ON THE RANK OVER ℚ OF SOME REAL MATRICES

Part 1.

Leonid A. Gutnik

To my daughters Tanya and Natasha

”While there is life there is hope,” said Sir Henry.

H.Rider Haggard. King Solomon’s mines.

Table of contents

§0. Introduction.

§1. Quasi-irrational matrices.

§2. On the zeros of some functions.

§3. Statement of the main results.

§4. Deduction of theorem 2 from theorem 1.

§5. Theorem 1 (the beginning of the proof).

§6. Upper bounds for the sums of certain series.

§7. On the arithmetic properties of the coefficients of some simple fractions.

§8. The first progress in our attempts to satisfy the conditions of Lemma 1;

the plan of the further proceedings.

§9. Linear independence over ℝ of certain sequences in ℚ^{ℕ}.

§10. Theorem 1 (the end of the proof).

§0. Introduction.

The arithmetic nature of the special values of the polylogarithm

F (z, k) = L_{k}(z) = ∑_{n=1}^{∞} n^{−k}z^{n}

in the rational points z = \frac{b}{a}, where b ∈ ℕ, a ∈ ℤ, b < |a| , and (a, b) = 1, has interested many mathematicians. An informative survey of the subject can be found in an interesting paper by Steffen Eckmann [11]. We cite:

”Neben vielen anderen Irrationalitätsaussagen konnte Maier [9] durch explicite Konstruktion von rationalen Approximationen zeigen, daß Dilogarithmusfunktion F (x, 2) an rationalen Stellen x_{0} = \frac{b}{a} mit |a| > b^{3}2^{8}e^{2} irrationale verte annimt. Chudnovski [1] entwickelte Maiers Methode weiter und bewies u.a.: Gilt a >> k^{k}(k + 1)^{−k−1} exp(k^{2}) für a ∈ ℕ, so ist F (a^{−1}, k) irrational. Gutnik [8] verallgemeinerte diese Ergebnisse und zeigte, daß die Werte der Polylogarithmusfunktionen

Typeset by AMS-TEX

an bestimmten Stellen linear unabhängig über dem Körper rationalen Zahlen (kurz: ℚ - l.u.) sind:

Satz 1a

Ist x_{0} = \frac{b}{a} mit (a, b) ∈ ℤ × ℕ, so sind für s ∈ ℕ die s + 1 Zahlen

1, F (x_{0}, 1), F (x_{0}, 2), …, F (x_{0}, s)

ℚ - l.u., falls |a| > b^{s+1}s^{−1} exp(s^{2}).

In [1] wurde ein ähnliches Resultat über obige Funktionen an gewissen Stellen a^{−1} mit a ∈ ℕ behauptet. Durch eine andere Beweismethode, welche Padé-Approximation benutzt, konnte Nikishin bereits in [10] eine etwas schwächere Version des Satzes 1a zeigen:

Satz 1b

Ist x_{0} = \frac{b}{a} mit (a, b) ∈ ℤ × ℕ,so sind für s ∈ ℕ die s + 1 Zahlen

1, F (x_{0}, 1), F (x_{0}, 2), …, F (x_{0}, s)

ℚ-l.u.,falls |a| > b^{s+1} exp{s(s − 1)(lns + 1 + log 4)}.

Man beachte hierbei, daß in [10] gegebene Vorraussetzung eine Korrektur erfahren musste, da dort auf seite 388 bei einer Abschätzung zwei Faktoren unberücksichtig geblieben waren.”

The proof of Theorem 1a mentioned above, which was given in [8], made use of Mejer’s function; later I have found an elementary proof of this result [15]. That elementary proof, with some generalisations, is presented here.

Acknowledgments. I am obliged to Prof. P. Bundshuh for a reprint of the paper [11], to Prof. B.Z. Moroz for translating this paper from the Russian, and to Prof. A.G. Aleksandrov for his constant support.

§1. Quasi-irrational matrices.

Let m ∈ ℕ, n ∈ ℕ. Let E_{m} denote the unit m × m− matrix.

Definition 1. A real m × n−matrix A = (a_{i,k}) is said to be quasi-irrational, if the columns of the matrix

(1)  (A, E_{m})

are linearly independent over ℚ.

Thus the real numbers

(2)  1, a_{1}, …, a_{n}

are linearly independent over ℚ if and only if the row (a_{1}, …, a_{n}) is quasiirrational.

Lemma 1. The following assertions are equivalent:

(i) for no column X in ℚ^{n}\{0} all the entries of the column AX are rational numbers;

(ii) there are m + n sequences of integers

α_{l}(ν), ν ∈ ℕ,

where l = 1 , …, m + n, satisfying the condition

(3)  lim_{ν→∞} |−α_{m+k}(ν) + ∑_{j=1}^{m} a_{j,k}α_{j}(ν)| = 0,

where k = 1 , …, n, and such that for each κ ∈ ℕ the m + n sequences

(4)  α_{l}(κ + μ), μ ∈ ℕ,

where l = 1 , …, m + n, are linearly independent over ℝ; (iii) the matrix A is quasi-irrational. Proof. Condition (i) is clearly equivalent to (iii). Suppose (ii) hold true, then there is a column X in ℚ^{n}\{0} such that all the

entries of the column AX are rational numbers. On choosing a suitable d ∈ ℕ and replacing X by dX, if necessary, we may assume that

X ∈ ℤ^{n}\{0} , AX ∈ ℤ^{m}.

Let x_{k}, for k = 1 , …, n, be the k−th entry of the column X. It follows from relation (3) that

(5)  lim_{ν→∞} ((− ∑_{k=1}^{n} x_{k}α_{m+k}(ν)) + ∑_{j=1}^{m} α_{j}(ν) ∑_{k=1}^{n} a_{j,k}x_{k}) =

∑_{k=1}^{n} x_{k} lim_{ν→∞} (−α_{m+k}(ν) + ∑_{j=1}^{m} a_{j,k}α_{j}(ν)) = 0.

Therefore there is such an integer κ ∈ ℕ that for any μ ∈ ℕ the absolute value of the integer

(6)  (− ∑_{k=1}^{n} x_{k}α_{m+k}(κ + μ)) + ∑_{j=1}^{m} α_{j}(κ + μ) ∑_{k=1}^{n} a_{j,k}x_{k}

is smaller than 1; consequently, the integer (6) is equal to 0. However the number (6) can not be equal to 0 for every μ in ℕ because sequences (4) are linearly independent over ℝ and x_{k} ≠ 0 for some k = 1 , …, n. Suppose (iii) hold true. Let us consider, for each i ∈ {1 , …, m}, a system of n linear forms

(7)  L_{i,k} = y_{0}a_{i,k} − y_{k}, k = 1 , …, n,

of n+1 independent variables y_{0} , …, y_{n}. Since the number (n+1) of the variables exceeds the number n of the linear forms, it follows from the well-known lemma (cf., for instance, [12, p.131]; the cited lemma is a consequence of the ”pigeon-hole” Dirichlet principle) that, given μ ∈ ℕ, for each k = 1 , …, n there are real numbers

(8)  θ_{k}(i + m(μ − 1)) ∈ [−1, 1],

and integers y_{0} , …, y_{n} such that

L_{i,k} = \frac{θ_{k}(i + m(μ − 1)}{i + mμ},

and y_{i} ≠ 0 for some i; therefore y_{0} ≠ 0. For these values of the variables y_{0} , …, y_{n}, we let

α_{i}(i + m(μ − 1)) = y_{0}, α_{m+k}(i + m(μ − 1)) = y_{k},

with k = 1 , …, n, and

α_{l}(i + m(μ − 1)) = 0

for l ∈ {1 , …, m}\{i}. We have thus constructed m + n sequences of integers

α_{l}(ν), ν ∈ ℕ,

with l = 1 , …, m + n, satisfying the following conditions:

α_{i}(ν) ≠ 0

for ν ≡ i (mod m) and i ∈ {1 , …, m},

α_{l}(ν) = 0

for ν ≡ i (mod m) and l ∈ {1 , …, m}\{i}, and

(9)  −α_{m+k}(ν) + ∑_{j=1}^{m} a_{j,k}α_{j}(ν) =

−α_{m+k}(ν) + a_{i,k}α_{i}(ν) = \frac{θ_{k}(ν)}{ν + 1},

for ν ≡ i (mod m) , k = 1 , …, n.

Suppose that for some κ ∈ ℕ the m+n sequences (4), arising from the sequences α_{l}(ν) we have just constructed when the first κ terms are removed, turn out to be linearly dependent over ℝ. Since α_{l}(ν) ∈ ℤ for l = 1 , …, m + n, ν ∈ ℕ, the sequences (4) are actually linearly dependent over ℚ; therefore one can find m + n integers t_{l}, l = 1 , …, m + n, such that

(10)  ∑_{i=1}^{m} t_{i}α_{i}(κ + μ) + ∑_{k=1}^{n} t_{k+m}α_{m+k}(κ + μ)) = 0

and t_{j} ≠ 0 for some j.

It follows from (9) and (10) that

(11) ∑_{i=1}^{m} α_{i}(κ + μ)(t_{i} + ∑_{k=1}^{n} t_{k+m}a_{i,k}) = −(κ + μ + 1)^{−1} ∑_{k=1}^{n} t_{k+m}θ_{k}(κ + μ).

For i ∈ [1, m] ∩ ℕ, let

(12)  μ ≡ −κ + i (mod m).

It follows from (9) that

(13)  |t_{i} + ∑_{k=1}^{n} t_{k+m}a_{i,k} |≤ (κ + μ + 1)^{−1} ∑_{k=1}^{n} |t_{k+m}| .

On passing to the limit μ → ∞, one obtains from (12) and (13) the following equation

t_{i} + ∑_{k=1}^{n} t_{k+m}a_{i,k} = 0,

with i = 1 , …, m, in contradiction to (iii). This proves the Lemma.

Remark. If the condition (iii) holds true, then it follows from [12, p. 131, Lemma 1] and the above construction that, for l = 1 , …, m+n, there is a sequence α_{l}(ν) satisfying (ii) and the extra condition

α_{l}(ν) = O(ν^{n+1}).

§2. On the zeros of some functions.

Suppose that 0 < α, 0 < δ, 0 ≤ η, 0 < D, and n ∈ ℕ. Let

f_{0} = f_{0}(η) = f_{0}(α, δ, D, η) = (η + δ)^{α+1} − (1 + δ)^{α+1}Dη,

(14)

f_{0,n}^{∧} = f_{0,n}^{∧}(η) = f_{0,n}^{∧}(α, δ, D, η) = ∑_{k=0}^{n−1}(η + δ)^{(α+1)k}((1 + δ)^{α+1}Dη)^{n−k}.

Since (\frac{∂}{∂η})^{2}f_{0} > 0, the function f_{0} is convex and has no more than two zeros, counted according to their multiplicity, in the interval [0, +∞). Consider the function

(15)  f_{1} = f_{1}(η) = f_{1}(α, δ, η) = (η + δ)^{α+1}η^{−1}(1 + δ)^{−α−1}

defined on the interval (0, +∞); since

(\frac{∂}{∂η} f_{1}) (α, δ, η) = (η + δ)^{α}η^{−2}(1 + δ)^{−α−1}(αη − δ),

the function f_{1} attains the global minimum

(16)  A_{1}(α, δ) = (\frac{α + 1}{1 + δ})^{α+1} (\frac{δ}{α})^{α} .

in the point η = \frac{δ}{α}. Thus

(17)  f_{1}(α, δ, η) ≥ A_{1}(α, δ)

for η ∈ (0, +∞); moreover, relation (17) is a strict inequality if and only if η ≠ \frac{δ}{α}. Since

(18)  f_{0}(α, δ, D, η) = η(1 + δ)^{α+1}(f_{1}(α, δ, η) − D),

(\frac{∂}{∂η} f_{0}) (α, δ, D, η) = \frac{f_{0}(α, δ, D, η)}{η} + η(1 + δ)^{α+1} (\frac{∂}{∂η} f_{1}) (α, δ, η)

for η ∈ (0, +∞), the function f_{0}(η) = f_{0}(α, δ, η) has no zeros in the interval [0, +∞) if D < A_{1}(α, δ) and has precisely two zeros in this interval, counted in accordance with their multiplicity, if D ≥ A_{1}(α, δ); let η_{1}(α, δ, D) and η_{2}(α, δ, D) stand for the smallest and the biggest of these zeros accordingly. Relations (17) and (18) show that if D > A_{1}(α, δ), then each of the zeros η_{k}(α, δ, D), k = 1, 2, is a simple zero and

(19)  η_{1}(α, δ, D) < \frac{δ}{α} < η_{2}(α, δ, D).

On the other hand, if D = A_{1}(α, δ), then

(20)  η_{1}(α, δ, D) = \frac{δ}{α} = η_{2}(α, δ, D),

and therefore it follows from (18) that η = \frac{δ}{α} is a double zero of the function f_{0}. Let

(21)  A_{0} = A_{0}(η) = η^{α}(1 + η)^{−α−1}

for η ∈ [0, +∞). Since

(\frac{∂}{∂η} A_{0}) (η) = η^{α−1}(1 + η)^{−α−2}(α − η)

for η ∈ (0, +∞), the function A_{0}(η) has a global minimum in that interval at the point η = α. Thus A_{0}(α) ≥ A_{0}(η) for η ∈ [0, +∞), where the equality is attained in the only point η = α. Since A_{1}(α, δ) = (A_{0}(δ))/A_{0}(α), it follows that

(22)  A_{1}(α, δ) ≤ 1,

the equality in (22) being attained in the only point δ = α.

Lemma 2. Suppose that 0 < δ ≤ α, 1 ≤ D. Then η_{1}(α, δ, D) is a simple root of the function f_{0} and η_{1}(α, δ, D) < δ/α ≤ 1 as soon as α − δ + D > 1, but η_{1}(α, α, 1) is a double root of f_{0} and η_{1}(α, α, 1) = 1.

Proof. If δ < α, then A_{1}(α, δ) < 1 ≤ D and it follows from (19) that η_{1}(α, δ, D) is a simple root of the function f_{0}(α, δ, D, η), satisfying the inequality

η_{1}(α, δ, D) < \frac{δ}{α} < 1.

If δ = α and D > 1, then A_{1}(α, δ) = 1 < D and it follows from (19) that η_{1}(α, δ, D) is a simple root of the function f_{0}(α, δ, D, η) satisfying the relation

η_{1}(α, δ, D) < \frac{δ}{α} = 1.

Finally, if α = δ and D = 1, then A_{1}(α, δ) = 1 = D and it follows from (20) that η_{1}(α, δ, D) = \frac{δ}{α} = 1 is a double root of the function f_{0}(α, δ, D, η). The lemma is proved.

Let η = uδ^{α+1}/D and let

(23)  f̃_{0}(u) = f_{0}(uδ^{α+1}/D) = δ^{α+1}((uδ^{α}/D + 1)^{α+1} − (1 + δ)^{α+1}u).

If

(24)  D ≥ δ^{α−1},

then f̃_{0}(1) ≤ 0 and

(25)  η_{1}(α, δ, D) ≤ δ^{α+1}/D.

§3. Statement of the main results.

For 0 < η ≤ 1, 0 < δ ≤ α, let

ρ̃(α, δ, η) = (\frac{η}{δ} )^{δ} (\frac{1 − η}{1 + δ} )^{α−δ}

if δ < α, and let

ρ̃(α, α, η) = (\frac{η}{α})^{α} .

For k ∈ ℕ, |x| < 1, let

L_{k}(x) = ∑_{n=1}^{∞} n^{−k}x^{n},

and let Λ_{m,s}(x) = (a_{i,j}(x)) denote the m × s− matrix, whose element at the intersection of its i-th row and its j-th column is equal to

a_{i,j}(x) = \binom{i + j − 2}{i − 1}L_{i+j−1}(x),

for each i = 1 , …, m and each j = 1 , …, s.

The following two theorems are the main results of this paper.

Theorem 1. Suppose that m ∈ ℕ, s ∈ ℕ, n ∈ s−1+ℕ, α = m/n, 0 < δ ≤ α, a ∈ ℤ, b ∈ ℕ, z = \frac{a}{b}, and D = |z|^{1/n} > 1. If

(26)  ρ̃(α, δ, η_{1}(α, δ, D)) exp (\frac{s}{n} + (α − \frac{1}{n}) max(δ, 1)) b^{1/n} < 1,

then the matrix Λ_{m,s} (\frac{b}{a}) is quasi-irrational.

Theorem 2. Let m ∈ ℕ, n ∈ ℕ, a ∈ ℤ, b ∈ ℕ. If

(27)  |a| > (\frac{m}{n})^{m} exp (\frac{n}{m} (n + (m − 1) max (\frac{m}{n}, 1))) b^{\frac{m+n}{m}} ,

then the matrix Λ_{m,n} (\frac{b}{a}) is quasi-irrational.

On letting m = 1, one observes that Theorem 1a in the Eckmann’s survey [11], cited in § 1, is a corollary of our Theorem 2.

§4. Deduction of theorem 2 from theorem 1.

Let us start by deducing theorem 2 from theorem 1 when δ = α and s = n. Since the inequality (25) follows from (24), the following inequality

(28)  η_{1}(α, α, D) ≤ α^{α+1}/D

is a consequence of the inequality D ≥ α^{α−1}, and therefore it follows from the inequality

(29)  |z| > (\frac{m}{n})^{m−n} .

Thus it follows from (29) that

ρ̃(α, α, η_{1}(α, α, D)) exp (\frac{s}{n} + (α − \frac{1}{n})max(δ, 1)) b^{1/n} ≤

(α^{α}/D)^{α} exp(n/n + (α − 1/n) max(α, 1))b^{1/n}.

Since the following inequality

(30)  (α^{α}/D)^{α} exp(n/n + (α − 1/n)) max(α, 1))b^{1/n} < 1

is equivalent to (27), it suffices to prove that the inequality (27) implies (29). But it follows from (27) that

|z| > (\frac{m}{n})^{m} e^{\frac{n}{m} (n+(m−1))} ≥ (\frac{m}{n})^{m} (\frac{n}{m})^{n} = (\frac{m}{n})^{m−n} .

Thus Theorem 2 is indeed a consequence of Theorem 1.

§5. Theorem 1 (beginning of the proof).

It follows from the relation (19) and Lemma 2 that

(\frac{∂}{∂η} f_{0}) (α, D, η_{1}(α, δ, D)) = \frac{α}{η_{1}(α, δ, D)}(η + δ)^{α}(η_{1}(α, δ, D) − δ/α)

and, consequently, that the function ρ̃(α, δ, η_{1}(α, δ, D)) is continuous at the point (α, δ, D) as soon as the following condition

(31)  0 < δ ≤ α, 1 ≤ D, α − δ + D > 1

is satisfied. Therefore, in view of the supposition of Theorem 1, it may be assumed that

(32)  δ ∈ ℚ, δ = d^{∨}/d^{∧},

where d^{∨} ∈ ℕ, d^{∧} ∈ ℕ.

10 LEONID A. GUTNIK

§6. Upper bounds for the sums of certain series.

For ν ∈ ℕ, let

R(t, ν) = \frac{∏_{k=1}^{νd^{∨}}(t − k)^{n}}{∏_{k=0}^{νd^{∧}+1}(t + k)^{m}} \frac{((νd^{∧} + 1)!)^{m}}{((νd^{∨})!)^{n}} .

Given μ ∈ ℕ − 1, let us consider the series

(33)  ∑_{t=1}^{∞} \frac{(−1)^{μ}}{μ!} ((\frac{∂}{∂t})^{μ} R) (t, ν)z^{−t}.

Since d^{∨}n ≤ d^{∧}m, the series (33) converges for |z| ≥ 1; let ψ_{μ}(z; ν) stand for the sum of this series. Clearly,

(34)  |ψ_{0}(z; ν)| ≤ ψ_{0}(|z|; ν).

Let

R_{1}(t, ν) = \frac{∏_{k=1}^{νd^{∨}}(t − k)^{n}}{∏_{k=0}^{νd^{∧}−1}(t + k)^{m}} \frac{((νd^{∧})!)^{m}}{((νd^{∨})!)^{n}} ,

(35)

R_{2}(t, ν) = \frac{(νd^{∧} + 1)^{m}}{(t + νd^{∧})^{m}(t + νd^{∧} + 1)^{m}} .

Since R_{1}(t, ν) = 0 for t = 1, , …, d^{∨}ν, it follows that

(36)  ψ_{0}(z; ν) = ∑_{t=1+νd^{∨}}^{∞} R_{1}(t, ν)z^{−t}R_{2}(t, ν),

and

∑_{t=1+νd^{∨}}^{∞} R_{2}(t, ν) ≤ (νd^{∧} + 1)^{m} ∑_{t=2}^{∞} \frac{1}{(t + νd^{∧})^{2m}} ≤

ON THE RANK OVER ℚ OF SOME REAL MATRICES 11

(νd^{∧} + 1)^{m} ∫_{1}^{∞} \frac{1}{(t + νd^{∧})^{2m}} = \frac{1}{(2m − 1)(1 + νd^{∧})^{m−1}} ≤ 1,

(38)  ψ_{0}(|z|; ν) ≤ sup_{t≥1+νd^{∨}} R_{1}(t, ν)|z|^{−t},
(39)  R_{1}(t + 1, ν)|z|^{−t−1} = R_{1}(t, ν)|z|^{−t}t^{m+n}(t − νd^{∨})^{−n}(t + νd^{∧})^{−m}z^{−1}.

Let t = ντd^{∧}. Then it follows from (39) that

(40) R_{1}(t + 1, ν)|z|^{−t−1} = R_{1}(t, ν)|z|^{−t}τ^{m+n}(τ − δ)^{−n}(τ + 1)^{−m}|z|^{−1}.

Let further

(41)  g(τ) = τ^{m+n}(τ − δ)^{−n}(τ + 1)^{−m}|z|^{−1} − 1

for δ < τ < +∞ and let

(42)  τ = \frac{δ + η}{1 − η}

with 0 ≤ η < 1. Let D = |z|^{1/n}. On substituting relation (42) in (41), one obtains

g(τ) = h(η) = \frac{(η + δ)^{m+n} − (1 + δ)^{m+n}(Dη)^{n}}{(1 + δ)^{m+n}(Dη)^{n}} =

\frac{f_{0}(α, δ, D, η)f_{0,n}^{∧}(α, δ, D, η)}{(1 + δ)^{m+n}(Dη)^{n}},

where D = |z|^{1/n}, and the functions f_{0}(α, δ, D, η), f_{0,n}^{∧}(α, δ, D, η) are given by the equalities (14). Let

τ_{1}(α, δ, D) = \frac{δ + η_{1}(α, δ, D)}{1 − η_{1}(α, δ, D)}.

Then

δ < τ_{1}(α, δ, D),

η_{1}(α, δ, D) = \frac{τ_{1}(α, δ, D) − δ}{τ_{1}(α, δ, D) + 1},

g(τ) < 0 for τ ∈ (δ, τ_{1}(α, δ, D)), and g(τ) > 0 for τ > τ_{1}(α, δ, D). Therefore R_{1}(t + 1, ν)|z|^{−t−1} ≤ R_{1}(t, ν)|z|^{−t}

12 LEONID A. GUTNIK

for t ≥ νd^{∧}τ_{1} > νd^{∨} and

R_{1}(t + 1, ν)|z|^{−t−1} ≥ R_{1}(t, ν)|z|^{−t}

for νd^{∨} + 1 ≤ t ≤ νd^{∧}τ_{1}. Consequently, one obtains

(43) sup_{t≥1+νd^{∨}} R_{1}(t, ν)|z|^{−t} = R_{1}(νd^{∧}τ_{1} + θ_{ν}; ν)|z|^{−t−1} ≥ R_{1}(t, ν)|z|^{−νd^{∧}τ_{1}−θ_{ν}}

for νd^{∧}τ_{1} + θ_{ν} ∈ ℕ and 0 ≤ θ_{ν} < 1. It follows from (35) that

R_{1}(t, ν)|z|^{−t} = \frac{(Γ(t))^{m+n}(Γ(1 + νd^{∧}))^{m}}{(Γ(t − νd^{∨}))^{n}(Γ(t + νd^{∧}))^{m}Γ(1 + νd^{∨})^{n}} |z|^{−t}.

To estimate that quantity for t = νd^{∧}τ, τ = τ_{1} + θ_{ν}/(νd^{∧}), |z| > 1, ν ≥ 2 we use the Stirling formula:

logΓ(x) = (x − \frac{1}{2})logx − x + O(1).

In view of the inequality the inequality τ − δ ≥ τ_{1} − δ > 0, it follows that

log(R_{1}(t, ν)|z|^{−t}) =

(m + n)((νd^{∧}τ − 1/2)log(νd^{∧}τ) − νd^{∧}τ)−
n((νd^{∧}(τ − δ) − 1/2) log(νd^{∧}(τ − δ)) − νd^{∧}(τ − δ))−
m((νd^{∧}(τ + 1) − 1/2) log(νd^{∧}(τ + 1)) − νd^{∧}(τ + 1))+
m((νd^{∧} + 1/2) log(νd^{∧}(τ + 1)) − νd^{∧})−

n((νd^{∨} + \frac{1}{2})(log(νd^{∨}) − νd^{∨})−

νd^{∧}τ log(|z|) + O(1) =

A(τ)ν log(ν) + B(τ)ν + C(τ) log(ν) + O(1),

where

A(τ) = (m + n)d^{∧}τ − nd^{∧}(τ − δ) − md^{∧}(τ + 1) − nd^{∨} + md^{∧} =
(m + n)d^{∧}τ − nd^{∧}τ − md^{∧}τ + nd^{∧}δ − nd^{∨} − md^{∧} + md^{∧} = 0,
B(τ) = (log(d^{∧}) − 1)×
((m + n)d^{∧}τ − nd^{∧}(τ − δ) − md^{∧}(τ + 1) + md^{∧} − nd^{∨})−

ON THE RANK OVER ℚ OF SOME REAL MATRICES 13

−nd^{∧}δlog(δ)+ (m + n)d^{∧}τ log(τ) − nd^{∧}(τ − δ) log(τ − δ)− −md^{∧}(τ + 1) log(τ + 1) − d^{∧}τ log(|z|) =

−nd^{∧}δ log(δ)+

(m + n) (d^{∧}τ_{1} + \frac{θ_{ν}}{ν} ) log (τ_{1} (1 + \frac{θ_{ν}}{d^{∧}τ_{1}ν} )) −

n (d^{∧}(τ_{1} − δ) + \frac{θ_{ν}}{ν} ) log ((τ_{1} − δ) (1 + \frac{θ_{ν}}{d^{∧}(τ_{1} − δ)ν} )) −

m (d^{∧}(τ_{1} + 1) + \frac{θ_{ν}}{ν} ) log ((τ_{1} + 1) (1 + \frac{θ_{ν}}{d^{∧}(τ_{1} + 1)ν} )) −

(d^{∧}τ_{1} + \frac{θ_{ν}}{ν} ) log(|z|) =

(m + n)d^{∧}τ_{1} log(τ_{1}) − nd^{∧}(τ_{1} − δ) log(τ_{1} − δ) − md^{∧}(τ_{1} + 1) log(τ_{1} + 1)−

nd^{∧}δ log(δ) − d^{∧}τ_{1} log(|z|) + O(1/nu),

and

C(τ) = −(m + n)/2 + n/2 + m/2 + m/2 − n/2 = (m − n)/2.

Thus

t = νd^{∧}τ, τ = τ_{1} + θ_{ν}/(νd^{∧}), |z| > 1, ν ≥ 2,
log(R_{1}(t, ν)|z|^{−t}) =

ν(m + n)d^{∧}τ_{1} log(τ_{1}) − νnd^{∧}(τ_{1} − δ) log(τ_{1} − δ) − νmd^{∧}(τ_{1} + 1) log(τ_{1} + 1)−

νnd^{∧}δ log(δ) − νd^{∧}τ_{1} log(|z|) + O(1)

for

t = νd^{∧}τ, τ = τ_{1} + θ_{ν}/(νd^{∧}), |z| > 1, ν ≥ 2.

It follows from (41) that

(44)  (m + n) log(τ_{1}) − n log(τ_{1} − δ) − m log(τ_{1} + 1) − log(|z|) = 0.

On multiplying (44) by

νd^{∧}τ_{1}

and then substracting the resulting equality from the previous one, we obtain the following equality:

ψ_{0}(|z|; ν) = ν^{(m−n)/2} (\frac{(τ_{1} − δ)^{δ}}{(τ_{1} + 1)^{α}(δ)^{δ}} )^{nνd^{∧}} exp(O(1)) =

ν^{(m−n)/2} ((\frac{η_{1}}{δ} )^{δ} (\frac{1 − η_{1}}{1 + δ} )^{α−δ})^{nνd^{∧}} exp(O(1)).

Since, for t ≥ νd^{∨} + 1, the product R(t, ν) consists of O(1)ν factors and since the quotient of each of the factors to its partial derivative with respect to t can be bounded above as O(1), it follows that

((\frac{∂}{∂t})^{μ} R) (t, ν) = ν^{μ}R(t, ν)O(1)

for μ ∈ ℕ − 1 and t ≥ νd + 1. Therefore

ψ_{μ}(|z|; ν) = ν^{μ}ψ_{0}(|z|; ν)O(1) =

(45)

ν^{s−1+(m−n)/2} ((\frac{η_{1}}{δ} )^{δ} (\frac{1 − η_{1}}{1 + δ} )^{α−δ})^{nνd^{∧}} O(1)

for μ = 0 , …, s − 1, where s ≤ n.

§7. On the arithmetic properties of the coefficients of some simple fractions.

Let

(46)  c_{1} = c_{1}(ν) = νd^{∨}, c_{2} = c_{2}(ν) = νd^{∧} + 1, u(ν) = c_{1}(ν) + c_{2}(ν).

It follows that

(47)  R(t, ν) = ∑_{k=0}^{c_{(}ν)} ∑_{λ=1}^{m} \frac{α_{k,λ}^{∼}(ν)}{(t + k)^{λ}} ,

with

α_{k,λ}^{∼} = \frac{1}{(m − λ)!} lim_{t→−k} (\frac{∂}{∂t})^{m−λ} ((t + k)^{m}R)(t, ν)

for k = 0 , …, c_{2}(ν) and λ = 1 , …, m. It follows from the well-known Leibniz differentiation rule that

(48)  α_{k,λ}^{∼} = (−1)^{nc_{1}(ν)+mk}\binom{c_{1}(ν) + k}{k}^{n}\binom{c_{2}(ν)}{k}^{m}×

∑_{\begin{matrix}λ_{1}+...+λ_{u(ν)}=m−λ \\ λ_{1}∈ℕ−1 ,... ,λ_{u(ν)}∈ℕ−1 \\ λ_{c_{2}(ν)−k+κ}≤n, \\ k=0 ,... ,c_{2}(ν),κ=k+1 ,... ,c_{1}(ν)+k \end{matrix}} Π_{k,λ}(ν; λ_{1} , …, λ_{u(ν)})

with

(49)  Π_{k,λ}(ν; λ_{1} , …, λ_{c_{1}(ν)+c_{2}(ν)}) =

(∏_{κ=1}^{k} \binom{m + λ_{κ} − 1}{λ_{κ}}κ^{−λ_{κ}}) ×

( ∏_{κ=1}^{c_{2}(ν)−k} \binom{m + λ_{k+κ} − 1}{λ_{k+κ}}(−κ)^{−λ_{k+κ}} ) ×

( ∏_{κ=k+1}^{c_{1}(ν)+k} \binom{n}{λ_{c_{2}(ν)−k+κ}}(−κ)^{−λ_{c_{2}(ν)−k+κ}} )

for k = 0 , …, c_{2}(ν) and λ = 1 , …, m. Let

c_{3}(ν) = min(c_{1}(ν), c_{2}(ν), c_{4}(ν) = max(c_{1}(ν), c_{2}(ν)),

(50)

c_{5}(ν) = u(ν) = c_{1}(ν) + c_{2}(ν)) = c_{3}(ν) + c_{4}(ν).

For a rational prime p, let v_{p} stand for the p-adic valuation on ℚ. Lemma 3. If p > c_{4}(ν), then

(51)  v_{p}(α_{k,λ}^{∼}) ≥ 0

for k = 0 , …, c_{2}(ν) and λ = 1 , …, m.

Proof. If p > max(c_{4}(ν), c_{1}(ν) + k), then the asserted inequality (51) is an immediate consequence of the equations (48) and (49). Suppose that

(52)  c_{4}(ν) < p ≤ c_{1}(ν) + k.

Then

(53)  c_{1}(ν) + c_{2}(ν)) ≤ 2c_{4}(ν) < 2p < p^{2},

and

(54)  v_{p} (\binom{c_{1}(ν) + k}{k}^{n}) = n.

Since the variable κ in the equation (49) satisfies the inequality

(55)  κ ≤ max(k, c_{2}(ν) − k, c_{1}(ν) + k)

and

\binom{n}{λ_{c_{2}(ν)−k+κ}} = 0

for λ_{c_{2}(ν)−k+κ} > n,then the assertion of the lemma now follows immediately from (48) and (49). The lemma is proved.

Let, as usual, π(x) stand for the number of rational primes p with p ≤ x, x ∈ ℝ, so that π(x) = 0 for x < 2. Let {α, β, γ} ⊂ ℝ. If π(α) < π(β) and 1 ≤ γ, let d_{α,β,γ} denote the smallest positive integer d satisfying the condition

(56)  v_{p} (\frac{d}{κ}) ≥ 0

for all primes p in the interval α < p ≤ β and all positive integers κ with κ ≤ γ. We let, by definition, d_{α,β,γ} = 1 if ether γ < 1, or π(β) ≤ π(α). Clearly,

(57)  d_{α,β,γ} = ∏_{α<p≤β} p^{[ln(γ)/ln(p)]} ≤ γ^{max(π(β)−π(α),0)}.

Lemma 4. If

(58)  1 ≤ α, 1 ≤ β, 1 ≤ γ ≤ γ_{0}α

for some γ_{0} > 0, then for every ε > 0 there exists C_{0}(γ_{0}, ε) > 0 such that

d_{α,β,γ} ≤ C_{0}(γ_{0}, ε) exp(α((max(β/α − 1, 0) + ε)),

(59)

d_{1,α,γ} ≤ C_{0}(γ_{0}, ε) exp α(1 + ε)).

Proof. We can assume that β ≤ γ. It follows from the relations (57) and (58) that

d_{α,β,γ} ≤

exp((log(γ_{0}/e) + log(αe))(max(β − α, 0)/ log(eα) + O(max(β − α, 0)/ log^{2}(eα)))) ≤

C_{0}(γ_{0}, ε) exp(α(max(β/α − 1, 0) + ε)),

(60)

d_{1,α,γ} ≤ exp((log(γ_{0}/e) + log(αe))(α/ log(eα) + O(α/ log^{2}(eα)))) ≤
C_{0}(γ_{0}, ε) exp(α(1 + ε)).

The lemma is proved.

Lemma 5.Each of the numbers

α_{k,λ}^{∼}(d_{1,c_{2}(ν),c_{5}(ν)})^{m−λ}(d_{c_{2}(ν),c_{1}(ν),c_{5}(ν)})^{min(m−λ,nc_{1}(ν))},

for k = 0 , …, c_{2}(ν), λ = 1 , …, m, is contained in ℤ.

Proof. Let us consider the equality

(61) Π_{k,λ}(ν; λ_{1} , …, λ_{u(ν)})(d_{1,c_{2}(ν),c_{5}(ν)})^{m−λ}(d_{c_{2}(ν),c_{1}(ν),c_{5}(ν)})^{min(m−λ,nc_{1})} =

(∏_{κ=1}^{k} \binom{m + λ_{κ} − 1}{λ_{κ}} (\frac{d_{1,c_{2}(ν),c_{5}(ν)}}{κ} )^{λ_{κ}}) ×

( ∏_{κ=1}^{c_{2}(ν)−k} \binom{m + λ_{k+κ} − 1}{λ_{k+κ}} (\frac{−d_{1,c_{2}(ν),c_{5}(ν)}}{κ} )^{λ_{k+κ}} ) ×

( ∏_{κ=k+1}^{c_{1}(ν)+k} \binom{n}{λ_{c_{2}(ν)−k+κ}} (\frac{−d_{1,c_{2}(ν),c_{5}(ν)}d_{c_{2}(ν),c_{1}(ν),c_{5}(ν)}}{κ} )^{λ_{c_{2}(ν)−k+κ}} ) ×

(d_{c_{2}(ν),c_{1}(ν),c_{5}(ν)}(ν))^{λ_{c_{5}(ν)+1}} ,

where

λ_{c_{5}(ν)+1} = min(m − λ, nc_{1}(ν)) − ∑_{κ=k+1}^{c_{1}(ν)+k} λ_{c_{2}(ν)−k+κ}

depends, of course, on λ and k and, in view of (48), λ_{c_{5}(ν)+1} ≥ 0. Further, c_{2}(ν) < c_{5}(ν); moreover, if 1 ≤ κ ≤ k, then κ and therefore any prime dividing κ do not exceed c_{2}(ν). Consequently,

v_{p} (∏_{κ=1}^{k} \binom{m + λ_{κ} − 1}{λ_{κ}} (\frac{d_{1,c_{2}(ν),c_{5}(ν)}}{κ} )^{λ_{κ}}) ≥ 0

for every prime number p,

v_{p} ( ∏_{κ=1}^{c_{2}(ν)−k} \binom{m + λ_{k+κ} − 1}{λ_{k+κ}} (\frac{−d_{1,c_{1}(ν),c_{5}(ν)}}{κ} )^{λ_{k+κ}} ) ≥ 0

for every prime number p, and

v_{p} ( ∏_{κ=k+1}^{c_{1}(ν)+k} \binom{n}{λ_{c_{2}(ν)−k+κ}} (\frac{−d_{1,c_{2}(ν),c_{5}(ν)}d_{c_{2}(ν),c_{1}(ν),c_{5}(ν)}}{κ} )^{λ_{c_{2}(ν)−k+κ}} ) ≥ 0

for every prime number p not exceeding c_{4}(ν). Therefore

v_{p} (α_{k,λ}^{∼}(d_{2,c_{1}(ν),c_{5}(ν)})^{m−λ}(d_{c_{2}(ν),c_{1}(ν),c_{5}(ν)})^{min(m−λ,nc_{1}(ν))}) ≥ 0

for every prime number p not exceeding c_{4}(ν). It follows now from Lemma 3 that

(62)  v_{p} (α_{k,λ}^{∼}(d_{1,c_{1}(ν),c_{5}(ν)})^{m−λ}(d_{c_{2}(ν),c_{1}(ν),c_{5}(ν)})^{min(m−λ,nc_{1}(ν))}) ≥ 0

for every prime number p. The lemma is proved.

§8. The first progress in our attempts to satisfy the conditions of Lemma 1; the plan of the further proceedings.

Since

∑_{t=1}^{∞} z^{−t}/(t + k)^{λ+μ} = z^{k}L_{λ+μ}(z^{−1}) − ∑_{l=1}^{k} z^{k−l}/l^{λ+μ}

with λ = 1 , …, m, μ ∈ ℕ − 1, it follows from (33) that

(64)  ψ_{μ}(z; ν) = −P_{m+1+μ}(z; ν) + ∑_{λ=1}^{m} P_{λ}(z; ν)\binom{μ + λ − 1}{μ}L_{λ+μ}(z^{−1}),

where

P_{λ}(z; ν) = ∑_{k=0}^{c_{2}(ν)} α_{k,λ}^{∼}z^{k}

P_{m+1+μ}(z; ν) = ∑_{l=0}^{c_{2}(ν)−1} z^{l} ∑_{k=l+1}^{c_{2}(ν)} ∑_{λ=1}^{m} \binom{μ + λ − 1}{μ}\frac{α_{k,λ}^{∼}}{(k − l)^{λ+μ}}

for λ = 1 , …, m, μ ∈ ℕ − 1.

Lemma 6. Let a ∈ ℤ, b ∈ ℕ, z = \frac{a}{b}, s ∈ ℕ, s ≤ n, ν ∈ ℕ, and

(66)  d_{s}^{∼}(ν) = (d_{1,c_{2}(ν),c_{5}(ν)})^{m+s−1}(d_{c_{2}(ν),c_{1}(ν),c_{5}(ν)})^{min(m−1,nc_{1}(ν))}.

Then

P_{λ}^{∗}(z; ν) = b^{c_{2}(ν)}d_{s}^{∼}(ν)P_{λ}(z; ν) ∈ ℤ, P_{m+1+μ}^{∗} = b^{c_{2}(ν)}d_{s}^{∼}(ν)P_{m+1+μ}(z; ν) ∈ ℤ,

(67)

P_{m+1+μ}^{∗} = b^{c_{2}(ν)}d_{s}^{∼}(ν)P_{m+1+μ}(z; ν) ∈ ℤ

where λ = 1 , …, m and μ = 0 , …, s − 1.

Proof. The assertion of the lemma is an immediate consequence of the equalities (65) and Lemma 5.

Lemma 7. For every ε > 0, there exists c(ε) > 1 such that

(b^{c_{2}(ν)}d_{s}^{∼}(ν))^{1/(nd^{∧}ν)} ≤

(68)

c(ε) (exp(s/n + (α − 1/n) max(δ, 1))b^{1/n})^{1+ε} ,

where α = m/n and δ = d^{∨}/d^{∧}.

Proof. Since c_{5}(ν)/c_{2}(ν) < 1 + δ, we may apply Lemma 4 with α = c_{2}(ν), β = c_{1}(ν), γ = c_{5}(ν), and γ_{0} = 1 + δ_{0}. By Lemma 4, for every ε > 0, there exists c_{0}(ε) > 1 such that

(d_{1,c_{2}(ν),c_{5}(ν)})^{m+s−1} ≤ (c_{0}(ε))^{m+s−1} exp(d^{∧}ν(1 + 1/d^{∧}ν)(1 + ε/2)(m + s − 1))

and

d_{c_{2},c_{1}(ν),c_{5}(ν)})^{m−1} ≤ (c_{0}(ε))^{m−1} exp(d^{∧}ν(1 + 1/d^{∧}ν)max(δ − 1, 0)(1 + ε/2)(m − 1))

since c_{1}(ν)/c_{2}(ν) < δ_{0}. Therefore it follows from (66) that

d_{s}^{∼}(ν) = (d_{1,c_{2}(ν),c_{5}(ν)})^{m+s−1}(d_{c_{2}(ν),c_{1}(ν),c_{5}(ν)})^{m−1} ≤

(d_{1,c_{2}(ν),c_{5}(ν)})^{s}(d_{1,c_{2}(ν),c_{5}(ν)}d_{c_{2}(ν),c_{1}(ν),c_{5}(ν)})^{m−1} ≤
(c_{0}(ε))^{2m−2+s} exp(d^{∧}ν(1 + 1/d^{∧}ν)(1 + ε/2)s)×
exp(d^{∧}ν(1 + 1/d^{∧}ν)(1 + max(δ − 1, 0))(1 + ε/2)(m − 1)) =
(c_{0}(ε)^{2m−2+s} exp(d^{∧}ν(1 + 1/d^{∧}ν)(1 + ε/2)s)×
exp(d^{∧}ν(1 + 1/d^{∧}ν) max(δ, 1)(1 + ε/2)(m − 1)).

Consequently,

(b^{c_{2}(ν)}d_{s}^{∼}(ν))^{1/(nd^{∧}ν)} ≤
b^{(1+1/d^{∧}ν)/n}(c_{0}(ε))^{(2m−2+s)/(nd^{∧}ν)} exp(d^{∧}ν(1 + 1/d^{∧}ν)(1 + ε/2)s/n)×
exp(d^{∧}ν(1 + 1/d^{∧}ν) max(δ, 1)(1 + ε/2)(α − 1/n)).

The last inequality implies the assertion of the lemma. The lemma is proved.

Making use of (64), we wish to apply Lemma 1 to the matrix Λ_{m,s} (\frac{b}{a}) of Theorem 1, taking m in (67) as ”m” in (3), s in (67) as ”n” in (3), λ in (67) as ”j” in (3), P_{λ}^{∗}(z; ν) = b^{c_{2}(ν)}d_{s}^{∼}(ν)P_{λ}(z; ν) in (67) as ”α_{j}(ν)” in (3), μ = 0 , …, s − 1 in (67) as ”k − 1” in (3), and P_{m+1+μ}^{∗}(z; ν) = b^{c_{2}(ν)}d_{s}^{∼}(ν)P_{m+1+μ}(z; ν) in (67) as ”α_{m+k}(ν)” in (3). To conclude the proof of Theorem 1, we have to prove that the relation

lim_{ν→∞} ψ_{μ}^{∗}(z; ν) = 0,

for ψ_{μ}^{∗}(z; ν) = b^{c_{2}(ν)}d_{s}^{∼}(ν)ψ_{μ}(z; ν) with ψ_{μ}(z; ν) from (64), follows from (26) and that for every κ ∈ ℕ the m + s sequences

{b^{c_{2}(ν)}d_{s}^{∼}(κ + ω)P_{l}(z; ν) : ω ∈ ℕ},

where l = 1 , …, m + s, are linearly independent over ℝ.

§9. Linear independence over ℝ of certain sequences in ℚ^{ℕ}.

For d ∈ ℕ − 1 and σ = 0 , …, d, let ν_{σ} ∈ ℤ. Let further

(69)  Π_{i,λ}^{∗}(d; ν_{0} , …, ν_{d}) = \binom{m + ν_{0} − 1}{iλ + ν_{0} − i} ∏_{κ=1}^{d} \binom{n}{ν_{κ}}κ^{−ν_{κ}}

and

(70)  h_{i,λ}(d) = ∑_{\begin{matrix}0≤ν_{1}≤n ,... ,0≤ν_{d}≤n \\ ν_{0} +... +ν_{d}=m−λ \end{matrix}} Π_{i,λ}^{∗}(d; ν_{0} , …, ν_{d}),

where λ = 1 , …, m and i = 0, 1. It follows from (69) that

(71)  h_{i,λ}(d) = ∑_{\begin{matrix}i−iλ≤ν_{0},0≤ν_{1}≤n ,... ,0≤ν_{d}≤n \\ ν_{0} +... +ν_{d}=m−λ \end{matrix}} Π_{i,λ}^{∗}(d; ν_{0}, ν_{1}, , …, ν_{d}).

where, as above, λ = 1 , …, m and i = 0, 1.

Lemma 8. We have

(72)  h_{1,λ}(0) = \binom{2m − λ − 1}{m − λ} = h_{0,λ}(0)

for λ = 1 , …, m. Moreover, if 2 ≤ λ ≤ m, 0 < d < m/n and ε^{3} = ε, then

(73)  h_{1,λ}(d) + εh_{0,λ}(d) > 0,

and

(74)  h_{1,1}(d) = h_{0,1}(d) > 0.

Let λ = 1 , …, m; the coefficient of z^{m−λ} in the Taylor’s series expansion of the function

(75)  S_{0}(z) = (1 − z)^{−m} ∏_{κ=1}^{d}(1 + z/κ)^{n}

is equal to h_{0,λ}(d) and, if 0 ≤ d < m/n, then the coefficient of z^{m−λ} in the Taylor’s series expansion of the function

(76)  S_{1}(z) = (1 − z)^{−m}(d!)^{−n} ∏_{κ=1}^{d}(κ + 1 − z)^{n}

is equal to h_{1,λ}(d).

Proof. Equalities (72) are a direct consequence of (70). It follows from (72) that h_{1,1}(0) = h_{0,1}(0) > 0. On rewriting (71), one obtains the following equations

(77)  h_{0,λ}(d) = ∑_{\begin{matrix}0≤ν_{0},0≤ν_{1}≤n ,... ,0≤ν_{d}≤n \\ ν_{0} +... +ν_{d}=m−λ \end{matrix}} Π_{i,λ}^{∗}(d; ν_{0} ≤ n , …, ν_{d}),

(78)  h_{1,λ}(d) = ∑_{\begin{matrix}1−λ≤ν_{0},0≤ν_{1}≤n ,... ,0≤ν_{d}≤n \\ ν_{0} +... +ν_{d}=m−λ \end{matrix}} Π_{i,λ}^{∗}(d; ν_{0} ≤ n , …, ν_{d}).

for λ = 1 , …, m. It follows from (77) that 0 ≤ ν_{0} ≤ λ + ν_{0} − 1 ≤ m + ν_{0} − 1 and ν_{0} ≤ m − λ. Therefore

(79)  \binom{m + ν_{0} − 1}{m − λ} = \binom{m + ν_{0} − 1}{λ + ν_{0} − 1} ≥ 1

and

(80)  \binom{m + ν_{0} − 1}{ν_{0}} ≤ \binom{m + ν_{0} − 1}{λ + ν_{0} − 1}.

Equality in (80) holds only in one of the following cases:

(i) if λ = 1,

or

(ii) if 2 ≤ λ ≤ m and ν_{κ} = 0 for all κ = 1 , …, m,

or

(iii) if m − 1 − ∑_{κ=1}^{d} ν_{κ} = ν_{0} + λ − 1 < 0.

If λ = 1, then ν_{0} ≥ 0; therefore relation (74), with d > 0, follows from the relations (77) - (79). Further, let 2 ≤ λ ≤ m and 0 < d < m/n, then condition (iii) does not hold and therefore relation (80) is a strict inequality for ν_{1} = n , …, ν_{d} = n. Thus the inequality (73) follows from the relations (77) - (80). The right hand side of the equation (70) for i = 0 clearly coincides with the coefficient of z^{m−λ} in the Taylor’s series expansion of the function (75). The function (76) is equal to

∑_{0≤ν_{1}≤n ,... ,0≤ν_{d}≤n} Π_{d}^{∼}(z; ν_{1} , …, ν_{d}),

where

Π_{d}^{∼}(z; ν_{1} , …, ν_{d}) = (1 − z)^{( ∑_{κ=1}^{d} ν_{κ})−m} ∏_{κ=1}^{d} \binom{n}{ν_{κ}}κ^{−ν_{κ}};

if d < m/n, then m− ∑_{κ=1}^{d} ν_{κ} > 0 in the last expression, and therefore the coefficient

of z^{m−λ} in the Taylor’s series expansion of the function Π_{d}^{∼}(z; ν_{1} , …, ν_{d}) is equal to

\binom{m + ν_{0} − 1}{λ + ν_{0} − 1} ∏_{κ=1}^{d} \binom{n}{ν_{κ}}κ^{−ν_{κ}}

with

ν_{0} = m − λ − ∑_{κ=1}^{d} ν_{κ}.

Thus the right hand side of the equation (70) for i = 1 coincides with the coefficient of z^{m−λ} in the Taylor’s series expansion of the function (76). This proves the lemma.

Corollary. If 0 ≤ d < m/n and |z| > 1, then

(81)  ±zh_{1,λ}(d) + h_{0,λ}(d) ≠ 0

for λ = 1 , …, m.

Lemma 9. Let p be a prime; suppose that

d ∈ ℕ − 1, r ∈ ℕ − 1, d_{1} ∈ ℕ − 1, d_{2} ∈ ℕ − 1, r_{1} ∈ ℕ − 1, r_{2} ∈ ℕ − 1,

and

(82)  max(r_{1}, r_{2}) < p.

Then

(83)  p^{−d}(dp + r)! ∈ (−1)^{d}d!r! + pℤ

and

(84)  \binom{(d_{1} + d_{2})p + r_{1} + r_{2}}{d_{1}p + r_{1}} ∈ \binom{d_{1} + d_{2}}{d_{1}}\binom{r_{1} + r_{2}}{r_{1}} + pℤ.

Proof. Since

(85)  p^{−d}(dp + r)! = (∏_{k=1}^{d} k ∏_{l=1}^{p−1}((k − 1)p + l)) ∏_{j=1}^{r}(dp + j),

relation (83) follows from Wilson’s theorem. Since

v_{p} ((∏_{j=1}^{r_{i}}(d_{i}p + j)) − r_{i}!) ≥ 1,

for i = 1, 2 and, in view of (82), v_{p}(r_{i}!) = 0 for i = 1, 2, it follows from (85) and Wilson’s theorem that

(86)  v_{p} (\binom{d_{1} + d_{2})p + r_{1} + r_{2}}{d_{1}p + r_{1}} − \binom{d_{1} + d_{2}}{d_{1}}\binom{r_{1} + r_{2}}{r_{1}}) > 0.

This proves the lemma.

24 LEONID A. GUTNIK

Lemma 10. Let p be a prime; suppose that

d ∈ ℕ, r ∈ ℕ, r < p, d^{∼} ∈ ℕ − 1,

and d^{∼} < d. Then

(87)  \binom{dp}{d^{∼}p + r} ∈ d\binom{d − 1}{d^{∼}}\binom{p}{r} + p^{2}ℤ.

Proof. Since

\binom{dp}{d^{∼}p + r} = \binom{(d − 1)p + p − 1}{d^{∼}p + r − 1}\frac{dp}{d^{∼}p + r} ,

v_{p}(d^{∼}p + r) = v_{p}(r) = 0, v_{p}((d^{∼}p + r)^{−1} − r^{−}) > 0,

\binom{(d − 1)p + p − 1}{d^{∼}p + r − 1} ∈ \binom{d − 1}{d^{∼}}\binom{p − 1}{r − 1} + pℤ

by Lemma 9, it follows that

v_{p} (\binom{(d − 1)p + p − 1}{d^{∼}p + r − 1}\frac{d}{d^{∼}p + r} − \binom{d − 1}{d^{∼}}\binom{p − 1}{r − 1}\frac{d}{r}) > 0,

v_{p} (\binom{(d − 1)p + p − 1}{d^{∼}p + r − 1}\frac{dp}{d^{∼}p + r} − \binom{d − 1}{d^{∼}}\binom{p − 1}{r − 1}\frac{dp}{r}) > 1,

and

\binom{d − 1}{d^{∼}}\binom{p − 1}{r − 1}\frac{dp}{r} = d\binom{d − 1}{d^{∼}}\binom{p}{r}.

This proves the lemma.

Lemma 11. Given a prime number p, let

r ∈ ℕ − 1, r < p, d ∈ ℕ − 1,

let

(88)  c_{2}(ν) = p, c_{1}(ν) = dp + r, c_{1}(ν) + c_{2}(ν) < p^{2}

for some ν ∈ ℕ, and let α_{k,λ}^{∼}(ν) be defined by the equations (48), (49). Then the following inequalities hold true:

(89)  v_{p}(α_{k,λ}^{∼}(ν)) ≥ λ,

where k = 1 , …, p − 1 and λ = 1 , …, m,

(90)  v_{p}(α_{sp,λ}^{∼}(ν)) ≥ λ − m,

where s = 0, 1 and λ = 1 , …, m,

(91)  v_{p} (α_{sp,λ}^{∼}(ν)(−1)^{spm+nc_{1}(ν)+(1−s)(m−λ)} − p^{λ−m}h_{s,λ}(d)) ≥ λ − m + 1,

where s = 0, 1 and λ = 1 , …, m.

Proof. The inequalities (89) and (90) follow directly from the equalities (48), (49). Let us prove (91). Since the parameters in the equations (48), (49) satisfy the following inequalities: 0 ≤ k ≤ c_{2}(ν) and 1 ≤ κ ≤ max(c_{1}(ν) + k, c_{2}(ν) − k) ≤ c_{1}(ν) + c_{2}(ν) < p^{2}, it follows that

(v_{p}(κ))^{2} = v_{p}(κ),

and that

v_{p} (Π_{sp,λ}(ν; λ_{1} , …, λ_{c_{1}(ν)+c_{2}(ν)})) =

(− ∑_{σ=1}^{s} λ_{pσ}) − (∑_{σ=1}^{1−s} λ_{p(s+σ)}) − ( ∑_{σ=s+1}^{d^{∼}+s} λ_{p(1−s+σ)}) = λ − m+

( ∑_{\begin{matrix}1≤κ≤^{s}p \\ (κ,p)=1 \end{matrix}} λ_{κ}) + ( ∑_{\begin{matrix}1≤κ≤c_{2}(ν)−sp \\ (κ,p)=1 \end{matrix}} λ_{sp+κ}) + ( ∑_{\begin{matrix}ps+1≤κ≤c_{1}(ν)+sp \\ (k,p)=1 \end{matrix}} λ_{c_{2}(ν)−k+κ}) .

Therefore

v_{p} (Π_{sp,λ}(ν; λ_{1} , …, λ_{u(ν)})) ≥ λ − m + 1

if at least one of the last three sums strictly positive. Thus

(−1)^{nc_{1}(ν)+msp}α_{sp,λ}^{∼} ∈ p^{λ−m+1}ℤ + \binom{c_{1}(ν) + sp}{sp}^{n}\binom{c_{2}(ν)}{sp}^{m}×

∑_{\begin{matrix}λ_{1}+...+λ_{u(ν)}=m−λ \\ λ_{1}∈ℕ−1 ,... ,λ_{c_{5}(ν)}∈ℕ−1 \\ ∑_{σ=1}^{1+d} λ_{pσ}=m−λ \end{matrix}} Π_{k,λ}(ν; λ_{1} , …, λ_{u(ν)}).

It follows from (49) that if the integers λ_{1} , …, λ_{u(ν)} lie in ℕ − 1 and satisfy the relations

λ_{1} + … + λ_{u(ν)} = m − λ,
λ_{1} ∈ ℕ − 1 , …, λ_{c_{5}(ν)} ∈ ℕ − 1,

26 LEONID A. GUTNIK

and

∑_{σ=1}^{1+d} λ_{pσ} = m − λ,

then

(92)  Π_{sp,λ}(ν; λ_{1} , …, λ_{u(ν)}) =

(∏_{σ=1}^{s} \binom{m + λ_{pσ} − 1}{λ_{pσ}}(pσ)^{−λ_{pσ}}) ×

(∏_{σ=1}^{1−s} \binom{m + λ_{p(s+σ)} − 1}{λ_{p(s+σ)}}(−pσ)^{−λ_{p(s+σ)}}) ×

( ∏_{σ=s+1}^{d+s} \binom{n}{λ_{p(1−s+σ)}}(−pσ)^{−λ_{p(1−s+σ)}}) =

p^{λ−m} (∏_{σ=1}^{s} \binom{m + λ_{pσ} − 1}{λ_{pσ}}(σ)^{−λ_{pσ}}) ×

(∏_{σ=1}^{1−s} \binom{m + λ_{p(s+σ)} − 1}{λ_{p(s+σ)}}(−σ)^{−λ_{p(s+σ)}}) ×

( ∏_{σ=s+1}^{d+s} \binom{n}{λ_{p(1−s+σ)}}(−σ)^{−λ_{p(1−s+σ)}})

for s = 0, 1 and λ = 1 , …, m. For s = 0, the last product (92) is equal to

(93)  p^{λ−m}\binom{m + λ_{p} − 1}{λ_{p}}(−1)^{−λ_{p}} (∏_{σ=1}^{d} \binom{n}{λ_{p(1+σ)}}(−σ)^{−λ_{p(1+σ)}}) .

For ν_{0} = λ_{p} and ν_{σ} = λ_{p(1+σ)}, with σ = 1, ..., d^{∼}, the product (93) is equal to (−p)^{λ−m}Π_{0,λ}^{∗}(d; ν_{0} , …, ν_{d^{∼}}). This proves relation (91) for s = 0. For s = 1, the last product (92) is equal to

(94)  p^{λ−m}\binom{m + λ_{p} − 1}{λ_{p}} (∏_{σ=2}^{d+1} \binom{n}{λ_{pσ}}(−σ)^{−λ_{p(1−s+σ)}}) .

For σ = κ + 1, ν_{κ} = λ_{p(κ+1)}, with κ = 1 , …, d, the product (94) is equal to p^{λ−m}Π_{2,λ}^{∗}(d; ν_{0} , …, ν_{d}), where

Π_{2,λ}^{∗}(d; ν_{0} , …, ν_{d}) = \binom{m + ν_{0} − 1}{ν_{0}} (∏_{κ=1}^{d} \binom{n}{ν_{κ}}(−(κ + 1))^{−ν_{κ}}) .

It follows from Lemma 9 that

\binom{c_{1}(ν) + p}{p}^{n}\binom{c_{2}(ν)}{p}^{m} =

\binom{(d + 1)p + r}{p + 0}^{n}\binom{p}{p}^{m} ∈ \binom{d + 1}{1}^{n}\binom{r}{0}^{n} + pℤ.

Therefore

(−1)^{nc_{1}(ν)+mp}α_{p,λ}^{∼} ∈ p^{λ−m+1}ℤ + h_{2,λ}(d),

where

h_{2,λ}(d) = (d + 1)^{n} ∑_{\begin{matrix}0≤ν_{1}≤n ,... ,0≤ν_{d}≤n \\ ν_{0} +... +ν_{d}=m−λ \end{matrix}} Π_{2,λ}^{∗}(d; ν_{0} , …, ν_{d}).

On rewriting (76) as follows:

(95)  S_{1}(z) = (1 − z)^{−m}(d + 1)^{n} ∏_{κ=1}^{d}(1 − z/(κ + 1))^{n},

one concludes that

h_{2,λ}(d) = h_{1,λ}(d),

with d ∈ ℤ, 0 ≤ d < m/n, λ = 1 , …, m. This proves the lemma.

Lemma 12. Suppose the assumptions of Lemma 11 be satisfied. Let z ∈ ℚ, λ ∈ ℕ, λ ≤ m and suppose that

v_{p}(z) = v_{p}(∑_{s=0}^{1}(−1)^{s}(h_{s,λ}(d)z^{s})^{2}) = 0.

Then the values of the polynomial P_{λ}(z; ν) defined by the equations (65) satisfy the following relation:

v_{p}(P_{λ}(z; ν)) = λ − m.

Proof. This lemma follows directly from (65),Lemma 11 and small Fermat theorem.

Lemma 13. Suppose the assumptions of Lemma 11 be satisfied. Let z ∈ ℚ, μ ∈ ℕ − 1, μ ≤ n − 1 and suppose that

v_{p}(z) = v_{p} (∑_{λ=1}^{m} h_{1,λ}(d)\binom{λ + μ − 1}{μ}) = 0.

Then the values of the polynomial P_{m+μ+1}(z; ν) defined by the equations (65) satisfy the following relation:

v_{p}(P_{m+μ+1}(z; ν)) = −μ − m.

Proof. It follows from Lemma 11 that the terms of the second sum in (65) with

l^{2} + (c_{2}(ν) − k)^{2} > 0

lie in p^{1−m}ℤ, while the sum of all the other terms lies in

p^{1−m−μ}ℤ + (−1)^{pm+nc_{1}(ν)}p^{−m−μ} ∑_{λ=1}^{m} h_{1,λ}(d)\binom{λ + μ − 1}{μ}.

This proves the lemma.

Lemma 14.Let z ∈ ℚ, |z| > 1, then for every κ ∈ ℕ the m + n sequences

{P_{i}(z, κ + ω) : ω ∈ ℕ},

with i = 1 , …, m + n, are linearly independent over ℝ.

Proof. Let

T_{i}(z, ν) = P_{m+1−i}(z, ν)

for i = 1 , …, m, and let

T_{i}(z, ν) = P_{i}(z, ν)

for i = m + 1 , …, m + n. Let us prove first that, for each κ ∈ ℕ, the m + n sequences

{T_{i}(z, κ + ω) : ω ∈ ℕ},

with i = 1 , …, m + n, are linearly independent over ℚ. Suppose it is not the case. Then there exist κ ∈ ℕ and a_{i} ∈ ℚ, with i = 1 , …, m+n, satisfying the conditions

∑_{i=1}^{m+n} |a_{i}| > 0

and

(96)  ∑_{i=1}^{m+n} a_{i}T_{i}(z; ν) = 0

for ν ∈ κ + ℕ. Let c be equal to the product of all the non-zero numbers a_{i} for i ∈ [1, m + n] ∩ ℕ. Let z = a/b, where a ∈ ℤ, b ∈ ℕ, and |a| > b. Finally, let d denote the biggest integer which is strictly smaller than δ = d^{∨}/d^{∧}, and let

F = (abc ∏_{λ=1}^{m}((h_{0,λ}(d))^{2} − (zh_{1,λ}(d))^{2})) ∏_{μ=0}^{n−1} (∑_{λ=1}^{m} h_{1,λ}(d)\binom{λ + μ − 1}{μ}) .

Since c_{1}(ν)/c_{2}(ν) < d^{∨}/d^{∧} and lim_{ν→+∞} c_{1}(ν)/c_{2}(ν) = d^{∨}/d^{∧} = δ, it follows that

lim_{ν→+∞}[c_{1}(ν)/c_{2}(ν)] = d,

where [y] stands for the integral part of y ∈ ℝ as usual. Therefore there is κ_{1} ∈ ℕ such that

[c_{1}(ν)/c_{2}(ν)] = d

for ν ∈ ℕ + κ_{1}. Since

d < d^{∨}/d^{∧} = δ ≤ α = m/n,

it follows from Lemma 8 and the Corollary to this lemma that F ≠ 0. Let ν ∈ (ℕ + κ_{1}) ∩ (ℕ + κ), and suppose that c_{2}(ν) = d^{∧}ν + 1 is equal to a prime p with v_{p}(F) = 0. Then c_{1}(ν) = dp+r, with r ∈ ℕ−1, r < p, d < δ ≤ α = m/n. Therefore the conditions of Lemmata 12 and 13 are satisfied and consequently

v_{p}(T_{i}(z; ν)) = −i + 1

for i = 1 , …, m + n. Let i_{0} = max{i : 1 ≤ i ≤ m + n, a_{i} ≠ 0}, then

v_{p} ( ∑_{i=1}^{m+n} a_{i}T_{i}(z; ν)) = −i_{0} + 1.

The last equation, however, contradicts (96). Therefore there is an invertible m × m− submatrix of the m × ℕ - matrix

𝔗 = (T_{i}(z, κ + ω))_{1≤i≤m+n, ω∈ℕ}.

The lemma is proved.

Corollary. Let z ∈ ℚ, |z| > 1. Then, for every κ ∈ ℕ, the m + n sequences

{P_{i}^{∗}(z, κ + ω) : ω ∈ ℕ},

where i = 1 , …, m + n, are linearly independent over ℝ.

Proof. Assuming the assertion does not hold, one can find κ ∈ ℕ and a_{i} ∈ ℝ, with i = 1 , …, m + n, satifying the conditions

∑_{i=1}^{m+n} |a_{i}| > 0

and

∑_{i=1}^{m+n} a_{i}P_{i}^{∗}(z; ν) = 0

for all ν ∈ ℕ + κ. Then

b^{c_{2}(ν)}d_{n}^{∼}(ν) ∑_{i=1}^{m+n} a_{i}P_{i}(z; ν) = ∑_{i=1}^{m+n} a_{i}P_{i}^{∗}(z; ν) = 0

and, since b^{c_{2}(ν)}d_{n}^{∼}(ν) > 0,

∑_{i=1}^{m+n} a_{i}P_{i}(z; ν) = 0

for all ν ∈ ℕ + κ, in cotradiction with Lemma 14. This proves the corollary.

§10. Theorem 1 (the end of the proof).

Suppose the inequality (26) be satisfied. Then

ρ̃(α, δ, η_{1}(α, δ, D))(e^{s/n+(α−1/n)max(δ,1)}b^{1/n})^{1+ε} < 1

for a small enough ε > 0. Let

(97) ψ_{j−1}^{∗}(z; ν) = −P_{m+j}^{∗}(z; ν) + ∑_{i=1}^{m} P_{i}^{∗}(z; ν)\binom{i + j − 2}{i − 1}L_{i+j−1}(z^{−1}),

where j = 1, ..., s with s ∈ ℕ, s ≤ n. It now follows from the relations (45),(64),(67), and (68) that

lim_{ν→+∞}[ψ_{j−1}^{∗}(z; ν)] = 0

for each j. In view of Lemmata 6 and 14, P_{l}^{∗}(z, ν) ∈ ℤ for l = 1 , …, m + s and, for each κ ∈ ℕ, the m + s sequences

{P_{i}^{∗}(z, κ + ω) : ω ∈ ℕ},

where i = 1 , …, m + s, are linearly independent over ℝ. Therefore the matrix Λ_{m,s}(b/a) satisfies condition (ii) of Lemma 1. This completes the proof of Theorem

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Moscow State Institute of Electronics and Mathematics Bolshoi Trekhsvyatitelskij pereulok, 1-3/12, str.8 Moscow 109028 Russian Federation E-mail: gutnik@gutnik.mccme.ru