C OMPOSITIO M ATHEMATICA
P. L. C IJSOUW
Transcendence measures of certain numbers whose transcendency was proved by A. Baker
Compositio Mathematica, tome 28, n
o2 (1974), p. 179-194
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
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179
TRANSCENDENCE MEASURES OF CERTAIN NUMBERS WHOSE TRANSCENDENCY WAS PROVED BY A. BAKER
P. L.
Cijsouw
Noordhoff International Publishing Printed in the Netherlands
1. Introduction
In the
subsequent
paper we continue theinvestigation
of transcen-dence measures of certain transcendental numbers 03C3, i.e.
positive
lowerbounds for
IP(u)1
in terms of thedegree
N andheight
H of P, where P is anarbitrary polynomial
withintegral
coefficient. For more informa- tion about transcendence measures and the type of transcendence mea- sures we will lookfor,
see the earlier paper[4];
see also the authors thesis[3],
which includes the results of the present paper.Let 03B11, ···, an be non-zero
algebraic
numbers suchthat,
for any(fixed)
values of the
logarithms, log
03B11, ···,log
03B1n arelinearly independent
overQ.
In this paper, transcendence measures are derived for numbers whichcan be written in one of the
following
ways:(i ) fi 1 log
a 1 + ···+ 03B2n log
anwith n ~ 2,
where03B21, ···, /3n
arealge-
braic
numbers,
not all zero(ii) e03B2o03B103B211 ··· 03B103B2nn with n ~
1 andwhere
03B20, 03B21, ···, Pn
arealgebraic
numbers such that at least one ofPl’ ..., Pn
is irrational. We prove the transcendence measurefor numbers of the form
(1),
and the transcendence measurefor numbers of the form
(2).
Here S =N+ log H,
a is anarbitrary posi.
tive number and
Ci
andC2
areeffectively computable numbers, depend- ing
on 8, n, the a’s and theirlogarithms
and thej8’s.
We
remark,
that these transcendence measures are the firstexplicit
ones to be
published
for these numbers in which both thedependence
on H and N is
expressed.
If one is interested merery in theheight H,
better results can be
given.
For numbers of the form(i),
N. I. FEL’DMAN[6 proved
the transcendence measureexp { - CS},
where C is apositive
number, depending
onN,
the a’s and theirlogarithms
and the03B2’s.
Fornumbers of the form
(ii),
a recent result of A. BAKER[2] implies
thetranscendence measure
exp { -
Clog
Hlog log H}
for H ?4,
where Cagain
is apositive
numberdepending
onN,
the a’s and theirlogarithms
and the
03B2’s.
An earlier result for thespecial
case of numbers of the forme03B2003B103B211
can be found in[7].
The method of
proof
of our transcendence measures is A. BAKER’S onewith some
improvements
introducedby
N. I. FEL’DMAN. In theproof
we
firstly
derive a measure for theapproximability
of numbers of thetypes
(i)
and(ii);
afterthat,
the transcendence measures aregiven.
The transcendence of numbers of the form
(i)
followsimmediately
from e.g. Theorem 1 of A. BAKER’S paper
[1].
The transcendence of numbers of the form(ii)
wasproved by
A.BAKER, distinguishing
thecases
flo =
0 and03B20 ~ 0;
see the same paper.2. Lemmas
We shall use the same notations
(especially
for thedegree; height
andsize)
as in[4].
Forshortness,
we use without reformulation the lemmas3, 6, 7,
8 and 9 of[4]. Further,
we need thefollowing
lemmas:LEMMA 1 : Let 03B11, ···, an be non-zero
algebraic
numbers suchthat, for any fixed
valuesof
thelogarithms, log
03B11, ···,log
an arelinearly indepen-
dent over
Q.
Let a bepositive
and let d be apositive integer.
Then thereexists an
effectively computable positive
numbersuch that
for
allalgebraic
numbersPo, 03B21, ···, 03B2n,
not all zero,of degrees
at mostd and
of heights
at most h.PROOF: See Theorem 1 of
[5].
LEMMA 2: Let 03B1
and P
bealgebraic numbers, 03B2
~ 0. ThenPROOF: From Lemma 3 of
[5] it
follows thatUsing d(03B1)+
1ed(03B1)
andd(P) +
1ed(P)
we getfrom which
(2)
follows.LEMMA 3:
Let fi
be analgebraic number,
and let k and ~ beintegers.
ThenPROOF: If
an zn + ···
+ al z + ao is the minimalpolynomial
of03B2,
thenis a constant
multiple
of the minimalpolynomial
ofk + ~03B2. Thus,
thecoefficient
of zi (i = 0, 1, ···, n)
of this minimalpolynomial
is in absolutevalue at most
From the obvious
inequality (m i) ~
2m -1 for allpositive integers
m, itfollows that
by
which theproof
iscompleted.
3. The case
fil log
ai + ...+ 03B2n log
anFirst we
give
a measure for theapproximability
for numbers of the form(i),
in thespecial
case in whichfi.
= - 1.THEOREM 1:
Let, for n ~ 2, 03B11, ···, 03B1n
and 03B31, ···, Yn-1 be non- zeroalgebraic
numbers suchthat, for
anyfixed
valuesof
theloga- rithms, log
03B11, ···,log
an arelinearly independent
overQ.
Let a be apositive
number. Then there exists aneffectively computable
numberS1
=81(8, log 03B11, ···, logrtn,
03B31, ···,1’n-l)such
thatfor
allalgebraic numbers 11 of degree
N and size S ~Sl.
PROOF: Put ô =
(2n3 + 4n2 + 3n + 7)-103B5.
Forabbreviation,
putand
It is sufficient to prove that
if S ~ Sl .
In thisproof
we may restrict ourselves to the case in which 03B4 is rather small.By
cl , c2, ··· we denotepositive
numbers whichdepend only
on n,log
03B11, ···,log
03B1n, yi , ..., 03B3n-1.Suppose
thatfor some
algebraic number 11
ofdegree
N and size S. We prove that this leads to a contradiction if S issufficiently large.
Choose the
following integers:
Put
where the numbers
Ck,
... knmv areintegers
of absolute values at mostC;
they
will bespecified
later.Fort =
0, 1, 2, ···
it iseasily
seen thatwhere
Put
We estimate the difference
as follows:
by |ez -1| ~ |z|e|z|,
one has fork.
=0, 1, ···, K-1
andthe
inequality
Hence,
Let
the
appropriate
way, such thatWe
apply
Lemma 6 of[4]
to thesepolynomials
in thespecified points.
If r, s and B
have the samemeaning
as in this lemma we haveHence, s ~
4rd where d =[Q(03B11, ···,
an , 03B31, ···,03B3n-1) : Q]. Further,
Hence,
theright
hand side of the second condition in Lemma 6 of[4]
isat most
It follows that the numbers
Ck1...
knmv can be chosen asintegers,
not allzero, of absolute values at most
C,
such that03A603C403C41 ... tn-1 (p)
= 0 forï, Ti, * ’ *1 ’r.-i =
0, 1, w,
T-1and p
=0, 1, w,
P-1.Doing
so, wecertainly
have03A603C403C41 ...03C4n-1(p)
= 0 for0 ~
03C4 + 03C41 + ...+ 03C4n - 1 ~
T- 1and
0 ~ p ~
P-1. From(8)
we now getfor 0 ~ 03C4+03C41+ ··· + 03C4n-1 ~ T-1 and 0 ~p~P-1.
Define
Tr
andPr
for r =0, 1, ...,
Rby
and
Remark that
LEMMA: For r =
0, 1, ···,
R theinequality
holds
for
allnon-negative integers
i, 03C41, ···, 03C4n - 1 and p withPROOF: We
proceed by
induction on r. For r = 0 the statement isproved
in(9).
Let r be aninteger
with 0 ~ r ~ R -1 for whichfor 0 03C4+03C41 + ···
+ 03C4n-1 ~ Tr -1
and0 ~ p ~ Pr -1.
Sincewe have for t =
0, 1, 2, ···
Together
with(12)
we obtainfor
From
(7)
we know thatfor 03C4+03C41+ ··· + 03C4n -1 ~ Tr+1-1. We apply Lemma 7 of [4] to F03C403C41...03C4n-1
with R =
Pr + 1, A
=6,
T =Tr + 1
and P =Pr.
From(13), (14)
and theinequality
we then obtain
In
particular,
From
(8)
and(10)
it follows thatfor
0 ~
03C4 + 03C41 + ···+ 03C4n-1 ~ Tr+1-1
and 0~ p ~ Pr + 1-1.
But forthese values of i, 03C41, ···, 03C4n-1
and p,
we canconsider 03A603C403C41 ... 03C4n-1(p)
as apolynomial in il,
03B11, ···, 03B1n , 03B31, ···, 1’n-l,of degree
less thanTr + 1 + N in il, KPr + 1
in 03B11, ···,an and Tr+1
in 03B31, ···, 03B3n -1. If B denotes thesum of the absolute values of the
coefficient,
then we haveAccording
to Lemma 3 of[4]
we thus have either03A603C403C41...03C4n-1(p) =
0 orfor
0 ~ 03C4+03C41+ ··· +tn-1 ~ Tr+1-1 and 0 ~p~Pr+1-1. Hence,
03A603C403C41 ...03C4n-1(p)
= o for these 1", 03C41, ···, zn-1and p. Again
from(8)
and(10)
we obtainfor 0 ~ 03C4 + 03C41 + ...
+ 03C4n-1 ~ Tr + 1-1
and 0 ~p ~ Pr+l -1.
The lem-ma has been
proved.
From
(11)
with r = R we getfor
0 ~
03C4 + 03C41 + ···+03C4n-1 ~ TR -1
and0 ~ p ~ PR-1.
From theirdefinitions we have
TR
= T’. Sincewe see that
Hence,
for
0 ~ 03C4 + 03C41 + ··· + tn-1 ~ T’ -1
and0 ~ p ~ P’ -1.
From(6)
it now follows that
fort =
0, 1, ..., T’ - 1 and p
=0, 1, ···, P’ -1.
We
apply
Lemma 8 of[4] to
F with Kreplaced by
K". Let 03A9 and co have the samemeaning
as in this lemma. Since the exponents of F have the formwe see that
With
this,
the conditionis
easily
checked.Further,
we know from Lemma1, applied
with 8 =03B4,
that
for all
integers k1, ···, kn ,
not all zero,with |k1| ~ K- 1, ···, |kn| ~
K-1. From
(21)
and(5)
it follows thatFrom
(20)
we haveFrom lemma 8 of
[4],
with(19), (22)
and(23)
we obtainfor k1, ···, kn
=0, 1, ···, K-1
and m =0, 1, ···, M-1.
But
according
to Lemma 3 of[4]
we have eitheror
for the same values of
k1, ..., kn
and m.Hence,
and m =
0, 1, ’ ’ ’,
M -1.Since 11
has thedegree N,
it follows that allintegers Ckl...
knmv are zero, in contradiction to their choice. This con-tradiction proves Theorem 1.
We have the
following
COROLLARY: Under the conditions
of
Theorem1,
there exists aneffec- tively computable,
numberC3
=C3 (8, log
03B11, ···,log
03B1n, 03B31, ···,03B3n - 1)
> 0 such thatfor
allalgebraic
numbers 11of degree
at most N and size at most S.PROOF: There are
only finitely
manyalgebraic
numbers 11 of sizes(~) S1.
ChooseC3 ~
1 such that(26)
holds for thesefinitely
many numbers.THEOREM 2:
Let, for
n >2,
03B11, ···, an andPl’ ..., Pn
be non-zeroalgebraic
numbers suchthat, for
anyfixed
valuesof
thelogarithms, log
03B11, ···,log
an arelinearly independent
overQ.
Let 8 be apositive
number. Then there exists an
effectively computable positive
numberC4 = C4 (8, log
03B11, ···,log
an,03B21, ···, fin)
such thatfor
allalgebraic numbers 03BE of degree
N and size S.We have
d(~) ~ c7N
with C7 =d(03B2n) and, by
Lemma2, s(~) ~ c8S
with c8 =
3d(03B2n)+s(03B2n).
From(26)
we now obtainfor some
effectively computable positive
numberC4.
THEOREM 3: Under the conditions
of
Theorem2,
there exists aneffec- tively computable
numberCl = el (8, log
03B11, ···,log
an,03B21, ···,03B2n)
>0,
such thatis a transcendence measure
of 03B21 log
al + ...+ 03B2n log
an .PROOF:
Apply
Lemma 9 of[4]
to the result of Theorem 2 and putCl = 6C4(1 +log 2)n+l+£.
4. The case
e03B2003B103B211 ··· oc On
THEOREM 4: Let n be a
positive integer.
Let03B20
bealgebraic
and let al, ..., an be non-zeroalgebraic
numbers suchthat, for
anyfixed
valuesof
thelogarithms, log
03B11, ···,log
an arelinearly independent
overQ.
Let03B21, ···, 03B2n
bealgebraic numbers,
not all rational. PutLet a be a
positive
number. Then there exists aneffectively computable
number
S2 = S2(s, log
03B11, ···,log
an,03B20, 03B21, ···, 03B2n)
> 0 such thatfor
allalgebraic 03BE of degree
N and size S ~S2 .
PROOF : Put 03B4 =
(2n3 + 8n2 + 10n +4)-1 03B5.
For the sakeof brevity,
putand
It is sufficient to prove that
if
S ~ S2;
in thisproof we
may assume that bis rather small.By
cl , c2, ···we shall denote
positive
numbers whichdepend
on n,log
03B11, ···,log
an,03B20, 03B21, ···, fi. only.
Suppose
thatfor some
algebraic number 03BE
ofdegree
N and size S. We prove that this isimpossible
if S’ issufficiently large.
Choose the
following integers:
Put
where the numbers
Ckl...
knlmv areintegers
of absolute values at mostC; they
will bespecified
later.where
Define
03A603C403C41 ... 03C4n by
For ~ = 0, 1, ···, L-1
andone has
Hence,
We
apply
Lemma 6 of[4] to
thepolynomials
chosen in the
appropriate
way such thatIf
r, s and B
denote the same numbers as in Lemma 6 of[4],
we haveand
From these
inequalities
it is easy to check the conditions of this lemma.Hence,
we can fix the numbersCkl ...
knlmv asintegers,
not all zero, ofabsolute values at most
C,
such that03A603C403C41 ... 03C4np
= 0 for r, Tl, ’ ’ ’, in=
0, 1, ···,
T-1and p
=0, 1, ···,
P-1. With(32)
thisimplies
for
0 ~
03C4+03C41+ ···+ 03C4n ~
T -1 and0 ~ p ~P-1.
Define Tr
andPr
for r =0, 1, ···, R by
and
Observe that
LEMMA: For r =
0, 1, ···,
R theinequality
holds
for
allnon-negative integers
i, 03C41, ···, in and p withPROOF: We use induction on r. For r = 0 the
inequality
hasalready
been
proved
in(33).
Let r be aninteger
with 0 ~ r ~ R-1 for whichfor 0 ~ 03C4 + 03C41 + ···
+ 03C4n ~ Tr -1
and0 ~ p ~ Pr-1.
Sinceit follows that for t =
0, 1, 2, ...
Hence,
(36) implies
For the same values of i, 03C41, ···, T, we obtain from
(31 )
We
apply
Lemma 7 of[4]
toF03C403C41 ... 03C4n
with R =Pr + 1,
A -6,
T =Tr + 1
and P =
Pr .
From(37), (38)
and(34)
we then obtainConsequently,
for 0 ~
03C4 + 03C41 + ···+ 03C4n ~ Tr + 1-1 and 0 ~ p ~ Pr + 1-1.
From(32)
and
(34),
it follows thatfor 0 ~ 03C4 + 03C41 + ··· + 03C4n ~ Tr+1-1 and 0 ~ p ~ Pr+1-1.
However, for these values of i, 03C41, ···, 03C4n and p, we can consider
03A603C403C41 ... 1:n(p)
as apolynomial in 03BE,
03B11, ···, 03B1n,/30’ 03B21, ···, 03B2n,
ofdegree
lessthan
LPr + 1 + N in 03BE, KPr + 1
in 03B11, ···, an andTr + 1
in03B20, 03B21, ···, /3n.
The sum of the absolute values of its coefficients is at most
According
to Lemma 3 of[4]
we have either03A603C403C41 ... 03C4n(p)
= 0 orHence,
From
(32)
and(34)
we seefor
0 ~ 03C4 + 03C41 + ··· +03C4n ~ Tr+1 -1
and0 ~ p ~ Pr + 1 -1,
whichproves the lemma.
From
(35)
with r = R we getThus,
for
0 ~
03C4 + 03C41 + ···+ 03C4n ~
T’ -1 and0 ~ p ~
P’ -1. From(30)
wenow obtain
fort=
0, 1, ..., T’ -1 and p =0,1,
... P’-1.The exponents of F have the foi m
Let 03A9 and co have the same
meaning
as in Lemma 8 of[4].
Thenfrom which the condition
follows
by
directcomputation.
The difference of two exponents of F is of the form
with
integral ki , ..., kn, l,
not all zero, and|ki| ~
K-1 for i =1, ···,
nand 1 fl
~ L -1.Moreover,
at least one of the numberski+tfli (i
=1,···, n)
is non-zero, since03B21, ···, 03B2n
are not all rational. Thedegrees
of1 po, k1 + ~03B21, ···, kn + ~03B2n
are constants. We estimate theirheights by
means of Lemma3;
we then see that theseheights
do notexceed
in which c. and C6 are upper bounds for the
heights
anddegrees
resp. of03B20, 03B21, ···, 03B2n.
From Lemma 1 with e = ô it follows thatHence,
the exponents of F are distinct andFrom Lemma 8 of
[4], using (41), (42)
and(43)
we obtain theinequality
According
to Lemma 3 of[4]
we have eitherfor the same values
of k1, ···, kn, ~
and m. It follows thatfor all of these values.
Since 03BE
has thedegree N,
thisimplies
that all in- tegersCk,
... knlmv are zero, in contradiction to their choice. The theorem has beenproved.
Using
thefact,
that there areonly finitely
manyalgebraic numbers 03BE
of size SS2,
andusing
Lemma 9 of[4],
oneimmedeately
obtains thefollowing
theorem:THEOREM 5: Under the conditions
of Theorem 4,
there exists aneffectively computable,
numberC2 = C2 (8, log
CX1, ...,log
oc,,,03B20, 03B21, ···, fi,,)
> 0 such thatis a transcendence measure
of e03B2003B103B211 ··· 03B103B2nn.
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[3] P. L. CIJSOUW: Transcendence measures, thesis. University of Amsterdam, 1972.
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(Oblatum 22-X-1973) Instituut voor Propedeutische Wiskunde Universiteit van Amsterdam
Roetersstraat 15 Amsterdam.