanged at every stage of a reduction. This not onlyhelps one check the arithmetic, but also allows one to deduce that anyirreducible sequence obtained by reducing will have all entriespositive, if , and all entries negative, if

. If , then the sequence with noentries is the only irreducible sequence obtainable from , since ifthe entries of are all non-zero and , then

must contain two adjacent entries of opposite sign and so thereduction can be carried a stage further.

(a1, … , an)a1 + … + an > 0

a1 + … + an < 0 a1 + … + an = 0(a1, … , an)

(b1, … , bn) b1 + … + bn = 0(b1, … , bn)

The key idea of our analysis is encapsulated in the following definition.Suppose at some stage in a reduction a sequence is formed from a longersequence by one application of the rules. Each non-zero entryin the shorter sequence is either a rewrite of an element, say, of

or has the (non-zero) value , where and haveopposite signs. In the first case we call in thepredecessor ofthe entry in the shorter sequence and in the second case we call whicheverof or has the same sign as thepredecessor of .Thus any non-zero element that occurs at any stage of a reduction has aunique predecessor at the previous stage (and hence at all previous stages),but it may have no successor at the next stage. Predecessors are not definedfor zero entries. Each non-zero element in the final stage thus has a uniqueultimate predecessor in the first.

(u1, … , ut)ui

(u1, … , ut) ui + ui + 1 ui ui + 1

ui (u1, … , ut)

ui ui + 1 ui + ui + 1 ui + ui + 1

The next proposition answers the first two questions raised in above.

Proposition: If , then can be reduced to asingle element if, and only if, there is an entry such that

and . One algorithm that achievesthis is to reduce (by application of appropriate rules) to asequence where each and to reduce to asequence say, where each ; then one successivelyamalgamates into . If , thenthe analogous condition and algorithm, with all inequalities reversed, apply.

a1 + … + an > 0 (a1, … , an)aj

a1 + … + aj −1 ≤ 0 aj +1 + … + an ≤ 0(a1, … , aj − 1)

(c1, … , cr) ck < 0 (aj + 1, … , an)(d1, … , ds) dk < 0

c1, … , cr, d1, … , ds ai a1 + … + an < 0

Proof: To see that the inequalities satisfied by imply reducibility to asingle element, we note that, by the first paragraph, the inequalities enablethe algorithm to be carried out. Conversely, suppose there is a reduction of

to the single element . Then we may taketo be the ultimate predecessor of under the reduction.

aj

(a1, … , an) (a1 + … + an) aj

a1 + … + an


An example of a sequence , which cannot be reduced to a singleelement . We have ,but none of the four entries such that , satisfy both the other twoinequalities required for the proposition.

S(−2,−3, 8,−5, 7,−3,−1, 7,−4, 10, −7) a1 + … + a11 = 7

ai aj ≥ 7

These ideas can be extended further. A necessary and sufficientcondition for a sequence to be reducible to some irreduciblesequence of length 2 is that there is a value of satisfying ,such that and each satisfy the conditions of theproposition and and both have the same sign.More generally, the sequence can be reduced to an irreduciblesequence with at least entries if, and only if, there exist suchthat and such that ,

, … , , … ,all have the same sign.

(a1, … , an)m 1 ≤ m < n

(a1, … , am) (am+ 1, … , an)a1, … , am am+ 1, … , an

(a1, … , an)r m1, … mr − 1

1 ≤ m1 < m2 < … < mr − 1 < n a1 + … + am1

am1 + 1 + … + am2ami + 1 + … + ami + 1

amr − 1 + 1 + … + an

For example, the sequence satisfies theproposition with and satisfies the propositionwith and the sum of the entries is positive for both these sequences.Thus can be reduced to an irreducible sequence of length 2, e.g. by

(−2, −3, 8, −5, 7, −3)j = 3 (−1, 7, −4, 10, −7)

j = 2S

(−2, −3, 8, −5, 7, −3, −1, 7, −4, 10, −7)→ (−2, 5, −5, 7, −3, −l, 7, −4, 10,) → (−2, 5, 2, −3, −1, 7, −4, 10, −7) → (−2, 5, 2, −3, 6, −4, 10, −7) →−7

(−2, 5, 2, −3, 6, −4, 3) → (3, 2, −3, 6, −4, 3) → (3, −1, 6, −4, 3) →

(3, −1, 6, −l) → (2, 6, −1) → (2, 5).Likewise, the reduction

(−2, −3, 8, −5, 7, −3, −1, 7, −4, 10, −7)→ (−2, 5, −5, 7, −3, −l, 7, −4, 10,) → (−2, 5, 2, −3, −1, 7, −4, 10, −7) → (−2, 5, 2, −3, 6, −4, 10, −7) →−7

(−2, 5, 2, −3, 6, −4, 3) → (3, 2, −3, 6, −4, 3) → (3, 2, 3, −4, 3) →

(3, 2, 3, −l) → (3, 2, 2)shows that 3 is a possible length for an irreducible sequence obtainable byreduction of . No reduction of can give an irreducible sequence of length5 or more, since the predecessors in of these 5 or more elements would allhave to be positive elements of , but has only 4 positive entries. To seethat length 4 also cannot be achieved we can use the criterion in the previousparagraph. Explicitly, the 8 must amalgamate in a reduction with thepreceding . For 8 to remain a predecessor (positive) the −5 mustamalgamate with the first 7. Likewise the 10 must amalgamate with thefollowing −7 and, for it to remain a predecessor the −4 must amalgamatewith the second 7. Now, whichever way −3 amalgamates, one of thepositive entries of ceases to be a predecessor.

S SS

S S

−2, −3

SM. H. EGGAR

Department of Mathematics and Statistics, University of Edinburgh EH9 3JZ

NOTES 101

84.18 An inductive proof of the arithmetic mean −geometric mean inequality

The aim of this note is to present a very simple proof of the arithmeticmean − geometric mean inequality:

a1 + a2 + … + an ≥ n(a1… an)1/n forn ∈n; ai ∈ r+, i = 1, 2, … , n, (1)with inequality if, and only if, for some .ai ≠ ai + 1 1 ≤ i < n

We use mathematical induction with the aid of the known inequality ofBernoulli:

(1 + x)n ≥ 1 + nx, for n ∈ n, x ∈ r andx > −1, (2)with equality if, and only if, or .n = 1 x = 0

Proof: If , inequality (1) reduces to , which is certainly true.Suppose that (1) is true (inductive hypothesis) for . Letting

, we have (by the inductive hypothesis)

n = 1 a1 ≥ a1

n = kGn = n a1… an

(a1 + … + ak) + ak + 1 ≥ kGk + ak + 1

with equality holding if, and only if, . Hencea1 = a2 = … = ak

(a1 + … + ak) + ak+1

≥ Gk+1( kGk

Gk+1+

ak+1

Gk+1)

= Gk+1(k( Gk

ak+1)1/(k+1)

+ (ak+1

Gk)k/(k+1))

= Gk+1( k

1 + x+ (1 + x)k) (by(ak+1/Gk)1/(k+1) = 1 + x > 0)

≥ Gk+1( k

1 + x+ 1 + kx) (by (2) with equality in the case )x = 0

≥ Gk+1(k + 1) ( is equivalent to )k1+x + 1 + kx ≥ k + 1 x2 ≥ 0

which is the case . Hence, by induction, the result is true for. On account of (2), the equality holds in the case , hence

. This implies that in the case ofequality. See [1, p. 68], [2] and [3-5].

n = k + 1n ≥ 1 x = 0Gn = an + 1 a1 = … = an = an + 1

References1. P. S. Bullen, D. S.Mitrinoviæ, and P. M.Vasiæ, Means and their

inequalities, Reidel, Dordrecht, Holland (1988).2. D. Rüthing, Proofs of the arithmetic mean − geometric mean

inequality, Int. J. Math. Educ. Sci. Technol. 13, (1) (1982) pp. 49-54.3. E. F. Beckenbach, R. Bellman, Inequalities, Springer (1971).


4. D. S. Mitrinoviæ, Analytic inequalities, Springer (1970).5. G. H. Hardy, J. E. Littlewood, G. Pólya,Inequalities, Cambridge

University Press (1967).ZBIGNIEW URMANIN

Otto-Hahn-Str. 8, 42897 Remscheid-Lennep, Germany

84.19 Weighted mean in a trapeziumIn this note we show how any line segment between and parallel to the

two parallel sides of a trapezium can be considered as a weighted mean oftheir lengths.

We recall that a weighted mean, , of two positive numbers and isgiven by , where and are the weights. Somespecial cases are worth noting. When , we obtain the arithmetic mean

. With as any nonzero constant and with and, we obtain the harmonic mean . With

and we obtain the geometric mean . Additional specialcases are given in [1].

w a bw = (bx + ay) / (x + y) x y

x = yw = (a + b) / 2 k x = kay = kb w = 2ab/ (a + b) x = k a

y = k b w = ab

For the proof, we let , , ,and for some arbitrary trapezoidABCD, as shown in Figure 1. Wenote that G is the intersection of the two nonparallel sides of the trapezoid.

a = DC < AB = b w = EF x = DE y = EAp = GE

A B

CD

E

G

F

a

w

b

x

y

p − x

FIGURE 1

Since , . HenceDC // EF // AB GDC ∼ GEF ∼ GAB

a

w=

p − x

pand

w

b=

p

p + y, respectively.

NOTES 103

By solving these two equations separately for we obtainp

p =wx

w − aandp =

wy

b − w.

Eliminating , we obtain the weighted mean, , asdesired.

p w = (bx + ay) / (x + y)

Reference1. Larry Hoehn, A geometrical interpretation of the weighted mean,

College Mathematics Journal 15 (March 1984) pp. 135-139.LARRY HOEHN

Department of Mathematics and Computer Science,Austin Peay State University, P.O. Box 4626, Clarksville, TN 37044 USA

84.20 A formula for integrating inverse functionsThe derivative of an inverse function is given in calculus textbooks (see,

for example, [1]) by the formula

(f −1) ′ (y) =1

f ′ (x) wherey = f (x) .

Wouldn't it be useful also to have the integration counterpart? In fact, asfar as we know, there is no calculus textbook that lists such a formula.Hence, the integration of inverse functions or expressions containing themcan represent a problem to mathematics students, as it implies exceptionalmemory or the availability of integral tables. Here we deduce a simpletheorem for integrating inverse functions based on a change of variable thatdoes not require prior knowledge of their antiderivatives. Namely, if thefunction has the integral , its inverse functioncan be easily integrated with the formula

y = f (x) ∫ f (x) dx x = f −1 (y)

∫ f −1 (y) dy = x f (x) − ∫ f (x) dx.

The proof of this theorem is straightforward. Let the functionand the differential be substituted in the integral

y = f (x)dy = f ′ (x) dx

∫ f −1 (y) dy = ∫ f −1 (f (x)) f ′ (x) dx

to obtain

∫ f −1 (y) dy = ∫ xf ′ (x) dx.

This expression can be integrated by parts to yield the required rule

∫ f −1 (y) dy = x f (x) − ∫ f (x) dx.

To appreciate its power, let us take, for example . Firstly,∫ cos−1 x dx


we introduce the change of variable with , andthen proceed:

x = cos(y) dx = − sin(y) dy

∫ cos−1 (x) dx = y cos(y) − ∫ cos(y) dy

∫ cos−1 (x) dx = y cos(y) − sin (y) + C.

By changing back the variable, and recalling a familiar trigonometricidentity, we obtain

∫ cos−1 (x) dx = x cos−1 (x) − 1 − x2 + C.

This procedure is not new (see [2]), as mathematicians seem to invoke itunconsciously when they need to integrate an inverse function of unknownantiderivative (see, for example, [3]); it is essentially a subtle substitutionbefore integrating by parts; but because of this simple origin, it has not beenformally stated, and thus newcomers have to discover it for themselves.

Finally, the present approach can be extended to integrate morecomplicated expressions containing inverse functions such as

∫ F (y, f −1 (y)) dy = F (f (x) , x) f (x) − ∫ f (x)d

dxF (f (x) , x) dx.

AcknowledgementsSS research has been funded by the JGH Award, ORS Award and

CONICIT. We thank E. Crampin and D. McInerney for their comments.

References1. Dale Varberg, Michael Sullivan and Edwin J. Purcell,Calculus with

analytical geometry, (7th edn.), Prentice Hall, (1996).2. R. M. Corless, G. H. Gonnet, D. E. G. Hare, D. J. Jeffrey and D. E.

Knuth, On the Lambert W function,Adv. Comput. Math.5 (1996)pp. 329-359.

3. K. B. Ranger, A complex variable integration technique for the 2-dimensional Navier-Stokes equations,Q. Applied Maths49 (1991)pp. 555-562.

S. SCHNELLCentre for Mathematical Biology, Mathematical Institute

24-29 St Giles', Oxford OX1 3LBe-mail: [email protected]

C. MENDOZACentro de Física, Instituto Venezolano de Investigaciones Científicas

(IVIC), PO Box 21827, Caracas 1020A, Venezuelae-mail: [email protected]

NOTES 105

84.21 Mathematician versus machineRecently, whilst devising examples of non-trivial functions for

reinforcing the standard techniques in calculus on the determination ofstationary points, I encountered two interesting properties of a functionwhich is also rich in other properties, which we leave readers to investigate.I shall highlight only the first two that interested me. There is also muchscope for investigation of the various properties of the function usingpackages such asMathematica, Derive and Matlab, or even a graphicscalculator. However, it must be stressed, such packages can only suggest aconjecture that will then require a formal proof. Indeed, sometimes thecalculations required are so detailed that a computer can fail to complete thetask and simply grind to a halt. The proofs of the two main properties givenhere involve recurrence relations and induction. Again, we leave readers toinvestigate other proofs of these results, and to make and prove otherconjectures.

Consider the function

ƒ (x) =x

1 + x2, x ≥ 0. (1)

We leave readers to determine the first few derivatives of usingappropriate techniques. For example, one can write

and compare the resulting series with thatobtained from MacLaurin's theorem .Evaluating these derivatives, up to and including the fifth, at and

gives the results shown in Table 1.

ƒ (x)x (1 + x2)−1 =

x (1 − x2 + x4 − x6 +… )ƒ(x) = ƒ(0) + ƒ′(0)x + 1

2!ƒ″(0)x2 +…x = 0

x = 1

n 0 1 2 3 4 5

ƒ(n) (0) 0 1 0 −6 0 120

ƒ(n) (1) 12 0 −1

232 −3 0

TABLE 1

N.B. denotes the th derivative of . These results suggest thatevery even derivative of , including itself, vanishes at .Moreover, the first and fifth derivatives vanish at , suggesting thatevery fourth derivative vanishes. Determining further derivatives directly,or using a package, provides additional evidence for these conjectures.However, it becomes increasingly more difficult to determine thederivatives, particularly using the popular and versatileMathematica.Therefore some mathematics is required, and the results will be conclusive.

ƒ(n) (x) n ƒ (x)ƒ (x) ƒ (x) x = 0

x = 1

We outline one approach involving recurrence relations andmathematical induction. Readers should check all the details, as well astrying their own proofs.

From (1) we have

(1 + x2) ƒ (x) = x, ƒ (0) = 0 ƒ (1) = 12. (2)with and


Differentiating (2) once yields

(1 + x2) ƒ′ (x) + 2xƒ (x) = 1, ƒ′ (0) = 1 ƒ′ (1) = 0. (3)with and

Differentiating (2) times, , yieldsn n ≥ 2

(1 + x2) ƒ(n) (x) + 2nxƒ(n − 1) (x) + n (n − 1) ƒ(n − 2) (x) = 0. (4)Considering first the case where , we have, from (4)x = 0

ƒ(n) (0) = −n (n − 1) ƒ(n − 2) (0) , n ≥ 2, (5)a recurrence relation with starting values and

, (5) can be simplified by writing , sothat

ƒ(0) (0) = ƒ (0) = 0ƒ(1) (0) = ƒ′ (0) = 1 an = ƒ(n) (0) / n!

an = −an − 2, n ≥ 2, a0 = 0, a1 = 1. (6)The solution of (6) is now trivial, with

, , , etc., i.e. and, . Therefore

a0 = 0, a1 = 1, a2 = −a0 = 0,a3 = −a1 = −1 a4 = −a2 = 0 a5 = −a3 = 1 a2k = 0a2k + 1 = (−1)k k ≥ 0

ƒ(2k)(0) = a2k(2k)! = 0 ƒ(2k+1)(0) = a2k+1(2k + 1)! = (−1)k(2k + 1)! (7)and

for , which can be proved by induction as an exercise.k ≥ 0For the second case, where , we have, from (4)x = 1

2ƒ(n) (1) + 2nƒ(n − 1) (1) + n (n − 1) ƒ(n − 2) (1) = 0. (8)a recurrence relation with starting values and

which can be simplified by writing , sothat

ƒ(0)(1) = ƒ(1) = 12

ƒ(1)(1) = ƒ′(1) = 0, bn = ƒ(n) (1) / n!

bn = −bn − 1 − 12bn − 2, n ≥ 2, b0 = 1

2, b1 = 0. (9)Generating the first few values of using (9) gives , ,

, , ,etc. Table 2 shows these and further values of .

bn b0 = 12 b1 = 0

b2 = −b1 − 12b0 = −1

4 b3 = −b2 − 12b1 = 1

4 b4 = −b3 − 12b2 = −1

8bn

n 0 1 2 3 4 5 6 7 8 9

bn12 0 −1

414 −1

8 0 116 − 1

16132 0

TABLE 2

From this we conjecture that

b4k =(−1)k

22k+1, b4k+1 = 0, b4k+2 =

(−1)k+1

22k+2 b4k+3 =

(−1)k

22k+2(10)and

for , which can readily be proved by induction, and thus there are fourdifferent cases that arise naturally from (10). These are, for ,

k ≥ 0k ≥ 0

f (4k) =(−1)k (4k)!

22k + 1, ƒ(4k + 2) (1) =

(−1)k + 1 (4k + 2)!22k + 2

ƒ(4k + 3) (1) =(−1)k (4k + 3)!

22k + 2and ƒ(4k + 1) (1) = 0.

NOTES 107

Table 3 is a continuation of Table 1 determined directly from thederivatives, and the values confirm the results obtained above.

n 6 7 8 9 10 11 12

ƒ(n) (0) 0 −5040 0 362880 0 −39916800 0

ƒ(n) (1) 45 −315 1260 0 −56700 623700 −3742200

TABLE 3

Even the most sophisticated mathematical software will find it difficultto generate the above formulae. Nevertheless, such packages can be usefulin suggesting such patterns, which can then be followed up with some solidmathematics.

P. GLAISTERDepartment of Mathematics, University of Reading RG6 2AX

84.22 On a conjecture of Paul ThompsonWe shall show that, for and real ,0 ≤ σ < 1 τ ≠ 0

arg(∑Nn=1n−σ − iτ) − arg(∑N−1

n=1 n−σ − iτ)arg(∑N

n=1(1 − 1n)σn−iτ) − arg(∑N−1

n=1 (1 − 1n)σn−iτ) → 1 asN → ∞, (1)

and that convergence is most rapid when . Here theexpression is taken to be the principal argument of , sothat its value lies between and . With , the function isthe famous Riemann zeta-function, and ‘most rapid’ will be made precise inthe following. Paul Thompson [1] had conjectured that, with , thelimit formula (1) holds if, and only if, .

ζ (σ + iτ) = 0arga − argb a/ b

−π π s = σ + iτ ζ (s)

σ = 12

ζ (σ + iτ) = 0

Some preliminary estimatesIt will be convenient to use order notation for asymptotic analysis which

we recall in the following. We write , where ,to mean that is bounded; in other words there is a numberKsuch that for all , and we usually apply this for

. Similarly, we also write as , where iscomplex and , to mean that is bounded in thepunctured neighbourhood of ; in other words, there are positive numbersKand such that for all satisfying .

f (N) = O (g(N)) g(N) > 0| f (N)/g(N) |

| f (N) | ≤ Kg(N) N ≥ 1N → ∞ f (z) = O (g(z)) z → ω z

g(z) > 0 | f (z)/g(z) |ω

r | f (z) | ≤ Kg(z) z 0 < |z− ω| < rLet us now write for the expression of which the

limit is stated in (1). We note that both the numerator and the denominatorfor are arguments of complex numbers in the form of a ratio

C (N) = C (N; σ, τ)

C (N)

∑Nn = 1 f (n; σ, τ)

∑N − 1n = 1 f (n; σ, τ)

= 1 +f (N; σ, τ)

∑N − 1n = 1 f (n; σ, τ)

, (2)

with an appropriate function . Thus, for the asymptotic expansion, we shallmake use of

f


11 + z

= 1 − z + O (| z |2) as z → 0, (3)

arg(1 + z) = I(z) + O (| z |2) as z → 0. (4)When applied to we shall also requireC (N)

(1 −1n)σ

= 1 −σn

+ O (n−2) as n → ∞, (5)

∑N − 1

n = 1

n−s =N1 − s

1 − s+ ζ (s) −

N−s

2+ O (N−σ − 1) as N → ∞. (6)

The reader can easily establish the elementary results (3), (4) and (5).Formula (6) is valid for and . Its derivation appears in [2],wherein much else on can be found.

σ > −1 s ≠ 1ζ (s)

Estimation of the denominator for C(N)From (6), we find that, as ,N → ∞

∑N − 1

n = 1

n−iτ =N1 − iτ

1 − iτ+ O (1) and ∑

N − 1

n = 1

n−1 − iτ = O (1) ,

and hence, by (5),

∑N−1

n=1(1 −

1n)σ

n−iτ = ∑N−1

n=1

n−iτ − σ ∑N−1

n=1

n−1− iτ + O( ∑N−1

n=1

n−2) =N1− iτ

1 − iτ+ O(1),

where we have used the fact that is absolutely convergent. Togetherwith (3) and (5) we deduce that

∑ n−2

∑Nn=1(1 − 1/n)σn−iτ

∑N−1n=1 (1 − 1/n)σn−iτ = 1 + (1 −

1N)σ 1 − iτ

N(1 + O(N −1))

= 1 +1 − iτ

N+ O(N−2),

and hence, by (4),

arg( ∑Nn = 1 (1 − 1 /n)σ n−iτ

∑N − 1n = 1 (1 − 1 /n)σ n−iτ) = −

τN

+ O (N −2) . (7)

Estimation of the numerator for C (N)A similar argument applied to the numerator for yields, for

and ,C (N)

σ > −1 s ≠ 1

∑Nn = 1 n−s

∑N − 1n = 1 n−s

= 1 +1 − s

N(1 + (1 − s) ζ (s) Ns − 1 + O (N−1))−1

.

We note that

| (1 − s)ζ (s)Ns−1 + O(N−1) |2

= O(N2σ −2) + O(Nσ −2) + O(N−2),and all three error terms are simply , if we restrict our attention toO(N2σ −2)

NOTES 109

. Thus, on applying (3), we now haveσ ≥ 0

∑Nn = 1 n−s

∑N − 1n = 1 n−s

= 1 +1 − s

N− (1 − s)2 ζ (s) Ns − 2 + O (N−2) + O (N2σ − 3) ,

and hence, by (4),

arg( ∑Nn=1n−s

∑N−1n=1 n−s) = −

τN

− I((1 − s)2ζ (s)Ns−2) + O(N−2) + O(N2σ −3). (8)

As a little aside, we remark that the argument also gives the followingformula for in the ‘critical strip’ ,ζ (s) 0 < σ < 1

ζ (s) = limN → ∞

( N2 − s

(1 − s)2 (1 −∑N

n = 1 n−s

∑N − 1n = 1 n−s) +

N1 − s

1 − s) .

Estimation of C (N)From (7) and (8), we see that both the numerator and the denominator

for have the asymptotic value as for comes from in (8)), so that as. In other words, we have established Thompson's limit formula

(1). Indeed, from (7) and (8), the expression itself has the asymptoticformula

C (N) −τ / N N → ∞ 0 ≤ σ < 1(σ < 1 Re(s − 2) < −1 C (N) → 1N → ∞

C (N)

C (N) = 1 +N

τI((1 − s)2 ζ (s) Ns − 2) + O (N−1) + O (N2σ − 2) .

Note that, if , then the second term on the right-hand side here willbe present, and it will not be for any as .On the other hand, if then this second term will be absent.Therefore, for ,

ζ (s) ≠ 0O (f (N) Nσ − 1) f (N) → 0 N → ∞

ζ (s) = 00 < σ < 1

C (N) =

1 + O (Nσ − 1) always

1 + O (N−1) + O (N2σ − 2) if, and only if, ζ (s) = 0.

The reason for imposing the condition is that, as it stands, the ‘onlyif’ part of the statement might not be valid. However, it is known thathas no zero on the line , so that no harm is done even if we write

. Readers who are not familiar with the theory of may beinterested to know that the distribution of the zeros of is closely relatedto the distribution of primes, and the fact that has no zero on , 1is used to prove the prime number theorem.

σ ≠ 0ζ (s)

σ = 00 ≤ σ < 1 ζ (s)

ζ (s)ζ (s) σ = 0

References1. P. Thompson, A conjecture with a prize,Math. Gaz. 82 (July 1998)

p. 309.2. E. C. Titchmarsh,The theory of the Riemann zeta function (2nd edn.),

Oxford University Press (1986).TIM JAMESON

13 Sandown Road, Lancaster LA1 4LN


84.23 Maximal volume of curved folding boxesIn this note, we consider the following problem.

Problem: Consider a square of paper which is identified with and, and cut off four corner regions which satisfy the inequalities

and , where is a continuous, piecewisefunction with , for every . Then find themaximal volume of the folding box which is made from the paper byattaching each pair of neighbouring cut edges.

| x | ≤ 1| y | ≤ 1| y | ≥ f (| x |) | x | ≥ f (| y |) f C2

f (t) f ′ (t) ∈ [0,1] t ∈ [0,1]V (f )

These are some results on the problem.(Fig. 1) If with , the volume takes the maximalvalue when , a typical exercise of differentialcalculus.

f (t) = a a ∈ [0,1] V (f )= 16 / 27 ≈ 0.5926 a = 2 / 3

(Fig. 2) If with , the volume takes themaximal value when and , a result of N.Lord [1].

f (t) b (t − a) = a a, b ∈ [0,1) V (f )≈ 0.6914 a ≈ 0.5356 b ≈ 0.5312

(Fig. 3) If with , the volume takes themaximal value when a , a result of N. Reed [2].

f (t) = a−1 sinat a ∈ (0, π / 2] V (f )≈ 0.7414 ≈ 1.2215

FIGURE 1 FIGURE 2 FIGURE 3

For the general problem, we obtain the following solution.

Solution: The volume takes the maximal valuewhen

V (f ) (4 / 3) α−2 ≈ 0.7757

f (t) = α−1sn(αt, 2)cn(αt, 2)

, α =π / 2

agm(1, 2)≈ 1.3110, (1)

where is the arithmetic-geometric mean of 1 and . The cutsquare is shown in Figure 4.

agm(1, 2) 2

In the above solution, we use the Jacobian elliptic functions, that is,

cn(u, k) = cos(am(u, k)) , sn(u, k) = sin (am(u, k)) , (2)where is the inverse function ofθ = am(u, k)

u = ∫0

dθ1 − k2 sin2 θ

. (3)

See §22.122 of [3], for example. In this note, we do not use other formulasfor elliptic functions to attain the above solution.

NOTES 111

FIGURE 4

Proof: We can assume that for every . Because,even if not so, we can take the function in place of .

0 ≤ f (t) ≤ t t ∈ [0,1]g(t) = mint, f (t) f

We put together the folding box, keep its mouth up, and cut it by ahorizontal plane of height . Then the positive -axis bends into an arcwhich lies in the centre of a side face. We denote by the length of the arcfrom the bottom to the section. Then the section is a square with edgelength . By Pythagoras' theorem, we have

. So we can give the volume of the folding box asfollows:

z xt

2f (t) dt2 = dz2 + df (t)2

= dz2 + f ′ (t)2 dt2 V (f )

V (f ) = 4 ∫ h

0f (t)2 dz = 4 ∫

1

0f (t)2 1 − f ′ (t)2 dt, (4)

whereh is the height of the folded box. The volume is a functionwhich has a function as a variable. So the problem becomes an exercise ofthe calculus of variations.

V (f )f

There is a formula to find the maximal value of (4), the Euler-Lagrangeequation. See §IV.3 of [4], for example. By using it, we obtain

d

dt ( ∂ L

∂ f ′(f (t) , f ′ (t))) −

∂ L

∂ f(f (t) , f ′ (t)) = 0, (5)

where . After a tedious calculation, we obtainL (f , f ′) = f 2 1 − (f ′)2

f (t) f ″ (t) + 2 (1 − f ′ (t)2) = 0. (6)By using (6), we obtain

d

dt(log (1 − f ′ (t)2) − 4 logf (t)) = −2 f ′ (t) f ″ (t)

1 − f ′ (t)2− 4 f ′ (t)

f (t)= 0. (7)

By integrating (7), and using fot , we obtain0 ≤ f (t) ≤ t t ∈ [0, 1]

1 − f ′ (t)2 = c4f (t)4 , f (0) = 0, (8)


where c is an integrating constant. We set , and , sof (t) = c−1h (ct) u = ct

1 − h′ (u)2 = h (u)4 , h (0) = 0. (9)The function attains the maximal value of under theboundary condition . So we must find a value ofc whichattains the maximal value of the function .

f (t) = c−1h (ct) V (f )f (1) = c−1h (c)

V0 (c) = V (f )By using (8) in (4), we obtain

V0 (c) =4c2 (1 − ∫

1

0f ′ (t)2 dt) . (10)

By an integration by parts, and by using (6), we obtain

∫ 1

0f ′ (t)2 dt = [f (t) f ′(t)]1

0 − ∫ 1

0f (t) f ″(t) dt

=1c h (c) h′ (c) + 2 (1 − ∫

1

0f ′ (t)2 dt) . (11)

So we obtain

∫ 1

0f ′ (t)2 dt =

23

+13c

h (c) h′ (c) . (12)

By putting (12) in (10), we obtain

V0 (c) =4

3c2 (1 −1c

h (c) h′ (c)) . (13)

By differentiating (13) once and twice with respect to , and by using(9) and its derivative to eliminate , we obtain

ch″ (t)

V′0 (c) =4c4

h′ (c) (h (c) − ch′ (c)) , (14)

V″0 (c) = −16c5

h′ (c)(h(c) − ch′ (c)) −8c4

h(c)3(h(c) − 2ch′ (c)). (15)

When , the volume is not maximal because. When , the volume is locally

maximal because . So the volume takes the maximumvalue when is the first positive zero of , say . Thus the maximalvolume is expressed by .

ch′ (c) = h(c) V0 (c)V″ (c) > 0 h′ (c) = 0 V0 (c) = (4 / 3) c−2

V″0 (c) < 0 V0 (c)c h′ α

V0 (α) = (4 / 3) α−2

Since is a first positive zero of , the function is monotoneincreasing on and . By separating the variables of (9), weobtain

α h′ h[0, α] h (α) = 1

α = ∫ 1

0

dh

1 − h4= ∫

π/2

0

dθcos2 θ + 2 sin2 θ

=π / 2

agm(1, 2), (16)

where . The proof of the last equality of (16) is found in [5].h = sinθBy separating the variables of (9), we obtain

u = ∫0

dh

1 − h4= ∫0

dφ1 − 2 sin2 φ

, (17)

where . So we obtain .h = tanφ h (u) = tan(am(u, 2))

NOTES 113

Remark. The function is equal to the lemniscate function, and the value is a quarter of the length of the lemniscate. So, by using the same

notations as [3], we can rewrite the equation (1) in the solution as

h (u)α

f (t) = α−1 sinlemn(αt) , α = ω

/ 2. (18)

The author thanks the referees for their valuable comments.

References1. N. Lord, The folding box problem, Math. Gaz. 74 (1990) pp. 361-365.2. N. Reed, A curved folding box, Math. Gaz. 76 (1992) pp. 275-277.3. E. T. Whittaker and G. N. Watson,A course of modern analysis (4th

edn.), Cambridge University Press (1927, reprint 1996).4. R. Courant and D. Hilbert,Methods of mathematical physics, Vol. I,

Interscience (1953, reprint John Wiley 1989).5. N. Lord, Recent calculations of : the Gauss-Salamin algorithm,Math.

Gaz. 76 (1992) pp. 231-242.π

KENZI ODANIDepartment of Mathematics, Aichi University of Education,

Kariya-shi, Aichi 448-8542, Japane-mail: [email protected]

84.24 More on a sine product formulaIn [1], Scott made the following conjecture, which he proved by

induction for positive integers of the form and .2n 3 × 2n

sin ( π2n) sin (2π

2n) sin (3π2n) … sin ((n − 1) π

2n ) = n1/22−n + 1.

We shall prove this conjecture without the use of induction. First, notethat, by the symmetry of the sine graph, this is equivalent to the statement

sin ( π2n) sin (2π

2n) sin (3π2n) … sin ((2n − 1) π

2n ) = n.2−2n + 2.

This now follows immediately from a slightly more general result.

TheoremFor any whole number greater than 1,m

sin ( πm) sin (2π

m ) sin (3πm ) … sin ((m − 1) π

m ) = m.2−m + 1.

ProofLet us write for the complex th root of unity given by .

The expression on the left-hand side is therefore equal toω 2m ω = eiπ/m


(ω − ω−1

2i ) (ω2 − ω−2

2i ) … (ωm− 1 − ω−m + 1

2i )= (1 − ω−2) (1 − ω−4) … (1 − ω−2m+ 2) ωm(m− 1)/2i−m + 12−m + 1.

Since , it remains to prove that. Consider the polynomial in an indeterminate given by

. This has roots equal to all the complexth roots of unity except 1. Since its first term is equal to , the

polynomial must coincide with

ωm/2 = i (1 − ω−2)(1 − ω−4)… (1 − ω−2m+2)= m X(X − ω−2)(X − ω−4)… (X − ω−2m+2)m Xm− 1

Xm − 1X − 1

= Xm− 1 + Xm− 2 + … + X + 1

which has these same roots. We take equal to 1 to complete the proof.X

Remark 1. Via we getsin kπ2n = eikπ/2n (1 − e−ikπ/n) / 2i

∏n − 1

k = 1

sinkπ2n

=1

(2i)n − 1ei π

2n(1 + 2 + … + (n − 1)) ∏n − 1

k = 1

(1 − e−ikπ/n) ,

i.e. (due to )1 + 2 + … + (n − 1) = 12n (n − 1)

∏n − 1

k = 1

(1 − e−ikπ/n) = in − 1e−i(n − 1)π/4 n = ei(n − 1)π/4 n.

Finally, conjugation yields

∏n − 1

k = 1

(1 − eikπ/n) = n (cos(n − 1) π

4− i sin

(n − 1) π4 ) .

Remark 2. The above proven formula conjectured by Scott is not new. Itcan be found in various tables, alongside many similar formulas such as

,sinπ

2n + 1sin

2π2n + 1

… sinnπ

2n + 1=

2n + 12n

cosπ2n

cos2π2n

… cos(n − 1) π

2n=

n

2n − 1,

cosπ

2n + 1cos

2π2n + 1

… cosnπ

2n + 1=

12n

,

but also

cot2π

2n + 1+ cot2

2π2n + 1

+ … + cot2nπ

2n + 1=

n (2n − 1)3

,

cosec2π2n

+ cosec23π2n

+ … + cosec2(2n − 1) π

2n= n2,

or even quite exotic ones as

.∑n

k= 1

cotxk ∏n

j =1j ≠k

cot(xk − xj) = sin(n + 1)π

2+ (−1)n+1∏

n

k= 1

cotxk

See, for example, [2, ch. 44].

NOTES 115

Remark 3. There are occasions when the inductive stepdoes prove useful, namely when it is obvious that for

. There is a standard proof of the AM-GM inequality that uses thisapproach.

P(n) ⇒ P(2n)P(n) ⇒ P(m)

m < n

References1. J. A. Scott, A conjecture and an unusual kind of inductive step,Math.

Gaz. 82 (July 1998) p. 277.2. A. P. Prudnikov et al,Integrals and series (elementary functions) [in

Russian], Nanka, Moscow (1981).WALTHER JANOUS

Ursulinegymnasium, Fürstenweg St, A 6020 Innsbruck, AustriaJEREMY KING

Tonbridge School, Tonbridge TN9 1JP

84.25 On a limit for prime numbersIn a recent Gazette article [1], the modified geometric mean

Q (n) = [a(1) a(2) … a (n)]1/a(n) (a (n) > 0)was considered for the sequence of prime numbers by Ruiz, who invokedthe Prime Number Theorem and the Stirling approximationfor to obtain the limit

p(n) ∼ n lognn!

Q = limn → ∞

Q (n) = e.

This limit may also be obtained by using the theorem of meansinstead of the Stirling formula, that is, by developing the inequality

H < G < A

n

∑ a(i)−1

n/a(n)

< Q (n) <∑ a(i)

n

n/a(n)

.

Let with (say) so thatby the Prime Number Theorem. Then we have, for ,

a (n) = n logn (n > 1) a (1) = log 2 a (n) ∼ p(n)n > 1

n1/ logn

1log2

+ ∑n

i =2

1i logi

−1/ logn

< Q(n) < n−1/ lognlog2+ ∑

n

i =2

i logi

1/ logn

.

Replacing by in the first expression and by in the last giveslog i log 2 logn

(n log2)1/ logn1 + ∑

n

i =2

1i

−1/ logn

< Q(n) < logn∑

n

i =1

i

n

1/ logn

.

Next, since , (Euler), and

if

n1/ logn = e ∑n

i = 2

1i

< logn ∑n

i = 1

i =12n (n + 1)

12 (n + 1) ≤ n n ≥ 1

log 2

1 + logn

1/ logn

<Q (n)

e< (logn)1/ logn .


Now denote by . Thenn1/n b (n)

(b (n + 1)b (n) )n(n + 1)

=(1 + 1

n)n

n< 1 n > 2,for

and clearly decreases to limit 1. A fortiori, for a positiveconstant. Thus follows.

b (n) a1/n → 1 aQ = e

It is interesting to note that the simpler expression

R = lim R(n) = lim [a(n)]n/a(n)

also has the limiting value for the sequence of prime numbers. Indeed, if where with , then

ea (n) = nf(n) f (n) → ∞ n

R = lim (nf(n))1/f (n) = lim n1/f (n)

whence ( ).f (n) ∼ logn / logR R > 1Note that the relationship between the limits and seems reminiscent

of ‘convergence in the mean’, that is, implies (see [2]).

Q Rsn → $

(s1 + s2 + … + sn) / n → $Finally, since for the convergent series where

or , but also for the divergent series givenby , it seems plausible to conjecture that issufficient for the divergence of the series of positive terms where

with .

R = 1 ∑ 1 /a (n)a (n) = np + 1 n (logn)p + 1 (p > 0)

a (n) = n logn log(logn) R > 1∑ 1 /a (n)

a (n) → ∞ n

References1. S. M. Ruiz, A result on prime numbers,Math. Gaz. 81 (491) p. 269

(1997).2. J. C. Burkill, A first course in mathematical analysis, Cambridge

University Press (1970).J. A. SCOTT

1 Shiptons Lane, Great Somerford, Chippenham, Wiltshire SN15 5EJ

84.26 SHM and projectionsMy first encounter with simple harmonic motion (SHM) was in the

motion of the projection of a point moving at constant speed in a circle ontoa diameter of the circle. If the circle has unit radius then the projectionvaries as , where is the angular speed of the point. Although thisapproach avoids studying differential equations, it is not an example fromwhich one can construct a simple physical demonstration to observe thevisual qualitative features of SHM. (Recall that, if a point moves atconstant (angular) speed in a circle of radius and centred onthe origin, then the projection onto the -axis satisfies

sinωt ω

Pω = dθ / dt R

x

x = Rcosθ, dx

dt= −Rsinθ

dθdt

= −Rω sinθ,

d2x

dt2= −Rω cosθ

dθdt

= −ω2Rcosθ = −ω2x.)and

NOTES 117

In this note I describe a common situation from which it is possible to makesuch observations. The example is again a projection and one that is morenatural and observable. I begin by reviewing the classical example oflinearSHM, by which I mean simple harmonic motion in a straight line.

A supported mass oscillating on the end of a spring exhibits SHM(assuming no resistance). The differential equation governing theseoscillations is

d2x

dt2= −

λml

x (1)

where represents the displacement of the mass, and and denote themass, and the modulus of elasticity and unextended length of the spring,respectively, and the solution is sinusoidal. Unfortunately, it is not easy toappreciate the qualitative features of SHM using such a model.

x m, λ l

An alternative example is the motion of a simple pendulum whosegoverning differential equation is

d2θdt2

= −g

lsinθ, (2)

where represents the angular displacement of the pendulum, anddenote the acceleration due to gravity and the length of the pendulum,respectively. Strictly speaking, the pendulum displays only approximateSHM. For small amplitudes and the differential equation (2) canbe approximated by

θ g, l

sinθ ≈ θ

d2θs

dt2= −

g

lθs, (3)

which is analogous to (1). The quantity denotes the small angleapproximation. The principal error made in making this approximation isthat the period, and hence the frequency, of the pendulum depends on theamplitude, in contrast to exact SHM, where it does not. Figure 1 shows thegraphs of the angular displacement for the pendulum as governed by (2),and the small angle approximation governed by (3). Figure 1 shows thatdespite the phase error in this approximation, the motion as observed issufficiently close to SHM to display the essential qualitative features.Unfortunately, it is difficult to appreciate the properties oflinear SHM asgoverned by (1), (even ignoring the fact that this is only an approximation),since it is theangular displacement that is exhibiting (approximate) SHM.One could observe the circular path traversed by the pendulum bob whosearc length is , but this is still not motion in a straight line. If one observesthe projection of the pendulum bob on a horizontal surface, however, thenan approximation to linear SHM is observed as I now show.

θs

θθs

lθ

The projection on the horizontal is and this variableexhibits approximate linear SHM. To see this observe that

, and already exhibits approximate SHM, or from theoriginal differential equation (2) with then ,so again exhibits approximate SHM.

x = l sinθ

x = l sinθ ≈ lθ θsinθ ≈ θ d2

dt2 (sinθ) = −gl sinθ

l sinθ


0.5 1.0 1.5 2.0 2.50

0.5

1.0

1.5

θs

θ

−1.5

−1.0

−0.5

FIGURE 1

From an observation point vertically above a pendulum one thereforesees (as a projection) an approximation to linear SHM. In Figure 2 theprojection value has been superimposed on the graph of shown inFigure 1, where is the exact angular displacement governed by (2).Similarly, in Figure 3, the projection value has been superimposed on

sinθ θθ

sinθs

0.5 1.0 1.5 2.0 2.50

0.5

1.0

1.5

−1.5

−1.0

−0.5

θ

sinθ

FIGURE 2

the graph of shown in Figure 1, where is the angular displacementdetermined by the small angle approximation governed by (3). In eithercase, it can be seen that the qualitative features of SHM are observed. (Asuperimposition of the graphs in Figure 2 onto those in Figure 3 confirms

θs θs

NOTES 119

this, although the graphs clearly become somewhat tangled.) Moreover,these features are observedregardless of the size of the amplitude. Theexamples given here have amplitudes of the order of .1 rad ≈ 57°

0.5 1.0 1.5 2.0 2.50

0.5

1.0

1.5

θs

−1.5

−1.0

−0.5

sinθs

FIGURE 3

So, where is the best place to observe this phenomenon? The idealplace is a playground when the sun is high in the sky. The linear movementof the shadow of a swing on the ground is sufficiently slow and exaggeratedto appreciate the qualitative features of linear SHM. It should be stressedthat this is only an approximation in the sense described above. Thetraditional mechanism for doing this is a pendulum with a pen inscribing atrace or an orifice letting out fine sand onto paper moving at constant speed.An alternative is to simulate linear SHM with a trace on a PC by integratingequations (2) or (3) and plotting the displacement against time, as shown inthe figures above. However, for me the playground seems much more fun!

P. GLAISTERDepartment of Mathematics, University of Reading RG6 2AX

84.27 Another cautionary chi-square calculationAlthough not as unsettling as my previous foray into this area [1], the

example discussed in this note highlights a point which tends to be passedover in introductory statistics texts. Consider the problem of testing whethera binomial distribution fits the observed data in the frequency tablebelow:

B(2, p)

x 0 1 2

Observed frequencyof (total 150)x

90 45 15

Conventional A level wisdom estimates by matching the observed andp


expected means: so that . Expected frequencies ( ) bothfor and are shown below, together with their correspondingvalues of .

2pˆ = 0.5 pˆ = 0.25 Epˆ p = 0.255

χ2

x 0 1 2 χ2

E (pˆ = 0.25) 84·38 56·25 9·38 6·00E (p = 0.255) 83·25 56·99 9·75 5·89

Since the 5% critical value for with 2 degrees of freedom is 5.991 wewould, at the 5% level, reject the ‘conventional wisdom’ value of butaccept the rival value . (This conclusion is unaltered by the factthat the test for is usually performed with 1 degree of freedom with thelower 5% critical value of 3.841.)

χ2

pp = 0.255

χ2

The moral of this example is that, although is a sensibleestimator of (indeed, it is the maximum likelihood estimator), there is noreason to suppose it will also be the minimum-chi-square estimator of(which in fact is 0.25786 with ). (See [2; chap. VII] for anextensive discussion of various types of estimator.)

pˆ = 0.25p

pχ2 = 5.876

It is worth noting that, now that values of the distribution function oftest statistics such as are available at the touch of a button, modernpractice among statisticians, though less so with scientists, is to quote exactprobabilities rather than use fixed significance levels, such as 5%. Ourseemingly dramatic contrast between (significant at the 5%level) and (not significant) evaporates in the newer vocabularywhere the probabilities are and respectively.

χ2

χ2 = 6.00χ2 = 5.89

0.0498 0.0526It is also worth pushing our original example further: if the observed

frequencies for 0, 1, 2 are , with , then , thevalue of arising from fitting , is given by:

a0, a1, a2 N = a0 + a1 + a2 f (p)χ2 B(2, p)

f (p) =a2

0

N (1 − p)2+

a21

2Np(1 − p)+

a22

Np2− N.

A little algebra shows that has a unique minimum on (0, 1) and that, with, has the same sign as .

fpˆ = (a1 + 2a2) / 2N f ′ (pˆ ) (a2 − a0) (a2

1 − 4a0a2)2

It follows that, apart from the intriguing special case where , is less that the minimum-chi-square estimator of when ; greater

when , and only coincides when or when .

a21 = 4a0a2

pˆ p a0 > a2

a0 < a2 a0 = a2 a21 = 4a0a2

It is a pleasure to acknowledge the perceptive suggestions of a refereewhich substantially improved the initial draft of this note.

References1. Nick Lord, A chi-square nightmare, Math. Gaz. 76 (July 1992) p. 274.2. A. M. Mood, F. A. Graybill, D. C. Boes,Introduction to the theory of

statistics (3rd edn.), McGraw-Hill (1974).NICK LORD


NOTES 121

84.28 More on dual Van Aubel generalisationsIn an article [1] by myself, the following two dual generalisations of

Van Aubel's theorem were presented:

Theorem 5If similar rectangles with centres and are erected externally

on the sides of quadrilateralABCD, then the segmentsEG andFH lie onperpendicular lines. Further, if and are the midpoints of thedashed segments shown, thenJL and KM are congruent segments,concurrent with the other two lines.

E, F, G H

J, K, L M

Theorem 6If similar rhombi with centres and are erected externally on

the sides of quadrilateralABCD as shown, then the segmentsEG andFH arecongruent. Further, if and are the midpoints of the dashedsegments shown, then JL and KM lie on perpendicular lines.

E, F, G H

J, K, L M

I am indebted to Hessel Pot from Woerden in the Netherlands whorecently, in a personal communication to me regarding these generalisations,pointed out that to Theorem 5 we can also add the following two properties:(a) the ratio of EG and FH equals the ratio of the sides of the rectangles.(b) the angle ofJL and KM equals the angle of the diagonals of the

rectangles.and to Theorem 6 the following corresponding duals:(a) the angle of EG and FH equals the angle of the sides of the rhombi(b) the ratio of JL and KM equals the ratio of the diagonals of the rhombi.

In fact, the latter four properties are contained in the following self-dualgeneralisation, which can be proved by using vectors or by generalising theapproach used in [1]:

Theorem 7If similar parallelograms with centres and are erected

externally on the sides of quadrilateralABCD as shown in Figure 1, then, and the angle ofEG andFH equals the angle of the sides of the

parallelograms. Further, if and are the midpoints of the dashedsegments shown, then , and the angle ofJL and KM equals theangle of the diagonals of the parallelograms (see Figure 1).

E, F, G H

FHEG = XY

XBJ, K, L M

KMJL = YA

XB

Reference1. M. de Villiers, Dual generalisations of Van Aubel's theorem,Math. Gaz.

495 (November 1998) pp. 405-412.


X

C

DA

B

Y

E

J

K F

M

L

H

GV U

FIGURE 1

MICHAEL de VILLIERSUniversity of Durban-Westville, South Africa

e-mail: [email protected]://mzone.mweb.co.za/residents/profmd/homepage.html

OBITUARY 123

ObituarySir Wilfred Cockcroft 1923-1999

Although Bill Cockcroft was by training a professional mathematician,it will be as a larger than life character who also made his mark in educationthat he will be remembered. But, first, a brief outline of his life.

Born the son of a plumber in Keighley, he attended Keighley GrammarSchool before going up to Balliol College in Oxford. The second world war,in which he served in the RAF in the far east, interrupted his career. He thenlectured in Aberdeen and Southampton before obtaining a chair at Hull in1961. His next move was to establish, as Vice-Chancellor, the NewUniversity of Ulster in Coleraine in 1976. He left Northern Ireland in 1982and, in 1983, became the chairman and chief executive of the newSecondary Examinations Council in London which, by the time he retired in1988, had developed the new GCSE. He was knighted in 1983, an event notunconnected with his Committee of Inquiry into the teaching ofmathematics which resulted in the highly praised reportMathematics Countspublished in 1982.

Although born Wilfred Halliday Cockcroft, he was always known asBill at home and later by everyone who knew him. The first impressionwhen meeting Bill Cockcroft was of a bluff Yorkshire man. He was a bigman, larger than life, friendly and abon viveur. He loved conversation,music, theatre, food and drink He always had a fund of amusing, and usuallytrue, stories. His flamboyant style carried with it a unique Falstaffianrumbustiousness. We remember the West Riding bellow of his voice, thegales of his laughter and his guffaws. Conversations were intense, full ofpauses, and the variety of ways he had for saying ‘I don't know’, expressingsorrow, disgust, ignorance, mystification and much else in different tones ofvoice. If you knew him at the height of his powers, you would marvel at hisstamina for work and play, notably his legendary capacity for strong drink.Such characteristics made him a vital presence in any gathering, political,academic or domestic.

Bill was totally unpompous and without malice. He made friendshipsunconstrained by class, race or wealth. He is probably the only Vice-Chancellor to play golf with the Head Porter and the Estates Manager of hisuniversity. Apocryphally, the fourth player was his driver with special dutiesfor the golfcart. On these forays at Portstewart Golf Club, it was alleged,Danny McGryllis's task was to shepherd enough bottles of Guinness aroundthe course to enable the winner of the hole either to be toasted at the point ofhis victory or to enjoy a bottle in solitary celebration.

But teachers will remember him forMathematics Counts which Bill,mischievously but not seriously, suggested should be entitledA Feeling ForFigures. For me, it was an exhilarating experience to be a member of thecommittee of inquiry. When the Secretary of State for Education, ShirleyWilliams, set up the committee in 1978, it was assumed that a quick reportwould ensue but it was two Secretaries of State later when Keith Joseph


received the report. Bill was unflappable. The government had to wait untilhe felt that the shape of the report was clear and that there was evidence tosupport all the statements. The rapport engendered within the committeewas such that, to mark each anniversary of the completion of the report, areunion was held at an Italian restaurant in London with, of course, largequantities of wine!

Members of the Cockcroft committee were allowed to discuss at lengthuntil a consensus was reached. It was rare to have a vote. It was interestingthat Keith Joseph was so completely satisfied with the academic rigour andpersuasive evidence that, from the publication of the Cockcroft Report inearly 1982, we saw a resurgence of confidence in the teaching ofmathematics. We had a blueprint which was widely accepted as the wayforward and the funding, including the advisory teachers known as‘Cockcroft missionaries’ and a significant boost to the Association'sDiploma courses, to help carry it out.

Again, Bill was to influence mathematics teachers when he oversaw theintroduction of the new GCSE. The Cockcroft Committee had, based onwork at Chelsea College and in particular Kath Hart's CSMS project,become convinced that the curriculum should be ‘bottom-up’, building onthe Foundation List rather than the traditional watering down of acurriculum for the most able. A particular concern, over which Sir KeithJoseph agonised, was the least able 40% who often left school with littlepositive achievement to show.

The principle enshrined in the bottom-up approach is that all pupilsshould be successful and have a sense of achievement. Too often pupilsobtain a ‘certificate of success’ knowing that they did not answer correctlymost of the questions on the examination paper. The notion of competencewas becoming important, for example, in reassuring an employer that acertificate was at least a minimal guarantee of attainment. The SECcontinued this work by attempting to identify the criteria of ‘Success’ atparticular grades.

Bill was such a colourful and energetic character that many more pagescould be filled by this obituary. I hope that others will write to the editorwith further tales of the unique Cockcroft qualities. He is sorely missed byeveryone that knew him.

AcknowledgementSome of the above material is taken from a memorial talk given at his

funeral by Hugh Sockett, a friend and colleague from Bill's days inColeraine.

PETER REYNOLDS6 Rosebery Road, Felixstowe IP11 7JR

CORRESPONDENCE 125

CorrespondenceDEAR EDITOR,

In ‘Moving the first digit of a positive integer to the last’ (Math. Gaz.83pp. 216-220), Braza and Tong's treatment of the problem is impressive, butthey fail to observe that there is an easy way to calculate by hand numberssuch that moving the first digit to the last is equivalent to a division.

Choose the first digit, say 8, and the divisor, say 4. We proceed usingthe usual paper method for division; but at each step the latest digit of thequotient provides the next digit for the dividend (together with any carrydigits).

→2 0 5

4 80 22 00

2 0 5 1

4 80 22 00 51

The process terminates when the latest digit of the quotient equals thefirst digit chosen and there is no remainder outstanding. The given examplewill terminate at 820512, with quotient 205128.

This method shows why there can only be one basic answer for a givenfirst digit. and why all other answers are concatenations of the basic answer.

Yours sincerely,JEREMY KING


DEAR EDITOR,I offer a couple of observations on items in the excellent July 1999

Gazette.1. Readers of Robert J. Clarke's article might also be interested in adiagram described by Ravi Vakil in his lively bookA mathematical mosaic

2

1T

x

x

y

y

x + y3

45°

30°

x − y

(enthusiastically reviewed by Andre Toom inthe January 1998American MathematicalMonthly—‘This is a book I would have like tohave read as a boy’). Page 87 features whatVakil calls the Ailles Rectangle, named afteran Ontario High School Teacher, Doug Ailles.

Here, from the 45°-triangles, and areimmediately seen to be and sothat the trigonometric ratios for 15° and 75°can be read off from triangle .

x y3 / 2 1 / 2

T

2. An alternative proof that J. A. Scott's recalcitrant seriesconverges for runs as follows: Fix a natural number with

, so that . Write with .Then, for all , we have:

∑ (n1/n − 1)p

p > 1 kk > p/ (p − 1) p > k/ (k − 1) n1/n = 1 + an an ≥ 0

n ≥ 2k


n = (1 + an)n > ( ) akn

nk

(ignoring all other terms ofthe binomial expansion)

= ∏k − 1

i = 0

n − i

1 + i ak

n

>(n

2)k

kk ak

n

and so .an <2k

n(k − 1)/k

Thus and converges by comparison with which is convergent because .

apn < (2k)p / np(k − 1)/k ∑ ap

n

∑ 1 /np(k − 1)/k p(k − 1) / k > 1

Keep up the fine editorial work!Yours sincerely,

NICK LORDTonbridge School, Tonbridge TN9 1JP

DEAR EDITOR,Themagic rectangles discussed in a recent Note [1] by Marián Trenkler

have a longer history than indicated there. Harmuth, in 1881, published twopapers [2, 3], establishing necessary and sufficient conditions for theexistence of magic rectangles in just this sense, that is, an arrangement ofthe integers 1, 2,… into an by rectangle where columns have thesame sum, as do rows (the natural generalisation of magic squares).

mn m n

Magic rectangles also have a more current interest than might begathered from [1]. For example, as with magic squares, they have foundsome favour in statistics; see [4] for a digest of the statistical uses of magicsquares, and [5] for some statistical work involving magic rectangles.Indeed, magic rectangles appeared in thisGazette in [6], in 1968, with anapplication of this sort in mind. The construction of magic rectanglescontinues to attract attention in research journals, as [7, 8, 9] attest.

Yours sincerely,D. G. ROGERS

Halewood Cottage, The Green, Croxley Green WD3 3HT

References1. M. Trenkler, Magic rectangles, Math. Gaz. 83 (1999) pp. 102-105.2. T. Harmuth, Über magische Quadrate und ähnliche Zahlenfiguren,

Arch. Math. Phys. 66 (1881) pp. 283-313.3. T. Harmuth, Über magische Rechteche mit ungeraden Seitenzahlen,

Arch. Math. Phys. 66 (1881) pp. 413-417.

CORRESPONDENCE 127

4. G. H. Freeman, Magic square designs,Encyclopedia of statisticalsciences, Vol. 5, (Wiley, New York, NY, 1985) pp. 173-174.

5. J. P. N. Phillips, Methods of constructing one way and factorial designsbalanced for trend, Appl. Statist., 17 (1968) pp. 162-170.

6. J. P. N. Phillips, A simple method of constructing certain magicrectangles of even order, Math. Gaz. 52 (1968) pp. 9-12.

7. T. Bier and A. Kleinschmidt, Centrally symmetric and magicrectangles, Discrete Math 176 (1997) pp. 29-42.

8. T. Bier and D. G. Rogers, Balanced magic rectangles,European J.Combin. 14 (1993) pp. 285-299.

9. M. A. Jacroux, A note on constructing magic rectangles,Ars Comb.36(1993) pp. 335-340.

Notices

On pp. 123-124 of this issue, you will see an obituary to Wilfred H.Cockcroft, a notable mathematician and figure in the mathematicalcommunity. Sadly, we have recently received notice a number ofother deaths that we feel it appropriate to record.

The Association has lost three of its past Presidents in the recentpast. Sir William McCrea (President from 1973−74), Bertha Jeffreys(President from 1969-70) and, on January 31st 2000, Mary Bradburn(President from 1994-95). Each of these contributed to the centenaryissue of theGazette (March 1996) and readers will find briefbiographical notes in that issue. There will be obituaries published inthe Gazette in due course.

In addition, theGazette itself has lost three overseas contributors.Folke Eriksson of Chalmers and Gotheburg University, Sweden, whowrote and refereed articles over several years, died in August 1999.Andrejs Dunkels of Luleå University of Technology, Sweden, who isthe author of Note 84.06 in this issue, died in December 1998.Finally, we have lost an extremely regular and long-servingcontributor with the passing in February 2000 of Dr. S.Parameswaran. His firstGazette article appeared in May 1946 andhe has recently provoked considerable interest in S·P numbers.


Problem CornerSolutions are invited to the following problems. They should be

addressed to Graham Hoare at 3 Russett Hill, Chalfont St. Peter, Bucks SL98JY and should arrive not later than 10 August 2000.

84.A (J. D. King)A trough of water with parabolic

cross-section is gently rolled over. Showthat the centre of mass of the waterremains a constant distance from , thepoint of contact.

P

P

84.B (Nick Lord)

(a) Find all rational numbers between 0 and 1 for which is an

integer. [One example, known to the Bernoullis, is , whichyields a sum of 26.]

x ∑∞

1

n3xn

x = 12

(b) Investigate the analogous question for ( is a fixed whole

number).

∑∞

1

nkxn k

84.C (C. F. Parry) is a scalene triangle with circumcentre and orthocentre . ,

are the reflections of the vertices in the corresponding sides ,, . If and meet at , then show that

ARC O H A′, B′C′ A, B, C BCCA AB B′C BC′ D(i) is a cyclic quadrilateral,OBCD(ii) is parallel to the Euler line .DA′ OH

84.D (Mark Stamp and Ruth Arumula)Consider the following map of consecutive one-way ‘roundabout’

roadways.n

finishstart . . .

Suppose that a car enters at the start, travels counterclockwise and eachtime the car can make a turn, a coin is flipped to decide which way to go. Awrong turn is any turn that results in a longer trip than necessary. What isthe expected number of wrong turns before the car arrives at the finish?

PROBLEM CORNER 129

Solutions and comments on 83.E, 83.F, 83.G, 83.H (July 1999).83.E (Nick Lord)

Let be a smooth,increasing function with

and . Characterise those

spirals with the ‘GP-property’,i.e. wherever the radial angle increases byconstant amounts, the corresponding areasbetween successive whorls of the spiralform a geometric progression : in thediagram, for constant , are inG. P.

f : R → Rlim

θ → −∞f (θ) = 0

limθ → ∞

f (θ) = ∞r = f (θ)

α A1, A2, A3

ααα

A1

A2

A3

Firstly, we require a function which maps Arithmeticprogressions to Geometric progressions. This is furnished by the functionalequation . Differentiating with respect to and thenwith respect to yields . This gives

f : r → r

f (x + y) = f (x) f (y) xy f ′ (x + y) = f ′ (x) f (y) = f (x) f ′ (y)

f ′ (x)f (x)

=f ′ (y)f (y)

= c (const),

so . Accordingly, we shall set the area function. Thus . So

f (x) = f (0)ecx A(θ) = a2e2k2θ

(a, k ≥ 0) ∫ 0−∞ f 2 (t) − f 2 (t − 2π) dt = 2a2e2k2θ

f 2 (θ) − f 2 (θ − 2π) = 4a2k2e2k2θ.In general

f 2 (θ − (2n − 2) π) − f 2 (θ − 2nπ) = 4a2k2e2k2(θ − (2n − 2)π), n ∈ z.Summing up, and noting that , we derivelim

θ → −∞f (θ) = 0

f 2 (θ) = 4a2k2l2k2θ (1 + e−4k2π + e−8k2π +… ) .

Hence, . (Since as , we have

). We conclude that the spirals we seek are necessarily logarithmic(= equiangular) with equation .

f (θ) =2akek2θ

1 − e−4k2πf (θ) → ∞ θ → ∞

a, k > 0r = Aeaθ

Correct solutions were also received from J. K. R. Barnett, J. M. Chick, R. P. C. Forman,M. Griffiths, G. Howlett, J. D. King, D. F. Lawden, S. N. Maitra, N. A. Routledge, I. F. Smith,H. B. Talbot, R. F. Tindall, G. B. Trustrum and the proposer, Nick Lord.

83.F (A. Robert Pargeter) is a quadrilateral such that is parallel to and .

Prove that .ABCD AD BC ∠ACD= ∠ABC

∠DBC ≤ 30°The preferred approach exploited trigonometry, often allied with


calculus. First we give the solution(s) of M. D. Fox.

c

A Q D

bS

Rh

TCa

PB

zx y a

aaa

bb

g

g

Given: .AD // BC, ∠ABC = ∠ACDDraw , and .AP, QC, DT⊥BT AR, QS⊥CDTo show that we shall prove the equivalent result that

.∠DBC ≤ 30°

BT2 ≥ 3h2

We need the following simple results:1. For all real , ⇒

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

x2 + y2 + z2 ≥ yz + zx + xy2. s are similar .ABC, DCA ⇒ bc = a × CD

3. In : .ABC ccosα + a cosγ = b4. s are similar .CDQ, CQS ⇒ CD × CS = CQ2

Then

BT2 = (x + y + z)2 = x2 + y2 + z2 + 2(yz+ zx+ xy) ≥ 3(yz+ zx+ xy) (by 1)

= 3(y × CDcosα + CDcosα × ccosβ + ccosβ × bcosγ)

= 3CD(ycosα + [ccosα + acosγ] cosβ (by 2)

= 3CD(RS+ bcosβ) (by 3)

= 3CD(RS+ CR) = 3CD× CS= 3CQ2 = 3h2 (by 4)

So , and .BT2 ≥ 3h2 ∠DBC ≤ 30°

We can do the same sort of thing by trigonometry, with the help of 1above. We need also:5. If , thenα + β + γ = 180°

cotγ = − cot(α + β) =(1 − cotα cotβ)(cotα + cotβ)

Let , then , and .Thus

∠DBC = ω BT = hcotω x + y + z = h(cotβ + cotγ + cotα)

PROBLEM CORNER 131

cot2ω = (cotα + cotβ + cotγ)2 ≥ 3(cotβcotγ + cotγ cotα + cotα cotβ)= 3[(cotα + cotβ) cotγ + cotα cotβ]= 3(1 − cotα cotβ + cotα cotβ) = 3 (by 5).

Therefore , so .cotω ≥ 3 ω ≤ 30°These results are closely related to the Brocard geometry of triangle, which has Brocard angle , one of its Brocard points being the

intersection of with the circumcircle of . It is (or perhapswas) astandard result that the Brocard angle cannot exceed 30°.

ABC ωDB DCA

For those who sought a purely geometrical solution without success, weoffer John Rigby's contribution.

Lemma. Let be a triangle whose height (the distance from to ) is, in which and . Then there exists a triangle

with , either of whose base angles or isequal to ; the height of this triangle is greater than or equal to .

PQR P QRh ∠PQR ≥ 60° ∠PRQ ≤ 60°P′QR ∠P′ = ∠P ∠P′QR ∠P′RQ

60° h

Q

P

P' M

R

Q

P

P' M

R

FIGURE 1(a) FIGURE 1(b)

The essence of the proof is indicated in Figure 1(a) or 1(b), dependingon whether the equal base angles of the isosceles triangle are less than

or greater than (the special case when is equilateral is easilydealt with); lies on the circle between and . We can obtain a triangle

in which the other base angle is by reflecting in thediameter through .

MQR60° 60° MQR

P′ P MP″QR 60° P′QR

MBy considering a triangle similar to or , but with height

, we obtain a corollary.XYZ P′QR P″QR

h

Corollary. Let be a triangle with height , in which and. Then there exists a triangle with height , with

, and with either of the base angles or equal to ;in this triangle .

PQR h ∠PQR ≥ 60°∠PRQ ≤ 60° XYZ h∠P = ∠X ∠XYZ ∠XZY 60°

YZ ≤ QR


A A' D' D

a

g

b

a

g b

a

60B C C'B'

FIGURE 2

Now let be a quadrilateral with parallel to and the anglesmarked equal, as in Figure 2. Since the angles are equal, then so are theangles . Since , , cannot be all greater than, or all less than, , wemay assume without loss of generality (since and play identical roles inthe figure) that one of is less than or equal to , the other greater thanor equal to . By the corollary there exists a triangle , as shown inFigure 2, with , , and . Then

, because triangles and are similar. Applying thecorollary to triangle , we obtain an equilateral triangle with

. Then . (The proofremains valid whatever the values of , and , although the relative orderof some of the points on the parallel lines will vary.)

ABCD AD BCβ γ

α α β γ 60°α β

β, γ 60°60° A′B′C

∠A′ = α ∠B′ = 60° B′C ≤ BC∠DCA′ = 60° DCA′ A′B′C

CDA′ C′D′A′D′A′ ≤ DA′ ∠DBC ≤ ∠DB′C ≤ ∠D′B′C = 30°

α β γ

Correct solutions were received from R. G. Bardelang, J. K. R. Barnett, R. L. Bolt,J. M. Chick, H. Martyn Cundy, R. P. C. Forman, S. Fowlie, M. D. Fox, M. Griffiths,G. Howlett, P. F. Johnson, J. D. King, D. F. Lawden, G. Leversha, N. Lord, S. N. Maitra,J. A. Mundie, C. F. Parry, J. Rigby, N. A. Routledge, I. F. Smith, H. B. Talbot, K. Thomas,R. F. Tindall, G. B. Trustrum and the proposer, A. Robert Pargeter.

83.G (Christian Turcu)The trapezium is given with and . is

the mid-point of , and lies on the line . Prove that:ABCD AB// CD AB = AC + CD E

BD BF // CE F AC(a) and are perpendicular to .AE DF BF(b) is the intersection of the angle bisectors of triangle if and only if

is perpendicular to .C DEFDA AB

(c) if and only if .EF // AD AB = 3.CD

A

M

E

C

F

BL

D

q

q

q

q

PROBLEM CORNER 133

Judicious constructions were the key to solving this problem. Althougha few used vectors, solvers predominantly favoured Euclid. H. MartynCundy's solution is selected to reflect this approach.

Given: parallel to , .AB DC |AB| = |AC| + |CD|The figure must clearly be as shown. Locate on such that

. Then, from the data, . is therefore aparallelogram, and will contain , the midpoint of . Let meetat .

L ABAL = AC LB = CD LBCD

LC E BD LC DFM

(a) Since and , is perpendicular to and thereforeto . The triangles , being similar, , and

. So ,say. Hence is parallel to and perpendicular to and .Therefore is the midpoint of , and bisects the angle

.

CE = EL AC = AL AE CLBF ALC ABF AB= AF CF = BL

= CD ∠DFC = ∠FDC = 12∠DCA= 1

2∠CAL = ∠CAE= ∠EAL = θDF AE BF CE

M DF DE = EF ECDEF

(b) will be the incentre of triangle if and only if also bisectsangle . This means that which is so if andonly if which is true if and only if is a right angle.

C DEF CDEDF ∠CDE = ∠EBA = θAE = EB= ED ∠DAB

(c) is equivalent to . From the similarity of thetwo triangles , this will be so if and only if ,i.e. if and only if is parallel to .

AB= 3CD AC = 2CD = 2CFAEC FME AE= 2FM = FD

EF ADCorrect solutions were received from J. K. R. Barnett, M. Bataille, R. L. Bolt, J. M. Chick,

H. Martyn Cundy, R. P. C. Forman, S. Fowlie, M. D. Fox, M. Griffiths, G. Howlett,D. F. Lawden, G. Leversha, N. Lord, S. N. Maitra, A. Robert Pargeter, C. F. Parry,N. A. Routledge, H.-J. Seiffert, I. F. Smith, H. B. Talbot, R. F. Tindall, G. B. Trustrum, ZhuHanlin and the proposer, Christian Turcu.

83.H (N. Gauthier and J. R. Gosselin)Given the Diophantine equation, , find all the

solutions for .qn+1 − mnq + mn − 1 = 0

q, m, n > 1

From the given equation we have and, since, we derive . But

since all but two of thebinomial coefficients exceed 1 and .

qn + 1 = mn (q − 1)q > 1 mn = qn + qn − 1 + … + q + 1qn < qn + qn − 1 + … + q + 1 < (q + 1)n

n ≥ 2Hence or , a contradiction since are integers.

qn < mn < (q + 1)n q < m < q + 1q, m

Some thought this was rather easy; J. M. Chick and J. A. Mundiesearched for rational solutions for , the latter coming within thedomain of ‘elliptic curves’. There is an infinity of solutions for both casesbut they found only one for satisfying , namely,

n = 2, 3

n = 3 m, q > 1

m =3454015799

, q =2679315799

.

Correct solutions were received from R. G. Bardelang, J. K. R. Barnett, M. Bataille,J. M. Chick, H. Martyn Cundy, R. P. C. Forman, M. Griffiths, G. Howlett, P. F. Johnson, J. D.King, D. F. Lawden, G. Leversha, N. Lord, S. N. Maitra, J. A. Mundie, N. A. Routledge, I. F.Smith, K. Thomas, R. F. Tindall, G. B. Trustrum, and the proposers, N. Gauthier and J. R.Gosselin.


CorrigendumIt has been pointed out by the proposer Dr S. Parameswaran, and two

other commentators, D. Desbrow and R. F. Tindall, that the condition, is even, is necessary but not sufficient for integral solutions to exist

for the equations , forgiven positive integers .(SeeMath. Gaz.83 (1999) pp. 318-320.) For example, forthere is only one solution, namely, . The correctnecessary and sufficient condition is that where is the largest integer not exceeding and . Indeed, as the

proposer indicates, can be uniquely determined from and by theequation .

n − mx + y = m x + y = n m, n

n − m = 2(x, y, n, m) = (4, 1, 5, 3)

n − m = a(a − 1) − r (r − 1)a n r = n − a2

m n am = (n − a2)2 + a

G.T.Q.H

The Turbulent UniverseThere is a time to cast stones away

=d

dt( )The Big Bang The stones of creation

And a time to gather stones together

= ∫ ∞

0( ) dtThe Cosmos The sands of time

Eccles 3 v 5A time when the cosmos evolved in a mighty explosionAnd nebulous mass scattered wide over the vastness of spaceTo coalesce in a myriad of galaxies, set in motionBy gravitational forces, and expanding space;

A time when order from chaos developed on EarthAnd the Sun and the Moon set in place in the skyWhen the passage of time and Adam received birth,Conceived in a twinkling of eternity's eye;

A time when raindrops and dew formed rivers and seasAnd rainbow colours blended to Sunlight,When embryos germinated into fauna and treesAnd dust clouds condensed into stars of the night;

And when the mountains and hills will be ground down to sandAnd creation returned to the palm of God's hand:

For above all this restless commotion and changeThere is permanence in the strong hand of God,A durability that defies both decay and change,Whence the footsteps of time have never yet trod.

A poem by ROY V. WHITLOCK

PROBLEM CORNER 135

Student ProblemsStudents up to the age of 19 are invited to send solutions to either of the

following problems toTim Cross, 75 Cardinal Crescent, Bromsgrove B61 7PR.

Two prizes will be awarded − a first prize of £25, and a second prize of£20 − to the senders of the most impressive solutions for either problem. Itis, therefore, not necessary to submit solutions to both. Solutions shouldarrive by May 20th 2000. Please give your school year, the name andaddress of your school or college, and the name of a teacher through whomthe award may be made. The names of all successful solvers will bepublished in the July 2000 edition.

Problem 2000.1A quartic polynomial is defined by

where are integers. Given that , and, determine the value of

p(x) p(x) ≡ x4 + ax3 + bx2 + cx+ da, b, c, d p(1) = 1999 p(2) = 2000

p(3) = 2001

p(2002) + p(−1998)2 × 2000

.

Problem 2000.2Prove that there are infinitely many pairs of positive integers for

whichx, y

x3 + y3 + 2x2 + y2 + 1

is twice a perfect square.

Solutions to 1999.5 and 1999.6I received a surprising number of submissions to these two tricky

problems, and the quality was exceptional with many students producingsome novel approaches to them; I'm sorry that I cannot pay adequate tributeto all ideas. Correct solutions to both problems were received from PeterAllen (Nottingham HS), Timothy Austin (Colchester RGS), Hannah Burton(City of London School for Girls), Tim Butler (Abingdon School), BrynGarrod (KE VI Camp Hill School for Boys), Daniel Lamy (NottinghamHS), Bruce Merry (Westerford HS, Cape Town, SA), Ahmed Asif Shaik(Woodhurst Secondary School, Durban, SA) and Koos van Zyl (OverkruinHS, Pretoria, SA). In addition to these, Philip Blakely (Barnard CastleSchool) and Owen Thompson (Abingdon School) solved 1999.5; andChristopher Deeks (Freman College, Buntingford), Soutrik Maitra (NationalDefence Academy, Pune, India), Pieter Mostert (Hudson Park HS, EastLondon, SA), Negsej T. K. (St. Joseph's College, Bangalore, India), JacobSteel (Colchester RGS) and Zhan Su (Nottingham HS) solved 1999.6.


Problem 1999.5Find, in the usual decimal form, the least positive integer which is

tripled in size when its final (right-most) digit is removed and placed in frontof its first (left-most) one.

N

Solution I Let be the required -digit number ( ),where is an -digit number (not beginning with a zero) and is asingle-digit number. Then

N = 10a + b n n ≥ 2a (n − 1) b

M = 3N = 10n−1b + a ⇒ 30a + 3b = 10n−1b + a ⇒ a =b

29(10n−1 − 3).

Now, since (being single-digit) is not divisible by 29 and is an integer,we must have that ; in other words, we want(mod 29).

b a29 | 10n − 1 − 3 10n − 1 ≡ 3

Examining powers of 10 modulo 29 (by multiplying by 10 anddiscarding remainders each time for instance), we find that the first such is28. [Note thatFermat's Little Theorem guarantees that (mod 29)

(mod 29) since .However, a little bit of work is needed to justify that this is the least forwhich this is so.]

n1028 ≡ 1

⇒ 1028 ≡ 30 = 3 × 10 ⇒ 1027 ≡ 3 hcf (10,29) = 1n

For the least solution we try etc. The first two casesgive leading zeroes for and to have the same number of digits. The case

gives

b = 1, 2, 3, …M N

b = 3

a =329

(1027 − 3) , N =3029

(1027 − 3) + 3.

In decimal form, N = 1 034 482 758 620 689 655 172 413 793.

Solution II Two or three submissions followed a much simpler constructiveapproach. For the least solution, we look for a number (of as yetundetermined length) which commences with the digit 1 the final digit isa 3 (or possibly a 4 or a 5, but we are looking for minimality). Since allremaining digits are simply shunted along one place, we simply continue bymultiplying by 3 as we go, carrying ‘tens’ units forwards as usual. In thisway, we find

⇒

3 × 3 = 9, 3 × 9 = 27, (7 carry2), 3 × 7 + 2 = 23(3 carry2), etc

All we need do now is to check our answer each time a digit 1 appears inour number. Indeed, the 1 would have to be followed (moving to the left) bya 3. The above value of is found in this way.N

Notice that the first example of a possible answer ending in a 4 ratherthan a 3 is , and similarly for a 5 in the units column. Notice also, assome did, that the suitable answers are simply re-orderings of the recurringdecimal cycle in the reciprocal of 29: viz

4N

129

= 0·034 482 758 62… .

PROBLEM CORNER 137

Problem 1999.6Prove that

cos 11° + cos 83° + cos 155° + cos 227° + cos 299° = 0.

Solution I Purely trigonometric approaches employed a variety of results,and most solutions were markedly different in some way or another. Onesuch example follows, and it is interesting that this is one of those caseswhen the specific result given is rather less obviously approached than thegeneral one lurking in the background.cosθ + cos(72° + θ) + cos(2×72° + θ) + cos(3×72° + θ) + cos(4×72° + θ)≡ cosθ + cos(72°+ θ) + cos(72°− θ) + cos(2×72°+ θ) + cos(2×72°− θ)

using the result cos(360° − A) = cosA≡ cosθ + 2 cos 72° cosθ + 2 cos 144° cosθ

using the result cosA + cosB = 2 cos12 (A + B) cos1

2 (A − B) using the result ≡ cosθ1+2 cos(2×36°)−2 cos36° cos(180°− A) = − cosA

upon writing and using thedouble-angle formula for cosine. We now show that is asolution of the equation , so that this last expression isidentically zero.

≡ cosθ 1 + 2 (2c2 − 1) − 2c c = cos 36°c = cos 36°

4c2 − 2c − 1 = 0

Note that ; that is,sin 36° = sin (180° − 36°) = sin (4 × 36°)sin 36° = sin (4 × 36°) = 2 sin(2 × 36°) cos(2 × 36°)

= 4 sin 36° cos 36° cos(2 × 36°) .Since , we have (writing )sin 36° ≠ 0 c = cos 36°

1 = 4c (2c2 − 1) ⇒ 0 = 8c3 − 4c − 1 = (2c + 1) (4c2 − 2c − 1) ,and since , we have the required result upon setting .cos 36° ≠ −1

2 θ = 11°

Solution II A second approach uses a complex argument.Let . Then the equation

implies for and so has rootsα = eiθ = cosθ + i sinθ z5 − α5 = 0

z5 = e i5θ ≡ ei(5θ + 360k°) k = 0, 1, 2, 3, 4

αk = cos(θ + 72k°) + i sin (θ + 72k°) (k = 0, 1, 2, 3, 4) .The sum of the roots of the given quintic equation is equal to the negative ofthe coefficient of in the left-hand-side, which is zero. Thus, both real andimaginary parts of this sum of roots are zero; i.e.

z4

∑4

k = 0

cos(θ + 72k°) = ∑4

k = 0

sin (θ + 72k°) = 0

and setting again gives the special case of the problem.θ = 11°


Solution III A rather lovely, and extremely concise and elegant, solutionappeared several times in submissions. As with Solution II, this result easilyextends to a whole family of similar problems with terms.n

Consider a regular pentagon resting with one vertex ( ) on the -axis,and slightly inclined to it (as shown).

A x

d

dd

d

A xθ

θ + 144°

θ + 72°

θ + 216°

θ + 288°

Then the projections, in turn, of the sides onto the horizontal axis form asequence of directed line segments whose sum is zero, since traversing thepentagon once leaves one returned to . These line segments are merelyA

d cosθ, d cos(θ + 72°) , … , d cos(θ + 288°) ,as in the given problem (with the cancellable factor of ). We thenimmediately see why the result holds for all values of , and in the case ofall regular -gons ( ), and why a similar result applies for the sums ofthe sines of angles equally spread throughout a 360° range (since theprojections onto the vertical axis must also sum to zero).

dθ

n n ≥ 3

As mentioned, the range of different ideas was outstandingly broad, andthis made the selection of prize-winners extremely difficult. However, I amawarding the first prize of £25 to Bruce Merry and the runners-up prize of£20 to Owen Thompson.

TIM CROSS

OTHER JOURNALS 139

Other JournalsWhat makes a great mathematics teacher? The case of Augustus De Morgan, byAdrian Rice, The American Mathematical Monthly 106 (6) pp. 534-552, 1999.

The name of Augustus De Morgan (1806-1871) is chiefly remembered forresearch in mathematical logic, geometric representation of complex numbers,convergence of series and some aspects of the history of mathematics. For almosthis entire career De Morgan was professor of mathematics at University College,London. Much of the present paper is devoted to describing his mathematics coursesand the surviving tracts by himself and by some of his students. The author claimsthat De Morgan deserves the adjective ‘great’ applied to his work as a teacher ofmathematics rather than for his original research. As a rearcher he was not in thesame league as his contemporaries Cauchy, Weierstrass, Klein, Maxwell andSylvester but De Morgan ‘possessed in abundance all the attributes of a memorableand effective educator’.

Putting the millennium in perspective, by Jim Hind,Mathematics Today35 (5)pp. 147-151, 1999.

On October 1 1949, probably for the first time, the whole world agreed on thedate: this was when Mao Tse-Tung announced that China would use the westerncalendar introduced by Pope Gregory in 1582. This article is a fascinatingdescription of the need for calendars, the historical development of calendars withrespect to the sun and moon's movements, and properties of the Jewish, Greek,Islamic, Mayan, Indian and Chinese calendars to name but a few. The present namesof days and months are somehow derived mostly from the Roman and Greek. Tointerchange dates between the different calendars presents a non-trivial mathematicalproblem.

Pentangram- a new puzzle, by Klaus Kühnie,The Mathematical Intelligencer21(2) pp. 15-17, 1999.

The well-known Chinese tangram is a puzzle consisting of seven pieces that canbe arranged into either one square of area 2 or two squares of area 1 each. The ratiobetween any two lengths occurring as side-lengths of the seven pieces is some powerof , which is the ratio between the side-length and the diagonal of a square.Equally well a puzzle could be designed where the pieces have to be arranged toregular hexagons and the magic number would be which is the length of thechord of a hexagon of side-length 1. The author gives details of pentangrams whichare based on regular pentagons whose ratio between chord-length and side-length isthe golden ratio.

2

3

Throwing elliptical shields on floorboards, by P. Glaister,Mathematical Spectrum32 (1) pp. 10-13, 1999.

The author describes a generalisation of the classical needle problem of Buffon.An elliptical shield is dropped onto a floor made of wooden planks (floorboards) andthe problem is to determine the probability that the shield crosses a crack betweenthe planks. As special cases there are the cases of a circle and a needle. Themathematical details involve complete and incomplete elliptic integrals which occurnaturally in the solution.

ANNE C. BAKER1 Carr Bank Close, Sheffield S11 7FJ


ReviewsA mathematical mystery tour: discovering the truth and beauty of the cosmos,by A. K. Dewdney. Pp. 218. £17.99. 1999. ISBN 0 471 23847 3 (Wiley).

This enjoyable odyssey, cast in the form of a travelogue in which the authorinterrogates four fictional characters at sites of epochal significance in the history ofmathematics, aims gently to explore the familiar philosophical questions:

• Is mathematics created or discovered?• What is the basis for the ‘unreasonable effectiveness’ of mathematics

when applied to the natural sciences?

Dewdney's first destination is Miletus for an introduction to early Greekmathematics, in particular a plausible route to Pythagoras' discovery of ‘his’ theoremand a description of the ‘crisis’ for Pythagoreans resulting from the discovery that

is irrational. His second stop is Jordan for an account of Islamic work on decimalnotation, algebra and the geometry/trigonometry of the celestial sphere with echoesof Omar Khayyám's, ‘And that inverted Bowl we call The Sky, Whereundercrawling coop'd we live and die,’. This is followed by a trip to Venice to probe therole of mathematics in modern science with a nicely motivated reconstruction ofBalmer's discovery of the formula giving the wavelengths of spectral lines forhydrogen and a confrontation with the possibility that a hydrogen atom may,perhaps, be no more than a solution of Schrödinger's equation. The tour ends inOxford with a discussion of twentieth century work on axiomatics andmetamathematics and the implications of Church's thesis and Gödel's theorem.

2

There are one or two signs of hasty editing, notably on page 74 where the ibnQurra-Fermat theorem should read, ‘The numbers and are amicable if

, and are all prime.’. Gazettereaders will find few surprises in the mathematical examples highlighted and somewill find the philosophical fare more soufflé than plum pudding, but I had a sneakingadmiration for Dewdney's willingness to fly the perhaps unfashionable Pythagorean-Platonist mathematics-as-independent-existence flag, even to the extent of coiningthe nameholos for the land over which the flag flies. The travelogue format isgimmicky but not, I think, too gimmicky and I would be happy to recommend thisbook to students wondering what mathematics is really all about, in particularwhether the mathematicians' worry blanket monogrammed ‘Mathematical Truth’ isknitted from massive self-delusion or woven from the fabric of the universe.

2npq 2nrp = 3.2n − 1 − 1 q = 3.2n − 1 r = 9.22n − 1 − 1

NICK LORDTonbridge School, Kent TN9 1JP

Women in science and engineering: choices for success, edited by Cecily CannanSelby. Pp. 263. £33.50. 1999. ISBN 1 57331 166 9 (New York Academy ofSciences).Notable women in mathematics: a biographical dictionary, edited by CharleneMorrow and Teri Perl. Pp. 302. £39.95. 1998. ISBN 0 313 29131 4 (GreenwoodPress).

These two books have similar aims but quite different styles. They address theconcern that the number of women active in science, engineering and mathematicsremains relatively small. Although much progress has been made in the last 50 yearsto tackle discrimination against women interested in science, there are very fewfields where women are represented in anything like equal numbers. Even in an arealike biotechnology, where 50% of research scientists are women, the senior positions

REVIEWS 141

are still dominated by men.The first book is based on a conference:Choices and Successes: Women in

Science and Engineering, held in New York in 1998. The New York Academy ofSciences had held conferences in 1972 and 1979 on a similar theme, and the latestconference was a conscious effort to review progress. An important finding was thatexplicit discrimination is now less common but that a bewildering morass of indirectbias and discrimination remains. Women continue to drop out of science andengineering courses at a proportionally greater rate than men, and the number of fullprofessorships offered to women remains small.

The Choices and Successes conference proceedings is divided into four mainparts:Changes, Choices, Successes andThe Future. The (relatively short) first partlooks at the changes that have taken place in the last 25 years and the (shorter) lastpart is a report of a panel discussion of the question:What have we learned and howcan it help? In between are two substantial sections, one looking atWhat keepswomen in science and engineering? and the other consideringWhich policies andpractices work?

Sue V. Rosser contributes an important paper on the impact of work climates onwomen. She identifies a five stage model to assess the extent of women's integrationand the adaptation of the workplace to women's needs. At Stage 1 the absence ofwomen is not noted. In Stage 2, women are seen as an add-on − tolerated as long asthey do not challenge the status quo. At Stage 3, women are seen as a problem,anomaly or deviant: they begin to challenge the ‘male-as-norm’ climate, for exampleby requesting child-friendly hours or by seeking more collaborative working styles.By Stage 4, there is a focus on women: both men and women begin to consider thepositive benefits of changes to the laboratory or workplace climate. Finally, at Stage5 the climate is redefined and reconsidered to include all − men and women. In asurvey she conducted of women who had received Professional Opportunities forWomen in Research and Education awards in 1997, Rosser categorised theworkplaces of the 56 respondents as follows:

Stage 1 2 3 4 5

Workplaces 9 9 34 4 0

Clearly, she considers that there is still some way to go!The conference seems not to have attracted speakers who disagree with the

fundamental premise that changes need to be made to accommodate women. Suchpeople doubtless exist.

The second book under review is intended to provide women with role-modelsin mathematics. It consists of 59 biographies of women in mathematics andmathematics education. These have been selected from a more comprehensive list ofwomen who could have been included if space had permitted. The desire to representwomen from a variety of nationalities and ethnic backgrounds has meant that somequite deserving cases have been omitted. On the other hand, a lack of publishedbiographical information on a candidate was not considered sufficient reason toleave anyone out: in many cases, the biography has been based on interviews heldspecifically for this book.

The examples of women born before 1900 are almost self-selecting: the listincludes Hypatia, Maria Agnesi, Sophie Germain, Sofya Kovalevskaya, CharlotteScott, Grace Chisholm Young and Emmy Noether. Included among those born thiscentury are Mina Rees, Olga Taussky-Todd, Julia Robinson, Anneli Lax, Cathleen


Morawetz, Mary Ellen Rudin, Dusa McDuff, Ingrid Daubechies and Karen Parshall.An obvious omission is Mary Lucy Cartwright. Despite the attempt to be inclusive,there are some obvious biases: the majority of the women live and work in theUnited States, and while there is a commendable concentration on the living, ratherthan on historical figures, only five of the women were born in the second half of thetwentieth century.

The biographies are designed to be accessible to readers who have a limitedmathematical background, though there are places where quite technical terms areused. In the biography of Marie-Louise Michelson, for example, we are told thatSaunders Mac Lane is ‘a well-known mathematician’ and that ‘topology is the studyof shapes and their properties—in particular those properties that remain unchangedeven if the shape is stretched and distorted’. However, later in the same biography,we learn that ‘Professor Michelson's achievements include research in Clifford andspinor cohomology, the geometry of spin manifolds and the Dirac operator’. This ispart of a deliberate policy in the book. The editors explain in the introduction: ‘Inthis volume we have tried to give a sense of the biographee's work in terms that canbe understood by the non-mathematician. [However] many mathematicians say theysimply cannot transmit any idea of their research, even to undergraduates. [If]mathematical terms are unclear, undefined, or unexplained, the reader is encouragedto use these terms as a starting point for conversations with someone who has greatermathematical training or to seek out reference works, such as mathematicaldictionaries.’

The purpose of the book is ‘not only to encourage more girls to become anintegral part of the next generation of mathematicians, but to spark the enthusiasm ofall students’. I feel sure that students who read about these mathematicians will findinspirational role models. However, in some ways I feel the book is toocomprehensive. On behalf of the Mathematical Association, I recently attended aDepartment of Trade and Industry conference on attracting girls into science,engineering and technology. One of the clear messages from participants who hadsucceeded in attracting young girls' attention was that ‘worthy but dull’ is not veryeffective. Instead the advised following the style of teen magazines, with ‘agonyaunts’ and lots of pictures of girls and young women having fun. Girls like the ideaof working in teams with other people like themselves—a message echoed by theNew York Academy of Sciences conference.

This book should be available in school and college libraries, but moreimportantly, someone needs to direct students to read it, perhaps by getting eachclass member, over the course of a year, to report back to the rest on the life of amathematician.

STEVE ABBOTTClaydon High School, Claydon, Ipswich IP6 0EG

Gnomon: from pharaohs to fractals, by Midhat J. Gazalé. Pp. 259. £17.95. 1999.ISBN 0 691 00514 1 (Princeton University Press).

To most of us, a gnomon is the spike on a sundial which shows the time (in thiscountry, outside the period of BST, and provided the sun is shining). Here the wordis used as defined by Hero of Alexandria: ‘a gnomon is that form that, when added tosome form, results in a new form similar to the original’. To a mathematician afamiliar example is the square which, when stuck on to the longer side of a rectanglewhose sides are in the golden ratio , generally denoted by ),produces a similar rectangle. Or, of course, a sheet of A4 paper is its own gnomon!This process, if continued indefinitely, results in a figure suitably selected vertices of

(= ( 5 + 1) / 2 φ

REVIEWS 143

which (in the case of the golden ratio rectangle) lie on an equiangular (orlogarithmic) spiral. This spiral also plays a large part in the book, and it may beregarded as a basic fractal, in that if enlarged it remains similar to the original (infact, congruent). Self-similarity, says the author, is the theme of the book.

There is a discussion of figurate numbers, which clearly lend themselves tognomonic representation. Much use is made throughout the book of continuedfractions: infinite periodic ones are gnomonic (in particular the familiar one for ,namely . As one might expect the Fibonacci sequence is prominentin this context, not, only the familiar one , , , but alsoa generalisation , , (where m is notnecessarily an integer), referred to as a Fibonacci sequence of order . The quantity

defined by is found to play an important part in thedevelopment of the theory. (In fact ). Applications of these generalisedFibonacci sequences are given to ‘ladders’ of transducers, resistances, and pulleys.The sequence of rectangles referred to above is an example of a ‘whorled figure’,which are exemplified in some detail. The ‘golden’ number satisfies

; the number which satisfies is called the ‘silver’number by the author (‘ ’ in honour of its discoverer Richard Padovan), and leads toseveral interesting constructions: in particular the ‘silver pentagon’ of sides

in order, with parallel to and at 60° to anequilateral triangle of side . Amusement is provided by twinkles, squinkles, andGolomb's rep-tiles. Various spirals are considered in detail. Positional numbersystems are described as a preliminary to the final chapter on fractals, which here areof the type not involving calculations with complex numbers, e.g. the Sierpinskigasket (based on Pascal's triangle), Cantor's ternary set, the Thue-Morse sequence,the Menger sponge, the Koch snowflake, and numerous fascinating figures, many in3-dimensions, arising from variations on these.

φ1 + 1

1 + 11 + … )

F0 = 0 F1 = 1 Fn+ 2 = Fn + Fn+ 1

Fm, 0 = 0 Fm, 1 = 1 Fm,n+2 = Fm,n + mFm,n+1

mΦm m = Φm + 1 / Φm

Φ1 = φ

φφ2 − φ − 1 = 0 p p3 − p − 1 = 0

p

1, p, p2, p3, p4 p4 p p3, whose gnomon isp4

The book is well illustrated (in black-and-white, but there is an interesting batchof colour plates). There is no deep mathematics, and most of it should be within thereach of a good 6th former, if not put off by some rather (at times) fearsome notation− lots of suffixes! (note: is used as a suffix, ). This is rather an unusualbyway of mathematics; of its importance it is difficult to judge; but it certainly hasits fascination, and I have much enjoyed reading about it. (But, you may ask, wheredo the Pharaohs come in? At the outset the author discusses, ultimately to dismiss,the idea that the Egyptian obelisks, of which few, sadly, remain on their originalsites, were intended as gnomons in the sundial sense. There are many suchinteresting historical interludes scattered throughout the book.)

i −1 = j

A. ROBERT PARGETER10 Turnpike, Sampford Peverell, Tiverton EX16 7BN

The mathematical tourist: new and updated snapshots of modern mathematics,by Ivars Peterson. Pp. 266. £13.95. 1998. ISBN 0 7167 3250 5 (W. H. Freeman).

This is an unusual book of its kind; in fact it is quite well named. Its subjectmatter is mainly modern developments and applications of quite recondite areas ofmathematics, but it is purely descriptive: there are extremely few formulae orequations. I quote from the preface: ‘Professional mathematicians, in formalpresentations and published papers, rarely display the human side of their work.Frequently missing among the rows of austere symbols… is the idea of what theirwork is all about − how and where their piece of the mathematical puzzle fits, theirfountains of inspiration, and the images that carry them from one discovery toanother. To most outsiders, modern mathematics represents unknown territory. Its


borders are protected by dense thickets of technical terms, its landscapes strewn withcryptic equations and inscrutable concepts. Few realize that the world of modernmathematics is rich with vivid images, provocative ideas, and useful notions.’ Thebook is very easy to read − one might almost call it light reading! These remarksmay tend to convey that the treatment is superficial: in terms of actual nitty-grittymathematics of course it is, but subject to this reservation it delves quite deeply intothe underlying concepts, their interconnections and applications, and the successes,failures, and hopes of some of the latest ideas.

A few of the topics are familiar classics, but most of more recent emergence. Abrief survey (which I do not pretend to be complete): map-colouring; unsymmetricaltiling; quasi-crystals; prime factoring (with its application to the RSA system ofcoding); topology; minimal surfaces; knots; manifolds in higher dimensions; stringtheory; fractals and fractal dimensions; the fractal simulation of natural objects;chaos and strange attractors; the emergence of chaos in Newton's method for solvingequations; cellular automata, including Conway's game of Life and Wolfram'smethod for generating patterns (several examples of these patterns are shown but Iwish there was a clearer explanation of the actual system); the Turing machine;automated theorem-proving. So it is by no means a dull book! It is well illustratedin black-and-white, and there are 16 colour plates. There are over 9 pages ofsuggestions for further reading, and a good index.

The present volume is a revised and enlarged version of that published in 1988:as I happen to have a copy of this I am able to make comparisons. The main thing isthat it has been brought well up-to-date in recent developments, such as quantumlogic, automated reasoning, virtual ants, and Wiles' proof of Fermat's ‘Last Theorem’(to mention a few), while revision elsewhere has not been neglected, e.g. the list ofMersenne primes has been extended by the 7 instances discovered since 1985. Ithink an owner of the earlier edition would need to browse through the new one todecide whether it is worth buying; to the new reader I would say have a go − youmay not learn much mathematics from it, but it will widen your horizons in apleasurable way.


Magic tricks, card shuffling and dynamic computer memories, by S. BrentMorris. £16.95. 1998. ISBN 0 883 85527 5 (Mathematical Association of America).

Brent Morris claims to be the only person in the world with a PhD in cardshuffling. He tells in the preface how his early interests in magic tricks andmathematics came to be combined in the mathematical papers on permutation groupsthat formed the basis of his doctoral dissertation. He also explains how he came toregister U.S. patent 4,161,036 (Method and apparatus for random and sequentialaccessing in dynamic memories).

Morris's book mixes quite sophisticated mathematics with practical instructionsfor carrying out card tricks. A prerequisite for mastery of the tricks is the ability toperform perfect shuffles to order and the first chapter deals with the history andmechanics of the such shuffles. There are two types of perfect shuffle, depending onwhether the original top card remains on top (the out-shuffle) or moves to the secondposition (the in-shuffle). These may be represented mathematically as permutations.Chapter two presents the surprising result is that perfect shuffles can be used to movethe top card to any desired position in the pack. If the top of the pack is labelledposition 0, the top card can be moved to the required position by performing in- andout-shuffles corresponding to the 1's and 0's of the binary representation of the

REVIEWS 145

position number. For example. the sequence in, out, out, in moves the top card toposition .10012 = 910

The mathematics gets steadily more involved in the third and fourth chapters,with the introduction of shuffle groups and permutation matrices. Many of themathematical results have quite recent proofs, some due to Morris himself. As withthe earlier chapters, the mathematics is used to justify the working of a number ofcarefully described card tricks.

The final chapter deals with the applications of the mathematics of shuffling todynamic computer memories (such as a hard or floppy disc), developed during thelate seventies. Though practical applications of the techniques became outdated bytechnical advances in data storage, the mathematics remains applicable tointerconnection networks in parallel processing.

Readers who have experienced Colin Wright's juggling lectures will appreciatethe power of combining performance with mathematics. I suspect that Brent Morrisis another master of the genre. His delightful book is highly recommended, but theperformance aspect has to be supplied by the reader. As Martin Gardner advises inhis introduction: ‘be sure you have on hand a deck of cards’.


Towing icebergs, falling dominoes, and other adventures in appliedmathematics, by Robert B. Banks. Pp. 329. £19.95. 1999. ISBN 0 691 05948 9(Princeton University Press).Slicing pizzas, racing turtles, and further adventures in applied mathematics, byRobert B. Banks. Pp. 284. £15.95. 1999. ISBN 0 691 05947 0 (Princeton UniversityPress).

The authors of most ‘popular’ mathematics books are at great pains to illustratethe vast sweep of mathematical thought and to counter the commonly held belief thatadvanced mathematicians involves ever harder sums. They seek to convey the bigideas, often at the expense of the mathematical details. Robert B. Banks is not thissort of mathematical writer: he likes to calculate.

Banks' first book,Towing icebergs, falling dominoes, and other adventures inapplied mathematics, was released in the United States in 1998. It is written in a‘popular’ style but includes much more mathematics than most of its competitors.Gazette readers will find many topics of interest, but I remain unconvinced that thisbook is suitable ‘for use as a text or a reference source for a first course inmathematical modelling’, to quote one of the snippets of ‘advance praise’ printed onits back cover.

The publishers supplied a pre-publication review copy of the sequel,presumably to give them time to incorporate favourable remarks for the officialpublication. Unfortunately, the features that disturbed me about the first book are, ifanything, exacerbated in the second, so I fear they will not quote this review.

Banks looks at a bewildering variety of problems and shows how they can bemodelled mathematically, mostly using mathematics that would be covered in anEnglish A level course. Apart from the problems alluded to in the two titles, theircombined 50 chapters include considerations of the trajectories of baseballs and golfballs, the number of people that have ever lived, how fast you should run in the rainand a better way to score the Olympics.

For my taste, there are too many asides and side issues that disturb the mainflow of the narrative. The use of insets and sidebars might have alleviated this minor


irritation. My main concern is Banks' tendency to pull rabbits from hats. He does thisin two ways: he introduces equations (and often their solutions) with insufficientjustification; and he offers very little discussion of the assumptions underlying hismathematical models or the validity of the solutions obtained.

For an example of the first type, in considering the towing of icebergs he raisesthe question of how thick the towing cables need to be.

‘The thrust force in the cable is N. Assuming that thesteel cable can tolerate a tension stress lb/N/m2, the cross-sectional area of the cable is ,and so m (20 inches).’

T = 50.8 × 106

σ = 35,000 in2 = 2.41 × 108

A = T / σ = 0.210 m2

d = 0.52The tension had been calculated earlier in the chapter, but the formula isplucked from the air. It might be argued that any reader with the mathematicalsophistication to follow the mathematics would know the necessary formula, and thatthe figure of lb/ can easily be looked up in a reference, but I feel uneasy.

A = T / σ

35,000 in2

While oversights of the first type are merely annoying, those of the second typeare inexcusable in a book that purports to show its readers how mathematicianswork. Not all sections omit to discuss the assumptions: the chapter Shotputs,Basketballs, Fountains (inTowing icebergs) has a section discussing when airresistance can be neglected, for example. However, the chapter Growth andSpreading Mathematical Analogies (in Slicing pizzas) begins:

‘How fast does a plant or person grow?… To help us obtain answers tothese and similar kinds of questions, we need to construct an appropriatemathematical framework. Such a framework is provided by the followingsimple differential equation:

dN

dt= aN(1 −

N

N∗) ,

in whichN is the magnitude of the growing or spreading quantity,t is thetime, a is the growth or spreading coefficient, and is the equilibriumvalue or carrying capacity.’

N∗

There is no discussion of the assumptions underlying this model (or thearctangent-exponential model introduced later). Later, numbers are substituted intosolutions of the differential equations without any discussion of the validity of theresults. The impression given of mathematics is that equations can be pulled from theshelf to suit any required purpose and uncritically applied to the problem in hand.

Despite my reservations, the books are often entertaining and certainlystimulating. I recommend readers to look before they buy.


Misused statistics(2nd edn.), by Herbert F. Spirer, Louise Spirer and A. J. Jaffe. .Pp. 263. $49.75. 1998. ISBN 0 8247 0211 5 (Marcel Dekker).

We are all continually being presented with ‘facts’, often in numerical form, thatwe accept uncritically; yet, if we thought more deeply about them, doubts wouldarise in our minds as to their meaning or validity, and further investigation mightreveal the ‘facts’ to be false. An important reason for this state of affairs is our useof the adversarial system whereby the presenter is trying to put the ‘facts’ favourableto their case, rather than provide a balanced judgement. Politicians are frequentoffenders but there is an increasing tendency for organisations such as Greenpeace tobias their statements. This is not the only reason for misleading statistics and the

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book under review contains many examples of misuse, listing four, major causes:1) lack of knowledge of the subject, 2) faulty interpretation, 3) flawed data,4) incorrect methodology. Most of the examples come from the media, not onlybecause the media are guilty of misuses − they are − but because their examples areexposed to view, whereas commercial organisations, for instance, sin in secret.

The examples range widely, though with an American bias, and are looselygrouped into similar types of error. It is often cruel fun to watch other people err,and the result is both entertaining and instructive. Furthermore, one can often learnfrom mistakes, so that to see a technique being misused or performed incorrectly canprevent your making similar errors. As a result, there is a strong case for this bookbeing in every school library so that pupils can better understand the pitfalls in‘facts’, not just in statistics but in all disciplines. For myself, I would have liked tohave seen, in addition to the excellent examples, more discussion of the principlesinvolved. For instance, the prosecutor's fallacy (in which the probability of , giventhat is true, is confused with that of , given the truth of , reversing the roles ofthe two events, and ) could have been developed more than it is. Incidentally, it ishard on lawyers to have the fallacy associated with them, for we are all capable ofmaking it. Even statisticians err whenever they perform a significance test, citing theprobability of data, given a hypothesis, rather than what is needed, the probabilityattached to the hypothesis on the basis of the data.

AB B A

A B

The book is not just destructive; it also provides sensible advice on how toseparate the genuine from the false. A favourite of mine is the recommendation torespond, when a speaker provides a ‘fact’, with the question ‘how do you knowthat?’ The technique is especially valuable when applied to the claims of manypractitioners of alternative medicines. A mild criticism of the book is that in dealingwith pictures, not enough attention has been paid by the publishers to goodpresentation. In this field, the three books by Edward R. Tufte, published by theGraphics Press, Cheshire, Connecticut, are masterpieces and certainly should be inevery library, not only to provide information, but to demonstrate the sheer beautypossible in a printed book. So buyMisused statistics to learn about mistakes and letTufte show you how to present facts graphically.

D. V. LINDLEY‘Woodstock’, Quay Lane, Minehead, TA24 5QU

Coordinating mathematics across the primary school, by Tony Brown. Pp. 236.£12.95. 1998 ISBN 0 7507 0687 2 (Falmer Press).

Taking on any role in the primary school can be a daunting task and, as stated inthe text, the primary teacher is asked to develop ‘numerous skills’. In doing so, it isbecoming increasingly difficult to master any fully and, since the curriculum is nolonger the sole responsibility of the headteacher, as it was in the past, so the need fora subject coordinator has become necessary. This book takes a methodical journeythrough the various stages of adopting this daunting role.

The text begins by looking at accepting the role of mathematics coordinator inthe teacher's own school, or moving to another, and the difficulties that may thenarise. The essential aspects involved in the settling-in period are clearly set out, asare the starting points for change and the acknowledgement of good practice.

The issues of good practice and change are examined thoroughly and soundpractical advice is given to assist effective implementation. The author is sensitiveto the fact that this is not easy and provides carefully-chosen case studies to illustratekey points. Also given are detailed background information about what thecoordinator should know and the ideas behind the strategies which are advocated.


All this helps to increase the reader's knowledge, allowing a new coordinator to gainconfidence in the decisions she or he makes.

Many professionals look upon change with suspicion, and decisions to changecurrent practice in a school have to be well thought out: this book will assist theprocess by setting out clearly ways forward. Questions that the coordinator shouldask him/herself about his/her own practice are suggested and also ways ofapproaching staff, and monitoring and auditing the materials and practices employedin the school. All information is set out in easy-to-read tables which can be usefullyreproduced.

Throughout, there is emphasis on the need for balance in everything, whether itbe in the need for change or in the needs of individuals or of the whole school. Thisis important because I think that peoplecan become too narrow-minded whenadopting a particular role.

The particular role of mathematics coordinator is explained clearly. It is amammoth one which people might rush into; setting aside time to read this bookwould aid any professional who is considering taking it on, settling into the role, oractually in such a post. Mathematics is a major element of the primary curriculum,so it is imperative that the mathematics coordinator is effective and confident in therole.

Although there are 236 pages in this book, there is text on only just over half ofthem. Each chapter is clearly labelled and the key points are all in the index, makingit easy for the reader to access the information required. I feel that this book wouldbe invaluable for anyone who is, or is considering being, a mathematics coordinator:it most definitely fulfils its claim to be a subject leader's handbook.

JACQUELINE G. ROBERTSONJohn Logie Baird Primary School, Helensburgh G84 9EP

Livewire maths, by Paul Harrison. Addition and subtraction to level 3, PupilBook, pp. 32. £2.99. ISBN 0 340 74908 3; Teacher's Resource Book, pp. 63. £12.99.ISBN 0 340 75397 8.Multiplication and division to level 3, Pupil Book, pp. 32.£2.99. ISBN 0 340 74909 1; Teacher's Resource Book, pp. 61. £12.99. ISBN0 340 75398 6.Multiplication tables to level 3, Pupil Book, pp. 24. £2.99. ISBN 0340 74911-3; Teacher's Resource Book, pp. 47. £12.99. ISBN 0 340 75399 4.Measures to level 3, Pupil Book, pp. 24. £2.99. ISBN 0 340 74910 5; Teacher'sResource Book, pp. 48. £12.99. ISBN 0 340 75400 1 (Hodder & Stoughton).

The pupil books are not write-on and so can be used many times. Althoughvery small, they do contain plenty of material. The teacher's resource books containadvice on using each section (relating it to the National Curriculum for England),answers and photocopiable sheets to support the work in the pupil books. A uniquefeature in each resource book is a photocopiable ‘buzzwords glossary’ which goesover the surprisingly large amount of vocabulary associated with each topic.

These books are aimed at pupils of age 11-14 who are underachieving in mathsworking at level 3 (approximately level C in Scotland). The content of the books(with the exception of Measures) has succeeded in being appropriate for this agegroup. The books attempt to tackle the basic concepts and ensure a realunderstanding of the topics. The material is good, with many excellent ideas tomake the work more enjoyable.

The writer assumes that you will be able to organise your class so that you canteach these pupils giving them a structured lesson of which the work from the bookis only a part. As there will only be a small group of pupils at this level in a mixed-

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ability class, this would be difficult to do in practice. Often the work in the pupilbook will require a lot of input from the teacher and, as there is very little work onany one page, many pupils will also need the supplementary photocopiable sheets aswell.

There are a few printing errors, which is a shame, as pupils can be confused bythem. Also, inAddition and subtraction, methods of mental working are given in thepupil book, which is unfortunate, as it is better to encourage them to use their ownmethods, and they lack the confidence to ignore the book's suggestions.

The only book about which I have serious reservations isMeasures. In thesection on litres and millilitres, it uses measuring jugs and water. My pupils wouldbe mortified to be given such a ‘baby thing’ to do in class.

Multiplication and Multiplication and division are very good, but they onlyconsider the 2, 3, 4, 5 and 10 times tables.

Despite the difficulties of using these books in a mixed-ability class I wouldrecommend them, as there is a lot of good material in them. The additionalresources in the resource books are excellent, both for class and homework.

Additional books were published in September 1999 to coverFractions, Time,Money and Number structure.

H. MASONMadras College, St Andrews, Fife KY16 9EJ

Maths workout: for homework and practice, by Bob Hartman and Mark Patmore.Book 1: pp. 64. £3.50. 1999. ISBN 0 521 63489 X.Book 2: pp. 64. £3.50. 1999.ISBN 0 521 63488 1.Book 3: pp. 64. £3.50. 1999. ISBN 0 521 63487 3.Teacher'sbook 1−3: 1999. ISBN 0 521 63483 0 (Cambridge University Press).

The whole series consists of six slim books, 64 pages each, and two teachers'books designed to provide homework and practice material for Years 7 to 9 as asupplement to any mainstream course. Books 1 to 3 are targeted at NC Levels 3 to 5but contain some harder, extension work. Each book is organised in 16 or 17 units,each unit having a common format beginning with ‘Key Ideas’ − a reminder of thebasic facts and skills needed for that section − followed by about 15 questions whichare split, in turn, into three sections ‘A’, ‘B’ and ‘C’. ‘A’ questions are described as‘straightforward and are intended to ensure confidence’, ‘B’ and ‘C’ questions ‘aremore challenging’. Most units also have a set of 3 or 4 extension questions at theend of the book intended to ‘challenge knowledge and understanding further’. Avery nice touch is a glossary of terms used at the back of each book.

The aim ofMaths workout is to provide a resource which is capable of beingused in a variety of ways according to the organisation and needs of the classes, withteachers being able to select work for the pupils to take home. The authors, in thepreface to the Teacher's Book, make much play of the modern need for greaterhome-school links and support, and ‘the intention has been to provide work thatcould involve the help and support of someone else at home. Currently, manyhomeworks tend be “more of the same” or “finish this off”; it is hoped that thiscollection will allow for more focussed and appropriate tasks to be given, manybeing set in a home-based context.’

Unfortunately, the Teacher's Book is simply a book of answers. The authorshave made a tilt at offering ‘Equipment Needed’ and ‘Teaching Points’ for eachsection but these are minimalist to say the least and do not offer any insight beyondsuch offerings as ‘There are opportunities for further discussion about differentshapes.’


The authors have tried to reflect the current trends for increased emphasis onnumeracy and numerical skills, and include many opportunities for practice andconsolidation. There has been no intention to include ‘teaching material’ − they arewhat they set out to be − extra homework-type questions arranged in topics. In fact,I could feel no sense of progression through the books 1 to 3 and think the publishershave simply divided the whole into six, more easily managed, units.

Would I buy a kit? I can't see the need as the texts in use at present have morethan enough questions to last for homework, or I can make up some practice/interesting/more challenging questions in direct response to the classes' immediategrasp and progress of each lesson. The lay-out of the three books does not lend itselfto snap judgements. The homeworks would have to be planned and the appropriateset booked out in advance. But there must be some super-teachers with advancedskills out there who plan homeworks as well as lessons.

DAVID WHETTONFerndown Upper School, Dorset BH22 9EY

On target for Key Stage 3 maths: summary and practice book, by Paul Hoganand Barbara Job. Pp. 168. £5. 1999. ISBN 0 7487 4453 3 (Stanley Thornes).

This summary and practice book has been designed to help students prepare forthe National KS3 test in Year 9. It covers the National Curriculum in 13 short topicsand includes plenty of practice questions similar to those in the tests. Each topic isorganised into levels 3-5, 6, 7 and 8 so that students can focus on their own level.Each topic provides a summary of ‘What you need to know’ followed by a ‘Testyourself’ exercise and a set of KS3−style practice questions, all with answers. Thefirst impression is very favourable − lots of colour and diagrams (not silly clip art).The book is well organised into the usual sections of Number, Algebra, Shape Spaceand Measures, and Handling Data, with four colours of pages in each section: yellowfor information, blue for simple self-tests, green for basic practice questions andmauve for KS3-style practice questions.

I can't fault this little gem and if the school capitation doesn't allow for newbooks next year, recommend it to doting parents as the perfect Easter present −cheap, cheerful and it'll last until July at least!


Transforming , by ATM Working Group. Pp. 22. £4.50 (£3.60 for ATM members).1998. ISBN 1 898611 02 5 (Association of Teachers of Mathematics).

The title of this book raises the question, ‘What is “Transforming”?’ The designof the front cover gives an enticing clue to the content, and the reader will not bedisappointed. The choice of title is then clearly explained in the introductoryparagraph.

Also in the introduction, the authors outline what they intend to cover in thetext. It is stressed that the suggested activities do not form a complete programme,and the additional resources and IT packages mentioned are listed comprehensivelyin the appendix. This list is a useful aid to the busy teacher, helping to avoid thefrantic search for appropriate resources, which can be off-putting and can oftendetermine whether or not a particular concept is taught.

The main areas covered are in chapters entitled Rotating, Reflecting, Enlargingand Combining Transformations which are themselves divided into subsections. Thismakes the text very user-friendly. Exact lesson plans are not given, but the key

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teaching points for delivering the various concepts are suggested and possibleactivities are described. The authors also give suggestions for further discussion andways to increase the pupils' knowledge, as well as extension work for groups withinthe class. There is a clearly identified progression through each of the sections. Thisis particularly noticeable in the last section, Combining Transformations, where thetext describes how the class teacher can expand and develop difficult ideas.

This book would encourage a teacher to enlarge his/her personal knowledge oftransformations. All aspects covered in the text are onlytouched upon in mostmathematics schemes at this level and, in my experience, for many children this isnot enough for them to acquire full understanding. This book gives a starting pointfor development of a complementary programme.

If teachers become familiar with the recommended software packages and timeallows, this text would be an ideal resource to aid delivery of this topic.

JACQUELlNE G. ROBERTSONJohn Logie Baird Primary School, Helensburgh G84 9EP

Calculator maths, by Alan Graham and Barrie Galpin.Foundations: Pp. 56. £8.95.1998. ISBN 0 9533 137 0 0.Number: Pp. 65. £8.95. 1998. ISBN 0 9533 137 1 9.Algebra: Pp. 60. £8.95. 1998 ISBN 0 9533 137 2 7.Shape: Pp. 50. £8.95. 1998ISBN 0 9533 137 3 5.Handling data: Pp. 65. £8.95. 1998 ISBN 0 9533 137 4 3.£33.50 for the series. (A + B Books).

The series consists of five slim books, between 50 and 65 pages each, designedfor the Texas Instruments range of graphical calculators (TI-80, TI-83, and TI-82)and is aimed at pupils aged 12 to 16.Foundations covers the basic calculator work −use of the various keys, list handling, graphing (sic) and simple programming;Number goes through the obvious numerical operations;Algebra includes formulas,graphs, equations (simple, simultaneous and quadratic) and inequalities;Shapeincorporates perimeters, areas and volumes, Pythagoras, similarity, loci, circles andelementary trig;Handling data uses the random number generator to look atprobability, frequencies, bar charts, histograms, line graphs, scatter diagrams andlines of best fit. Each area of mathematics is covered by several worked exampleswith explicit and very clear instructions for using the TI calculator key stroke by keystroke. The books are a perfect marriage of showing how easily mathematics can bedone with a suitable calculator and, at the same time, demonstrating the full range ofuse of the TI graphical calculators. The only drawback is that they are so model-dependent. I tried a few pages out on my trusty Casio fx-something-or-other andcouldn't get past ‘Set the Mode button to 4’ or ‘6 STO> ALPHA A’. The seriescannot be used for anything other than the TI-80, TI-83, and TI-82 as the instructionsin every question are so specific.

If you do have the right kit, then the series certainly covers all the functionsalthough many of them are a bit contrived − for example, simple equations are notexactly equations − you put in the number first and do the operations then try toguess what number you entered in the first place! Simultaneous equations are donegraphically, of course, as are quadratics; similarity of shapes is simply an exercise inplotting points. However, a graphical calculator makes short work of all aspects ofhandling data and number work, and the books do them well.

I don't see the series being used toteach all the aspects of the nationalcurriculum but more as explorations exciting pupils' interest preceding more explicitwork. Ideally, calculator work should encourage pupils to ask more questions ratherthan provide answers and, in this sense, the books are too prescriptive in their


instructions, killing adventure rather than encouraging it. Of course, the authors havehad an up-hill struggle anyway, as any of us trying to use graphics calculators in theclassroom will tell − they're damn hard, rarely give the right result and actuallydetract from the mathematics in question! Now, where's the chalk?


Basic mathematics skills: revision and practice, by A. Ledsham and M. E.Wardle. Pp. 376. £10.00. 1998. ISBN 0 19 914704 3 (Oxford University Press).

Now this is the sort of mathematics textbook I always wanted to write. Millionsof examples, all of a practical and realistic nature covering everything in basicnumber work including handling data and shape and space set in ‘everyday’applications.

‘Basic mathematics skills has been written to provide a sound foundation inbasic skills specifically aimed at those students who wish to develop the numeracyskills that are particularly applicable to everyday life.’ Whilst the authors go on tostate that the book is aimed at students working towards City and Guilds Numeracyand RSA Number Skills Certificates etc., I think it is also ideally suited forFoundation and ‘Basic Skills’ students in school, both mainstream and those takingGNVQ courses.

Each topic is introduced with a mixture of teaching points and worked examplesfollowed by structured and progressive exercises. Although the introductoryexplanations are brief, I think the book could equally be used by the independentlearner either as a source book or as a very useful revision guide. There is themandatory answers section at the back, something I personally loathe, and there areno past exam-type questions, but I can't imagine a better basic skills (née arithmeticand mensuration) book for both individual and class use. When I reviewed one ortwo of the other books in the seriesGCSE mathematics: revision and practice, I hadthe impression that they had been quickly cobbled together by Oxford from oldertexts, butBasic mathematics skills has a fresh, new feel to it that is immediatelyengaging. Highly recommended.


Oxford mathematics: Foundation GCSE, by Jim Kirkby, Peter McGuire, DerekPhilpott and Ken Smith. Pp. 400. £13.00. 1998. ISBN 0 19 914717 5.Oxford mathematics: Intermediate GCSE, by Sue Briggs, Peter McGuire, DerekPhilpott, Susan Shilton and Ken Smith. Pp. 414. £13. 1997. ISBN 0 19 914694 2.Oxford mathematics: Higher GCSE, by Sue Briggs, Peter McGuire, DerekPhilpott, Susan Shilton and Ken Smith. Pp. 416. £13.00; 1998 ISBN 0 19 914707 8(Oxford University Press).

Each of the threeOxford mathematics GCSE books provides a self-contained,two-year course for GCSE mathematics aimed at Years 10 and 11. They form thefinal part of theOxford mathematics series for Years 7 to 11 which includes thetwenty-four Year 7 and 8 topic books and the Year 9 Link books.

Each of the books is written with a view toorganised learning − a fairly rareand very refreshing approach − and uses some innovative ideas. Firstly, the index isat the front and for each topic there are four references − the main section, its review,practice and, finally, where it appears in any exam questions at the end of the book.The second most striking feature is the intelligent use of full colour − blue panels for

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‘Starting Points’ (information and skills needed before starting a new section),yellow panels for learning, occasional red text that links with other sections and bluetext for harder work.

Each section starts with a ‘need-to-know’ panel followed by highlightedinformation and worked examples, a fair number of exercises and, finally, a reviewin which each question has its skill cross-referenced. From time to time there is a‘Skills Break’ − an extended piece of work based on a single, real life situation (forexample, a day trip to Paris) utilising all the different skills learnt so far − and ‘EndPoints’ − a longer, basic revision exercise which is, again, cross-referenced. And ifthat is not enough practice, there are about five, straightforward, major revisionsections with about 50 questions each!

There are one or two little niggles of course − irritating typing errors in algebraand fractions; trigonometrical ratios being initially defined as OPP/HYP etc. thenhaving to be redefined for ratios of all angles; and some foundation work missingfrom the Higher book which often jumps in with only higher work. But on thewhole this is an excellent set of books which should be attractive and interesting forpupils without having fallen into the popularist trap.

On the flyleaf there is a free hotline number (0800 318245), a call to whichresulted in the overnight delivery of a very useful pack giving details of how thewhole series links together and even an analysis of every exam board's syllabusstatements against page numbers in the GCSE books. Organised learningandteaching − now there's a winning formula!


Improve your maths! A refresher course, by Gordon Bancroft and Mike Fletcher.Pp. 206. £11.95. 1998. ISBN 0 201 331306 (Addison-Wesley).

I quote from the preface: ‘This book aims to help all students who intend to startor who have already started a course in business studies, the social sciences or othersubject areas that require a student to be reasonably numerate…. It can be used bothas an introductory text before a course begins and throughout a course wherever aparticular mathematical skill or concept causes difficulty. Individuals who are notstudents, but who wish to develop numeracy skills vital to everyday life, may alsofind this text a valuable aid.’ The chapter headings are: Arithmetic Operations;Fractions and Decimals; Percentages and Ratios; Powers and Roots; Tables; Charts;Co-ordinates; Graphs; Averages; Spread; Correlation; Probability; Simple Algebra;Linear Equations; Simultaneous Equations; Quadratic Equations. Each chapterbegins with a statement of objectives, e.g. Chapter 2: ‘After reading this chapter youshould be able: to express a fraction in its simplest form; to add, subtract, multiplyand divide fractions;… (etc)…’. The text follows the usual routine of introductoryexplanation, worked examples, and exercises. Basic definitions, formulae, andtechniques are labelled ‘Key Point’ and enclosed in a box.

All is very clearly set out in much detail. Although the book is intended for self-study, there are places where a guiding hand might be helpful − e.g. to thoseunfamiliar with it, the use of notation in the statistics sections could be off-putting.The book is purely practical: virtually no attempt is made to prove or justifyanything: e.g. is taken almost for granted, some attempt to explain thenotation is made, but (say) the formulae for the roots of a quadratic or forcalculating standard deviations are just stated. It is assumed that a calculator will beused for all but the simplest arithmetical operations, and moreover that a fairly

Σ

a0 = 1a−n


sophisticated one is available, as reference is made to the use of an key forcalculations with vulgar fractions and to and keys for powers and roots. Thereis no geometry at all, nor any trigonometry − in fact a better title for the book wouldhave been ‘Improve your numeracy!’! Answers are given to all the exercises, andthere is an index. Although we read in the preface: ‘It is important to realize that tounderstand a piece of mathematics fully requires considerable application byattempting and successfully completing a number of examples and exercises’, Iconsider that, as with so many modern texts, compared with those of earlier days, theprovision of examples is rather meagre. The text is enlivened with a few reasonablyamusing cartoons.

abc

xy x1/y

It will be clear from my description that this is not a book for the would-bemathematician, but for those who need to know about and use basic techniques andare scared of the subject or lacking in confidence, it could be quite a useful guide.


STP National Curriculum mathematics 10B, by L. Bostock, S. Chandler, A.Shepherd, E Smith. Pp. 426. £12.00. 1999. ISBN 0 7487 3191 1 (Stanley Thornes).

The STP National Curriculum mathematics series is designed for secondaryschool students working at Higher and Intermediate levels. The books are helpfullynumbered by year group, starting at book 7, and in two, parallel series, A for Higherand B for Intermediate students. That, however, is about as good as it gets.

It might be that I was in poor humour at the end of a very busy term. It mightbe that I had just reviewed one of the best of the new breed of school texts (theOxford mathematics series). Whatever the reason, this book just doesn't appeal andis exactly the sort of mathematics text book that has put off generations of ‘normal’people (defined as those not being totally mad about mathematics). It presents a dry,featureless landscape with vast acres devoid of scenery to trudge through but one inwhich the traveller fears he or she will soon get bogged down. Even the words arenegative − on the random page open at the moment introducing an exercise onPythagoras' Theorem we read ‘For theproblems in this exercise… in most cases thiswill involve adding a line. You will also need to know how to find the areas ofsquares, rectangles, parallelograms and trapeziums.’ (my italics). And this beforeyou start. I can just picture Marcus telling me to ‘get a life Mr. W.’

From a teaching point of view I could see little organisation of learning in thelayout of the chapters. For example, we have this sequence of topics: formulas −length, area and volume − enlargement − straight line graphs − similar figures −changing the subject of a formula; and there seems no logic to the placing ofsummary sections throughout the book which, in themselves, imply there is asequence to be followed. Sadly, in the introduction, the authors write to the pupil‘Mathematics is an exciting and enjoyable subject.’ Not in this book it ain't.


Oxford revision guides GCSE : mathematics through diagrams, by AndrewEdmondson. Pp. 128. £9.00. 1998. ISBN 0 19 914708 6 (Oxford University Press).

The introductory pages of this revision guide give a very full explanation of thecourses and examining bodies that pupils may encounter within the GCSE structure.The introduction also has four pages that offer excellent advice and tips on how tostudy, plan and prepare for examinations at all levels.

REVIEWS 155

The revision notes and questions in the book aim to cover all syllabuses,examining bodies and levels within the structure. Notes and questions aimed atpupils sitting the Higher level of the GCSE are marked with an , but pupils whoare sitting exams at a Foundation or Intermediate tier may require assistance from ateacher to highlight appropriate course content and suitable revision questions.

H

The main contents of the textbook are well-structured and presented with clearexplanations and examples. It must be emphasised however, that other revision textswill be required to give full practice at examination questions as there are not enoughin this book. Fully worked solutions for the examination questions can be found atthe back, together with a formula sheet and a full index for quick topic and subjectreference.

Although this is a revision guide for pupils, I feel that they would gain mostbenefit from using the book if they first have the aid of a teacher at a ‘study skills’class, ensuring it is not read from cover to cover but used for reference to strengthenknowledge. The book may therefore be better as a school resource given out beforeexam time rather than as an individual pupil purchase.

CHRISTlNE HUNTERGarnock Academy, Kilbirnie, Ayrshire KA25 7AX

Practice for advanced mathematics: pure mathematics, by K. M. Morley.Pp. 244. £9.99. 1999. ISBN 0 340 70167 6 (Hodder & Stoughton).

Practice for advanced mathematics provides a bank of graded questions whichcovers most of the post-2000 A level common core. Like its companion volumes inmechanics and statistics, it is envisaged that the questions will mainly be used toconsolidate classroom understanding rather than for unsupported self-study, althoughkey points and worked examples are provided for those wanting to use it for revisionpurposes. The exercises range from routine to A level standard: it is worth notingthat all are newly written rather than recycled and that there is a good blend ofcalculator/non-calculator questions. The twelve chapters are organised thematically:Algebra − Trigonometry − Coordinate Geometry and Functions − Differentiation −Integration − Numerical Methods. This is a sensible response to what seems to be anaxiomatic fact of life: given any teacher, any textbook and any class, teaching order≠ chapter order≠ revision order! (When challenged about this by students, I mimicEric Morecombe's immortal reply to André Previn, ‘I am playing the right notes, butnot necessarily in the right order!’)

There are a few typos (such as transpositions of letters and non-italicisedvariables) and a bizarre misprint on page 10 which seems to imply that all logarithmsare rational numbers. From a personal standpoint, I found the author's style ratherstaccato and the pace, in places, uneven. In particular, while recognising that this isnot a conventional textbook, I did miss some of the commentary from the classroomthat supplies the glue to make the key points and methodology of the workedexamples really stick. I also had a few qualms about the syllabus coverage: there aresome variations in the content of the core-covering pure modules of the various post-2000 specifications but no questions included here on the factor formulae, the form

, the general binomial expansion, harder curve sketching (withasymptotes and behaviour at infinity) and vectors. These though are minor quibbles:this book merits a wide circulation − it is the sort of no-nonsense course companionthat I suspect a lot of us A level teachers would have liked to have got around towriting!

a cosθ + b sinθ



Practice for advanced mathematics: statistics, by Alan Smith. Pp. 188. £9.99.1998. ISBN 0 340 70165 X (Hodder & Stoughton).

This book is primarily an extensive bank of practice questions (with answers)serving the needs of AS/A level candidates either as a course support text or as arevision aid. It is worth emphasising straightaway that the exercises are the book'sgreat strength: they are inventive, well-structured (being graded A, B, C where A aresingle-skill questions and C are essentially A level questions) and notably fresh − theauthor has resisted any urge to recycle old examination questions. Coverage extendsas far as hypothesis tests for two means, -tests and the non-parametric sign andWilcoxon tests. Perhaps it is ‘top of the syllabus heavy’: I was surprised not to findcoding/pooling for means/variances, counting problems, or non-polynomialprobability density function questions. Each set of exercises is prefaced by a well-chosen collection of worked examples: these are clearly set out in a friendly styleusing ‘bubbles’ for reminders. Less successful to my eyes were the ‘Key point’summaries: these are very thin and their very baldness tends to magnify lapses andhighlight omissions. For example, taking variance as a definitionstruck me as eccentric; the probability summary only really covers conditionalprobability; there is an unfortunate discrete/continuous transposition near the top ofp. 28; and both the statement of the Central Limit Theorem (p. 86) and the definitionof a confidence interval (p. 90) are much too sloppy. And I always like to mentionthe equation of a Normal distribution (to bring home that it is not any old bell-shapeddistribution!) and the fact that Spearman's rank is precisely the product momentcorrelation coefficient for the rankings. To balance these gripes, the book isexcellent on protocols for hypothesis testing, the ‘big three’ distributions and theirmutual approximations, niggly details (such as continuity corrections, Yates, andmodelling assumptions for bivariate analysis) and the summary on Data Presentation(including an inventive ‘spot the errors’ exercise) has gone straight into my teachingfile.

χ2

= Σx2 / n − x¯ 2

Overall then, a well-organised compendium of fresh, bread-and-butter statisticsquestions which I could see as being particularly valuable in conjunction with amodular series of texts where (with zealous or resitting candidates) it is all too easyto run out of questions. But do check carefully against your own syllabusrequirements: you may find some surprises.


Practice for advanced mathematics: mechanics, by Peter Nunn and DavidSimmons. Pp. 212. £9.99. 1998. ISBN 0 340 70166 8 (Hodder & Stoughton).

This book supplies a comprehensive bank of practice questions to support anAS/A level course in mechanics. With the possible exceptions of light frameworksand dimensional analysis, all topics on current ‘pure with mechanics’ syllabuses arecovered, including variable acceleration, non-uniform circular motion, obliqueimpact, calculus methods for Centre of Mass and SHM; the coverage is notablythorough on motion in 2 dimensions and energy, work, power, momentum, impulseand impacts. The questions are crisp, sharply focused and workman-like; they areusefully graded A, B, C in terms of increasing difficulty and sophistication. Inparticular, the C grade questions are not just recycled past examination questions, butsome arguably do have a rather dated feel relative to some recent syllabusinnovations.

The exercise sets are supplemented by worked examples and summaries of keypoints: these are brisk and contribute to my personal hunch that some of the text may

REVIEWS 157

be a shade ambitious for the average student, for whom arguably the greatest hurdlesare the initial stages such as coping with vectors, setting up force diagrams,formulating equations and matching mathematics to physical intuition. I noticed oneor two oddities: a rearranged equation on p. 7 should read and thehighlighted tangential acceleration on p. 118 should be ; also is notmentioned but is, although the scalar product is nowhere defined orused!

v − u = atrθ s = vt − 1

2at2

2a.r = v2 − u2

The teaching of mechanics tends to excite strong passions andGazette readersmay have views about some of the authors' decisions such as to treat projectilesentirely vectorially, to tend to leaveg in equations until numerical evaluation at theend, to use exclusively modulus of elasticity (and not stiffness) in Hooke's Law and,on occasions, to suppress forces from force diagrams if they are not needed in theanalysis. But the book's undoubted strength is in its outstanding collection of well-constructed exercises: these will almost certainly cover your syllabus needs andusefully augment areas in which some of the recent modular texts are apt to be ratherthin.


Calculus mysteries and thrillers, by R. Grant Woods. Pp. 131. £16.95. 1999. ISBN0 883 85711 1 (Mathematical Association of America).

Calculus mysteries and thrillers is a collection of 11 problems suitable forstudent projects. Each problem is introduced by means of a short story and theprojects take the form of a report to be prepared for a particular purpose. In effect,the student is put in the position of a mathematical consultant reporting to a client.

The projects are classified as easy, moderate or difficult, though the term isrelative. For example, ‘The Case of the Swivelling Spotlight’ is described as difficult‘because it is likely to be assigned early in a course’, though the mathematicsinvolved (tangents, normals and Newton-Raphson applied to a cubic) is within thescope of an A level student. Indeed, most of the projects can be solved with A levelmathematics, the exceptions being three that require knowledge of arc-length orsurface of revolution and one that involves an application of the intermediate valuetheorem. However, given the difficulties that students tend to experience when askedto apply recently learned knowledge, some might prefer to use the problems in thefirst year of a university course.

Model solutions, written in the style appropriate to the context, are provided forall eleven problems. The publishers give permission for purchasers to copy theprojects for their students, so a mathematics department could legitimately buy asingle copy. This excellent classroom resource represents excellent value for money.


Understanding statistics, by Graham Upton and Ian Cook. Pp. 657. £18.50. 1997.ISBN 0 19 914391 9 (Oxford University Press).

Understanding statistics is designed to cover all A level syllabi and also to besuitable for introductory statistics courses at university level. It is a large book and isvery comprehensive in its content. It has plenty of exercises, both single topic andmiscellaneous exercises at the end of each section. Answers to the exercises areprovided.

The book is well written and easy to read, with lots of helpful asides and notes.


There is a useful glossary of notation at the front. The layout is clear and easy tounderstand. However, the authors aimed to produce more than just a comprehensiveintroductory statistics course. In their introduction they say: ‘The purpose of thisbook is to present a wide range of essential statistical ideas in a simple and (we hope)enjoyable fashion. If a user of the book does not derive the tiniest bit of pleasurefrom some part of the book then we will be disappointed (but there will be no cashrefunds).’ In order to fulfil this aim, they have provided historical notes and usedhistorical examples in some of the exercises. These are interesting. I enjoyedlooking at historical examples of graphs. These included graphs by FlorenceNightingale and a graphical record of Bonaparte's march to Moscow. I also foundout why a regression line is called a ‘regression’ line. Where historical examples arenot readily available for the exercises, the writers have introduced some light-heartedquestions to provide some variety. For example, one of the basic probabilityquestions begins: ‘A class of 100 students comprises a group of 40 people called“idiots” and a group of 60 called “complete idiots” . . . . . .’ !

I am finding this a very useful book to have. The measure of enjoyment I getfrom reading the book is to be found in how often I am distracted from the section Iam supposed to be looking at, because something interesting has caught my eye.This happens quite a lot!

I think every statistics teacher should have a copy of this book. For those whocannot afford this book for their pupils, the authors have produced a slightly shorterversionIntroducing statistics for only £12.50 which was reviewed in theGazette83(November 1999) p. 550.

H. MASONMadras College, St Andrews, Fife KY16 9EJ

The art and craft of problem solving, by Paul Zeitz. Pp. 280. £19.99. 1999. ISBN0 471 13571 2 (Wiley).

Those who enjoy good problems − either for themselves or for their students −will welcome this book. The first half contains four chapters on generalities (‘Whatthis book is about and how to read it’, ‘Strategies for investigating problems’,‘Fundamental tactics for solving problems’, ‘Three important crossover tactics’); thesecond half has four topic oriented chapters (on Algebra, Combinatorics, Numbertheory, and Calculus). Each chapter contains dozens of problems suitable forinterested students in their last two years at school, or in undergraduate courses witha problem solving theme.

However the book claims not only to contain lots of good problems, but tointroduce the reader to ‘The art and craft of problem solving’. This is a bold claim,and I shall examine it on two levels.

First, any mathematician who loves good problems and who sees how they canbe used to catch the imagination and to motivate hard work in their students is boundto look for ways of sharing such problems with a wider audience.

In a culture with a rich problem solving tradition among students and teachers itmay sometimes suffice simply to print lists of problems − structured by topic, or byapproximate level of difficulty. ‘But most people don't grow up in [such a] problemsolving culture’ (Preface, page x). Lacking such a culture, the prospective authorneeds a framework within which this rich material can be presented, and throughwhich students and teachers can be tempted to try more of the problems than theyotherwise might. Hence, some such context is certainly needed. In this instance, theapproach chosen by the author has allowed him to structure the problems he

REVIEWS 159

presents, and so to improve the likely impact of the book. As long as readers expectno more than this, and are prepared to work on a relatively high level, they will notbe disappointed.

However, the title appears to promise rather more than this. As the new centurydawns it may be worth taking this opportunity to look back over recent attempts tomake the ‘craft of problem solving’ something which can be explicitly taught.

Every good mathematics teacher has to coordinate two conflicting features ofmathematics. First, most worthwhile activity depends on using good problems − thatis, problems that can be relatively easily understood, but whose solution is far fromobvious, leading the would-be solver into unexpected territory. Second, goodproblems of this kind are by their nature unpredictable, and so are generallyperceived as being ‘too hard’ for most students.

One way of resolving this conflict is to restrict the mass of ordinary students toa more predictable diet of routine ‘exercises’. This has the unfortunate effect ofpresenting mathematics as a kind of over-processed mental pap − with all theinteresting (and hence problematic) tastes, spices and lumps removed, and withnothing left for students to really chew on. When such students meet real food, theirmental digestive systems cannot adapt, and they are left to pick at the pieces in anunsatisfying and ineffective way.

In recent years there have been, at all levels, attempts whose declaredmotivation was to break with this perceived tradition. Primary schools haveencouraged exploration; secondary curricula and examinations have includedinvestigation and problem solving − both as part of the routine diet and as assessedcoursework − and universities and teacher training courses have developed moduleswhich focus on the ‘process’ of solving problems.

At first sight such moves appear attractive. However, much of this work hasbeen rooted in optimism or ideology (e.g. ‘key skills’) rather than in any seriouspedagogical or didactical framework. Rhetoric has regularly outpaced reality, andthe outcome has often been to reduce potentially rich problem solving activity tosomething strangely similar to (but less useful than) the ‘predictable diet of routineexercises’ it was meant to replace. Faced with a very difficult task, and lacking anyeffective didactical framework, students, teachers and examiners have naturally beenforced to make the task more manageable. In the process they have oftenmisrepresented the mathematical character of the problems used, and have replacedthe elusive nature of the problem solving process by a more predictable ritual (‘Makea table; try a few simple cases; etc.’) whose main virtue is that it can be taught andassessed.

Such degeneration is not necessary. Videos of ordinary mathematics lessons inother countries show how one can develop a didactically precise way of using hardproblems to motivate the development and application of routine techniques.Lacking such a framework, the best we can often do to keep the flame ofmathematics alive is to ensure that students have the experience of struggling tosolve good problems.

In mathematics, insight and confidence can only be bought at a price. Euclid,when asked to recommend a short-cut to wisdom, is credited with the reply: ‘there isno royal road to geometry’. And when Gauss was asked how he achieved hisprofound insights, he replied that if others would think on things as long and as hardas he did himself, they could not fail to achieve similar insights. Most of theproblems presented and discussed in this book are entirely consistent with thismessage. And as long as the author's chosen framework is perceived as a light-


hearted way of encouraging his readers to travel further along this path than theyotherwise would, little harm will be done. But it is important to realise that the titlemeans no more than this.

The level of pedagogical sophistication is visible already in the Preface (pageix): ‘The average (non-problem solver) math student is like someone who goes to thegym three times a week to do lots of repetitions with low weights on various exercisemachines. In contrast the problem solver goes on a long, hard backpacking trip.Both people get stronger. The problem solver gets hot, cold, wet, tired and hungry.The problem solver gets lost, and has to find his or her way. The problem solvergets blisters ... .’; and again on page 5 ‘Some branches of mathematics have verylong histories, with many standard symbols and words. Problem solving is not oneof them. We use the terms “strategy”, “tactics” and “tools” to denote three differentlevels of problem solving.’ Such details are not only harmless, but − for those whoare not already committed problem solvers − can provide welcome packagingbetween successive bouts of exertion. However, when the author seeks to develophis private psychology of solving problems as part of some universal ‘art and craft ofproblem solving’, this reviewer regularly lost patience. The overall strategy is toooften naive (‘Anything that stimulates investigation is good’ (page 29)); and whenthe author attempts to motivate solutions which might elude the beginner, themotivation is sometimes harder to swallow than the solution being motivated (e.g.page 7).

Such idiosyncratic details are an inevitable feature of courses taught byindividuals − and can contribute to their appeal. But a book whose main titlepromises so much should be more than a printed version of such a course. So buythe book for problems to solve − but be prepared to take the advertised ‘art and craftof problem solving’ with a pinch of salt.

TONY GARDINER77 Farquhar Rd, Birmingham B15 2QP

A primer of abstract algebra, by Robert B. Ash. Pp. 181. £19.95. 1998. ISBN 0883 85708 1 (Mathematical Association of America).

It has often been remarked that there is a considerable difference between themathematics studied at school and that met in an undergraduate course. Manystudents find it difficult to cope with the abrupt leap in abstraction. This book isdesigned to precede first courses in abstract algebra and analysis, and so ease thetransition.

The material included is well-chosen, though the title suggests a much broaderrange of topics than is actually included. Vector spaces are the only algebraicstructure covered in any depth, for example. The exposition follows the definition-theorem-proof style typical of more advanced texts, but includes plenty of discussionof ideas to sweeten the pill. Each chapter has a number of exercises to whichsolutions are provided in an appendix.

There are six chapters, beginning with Logic and Foundations, which introducestruth tables, quantifiers, proofs, sets, functions and relations. The next two chaptersdeal with numbers: first in the sense of counting and countability, then via somebasic number theory. The latter chapter includes the Euclidean algorithm, uniquefactorisation, congruences and Diophantine equations, including the Chineseremainder theorem. The opportunity is taken to define algebraic structures such asgroups, rings and fields, with the integers modulon serving as examples. Fermat'slittle theorem is given as a corollary of Euler's theorem. The chapter ends with a

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section defining the Möbius function and linking it with Euler's φ function.

Chapter 4 returns to the theme of sets and the thorny issues of the well-orderingprinciple, Zorn's lemma and the axiom of choice. The last two chapters are devotedto a more extensive study of vector spaces and linear operators. Though thedefinitions are given in terms of arbitrary fields, the examples are mostly set inrn.Chapter 5 covers linear independence, change of basis, inner product spaces andeigenvalues. By the last chapter we are well beyond anything a student is likely tohave met at school. Jordan canonical form, normal and self-adjoint operators areintroduced leading to proofs of the Cayley-Hamilton Theorem and the spectraltheorems. The existence proof for Jordan canonical form is given in the last section,but readers are advised to skip it on a first reading.

Although it was written with American students in mind, this primer follows onrather neatly from English A level mathematics courses, and is therefore well wortha look. The book could be used as the basis of an introductory course for first yearundergraduates or for pre-course reading. Very able school students might find thefirst four or five chapters useful in preparation for Olympiads.


Probability and random variables: a beginner's guide, by David Stirzaker.Pp. 368. £16.95 (paperback) £45.00 (hardback). 1999. ISBN 0 521 64445 3paperback) ISBN 0 521 64297 3 (hardback) (Cambridge University Press).

‘Probability is the only branch of mathematics in which good mathematiciansfrequently get results which are entirely wrong.’ This quote, from C. S. Pierce, istypical of the many true and entertaining remarks that occur in this fine book. Thecase of PaulErdõs, who was wrong about the Monty Hall problem, discussed onpage 84, might have been mentioned (The man who loved only numbers, PaulHoffman. 1998 London: Fourth Estate). Wrong results would be reduced if themathematician had read this clear exposition of the basic concepts of mathematicalprobability and random variables. Continuous distributions naturally need somecalculus, but otherwise the prerequisites are modest and clearly explained at thebeginning of each chapter. After introductory material concerned with the nature ofprobability, rather than its mathematics, there is a chapter on the rules of probability,which are then applied to counting and gambling. The first part concludes with achapter on distributions. Part two deals with random variables in both discrete andcontinuous cases and concludes with a chapter on generating functions. There aremany examples, most of which are both mathematically interesting and of practicalrelevance. There are numerous exercises, with hints and solutions provided formost.

The most impressive feature is the clarity of the writing. The author has theability to proceed carefully and at length yet, at the same time, is not boring. It isdifficult to introduce humour into a mathematical text, yet here there are severaldelightful touches that enliven what could easily become dull, repetitive logic.Historical references also add to the interest. It is one of the most impressiveintroductions to the theory of probability, explaining the basic concepts anddeveloping them to the point where the results are interesting, yet not to where theybecome obscure or difficult. As an introduction for a mathematician to probability,it has rarely been bettered.

It does have its limitations, for its primary, and intended, emphasis is on thetheory, so that the practice is principally confined to the many excellent examples


and exercises. For example, suppose you are presented with the usual die with sixfaces; then there is a difference between your belief that it will show a 6, say, whenreasonably tossed, and the frequency of 6's in a long series of similar tosses. In aclassic experiment, Weldon had a belief of 1/6 but the frequency exceeded thisvalue. In presenting probability as frequency, the author does restrict the applicationof his topic, in particular to statistical questions. He fortunately does not show thesnobbishness towards statistics that is common amongst probabilists, except when hesuggest we statisticians are labourers (page 214); this in connection with the aptly-named law of the unconscious statistician. It is unreasonable to criticise an authorfor omitting something it was not intended to cover, but some criticism might belevelled at his failure to distinguish between continuity and its stricter form, absolutecontinuity. The treatment of independence for more than two events is cursory,though at least the definition is correct. Although he quotes Jeffreys ‘probability is afunction of two propositions’ he often fails to recognize this. The justification forexpectation is not convincing.

These are but minor blemishes to set against the genuine achievement of writinga clear, and interesting, introduction to mathematical probability that all involved inteaching at sixth-form, or first-year undergraduate, levels should look at.Remarkably, even the ‘blurb’ does not exaggerate.

D. V. LINDLEYWoodstock, Quay Lane, Minehead TA24 5QU

Bernhard Riemann 1826 − 1866: turning points in the conception ofmathematics, by Detlef Laugwitz, translated by Abe Shenitzer. Pp. 357. SFr148.1999. ISBN 3 7643 4040 1 (Birkhäuser).Riemann, topology and physics (2nd edition), by Michael Monastyrsky, translatedby Roger Cooke. Pp. 215. SFr98. 1999. ISBN 3 7643 3789 3 (Birkhäuser).

In his short mathematical career, Riemann made important contributions to realand complex analysis, analytic number theory, geometry, topology and mathematicalphysics. His name is associated with the Riemann Hypothesis, the Riemann integraland Riemannian geometry. Many developments in modern mathematics are relatedto Riemann's mathematics.

The two books under review complement each other nicely. In his book, DetlefLaugwitz describes the work of Riemann and of many other nineteenth centurymathematicians and argues that Riemann was responsible for a major change inmathematical culture. Michael Monastyrsky devotes more space to Riemann's life,but then switches to an account of twentieth century developments in topology andphysics that can be traced back to Riemann.

Laugwitz has structured his book to bring out the revolutionary nature ofRiemann's contributions to mathematics. The first three chapters (numbered 0 to 2)cover Riemann's life, his work on complex analysis and number theory, and hiscontributions to real analysis. Chapter three has sections on Riemann's geometry,physics and philosophy. In each of these chapters the author describes at some lengththe situation in the years before and after Riemann's own work. In this way, hethrows Riemann's efforts into relief, allowing the reader to appreciate the scale of thechanges in mathematical thinking in the mid-nineteenth century.

In the final chapter, Turning Points in the Conception of Mathematics, Laugwitzdiscusses some of the ‘paradigm shifts’ in mathematical thinking and suggests thatRiemann's importance in this respect has hitherto been overlooked by manycommentators. He constructs an argument based on historical evidence from

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Riemann's own papers and the writings of others such as Dedekind, Cantor andHilbert. He identifies a number of turning points where he considers Riemann's ideasto have been influential and proposes a unifying theme: that Riemann workedconceptually, seeking to understand ideas in preference to the algorithmic thinking ofhis contemporaries. In this respect Riemann helped change the way mathematiciansoperate.

The extensive background material that Laugwitz has marshalled to support hismain thesis also serves as an excellent general account of the development ofmathematics in the nineteenth century.

Monastyrsky first wrote a biography of Riemann because there was nothingavailable in Russian. He also wrote a separate book on topological themes in modernphysics. Both were so well received that they were translated into English andpublished together in 1987, with some updating of the work on topology andphysics. The primary purpose of the second edition is to further update this secondsection. The biographical section of around 80 pages is largely unchanged, andremains an excellent account of Riemann's life and mathematics.

Since the Russian edition was published, mathematical physics has moved on.Monastyrsky can (and does) congratulate himself on foreseeing the continued vitalityof this field, and now updates his account. He describes a variety of ways in whichtopology and physics have developed through the interaction of shared concepts. Hedeals with symmetry breaking, liquid crystals, gauge fields, instantons, solitons,knots and braids, and the future. The whole area is noted for the difficulty of themathematics and resists attempts to give a description for the lay-person. However,Monastyrsky does achieve a description that can be appreciated by mathematicallywell-informed readers such as the readers of the Gazette.

Both books are important for anyone who wants to understand how mathematicschanges its nature over the years.


La correspondance entre Henri Poincaré et Gösta Mittag-Leffler, Présentée etannotée par Philippe Nabonnand. Pp. 421. SFr228. 1999. ISBN 3 7643 5992 7(Birkhäuser).

Mittag-Leffler (1846-1927) and Poincaré (1854-1912) corresponded regularlyfrom 1881 to 1911, when Mittag-Leffler retired. Following his graduation, in 1872,from Uppsala University in Sweden, Mittag-Leffler studied with Hermite in Parisand attended lectures by Weierstrass in Berlin before settling in Helsingfors. In 1882,newly installed as a Professor at the Högskola in Stockholm, he founded the journalActa Mathematica. Hermite alerted Mittag-Leffer to the talents of his studentsAppell, Picard and Poincaré, with the result that Poincaré contributed a number oflengthy papers to the first few issues of the journal.

Poincaré had attended the Ecole Polytechnique and the Ecole des Mines beforeachieving his doctorate in 1879. He spent two years at the University of Caen beforemoving to the University of Paris in 1881. He is considered by some to be the lastman whose interests encompassed the whole of the mathematics of his time.

This book is a publication of the Archives Henri-Poincaré, the first in a plannedseries of four volumes of Poincaré's correspondence. It collects 259 letters thatpassed between the two mathematicians. Most are held at the Mittag-Leffler Instituteand all are written in French. Many of them are quite short, particularly in the lateryears, but some have a substantial mathematical content. The editor, Phillipe


Nabonnand, has annotated the letters to provide much background detail and manyreferences to the works of other mathematicians of the period. In many cases theannotations are longer than the letters. A few letters that passed between Poincaréand Fredhom, Gyldén and Phragmén are collected in appendices.

In the first letter of this collection Mittag-Leffler seeks information aboutPoincaré's work on automorphic functions. Poincaré's first two papers inActaMathematica concerned Fuchsian functions, a particular class of automorphicfunctions.

The book contains a substantial bibliography and a list of names, andNabonnand has written a 26 page introduction which includes much biographicalinformation about Mittag-Leffler as well as Poincaré. However, for most historiansof mathematics the interest will lie in the letters themselves. As well as beingprimary historical sources for those interested in the two correspondents and theirfields of mathematics, the letters shed light on other aspects of scientific life at theturn of the twentieth century.

The second volume in the series will cover Poincaré's correspondence withphysicists, the third his correspondence with other mathematicians and the fourth hisprivate and administrative letters.


Mathematics and mathematicians: mathematics in Sweden before 1950, by LarsGårding. Pp. 288. 1998. $75. ISBN 0 8218 0612 2 (American Mathematical Society/London Mathematical Society).

In this book, which he dedicates to Swedish mathematicians, the distinguishedanalyst Lars Gårding gives an in-depth chronological account of the history ofmathematics in Sweden. This is largely a tale of three university centres Uppsala(founded in 1477), Lund (1668) and Stockholm (1880); (moreover, with just half-a-dozen mathematics professors between them even early in this century, it is a storylargely driven by the talents and charisma of individuals.

Gårding only cites the work of two pre-nineteenth century mathematicians(Klingenstierna and Bring) and the bulk of the book centres on the period 1860 to1950 − from the birth of modern Sweden to the post-war reorganisation of itsuniversities. He supplies a crisp, urbane commentary with neat summaries ofbackground mathematical themes (in modern gloss) and an insider's sympathy forpeculiarities of Swedish academic life. One of these concerns the perils ofappointments by open committees: we learn that in 1873, Björling was preferred toBäcklund (of transformations fame) and that, in 1923, the experts were unable todecide between the merits of Marcel Riesz and one Nils Zeilon! Although Gårding'sprimary aim is ‘to write about the work of mathematicians for readers interested inmathematics’, he fleshes out the mathematics with historical and biographicaldetails. He is not afraid to express strong opinions and his witty one-liners, togetherwith anecdotal quotations from archival sources, and the 40 photographs at the endof the text serve to bring the pen-portraits to life.

Sweden was rather a mathematical backwater until the 1880s and theappointment of Gösta Mittag-Leffler as the first professor of mathematics at thenewly-founded Stockholm University. Fired with Weierstrassian enthusiasm for thetheory of analytic functions, it was not so much his own mathematical achievementsas his energy, vision and initiative which led to him becoming an internationallyrespected figure with the clout to put Swedish mathematics on the map. This was

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reflected in his founding of the journalActa Mathematica (in the early volumes ofwhich Poincaré published on Fuchsian groups and Cantor on set theory), in hispromotion of the 1885 King Oscar II Prize Competition (one of the winners of whichwas Poincaré on the three body problem), in his securing of the appointment ofSonya Kovalevski to a post at Stockholm, and in his bequest of what later becamethe Mittag-Leffler Institute. (There is even a linkage to more recent times: AndréWeil, who died in August 1998, testified to Mittag-Leffler's nose for youngmathematical talent when he recalled how the latter, just before his death in 1927,promised to publish his thesis (unseen and then unfinished!) inActa.) Since then, asGårding puts it, Sweden has not been lacking in good mathematicians: I'll list a fewof those described together with brief reminders of their work.

Bendixon (Cantor-Bendixon theorem)Beurling (inner/outer functions, spectral analysis)Carleman (Carleman kernels, quasianalytic functions)Cramer (modern probability theory) Fredholm (integral equations)Frostman (potential theory) von Koch (snowflake curve)Nagell (number theory) Nörlund (difference equations)M. Riesz (conjugate functions, Riesz-Thorin theorem).

But, as Gårding rather elegiacally reflects in a postscript, the attraction of a booksuch as this is not so much in the rehearsing of those names which Posteritysanctifies but in the dusting-off of those which Time has forgotten: who now hasheard of Björling, Dillner, Falk, Malmsten, Wiman, or Gullstand (who won the 1911Nobel Prize for Medicine)? Indeed, to share a personal reminiscence, the nameEdvard Phragmén rang a rusty bell with me from the Phragmén-Lindelöf principle(an extension of the maximum modulus theorem to sectors). He now emerges fromthe shadows first as the eagle-eyed proof-reader who detected Poincaré's seriousmistake in his initial submission for the King Oscar II Prize (the rectification ofwhich led Poincaré to his intimation of the possibility of chaotic behaviour ofsolutions to the three body problem, [1]). He then succeeded Kovalevski asprofessor at Stockholm in 1892 but (as Erdös would have put it) he ‘died’ in 1903 tobecome a highly successful chief inspector of insurance!

I found this book full of such surprises and fresh insights: I can warmlyrecommend both it, and the preceding twelve volumes (such as [1]) of the jointAMS/LMS History of Mathematics series in which it takes it place.

Reference1. June Barrow-Green, Poincaré and the three body problem, AMS/LMS (1997),

reviewed in Math. Gaz. 83 (July 1999), p. 343.NICK LORD

Tonbridge School, Kent TN9 1JP

The mathematics of Plato's Academy: a new reconstruction(2nd edn.), byD. H. Fowler. Pp. 441. £60. 1999. ISBN 0 19 850258 3 (Clarendon Press).

The first edition of this book caused considerable interest, and even controversy,from its appearance in 1987, for the author really did offer a new interpretation ofGreek mathematics, especially Euclid'sElements. The principal theses, whichconcern both the mathematics itself and the available sources, may be summarised asfollows:

1) Reliable historical evidence for all ancient Greeks is scanty, and not only for


the deep past such as the Pythagoreans; for example, the provenance of the statementover Plato's Academy ‘let no one unskilled in geometry enter’ dates only from the+6th century, and even the reading of the word ‘geometry’ needs pondering (ch. 6).

2) The emphasis on geometry was indeed strong; the objects treated were, forexample, lines, rather than lengths upon which a manner of measuring and therebyarithmetic has been imposed. In particular, ratios of lines have to be distinguishedfrom rational numbers (chs. 1-4).

3) The Euclidean algorithm led to a sophisticated theory of (say) geometricalmagnitudes and their parts (ch. 2), with especial attention in Euclid's Book 10 toproperties (including ratios) of the forms and for givenmagnitudes and ; relationships between squares and their sides was an importantmotivation (ch. 5). This theory resembles that of continued fractions in arithmetic,though not to be identified with it; but features of the latter, a curiously fugitive topicin mathematics, can be examined (ch. 9) to improve understanding.

( a + b) ( a + b)a b

4) Much doubt is to be cast upon the well-known claim that the discovery ofirrational numbers caused a crisis in Greek mathematics. Apart from the distinctionbetween such numbers and incommensurable ratios anyway, there are good reasonsto think that the properties of the algorithms, including periodicity of the residues inmany cases, led to exciting developments (ch. 8).

5) Many of the oldest texts are written on materials such as papyrus or wood, sothat their layout and state needs to be considered in detail (ch. 6 and several plates).So do the systems of numerals deployed, especially concerning fractions (ch. 7).

The chapter numbers given above are still valid, for the new edition does notexhibit major remodelling. Some further plates are provided, and ch. 5 has beenrewritten; but the main addition is a new ch. 10, which focuses mainly upon theses 3)and 4) above. The reader is now advised to start with the opening section of this newchapter before proceeding to its predecessors; but this is not very satisfactory if (as islikely) he is unfamiliar with the kinds of argument deployed.

Two features of the first edition have been retained where change would havebeen appreciated. First, the textual references to items in the substantialbibliography are given as, say, 'Hogendijk HTATGIG', which is an unwelcome useof acronyms. Second and more important as an example of the dominance ofgeometry, these Greek mathematicians spoke of, for example, ‘the squareon theside’ and not ‘the squareof the side’, which is redolent of the multiplication oflengths. The author's stress on this point is sabotaged by his denoting a square online by the notation ‘ ’, which has been read in the other sense for centuries underthe influence of common algebra. Preferable alternative notations include ‘ ’,the tetragon (or square) on , which was adopted by E. J. Dijksterhuis in his Dutchedition of theElements in 1929-1930, along with similar symbols for cubes, circles,and so on.

b b2

T (b)b

The summary above shows that this book is a research monographparexcellence, and so not directly usable for mathematics teaching before a late stage ina first-degree course. But the issues raised are of major importance in order not tomisunderstand ancient Greek mathematics, and the book deserves to play animportant role on the rewriting of general and introductory histories of mathematics.

I. GRATTAN-GUlNNESSMiddlesex University at Enfield EN3 4SF

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Geometry civilized: history, culture and technique, by J. L. Heilbron. Pp. 309.£35. 1998. ISBN 0 19 850078 5 (Oxford University Press).

As the subtitle indicates, this beautifully produced and sumptuously illustratedbook is a multi-layered work. The mathematical canvas consists of a faithfulaccount of Euclid'sElements (Book I-IV and some of VI) but Heilbron embroidersthis to make a rich cultural and historical tapestry depicting the place of geometry inthe natural and man-made worlds, its traditional role in a liberal education and thepure pleasure of geometrical problem solving. He touches on a wide spectrum ofapplications − from Vedic altars to manhole covers, from native American patternsto Gothic tracery, from the ovals used in the lay-out of St.Peter's Square, Rome to theoctagons favoured by Thomas Jefferson in domestic architecture. And he integratesan impressive range of early sources: among many such gobbets, we learn that thereis a reference to the ‘angle in a semicircle theorem’ in Dante'sDivine Comedy andthat the surprising observation that, in a walk round the Equator, the extra distanceyour head travels compared with your feet is independent of the size of the Earth,goes back at least as far as a problem posed in 1715.

After an introductory chapter, the next four chapters take geometry from basicdefinitions, postulates and common notions as far as familiar angle, triangle,Pythagoras and circle theorems; chapter 6 tackles some harder applications. Twosamples will, I hope, convey the flavour ofGeometry civilized. First, in betweenpages 152 and 153 there are 8 colour plates; these depict Islamic geometricalmosaics, the great octagonal dome of Santa Maria del Fiore, Florence, a mosaic ofthe death of Archimedes (one of several Archimedes episodes mentioned), twolovely illustrations ‘Geometria’ and ‘Surveying’ from Reisch'sMargaritaphilosophica (1512), contrasting pedagogic uses of colour in Byrne'sEuclid (1847)and in the Chinese dissection method of ‘colour and cut’, and Raphael's famousSchool of Athens fresco (1510-11) − with a convincing identification of the theoremabout star hexagons that Euclid appears to be enunciating there. Second, in chapter 5‘From Polygons to Pi’, which is ostensibly about circle theorems, we encounter enroute Stonehenge, Descartes' explanation of the rainbow, inscribed/circmscribed/escribed triangles, regular -gons (with exact constructions for = 3, 4, 5, 6, 8, 15and Renaldini's approximation, [1]), Gothic windows, Archimedes' bounds for(and related Egyptian and Chinese calculations), a table of specific cross-referencesto Euclid, and 35 exercises.

n nπ

Some proofs are given with full Euclidean pomp (including, pleasingly, severalof the reductio ad absurdum arguments) but Heilbron sensibly uses alternativearguments and methods either for clarity or for contrast or, especially in his use ofalgebra and trigonometry, for the comfort of modern readers. Several laterdevelopments are included, notably the theorems of Ceva, Menelaus, Ptolemy, thelines of Euler and Simson, Hero's area formula and Descartes' algebraic solution ofApollonius' three-circle tangency problem [2], charmingly reconstructed from hiscorrespondence with Princess Elisabeth of Bohemia. There are also somefascinating cameo digressions; Eratosthenes on the size of the Earth and Columbus'planning for his New World voyages, surveying methods for inaccessible heights,the dangers of misleading diagrams, refraction (rainbows and burning-mirrors) andHuygens' acceleration of Archimedes' method for calculating . Raw theory isfurther consolidated by an excellent collection of some 150 solved exercises fromsources as diverse as ancient Chinese texts, the 1802 Cambridge Tripos papers and(especially) the pages of theLadies' Diary. Heilbron has a connoisseur's eye for anattractive problem − indeed, he cites his own delight in solving geometry problemsas a major motivation for composing this book. My only quibble concerns his

π


solutions: I noticed several algebraical typos and a number which could be tidied-up.For example, Exercise 5.2.4 (p. 205) features three circles radii sharing twocommon tangents with the middle circle touching the other two. The published

solution leads to the expression which the author fails to cancel to

which, in fact, would have fallen out immediately from a different, shorterargument. Also, although I can understand Heilbron's desire not to range beyondalgebra and basic trigonometry in his repertoire of techniques, it was a pity not tofind references to alternative methods such as transformation geometry, [3], for thetheorems of Aubel and Napoleon (4.2.26, 4.2.27), inversion (as in [4]) for the radii ofnested circles in Gothic tracery (6.5), or to previous work on what the author callsthe ‘Tantulus problem’. The latter seems to have caused quite a stir when it surfacedin the columns of theWashington Post in 1995, but will be all too familiar toGazettereaders as the ‘adventitious angles’ configuration of [5, 6].

a, x, b

x =a b + b a

a + bx = ab

The teaching of geometry was, of course, the contentious issue which led to theformation of the Association for the Improvement of Geometrical Teaching − theprecursor of the Mathematical Association − in 1871. Geometry civilized is a timelyreminder of just what we lose in mathematical, pedagogical and cultural terms if wefail to teach an honest dose of geometry, ‘... geometry exercises, whereas algebrarelaxes, the mind’ (p. 27l). With the readability, leap-off-the-page enthusiasm andwide-ranging scope of one of David Attenborough's Natural History blockbusters,perhaps the OUP could be persuaded to produce a more attractively priced paperbackedition − even if we cannot have an associated television series!

References1. D. Bousfield, The construction of an approximately regular -gon,Math. Gaz.

66 (Oct. 1982) pp. 229-30.n

2. H. Dörrie, 100 great problems of elementary mathematics, Dover (1965)pp. 154-60.

3. J. Rigby, Aubel & Thébault's theorems,SymmetryPLUS9 (Summer 1999)pp. 4-5.

4. M. Harvey, Ever decreasing circles and inversion,Math. Gaz. 82 (Nov. 1998)pp. 472-475.

5. C. Tripp, Adventitious angles, Math. Gaz. 59 (June 1975), pp. 98-106.6. D. A. Quadling, Last words on adventitious angles,Math. Gaz. 62 (Oct. 1978),

pp. 174-183.NICK LORD

Tonbridge School, Kent TN9 1JP

A history of algorithms: from the pebble to the microchip, by Jean-Luc Chabert(ed.), tr. Chris Weeks. Pp. 524. 1999. ISBN 3 540 63369 3 (Springer-Verlag).History of mathematics, histories of problems, Inter-IREM Commission, tr. ChrisWeeks. Pp. 429. 1997. ISBN 2 7298 4730 8 (Ellipses, Paris).

In recent years French readers have benefited from the astonishing energies of agroup of French historians and mathematics teachers, many associated with thehistory and epistemology section of the nationwide ‘IREM’ movement (Institutes forResearch into Mathematics Education). A large number of collaborative works haveappeared in print, in which mathematical texts from the past are studied and re-presented as resources for mathematics teachers today. The combination of

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experienced mathematics teachers, mathematics educators and historians ofmathematics has generated reliable and well-judged books in which the riches of themathematical past are made available and contextualized in a historical andeducational narrative, thus providing material for teachers to use in theirmathematics lessons in a variety of ways.

Only a small proportion of this cornucopia has hitherto been translated intoEnglish. An early such venture was the set of essays published by the MathematicalAssociation in 1990 under the titleHistory in the mathematics classroom: the IREMpapers. Two years ago there appeared a volume of fifteen essays,History ofmathematics, history of problems, containing material on a range of topics to appealto upper secondary school pupils such as infinity, prime numbers, ruler and compassconstructions, and non-Euclidean geometry. Since some thirty IREM members wereinvolved in working on the book, somewhat like an Open University Course Team, ithas modestly appeared under the authorship of ‘The Inter-IREM Commission’,though one of the driving forces was evidently the Commission's chair, EvelyneBarbin.

Any of these essays serves to illustrate the style and quality of this production.‘How may pictures appear to be real?’ by Didier Bessot and Jean-Pierre Le Gaff,from the Caen IREM, is a well-illustrated survey of perspective and projection fromEuclid to Lambert, through Brunelleschi, Alberti, Piero della Francesca, Dürer, andDesargues, with a number of in-text exercises and their answers (called ‘How didyou get on?’). Jean-Pierre Friedelmayer's ‘A desperate search’ is a clear and livelytour through the history of equation-solving from the Rhind Papyrus to Galois,impressive in the way it combines a strong narrative flow with sufficientmathematical exposition for students to get their teeth into. And the other essays areequally helpful and accessible.

History of mathematics, histories of problems was translated into English byChris Weeks, whose growing reputation as a skilful translator of material like this iswell-deserved, built on his experience as a teacher and teacher educator over manyyears. Now his translation has appeared of the splendidHistoire d'algorithmes: ducaillou à la puce (Berlin 1994). This work, written by a similar team and indeedsome of the same writers, under the overall leadership of Jean-Luc Chabert of theUniversity of Picardy, is one of the most enthralling of these collective works, in partbecause it has a single theme richly illustrated and built up over several hundredpages. To ensure the highest standards of scholarship, the team of seven included aleading historian of Arab mathematics (Ahmed Djebbar) and of Chinesemathematics (Jean-Claude Martzloff). The result is a delight and should be on everyteacher's bookshelf. The book can serve as a resource for teaching numericalanalysis, but more than that presents a view of current mathematics as the inheritorof its history, and of mathematics teaching as something that can be enriched andstrengthened by knowledge of that history. As history books go, it has a largequantity of original sources (in translation), so is well suited for a number ofdifferent classroom approaches, styles and needs.

All concerned with this production, from the authors and translator to thepublishers and their designers are to be congratulated on an excellent addition toeducational and historical resources for English-speaking mathematicians.

JOHN FAUVELFaculty of Mathematics, The Open University, Milton Keynes MK7 6AA


The emergence of the American mathematical research community 1876-1900:J. J. Sylvester, Felix Klein and E. H. Moore,by Karen Hunger Parshall and DavidE. Rowe. Pp. 500. £25.00. 1997. ISBN 0 8218 9004 2 (American MathematicalSociety / London Mathematical Society).

As International Exhibitions go, the one which took place in Chicago in 1893was small beer. Only 2.5 million attended compared with 32 million who had turnedup to marvel at the new Eiffel Tower in Paris four years earlier, and, when theExhibition returned to Paris in 1900, no fewer than 50 million passed through theturnstiles (these attendance figures according to the latest edition (2000) ofLe petitLarousse).

The authors of this book on the beginning of American mathematical researchsee the World’s Columbian Exposition of 1893—to celebrate 500 years postChristopher Columbus—and the associated Mathematical Congress as a pivotalpoint in the growth of American mathematical research. The city of marble (‘WhiteCity’) which arose on the marshlands on the south side of Chicago was host to theEuropeans, and in particular a strong German mathematical presence. The Germanscame to the New World with their latest mathematical wares (abstruse theses, eruditebooks, geometrical models), led by Felix Klein and Göttingen mathematics. By thistime many young American post-graduates had visited Germany to sample theeducational system at first-hand, and were receptive to a new mood and thirst forresearch.

The conjunction of the old world meeting the new is nicely captured on the frontcover with a picture of the ‘White City’, flags fluttering around its mock classicalarchitecture, and combined with a sepia insert of a gathering of the Germanmathematical establishment. In a rapidly constructed venue linked by waterways tothe industrial exhibits and showcases, American mathematics received an definiteimpulse by way of the mathematical exhibits. It was only technically inter-nationalsince there were only four non-American mathematicians present, but many otherssent good wishes and had their papers readin absentia. David Hilbert, HeinrichWeber, Eugen Netto and Adolf Hurwitz were in absentia on the opening dayfollowed later by Hermann Minkowski, Charles Hermite and Arthur Schönflies, withthe show rounded off by Felix Klein in person. After the Exposition, Klein gave hisfamous Evanston Colloquium Lectures.

The structure of the book gives a well-rounded picture. An overview of the‘Emergence’ is followed by an outline of American mathematics immediately priorto Chicago. Sylvester’s contribution was a shock to the American way such as itexisted at that time. He encouraged the callow youth to abandon a view ofmathematics as a body of knowledge carved in stone waiting to be learnt off byheart. Sylvester, as no other could do, demonstrated that mathematics was a livingsubject. During 1876-1882 Sylvester showed his small band of students at JohnsHopkins University that they too could do research, and that it was the real businessof mathematicians to make new discoveries. As the authors point out, Sylvester’sclassroom was no place for the student who placed a premium on taking away a clearset of lecture notes—he never even had a set himself. Lectures from Sylvester werelaunched from his latest mathematical thought and the whole class, which in a veryreal sense included himself, were involved with speculation, inductive reasoning,wild guesses and all those artifices which mathematicians are not supposed to use.

Meanwhile some students were travelling abroad, and by the late 1880s, asteady stream of post-graduates had shiny new German PhDs to show theirprospective employers. (It is surprising to learn that the much heralded German PhDcould be obtained for as little as a semester in residence, writing a thesis followed by

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an oral examination.) The experience of each traveller varied. Those who went toBerlin, for example, received a wide exposure to mathematical ideas, unlike the oneswho took part in Sylvester’s seminar, where the fare was limited to his own subjects:higher algebra, analytical geometry and specialised questions in the theory ofnumbers. In Göttingen Klein soared above the details and exposed his students to thevastness of mathematics. The section of the book devoted to Klein teases out hischaracter. At root a highly competitive man, Klein ran his Göttingen Seminar onautocratic lines. He did not tolerate fools gladly, as the hapless James Boyddiscovered. Boyd made the fatal mistake of not learning the basics of mathematicsfirst!

Towards the end of the nineteenth century researchers in mathematics werebeginning to be grown at home. The principal schools emerging in the USA wereChicago, Harvard and Princeton. In Chicago, E. H. Moore was the main drivingforce. The young group of mathematicians to emerge from this school at thebeginning of the new century included L. E. Dickson, Oswald Veblen, R. L. Moore(of the ‘Moore method’ of teaching mathematics) and George David Birkhoff. Eachone made a mark in his own way, although Dickson is perhaps undervalued today.

To specialists in the history of nineteenth century mathematics this will be anindispensable source. Throughout the book factual material is supplied, whichotherwise might be scattered or difficult to obtain. There is a list of lectures given atKlein’s Seminar between 1881-1896 (the Leipzig and Göttingen years) and thelecture programme of the Chicago Mathematical Congress is carefully documented,to give just two examples. The authors are expert on the mathematicians in the titleof the book, Parshall (on Sylvester and Moore) and Rowe (on Klein and Germanmathematics). Although two authors were involved, they have succeeded in writinga seamless book and one which reads well. It is fully footnoted and there is anextensive bibliography. Photographs enliven the presentation and especiallywelcome are the photographs of the lesser known American mathematicians wholaid the groundwork for the research schools of America which exist today.

TONY CRILLYMiddlesex Business School, The Burroughs, Hendon, London NW4 4BT


Introduction to cardinal arithmetic , by M. Holz, K. Steffens and E. Weitz.Pp. 304. SFr88. 1999. ISBN 3 7643 6124 7. (Birkhäuser).

It is not uncommon forThe Mathematical Gazette to receive a paper on cardinalarithmetic, but very few are published. The reason for rejection is usually that theauthor has misunderstood some aspect of Cantor's work. For example, a recent paperclaimed to show that the set of real numbers between 0 and 1 is countable. In fact itshowed only that the set of numbers with finite decimal fraction representations iscountable. This will not surprise anyone who knows that the rational numbers arecountable.

I doubt that these misguided authors would appreciate a text book on cardinalarithmetic, for their minds are made up: they cannot accept the concept of a biggerinfinity than the cardinality of the natural numbers. For the rest of us however, thistextbook provides a valuable, if fairly intense introduction to the classical ideas andmore recent work on cardinal arithmetic.

Chapter 1 begins with the Zermelo-Fraenkel axioms of set theory and the axiomof choice (the ZFC theory of sets) and develops the classical theory of Bernstein,Cantor, Hausdorff, König and Tarski from the period 1870-1930. Even this first


chapter is quite hard going for anyone starting from scratch: there are dozens ofdefinitions to absorb before one can make sense of the considerable number oflemmas and theorems. In addition, there is much discussion of what can beestablished with various different axiomatisations of set theory. Indeed, in theintroduction, readers are warned if they want to accept the generalised continuumhypothesis (that is the smallest infinite cardinal greater than ) they can stopreading immediately, since the exponentiation of cardinals then obeys very simplerules. More interesting is the study of exponentiation using the ZFC axioms alone.Note that, between them, Kurt Gödel and Paul Cohen proved that the consistency ofZFC implies the consistency of ZFC with the generalised continuum hypothesis(Gödel) and with its negation (Cohen).

2κ κ

The second chapter is based on the work of Galvin, Hajnal and Silver from the1970s. Tarski and Gödel had already noted the importance of the gimel function

, where is the cofinality of the infinite cardinal . The Galvin-Hajnal formula provides an estimate for the possible size of the value of the gimelfunction when the argument is one of the successor cardinals of ¼0, the cardinality ofthe natural numbers. From this formula springs a wealth of other results on cardinals.

κ → 2ck(κ) ck (κ) κ

The first two chapters make up almost half of the book. The remaining sevenchapters are devoted to a presentation of pcf theory, which was developed by S.Shelah in the 1990s to deal with many of the open questions that remained from theseventies. This later material brings the determined reader almost to the limit ofpresent knowledge of cardinality results provable in ZFC.

The press release claims that the book is aimed at ‘undergraduates, and also atpostgraduate students and researchers who want to broaden their knowledge ofcardinal arithmetic’. I doubt the validity of including of undergraduates in theaudience for the book. Perhaps things are different in Germany, the authors' homecountry, but in Britain, I would be surprised if even final year students were able toget far beyond the first chapter.


Logic as algebra, by Paul Halmos and Steven Givant. Pp. 141. $27. 1998. Vol.21. ISBN 0 88385 327 2 (Mathematical Association of America).

The algebraization of logic, envisaged first possibly by Leibniz, received its firstsignificant boost with the work of George Boole. Recognising the inadequacies ofAristotelian syllogistic reasoning as a basis for a general theory of inference, Boole,by restricting numerical algebra to two values 1 and 0, developed Boolean algebra.Others, notably De Morgan, W. S. Jevons, C. S. Peirce and E. Shröder, by theirelaboration of the algebra of logic based on Boole's work, edged ever closer tofulfilling Leibniz's dream of a Characteristica Universalis.

This slim volume is the 21st in the seriesDolciani Mathematical Expositions, asignificant proportion of which have been written by Ross Honsberger. What does itprovide? As the title suggests it is intended ‘to show that logic can (and perhapsshould) be viewed from an algebraic perspective…. Moreover, the connectionbetween the principal theorems of the subject and well-known theorems in algebrabecome clearer.’ Readers anticipating arguments based on truth tables or diagramsof switching circuits will be disappointed. In compensation they will be entertainedby a rich array of algebraic concepts such as prime and maximal ideals, filters,homomorphisms, equivalence classes, kernels, quotient algebras and duality, all inthe service of logic. As the authors state, ‘propositional logic and monadic predicate

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calculus − predicate logic with a single quantifier, are the principal topics treated’.The propositional calculus and Boolean algebra are introduced in a gentle and clearmanner and their relationship explained. This leads to a section on Booleanuniversal algebra which culminates in the maximal ideal theorem and the powerfulrepresentation theorem of M. H. Stone which asserts that every Boolean algebra isisomorphic to a field of sets, a field of sets being a collection of subsets of a setclosed under complementation, union and intersection. (A theorem that asserts that,for a given axiomatic theory , a distinguished subset of the set of all models has theproperty that every model is isomorphic to some member of this subset, is arepresentation theorem for ).

T

TPre-Boolean algebras, quotient algebras and Boolean logics feature in the next

section, Logic via Algebra, where propositional logic receives a more thoroughgoingalgebraic treatment than hitherto. This section concludes with the deductiontheorem, the strong soundness theorem, (every theorem is valid), the strongcompleteness theorem, (every valid formula is a theorem) and, finally, thecompactness theorem for this system. The short section that follows this adumbratesthe concept of a lattice which constitutes a generalisation of Boolean algebra.

In the final chapter, Monadic Predicate Calculus, quantifiers (‘there exists’ and‘for all’) now appear. It should be recalled that first order predicate calculus enabledmathematicians to formalise set theory, which in turn provided them with anadequate foundation for formulating all other mathematical objects and structures.Now the algebraization of this area of logic seems to have been realised first by S.M. Ulam and C. J. Everett in theirProjective algebra I (1945), where the authorssought to express abstractly the Boolean algebra of subsets of a plane and theirprojections on the two coordinate axes, thus facilitating an algebraic treatment oflogical quantifiers. Subsequently, two further generalisations of Boolean algebrawere created, namely, the cylindric algebras of Tarski and his collaborators and thepolyadic algebras of one Paul Halmos. The last chapter of this book, however, dealswith monadic algebra which is defined as a Boolean algebra together with anexistential qualifier. The authors proceed ‘to show how the theory of syllogismsfinds a simple and natural expression in the framework of monadic algebras.’ Notuntil the final sentence of this tract does the creator of polyadic algebras refer to hisown exposition on the subject inAlgebraic logic. Accordingly, we can conceive ofthis last chapter as an an introduction to more profound matters. Indeed, the wholewill serve as a neat, succinct, introduction to logic particularly for readers very muchat home with algebraic concepts.

GRAHAM HOARE3 Russett Hill, Chalfont St. Peter SL9 8JY

Introduction to set theory (3rd edn.) by Karel Hrbacek and Thomas Jech. Pp. 291.$69.75. 1999. ISBN 0 8247 7915 0 (Dekker).

This very nicely constructed little book really needs two reviews. The first oneis all praise. The aim of the authors is to provide a final undergraduate yearintroduction to set theory as complete as possible without requiring the reader tomaster any symbolic logic. This aim is completely met and in a very readable form,especially because of the way in which the numerous straightforward but oftenlengthy proofs are set as examples (with hints), keeping only the important ones tobe set out in the text. They begin with the intuitive notion of a set, introducingRussell's Paradox as early as page 2, to motivate the need for care. The whole bookprovides a reasoned argument for Zermelo-Fraenkel set theory with (after Chapter 8)the Axiom of Choice. In addition, the uses of set theory in the construction of the


natural numbers, the reals, cardinal arithmetic and ordinal arithmetic are fully dealtwith. This third edition adds later chapters on filters and ultra filters, partitions foruncountable cardinals, Suslin's problem, large cardinals and the axiom of foundation.These later chapters do not prove everything − this would hardly have been possiblewith the absence of logic − but they do give enough to give the reader a good idea ofthe notions and the general approach of proofs. In this way the reader is brought intocontact with the latest work in an active field, although without full proofs.

The second review is slightly more critical. As well as teaching the young how‘to do the sums’ in set theory, they should be told what they are doing; the authors donot shirk this task. On page 1 they nail their colours to the mast: ‘Sets are notobjects of the real world, like tables or stars, they are created by our minds, not byour hands.’ What is being said here? It is clear from later on that the first clause isnot to be read as denying sets are objects, only the sort of object. So it seems that thedistinction being made is between a box of 144 apples and a set of 144 apples. Butdoes the box then contain the set (as well as the apples)? Does it have 145 objectsalbeit of different kinds in it? These are not quibbles; they have exercised manyminds. The question of whether the third member of the Holy Trinity was really theset of the other two divided the Western and Eastern churches. So it seems to methat there are two questions about their statement: firstly, to explore its meaning inthe way it is used; secondly, to see whether it pays off in the later understanding ofthe theory.

As I noted above, the early motivation for axiomatic set theory comes fromRussell's Paradox. They draw the lesson that ‘by merely defining a set we do notprove its existence (similarly as by defining a unicorn we do not prove that unicornsexist).’ Surely something is wrong here? When we define a unicorn, there issomething created by our minds, and it is hard to see how this does not exist (thoughnot like tables or stars, but like sets)! It would have helped here to have hadsomething more on the relation between existence and consistency. By page 39 wereach the usual ZF definition of the natural numbers as sets; so we have to concludethat the natural numbers also are not objects of the real world but created by ourminds. This has the advantage of brushing away any questioner like Frege, who maywant to know what the natural numbers are, though the shoe pinches in other places.Is it only a creation of our minds that 2 + 2 = 4? Wittgenstein would have it that thiswas a grammatical proposition; Quine that it was no different from an empirical one,accepted to ‘expedite our dealings with sense experience’. Where do the authorsstand between these two? Certainly they put some emphasis on being concerned‘only with sets ofmathematical objects, such as numbers, points of space, functionsor sets. Actually, the first three concepts can be defined ... as sets ... So the onlyobjects with which we are concerned from now on are sets.’ (The apparentconjuring trick of climbing up with no objects at all is achieved, of course, bystarting with the null set.)

What of the pay-off for this rather specific notion of reality? Here it seems tome that not as much use has been made of it as might be. To take only one example:the axiom of choice is introduced in a fairly standard way. It is shown that a finitesystem has a choice function, and pointed out that the same proof will not do for acountable system. Then the axiom is described as a new principle of set formation,different because non-effective. Gödel and Cohen and their results are mentioned. Iwould have thought that a useful pay-off would be to see Cohen's results as showingthat the mind has the power to construct different systems of sets − to remove, as itwere, any philosophical cramp about the axiom of choice or equally about theContinuum Hypothesis.

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Such nit-picking should not be read as saying that the book is other than anexcellent introduction to the techniques of set theory. Besides, I cannot but warm toa book which gives me such a quotable paragraph as: ‘Some mathematicians objectto the Axiom of Infinity on the grounds that a collection of objects produced by aninfinite process ... should not be treated as a completed entity. However, mostpeople with some mathematical training have no difficulty visualising the collectionof natural numbers that way.’ Note the implied threat in the second sentence;perhaps it can be read as a warning against ‘some mathematical training’.

C. W. KILMISTERRed Tiles Cottage, High Street, Barcombe BN8 5DH

Practical foundations of mathematics, by Paul Taylor. Pp. 572. £50. 1999. ISBN0 521 63107 6 (Cambridge University Press).

Whereas the motivation for logicians in medieval times was theology, today it ismore likely to be mathematics and programming. The book offers an understandingof foundations of mathematics and informatics, beyond the mere codification ofmathematics in logical scripture. The subject matter is becoming important,especially because the possibility of automated deduction may help with computer-assisted construction, verification and dissemination of proofs. Take, for example,the concept of ‘types’. Many readers will be at least vaguely aware of the work ofGeorg Cantor on the magnitude of sets, and the subsequent work of Bertrand Russellon the theory of types in his attempt to resolve paradoxes in set theory. Morerecently, in the development of high level computer languages, it was found that adistinction between integer and real data has to be made if only because they havedifferent storage requirements. Nowadays, the software industry has learned frominformatics that the type discipline helps to make programs more reliable.

The first part of this research-level book is well worth reading by anymathematician or computer scientist, treating it as a text on ‘discrete math for grown-ups’. There is a well written account on the difference between object-language andmeta-language, and also between classical and intuitionistic logic, and why the latteris used in the current context. Order structures are introduced as tools forinvestigating semantics, and they serve to describe systems of propositions, and alsoas the substance of individual types. The wide-ranging examples transcenddisciplinary boundaries between universal algebra, type theory, category theory, settheory, sheaf theory, topology and programming. No part of the book can bedescribed as easy reading, and the more mathematics, logic and informatics youknow, the more you will benefit from reading it. If you already know some, or areparticularly keen to find out about, category theory then much of the book is veryinteresting and useful, but it will be tough going. In any case, unless one isreasonably familiar with ZF theory and categorical type theory, the second half ofthe book will be quite impenetrable.

The following chapter titles are not very informative, but they may convey atleast an impression of what the book is about: 1 First Order Reasoning; 2 Types andInduction; 3 Posets and Lattices; 4 Cartesian Closed Categories; 5 Limits andColimits; 6 Structural Recursion; 7 Adjunctions; 8 Algebra with Dependent Types;9 The Quantifiers. Each chapter opens with a well considered preliminary sectionwhich describes in broad terms what is installed in the chapter. For example, thefollowing is the opening paragraph for Chapter 4.Category theory unifies thesymbolic (Formalist) and model-based (Platonist) views of mathematics. Inparticular, it offers an agnostic solution to the question we raised in Section 1.3 ofwhether a function is an algorithm or an input-output relation.


There are some 50 exercises at the end of each chapter, but only a few of themare really exercises as such; the rest of them seem to be miscellaneous related factsfor which the author could not spare extra space. Several hundred items are listed inthe bibliography, and there is a good index. The book is a beautiful product of thelabour of the author, who composed it using Emacs, typesetting it in TEX using hisown commutative diagram and design macros. Students and teachers of computing,mathematics and philosophy will find it good value as a reference work.

P. SHIUDepartment of Mathematical Sciences, Loughborough University LE11 3TU

The mathematics of ciphers: number theory and RSA cryptography, byS. C. Coutinho. Pp. 196. £19.00. 1999. ISBN 1 56881 082 2 (A. K. Peters).

I quote from the Preface: ‘This book will take you on a journey whose finaldestination is the celebrated Rivest, Shamir, and Adleman (RSA) public keycryptosystem. In fact, the book is more concerned with mathematics than withcryptography. Although the working of the RSA cryptosystem is described in detail,we will not be concerned with details of its implementation. Instead, we concentrateon the mathematical problems it poses, which are related to the factorization ofintegers, and to determining whether a given integer is prime or composite. ... Theway number theory is presented in this book differs in some important respects fromthe classical treatment of some older books. Thus we emphasize the algorithmicaspects everywhere, not forgetting to give complete mathematical proofs of all thealgorithms that appear in the book. ... Hence this is really a book about algorithmicnumber theory and its applications to RSA cryptography. ...’

This is certainly the most readable book about the RSA system that I have seen.Perhaps ‘about’ is not quite the right word: the description of the RSA cryptosystemcomes only in the last chapter, of 9 pages! − but of course this brevity is the result of,and is justified by, the detailed and thorough build-up of the underlying ideas,through fundamental algorithms which include the Euclidean algorithm for the gcdof 2 integers (which, from my school-days, I still prefer to call the hcf!), primenumbers (including Mersenne and Fermat primes), unique factorisation, the sieve ofEratosthenes, modular arithmetic, Fermat's little theorem, pseudoprimes, Carmichaelnumbers, systems of congruences, the Chinese remainder theorem, groups (up toLagrange's theorem), the Lucas-Lehmer and other tests for primes. Although thepreface states that the required previous knowledge of mathematics does not gobeyond geometric progressions and the binomial theorem, the proofs in the bookrequire close attention and are sometimes rather abstract − more in the way ofnumerical instances might have been helpful. Each algorithm, once established, isdisplayed as an outline program, which the reader can then implement for himself onhis own equipment. The historical comments are smoothly interwoven into the text,adding much to the attractiveness and readability of the book. There is a set ofexercises at the end of each chapter − but as these are by no means trivial, it is a pitythat no answers are given. There is a bibliography, and indexes of algorithms and‘main results’ as well as a general index.

For anyone interested in this branch of number theory and its application tocryptography, who is prepared to study the text seriously, this book is to be highlyrecommended.


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Introduction to abstract algebra (2nd edn.), by W. Keith Nicholson. Pp. 599.£58.50. 1999. ISBN 0 471 33109 0 (Wiley).

At 599 pages this is a big book, mainly because nearly everything is provedfully and carefully, there are lots of exercises and lots of examples to illuminate andconsolidate the theory, and the overall style is fairly discursive. All this should, anddoes, make it a text suitable for readers working on their own. The only disincentiveto this is the price tag − students able or willing to invest £58.50 on a single bookwill be hard to find.

There are of course many books with this title, contents and emphasis, and at asimilar level. A text which has been around for a long time, and with which manyreaders will be familiar, is Fraleigh'sAbstract algebra. I would suggest this as asuitable comparison if you are thinking of buying Nicholson's book.

The author claims that different selections from the contents make a one- ortwo-semester course. This is certainly true, though my impression is that two ormore one-semester courses in different years would be most appropriate, due to therising level of treatment. The early chapters certainly lead the reader by the hand,leaving little to chance, whereas later chapters leave far more for the reader to fill infrom his or her own initiative, and depend more on the assumption that mathematicalmaturity has had time to develop.

The starting level is very basic: chapter 0 is a careful explanation of thenecessary preliminaries on types of proof, sets, mappings and equivalence relations.I then turned to some of the early topics which I know from experience can bedifficult to teach − induction, well-defined operations, conservation of parity ofpermutations. The treatment of these seemed excellent.

At a higher level of abstractness, I turned to the material on field theory, inparticular the existence of splitting fields. Here too the treatment was thorough, tookadvantage of the power of abstractness, yet kept the reader's feet on the ground withplenty of concrete illustrations.

To promulgate the message that abstract algebra is not just for puremathematicians the more applicable aspects of the subject get a decent exposure −cryptography, linear codes, cyclic and BCH codes, Polya counting, classicalconstruction problems and Galois theory.

The end level of the book is a survey of work on algebra associated with thenames of Wedderburn, Jacobson, Artin, Brauer, Chevalley and Noether, including aproof of Wedderburn's theorems and some of its extensions. It is written in anappetite-whetting way, which is of course how all books should end.

So the panorama is wide, and Nicholson is a good guide, definitely worthconsidering.

JOHN BAYLISDepartment of Mathematics, The Nottingham Trent University, Burton Street,

Nottingham NG1 4BU

Classical invariant theory, by Peter J. Olver. LMS Student Texts 44. Pp. 280.£13.95 (pb), £37.50 (hb). 1999. ISBN 0 521 55821 2 (pb), 0 521 55243 5 (hb)(Cambridge University Press).

Very recently Gian-Carlo Rota wrote of invariant theory (the ‘classical’invariant theory largely concerned with polynomials) as ‘the great Romantic story ofmathematics’ [1]. It was one of his principal research interests—and indeed he wasa modern champion of the theory and one who stood in direct line from GeorgeBoole, Arthur Cayley and a whole gamut of algebraists down the last 150 years.


Sadly Rota died on 19 April 1999, but he would have seen this work under reviewwritten by one of his former students at MIT. Indeed Peter Olver paid due respect tohis former teacher and the ‘wonderful lectures [which] opened my vistas’ (p. xxi).

Invariant theory aims to bring out intrinsic properties. In the simple case ofpolynomials of one variable there would be no point in treating say,and separately, since one is a linear transformation of theother: they are therefore equivalent in an algebraic sense. In the language of invarianttheory both have the same set of invariants and covariants. In practice invarianttheory does not deal with numerical polynomials but with symbolic ones. Thesimplest example of an invariant is the discriminant for the ordinaryquadratic polynomial . In projective geometry, invariant theory isconcerned with the intrinsic properties which do not depend on choice of coordinateaxes.

x3 + 2x − 38x3 − 12x2 + 10x − 6

b2 − 4acax2 + bx + c

The history of invariant theory reminds us that mathematics is subject to thevagaries of fads and fashions just as any other human enterprise. While a topic inmathematics comes back into fashion at intervals it invariably comes back wearingdifferent clothes. Modern invariant theory hardly follows the lines of Cayley’s andSylvester’s research programme. Rota erected an abstract scheme for the theory in away never dreamed of by the pioneers of the 1840s and 1850s though his work builton theirs. In any explanation of invariant theory, it is significant that the commondiscriminant of the quadratic remains the standard exemplar forall thealgebraic versions of the subject which have arisen. The newness of present daymodern invariant theory is due to the central role which group theory plays.

b2 − 4ac

The present book brings much of classical invariant theory up-to-date. Grouptheory is introduced early, and there is the recognition of the part David Hilbertplayed in the 1880s. However, the modern theory is far from being a continuationfrom the point at which Hilbert left off. To be sure, Hilbert’s Basis Theorem, theSyzygy theorem and the Nullstellensatz are still central planks of the theory, butthere has been a reorganisation too.

It might be well to compare the present text with a text which appeared acentury ago that was designed for the same purpose as the book under review: i.e. tobring invariant theory within the compass of students and to bring it up to date.Thealgebra of quantics was written by Edwin Bailey Elliott, who taught the subject atOxford when Sylvester was there in the 1880s. The work is highly derivative of theEnglish approach to invariant theory, as we might expect, with only a brief nod inthe direction of the German mathematicians Paul Gordan and Alfred Clebsch, thoughit does contain a summary of Hilbert’s work [2].

A good portion of Elliott’s text is bound up with combinatorial techniques in theshape of Eulerian generating functions. This was quite natural for Elliott sincegenerating functions could be used not only to count the number of irreducibleinvariants and covariants but to discover them and the linear dependencies (syzygies)between them. Elliott, influenced very heavily by the English school, used hisintroductory chapters for an amplification of material found in George Salmon’stexts. It adopts the English terminology and it contains a thorough discussion ofCayley’s twin differential operators and . Elliott gives the complete listing of theirreducible invariants and covariants for the polynomials of degree five and six(already established by the ‘King of the invariants’ Paul Gordan, using a compressednotation) and goes on to show the connection between invariant theory and analyticalgeometry. (Olver decided to leave out the combinatorial aspect of invariant theory inhis book because of pressure of space, so he refers the reader elsewhere [3]).

Ω Ο

REVIEWS 179

As mentioned above, the modern view of invariant theory makes transformationgroups as the unifying idea. From this there is an easy extension from polynomials tothe study of differential invariants of Lie groups. It also lends itself to an easierappreciation of Sophus Lie’s attempt to construct a ‘Galois theory’ for differentialequations. Olver’s book gains added interest since his own background is indifferential equations and mathematical physics. An interesting feature of the book isthe reappearance of Sylvester’s ‘chemical’ viewpoint, a way of representinginvariants and covariants by formulae expressed graphically.

Invariant theory is a highly technical subject. To study it will never be easy butthe present text presents the leading modern ideas in a highly cogent andunderstandable form.

References1. Gian-Carlo Rota, Two turning points in invariant theory,The Mathematical

Intelligencer 21 (1999), pp. 20-27.2. E. B. Elliott, An introduction to the algebra of quantics, 2nd edn. Reprint.

(1913).3. Bernd Sturmfels, Algorithms in invariant theory, Springer (1993).



Abelian groups and modules, edited by Paul Eklof and Rüdiger Göbel. Pp. 373.SFr168. 1999. ISBN 3 7643 6172 7. (Birkhäuser).Analysis and geometry in several complex variables, edited by Gen Komatsu andMasatake Kuranishi. Pp. 314. SFr158. 1999. ISBN 3 7643 4067 3. (Birkhäuser).

These books, published in the Trends in Mathematics series, contain conferenceproceedings. Naturally, the papers deal with matters at the forefront of research.

The first book arises from the International Conference on Abelian Groups andModules that was held in Dublin during August 1998. Some of the papers deal withmethods borrowed from other areas of mathematics and applied to abelian groupsand modules, including model theory, category theory, infinite combinatorics,classical algebra and geometry. Other papers use abelian group theory in the study ofmodule theory and non-commutative groups.

The second book is a collection of papers from the 40th Taiguchi SymposiumAnalysis and Geometry in Several Complex Variables, held in Kataka, Japan in June1997. Several of the papers cover recent applications of complex analysis to otherareas, such as partial differential equations, differential geometry, quantummechanics and algebraic geometry.

These books will have limited appeal beyond the respective researchcommunities, but both contain papers which may interest workers in other fields.


Combinatorics: a problem oriented approach, by Daniel A. Marcus. Pp. 136.£16.95. 1999. ISBN 0 883 85710 3 (Mathematical Association of America).

As the title might suggest, the greater part of this attractive little book consistsof problems. The four sections that make up Part I cover Strings, Combinations,Distributions and Partitions. Part II covers more advanced methods of counting, with


sections on Inclusion and Exclusion, Recurrence Relations, Generating Functionsand the Pólya-Redfield Method. Within each section there are introductory problemsthat build towards one of the nineteen Standard Problems (for example, #9 is ‘Findthe number of distributions of a given set of identical balls into a given set of distinctboxes’). These are generally followed by further problems that can be solved bysuitable adaptation of the Standard Problem. The problems are connected by fairlyshort sections of text which include examples and any definitions that are required.

The first 6 sections are found in most elementary books on combinatorics, andthe treatment of generating functions is quite short, so as far as content is concerned,it is the final section that distinguishes the book from its competitors. The Pólya-Redfield Method is useful for solving counting problems where there is an elementof symmetry. One example of this type of problem is to find the number of ways ofcolouring the squares of a 3 by 3 grid using two colours, two colourings beingconsidered the same if one can be obtained by rotating the other. Not surprisingly,this final section contains a dose of group theory.

The book is based on the author's problem-led course on combinatorics to‘mathematics and computer science majors… generally third and fourth year’ atCalifornia State Polytechnic University. The prerequisites are few however, and thebook could form the basis of a first-year undergraduate course. It would also besuitable for independent study, for example by a student preparing for the Olympiad.


Discrete mathematics using latin squares, by Charles F. Laywine and Gary L.Mullen. Pp. 305. £51.95. 1998. ISBN 0 471 24064 8 (Wiley-Interscience).

A latin square of order is an array in which distinct symbols arearranged so that each symbol occurs once in each row and column. Readers mayhave come across such objects in statistical designs used to determine whethersignificant differences in some variable exist between various samples. The subjectitself is rich with unsolved problems, and methods employed on obtaining generalresults touch on a variety of other mathematical areas, especially in combinatorics,finite geometry and coding theory.

n n × n n

The book introduces many basic properties and results of latin squares togetherwith diverse applications. The sixteen chapters are divided into four parts, with thefirst two parts devoted to the introduction to latin squares and generalisations such aspermutation cubes, orthogonal hypercubes and frequency squares. Relatedmathematics, such as the sieve principle, groups and graphs, are dealt with in thethird part. The nine chapters on applications are given in the last part, whichconstitutes half of the book. There is a useful chapter on ‘nets’, which are point setswith a very uniform distribution in a high dimensional cube, and can be used toovercome the problem of generating truly random sequences in numerical techniquessuch as Monte Carlo methods. Other topics include affine designs, statistics, error-correcting codes and cryptology, with some topics being discussed in quiteconsiderable detail. Duplicate bridge players may be interested in a short chapter on‘Room squares’, a topic based on the article [1] in theGazette by T. G. Room, whichis related to the construction of Howell movements. The last chapter gives shortintroductions to more applications including conflict-free access to parallelmemories, broadcast squares and tournaments. Thus, although many readers will beable to construct solutions to round-robin tournaments, perhaps few will be able totackle a mixed-doubles tournament with a spouse-avoiding condition.

REVIEWS 181

The well-written book, which can be read by undergraduates, contains veryuseful notes and references at the end of each chapter. Many of the exercises havehints or partial solutions given in an appendix. Unfortunately, I am afraid that thevery high price for such a text means that it may only be bought by libraries.

Reference1. T. G. Room, A new type of magic square, Math. Gaz. 39 (1955), p. 307.

P. SHIUDepartment of Mathematical Sciences, Loughborough University LE11 3TU

Graph theory as I have known it, by W. T. Tutte. Pp. 156. £27.50. 1998. ISBN0 19 850251 6 (Oxford University Press).

W. T. Tutte is one of the principal pioneers in the field of graph theory, wherehe has exerted a major influence for over 60 years. This book is an account of a pre-retirement series of lectures that Tutte gave in 1984 in which he reflected on his life'swork and disclosed many of his lines of thought, creative processes, triumphs andfrustrations. As he has also memorably described elsewhere in [1], Tutte'sintroduction to graph theory occurred while he was an undergraduate at Cambridgein the 1930s through his involvement with the Trinity College ‘Team of Four’(Brooks, Smith, Stone, Tutte). Initially fired by one of H. E. Dudeney'sCanterburypuzzles, they eventually disproved Lusin's conjecture, that there is no dissection of asquare into a finite number of unequal smaller squares, and Tait's conjecture (whichimplies the Four Colour Theorem, FCT), that there is a Hamiltonian circuit on theedges of any convex polyhedron. ‘Squaring the square’ led to generalisations,involving triangulations of triangles and parallelograms, and to work on rotationalsymmetries of graphs (including the search for highly symmetrical graphs) and ongraphs on spheres: here Tutte whisks us from Brooks' Theorem via Hadwiger'sConjecture (which generalises the FCT and is still unproved) to the theory of bridgesof bonds, ‘a beautiful theory needing applications’. Apart from the FCT, two othernineteenth century precursors of twentieth century graph theoretical concerns wereCayley's famous enumeration formula for the number of labelled trees onvertices and Kirchhoff's ‘Matrix-Tree’ Theorem which asserts that the singular

matrix associated to the graph by:

nn − 2 n

n × n K = (cij) G = v1, … , vncii = vi,valency of

cij = −1 if vi, vj are adjacent,

= 0 otherwise,

has constant cofactors, the constant being the number of spanning trees of .GTutte describes his own work on the enumeration of various types of

triangulations: this is a fiendishly difficult task which put me in mind of Piet Hein's‘grook’, ‘Problems worthy of attack prove their worth by hitting back.’. Kirchhoff'sideas spurred Tutte's interest in subgraphs which culminated in hisf-Factor Theorem.Another important enumerative tool is thechromatic polynomial, , of a(loopless) graph . This has degree equal to the number of vertices of and, forany whole number , is the number of colourings of the vertices of using

colours in which no edge has both its ends the same colour. Much of Tutte's laterwork has dealt with specific properties of the roots of orchromaticeigenvalues, including reasons for the ubiquitous appearance of the golden ratio andother Beraha numbers of the form , and also with generalisations such as

P(G, x)G G

m P(G, m) Gm

P(G, x)

4 cos2 (π / k)


the 2-variable Tutte polynomial which mimics the behaviour of the chromaticpolynomial with respect to contraction and formation of subgraphs. His paper withthe catchy title, ‘All the King's horses’ demonstrated, among other things, that

can be reconstructed from a knowledge of for, although Ulam's full reconstruction conjecture − that is determined

up to isomorphism by the isomorphism classes of the subgraphs − remainsstubbornly unresolved. More abstract algebraic techniques also feature: Tutte tracesthe impact of some homological ideas from combinatorial topology, and the routethat led to his 1959 characterisation of which abstract matroids are graphic.

P(G, x) P(G − vi , x)i = 1, … , n G

G − vi

This is unusual and enchanting book which will be accessible to anyone who iscomfortable with the contents of [2]. Tutte modestly shows us ‘mathematics in themaking’ and portrays in a delightfully whimsical, personal manner the multi-dimensional warps and wefts, cul de sacs, wishful thinking, pleasures andsatisfaction associated with a fulfilled life in mathematics.

References1. Martin Gardner,More mathematical puzzles and diversions. Penguin. 1980.

Chapter 17.2. Robin J. Wilson, Introduction to graph theory. Longman. 1975.


Graph theory and its applications, by Jonathan Gross and Jay Yellen. Pp. 585.£47.50. 1999. ISBN 0 8493 3982 0 (CRC Press)*.

Graph theory is a relatively new area of mathematics with many applications inoptimisation, scheduling, communication networks, computer architecture and evenbiology. This comprehensive text assumes little background and the authors claim itcan be used as the basis of advanced undergraduate or beginning graduate courses ingeneral graph theory, data structures and algorithms, or operations research andoptimisation.

The first few chapters cover the fundamentals of graph theory such as types ofgraph, matrix representation, spanning trees, Eulerian trails, Hamiltonian cycles andtravelling salesman problems. Later chapters introduce more advanced ideas andapplications such as drawing graphs on various surfaces, planarity, graph colourings,digraph applications, network flows, enumeration, voltage graphs and non-planarlayouts. The variety of courses is possible by being selective in the coverage of thelater chapters in particular.

Although the book caters in theory for readers with minimal background, inpractice such readers would have to absorb an awful lot. Taking Chapter 8 (DrawingGraphs and Maps) as a typical example, there are 72 definitions, 17 remarks and 20propositions, theorems and corollaries. The first section is particularly tough foranyone without any knowledge of topology: there are 20 definitions to be masteredin order to state (but not prove) the Jordan Curve Theorem. To be fair though, anadvanced undergraduate who has not studied the topology of surfaces, and thereforehas to work hard in this chapter, will probably have studied topics which make otherchapters more straightforward.

As well as being comprehensive, the book has several other useful features.There are hundreds of illustrations ‘to strengthen intuition’ and over 1600 exercises,

* CRC Press has become part of the same group as Springer-Verlag.

REVIEWS 183

some to secure understanding and others to challenge. In many sections explicitalgorithms are given for those who wish to use computers to solve particularproblems. Finally, a large number of applications are described, some inconsiderable detail.

The Chinese postman problem (in Chapter 6) provides one example of the‘algorithm and applications’ approach. The problem, proposed in 1962 by theChinese mathematician Meigu Guan, is to find the shortest closed walk that traversesevery edge of a graph at least once. It corresponds to the postman (or woman)seeking the shortest route that allows all the mail to be delivered, starting andfinishing at the sorting office. The authors describe an algorithm for solving theproblem and give five examples of applications: street sweeping, mechanical graphplotters, arranging a sequence of two-person meetings, determining an RNA chainfrom its fragments, and information encoding. For the RNA example, they devotethree pages to providing enough information about RNA fragmentation for readers toappreciate the value of the solution.

This book succeeds in its aim of comprehensive coverage. It will be useful foranyone needing to learn about algorithms, applications, specific topics or aboutgraph theory in general.


Theory of differentiation: a unified theory of differentiation via new derivatetheorems and new derivatives, by Krishna M. Garg. Pp. 525. £80.95. 1998. ISBN0 471 25387 1 (Wiley).

This imposing research monograph is the outcome of a life-time's research intothe minutiae of the theory of differentiation. In it, Garg heroically strives to collate,unify, systematize and, in several instances, improve the various results aboutgeneralised derivatives that have sporadically appeared in the literature since thedays of the early pioneers such as Dini, Peano, Denjoy, Perron, G. C. and W. H.Young, Lusin, Banach and Saks.

The archetypal unilateral ‘derivates’ are the upper and lower Dini derivates:

D+f (x) = limsuph ↓ 0

f (x + h) − f (x)h

, D−f (x) = liminfh ↑ 0

f (x + h) − f (x)h

(with analogous definitions for and ).D+f (x) D−f (x)Notable early theorems were those of Denjoy-Young (Almost everywhere, either

is differentiable or one of the upper derivates is and one of the lower derivates.) and of Denjoy-Young-Saks (An arbitrary function is differentiable at almost

every point at which it has a unilateral derivative.). Garg bases his development onwhat he calls upper and lower new derivatives defined − where they exist − as theset-valued functions and ; the setsinvolved are, in fact, singletons nearly everywhere. (These link with thesubgradients of convex analysis and the notions of sub/super/semi-differentiability innon-smooth analysis.)

f +∞−∞

x [D+f (x) , D−f (x)] x [D−f (x) , D+ f (x)]

What might be hoped for of a generalised derivative?

• Versions of the familiar manipulative devices such as the product,quotient, chain and l'Hôpital's rules.

• Versions of characteristic results such as the mean value theorem andthe ‘monotonicity theorem’, guaranteeing that a function is non-decreasing if its derivative is non-negative.


• Versions of familiar structural properties of derivatives such as theDarboux (intermediate value) property, the Denjoy property (that

is either empty or non-null), and the Lusin property ( isnull whenever is null).(f ′)−1 (a, b) f (E)

E• Reconstructibility of a function from its derivative by a suitably

generalised Denjoy-Perron integration process.• Analogues of the Banach-Saks theorem, which characterises those

continuous functions that are differentiable (everywhere) on almostall of their level sets .

ff −1 (y)

• Although since Weierstrass it has been known that continuous, nowheredifferentiable functions exist, a continuous function might have ageneralised derivative on, hopefully, a rich set of points. Again, the setof somewhere differentiable functions is meagre in the space ofcontinuous functions: does the same hold for generalised derivatives?

These constitute some of the recurring themes explored in this book whichculminates in an axiomatic framework which, building on work with newderivatives, explains the differences in behaviour of the various notions ofgeneralised derivative that have appeared in the literature (Garg lists over a dozensuch!). Throughout, the author's canvas is that of real-valued functions defined onsubsets ofr − indeed, he is able to use the symbol to denote the spaces ofcontinuous functions on [0, 1]!

C

This is a dauntingly technical work (the index of symbols runs to 5 pages!)which demands close concentration on the part of the reader. But, as a notinconsiderable feat of presentation of a frustratingly diffuse area of analysis, it willrepay such attention by real-variable aficionados.


Functional analysis and differential equations in abstract spaces, by S. Zaidman.Pp. 226. £36.00. 1999. ISBN 1 58488 011 2 (Chapman & Hall/CRC).

In form and content, this is very much a book of two contrasting halves. Thefirst half consists of a carefully written, smoothly organised introduction to classicallinear functional analysis. The route chosen to the three big theorems (uniformboundedness, closed graph, Hahn-Banach) is a very familiar one, but there are somenice pedagogical flourishes (such as the easy proof of Hahn-Banach for Hilbertspaces: extend defined on to by uniform continuity and thence to the wholespace by setting on ). After that, the topics presented reflect the lessstandard prerequisites for the second half of the text with material on unbounded/closed/closable operators, operator semigroups and their infinitesimal generators,compact operators, symmetric operators (with their square roots obtained by a prettyiterative method) and a soupçon of spectral theory.

f Y Y

f = 0 Y⊥

In the more technical second half of the book, these tools are shown in action inthe context of results taken from the recent research literature on differentialequations in Hilbert and Banach spaces. As ever, the devil is in the details, but wecan glimpse something of what is involved from the simplest first order problem:solve with given. Here, is a Banach space-valuedfunction, is the strong derivative (obtained by replacing by in the usualdefinition of derivative) and is a closed densely defined operator. If the problem iswell-posed in a strong enough sense, the prescription gives rise to anoperator semigroup with infinitesimal generator , i.e.

u′ = Au u0 u : [0, ∞) → Xu′ | . | .

Au (t) = T (t) u0

T (t) : X → X A

REVIEWS 185

, and the solution of the related non-homogeneousequation ( given) isAx = lim t ↓ 0 (T (t) x − x) / t

u′ = Au + f u0

u (t) = T (t) u0 + ∫t

0T (t − s) f (s) ds.

The majority of Zaidman's later theorems relate to the question of uniqueness ofsolutions. He introduces the concepts of weak and ultraweak solutions (which relyon test functions in much the same way that these are used in the theory ofdistributions) and, via delicate differential inequalities, proves restricted uniquenessresults in this situation for ( symmetric on Hilbert space) and

closed and densely defined on a reflexive Banach space). He is also able totransfer well-known results such as the resolvent representation formula

( as above, large) to the ultraweak setting.Finally, on a different tack, he discusses the striking theorem that, for solutions ofequations such as (with the infinitesimal generator of an almostperiodic (a.p.) group of operators and a.p.) the concepts of weakly a.p and stronglya.p. coincide with a simple connection between the spectrums of and : here, isweakly a.p. if is a.p. in the classical sense for all in and strongly a.p. ifmimics the classical definition but with replacing .

u″ = Mu M u′ = Au(A

∫∞0 e−λtT (t) x dt = (λI − A)−1 x T, A Reλ

u′ = Au + f Af

f u ux′u x X′ u

. | . |There are no exercises and, to my mind, a dearth of illustrative and motivational

examples in the text. The juxtaposition of the elementary and rather standardcontents of the first half with the specialised and much more rarefied concerns of thesecond half certainly give the book an unusual and very distinctive flavour but may,I fear, serve to split its potential readership.


Ordinary differential equations and applications: mathematical methods forapplied mathematicians, physicists, engineers, bioscientists, by Werner S.Weigelhofer and Kenneth A. Lindsay. Pp. 215. £14. 1999. ISBN 1 898563 57 8(Horwood Publishers).

This book, according to the authors' preface, is based on lecture notes for a thirdyear course on mathematical methods at Glasgow University. The book opens withthree chapters entitled respectively: Differential Equations of First Order; ModellingApplications; and Linear Differential Equations of Second Order. The content of thefirst and third is fairly conventional, but the second deals with a variety of modellingtopics coming from fields not usually touched on in differential equation textbooks,such as Newton's law of cooling, the Gompertz population law and pursuit curves.The fourth chapter contains work on a variety of methods which have proved usefulin the solution of linear second order equations. The fifth chapter on oscillatorymotion includes a discussion on the concept of resonance.

The later chapters deal with more advanced work. The sixth chapter gives anintroduction to the use of Laplace transforms for solving linear differential equationswith constant coefficients. The seventh chapter deals with higher order initial valueproblems (the marching problems of numerical analysis) and introduces the idea ofthe Wronskian. The eighth chapter discusses systems of first order linear equationsand shows how such systems can be dealt with using the aid of matrices. The twofinal chapters deal with more sophisticated material. The ninth introduces theconcept of eigenvalues and eigenfunctions and Sturm-Liouville theory, and the tenthgives an introduction to the calculus of variations and considers some elementaryexamples. An appendix gives a number of self-study projects from a variety of


fields; for example, rockets, bridges and snowploughs. Examples with solutions arescattered throughout each chapter and at the end of each chapter there are a numberof tutorial examples, the answers to which (taking up about a quarter of the book) aregiven in a second appendix.

This is an interesting book and reads easily. The topic mix may not suit everycourse, but many will find it a useful textbook. Parts of chapter 1 and chapters 2, 3and 5 could in fact be useful to A level students. (Might one suggest a pamphlet onthese lines by the authors?) There are one or two places where I feel the book couldbe improved. In the first chapter there is a mention of singular solutions but nomention of envelopes, and some discussion could have been given in the ninth andtenth chapters of the connection between eigenvalues and the calculus of variations.However, the appearance of the book is pleasing and the price is, for these days,reasonable. I have no hesitation in recommending it.

Ll. G. CHAMBERSSchool of Mathematics, University of Bangor LL57 1UT

Elementary Lie group analysis and ordinary differential equations, by Nail H.Ibragimov (Mathematical Methods in Practice 4). Pp. 347. £55. 1999. ISBN 0 47197430 7 (Wiley).

The seriesMathematical Methods in Practice is intended to provide a one-stop-shop for applied scientists who wish to use mathematics in their work. The idea is tocombine some of the theory traditionally taught in pure mathematics courses withapplications that are sometimes taught in the absence of rigour. Lie originallydeveloped his theory as a way of unifying the treatment of many types of differentialequations, analogous to Galois's theory for algebraic equations. Many books on Liegroups develop the theory in the most general form, requiring the reader to beacquainted with topological groups and manifolds, which are difficult ideas forbeginners. This book is intended for students who wish to understand Lie groupanalysis specifically in the context of differential equations.

In order that the book be self-contained, the first part gives a brief but fairlycomprehensive treatment of the classical approach to differential equations. The firstchapter gives many excellent examples of the use of differential equations inmathematical modelling. The next three cover methods for various ordinaryequations, general properties of solutions and first order partial differentialequations. The second part develops the fundamental ideas of Lie group analysis,beginning with an interesting historical survey of Lie theory from Lie's originalwork, to the resurrection of applied group analysis led by L. V. Ovsyannikov in thesixties and seventies. The other chapters cover transformation groups, infinitesimaltransformations and local groups, differential algebra, symmetry of differentialequations and invariants. Having established the machinery of Lie group analysis inpart two, the remaining section is devoted to showing how it brings unity to theadhoc methods for differential equations presented in the first part.

This book presents Lie groups in a way that will appeal to two groups: thosewho want an accessible introduction to the theory and those who primary focus is ondifferential equations. Each chapter is supplemented with numerous historical notesand references and a collection of problems, graded in difficulty. Answers or hintsare collected in an appendix.


REVIEWS 187

Beginning partial differential equations, by Peter V. O’Neil. Pp. 500. £51.95.1999. ISBN 0 471 23887 2 (Wiley).

This is a self-confessed bread-and-butter book on partial differential equations.Since the subject is a difficult one for students, the appearance of this book is verywelcome, although the price will not be conducive to large sales. In fact, itsexorbitant price will ensure its exclusion from all but the most well-heeled libraries.

The topics chosen are first and second order differential equations, Fourieranalysis, the treatment of the wave and heat equations (the latter now used as a toolin financial mathematics) and finally, a long section on the Dirichlet and Neumannproblems. To his credit the author makes certain prerequisites clear at the beginning,and it is not one of those books with lofty ideas (not to mention blurbs written by themarketing department) about teaching the latest research to students without GCSE.The reader is not asked to plunge into partial differential equations without a facilitywith the standard properties of real-valued functions ofn real variables, vectorcalculus (theorems of Green and Gauss), a post-calculus course on ordinarydifferential equations and the convergence of series and improper integrals. Accessto the computer softwareMAPLE would also be handy, but is not essential for aproper study of the mathematical contents of this book.

The very sparse historical comments are too incomplete to be of much help buttheir inclusion is a nod in the right direction. A few afternoons in a reasonable librarywould have offered much more to bolster the meagre comments made here. Why notgive us a bit more (than nothing) on say, Jean Marie Constant Duhamel (1797-1872),who crops up in various places in the book. And which Neumann of the Neumannproblem is the author not talking about? A curious diversion in a technical book isthe long historical digression on ‘The Great Debate Over the Age of the Earth’ (pp.317-320). One can only guess that this is still a popular topic in Alabama, where theauthor is based. In this portion, which is presented well, the author summarises JoeBurchfield’s Lord Kelvin and the age of the Earth (1875) and shows the wayWilliam Thomson used the heat equation to gain his estimate of the Earth’s age asbetween 100 and 400 million years.

The book will appeal mostly to applied mathematicians and those engineerswho are relatively strong mathematically. The presentation of surfaces in threedimensions in these days of computer graphics leaves something to be desired. It is auseful book but the same information can be found in books which are morereasonably priced.



Numerical solution of partial differential equations in science and engineering,by Leon Lapidus and George F. Pinder. Pp. 677. £41.95. 1999. ISBN 0 471 35944 0(paperback) (Wiley).

Originating as a support text for courses given by both authors at PrincetonUniversity, this book is ideally suited to students from a variety of academicdisciplines. It is virtually free from jargon or other nomenclature which may detercertain students; however it provides references to applications in a diversity ofsubject areas.

It is very sad to report the sudden death (5 May 1977) of co-author, LeonLapidus, whilst at work in the Department of Chemical Engineering at PrincetonUniversity. This tragedy undoubtedly contributed to the delay in publication ....


which in no way devalues this work. The objectives of providing a balancedtreatment of finite differences and finite element methods which can be read eitherby equation-type or by numerical approximation have been fully achieved.

The first quarter of the text consists of introductory material relating to partialdifferential equations of the first and second orders including their classification,together with the method of characteristics. Basic concepts of finite differences andfinite elements follow, the latter being extended to include triangular, isoparametricand three-dimensional elements. There are limited references here covering bothstandard texts such as R. Courant and D. HilbertMethods of mathematical physics(Interscience 1962) and some less readily available publications!! For example B. G.Galerkin, Vestn. Inzh. Tekh (USSR), 19, pp. 897-908 (1915).

As expected the majority of the book is trisected to cover parabolic, elliptic andhyperbolic partial differential equations. Each section concludes with a verysubstantial list of references. The absence of sets of exercises for the reader is onlypartially offset by the inclusion of selected ‘Example Problems’ within the body ofthe text. However the style of presentation is such that almost all readers would beexpected to go away and practice what they have assimilated; either in ‘real world’applications or in sets of exercises taken from elsewhere.

The three types of PDE each receive a thorough treatment, carefully balancedbetween finite differences and finite elements, as well as between one, two and threespatial dimensions.

Authors of the twenty-first century attempting to supersede this work, possiblyby the inclusion of extensive material relating to digital computer software, will dowell to ask themselves ‘Which came first, the chicken or the egg?’

Whilst the publication price of £41.95 for a paperback edition of approximately700 pages may be considered substantial it must be emphasised that − despite itsorigins − the work is lecturer independent. Thus the reader has a readily accessibleself-contained reference work for the subject of the numerical solution of PDEs −itself central to all applicable mathematics. This book is strongly recommended forall students with significant involvement in this subject area.

MICHAEL R. MUDGE23, Gors Fach, Pwll-Trap, St Clears SA33 4AQ

Evaluation and optimisation of electoral systems, by Pietro Grilli di Cortona,Cecilia Manzi, Aline Pennisi, Federica Ricca and Bruno Simeone. Pp. 230. $53(SIAM members $42.40). 1999. ISBN 0 89871 422 2 (Society for Industrial andApplied Mathematics).

The problem of determining the best electoral system has occupied many mindsin this country recently. New assemblies in Scotland and Wales and the currentlysuspended assembly for Northern Ireland have been established and the system forelecting Members of the European Parliament has been changed. Reform of theHouse of Lords and the possibility of proportional representation for the Commonswill keep the subject of electoral systems in the public mind.

This book is the result of a collaboration involving experts from operationalresearch, statistics, decision sciences and political sciences. It has four sections, threepresenting mathematical treatments of various aspects of electoral systems and afinal section taking the political view. The mathematical treatment has been kept asaccessible as possible in the hope of attracting readers from beyond the mathematicalcommunity. Though the book is essentially mathematical, the consequences of thevarious equations and formulas have been interpreted in words for thoseuncomfortable with mathematical notation.

REVIEWS 189

Previous analyses of electoral systems have taken various points of view; forexample, axiomatic, statistical, game-theoretic and geometric approaches have allbeen used. The present authors consider the optimisation of various criteria.

In Chapter 2, Cecilia Manzi proposes a general set-theoretic model for electoralsystems that encompasses quotient methods and first-past-the-post, double ballot,alternative transferable vote and single transferable vote systems. This model enablesa classification of the electoral systems of many countries. These systems can becompared using a variety of criteria and indicators that are dealt with in thefollowing chapter. The next five chapters, by Aline Pennisi, consider the design ofelectoral systems. Different electoral formulas are treated as algorithms thatminimise certain measures of unfairness, with the surprising result that the sameformula may minimise more than one measure.

The third section, by Fedrica Ricca, considers the problem of designingelectoral districts. An artificial example is given of a territory consisting of 45 wardsto be grouped into 9 equal constituencies. Although 24 wards favour party C and 21favour party P, it is possible to assign wards to constituencies to achieve 8 seats to 1victories for either party. The political sensitivity of deciding constituencyboundaries could not be better illustrated.

In the final section, Pietro Grilli di Cortona analyses the benefits, limitations andpolitical implications of the methodology presented in the first three parts.

This is a fascinating study that deserves to be widely read.STEVE ABBOTT

Claydon High School, Claydon, Ipswich IP6 0EG

Perplexing problems in probability: Festschrift in honor of Harry Kesten, ed.Maury Bramson and Rick Durrett. Pp. 398. DM158. 1999. ISBN 3 7643 4093 2(Birkhauser).

It has become fashionable recently for academics to honour a distinguishedcolleague on reaching a certain age by publishing aFestschrift, being a collection, inbook form, of papers, each on a topic on which the person honoured has worked.Often the contributors are former students. Cynics say this is a device for the authorsto get another paper to add to their c.v. Others would say that it provides anopportunity both to honour a major contributor and to present a view of the state ofknowledge in a particular field. The book under review is such a Festschrift inhonour of Harry Kesten, consisting of a paper by Rick Durrett, which summarizeshis work, and twenty papers of original material on topics on which Kesten hasworked. The honorand has made major contributions to mathematical probabilityand has gained a deserved reputation as a superb solver of problems and, as such, hiswork is not confined to a single topic.

The papers therefore range widely and all require substantial background fortheir appreciation. One class of problems concerns percolation processes. Take asquare, integer lattice and add edges connecting adjacent sites. Imagine the edges tobe channels conveying a fluid which are open (or closed) with probability

independently of other channels. A question of physical interest is the setof sites that can be reached from the origin by the use of open channels. It turns outthat there is a critical value at and the number of such sites is infinite withpositive probability only if exceeds that value. Kesten has extended enormouslyour knowledge of the behaviour for values of near to . Another problem, again onthe lattice, is the self-avoiding random walk, which is the usual walk except that awalk can never visit the same site twice. This is an especially difficult scenario

p(or 1 − p)

p = 12

pp 1

2


because the process has to remember all its past and cannot forget most of it, unlike aMarkov process.

These are difficult papers of limited appeal. When a writer says ‘well-known’he likely means ‘well-known to 500 people’. But nevertheless they represent realadvances in difficult problems and form a fine testimonial to one of the world'sleading probabilists.

D. V. LINDLEYWoodstock, Quay Lane, Minehead TA24 5QU

Resampling methods: a practical guide to data analysis, by Phillip I. Good.Pp. 269. SFr118. 1999. ISBN 3 7643 4091 6 (Birkhäuser).

This book has been written for a diverse audience: medical students, healthworkers, business people, biologists, social scientists, industrial statisticians andstatistical consultants are all mentioned in the preface as potential beneficiaries. Asthe title suggests, it is not a traditional statistics text, preferring to emphasise ‘table-free’ resampling methods (bootstrap, density estimation and permutations) ratherthan methods based on tables of standard distributions.

Much of the first part of the book is aimed at those with minimal experience ofstatistics. The first two chapters include such material as types of data, measures ofaverage, statistical diagrams and the binomial distribution. However, the resamplingapproach is introduced as early as chapter one, where the bootstrap method is used toestablish the precision of a sample median. This involves repeatedly sampling withreplacement from the original sample of size 22, each case yielding another sampleof size 22, usually involving repetitions. The sample medians of 50 ‘resamples’ arecomputed. These are illustrated on a number line to ‘provide a feel for what mighthave been had we sampled repeatedly from the original population’. Clearly such amethod has drawbacks, and these are discussed in a section headed ‘Caveats’.

Permutation tests are introduced in the third chapter, Testing Hypotheses. Twosamples are compared: one has been subjected to a treatment that the other has not.The results of the treated sample are 121, 118, 110; those of the untreated sampleresults are 34, 22, 12. The total of the treated sample is compared with the totals ofevery possible choice of 3 results. Since it is the highest of the 20 possible resultsone can assert at the 5% significance level that the treatment was effective.

Density estimation is not introduced until chapter 10, Classification andDiscrimination, which considers the problem of deciding how many distinctpopulations are represented in a sample. Suppose that a sample containing data fromn distinct populations is used to construct a histogram. Whether or not the histogramshows then sub-populations depends on the number of intervals used and the extentto which the populations ‘overlap’. The histogram can be smoothed by replacingeach of its blocks by a normal distribution curve of corresponding area, centred onthe block, and then summing the results. The resulting curve is likely to have anumber of modes (maxima). The process can be repeated for many different intervalwidths to determine the smallest interval width that gives k modes. For each ofthese critical interval widths, bootstrap resampling can be used with the smoothingprocess to estimate the proportion of times that more thank modes appear. Thisproportion will be close to 1 if and close to 0 if , thus allowing us toestimate the value of n.

hk

k < n k ≥ n

This book illustrates many applications of resampling methods. In each case, theauthor discusses the assumptions that have been used and the correspondinglimitations. He writes in a conversational style and makes effective use of concrete

REVIEWS 191

examples to introduce the various methods. Each chapter has a chapter summary, asection called ‘To Learn More’ guiding readers to appropriate references, and a setof exercises. Overall this is a helpful introduction, but one that will stretch many ofthe target groups mentioned in the preface. His claim that readers need only ‘high-school algebra’ is stretching the truth: an expression like

eff [w∗, F] =1B ∑

B

b = 1(1n ∑

n

i = 1

[yi, η [xi, w∗b]])would frighten the life out of most ‘Physicians and physicians in training, nurses andnursing students’ of this reviewer's acquaintance!


Epidemiology: study design and data analysis, by Mark Woodward. Pp. 699.£39.95. 1999. ISBN 1 584 88009 0. (Chapman & Hall/CRC).

This book has been aimed at statisticians who want to see how their subject canbe applied to epidemiology and to medical researchers who need a betterunderstanding of statistics. The first two chapters reflect this dual aim: thestatisticians need Chapter 1 to get a basic understanding of the fundamental issues ofepidemiology and the researchers need Chapter 2 to remind them of the basic toolsof statistics.

Epidemiologists investigate the causes of disease as well as modelling thespread of disease. The emphasis in this book is on the first category of questions,including the estimation of risk, the effect of confounding variables and theinteraction among risk factors. A good example of a confounding variable is thepresence of grey hair among stroke victims: the important risk factor is age, but greyhair increases with age. An example of interaction between two risk factors occurs inthe study of lung disease among porcelain painters: lung function is affected byexposure to cobalt, but this is much worse among painters who smoke.

Necessarily Woodward pays a good deal of attention to the design of studies toinvestigate risk factors, with chapters on cohort studies, case-control studies andintervention studies.

The last four chapters present statistical methods such as hypothesis testing,analysis of variance for one or more explanatory variables, various forms ofregression analysis and probability models for hazard and survival functions, such asthe exponential and Weibull distributions

The book contains many examples and exercises based on real epidemiologicaldata. In some cases, where the data set is large, the information is also providedelectronically on an associated web site (http://www.reading.ac.uk/AcaDepts/sn/wsn1/publications99.html).

The use of real data sets to illustrate ideas, the emphasis on practicality and theomission of proofs are all features that will appeal to those interested in applyingstatistical methods to the study of disease. The book looks particularly suitable forstudents.



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