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FRACTIONAL PARTS OF POWERS AND RELATED TOPICS

By

Michael A. Bennett

B. Sc. (Mathematics) Dalhousie University, 1987

M. Sc. (Mathematics) University of British Columbia, 1989

A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF

THE REQUIREMENTS FOR THE DEGREE OF

DOCTOR OF PHILOSOPHY

in

THE FACULTY OF GRADUATE STUDIES

MATHEMATICS

We accept this thesis as conforming

to the required standard

THE UNIVERSITY OF BRITISH COLUMBIA

March 1993

© Michael A. Bennett, 1993

In presenting this thesis in partial fulfilment of the requirements for an advanced

degree at the University of British Columbia, I agree that the Library shall make it

freely available for reference and study. I further agree that permission for extensive

copying of this thesis for scholarly purposes may be granted by the head of my

department or by his or her representatives. It is understood that copying or

publication of this thesis for financial gain shall not be allowed without my written

permission.

(Signature)

Department of tla +6 ernoA(sThe University of British ColumbiaVancouver, Canada

Date ^Al 61 2c ) /913

DE-6 (2/88)

Abstract

In this thesis, we employ a variety of explicit approximations to tackle some problems

in Diophantine analysis. We are generally concerned with constructing rational function

approximations 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 and

known results from the literature. The situation for algebraic numbers in general is briefly

surveyed. In Chapter 1, we utilize Pade approximation in the manner of F. Beukers to

generalize and sharpen results of Beukers, D. Easton and G. Xu about lower bounds for

fractional parts of powers of rationals. Some theorems of a "semi-effective" nature are

discussed 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 Chapter

1 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, in

particular 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 prove

a version of the Ideal Waring problem with restricted summands.

ii

Table of Contents

Abstract^ ii

List of Tables^ v

Acknowledgements^ vi

0 Introduction 1

0.1 Preamble ^ 1

0.2 Lower bounds on 11011 for 0 rational ^ 2

0.3 Lower bounds on Pkii for 0 algebraic ^ 3

0.4 Relations to Additive Number Theory ^ 4

0.5 Lower bounds on 110 k 11 for 0 transcendental ^ 4

1 Fractional Parts of Powers of Rationals 6

1.1 Introduction ^ 6

1.2 Some Preliminary Definitions ^ 11

1.3 Generating Our Approximants ^ 12

1.4 The Main Theorem ^ 15

1.5 Bounding the Approximants ^ 17

1.6 A More Explicit Treatment of Q(s) and E(s) ^ 20

1.7 Common Factors of Coefficients of Pri and Q, ^ 22

1.8 The Proof of the Main Theorem ^ 25

1.9 The case 11(1 + )3/N) k 11 ^ 27

iii

1.10 Semi-Effective Results ^ 37

1.11 Density Results ^39

2 Contents of Pade Approximants to (1 — z) k^43

2.1 Basic Behaviour of L(a, b, s) ^43

2.2 The Case L(1,1,3) ^ 46

2.3 Limiting Behaviour of L(a,b,^) ^ 58

3 Connections to Waring's Problem^ 60

3.1 Introduction ^60

3.2 Asymptotic Theory: Notation and Definitions ^ 62

3.3 Necessary Lemmas ^ 65

3.4 The Contribution of the Major Arcs ^ 70

3.5 Minor Arc Estimates ^ 78

3.6 Dickson's Ascent Argument ^ 82

3.7 Proof of Theorem 3.1.3^ 86

3.8 Concluding Remarks ^ 89

4 Bounds for IIe k Il^ 90

4.1 Multi-point Bounds ^ 90

4.2 A Single-point Bound ^ 90

( IV' + 1 VcNb )

Appendix: Special cases of lower bounds for 95

Bibliography^ 100

iv

(N + N1 2 )k

List 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) ^ 99

Acknowledgements

I would like to thank my supervisor, Dr. David Boyd for many helpful discussions and

admirable patience. Thanks are also due to Mark MacLean, Dr. Larry Roberts, Joan de

Niverville and the many others who've influenced me over the last five years, especially

the 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-god

E. David Robinson and Dr. Rajiv Gupta for periodic advice.

vi

Chapter 0

Introduction

0.1 Preamble

The behaviour of the fractional parts of real sequences has interested mathematicians

for 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 promises

continued interest in coming years. In the following work, we restrict our attention to

sequences of the form O k where 0 >1 is a fixed real number and k is a positive integer.

If we define

oil _--, minook — MI)MEZ

and

{0kfoki = min (0k — M)} ma \Af<ek

then 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 every

pair (a,b), with 0 < a < b <1, one has

1lirn —„,1{k<N:a<fxkl<b}1=b—a

N-4.00 IV

where {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 being

clearly the weaker one). While a result of Koksma [34] ensures that this is in fact the

case for almost all 0 > 1 (in the sense of Lebesgue measure), it is difficult to find such

1

Chapter 0: Introduction^ 2

0 — indeed this appears to be the unfortunate situation for any exponentially growing

sequence.

0.2 Lower bounds on ilO k li for 0 rational

In Chapter 1, we discuss the case of fractional parts of powers of rational numbers. By

applying Pade approximation techniques a la F. Beukers [9], we are able to sharpen

effective lower bounds for certain infinite classes of rationals, providing some advantages

over stronger, ineffective bounds of K. Mahler [41] and weaker, more general effective

bounds due to A. Baker and J. Coates [4]. To do this, we find approximating polynomials

PP (z) and Q,,(z) of degree n in Z[z] which satisfy

pn (z ) — ( 1 — z )k Q n ( z ) = z2n+ 1 En ( z )

or more generally

Pn(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. By

estimating the size of 1Q 7,(z)! and lEn (z)I for specific rational choices of z and using the

fact that a nonzero integer has modulus at least one (a trivial fact, but of undeniable

utility in transcendence theory!) we obtain a number of bounds, including

0.2.1 If 4 < N < k • 3 k then^ + --(

1^1 yc

and

> 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 )


We 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 of

the coefficients of the polynomials Pn (z) and Qn (z). This idea has been exploited with

relation to fractional parts of powers of rationals by F. Beukers [9], D. Easton [26] and

A. 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 coefficients

and discuss its limiting behaviour. All of the aforementioned improved bounds hinge

upon evaluation of this function, so its asymptotics are of clear interest.

0.3 Lower bounds on IjO k ii for 0 algebraic

Results 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 and

transcendence theory may be applied in both settings. In general, for algebraic 0, it is

not possible to find better than exponential bounds upon 1104, due to the existence of

the Pisot (or Pisot-Vijayaraghavan) numbers. These are defined to be algebraic integers

all of whose other conjugates lie strictly inside the unit circle. It is easy to show that if

0 is a Pisot number, then

(0.1) k—■oolim 110k11— 0

and in fact (see Salem [51]), a partial converse exists — if (0.1) holds for 0 algebraic then

0 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 (as

is in fact the case for all Pisot numbers). An interesting open problem is whether there

are any transcendental 0 satisfying (0.1) with even, say, logarithmic decrease.


Another class of algebraic integers 0 for which we have some information regarding

ii°kii is that of the Salem numbers. These are algebraic integers, all of whose conjugates

lie 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 uniformly

distributed modulo 1.

0.4 Relations to Additive Number Theory

Lower bounds upon II (3/2) k II have long been known to be related to the number g(k) in

Waring's problem (and vice versa). In fact, one has

(0.2) g(k) = 2 k + [(3/2) k ] — 2

provided 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 obtain

this, we generalize a by now classical bound due to I. Vinogradov [57] on G(k) in Waring's

problem and apply an ascent argument of L. E. Dickson [21]. We are unfortunately not

able 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 11 for 0 transcendental

Few results in this area have been stated explicitly in the literature and what ones exist

seem to appear as byproducts of irrationality measures for In n or ir. Through the use

of 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 by

Mahler [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,


we apply single-point diagonal Pade approximation to ez to deduce a lower bound which

is weaker than those already known, but simple to prove. The same techniques, one

may hope, could be utilized in many other settings where closed form approximants of

Pade type are available. It would be of genuine interest to determine when this is in fact

possible and to compare the induced bounds to those available by other methods.

Chapter 1

Fractional Parts of Powers of Rationals

1.1 Introduction

There 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 related

approximations 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 these

fractional parts and discuss their consequences.

If we suppose (here and subsequently) that a > b > 2 are relatively prime integers

and k a positive integer, then our starting point is the trivial "Liouville" bound

11(a/b) k il .? b - k .

A fair indication of some of the difficulties involved with these problems is that an

improvement of (1.1.1) to strict inequality (for k > 2) would imply the truth of Catalan's

conjecture — 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] showed

1.1.2 If e > 0 is given, then there exists ko^ko (a, b, e) such that for all k > ko , we have

ii(a/101 > e -Ek

6

Chapter 1: Fractional Parts of Powers of Rationals^ 7

The 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 theorem

merely asserts its existence. Much of what follows deals with finding constructive, that is

effective, 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 and

thus would be considerably more desirable than an ineffective result, though usually much

weaker. 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 first

and still essentially best results are due to Baker and Coates [4] who proved

1.1.3 There exist effectively computable numbers 77 and k o , dependent on a and b, such

that 0 < <1 and if k > ko , then

IRa I b) k II >

This follows as a p-adic corollary to Baker's work on linear forms in logarithms and in

practice they turn out to be extremely close to 1 — for a = 3, b 2 one finds, with

present estimates, 77 ti 1 —10'. It seems unlikely that further research along these lines

will 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 the

polynomial (1 — z) k and related functions. In 1981, F. Beukers [9] used this approach to

prove

1.1.4 If k > 5000, then 11(312)11>

and


1.1.5 If N and k are positive integers, N > 2, then

(1 + 7\71 )k > N-3/2(8.4)-k.

4

The basic notion involved here is that while it may be difficult to derive arithmetic

information directly from hypergeometric functions such as the binomial, very regular

approximants exist which can be more readily handled. This usually entails finding upper

bounds for these approximants (or sometimes asymptotics) and then connecting this to

other, often trivial, lower bounds for related quantities. By sharpening some of these

estimates, D. Easton [26] improved (1.1.4) and (1.1.5) to

1.1.6 There exists an effectively computable ko such that if k > ko , we have

11(3/2) k ii > 2 -0.82k .

and

1.1.7 If N > 4 is an integer, then there exists an effective k ip ,---- ko (N), such that if

k > 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 prove

1.1.8 There exists an effectively computable k o such that if k > ko , we have

ii(3 / 2 )kil > 2 -0.794k .

It is not obvious how this result can be much improved without, to quote from Baker

and Coates, the introduction of "fundamentally new ideas". We will see later exactly

which effective bound is desirable and note here only that the coefficient involved in the

exponent of 2 would need to be much reduced.

On somewhat more general lines, we mention further results by Easton, who from a

technical theorem similar to our Theorem 1.4.1, deduced the following corollaries

1.1.9 If N > 4 is an integer, then there exists effective ko = ko (N) such that if k > ko,

then

(N^)k > (3.15) -k .

and

1.1.10 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 quite

dramatic in some cases!), G. Xu [61] employed Beukers' techniques and some facts about

Legendre polynomials to prove

1.1.11 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 mind

that the above is only an improvement upon the trivial bound (1.1.1) when 362a0 2 < N

and hence is still subject to fairly strong restrictions.

In what follows, the author will derive a number of bounds similar in nature to those

mentioned above. In particular, (1.1.5) and (1.1.7) are sharpened in respectively, explicit

and effective versions to

1.1.12 If N and k are positive integers with 4 < N < k • 3k , then

> 3-k(

+ k


and


then

(

7, -v-1 + )k

> (2.85) -k .

By generalizing Dubitskas' lemma on divisibility of binomial coefficient products it is

possible to also improve (1.1.9) and (1.1.10), to obtain


then

(N Tv--+ )k> (2.65) -k .

and


then

N + 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 result

01.1.16 If a, 13, N are integers with N > 101 and a + —N

> 1, then there is an effective

ko = ko la, /3,N) such that if k > ko,

,^/3 \a + WO

k

> (4a3 2 ) -k .

The constant 4 can be improved somewhat, but already is rather preferable to the 362

occuring in (1.1.11).

1.2 Some Preliminary Definitions

Since we will be dealing frequently with the binomial function, we define, in analogy to

the 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 = 1.0

Some straightforward properties of this are

ax(1.4) for real a > 0

ay_

x^y —1

(1.5) = max (1 — t)x0

xy tom.)

(1.6)y

< 2x^with equality exactly when x = 2y

and

(1.7)x + ay + a y

for 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 our

approximation. If we set

6-1 a(1.8)^zb H (a, b, z) = (1 — z)a+ b — E^(-zy

i.0 ( I

then we see that it is possible to view this function as a "truncated" binomial. The above

definition is motivated with respect to fractional parts of powers of rationals by

Lemma 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)11

NP j Mem ' ceNr ) II .II


Proof: 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—i

i=(rq—p)cm

while

((aNr 13)9vcm rqcm (rqcm\

^NP ) = E^)j=0rqcm (rgern\

^= E^)1=0

and hence we have

(aN(r -piq))(rqcm -i) ( N yNP/q)

ar qcm-i Nr((rq—p)cm—i) oi

rqcm

apcm iaqrq-ocniH(pcm, (rq — p )cm, ^EceNr i=(rq—p)cm

so that, since

frqCrn _rqcm—iNr((rq—p)—i)pi

i ) u

(rq—p)cm-1 „cm arqcm_2 Nr((rq_ocin_z) 3i

1=0

is an integer, we obtain the required equality.^ ■

Throughout the following chapter, we will assume that a, f3 , c, m, etc. are as in the

statement of the above lemma and add the constraint that

(ceNr #) q^1NP

to avoid trivialities.

1.3 Generating Our Approximants

The following method for construction of approximants to H(a, b, z) was suggested to

the author by F. Beukers. We suppose that A, B, and C are nonnegative integers with

A < C and that^< 1. Writing

fo 1 0(1 — 013(z — t) Ct) dt f tA (1 — t) B (z — t) c dt + f tA(1 _^—^di

and making the changes of variable t^zt and t^1 — t zt in the first and second

integrals on the right hand side respectively, we can conclude

tA(1 — t) B (z — t)c dt

(1.9)^= Fin]. — zr+c-FilltBq —^— •(1^t^zt)Ao

1zA+C +1 I tA^t)C ^zt)B di.

Applying definition (1.8), this last equation becomes

Jo 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) A

zA+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 hand

sideand so if we let

z A + B + C + )PA(z) = A-c(

A! B! C!^l ' ! (1 1 tA (1 ^— t) B (z — t)c dt

c-A-1(B + c +^fl tB(1 - t)c(1 -^zt)A dt)+ Hoc,. (-z)10i=0


(1.12)^QA(z) =(-1)c (A B C 1)! f l tB — 1)C (1 — t + zt)A dt

A! B! C!

(1.13)ÊA(z) =^B(A++ C+1)7 1 A

(1—t)c (1— zt) BA! B! C!ô

then we have from (1.10),

(1.14)^PA(z) — H(A + B +1,C — A, z)Q A(z) = z 2A+1 EA(z)•

Now

fo^1 Â

^

tB (1 — t) c (1 — t + zt) A dt =^1 B (1 — t) C 2 (A )(1 - t)A-i (zt) i dti=0^6

A

^

=^() Z 2 tB+2(1 t)A + C _ i dt

^

i=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^)zZ0

which belongs to Z[z] (of degree A). Similarly,

B (A + i) ( A B C 1) (— )^(1.16)ÊA(z) = E

i=0^)^1

which is also in Z [z] (of degree B). It is clear from (1.11) that PA (z) is a polynomial of

degree A in z. The coefficients of PA (z) in general can be fairly complicated, so at this

juncture 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 polynomials

PA (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 only

the following:


Lemma 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 find

PA (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+ 1

since 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). Since

EA(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 Theorem

Before we proceed, we must introduce some more terminology. Let us define, for s > l/p,

(1.17)

and

(1.18)

where

cd(s)= (14r,,)0Q(s, 01) •

rqs +1(rq — p)s + 2

E(s). ( tiAtici lE(s,t)l) • -(rq — p)s + 2_

(rq — p)s +1

cd (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 + a

13 tr 1Nr^•

Further, we set

and

(1.19)^L(s) = L(rq, p, s) = exp(t

rqs +1t^

e(s,t)))

where

0(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 +1

and the summation is over all t such that

t ^[(ps — 1)1^[((rq — p)s + 1)1[rqs +1^rqs +1^rqs + 1^= t — 2.

Under these assumptions, we have

Theorem 1.4.1 Ifs is rational, s > 11p, N > 2, e > 0, and

rqs +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 Onk

NP )

(1.21) > (aNr Q(s)L(s) -1 +e • (rq — p)s + 2_(rq — p)s + 1

To 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 )^NaNr

where 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 multiples

of m. We have, from (1.14), that for n = din or din — 1

Qn ( 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 p

n —0^m N-pe5

^aNr^n aNr I

and 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 upon

Qn( - 13 )aNr and En ( 13

aNr) , while bounding the right hand side of the inequality (1.23)

from below.

1.5 Bounding the Approximants

We first deal with the associated binomial coefficients, proving if A, B, C, and m are

positive 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^AABBCc

Proof: 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 l

This directly yields the above inequality with the RHS multiplied by eX , where

X =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 or

dm — 1, where d is a positive integer satisfying d < pc, we show

(rq — p)c + 2d(rq — p)c+ d

Lemma 1.5.2 Q.( --/ )aNr< (Q(c/d) d •

)-

Proof: Before proceeding, we note that the implied constant in the above inequality

depends upon c, d, p, q, r, a, 9, N and the choice of n as either dm or dm — 1, but not

and

- (1 + ° \ t\n dt.aNr ) )I = 1 1 ( 1 — tyrq—p)cm-En tpern—n-1 (10 k

11..11 =


upon 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. 1

and 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 • (

where

D1=

1^(rqc+ d)(pc - d)27r \ ((rq - p)c + d) • d

1^((rq - p)c -I- d) • d27r \ (rqc+ d)(pc - d)

if n = dm

if n = dm - 1

Now we have

1 ^, 11[n/ ml1 = 1 (Q (C I CI, trr - 1 (1 — t) ("—P)c+ rnimitPc—( rnirni +1) (' — I1^0+ 1'. )t)^dt

o^ aNr

and so

1/1< (max1Q(c/d,t) d ir 1 • /1, tE[0,11

where(1 — t)(rq—p)c-i-in/mI tpc—(rnim]+1) (1 — ( 1 + /3 )tyrtimi dt

aNr I

Combining 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 prove

En( --° )aNr

rqc+ d(rq — p)c + 2d

Lemma 1.5.3

< (E (c/d) d •

Proof: Again we may note that the implied constant is independent of m (and dependent

as 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-1dt

a Nr ifoand hence a further application of Lemma 1.5.1 yields

(En \

—13aNr )i

rqc + 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) and

1J = i (1 — t) (rq—p)cm-l-ntn (1 + ^ tYcm—n-1 dt.

aNr

Since

1J = f (E(c/d,t) d ) m-1 (1 — (rq—P)c+[ninli t [n/mi a + 13 0"-(inimi+1) dtJo^ aNr

we have, as before,

IJI < (maxe[cui1E(c/d, t) dr 1 • Jl,t

Ji. = fo


where1(1 — t)

(r q — p) c-F[nim]t[n i nil (1 + ^ )l3r t \Pc— ([nim]+ 1 ) dt

aN

From (1.28) we conclude

En( 3 )aNr

rqc + 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) were

defined as they were. To see the underlying motivation, we prove the following

Lemma 1.6.1 Ifs E R is such that

(1.30) pi(rq—p)s-1-2 ceps-1 rqs +1< NT

(rq — p)s + 2_

1 if/3 >0then we have Q(s) < 1 5/4 if #<0

aNrfor some t i E ( (IN, + 0 ,1). It follows that IQ(s, 01 is maximal at either to or t 1 . Now

aNr 13 +

^13 r

t1 > 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^aNr

aNrSince 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 value

rqs +1

rqs +1(rq — p)s + 2

TProof: 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 + 0

rqs +1?

-^( /3 )(r-p).51-2

(rq — p)s + 2^aNr

-1

> 1(2(3, ti) I

1Q (s ) < 1+ 1131(7. g -O s+ 1 aps(1.31)


implies

1Q(s,to)1 <

rqs +1(rq — p)s + 2

0^

Nr

(r q-p)s-F2 aps-1Now by (1.30), we have^ <

_rqs +1

(rq — p)s + 2

-1and hence

so thatrqs + 1

-max1Q(8,01 <te[om^(rq — p)s + 2

and thus Q(s) < 1 as required. If, however, # < 0, we note that

max 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 +1

and since (1.30) yields 101 <aNr^1#^

1 NI

rqs + 1 - -1we have(rq — p)s + 2

rqs +1(rq — p)s +1

is a monotone increasing function in x (for y constant) and from (1.6), we

rqs +1(rq — p)s +1

xSince

,L Y

have that > 4, and the result obtains.^ ■

Lemma 1.6.2 Ifs E IR satisfies (1.30), then

E(s) <{

23/20 if 13 > 0

1 if 13 < 0

Proof: If # > 0, then

max1E(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 )=

p )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 Ces

rqs +1(rq — p)s + 2

and thus

(1 + ^((rq — p)s + 2) ((rq—P)s+2) • (ps — 1)(P8-173 -1

T_8-1 5 (1 +

^

aNr^ (rqs + 1)( r q 3+1)

which, by (1.7) is(ps ^1yps-1) \ps -1

5_ (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) and

E(s) < 1 (/3 < 0) in Lemmas 1.6.1 and 1.6.2 respectively are sharp. Subject to the

constraint (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 s

we obtain stronger results still.

1.7 Common Factors of Coefficients of Pr, and Q„

The arithmetic behaviour of the Pade approximants to various hypergeometric functions

frequently displays a striking regularity. In our case, the binomial coefficients associated

to /3, and Q„ (see e.g. (1.15)) contain large common integer factors. It is, in some

circumstances, these factors that are necessary to attain any nontrivial result.

Lemma 1.7.1 Suppose t is a positive integer satisfying

r ^At ^r ^Bt ^r(1.32)

LAS-B+Ci+ Leld-B+Ci + [A+Bct

+Ci —t 2

and that we define

^/ = max { Ât 1 + ^Bt 1 + r ct^ +IT

LA+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.

C

AtProof: Since x > I, we have A/x < LA + 1, while x < (A + B + C)/t implies

Aix > At/(A B C) whence {A/x} > {At/(A B C)} (where {y} denotes the

fractional 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 Ât^Bt^Ct1 A-I-B-FC }+{A+B-FC } 1/4A+B+C

= 2

and hence we havef_A-t+ f/3-t + LEI > 2Ix' Ix' Ix'

and thus, if i is an integer, 0 < i < A, either

A

(1.33)

Or

(1.34)

In the first instance,

{-B }+{}?1x^x

C+^1 +X^X^X

[B : ii LB]

{xi] =1

so that x divides (B + i

. ) (since ord x (n!) = E[nIxi]). If, however, (1.34) holds, we mustzî=1

have {A/x} {i/x} and thus

or equivalently

{ A^ + fel =

x

which is > 1 by (1.34). It follows that

whence x divides

A 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 n

We use Lemma 1.7.1 to prove

Lemma 1.7.2 Let e > 0 be given. There exists an effectively computable mo such that

if m > mo we have

G(c, d) > L(cl d) (1- E)dm

Proof: 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 0 and mo such that if

m > 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 xEIt

frqc+ d>flexp ^ dO(c d, t) — •E Y7 .

2tThis is

> 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 Theorem

The 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-1 rqs +1(rq — p)s + 2

E(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 implying

that Pn and Qn are polynomials in z whose integer coefficients contain a common factor

exceeding L(s)(1-el)dm. Hence we may write, for such m,

En(a—)3 )Nr

Q„ ( 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 have

rqs +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 that

En(-13 )aNr1

<2a'N-rn-P8L(s)(1-eodm,(1 . 39) 101(r9-0cm-1-2n-l apcm-2n-1N-r(2n+1)

for 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-

)—din

where C1 is a computable constant, m > m 2 and M is an integer ("translated" from M((aNr

P

+ constant,

by the difference^- a'''. 13 ( rq- P )cm H (pcm, (rq - p)cm, - 0 / (aNT))). NowN

taking 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 that

k) 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 there

N


1.9 The case 11(1 --1-,3/N) k il

In the next few sections, we will be primarily concerned with utilizing somewhat simpler

forms of Theorem 1.4.1 obtained by specifying certain parameters. With the situation

above in mind, we let

ln 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, proving

Lemma 1.9.1 Under the conditions in (144 we have

E(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)-1

E(s) < (1 + --)^< 1

ln NSince (1.41) gives , < ^

/V ln 4N1

2N^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)^ l

ln Nn 4N

V2N > 50

i.e., for all N > 861. Now, if N < 860, then we check the first inequality in (1.42) and

find that it holds unless /3 = 1 and 7 < N < 49. Noting that E(s, t) is increasing in

s, we choose s =^,32

and compute the remaining cases from (1.18). Since each has

E(s) < 8/e 2 , the result follows as stated.^ ■

We use this lemma to prove

Theorem 1.9.2 If and N are positive integers satisfying 13 <In 4N

is an effective constant ko = ko (#,N) such that for all k > k o , we have

k(1

N > (4N) -0 1"k •

In N

Proof: If we define

.R8)=1[3+2111/2'92 — 83(c2s(s+ 1)18+)81- i

then differentiating with respect to s, we find that f is maximal for s satisfying

(S + 1 y = e 2

—1s

8 8 22 ^1^ 1

i.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)/4

and 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 >^N2/32

and (1.44). It follows, then, from Theorem 1.4.1, Lemma 1.6.2, and L(s) > 1, that for

k > ko , effectively computable,

(1.45) > (4N) -kis

which, 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 ensures

that 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 oo

and so if e > 0 is given, we can achieve a bound of the form 1.1.2, for large enough N. We

will discuss the ramifications of this in later section, but mention only that the methods1 )k

of Beukers and Easton yield just a lower bound for (1 + N— of e -k as N —> oo. We

next modify Theorem 1.9.2 so as to provide a bound independent of N, akin to the results

of Beukers and Easton (1.1.4 and 1.1.6 respectively). Along these lines, we have

Theorem 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) - k

Proof: 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 largest

values for s such that (1.20) holds. In each case, we have

(1.47)^ (4NQ(s)L(s)-14-')lis < 2.85

for sufficiently small s, and the result obtains. For example, in the worst situation, we

have 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 with

4 < N < 51.^ ■

To obtain explicit rather than just effective bounds we must bound a number of constants

and perform a sizeable amount of computation. We have

Theorem 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) holds

for all k > N/2. If k < N/2 then (1 + N—1 )k < e 1 / 2 and so

(1 + ...N )k

or

1 +(^7v- ) k 1 )kkN

= (1 + -N- — 1 > — > 3-k

> 0.3512

whence (1.49) holds for these k also. For N with 4 < N < 728 we compute explicit

ko = ko (N) beyond which the desired bound obtains and then, by checking the remaining

values of k, complete the proof.

We take, as before, M to be an arbitrary integer, c and d relatively prime integers

with c > d > 1, s = c/d, 6 an integer with 0 < 6 < c, m a positive integer and n = dm

or dm —1. If we define k = cm — 6 and

^

(1.50)^ lE72(-1 / MI < 2G(c, d)Nn -s+ 1

then from the proof of Theorem 1.4.1 (where we do not apply Lemma 1.7.2 as is done in

that proof) we may write

^

(1.51)^(1 + Tv-1 )k — M(N + 1) -6 > D4 G(c, d)(4Q(s)N) -k/ s

where D4 = (2D2((N + 1 )(4Q(S) • N) 1 i s ) 81 1 . The inequality (1.49) then obtains if the

righthand side of (1.51) exceeds ric (since the choice of M was arbitrary). Suppose

N > 729, c = [NV3] and d = 1. We show that (1.50) is satisfied for k > N/2. Now the

first inequality in (1.44) yields

lEn (- 1/N )I < D3 (2s 2 )m.

To bound D3, we suppose n = 771 and hence that

1Ji = I t(1 — t)(1 + t/nr 2 dt.I

Now

E(s, t) = t(1 — t)(1 + t/N)- 1

which implies J1 imax1E(s, 01 < 1/(1 + t'/N) < 1 where t' maximizes the functiontE[om

t(1 — t)(1 + t/N)' on [0,1]. It follows that D3 < D1 < c/(27). The case n = m — 1 is

similar, with a sharper bound for D3. We may conclude, therefore, that

jEn (-1/N)I < Ter(2c2)m.

Since G(c, d) > 1, the above implies (1.50) provided m is such that

N > c Nc_ i

2c2^— 7r

or equivalently,

(1.52) m > ln( cNc-1 Win (—N ).—^7r^2c2

Now 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 bounded

above by 9c/2 and so if k > N/2, it follows that m > N/(2c) > 9c/2.

Now by Lemmas 1.5.2 and 1.6.1

Pn (-11N)1 < D2 • en

and we wish to bound D2. If n m, we have

\/1 =i

o to-2 (1 — t)(1

(N + 1 )t) dt

2 + (1 — c)/N— c

2c3 — c

and

max IQ (c, t)

This last quantity is

Q(c, cc + 11 )

181 [c-F 111>

182^2

( 2 ^(c ^( 2c ^— 1cl-lc+1)^c+1^N(c+1)) .

since c — 1 < N/91, whence

1tmepaxIQ(c,t)I>

2c2 .,i]

Thus

1 r D2 <^VC2 1•

(^ 2C2 =

27c3 2

c)— ^7r

Similarly if n m — 1, we have

c2

c2 — 1 •

1D2 < 27r

1^112c2 =

c2 — 1 c^7rc2

c2 — 1

and in either case, we can write

IQ.(-1/N)I < 4'n.

The result, then, will follow from (using Lemma 1.6.1)

(3/(4N)'/c) k > 8N(N +1)c-1

or equivalently

k > ln(8N(N + 1)a -1 )/1n(3/(4N) 1 /c).

Now since c = [N 1 /3] and N > 729, we have

ln(3/(4N) 1 /e) > ln(3/(3996) 1 /9 ) > 0.17716

and

ln(8N(N + 1)e -1 ) < ln 8 + c ln(N + 1)

so that

ln(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 intervals

containing primes dividing G(c, d). To estimate the contribution of these primes, we apply

upper and lower bounds on the Chebyshev function 0(x) = Ep<x lnp from Theorem 10

of 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 deduce

explicit ko = k0 (N) for each such N beyond which (1.49) holds and tabulate the results in

Table 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 complete

the 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. ■

Following the same lines, we can now sharpen a pair of results of Easton [26], proving

Theorem 1.9.5 If N > 2 is an integer, then there exists an effective ko = ko (N) such

that if k > ko , then

N + 1cN

) > (2 .65) -k

and

Theorem 1.9.6 If N > 2 is an integer, then there exists an effective ko = ko (N) such

that if k > ko , then

(N + v2-1 )k > (3 .0].) -k

Proof of Theorem 1.9.5: Firstly, for N > 80, we choose s to be some rational in the

interval (1.3791n N,1.380 In N) and note that

2s + 1s + 2

< N2

SO

2 + 11-1 s-iE(s) < (1+ —N1 2 ) 8-I < (1 +I[ 8

s + 2

Since this last function is decreasing for s in the intervals above, we have

E(s) < (1 + r3riy. 1.0008658 .8

and we may check thatir

s + 22s + 111

• E(s) < N2

for 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 + 111

which, by the choice of s, yields the desired result.

If 3 < N < 79, we choose rational s as in our Appendix (see Table 3), estimate the

contributions 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 this

situation 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 that

and hence

1138 + 111ii. s + 2 11 < N3

(1 +1 \ 2s-1^i,^ri3S + 111 -1 )2s-1 .V^< Y + 11. s + 2 J1

Arguing 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 + 1

1/(3s)—k> (3.01) —k

for the values of N in question, k > ko = ko (N). If 2 < N < 25, we proceed as above

with the weakest bound obtaining for N = 2, s = 1.0725, where for k > ko effectively

computable, we have

(2 +4/\k > (3.000353 ...)-k .

E(s) <

■

Regarding 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 much

improved.

By way of another example in this area, we consider the case of 11(5/4) k II. From

Theorem 1.9.3, we have for k > ko , effectively computable

^

(1.53)^ 11(5/4)kll > (2.85) -k

and careful consideration of Theorem 1.4.1 with a — /3 — p — q — r — 1, N = 4 and

s = 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. If

we choose a = q =1, 13 = —3, p = 6, r = 7, N = 2, and s = 0.436, then Theorem 1.4.1

implies, 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 of

obtaining a strong bound from the latter form. For arbitrary a/b, though, finding such

an m is often difficult, sometimes impossible.

Without such a form, the best we can do is the following strengthening of a result of

G. 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 that

k > ko implies

Ra + 11'—,f)k

> (4a1,31) -k •

Proof: We take p = q = r = 1 in Theorem 1.4.1. If a = /3 = 1 the conclusion follows

from Theorem 1.9.3, so we may suppose that la,31 > 2. Now, as in the proof of Lemma

1.6.2, we have

(1.56) E(s) < IQIaN

ys

and

(1.57)Ê(s)Ê(s, 1/2) • 4 = (1 + 13.^1 .2aN

This implies that E(2) < 9/8 and since L(2) = (3/2)1n 3 — it//6 = 2.09807..., we may

note02a 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 have

lim^11-s + 111 E(s)

11- 2 11 L(s ) = 00

(since /3 < 0 implies a 2). By the continuity of 1[s 1

2Ê(s) and L(s) (see Chapter

2, Lemma 2.1.2 for the last), given e > 0, we can choose s E Q such that .s > 2 and

N = (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 > ko

implies

> (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 .

■

As a corollary, we have

Corollary 1.9.8 There is an effective ko = ko (a,13,N) such that for k > ko,

icx ^

k

N )

> (40 2 ) -k .

Proof: If 40 2 < N, this follows from Theorem 1.9.7, while 40 2 > N implies the

inequality is trivial in the Liouville sense.^ ■

It should be noted that a more careful analysis of L(s) can enable a sharpening of

the constant 4 in the above results, probably to somewhat less than 3.

1.10 Semi-Effective Results

If, for a given 7/ > 0, we have a rational alb satisfying

(1.58)^ < b-nm

for some m, then, as mentioned earlier, we can use this information to deduce lower

bounds for 11(0)4 for k > ko = ko (a, b,q,m). In particular, we can prove

Theorem 1.10.1 Suppose I/ E R is such that b < a m o satisfies (1.58), then we can find a

constant ko = ko (a, b, , m) with

II(a/b) k I I > b- nk

for all k > ko.

We call this a "semi-effective result" in that it depends upon the existence (and size) of

m — which is by no means guaranteed. This can be generalized if we have

b < a < Ms-2+271)/(s-1)(1.59)

3211

a#2


for 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))k

for 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 > 1

implies (1.58) has no solutions while ri < 1/2 contradicts b < a 0 such

that

^

(1.61)^ a < b271 ' and 77 > 1 + E

2

and mo with

^

(1.62)^

b€"1° > 100.

Suppose that m is minimal with in > mo and (1.58) and write

(a/b)m = a + —N

so 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) yields

8a8 2 < 16b(1- e )m < brn = N.

Since L(s) > 1, we may apply Lemma 1.6.2 to conclude

• E(2) < L(2) • N

and 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 \k

f2 " - 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 to

conclude that there is a k o = ko (a,b,ii, in) with

Il(a/b) k il > b- '71c

for all k > ko , as desired.^ ■

While the statement of the above result may be somewhat confusing, we note that if

one has an upper bound upon the smallest "suitably large" solution in to (1.58), then

Theorem 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 only

finitely many rationals a/b in any finite interval which do not satisfy b < a < b27?. It

follows, then, that Theorem 1.10.1 may be applied to "almost all" rationals in (1, N) for

N a positive integer. For example, if I/ = 0.7925, we may utilize Theorem 1.10.1 for all

rationals in (1,2), in particular for a/b = 3/2.

1.11 Density Results

In 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, effective

bounds. We first prove

Lemma 1.11.1 Let N > 2 be an integer and e > 0 be given. Then there is an effectively

computable 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 -ek

for 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)p

So with .3 = In pi, it follows that

1 Nps—i^/^1 \plopE(s) < (1 + Np+ i^ )^ 32, (1.64) obtains for such ^Applying

Theorem 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)1c

for all k > ko . Since .--- 0(s), we have

fi lm (Nils [is + 211 iMP -Fl)s))

P-000^11.8 ^1.11^= 1

(remembering that cs = [lnp]) which concludes the proof.^ ■


We 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 prove

Theorem 1.11.2 Let e > 0 be given. Then there exists a constructible set of rationals

S,, dense in the interval (1,00), such that if alb is in S,, there is an effective constant

ko = ko la, b, e) with

11(a/b)kil > b-ek

for all k > ko.

Proof: Take ai/bi > 1 to be any rational and e, S > 0 to be arbitrary. We construct a

rational alb in a deleted 6-neighbourhood of a libi which effectively satisfies the above

bound. From Lemma 1.11.1, we find a computable p o with

(al + -71 k

a l )

(1.66) > brk

for p > Po and all k > k1 = ki(P, a l , b1 , e). Let

pi = max{po,In b1

+1, ln(biS)

} 2}.e In a i^ln a i

Then there is an effectively computable constant ko = ko (a i ,b i ,E, 8) such that k > ko

implies (1.66) with p replaced by pi . It follows that

( D + 1 \k

bi^biar )

> (bli + c )- k

ln bifor all such k. We take a = 41+1 +1 and b = biar and note that pi > ^ + 1 impliesE ln a l

II(a/b)kII > b'k

ln(bi 6)for k > ko . Also since pi >

ln a^ + 2, we have alb in a deleted 6-neighbourhood of

i— a l ibi , as required. Since the choices of E and 6 were arbitrary, the result obtains.^■


There are certainly many different ways to construct sets like the above — here weaP+ 1 + 1

have just considered (for a/b E Q) rationals of the form baP

^with large p. It would

be of some interest to find a dense set in (1, oo) for a given e > 0 with the property that

if 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 2

Contents of Pacle Approximants to (1 — z) k

2.1 Basic Behaviour of L(a,b, s)

As we have seen in the preceding chapter, a careful analysis of the coefficients of the Pade

approximants 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 and

s > 1/b is a real number) defined by (see (1.19) for comparison)

L(a,b, s) expi(E as + 1 0(s, t))

where

0(s, t) = max{e i (s, t), 0 2 (s, t), 03 (s, t)},1^ bs —1^ (a — b)s +1

0 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 + 1

and 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 a

variety of reasons suggested previously. To begin, we prove

Lemma 2.1.1 If a > b are positive integers and s >11b, then L(a,b,^) < oo.

Proof: Since

we have

L(a, b, s) exp(as +1

t ^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 conclude

Lemma 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)^

tê(s,t)). Let E > 0 and s be given

and suppose that Is — 8 1 1 < 1. Then by the proof of Lemma 2.1.1, we can find t o with

tE as + 1 e(s,t)) <612>to^t

and(asi

t +1

0(s i ,t)) <E12.i>to

Consider

tbs — 1 t l + f as +1(a — 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 t

1 Oa — b)s,+ 1

t} = 0}as +1^t as +1âs + 1

Then

tETi(s,..)

În L(a, b, s i ) =Ê (as i

t

+ 1ô(si,t))

and

In L(a,b,^). E ( as +1 t^0(s, t))

tETi (Si 00 )

Chapter 2: Contents of Pade Approximants to (1 — z) k^45

and

IlnL(a,b,^) — lnL(a, b, 8 1 )1(asi + 1 z (as + 1

O(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 that

Ti(s, to) C TI(si, to) c Ti(s, to) U T2(s, to)

z

and so

as + 1 0(8,0 )^z as i + 10(3 1 ,0)

t^ ttE_L (s,to)^ teTi(si,.

4 o)

=tET1(s,to )

as + 1

0(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)^(asi

t 1^0(s i ,t))

t= (as + 1^as i +

1 )^(et(s,t) — Oi(s i ,t))

< 4to

for 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 that

Is — s i l < St• impliesasi + 1

Cl(si ' t) < 4T0

and thus if 6. = min{1, 61 , 8t (t < to ), (5t.(t < to )} and Is — s 1 I < S, we have

Iln L(a,b,^) — ln L(a, b,^< e

as desired.^ ■

Chapter 2: Contents of Pade Approximants to (1 - z) k^46

2.2 The Case L(1,1,^)

The behaviour of L(a, b, s) is perhaps a little more predictable than may be obvious from

a 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+

have

Theorem 2.2.1 L(s) satisfies

lim L(s) = /e11 = 1.76387 • • •8-4•00

and

Theorem 2.2.2 If 2 < .5 < oo, then

1.66303 • • • = 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 the

related 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 bound

to approximately V1.048/N. To prove the above theorems, we perform a fairly careful

analysis 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^t

where

1 ^e(s, t) = max^ s-1. tj+i { t i +11Ls+1^s+iand t satisfies

(2.68)^2 Ls + LI

t 1 + Ls + t

.j. t_ 2.

s - 1^1

for

e(s,t) = {A(s —1) + B —1s — 1

if B < s

1if B > s

A + 1


Clearly (2.68) is equivalent to {s + 1

} > 1/2, and

0(s, t) =

t if

ts+1 1— s+1

tif

ls+1 1> s+1

Write t = A(s + 1) + B with A E Z and 0 1, A > 0, and 0 < j < A are integers, then

i) 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)]•

Proof: i) Suppose .5 E (2n + j/(A + 1), 2n + (2j + 1)/(2A + 1)). Then the values of t

contributing to LA(s) are

t=(2n+1)Ad-n+j+1

t.(2n+1)A+n+j+2

t = (2n + 1)A + 2n + j + 1

and 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 gives

Yi 1^Y2 1LA(s ) =

^t=si ^t=s2

where= (2n + 1)A + j + n + 1

Yi = (2n + 1)A + j + 2n + 1

x 2 = (2n — 1)A + j + n

Y2 =-- (2n — 1)A + j + 2n — 1.

Applying Euler-Maclaurin summation yields

A +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)Y2

Since

— 1= (2n + 1)A + j + n > (2A + 1)n

and

Y2 = (2n — 1)(A +1) + j (2n — 1)(A + 1)

Chapter 2: Contents of Pacie Approximants to (1 — z) k^49

we can write

2n(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)Y2

n + 12n — 1

(x 2 — 1 )(Yi)

and hence L'A (s) < 0 provided n > 3 (remembering that A + 1 — j > 1). If n = 2, then

4A + 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. If

n = 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) < 0

while A = 0 implies j = 0 and

1 ^1^1 ^1^1^n

L'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)],

that

S + 1^1^2n-1/^S + 1^S — 1

LA(s) (2n + 1)(A + 1) + j A +^(2n + 1)A + j + + 1 (2n-1)A+ j + il

and hencev2

L'4 (s) = E 1 -^

1

t..1+1 t^t=x2+1

where x1, x2, Yi, Y2 are as previously. From Euler-Maclaurin, we can write

VA (s) > ln(1 +A + j +1) 1(2n(n +1)(2A +1)(A +1) (A +1 + j) 2 )

x1Y2^2^xlylx2y2

and the second term on the right here satisfies

2n(n + 1)(2A + 1)(A + 1) — (A + 1 + j) 2^n(n 1)(2A + 1)(A + 1)

2xiyix2y2^ x1y1x2Y2

Chapter 2: Contents of Pade Approximants to (1 — z) 1'^ 50

This in turn is less than2n(n + 1)

(2n — 1)(2n + 1) x1Y2

which implies that L'A (s) > 0 provided n > 2. If n = 1, L'A(s)^13A +3+i > 0 as

required.^ ■

For the intervals of the form [2n + 1, 2n + 2], we have

Lemma 2.2.4 If n > 1, A > 0 and 0 < j < A are integers, then

i) 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 are

t =-- (2n +2)A+n+j+1

t.(2n+2)A+n+j+2

t = (2n + 2)A + 2n + j + 2

and thus

s + 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 the

second. ■

The preceding results lead us to


Lemma 2.2.5 If n > 1 then

i) L(s) is maximal on [2n, 2n + 1] for s = 2n

ii) L(s) is minimal on [2n, 2n + 1] for s = 2n + 1

iii) L(s) is maximal on [2n + 1, 2n + 2] for s = 2n + 2

iv) 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 similar

fashion. By the definition of LA(s) and Lemma 2.2.3, we will be done if we can show

that

(2.69)^LA (2n + A 1

^LA (2n + A +1 )

holds for all j = 0, 1,^, A. Now

j 1 (2n + 1)(A + 1) + j (2n — 1)(A + 1) + jLA (2n + A + 1 )

while

A + 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 1

t=x+1 t^y t=y+1

where x = (2n + 1)A + j+ n +1 and y (2n — 1)A + j+ n. Define the auxiliary variables

u 2A + 1

v = A + j + 1

and consider the function

(u+1)n+v 1E •un + v t=und-v-1-1

f (n)

Chapter 2: Contents of Pade Approximants to (1 — z)k^ 52

Apply the Euler-Maclaurin summation formula to yield

n1 ( 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) where

ii (n ) =-. n n^1( 1^ 1 ln(l: +

un + v )+ 2 un + v (u + 1)n + v )un + v

and f2(n) < 0 with

1 if2( 7 )1 .^1 ( 112 (un + v) 2^((u + 1)n + v) 2 1

)

.

It follows that

f (n) — f (n — 1) = fi(n) — fi(n — 1) + h(n) — f2(n — 1)

, in faZ) dz + f2(n) — f2(n — 1)n--1

and 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 ) 2

SO

(2v — u)(u + 1)(n 1) 2 + 2v 2 (n 1) + v 2min fl(z) >

zE74^ 2((u + 1)n + v) 2 (un + v) 2

Since1(^1 1^) 1 (^(2u + 1)n 2 + 2vn

12^(un + v) 2 ((u + 1)n + v) 2 ) — 12 (un + v) 2 ((u + 1)n + v) 2 )

from (2.71), we conclude that

f(n) — f(n — 1) > 0

and hence (2.70) obtains, provided n > 2 (remembering that our definitions of u and v

ensure 2u < v < u). If n = 1, (2.70) is just

1^1>

x x + 1 — 0

which 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 that

lim L(2n) = lim L(2n + 1) = ir/e.n.--.co^n-4.co

We 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 1

B=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)

=2

and


so that

We have

2n 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)^n

and 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)^n

Now

ln ((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().= n

From the preceding, it follows that

A +1 1 +0 ( n1)LA (2n) = 21n( 22AA ++ 21)


and hence

Now

n—colim L(2n) = exp(E (21n( 2A + 2 )^1^)).

2A+1^A+1

00

A=0

2 E lrq2A + 1

N 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 that

N

A d=0 A + 1=1nN+-y + 0()N

N^2A + 2^1^1E (2111( 2A + 1) A+1 ) -- 1n7r —7+0(y ).A=0

Thusct° (21n( 2A + 2 ) ^1 )

= lnr _

7A=0^2A + 1 ) A + 1

whence lira L(2n) = 7r/e, as desired. If, however, ,s = 2n + 1 thenn---*oo

2n1-1^2n + 2^2n (PLA(s) = E

B=n+2 n + 2)A + B 2nA + B —1))

^2n-I-1^ 2n-I-1^1^12

1

^

= 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 situation

1LA(2n +1) = 21n

(22AA ++ 21) A +1 4. 0(771.).

=1n

Hence lim L(2n + 1) = 7r/0 and thus also lim L(s) = 7r/ell (by Lemma 2.2.5).^■n--).00^ s-400


We may also now prove Theorem 2.2.2.

Proof of Theorem 2.2.2: Again, we will handle the case

L(2) ? L(s)

for s > 2 and note that the lower bound on L(s) obtains from a similar analysis. By

Lemma 2.2.5, it suffices to show that

L(2n) > L(2n + 2)

for n = 1, 2, ... which follows from the like inequality with L replaced by LA. Now,

^2n^2n + 1^2n — 1^LA (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 that

LA(2n) — LA(2n + 2)

x2Y3^ )= (2n + 1)1n( Y1x2x3Y4 ) + 21n( Y2X3

x1y2y3x4_ _ 12 )

— [(2n + 1)(— — —1 ) — (2n —1)(1x i^y 1^

x2^y

1 _ 1^X4 ^Y4

— (2n + 3)(—^) + (2n + 1)(-1 — 1 )1

+ u [(2n +

x3 y3

1)(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/ 4

Chapter 2: Contents of Pads Approximants to (1 — z) k^

57

wherex l = (2n + 1)A + n

x 2 = (2n — 1)A + n — 1

x3 = (2n + 3)A + n + 1

x4 = (2n + 1)A + n

yl = (2n + 1)A + 2n

Y2 = (2n — 1)A + 2n — 1

y3 = (2n + 3)A + 2n + 2

y4= (2n + 1)A + 2n + 1.

Since, if we expand the arguments of the above logarithms, we verify that

Y1X2X3Y4 < 1X1Y2Y3X4

and Y2x3 > 1x 2y3

we can find positive z 1 , z2 with

yix2x3y4 =x1y2y3x4

2X 3 and^_ 1 + Z2.x2y3

It is thus possible to bound ln(Y1X2X3Y4) below by—z 1 — 34/5 (assuming that z 1 < 1/6xi.y2y3x4

which follows from A > 1 and n > 2) and ln( Y2x3 ) by z2 — 4/2. By the properties of

Euler-Macaurin summation, then, we can write

LA (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)(-1

X4 Y4 )]

^1^1^ 1+ -2- [(2n +^1

—^— (2n — 1)(A- — 12 )1^y1

/ ^1—2- \— (2n + 3)L — ) + (2n +^— )1

^

x3 y3^X4^Y4

^

1^/ ^1 \--6-0-(2n +4^y4

Now expanding the right hand side of this inequality, with the aid of Maple V, yields

^

a 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^58

Moreover, collecting the 225 terms involved in f (A, n) and noting when the leading term

dominates, we have that

f (A, n) > 589824A 14 n 14

provided A > 1 and n > 2, unless (A, n) = (1,2).

If A = 01

Lo (2n) — Lo (2n + 2) =2n(n + 1)(2n + 1) > 0

while if n = 1,

LA(2) — L A (4) = ^5A + 2 > O.(3A + 2)(5A + 3)(5A + 4)

Supposing n = 2 and A = 1, then

19 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 unbearably

so. One can obtain upper and lower bounds in much the same manner as in the previous

section and in fact prove that

aa/(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 respectively

lira L(2,1,8)=2 3/ 2 • 3 -3/4 .17r/e'Y = 1.64792 • • •s.00

and

lira 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^ 59

For 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 distinct

from those with a > b > 1, in that the latter is always an algebraic multiple of \br / 67

rather than of 'r/e1'. The explanation for this, however, is by no means readily apparent.

Chapter 3

Connections to Waring's Problem

3.1 Introduction

It is well known from work of Dickson [22], Pillai [46], Rubugunday [50], Niven [45], et. al.

that if we define

g(k) = min{ m : all positive integers are sums of at most

m k-th powers of 1, 2, 3, ... }

then if

(3.1)^ 3k — 2 k [ ( 3 /2) k ] < 2k — [ ( 3 / 2)1

we 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 opposite

of (3.1) is true, that is

3k — 2 k [ (3 / 2 ) k I > 2k — [(3/2) k ] .

Then we have

0 < ([(3/2) k ] + 1) — (3/2) k < [(3/2) k ]/2k < (3/4) k

so that 11(3/2) k li < (3/4) k . From Mahler's result (1.1.2), this implies that only finitely

many 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).

60

Chapter 3: Connections to Waring's Problem^ 61

To place the Ideal Waring Problem in a somewhat more general context we consider

the following.

Suppose S is an increasing sequence of positive integers. We call S a basis of order

h if every positive integer can be written as a sum of at most h elements of S and at

least one positive integer requires the full h terms. If we further define the Schnirelmann

density a(S) by

o-(S) = inf {#{ s : s E S, s 5_ n }/n}riEN

then 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 that

3.1.2 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 Waring

problem, 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-th

powers. Since a(S) = 1/(N — 1) > 0, we have by (3.1.2) that Si(\/; ) form a basis and it is

a natural question as to its order (the case N = 2 of course corresponds exactly to the

standard Waring problem). If, in analogy to the above, we denote this order by gN(k),

we prove

Theorem 3.1.3 If N and k are positive integers with 4 < N < (k +1)(k-1 )/k —I then

gN(k) = Nk + I NN + 1 \k1 — 9\^I J^`•

This is achieved through use of the Hardy-Littlewood-Vinogradov method, bounding

certain exponential sums and an ascent argument due to Dickson.

3.2 Asymptotic Theory: Notation and Definitions

In this section, we concern ourselves with proving a slight generalization of a result of

Vinogradov's. To be precise, we show

Theorem 3.2.1 If k > 6 and M > e446/c6 are positive integers, then there exist s integers

x i , x 2 , ... , x s with

i) s < 6k In k (31n 6 + 4)k

ii) xi > miAsk3) for i = 1, 2, ..., s

iii) M = xi + 4 + + X .sk

The upper bound for s can be strengthened to order ti 3k In k or ti 2k In k, but, in

both cases, these induce a lower bound for M which is too large for our purposes. The

difficulty chiefly arises from estimating the size of the implied constant in

y(a) < qe

where q(a) is the number of solutions to the congruence V k^a (mod q) for v and a

integers in [0, q).

In what follows, we will utilize the Hardy-Littlewood method a la Vinogradov and

attempt to keep notation as "standard" as possible. In particular, we use as much

technical 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)t

and adopt the convention that 0 (with or without subscripts or superscripts) is a complex

number satisfying 101 < 1. Let

9)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 constitute

our 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 z

with 1z1 < 1/(qT), q E (P 1 / 2 ,4

We now construct a pair

terms in our upper bound

(3.5)^P1 =

of

for

4

"exceptional

s in Theorem

P2 =2

1i'

[-2P

1

sets"

3.2.1.

which will

Let

• • • , Pe =—

contribute

1^_[-2 P

i vt-1

the majority of

1 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 elements

of 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,


With these definitions in mind, we further defineY

^(3.10)Û(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=N

the principal object of our study:

fr(M) =LT-1

l-T-1

(A(x))4k (U(°)) 21/ (a)e( --aill)(3.11)

= 1.931 +Im

By expanding the integral for r(M) and making use of

f 1-1/1-^1 if M = 0(3.12) e(aM) da =

0 if M 0 an integer

we have that r(M) is the number of representations of M as

X +... X zik +72+U l + y k V

where N < x i < P, u and u' E U, 1 < y < Y, v E V. If we therefore show that

1.19111 > 1 1m1 and hence that r(M) 0 for all M sufficiently large, if Re > N then we will

have in the notation of Theorem 3.2.1

s < 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)=---ê(axkÂ(q) =^(S(a, 0/04ke( amiox=1^ a=1

(a,q)=1

and

6 = E A(q)q=1

(3.14)^ /(Z)= f e(zxk ) dx

J(M) = 1:(I(z)) 4k e(— zM) dz

3.3 Lemmas on Exponential Sums, Integrals and Other Topics

Throughout 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 the

use of the Vinogradov symbol < in order to make our treatment more concrete and to

motivate 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 applying

Rosser and Schoenfeld [48, Corollary 2 to Theorem 1] to get

5(k _ 2)k2k/(k-2)ln^< (k — 1) ln k ^

8k

Remembering that k > 6, the result obtains.^ •

Lemma 3.3.2 E A(q) =^c2p-1 where 1C2I < e i•ook3 C3 > e-4.29k3

Proof: By Landau [36, Theorems 325-326], we have

6 > C3 = fJ b(p) • H (1 - 7)-3/2)1p<C^p>C

whereCO

b(p) = EA(P1)j=1

From James [33], we may write

and C = (1 + k4k )2/(4k-5) .

Hence

b(p) > 1 2—(ord2(k)+2)(4k-1) , p = 2

P-(ordp(k)+1)(2k-1) p > 2

ln(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 that

ln(1/C3) < 4.29k3 .

Supposing that k > 9, we note that

E (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<C

Applying the bound upon the Chebyshev function

0(x) < 1.000081x for all x > 0

from Schoenfeld [52], we conclude that

(2k — 1) E lnp < 1.000081(2k — 1)C.p<C

It 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 3

and 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. Then

M'E e(ax)

x=M

1,

- 2lia l l

where 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. Then

m,^M'E e(+ f (x)) — I m e(+ f (x)) dx

x=M

Proof: 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) =q

where (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 between

the greatest and least values of the function 0(y) does not exceed A > O. If we define

m+x-i N+Y-1

S = E > 7(x)q(Y)e(x(Y))s=m y=N

and putM+X-1^ N+Y-1

E I7(x)1 2 = Xo ,^E 19(01 = Yo, max19(y)1= 9x=M^ y=N

then

< 2.

ISM < (X0 Y0 77((2A + 6)X + 3q)[Yq -1 + 101/2.

Proof: See Vinogradov [58, p.30, Lemma 104^ ■

if 1z1 P"Lemma 3.3.7 1/(z)1 < Z :=

Azi - vîf 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 obtain

2zPk^ 2zPk/(z) = f2z(N 1), O(,0) cos ri3d0 +63) sin 7 r,3 cl,3

i 124N-1)k 0

where 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 write

2zPk^1/2^ 3/2f2z(N 1)k OP) COS 7r# c113ÔP) cos 7 )3 +ÔP) cos r clf3

2z(N -1)k^ 1/22zPk

+ • • • +O P) cos^d3.f[2zPk +112]-1 / 2

Since 0(0) is positive and decreasing, the first integral here is positive and the remainder

alternate between negative and positive while decreasing in modulus. Hence

f2zPk11(,3) cos R- 13 d,3

J2z(N -1)kmax{101/2 O(/3) d,3, fi3:22 2(0) d)3}

f1

o O(Q) di(3 (2z)'

Similarly, after writing

1.2zPk^ 2zPk^2zPkAzov_ i) , ,(#) sin 71- 13 cl13 = 12z(N_nk + 11 2 . . .ÎÔP) sin ir/30

[2zPk]

the above argument yields

1 2zPkOP) sin 70 di3

i2z(N-1)k

(2z)-11

and thus

1/(4 < ((2z) -2 ' (2z) -2/ 1 / 2

5_ \(2z) -11

With 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 the

inequality

Xi^. . .^X4^k ^.31

in integers x i > N. Then

ir(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 of

solutions to

in positive integers x i , ..., x r while krN (M) denotes the number of solutions to the same

inequality subject to the constraint xi > N, i^1, 2, ..., r.

Clearly, we have li.v(M) < kr (M), but also

N -1(3.15)^k,NM) ?_ kr (M) — r E kr_ i (m — sk).

3=1

Now by Vinogradov [58, p.22, Lemma 3], we have

(3.16)^kr(M) Tr M" — OrM"' , 0 < 0 < 1

(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 hence

TrM" - 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 Arcs

We seek a lower bound upon 1/9A1, the contribution of the major arcs to r(M). In all

of what follows in this section we will assume that a = al q + z E 931(a, q). By direct

substitution, we can write

(3.17)^ 1-931 = E f^e(f(a))4ke( aMi )dau,tt ,v,y

for MI = M — u — u' — vy k , where u, u' E U, v E V, 1 < y < Y and so from (3.9) it

follows 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 in

order to obtain an asymptotic formula. We will show first

Lemma 3.4.1 f(a) = S (a, q)

I(z)d- C5 q, where IC5 1 < 4.q

Proof: If we write x = qt + s for s = 0, 1, ..., q — 1 and t running through all integers

in the interval ((N —1 — s)Iq,(P — s)/q], then

Pf(a) = E e(axk )

x=Nq--1

. E e(ask I q)Fs (z)s=o

with 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 Lemma

3.3.5 to get

(3.19)^F8(z) =ê(z(qt + 3) k ) dt + 40, 101 < 1.1(1(P-3)1q

V-1-01q

(the constant 4 occurs because the endpoints of the interval defining t need not be

integers). Hence

Fs(z)Î(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 prove

k^(S(a,q)^co-34-uz^1.îf, <Lemma 3.4.2 (f (a)) 4 =^1-(z))4k^4k-i, wh ereê1100k3 .q

Proof: From Lemma 3.4.1, we deduce that

,q)^(Z))4k(f(a ))4k (41c)(S(a,q)

I(z)) (C50 4"^(S(a

^41t1qî=0

and so from Lemmas 3.3.1 and 3.3.7, if we let i = 0, 1,^, 4k — 1

( c1

di = 4.kî 1 c5 14k-tztok-i-ivz^1

we have4.tic-,-1 (4k) ( S(a,^\i^4k-i2_,^•^(z)) (C5q)1.0^q

4k_i< E di .

i=o

Sincedt+i^4k — ^Zq-(1+v) >^Zq-(1+0diî^1 105 1^— 16k

and q 1 for all i = 0, 1,^, 4k — 2. If, on the other

hand, Z \12- 1z1 -v, then Z >^rye > P'qv and again since q < P112 , we conclude

that di+i /di > 1, i = 0, 1,^, 4k — 2. This implies that4k-1E di < 4kd4k_1 <i=o

and from the preceding, the result follows with IC61

We are now ready to estimate the contribution of the integral associated with a single

major 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))4kq

for IC71 < en.o4k3

Proof: 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^ q

cr647-3-1-vz4k-le(_am1) dz

(S(a,q)\4kq • J(Mi)e(—aMi/q)

liqrC6q -3-Pv z4k-l e(_aMi)dz

fl/ qr

0.0 S (a, q) I (z))4k e(— a MI ) dzfv q, q

fol/Tr (S(a,q) I (z))4k e (_ami ) dz.q

< e l1 .00k3 .^ ■

Now

f1/QT liqr z4k-1 dzCoq -3+ 'Z4k-l e(-ciMi )dz <2106 1q -3+' f

<21Colq-3+u (10P-k

P4k-1 dz fp_k(\fi z-v)4k-1 dz

22k-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 3.3.7

and (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 <F

Proof: Let M0 be an integer satisfying 0 < M1 - Mo < 2Pk'• Then

IJ(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-k

e lk .

and this is equal to 2-V- 7r(4k1-1)(Mi -Mo )P2k . Applying Lemma 3.4.3 with Mo in place

of Mi., a = 0, q = I yields

L iir (i(z))4k e(—zMo) dz = J(M0) C7P3k-1

and from IJ(Mi) - AMOI < 21.71- (4k 1)(Mi - Mo)P2k with IA - Mo l < 2Pk-u we

have

[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- y

It follows, therefore, that

LT ( f (a)) 4k e(-aMo) da^(MI) fiiir i /r(f(a)) 41c e( -aMo) da(3.20)

+60(C7P'l 412- 7r(4k 1))P3k-v

where a E 97t. Fixing H^[Pk-"], we let M' and M" range over 1, 2, ... , H and set

Mo = M1 -^- M" (so that M1 - Mo < 2Pk- '1 holds as required). We sum the above

equation (3.20) over M' and M" to yieldH H

E E ki ( f ( a )) 4k e( a (Mi - - M")) daAv=i m"=1

H2 j (mo^H^(f (a))4k e(_ct(mi^M")) da

mi=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")) da

m"=1 ir

1-1/1-^1^2<^p4k ^

^

2iiaii^da

= 1 p4k I 1/2 a -2 da21

= -2P4k (3- - 2) < kP5k-1

and since H = [Pk-"] > 1 Pk', we have that the above is < 2H2 P3k-1+21/ . This

implies that

H H

E 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-v

Now the above is just the number of representations of M i. as

M' + 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=i

where k(M) has the same meaning as in Lemma 3.3.8. From the lemma, we have

Fo. v yik^c4w4k(W) ^r(5)

and (3.22) becomes

1 F(1 -I- v)4k ((A^_ 1)4 _ (4-1^— H — 1) 4 )M'=1

H^(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 to

4 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 obtain

j

^F(1 + v)4k^+ ■v8p3k-v

(M1) - r(4)^Ml

f (f(a))4k

e( —"I ) da = E E f (f(a))4ke(—"1)dz

q^liqr

931^ q<p1/2 a=1^-1/9T(c 1,0=1

( s(a, oyik r( i + vylk m.3)^F(4)^e(—aMi/q)= E

q<pi,2(a , q)= 1

q)=1


where

!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 in

f(f(a)) 4ke( —cf-M) da.

Without further ado

Lemma 3.4

where 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)=1

q

+ E E cr7q-lp3k-1.

Now by Lemma 3.4.4, this

q<pi/2 a=1(a,q)=1

(S (a, q) )4k c8p3k , e( a mi / 0a=1^q

q))4k

(a,q)=1

q

+ E E

+ Eq<pi,2

c77 -1p3k-1 .

q<P 1 / 2 a=1(a,q)=1

The 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)=1

5_ ci4kIc81 E c 3P3k- vq<pl /2

< C1k IC8 I( (3)P3k- v

while the third is

< ^nip., < n1p3k_1 /2 .q<pi,2

The first is justjr(1+ v)4k v > A(q)

r(4) q<pi,and hence by Lemma 3.3.2, is equal to

ro. + 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(where

1C91<ci4k 1GIC(3)+ Ic71Pu-1 / 2 + 1c2Ir(1r+(4)v)4k P-1

< e l1.06k3, for k > 6

■


Putting 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,y

and Lemma 3.3.3 yields

F(1 + v)4k1/9311>^1

2

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)k4k

where Co = C3 F(1co) / , provided P >^Since F(1 x) > 0.885 for all4^ Cio

positive 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 Estimates

Most of the bounds utilized in the minor arc case are in fact trivial, but that is not to

say that the estimates are in any sense simpler. In fact, the deepest result on exponential

sums that appears in this proof, namely that of Lemma 3.3.6, plays a crucial role. We

u,u 1 ,v,y

have

liml = fm (f (a)) 4k U(o) 2 V(a)e( — °M)da

Lif(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)

ctE

and 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] ,

and

Y'(Y) = oY =ây + zqy

q

so 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 take

ii(y) = ^1îf y is a k-th power

0ôtherwise

and

^y(x) = 1îf x E V

^0 ôtherwise

then 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)1/2

and again utilizing that a E m (and so P 1 / 2 < q < 2kPk-1 )

117(a)1 < V(0) 1/2pk/2-1/4.

Therefore,

and so

(3.29)

pk+k/2-1/4

C10 V(0) 1 /2 U(0) .

Now, by our construction of U and V, and the uniqueness of the elements so derived, we

have

U(0) = P1 • P2 • • • Pe

V(0)^• R2 ... Re • [P 1 /2k ]

and we can readily prove by induction that, for j^1, 2, ... , t,

(3.30)^ > 3—k p(1-03-1

and

(3.31)^ Rj > 3 -kR(1-03-1

(this last necessitates the first inequality in (3.3)). It follows that

U(0) > 3 —"Pk—cr

and

V (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/ 4

V(0) 112U(0)P4k+ki 2-1 / 4

1-19A1

< ^CioV(0)U(0)2P3k

1^1< <9.5k^6k

(3.32)

so that

(3.33)^

U(0) > C11 kp -46

V(0) > Ci2pk-112-46k-Fav14-1-1.12

where

= 3-kt > e-k3

and

C12 = 3 -kt • 246-k-1 > e -k3

Hence

Ilml^pk+k/2-1/4-k+1/6k-k/2+1/44-1/12k-o/8k-1/4k

-19311^ Cl0*C111•C122P-a/8k

ryV2Cm•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 from

9R

P > e445.36k5

and hence from (3.3).

lIm l

3.6 Dickson's Ascent Argument

In this section, we will study the representations of integers as sums of elements of the

set

^

(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 N

Suppose N < (k + 1)( k-1) / k — 1 and write [a, E 4 ) (m) (or (a, b) E S (A,) (m)) if every

integer 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 of

SN (k) required for representations of "small" integers before applying an ascent argument

to 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 =1

has only the solution N M k = 2.

Proof: Suppose that

^

(3.35)^ (N 1)k MNk + 1

where 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) MNk

and 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 4

or

^

(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) implies

that N = 2. Since 3 belongs to the exponent 2' modulo 2 k , we must have 2' dividing

k, 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 priori

upper bound for k. From (3.39), we clearly have N I k and if we suppose that Na k,

then since

ordp ( i ) > ordp k — ordp i (p prime)

we have

It follows that

and so if i > 2,

ordN (k) > a — max(ordP i) .•

ordN ((k)

Ni^m> a — ax(ordp i) + i•Pli

p odd

p odd

ord 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^84

We will also use

Lemma 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 t + Nk and choose i such that i k < n — £ < (i + 1) k . Thenik E Si(N" ) and since, by calculus

n — £ — ik < k(n — t)(k-1)/ k < kn (k-1)/ k

we have (3.40). If, however, n < .e + Nk , take i = 1 and write n = £ + m (so that1 < m < Nk ). We conclude

k(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 have

Lemma 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 ) .

■and

Lemma 3.6.4 (aNk, (a + 1)Nk) E SP(E) where E = max{a + # — 1, Nk —,3}.

Proof: 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 in

Lemma 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 satisfying

L + 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 obtain

Lemma 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 i

k+i ki. /\T-F 1) i

, 1.

k and n =^ , t follows that

r ,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)

Our main ascent relies upon the following result, which is essentially a variant of a

theorem of Dickson [19, Theorem 12].

Proposition 3.6.9 Let £ and Lo be integers with

Lo > t > (N +1) k , v = (1 —11L0 )1k and vk Lo > 1

Then if for t E N we define L i by

(3.41)^In L t = (k k

1 )t (1n Lo + k In v) — klnv

and 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) is

equivalent to

Lt = (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). In

general, (3.41) yields

Lt+i = (v_L t ) ki (k-1)

and the result obtains by induction upon t.^ ■

3.7 Proof of Theorem 3.1.3

Assume 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 to

N < (k +1) (k-1)1k —1.

If we choose t large enough that

(3.42)^ Lt > max1N8k3, e446k6 1 = e446k6

then 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 write

ln 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 this

choice of t that (1, L i ) E 4 1 (Irk) ) where

4)= 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 + 1

and need

(^< pk) = Nk + a _ 2 .(3.43)^E-1-a+t-1-(k—N)rN+2)k1../V+ 1 i ^N

If E = a + /3 — 1, then (3.43) becomes

^

(3.44)^E+a+t+(k—N)1N+21 —N k <-11. ( 1 \ r + 1

while E = Nk — )3 implies the inequality

^

(3.45)^ t + (k — N)[( NN ++ 21 )kJ — # < —2.

To prove that (3.44) and (3.45) obtain for all N and k satisfying

4 < N < (k + 1)(k-1) /k — 1

we employ Theorem 1.9.4 to deduce

3-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)1c

N )^3^ N + 1 )

and hence is < —1 for N and k unless

i) N= 4, 6 < k < 34

Or

ii) N= 5, 8 < k < 11

Additionally, we bound the left hand side of (3.45) by

6k ln k + 7k + (k — Ne + 2 )kN + 1

( N3 )k

which is < —2 for all values of N and k under consideration except

iii) N= 4, 6 < k < 32

and

iv) N --= 5, 8 < k < 11

Checking that (3.44) and (3.45) hold for the cases i), ii) and iii), iv) respectively, we

conclude the proof of the theorem by noting that M = aNk — 1 ct sr (Nk + a — 3) and

thusNk + [ ( NN+ 11

2 < gN (k) < Nk + [ ( NN+ V] _ .

2

3.8 Concluding Remarks

2(N N^

>+ 1 )k (

N N+ 1 )-kwhich is rather weaker than Theo-In general, we require only

rem 1.9.4. It appears though, that effectively proving the bound

11(4 /3 )4 > (9 /4 ) -k

is as difficult as that involved in the classical Ideal Waring problem.

Chapter 4

Bounds for Ilek li

4.1 Multi-point Bounds

Simultaneous or multi-point Pade and Pade-type approximation to In x and ex has a long

history dating back to Hermite's proof of the transcendence of e. By application of these

techniques, in what amounts to two essentially distinct ways, K. Mahler [39, 42] was able

to deduce

Theorem 4.1.1 (Mahler) There are constants c and k o such that

He il > k- ck

for 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], where

one finds that

'Irk 11 > k- ck2lnk

for all k sufficiently large and c an absolute positive constant.

4.2 A Single-point Bound

The techniques applied in Chapter 1 can be utilized in a variety of situations where closed

forms are available for the relevant Pade approximations. In the following section, we

deduce the lower bound

90

Chapter 4: Bounds for !le i'II^ 91

Theorem 4.2.1 There is an effective constant c such that

IIekII > c • (2k2 )- k2

for all k E N.

While this bound is weaker than that obtained via multi-point Pade approximation, it

is easier to produce.

Proof of Theorem 4.2.1: From Luke [37, p. 192, (19)-(22)], we can write

e_ z = Gm (-z) + em (z)Gm (z)

valid for Rez > 0, where Gm (z) is an m-th degree polynomial in z with integer coefficients

and 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!) 2

It follows that if k is a positive integer

(4.47) ek^Gm(k)^k Grn (k)Gm(-k) e Gm (-k) EniV

We now seek a bound upon Gm (k), proving

Gm+ i(k) Lemma 4.2.2 Gm(k))

< k + 4m + 2, m = 0, 1, 2, ....

Proof: We proceed via induction, proving that

Gni+i (k) k <^<k+4m+ 2.Gm (k)

Since Go (k) = 1 and G 1 (k) = k + 2, we suppose that

Gi+1(k) < k + 4i + 2G2(k)

Chapter 4: Bounds for He'll^ 92

for 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)

and

thus follows from

Gi+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) yields

kGm_ i (k) > (4in — 2)Gm_ i (k) + k 2Gm _2 (k)

Or

Gm _1(k) > ^k2Gm_2 (k) k — 4m + 2

But our induction hypothesis implies

Gm -1(k) <k+4m-6Gm,-2(k)

which gives

< k 4rn — 6k —4m-I-2

Or

—16m2 + 32m — 4k — 12 > 0

a contradiction (since G2(1)/G1 (1) = 19/3, we may assume (m, k) (1, 1)).^■

By the preceding lemma, then, we have that

k2 -1

Gk2(k) <^(k + 4m + 2)m=o

k2

Chapter 4: Bounds for 11 ek ii^ 93

and since1 k2-1_k2

E (k + 4m + 2) = 2k 2 + k

the 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 + 1

Gm (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 inequality

is 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 similar

with 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 e2^2

lE k2(101 = 4kek+1/(6k2 )+1/(2e)^16k2)k

[1 + o(k-2)] .

Thus

ek Ck2(k)jek2(k)1 < ( e2(2k

16k2

2 + k)) k2

< (0.99) k2

for k > ko effectively computable. Now

Gk2 (k) < (2k2l/k

+ k \ k2^) < (2k 2 ) k2

^

k —^e

m=0

Chapter 4: Bounds for He il^ 94

and hence (4.47), (4.50) and the last result imply that there is an effective constant c

(independent of k) such that

Hell > c • (2k2 ) — k2

as desired.^ ■

The above yields a bound of order

Ilek II > (f(k)) -1

where In f (k) = 0(k 2 in k) and hence is no match in strength for Mahler's

He ll > (g(k))-1

with g satisfying In g(k) = 0(k In k).

Appendix

( Na + 1)k

Nb )

Special cases of lower bounds for

In 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^N

and 4 include values of N with admissable s (in the notation of Theorem 1.4.1) and the

induced 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 involved

Pade approximants (as given in section 1.3). In all cases, the values of s = s(N) are not

claimed to be in any sense best possible, but are adequate for our purposes.

95

(Na +1 \k

Nb )

Appendix : Special cases of lower bounds for 96

( 1 + 1\77. )k

Table 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 2.0584

10 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 )


Table 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 1.40286

19-21 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 )


(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 2.4258

10 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 )


1 \ k

N2

Table 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 2.3672

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