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Fractional parts of powers and related topics Bennett, Michael A. 1993

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FRACTIONAL PARTS OF POWERS AND RELATED TOPICSByMichael A. BennettB. Sc. (Mathematics) Dalhousie University, 1987M. Sc. (Mathematics) University of British Columbia, 1989A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFDOCTOR OF PHILOSOPHYinTHE FACULTY OF GRADUATE STUDIESMATHEMATICSWe accept this thesis as conformingto the required standardTHE UNIVERSITY OF BRITISH COLUMBIAMarch 1993© Michael A. Bennett, 1993In presenting this thesis in partial fulfilment of the requirements for an advanceddegree at the University of British Columbia, I agree that the Library shall make itfreely available for reference and study. I further agree that permission for extensivecopying of this thesis for scholarly purposes may be granted by the head of mydepartment or by his or her representatives. It is understood that copying orpublication of this thesis for financial gain shall not be allowed without my writtenpermission.(Signature) Department of  tla +6 ernoA(sThe University of British ColumbiaVancouver, CanadaDate ^Al 61 2c ) /913DE-6 (2/88)AbstractIn this thesis, we employ a variety of explicit approximations to tackle some problemsin Diophantine analysis. We are generally concerned with constructing rational functionapproximations to certain polynomials and to the exponential and other related functions.From these systems, we deduce arithmetic information about the original problem.Chapter 0 is an introduction to the field and contains background information andknown results from the literature. The situation for algebraic numbers in general is brieflysurveyed. In Chapter 1, we utilize Pade approximation in the manner of F. Beukers togeneralize and sharpen results of Beukers, D. Easton and G. Xu about lower bounds forfractional parts of powers of rationals. Some theorems of a "semi-effective" nature arediscussed and density results for certain constructed sets are proved.In Chapter 2, we utilize Euler-Maclaurin summation to describe the content of Pade-type approximants to the binomial function. These results form the basis for the afore-mentioned improved bounds. Through similar techniques to those employed in Chapter1 we derive lower bounds upon fractional parts of values of the exponential at integers.These are given in Chapter 4.Chapter 3 contains an application of the above theory to additive number theory, inparticular to Waring's problem and related questions about additive bases. The Hardy-Littlewood-Vinogradrov circle method is used with a theorem from Chapter 1 to provea version of the Ideal Waring problem with restricted summands.iiTable of ContentsAbstract^ iiList of Tables vAcknowledgements^ vi0 Introduction 10.1 Preamble ^ 10.2 Lower bounds on 11011 for 0 rational ^ 20.3 Lower bounds on Pkii for 0 algebraic 30.4 Relations to Additive Number Theory ^ 40.5 Lower bounds on 110 k 11 for 0 transcendental 41 Fractional Parts of Powers of Rationals 61.1 Introduction ^ 61.2 Some Preliminary Definitions ^ 111.3 Generating Our Approximants 121.4 The Main Theorem ^ 151.5 Bounding the Approximants ^ 171.6 A More Explicit Treatment of Q(s) and E(s) ^ 201.7 Common Factors of Coefficients of Pri and Q, 221.8 The Proof of the Main Theorem ^ 251.9 The case 11(1 + )3/N) k 11 ^ 27iii1.10 Semi-Effective Results ^  371.11 Density Results  ^392 Contents of Pade Approximants to (1 — z) k^432.1 Basic Behaviour of L(a, b, s)  ^432.2 The Case L(1,1,3) ^  462.3 Limiting Behaviour of L(a,b,^)  583 Connections to Waring's Problem^ 603.1 Introduction  ^603.2 Asymptotic Theory: Notation and Definitions ^  623.3 Necessary Lemmas ^  653.4 The Contribution of the Major Arcs ^  703.5 Minor Arc Estimates ^  783.6 Dickson's Ascent Argument ^  823.7 Proof of Theorem 3.1.3^ 863.8 Concluding Remarks  894 Bounds for IIe k Il^ 904.1 Multi-point Bounds ^  904.2 A Single-point Bound  90 ( IV' + 1 VcNb )Appendix: Special cases of lower bounds for 95Bibliography^ 100iv(N + N1 2 )kList of Tables> B -k for all k > ko (N) ^ 96> 3 — k for all k > ko (N) 97> B—k for all k > ko (N) ^ 98> B -k for all k > ko (N) 99AcknowledgementsI would like to thank my supervisor, Dr. David Boyd for many helpful discussions andadmirable patience. Thanks are also due to Mark MacLean, Dr. Larry Roberts, Joan deNiverville and the many others who've influenced me over the last five years, especiallythe other graduate students including Big Vaughn Anderson and Mighty Djun Kim.Lastly, I must acknowledge the Math Club (and all who sail in her), noted TX-godE. David Robinson and Dr. Rajiv Gupta for periodic advice.viChapter 0Introduction0.1 PreambleThe behaviour of the fractional parts of real sequences has interested mathematiciansfor much of this century and remains a popular topic of research. A number of impor-tant open problems remain and the field in general displays a diversity which promisescontinued interest in coming years. In the following work, we restrict our attention tosequences of the form O k where 0 >1 is a fixed real number and k is a positive integer.If we defineoil _--, minook — MI)MEZand{0kfoki = min (0k — M)} ma \Af<ekthen we wish to study the size of 110k11 or, more generally, the distribution of {Ok }.Recall that a sequence (x k ) is said to be uniformly distributed modulo 1 if for everypair (a,b), with 0 < a < b <1, one has1lirn —„,1{k<N:a<fxkl<b}1=b—aN-4.00 IVwhere {xk }, as above, denotes the fractional part of x k . It might be asked if the sequence(Ok ) is uniformly distributed modulo 1 or even dense modulo 1 (the latter property beingclearly the weaker one). While a result of Koksma [34] ensures that this is in fact thecase for almost all 0 > 1 (in the sense of Lebesgue measure), it is difficult to find such1Chapter 0: Introduction^ 20 — indeed this appears to be the unfortunate situation for any exponentially growingsequence.0.2 Lower bounds on ilO k li for 0 rationalIn Chapter 1, we discuss the case of fractional parts of powers of rational numbers. Byapplying Pade approximation techniques a la F. Beukers [9], we are able to sharpeneffective lower bounds for certain infinite classes of rationals, providing some advantagesover stronger, ineffective bounds of K. Mahler [41] and weaker, more general effectivebounds due to A. Baker and J. Coates [4]. To do this, we find approximating polynomialsPP (z) and Q,,(z) of degree n in Z[z] which satisfypn (z ) — ( 1 — z )k Q n ( z ) = z2n+ 1 En ( z )or more generallyPn(z) — H (a , b, z)Q ri(z) = z 2 n -l1 En (z)where6-1^ _i_ I, (zb H (a, b, z)^(1^zr^( n+ b^v• .-,, i ,-,), —ZY .L'd^ii=oHere, En (z) is also a polynomial in z with integer coefficients and degree k — n — 1. Byestimating the size of 1Q 7,(z)! and lEn (z)I for specific rational choices of z and using thefact that a nonzero integer has modulus at least one (a trivial fact, but of undeniableutility in transcendence theory!) we obtain a number of bounds, including0.2.1 If 4 < N < k • 3 k then^ + --( 1^1 ycand > 3 -k .0.2.2 If a, 0, and N are integers with a > 1, 0 <101 < N and N > 2, then there is an> (4a02)-k .effective ko = ko la, 0, N) such that k > ko implies,^,^k0 N(ct + Ki )Chapter 0: Introduction^ 3We are also able to construct dense sets (in the interval (1, oo)) of "poorly approximated"rationals by a like approach, where the resulting bounds are again effective.To improve the situation above, we often need to rely upon arithmetic properties ofthe coefficients of the polynomials Pn (z) and Qn (z). This idea has been exploited withrelation to fractional parts of powers of rationals by F. Beukers [9], D. Easton [26] andA. K. Dubitskas [25] and in other settings by G. V. Chudnovksy [14] and M. Hata [30, 31].In Chapter 2, we examine, in some detail, a function associated with these coefficientsand discuss its limiting behaviour. All of the aforementioned improved bounds hingeupon evaluation of this function, so its asymptotics are of clear interest.0.3 Lower bounds on IjO k ii for 0 algebraicResults about 1104 can be seen as being somewhat analogous to irrationality measures,differing only in that one is concerned with approximation by integers rather than ra-tionals. It is unsurprising, then, that many of the techniques from irrationality andtranscendence theory may be applied in both settings. In general, for algebraic 0, it isnot possible to find better than exponential bounds upon 1104, due to the existence ofthe Pisot (or Pisot-Vijayaraghavan) numbers. These are defined to be algebraic integersall of whose other conjugates lie strictly inside the unit circle. It is easy to show that if0 is a Pisot number, then(0.1) k—■oolim 110k11—  0and in fact (see Salem [51]), a partial converse exists — if (0.1) holds for 0 algebraic then0 is a Pisot number. In the case of quadratic Pisot numbers (0.1) becomes more explicit.For example, if 0 = f+1, then 11 0k 11 = 2 — 1) k and hence 0 exponentially (asis in fact the case for all Pisot numbers). An interesting open problem is whether thereare any transcendental 0 satisfying (0.1) with even, say, logarithmic decrease.Chapter 0: Introduction^ 4Another class of algebraic integers 0 for which we have some information regardingii°kii is that of the Salem numbers. These are algebraic integers, all of whose conjugateslie inside or on the unit circle (with at least one in the latter category). From Salem [51],we know that if 0 is a Salem number, then {0 1 } is dense modulo 1, but not uniformlydistributed modulo 1.0.4 Relations to Additive Number TheoryLower bounds upon II (3/2) k II have long been known to be related to the number g(k) inWaring's problem (and vice versa). In fact, one has(0.2) g(k) = 2 k + [(3/2) k ] — 2provided II (3/2) k II > (3/4) k . In Chapter 3, we treat a slight generalization of this result,proving that an analogous equality to (0.2) for the order of {1, Nk , (N +1) k ,...} (N > 4)as an additive basis of the positive integers follows from a theorem in Chapter 1. To obtainthis, we generalize a by now classical bound due to I. Vinogradov [57] on G(k) in Waring'sproblem and apply an ascent argument of L. E. Dickson [21]. We are unfortunately notable to improve the known effective bound of II (3/2) k II so as to imply (0.2) for all k.0.5 Lower bounds on 110 1  for 0 transcendentalFew results in this area have been stated explicitly in the literature and what ones existseem to appear as byproducts of irrationality measures for In n or ir. Through the useof multi-point Pade approximants or Pade-type approximants to 1, e, e 2 , ... , en or 1,In x, 1n2 x, ... , lnn x, lower bounds upon the forms iiekii and Ii7k11 have been obtained byMahler [39, 40, 42] and can be inferred from more recent work on the approximation of ir,en and Inn (see e.g. G. V. Chudnovsky [14]). By direct analogy to Chapter 1, however,Chapter 0: Introduction^ 5we apply single-point diagonal Pade approximation to ez to deduce a lower bound whichis weaker than those already known, but simple to prove. The same techniques, onemay hope, could be utilized in many other settings where closed form approximants ofPade type are available. It would be of genuine interest to determine when this is in factpossible and to compare the induced bounds to those available by other methods.Chapter 1Fractional Parts of Powers of Rationals1.1 IntroductionThere are two natural reasons for interest in questions regarding fractional parts of pow-ers of rationals. The first is as an illustrative example of the utility of Pade and relatedapproximations in number theory, particularly in irrationality and transcendence mea-sures. The second and more frequently cited is the topic's well known connection to g(k)in Waring's problem. In the following, we will derive various lower bounds upon thesefractional parts and discuss their consequences.If we suppose (here and subsequently) that a > b > 2 are relatively prime integersand k a positive integer, then our starting point is the trivial "Liouville" bound11(a/b) k il .? b - k .A fair indication of some of the difficulties involved with these problems is that animprovement of (1.1.1) to strict inequality (for k > 2) would imply the truth of Catalan'sconjecture — i.e., that the equation ax — by 1 has only the solution a = 3,b x = 2 in positive integers (with x, y > 2).In 1957, Mahler [41] showed1.1.2 If e > 0 is given, then there exists ko^ko (a, b, e) such that for all k > ko , we haveii(a/101 > e -Ek6Chapter 1: Fractional Parts of Powers of Rationals^ 7The proof relies upon Ridout's extension of Roth's theorem and is unfortunately ineffec-tive i.e., it is not possible to construct the constant k o given a, b and e — the theoremmerely asserts its existence. Much of what follows deals with finding constructive, that iseffective, proofs of results similar to the above. Any effective improvement upon (1.1.1)enables one to bound the size of possible solutions to certain Diophantine equations andthus would be considerably more desirable than an ineffective result, though usually muchweaker. In fact, however, it appears very difficult to find effective bounds which resemble(1.1.2) either in strength or in universality.If one wishes to find effectively computable bounds for all rationals alb, then the firstand still essentially best results are due to Baker and Coates [4] who proved1.1.3 There exist effectively computable numbers 77 and k o , dependent on a and b, suchthat 0 < <1 and if k > ko , thenIRa I b) k II >This follows as a p-adic corollary to Baker's work on linear forms in logarithms and inpractice they turn out to be extremely close to 1 — for a = 3, b 2 one finds, withpresent estimates, 77 ti 1 —10'. It seems unlikely that further research along these lineswill lead to results of the form of (1.1.3) with 77 arbitrarily close to zero. In section 1.11,however, we construct a dense set of rationals with this property.To obtain stronger bounds than those given by (1.1.3), various authors have consid-ered restricted sets of rationals and applied the method of Pade approximation to thepolynomial (1 — z) k and related functions. In 1981, F. Beukers [9] used this approach toprove1.1.4 If k > 5000, then 11(312)11>andChapter 1: Fractional Parts of Powers of Rationals^ 81.1.5 If N and k are positive integers, N > 2, then(1 + 7\71 )k > N-3/2(8.4)-k.4The basic notion involved here is that while it may be difficult to derive arithmeticinformation directly from hypergeometric functions such as the binomial, very regularapproximants exist which can be more readily handled. This usually entails finding upperbounds for these approximants (or sometimes asymptotics) and then connecting this toother, often trivial, lower bounds for related quantities. By sharpening some of theseestimates, D. Easton [26] improved (1.1.4) and (1.1.5) to1.1.6 There exists an effectively computable ko such that if k > ko , we have11(3/2) k ii > 2 -0.82k .and1.1.7 If N > 4 is an integer, then there exists an effective k ip ,---  ko (N), such that ifk > ko , then(1 + TvI--)k > (3.87) -k . By further strengthening a technical lemma on primes dividing binomial coefficients,A. K. Dubitskas [25] was able to prove1.1.8 There exists an effectively computable k o such that if k > ko , we haveii(3 / 2 )kil > 2 -0.794k .It is not obvious how this result can be much improved without, to quote from Bakerand Coates, the introduction of "fundamentally new ideas". We will see later exactlywhich effective bound is desirable and note here only that the coefficient involved in theexponent of 2 would need to be much reduced.Chapter 1: Fractional Parts of Powers of Rationals^ 9On somewhat more general lines, we mention further results by Easton, who from atechnical theorem similar to our Theorem 1.4.1, deduced the following corollaries1.1.9 If N > 4 is an integer, then there exists effective ko = ko (N) such that if k > ko,then (N^)k > (3.15) -k .and1.1.10 If N > 2 is an integer, then there exists effective k o ko (N) such that if k > ko,then(N + )k > (3.27) -k .Along a more explicit vein (the distinction between explicit and effective being quitedramatic in some cases!), G. Xu [61] employed Beukers' techniques and some facts aboutLegendre polynomials to prove1.1.11 If N > i3 and a are positive integers, then> (18(a + 1)N(36202 ) k ) -1 .Such a statement can be somewhat deceptive in its simplicity. One must bear in mindthat the above is only an improvement upon the trivial bound (1.1.1) when 362a0 2 < Nand hence is still subject to fairly strong restrictions.In what follows, the author will derive a number of bounds similar in nature to thosementioned above. In particular, (1.1.5) and (1.1.7) are sharpened in respectively, explicitand effective versions to1.1.12 If N and k are positive integers with 4 < N < k • 3k , then> 3-k(+ kChapter 1: Fractional Parts of Powers of Rationals^ 10and1.1.13 If N > 4 is an integer, then there exists effective ko = ko (N) such that if k > ko,then(7, -v-1 + )k > (2.85) -k .By generalizing Dubitskas' lemma on divisibility of binomial coefficient products it ispossible to also improve (1.1.9) and (1.1.10), to obtain1.1.14 If N > 3 is an integer, then there exists effective ko = ko (N) such that if k > ko,then(N Tv--+ )k> (2.65) -k .and1.1.15 If N > 2 is an integer, then there exists effective ko = ko (N) such that if k > ko,thenN + v2-( ^)k> (3.01) -k .The latter result yields (1.1.8), essentially as a direct corollary, with N = 2 (noting that(3/2) 2 = 2 + 1/4).Finally, by way of strengthening (1.1.11), we have the following effective result01.1.16 If a, 13, N are integers with N > 101 and a + —N > 1, then there is an effectiveko = ko la, /3,N) such that if k > ko,,^/3 \a + WOk> (4a3 2 ) -k .The constant 4 can be improved somewhat, but already is rather preferable to the 362occuring in (1.1.11).Chapter 1: Fractional Parts of Powers of Rationals^ 111.2 Some Preliminary DefinitionsSince we will be dealing frequently with the binomial function, we define, in analogy tothe binomial coefficients,(1 .3) X_Y= xx for real x > y > 0yY(x — y)x- Y where we adopt the convention l[x^ 1{ x JJ = 1.0Some straightforward properties of this areax(1.4) for real a > 0ay_x^y —1(1.5) = max (1 — t)x0xy tom.)(1.6)y< 2x^with equality exactly when x = 2yand(1.7)x + ay + a yfor real a > 0, x > y > O.The proofs of these follow directly from the definition and/or calculus.We'll further define a function H (a, b, z) which will be the primary object of ourapproximation. If we set6-1 a(1.8)^zb H (a, b, z) = (1 — z)a+ b — E^(-zyi.0 ( Ithen we see that it is possible to view this function as a "truncated" binomial. The abovedefinition is motivated with respect to fractional parts of powers of rationals byLemma 1.2.1 If a, c, m, N, p, q, and r are positive integers, 0 L 0 an integer, N > 2,p < rq and (p, q) = 1, then(aNr + ycrn = 1aPc1 n 13(rq-P)cm I I (Pcm ' (rq)11NP j Mem ' ceNr ) II .IIChapter 1: Fractional Parts of Powers of Rationals^ 12Proof: From the definition (1.8),H(pcm,(rq — p)cm,  —#aNr(rq—p)cm-1 (rqcm)(^\i)# rqcm(--CeNr )(rq—p)cm /(1 aNr 1=0^ct/Vrrqcm= (-1) ("—P)cm^E^(rqcrn) ( CeNr )(rq—p)cm—ii=(rq—p)cm while((aNr 13)9vcm rqcm (rqcm\^NP ) = E^)j=0rqcm (rgern\^= E^)1=0and hence we have(aN(r -piq))(rqcm -i) (  N  yNP/q)ar qcm-i Nr((rq—p)cm—i) oirqcmapcm iaqrq-ocniH(pcm, (rq — p )cm, ^EceNr i=(rq—p)cmso that, sincefrqCrn _rqcm—iNr((rq—p)—i)pii ) u(rq—p)cm-1 „cm arqcm_2 Nr((rq_ocin_z) 3i1=0is an integer, we obtain the required equality.^ ■Throughout the following chapter, we will assume that a, f3 , c, m, etc. are as in thestatement of the above lemma and add the constraint that(ceNr #) q^1NPto avoid trivialities.1.3 Generating Our ApproximantsThe following method for construction of approximants to H(a, b, z) was suggested tothe author by F. Beukers. We suppose that A, B, and C are nonnegative integers withChapter 1: Fractional Parts of Powers of Rationals^ 13A < C and that^< 1. Writingfo 1 0(1 — 013(z — t) C) dt f tA (1 — t) B (z — t) c dt + f tA(1 _^—^diand making the changes of variable t^zt and t^1 — t zt in the first and secondintegrals on the right hand side respectively, we can concludetA(1 — t) B (z — t)c dt(1.9)^= Fin]. — zr+c-FilltBq —^— •(1^t^zt)Ao1zA+C +1 I tA^t)C ^zt)B di.Applying definition (1.8), this last equation becomesJo 1 tA (1 — t) B (z — t) c dt(1.10)^ i=0 (13 i+ C + 1)(—z) fo 1 tB (1 — t)c (1 — t + zt) A dt(-1)c zc-A H (A + B + 1, C — A, z) 10 1 tB (1 — t) c (1 — t + zt) AzA+c+ f tA (1 — t)c (1 — zt)B dt.Since z' divides the right hand side of (1.10), it must also divide the left handsideand so if we letz A + B + C + )PA(z) = A-c(A! B! C!^l ' ! (1 1 tA (1 ^— t) B (z — t)c dtc-A-1(B + c +^fl tB(1 - t)c(1 -^zt)A dt)+ Hoc,. (-z)10i=0Chapter 1: Fractional Parts of Powers of Rationals^ 14(1.12)^QA(z) =(-1)c (A B C 1)! f l tB — 1)C (1 — t + zt)A dtA! B! C!(1.13)^EA(z) =^B(A++ C+1)7 1 A (1—t)c (1— zt) BA! B! C!^othen we have from (1.10),(1.14)^PA(z) — H(A + B +1,C — A, z)Q A(z) = z 2A+1 EA(z)•Nowfo^1 ^A^tB (1 — t) c (1 — t + zt) A dt =^1 B (1 — t) C 2 (A )(1 - t)A-i (zt) i dti=0^6A=^() Z 2 tB+2(1 t)A + C _ i dti=o^°( zi (B i)! (A + C — i)!i=o i^(A+B+C+1)!and henceB+i)^(1.15)^QA(z) = ( -1 ) c 2^C^)zZ0which belongs to Z[z] (of degree A). Similarly,B (A + i) ( A B C 1) (— )^(1.16)^EA(z) = Ei=0^)^1which is also in Z [z] (of degree B). It is clear from (1.11) that PA (z) is a polynomial ofdegree A in z. The coefficients of PA (z) in general can be fairly complicated, so at thisjuncture we will only note that if a prime divides every coefficient of QA (z), then from(1.14) it must divide every coefficient of PA(z).From comparison with, for example, Beukers [9], one may note that the polynomialsPA (z) and QA (z) coincide with the diagonal Pade approximations to H(A + B 1, C —A, z), with error terms EA(z). Of the properties of Pade approximants, we will need onlythe following:Chapter 1: Fractional Parts of Powers of Rationals^ 15Lemma 1.3.1 PA (z)QA+1 (z) — QA(z)PA-Fi(z) = cz 2A+ 1 where c is a non-zero constant(dependent upon A, B and C ).Proof: Applying (1.14) twice, we findPA (Z)QA-F1(Z) — QA(z)PA+1(z) = z2A+1 (EA(Z)QA+1(Z) — Z 2 EM-1(Z)QA(Z))•Now the term on the left is a polynomial in z of degree at most 2A +1, divisible by z 2A+ 1since the right hand side is. This implies that the term E A(Z)( 2 A-Fi(Z) - Z 2 E A.4_1(Z)Q A(Z)must be constant and hence equal to EA(0)QA +1(0). SinceEA(0)(A + B B+ C +1)=^0 and Q A-H.(0) = ( -1 )c (A + cyC + 1) °we have the result with c = EA(0)QA +1 (0). ■1.4 The Main TheoremBefore we proceed, we must introduce some more terminology. Let us define, for s > l/p,(1.17)and(1.18)wherecd(s)= (14r,,)0Q(s, 01) • rqs +1(rq — p)s + 2E(s). ( tiAtici lE(s,t)l) • -(rq — p)s + 2_(rq — p)s +1cd (s,t). (1 - orq -os+ies-1 (1 — (1 + a N3 r ) t)E (8 ,t) = (1 — t) (r q-P)s+1 t (1 + a13 tr 1Nr^•Further, we setand(1.19)^L(s) = L(rq, p, s) = exp(trqs +1t^e(s,t)))Chapter 1: Fractional Parts of Powers of Rationals^ 16where0(s,t) = max 1^^ps —1^(rq — p)s +1 /[ ^t ^+^r(ps — 1)ti +1 ' 1((rq — p)s + 1)ti +1{rqs + 1 rqs +1 j^rqs +1and the summation is over all t such thatt ^[(ps — 1)1^[((rq — p)s + 1)1[rqs +1^rqs +1 rqs + 1^= t — 2.Under these assumptions, we haveTheorem 1.4.1 Ifs is rational, s > 11p, N > 2, e > 0, andrqs +1(rq — p)s + 2_then there exists an effective constant k o = ko (ex, 0,p, q, r, N,s,e) such that for all k > ko(1.20) 101(rq-p)s-1-2aps-1 • E(s) < L(s) • NT(aN r OnkNP )(1.21) > (aNr Q(s)L(s) -1 +e • (rq — p)s + 2_(rq — p)s + 1To prove this, we write s = c/d for c and d relatively prime integers, let M be anyinteger and define(1.22) A apern /3(r q -p)cm H (pern (rq p)crn, —0 )^NaNrwhere 6 E Z, 0 < S < rc. The term N -P5 is appended to M to account for the fact thatwe initially consider not arbitrary k as exponents in the above, but rather strict multiplesof m. We have, from (1.14), that for n = din or din — 1Qn ( CVN13r(1.23 )+ 10 rg-p)cm-E2n- 1 apern-2n-1 N-r(2n+1)^(  — 13  )1\ar^apC771 i3 ( 7- q — p)crn pn —0^m N-pe5N n aNr IChapter 1: Fractional Parts of Powers of Rationals^ 17and by Lemma 1.3.1 can choose n = dm or dm — 1 such that the last quantity is nonzero.By Lemma 1.2.1, the problem is reduced to finding suitably good upper bounds uponQn( - 13 )aNr and En ( 1^3aNr) , while bounding the right hand side of the inequality (1.23)from below.1.5 Bounding the ApproximantsWe first deal with the associated binomial coefficients, proving if A, B, C, and m arepositive integers,((A + B + C)m)!^1A + B + C^ ((A + B + C)A+B+c ILemma 1.5.1^<^(Am)! (Bm)! (Cm)! 271- m .1 ABC AABBCcProof: We use the following explicit version of Stirling's formula (see, for example,Stromberg [55])/(12n+1/4) < n ! < rtn e-norn evi2n .if n E N, nne -nV2rn e lThis directly yields the above inequality with the RHS multiplied by eX , whereX =12(A + B + C)m 12Am + 1/4 12Bin + 1/4 12Cm + 1/4 .If A, B, and C are positive integers then we have X < 0 and so eX < 1.^■We may now set about finding upper bounds for 1Q n 1 and 'En '. Taking n = dm ordm — 1, where d is a positive integer satisfying d < pc, we show (rq — p)c + 2d(rq — p)c+ dLemma 1.5.2 Q.( --/  )aNr< (Q(c/d) d •)-Proof: Before proceeding, we note that the implied constant in the above inequalitydepends upon c, d, p, q, r, a, 9, N and the choice of n as either dm or dm — 1, but notand- (1 + ° \ t\n dt.aNr ) )I = 1 1 ( 1 — tyrq—p)cm-En tpern—n-1 (10 k11..11 =Chapter 1: Fractional Parts of Powers of Rationals^ 18upon m itself. From (1.12), we may write (rqcm + n)! =(pcm - n - 1)! ((rq - p)cm + n)! n!Qn (.1\r , )(1.24)rqc+ d -(rq - p)c + 2d(rq - p)c + 2d(rq - p)c + d(1 — t) (rq—p)cm+n tpcm—n-1 (-. — a aNr+ 13^ )t)n1^ dt1. 1and thus, applying Lemma 1.5.1 with A = pc - d, B = (rq - p)c + d and C = d yields(using (1.3))(1.25)^Qn(aNi3r) < D 1 • (whereD1=1^(rqc+ d)(pc - d)27r \ ((rq - p)c + d) • d1^((rq - p)c -I- d) • d27r \ (rqc+ d)(pc - d)if n = dmif n = dm - 1Now we have1 ^, 11[n/ ml1 = 1 (Q (C I CI, trr - 1 (1 — t) ("—P)c+ rnimitPc—( rnirni +1) (' — I1^0+ 1'. )t)^dto^ aNrand so1/1< (max1Q(c/d,t) d ir 1 • /1, tE[0,11where(1 — t)(rq—p)c-i-in/mI tpc—(rnim]+1) (1 — ( 1 +  /3  )tyrtimi dtaNr IChapter 1: Fractional Parts of Powers of Rationals^ 19Combining this with (1.25) gives Qn (aN/3r )-^_(rq — p)c +2d(rq — p)c + d _ )-(1.26) < D2 (Q(C/CO d •for D2 = D1 • /1/iTtriCi lQ(C/C1 , WI'^ ■To bound 'En ', we proveEn( --° )aNrrqc+ d(rq — p)c + 2dLemma 1.5.3 < (E (c/d) d •Proof: Again we may note that the implied constant is independent of m (and dependentas in Lemma 1.5.2). From (1.6), it follows that(rqcm + n)!=(1.27)(pcm — n — 1)! ((rq — p)cm + n)! n!1 0 — tyrq—p)cm-En in (1 + ^ tycm—n-1 dta Nr ifoand hence a further application of Lemma 1.5.1 yields(En \ —13aNr )irqc + d -(rq — p)c+ 2d(rq — p)c + 2d -r(rq — p)c + d -(1.28) < Di - (where D 1 is as in (1.25) and1J = i (1 — t) (rq—p)cm-l-ntn (1 + ^ tYcm—n-1 dt.aNrSince1J = f (E(c/d,t) d ) m-1 (1 — (rq—P)c+[ninli t [n/mi a + 13 0"-(inimi+1) dtJo^ aNrwe have, as before,IJI < (maxe[cui1E(c/d, t) dr 1 • Jl,t Ji. = foChapter 1: Fractional Parts of Powers of Rationals^ 20where1(1 — t)(r q — p) c-F[nim]t[n i nil (1 + ^ )l3r t \Pc— ([nim]+ 1 ) dtaNFrom (1.28) we conclude En(  3 )aNrrqc + d(rq — p)c + 2d )rn(1.29) < D3 • (E(C/CO d •for D3 = D1 • Ji /max IE(c/d, t) d I.^ ■te[o,i]1.6 A More Explicit Treatment of Q(s) and E(s)At first glance, it is not immediately obvious why the quantities Q(s) and E(s) weredefined as they were. To see the underlying motivation, we prove the followingLemma 1.6.1 Ifs E R is such that(1.30) pi(rq—p)s-1-2 ceps-1rqs +1< NT(rq — p)s + 2_1 if/3 >0then we have Q(s) < 1 5/4 if #<0aNrfor some t i E ( (IN, + 0 ,1). It follows that IQ(s, 01 is maximal at either to or t 1 . NowaNr 13 + ^13 rt1 > ctivr +^  implies that 1 t i. < aNr^and 1 (1 + aNr )t i < aN , so that#^ #1Q( 8 ,ti)l < ( aNr + 0 ) (rq-P)s+1 ( 13 ) < ( 13 ) ( rq—P)s+2aNr^aNraNrSince to < aN, + 0 , we have Q(s, to ) < (1 — to) ("—P)s+2 48-1 and the latter function is-1. Thismaximal on [0,1] for t = ps — 1 where it assumes the valuerqs +1rqs +1(rq — p)s + 2TProof: If i3 > 0, then since Q(s,0) = Q(3,1). Q (s, aNa rN-F ,3) = 0, we have that Q(s,t)Tis maximal (and positive) on [0,1] for some t o E (0, aN  ) and minimal (negative)aNr + 0rqs +1?-^( /3 )(r-p).51-2(rq — p)s + 2^aNr-1> 1(2(3, ti) I1Q (s ) < 1+ 1131(7. g -O s+ 1 aps(1.31)Chapter 1: Fractional Parts of Powers of Rationals^ 21implies 1Q(s,to)1 <rqs +1(rq — p)s + 20^Nr(r q-p)s-F2 aps-1Now by (1.30), we have <_rqs +1(rq — p)s + 2-1and henceso thatrqs + 1 -max1Q(8,01 <te[om^(rq — p)s + 2and thus Q(s) < 1 as required. If, however, # < 0, we note thatmax 1Q(3, t)1 < max 1(1 — 0 (7.g-03+2es-1i +  I °I. ‘: max 1(1 — t) ( rq -P )s+ l tPs1^tE[co^tE[o,i)l aNr tE[0 ,11,^-1^-rqs +1^1)31^rqs +1^- -1..-..(rq — p)s + 2^+ aNr (rq — p)s +1and since (1.30) yields 101  <aNr^1#^1 NIrqs + 1 - -1we have(rq — p)s + 2rqs +1(rq — p)s +1is a monotone increasing function in x (for y constant) and from (1.6), werqs +1(rq — p)s +1xSince,L Yhave that > 4, and the result obtains.^ ■Lemma 1.6.2 Ifs E IR satisfies (1.30), thenE(s) <{23/20 if 13 > 01 if 13 < 0Proof: If # > 0, thenmax1E(s,t)1< max 1(1 — t) (rq-P)s+1 ti • (1 + 13 Ys -1tE[0,1]^tE[0,1]^ (IN'.(rq — p)s + 2 -1 t^13 vs-1(rq — p)s +1^0 + aNr )=Chapter 1: Fractional Parts of Powers of Rationals^ 22p )ps_iso that E(s) < (1 aNr^. Now from (1.30) we know that rqs +1^-1(rq — p)s + 2^5-^<^aNr^13(r q -1)) '9+1 Cesrqs +1(rq — p)s + 2and thus(1 + ^((rq — p)s + 2) ((rq—P)s+2) • (ps — 1)(P8-173 -1T_8-1 5 (1 +^aNr (rqs + 1)( r q 3+1)which, by (1.7) is(ps ^1yps-1) \ps -15_ (1 + 4(ps +1)(7'8+ 1 ) )and hence < 23/20 by calculus.If we have < 0, then 1 +aNr^ t < 1 and we reach the desired result immediately.■It should be remarked at this juncture that the bounds Q(s) < 1 (/3 > 0) andE(s) < 1 (/3 < 0) in Lemmas 1.6.1 and 1.6.2 respectively are sharp. Subject to theconstraint (1.30), however, the upper bounds of 5/4 and 23/20 may both be improved(in the direction of unity) and if we add further conditions upon a, /3, N, p, q, r and swe obtain stronger results still.1.7 Common Factors of Coefficients of Pr, and Q„The arithmetic behaviour of the Pade approximants to various hypergeometric functionsfrequently displays a striking regularity. In our case, the binomial coefficients associatedto /3, and Q„ (see e.g. (1.15)) contain large common integer factors. It is, in somecircumstances, these factors that are necessary to attain any nontrivial result.Lemma 1.7.1 Suppose t is a positive integer satisfyingr ^At ^r ^Bt ^r(1.32)LAS-B+Ci+ Leld-B+Ci + [A+Bct +Ci —t 2Chapter 1: Fractional Parts of Powers of Rationals^ 23and that we define^/ = max { ^At  1 + ^Bt 1 + r ct^ +ITLA+B+C^LAA-B-ECJ^[A+B-I-C]If x is a prime such that I < x < (A + B C)/t, then we may conclude that x divides(A+C—iVB-ki) for all i = 0, 1, ..., A.CAtProof: Since x > I, we have A/x < LA^  + 1, while x < (A + B + C)/t impliesAix > At/(A B C) whence {A/x} > {At/(A B C)} (where {y} denotes thefractional part of y). Similarly, {B Ix} > {Bt/(A B C)} and additionally {C/x} >{CtI(A+B+ C)}. Now (1.32) is readily seen to be equivalent to^f ^At^Bt^Ct1 A-I-B-FC }+{A+B-FC } 1/4A+B+C = 2and hence we havef_A-t+ f/3-t + LEI > 2Ix' Ix' Ix'and thus, if i is an integer, 0 < i < A, eitherA(1.33)Or(1.34)In the first instance,{-B }+{}?1x^xC+^1 +X^X^X[B : ii LB] {xi] =1so that x divides (B + i. ) (since ord x (n!) = E[nIxi]). If, however, (1.34) holds, we mustz^i=1have {A/x} {i/x} and thusChapter 1: Fractional Parts of Powers of Rationals^ 24or equivalently{ A^ + fel =xwhich is > 1 by (1.34). It follows thatwhence x dividesA C — i [Ax—x+ cC — ■The above lemma is a generalization of a result of Dubitskas [25]. Let us define((rq — p)cm + 2n — i (pcm — n — 1 +G(c, d) =^min^gcd^n=dm or dm-1 1=0,1,...,n^(rq — p)cm nWe use Lemma 1.7.1 to proveLemma 1.7.2 Let e > 0 be given. There exists an effectively computable mo such thatif m > mo we haveG(c, d) > L(cl d) (1- E)dmProof: Firstly, we take A = n, B pcm — n — 1 and C = (rq — p)cm+ n in Lemma 1.7.1.Then via Rosser and Schoenfeld [48] (which quantifies the statement E In x ti b — a,a<x<bwhere the sum is over prime x), we can find effective Ei E2 > 0 and mo such that ifm > mo and if we define, for t satisfying (1.32)(1.35)^It = (0(cl cl,t)dm,(rqc d)m/t]then)1-E,^G(c,d)>11(11 x^(x prime)t xEItfrqc+ d>flexp ^ dO(c d, t) — •E Y7 .2tThis isChapter 1: Fractional Parts of Powers of Rationals^ 25> exp((rqct+ d de(cld,t))m — e2m)= L(c1d) dm exp( —E2m).If we choose e 2 suitably small relative to e, the result obtains.^ ■1.8 The Proof of the Main TheoremThe machinery and notation is now in place to prove Theorem 1.4.1.Proof of Theorem 1.4.1: Suppose that (1.20) is satisfied. Since it is a strict inequality,we can find an e i with 0 < E l < E/2 so that(1.36) loi(rq-p)s-F2aps-1rqs +1(rq — p)s + 2E(s) < L(s) 1-2e1Nr. Applying Lemmas 1.7.1 and 1.7.2 with E i yields an effective mo with m > m o implyingthat Pn and Qn are polynomials in z whose integer coefficients contain a common factorexceeding L(s)(1-el)dm. Hence we may write, for such m,En(a—)3 )NrQ„ ( Th  ^)aNr + 101(rq-p)cm-F2n-lapcm-2n-1N-r(2n+1)(1.37)> ce- riN-rn-P5 L (s)( 1 -61)dm.Now by Lemma 1.5.3, we haverqs +1(rq — p)s + 2)drn(1.38) En( )aNr < (E(s)and since L(s) > 1, from (1.36) there exists an effectively computable m i such thatEn(-13 )aNr1<2a'N-rn-P8L(s)(1-eodm,(1 . 39) 101(r9-0cm-1-2n-l apcm-2n-1N-r(2n+1)Chapter 1: Fractional Parts of Powers of Rationals^ 26for in > m i . We conclude, therefore, that if m > m 2 = max{mo , m1}, > —1 a—nN—rn—P8 L(8) (1-61)dm•2(1.40)Remembering (1.37), we now apply Lemmas 1.2.1 and 1.5.2 so that (1.40) yields( (CeN r + O) q ' cm^-114--NP^)^NPs> C1 (ceNrQ(s)L(s) -14-e 1 •(rq - p)s + 2-(rq - p)s + 1-)—dinwhere C1 is a computable constant, m > m 2 and M is an integer ("translated" from M((aNr P+ t t,by the difference^- a'''. 13 ( rq- P )cm H (pcm, (rq - p)cm, - 0 / (aNT))). NowNtaking k = rcm - 8 gives—^-^k(rq - p)s + 2 ) TS(rq - p)s +1((aNr + 13)q  )k ^M NP ) (ceNr + OW > C2 (aN r Q(S)L(S) -1+el •where C2 is effective and independent of k. Again, since L(s) > 1 and 0 < e l < E/2,there is a computable ko such thatk) rs((aNr + ,6)q )k ^M NP^) ((INT' + OW> (aN"Q(s)L (s) -1 +e (rq - p)s + 2(rq - p)s +1_for all k > 14) . Since the choice of M was arbitrary, the result obtains.^■2 then thereNChapter 1: Fractional Parts of Powers of Rationals^ 271.9 The case 11(1 --1-,3/N) k ilIn the next few sections, we will be primarily concerned with utilizing somewhat simplerforms of Theorem 1.4.1 obtained by specifying certain parameters. With the situationabove in mind, we letln NN(1.41)^p=g=r.a=1,0<#1n 4N2 and s < 2/32 real.Under these conditions, we first sharpen Lemma 1.6.2, provingLemma 1.9.1 Under the conditions in (144 we haveE(s) < 8/e2 .Proof: First we note that as in the proof of Lemma 1.6.2, we have from (1.41)(1.42)\\/NA202) -1^(^)VN/(20)-1E(s) < (1 + --)^< 1 ln NSince (1.41) gives , < ^/V ln 4N12N^1 \A/2-1)and the function (1+ -^< 8/e 2 for all x > 50,the desired result obtains for all N satisfying(1.43)^ lln Nn 4N V2N > 50i.e., for all N > 861. Now, if N < 860, then we check the first inequality in (1.42) andfind that it holds unless /3 = 1 and 7 < N < 49. Noting that E(s, t) is increasing ins, we choose s =^,32 and compute the remaining cases from (1.18). Since each hasE(s) < 8/e 2 , the result follows as stated.^ ■We use this lemma to proveTheorem 1.9.2 If and N are positive integers satisfying 13 <In 4Nis an effective constant ko = ko (#,N) such that for all k > k o , we havek(1N > (4N) -0 1"k •In NChapter 1: Fractional Parts of Powers of Rationals^ 28Proof: If we define.R8)=1[3+2111/2'92 — 83(c2s(s+ 1)18+)81- ithen differentiating with respect to s, we find that f is maximal for s satisfying(S + 1 y = e 2—1s 8 8 22 ^1 1i.e., that f(s) < ^ e2 < e 2 /8. Now this implies that /32 1 3 2+ }I E(s) < #2 e 2s 2 E(s)/4and so Lemma 1.9.1 gives(1.44)^ 132 r +2 1 11 E(s) < 20 2 3 2 < N.\ISince the first inequality in (1.44) is strict, we can find an s E Q satisfying s >^N2/32and (1.44). It follows, then, from Theorem 1.4.1, Lemma 1.6.2, and L(s) > 1, that fork > ko , effectively computable, (1.45) > (4N) -kiswhich, by the choice of s, yields the required result.^ ■It may be noted that the bound upon /3 in the statement of the theorem above ensuresthat the lower bound on II(1 + /3/N) k II is nontrivial in the Liouville sense. Furthermore,if we fix 0, and consider the limiting case, we have(1.46)^ lim (4N)V2/N i3 =-- 1N ooand so if e > 0 is given, we can achieve a bound of the form 1.1.2, for large enough N. Wewill discuss the ramifications of this in later section, but mention only that the methods1 )kof Beukers and Easton yield just a lower bound for (1 + N— of e -k as N —> oo. Wenext modify Theorem 1.9.2 so as to provide a bound independent of N, akin to the resultsof Beukers and Easton (1.1.4 and 1.1.6 respectively). Along these lines, we haveChapter 1: Fractional Parts of Powers of Rationals^ 29Theorem 1.9.3 If N > 4 is an integer, then there exists an effectively computable ko =ko (N) such that if k > ko , then(1 + -iv1 )k > (2 .85) - kProof: Taking # = 1, Theorem 1.9.2 implies the above for all N > 52. For N = 4, 5,... 51, we apply Theorem 1.4.1 directly, choosing rational approximations to the largestvalues for s such that (1.20) holds. In each case, we have(1.47)^ (4NQ(s)L(s)-14-')lis < 2.85for sufficiently small s, and the result obtains. For example, in the worst situation, wehave N =10, s = 2.936 and(1.48)^(40Q(2.936)L(2.936)-°.99)1/2.936 < 2.8455.The reader may consult Table 1 of our Appendix for precise results for each N with4 < N < 51.^ ■To obtain explicit rather than just effective bounds we must bound a number of constantsand perform a sizeable amount of computation. We haveTheorem 1.9.4 If N and k are positive integers with 4 < N < k • 3 k , then(1.49) ( 1 + 7\7 )k > 3-k .Proof: To begin, we consider two cases. Firstly if N > 729, we prove that (1.49) holdsfor all k > N/2. If k < N/2 then (1 + N—1 )k < e 1 / 2 and so(1 + ...N )kor1 +(^7v- ) k 1 )kkN= (1 + -N- — 1 > — > 3-k> 0.3512Chapter 1: Fractional Parts of Powers of Rationals^ 30whence (1.49) holds for these k also. For N with 4 < N < 728 we compute explicitko = ko (N) beyond which the desired bound obtains and then, by checking the remainingvalues of k, complete the proof.We take, as before, M to be an arbitrary integer, c and d relatively prime integerswith c > d > 1, s = c/d, 6 an integer with 0 < 6 < c, m a positive integer and n = dmor dm —1. If we define k = cm — 6 and^(1.50)^ lE72(-1 / MI < 2G(c, d)Nn -s+ 1then from the proof of Theorem 1.4.1 (where we do not apply Lemma 1.7.2 as is done inthat proof) we may write(1.51)^(1 + Tv-1 )k — M(N + 1) -6 > D4 G(c, d)(4Q(s)N) -k/ swhere D4 = (2D2((N + 1 )(4Q(S) • N) 1 i s ) 81 1 . The inequality (1.49) then obtains if therighthand side of (1.51) exceeds ric (since the choice of M was arbitrary). SupposeN > 729, c = [NV3] and d = 1. We show that (1.50) is satisfied for k > N/2. Now thefirst inequality in (1.44) yieldslEn (- 1/N )I < D3 (2s 2 )m.To bound D3, we suppose n = 771 and hence that1Ji = I t(1 — t)(1 + t/nr 2 dt.INowE(s, t) = t(1 — t)(1 + t/N)- 1which implies J1 imax1E(s, 01 < 1/(1 + t'/N) < 1 where t' maximizes the functiontE[omt(1 — t)(1 + t/N)' on [0,1]. It follows that D3 < D1 < c/(27). The case n = m — 1 issimilar, with a sharper bound for D3. We may conclude, therefore, thatjEn (-1/N)I < Ter(2c2)m.Chapter 1: Fractional Parts of Powers of Rationals^ 31Since G(c, d) > 1, the above implies (1.50) provided m is such thatN > c Nc_ i2c2^— 7ror equivalently,(1.52) m > ln( cNc-1 Win (—N ).—^7r^2c2Now ln(cNc-1 /7r) < clnN by our choice of c and N while ln(N/(2c2 )) > ln(N 113 /2) >(2/9)1n N (since N > 729). We conclude that the right hand side of (1.52) is boundedabove by 9c/2 and so if k > N/2, it follows that m > N/(2c) > 9c/2.Now by Lemmas 1.5.2 and 1.6.1Pn (-11N)1 < D2 • enand we wish to bound D2. If n m, we have\/1 =io to-2 (1 — t)(1 (N + 1  )t) dt2 + (1 — c)/N— c2c3 — candmax IQ (c, t)This last quantity isQ(c, c + 11 )181 [c-F 111>182^2( 2 ^(c ^(  2c ^— 1cl-lc+1)^c+1^N(c+1)) .since c — 1 < N/91, whence1tmepaxIQ(c,t)I> 2c2 .,i]  Thus1 r D2 <^VC2 1• (^ 2C2 =27c3 2 c)— ^7rSimilarly if n m — 1, we havec2c2 — 1 •1D2 < 27r1^112c2 =c2 — 1 c^7rc2c2 — 1Chapter 1: Fractional Parts of Powers of Rationals^ 32and in either case, we can writeIQ.(-1/N)I < 4'n.The result, then, will follow from (using Lemma 1.6.1)(3/(4N)'/c) k > 8N(N +1)c-1or equivalentlyk > ln(8N(N + 1)a -1 )/1n(3/(4N) 1 /c).Now since c = [N 1 /3] and N > 729, we haveln(3/(4N) 1 /e) > ln(3/(3996) 1 /9 ) > 0.17716andln(8N(N + 1)e -1 ) < ln 8 + c ln(N + 1)so thatln(8N(N -I- 1)c -1 )/1n(3/(4N) l ik ) < 11.74 + 5.65N 1 /3 1n(N + 1).Since this is less than N/2, the result obtains.For 4 < N < 728, we choose values c and d and use Lemma 1.7.1 to find intervalscontaining primes dividing G(c, d). To estimate the contribution of these primes, we applyupper and lower bounds on the Chebyshev function 0(x) = Ep<x lnp from Theorem 10of Rosser and Schoenfeld [48], the corollary to Theorem 6 of Rosser and Schoenfeld [49],Corollary 2 (9.8) of Schoenfeld [52] and the closing remarks to this last paper. We deduceexplicit ko = k0 (N) for each such N beyond which (1.49) holds and tabulate the results inTable 2, together with the choices of c and d, and the lower bound derived for G(c, d) 1 /( dm)(denoted by G). By checking the required inequality for smaller values of k, we completethe proof. This calculation utilizes Fortran code which computes the N-ary expansion of(N + 1) k and searches for long strings of 0's or (N — 1)'s. ■Chapter 1: Fractional Parts of Powers of Rationals^ 33Following the same lines, we can now sharpen a pair of results of Easton [26], provingTheorem 1.9.5 If N > 2 is an integer, then there exists an effective ko = ko (N) suchthat if k > ko , thenN + 1cN) > (2 .65) -k andTheorem 1.9.6 If N > 2 is an integer, then there exists an effective ko = ko (N) suchthat if k > ko , then(N + v2-1 )k > (3 .0].) -kProof of Theorem 1.9.5: Firstly, for N > 80, we choose s to be some rational in theinterval (1.3791n N,1.380 In N) and note that2s + 1s + 2 < N2SO2 + 11-1 s-iE(s) < (1+ —N1 2 ) 8-I < (1 +I[ 8s + 2Since this last function is decreasing for s in the intervals above, we haveE(s) < (1 + r3riy. 1.0008658 .8and we may check thatir s + 22s + 111• E(s) < N2for N > 80. Applying Theorem 1.4.1 and Lemma 1.6.1, then, implies, for k > ko,effectively computable(N + -N-1 ) k s^211 1 /( 2s)) — ks + 111which, by the choice of s, yields the desired result.Chapter 1: Fractional Parts of Powers of Rationals^ 34If 3 < N < 79, we choose rational s as in our Appendix (see Table 3), estimate thecontributions of primes dividing the associated binomial coefficients via Lemma 1.7.1,compute E(s) from the definition and conclude as above. The "worst case" in thissituation occurs for N = 4 where, taking s = 2.39, we have the bound(4 + l\k4) > (2.64917 ...) -k for k sufficiently large.^ ■Proof of Theorem 1.9.6: Suppose N > 26 and take .s rational in (1.3491n N, 1.3501n N).It follows thatand hence1138 + 111ii. s + 2 11 < N3(1 +1 \ 2s-1^i,^ri3S + 111 -1 )2s-1 .V^< Y + 11. s + 2 J1Arguing as before, we have E(s) < 1.0008... and again,3s + 1s + 2• E(s) < N 3 .We conclude via Theorem 1.4.1 that> (Nl is i[ s + 2-s + 11/(3s)—k> (3.01) —kfor the values of N in question, k > ko = ko (N). If 2 < N < 25, we proceed as abovewith the weakest bound obtaining for N = 2, s = 1.0725, where for k > ko effectivelycomputable, we have(2 +4/\k > (3.000353 ...)-k .E(s) <■Chapter 1: Fractional Parts of Powers of Rationals^ 35Regarding this last result, one may note that this argument implies Dubitskas's bound(1.1.8) for 11(3/2)4 (since (3/2) 2 = 2 + 1/4). Again, it not clear how this can be muchimproved.By way of another example in this area, we consider the case of 11(5/4) k II. FromTheorem 1.9.3, we have for k > ko , effectively computable^(1.53)^ 11(5/4)kll > (2.85) -kand careful consideration of Theorem 1.4.1 with a — /3 — p — q — r — 1, N = 4 ands = 2.044, yields for k > ko^(1.54)^ II(5/4)kII > (2.60) -k .Now the observation that (5/4) 3 = 2 — 3/64 enables us to obtain a better bound. Ifwe choose a = q =1, 13 = —3, p = 6, r = 7, N = 2, and s = 0.436, then Theorem 1.4.1implies, for k > ko effective(1.55)^ 11(5/4)4 > (1.93) -k .It appears that, as a general rule of thumb, if we know an m such that (a/b)m =(aNr + 13) 9 /NP for a and /3 "small" and rq > p, then we'll stand a better chance ofobtaining a strong bound from the latter form. For arbitrary a/b, though, finding suchan m is often difficult, sometimes impossible.Without such a form, the best we can do is the following strengthening of a result ofG. Xu [61] (assuming as previously that a + ii/N > 1).Theorem 1.9.7 If 4a/3 2 < N, then there is an effective ko = ko (a, /3, N) such thatk > ko impliesRa + 1 '—,f)k > (4a1,31) -k •Proof: We take p = q = r = 1 in Theorem 1.4.1. If a = /3 = 1 the conclusion followsfrom Theorem 1.9.3, so we may suppose that la,31 > 2. Now, as in the proof of LemmaChapter 1: Fractional Parts of Powers of Rationals^ 361.6.2, we have(1.56) E(s) < IQIaNysand(1.57)^E(s)^E(s, 1/2) • 4 = (1 + 13.^1 .2aNThis implies that E(2) < 9/8 and since L(2) = (3/2)1n 3 — it//6 = 2.09807..., we maynote02a ir -11 E ( 2 ) < 402 N.L21 L(2)From (1.57) and the fact that lim L(s) = it/e1' (see Chapter 2, Theorem 2.2.1), we havelim^11-s + 111 E(s)11- 2 11 L(s )= 00(since /3 < 0 implies a 2). By the continuity of 1[s 12^E(s) and L(s) (see Chapter2, Lemma 2.1.2 for the last), given e > 0, we can choose s E Q such that .s > 2 andN = (1 + E0),3 2as-1 r +2 1 11 E(s)/ L(s)where 0 < 0 < 1. Applying Theorem 1.4.1, we have a computable /c o such that k > koimplies> (4(1 + 6) a s 132 3 +2 11 Q(s)E(s)L(s) -2+e)= (h(s) • ceii3 1 2/s)-k—k/s i/swhere h(s) = (4(1 + 6) r +2 1 1) Q(s)E(s)L(s) -2+e)Now, since s > 2, we have L(s) > 1.66303... (again, see Chapter 2, Theorem 2.2.2)and so Lemmas 1.6.1 and 1.6.2 imply, taking E > 0 small enough, that h(s) < 4 and so> (4431 2)s) —k^(4431) —k .■Chapter 1: Fractional Parts of Powers of Rationals^ 37As a corollary, we haveCorollary 1.9.8 There is an ef ective ko = ko (a,13,N) such that for k > ko, icx ^kN ) > (40 2 ) -k .Proof: If 40 2 < N, this follows from Theorem 1.9.7, while 40 2 > N implies theinequality is trivial in the Liouville sense.^ ■It should be noted that a more careful analysis of L(s) can enable a sharpening ofthe constant 4 in the above results, probably to somewhat less than 3.1.10 Semi-Effective ResultsIf, for a given 7/ > 0, we have a rational alb satisfying(1.58)^ < b-nmfor some m, then, as mentioned earlier, we can use this information to deduce lowerbounds for 11(0)4 for k > ko = ko (a, b,q,m). In particular, we can proveTheorem 1.10.1 Suppose I/ E R is such that b < a < b 2n . Then there is a computableconstant mo = mo (a,b,ii) such that if m > m o satisfies (1.58), then we can find aconstant ko = ko (a, b, , m) withII(a/b) k I I > b- nkfor all k > ko.We call this a "semi-effective result" in that it depends upon the existence (and size) ofm — which is by no means guaranteed. This can be generalized if we haveb < a < Ms-2+271)/(s-1)(1.59)3211a#2Chapter 1: Fractional Parts of Powers of Rationals^ 38for some s > 1 and a large enough m satisfying (1.58) to^(1.60)^ 11(a/b)111 > b -(( s -2+271) /3(8-1))kfor k > ko = ko (a, b, s, ri , m). The following is just the case s = 2.Proof of Theorem 1.10.1: Let us first note that we may assume 1/2 < ri < 1, since y > 1implies (1.58) has no solutions while ri < 1/2 contradicts b < a < b 27). Choose e > 0 suchthat(1.61)^ a < b271 ' and 77 > 1 + E2and mo with(1.62)^b€"1° > 100.Suppose that m is minimal with in > mo and (1.58) and write(a/b)m = a + —Nso that(1.63)^a 5_ [(a/b)m ] + 1, lot < 0-0. , N = bm.Now by (1.61) and (1.63)8a/3 2 < 8b (2-2" )m(b (2"- e ) mb-m + 1)and hence (1.61) and (1.62) yields8a8 2 < 16b(1- e )m < brn = N.Since L(s) > 1, we may apply Lemma 1.6.2 to conclude• E(2) < L(2) • NChapter 1: Fractional Parts of Powers of Rationals^ 39and thus by Theorem 1.4.1, we find a k i = ki (a,b,ii,m) such that for k > ki. (a + Pk> (4Q(2 ) • aN) -kl2> (5aN ) -kl2> (5(am + bm)) -kl2> (fldb (7)-e/2)m) - k .From (1.62), this gives/^13 \kf2 " - Tr ) > (b6-e/4)"1 - 1'.Since (a/b)m = a + /3/N, we may modify the argument in the proof of Theorem 1.4.1 toconclude that there is a k o = ko (a,b,ii, in) withIl(a/b) k il > b- '71cfor all k > ko , as desired.^ ■While the statement of the above result may be somewhat confusing, we note that ifone has an upper bound upon the smallest "suitably large" solution in to (1.58), thenTheorem 1.10.1 implies a bound upon the size of all possible solutions.As a final comment in this section, we mention that for 1/2 < /7 < 1, there are onlyfinitely many rationals a/b in any finite interval which do not satisfy b < a < b27?. Itfollows, then, that Theorem 1.10.1 may be applied to "almost all" rationals in (1, N) forN a positive integer. For example, if I/ = 0.7925, we may utilize Theorem 1.10.1 for allrationals in (1,2), in particular for a/b = 3/2.1.11 Density ResultsIn what follows, we will construct a set of rationals alluded to in the introduction —namely a dense one (in the interval (1, oo)) for which we obtain quite strong, effectivebounds. We first proveChapter 1: Fractional Parts of Powers of Rationals^ 40Lemma 1.11.1 Let N > 2 be an integer and e > 0 be given. Then there is an effectivelycomputable po = po (N, e) such that if p > po , then there exists an effective ko = ko (p, N, e)with 1 )k(N+ N P> e -ekfor all k > ko .Proof: We take, in Theorem 1.4.1, ce= i3=q= 1 and r=p+1. Then (1.20) becomes(p + 1)s + 1(1.64)^ • E(s) < L(s)NP+ 1 .s + 2 -Now(p + 1)s + 1^((PI+ 1)s + 1)(0)+ 1 ) 8+ 1 )=. 5 + 2^(Ps — 1)(ps-- 1 )( s + 2)(s+2) < (p + 1)pSo with .3 = In pi, it follows that1 Nps—i^/^1 \plopE(s) < (1 + Np+ i^ ) < i + Np+1^ )and henceSin( (p + 1)s2 + 1 •p + 1 7^pin p• E(s)) < 21n(p+ 1) + lnpins +^ PSince this is < (p+ 1)1n 2 < (p + 1)1n N for p > 32, (1.64) obtains for such ^ApplyingTheorem 1.4.1 then yields an effectively computable ko = ko (p, N) with> (Np+ , ir s + 2 -18 + 1) k/(03+1)s)(1.65) (N + NP)1cfor all k > ko . Since .--- 0(s), we havefi lm (Nils [is + 211 iMP -Fl)s))P-000^11.8 ^1.11 = 1(remembering that cs = [lnp]) which concludes the proof.^ ■Chapter 1: Fractional Parts of Powers of Rationals^ 41We note that the above may be generalized to replace 1 by any fixed nonzero integer/3, with the same conclusion, though for our purposes 3 = 1 suffices.We use the preceding lemma to proveTheorem 1.11.2 Let e > 0 be given. Then there exists a constructible set of rationalsS,, dense in the interval (1,00), such that if alb is in S,, there is an effective constantko = ko la, b, e) with11(a/b)kil > b-ekfor all k > ko.Proof: Take ai/bi > 1 to be any rational and e, S > 0 to be arbitrary. We construct arational alb in a deleted 6-neighbourhood of a libi which effectively satisfies the abovebound. From Lemma 1.11.1, we find a computable p o with (al + -71 ka l )(1.66) > brkfor p > Po and all k > k1 = ki(P, a l , b1 , e). Letpi = max{po,In b1 +1, ln(biS) } 2}.e In a i^ln a iThen there is an effectively computable constant ko = ko (a i ,b i ,E, 8) such that k > koimplies (1.66) with p replaced by pi . It follows that( D + 1 \kbi^biar )> (bli + c )- kln bifor all such k. We take a = 41+1 +1 and b = biar and note that pi > ^ + 1 impliesE ln a lII(a/b)kII > b'kln(bi 6)for k > ko . Also since pi >ln a^ + 2, we have alb in a deleted 6-neighbourhood ofi—a l ibi , as required. Since the choices of E and 6 were arbitrary, the result obtains.^■Chapter 1: Fractional Parts of Powers of Rationals^ 42There are certainly many different ways to construct sets like the above — here weaP+ 1 + 1have just considered (for a/b E Q) rationals of the form baP^with large p. It wouldbe of some interest to find a dense set in (1, oo) for a given e > 0 with the property thatif a/b were in the set, then for k effectively large,(1.67)^ 11(a/b)kil > e -ek .Such a result appears to be somewhat more difficult than Theorem 1.11.2, however.Chapter 2Contents of Pacle Approximants to (1 — z) k2.1 Basic Behaviour of L(a,b, s)As we have seen in the preceding chapter, a careful analysis of the coefficients of the Padeapproximants to (1 — z)k can provide rewarding arithmetic information. In this chapter,we study in greater detail the function L(a , b, s) (where a > b are positive integers ands > 1/b is a real number) defined by (see (1.19) for comparison)L(a,b, s) expi(E as + 1 0(s, t))where0(s, t) = max{e i (s, t), 0 2 (s, t), 03 (s, t)},1^ bs —1^ (a — b)s +10 1 (8,0=[^t,- 1as+ 11 ++ibse2(8't)^03(8 ' t),tJ +1 = i(a — b)s +1 tlJ+11 as 1 I.^as + 1and t satisfies{ast+ 1]rbs — 1^[  as +1(a — b)s +1 Las + 1t +  ^= t — 2.Upper and lower bounds as well as asymptotics for this function are of interest for avariety of reasons suggested previously. To begin, we proveLemma 2.1.1 If a > b are positive integers and s >11b, then L(a,b,^) < oo.Proof: Sincewe haveL(a, b, s) exp( as +1t ^0(s, t))43^Chapter 2: Contents of Pacle Approximants to (1 — z)k^ 44( as +1^1^ln L(a,b,^) <t°^t^[t1(as +1)] +1)t=ix_,°° ( as +1^1< 2-,^t^t1(as +1) + 1)t=ix_, (as +1) 2  < (as + 1)2 • Li .'— tL2it(t 4. as + 1 )■We may also concludeLemma 2.1.2 L(a, b, s) is a continuous function in s.E as + 1= tProof: Again we consider In L(a, b, s)^t^e(s,t)). Let E > 0 and s be givenand suppose that Is — 8 1 1 < 1. Then by the proof of Lemma 2.1.1, we can find t o withtE a^s + 1 e(s,t)) <612>to^tand(asi t +10(s i ,t)) <E12.i>toConsidertbs — 1 t l + f as +1(  — b)s +1 Ti (s,k).= ft E N, t < k : {as +1} + {as +1 I t^t} = 2}T2 (s,k)-- qt E N, t<k:{  t 1 f bs —1 t1 Oa — b)s ,+ 1 t} = 0}as +1^t as +1^as + 1ThentETi(s,..)^In L(a, b, s i ) =^E (as it+ 1^o(si,t))andIn L(a,b,^). E ( as +1 t^0(s, t))tETi (Si 00 )Chapter 2: Contents of Pade Approximants to (1 — z) k^45andIlnL(a,b,^) — lnL(a, b, 8 1 )1(asi + 1 z (as + 1O(s, t))^t^0(81,0)tovs,to) + e / 2.Now we can choose (5 1 > 0 such that s i with Is — 3 1 1 < Si implies thatTi(s, to) C TI(si, to) c Ti(s, to) U T2(s, to)zand soas + 1 0(8,0 )^z as i + 10(3 1 ,0)t^ ttE_L (s,to)^ teTi(si,.4 o)=tET1(s,to ) as + 10(3,0 )^(as i + 1^o(81,0)]as1 + 1 ^0(Si, t))ttET1(si,to)nT2(s,to)for such s i . If t E Ti (s,t0 ), then we can choose St > 0 with(as + 1^0(3,0)^(asit 1^0(s i ,t))t= (as + 1^as i + 1 )^(et(s,t) — Oi(s i ,t))< 4tofor s i satisfying Is — 8 1 1 < St . If t E Ti (s i , t0 ) fl T2 (s, to ), then we can find St* such thatIs — s i l < St• impliesasi + 1Cl(si ' t) < 4T0and thus if 6. = min{1, 61 , 8t (t < to ), (5t.(t < to )} and Is — s 1 I < S, we haveIln L(a,b,^) — ln L(a, b,^< eas desired.^ ■Chapter 2: Contents of Pade Approximants to (1 - z) k^462.2 The Case L(1,1,^)The behaviour of L(a, b, s) is perhaps a little more predictable than may be obvious froma cursory glance In the special case L(s) = L(1,1, s), related to bounds upon forms like(1 + Tv-ilk  or (a^)k (see Chapter 1, Theorems 1.9.2, 1.9.3, 1.9.4 and 1.9.7), we+haveTheorem 2.2.1 L(s) satisfieslim L(s) = /e11 = 1.76387 • • •8-4•00andTheorem 2.2.2 If 2 < .5 < oo, then1.66303 • • • = L(3) < L(s) < L(2) = 2.09807 • •Recall that L(c/d) provides a lower bound for the dm-th roots of the content of therelated Pade approximants. With this in mind, we may use the above results to sharpen,for instance, Theorem 1.9.2 by reducing the J2/N involved in the exponent of the boundto approximately V1.048/N. To prove the above theorems, we perform a fairly carefulanalysis of certain finite sums via the method of Euler-Maclaurin summation (see e.g.Bromwich [10]). Before we proceed, however, we need some preliminary results. Now,In L(s) =E ts +1 , (s, t) )t^twhere1 ^e(s, t) = max^ s-1. tj+i {  t i +11Ls+1^s+iand t satisfies(2.68)^2 Ls + LIt  1 + Ls +^ t.j. t_ 2.s - 1^1fore(s,t) = {A(s —1) + B —1s — 1if B < s1if B > sA + 1Chapter 2: Contents of Pacle Approximants to (1 — z)k^ 47Clearly (2.68) is equivalent to {s + 1} > 1/2, and0(s, t) =t if ts+1 1— s+1tifls+1 1> s+1Write t = A(s + 1) + B with A E Z and 0 < B < s + 1. Then00lnL(s) = E LA (8)A=0whereLA(S) = E ^+ 1^e(S, t))B A(S+1)+Band B is such thati) A(s + 1) + B E Nii) (s + 1)/2 < B < s + 1.We haveLemma 2.2.3 If n > 1, A > 0, and 0 < j < A are integers, theni) LA (s) is decreasing on (2n + j 1(A + 1), 2n + (2j + 1)/(2A + 1))ii) LA (s) is increasing on [2n + (2j + 1)/(2A + 1), 2n + (j + 1)/(A 1)]•Chapter 2: Contents of Pade Approximants to (1 — z) k^48Proof: i) Suppose .5 E (2n + j/(A + 1), 2n + (2j + 1)/(2A + 1)). Then the values of tcontributing to LA(s) aret=(2n+1)Ad-n+j+1t.(2n+1)A+n+j+2t = (2n + 1)A + 2n + j + 1and hence+ 1^1^2n-1 (^+ 1^s — 1LA(s) 7.= (2n + 1)(A + 1) + j A + 1 + E=n (2n + 1)A + j + i + 1 (2n — 1)A + j + i) .Differentiating with respect to s givesYi 1^Y2 1LA(s ) =^t=si ^t=s2where= (2n + 1)A + j + n + 1Yi = (2n + 1)A + j + 2n + 1x 2 = (2n — 1)A + j + nY2 =- (2n — 1)A + j + 2n — 1.Applying Euler-Maclaurin summation yieldsA +1 — j \ 1 /2n(n + 1)(2A + 1)(A + 1) — (A +1 — j) 2 ) .L'A (s) < ln + (x2 — 1)yi + 2 C(x1 — nyi(x2 — 1)Y2Since— 1= (2n + 1)A + j + n > (2A + 1)nandY2 = (2n — 1)(A +1) + j (2n — 1)(A + 1)Chapter 2: Contents of Pacie Approximants to (1 — z) k^49we can write2n(n + 1)(2A + 1)(A + 1) — (A + 1 j) 2 < n(n + 1)(2A + 1)(A + 1) <2(x i — 1)yi(x2 — 1)Y2^(x1 — 1 )yi(x2 — 1)Y2n + 12n — 1 (x 2 — 1 )(Yi)and hence L'A (s) < 0 provided n > 3 (remembering that A + 1 — j > 1). If n = 2, then4A + 2n(n + 1)(2A + 1)(A + 1) <  5A + 2 — 1)yi (x2 — 1)y2^(x2 — 1)(Yi)which implies LA(s) < 0 if A > 1. It remains to check the cases n = 1 and A = O. Ifn = 1,1^1^1 L'A (3)=-. 3A +3+ j + 3A + 2 +:7 A+ 1 +j=—(2A + 1) (3A+3+j)(3A+2+j)(A+1+j) < 0while A = 0 implies j = 0 and1 ^1^1 ^1^1^nL'A(s) — 2n + 1 + 2n n — 2n + 1 2n <ii) Proceeding as above, we have for s E [2n + (2j + 1)/(2A + 1), 2n + (j + 1)/(A + 1)],thatS + 1^1^2n-1/^S + 1^S — 1LA(s)  (2n + 1)(A + 1) + j A + (2n + 1)A + j + + 1 (2n-1)A+ j + iland hencev2L'4 (s) = E 1  -^1t..1+1 t^t=x2+1where x1, x2, Yi, Y2 are as previously. From Euler-Maclaurin, we can writeVA (s) > ln(1 +A + j +1) 1(2n(n +1)(2A +1)(A +1) (A +1 + j) 2 )x1Y2^2^xlylx2y2and the second term on the right here satisfies2n(n + 1)(2A + 1)(A + 1) — (A + 1 + j) 2^n(n 1)(2A + 1)(A + 1)2xiyix2y2^ x1y1x2Y2Chapter 2: Contents of Pade Approximants to (1 — z) 1'^ 50This in turn is less than2n(n + 1) (2n — 1)(2n + 1) x1Y2which implies that L'A (s) > 0 provided n > 2. If n = 1, L'A(s)^13A +3+i > 0 asrequired.^ ■For the intervals of the form [2n + 1, 2n + 2], we haveLemma 2.2.4 If n > 1, A > 0 and 0 < j < A are integers, theni) LA (s) is decreasing on (2n + 1 + j 1(A + 1), 2n + 1 + 2j/(2A + 1))ii) LA(s) is increasing on [2n + 1 + 2j/(2A + 1), 2n + 1 + (j + 1)/(A + 1)].Proof: If s E (2n + 1 + j 1(A + 1), 2n + 1 + 2j/(2A + 1)), the relevant values of t aret =-- (2n +2)A+n+j+1t.(2n+2)A+n+j+2t = (2n + 2)A + 2n + j + 2and thuss + 1^1^2n s^+1^S — 1LA(S) = (2n + 2)(A + 1) + j A + 1 +^(2n + 2)A + j + + 1 2nA+j+ i)If, however, s E [2n + 1 + 2jA2A + 1), 2n + 1 + (j + 1)/(A + 1)],s + 1^1^2n S + 1^S — 1 )LA(S) =     E ^(2n + 2)(A + 1) + j A + 1 + ,...„((2n + 2)A + j + i + 1 2nA + j + i) •Arguing as in Lemma 2.2.3, we have L'As) < 0 in the first case, while L'A (s) > 0 in thesecond. ■The preceding results lead us toChapter 2: Contents of Pacle Approximants to (1 — z)k^ 51Lemma 2.2.5 If n > 1 theni) L(s) is maximal on [2n, 2n + 1] for s = 2nii) L(s) is minimal on [2n, 2n + 1] for s = 2n + 1iii) L(s) is maximal on [2n + 1, 2n + 2] for s = 2n + 2iv) L(s) is minimal on [2n + 1,2n + 2] for s = 2n + 1.Proof: We will prove the first case and note that the other three follow in a similarfashion. By the definition of LA(s) and Lemma 2.2.3, we will be done if we can showthat(2.69)^LA (2n + A 1^LA (2n + A +1  )holds for all j = 0, 1,^, A. Nowj 1 (2n + 1)(A + 1) + j (2n — 1)(A + 1) + jLA (2n + A + 1 )whileA + 1 (2n + 1)(A + 1) + j — i (2n — 1)(A + 1) + j + 1 — i)j + 1 =1^÷',( (2n + 1)(A + 1) + j + 1 (2n — 1)(A + 1) + j + 1LA(2n+A + 1 A + 1 + 1)(A + 1) + j+ 1 — i (2n — 1)(A + 1) + j + 2 — i)and it follows that (2.69) is equivalent to(2.70) —n — E+n 1 > n— 1 v+n-1 1t=x+1 t^y t=y+1where x = (2n + 1)A + j+ n +1 and y (2n — 1)A + j+ n. Define the auxiliary variablesu 2A + 1v = A + j + 1and consider the function(u+1)n+v 1E •un + v t=und-v-1-1f (n)Chapter 2: Contents of Pade Approximants to (1 — z)k^ 52Apply the Euler-Maclaurin summation formula to yieldn1 ( 1f(n) =^un + v In \.' + un + v ) + 2 g.tn + v (u + 1)n + v)1 ( ^1^1 12 (un + v) 2^((u + 1)n + 0 2 ) + . • •We write f (n) = fi (n) + f2 (n) whereii (n ) =-.n n^1( 1^ 1 ln(l: +un + v )+ 2 un + v (  1)n + v )un + vand f2(n) < 0 with1 if2( 7 )1 .^1 ( 1^12 (un + v) 2^((u + 1)n + v) 2 1).It follows thatf (n) — f (n — 1) = fi(n) — fi(n — 1) + h(n) — f2(n — 1), in faZ) dz + f2(n) — f2(n — 1)n--1and hence(2.71) f (n) — f (n — 1) ? min faz) + f2 (n)zE[n-1,n](since f2(n — 1) < 0). Now(2v — u)(u + 1)z 2 + 2v 2 z + v 2f; (z ) = 2^(( u + 1)z + v) 2 (uz + v ) 2SO(2v — u)(u + 1)(n 1) 2 + 2v 2 (n 1) + v 2min fl(z) >zE74^ 2((u + 1)n + v) 2 (un + v) 2Since1(^1 1^) 1 (^(2u + 1)n 2 + 2vn12^(un + v) 2 ((u + 1)n + v) 2 ) — 12 (un + v) 2 ((u + 1)n + v) 2 )from (2.71), we conclude thatf(n) — f(n — 1) > 0Chapter 2: Contents of Pade Approximants to (1 — z)k^ 53and hence (2.70) obtains, provided n > 2 (remembering that our definitions of u and vensure 2u < v < u). If n = 1, (2.70) is just1^1>x x + 1 — 0which follows from x > 0.We are now in a position to prove Theorem 2.2.1.Proof of Theorem 2.2.1: By Lemma 2.2.5, it suffices to show thatlim L(2n) = lim L(2n + 1) = ir/e.n.--.co^n-4.coWe first show that lira L(2n) = r/e1'. Considern---■oo■^2n^2n + 1^2n — 1^LA(2n) , E B=n+1 ((2n + 1)A + B (2n — 1)A + B — 1)2n^1(2n +1)A + BB.n-Fi 2n^1^ 1— (2n — 1) E B=n+1 ((2n — 1)A + B — 1 (2n + 1)A +^ B)By Euler-Maclaurin summation,2n^1^(2n+1)A1-2n 1B=n-1-1 (2n + 1)A + B = t=(2nd-Eim+n+1 -t= ln (  (2n + 1)A +^1^12n ) 1 ((2n + 1)A + n ) 2 (2n + 1)A + n (2n + 1)A + 2n)1 ^1 ^1 + 12 (((2n + 1)A + n) 2 ((2n + 1)A + 2n) 2 )= ln ( (2n + 1)A + 2in n ^(1 )+ 0^(2n + 1)A + n } 2((2n + 1)A + n)((2n + 1)A + 2n) n(2n + 1)A + 212) = In (2A + 2) + 0 (1)ln((2n + 1)A + n i^UA + li^i-i)=2andChapter 2: Contents of Pade Approximants to (1 — z) k^54so thatWe have2n 1^  =ln(2A +2) +0(1)B=En+i (2n + 1)A + B^2A +1 )^).2n^1^ 1(2n — 1) E B=n+1 ( (2n — 1)A + B — 1 (2n + 1)A + B)—1)A + 2n — 1 ) in ((2n + 1)A + 2n))= (2n — 1) (ln (2n^(2n — 1)A + n — 1^(2n + 1)A + n l)(2n — 1)n+ 2((2n + 1)A + n)((2n + 1)A + 2n)(2n — 1)n ^+ 0(1 )2((2n —1)A + n —1)((2n — 1)A + 2n — 1)^nand the last two terms here sum to(2n — 1)n(8nA 2 + (10n — 2)A + 3n — 1)^= o( 1 ).2((2n + 1)A + n)((2n + 1)A + 2n)((2n — 1)A + n — 1)((2n — 1)A + 2n — 1)^nNowln ((2n — 1)A + 2n — 1)) ln ((2n + 1)A + 2n)(2n — 1)A + n — 1^(2n + 1)A + n i=141 +^(2A + 1)n ((2n + 1)A + 2n)((2n — 1)A + n — 1))and so— 1)A + 2n — 1) \^2n+ 1)A + 2n))(2n — 1) - (ln ( (2n(2n — 1)A + n — 1 ) in(((2n+ 1)A + n )=(2n — 1)(2A + 1)n ^1((2n + 1)A + 2n)((2n — 1)A + n — 1) + 0(-0)1 A + 1 + o().= nFrom the preceding, it follows thatA +1 1 +0 ( n1)LA (2n) = 21n( 2A ++ 21)Chapter 2: Contents of Pade Approximants to (1 — z) k^55and henceNown—colim L(2n) = exp(E (21n( 2A + 2 )^1^)).2A+1 A+100A=02 E lrq2A + 1N i 2A + 2) =21n (TNT(2A+2))A=0^ .11k2A+1J)A=0( TTN ( 2A + 2 )2)1-10 k2A+1/2 • 2 • 4 • 4. • • (2N)(2N + 2) )=ln(2N + 2) +1n(1 • 3 - 3 • 5 • • • (2N + 1)(2N + 1))1 \=ln N + In 7r + 0 t—N)^(by Wallis' formula)Also,so thatNA d=0 A + 1=1nN+-y + 0()NN^2A + 2^1^1E (2111( 2A + 1) A+1 ) - 1n7r —7+0(y ).A=0Thusct° (21n( 2A + 2 ) ^1  ) = lnr _7A=0^2A + 1 ) A + 1whence lira L(2n) = 7r/e, as desired. If, however, ,s = 2n + 1 thenn---*oo2n1-1^2n + 2^2n (PLA(s) = EB=n+2 n + 2)A + B 2nA + B —1))^2n-I-1^ 2n-I-1^1^121 ^= E  2n E B=n+2 (2n + 2)A + B B=n-I-2 (2nA + B — 1 (2n + 2)A +^ B)and applying Euler-Maclaurin yields, as in the previous situation1LA(2n +1) = 21n(22AA ++ 21) A +1 4. 0(77 .).=1nHence lim L(2n + 1) = 7r/0 and thus also lim L(s) = 7r/ell (by Lemma 2.2.5).^■n--).00^ s-400Chapter 2: Contents of Pade Approximants to (1 — z)k^ 56We may also now prove Theorem 2.2.2.Proof of Theorem 2.2.2: Again, we will handle the caseL(2) ? L(s)for s > 2 and note that the lower bound on L(s) obtains from a similar analysis. ByLemma 2.2.5, it suffices to show thatL(2n) > L(2n + 2)for n = 1, 2, ... which follows from the like inequality with L replaced by LA. Now,^2n^2n + 1^2n — 1LA (2n) . E B=n-I-1 ((2n + 1)A + B (2n — 1)A + B — 1)and so, once again applying Euler-Maclaurin summation, we have thatLA(2n) — LA(2n + 2)x2Y3^ )= (2n + 1)1n( Y1x2x3Y4 ) + 21n( Y2X3x1y2y3x4_ _ 12 )— [(2n + 1)(— — —1 ) — (2n —1)(1x i^y 1^x2^y1 _ 1^X4 ^Y4— (2n + 3)(—^) + (2n + 1)(-1 — 1 )1+ u [(2n +x3 y31)(1^1 ^1-2- — --) — (2n — 1)(72- — --)xi^Yi^X2^Y2—(2n + 3)(7-2-1 — —2-1 ) + (2n + 1)(-2-1 — —2-1 )1x3^y3^\ x^V4^4 ,1^,^ ,^1^1^,,,^,^1^1)– — k272 + 1) (-71 – --i) – (2n —1)(- 4 — 7-4120 xi ^Yi X^il2 ,2—(2n + 3)(- 11.X3— —4-1 ) + (2n + 1)(-71 — —4-1 )] + • •y3 x 4 ,7/ 4Chapter 2: Contents of Pads Approximants to (1 — z) k^57wherex l = (2n + 1)A + nx 2 = (2n — 1)A + n — 1x3 = (2n + 3)A + n + 1x4 = (2n + 1)A + nyl = (2n + 1)A + 2nY2 = (2n — 1)A + 2n — 1y3 = (2n + 3)A + 2n + 2y4= (2n + 1)A + 2n + 1.Since, if we expand the arguments of the above logarithms, we verify thatY1X2X3Y4 < 1X1Y2Y3X4and Y2x3 > 1x 2y3we can find positive z 1 , z2 withyix2x3y4 =x1y2y3x42X 3 and^_ 1 + Z2.x2y3It is thus possible to bound ln(Y1X2X3Y4) below by—z 1 — 34/5 (assuming that z 1 < 1/6xi.y2y3x4which follows from A > 1 and n > 2) and ln( Y2x3 ) by z2 — 4/2. By the properties ofEuler-Macaurin summation, then, we can writeLA (2n) — LA (2n + 2)> —(2n + 1)(zi + 34/5) + 2z 2 — 41 1 1 \ /1^1 \—^[(2n + 1)(— — --) — (2n —X2 — --Y2)/ 1 1 \1— (2n + 3) x3 — --)y3 + (2n + 1)(-1X4 Y4 )]^1^1^ 1+ -2- [(2n +^1—^— (2n — 1)(A- — 12 )1^y1/ ^1—2- \— (2n + 3)L — ) + (2n +^— )1^x3 y3^X4^Y41^/ ^1 \--6-0-(2n +4^y4Now expanding the right hand side of this inequality, with the aid of Maple V, yieldsa rational function f (A ' n) with degA f = deg as f = 14, degA g^16 and deg as g = 17.g(A,n)Explicitly,g(A,n) = 60(A + 1)(2n — 1) 2 xly?x4yM.Chapter 2: Contents of Pad6 Approximants to (1 — z) k^58Moreover, collecting the 225 terms involved in f (A, n) and noting when the leading termdominates, we have thatf (A, n) > 589824A 14 n 14provided A > 1 and n > 2, unless (A, n) = (1,2).If A = 01 Lo (2n) — Lo (2n + 2) = 2n(n + 1)(2n + 1) > 0while if n = 1,LA(2) — L A (4) = ^5A + 2> O.(3A + 2)(5A + 3)(5A + 4) Supposing n = 2 and A = 1, then19 L i (4) — L i (6) 6435 > °and the conclusion obtains as desired.^ ■2.3 Limiting Behaviour of L (a , b, s)In general, the situation is slightly more complicated when a > b > 1, but not unbearablyso. One can obtain upper and lower bounds in much the same manner as in the previoussection and in fact prove thataa/(2*-0)lim L(a,b,^) = bb/(2a(a-b))(a + bya-1-01(2ab) /27r/CY .s..It follows that the limits associated with the forms (N + Tv-1 )k and (N + 1 tN2 (see Theorems 1.9.4 and 1.9.5) are respectivelylira L(2,1,8)=2 3/ 2 • 3 -3/4 .17r/e'Y = 1.64792 • • •s.00andlira L(3, 2, 8) = 2 1 /6 • 33/4 - 5 -5/ 12 V7r/CY = 1.73783 • • •s..Chapter 2: Contents of Pad Approximants to (1 — z)k^ 59For details of these results, the reader is directed to the author's forthcoming paper(M. Bennett [5]). It seems to be somewhat curious that the case a = b = 1 is so distinctfrom those with a > b > 1, in that the latter is always an algebraic multiple of \br / 67rather than of 'r/e1'. The explanation for this, however, is by no means readily apparent.Chapter 3Connections to Waring's Problem3.1 IntroductionIt is well known from work of Dickson [22], Pillai [46], Rubugunday [50], Niven [45], et. al.that if we defineg(k) = min{ m : all positive integers are sums of at mostm k-th powers of 1, 2, 3, ... }then if(3.1)^ 3k — 2 k [ ( 3 /2) k ] < 2k — [ ( 3 / 2)1we have(3.2)^ g(k) = 2 k + [(3/2) k ] — 2.The Ideal Waring Problem states that the equality above holds for all k > 2.To see the connection to fractional parts of powers of rationals, suppose the oppositeof (3.1) is true, that is3k — 2 k [ (3 / 2 ) k I > 2k — [(3/2) k ] .Then we have0 < ([(3/2) k ] + 1) — (3/2) k < [(3/2) k ]/2k < (3/4) kso that 11(3/2) k li < (3/4) k . From Mahler's result (1.1.2), this implies that only finitelymany k fail to satisfy (3.2) and hence it would be desirable to have an effective lowerln 3bound upon j1(3/2)/1 of the form in (1.1.8) with 0.794 replaced by 2 — ln 2 (which is--, 0.415).60Chapter 3: Connections to Waring's Problem^ 61To place the Ideal Waring Problem in a somewhat more general context we considerthe following.Suppose S is an increasing sequence of positive integers. We call S a basis of orderh if every positive integer can be written as a sum of at most h elements of S and atleast one positive integer requires the full h terms. If we further define the Schnirelmanndensity a(S) byo-(S) = inf {#{ s : s E S, s 5_ n }/n}riENthen it is straightforward to show that (see e.g. Ellison [27])3.1.1 If u(S) > 0 then S is a basis (of some order).If S(k) denotes the set of k-th powers of elements of S, then Rieger [47] proved that3.1.2 If a(S) > 0, then S( k) is a basis.Waring's problem is the special case S = N (so that a(S) = 1). The ideal Waringproblem, then, is to show that the order of the basis S(k) is given by (3.2).In the following, we consider the set SN = {i, N, N +1, N + 2, ... } and its k-thpowers. Since a(S) = 1/(N — 1) > 0, we have by (3.1.2) that Si(\/; ) form a basis and it isa natural question as to its order (the case N = 2 of course corresponds exactly to thestandard Waring problem). If, in analogy to the above, we denote this order by gN(k),we proveTheorem 3.1.3 If N and k are positive integers with 4 < N < (k +1)(k-1 )/k —I thengN(k) = Nk + I  NN + 1  \k1 — 9\^I J^`•This is achieved through use of the Hardy-Littlewood-Vinogradov method, boundingcertain exponential sums and an ascent argument due to Dickson.Chapter 3: Connections to Waring's Problem^ 623.2 Asymptotic Theory: Notation and DefinitionsIn this section, we concern ourselves with proving a slight generalization of a result ofVinogradov's. To be precise, we showTheorem 3.2.1 If k > 6 and M > e446/c6 are positive integers, then there exist s integersx i , x 2 , ... , x s withi) s < 6k In k (31n 6 + 4)kii) xi > miAsk3) for i = 1, 2, ..., siii) M = xi + 4 + + X .skThe upper bound for s can be strengthened to order ti 3k In k or ti 2k In k, but, inboth cases, these induce a lower bound for M which is too large for our purposes. Thedifficulty chiefly arises from estimating the size of the implied constant iny(a) < qewhere q(a) is the number of solutions to the congruence V k^a (mod q) for v and aintegers in [0, q).In what follows, we will utilize the Hardy-Littlewood method a la Vinogradov andattempt to keep notation as "standard" as possible. In particular, we use as muchtechnical machinery as we can from Vinogradov [58]Let us suppose that N > 2 and k > 6 are integers and that M is an integer, satisfying(3.3)^ M > max{N8k3, e446k6 }.We define(3.4)1v = —k' P = [MS R = [P' - '112], Y [P'12 ]T = 2kP",^= [2k In k kln 6] and a = k(1 — v)tChapter 3: Connections to Waring's Problem^ 63and adopt the convention that 0 (with or without subscripts or superscripts) is a complexnumber satisfying 101 < 1. Let9)1(a,q)={aER:a= a /q z, (a, q) = 1, 0 <a<q<P1)2 and 1z1< 1/ }and set TR = 9)2(a, q), which is easily seen to be a disjoint union. These will constituteour major arcs. Further, we take our minor arcs m to be the complement of fit in[—T -1 , 1 — T -1 ] so that, by a theorem of Dirichlet, if a E m we can write a = a/q zwith 1z1 < 1/(qT), q E (P 1 / 2 ,4We now construct a pairterms in our upper bound(3.5)^P1 =offor4"exceptionals in TheoremP2 =21i'[-2P1sets"3.2.1.which willLet• • • , Pe =—contribute1^_[-2 Pi vt-1the majority of1 1(3.6)^=and set[-1 R]4, R2 = k-R1 -11, ,^=2 "-(3.7)^U ={ u^u =-- xi^. . .^x l; , Pt < x i < 2P1 }and(3.8)^V = {v : v^yi; , Ri < yi < 2Ri}We may note that a lemma of Vinogradov [58, p.63, Lemma 1] gives that the elementsof U and V are all distinct, satisfying(5p)k < u < ‘2Llp\k^for all u E U;(3.9)for all v E V.f91,Chapter 3: Connections to Waring's Problem^ 64With these definitions in mind, we further defineY^(3.10)^U(a) = >2 e(au), V(a) = >2 >2 eofyykouEU^ uo.where here and henceforth, e(x) = e2irix . Taking f(a) = E e(axk ), we now introducex=Nthe principal object of our study:fr(M) =LT-1l-T-1 (A(x))4k (U(°)) 21/ (a)e( --aill)(3.11)= 1.931 +ImBy expanding the integral for r(M) and making use off 1-1/1-^1 if M = 0(3.12) e(aM) da =0 if M 0 an integerwe have that r(M) is the number of representations of M asX +... X zik +72+U l + y k Vwhere N < x i < P, u and u' E U, 1 < y < Y, v E V. If we therefore show that1.19111 > 1 1m1 and hence that r(M) 0 for all M sufficiently large, if Re > N then we willhave in the notation of Theorem 3.2.1s < 4k + 3e < 6k In k + (In 216 + 4)k.To accomplish this, we will attempt to approximate and bound the various terms asso-ciated with r(M) by objects familiar from the theory of exponential sums.With this in mind, we further define for 0 < a < q, (a, q) = 1q^ q^(3.13)^S(a,q)=---^e(axk^A(q) =^(S(a, 0/04ke( amiox=1 a=1(a,q)=1Chapter 3: Connections to Waring's Problem^ 65and6 = E A(q)q=1(3.14)^ /(Z)= f e(zxk ) dxJ(M) = 1:(I(z)) 4k e(— zM) dz3.3 Lemmas on Exponential Sums, Integrals and Other TopicsThroughout this chapter, we will denote by CT, for i = 1, 2, ... quantities which depend(in a definite and explicit fashion) upon k (usually) and N (occasionally). We avoid theuse of the Vinogradov symbol < in order to make our treatment more concrete and tomotivate the derivation of our bounds upon M.The following preliminary lemmas are necessary.Lemma 3.3.1 IS (a, q)j <^where C1 < e 2.75k2 .Proof: One may see Landau [36, Theorem 315]. The bound for C 1 follows by applyingRosser and Schoenfeld [48, Corollary 2 to Theorem 1] to get5(k _ 2)k2k/(k-2)ln^< (k — 1) ln k ^8kRemembering that k > 6, the result obtains. •Lemma 3.3.2 E A(q) =^c2p-1 where 1C2I < e i•ook3< p l /2Proof: This is essentially a corollary of Lemma 3.3.1 (see Dickson [23, Lemma B]). Wehave 1C2 1 < Cik < e l1.00k3 ■Lemma 3.3.3 6 > C3 > e-4.29k3Chapter 3: Connections to Waring's Problem^ 66Proof: By Landau [36, Theorems 325-326], we have6 > C3 = fJ b(p) • H (1 - 7)-3/2)1p<C^p>CwhereCOb(p) = EA(P1)j=1From James [33], we may writeand C = (1 + k4k )2/(4k-5) .Henceb(p) > 1 2—(ord2(k)+2)(4k-1) , p = 2P-(ordp(k)+1)(2k-1) p > 2ln(1/C3 ) < (4k — 1)(ord 2 (k) + 2) In 2 + E (2k — 1)(ordp (k) + 1) In p ln(C(3/2))3<p<C(noting that 11(1—p -312 ) = C(3/2) -1 > 1/3). If 6 < k < 8, from the preceding inequality,we check thatln(1/C3) < 4.29k3 .Supposing that k > 9, we note thatE (2k — 1)(ordp(k) 1)1np= E(2k — 1) ord p (k)lnp E (2k — 1)1npp<C^ p<k^ p<C= (2k — 1)ln k + (2k — 1) E lnpp<CApplying the bound upon the Chebyshev function0(x) < 1.000081x for all x > 0from Schoenfeld [52], we conclude that(2k — 1) E lnp < 1.000081(2k — 1)C.p<CChapter 3: Connections to Waring's Problem^ 67It follows that (using ord 2 k < In k/ln 2)ln(1 /C3 ) < (2k — 1)1.000081C + (4k — 1) In k + (6k — 1) ln 2 + ln 3and since k > 9, the right hand side of this is < 4.29k3 , as required.^■Lemma 3.3.4 Let M and M' be integers with M < M' and let a E R \ Z. ThenM'E e(ax)x=M1,- 2lia l lwhere Hail is the distance from a to the nearest integer.Proof: See Vinogradov [58, p.23, Lemma 6].^ ■Lemma 3.3.5 Let M and M' be integers with M < M' and let f(x) be a twice differ-entiable function on [M, M'] satisfying 0 < 1(x) < 1/2 and f"(x) > O. Thenm,^M'E e(+ f (x)) — I m e(+ f (x)) dxx=MProof: See Vinogradov [58, p.34, Lemma 13].^ ■ay + 0(Y) Lemma 3.3.6 Let M, X , M', Y be integers with X , Y > O. Let c,o(y) =qwhere (a, q) = 1, q > 0 and y runs through the values y=N,...,N+Y — 1. Further,suppose that when y runs through any q successive values in this set, the difference betweenthe greatest and least values of the function 0(y) does not exceed A > O. If we definem+x-i N+Y-1S = E > 7(x)q(Y)e(x(Y))s=m y=Nand putM+X-1^ N+Y-1E I7(x)1 2 = Xo ,^E 19(01 = Yo, max19(y)1= 9x=M y=Nthen< 2.ISM < (X0 Y0 77((2A + 6)X + 3q)[Yq -1 + 101/2.Chapter 3: Connections to Waring's Problem^ 68Proof: See Vinogradov [58, p.30, Lemma 104^ ■if 1z1 P"Lemma 3.3.7 1/(z)1 < Z :=Azi - v^if izi >Proof: Since N > 1 and /(z) = f e(zx k ) dx, the first of these assertions follows easily.Suppose that z > P' and make the change of variables = 2zx k to obtain2zPk^ 2zPk/(z) = f2z(N 1), O(,0) cos ri3d0 +63) sin 7 r,3 cl,3i 124N-1)k 0where OP) = v(2z) -i'fi'. Now by (3.3) we have that N < -2-1 P 1 ', and so 2z(N- 1) k <1-2 and 2zPk > 2 and thus we can write2zPk^1/2^ 3/2f2z(N 1)k OP) COS 7r# c113^OP) cos 7 )3 +^OP) cos r clf32z(N -1)k 1/22zPk+ • • • +OP) cos^d3.f[2zPk +112]-1 / 2Since 0(0) is positive and decreasing, the first integral here is positive and the remainderalternate between negative and positive while decreasing in modulus. Hence f2zPk11(,3) cos R- 13 d,3J2z(N -1)kmax{101/2 O(/3) d,3, fi3:22 2(0) d)3}f1o O(Q) di(3 (2z)'Similarly, after writing1.2zPk^ 2zPk^2zPkAzov_ i) , ,(#) sin 71- 13 cl13 = 12z(N_nk + 11 2 . . .^I^OP) sin ir/30[2zPk]the above argument yields1 2zPkOP) sin 70 di3i2z(N-1)k(2z)-11Chapter 3: Connections to Waring's Problem^ 69and thus1/(4 < ((2z) -2 ' (2z) -2/ 1 / 25_ \(2z) -11With minor modifications, the same procedure works for z < 0 and the result obtains.■Lemma 3.3.8 Suppose M > 0 and let k(M) denote the number of solutions to theinequalityXi^. . .^X4^k ^.31in integers x i > N. Thenir(i +0 ,4kk(M)^r(5) " M4 - c4 m4—.Proof: In general, we let r be any positive integer and let kr (M) denote the number ofsolutions toin positive integers x i , ..., x r while krN (M) denotes the number of solutions to the sameinequality subject to the constraint xi > N, i^1, 2, ..., r.Clearly, we have li.v(M) < kr (M), but alsoN -1(3.15)^k,NM) ?_ kr (M) — r E kr_ i (m — sk).3=1Now by Vinogradov [58, p.22, Lemma 3], we have(3.16)^kr(M) Tr M" — OrM"' , 0 < 0 < 1Chapter 3: Connections to Waring's Problem^ 70(F(1+ v))rwhere Tr = r(i + rv)^ and so, using Tr <1, it follows that kr _ i (M — sk) < kr_ i (M) <Mir-1)" and henceTrM" - OrM (r-1)" - r(N — 1)M (1- 1>v < krN (M) < kr (M)which implies the stated result with 0 < C4 < 4kN.^ ■3.4 The Contribution of the Major ArcsWe seek a lower bound upon 1/9A1, the contribution of the major arcs to r(M). In allof what follows in this section we will assume that a = al q + z E 931(a, q). By directsubstitution, we can write(3.17)^ 1-931 = E f^e(f(a))4ke( aMi )dau,tt ,v,yfor MI = M — u — u' — vy k , where u, u' E U, v E V, 1 < y < Y and so from (3.9) itfollows that(3.18)^ (1 — __3 )pk < mi. < Pk.‘^2k )^—We will now attempt to come to grips with 1931 (f(a)) 4k e(—aMi )da, approximating inorder to obtain an asymptotic formula. We will show firstLemma 3.4.1 f(a) = S (a, q) I(z)d- C5 q, where IC5 1 < 4.qProof: If we write x = qt + s for s = 0, 1, ..., q — 1 and t running through all integersin the interval ((N —1 — s)Iq,(P — s)/q], thenPf(a) = E e(axk )x=Nq--1. E e(ask I q)Fs (z)s=oChapter 3: Connections to Waring's Problem^ 71with Fs (z) = E e(z(qt s) k )• Consider, then, f (t) =1z1(qt 3)k. Now, f"(t) > 0 and,remembering that a E 9n(a,q), 0 < f'(t) < klzIqPk-1 < 1/2 and so we can apply Lemma3.3.5 to get(3.19)^F8(z) =^e(z(qt + 3) k ) dt + 40, 101 < 1.1(1(P-3)1qV-1-01q(the constant 4 occurs because the endpoints of the interval defining t need not beintegers). HenceFs(z)^I(z) -1- 40(returning to our original variables) and thus^ /(z) C5qf(a) = S(a,with 105 1 < 4.^ ■From this result, it is straightforward to provek^(S(a,q)^co-34-uz^1.^if, <Lemma 3.4.2 (f (a)) 4 = 1-(z))4k^4k-i, wh ere^e1100k3 .qProof: From Lemma 3.4.1, we deduce that,q)^(Z))4k(f(a ))4k(41c)(S(a,q)I(z)) (C50 4"^(S^41t1q i=0and so from Lemmas 3.3.1 and 3.3.7, if we let i = 0, 1,^, 4k — 1( c1di = 4.k^i 1 c5 14k-tztok-i-ivz^1we have4.tic-,-1 (4k) ( S(a,^\i^4k-i2_,^• (z)) (C5q)1.0 q4k_i< E di .i=o Sincedt+i^4k — ^Zq-(1+v) >^Zq-(1+0di^i^1 105 1^— 16kChapter 3: Connections to Waring's Problem^ 72and q < P 1 / 2 , if Z = P, then di+i /di > 1 for all i = 0, 1,^, 4k — 2. If, on the otherhand, Z \12- 1z1 -v, then Z >^rye > P'qv and again since q < P112 , we concludethat di+i /di > 1, i = 0, 1,^, 4k — 2. This implies that4k-1E di < 4kd4k_1 <i=oand from the preceding, the result follows with IC61We are now ready to estimate the contribution of the integral associated with a singlemajor arc.Lemma 3.4.3 If a E Tt(a, q) then11fTr j(moeHami/q) c7q-i 3k-1pI^f (a)) 4k e ,_( aMi )dz = (S(a,q))4kqfor IC71 < en.o4k3Proof: From Lemma 3.4.2, it is clear that^llqr^ ihr S(a q) (f())4k e(—aMi ) dz =^' I (z))4k e(—ctMi ) dz^f liqr qcr647-3-1-vz4k-le(_am1) dz(S(a,q)\4kq • J(Mi)e(—aMi/q)liqrC6q -3-Pv z4k-l e(_aMi)dzfl/ qr0.0 S (a, q) I (z))4k e(— a MI ) dzfv q, qfol/Tr (S(a,q) I (z))4k e (_ami ) dz.q< e l1 .00k3 .^ ■Chapter 3: Connections to Waring's Problem^ 73Nowf1/QT liqr z4k-1 dzCoq -3+ 'Z4k-l e(-ciMi )dz <2106 1q -3+' f<21Colq-3+u (10P-kP4k-1 dz fp_k(\fi z-v)4k-1 dz22k-12 1 C6 1q -34-11 (P3k-1 + 3 - 1 /2/k p3k-1)which in turn is less than e11.036k3q-31-vp3k-1 The sum in modulus of the integrals/ 9 T^q(S(a,q) 1 (z))4ke( cali )dzq is, again using Lemmas 3.3.1 and 3.3.7and (S(a, q) I(z))4ke(_ami)dzq< 2 I c r-y4k —4 f °° —4^2^k^ui^z dz = - A '''z k-4i1 \34 Ul^WT)1/qT^3= —136 k34k C1kg -1 P3k-3 •Lemma 3.4.3 then follows as stated, with IC71 < en.o4k3 (noting that q2-1'13-2 5 P-1-42 <e -446k5 ) .•It remains only to find an asymptotic estimate for J(Mi ) and to sum up all the majorarcs. For the former of these, we prove^Lemma 3.4.4 J(Mi) = (11(1 +(4)1i))4k^C8P3k-' with IC81 <FProof: Let M0 be an integer satisfying 0 < M1 - Mo < 2Pk'• ThenIJ(Mi ) - J(Mo)I = f:(I(z))4k (e(-zMi ) - e(-zMo))dzI<^1°° Z4kz(Mi - Mo ) dz-k 00= 4Afir(Mi - M0) (f p P4k z dz f 4 k z -3 dz)P-ke lk .Chapter 3: Connections to Waring's Problem^ 74and this is equal to 2-V- 7r(4k1-1)(Mi -Mo )P2k . Applying Lemma 3.4.3 with Mo in placeof Mi., a = 0, q = I yieldsL iir (i(z))4k e(—zMo) dz = J(M0) C7P3k-1and from IJ(Mi) - AMOI < 21.71- (4k 1)(Mi - Mo)P2k with IA - Mo l < 2Pk-u wehave[111-r ( f (Z))4k e( - z Mo) dz = J(M1 ) C7P3k-1 0(4‘fir(4k 1)P3k- u)=J(Mi )-1- 0'(C7 Pu -1 4V2- R- (4k 1))P3k- yIt follows, therefore, thatLT ( f (a)) 4k e(-aMo) da^(MI) fiir i /r(f(a)) 41c e( -aMo) da(3.20)+60(C7P'l 412- 7r(4k 1))P3k-vwhere a E 97t. Fixing H^[Pk-"], we let M' and M" range over 1, 2, ... , H and setMo = M1 -^- M" (so that M1 - Mo < 2Pk- '1 holds as required). We sum the aboveequation (3.20) over M' and M" to yieldH HE E ki ( f ( a )) 4k e( a (Mi - - M")) daAv=i m"=1H2 j (mo^H^(f (a))4k e(_ct(mi^M")) dami=i mn=iO'H2 (C7P/ -1 4\12- 7r(4k 1))P3k- ')and by applying Lemma 3.3.4 to the summation on the RHS, we findH H  i_i / TIE E^f (a)) 4k e(-a(Mi - M' - M")) dam"=1 ir1-1/1-^1^2<^p4k ^^2iiaii^da= 1 p4k I 1/2 a -2 da21= -2P4k (3- - 2) < kP5k-1Chapter 3: Connections to Waring's Problem^ 75and since H = [Pk-"] > 1 Pk', we have that the above is < 2H2 P3k-1+21/ . Thisimplies thatH HE E frin(i(a))4ke(—a(Mi — M' — M")) da^(3.21)^m"=1 = H2 j (mi) on (c-vv-i 41fi r(4k + 1) + 2 p3u-1)H2p3k-vNow the above is just the number of representations of M i. asM' + M" + + + xnkk(where N < x i < P, i = 1, 2, ... , 4k) and so, summing over M", it is justH^(3.22)^E (k(mi — M' —1) — k(Mi — M' — H — 1))itv=iwhere k(M) has the same meaning as in Lemma 3.3.8. From the lemma, we haveFo. v yik^c4w4k(W) ^r(5) and (3.22) becomes1 F(1 -I- v)4k ((A^_ 1)4 _ (4-1^— H — 1) 4 )M'=1H^(3.23)^-C4 E ((mi —^— 04--- — (m, —^— H — 1) 4')ivr=1= F(1 + 11)4k H2 (4M1 + 0 1 • 121/M12) + 0 2 • 4 • C4H2 MT-11F(5)Moreover, this expression is equal to4 r(1 + V)4k^2 3 + 03 (12F(1^vrk 4C4H-1M1^(3.24)^ ') H31WF(5^H) F(5)Substituting this into (3.21), and dividing by H 2 , we obtainj^F(1 + v)4k^+ ■v8p3k-v(M1) - r(4)^Mlf (f(a))4k e( —"I ) da = E E f (f(a))4ke(—"1)dzq^liqr931^ q<p1/2 a=1^-1/9T(c 1,0=1( s(a, oyik r( i + vylk m.3)^F(4)^e(—aMi/q)= Eq<pi,2(a , q)= 1q)=1Chapter 3: Connections to Waring's Problem^ 76where!GI < 3F(1 + iirk 41^-1^1-11F(4)^+ 1C41 1/ Mi. + IC7IPv -1 +4V271- (4k + 1) + 2p3v-1 .Using (3.3), 0 < F(1 + 0 < 1 and the inequalities in (3.18), the result obtains for k > 6.■We are finally ready to estimate the contribution of all the major arcs inf(f(a)) 4ke( —cf-M) da.Without further adoLemma 3.4where IC9 1 <F(1 +4)v)4k. 5 f (f(a)) 4ke( — ceM1) da =^Mi36 + C9P3k- ufit^ F(ii.o6k3e Proof: Well,which by Lemma 3.4.3= E^cE (S(a,q) \4k jfmoe( ami/q)g<P 1 / 2 a=1^q^)(a,q)=1q+ E E cr7q-lp3k-1.Now by Lemma 3.4.4, thisq<pi/2 a=1(a,q)=1(S (a, q) )4k c8p3k , e( a mi / 0a=1^q q))4k(a,q)=1q+ E E+ Eq<pi,2c77 -1p3k-1 .q<P 1 / 2 a=1(a,q)=1Chapter 3: Connections to Waring's Problem^ 77The second sum in modulus is less than (by Lemma 3.3.1)qE E qlkic 8 1 q - 4 p3k-uq<pi.12 a=1(a,q)=15_ ci4kIc81 E c 3P3k- vq<pl /2< C1k IC8 I( (3)P3k- vwhile the third is< ^nip., < n1p3k_1 /2 .q<pi,2The first is justjr(1+ v)4k v > A(q)r(4) q<pi,and hence by Lemma 3.3.2, is equal toro. + 0 4kr(4)^v(e + c2p 1)and by (3.18),ro + vrk 1113r2 ip_ i < lc2 1 r(1+ vrk p3k_ i.r(4)^1'^' ^'^'^r(4)Hence(3.25)^/931(f(a))4ke(—aMi) da = 11(1+4)v )4k ve + Gp3k,r(where1C91<ci4k 1GIC(3)+ Ic71Pu-1 / 2 + 1c2Ir(1r+(4)v)4k P-1< e l1.06k3, for k > 6■Chapter 3: Connections to Waring's Problem^ 78Putting the results of this section together, then, we have(3.26)I9) = E 9X `j,f(ayke(_amoda,v,y=^(ro vyik■ F(4)^/We + c9p3k-v)(where M1 = M — u — u' — vy k). This implies that(3.27)^> E ir(1 (4)v)4k(0.9M) 38— E ic9 1/33k—)u,u ,^F,v,y u,u1,v,yand Lemma 3.3.3 yieldsF(1 + v)4k1/9311>^12 copu(0) 2 v(0) _ ro3k-uu(0)2V(0)r(4)c3 r(1 +4)v)4k >^P3ku(o)2v(o) - IC9 IP3k-vU(0) 2 V(0)2^F(> Ci0P3kU(0) 2 V(0)P(4j) ^(IC91)k4kwhere Co = C3 F(1co) / , provided P >^Since F(1 x) > 0.885 for all4^ Ciopositive x, we have> C3 4F(4)(0 . 885)4k> e -4.36k3 for k > 6.The desired bound upon P, then, will follow from P > e15A2k3 and hence from (3.3).3.5 Minor Arc EstimatesMost of the bounds utilized in the minor arc case are in fact trivial, but that is not tosay that the estimates are in any sense simpler. In fact, the deepest result on exponentialsums that appears in this proof, namely that of Lemma 3.3.6, plays a crucial role. Weu,u 1 ,v,yChapter 3: Connections to Waring's Problem^ 79have liml = fm (f (a)) 4k U(o) 2 V(a)e( — °M)daLif(a)1 4k 1U(4 2 1 17 (o)1da< i--4k mcIV(a)i fm 1U(a)1 2 da(3. 28)< P4k maMx1V(a)1U(0) (via Parseval's identity)ctEand it remains to investigate 117(a)1 using Lemma 3.3.6.In the notation of that lemma, we take M = M' = 1, X = [2—kpk-1/2] , y = [p1/2] ,andY'(Y) = oY =^ay + zqyqso that 0(y) = zqy. Since a E m, we have IzI < ^1 ^for P 1 / 2 < q < 2kPk-1 and2kqPk -1-thus Izi < 1/q 2 which implies that 0 < A < 1 in our case. If we further takeii(y) = ^1^if y is a k-th power0^otherwiseand^y(x) = 1^if x E V0 ^otherwisethen we have Xo = V(0)/Yo , Yo = [P42] and 77 = 1 and by the conclusion of Lemma 3.3.6,IV(a)l _< (V(0)(8 • 2-k pk-112 + 30(p112 q-1 + 0)1/2and again utilizing that a E m (and so P 1 / 2 < q < 2kPk-1 )117(a)1 < V(0) 1/2pk/2-1/4.Chapter 3: Connections to Waring's Problem^ 80Therefore,and so(3.29)pk+k/2-1/4C10 V(0) 1 /2 U(0) .Now, by our construction of U and V, and the uniqueness of the elements so derived, wehaveU(0) = P1 • P2 • • • PeV(0)^• R2 ... Re • [P 1 /2k ]and we can readily prove by induction that, for j^1, 2, ... , t,(3.30)^ > 3—k p(1-03-1and(3.31)^ Rj > 3 -kR(1-03-1(this last necessitates the first inequality in (3.3)). It follows thatU(0) > 3 —"Pk—crandV (o) > 3-k€Rk— [pv/ 2 ]Since = [2k In j k In 6] and a = k(1 — 0', we deduce that for k > 6,< v(o) 1/ 2u(o)p4k+ki 2- 1/ 4V(0) 112U(0)P4k+ki 2-1 / 41-19A1 < ^CioV(0)U(0)2P3k1^1< <9.5k^6k(3.32)Chapter 3: Connections to Waring's Problem^ 81so that(3.33)^U(0) > C11 kp -46V(0) > Ci2pk-112-46k-Fav14-1-1.12where= 3-kt > e-k3andC12 = 3 -kt • 246-k-1 > e -k3HenceIlml^pk+k/2-1/4-k+1/6k-k/2+1/44-1/12k-o/8k-1/4k-19311 Cl0*C111•C122P-a/8kryV2Cm•CH•u n/m 1and so j III < 1 if P-a/8k < C10 • C11 • Cg 2 . Now C10 • C11 • C122 > e_5.86k3, so we have1III1 < 1 if pa 1(80 < e5.86k3 From (3.32), this follows from9RP > e445.36k5and hence from (3.3).lIm lChapter 3: Connections to Waring's Problem^ 823.6 Dickson's Ascent ArgumentIn this section, we will study the representations of integers as sums of elements of theset^(3.34)^ S.1■1;) = {1, Nk ,(N +1) k ,...}where we assume, as before, N > 2 and k > 6.We adopt the notation= [(NN+ 1)k] Nk 1(N + 1  )1and^(N +1) k Lk NSuppose N < (k + 1)( k-1) / k — 1 and write [a, E 4 ) (m) (or (a, b) E S (A,) (m)) if everyinteger in [a, b] (respectively (a, b)) can be written as a sum of at most m elements of 4 )(where we allow repetitions). Following Dickson [21], we count the number of elements ofSN (k) required for representations of "small" integers before applying an ascent argumentto enable the use of Theorem 3.2.1.Before we begin, we need a pair of preliminary lemmas.Lemma 3.6.1 If N, M and k > 2 are positive integers then(N +1) k — MNk =1has only the solution N M k = 2.Proof: Suppose that(3.35)^ (N 1)k MNk + 1where N > 2 and k > 2 (but not N k = 2). If k is even, then we may write(3.36)^((N 1) k/2 — 1)((N 1) ki2  1) MNkChapter 3: Connections to Waring's Problem^ 83and so conclude if N is odd that Nk divides (N+1) kl2 —1. Since this implies N2 < N+ 1,it contradicts N > 2. If, however, N is even, then we have^(3.37)^Nk 2((N + 1) 142 — 1) if N 0 mod 4or(3.38)^Nk 2k ((N + 1) 142 — 1) if N a 2 mod 4.From (3.37), we have N 2 < 2(N +1) which contradicts N 0 mod 4 while (3.38) impliesthat N = 2. Since 3 belongs to the exponent 2' modulo 2 k , we must have 2' dividingk, so that k < 4.It remains only to consider odd k. We can write, from (3.35)(3.39)k k) Ni MNki=1^)and proceed via induction, proving that ord N (k)^oo, thus contradicting any a prioriupper bound for k. From (3.39), we clearly have N I k and if we suppose that Na k,then sinceordp ( i ) > ordp k — ordp i (p prime)we haveIt follows thatand so if i > 2,ordN (k) > a — max(ordP i) .•ordN ((k) Ni^m> a — ax(ordp i) + i•Plip oddp oddord N^N') ?_ a + 2.We conclude, then that Na+l I k as required and hence (3.39) has no solutions for k odd.■Chapter 3: Connections to Waring's Problem^84We will also useLemma 3.6.2 If n and £ are integers with n > £ > (N + 1)k, then there is an elementof 4 ) , say ik, such that(3.40)^ £ < n — i k < i + kn (k-1 )/k.Proof: Suppose first that n > t + Nk and choose i such that i k < n — £ < (i + 1) k . Thenik E Si(N" ) and since, by calculusn — £ — ik < k(n — t)(k-1)/ k < kn (k-1)/ kwe have (3.40). If, however, n < .e + Nk , take i = 1 and write n = £ + m (so that1 < m < Nk ). We concludek(e + rn)(k-1)/k > k(N + 1) k-1 = kN+ 1 (N + W .Since k > N + 1, this is at least (N + 1)k and hence greater than m, as desired.^■Let us now begin to consider representations of comparatively small integers as sumsof elements of SP. We haveLemma 3.6.3 [1, aNk] E SP(Nk + a — 2).Proof: If M < aNk — 1, then we can write M = Nkx + y with 0 < y < Nk — 1 andx < a. It follows that M is a sum of x + y < Nk + a — 2 elements of 4 ) . If, however,M = aNk, clearly M E S$ (a ) .■andLemma 3.6.4 (aNk, (a + 1)Nk) E SP(E) where E = max{a + # — 1, Nk —,3}.Chapter 3: Connections to Waring's Problem^ 85Proof: The integers aNk , aNk + 1, ..., aNk + # — 1 are in S1V(a + /3 — 1) whileaNk + # = (N +1) k , ..., aNk + Nk —1 = (N +1)k — 13+ Nk — 1 belong to Sr(Nk — /3).Since (a + 1)N k E S'Al; ) (a + 1) and /3 > 2 via Lemma 3.6.1, we are done. ■The beginning of our ascent argument, following Dickson [21], lies inLemma 3.6.5 If p and L are positive integers with p > N and (L,L + p k ) E 4) (rn),then (L, L + 2pk ) E Snm + 1).Proof: Let M be an integer satisfyingL + pk < M < L + 2pk .Then M - Pk E 4 ) (m) and so M E SV(m + 1). If M E (L, L + p k ), the result is trivial.■By induction on n, we readily obtainLemma 3.6.6 If p, n and L are positive integers with p > N and (L, L +pk) E 4) (m),then (L, L + pk (n + 1)) E SV(m + n).Taking L = aNk, p = N, n = a + 1 and applying Lemmas 3.6.4 and 3.6.6, sincenNk > (N + 1)k ,Lemma 3.6.7 We have(aNk , aNk + (N + 1) k ) E Sr(E + a)If we now successively apply Lemma 3.6.7 and Lemma 3.6.6 with p = N +1, N + 2, ...,r,N+2,11 1-,,T+2)N + 3 )11.1'^[r..^(  k  ) i ik+i ki. /\T-F 1) i, 1.k and n =^ , t follows thatr ,N+ 2 \k J1 + ... + r i k +Lemma 3.6.8 (aNk , aN k + (k + 1) k ) E SN P (E + a +[4 + 1 ) ^)1,1). i)Chapter 3: Connections to Waring's Problem^ 86Our main ascent relies upon the following result, which is essentially a variant of atheorem of Dickson [19, Theorem 12].Proposition 3.6.9 Let £ and Lo be integers withLo > t > (N +1) k , v = (1 —11L0 )1k and vk Lo > 1Then if for t E N we define L i by(3.41)^In L t = (k k 1 )t (1n Lo + k In v) — klnvand if (1, Lo) E SV(m), then we conclude that (t, L i ) E S (Airc) (m + t).Proof: We suppose (f, Lo) E 4 ) (m) and that n E (t, L1). Now for t = 1, (3.41) isequivalent toLt = (vLo )kk- 1)and hence we may use Lemma 3.6.2 to find i k E SN ) such that£ < n — ik < t + kn (k-1)I k < t + kVL0.Since v = (1 — t/Lo)/k, we have t < n — i k < Lo, whence (f, L i ) E 4 ) (m + 1). Ingeneral, (3.41) yieldsLt+i = (v_L t ) ki (k-1)and the result obtains by induction upon t.^ ■3.7 Proof of Theorem 3.1.3Assume N > 4. To apply the preceding proposition, we let £ =- (N + 1)k and Lo = (k-1-1) k .The condition that vkL o > 1 is then equivalent toN < (k +1) (k-1)1k —1.Chapter 3: Connections to Waring's Problem^ 87If we choose t large enough that(3.42)^ Lt > max1N8k3, e446k6 1 = e446k6then Theorem 3.2.1 gives [L i , oo) E 41,` ) (6k In k+ (3 ln 6+4)k). Now from v = (1 —e/L0 )/k,we may writeln Lt = ( k k i ) t (k ln(k + 1) — k ln v) — k in v> ( k k i^ ) t (kln( k+k 1 )).Since In (k +k 1)^1 _ 1^11>^>k^2k2 -- 12k for k > 6, this implies- 11 (  k  ) tIn L t >12 k — 1 ) 'k  )If we note that ln(^> 1^ 11^ >we obtain (3.42) providedk —1^k —1^2(k —1) 2^k't > k(61nk+1n (5352 \ \ .11 1)Taking t = [6k In k + 7k}, then, yields the desired conclusion. It remains to show for thischoice of t that (1, L i ) E 4 1 (Irk) ) where4)= Nk + {( NN+ 1 )1 — 2(we have [L i , oo) E Sr(i )) because 6k In k + (31n 6 + 4)k < 4 ) for 4 < N < (k +i)(k-i)/k — 1).By Lemma 3.6.8 and Proposition 3.6.9, we have(1,L t ) E Sr (E + a + t+ (k —N)[(N+ 2)k})N + 1and need(^< pk) = Nk + a _ 2 .(3.43)^E-1-a+t-1-(k—N)rN+2)k1../V+ 1 i ^NChapter 3: Connections to Waring's Problem^ 88If E = a + /3 — 1, then (3.43) becomes^(3.44)^E+a+t+(k—N)1N+21 —N k <-11. ( 1 \ r + 1while E = Nk — )3 implies the inequality(3.45)^ t + (k — N)[( N + 21 )kJ — # < —2.To prove that (3.44) and (3.45) obtain for all N and k satisfying4 < N < (k + 1)(k-1) /k — 1we employ Theorem 1.9.4 to deduce3-k < fork < 1 - 3 -k.The left hand side of (3.44) is then bounded above by( N + 1\k (3 )\k+6klnk +7k +(k N)( N +2)1cN )^3 N + 1 )and hence is < —1 for N and k unlessi) N= 4, 6 < k < 34Orii) N= 5, 8 < k < 11Additionally, we bound the left hand side of (3.45) by6k ln k + 7k + (k — Ne + 2 )kN + 1( N3 )kwhich is < —2 for all values of N and k under consideration exceptiii) N= 4, 6 < k < 32Chapter 3: Connections to Waring's Problem^ 89andiv) N --= 5, 8 < k < 11Checking that (3.44) and (3.45) hold for the cases i), ii) and iii), iv) respectively, weconclude the proof of the theorem by noting that M = aNk — 1 ct sr (Nk + a — 3) andthusNk + [ ( NN+ 112 < gN (k) < Nk + [ ( NN+ V] _ .23.8 Concluding Remarks 2(N N^>+ 1 )k ( N N+ 1 )-k which is rather weaker than Theo-In general, we require onlyrem 1.9.4. It appears though, that effectively proving the bound11(4 /3 )4 > (9 /4 ) -kis as difficult as that involved in the classical Ideal Waring problem.Chapter 4Bounds for Ilek li4.1 Multi-point BoundsSimultaneous or multi-point Pade and Pade-type approximation to In x and ex has a longhistory dating back to Hermite's proof of the transcendence of e. By application of thesetechniques, in what amounts to two essentially distinct ways, K. Mahler [39, 42] was ableto deduceTheorem 4.1.1 (Mahler) There are constants c and k o such thatHe il > k- ckfor all k > ko.The first of Mahler's methods gives the constant c = 40, while the second yields c = 33.A similar approach to the problem of bounding Prkll is mentioned in Mahler [40], whereone finds that'Irk 11 > k- ck2lnkfor all k sufficiently large and c an absolute positive constant.4.2 A Single-point BoundThe techniques applied in Chapter 1 can be utilized in a variety of situations where closedforms are available for the relevant Pade approximations. In the following section, wededuce the lower bound90Chapter 4: Bounds for !le i'II^ 91Theorem 4.2.1 There is an effective constant c such thatIIekII > c • (2k2 )- k2for all k E N.While this bound is weaker than that obtained via multi-point Pade approximation, itis easier to produce.Proof of Theorem 4.2.1: From Luke [37, p. 192, (19)-(22)], we can writee_ z = Gm (-z)^ + em (z)Gm (z)valid for Rez > 0, where Gm (z) is an m-th degree polynomial in z with integer coefficientsand e nz (z) satisfies (see Luke [37, p. 74, (35)])(4.46)^Em(z) = (-1)rn-E'rz2m+1e-z [1 + 0(M -1 )].24m+ 1 (n2!) 2It follows that if k is a positive integer(4.47) ek^Gm(k)^k  Grn (k)Gm(-k) e Gm (-k) EniVWe now seek a bound upon Gm (k), provingGm+ i(k) Lemma 4.2.2 Gm(k)) < k + 4m + 2, m = 0, 1, 2, ....Proof: We proceed via induction, proving thatGni+i (k) k <^<k+4m+ 2.Gm (k)Since Go (k) = 1 and G 1 (k) = k + 2, we suppose thatGi+1(k) < k + 4i + 2G2(k)Chapter 4: Bounds for He'll^ 92for i = 0, 1,^, m — 1, whileGi+i(k) > kGi(k) —for i = 0, 1, ..., m — 2. From Luke [37, p. 92, (22)] we have(4.48)andthus follows fromGi+i(k) = (4i + 2)Gi(k) k 2 Gi_ 1 (k)Gm+i (k) < (k + 4m + 2)Gm (k)Gm (k) > kGm_ i (k).If Gm (k) < kGm _ 1 (k), then (4.48) yieldskGm_ i (k) > (4in — 2)Gm_ i (k) + k 2Gm _2 (k)OrGm _1(k) > ^k2Gm_2 (k) k — 4m + 2But our induction hypothesis impliesGm -1(k) <k+4m-6Gm,-2(k)which gives< k 4rn — 6k —4m-I-2Or—16m2 + 32m — 4k — 12 > 0a contradiction (since G2(1)/G1 (1) = 19/3, we may assume (m, k) (1, 1)).^■By the preceding lemma, then, we have thatk2 -1Gk2(k) <^(k + 4m + 2)m=ok2Chapter 4: Bounds for 11 ek ii^ 93and since1 k2-1_k2 E (k + 4m + 2) = 2k 2 + kthe arithmetic-geometric mean inequality implies that^(4.49)^ Gk2(k) < (2k 2 + k)k2 .From (4.47), we write, for m = k 2 or k 2 + 1Gm (k) — ek Gm (—k) = ek G,i (k)Em (k)and so if N is an arbitrary integer,(4.50)^lek — NIG,(—k)+ ekGm(k)IEm(k)I >IG,(k) — NG,,(—k)j.Now we can choose in = k 2 or k 2 + 1 such that the right hand side of the above inequalityis nonzero (a rather general property of Pade approximants, analogous to Lemma 1.3.1)and hence > 1. We assume that this occurs for m = k 2 , the other case being similarwith the same result obtaining. By (4.46),rk2k2+1 6-k1 6 1c 2 (k )1 = 24k2 +102)0 2 [ 1 + °(k 2 )]and Stirling's formula yields (for some 0 < 0 < 1)^1 ^t  e2^2lE k2(101 = 4kek+1/(6k2 )+1/(2e)^16k2)k [1 + o(k-2)] .Thusek Ck2(k)jek2(k)1 < ( e2(2k16k22 + k)) k 2< (0.99) k2for k > ko effectively computable. NowGk2 (k) < (2k2l/k+ k \ k2^) < (2k 2 ) k2k —^em=0Chapter 4: Bounds for He il^ 94and hence (4.47), (4.50) and the last result imply that there is an effective constant c(independent of k) such thatHell > c • (2k2 ) — k2as desired.^ ■The above yields a bound of orderIlek II > (f(k)) -1where In f (k) = 0(k 2 in k) and hence is no match in strength for Mahler'sHe ll > (g(k))-1with g satisfying In g(k) = 0(k In k).Appendix ( Na + 1)kNb )Special cases of lower bounds forIn the following, we tabulate the results for small values of N as cited in Theorems 1.9.3,1.9.4, 1.9.5 and 1.9.6 for forms (1 + 1 )k^(N + 1 )k and (N + NZ )k . Tables 1, 3N Nand 4 include values of N with admissable s (in the notation of Theorem 1.4.1) and theinduced lower bound (for exponent k > ko = ko (N) effective). Table 2 has, additionally,explicit ko (N) and derived lower bounds for the nth root of the contents of the involvedPade approximants (as given in section 1.3). In all cases, the values of s = s(N) are notclaimed to be in any sense best possible, but are adequate for our purposes.95 (Na +1 \kNb )Appendix : Special cases of lower bounds for 96( 1 + 1\77. )kTable 1: > B-k for all k > ko (N)N s B N s B N s B4 2.044 2.5928 21 4.35 2.3809 38 5.88 2.10725 2.211 2.7548 22 4.45 2.3612 39 5.98 2.08786 2.381 2.8054 23 4.55 2.3423 40 6.05 2.07977 2.548 2.8058 24 4.64 2.3282 41 6.12 2.07228 2.689 2.8229 25 4.72 2.3178 42 6.18 2.06669 2.818 2.8354 26 4.81 2.3033 43 6.25 2.058410 2.936 2.8452 27 4.89 2.2925 44 6.32 2.050411 3.070 2.8111 28 4.97 2.2816 45 6.39 2.042012 3.230 2.7298 29 5.06 2.2640 46 6.46 2.033513 3.38 2.6640 30 5.16 2.2415 47 6.53 2.025314 3.53 2.6005 31 5.25 2.2232 48 6.59 2.019515 3.66 2.5602 32 5.35 2.2021 49 6.66 2.011416 3.80 2.5079 33 5.44 2.1850 50 6.72 2.005617 3.96 2.4403 34 5.53 2.1681 51 6.78 1.999718 4.07 2.4183 35 5.62 2.152319 4.16 2.4098 36 5.70 2.139420 4.26 2.3935 37 5.79 2.1233 (Na + 1 vcNb )Appendix : Special cases of lower bounds for97Table 2: > 3 -k for all k > ko (N)N c d ko G4 25 13 28,375 1.736685 17 8 66,045 1.746516 30 13 162,600 1.709347 37 15 127,391 1.689588 60 23 177,060 1.649599 52 19 219,180 1.6182310 20 7 269,480 1.5865411 86 29 359,050 1.5529912 46 15 170,890 1.5212513 79 25 63,437 1.4897914 13 4 36,491 1.4680115 10 3 19,900 1.4407416 17 5 16,728 1.4239117 7 2 10,276 1.4042118 25 7 10,325 1.4028619-21 11 3 < 12,749 1.4220222-28 4 1 < 11,288 1.4522629-31 13 3 < 11,466 1.3402132-37 9 2 < 8703 1.3064538-77 5 1 < 6175 1.3067878-135 6 1 < 1359 1136-274 7 1 < 647 1275-545 8 1 < 422 1546-728 9 1 < 382 1 (Na +1\kk Nb )Appendix : Special cases of lower bounds for 98(N + -1k^N )Table 3: > B -k for all k > ko (N)N s B N s B N s B N s B3 2.10 2.5477 23 4.68 2.5082 43 5.55 2.4581 63 6.09 2.42564 2.39 2.6492 24 4.75 2.4970 44 5.58 2.4567 64 6.11 2.42605 2.71 2.5719 25 4.82 2.4828 45 5.61 2.4552 65 6.13 2.42636 3.05 2.4192 26 4.89 2.4677 46 5.65 2.4496 66 6.15 2.42647 3.19 2.4899 27 4.96 2.4504 47 5.68 2.4477 67 6.17 2.42638 3.33 2.5246 28 5.02 2.4409 48 5.71 2.4461 68 6.19 2.42619 3.45 2.5535 29 5.05 2.4493 49 5.73 2.4477 69 6.21 2.425810 3.58 2.5609 30 5.09 2.4531 50 5.76 2.4454 70 6.23 2.425411 3.69 2.5686 31 5.13 2.4556 51 5.79 2.4428 71 6.25 2.425112 3.79 2.5773 32 5.17 2.4572 52 5.82 2.4400 72 6.27 2.424313 3.89 2.5799 33 5.21 2.4578 53 5.85 2.4370 73 6.28 2.426514 3.98 2.5822 34 5.24 2.4614 54 5.88 2.4338 74 6.30 2.425615 4.07 2.5777 35 5.28 2.4607 55 5.91 2.4302 75 6.32 2.424516 4.16 2.5687 36 5.32 2.4593 56 5.94 2.4264 76 6.34 2.423417 4.24 2.5624 37 5.35 2.4613 57 5.97 2.4222 77 6.36 2.422618 4.32 2.5530 38 5.38 2.4631 58 6.00 2.4172 78 6.37 2.424619 4.40 2.5438 39 5.42 2.4610 59 6.02 2.4191 79 6.39 2.423520 4.47 2.5382 40 5.45 2.4622 60 6.03 2.423221 4.54 2.5300 41 5.48 2.4632 61 6.05 2.424322 4.62 2.5146 42 5.51 2.4633 62 6.07 2.4251 (Na + 1 VcNb )Appendix : Special cases of lower bounds for 991 \ kN2Table 4: (N + > B -k for all k > ko (N)N s B2 1.0725 3.00043 1.62 2.67084 2.13 2.36575 2.35 2.48606 2.56 2.52867 2.72 2.58478 2.86 2.62519 2.99 2.658110 3.16 2.606311 3.31 2.568312 3.48 2.504713 3.59 2.496914 3.70 2.483815 3.82 2.458016 3.92 2.441217 4.02 2.422618 4.10 2.419819 4.18 2.413720 4.26 2.405321 4.33 2.401322 4.41 2.387723 4.49 2.372724 4.55 2.370525 4.61 2.3672Bibliography[1] K. Alladi and M.L. Robinson. Legendre polynomials and irrationality. J. ReineAngew. Math., 318:137-155, 1980.[2] A. Baker. Rational approximation to 2 1 /3 and other algebraic numbers. Quart. J.Math. Oxford, 15:375-383, 1964.[3] A. Baker. Rational approximations to certain algebraic numbers. Proc. LondonMath. Soc., 4:385-398, 1964.[4] A. Baker and J. Coates. Fractional parts of powers of rationals. Math. Proc. Cam-bridge Philos. Soc., 77:269-279, 1975.[5] M. Bennett. The content of Pade approximants to the binomial function. In prepa-ration.[6] M. Bennett. Fractional parts of powers of rational numbers. Math. Proc. CambridgePhilos. Soc. To appear.[7] M. Bennett. An Ideal Waring problem with restricted summands. To appear.[8] F. Beukers. Pade-Approximations in Number Theory. Seminaire de Theorie desNombres, 1980-1981. expose n° 10.[9] F. Beukers. Fractional parts of powers of rationals. Math. Proc. Cambridge Philos.Soc., 90:13-20, 1981.[10] T.J. Bromwich. Introduction to the theory of infinite series. London: MacMillanCo., 1926.[11] J.W.S. Cassels. An introduction to Diophantine analysis. Cambridge UniversityPress, 1957.[12] S. Chowla. On g(k) in Waring's problem. Proc. Lahore Philos. Soc., 6:16-17, 1944.[13] D. Chudnovsky and G. Chudnovsky. Pade and rational approximations to systemsof functions and their arithmetic applications. In Number Theory, volume 1052,pages 37-84. Springer, Lecture Notes in Mathematics.100Bibliography^ 101[14] G. Chudnovsky. Hermite-Pade approximations to exponential functions and elemen-tary estimates of the measure of irrationality of 7r. In The Riemann Problem, Com-plete Integrability and Arithmetic Applications, volume 925, pages 229-322. Springer,Lecture Notes in Mathematics.[15] G. Chudnovsky. On the method of Thue-Siegel. Ann. of Math., 117:325-382, 1983.[16] H. Davenport. Analytic methods for diophantine equations and approximations. AnnArbor Publishers, 1962.[17] J. M. Deshouillers and F. Dress. Sums of 19 biquadrates: On the representation oflarge integers. Ann. Scuola Norm. Sup. Pisa, XIX.1:113-153, 1992.[18] L. E. Dickson. Proof of Waring's theorem for fifth powers. Bull. Amer. Math. Soc.,37:549-553, 1931.[19] L. E. Dickson. Recent Progression on Warings's Theorem and its Generalizations.Bull. Amer. Math. Soc., 39:701-702, 1933.[20] L. E. Dickson. On Waring's problem and its generalizations. Ann. of Math., 37:293-316, 1936.[21] L. E. Dickson. Proof of the ideal Waring's theorem for exponents 7-180. Amer. J.Math., 58:521-529, 1936.[22] L. E. Dickson. Solution of Waring's problem. Amer. J. Math., 58:530-535, 1936.[23] L. E. Dickson. Waring's problem and its generalizations. Ann. of Math., 37:833-842,1936.[24] A. K. Dubitskas. Approximation of 7r/0 by rational fractions. Vestnik MoskovUniv. Ser. 1. Mat. Mekh., 6:73-76, 1987.[25] A.K. Dubitskas. A lower bound for the quantity 11(3/2)4. Russian Math. Surveys,45:163-164, 1990.[26] D. Easton. Applications of the hypergeometric function in effective Diophantineapproximation. PhD thesis, University of Waterloo, 1986.[27] W. J. Ellison. Waring's problem. Amer. Math. Monthly, 78:10-36, 1971.[28] G. H. Hardy. Some famous problems in the theory of numbers and in particularWaring's problem. Oxford University Press, London, 1920.[29] G. H. Hardy and E. M. Wright. An introduction to the theory of numbers. NewYork: Oxford University Press, 4th edition, 1960.Bibliography^ 102[30] M. Hata. On the linear independence of the values of polylogarithmic functions. J.Math. Pures Appl., 69:133-173, 1990.[31] M. Hata. A lower bound for rational approximations to ir. J. Number Theory,43:51-67, 1993.[32] L. K. Hua. Additive theory of prime numbers. In Translations of MathematicalMonographs, vol. 13. Cambridge, 1966.[33] R. D. James. The value of the number g(k) in Waring's problem. Trans. Amer.Math. Soc, pages 395-444, 1934.[34] J. K. Koksma. Ein mengentheoretischer satz iiber die gleichverteilung modulo eins.Compositio Math., 2:250-258, 1935.[35] J. M. Kubina and M. C. Wunderlich. Extending Waring's conjecture to 471,600,000.Math. Comp., 55:815-820, 1990.[36] E. Landau. Vorlesungen I. Hirzol, Leipzig, 1927.[37] Y. Luke. The Special Functions and their Approximations, volume II. AcademicPress, 1969.[38] K. Mahler. Zur Approximation der exponential Funktion and des Logarithms. J.Reine Angew. Math., 166:118-150, 1932.[39] K. Mahler. On the approximation of logarithms of algebraic numbers. Philos. Trans.Roy. Soc. London Ser. A., 245:371-398, 1952/3.[40] K. Mahler. On the approximation of ir. Proc. Ned. Acad. Wet., 56:30-42, 1953.[41] K. Mahler. On the fractional parts of the powers of a rational number: II. Mathe-matika, 4:122-124, 1957.[42] K. Mahler. Applications of some formulae by Hermite to the approximation ofexponentials and logarithms. Math. Ann., 168:200-227, 1967.[43} K. Mahler. Lectures on transcendental numbers, volume 546 of Lecture Notes inMath. Springer, 1976.[44] W. Narkiewicz. Classical problems in number theory. Polish Scientific Publications,1986.[45} I. Niven. An unsolved case of the Waring problem. Amer. J. Math., 66:137-143,1944.Bibliography^ 103[46] S. S. Pillai. On Waring's problem. J. Indian Math. Soc., 2:16-44, 1933.[47] G. J. Rieger. Uber eine Verallgemeinerung des Waringsches Problems. Math. Z.,58:281-283, 1953.[48] J. Rosser and L. Schoenfeld. Approximate formulas for some functions of primenumbers. Illinois J. Math., 6:64-94, 1962.[49] J. Rosser and L. Schoenfeld. Sharper bounds for the Chebyshev functions 0(x) andzk(x). Math. Comp., 29:243-269, 1975.[50] R. K. Rubugunday. On g(k) in Waring's problem. J. Indian Math. Soc., 6:192-198,1942.[51] R. Salem. Algebraic numbers and Fourier analysis. D.C. Heath and Co., 1963.[52] L. Schoenfeld. Sharper bounds for the Chebyshev functions 0(x) and '(x), II. Math.Comp., 30:337-360, 1976.[53] C. Small. Waring's problem. Math. Mag., 1:12-16, 1977.[54] R. M. Stemmler. The ideal Waring's theorem for exponents 401-200,000. Math.Comp., 18:144-146, 1964.[55] K. Stromberg. An Introduction to Classical Real Analysis. Wadsworth InternationalMathematical Series, 1981.[56] E. Trost. Eine Bemerkung zum Waringschen Problem. Elem. Math., 13:73-75, 1958.[57] I. M. Vinogradov. On Waring's problem. Ann. of Math., 36:395-405, 1935.[58] I. M. Vinogradov. The method of trigonometric sums in the theory of numbers.Interscience, London, 1954.[59] I. M. Vinogradov. On an upper bound for G(n). Izv. Akad. Nauk. SSSR Ser. Mat.,23:637-642, 1959.[60] I. M. Vinogradov. Special variants of the method of trigonometric sums. Navka,Moscow, 1977.[61] G. Xu. Fractional parts of powers of rationals. Journal of Mathematical Researchand Exposition, Beijing, China, 8(2):207-213, 1988.

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