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TRANSCRIPT
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 )
Chapter 0: Introduction^ 3
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.
Chapter 0: Introduction^ 4
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,
Chapter 0: Introduction^ 5
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
Chapter 1: Fractional Parts of Powers of Rationals^ 8
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.
Chapter 1: Fractional Parts of Powers of Rationals^ 9
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
Chapter 1: Fractional Parts of Powers of Rationals^ 10
and
1.1.13 If N > 4 is an integer, then there exists effective ko = ko (N) such that if k > ko,
then
(
7, -v-1 + )k
> (2.85) -k .
By generalizing Dubitskas' lemma on divisibility of binomial coefficient products it is
possible to also improve (1.1.9) and (1.1.10), to obtain
1.1.14 If N > 3 is an integer, then there exists effective ko = ko (N) such that if k > ko,
then
(N Tv--+ )k> (2.65) -k .
and
1.1.15 If N > 2 is an integer, then there exists effective ko = ko (N) such that if k > ko,
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).
Chapter 1: Fractional Parts of Powers of Rationals^ 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
Chapter 1: Fractional Parts of Powers of Rationals^ 12
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
Chapter 1: Fractional Parts of Powers of Rationals^ 13
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
Chapter 1: Fractional Parts of Powers of Rationals^ 14
(1.12)^QA(z) =(-1)c (A B C 1)! f l tB — 1)C (1 — t + zt)A dt
A! B! C!
(1.13)^EA(z) =^B(A++ C+1)7 1 A
(1—t)c (1— zt) BA! B! C!^o
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 ^A
^
tB (1 — t) c (1 — t + zt) A dt =^1 B (1 — t) C 2 (A )(1 - t)A-i (zt) i dti=0^6
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)^EA(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:
Chapter 1: Fractional Parts of Powers of Rationals^ 15
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)))
Chapter 1: Fractional Parts of Powers of Rationals^ 16
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
Chapter 1: Fractional Parts of Powers of Rationals^ 17
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 =
Chapter 1: Fractional Parts of Powers of Rationals^ 18
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
Chapter 1: Fractional Parts of Powers of Rationals^ 19
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
Chapter 1: Fractional Parts of Powers of Rationals^ 20
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)
Chapter 1: Fractional Parts of Powers of Rationals^ 21
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 )=
Chapter 1: Fractional Parts of Powers of Rationals^ 22
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
Chapter 1: Fractional Parts of Powers of Rationals^ 23
and that we define
^/ = max { ^At 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 ^At^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 mustz^i=1
have {A/x} {i/x} and thus
Chapter 1: Fractional Parts of Powers of Rationals^ 24
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<b
where the sum is over prime x), we can find effective Ei E2 > 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
Chapter 1: Fractional Parts of Powers of Rationals^ 25
> exp((
rqct+ d
de(cld,t))m — e2m)
= L(c1d) dm exp( —E2m).
If we choose e 2 suitably small relative to e, the result obtains.^ ■
1.8 The Proof of the Main 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)
Chapter 1: Fractional Parts of Powers of Rationals^ 26
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
Chapter 1: Fractional Parts of Powers of Rationals^ 27
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
Chapter 1: Fractional Parts of Powers of Rationals^ 28
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
Chapter 1: Fractional Parts of Powers of Rationals^ 29
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
Chapter 1: Fractional Parts of Powers of Rationals^ 30
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.
Chapter 1: Fractional Parts of Powers of Rationals^ 31
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
Chapter 1: Fractional Parts of Powers of Rationals^ 32
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. ■
Chapter 1: Fractional Parts of Powers of Rationals^ 33
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.
Chapter 1: Fractional Parts of Powers of Rationals^ 34
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) <
■
Chapter 1: Fractional Parts of Powers of Rationals^ 35
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
Chapter 1: Fractional Parts of Powers of Rationals^ 36
1.6.2, we have
(1.56) E(s) < IQIaN
ys
and
(1.57)^E(s)^E(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
2^E(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 .
■
Chapter 1: Fractional Parts of Powers of Rationals^ 37
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 < b 2n . Then there is a computable
constant mo = mo (a,b,ii) such that if m > 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
Chapter 1: Fractional Parts of Powers of Rationals^ 38
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 < b 27). Choose e > 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
Chapter 1: Fractional Parts of Powers of Rationals^ 39
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
Chapter 1: Fractional Parts of Powers of Rationals^ 40
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^ )^< i + Np+1^ )
and hence
S
in( (p + 1)s2
+ 1 •
p + 1 7^pin p• E(s)) < 21n(p+ 1) + lnpin
s +^ P
Since this is < (p+ 1)1n 2 < (p + 1)1n N for p > 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.^ ■
Chapter 1: Fractional Parts of Powers of Rationals^ 41
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.^■
Chapter 1: Fractional Parts of Powers of Rationals^ 42
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)^
t^e(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 +1^as + 1
Then
tETi(s,..)
^In L(a, b, s i ) =^E (as i
t
+ 1^o(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
Chapter 2: Contents of Pacle Approximants to (1 — z)k^ 47
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 < B < s + 1. Then
00
lnL(s) = E LA (8)A=0
where
LA(S) = E ^+ 1^e(S, t))B A(S+1)+B
and B is such that
i) A(s + 1) + B E N
ii) (s + 1)/2 < B < s + 1.
We have
Lemma 2.2.3 If n > 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)]•
Chapter 2: Contents of Pade Approximants to (1 — z) k^48
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
Chapter 2: Contents of Pacle Approximants to (1 — z)k^ 51
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
Chapter 2: Contents of Pade Approximants to (1 — z)k^ 53
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
Chapter 2: Contents of Pade Approximants to (1 — z) k^54
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)
Chapter 2: Contents of Pade Approximants to (1 — z) k^55
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
Chapter 2: Contents of Pade Approximants to (1 — z)k^ 56
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.
Chapter 3: Connections to Waring's Problem^ 62
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
Chapter 3: Connections to Waring's Problem^ 63
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,
Chapter 3: Connections to Waring's Problem^ 64
With these definitions in mind, we further defineY
^(3.10)^U(a) = >2 e(au), V(a) = >2 >2 eofyykouEU^ uo.
where here and henceforth, e(x) = e2irix . Taking f(a) = E e(axk ), we now introducex=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)=---^e(axk^A(q) =^(S(a, 0/04ke( amiox=1^ a=1
(a,q)=1
Chapter 3: Connections to Waring's Problem^ 65
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< p l /2
Proof: This is essentially a corollary of Lemma 3.3.1 (see Dickson [23, Lemma B]). We
have 1C2 1 < Cik < e l1.00k3 ■
Lemma 3.3.3 6 > C3 > e-4.29k3
Chapter 3: Connections to Waring's Problem^ 66
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
Chapter 3: Connections to Waring's Problem^ 67
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.
Chapter 3: Connections to Waring's Problem^ 68
Proof: See Vinogradov [58, p.30, Lemma 104^ ■
if 1z1 P"Lemma 3.3.7 1/(z)1 < Z :=
Azi - v^if izi >
Proof: Since N > 1 and /(z) = f e(zx k ) dx, the first of these assertions follows easily.
Suppose that z > P' and make the change of variables = 2zx k to 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# c113^OP) cos 7 )3 +^OP) 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 . . .^I^OP) sin ir/30
[2zPk]
the above argument yields
1 2zPkOP) sin 70 di3
i2z(N-1)k
(2z)-11
Chapter 3: Connections to Waring's Problem^ 69
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
Chapter 3: Connections to Waring's Problem^ 70
(F(1+ v))rwhere Tr = r(i + rv)^ and so, using Tr <1, it follows that kr _ i (M — sk) < kr_ i (M) <
Mir-1)" and 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
Chapter 3: Connections to Waring's Problem^ 71
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) =^e(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)^I(z)
-1- 40
(returning to our original variables) and thus
^ /(z) C5qf(a) = S(a,
with 105 1 < 4.^ ■
From this result, it is straightforward to prove
k^(S(a,q)^co-34-uz^1.^if, <Lemma 3.4.2 (f (a)) 4 =^1-(z))4k^4k-i, wh ere^e1100k3 .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
^41t1q^i=0
and so from Lemmas 3.3.1 and 3.3.7, if we let i = 0, 1,^, 4k — 1
( c1
di = 4.k^i 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+0di^i^1 105 1^— 16k
Chapter 3: Connections to Waring's Problem^ 72
and q < P 1 / 2 , if Z = P, then di+i /di > 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 .^ ■
Chapter 3: Connections to Waring's Problem^ 73
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 .
Chapter 3: Connections to Waring's Problem^ 74
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
Chapter 3: Connections to Waring's Problem^ 75
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
Chapter 3: Connections to Waring's Problem^ 76
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
Chapter 3: Connections to Waring's Problem^ 77
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
■
Chapter 3: Connections to Waring's Problem^ 78
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
Chapter 3: Connections to Waring's Problem^ 79
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 =^ay + 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) = ^1^if y is a k-th power
0^otherwise
and
^y(x) = 1^if x E V
^0 ^otherwise
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.
Chapter 3: Connections to Waring's Problem^ 80
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)
Chapter 3: Connections to Waring's Problem^ 81
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
Chapter 3: Connections to Waring's Problem^ 82
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
Chapter 3: Connections to Waring's Problem^ 83
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 < i + kn (k-1)/k.
Proof: Suppose first that n > t + Nk and choose i such that i k < n — £ < (i + 1) k . Thenik E Si(N" ) and since, by 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}.
Chapter 3: Connections to Waring's Problem^ 85
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)
Chapter 3: Connections to Waring's Problem^ 86
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.
Chapter 3: Connections to Waring's Problem^ 87
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
Chapter 3: Connections to Waring's Problem^ 88
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
Chapter 3: Connections to Waring's Problem^ 89
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 )
Appendix : Special cases of lower bounds for 97
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 )
Appendix : Special cases of lower bounds for 98
(N + -1k^N )
Table 3: > B -k for all k > ko (N)
N s B N s B N s B N s B3 2.10 2.5477 23 4.68 2.5082 43 5.55 2.4581 63 6.09 2.42564 2.39 2.6492 24 4.75 2.4970 44 5.58 2.4567 64 6.11 2.42605 2.71 2.5719 25 4.82 2.4828 45 5.61 2.4552 65 6.13 2.42636 3.05 2.4192 26 4.89 2.4677 46 5.65 2.4496 66 6.15 2.42647 3.19 2.4899 27 4.96 2.4504 47 5.68 2.4477 67 6.17 2.42638 3.33 2.5246 28 5.02 2.4409 48 5.71 2.4461 68 6.19 2.42619 3.45 2.5535 29 5.05 2.4493 49 5.73 2.4477 69 6.21 2.4258
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 )
Appendix : Special cases of lower bounds for 99
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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