of Number TT by Algebraic from Special Fields N. I.
Numbers
FELDMAN
Department of Mathematics, Faculty of Mathematics and Mechanics, University of Moscow, Moscow, V234 USSR Communicated by H. Zassenhaus Received
DEDICATED
The algebraic
estimate numbers
December
TO
from below from the
of fields
26,
PROFESSOR
the modulus generated
1973
K.
MAHLER
of the difference by the roots of unity
between is made.
v and
1
LetP(z)=a,z”+...+a,, n(m))
= n,
H(p(z))
=
oFk%%
1 uk
1,
W(z))
=
I ql
I +
***
+
I a,
I. (1)
If 5 is an algebraic number and P(z) is the irreducible polynomial with relatively prime integer coefficients, we let 40
= W(z)), H(5) = H(m),
WI
for 4
= W(z)).
(2)
Estimates from below for the modulus of the difference between rr algebraic numbers have been obtained by various authors: Popken Siegel [2], Mahler [3,4] and others. The most precise are Mahler’s equalities (see [4]) I 3-f- P/4 I > q-42, 1VT- 5 1 > H-‘I”,
and the inequalities
(3) (4)
obtained in papers [5,6]
1T - 5 1 > exp(-c&n 1n- - 5 1 > exp(-c,n
In n + 1 + In H) ln(n Inn + 2 + In H)), In(n + 2) In H),
n2 ln4(n + 2) -C In H,
48 Copyright All rights
and [l], in-
Q 1977 by Academic Press, Inc. of reproduction in any form reserved.
ISSN
0022-314X
(5) (6)
49
APPROXIMATION OF rr
where p and q are natural numbers, c1 , c2 , c, , as well as co, c4 ,... occurring below are positive constants n = n(5), H = H(5). It is natural to compare the inequalities (3~(6) with the best possible estimate exp(-c,n In H), which it is impossible to improve (see [13]). It was shown in [7], that if m is natural, v = q(m), the Euler’s function, K = Q(exp(2+n)), 5 E K, and its degree over Q is equal to n, No = v + n-lv In(H + 2)/lnln(H + 2) E > 0, In m > E In No, then / 7T _ 5 1 > e-w&JnNo, c-5= Cs(E). (7) For n w c,v inequality (7) gives us an estimate of the form exp(-c,n In n), but as can be seen from the hypotheses of the theorem, only for 5, belonging to the fields of a special type. In the present paper a proof of some stronger result is given (in particular, the hypothesis In m > EIn No is excluded and for large v, for v M c,ZV, the v2 In v is replaced by v2). Namely, the following theorem is proved. THEOREM. Let m be integer, m 3 3, p = exp(2G/m), K = Q
N = v + vn-l In L.
5 1 > emcovN,
(8)
Note 1. It is evident that v = q(m) or 2v(m). Note 2.
y(m) > c,m(hrln m)-l (see [ll, p. 2161).
Note 3. For m bounded the inequality (8) is of the same type as inequalities (4) and (6). If n In n < v then it is worse than (6) but if n In n > v it is sharper already. It is necessary to take into consideration that there are no additional requirements to 5 in inequality (6) while inequality (8) concerns only 5 from special fields. 2
The following lemmas are needed for what follows. LEMMA 1 [5, Lemma 11. Let X be a natural nu’mber and let ai,j be real numbers, where 9i I afsj I G A
i = l,..., r;
r < t.
There exists a nontrivial collection of rational integers x1 ,..., xt for which I ais1
+ I-* + aiStx, 1 < ZAX((X I x, I < x
+ I)$/, - 2)-l,
j = l,..., t.
i = I,..., r, (9)
50
N. 1. FELDMAN
LEMMA 2 [S]. **.+a,,a,>O.
Let & ,..., 5, be all the zeros of the polynomial Then
a,zm +
a, tnl max(L i 5t I) < L(P). LEMMA 3. If p is a root of unity, the field K = Q(p, i) has the degree v, 5 E K, n(c) = n, and if the rational integers At,tena satisfy the inequality T
L
M
then either 6 = 0 or I 6 I = / t i f A,~,,i’~i~p~ t=o LO m=o
1 >, B’-uL(&‘TI”.
(10)
Proof Let K, ,..., K, be the fields conjugate to K = K1 , and &, aj be the elements of the field Kj which are conjugate to 4 = 5, and 6 = 6, . Let a, be the leading coefficient of the fundamental polynomial of 5. If 6 # 0, then
hence, applying Lemma 2, we find that 18 1 > aiT fi [ 6j 1-l > aiT ?fi (B max(1, ( [j I’))-’ j=2
3 B1-“L(ZJ-‘T’“,
for each of the roots <(I),..., C (In p/p) > In x - (l/2 In x) + E, x > 1, E = -I,3325 P
C In p -c xfl + l/(2 In x>),
x > 1.
e<=
. . ..
(11)
LEMMA 5. Suppose that s and a b 3(s + 1) are nonnegative integers. Let A, be the set of all integers obtainedfrom theproduct N(x) = (x + 1) a** (x + a) by crossing out in all possible ways no more than s of the terms in parentheses, x = 0, 3~1, &2,.... Then there is a common divisor d, of aN the numbers in A, , which satisfies the inequality a=(s + l)-Q e3*250 < d, < aa(s + l)-“.
(12)
APPROXIMATION
51
OF TT
Proof. Let p be a prime number, p < u/(s + 1). It enters in no less than e, = [u/p] parentheses of N(x). We have crossed out no more than s of them, hence the obtained number divides by peg-*. Let
d, =
n prww. P(a/(s+l)
Using Lemma 4 and the obvious inequality a!ea > CP, we obtain d, >