donald j newman analytic number theory gradu

7/29/2019 Donald J Newman Analytic Number Theory Gradu

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GraduateTexts in Mathematics 177Editorial Board

S. Axler F.W. Gehring K.A. Ribet

SpringerNew York

BerlinHeidelbergBarcelonaHongKongLondonMilanParisSingaporeTokyo


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Donald J. Newman

Analytic Number Theory

13


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Donald J. NewmanProfessor EmeritusTempleUniversityPhiladelphia, PA 19122

USA

Editorial Board

S. Axler F.W. Gehring K.A. RibetDepartment of Departmentof Department of

Mathematics Mathematics MathematicsSan Francisco StateUniversity University of Michigan University of California

San Francisco, CA 94132 Ann Arbor, MI 48109 at BerkeleyUSA USA Berkeley, CA 94720-3840USA

Mathematics Subject Classification (1991): 11-01, 11N13, 11P05, 11P83

Library of CongressCataloging-in-PublicationData

Newman, Donald J., 1930Analytic number theory / Donald J. Newman.

p. cm. (Graduatetexts in mathematics; 177)Includes index.ISBN 0-387-98308-2 (hardcover: alk. paper)

1. Number Theory. I. Title. II. Series.QA241.N48 1997512.73dc21 97-26431

1998 Springer-Verlag New York, Inc.All rights reserved. This work may not betranslated or copied in whole or in part without thewrittenpermissionof thepublisher (Springer-VerlagNewYork, Inc.,175FifthAvenue, NewYork, NY 10010,USA),exceptforbrief excerptsinconnectionwithreviewsorscholarlyanalysis.Useinconnectionwithanyformof informationstorageandretrieval,electronicadaptation,computersoftware,orbysimilarordissimilar methodologynowknownorhereafter developedisforbidden.Theuseof general descriptivenames,tradenames,trademarks,etc.,inthispublication,evenif theformerarenotespeciallyidentified,isnottobetakenasasignthatsuchnames, asunderstoodby theTradeMarksandMerchandiseMarksAct, may accordingly beused freely by anyone.

ISBN 0-387-98308-2 Springer-Verlag New York Berlin Heidelburg SPIN 10763456


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Contents

Introduction and Dedication vii

I. The Idea of Analytic Number Theory 1

Addition Problems 1ChangeMaking 2Crazy Dice 5Can r(n) beconstant? 8A Splitting Problem 8An Identity of Eulers 11Marks on aRuler 12Dissection into Arithmetic Progressions 14

II. The Partition Function 17TheGenerating Function 18TheApproximation 19Riemann Sums 20TheCoefficients ofq(n) 25

III. The ErdosFuchs Theorem 31

ErdosFuchs Theorem 35

IV. Sequences without Arithmetic Progressions 41

TheBasic Approximation Lemma 42

v


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vi Contents

V. The Waring Problem 49

VI. A Natural Proof of the Nonvanishing ofL-Series 59

VII. Simple Analytic Proof of the Prime NumberTheorem 67

First Proof of thePrimeNumber Theorem. 70Second Proof of thePrimeNumber Theorem. 72

Index 77


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Introduction and Dedication

This book is dedicated to Paul Erdos, the greatest mathematician Ihave ever known, whom it has been my rare privilege to considercolleague, collaborator, and dear friend.

I liketothink thatErdos, whosemathematicsembodiedtheprinci-ples whichhaveimpressed themselves uponmeas defining thetruecharacter of mathematics, would have appreciated this little book

and heartily endorsed its philosophy. This book proffers the thesisthat mathematicsisactually aneasy subjectandmany of thefamousproblems, even thosein number theory itself, which havefamouslydifficult solutions, can beresolved in simpleand moredirectterms.

There is no doubt a certain presumptuousness in this claim. Thegreat mathematicians of yesteryear, those working in number the-ory and related fields, did not necessarily striveto effect thesimple

solution. Theymayhavefeltthatthestatusandimportanceof mathe-maticsasanintellectual disciplineentailed, perhapsindeedrequired,a weighty solution. Gauss was certainly a wordy master and Euleranother.Theybelongedtoatraditionthatundoubtedly reveredmath-ematics, but as a discipline at some considerable remove from thecommonplace. Inkeepingwithamoredemocraticconceptof intelli-

genceitself,contemporarymathematicsdivergesfromthissomewhatelitist view. The simple approach implies a mathematics generallyavailable even to those who have not been favored with the naturalendowments, nor thecareful cultivation of an Euler or Gauss.

vii


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viii Introduction and Dedication

Such an attitude might prove an effective antidote to a generallydeclininginterestinpuremathematics. Butitisnotsomuchasincen-tivethat weproffer what might best be called the fun and gamesapproach to mathematics, but as a revelation of its true nature. Theinsistence on simplicity asserts a mathematics that is both magi-cal and coherent. Thesolution that strives to master thesequalitiesrestores to mathematics that element of adventure that has alwayssupplied its peculiar excitement. That adventureis intrinsic to eventhemost elementary descriptionof analytic number theory.

The initial step in the investigation of a number theoretic itemis the formulation of the generating function. This formulationinevitably movesusaway fromthedesignatedsubjectto aconsider-ationof complexvariables. Havingwanderedawayfromour subject,itbecomesnecessary toeffectareturn. TowardthisendTheCauchyIntegralprovestobeanindispensabletool.Yetitleadsus,inevitably,

further afieldfromall theintricaciesof contour integrationandthey,in turn entail thefamiliar processes, thedeformation and estimationof thesecontour integrals.

Retracingour stepswefindthatwehavegonefromnumber theorytofunctiontheory, andbackagain. Thejourney seemscircuitous, yetin its wakea pattern is revealed that implies a mathematics deeplyinter-connected andcohesive.


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I

The Idea of Analytic NumberTheory

ThemostintriguingthingaboutAnalyticNumber Theory (theuseofAnalysis, orfunction theory, innumber theory) is itsvery existence!How could oneuseproperties of continuous valued functions to de-terminepropertiesof thosemostdiscreteitems,theintegers.Analyticfunctions? What has differentiability got to do with counting? Theastonishment mounts further when welearn that thecomplex zeros

of acertain analytic function are the basic tools in theinvestigationof theprimes.The answer to all this bewilderment is given by the two words

generating functions. Well, thereareanswers and answers. To thoseof uswhohavewitnessedtheuseof generatingfunctionsthisisakindof answer, buttothoseof uswhohavent, thisissimplyarestatementof the question. Perhaps the best way to understand the use of the

analytic method, or the use of generating functions, is to see it inaction in a number of pertinent examples. So let us take a look atsomeof these.

Addition Problems

Questionsaboutadditionlendthemselvesvery naturally totheuseofgenerating functions. Thelink is thesimpleobservation that addingm and n is isomorphic to multiplying zm and zn. Thereby questionsabout the addition of integers are transformed into questions aboutthemultiplication of polynomials or power series. For example, La-granges beautiful theoremthat every positive integer is the sumof

1


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2 I. TheIdeaof Analytic Number Theory

four squares becomes thestatement that all of thecoefficientsof the

powerseriesfor1 + z + z4 + + zn2 +

4

arepositive.How

oneproves such a fact about thecoefficients of such a power seriesis another story, but at least one begins to see how this transitionfromintegersto analytic functionstakesplace. But now lets look atsomeadditionproblemsthatwecan solvecompletelybytheanalyticmethod.

Change Making

How many ways can one make change of a dollar? The answer is293, but theproblemisbothtoo hardandtoo easy. Too hardbecausethe available coins areso many and so diverse. Too easy becauseitconcernsjust onechangee, adollar. Morefittingto our spirit isthe

followingproblem: Howmanywayscanwemakechangeforn if thecoins are 1, 2, and 3? To formthe appropriate generating function,let us write, for |z| < 1,

1

1 z 1 + z + z1+1 + z1+1+1 + ,

1

1 z2

1

+z2

+z2+2

+z2+2+2

+ ,

1

1 z3 1 + z3 + z3+3 + z3+3+3 + ,

and multiplying thesethreeequations to get

1

(1

z)(1

z2)(1

z3)

(1 + z + z1+1 + )(1 + z2 + z2+2 + ) (1 + z3 + z3+3 + ).

Now we ask ourselves: What happens when we multiply out theright-handside? Weobtaintermslikez1+1+1+1 z2 z3+3. Ontheonehand, this term is z12, but, on the other hand, it is zfour1

s+one2+two3s

and doesnt this exactly correspond to the method of changing theamount 12 into four 1s, one 2, and two 3s? Yes, and in fact we


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ChangeMaking 3

see that every way of making change (into 1s, 2s, and 3s) forevery n will appearinthismultiplyingout. Thusif wecall C(n) thenumber of waysof makingchangefor

n, then

C(n)will betheexact

coefficient of zn when the multiplication is effected. (Furthermoreall isrigorousandnot just formal, sincewehaverestrictedourselvesto |z| < 1 wherein convergenceis absolute.)

Thus

C(n)zn 1

(1 z)(1 z2

)(1 z3

)

, (1)

and the generating function for our unknown quantity C(n) isproduced. Our number theoretic problem has been translated intoa problem about analytic functions, namely, finding the Taylorcoefficients of thefunction 1

(1z)(1z2)(1z3) .Fine.A well definedanalyticproblem,buthowtosolveit?Wemust

resist the temptation to solve this problem by undoing the analysiswhichledto itsformulation. Thusthethingnotto do isexpand 1

1z ,1

1z2 ,1

1z3 respectively into

za,

z2b,

z3c andmultiply only todiscover thatthecoefficientisthenumber of waysof makingchangefor n.

The correct answer, in this case, comes from an algebraic tech-niquethatweall learnedincalculus, namely partial fractions. Recallthat this leads to terms like A

(1z)k for which we know the expan-sionexplicitly (namely, 1

(1z)k is just aconstant timesthe(k 1)thderivativeof 1

(1z)

nzn).Carrying out the algebra, then, leads to the partial fractional

decomposition which wemay arrangein thefollowingform:

1

(1 z)(1 z2)(1 z3)

16

1

(1 z)3 +1

4

1

(1 z)2 +1

4

1

(1 z2) +1

3

1

(1 z3) .

Thus, since

1

(1 z)2 d

dz

1

1 z d

dz zn (n + 1)zn


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and

1

(1 z)3 d

dz

1

2(1 z)2 d

dz n + 1

2

zn

(n + 2)(n + 1)

2zn,

C(n) (n + 2)(n + 1)12

+ n + 14

+ 1(n)4

+ 2(n)3

(2)

where 1(n)

1 if 2

|n and

0 otherwise; 2(n)

1 if 3

|n

and 0else. A somewhat cumbersomeformula, but onewhichcanbeshortened nicely into

C(n)

n2

12+ n

2+ 1

; (3)

wheretheterms in thebracketsmean thegreatest integers.

A nice crisp exact formula, but these are rare. Imagine the messthatoccursif thecoinsweretheusual coinsof therealm, namely1, 5,10, 25, 50, (100?). Theright thingto ask for then isan asymptoticformularather than an exact one.

Recall that anasymptotic formulaF(n) for afunctionf(n) isonefor which limn

f (n)

F(n) 1. In thecolorful languageof E. Landau,

the relative error in replacing f (n) by F(n) is eventually 0%. Atany rate, wewritef (n) F(n) whenthisoccurs. Onefamoussuchexampleis Stirlings formula n!

2 n( n

e)n. (Also notethat our

result (3) can beweakened to C(n) n212

.)Soletusassumequitegenerally thattherearecoinsa1, a2, a3, . . .,

ak, whereto avoid trivial congruenceconsiderations wewill requirethat therebeno commondivisiorsother than1. Inthisgenerality we

ask foranasymptoticformulaforthecorrespondingC(n). Asbeforewefind that thegenerating function is given by

C(n)zn 1(1 za1)(1 za2) (1 zak ) . (4)

But the next step, explicitly finding the partial fractional decompo-sition of this function is the

hopelesstask. However, let us simply

look for one of the terms in this expansion, the heaviest one. Thus


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Crazy Dice 5

at z 1 the denominator has a k-fold zero and so there will be aterm c

(1z)k . All the other zeros are roots of unity and, because weassumed no common divisiors, all will beof order lower than

k.

Thus, although the coefficient of the term c(1z)k is c

n+k1

k1, the

coefficients of all other terms a(1z)j will be a

j

n+jj1. Since all of

thesej arelessthank, thesumtotal of all of thesetermsisnegligiblecompared to our heavy termc

n+k1

k1. In short C(n) cn+k1

k1, or

even simpler,

C(n) c nk

1

(k 1)! .

But, what is c? Although we have deftly avoided the necessity offindingall of theother terms, wecannotavoidthisone(itsthewholestory!). So let us write

1

(1 za1)(1 za2) (1 zak ) c

(1 z)k + other terms,multiply by (1 z)k to get

1 z1 za1

1 z1 za2

1 z1 zak c + (1 z)

k other terms,

andfinally letz

1. ByLHopitals rule, for example, 1z

1

zai

1ai

whereas each of theother terms times (1 z)k goes to 0. Thefinalresult is c 1

a1a2ak , and our final asymptotic formulareads

C(n) nk1

a1a2 ak(k 1)!. (5)

Crazy Dice

An ordinary pair of dice consist of two cubes each numbered 1through 6. When tossed together there are altogether 36 (equallylikely) outcomes. Thus the sums go from 2 to 12 with variednumbers of repeats for these possibilities. In terms of our ana-lytic representation, each die is associated with the polynomialz + z2 + z3 + z4 + z5 + z6. The combined possibilities for the


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sums then aretheterms of theproduct

(z

+z2

+z3

+z4

+z5

+z6)(z

+z2

+z3

+z4

+z5

+z6)

z2 + 2z3 + 3z4 + 4z5 + 5z6 + 6z7

+ 5z8 + 4z9 + 3z10 + 2z11 + z12

Thecorrespondence, for example, says that thereare3 ways for the10to show up, thecoefficientsofz10 being3, etc. Thequestionis: Isthereanyotherwaytonumber thesetwocubeswithpositiveintegers

so as to achievethevery samealternatives?Analytically, then, the question amounts to the existence of

positiveintegers, a1, . . . , a6; b1, . . . , b6, so that

(za1 + + za6)(zb1 + + zb6) z2 + 2z3 + 3z4 + + 3z10 + 2z11 + z12.

Thesewould betheCrazy Dice referred to in thetitleof this sec-tion. They look totally different fromordinary dicebut they produceexactly thesameresults!

So, repeatingthequestion, can

(za1 + + za6)(zb1 + + zb6) (z + z2 + z3 + z4 + z5 + z6) (6)

(z + z2 + z3 + z4 + z5 + z6)?

To analyze this possibility, let us factor completely (over the ratio-nals) thisright-handside. Thusz + z2+ z3+ z4+ z5+ z6 z 1z6

1z z(1

+z

+z2)(1

+z3)

z(1

+z

+z2)(1

+z)(1

z

+z2).Weconclude

from(6) thatthea-polynomial andb-polynomialmustconsistofthesefactors. Also therearecertainsiderestrictions. Theas andbsareto bepositive andso az-factor must appear inbothpolynomials.Thea-polynomial mustbe6atz 1andso the(1+ z + z2)(1+ z)factor must appear in it, and similarly in the b-polynomial. All thatisleft todistributearethetwofactorsof 1 z + z2. If oneapiecearegiven to the a- and b-polynomials, then we get ordinary dice. Theonly thing left to try is putting both into thea-polynomial.


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Crazy Dice 7

This works! Weobtain finally

za z(1 + z + z2)(1

+z)(1

z

+z2)2

z + z3 + z4 + z5 + z6 + z8

and zb z(1 + z + z2)(1 + z) z + 2z2 + 2z3 + z4.

Translating back, thecrazy diceare1,3,4,5,6,8 and 1,2,2,3,3,4.

Now we introduce the notion of the representation function. So,supposethereis aset A of nonnegativeintegers and that wewish toexpressthenumber of waysinwhichagiveninteger n canbewrittenas the sum of two of them. The trouble is that we must decide onconventions. Does order count? Can the two summands be equal?

Thereforeweintroducethree representation functions.

r(n) #{(a, a) : a, a A, n a + a};

So hereorder counts, and they can beequal;

r+(n) #{(a, a) : a, a A, a a, n a + a},

order doesnt count, and they can beequal;

r(n) #{(a, a) : a, a A, a < a, n a + a},

order doesnt count, andthey cant beequal. Intermsof thegenerat-ingfunctionfor theset A, namely, A(z)

aA z

a, wecanexpressthegenerating functions of theserepresentationfunctions.

Thesimplest is that ofr(n), whereobviouslyr(n)zn A2(z). (7)

To deal withr(n), wemust subtractA(z2) fromA2(z) toremovethecaseofa a andthendivideby 2to removetheorder. So here

r(n)zn 12

[A2(z) A(z2)]. (8)


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Finally for r+(n), we must add A(z2) to this result to reinstate thecaseofa a, and weobtain

r+(n)zn 1

2[A2(z) + A(z2)]. (9)

Can r(n) be constant?

Isitpossibletodesignanontrivial setA,sothat,say,r+(n) isthesamefor all n? Theanswer is NO, for wewould haveto have0

A. And

then 1 A, else r+(1) r+(0). And then 2 / A, else r+(2) 2.And then 3 A, elser+(3) 0 (whereas r+(1) 1), then 4 / A,elser+(4) 2. Continuing in this manner, wefind 5 A. But nowwearestymied sincenow 6 1 + 5, 6 3 + 3, and r+(6) 2.

The suspicion arises, though, that this impossibility may just beaquirk of small numbers. Couldnt A bedesigned so that, except

for somemisbehavior at thebeginning, r+(n) constant?We will analyze this question by using generating functions. So,

using (9), the question reduces to whether thereis an infinite set Afor which

1

2[A2(z) + A(z2)] P(z) + C

1 z , (10)

P(z) is apolynomial.Answer:No.Justlookwhathappensif weletz (1)+. Clearly

P(z) and C1z remain bounded, A

2(z) remains nonnegative, andA(z2) goes to A(1) , acontradiction.

A Splitting Problem

Can we split the nonnegative integers in two sets A and B so thatevery integer n is expressible in the same number of ways as thesumof twodistinctmembers ofA, as it is as thesumof two distinctmembers of B?

If weexperiment abit, beforeweget downto business, andbeginby placing 0

A, then 1

B, else 1 would be expressible as


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A Splitting Problem 9

a + a but not as b + b. Next 2 B, else 2 would be a + a butnot b + b. Next 3 A, else 3 would not be a + a whereas itis

b + b 1

+2. Continuing in this manner, we seem to force

A {0, 3, 5, 6, 9, } andB {1, 2, 4, 7, 8, }. Butthepatternisnotclear,noristheexistenceoruniquenessof thedesiredA, B. Wemust turn to generating functions. So observe that wearerequiringby (8) that

1

2

[A2(z)

A(z2)]

1

2

[B2(z)

B(z2)]. (11)

Also, because of the condition that A, B be a splitting of thenonnegatives, wealso havetheconditionthat

A(z) + B(z) 11 z . (12)

From(11) weobtain

A2(z) B2(z) A(z2) B(z2), (13)

and so, by (12), weconcludethat

[A(z) B(z)] 11 z A(z

2) B(z2),

or

A(z) B(z) (1 z)[A(z2) B(z2)]. (14)

Now this is arelationship that can beiterated. Weseethat

A(z2) B(z2) (1 z2)[A(z4) B(z4)],

so that continuing gives

A(z) B(z) (1 z)(1 z2)[A(z4) B(z4)].

And, if wecontinueto iterate, weobtain

A(z) B(z) (1 z)(1 z2) (1 z2n1)

A(z2n

) B(z2n )

,

(15)


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and so, by letting n , since A(0) 1, B(0) 0, wededucethat

A(z) B(z) i0

(1 z2i ). (16)

And this product is easy to multiply out. Every term zn occursuniquely sinceevery n is uniquely the sumof distinct powers of 2.Indeedzn occurswithcoefficient+1ifn isthesumof aneven numberof distinct powersof 2, and it has coefficient

1, otherwise.

Wehaveachievedsuccess! ThesetsA and B do exist, areunique,andindeedaregivenby A Integers, whicharethesumof anevennumber of distinctpowersof 2, andB Integers,whicharethesumof an odd number of distinct powers of 2. This is not one of thoseproblems where, after theanswer is exposed, oneproclaims, oh, ofcourse. It isnt really trivial, even in retrospect, why the A and B

havethesamer(n), or for that matter, to what this common r(n)is equal. (Seebelow whereit is proved that r(22k+1 1) 0.)A Integers with an even number of 1s in radix 2. Then and

only then

2k+1

111 1 22k+1 1

is not thesumof two distinct As.

Proof. A sumof two As, with no carries has an even number of

1s(so it wont give

odd

111 1), elselook at thefirstcarry. Thisgives

a0 digit so, again, its not 11 1.So r

(22k+1

1)

0. We must now show that all other n have

a representation as the sum of two numbers whose numbers of 1digits areof likeparity. First of all ifn contains 2k 1s then it is thesum of the first k and the second k. Secondly if n contains 2k + 11s but also a 0 digit then it is structured as 111

m

A where A

contains 2k + 1 m 1s and, say, is of total length L then it can be

expressed as 111 1

m1 00 00

2

plus 1A and these two numbers


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An Identity of Eulers 11

haverespectively m 1s and 2k + 2 m 1s. Theseareagain of likeparity so wearedone.

An Identity of Eulers

Consider expressing n as the sum of distinct positive integers, i.e.,whererepeatsarenotallowed.(SoForn 6,wehavetheexpression1 + 2 + 3 and also 2 + 4, 1 + 5, and just plain 6 alone.)

Also consider expressing n as the sum of positiveodd numbers,butthistimewhererepeatsare allowed. (Sofor n 6, weget1+ 5,3 + 3, 1 + 1 + 1 + 3, 1 + 1 + 1 + 1 + 1 + 1.) In both cases weobtained four expressions for 6, and a theoremof Eulers says thatthis is no coincidence, that is, it says thefollowing:

Theorem. The number of ways of expressingn as the sum of distinct

positive integers equals the number of ways of expressing n as thesum of (not necessarily distinct) odd positive integers.

To provethis theoremweproducetwo generating functions. Thelatter is exactly the coin changing function where the coins havethedenominations 1, 3, 5, 7, . . . . This generating function is givenby

1

(1 z)(1 z3)(1 z5) . (17)

The other generating function is not of the coin changing varietybecauseof thedistinctnesscondition. A momentsthought,however,showsthat this generating function is given as theproduct of 1 + z,

1+ z2

, 1+ z3

, . . . . For, whenthesearemultipliedout,eachzk

factoroccurs at most once. In short, theother generating function is

(1 + z)(1 + z2)(1 + z3) . (18)Eulers theoremin its analytic formis then just theidentity

1

(1

z)(1

z3)(1

z5)

(1 + z)(1 + z2)(1 + z3) throughout |z| < 1. (19)


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Another way of writing (19) is

(1

z)(1

z3)(1

z5)

(1

+z)(1

+z2)(1

+z3)

1 (20)

which is the provocative assertion that, when this product ismultiplied out, all of theterms (asidefromthe1) canceleach other!

To prove(2) multiply the1 z by the1 + z (to get 1 z2) anddo the same with 1 z3 by 1 + z3, etc. This gives the new factors1 z2, 1 z6, 1 z10, and leaves untouched the old factors1

+z2, 1

+z4, 1

+z6,

. These rearrangements are justified by

absoluteconvergence, and so weseethat theproduct in (20), call itP(z), is equal to

(1 z2)(1 z6)(1 z10) (1 + z2)(1 + z4) which just happens to beP (z2)! So P(z) P (z2) which of coursemeans that there cant be any terms azk, a 0, k 0, in theexpansionof P(z), i.e., P(z) is just its constant term1, as asserted.

Marks on a Ruler

Suppose that a 6 ruler is marked as usual at 0, 1, 2, 3, 4, 5, 6.Using this ruler wemay of coursemeasureany integral length from1 through 6. But wedont need all of thesemarkings to accomplishthese measurements. Thus we can remove the 2, 3, and 5, and themarksat 0, 1, 4, 6 aresufficient. (The2 can bemeasured between 4and 6, the3 can begotten between 1 and 4, and the5 between 1 and6.) Since

42

6, thisisaperfect situation. Thequestionsuggestsitself then, are there any larger perfect values? In short, can therebe integers a1 < a2 < < an such that the differences ai aj,i > j, takeon all thevalues 1, 2, 3, . . . ,

n2

?If we introduce the usual generating function A(z) ni1 zai ,

thenthedifferencesareexposed, notwhenwesquareA(z), butwhenwemultiply A(z) by A( 1

z). Thus A(z) A( 1

z) ni,j1 zaiaj and

if wesplit this (double) sumas i > j, i j, and i < j, weobtain

A(z)

A 1z n

i,j1

i>j

zaiaj+

n+

n

i,j1

i


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Dissection into Arithmetic Progressions 13

Our perfectruler, by hypothesis, thenrequiresthat thefirst sumbeequal to

N

k1 zk, N

n

2, and since the last sum is the same as

first, with 1z

replacingz

, our equation takes thesimpleform

A(z) A

1

z

NkN

zk + n 1, N

n

2

,

or, summing this geometric series,

A(z) A 1z zN+1

zN

z 1 + n 1, N n2. (21)In search of a contradiction, we let z lie on the unit circle z ei,so that the left side of (21) becomes simply |A(ei)|2, whereas theright-hand sideis

zN+12

z(N+

12 )

z12 z 12 + n 1

sin(N+

12

)

sin 12

+ n 1

and (21) reduces to

A(ei)

2

sinn2n+1

2

sin 12

+ n 1. (22)

A contradiction will occur, then, if wepick a which makes

sin n2n+1

2

sin 12

< (n 1). (23)

(And we had better assume that n 5, since wesaw the perfectruler for n

4.)

A good choice, then, is to make sin n2n+12 1, for exam-ple by picking 3

n2n+1 . In that case sin

2<

2, 1

sin 2> 2

,

1sin 2

< 2

2n22n+23

. and so the requirement (23) follows

from 2n22n+23

< (n 1) or 2n2 2n + 2 > 3(n 1). But2n2 2n + 2 3(n 1) > 2n2 2n + 2 10(n 1) 2(n

3)2

6

2

22

6

2, for n

5. Thereareno perfect

rulers!


21/80


Dissection into Arithmetic Progressions

It is easy enough to split the nonnegative integers into arithmeticprogressions. For example they split into the evens and the odds orinto the progressions 2n, 4n + 1, 4n + 3. Indeed there are manyother ways, but all seemto require at least two of the progressionstohavesamecommondifference(theevensandoddsbothhave2asa common difference and the 4n + 1 and 4n + 3 both have 4). SothequestionarisesCanthepositiveintegersbesplit into at least two

arithmetic progressions any two of which have a distinct commondifference?

Of course we look to generating functions for the answer. Theprogression an + b, n 0, 1, 2, . . . will be associated with thefunction

n0 z

an+b. Thus the dissection into evens and odds cor-responds to the identity

n0 z

n

n0 z

2n +

n0 z

2n+1, andthe dissection into 2n, 4n

+1, 4n

+3 corresponds ton0 zn

n0 z2n +n0 z4n+1+n0 z4n+3, etc.Sinceeachof theseseries

isgeometric, wecanexpresstheir sumsby

n0 zan+b zb

1za . Ourquestion then is exactly whether therecan bean identity

1

1 z zb1

1 za1 +zb2

1 za2 + +zbk

1 zak ,

1 < a1 < a2 < . . . < ak. (24)Well, just as the experiment suggested, there cannot be such a dis-section, (24) is impossible. To see that (24) does, indeed, lead to a

contradiction, all weneed do is let z e 2 iak and observethat thenall of the terms in (24) approach finite limits except the last term

zbk

1zak which approaches.Hopefully, then, this chapter has helped takethesting out of the

preposterous notionof using analysis in number theory.


22/80

Problems for Chapter I 15

Problems for Chapter I

1. Produceaset A such that r(n) > 0 for all n in 1

n

N, but

with |A| 4N + 1.2. Show that every set satisfying the conditions of (1) must have

|A|

N.

3. Showdirectly, withnoknowledgeof Stirlingsformula, thatn! >( n

e)n.


23/80

II

The Partition Function

Oneof thesimplest,mostnatural, questionsonecanask inarithmeticishow to determinethenumber of waysofbreaking up agiveninte-

ger. Thatis, weask aboutapositiveinteger n: Inhowmany wayscan

it bewritten as a + b + c + wherea , b , c , . . . arepositiveinte-gers?Itturnsoutthattherearetwodistinctquestionshere, depending

on whether we elect to count the order of the summands. If wedo

chooseto let theorder count, then theproblembecomes too simple.The answer is just 2n1 and the proof is just induction. Things areincredibly different and morecomplicated if order is not counted!

Inthiscasethenumber of breakupsor partitions is1for n 1,2 for n 2, 3 for n 3, 5 for n 4, 7 for n 5, e.g., 5 has therepresentations1+ 1+ 1+ 1+ 1, 2+ 1+ 1+ 1, 3+ 1+ 1, 4+ 1,5, 3

+2, 2

+2

+1, and no others. Remember such expressions

as 1 + 1 + 2 + 1 are not considered different. The table can beextended further of course but no apparent pattern emerges. There

is a famous story concerning the search for some kind of pattern in

this table. This is told of Major MacMahon who kept alist of these

partitionnumbers arranged oneunder another up into thehundreds.

It suddenly occurredto himthat, viewedfromadistance, theoutline

of the digits seemed to forma parabola! Thus the number of digitsinp(n), thenumber of partitionsofn, isaroundC

n, or p(n) itself

is very roughly e

n. Thefirst crudeassessment of p(n)!

Among other things, however, this does tell us not to expect any

simpleanswers.Indeedlaterresearchshowedthatthetrueasymptotic

formulafor p(n) is e

2n/3

4

3n, certainly not aformulato beguessed!

17


24/80

18 II. ThePartition Function

Now we turn to the analytic number theory derivation of this

asymptotic formula.

The Generating Function

To put into sharp focus the fact that order does not count, we may

view p(n) as thenumber of representations ofn as asumof 1s and

2s and 3s . . . , etc. But this is just the change making problem

wherecoinscomeinall denominations. Theanalysisinthatproblemextends verbatim to this one, even though we now have an infinite

number of coins, So weobtain

n0

p(n)zn

k1

1

1 zk (1)

valid for |z| < 1, whereweunderstand that p(0) 1.Having thus obtained thegenerating function, weturn to thesec-

ond stage of attack, investigating the function. This is always the

tricky(creative?) partof theprocess. Weknow pretty well what kind

of information wedesireabout p(n): an estimateof its growth, per-

hapsevenanasymptotic formulaif wearelucky. Butwedont know

exactly how this translates to the generating function. To grasp the

connectionbetweenthegeneratingfunctionanditscoefficients,then,

seems to be the paramount step. How does one go from one to the

other? Mainly how does onego fromafunction to its coefficients?

It is here that complex numbers really play their most important

role. Thepoint isthat thereareformulas(for saidcoefficients). Thus

we learned in calculus that, if f(z)

anz

n, then an

f(n)(0)

n!,

expressingthedesiredcoefficientsintermsof highderivativesof thefunction. But this a terrible way of getting at the thing. Except for

raremadeup examplesthereisverylittlehopeof obtainingthenth

derivativeof a given function and even estimatingthesederivatives

isnot atask withvery goodprospects. Faceit, thecalculusapproach

is aflop.

Cauchystheoremgivesadifferentandmorepromisingapproach.

Thus, again with f(z) anzn, this time we have the formula


25/80

TheApproximation 19

an 12 i

C

f(z)

zn+1 dz, an integral rather than a differential operator!

Surely thisisamoresecureapproach, becauseintegral operatorsare

bounded, anddifferential operatorsarenot. Thepricewepay is thatof passingto thecomplex numbersfor our zs. Not abadprice, is it?

So let us get under way, but armed with the knowledge that the

valuableinformationabout f(z) will help in getting agood approx-

imation to

C

f(z)

zn+1 dz. But a glance at the potentially explosive1

zn+1

showsus that C hadbetter stay as far away fromtheoriginas it can,

i.e., it must hugtheunit circle. Again, alook at our generatingfunc-

tion

p(n)zn shows that its biggest when z is positive(since thecoefficientsarethemselvespositive). All inall, weseethatweshould

seek approximations to our generating function which aregood for

|z| near 1 with special importance attached to those zs which arenear +1.

The Approximation

Starting with (1), F(z) k1 11zk , and taking logarithms, weobtain

logF(z)

k1

log1

1 zk

k1

j1

zkj

j

j1

1

j

k1

zj k

j1

1

j

zj

1 zj . (2)

Now write z ew so that w > 0 and obtain logF (ew)

k11k

1ekw

1

. Thus noticing that theexpansion of 1ex

1

begins with

1x 12 + c1x + or equivalently (near 0) 1x e

x

2 + cx + ,werewritethis as

logF (ew) 1

k

1

kw e

kw

2

+ 1k 1ekw 1 1kw + ekw

2

(3)


26/80


2

6w+ 1

2log(1 ew)

+

1k

1

ekw 1 1

kw+ e

kw

2

.

Theformof thisseriesisverysuggestive. Indeedwerecognizeany

series

1k

A(kw) A(kw)kw

w as aRiemann sum, approximating

theRiemannintegral

0A(t)

tdt for smallpositive w. Itshouldcome

as no surprise then, that such series are estimated rather accurately.

So let us review theRiemann sumstory.

Riemann Sums

Suppose that (x) is a positive decreasing function on (0, ) andthat h > 0. The Riemann sumk1 (kh)h is clearly equal to thearea of the union of rectangles and so is bounded by the area undery (x). Hencek1 (kh)h 0 (x)dx. On theother hand,theseries

k0 (kh)h canbeconstruedastheareaof thisunionof

theserectangles and, as such, exceeds theareaunder y (x). Sothis timeweobtain

k0 (kh)h

0

(x)dx.

Combining these two inequalities tells us that the Riemann sum

lieswithinh

(0) of theRiemannintegral. Thisisall veryniceand

ratheraccuratebutitrefersonlytodecreasingfunctions.However,we

may easily remedy thisrestrictionby subtractingtwosuchfunctions.

Thereby weobtain

k1

[(kh) (kh)]h

0

[(x) (x)] h[ (0) + (0)].

Calling(x) (x) F(x) andthenobservingthat (0) + (0)is thetotal variation V ofF(x) wehavetherather general result

k1

F(kh)h

0

F(x) h V (F ). (4)

To be sure, we have proven this result only for real functions but

in fact it follows for complex ones, by merely applying it to thereal

and imaginary parts.


27/80

Riemann Sums 21

To modify this result to fit our situation, let us write w hei,h > 0, /2 < < /2, and concludefrom(4) that

k1

F(khei)h

0

F(xei)dx h V(F )

(V is thevariation along theray of argument ), so that

k1

F(kw)w

0

F(xei)d(xei)

w

V(F).

Furthermore, inour caseof an analytic F, this integral is actually

independentof . (Simply apply Cauchys theoremandobservethat

at F falls off like 1x2

). We also may use the formula V(F )

0|F(xei)|dx and finally deducethat

k1

F(kw)w

0

F(x)dx w

0

|F(xei)|dx.

Later on weshow that

0

1

ex

1

1x

+ ex

2

dx

x log 1

2, (5)

and right now wemay notethat the(complicated) function

F(xei) 2x3e3i

exei

2x2e2i e

xei

2xei

1x2e2i(exe

i

1)

exei

xei(exei

1)2

is uniformly bounded by M(x+1)2 in any wedge || < c < /2(m +

M(c)), so that weobtain

k11

k

1

ekw 1 1

kw+ e

kw

2

log 1

2 Mw (6)

throughout | argw| < c < /2.


28/80


The Approximation. Wehave prepared theway for the useful ap-

proximationtoourgeneratingfunction. All weneedtodoiscombine

(1), (3), and (6), replacew by log

1

z , and exponentiate. Theresult is

k1

1

1 zk

1 z2

exp

2

6log 1z

[1 + O(1 z)]

in|1 z|1 |z| c.

But we perform one more neatening operation. Thus log 1z

is

an eyesore! It isnt at all analytic in the unit disc, we must replace

it (before anything good can result). So note that, near 1, log 1z

(1 z) +(1

z)2

2 +(1

z)3

3 + 21

z

1+z + O((1 z)3

), or1

log

1z

12

1+z1z + O(1 z). Finally then,

k1

1

1 zk

1 z2

exp

2

121 + z1 z

[1 + O(1 z)] (7)

in|1 z|1 |z| c.

This is our basic approximation. It is good near z 1, which

we have decided is the most important locale. Here we see thatwecan replace our generating function by the elementary function1z2

exp

2

121+z1z

whosecoefficientsshouldthenproveamenable.

However, (7) is really of no use away from z 1, and, sinceCauchys theorem requires values of z all along a closed loop sur-

rounding 0, weseethat something elsemust besupplied. Indeed we

will show that, away from1, everythingisnegligibleby comparison.


29/80

Riemann Sums 23

To seethis, let us return to (2) and concludethat

logF(z) 1

1 z

j2

1

j |z

|j

1 |z|j 1

1 |z|

j2

1

j

1

j

11 |z|

2

6 1

,

or

F(z) exp 1|1 z| + 2

6 1 11 |z| , (8)

an estimatewhich is just what weneed. It shows that, away from1,

where 1|1z| is smaller than1

1|z| , F(z) is rather small.Thus, for example, weobtain

F(z) exp1

|1 z| when |1

z

|1 |z| 3. (9)Also, in this sameregion, setting

(z)

1 z2

exp

2

12

1 + z1 z

n0

q(n)zn, (10)

(z)

22

exp

2

122

1 z

exp

2

122

3(1 |z|)

so that

(z) exp

1

1 |z|

when

|1 z|1 |z| 3. (11)

The Cauchy Integral. Armed with these preparations and the

feelingthatthecoefficientsof theelementaryfunction(z) areacces-

sible, welaunchour major Cauchy integral attack. So, to commence

thefiring, wewrite

p(n)

q(n)

1

2 i

C

F(z) (z)

zn

+1

dz (12)


30/80


and wetry C acirclenear theunit circle, i.e.,

C is

|z

| r, r < 1. (13)

Next we break up C as dictated by our consideration of |1z|1|z|| ,

namely, into

A is thearc |z| r, |1 z|1 |z| 3,

and (14)

B is thearc |z| r, |1 z|1 |z| 3.

So,

p(n) q(n) (15)

1

2 i

A

F(z) (z)

zn+

1dz

+

1

2 i

B

F(z) (z)

zn+

1dz,

and if we use (7) on this first integral and (9), (11) on this second

integral wederivethefollowing estimates:

1

2 i

A

F(z) (z)zn+1

dz

M

r n+1 (1 r)3/2

exp 2

6

1

1 r thelength of A.

(M is theimplied constant in theO of (7) when c 3).As for thelength ofA, elementary geometry gives theformula

4r arcsin

2(1 r)

r

and this is easily seen to beO(1 r). Wefinally obtain, then,1

2 i

A

F(z) (z)zn+1

dz

M (1 r)5/2.

r nexp

2

6

1

1

r

, (16)

whereM is an absoluteconstant.


31/80

TheCoefficients ofq(n) 25

For thesecond integral,

1

2 i

B

F(z)

(z)

zn+1 dz 1

2 r n+1 2exp1

1 r 2 r 2

r nexp

1

1 r

.

And this is even smaller than our previous estimate. So combining

thetwo gives, by (15),

p(n) q(n) M (1 r)5/2

r nexp

2

6

1

1 r

. (17)

But what is r? Answer: anything we please (as long as 0 < r 0),

e(atb)2

dt

a,


32/80


or

ea

2t2e+2abtdt

a

eb2

.

Thus if we set b2 26

11z and a

2 1 z (thinking of z as real(|z| < 1) for now), weobtain

ezt

2

e+

23 tt2dt

1 z exp

2

6

1

1 z

,

whichgives, finally,

(z) e 2/12

2(1 z)

ezt2

e

23 tt2dt. (19)

Equating coefficientsthereforeresultsin

q(n) e

2/12

2

t2n

n! t2n2

(n 1)!

e

2/3t

t2

dt (20)

theformula for q(n) fromwhichwecan obtain asymptotics.

Reasoningthatthemaximumof theintegrandoccursnear t nwechangevariables by t s + n, and thereby obtain

q(n) Cn

Kn(s)2se

2s 262

ds, (21)

where

Cn e

2n/3

2n

nn+12

enn!,

Kn(s) 1

+s

2

n1 + s

n

2

1 +s

n es

n

+s2

2n 2n.

Since Kn(s) 1, wesee, at least formally, that the above integralapproaches

2se

2

s

2

62

ds

u +

23eu

2

du,


33/80


where we have set s u2

+ 2

6. Furthermore, since ueu

2

is

odd, it is equal to 2

3 e

u2du

2

3. Thus (21) formally

becomes

q(n) e

2n/3

4

3n

2 nnn

enn!. (22)

And score another one for Stirlings formula, which in turn

gives

q(n) e

2n/3

4

3n, (23)

and our earlier estimate(18) allowsus thereby to concludethat

p(n) e

2n/3

4

3n. (24)

Success! We have determined the asymptotic formula for p(n)!

Well, almost. We still have two debts outstanding. We must justify

our formal passage to the limit in (21), and we must also prove our

evaluation(5). So first weobservethat xex ismaximizedat x

1,

so wededucethat 1 + s

n

e

sn 1 (25)

(using x (1 + sn

)) and also

1 + sn e sn e s2

2n (26)

(using x (1 + sn

)2).

Thus using (25) for positives, by (21),

Kn

(s)

es2

for s

0,


34/80


and using (26) for negatives gives us

|Kn(s)| (1 s)es2 2s

n 1 +

s

n es

n 2n2

(1 s)es2 2sn e n1n s2

(1 s)e2s2+1(1+s/

n)2

or

|K

n(s)

| (1

s)e2s

2+1 for s < 0. (28)

Thus (27) and (28) givethebound for our integral in (21) of

2ses22

s

2

6

2for s 0,

and

2s(s 1)e1

+

2/3s

for s < 0.

This bound, integrable over (, ), gives us the requireddominated convergence, and the passage to the limit is indeed

justified.

Finally wegivethefollowing:

Evaluation of our Integral (5). Toachievethislet usfirst notethatas N our integral is thelimit of theintegral

0

(1 eNx )

1

ex 1 1

x+ e

x

2

dx

x

(by dominated convergence, e.g.). But this integral can besplit into

0

(1 eNx ) 1ex 1 1x dxx +

0(1 eNx )

e

x

2xdx

N

k1

0

ekx1 + x ex

x2dx + 1

2

0

ex e(N+1)xx

dx.

Next notethat

1+

x

ex

x2 1

0t e(1t)x dt


35/80


and

ex e(N+1)x

x N+1

1

esx ds.

Hence,byFubini,wemayinterchangeandobtain,forourexpression,

theelementary sum

N

k1

10

t

k + t 1dt +1

2

N+11

ds

s

Nk1

(k 1) log

k

k 1

1

+ 12

log(N + 1)

N

k1(k 1) logk (k 1) log(k 1) N

+1

2log(N

+1)

N logN logN log(N 1) log1 N

+ 12

log(N + 1)

N logN logN! N + 12

log(N + 1).

What luck! This is equal to logN+1(N/e)N

N!and so, by Stirlings

formula, indeed approaches log 12

.

(Stirlings formula was used twice and hence neednt have been

used at all! Thus we ended up not needing the fact that C

2

in theformulan! Cn(n/e)n sincetheC cancels against aC inthe denominator. The n! formula with C instead of

2 is a much

simpler result.)


36/80


Problems for Chapter II

1. ExplaintheobservationthatMacMahonmadeof aparabola when

he viewed the list of the (decimal expansions) of the partitionfunction.

2. Provethesimple factthat, if order counts(e.g., 2+ 5isconsid-ered a different partition of 7 than 5 + 2), then the total numberof partitions on n would be2n1.

3. Explain theapproximation near 1 of log 1z

as 21z1+z +

O(1 z)3. Why does this lead to1

log 1z

12

1 + z1 z + O(1 z)?

4. Why is the Riemann sum such a good approximation to the in-

tegral when the function is monotone and the increments are

equal?


37/80

III

The ErdosFuchs Theorem

Therehas always been somefascination with thepossibility ofnearconstancyof therepresentationfunctionsri (n) (of I (7), (8) and(9)).In Chapter I wetreated thecaseofr+(n) and showed that this couldnoteventuallybeconstant.Thefactthatr(n) cannotbeconstantforaninfiniteset is really trivial sincer(n) is odd for n 2a, a A, andevenotherwise. Thecaseofr(n) ismoredifficult, andwewill treat

itinthischapter asanintroductiontotheanalysisintheErdosFuchstheorem.TheErdosFuchstheoreminvolvesthequestionof justhownearly

constant r(n) can beon average. Historically this all began with thesetA {n2 : n N0},thesetof perfectsquares,andtheobservationthat then r(0)+r(1)+r(2)++r(n)

n+1 , the average value, is exactly equal to1

n+

1times thenumber of latticepoints in thequarter disc x, y

0,

x2 + y2 n. Consideration of the double Riemann integral showsthatthisaverageapproachestheareaof theunitquartercircle,namely/4, and so for this set A, r(0)+r(1)+r(2)++r(n)

n+1 4 (r(n) is onaverage equal to theconstant/4.)

Thedifficultquestionishowquicklythislimitisapproached. Thusfairly simplereasoning shows that

r(0) + r(1) + r(2) + + r(n)n + 1

4+ O

1

n

,

whereas moreinvolved analysis shows that

r(0) + r(1) + r(2) + + r(n)n

+1

4

+ O 1

n2/3 .31


38/80

32 III. TheErdosFuchs Theorem

Very deep arguments have even improved this to o

1n2/3

, for ex-

ample, and the conjectureis that it is actuallyO 1

n 34 for every

> 0. On theother hand, further difficult arguments show that it is

not O

1

n34

+

.

Now all of these arguments were made for the very special caseofA the perfect squares. What a surprise then, when Erdos andFuchs showed, bysimple analytic number theory, thefollowing:

Theorem. For any setA, r(0)+r(1)+r(2)++r(n)n+1 C + O

1

n34

+

is

impossible unless C 0.

This will be proved in the current chapter, but first an appetizer.Weprovethat r(n) cant eventually beconstant.

So let us assumethat

A2(z) A(z2) P(z) + C1 z , (1)

P isapolynomial, and C isapositiveconstant. Now look for acon-

tradiction. The simple device of letting z (1)+ which workedso nicely for the r+ problem, leads nowhere here. The exercises inChapterI were,afterall,handpickedfortheirsimplicityandinvolvedonlythelightest touch of analysis.Hereweencounteraslightlyheav-ier dose. We proceed, namely, by integrating the modulus around acircle. From(1), weobtain, for 0 r < 1,

|A2(rei)|d

|A(r 2e2i)|d +

|P(rei)|d (2)

+ C

d

|1 rei| .


39/80

III. TheErdosFuchs Theorem 33

Certain estimates arefairly evident. P(z) is apolynomial and so

|P(rei)

|d

M, (3)

independent of r (0 r < 1).We can also estimate the (elliptic) integral

d

|1rei| 2

0d

|1rei| by the observation that if z is any complex number inthefirst quadrant, then |z| z + z. Thus sincefor 0 ,1 rei is in the first quadrant, ie i

eir i1rei also is, and1|1rei|

ieieir

( + ) ie ieir . Hence0

d

|1 rei| ( + )

0

iei

ei r d

( + )

log(ei r)

0

( + ) log 1 + r1 r + log

1 + r1 r

.

Thebound, then, is

d

|1 rei| 2 + 2log1

+r

1 r . (4)Theintegral

|A(rei)|2d is adelight. It succumbs to Parsevals

identity. This is theobservation that

|

anein|2d

ane

in

ameimd

m,n

anamei(nm)d

n,m

anam

ei(nm)d

and these integrals all vanish except that, when n m, they areequal to 2 . Hencethisdoublesumis2 |an|2. Thederivationisclearly valid for finite or absolutely convergent series which covers


40/80


our caseof A(rei) (but it even holds in muchgreater miraculousgeneralities).

At any rate, Parsevals identity gives us

|A(rei)|2d 2aA

r2a 2A(r2). (5)

The last integral we must cope with is

|A(r 2e2i)|d, and,unlikeintegrals of |f|2, thereis no formulafor integrals of|f|.ButthereisalwaystheSchwarz inequality|f| (1 |f|

2)1/2, and

so at least we can get an upper bound for such integrals, again byParseval. Theconclusion is that

|A(r 2e2i)|d 2

A(r 4). (6)

All four of theintegralsin(2) havebeenspokenfor andso, by (2)through (6), weobtain

A(r 2)

A(r4) + M2

+ C + C

log

1 + r1 r

. (7)

It is a nuisance that our function A is evaluated at two differentpoints, but wecan alleviate that by the obvious monotonicity of A,A(r 4)

A(r 2), and obtain

A(r2)

A(r 2) + M + C

log

1 + r1 r

. (8)

Is somethingbounded interms of its ownsquareroot? But ifx x + a, weobtain(x 1

2)2 a + 1

4,

x

a + 14

+ 12

, x a

+12

+ a +14

. This yields apure bound onx. Then

A(r 2) M+ C

log

1 + r1 r

+

M + C

log

1 + r1 r

. (9)

But, so what? This says that A(r 2) grows only at the order oflog 1

1r as r 1, but it doesnt say that A(r 2) remains bounded,does it? Wherein is the hoped contradiction? We must revisit (1)for this. Thereby we obtain, in turn A2(r 2) A(r 4) P (r 2) +


41/80


C

1r 2 , A2(r 2) P (r 2) + C

1r 2 , A2(r2) M + C

1r 2 , and finally

A(r2

) M + C1 r2 , (10)arateof growthwhichflatly contradicts(9) andso givesour desiredcontradiction.

If this proof seems like just so much sleight of hand, let us ob-serve what is really going on. We find ourselves with a set Awhose ri (n) is almost constant and this means that A

2(z)

C

1

z

.

On the one hand, this forces A(z) to be large on the positive axisA(r2) > C

1r 2

, and, on the other hand Parseval says that the

integral of|A2(z)| isA(r 2) and C1z

(beingfairly small except near1) has asmall integral, only O(log 1

1r ). (So A(r2) < C log 1

1r ).In cruder terms, Parseval tells us that A2(z) is large on average,

so it must belargeelsewherethan just near z 1, and so it cannotreally be like C

1z . (Note that the elsewhere in the earlier r+(n)problemwas the locale of1, and so even that argument seems tobein this spirit.)

So let us turn to theErdosFuchs theoremwiththesamestrategyin mind, viz., to bound A(r 2) below by C

1r 2

for obvious reasons

and then to bound it aboveby Parseval considerations.

ErdosFuchs Theorem

WeassumetheA is aset for which

r(0) + r(1) + + r(n) C(n + 1) + O(n

), C > 0, (11)and we wish to deduce that 1

4. As usual, we introduce the

generating function A(z) aA za, so that A2(z) r(n)zn,and therefore 1

1z A2(z) [r(0) + r(1) + + r(n)]zn. Since

(n + 1)zn 1(1z)2 our hypothesis (11) can bewritten as

1

1 z A2

(z) C

(1 z)2 +

n0anz

n

, an O(n

),


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or

A2(z)

C

1 z +(1

z)

n0

anzn, an

O(n). (12)

Of course we may assume throughout that < 1. Thereby (12)yieldstheboundM(1r 2)1 for anr 2n,sothatweeasilyachieveour first goal namely,

A(r 2) >C

1 r2

, C > 0. (13)

As for the other goal, the Parseval upper bound on A(r 2), againwe wish to exploit the fact that A2(z) is near C

1z , but this takessomedoing. Fromthelook of (12) unlike(1), this nearness seemsto occur only where(1 z) anzn is relatively small, that is, onlyinaneighborhoodofz 1. Wemustenhance thislocaleif wearetoexpectanythingfromtheintegration, andwedosoby multiplyingby a function whose heft or largeness is all near z 1. A handysuch multiplier for us is thefunction S2(z) where

S(z) 1 + z + z2 + + zN1, N large. (14)Themultiplication ofS2(z) by (12) yields

[S(z)A(z)]2 CS2(z)1 z + (1 z

N)S(z)

anzn, (15)

whichgives

|S(z)A(z)|2 CN2

|1 z| + 2|S(z)

anzn|, (16)

and integration leads to

|S(rei)A(rei)|2d

CN2

d

|1 rei| (17)

+ 2 |S(rei) an(rei)n|d.


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As before, wewill useParseval on thefirst of theseintegrals, (4)on the second, and Schwarzs inequality together with Parseval onthethird.

So write S(z)A(z) cnzn, and conclude that |S(rei)A(rei)|2d 2 |cn|2r2n. Since the cn are integers, |cn|2 c2n cn andsothisis,furthermore, 2

cnr

2n 2S(r 2)A(r 2).(The general fact then is that, if F(z) has integral coefficients,

|F(rei)|2d 2F(r 2).)Now we introduce a side condition on our parameters r and N

whichweshall insist on henceforth namely that

1

1 r 2 N. (18)

Thus,by(14), S(r 2) > N r2N N (1 1N

)N N (1 12

)2 N4

,and by (13), A(r2) > C

1

r 2

, and weconcludethat

|S(rei)A(rei)|2d > CN

1 r 2, C > 0. (19)

Next, (4) gives

CN2

d

|1 rei

| MN2 log

e

1 r2

(20)

and our last integral satisfies

S(rei) an(rei)n d

S(re

i

)2

d

an(re

i

)

n2d

2

k


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At last, combining (19), (20), and (21) allowstheconclusion

C

M N1 r 2 log

e

1 r 2 +1

N (1 r2). (22)

Once again we are masters of the parameters (subject to (18)),and so we elect to choose r, so that N

1 r2 1

N (1r 2) . Thus

our choiceis to make 11r 2 N

32+1 and notehappily that our side

condition(18) issatisfied.Alsopluggingthischoiceinto(22) gives

C

M N

4

14+2 (2 + 3logN ). (23)

Well, success is delicious. We certainly see in (23) the fact that 1

4. (If the exponent of N, 41

4+2 , werenegative then this right-handsidewouldgoto0,2+3logNnotwithstanding,and(23) wouldbecomefalsefor largeN.)


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Problems for Chapter III 39

Problems for Chapter III

1. Showthatthenumber of latticepointsinx2

+y2

n2, x, y

0,

is 4 n2. By theRiemannintegral methodshow that it is, infact

4n2 + O(n).

2. Ifx is bounded by its ownsquareroot (i.e., by

x + a), thenwefind that it has a pure bound. What if x, instead, is bounded byx2/3 + ax 1/3 + b? Does this insureabound on x?

3. Supposethat aconvex closed curvehas itscurvaturebounded by. Show that it must comewithin 2 of somelatticepoint.

4. Produceaconvexclosedcurvewithcurvatureboundedby whichdoesnt comewithin

1200of any latticepoint.


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IV

Sequences without ArithmeticProgressions

The gist of the result of Chapter IV is that a sequence of integerswithpositivedensity mustcontainanarithmetic progression(of atleast threedistinct terms).

Moreprecisely andinsharper, finitizedform, thisisthestatementthat, if > 0, thenfor largeenoughn, any subsetof thenonnegativeintegers below n withat least n members must contain threeterms

a, b, c wherea < b < c anda + c 2b. Thisisashock tonobody.If a set is fat enough, itshould contain all sorts of patterns. Theshock is that this is so hardto prove.

Atany ratewebeginwithavastly moregeneral consideration, thenotionof anaffinepropertyof finitesetsof integers. Soletusagreetocall aproperty P anaffine property if itsatisfiesthefollowingtwoconditions:

1. For each fixed pair of integers, with 0, theset A(n) hasP if and only ifA(n) + has P.

2. Any subset of aset, which has P, also has P.

Thus, for example, the property PA of not containing any arith-

metic progressions is an affine property. Again the trivial propertyP0 of just being any set is an affineone.Now wefix an affineproperty P and consider a largest subset of

the nonnegative integers below n, which has P. (Thus we requirethat this set hasthemost memberspossible, not just to bemaximal.)Theremaybeseveral suchsetsbutwechooseoneof themanddenoteit by S(n

;P ). Wealso denotethenumber of elements of this set by

41


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42 IV. Sequences without Arithmetic Progressions

f (n;P ). So, for example, for thetrivial property, f (n;P0) n, andfor PA, f (3;PA) 2, f (5;PA) 4.

It follows easily from conditions 1 and 2 that thisf (n)

is sub-additive, i.e., f (m + n) f (m) + f (n). If we recall the factthat subadditive functions enjoy the property that limn

f (n)

nex-

ists (in fact limnf (n)

n inf f (n)

n), we are led to define CP

limnf (n;P )

n. This number is a measure of how permissive the

property P is. Thus CP0 1, becauseP0 is totally permissive. Theannouncedresult about progression=freesequencesamountstothe

statement that CPA 0, so that PA is, in this sense, totally unper-missive. At any rate, wealways have0 CP 1, and wemay dubCP thepermission constant.

The remarkable result proved by Szemeredi and then later byFurstenberg is that, except for P0, CP is always 0. Their proofs arebothrather complicated, andweshall contentourselveswiththecase

ofPA, which was proved by Roth.

The Basic Approximation Lemma

It turns out that the extremal sets S(n;P ) all behave very much asthoughtheir elementswerechosenat random. For example, wenotethat such a set must contain roughly the same number of evensas odds. Indeed if 2b1, 2b2, . . . , 2bk were its even elements, thenb1, b2, . . . , bk wouldbeasubset of

0, n

2

andso wecouldconclude

that k f

n

2

. Similarly the population of the odd elements of

S would satisfy this same inequality. Since n2

12

f(n), we con-

clude that both the evens and the odds contain not much more thanhalf thewholeset. Thereby theevens and theodds must beroughlyequinumerous. (Thus, two upper bounds imply thelower bounds.)

Delaying for the moment the precisestatement of this random-ness, let us just note how it will prove useful to us with regard toour arithmetic progression considerations. Thepoint is simply that,if integers were chosen truly at random with a probability C > 0,there would automatically bea huge number of arithmetic progres-


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sions formed. So we expect that even an approximate randomnessshould produceat least onearithmetic progression.

Thepreciseassertion is that of thefollowing lemma.

Lemma.

aS(n;P ) za CP

kn z

k + o(n), uniformly on |z| 1.

Remark. In terms of the great SzemerediFurstenberg result thatCP 0(exceptfor P P0), thisisatotal triviality. Weareprovingwhat in truth is an empty result. Nevertheless we are not preparedto givethelengthy and complex proofs of this general theorem, andso we must prove the Lemma. (We do what we can.) The proof, infact, isreallyjustanelaborationof theoddsandevensconsiderationsabove.

Proof. The basic strategy is to estimate qn(z) aS za CP

k


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So let N be chosen, and let be any th root of unity, i.e., 1. To estimateqm(), let us writeit as

1

aSa


50/80


Substituting (4) in (3) gives

qm()

2 fm CP

m

+(f (nm)CP(nm)). (5)

Now wefind it useful to replace the function f(x) CPx by itsmonotonemajorantF(x) maxtx (f(t)CPt ) andnotethatthisF(x) isnondecreasingandsatisfiesF(x) o(x) sincef(x) CPxsatisfies thesame. So (5) can bereplaced by

qm() 2Fm

+ F (n m) 2F

n

+ F (n) (6)

(a bound independent of m).So choosen0 so that x n0 impliesF(x) x, andthenchoose

n1 so that x n1 implies F(x) n0 x. Fromnow on wewill pickn n1 and also will fix N [ nn0 ].

Dirichlets theorem1 on approximation by rationals now tells us

that the totality of arcs surrounding these with length 2 2(N+1)covers the whole circle. Thus using (2) for q(z), and 2 2

(N+1) 22 n0n gives

q(z) [2F

n

+ F(n)]

1 + 2 n0

. (7)

Weseparatetwo cases:CaseI: n0. HereweuseF ( n ) F(n) and obtain [2F ( n ) +F(n)](1+ 2 n0

) (2+1)(1+ 2 n0

)F(n) 3(1+ 2 n0

)F(n)

(3 + 6 n0)F(n) (6 + 3)n0F(n) (6 + 3)n0 n0 n 22n.CaseII: > n0. Here[2F (

n

) + F(n)](1+ 2 n0

) [2F ( n

) +

F(n)](1 + 2 ). But still nn0

, or n

n0. So F ( n ) n , andtheaboveis

(2n

+n)(1

+2 )

(3

+6)n < 22n.

In either caseDirichlets theoremyields our lemma.So let P be any affineproperty, and denotebyA A(n;P ) the

number of arithmetic progressionsfromS(n;P )(whereorder counts

1Dirichletstheoremcanbeprovedbyconsideringthepowers1,z, z2, , zN for zany pointontheunitcircle. SincetheseareN+ 1pointsonthecircle, twoof themz

i

,jj

mustbewithinarclength2

N+1 of oneanother.Thismeans| argzi

j

| 2

N+1and calling|i j| gives theresult.


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and equality is allowed). Weshow that

A(n;P )

C3P

2 n2

+ o(n2

). (8)

The proof is by contour integration. If we abbreviate

aS za

g(z), then werecognizeA as the constant termin g(z)g(z)g(z2),and so wemay write

A 12 i |z|1

g2(z)g(z2)dz

z. (9)

Now writing G(z) k


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Proof. By the definition of PA, the only arithmetic progressionsin S(n;PA) are the trivial ones, three equal terms, which number isat most

n. ThusA

(n;PA) n, and so, by (8),

C

3

PA

n2

2 + o(n2

) n.

ThereforeCPA 0.


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Problems for Chapter IV

1. Attach a positiverational to each integer from1 to 12 so that all

A.P.s with common difference d up to 6 obtain their correctmeasure 1

d.

2. Provethat, if weask forageneralizationof this, thenwecanonlyforcethecorrect measure 1

dfor all A.P.s of common difference

d, by attaching weights onto 1, 2, . . . , n, ifd O(n).3. If we insist only on approximation, however, show that we can

always attach weights onto 1, 2, . . . , n such that the measuregivento every A.P. withcommondifference m is within en/mof 1

a.


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V

The Waring Problem

In afamous letter to Euler, Waring wrotehis great conjectureaboutsums of powers. Lagrange had already proved his magnificent the-orem that every positive integer was the sum of four squares, andWaring guessed that this was not just aproperty of squares, but that,infact, thesumof afixednumber of cubes, fourthpowers, fifthpow-ers, etc., also worked. He guessed that every positive integer was

thesumof 9 cubes, 19 fourth powers, 37 fifth powers, and so forth,and although no serious guess was made as to how the sequence 4(squares), 9, 19, 37, . . . went on, he simply stated that it did! Thatis what weproposeto do in this chapter, just to provetheexistenceof therequisitenumber of thecubes, fourth powers, etc. Wedo notattempt to find thestructureof the4, 9, 19, . . . , but just to proveitsexistence.

So let us fix k and view the kth powers. Our aim, by Schnirel-manns lemmas below, need be only to producea g g(k) and an (k) > 0 such that the sumof g(k) kth powers represents atleast thefraction (k) of all of theintegers.

Oneof thewonderful thingsabout thisapproachisthat it requiresonly upper bounds, despite the fact that Warings conjecture seems

to require lower bounds, something seemingly totally impossiblefor contour integrals to produce. But theadequateupper bounds areobtained by theso called Weyl sums given below.

So first we turn to our three basic lemmas which will eventuallyyield our proof. These are A, the theorem of Dirichlet, B, that ofSchnirelmann, and finally C, theevaluation of theWeyl sums.

49


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50 V. TheWaring Problem

A. Theorem (Dirichlet). Given a real x and a positive integer M,

there exists an integer a and a positive number b M such that

|x a

b | 1

(M+1)b .Proof. Consider the numbers 0, x, 2x, 3x, . . . , Mx all reduced(mod 1). Clearly, two of these must be within 1

M+1 of each other.If these two differ by bx, then 1 b M and bx (mod 1) is, inmagnitude, 1

M+1 . Nextpick aninteger a that makesbx a equaltobx (mod1). So|bx a| 1

(M+1) whichmeans|x ab | 1(M+1)b ,

as asserted. Q.E.D.

We also point out that this is a best possible result as the choicex 1

M+1 showsfor every M. (Again, wemay assumethat (a, b) 1for, if they haveacommondivisior, thiswouldmaketheinequality|b| M even truer).

B. Schnirelmanns Theorem. IfS is a set of integers with positive

Schnirelmann density and0 S, then every non-negative integer isthe sum of at mostk members ofS for some k 1.

Lemma 1. LetS have density and0 S. Then S S has densityat least2

2.

Proof. All thegapsinthesetSarecoveredinpartbythetranslationofS by thetermofS just beforethisgap. Hence, at least thefraction of this gap gets covered. So fromthis covering wehavedensity fromS itself and times thegaps. Altogether, then, weindeed have + (1 ) 2 2, as claimed.

Lemma 2. IfS has density > 12, then S S contains all the

positive integers.

Proof. Fix an integer n which is arbitrary, let A be the subset ofS which lies n, and let B be the set of all n minus elements ofS. SinceA contains morethan n/2 elementsand B contains at leastn/2elements, thePigeonholeprincipleguaranteesthatthey overlap.So supposethey overlap at k. Sincek A, weget k S, and since


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V. TheWaring Problem 51

k B, we get n k S. These arethe two elements of S whichsumto n.

RepeatingLemma1j

times, then, leadstoasummingof 2j copiesofSandadensity of 1 (1 )2j ormore. Sincethislatter quantity,for large enough j, will become bigger than 1

2, Lemma 2 tells us

that 2j+1 copies of S giveus all theintegers, just as Schnirelmannstheoremclaims. Q.E.D.

C. Evaluation of Weyl Sums. Let bZ, b

0 and k

N,

P(n) be a polynomial of degree k with real coefficients and leading

coefficient integral and prime to b, and letI be an interval of length

N. ThennI

e

P(n)

b

N1+o(1)b21k

where the bound depends on k.

Here as usual wedenotee(x) e2 ix .We proceed by induction on k, which represents the degree of

P(n). It is clearly truefor k 1, and generally wemay write

S nI

e

P(n)

b

and may assumew.l.o.g. that I {1, 2, 3, . . . , N}. Thereby

|S

|2

N1

jN+1 n{1,2,...,N}

n{j+1,j+2,...,j+N}

e P(n) P (n j )

b .

This inner sum involves a polynomial of degree (k 1) but has aleading coefficient which varies with j. If we count those j whichproduceadenominator ofd, whichof coursemust divideb, thenweobservethatthismustappear roughly d timesinaninterval of lengthb. Sothisnumber ofj inthefull interval of length2N

+1isroughly

(2N+1)b

d.


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Thefull estimate, then, by theinductivehypothesis is

|S|2

d|b

N

b dN1

+o(1)

d1

2k

2

N2+o(1)

b b

1

1

2k

2 d|b 1

N2+o(1)b 12k2 bo(1).So weobtain

S N1+o(1)b 12k1 ,

and theinductionis complete.Now wecontinueas follows:

Lemma 3. Letk > 1be a fixed integer. There exists a C1 such that,for any positive integers N, a, b with (a, b) 1,

N

n1e

a

bnk C1N1+o(1)b21k .

Our endpoint will bethefollowing:

Theorem. If, for each positive integers, we write

rs (n)

nk1++nks n

ni 0

1,

then there exists g andC such thatrg(n) Cng/ k1for alln > 0.

Thepreviously citednotionsof Schnirelmannallow deducing, thefull Waring result fromthis theorem:

There exists a G for which rG(n) > 0for alln > 0.

To proveour theorem, since

rs (n) 1

0

mn1/ k e(xm

k

)

s

e(nx)dx,


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it suffices to provethat thereexists g and C for which

10

N

n1e(xnk)

g

dx CNgk for all n > 0. (1)

First someparenthetical remarks about this inequality. Supposeit isknowntoholdfor someC0 andg0. Then, since|

Nn1 e(xn

k)| N,it persists for C0 andany g g0. Thus (1) is aproperty of largegs,in other words, it is purely a magnitude property. Again, (1) is a

best possibleinequality in that, for each g, thereexists ac > 0 suchthat

10

N

n1e(xnk)

g

dx > cNgk for all n > 0. (2)

To see this, note thatN

n1 e(xnk

) has a derivative bounded by2 Nk+1. Hence, in theinterval (0, 14 Nk

),

N

n1e(xnk)

N 2 Nk+1 14 Nk N

2,

and so (2) follows with c 1

4 2g .Theremainder of our paper, then, will bedevotedtothederivationof (1) from Lemma 3. Henceforth k is fixed. Denote by Ia,b,N thex-interval |x a

b| 1

bNk1/2 , and call J Nk|x ab |, j [J],wherea, b, N, j areintegers satisfying N > 0, b > 0, 0 a < b,(a, b) 1, b Nk 12 .

By Dirichletstheorem, theseintervalscover (0, 1). Our maintool

is thefollowinglemma:

Lemma 4. There exists > 0 and C2 such that, throughout anyintervalIa,b,N,

N

n1 e(xnk

)

C2N

(b + j ) .


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Proof. This is almost trivial ifb > N2/3, for, sincethederivativeof|

N

n1 e(xnk)| is bounded by 2 Nk+1,

N

n1e(xnk)

Nn1

e(a

bnk)

+x ab

2 Nk+1

N1+o(1)

b1

2k1+ 2 N

3/2

b N

1+o(1)

b1

2k1+ 2 N

b1/4,

by C, which gives the result, since j 0 automatically. Assumethereforethatb N

2/3

, andnotethefollowingtwosimplefacts(A)and(B). Fordetailssee[K. Knopp, Theory and Application of InfiniteSeries,Blackie& Sons,Glasgow,1946.] and[G.PolyaundG. Szego,

Aufgaben und Lehrsatze aus der Analysis, Dover Publications, NewYork 1945, Vol. 1, Part II, p. 37]. Q.E.D.

(A)IfM isthemaximumof themoduli of thepartial sumsmn1

an,V the total variation of f(t) in 0 t N, and M the maximumof themodulus off(t) in 0 t N, then

Nn1

anf (n)

M(V + M).(B) IfV is thetotal variation off(t) in 0

t

N, then

Nn1

f(n) N

0

f(t)dt

V .Now write 1

b

bn1 e(

a

bnk) and

N

n1

e(xnk)

S1 +

S2

, (3)

where

S1 N

n1

e

a

bnk

e

x a

b

nk

,

S2 N

n1e

x a

b

nk

.


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Weapply (A) to S1. To do so, wenotethat

m

n1

e

ab

nk

0 +

b[m/b]


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Proof of (1). Chooseg 4

, givenasabove. By Lemma4, sincethelength of each Ia,b,N is at most 2N

k,

Ia,b,N

N

n1e(xnk)

g

dx C7Ng

(b + j )41

Nk.

Summing over all a, b, j gives theestimate

C7Ngk

b,j

b

(b + j )4 CNgk

since

b1

j01

(b+j )3 < , and theproof is complete.


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Problems for Chapter V 57

Problems for Chapter V

1. If wepermitpolynomialswitharbitrary complex coefficientsand

ask the Waring problem for polynomials, then show that x isnot thesumof 2 cubes, but it is thesumof 3 cubes.

2. Show that every polynomial is thesumof 3 cubes.

3. Show, in general, that thepolynomial x is pivotal, that is ifx isthe sumof g nth powers, then every polynomial is the sumof gnthpowers.

4. Show that if max(z, b) > 2c, wherec isthedegreeofR(x), thenPa + Qb R is unsolvable.

5. Show that theconstant polynomial 1canbewrittenasthesumof4n + 1 nthpowers of nonconstant polynomials.


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VI

A Natural Proof of theNonvanishing of L-Series

Rather than the usual adjectives of elementary (meaning not in-volving complex variables) or simple (meaning not having too

manysteps) whichrefertoproofs, weintroduceanewone, natural.

Thisterm, whichisjust asundefinableastheothers, isintroducedto

meannothavinganyadhocconstructionsorbrilliancies.A natural

proof, then, isonewhichprovesitself, oneavailabletothecommon

mathematician in thestreets.A perfect example of such a proof and one central to our whole

constructionisthetheoremof PringsheimandLandau. Herethecru-

cial observation is that aseries of positiveterms (convergent or not)

canberearrangedatwill. Additionremainsacommutative operation

whenthetermsarepositive. Thisisasumof asetof quantitiesrather

than thesumof asequence of them.

Theprecisestatement of thePringsheimLandau theoremis that,foraDirichletserieswithnonnegativecoefficients, thereal boundary

point of its convergenceregion must beasingularity.

Indeed this statement proves itself through the observation that

naz

k(az)k

k!(logn)k is a power series in (a z) with non-

negativecoefficients. Thusthe(unique) power seriesfor annz ann

a

naz

has nonnegative coefficients in powers of (a z).So let b be the real boundary point of the convergence region of

annz, andsupposethat b isaregular point andthat b < a. Thus

the power series in (a z) continues to converge a bit to the left

ofb and, by rearranging terms, the Dirichlet series converges there

also,contradictingthemeaningofb. A natural proof of anatural

59


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60 VI. A Natural Proof of theNonvanishing ofL-Series

theoremfollows, onewithavery nicecorollary whichwerecordfor

futureuse.

(1) If a Dirichlet series with nonnegative coefficients represents afunctionwhichis(canbecontinuedtobe)entire,thenitiseverywhere

convergent.

Our ultimate aim is to prove that the L-series have no zeros on

the line z 1. This is the nonvanishing of the L-series that we

referred to in the chapter title. So let us begin with the simplest of

all L-series, the -function, (z) 1nz . Our proof, in fact, wasnoticedby Narasimhanandisasfollows: Assume, par contraire, that(z) hadazeroat1+ia,a real.Then(sic!)thefunction(z)(z + ia)

would beentire. (Seetheappendix, pageno. 63).

Theonly troublepointscouldbeat z 1or at z 1 ia where

oneof thefactorshasapole, butthesearethencancelledby theother

factor, which, by our assumption, has azero.

A bizarreconclusion, perhaps, that theDirichlet series (z)(z +ia) is entire. But how to get acontradiction? Surely thereis no hint

fromitscoefficients, they arentevenreal. A naturalstepthenwould

be to make them real by multiplying by the conjugate coefficient

function, (z)(z ia), which of course is also entire. We areled,

then, to form2(z)(z + ia)(z ia).

This function is entire and has real coefficients, but arethey pos-

itive? (We want themto be so that we can use (1).) Since these arecomplicated coefficients dependent on sums of complex powers of

divisiors, we pass to the logarithm, 2log(z) + log (z + ia) +

log (z ia), which, by Eulersfactorization of the-function, has

simplecoefficients. A dangerousroute, passingtothelogarithm, be-

cause this surely destroys our everywhere analyticity. Nevertheless

let us brazen forth (faint heart fair maiden never won).By Eulersfactorization, 2log(z) + log (z + ia) + log (z

ia)

p

2log 1

1pz+ log 1

1pzia+ log 1

1pz+ia

p,v1

vpvz

(2 + piva + p+iv a ), and indeed these coefficients are nonnega-

tive! The dangerous route is now reversed by exponentiating. We

return to our entire function while preserving the nonnegativity of

thecoefficients. All in all, then,


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VI. A Natural Proof of theNonvanishing ofL-Series 61

(2) 2(z)(z + ia)(z ia) isanentireDirichletserieswithnonneg-

ative coefficients. Combining this with (1) implies the unbelievable

fact that(3) the Dirichlet series for 2(z)(z + ia)(z ia) is everywhere

convergent.

The falsity of (3) can be established in may ways, especially if

werecall that the coefficients areall nonnegative. For example, the

subseries corresponding to n power of 2 is exactly equal to1

(12z)2 1

12zia 1

12z+iawhich exceeds 1

(12z)2 1

4along the

nonnegative (real) axis and thereby guarantees divergence at z 0. Q.E.D.

Andsowehavethepromisednatural proof of thenonvanishingof

the-function which can then lead to thenatural proof of thePrime

Number Theorem. Wemust turntothegeneral L-serieswhichholds

the germof the proof of the Prime Progression Theorem. Dirichlet

pointed out that the natural way to treat these progressions is notone progression at a time but all of the pertinent progressions of a

givenmodulussimultaneously, forthisleadstotheunderlyinggroup

and henceto its dual group, thegroup of characters. Let us look, for

example, atthemodulus10. Thepertinent progressionsare10k + 1,

10k + 3, 10k + 7,10k + 9, so that the group is the multiplicative

group of 1,3,7,9 (mod 10). Thecharacters are

1 : 1(1) 1, 1(3) 1, 1(7) 1, 1(9) 1,

3 : 3(1) 1, 3(3) 1, 3(7) 1, 3(9) 1,

7 : 7(1) 1, 7(3) 1, 7(7) 1, 7(9) 1,

9 : 9(1) 1, 9(3) 1, 9(7) 1, 9(9) 1,

and so theL-series are

L1(z) p1

1

1 pz

p3

1

1 pz

p7

1

1 pz

p9

1

1 pz,

L3(z) p1

1

1 pzp3

1

1 ipzp7

1

1 + ipzp9

1

1 + pz ,


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62 VI. A Natural Proof of theNonvanishing ofL-Series

L7(z) p1

1

1 pz

p3

1

1 + ipz

p7

1

1 ipz

p9

1

1 + pz,

and

L9(z) p1

1

1 pz

p3

1

1 + pz

p7

1

1 + pz

p9

1

1 pz.

(Here z > 1 to insure convergence and the subscripting of the

characters is used to reflect the isomorphismof the dual group and

theoriginal group.)The generating function for the primes in the arithmetic pro-

gressions ((mod 10) in this case) are then linear combinations of

the logarithms of these L-series. And so indeed the crux is the

nonvanishing of theseL-series.

What could bemorenatural or morein thespirit of Dirichlet, but

to prove these separate nonvanishings altogether? So we are led to

taketheproduct of all theL-series! (Landauusesthesamedeviceto

provenonvanishing of theL-series at point 1.)

Theresult is theDirichlet series

Z(z) p1

1

(1 pz)4

p3

1

(1 p4z)

p7

1(1 p4z)

p9

1(1 p2z)2

,

andtheproblemreducestoshowingthatZ(z) iszero-freeonz 1.

Of course, this is equivalent to showing that

p11

1pzis zero-

freeon z 1, which seems, at first glance, to beamoreattractive

formof the problem. This is misleading, however, and we are bet-

ter off with Z(z), which is the product of L-series and is an entirefunction except possibly for a simple pole at z 1. (See the

appendix.)

Guidedby thespecial caseslet usturnto thegeneral one. Solet A

be a positive integer, and denote by GA the multiplicativegroup of

residue classes (mod A) which are prime to A. Set h (A), and

denote the group elements by 1 n1

, n2

, . . . , nh

. Denote the dual

group of GA by GA and its elements by 1, n2, . . . , nh arranged


67/80

VI. A Natural Proof of theNonvanishing ofL-Series 63

so that ni ni is an isomorphism of G and G. Next, for z >

1, write Lni (z) nj pnj1

1ni (nj)pz and finally set Z(z)

ni Lni (z). As in the case A 10, elementary algebra leads toZ(z)

nj

pnj

1

(1phjz)

h/ hj, wherehj is the order of the group

element nj.

Asbefore,Z(z) isentireexceptpossiblyforasimplepoleatz 1,

andweseekaproof thatZ(1+ia) 0forreal a.Soagainweassume

Z(1 + ia) 0, formZ2(z)Z(z + ia)Z(z ia), and concludethat

it is entire. We note that its logarithm and hence that it itself has

nonnegativecoefficientsso that (1) is applicable.

So, with dazzling speed, weseethat azero of any L-series would

lead to the everywhere convergence of the Dirichlet series (with

nonnegativecoefficients) Z2(z)Z(z + ia)Z(z ia).

The end game (final contradiction) is also as before although 2

may not be among the primes in the resultant product, and wemay

haveto takesomeother prime . Nonetheless again weseethat thesubseriesof powersof divergesat z 0whichgivesusour QED.

Appendix. A proof that theL-series areeverywhereanalytic func-

tions with the exception of the principal L-series, L1 at the single

point z 1, which is asimplepole.

Lemma. For any in [0,1), define f(z)

n11

(n )z 1

z1for

z > 1. Then f (z) is continuable to an entire function.

Proof. Since, for z > 1,

0ente ttz1dt 1

(n )z

0et

tz1dt (z)(n )z

, by summing, weget

1(n

donald j newman analytic number theory gradu

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