ON THE DIMENSION OF SOME LINEAR SPACES OVER ℚ (Part 5)

L.A.Gutnik

Table of contents

§0. Foreword. §1. Some simple properties of the binomial coefficients. §2. Chudnovsky-Hata’s multipliers. §3. On Diophantine Approximations of logarithms in cyclotomic fields. Part 1.

§0. Foreword.

The section 1 has a preparatory character. In the section 2 I describe the theory of Chudnovsky-Hata’s multipliers. In the section 3 I begin the proof of the Theorem 1 in [25], described also as

Theorem 2.4.3 in [34].

Acknowledgments. I express my deepest thanks to Professors B.Z.Moroz, I.I. Piatetski-Shapiro, A.G.Aleksandrov, P.Bundshuh and S.G.Gindikin for help and support.

§1. Some simple properties of the binomial coefficients.

For each prime p ∈ ℕ let v_{p} denotes the p-adic valuation on ℚ. Lemma 1.Let p ∈ ℕ + 2 is a prime number, {d, r} ⊂ ℕ − 1, r < p. Then v_{p}((dp + r)!/((−p)^{d}d!r!) − 1) ≥ 1. Proof. The assertion of the Lemma is the direct corollary of the Wilson’s

theorem. ■.

Lemma 2. Let p ∈ ℕ + 2 is a prime number, {d, d_{1}} ⊂ ℕ − 1,

(1)  {r, r_{1}} ∈ [0, p − 1] ∩ ℤ, d_{1}p + r_{1} ≤ dp + r.

Then

(2)  v_{p} (\binom{dp + r}{d_{1}p + r_{1}}) = v_{p} (\binom{d}{d_{1}}) ,

if r_{1} ≤ r,

(3)  v_{p} (\binom{dp + r}{d_{1}p + r_{1}} (\binom{d}{d_{1}}\binom{r}{r_{1}})^{−1} − 1) ≥ 1,

Typeset by AMS-TEX

if r_{1} ≤ r,

(4)  v_{p} (\binom{dp + r}{d_{1}p + r_{1}}) = 1 + v_{p} ((d − d_{1})\binom{d}{d_{1}}) ,

if r < r_{1},

(5) v_{p} (\binom{dp + r}{d_{1}p + r_{1}}\binom{r_{1}}{r}(r_{1} − r) (p\binom{d}{d_{1}}(d − d_{1}))^{−1} + (−1)^{r_{1}−r}) ≥ 1,

if r < r_{1}.

Proof. Clearly, d_{1} ≤ d. If r_{1} ≤ r, then let r_{2} = r − r_{1}, d_{2} = d − d_{1}. If r_{1} > r, then, in view of (1), d ≥ d_{1} + 1; in this case we let

(6)  r_{2} = p + r − r_{1}, d_{2} = d − d − 1.

Then d = d_{1} + d_{2}, r = r_{1} + r_{2},

\binom{dp + r}{d_{1}p + r_{1}} = ((dp + r)!)/((d_{1}p + r_{1})!(d_{2}p + r_{2})!).

Accordindg to the Lemma 1,

(7)  v_{p} (\binom{dp + r}{d_{1}p + r_{1}}(−p)^{−d+d_{1}+d_{2}}(d_{1})!(r_{1})!(d_{2})!(r_{2})!/(d!r!) − 1) ≥ 1,

(8)  v_{p} (\binom{dp + r}{d_{1}p + r_{1}}) = d − d_{1} − d_{2} + v_{p} (d!r!)/((d_{1})!(r_{1})!(d_{2})!(r_{2})!)) .

The equality (2) and the inequality (3) diectly follow from (7) and (8). If the inequality r < r_{1} holds, then in view of (6) – (8),

r_{2}! ∏_{j=1}^{r_{1}−r−1}(p + r − r_{1} + j) = (p − 1)!, v_{p}(r_{2}!(r_{1} − r − 1)!(−1)^{r_{1}−r} − 1) ≥ 1.

Therefore (4) and (5) hold. ■.

Corollary 1. Let p ∈ ℕ + 2 is a prime number,

{d, d_{1}, d_{2}, r, r_{1}, r_{2}} ⊂ ℕ − 1, max(r_{1}, r_{2}) < p.

Then

p^{−d}(dp + r)! ∈ (−1)^{d}d!r! + pℤ,

\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. This is direct corollary of the Lemmata 1 and 2.

See also Lemma 9 in [32]. ■

Corolary 2. Let p ∈ ℕ + 2 is a prime number,

{d, r_{1}} ⊂ ℕ, r_{1} < p, d_{1} ∈ ℕ − 1, d_{1} < d.

Then

(9)  v_{p} (\binom{dp}{d_{1}p + r_{1}} (d\binom{d − 1}{d_{1}}\binom{p}{r_{1}})^{−1} + 1) ≥ 1.

Proof. Since,

d\binom{d − 1}{d_{1}} = (d − d_{1})\binom{d}{d_{1}}, v_{p} (\binom{p}{r_{1}}r_{1}/p − (−1)^{r_{1}}) ≥ 1,

the equality (9) directly follows from (5). ■

Corolary 3. Let p ∈ ℕ + 2 is a prime number,

{d, r_{1}} ⊂ ℕ, r_{1} < p, d_{1} ∈ ℕ − 1, d_{1} < d.

Then

\binom{dp}{d_{1}p + r_{1}} ∈ d\binom{d − 1}{d^{∼}}\binom{p}{r_{1}} + p^{2}ℤ.

Proof. This is a corollary of the Corrolary 2. See also Lemma 10 in [32] . ■

Lemma 3. Let p ∈ ℕ+2 is a prime number, α ∈ ℕ−1, {r_{i}, ρ_{i}} ⊂ [0, p−1]∩ℤ, where i = 0, …, α,

(10)  m = ∑_{i=1}^{α} ρ_{i}p^{i}, n = ∑_{i=1}^{α} r_{i}p^{i}, m ≤ n.

Let h_{i}(m, n) = 0, if ρ_{i} ≤ r_{i} for some i ∈ [0, α] ∩ ℤ (in particular, for i = α). But, if ρ_{i} > r_{i} for some i ∈ [0, α) ∩ ℤ, then {h ∈ [1, α − i] ∩ ℤ: ρ_{i+h} ≠ r_{i+h}} ≠ ∅ (in view of (10)), and we put h_{i}(m, n) = inf({h ∈ [1, α − i] ∩ ℤ: ρ_{i+h} ≠ r_{i+h}}). Then

(11)  v_{p} (\binom{n}{m}) = ∑_{i=1}^{α} h_{i}(m, n).

Proof. We use induction on α. Clearly, for α = 0 the assertion of the Lemma is obvious. Let α > 0 and the assertion of the Lemma is true for all the α′ ∈ [0, α)∩ℤ. If h_{i}(m, n) = 0 for i = 0, …, α, then applying (3) sequentially, we see that

v_{p} (\binom{n}{m}) = 0 = ∑_{i=1}^{α} h_{i}(m, n).

If {i ∈ {0, …, α}: h_{i}(m, n) ≠ 0} ≠ ∅, then let

β = inf({i ∈ {0, …, α}: h_{i}(m, n) ≠ 0}).

In view of (10), β < α. Let α^{∗} = α − β,

ρ_{i}^{∗} = ρ_{β+i}, r_{i}^{∗} = r_{β+i},

where i = 0, …, α^{∗}. Let further

m^{∗} = ∑_{i=β}^{α} ρ_{i}p^{i−β} = ∑_{i=0}^{α^{∗}} ρ_{i}^{∗}p^{i}, n^{∗} = ∑_{i=β}^{α} r_{i}p^{i−β} = ∑_{i=0}^{α^{∗}} r_{i}^{∗}p^{i}

Since n^{∗}−m^{∗} = n−m, it follows from (10) that n^{∗} ≥ m^{∗}. Applying (3) sequentially, we see that

(12)  v_{p} (\binom{m}{n}) = v_{p} (\binom{m^{∗}}{n^{∗}}) = v_{p} (\binom{m^{∗}}{n^{∗}}) + ∑_{i=0}^{β−1} h_{i}(m, n).

When i ∈ [0, α^{∗}] ∩ ℤ, then the iequalities ρ_{i}^{∗} ≤ r_{i}^{∗} and ρ_{β+i} ≤ r_{β+i} are equivalent; in this case h_{i}(m^{∗}, n^{∗}) = h_{β+i}(m, n) = 0. When i ∈ [0, α^{∗}) ∩ ℤ, then ρ_{i}^{∗} > r_{i}^{∗} if and only if ρ_{β+i} > r_{β+i}, h ∈ [0, α^{∗} − i) ∩ ℤ if and only if h ∈ [0, α − β − i) ∩ ℤ, the inequalities ρ_{i+h+1}^{∗} ≠ r_{i+h+1} and ρ_{β+i+h+1}^{∗} ≠ r_{β+i+h+1} are equivalent; therefore, if ρ_{i} > r_{i} for some i ∈ [0, α^{∗}) ∩ ℤ, then h_{i}(m^{∗}, n^{∗}) = h_{β+i}(m, n). We take now on the role of r, r_{1}, d, d_{1} in the Lemma 2 respectively the numbers

ρ_{β}, r_{β}, m′ = (m^{∗} − ρ_{β})/p = ∑_{i=0}^{α′} ρ′_{i}p^{i}, n′ = ∑_{i=0}^{α′} r′_{i}p^{i},

where α′ = α^{∗} − 1, ρ′_{i} = ρ_{1+i}^{∗}, r′_{i} = r_{1+i}, i = 0, …, α′. In view of (5),

(13)  v_{p} (\binom{n^{∗}}{m^{∗}}) = 1 + v_{p}(n′ − m′) + v_{p} (\binom{n′}{m′}) = h_{0}(m^{∗}, n^{∗})+

v_{p} (\binom{n′}{m′}) = h_{β}(m, n) + v_{p} (\binom{n′}{m′}) = v_{p} (\binom{n′}{m′}) + ∑_{i=0}^{β} h_{i}(m, n).

Clearly, h_{i}(m′, n′) = h_{1+i}(m^{∗}, n^{∗}) = h_{1+β+i}(m, n), where i = 0, …, α′. Since α′ = α^{∗} − 1 < α, it follows that, according to the assumption of induction,

(14)  v_{p} (\binom{n′}{m′}) = ∑_{i=0}^{α′} h_{i}(m′, n′) = ∑_{i=1}^{α^{∗}} h_{i}(m^{∗}, n^{∗}) = ∑_{i=β+1}^{α} h_{i}(m, n).

In view of (14), (13) and (12), the equality (11) holds. ■.

§2. Chudnovsky-Hata’s multipliers.

Let ν ∈ ℕ, Δ ∈ ℕ, Δ ≥ 2. Let us consider the numbers

(15)  \binom{νΔ}{k}\binom{νΔ + k}{νΔ − ν},

where k = 0, …, νΔ. How small may be chosen the number r(ν) ∈ (0, +∞) ∩ ℚ such that r(ν)\binom{νΔ}{k}\binom{νΔ+k}{νΔ−ν} ∈ ℤ for all the k = 0, …, νΔ? Probably G.V. Chudnovsky was the first man, who discovered, that r(ν) may be chosen sufficiently small; Hata [16] in details studied this effect. Therefore I name such r(ν) by Chudnovsky-Hata’s multiplier. In this section I follow Hata’s paper [16], making use also some results of the section 1. Since ℤ contains the numbers (15), it follows that we can seek r(ν) in the form

(16)  r(ν) = r^{∗}(ν) = 1/d_{1}^{∗}(Δ, ν),

where d_{1}^{∗}(Δ, ν), is some common divisor of the numbers (15). According to the Lemma 2, the prime number p ∈ ℕ is a divisor of all the numbers (15) if and only if, when

(17)  0 < max ({k/p} − {νΔ/p}, {ν(Δ − 1)/p} − {(k + νΔ)/p},)

where k = 0, …, νΔ. Let us to prove that the condition (17) is equivalent to the inequality

(18)  2{νΔ/p} < {ν(Δ − 1)/p}.

In fact, if, in contrary to (18),

(19)  {νΔ/p} ≥ k/p = max({ν(Δ − 1)/p} − {νΔ/p}, 0),

then {k/p} = k/p, 1 > max({ν(Δ − 1)/p}, {νΔ/p}) = {νΔ/p} + {k/p} and therefore {νΔ/p} + {k/p} = {(νΔ + k)/p},

{(νΔ + k)/p} = max({ν(Δ − 1)/p}, {νΔ/p}) ≥ {ν(Δ − 1)/p}

with k ∈ [0, νΔ] ∩ ℤ; but the last inequality, in view of (23), implies the inequality {νΔ/p} < {k/p}, which contradicts to (19). Let the inequality (18) holds and let {νΔ/p} ≥ {k/p} for some k ∈ [0, νΔ] ∩ ℤ. Then, in view of (18),

{νΔ/p} + {k/p} ≤ 2{νΔ/p} < {ν(Δ − 1)/p} ≤ (p − 1)/p < 1,
{(νΔ + k)/p} = {νΔ/p} + {k/p} < {ν(Δ − 1)/p},

and therefore (17) holds. So, if (18) holds, then p is divisor of all the numbers (15). We will study the set

(20)  X_{Δ}^{∗} = {x ∈ ℝ: 2{xΔ} < {x(Δ − 1)} }.

Clearly, X_{Δ}^{∗} ∩ ℤ = ∅ and m + X_{Δ}^{∗} = X_{Δ}^{∗}, for any m ∈ ℤ. Therefore it is suffiient to study the set X_{Δ}^{∗} ∩ (0, 1). Let

(21)  ξ ∈ X_{Δ}^{∗} ∩ (0, 1),

and let D^{∗} ∈ Δ(Δ − 1)ℕ. Let

(22)  ξ ∈ [κ/D^{∗}, (κ + 1)/D^{∗})

for some κ ∈ {0, …, D^{∗} − 1}. Then

κ/(D^{∗}/(Δ − ϵ)) ≤ ξΔ < (κ + 1)/(D^{∗}/(Δ − ϵ)),
[κ/(D^{∗}/(Δ − ϵ))] ≤ [κ/(D^{∗}/(Δ − ϵ))] + {κ/(D^{∗}/Δ)} ≤ ξΔ <
[κ/(D^{∗}/(Δ − ϵ))] + {κ/(D^{∗}/(Δ − ϵ))} + 1/(D^{∗}/(Δ − ϵ)) ≤
[κ/(D^{∗}/(Δ − ϵ))] + (D^{∗}/(Δ − ϵ) − 1)/(D^{∗}/Δ) + 1/(D^{∗}/(Δ − ϵ) =
[κ/(D^{∗}/(Δ − ϵ))] + 1,

where ϵ^{2} = ϵ. Consequently, [κ/(D^{∗}/(Δ − ϵ))] = [ξ(Δ − ϵ)],

ξ(Δ − ϵ) = ξ(Δ − ϵ) − [ξ(Δ − ϵ)] = ξ(Δ − ϵ) − [κ/(D^{∗}/(Δ − ϵ))],

Therefore, the existence of κ ∈ [0, D^{∗} − 1) ∩ ℤ such that the inclusion (22) and the inequality

(23)  2ξΔ − 2[κ/(D^{∗}/Δ)] < ξ(Δ − 1) − [κ/(D^{∗}/(Δ − 1))]

hold is equivalent to the condition (21). The inequality (23) is equivalent to the inequality

(24)  ξ < κ/D^{∗} + (({κ/(D^{∗}/(Δ − 1))} − 2{κ/D^{∗}/Δ)]})(Δ + 1)^{−1}.

Let us note that for κ = 0 the system (22) – (24) has no solutions. Let

η^{∗}(κ) = +({κ/(D^{∗}/(Δ − 1))} − 2{κ/(D^{∗}/Δ)]})(Δ + 1)^{−1},
η_{0}^{∗}(κ) = max(0, min(η^{∗}(κ), 1/D^{∗})) = max(0, η^{∗}(κ)),

where κ ∈ {0, …, D^{∗} −1}. Then the system of conditions (22) – (24) is equivalent to the condition

(25)  ξ ∈ [κ/D^{∗}, κ/D^{∗} + η_{0}^{∗}(κ)).

So, if ν/p = m + ξ, where

m ∈ [0, (ν/(2Δ))^{1/2} − 1] ∩ ℤ, κ ∈ ℤ, 1 ≤ κ ≤ D − 1,
ξ ∈ (0, 1), κ/D^{∗} ≤ ξ < κ/D^{∗} + η_{0}^{∗}(κ),

i.e. if

(26)  ν/(m + κ/D^{∗} + η_{0}^{∗}(κ)) < p ≤ ν/(m + κ/D^{∗}),

where

(27)  m ∈ [0, (ν/(Δ + 1))^{1/2} − 1] ∩ ℤ, κ ∈ ℤ, 1 ≤ κ ≤ D^{∗} − 1,

then the condition p > (2νΔ)^{1/2} and the condition (24) are fulfilled. We take on the role of d_{1}^{∗}(ν) the product of all the prime nubers p, which satisfy to the conditions (26) – (27). It is well known that there exists a > 0, such that

(28)  ψ(x) = ∑_{p≤x} = x(1 + O(exp(−a(log(x))^{1/2})),

where x ≥ 2. Therefore

(29)  log(d_{1}^{∗}(ν)) =

∑_{κ=1}^{D^{∗}−1} ∑_{m=0}^{[(ν/(Δ+1))^{1/2}−1]}(ψ(ν/(m + κ/D^{∗})) − ψ(ν/(m + κ/D^{∗} + η_{0}^{∗}(κ))).

In view of (26) – (28) we have

(30)  ψ(ν/(m + κ/D^{∗} + η_{0}(κ))) = ν/(m + κ/D^{∗} + η_{0}^{∗}(κ))+
O((ν/(m + 1))(exp(−(a/\sqrt{2})(log(ν))^{1/2}),

(31)  ψ(ν/(m + κ/D^{∗})) = ν/(m + κ/D^{∗})+
O((ν/(m + 1)) exp(−(a/\sqrt{2})(log(ν))^{1/2}).

But

∑_{m=0}^{[(ν/(Δ+1))^{1/2}−1]}(1/(m + κ/D)) − 1/(m + κ/D^{∗} + η_{0}^{∗}(κ)) =

(Γ′(κ/D^{∗} + η_{0}^{∗}(κ)))/(Γ(κ/D^{∗} + η_{0}^{∗}(κ))) − (Γ′(κ/D^{∗}))/(Γ(κ/D^{∗}))) + O(ν^{−1/2}), where κ ∈ [1, D^{∗} − 1] ∩ ℤ. Consequently,

log(d_{1}^{∗}(ν)) = ν ∑_{κ=1}^{D−1}(Γ′(κ/D^{∗} + η_{0}^{∗}(κ)))/(Γ(κ/D^{∗} + η_{0}^{∗}(κ)))−

ν ∑_{κ=1}^{D^{∗}−1}(Γ′(κ/D^{∗}))/(Γ(κ/D^{∗})) + O(ν log(ν) exp(−(a/\sqrt{2})(log(ν))^{1/2}).

Let δ_{0} = 1/Δ, η^{∨}(x) = {(1 − δ_{0})x} − 2{x}, η_{0}^{∨}(x) = max(0, η^{∨}(x)), where x ∈ ℝ. Let further χ^{∨} = χ^{∨}(x) denotes the characteristic function of the set {x ∈

ℝ: 0 < η_{0}^{∨}(x)}. If x ∈ (0, 1), then χ^{∨}(x) = 0. Therefore the integral ∫_{0}^{∞}χ^{∨}(x)x^{−2} dx

converges. We wish to finde the value Δ ∫_{0}^{∞}χ^{∨}(x) dx = ∫_{0}^{∞}χ^{∨}(Δx)x^{−2} dx. Clearly,

∫_{0}^{∞}χ^{∨}(Δx)x^{−2} dx = ∑_{m=0}^{∞} ∫_{m}^{m+1}χ^{∨}(Δx)x^{−2} dx = ∑_{m=0}^{∞} ∫_{0}^{1} χ^{∨}(Δξ)(m + ξ)^{−2} dξ.

In view of (20), X_{Δ}^{∗} = {x ∈ ℝ: 0 < η_{0}^{∨}(Δx)}. Clearly, the condition (21) is equivalent to the equality χ^{∨}(Δξ) = 1. Therefore, in view of (25),

∫_{0}^{1} χ^{∨}(Δξ)(m + ξ)^{−2} dξ =

∑_{κ=1}^{D^{∗}−1}(1/(m + κ/D^{∗})) − 1/(m + κ/D^{∗} + η_{0}^{∗}(κ))

Consequently, in view of (29) – (31),

(32)  log(d_{1}^{∗}(ν)) = νΔ ∫_{0}^{∞}χ^{∨}(x) dx + O(ν log(ν) exp(−(a/\sqrt{2})(log(ν))^{1/2}).

Let

(33)  x = mΔ + κ + ξ,

where m ∈ ℕ − 1, κ = 0, …, Δ − 1, 0 ≤ ξ < 1. Then

∫_{0}^{∞}χ^{∨}(Δx)x^{−2} dx = ∑_{m=0}^{∞} ∑_{κ=0}^{Δ−1} ∫_{0}^{1} χ^{∨}(mΔ + κ + ξ)(mΔ + κ + ξ)^{−2} dx.

If {x} ≥ {x/Δ}, then {x − x/Δ} = {{x} − {x/Δ}} = {x} − {x/Δ} and the inequality {x−x/Δ} > 2{x} is impossible; on the other hand, clearly, the inequality {x} < {x/Δ}, implies the equality

{x − x/Δ} = {{x} − {x/Δ}} = {x} − {x/Δ} + 1

and therefore in this case the inequality {x − x/Δ} > 2{x} turns into the the inequality 1 − {x} > {x/Δ}; so, the inequality {x − x/Δ} > 2{x} is equivalent to the condition

(34)  1 − {x} > {x/Δ} > {x}.

In view of (33),

(35)  {x} = ξ, {x/Δ} = (κ + ξ)/Δ,

and the condition (34) may be rewritten in the form

1 − ξ > (κ + ξ)/Δ > ξ,

i.e. in the form

0 ≤ ξ < min(κ/(Δ − 1), (D − κ)/(Δ + 1)),

where κ = 0, …, Δ − 1. Clearly,

min(κ/(Δ − 1), (Δ − κ)/(Δ + 1)) = κ/(Δ − 1), ,

if κ ≤ (Δ − 1)/2,

min(κ/(Δ − 1), (D − κ)/(Δ + 1)) = (D − κ)/(Δ + 1),

if κ ≥ (Δ − 1)/2. Therefore

∫_{0}^{1} χ^{∨}(mΔ + κ + ξ)(mΔ + κ + ξ)^{−2} dx = (mΔ + κ)^{−1}−

(mΔ + κ + κ/(Δ − 1))^{−1} = Δ^{−1}(m + κ/Δ)^{−1} − (m + κ/(Δ − 1))^{−1},

if κ ≤ (Δ − 1)/2,

∫_{0}^{1} χ^{∨}(mΔ + κ + ξ)(mΔ + κ + ξ)^{−2} dx = (mΔ + κ)^{−1}−

(mΔ + κ + (Δ − κ)/(Δ + 1))^{−1} = Δ^{−1}(m + κ/Δ)^{−1} − (m + (κ + 1)/(Δ + 1))^{−1},

10 L.A.GUTNIK

if κ ≥ (Δ − 1)/2. Consequently,

(36)  ∫_{0}^{∞}χ^{∨}(Δx)x^{−2} dx = Δ^{−1} ∑_{1≤κ<Δ/2} Γ′(κ/(Δ − 1))/Γ(κ/(Δ − 1))−

Δ^{−1} ∑_{1≤κ<Δ/2} Γ′(κ/Δ)/Γ(κ/Δ)+

Δ^{−1} ∑_{Δ/2≤κ≤Δ−1} Γ′((κ + 1)/(Δ + 1))/Γ((κ + 1)/(Δ + 1))−

Δ^{−1} ∑_{Δ/2≤κ≤Δ−1} Γ′(κ/Δ)/Γ(κ/Δ) = Δ^{−1} ∑_{1≤κ<Δ/2} Γ′(κ/(Δ − 1))/Γ(κ/(Δ − 1))+

Δ^{−1} ∑_{Δ/2≤κ≤Δ−1} Γ′((κ + 1)/(Δ + 1))/Γ((κ + 1)/(Δ + 1))−

Δ^{−1} ∑_{1≤κ≤Δ−1} Γ′(κ/Δ)/Γ(κ/Δ) =

Δ^{−1} ∑_{1≤κ≤(Δ−1)/2} Γ′(κ/(Δ − 1))/Γ(κ/(Δ − 1))+

Δ^{−1} ∑_{(Δ−1)/2≤κ≤Δ−1} Γ′((κ + 1)/(Δ + 1))/Γ((κ + 1)/(Δ + 1))−

(1 − (−1)^{Δ})(2Δ)^{−1}Γ′(1/2)/Γ(1/2) − Δ^{−1} ∑_{1≤κ≤Δ−1} Γ′(κ/Δ)/Γ(κ/Δ) =

Δ^{−1} ∑_{1≤κ≤(Δ−1)/2} Γ′(κ/(Δ − 1))/Γ(κ/(Δ − 1))+

Δ^{−1} ∑_{(Δ+1)/2≤κ≤Δ} Γ′(κ)/(Δ + 1))/Γ(κ/(Δ + 1))−

(1 − (−1)^{Δ})(2Δ)^{−1}Γ′(1/2)/Γ(1/2) − Δ^{−1} ∑_{1≤κ≤Δ−1} Γ′(κ/Δ)/Γ(κ/Δ) =

Δ^{−1} ∑_{1≤κ≤(Δ−1)/2} Γ′(κ/(Δ − 1))/Γ(κ/(Δ − 1))+

Δ^{−1} ∑_{(Δ+1)/2≤κ≤Δ} Γ′(κ/(Δ + 1))/Γ(κ/(Δ + 1))−

(1 − (−1)^{Δ})(2Δ)^{−1}Γ′(1/2)/Γ(1/2) − Δ^{−1} ∑_{1≤κ≤Δ/2} Γ′(κ/Δ)/Γ(κ/Δ)−

ON THE DIMENSION OF SOME LINEAR SPACES OVER ℚ (PART 5) 11

Δ^{−1} ∑_{Δ/2≤κ≤Δ−1} Γ′(κ/Δ)/Γ(κ/Δ) + (1 + (−1)^{Δ})(2Δ)^{−1} \frac{Γ′(1/2)}{Γ(1/2)}.

Let p ∈ ℕ, q ∈ ℕ + p. According to well-known theorem of K.F.Gauß

(37)  \frac{Γ′(p/q)}{Γ(p/q)} = −γ − log(q) − (π/2) cot(pπ/q)+

∑_{κ=1}^{q−1}(cos(2kpπ/q)) log(|1 − exp(−2kπ/q)|),

where γ is the Euler’s constant. Let σ(p, q) = ∑_{κ=1}^{[q/2]} cos(kpκπ/q).

Since cos(2p(q − kκ)π/q) = cos(2kκπ/q) for k = 1, …, [(q − 1)/2], it follows that 2σ(p, q) + 1 = 0, if q ∈ 2ℕ + 1, and 2σ(p, q) + 1 − (−1)^{p} = 0, if q ∈ 2ℕ. So,

(38) σ(p, q) = −1/2 + (−1)^{p}((−1)^{q} + 1)/4 = 0, σ(p, q) = ∑_{[q/2]}^{q−1} cos(2kpπ/q).

If q ∈ 2ℕ, then

∑_{κ=1}^{q−1}(−1)^{k} log(|1 − exp(−2kπ/q)|) =

log ( lim_{x→1} (|\frac{∏_{κ=1}^{q}(x − exp(−2kπ/q)}{(x − 1) ∏_{κ=1}^{q/2}(x − exp(−(2k − 1)π/(q/2))}|)) −

log (|∏_{κ=1}^{q/2}(x − exp(−(2k − 1)π/(q/2))|_{x=1}|) =

log( lim_{x→1}(|((x^{q/2} − 1)//(x − 1)|))−

log(| (x^{q/2} + 1) |_{x=1}|) = log(q) − 2 log(2).

If q ∈ 2ℕ + 1, then k + (q − k) ∈ 2ℤ + 1 and

∑_{κ=1}^{q−1}(−1)^{k} log(|1 − exp(−2kπ/q)|) =

∑_{κ=1}^{(q−1)/2}(−1)^{k}(log (\frac{|1 − exp(−2kπ/q)|}{|1 − exp(−2(q − k)π/q)|}) = 0.

12 L.A.GUTNIK

Therefore

(39)  ∑_{κ=1}^{q−1}(−1)^{k} log(|1 − exp(−2kπ/q)|) = (log(q) − log(4))(1 + (−1)^{q})/2.

Clearly,

(40)  ∑_{κ=1}^{q−1}(−1)^{k} log(|1 − exp(−2kπ/q)|) = log(q),
(41)  [q/2] + 1/2 − (1 + (−1)^{q})/4 = q/2.

Therefore, in view of (36) – (41),

(42)  ∑_{1≤κ≤(Δ−1)/2} Γ′(κ/(Δ − 1)/Γ(κ/(Δ − 1)) =

−γ[(Δ − 1)/2] − [(Δ − 1)/2] log(Δ − 1)) − (π/2) ∑_{1≤κ≤(Δ−1)/2} cot(πκ/(Δ − 1))+

∑_{1≤κ≤Δ−1} σ(k, Δ − 1) log(|(1 − exp(−2ikπ/(Δ − 1)) =

−γ[(Δ − 1)/2] − [(Δ − 1)/2] log(Δ − 1) − (π/2) ∑_{1≤κ≤(Δ−1)/2} cot(πκ/(Δ − 1))+

∑_{1≤κ≤Δ−1} (−1/2 + (−1)^{k} ((−1)^{Δ−1} + 1) /4) log(|(1 − exp(−2ikπ/(Δ − 1)) =

−γ[Δ − 1)/2] − [Δ − 1)/2] log(Δ − 1) − (π/2) ∑_{1≤κ≤(Δ−1)/2} cot(πκ/(Δ − 1))+

(−1/2) log(Δ − 1)) + (log(Δ − 1) − log(4)) (1 + (−1)^{Δ−1}) /4 =

−γ[Δ − 1)/2] − (Δ − 1)/2) log(Δ − 1/2) − (π/2) ∑_{1≤κ≤(Δ−1)/2} cot(πκ/(Δ − 1))+

(− log(4)) (1 + (−1)^{Δ−1}) /4,

∑_{(Δ+1)/2≤κ≤Δ} Γ′(κ/(Δ + 1))/Γ(κ/(Δ + 1)) =

−γ[Δ + 1)/2] − [Δ + 1)/2] log(Δ − 1)) + (π/2) ∑_{1≤κ≤(Δ+1)/2} cot(πκ/(Δ + 1))+

∑_{1≤κ≤Δ−1} σ(k, Δ + 1) log(|(1 − exp(−2ikπ/(Δ + 1)) =

ON THE DIMENSION OF SOME LINEAR SPACES OVER ℚ (PART 5) 13

−γ[Δ + 1)/2] − [Δ + 1)/2] log(Δ − 1)) + (π/2) ∑_{1≤κ≤(Δ+1)/2} cot(πκ/(Δ + 1))+

∑_{1≤κ≤Δ−1} (−1/2 + (−1)^{k}((−1)^{Δ+1} + 1)/4) log(|(1 − exp(−2ikπ/(Δ + 1)) =

−γ[Δ + 1)/2] − [Δ + 1)/2] log(Δ + 1)) + (π/2) ∑_{1≤κ≤(Δ+1)/2} cot(πκ/(Δ + 1))+

(−1/2) log(Δ + 1) + (log(Δ + 1) − log(4)) ((−1)^{Δ+1} + 1) /4 =

−γ[Δ + 1)/2] − (Δ + 1)/2) log(Δ + 1)) + (π/2) ∑_{1≤κ≤(Δ+1)/2} cot(πκ/(Δ + 1))+

(− log(4)) ((−1)^{Δ+1} + 1) /4,
(43)  ∑_{1≤κ≤Δ/2} Γ′(κ/Δ)/Γ(κ/Δ) + ∑_{Δ/2≤κ≤Δ−1} Γ′(κ/Δ)/Γ(κ/Δ) =
−2γ[Δ/2] − 2[Δ/2] log(Δ)+
(− log(Δ)) + (log(Δ) − log(4)) ((−1)^{Δ} + 1) /2 =
−2γ[Δ/2] − Δ log(Δ) + (− log(4)) ((−1)^{Δ} + 1) /2,
(− log(4)) (−(−1)^{Δ} + 1) /2 − (− log(4)) ((−1)^{Δ} + 1) /2−
(1 − (−1)^{Δ})Γ′(1/2)/(2Γ(1/2)) + (1 + (−1)^{Δ})Γ′(1/2)/(2Γ(1/2)) =
(−1)^{Δ} log(4) + (−1)^{Δ}Γ′(1/2)/Γ(1/2) =
(−1)^{Δ}(log(4) − γ − log(2) − log(2)) = −(−1)^{Δ}γ,
(44)  −γ([(Δ − 1)/2] + [(Δ + 1)/2] − 2[Δ/2] + (−1)^{Δ}) = 0.

Consequently, in view of (36), (42) – (44),

(45)  ∫_{0}^{∞}χ^{∨}(x)x^{−2} dx =

log(Δ) − ((Δ − 1)/(2Δ)) log(Δ − 1) − ((Δ + 1)/(2Δ)) log(Δ + 1)−

(π/(2Δ)) ∑_{1≤κ≤(Δ−1)/2} cot(πκ/(Δ − 1))+

(π/(2Δ)) ∑_{1≤κ≤(Δ+1)/2} cot(πκ/(Δ + 1)).

Hata reasons somewhat differently. He uses the equality.

\binom{νΔ}{k}\binom{νΔ + k}{νΔ − ν} = (νΔ)!(νΔ + k)!/(k!(νΔ − k)!(νΔ − ν)!(ν + k)!) =

((νΔ)!)^{2}/((νΔ + ν)!(νΔ − ν)!)\binom{νΔ + ν}{ν + k}\binom{νΔ + k}{k},

where k = 0, …, νΔ, and making use that

(46)  \binom{νΔ + ν}{ν + k}\binom{νΔ + k}{k} ∈ ℕ,

for k = 0, …, νΔ, he seeks r(ν) in the form

(47)  r(ν) = r^{∗∗}(ν) = ((νΔ + ν)!(νΔ − ν)!)/(((νΔ)!)^{2}d_{1}^{∗∗}(Δ, ν)),

where d_{1}^{∗∗}(Δ, ν), is some common divisor of the numbers (46). According to the Lemma 2, the prime number p ∈ ℕ is a divisor of all the numbers (46) if and only if, when

(48)  0 < max({(ν + k)/p} − {ν(Δ + 1)/p}, {νΔ/p} − {(k + νΔ)/p}),

where k = 0, …, νΔ. Let us to prove that the condition (48) is equivalent to the inequality

(49)  {ν(Δ − 1)/p} + {ν(Δ + 1)/p} < {νΔ/p}.

Let the inequality (49) holds and let {(ν + k)/p} ≤ {ν(Δ + 1)/p} for some k ∈ [0, νΔ] ∩ ℤ. Then, in view of (49),

{ν(Δ − 1)/p} + {(ν + k)/p} ≤
{ν(Δ − 1)/p} + {ν(Δ + 1)/p} < {νΔ/p} ≤ (p − 1)/p < 1,
{(νΔ + k)/p} = {ν(Δ − 1)/p} + {(ν + k)/p} < {νΔ/p},

and therefore (48) holds. If (63) holds, then, taking in the inequality (48) k = 0, we see that

{ν/p} > {ν(Δ + 1)/p} = {{νΔ/p} + {ν/p}},

and therefore

(50)  {νΔ/p} + {ν/p} ≥ 1, {νΔ/p} + {ν/p} − 1 = {ν(Δ + 1)/p}.

If

(51)  {ν/p} > {νΔ/p},

ON THE DIMENSION OF SOME LINEAR SPACES OVER ℚ (PART 5) 15

then k = p − p{ν/p} > 0,, in view of (50),

{νΔ/p} ≥ 1 − {ν/p} = k/p = {k/p},

and {(ν + k)/p} = 0 ≤ {ν(Δ + 1)/p}. Therefore, in view of (48),

(52)  {νΔ/p} > {(k + νΔ)/p}) = {{k/p} + {(νΔ)/p}}.

But the inequality (52) is equivalent to the inequality

1 ≤ {k/p} + {(νΔ)/p} = 1 − {ν/p} + {(νΔ)/p},

what contradicts to (51). Therefore the inequality

{ν/p} ≤ {νΔ/p}

follows from (48). In view of (50),

{νΔ/p} = {ν(Δ − 1/p} + {ν/p}} = {ν(Δ − 1)/p} + {ν/p} >

{ν(Δ − 1)/p} + {ν/p} + {νΔ/p} − 1 = {ν(Δ − 1)/p} + {ν(Δ + 1)/p}.

So, if (49) holds, then p is divisor of all the numbers (46). We will study the set

(53)  X_{Δ}^{∗∗} = {x ∈ ℝ: {x(Δ − 1)} + {x(Δ + 1)} < {xΔ} }.

Clearly, X_{Δ}^{∗} ∩ ℤ = ∅ and m + X_{Δ}^{∗∗} = X_{Δ}^{∗∗}, for any m ∈ ℤ. Therefore it is suffiient to study the set X_{Δ}^{∗∗} ∩ (0, 1). Let

(54)  ξ ∈ X_{Δ}^{∗∗} ∩ (0, 1),

and let D^{∗∗} ∈ Δ(Δ^{2} − 1)ℕ. Let

(55)  ξ ∈ [κ/D^{∗∗}, (κ + 1)/D^{∗∗})

for some κ ∈ {0, …, D^{∗∗} − 1}. Then

κ/(D^{∗∗}/(Δ − ϵ)) ≤ ξ(Δ − ϵ) < (κ + 1)/(D^{∗∗}/(Δ − ϵ)), [κ/(D^{∗∗}/(Δ − ϵ))] ≤ [κ/(D^{∗∗}/(Δ − ϵ))] + {κ/(D^{∗}/(Δ − ϵ))} ≤ ξ(Δ − ϵ) <

[κ/(D^{∗∗}/Δ)] + {κ/(D^{∗∗}/Δ)} + 1/(D^{∗∗}/(Δ − ϵ)) ≤
[κ/(D^{∗∗}/Δ)] + (D^{∗∗}/(Δ − ϵ) − 1)/(D^{∗∗}/(Δ − ϵ)) + 1/(D^{∗∗}/(Δ − ϵ) =
[κ/(D^{∗∗}/(Δ − ϵ))] + 1,

where ϵ^{3} = ϵ. Consequently,

[κ/(D^{∗∗}/(Δ − ϵ))] = [ξ(Δ − ϵ)],
ξ(Δ − ϵ) = ξ(Δ − ϵ) − [ξ(Δ − ϵ)] = ξ(Δ − ϵ) − [κ/(D^{∗∗}/(Δ − ϵ))],

where ϵ^{3} = ϵ. Therefore, the existence of κ ∈ [0, D^{∗} −1)∩ℤ, for which the inclusion (55) and the inequality

(56)  2ξΔ − [κ(Δ − 1)/D^{∗∗}] − [κ(Δ + 1)/D^{∗∗}] < ξΔ − [κΔ/D^{∗∗}],

hold, is equivalent to the condition (54). The inequality (56) is equivalent to the inequality

(57) ξ < η^{∗∗}(κ) := κ/D^{∗∗} + ({κΔ/D^{∗∗}} − {κ(Δ − 1)/D^{∗∗}} − {κ(Δ + 1)/D^{∗∗}})/Δ.

Let us note that for κ = 0 the system (55) – (57) has no solutions. Let

η_{0}^{∗∗}(κ) = max(0, min(η^{∗∗}(κ), 1/D^{∗∗})) = max(0, η^{∗∗}(κ)),

where κ ∈ {0, …, D^{∗∗}−1}. Then the system of conditions (55) – (57) is equivalent to the condition

(58)  ξ ∈ [κ/D^{∗∗}, κ/D^{∗∗} + η_{0}^{∗∗}(κ)).

So, if ν/p = m + ξ, where

m ∈ [0, (ν/(2Δ))^{1/2} − 1] ∩ ℤ, κ ∈ ℤ, 1 ≤ κ ≤ D − 1,
ξ ∈ (0, 1), κ/D^{∗∗} ≤ ξ < κ/D^{∗∗} + η_{0}^{∗∗}(κ),

i.e. if

(59)  ν/(m + κ/D^{∗∗} + η_{0}^{∗∗}(κ)) < p ≤ ν/(m + κ/D^{∗∗}),

where

(60)  m ∈ [0, (ν/(Δ + 1))^{1/2} − 1] ∩ ℤ, κ ∈ ℤ, 1 ≤ κ ≤ D^{∗∗} − 1,

then the condition p > (2νΔ)^{1/2} and the condition (64) are fulfilled. We take on the role of log(d_{1}^{∗∗}(ν)) the product of all the prime nubers p, which satisfy to the conditions (59) – (60). In view of (59) – (60) and (28),

(61)  log(d_{1}^{∗∗}(Δ, ν)) =

∑_{κ=1}^{D^{∗∗}−1} ∑_{m=0}^{[(ν/(Δ+1))^{1/2}−1]}(ψ(ν/(m + κ/D^{∗∗})) − ψ(ν/(m + κ/D^{∗∗} + η_{0}^{∗∗}(κ))),

(62)  ψ(ν/(m + κ/D^{∗∗} + η_{0}^{∗∗}(κ))) = ν/(m + κ/D^{∗∗} + η_{0}^{∗∗}(κ))+
O((ν/(m + 1))(exp(−(a/\sqrt{2})(log(ν))^{1/2}),

ψ(ν/(m + κ/D^{∗∗})) = ν/(m + κ/D^{∗∗}) + O((ν/(m + 1)) exp(−(a/\sqrt{2})(log(ν))^{1/2}).

But

∑_{m=0}^{[(ν/(Δ+1))^{1/2}−1]}(1/(m + κ/D^{∗∗})) − 1/(m + κ/D^{∗∗} + η_{0}^{∗∗}(κ)) =

Γ′(κ/D^{∗∗} + η_{0}^{∗∗}(κ))/Γ(κ/D^{∗∗} + η_{0}^{∗∗}(κ)) − Γ′(κ/D^{∗∗})/Γ(κ/D^{∗∗}) + O(ν^{−1/2}),

where κ ∈ [1, D^{∗∗} − 1] ∩ ℤ. Consequently,

log(d_{1}^{∗∗}(Δ, ν)) = ν ∑_{κ=1}^{D^{∗∗}−1} Γ′(κ/D^{∗∗} + η_{0}(κ))/Γ(κ/D^{∗∗} + η_{0}(κ))−

ν ∑_{κ=1}^{D^{∗∗}−1} Γ′(κ/D^{∗∗})/Γ(κ/D^{∗∗}) + O(ν log(ν) exp(−(a/\sqrt{2})(log(ν))^{1/2}).

Let η^{∧}(x) = {x} − {(1 − δ_{0})x} − {(1 + δ_{0})x}, η_{0}^{∧}(x) = max(0, η^{∧}(x)), where x ∈ ℝ. Let further χ^{∧} = χ^{∧}(x) denotes the characteristic function of the set {x ∈ ℝ: 0 < η_{0}^{∧}(Δx)}. If x ∈ (0, 1/(1 + δ_{0})), then χ^{∧}(x) = 0. Therefore the

integral ∫_{0}^{∞}χ^{∧}(x)x^{−2} dx converges. We wish to finde the value Δ ∫_{0}^{∞}χ^{∧}(x) dx =

∫_{0}^{∞}χ^{∧}(Δx)x^{−2} dx. Clearly,

∫_{0}^{∞}χ^{∧}(Δx)x^{−2} dx = ∑_{m=0}^{∞} ∫_{m}^{m+1}χ^{∧}(Δx)x^{−2} dx = ∑_{m=0}^{∞} ∫_{0}^{1} χ^{∧}(Δξ)(m + ξ)^{−2} dξ.

In view of (53), X_{Δ}^{∗∗} = {x ∈ ℝ: 0 < η_{0}^{∧}(Δx)}. Clearly, the condition (54) is equivalent to the equality χ^{∧}(Δξ) = 1. Therefore, in view of (58),

∫_{0}^{1} χ^{∧}(Δξ)(m + ξ)^{−2} dξ = ∑_{κ=1}^{D^{∗∗}−1}(1/(m + κ/D^{∗∗})) − 1/(m + κ/D^{∗∗} + η_{0}^{∗∗}(κ)).

Consequently, in view of (60) – (62),

(63)  log(d_{1}^{∗∗}(ν)) = νΔ ∫_{0}^{∞}χ^{∧}(x) dx + O(ν log(ν) exp(−(a/\sqrt{2})(log(ν))^{1/2}).

In view of (33),

∫_{0}^{∞}χ^{∧}(Δx)x^{−2} dx = ∑_{m=0}^{∞} ∑_{κ=0}^{Δ−1} ∫_{0}^{1} χ^{∧}(mΔ + κ + ξ)(mΔ + κ + ξ)^{−2} dx.

If

(64)  {x} > {x + x/Δ} + {x − x/Δ},

then

{x} > {x + x/Δ} + {x − x/Δ} = {{x + x/Δ} + {x − x/Δ}} =
{2x} = {2{x}}, 1 ≥ 2{x},
{x} > {x + x/Δ} = {{x} + {x/Δ}}, {x} + {x/Δ} ≥ 1, {x/Δ} > 0,
{x} > {x − x/Δ} = {{x} + {−x/Δ}}, {x} ≥= 1 − {1 − {x/Δ}} = {x/Δ},

because {x/Δ} > 0. If

(65)  {x} ≥ {x/Δ} ≥ 1 − {x},

then

2{x} ≥ 1, {x + x/Δ} + {x − x/Δ} =
{{x} + {x/Δ} + {x} − {x/Δ}} = {2{x}} = {2x} < {x}.

So, the conditions (64) and (65) are equivalent.

According to (33) and (35), {x} = ξ, {x/Δ} = (κ + ξ)/Δ and the condition (65) takes form ξ ≥ (κ+ξ)/Δ ≥ 1−ξ, i.e. the form ξ ≥ max(κ/(Δ−1), (Δ−κ)/(Δ+1)), where κ = 0, …, Δ − 1. Clearly,

max(κ/(Δ − 1), (D − κ)/(Δ + 1)) = κ/(Δ − 1), if κ ≥ (Δ − 1)/2,

max(κ/(Δ − 1), (D − κ)/(Δ + 1)) = (D − κ)/(Δ + 1), if κ ≤ (Δ − 1)/2.

Therefore the integral

∫_{0}^{1} χ^{∧}(mΔ + κ + ξ)(mΔ + κ + ξ)^{−2} dx

is equal to

(mΔ + κ + κ/(Δ − 1))^{−1} − (mΔ + κ + 1)^{−1} =
Δ^{−1}(m + κ/(Δ − 1))^{−1} − Δ^{−1}(m + (κ + 1)/Δ)^{−1},

if κ ≥ (Δ − 1)/2, and it is equal to

(mΔ + κ + (Δ − κ)/(Δ + 1))^{−1} − (mΔ + κ)^{−1} =

Δ^{−1}(m + (κ + 1)/(Δ + 1))^{−1} − Δ^{−1}(m + (κ + 1)/Δ)^{−1}),

if κ ≤ (Δ − 1)/2. Consequently,

(66)  ∫_{0}^{∞}χ^{∧}(Δx)x^{−2} dx = Δ^{−1} ∑_{0≤κ<Δ/2} Γ′((κ + 1)/Δ)/Γ((κ + 1)/Δ)−

Δ^{−1} ∑_{0≤κ<Δ/2} Γ′((κ + 1)/(Δ + 1))/Γ((κ + 1)/(Δ + 1))+

Δ^{−1} ∑_{Δ/2≤κ≤Δ−1} Γ′((κ + 1)/Δ)/Γ((κ + 1)/Δ)−

Δ^{−1} ∑_{Δ/2≤κ≤Δ−1} Γ′(κ/(Δ − 1))/Γ(κ/(Δ − 1)) =

Δ^{−1} ∑_{0≤κ≤Δ−1} Γ′((κ + 1)/Δ)/Γ((κ + 1)/Δ)−

Δ^{−1} ∑_{0≤κ<Δ/2} Γ′((κ + 1)/(Δ + 1))/Γ((κ + 1)/(Δ + 1))−

Δ^{−1} ∑_{Δ/2≤κ≤Δ−1} Γ′(κ/(Δ − 1))/Γ(κ/(Δ − 1)) =

Δ^{−1} ∑_{0≤κ≤Δ−1} Γ′((κ + 1)/Δ)/Γ((κ + 1)/Δ)+

Δ^{−1} ∑_{1≤κ≤Δ−1} Γ′(κ/Δ)/Γ(κ/Δ)−

Δ^{−1} ∑_{0≤κ≤Δ−1} Γ′((κ + 1)/(Δ + 1))/Γ((κ + 1)/(Δ + 1))−

Δ^{−1} ∑_{1≤κ≤Δ−1} Γ′(κ/(Δ − 1))/Γ(κ/(Δ − 1))−

Δ^{−1} ∑_{1≤κ≤Δ−1} Γ′(κ/Δ)/Γ(κ/Δ)+

Δ^{−1} ∑_{Δ/2≤κ≤Δ−1} Γ′((κ + 1)/(Δ + 1))/Γ((κ + 1)/(Δ + 1))+

Δ^{−1} ∑_{1≤κ<Δ/2} Γ′(κ/(Δ − 1))/Γ(κ/(Δ − 1)).

In view of (43) and (37),

∑_{1≤κ≤Δ−1} Γ′(κ/Δ)/Γ(κ/Δ) =

−2γ[Δ/2] − Δ log(Δ) + (− log(4)) ((−1)^{Δ} + 1) /2−
(Γ′(1/2)/Γ(1/2))(1 + (−1)^{Δ})/2 = −2γ[Δ/2] − Δ log(Δ)+
(− log(4)) ((−1)^{Δ} + 1) /2 − (−γ − log(4)) (1 + (−1)^{Δ}) /2 =
−2γ[Δ/2] − Δ log(Δ) + γ (1 + (−1)^{Δ}) /2.

Therefore

∑_{0≤κ≤Δ−1} Γ′((κ + 1)/(Δ + 1))/Γ((κ + 1)/(Δ + 1)) =

−2γ[(Δ + 1)/2] − γ − (Δ + 1) log(Δ + 1) + γ (1 + (−1)^{(Δ+1)}) /2,

∑_{1≤κ≤Δ−1} Γ′((κ + 1)/(Δ + 1))/Γ((κ + 1)/(Δ + 1)) =

−2γ[(Δ − 1)/2] − (Δ − 1) log(Δ − 1) + γ (1 + (−1)^{(Δ−1)}) /2,

(67)  Δ^{−1} ∑_{0≤κ≤Δ−1} Γ′((κ + 1)/Δ)/Γ((κ + 1)/Δ)+

Δ^{−1} ∑_{1≤κ≤Δ−1} Γ′(κ/Δ)/Γ(κ/Δ)−

Δ^{−1} ∑_{0≤κ≤Δ−1} Γ′((κ + 1)/(Δ + 1))/Γ((κ + 1)/(Δ + 1))−

Δ^{−1} ∑_{1≤κ≤Δ−1} Γ′(κ/(Δ − 1))/Γ(κ/(Δ − 1)) =

−γ(4[Δ/2] − 2[(Δ + 1)/2] − 2[(Δ − 1)/2] − 2(−1)^{Δ}+ (Δ + 1) log(Δ + 1) + (Δ − 1) log(Δ − 1) − 2Δ log(Δ) = (Δ + 1) log(Δ + 1) + (Δ − 1) log(Δ − 1) − 2Δ log(Δ).

In view of (36), (66), (67), (45),

(68)  ∫_{0}^{∞}χ^{∧}(Δx)x^{−2} dx =

− log(Δ) + \frac{Δ − 1}{2Δ} log(Δ − 1) + \frac{Δ + 1}{2Δ} log(Δ + 1)−

\frac{π}{2Δ} ∑_{1≤κ≤(Δ−1)/2} cot(πκ/(Δ − 1))+

\frac{π}{2Δ} ∑_{1≤κ≤(Δ+1)/2} cot(πκ/(Δ + 1)).

The equality (93) was obtained in Hata’s paper [16]. Applying the Stirling’s formula to 2 log((νΔ)!)−log((νΔ+ν)!)−log((νΔ−ν)!) and making use of (16), (47), (32),

(63) and (68), we obtain the equality

log(r^{∗}(ν)) − log(r^{∗∗}(ν)) = O(ν log(ν) exp(−(a/\sqrt{2})(log(ν))^{1/2}).

§3. On Diophantine Approximations of logarithms in cyclotomic fields. Part 1.

Here I use the same method, as in [21] – [40] and apply also the results of the section 2. I work on the Riemann surface 𝔉 of the function Log (z) and identify it with the direct product of the multiplicative group ℝ_{+}^{∗} = {r ∈ ℝ: r > 0} of all the positive real numbers with the operation ×, not to be written down explicitly as usual, and the additive group ℝ of all the real numbers, so that z_{1}z_{2} = (r_{1}r_{2}, ϕ_{1}+ϕ_{2}) for any two points z_{1} = (r_{1}, ϕ_{1}) and z_{2} = (r_{2}, ϕ_{2}) on 𝔉. I will illustrate the appearing situations on the half plain (ϕ, r), where r > 0. For any z = (r, ϕ) ∈ 𝔉, α ∈ ℝ let

θ_{0}(z) = r exp iϕ, Log (z) = ln(r) + iϕ, η_{α}^{∗}(z) = (r, ϕ − α).

Then θ_{0}(z) = exp(Log (z)), Log (z_{1}z_{2}) = Log(z_{1})+Log (z_{2}) for any {z, z_{1}, z_{2}} ⊂ 𝔉. Let further ρ(z_{1}, z_{2}) = | Log (z_{1}) − Log (z_{2})|, where {z_{1}, z_{2}} ⊂ 𝔉; clearly, (𝔉, ρ) is a metric space, ρ(zz_{1}, zz_{2}) = ρ(z_{1}, z_{2}) for any {z, z_{1}, z_{2}} ⊂ 𝔉. For any α ∈ ℝ the map z → η_{α}^{∗}(z) is the bijection of 𝔉 onto 𝔉 and θ_{0}((η_{α}^{∗})^{m}(z)) = exp(−miα)θ_{0}(z) for each z = (r, ϕ) ∈ 𝔉, α ∈ ℝ and m ∈ ℤ. The group 𝔉 may be considered as ℂ-linear space, if for any z ∈ 𝔉 and any s ∈ ℂ we let z^{s} = (| exp(s Log (z))|, Im (s Log (z)). Let D be a domain in 𝔉, and f(z) = f^{∧}(r, ϕ) is a complex-valued function f(z) on the domain D. It is well known that f(z) is holomorphic in D if the complex-valued function f^{∧}(r, ϕ) of two real variables r and ϕ has continuous partial derivatives in the domain D, and the Cauchy-Riemann conditions

(69)  (δf)(z) := θ_{0}(z)(((∂/∂z)f)(z)) :=
r(((∂/∂r)f^{∧})(r, ϕ)) = −i(((∂/∂ϕ)f^{∧})(r, ϕ))

are satisfied for every point z = (r, ϕ) ∈ D. The equality (69) determines a differentiations ∂/∂z and δ = θ_{0}(z)∂/∂z on the ring of all the holomorphic in the domain D functions. Clearly, the function Log (z) is holomorphic on 𝔉 and, moreover, ((∂/(∂z) Log )(z) = θ_{0}(z^{−1}), (δ Log )(z) = 1. I use the functions of C.S.Mejer. Let Δ ∈ ℕ + 1, δ_{0} = 1/Δ, γ_{1} = (1 − δ_{0})/(1 + δ_{0}), d_{l} = Δ + (−1)^{l}, l = 1, 2. I use the auxiliary set Ω_{0} = {z ∈ 𝔉: |z| ≤ 1} to introduce the first of my auxiliary function f_{1}(z, ν), and prove that, for each ν ∈ ℕ, the function f_{1}(z, ν) belongs to the ring ℚ[θ_{0}(z)]; therefore using the principle of analytic continuation we may regard it as being defined in 𝔉. For ν ∈ ℕ, let

(70) −(−1)^{ν(Δ+1)}f_{1}(z, ν) = G_{2,2}^{(1,1)} (z| \begin{matrix}−νd_{1}, & 1 + νd_{2} \\ 0, & ν \end{matrix} ) = \frac{1}{2πi} ∫_{L_{1}} g_{2,2}^{(1,1)}(s)ds,

where g_{2,2}^{(1,1)}(s) = θ_{0}(z^{s})Γ(−s)Γ(1 + d_{1}ν + s)/(Γ(1 − ν + s)Γ(1 + d_{2}ν − s)), and the curve L_{1} passes from +∞ to +∞ encircling the set ℕ − 1 in the negative direction, but not including any point of the set −ℕ. So, for the parameters of the Meyer’s functions we have

p = q = 2, m = n = 1, a_{1} = −νd_{1}, a_{2} = 1 + νd_{2}, b_{1} = 0, b_{2} = ν,

Δ^{∗} = (∑_{k=1}^{q} b_{k}) − ∑_{j=1}^{p} a_{j} = −ν − 1 < −1,

and, since we take |z| ≤ 1, convergence conditions of the integral (70) hold. To compute the function f_{1}(z, ν), we use the following formula

(71)  G = (−1)^{k} ∑_{s∈S_{k}} Res (g; s),

where k = 1, G denotes the integral (70) with L = L_{k}, g denotes the integrand of the integral (70), S_{k} denotes the set of all the unremovable singularities of g encircled by the curve L_{k}, and Res (g; s) denotes the residue of the function g at the point s. Then we obtain the equlity

f_{1}(z, ν) = (νd_{1})!/(νΔ)!z^{ν}(−1)^{νΔ} ∑_{k=0}^{νΔ}(−θ_{0}(z))^{k}\binom{νΔ}{k}\binom{νΔ + k}{νd_{1}}.

Therefore, as it has been already remarked, using the principle of analytic continuation we may regard it as being defined in 𝔉. Let Ω_{1} = {z ∈ 𝔉: |z| ≥ 1}. Now, let me introduce my second auxiliary function defined for z ∈ Ω_{1}. For ν ∈ ℕ, let

(72) −(−1)^{ν(Δ+1)}f_{2}(z, ν) = G_{2,2}^{(2,1)} (z| \begin{matrix}−νd_{1}, & 1 + νd_{2} \\ 0, & ν \end{matrix} ) = \frac{1}{2πi} ∫_{L_{2}} g_{2,2}^{(2,1)}(s)ds,

where g_{2,2}^{(2,1)}(s) = θ_{0}((η_{π}(z))^{s})Γ(−s)Γ(ν − s)Γ(1 + d_{1}ν + s)/Γ(1 + d_{2}ν − s), and the curve L_{2} passes from −∞ to −∞ encircling the set −ℕ in the positive direction, but not including any point of the set ℕ − 1. So, for the parameters of the Meyer’s functions we have

p = q = m = 2, n = 1, a_{1} = −νd_{1}, a_{2} = 1 + νd_{2}, b_{1} = 0, b_{2} = ν,

Δ^{∗} = (∑_{k=1}^{q} b_{k}) − ∑_{j=1}^{p} a_{j} = −nu − 1 < −1,

and, since we take |z| ≥ 1, convergence conditions of the integral (72) hold. To compute the function f_{2}(z, ν), we use the formula (71), where k = 2, G denotes the integral (72) with L = L_{k}, g denotes the integrand of the integral (72), S_{k} denotes the set of all the unremovable singularities of g encircled by L_{k}, and Res (g; s) denotes the residue of the function g at the point s. Then we obtain the equality

(73)  f_{2}(z, ν)(νΔ)!/(νd_{1})! = (−1)^{ν} ∑_{t=ν+1}^{∞} R_{0}(t; ν)θ_{0}(z^{−t+ν}),

ON THE DIMENSION OF SOME LINEAR SPACES OVER ℚ (PART 5) 23

where R_{0}(t; ν) = (νΔ)!/(νd_{1})! ∏_{κ=ν+1}^{νΔ}(t − κ) ( ∏_{κ=0}^{νΔ}(t + κ))^{−1} . Let further

(74)  f_{k}^{∗}(z, ν) = f_{k}(z, ν)(νΔ)!/(νd_{1})!,

where k = 1, 2. Expanding the function R_{0}(t; ν) into partial fractions, we obtain the equality

R_{0}(t; ν) = ∑_{k=0}^{νΔ} α_{ν,k}^{∗}/(t + k)

with

(75)  α_{ν,k}^{∗} = (−1)^{ν+νΔ+k}\binom{νΔ}{k}\binom{νΔ + k}{νΔ − ν},

where k = 0, …, νΔ. It follows from (73) – (75) that

f_{2}^{∗}(z, ν) = (−θ_{0}(z))^{ν} ∑_{t=1+ν}^{+∞}(θ_{0}(z))^{−t}R_{0}(t; ν) =

(−θ_{0}(z))^{ν} ∑_{t=1+ν}^{+∞}(θ_{0}(z))^{−t−k+k} ∑_{k=0}^{νΔ} α_{ν,k}^{∗}/(t + k) =

(−θ_{0}(z))^{ν} ∑_{t=1+ν}^{+∞}((θ_{0}(z))^{−t−k}/(t + k)) ∑_{k=0}^{νΔ} α_{ν,k}^{∗}(θ_{0}(z))^{k} =

(76)  (−θ_{0}(z))^{ν} ∑_{k=0}^{νΔ} α_{ν,k}^{∗}(θ_{0}(z))^{k} ∑_{τ=1+ν+k}^{+∞}((θ_{0}(z))^{−τ} /τ)) =
= α^{∗}(z; ν)(− log(1 − 1/θ_{0}(z))) − ϕ^{∗}(z; ν),

where log(ζ) is a branch of Log (ζ) with | arg(ζ)| < π,

(77)  α^{∗}(z; ν) = (−(θ_{0}(z))^{ν} ∑_{k=0}^{νΔ} α_{ν, k}^{∗}(θ_{0}(z))^{k} = f_{1}^{∗}(z; ν),
(78)  ϕ^{∗}(z; ν) = (−θ_{0}(z))^{ν} ∑_{k=0}^{νΔ} α_{ν,k}^{∗}(θ_{0}(z))^{k} ∑_{τ=1}^{ν+k}((θ_{0}(z))^{−τ} /τ)) =

(−θ_{0}(z))^{ν} ∑_{τ=1}^{ν}((θ_{0}(z))^{−τ} α^{∗}(z; ν)/τ+

(−θ_{0}(z))^{ν} ∑_{k=0}^{νΔ} α_{ν,k}^{∗}(θ_{0}(z))^{k} ∑_{τ=1+ν}^{ν+k}((θ_{0}(z))^{−τ} /τ)).

The change of order of summation by passage to (76) is possible, because the series in the second sum in (76) is convergent, if |z| ≥ 1 and θ_{0}(z) ≠ 1. We see that in the domain D_{0} = {z ∈ 𝔉: |z| > 1} the funcion f_{2}^{∗}(z, ν) coincides with the function

(79)  f_{0}^{∗}(z, ν) = α^{∗}(z; ν)(− log(1 − 1/θ_{0}(z))) − ϕ^{∗}(z; ν),

The form (79) may be used for various applikations. The following case is interesting for us:

m ∈ ℕ + 1, k ∈ ℤ, (|k|, m) = 1, θ_{0}(z)) = exp(2ikπ/m).

If θ_{0}(z) = −r exp(iψ) with r > 0 and −π < ψ < π, then Im (1 − 1/θ_{0}(z)) = 0 only, when Re (1 − 1/θ_{0}(z)) > 0 and we have no problems with log(1 − 1/θ_{0}(z)). So, we take z = (r, ψ − π) = (r, θ) , where r > 0, |ψ| < π and −2π < θ = ψ − π < 0; clearly, the function (79) is analytic in the domain

D_{1} = {z = (r, ψ − π)): r > 0, −π < ψ < π} = {z = ((r, θ)): r > 0, −2π < θ < 0}.

Let

(80)  D_{2}(δ_{0}) = {z ∈ 𝔉: |z| > 1 + δ_{0}/2, D_{3} = D_{2}(δ_{0}) ∪ D_{1}.

Since D_{2}(δ_{0}) ∩ D_{1} ≠ ∅, it follows that D_{3} = D_{2}(δ_{0}) ∪ D_{1} is a domain in 𝔉, and the function (79) is analytic in this domain and coincides with the function f_{2}^{∗}(z, ν) in the domain D_{2}(δ_{0}) ⊂ D_{0}. The conditions, which imply the equality

(81)  (−1)^{m+p−n} exp(−iα)θ_{0}(z)×

((∏_{j=1}^{p}(δ + 1 − a_{j})) (G ∘ η_{α}^{∗})) (z) = ((∏_{k=1}^{q}(δ − b_{k})) (G ∘ η_{α}^{∗})) (z)

hold for z ∈ D_{0} for the Mejer’s function

G = G_{p,q}^{(m,n)} (z| \begin{matrix}a_{1}, & …, & a_{p} \\ b_{1}, & …, & b_{q} \end{matrix} )

in our case (see [30], section 1, Corollary of the Lemma 2.1.5, the case C_{2}.) In fact we have p = q = 2, m = n = 1, α = 0 for the function f_{1}(z, ν) and |z| > 1 in D_{0}; therefore the equation (81) holds and takes the form

θ_{0}(z)((δ + 1 + d_{1}ν)(δ − d_{2}ν)f_{1})(z, ν) = (δ(δ − ν)f_{1})(z, ν)

We have p = q = m = 2, n = 1, α = π for the function f_{2}(z, ν) and |z| > 1 in D_{0}; terefore the equation (115) holds and takes the form

θ_{0}(z)((δ + 1 + d_{1}ν)(δ − d_{2}ν)f_{2})(z, ν) = (δ(δ − ν)f_{2})(z, ν).

We see that both the functions f(z, ν) = f_{k}^{∗}(z, ν), where k = 1, 2 satisfy to the same differential equation

(82)  θ_{0}(z)(δ + 1 + d_{1}ν)(δ − d_{2}ν)f(z, ν) = (δ(δ − ν)f)(z, ν).

in the domain D_{0}. According to the general properties of the Mejer’s functions we have the equality

(83)  ( ∏_{κ=1}^{Δ−1}(ν(Δ − 1) + κ)) ∏_{κ=1}^{d_{2}}(δ − d_{2}ν − κ)f_{k}^{∗}(z, ν + 1) =

(∏_{κ=1}^{Δ}(νΔ + κ)) (δ − ν) ∏_{κ=1}^{d_{1}}(δ + d_{1}ν + κ)f_{k}^{∗}(z, ν),

where k = 1, 2 and z ∈ D_{0}. Since f_{0}^{∗}(z, ν) and polynomial f_{1}^{∗}(z, ν) are analytic in the domain D_{0} ∪ D_{1}, and f_{0}^{∗}(z, ν) coincides with f_{2}^{∗}(z, ν) in D_{0}, it follows that the equations (82) and (83) hold in D_{0} ∪ D_{1} for k = 0, 1. Let

(84)  D^{∨}(w, η) = (η + 1)(η + γ_{1}) − 2(1 + γ_{1})wη,

(85)  D^{∧}(z, η) = D^{∨}(θ_{0}(z), η),

where

(86)  w = θ_{0}(z) = −r exp(iψ), r > 0, |ψ| < π.

In view of (86), the polynomials (1) in [36] and (85) coincide.

As in [30], we consider ν^{−1} as an independent variable taking its values in ℂ including 0. Let F be a bounded closed subset of 𝔉 (in particular, compact F may be an one-point set). Let ℌ_{0}(F) be the ring of all the functions in ℚ(w) well defined for every w ∈ θ_{0}(F). For ε ∈ (0, 1), let ℌ(F, ε) be the ring of all the functions in ℚ(w, ν^{−1}) well defined for every (w, ν^{−1}) with w ∈ θ_{0}(F), |ν^{−1}| ≤ ε_{0}. Let further h^{∼}(η) = (η − 1)(1 − δ_{0})^{−d_{1}}(η + 1)2^{−2}η^{d_{1}}.

Lemma 4. Let F be a closed bounded subset of D_{0} ∪ D_{1} (in particular, F may be an one-point set). Let for any z ∈ F the polynomial (85) has only simple roots and on the set of all the its roots η the map

(87)  η → h^{∼}(η)

is injective. Then there is ε ∈ (0, 1) such that, for any z ∈ F, ν ∈ ℕ + [1⧸ε], the functions f_{0}^{∗}(z, ν), f_{1}^{∗}(z, ν) = α^{∗}(z; ν) and ϕ^{∗}(z; ν) are solutions of the difference equation

(88)  x(z, ν + 2) + ∑_{j=0}^{1} q_{j}^{∗}(z, ν^{−1})x(z, ν + j) = 0,

moreover, q_{j}^{∗}(z, ν^{−1}) ∈ ℌ(F, ε) for j = 0, 1, and trinomial w^{2} + ∑_{j=0}^{1} q_{j}^{∗}(z, 0)w^{j}

coincides with

∏_{k=0}^{1}(w − h(η_{k})), 89

if ∏_{k=0}^{1}(w − η_{k}) coincides with D^{∨}(w, η) from (84).

Proof. Proof may be found in [30]. ■

This Lemma shows the importance of the properties of the roots of the polynomial (85); all the its roots η_{k}^{∧}(r, ψ, δ_{0}), where k = 0, 1; 0 < r; , ψ ∈ ℝ, are presented in the equality (65) of the paper [36].

Let δ_{0} > 0, 2r ≥ δ_{0}. It follows from the Lemma 3.1.3 in [35] and its proof that

(90)  1 ≤ |η_{0}^{∧}(r, π, δ_{0})| /|η_{1}^{∧}(r, π, δ_{0})| ≤ |η_{0}^{∧}(r, ψ, δ_{0})| /|η_{1}^{∧}(r, ψ, δ_{0})|,

and if 2r ≥ δ_{0}, then

(91) 1 ≤ |η_{0}^{∧}(r, π, δ_{0}) + 1| /|η_{1}^{∧}(r, π, δ_{0}) + 1| ≤ |η_{0}^{∧}(r, ψ, δ_{0}) + 1| /|η_{1}^{∧}(r, ψ, δ_{0}) + 1|,

where ψ ∈ ℝ; moreover, if 2r > 1 + \sqrt{1 − δ_{0}^{2}}, then left inequality in (90) and left inequality in (91) are strong; on the other hand, when r ≥ δ_{0}/2 and ψ ∈ [0, π), then right inequality in (90) and right inequality in (91) are strong. Let δ_{0} > 0, 2r ≥ δ_{0},

(92)  μ_{0}(r, δ_{0}) =
(|η_{1}^{∧}(r, π, δ_{0})| /|η_{0}^{∧}(r, π, δ_{0})|)^{(1/δ_{0}−1)}|η_{1}^{∧}(r, π, δ_{0}) + 1| /|η_{0}^{∧}(r, π, δ_{0}) + 1|,

(93)  c_{0}(r, δ_{0}) = (μ_{0}(r, δ_{0}))^{2} + 1)/(2μ_{0}(r, δ_{0})).

According to (3.1.43) in [34] and (90) – (92), if 2r ≥ δ_{0}, then

(94)  0 < μ_{0}(r, δ_{0}) ≤ 1,

(95)  1/μ_{0}(r, δ_{0})) ≤

(|η_{0}^{∧}(r, ψ, δ_{0})| /|η_{1}^{∧}(r, ψ, δ_{0})|)^{1/δ_{0}−1}|η_{0}^{∧}(r, ψ, δ_{0}) + 1| /|η_{1}^{∧}(r, ψ, δ_{0}) + 1|,

and if 2r > 1 + \sqrt{1 − (δ_{0})^{2}}, then right inequality in (94) is strong.

Lemma 5. Let μ > 0 and all roots of quadratic trinomial x^{2} − px + q are positive. Let further x_{max} and x_{min} denote respectively the maximal and minimal of these roots. Then the inequality

(96)  x_{min}/x_{max} > μ^{2}

is equivalent to the inequality

(97)  p/\sqrt{q} < (μ^{2} + 1)/μ.

Proof. We have

x_{min}/x_{max} = (x_{min})^{2}/q = (x_{min}/\sqrt{q})^{2} = (p/\sqrt{q} − ((p/\sqrt{q})^{2} − 4)^{1/2})^{2}/4.

Therefore (96) is equivalent to the inequality p/\sqrt{q} − 2μ > (p/\sqrt{q})^{2} − 4)^{1/2}, which is equivalent to (97) ■.

Let

(98)  s = s_{0}(r, ψ) = |r exp(iψ) + 1| /2 = ((r − 1)^{2}/4 + r cos^{2}(ψ/2))^{1/2},

where ψ ∈ [−π, π].

Lemma 6. Let 0 < δ_{0} < 1 and at least one of the following conditions is fulfilled:

A) r ≥ 1 + δ_{0}/2, |r − 1| ≤ 2s ≤ (r + 1);

B) max(2|r − 1|, δ_{0}) < 4s ≤ 2(r + 1), δ_{0} ≤ 2r

C) 2|r − 1| < δ_{0} = 4s ≤ (r + 1), δ_{0} ≤ 2r,

D) 0 < |r − 1|/2 ≤ s ≤ (r + 1)/2, r > (1 + \sqrt{1 − (δ_{0})^{2}})/2,

E)

(99)  1 − δ_{0}/2 < r < 1 + δ_{0}/2, |r − 1|/2 ≤ s ≤ δ_{0}/4

and

(100) s > σ_{0}(r, δ_{0}) = δ_{0}((1 + δ_{0}/2)^{2} − r^{2})/(4(c_{0}(r, δ_{0})(1 + δ_{0}) − δ_{0}/2 − r)).

Then (101)

|η_{0}^{∧}(r, ψ, δ_{0})|^{1−δ_{0}}|(η_{0}^{∧}(r, ψ, δ_{0}))^{2} − 1|^{δ_{0}} > |η_{1}^{∧}(r, ψ, δ_{0})|^{1−δ_{0}}|(η_{1}^{∧}(r, ψ, δ_{0}))^{2} − 1|^{δ_{0}}.

Proof. In the cases A), C) and D), according to the Lemma 4.4 in [36],

|η_{0}^{∧}(r, ψ, δ_{0})|^{1−δ_{0}}|η_{0}^{∧}(r, ψ, δ_{0}) + 1|^{δ_{0}} > |η_{1}^{∧}(r, ψ, δ_{0})|^{1−δ_{0}}|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{δ_{0}},

but, according to the Lemma 4.6 in [36], and in view of (3.1.71) in [35],

|η_{0}^{∧}(r, ψ, δ_{0}) − 1|^{δ_{0}} ≥ |η_{1}^{∧}(r, ψ, δ_{0}) − 1|^{δ_{0}} > 0,

So, in the cases A), C) and D) the inequality (100) holds.

In the case B, in view of the Lemma 4.4 in [36] and (3.1.43) in [35],

|η_{0}^{∧}(r, ψ, δ_{0})|^{1−δ_{0}}|η_{0}^{∧}(r, ψ, δ_{0}) + 1|^{δ_{0}} ≥ |η_{1}^{∧}(r, ψ, δ_{0})|^{1−δ_{0}}|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{δ_{0}} > 0,

but, according to the Lemma 4.6 in [36], |η_{0}^{∧}(r, ψ, δ_{0}) − 1|^{δ_{0}} > |η_{1}^{∧}(r, ψ, δ_{0}) − 1|^{δ_{0}}. So, in the case B) the inequality (100) holds. In the case E) (see the proof of the Lemma 4.9 for the case E) in [36] ) the inequality (101) is equivalent to the inequality

(|η_{0}^{∧}(r, ψ, δ_{0})| /|η_{1}^{∧}(r, ψ, δ_{0})|)^{(1/δ_{0}−1)}(|η_{0}^{∧}(r, ψ, δ_{0}) + 1| /|η_{1}^{∧}(r, ψ, δ_{0}) + 1|)×
|η_{0}^{∧}(r, ψ, δ_{0}) − 1| > |η_{1}^{∧}(r, ψ, δ_{0}) − 1|,

which, in view of (95), will be fulfilled, if the inequality

(102)  |η_{0}^{∧}(r, π, δ_{0}) − 1|^{2} > |η_{1}^{∧}(r, π, δ_{0}) − 1|^{2}(μ_{0}(r, δ_{0}))^{2}

will be fulfilled. Since we are found in the case h) of the Lemma 4.6 in [36], it follows that x_{max} = |η_{1}^{∧}(r, π, δ_{0}) − 1|^{2}, x_{min} = |η_{0}^{∧}(r, π, δ_{0}) − 1|^{2} and (102) will be fulfilled, when (97) will be fulfilled with p = p_{0}, q = q_{0}, where p_{0} and q_{0} are taken from (70) – (71) in [36]. The equality (3.1.69) in [35] may be rewritten in the form

(103)  |R_{0}(r, ψ, δ_{0})|^{4} = |D_{0}(r, ψ, δ_{0})|^{2} = r^{2} + (r^{2} − (δ_{0}/2)^{2})^{2}+
2r(r^{2} + (δ_{0}/2)^{2})t + r^{2}(δ_{0})^{2}t^{2} = r^{2}(δ_{0})^{2}((4s^{2} − r^{2} − 1)/2r)^{2}+
2r(r^{2} + (δ_{0}/2)^{2})((4s^{2} − r^{2} − 1)/2r) + r^{2} + (r^{2} − (δ_{0}/2)^{2})^{2} =
4(δ_{0})^{2}s^{4} + (4r^{2} − (δ_{0})^{2}(2r^{2} + 1))s^{2} + (δ_{0}/2)^{2}r^{2}(r^{2} − 1) + (δ_{0}/2)^{4} =
(2rs + (δ_{0}/2)^{2})^{2} + (δ_{0})^{2}α(s),

where

α(s) = 4s^{4} − (2r^{2} + 1)s^{2} + r^{2}(r^{2} − 1)/4 − rs = (2s^{2} − r^{2}/2)^{2} − (s + r/2)^{2} =

(2s^{2} + s + r/2 − r^{2}/2)(2s^{2} − s − r/2 − r^{2}/2) =

4(s − (r − 1)/2)(s + r/2)^{2}(s − (r + 1)/2)) = (2s + r)^{2}(s − (r − 1)/2)(s − (r + 1)/2)).

Since α(s) ≤ 0, when \frac{r−1}{2} ≤ s ≤ \frac{r+1}{2}, it follows, in view of (3.1.70) - (3.1.71) in [35] and (103), that

\frac{p_{0}}{2\sqrt{q_{0}(r, ψ, δ_{0})}} ≤ (2(1 + δ_{0})s)^{−1}(2(2 + δ_{0})s^{2} − δ_{0}(r^{2} − 1) + (δ_{0}/2)^{2} + 2rs + (δ_{0}/2)^{2}).

ON THE DIMENSION OF SOME LINEAR SPACES OVER ℚ (PART 5) 29

Therefore (97) will be fulfilled, if the inequality

c_{0}(r, δ_{0}) > (2(1 + δ_{0})s)^{−1}(2(2 + δ_{0})s^{2} − δ_{0}(r^{2} − 1) + (δ_{0}/2)^{2} + 2rs + (δ_{0}/2)^{2}), will be fulfilled; according to (99),

2(2 + δ_{0})s^{2} ≤ (2 + δ_{0})(δ_{0}/2)s ≤ δ_{0}s + (δ_{0}/2)^{3}.

Therefore (97) will be fulfilled, if the inequality

2(c_{0}(r, δ_{0})(1 + δ_{0}) − δ_{0}/2 − r)s > δ_{0}(1 + δ_{0}/2)^{2} − r^{2})/2,

will be fulfilled; the last inequality is equivalent to (100). ■

According to the Remark 1 in [36] (see also the section 4 below), if ε^{3} = ε, k^{2} = k, |2r − 1| ≥ \sqrt{1 − δ_{0}^{2}}, then

(104)  η_{k}^{∧}(r, π, δ_{0}) + ε =

(2(r − 1/2 + ε(1 + δ_{0})/2 + (−1)^{k} sign (2r − 1)(r^{2} − r + (δ_{0}/2))^{2})^{1/2})/(1 + δ_{0}). and if ε^{3} = ε, k^{2} = k, |2r − 1| ≤ \sqrt{1 − δ_{0}^{2}}, then η_{k}^{∧}(r, π, δ_{0}) + ε = \overline{η_{1−k}^{∧}(r, π, δ_{0}) + ε}. If ε^{2} = ε, k^{2} = k and r ≥ (1 + (1 − δ_{0}^{2})^{1/2})/2, then (104) implies the relations

η_{k}^{∧}(r, π, δ_{0}) > 0,

(105) |η_{k}^{∧}(r, π, δ_{0}) + ε| = η_{k}^{∧}(r, π, δ_{0}) + ε = \frac{2r − 1 + 2(−1)^{k}\sqrt{r^{2} − r + (\frac{δ_{0}}{2} )^{2}}}{1 + δ_{0}} + ε.

Therefore, if ε^{2} = ε, k^{2} = k and r ≥ (1 + \sqrt{1 − δ_{0}^{2}}) /2, then

(106)  (d/dr)|η_{k}^{∧}(r, π, δ_{0}) + ε| = (d/dr)|η_{k}^{∧}(r, π, δ_{0})|.

In view of (105), if r > (1 − δ_{0}^{2})^{1/2}/2, then

(107)  (d/dr)|η_{0}^{∧}(r, π, δ_{0})| > 0,

and in view of the first of the equalities (3.1.43) in [35], if r > ^{\frac{1+\sqrt{1−δ_{0}^{2}}}{2}} , then

(108)  (d/dr)|η_{1}^{∧}(r, π, δ_{0})| < 0.

In view of (106) – (108), if ε^{2} = ε, k^{2} = k and r ≥ (1 + (1 − δ_{0}^{2})^{1/2})/2, then

(109)  (−1)^{k}(d/dr)|η_{k}^{∧}(r, π, δ_{0}) + ε| > 0.

If r ≥ (1 + \sqrt{1 − δ_{0}^{2}}) /2, then, in view of (113) – (114) in [36], and, in view of (92) and (109), 1/μ_{0}(r, δ_{0}) > 1, (d/dr)(1/μ_{0}(r, δ_{0})) > 0. Therefore, according to (93) and (100), if r > (1 + (1 − δ_{0}^{2})^{1/2})/2, then (d/dr)c_{0}(r, δ_{0}) > 0, (d/dr)σ_{0}(r, δ_{0}) < 0.

Lemma 7. If r = 1, Δ ∈ 2 + ℕ, then σ_{0}(r, δ_{0}) < (1/51)(26/27).

Proof. See the Proof of the Corollary of the Lemma 4.9 in [36]. ■

Lemma 8. Let Δ ∈ 2 + ℕ, δ_{0} = 1/Δ. Then ((1 + (1 − δ_{0}^{2})^{1/2})/2, 1) contains the number ρ_{0}(δ_{0}) such that σ_{0}(r, δ_{0}) < 1/51 for r ∈ (ρ_{0}(δ_{0}), 1 + δ_{0}/2).

Proof. Since c_{0}(r, δ_{0}) ≥ 1, it follows that the assertion of the Lemma is direct corollary of (100), the Lemma 7 and the Lemma 6. ■

Clearly, for ρ_{0}(δ_{0}) of the Lemma 8 we have the inequality

1−ρ_{0}(δ_{0}) < (1−(1−(δ_{0}^{2}))^{1/2})/2 = (δ_{0}^{2}/2)/(1+(1−(δ_{0})^{2})^{1/2}) ≤ δ_{0}/(6+4\sqrt{2}) < δ_{0}/2.

If Δ ∈ [3, 16] ∩ ℕ],then 1/51 < δ_{0}/4. Therefore, if Δ ∈ [3, 16] ∩ ℕ], then

(110)  D_{4}(δ_{0}) =

\{\{r, s\} ⊂ ℝ: ρ_{0}(δ_{0}) < r < 1 + \frac{δ_{0}}{2} , max (\frac{|r − 1|}{2}, \frac{1}{51}) < s < \frac{δ_{0}}{4} \} ≠ ∅;

so the domain D_{4}(δ_{0}) consists of all the points (r, ψ − π) ∈ 𝔉, where, in correspondence with the equality (98), r and ψ ∈ (−π, π) satisfy to the condition (110). Let D_{4}^{∗}(δ_{0}) denotes the closure of D_{4}(δ_{0}). According to E) of the Lemma 6, the inequality (101) holds for any (r, ψ − π) ∈ D_{4}^{∗}(δ_{0}). Clearly,

D_{4}(δ_{0}) ⊂ D_{5}(δ_{0}) =

\{\{r, s\} ⊂ ℝ: ρ_{0}(δ_{0}) < r < 1 + \frac{δ_{0}}{2} , max (\frac{|r − 1|}{2}, \frac{1}{51}) < s < (r + 1)/2\} .

According to B), C) and E) of the Lemma 6, if Δ ∈ [3, 16] ∩ ℕ], then (101) holds for any (r, ψ − π) ∈ D_{5}(δ_{0}). Let

D_{6}(δ_{0}) = {{r, s} ⊂ ℝ: δ_{0}/2 < r, max(2|r − 1|, δ_{0}) < 4s ≤ 2(r + 1)},

and let D_{6}^{∗}(δ_{0}) denotes the closure of the domain D_{6}(δ_{0}). According to B) of the Lemma 6, the inequality (101) holds for any (r, ψ−π) ∈ D_{6}^{∗}(δ_{0}). Let further D_{2}^{∗}(δ_{0}) denotes closure of the domain D_{2}(δ_{0}). In view of (80) and according to A) of the Lemma 5, the inequality (101) holds for the arbitrary point (r, ψ − π) ∈ D_{2}^{∗}(δ_{0}). Let D_{7}(δ_{0}) = D_{5}(δ_{0}) ∪ D_{6}(δ_{0}) ∪ D_{2}(δ_{0}) and D_{7}^{∗}(δ_{0}) denotes the closure of D_{7}(δ_{0}). Since D_{5}(δ_{0}) ∩ D_{6}(δ_{0}) ≠ ∅, D_{2}(δ_{0}) ∩ D_{6}(δ_{0}) ≠ ∅, it follows that D_{7}(δ_{0}) is a domain in 𝔉, and the inequality (101) holds for any point (r, ψ − π) ∈ D_{7}^{∗}(δ_{0}).

Let D is a domain in ℂ or 𝔉. and D^{∗} is closure of D. Let

(111)  a_{0}^{∼}(z) , …, a_{n}^{∼}(z)

which are functions continuous on D^{∗} and analytic in D. Let a_{n}^{∼}(z) = 1 for any z ∈ D^{∗}. Let

(112)  T (z, λ) = ∑_{i=0}^{n} a_{i}^{∼}(z)λ^{k}.

Let s ∈ ℕ, n_{i} ∈ ℕ − 1, where i = 1, …, s and ∑_{i=1}^{s} n_{i} = n. We say that T (z, λ) has

(n_{1}, …, n_{s})-disjoint system of roots on D^{∗}, if for any z ∈ D^{∗} the set of all the roots λ of the polynomial T (z, λ) splits in s klasses 𝔎_{1}(z), …, 𝔎_{s}(z) with following properties:

a) for i = 1, …, s the sum of the multiplicities of the roots of the klass 𝔎_{i}(z) is equal to n_{i};

b) if i ∈ [1, s] ∩ ℕ, j ∈ (i, s] ∩ ℕ, n_{i}n_{j} ≠ 0, then the absolute value of each root of the klass 𝔎_{i}(z) is greater than absolute value of the each root of the klass 𝔎_{j}(z).

If the polynomial (112) has (n_{1}, …, n_{s})-disjoint system of roots on D^{∗}, then for each i = 1, …, s we denote by ρ_{i,0}^{∗}(z) and ρ_{i,1}^{∗}(z) respectively the maximal and minimal absolute value of the roots of the klass 𝔎_{i}(z). Let D is bounded domain in 𝔉 such that D^{∗} ∈ D_{7}(δ_{0}). Let

(113)  F^{∧}(z, η) = ∏_{i=1}^{2}(θ_{0}(z) − h(η_{i−1}(r, ψ, δ_{0}))),
n = s = 2, n_{1} = n_{2} = 1, 𝔎_{i}(z) = {h(η_{i−1}(r, ψ, δ_{0}))}
ρ_{i,0} = ρ_{i,1} = |h(η_{i−1}(r, ψ, δ_{0}))|,

where i = 1, 2.

Lemma 9. The polynomial F^{∧}(z, η) in (113) has (1, 1)-disjoint system of roots on D^{∗}.

Proof. According to the previous considerations, the more strong assertion holds: the polynomial (113) has (1, 1)-disjoint system of roots on D_{7}^{∗}(δ_{0}). ■

Corollary. Let D is bounded domain in 𝔉 such that D^{∗} ⊂ D_{7}(δ_{0}). Then the map (87) is injective for every z ∈ D^{∗}; the functions f_{0}^{∗}(z, ν), α^{∗}(z, ν) and ϕ^{∗}(z, ν) in (79), (77) and (78) satisfy to all the conditions of the Lemma 2 for every z ∈ D^{∗}; therefore for every z ∈ D^{∗} these functions are solutions of the difference equation of Poincaré type (88), and characteristical polynomial of this equation coincides with the polynomial (89). ■

Let for each ν ∈ ℕ − 1 are given continuous on D^{∗} functions

(114)  a_{0}(z; ν), …, a_{n}(z, ν),

which are analytic in D. Let a_{n}(z : ν) = 1 for any z ∈ D^{∗} and any ν ∈ ℕ − 1. Let for any i = 1, …, n − 1 the sequence of functions a_{i}(z; ν) converges to a_{i}^{∼}(z) uniformly on D^{∗}, when ν → ∞. Let us consider now the difference equation

(115)  a_{0}(z; ν)y(ν + 0) + … + a_{n}(z; ν)y(ν + n) = 0,

i.e. we consider a difference equation of the Poincaré type, coefficients (114) of this equation are continuous on D^{∗} and analytic in D, and they uniformly converge to limit functions (111), when ν → ∞.

Lemma 10. Let the equation (115) has a solution y(z, ν), which is continuous on D^{∗} and analytic on D. Let polynomial (112) has (n_{1}, …, n_{s})-disjoint system of roots on D^{∗}. Let i ∈ [1, s] ∩ ℤ. Let us consider the set of all the z ∈ D, for which the following inequality holds

(116)  lim sup_{ν∈ℕ, ν→∞} |y(z, ν)|^{1/ν}) < ρ_{i,1}(z);

if this set has a limit point in D, then the inequality (116) holds in D^{∗}. Proof. The proof may be found in [20] (Theorem 1 and its Corollary). ■ Lemma 11. Let D is bounded domain in 𝔉 such that D^{∗} ∈ D_{3}. Then

(117)  lim sup_{ν∈ℕ, ν→∞} |f_{0}^{∗}(z, ν)|^{1/ν}) < ρ_{1,1}(z) = |h^{∼}(η_{0}(r, ψ, δ_{0}))|

for any z ∈ D^{∗}.

Proof. In view of (80), expanding the domain D, if necessary, we can suppose that {(r, ϕ): r ∈ [2, 3], ϕ = 0} ⊂ D. Making use the same arguments, as in [33], Lemma 4.2.1, we see that (117) holds for any z = (r, ϕ) ∈ {r ∈ [2, 3], ϕ = 0}. According to the Lemma 5, (117) holds for any z ∈ D^{∗}. ■

Let K/ℚ be the finite extension of the field ℚ, [K : ℚ] = d. Let K has r_{1} real places and r_{2} complex places. Each such place is the monomorphism of the field K in ℝ, if a place is real, or in ℂ, if a place is not real; we will denote these monomorphisms respectively by σ_{1} , … σ_{r_{1}+r_{2}}. Then d = r_{1} + 2r_{2}. Let 𝔅 be the fixed integer basis

(118)  ω_{1} , …, ω_{d}

of the field K over ℚ. Clearly,K is an algebra over ℚ. With extension of the ground field from ℚ to ℝ appears an isomorphism of the algebra 𝔎 = K ⊗ ℝ onto direct sum \underbrace{ℝ ⊕ … ⊕ ℝ}_{r_{1} times} ⊕ \underbrace{ℂ ⊕ … ⊕ ℂ}_{r_{2} times} of r_{1} copies of ℝ and r_{2} copies of ℂ. We identify by

means of this isomorphism the aIgebra 𝔎 with the specified direct sum. We denote below by π_{j}, where j = 1 , …, r_{1} + r_{2}, the projection of 𝔎 onto its j−th direct summand and also the extension of this projection onto all the kinds of matrices which have all the elements in 𝔎. So, π_{j}(𝔎) = ℝ for j = 1 , …, r_{1} and π_{j}(𝔎) = ℂ for j = r_{1} + 1 , …, r_{1} + r_{2}. We denote by 𝔦_{𝔎} the embedding of ℝ in 𝔎 in diagonal way and also the extension of this embedding onto all kinds of the real matrices. So, ℝ is imbedded by means of 𝔦_{𝔎} in 𝔎 in diagonal way. Each element Z ∈ 𝔎 has

ON THE DIMENSION OF SOME LINEAR SPACES OVER ℚ (PART 5) 33

a unique representation in the form:

(119)  Z = ( z_{1} ; ⋮ ; z_{r_{1}+r_{2}} ; \overline{z_{r_{1}+1}} ; ⋮ ; \overline{z_{r_{1}+r_{2}}} ) ,

with z_{j} = π_{j}(Z) for any j = 1 , …, r_{1} + r_{2}. We denote by Tr_{𝔎}(Z) the sum

∑_{j=1}^{r_{1}} π_{j}(Z) + ∑_{j=r_{1}+1}^{r_{1}+r_{2}} 2 Re (π_{j}(Z)),

and by q_{∞}^{(𝔎)}(Z) we denote the value max(|π_{1}(Z)| , …, |π_{r_{1}+r_{2}}(Z)|). Clearly,

q_{∞}^{(𝔎)}(Z_{1}Z_{2}) ≤ q_{∞}^{(𝔎)}(Z_{1})q_{∞}^{(𝔎)}(Z_{2}),

q_{∞}^{(𝔎)}(Z_{1} + Z_{2}) ≤ q_{∞}^{(𝔎)}(Z_{1}) + q_{∞}^{(𝔎)}(Z_{2}), q_{∞}^{(𝔎)}(𝔦_{𝔎}(λ)Z) = |λ|q_{∞}^{(𝔎)}(Z)

for any Z_{1} ∈ 𝔎, Z_{2} ∈ 𝔎, Z ∈ 𝔎 and λ ∈ ℝ. The natural extension of the norm q_{∞}^{(𝔎)} on the set of all the matrices, which have all the elements in 𝔎 (i.e. the maximum of the norm q_{∞}^{(𝔎)} of all the elements of the matrix) also will be denoted by q_{∞}^{(𝔎)}.

If Z = ( z_{1} ; ⋮ ; z_{d} ) ∈ K ⊂ 𝔎, then z_{j} = σ_{j}(Z) = π_{j}(Z), where j = 1 , …, r_{1} + r_{2},

and z_{r_{1}+r_{2}+j} = \overline{σ_{r_{1}+j}(Z)}, where j = 1 , …, r_{2}. In particular,

ω_{k} = ( σ_{1}(ω_{k}) ; ⋮ ; σ_{r_{1}+r_{2}}(ω_{k}) ; \overline{σ_{r_{1}+1}(ω_{k})} ; ⋮ ; \overline{σ_{r_{1}+r_{2}}(ω_{k})} ) ,

The ring of all the integer elements of the field K will be denoted by ℤ_{K}. On 𝔎 there is the involution which turns each element (119) in 𝔎 into \overline{Z} = ( \overline{z_{1}} ; ⋮ ; \overline{z_{d}} ) ∈ 𝔎. Every element Z of the algebra 𝔎 has a unique representation in the form

(120)  Z = ∑_{j=1}^{d} x_{j}ω_{j}

with x_{j} ∈ ℝ and ω_{j} in (118). Let D ∈ Mat_{d}(ℂ), and ω_{j} in (118) is its j-th column

for j = 1, …, d. Let further X_{Z}^{(𝔅)} denotes the column ( x_{1} ; ⋮ ; x_{d} ) . Then (120) may

be rewritten in the form Z = DX_{Z}^{(𝔅)}. Clearly,

K = {∑_{j=1}^{d} x_{j}ω_{j}: x_{j} ∈ ℚ, j = 1, …, d} , ℤ_{K} = {∑_{j=1}^{d} x_{j}ω_{j}: x_{j} ∈ ℤ, j = 1, …, d} .

The ring ℤ_{K} is embedded in the ring 𝔎 as discrete lattice. Moreover, if Z ∈ ℤ_{K}\{0},

then ( ∏_{i=1}^{r_{1}} |σ_{j}(Z)|) ∏_{i=1}^{r_{2}} |σ_{r_{1}+i}(Z)|^{2} = | Nm_{K/ℚ}(Z)| ∈ ℕ and therefore

(121)  q_{∞}^{(𝔎)}(Z) ≥ 1.

for any Z ∈ ℤ_{K}\{0}. For each Z ∈ 𝔎 let ∥ℤ∥_{K} = inf_{W ∈ℤ_{K}}{q_{∞}^{(𝔎)}(Z − W)}. The

elements of ℤ_{K} we name below by K-integers.

Lemma 12. If Z ∈ 𝔎 and q_{∞}^{(𝔎)}(Z) ≤ 1/2 then ∥Z∥_{K} = q_{∞}^{(𝔎)}(Z).

Proof. If q_{∞}^{(𝔎}(Z − W) < q_{∞}^{(𝔎)}(Z) with W ∈ ℤ_{K}, then

q_{∞}^{(𝔎)}(W) ≤ q_{∞}^{(𝔎)}(Z) + q_{∞}^{(𝔎)}(Z − W) < 2q_{∞}^{(𝔎)}(Z) ≤ 1.

In view (121), W = 0. ■

Let m ∈ ℕ, n ∈ ℕ, a_{i,k} ∈ 𝔎 for i = 1 , …, m, k = 1 , …, n, let α_{j}^{∧}(ν) ∈ ℤ_{K}, where j = 1, …, m + n and ν ∈ ℕ. Let there are γ_{0}, r_{1}^{∧} ≥ 1, …, r_{m}^{∧} ≥ 1 such that

(122)  q_{∞}^{(𝔎)}(α_{i}(ν)) < γ_{0}(r_{i}^{∧})^{ν}

where i = 1 , …, m and ν ∈ ℕ. Let further y_{k}(ν) = −α_{m+k}^{∧}(ν) + ∑_{i=1}^{m} a_{i,k}α_{i}^{∧}(ν)

where k = 1 , …, n and ν ∈ ℕ. If X = ( Z_{1} ; ⋮ ; Z_{n} ) ∈ 𝔎^{n}, then let ϕ_{i}(X) = ∑_{k=1}^{n} a_{i,k}Z_{k}

for i = 1 , …, m, let y^{∧}(X) = y^{∧}(X, ν) = ∑_{k=1}^{n} y_{k}^{∧}(ν)Z_{k} for ν ∈ ℕ, and let

(123)  α_{0}^{∧}(X, ν) = ∑_{k=1}^{n} α_{m+k}^{∧}(ν)Z_{k}

for ν ∈ ℕ. Clearly, y^{∧}(X, ν) = −α_{0}^{∧}(X, ν) + ∑_{i=1}^{m} α_{i}^{∧}(ν)ϕ_{i}(X) for X ∈ 𝔎^{n}, ν ∈ ℕ;

moreover, α_{0}^{∧}(X, ν) ∈ ℤ_{K} for X ∈ (ℤ_{K})^{n} and ν ∈ ℕ.

ON THE DIMENSION OF SOME LINEAR SPACES OVER ℚ (PART 5) 35

Lemma 13. Let l ∈ ℕ, γ_{1} > 0, let

(124)  γ_{2} > \frac{1}{2}, R_{1} ≥ R_{2} > 1, α_{i} = (log(r_{i}^{∧}R_{1}/R_{2}))/ log(R_{2}),

let where i = 1 , …, m, let γ_{3} = γ_{1}(R_{1})^{(− log(2γ_{2}R_{2}))/ log(R_{2})},

γ_{4} = γ_{3} (∑_{i=1}^{m} γ_{0}(r_{i}^{∧})^{(log(2γ_{2}))/ log(R_{2})+l})^{−1}

(188)  X ∈ (ℤ_{K})^{n}\{(0)},

and let for each ν ∈ ℕ − 1 hold the inequalities

(126)  q_{∞}^{(𝔎)}(y^{∧}(X, ν)) ≤ γ_{2}(R_{2})^{−ν}q_{∞}^{(𝔎)}(X),
γ_{1}(R_{1})^{−ν}q_{∞}^{(𝔎)}(X) ≤ sup{q_{∞}^{(𝔎)}(y^{∧}(X, κ)): κ = ν, …, ν + l − 1}.

Then

(127)  sup{∥ϕ_{i}(X)∥_{K}(q_{∞}^{(𝔎)}(X))^{α_{i}}: i = 1, …, m} ≥ γ_{4}.

Proof. According to (124), (125) and (121), 2γ_{2}q_{∞}^{(𝔎)}(X) > 1. Let

(128)  ν_{1} = inf{ν ∈ ℕ − 1: γ_{2}q_{∞}^{(𝔎)}(X)(R_{2})^{−ν} ≤ 1/2} =
inf{ν ∈ ℕ − 1: ν ≥ log(2γ_{2}q_{∞}^{(𝔎)}(X))/ log(R_{2}).

Clearly, ν_{1} = log(2γ_{2}q_{∞}^{(𝔎)}(X))/ log(R_{2}) + θ, where θ ∈ [0, 1),

(129)  γ_{1}(R_{1})^{−ν_{1}}q_{∞}^{(𝔎)}(X) =
γ_{1} exp(−(log(R_{1}))(log(2γ_{2}q_{∞}^{(𝔎)}(X))/ log(R_{2}) + θ))q_{∞}^{(𝔎)}(X) =

γ_{1}(q_{∞}^{(𝔎)}(X))^{−(log((R_{1})/R_{2}))/ log(R_{)}} exp(−(log(R_{1}))((log(2γ_{2}))/ log(R_{2}) + θ)) ≥ γ_{1}(exp(−(log(R_{1}))(log(2γ_{2}R_{2}))/ log(R_{2})))(q_{∞}^{(𝔎)}(X))^{−(log((R_{1})/R_{2}))/ log(R_{2})} =

γ_{3}(q_{∞}^{(𝔎)}(X))^{−(log((R_{1})/R_{2}))/ log(R_{2})}.

In view of (128), γ_{2}q_{∞}^{(𝔎)}(X)(R_{2})^{−ν} ≤ 1/2 for ν ∈ ν_{1} + ℕ − 1. Let q_{∞}^{(𝔎)}(y^{∧}(X, ν_{2})) = sup{q_{∞}^{(𝔎)}(y^{∧}(X, κ)): κ = ν_{1}, …, ν + l − 1} with ν_{2} ∈ {ν_{1} , …, ν_{1} +l−1}. Then, in view of (126), (129), (125), (128) and (122),

\frac{1}{2} ≥ q_{∞}^{(𝔎)}(y^{∧}(X, ν_{2})) ≥ γ_{1}(R_{1})^{−ν_{1}}q_{∞}^{(𝔎)}(X) ≥ γ_{3}(q_{∞}^{(𝔎)}(X))^{−(log((R_{1})/(R_{2})/ log(R_{2})}.

Let be fixed f_{1} ∈ ℤ_{K}, …, f_{m} ∈ ℤ_{K} such that

(130)  q_{∞}^{(𝔎)}(ϕ_{i}(X) − f_{i}) = ∥ϕ_{i}(X)∥_{K}.

According to (123), (130) and Lemma 12, g = −α_{0}^{∧}(X, ν_{2}) + ∑_{i=1}^{m} α_{i}^{∧}(ν_{2})f_{i} ∈ ℤ_{K}

and

γ_{3}(q_{∞}^{(𝔎)}(X))^{−(log((R_{1})/R_{2}))/ log(R_{2})} ≤ q_{∞}^{(𝔎)}(y^{∧}(X, ν_{2}) =

∥q_{∞}^{(𝔎)}(y^{∧}(X, ν_{2}))∥_{K} ≤ q_{∞}^{(𝔎)}(y^{∧}(X, ν_{2}) − g) ≤ ∑_{i=1}^{m} q_{∞}^{(𝔎)}(α_{i}^{∧}(ν_{2}))q_{∞}^{(𝔎)}(ϕ_{i}(X) − f_{i}) =

∑_{i=1}^{m} q_{∞}^{(𝔎)}(α_{i}^{∧}(ν_{2}))∥ϕ_{i}(X)∥_{K} ≤ ∑_{i=1}^{m} ∥ϕ_{i}(X)∥_{K}γ_{0}(r_{i}^{∧})^{ν_{1}+l−1}.

But

(r_{i}^{∧})^{ν_{1}} = exp((log(r_{i}^{∧}))((log(2γ_{2}q_{∞}^{(𝔎)}(X)))/ log(R_{2}) + θ)) =
(q_{∞}^{(𝔎)}(X))^{(log(r_{i}^{∧}))/ log(R_{2})}(r_{i}^{∧})^{(log(2γ_{2}))/ log(R_{2}))+θ))},

where i = 1, …, m. Therefore

(131)  γ_{3}(q_{∞}^{(𝔎)}(X))^{−(log((R_{1})/(R_{2})/ log(R_{2})} ≤ q_{∞}^{(𝔎)}(y^{∧}(X, ν_{2}) ≤
sup{∥ϕ_{i}(X)∥_{K}(q_{∞}^{(𝔎)}(X))^{(log(r_{i}^{∧}))/ log(R_{2})}: i = 1, …, m}×

∑_{i=1}^{m} γ_{0}(r_{i}^{∧})^{(log(2γ_{2}))/ log(R_{2})+l}.

¿From (131) follows (127). ■

Corollary. Let a ∈ 𝔎, α_{1}^{∧}(ν) ∈ ℤ_{K}, α_{2}^{∧}(ν) ∈ ℤ_{K}, y(ν) = −α_{2}^{∧}(ν) + aα_{1}^{∧}(ν) where ν ∈ ℕ.

Let there are γ_{0}, r_{1}^{∧} ≥ 1 such that q_{∞}^{(𝔎)}(α_{1}(ν)) < γ_{0}(r_{1}^{∧})^{ν}, where ν ∈ ℕ. Let

l ∈ ℕ, γ_{1} > 0, γ_{2} > \frac{1}{2}, R_{1} ≥ R_{2} > 1, α_{1} = (log(r_{1}^{∧}R_{1}/R_{2}))/ log(R_{2}),

γ_{3} = γ_{1}(R_{1})^{(− log(2γ_{2}R_{2}))/ log(R_{2})}, γ_{4} = γ_{3} (γ_{0}(r_{1}^{∧})^{(log(2γ_{2}))/ log(R_{2})+l})^{−1} .

Let X ∈ ℤ_{K} and let for each ν ∈ ℕ − 1 hold the inequalities

γ_{1}(R_{1})^{−ν}q_{∞}^{(𝔎)}(X) ≤ sup{q_{∞}^{(𝔎)}(y_{1}(κ)X): κ = ν , …, ν + l − 1))},

q_{∞}^{(𝔎)}(y(ν)X) ≤ γ_{2}(R_{2})^{−ν}q_{∞}^{(𝔎)}(X).

Then ∥aX∥_{K}(q_{∞}^{(𝔎)}(X))^{α} ≥ γ_{4}.

Proof. This Corrolary is the Lemma 13 for m = n = 1. ■

Let l ∈ ℕ, let p be prime in (2, +∞), let K be a finite extension of ℚ let 𝔭 be a prime ideal in ℤ_{K} and p ∈ 𝔭, let f be the degree of 𝔭, let (p) = 𝔭^{e}𝔮, where 𝔮 is a entire ideal, not contained in 𝔭, let v_{𝔭} be additive 𝔭-valuation, which prolongs v_{p}; so, if π is a 𝔭-prime number, then v_{𝔭}(π) = 1/e. If f is the degree of the ideal 𝔭 then

(132)  v_{𝔭} (w^{p^{β}} − w) ≥ 1,

where β ∈ ℕf, w ∈ K and v_{𝔭}(w) ≥ 0. In viw of (132), (77), and (75),

v_{𝔭}(α^{∗}(z; p^{β}l) − α^{∗}(z; l)) > 1/e,

if β ∈ ℕf, θ_{0}(z) ∈ K and v_{𝔭}(θ_{0}(z)) ≥ 0. In view of (78),

(133)  ϕ^{∗}(z; ν) =

(−θ_{0}(z))^{ν} ∑_{k=0}^{νΔ} α_{ν,k}^{∗}(θ_{0}(z))^{k} ∑_{τ=1}^{ν+k}((θ_{0}(z))^{−τ} /τ)) =

(−θ_{0}(z))^{ν} ∑_{τ=1}^{ν}((θ_{0}(z))^{−τ} α^{∗}(z; ν)/τ+

(−θ_{0}(z))^{ν} ∑_{k=0}^{νΔ} α_{ν,k}^{∗}(θ_{0}(z))^{k} ∑_{τ=1+ν}^{ν+k}((θ_{0}(z))^{−τ} /τ)) =

(−1)^{ν} ∑_{τ=1}^{ν(Δ+1)} \frac{1}{τ} ∑_{k=max(0, τ−ν)}^{νΔ} α_{ν,k}^{∗}(θ_{0}(z))^{ν−τ+k};

therefore, if ν = p^{β}l, f = 1, β ∈ ℕf, p > l(Δ + 1), θ_{0}(z) ∈ K and v_{𝔭}(θ_{0}(z)) ≥ 0, then, according to the Lemma 2,

(134)  1 − β ≤

v_{𝔭} (ϕ^{∗}(z; ν) − ∑_{\begin{matrix}η∈[1, Δ+1]∩ℤ \\ k∈[p^{β}(η−l), p^{β}lΔ]∩ℤ \\ k≥0, v_{𝔭}(k)>0 \end{matrix}} \frac{(−1)^{pl}}{p^{β}η} (θ_{0}(z))^{p^{β}(l−η)+k}α_{ν,k}^{∗}) ,

(135)  1/e − β ≤

v_{𝔭} (ϕ^{∗}(z; ν) − ∑_{\begin{matrix}η∈[1, Δ+1]∩ℤ \\ k∈[p^{β−1}(η−l), p^{β−1}lΔ]∩ℤ \\ k≥0 \end{matrix}} \frac{(−1)^{pl}p}{p^{β}η}(θ_{0}(z))^{p^{β−1}(l−η)+k}α_{ν/p,k}^{∗}) .

We make the pass (134) → (135) β times and obtain the inequality

(136)  1/e − β ≤ v_{𝔭} (ϕ^{∗}(z; p^{β}l) − p^{−β}ϕ^{∗}(z; l)) ,

where β ∈ ℕ, p > l(Δ + 1), p ∈ 𝔭 and 𝔭 is ideal of the first degree.

Lemma 14. If m ∈ ℕ + 1, K = ℚ[exp(2pi/m)], α^{∗}(z; l_{1})ϕ^{∗}(z; l_{2})) ≠ 0 for some z ∈ K⧹{0}, l_{1} ∈ ℕ, l_{2} ∈ ℕ, then for any μ ∈ ℕ the sequenses

(137)  α^{∗}(z; ν), ϕ^{∗}(z; ν),

where ν ∈ μ + ℕ form a linear independent system over K.

Proof. There exists d^{∗} ∈ ℕ such that

d^{∗}z ∈ ℤ_{K}, d^{∗}zα^{∗}(z; l_{1}) ∈ ℤ_{K}, d^{∗}zϕ^{∗}(z; l_{2}) ∈ ℤ_{K}.

Let a prime p ∈ ℕm + 1 satisfies to the inequality

p > | Nm_{K/ℚ}(d^{∗}zα^{∗}(z; ν))| + | Nm_{K/ℚ}(d^{∗}zϕ^{∗}(z; ν))+
| Nm_{K/ℚ}(d^{∗}z)| + | Nm_{K/ℚ}(d^{∗}) + (Δ + 1)(l_{1} + l_{2}).

Let 𝔭 is a prime ideal containing p. Then v_{𝔭} (α^{∗}(z; l_{1})) = v_{𝔭} (ϕ^{∗}(z; l_{2})) = 0, and, in view of (136), v_{𝔭} (ϕ^{∗}(z; p^{β}l_{2})) = −β, but v_{𝔭} (α^{∗}(z; p^{β}l_{1})) = 0 with β ∈ ℕ. ■

We choose for each m = 2, …, 37 the corresponding Δ = Δ(m) and check for it and z = (1, 2π/m) the filfilment of the condition of the Lemma 14. For m ∈ [2, 4] let Δ = Δ(m) = 7, l_{1} = l_{2} = 1.

If m = 2, then θ_{0}(z) = −1, and in view of (75), (77), (78),

α^{∗}(z; ν) = ∑_{k=0}^{7} \binom{7}{k}\binom{7 + k}{6} > 0.

If m ∈ ℕ, Δ = Δ(m) = 7 then, in view of (75), (77), (79),

(1/7)α^{∗}(z; 1) = (1/7)(−θ_{0}(z)) ∑_{k=0}^{7} \binom{7}{k}\binom{7 + k}{6}(−θ_{0}(z))^{k} ≡

(−(θ_{0}(z))(1 + (θ_{0}(z))^{6} − 2(θ_{0}(z))^{7}) ( mod 7).

If m = 3, then (1/7)α^{∗}(z; ν)/(−θ_{0}(z)) ≡ 2(1 − θ_{0}(z)) ≢ 0 ( mod 7).

If m = 4, then (1/7)α^{∗}(z; ν)/(−θ_{0}(z)) ≡ −2i ≢ 0 ( mod 7). Let 𝔮 be a prime ideal containing the number 2, 𝔮 = (2), if m = 2, 3, 𝔮 = (1 + i), (2) = 𝔮^{2}, if m = 4. Then in view of (75), v_{𝔮}(α_{1, 7}^{∗}) = v_{𝔮} (−\binom{7}{7}\binom{14}{6}) = 0 and in view of (133),

v_{𝔮} (ϕ^{∗}(z, 1) + (1/8)α_{1, 7}^{∗}) ≥ −2.

If m ∈ [5, 12] ∩ ℕ, we take Δ = Δ(m) = 5, l_{1} = l_{2} = 1. If m ∈ ℕ, Δ = Δ(m) = 5 then in view of (75), (77), (78),

(1/5)α^{∗}(z; 1) = (−(θ_{0}(z)) ∑_{k=0}^{5} \binom{5}{k}\binom{5 + k}{4}(−θ_{0}(z))^{k} ≡

(−(θ_{0}(z))(1 + (θ_{0}(z))^{4} − 2(θ_{0}(z))^{5})( mod 5).

If m = 5, then (1/5)α^{∗}(z; 1)/(−(θ_{0}(z)) ≡ −1 + exp(−2πi/5) ≢ 0( mod 5).

If m = 6, then

(1/5)α^{∗}(z; 1)/(−(θ_{0}(z)) ≡
1 + exp(4πi/3) − 2 exp(−πi/3) = exp(2πi/3) ≢ 0 ( mod 5).

If m ∈ {7, 9, 11}, then the numbers exp(2πk/m), where k = 0, …, ϕ(m) compose an integer basis of the field ℚ[exp(2π/m)], α_{1, Δ}^{∗} ≠ 0, ϕ(m) ≥ 6 and

α^{∗}(z; 1) = − exp(2π/m) ∑_{k=0}^{Δ} α_{1, k}^{∗} exp(2kπ/m) ≠ 0.

If m = 8, then \frac{1}{5}α^{∗}(z; 1) ≡ − exp(π/4)2 exp(π/4) ( mod 5). If m = 10, then

(138)  ∏_{k=1}^{4}(2 − exp(2kπ/5) = 31

(1/5)α^{∗}(z; 1) ≡ (−(θ_{0}(z))(1 + (θ_{0}(z))^{4} − 2(θ_{0}(z))^{5}) ≡
− exp(π/5)(−2 + exp(4π/5)) ≢ 0( mod 5).

If m = 12, then Nm_{K/ℚ}(−2 − i) = Nm_{K/ℚ}(−2 + i) = 25, the ideal (5) is equal to the product of ideals (−2 − i), (−2 + i), and therefore

(1/5)α^{∗}(z; 1) ≡ (−(θ_{0}(z))(1 + (θ_{0}(z))^{4} − 2(θ_{0}(z))^{5}) ≡
− exp(iπ/6)(1 + exp(2iπ/3)) − 2 exp(5iπ/6)) =
− exp(iπ/6)(− exp(8iπ/6)) − 2 exp(5iπ/6)) = −2 − i ≢ 0( mod 5).

Let 𝔮 be a prime ideal containing the number 5. Then, in view of (133),

v_{𝔮} (ϕ^{∗}(z, 1) − \frac{−1}{5}(α_{1, 4}^{∗} + α_{1, 5}^{∗}θ_{0}(z)) ≥ 1,

40 L.A.GUTNIK

v_{𝔮} (ϕ^{∗}(z, 1) − (1 − 2θ_{0}(z)) ≥ 1.

If m = 5, then in view of (138),

v_{𝔮}(1 − 2 exp(2iπ/5)) = v_{𝔮}(− exp(2iπ/5)(2 − exp(8iπ/5)) = 0.

If m = 6, then (2 − exp(10iπ/6)(2 − exp(2iπ/6) = 3, and

v_{𝔮}(1 − 2 exp(2iπ/6)) = v_{𝔮}(− exp(2iπ/6)(2 − exp(10iπ/5)) = 0.

If m ∈ {7, 9, 11}, then the numbers exp(2πk/m), where k = 0, …, ϕ(m) compose an integer basis of the field ℚ[exp(2iπ/m)], α_{1, 5}^{∗} ≠ 0, ϕ(m) ≥ 6, in view of (78),

(139)  ϕ^{∗}(z; ν) = (−1)^{ν} ∑_{k=0}^{νΔ} α_{ν,k}^{∗} ∑_{τ=1}^{ν+k}((θ_{0}(z))^{ν+k−τ} /τ)) =

(−1)^{ν} ∑_{k=0}^{νΔ} α_{ν,k}^{∗} ∑_{s=0}^{ν+k−1}((θ_{0}(z))^{s}/(ν + k − s) =

(−1)^{ν} ∑_{k=0}^{νΔ−1} α_{ν,k}^{∗} ∑_{s=0}^{ν+k−1}((θ_{0}(z))^{s}/(ν + k − s)+

(−1)^{ν}α_{ν,νΔ}^{∗} ∑_{s=0}^{ν(Δ+1)−1}((θ_{0}(z))^{s}/(ν(Δ + 1) − s),

and, since Δ = 5 now, it follows that

ϕ^{∗}(z; 1) = − (∑_{k=0}^{4} α_{1,k}^{∗} ∑_{s=0}^{k} \frac{exp}{(}2isπ/m)1 + k − s) −

α_{1,5}^{∗} ∑_{s=0}^{5} \frac{exp}{(}2isπ/m)6 − s ≠ 0.

If m = 8, then ∏_{k=1} 4(2 − exp(2i(2k − 1)π/8)) = 17, and

v_{𝔮}(1 − 2 exp(2iπ/8)) = v_{𝔮}(− exp(2iπ/8)(2 − exp(14iπ/8)) = 0.

If m = 10, then ∏_{k=1} 2(2 − exp(2i(2k − 1)π/10))(2 − exp(−2i(2k − 1)π/12)) = 11, and

v_{𝔮}(1 − 2 exp(2iπ/10)) = v_{𝔮}(− exp(2iπ/10)(2 − exp(18iπ/12)) = 0.

If m = 12, then ∏_{k=1} 2(2 − exp(2i(4k − 3)π/12))(2 − exp(−2i(4k − 3)π/12)) = 13, and

v_{𝔮}(1 − 2 exp(2iπ/12)) = v_{𝔮}(− exp(2iπ/12)(2 − exp(22iπ/12)) = 0.

If m ∈ [13, 37] ∩ ℕ, we take Δ = Δ(m) = 3, l_{1} = l_{2} = 1. Then ϕ(m) ≥ 6, α_{1,3}^{∗} ≠ 0, and

α^{∗}(z; 1) = − exp(2π/m) ∑_{k=0}^{3} α_{1, k}^{∗} exp(2kπ/m) ≠ 0,

moreover, in view of (139), ϕ^{∗}(z; 1) ≠ 0 in this case. So, the condition of the Lemma 14 holds for all the m = 2, …, 37. We have proved the following

Lemma 15. For each μ ∈ ℕ the two sequences (137) are linearly independent over ℂ. ■

References

[1] R.Apéry, Interpolation des fractions continues et irrationalite de certaines constantes No 3 (1981), Bulletin de la section des sciences du C.T.H.S, 37 – 53.

[2] F.Beukers, A note on the irrationality of ζ(2) and ζ(3) 11 (1979), Bull. London Math. Soc., 268 – 272.

[3] W. Maier, Potenzreihen irratonalen Genzwertes. 156 (1927), J. Reine Angew.Math, 93 – 148..

[4] E.M. Nikis̆in, On irrationality of the values of the functions F(x,s) (Russian), Mat.Sb. 109 (1979), 410 – 417; English transl. in Math. USSR Sb. 37 (1980), 381 – 388..

[5] G.V. Chudnovsky, Pade approximations to the generalized hyper-geometric functions I, J.Math.Pures Appl. 58 (1979), 445 – 476..

[6] , Transcendental numbers 751 (1979), Number Theory,Carbondale,Lecture Notes in Math,,Springer-Verlag, 45 – 69..

[7] , Approximations rationelles des logarithmes de nombres rationelles, C.R.Acad.Sc. Paris,Série A t.288 (1979), 607 – 609.

[8] , Formules d’Hermite pour les approximants de Padé de logarithmes et de fonctions binômes, et mesures d’irrationalité, C.R.Acad.Sc. Paris,Série A t.288 (1979), 965 – 967.

[9] , Un systḿe explicite d’approximants de Padé pour les fonctions hypérgéometriques généraliesées,avec applications a l’arithmétique, C.R.Acad.Sc. Paris, Série A t.288 (1979), 1001 – 1004.

[10] , Recurrenses defining Rational Approximations to the irrational numbers, Proceedings of the Japan Academie, Ser. A, 58 (1982), 129 – 133.

[11] , On the method Thue-Siegel, Annals of Mathematics 117 (1983), 325 – 382.

[12] K.Alladi and M. Robinson, Legendre polinomials and irrationality 318 (1980), J. Reine Angew.Math, 137 – 155..

[13] A. Dubitskas, An approximation of logarithms of some numbers (1986), Diophantine approximations II,Moscow, 20 – 34.

[14] ,1987., On approximation of π/\sqrt{3} by rational fractions, Vestnik MGU, series 1 No 6 (1987), 73 – 76.

[15] S.Eckmann, Über die lineare Unaqbhangigkeit der Werte gewisser Reihen, Results in Mathematics 11 (1987), 7 – 43.

[16] M.Hata, Legendre type polinomials and irrationality mesures, J. Reine Angew. Math. 407 (1990), 99 – 125.

[17] A.O. Gelfond., Transcendental and algebraic numbers (Russian), GIT-TL, Moscow, 1952.

[18] H.Bateman and A.Erdélyi, Higher transcendental functions, New-York – Toronto – London, Mc. Grow-Hill Book Company, Inc., 1953.

[19] H.WEYL, Algebraic theory of numbers, 1940, Russian translation by L.I.Kopejkina.

[20] L.A.Gutnik, On the difference equations of Poincaré type in normed algebras(in Russian), , 668 -B94 (1994), VINITI, Moscow, 1 – 44..

[21] , On the Irrationality of Some Quantyties Containing ζ(3) (in Russian), Uspekhi Mat. Nauk 34, 3(207) (1979), 190.

[22] , On linear forms with coefficients in ℕζ(1 + ℕ) 3 (2000), Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 1 – 13.

[23] , Eleven papers translated from the Russian, vol. 140, American Mathematical Society, 1988, pp. 45 - 56.

[24] , About linear forms, whose coefficients are logarithms of algebraic numbers (in Russian) No 135-B95 (1995), VINITI, 1 – 149.

[25] , On systems of vectors, whose coordinates are linear combinations of logarithms of algebraic numbers with algebraic coefficients (in Russian) 3122-B94 (1994), VINITI, 1 – 158.

[26] , On linear forms, whose coefficients are 𝔸- linear combinations of logarithms of 𝔸numbers (in Russian) No 1617-B96 (1996), VINITI, 1 – 23.

[27] , On systems of linear forms, whose coefficients are 𝔸- linear combinations of logarithms of 𝔸- numbers (in Russian) No 2663-B96 (1996), VINITI, 1 – 18.

[28] , On linear forms with coefficients in ℕζ(1 + ℕ) 3 (2000), Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 1 – 13.

[29] , On linear forms with coefficients in ℕζ(1 + ℕ) (the detailed version,part 1) 15 (2001), Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 1 – 20.

[30] , On linear forms with coefficients in ℕζ(1 + ℕ) (the detailed version,part 2) 104 (2001), Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 1 – 36.

[31] , On linear forms with coefficients in ℕζ(1 + ℕ) (the detailed version,part 3) 57 (2002), Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 1 – 33.

[32] , On the rank over ℚ of some real matrices 27 (2002), Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 1 – 32.

[33] , On linear forms with coefficients in ℕζ(1 + ℕ) (the detailed version,part 4) 142 (2002), Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 1 – 27.

[34] , On the dimension of some linear spaces over finite extension ℚ (part 2) 107 (2002), Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 1 – 37.

[35] , On the dimension of some linear spaces over finite extension ℚ (part 3) 16 (2002), Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 1 – 45.

[36] , On the dimension of some linear spaces over ℚ (part 4) 73 (2003), Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 1 – 37.

[37] , On linear forms with coefficients in ℕζ(1 + ℕ) (the detailed version,part 5) 83 (2003), Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 1 – 13.

[38] , On linear forms with coefficients in ℕζ(1 + ℕ) (the detailed version,part 6) 99 (2003), Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 1 – 33.

[39] , On linear forms with coefficients in ℕζ(1 + ℕ),, Bonner Mathematishe Schriften Nr. 360, 360 (2003).

[40] , On linear combinations of logarithms of algebraic numbers with algebraic coefficients. (2003), International conference ”Diophantine analysis, uniform distributions and applications” August 25-30, 2003, Minsk, Belarus Abstracts, pp. 16 – 17..

[41] , On the difference equation of Poincaré type (Part 1). 52 (2003), Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 1 – 43.

[42] , On the difference equation of Poincaré type (Part 2) 107 (2003), Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 1 – 25.

[43] , On the difference equation of Poincaré type (Part 3) 9 (2004), Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 1 – 33.

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