On the irrationality measure of certain numbers

0.12 MB·English

by  Alexandr Polyanskii

On the irrationality measure of certain numbers
A. Polyanskii1
Abstract
The paper presents upper estimates for the irrationality measure and the non-
quadraticity measure for the numbers α = √2k+1ln √2k+1 1, k N.
k √2k+1+−1 ∈
Forany irrational α, the irrationality measure canbedefined as theexact upper bound
5
1 on the numbers κ such that the inequality
0
2
p
α < q κ
n −
− q
a (cid:12) (cid:12)
J (cid:12) (cid:12)
has infinitely many rational solutions(cid:12)p. By(cid:12)µ(α) denote the irrationality measure of α.
7 (cid:12)q (cid:12)
2 The non-quadraticity measure can be defined for any real α which isn’t a root of a

quadratic equation as the exact upper bound on the numbers κ such that the inequality
]
T
N α β < H κ(β)
−
| − |
.
h
has infinitely many rational solutions in quadratic irrationalities β. Here H(β) is the
t
a
height of the characteristic polynomial of β (taking an irreducible integer polynomial with
m
one of the roots equal to β, H the largest absolute value of this polynomial’s coefficients).
[
1 By µ2(α) denote the non-quadraticity measure of α.
v We are going to preset improved bounds on the irrationality measure and the non-
2 quadraticity measure for the numbers
5
7
√2k +1 1
6 α = √2k +1ln − , where k N. (1)
0 k √2k +1+1 ∈
.
1
0 This paper is, in a sense, a continuation of the paper [2] by M. Bashmakova: the
5
same integral is considered, but the denominator is estimated more accurately by using a
1
: certain coefficient symmetry. Some earlier estimates for the irrationality measure of the
v
i numbers investigated by the author have been obtained by A. Heimonen, T. Matala-Aho,
X
and K. V¨a¨an¨anen [6], M. Hata [4], G. Rhin [10], E. Salnikova [11], M. Bashmakova [1],
r
a [2].
Some of the numerical results obtained in this paper have been summarized in the
table below:
k µ(α ) 6 µ (α ) 6 k µ(α ) 6 µ (α ) 6 k µ(α ) 6 µ (α ) 6
k 2 k k 2 k k 2 k
3 6.64610... — 7 5.45248... — 10 3.45356... 10.0339...
5 5.82337... — 8 3.47834... 10.9056... 11 5.08120... —
6 3.51433... 12.4084... 9 5.23162... — 12 3.43506... 9.46081...
Let n be an odd positive integer, and let a,b be fixed positive integers (where b is odd)
such that b > 4a. Define a polynomial as follows:
x+(b 2a)n x+(b a)n x+bn
A(x) = − − =
(b 4a)n (b 2a)n bn
(cid:18) − (cid:19)(cid:18) − (cid:19)(cid:18) (cid:19)
1Moscow Institute of Physics and TechnologyE-mail: [email protected]. The research
was partially supported by the Russian Foundation for Basic Research, grant №09-01-00743.
1
(x+2an+1)...(x+(b 2a)n) (x+an+1)...(x+(b a)n) (x+1)...(x+bn)
= − − .
((b 4a)n)! · ((b 2a)n)! · (bn)!
− −
Consider an integral
z bn+1 π 3
I(z) = − 2 A(ζ) ( z) ζdζ,
−
2πi sinπζ −
ZL (cid:18) (cid:19)
where z = 0, the vertical line L is given by the equation Re ζ = C, where (b 2a)n <
6 − −
C < 2an 1, and this line is traversed from bottom to top. We also suppose that
− −
( z) ζ = e ζln( z), where the branch of the logarithm ln( z) = ln z + iargz + iπ is
− − −
− − | |
chosen so that argz < π.
| |
Statement 1. For all z C such that 0 < z < 1 we have
∈ | |
1 1
I(z) = U(z)ln2z +V(z)lnz W(z) iπ(U(z)lnz V(z)), (2)
−2 − 2 − −
where the functions U(z),V(z),W(z) Q(z) are defined for z < 1 by the following
∈ | |
equations:
U(z) = z−bn2+1 ∞ A( k)zk,
− −
k=bn+1
X
V(z) = z−bn2+1 ∞ A′( k)zk, (3)
− −
k=(b a)n+1
X−
W(z) = z−bn2+1 ∞ A′′( k)zk.
− −
k=(b 2a)n+1
X−
The proof of this statement (in a somewhat different form) has been given by Yuri
Nesterenko [7]. He has also proved the following lemma (see Lemma 1 in [7]).
Lemma 1. Let P(x) C[x] be a polynomial of degree d. Then for all z, z < 1, we have
∈ | |
∞ d z j+1
P( k)zk = c ,
− − j z 1
k=1 j=0 (cid:18) − (cid:19)
X X
where
k=j+1
j
c = ( 1)k 1P( k) .
j −
− − k 1
k=1 (cid:18) − (cid:19)
X
It has also been shown (see Statement 2 and Lemma 1 in [2]) that
U(z) = U 1 = Uˆ z + 1 ,V(z) = V 1 = z 1 Vˆ z + 1 ,
z z − z − z z (4)
W(z) = W 1 = Wˆ z + 1 , where Uˆ(z),Vˆ(z),Wˆ (z) Q(z).
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
z z ∈
If the number x satisfies(cid:0)x+(cid:1) 1 Q(cid:0), then(cid:1)we obtain U(x), V(x) ,W(x) Q.
x ∈ x 1/x ∈
Now consider the values of the integral at the following −points:
k +1 √2k +1 √2k +1 1
x = − = − , where k N. (5)
k
k √2k +1+1 ∈
2
In this case we have x + 1 = 2k+2 and x 1 = 2√2k+1. This easily leads to
k xk k k − xk − k
U(x ),√2k +1V(x ),W(x ) Q. (6)
k k k
∈
Let us define Ω as the set of 0 6 y < 1 such that for all x R the inequality
∈
([x 2ay] [x (b 2a)y] [(b 4a)y])+
− − − − − − (7)
+([x ay] [x (b a)y] [(b 2a)y])+([x] [x by] [by]) > 1
− − − − − − − − −
is satisfied.
The set Ω is a union of points, as well as closed, half-open and open intervals. Clearly,
the points of the interval 0; 1 do not belong to Ω, and thus the set of primes p > √bn
b
such that n Ω is finite. Denote the product of these primes as ∆, and denote as ∆
p ∈ (cid:2) (cid:1) 1
the producnt oof the primes p > (b 2a)n satisfying n Ω.
− p ∈
Let dn be the least common multiple of 1,2,...n,n.o
Let us introduce the following rational numbers:
bn+1
m− 2 , if k = 2m;
R =
k,n 3(b−2a)n+1 bn+1
(2 2 k− 2 , if k = 2m 1.
−
(b−2a)n+1
m− 2 , if k = 2m;
S =
k,n 3(b−2a)n+1 (b−2a)n+1
(2 2 k− 2 , if k = 2m 1.
−
(b−4a)n+1
m− 2 , if k = 2m;
T =
k,n 3(b−2a)n+1 (b−4a)n+1
(2 2 k− 2 , if k = 2m 1.
−
Lemma 2. Let x be defined by (5), then we have
k
d d
A = R U(x ) Z, B = S bnV(x )√2k +1 Z, C = T d ∆ bnW(x ) Z.
k,n k k,n k k,n (b 2a)n 1 k
∈ ∆ ∈ − ∆ ∈
Proof. Let us prove the statement for B. The statements for A and C can be proved by
the same argument.
It suffices to show that B2 K, where K is the ring of algebraic integers, which would
∈
imply that B is also an algebraic integer. Then from the property (6) we could say that
B Q, and consequently B Z.
∈ ∈
Let A (z) be a polynomial of degree 3(b 2a)n such that A (z) = A(z an). Then
1 1
− −
it’s easy to see that
A ( 1) = = A ( (b 2a)n) = A( an 1) = = A( (b a)n) = 0. (8)
′1 − ··· ′1 − − ′ − − ··· ′ − −
Let us rewrite (3). Defining l as l = k an and applying (8), we obtain
−
+ +
V(z) = z−bn2+1 ∞ A′( k)zk = z−bn2+1 ∞ A′( k)zk =
− − − −
k=(b a)n+1 k=an+1
X− X
+ + +
= −z−bn2+1+an ∞ A′(−k)zk−an = −z−bn2+1+an ∞ A′(−l−an)zl = −z−bn2+1+an ∞ A′1(−l)zl.
k=an+1 l=1 l=1
X X X
3
Then by Lemma 1 we have
3(b−2a)n−1 z j+1 j+1 j
V(z) = z−bn2+1+an bj z 1 , where bj = (−1)k−1A′1(−k) k 1 .
j=0 (cid:18) − (cid:19) k=1 (cid:18) − (cid:19)
X X
Lemma 4 of [7] states that
d d
bnA ( k) Z, where 1 k 3(b 2a)n, i.e. bnb Z
∆ ′1 − ∈ ≤ ≤ − ∆ j ∈
From (8) it follows that
3(b−2a)n−1 z j+1
V(z) = z−bn2+1+an bj z 1 =
j=(b 2a)n (cid:18) − (cid:19)
X−
= z−bn2+1+an z z 1 (b−2a)n+13(b−2a)n−1bj z z 1 j−(b−2a)n.
(cid:18) − (cid:19) j=(b 2a)n (cid:18) − (cid:19)
X−
It is true that xk = 1 √2k+1 = t and 1 = 1+√2k+1 = t are the solutions of the
xk 1 − 2 1 1 xk 2 2
equation t2 t −k = 0. Clearly, they must−be algebraic numbers.
− − 2
Applying (4) yields that
3(b 2a)n 1 3(b 2a)n 1
1 − − − −
−(V(xk))2 = V(xk)V x = (t1t2)(b−2a)n+1 bjt1j−(b−2a)n bjt2j−(b−2a)n =
(cid:18) k(cid:19) j=(b 2a)n j=(b 2a)n
X− X−
= k2 (b−2a)n+13(b−2a)n−1bjt1j−(b−2a)n3(b−2a)n−1bjt2j−(b−2a)n. (9)
(cid:18) (cid:19) j=(b 2a)n j=(b 2a)n
X− X−
If k is even, then for all positive integers N we have tN K since t K.
i ∈ i ∈
If k is odd, then for all positive integers N we can write 2[N2+1]tNi ∈ K since (ti)2N =
N N
k+1 √2k+1 and (t )2N+1 = 1 √2k+1 k+1 √2k+1 .
±2 i ± 2 ±2
(cid:16) Let us c(cid:17)onsider the following two ca(cid:16)ses: (cid:17)
1. k = 2m. Then by (9) we have
3(b 2a)n 1 3(b 2a)n 1
− − d − − d
B2 = − ∆bnbj t1j−(b−2a)n ∆bnbj t2j−(b−2a)n(2k +1) ∈ K.
j=(b 2a)n (cid:18) (cid:19) j=(b 2a)n (cid:18) (cid:19)
X− X−
  
2. k = 2m 1. Then (9) yields that
−
3(b 2a)n 1
− − d
B2 = − ∆bnbj 2(b−2a)nt1j−(b−2a)n×
j=(b 2a)n (cid:18) (cid:19)
X−
 
3(b 2a)n 1
− − d
× ∆bnbj 2(b−2a)nt2j−(b−2a)n(2k+1) ∈ K.
j=(b 2a)n (cid:18) (cid:19)
X−
 
This concludes the proof of the lemma.
4
We proceed by formulating several known results, which have been stated in [2] as
Lemmas 5 and 6.
Statement 2. Let x R, 0 < x < 1. If the equation
∈
z(z a)(z 2a) 1
− − =
(z (b 2a))(z (b a))(z b) x
− − − − −
has a unique solution z > b, then
0
1 (z (b 2a))b 2a(z (b a))b a(z b)b b
lim ln U(x) = M = ln 0 − − − 0 − − − 0 − lnx.
n→+∞ n | | 1 (cid:18)(z0 −2a)2a(z0 −a)a(b−4a)b−4a(b−2a)b−2abb(cid:19)− 2
Statement 3. Denote
z +(b 2a) b 2a z +(b a) b a z +b b b
M = ln | 1 − | − | 1 − | − | 1 | lnx,
1 z +2a 2a z +a a(b 4a)b 4a(b 2a)b 2abb − 2
1 1 − −
| | | | − −
where z is the complex root of the equation
1
(z +2a)(z +a)z 1
=
(z +(b 2a))(z +(b a))(z +b) x
− −
satisfying the condition Im z > 0. Then we have
1
1
limsup ln I(x) 6 M .
n | | 2
n +
→ ∞
To compute
1 d 1 d
N = lim ln bn and N = lim lnd ∆ bn, (10)
1 n n ∆ 2 n n (b−2a)n 1 ∆
→∞ →∞
we are going to use Lemma 6 from [7].
Lemma 3. Let u,v be real numbers such that 0 < u < v < 1. Then
1
lim lnp = ψ(v) ψ(u),
n n −
→∞ u≤{Xnp}<v
where ψ(x) = Γ′(x) is the logarithmic derivative of the gamma function, and the sum is
Γ(x)
taken over all primes p such that the fractional part n lies in the given range.
p
n o
Let us formulate a lemma by Hata (see [3], Lemma 2.1), which will allow us to prove
the principal theorem of this paper.
Lemma 4. Let n N, α R, let α be an irrational number, and let l = q α+p , where
n n n
∈ ∈
q ,p Z and
n n
∈
1 1
lim ln q = σ, limsup ln l τ, σ,τ > 0.
n n | n| n n | n| ≤ −
→∞ →∞
Then
σ
µ(α) 1+ .
≤ τ
5
Now we can state the principal theorem, which will be proved by applying the Hata’s
lemma to the sequence
d Im(I(x )) d d
E = S bn k √2k +1 = S bnU(x )α S bnV(x )√2k +1 = P α +Q .
n k,n k,n k k k,n k n k n
∆ − π ∆ − ∆
(cid:18) (cid:19)
By Lemma 2, we have P ,Q Z. Clearly, we can also write
n n
∈
b 2a
− lnm, if k = 2m;
1 − 2
K = lim ln(S ) =
1 n n k,n  b 2a 3(b 2a)
→∞  − lnk + − ln2, if k = 2m 1.
− 2 2 −
Theorem 1. Assume that a,b N satisfy b > 4a, x is defined by (5), the numbers M
1
∈
and M are defined by Statements 2 and 3, the set Ω is defined by (7), and N is defined
2 1
by (10).
If M +K +N < 0, we have
2 1 1
lim P
n M +K +N
µ(αk) ≤ 1− limn→s∞upE ≤ 1− M1 +K1 +N1.
n 2 1 1
n
→∞
We are going to formulate another lemma by Hata (see [5], Lemma 2.3), which will
allow us to prove another theorem.
Lemma 5. Let n N, and assume that α R is not a quadratic irrationality (i.e., not
∈ ∈
a root of a quadratic integer polynomial). Take l = q α + p , m = q α2 + r , where
n n n n n n
q ,p ,r Z, and assume
n n n
∈
1 1 1
lim ln q = σ, max limsup ln l ,limsup ln m τ, σ,τ > 0.
n→∞ n | n| (cid:26) n→∞ n | n| n→∞ n | n|(cid:27) ≤ −
Then we have
σ
µ (α) 1+ .
2 ≤ τ
Now we can easily formulate our second theorem by applying lemma 5 to the following
sequences:
d Im(I(x ))
F = T d ∆ bn k √2k +1 =
n k,n (b−2a)n 1 ∆ − π
(cid:18) (cid:19)
d d
= T d ∆ bnU(x )α T d ∆ bnV(x )√2k +1 = X α +Y ;
k,n (b 2a)n 1 k k k,n (b 2a)n 1 k n k n
− ∆ − − ∆
d Im(I(x ))
G = T d ∆ bn(2k +1) 2Re(I(x )) 2 k =
n k,n (b−2a)n 1 ∆ k − π
(cid:18) (cid:19)
d d
= T d ∆ bnU(x )α2 T d ∆ bn(2k +1)W(x ) = X α2 +Z .
k,n (b−2a)n 1 ∆ k k − k,n (b−2a)n 1 ∆ k n k n
By Lemma 2, we have X , Y , Z Z. We can also see that
n n n
∈
b 4a
− lnm, if k = 2m;
1 − 2
K = lim ln(T ) =
2 n n k,n  b 4a 3(b 2a)
→∞  − lnk + − ln2, if k = 2m 1.
− 2 2 −


6
Theorem 2. Assume that a,b N satisfies b > 4a, x is given by (5), the numbers M
k 1
∈
and M are defined by Statements 2 and 3, the set Ω is defined by (7), and N is defined
2 2
by (10).
If M +K +N < 0, then we have
2 2 2
lim X
n M +K +N
µ2(αk) 1 n→∞ 1 1 2 2.
≤ − max limsupF , limsupG ≤ − M +K +N
n n 2 2 2
{ }
n n
→∞ →∞
Remark 1. In conclusion, let us give the parameter values that have been used to
obtain the results presented in the beginning of this article. For each of the given values of
k, the irrationality measure µ(α ) were obtained by taking a = 1, b = 7. Non-quadraticity
k
measures µ (α ) have been derived by taking a = 2, b = 23 for k = 6 and a = 1, b = 13
2 k
for k = 8,10,12. Note that we couldn’t estimate the quadratic irrationality for odd values
of k because of the high growth rate of the “denominators” denoted as q in Lemma 5.
n
Remark 2. Using the same approach the author proved a few theorems. If one takes
y = k 1 i√2k 1 in (2) instead of x and uses the sketch of the proof Theorem 1, one can
k − −k − k
prove
Theorem 3. Let β = √2k 1arctan √2k 1. Then we have
k − k −1
−
k µ(β ) 6 µ (β ) 6 k µ(β ) 6 µ (β ) 6
k 2 k k 2 k
2 4.60105... — 8 3.66666... 14.37384...
4 3.94704... 44.87472... 10 3.60809... 12.28656...
6 3.76069... 19.19130... 12 3.56730... 11.11119...
One can read the proof of Theorem 3 in [8].
If onetakes y = 1 i√3 in(2)insteadofx anduses thesketch oftheproofTheorem 1
k 2− 6 k
(one needs Lemma 4 in a somewhat different form), one can prove
Theorem 4. For any ε > 0 there exists 0 < q(ε) Z such that for any q > q(ε),p ,p Z
1 2
∈ ∈
we have
p π p
max ln3 1 , 2 > q 3.86041 ε.
− ···−
− q √3 − q
(cid:18)(cid:12) (cid:12) (cid:12) (cid:12)(cid:19)
(cid:12) (cid:12) (cid:12) (cid:12)
One can read the proof of Theorem 4 in [8].
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
If one considers the integral (2), where
x+10n+1 x+9n+1 x+8n+1 x+7n+1
A(x) = ,
10n+1 8n+1 6n+1 4n+1
(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
takes x = √2k +1ln k+1 √2k+1, then using the sketch of the proof Theorem 2 one can
k −k
prove
Theorem 5. Let α = √2k +1ln √2k+1 1. Then we have
k √2k+1−+1
k µ (α ) 6 k µ (α ) 6 k µ (α ) 6 k µ (α ) 6 k µ (α ) 6
2 k 2 k 2 k 2 k 2 k
2 18.5799... 8 10.3786... 13 245.5913... 16 9.03034... 19 55.9694...
4 12.8416... 10 9.86485... 14 9.23973... 17 71.3960... 20 8.7192...
6 11.2038... 12 9.50702... 15 105.4297... 18 8.86054... 21 47.1243...
One can read the proof of Theorem 5 in [9].
The author would like to express his deep gratitude to his scientific advisor Yuri
Nesterenko and to Nikolai Moshchevitin for their interest in the research and their timely
advice.
7
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