ON THE DIMENSION OF SOME LINEAR SPACES

OVER ℚ (Part 4) L.A. Gutnik

Table of contents

§0. Foreword. §1. Several properties of the roots of the polynomial (1). §2. Some corrections to [53].

§0. Foreword.

For some problems studied in [40]-[58], the properties of the roots η of the polynomial

(1)  D^{∨}(z, η) = (η + 1)(η + γ_{1}) + 2r exp(iψ)(1 + γ_{1})η,

where

(2)  −1 < δ_{0} < 1, 0 < r, γ_{1} = (1 − δ_{0})/(1 + δ_{0}), ψ ∈ ℝ,

play an important role.

The zeros of the polynomial (1) are partially studied in the papers [55], [53], [58] and in more detail in [47]. In Section 1, I continue the study of the zeros of the polynomial (1). In Section 2, I make some corrections to [53].

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

An essential part of this paper has been written during the breaks between the lectures of the Arbeitstagung 2003 at Max-Planck-Institut für Mathematik (Bonn). I thank the MPI for the invitation to the Arbeitstagung and excellent working conditions.

L.A.Gutnik, On the dimension of some linear spaces over ℚ 2

”There is one point that you haven’t noticed,” said the wizad, ”and that is a secret entrance...”

J.R.R. Tolkien

§1. Several properties of the roots of the polynomial (1).

Let η = θ(1 + γ_{1}), θ^{∨} = − exp(−iψ)θ. Then

(3)  D^{∨}(z, η) = (1 + γ_{1})^{2}D_{1}^{∨}(z, θ),

where

(4) D_{1}^{∨}(z, θ) = θ^{2} + 2 (\frac{1}{2} + r exp(iψ)) θ + \frac{γ_{1}}{(1 + γ_{1})^{2}} = exp(2iψ)D_{2}^{∨}(z, θ^{∨})

with

(5)  D_{2}^{∨}(z, θ^{∨}) = (θ^{∨})^{2} − 2 (\frac{exp(−iψ)}{2} + r) θ^{∨} + \frac{γ_{1} exp(− 2iψ)}{(1 + γ_{1})^{2}}.

Let

(6)  D_{0}(r, ψ, δ_{0}) = r^{2} + r exp(−iψ) + (δ_{0}/2)^{2} exp(− 2iψ).

If we put t = cos(ψ), then

(7)  Im(D_{0}(r, ψ, δ_{0})) = −((δ_{0})^{2}/2)t + r)sin(ψ),

(8)  Re(D_{0}(r, ψ, δ_{0})) = (δ_{0})^{2}t^{2}/2 + rt + r^{2} − (δ_{0}/2)^{2},

where t = cos(ψ), ψ ∈ ℝ,

(9)  Re(D_{0}(r, 0, δ_{0})) = r^{2} + r + (δ_{0}/2)^{2},

(10)  Re(D_{0}(r, π, δ_{0})) = r^{2} − r + (δ_{0}/2)^{2},

(11)  Im(D_{0}(r, kπ, δ_{0})) = 0,

where k ∈ ℤ. Clearly, for t = cos(ψ) holds the equality

(12)  |D_{0}(r, ψ, δ_{0})|^{2} = r^{4} + r^{2} + (δ_{0}/2)^{4}+
2r^{2}(δ_{0}/2)^{2}(2t^{2} − 1) + 2r(r^{2} + (δ_{0}/2)^{2})t =
r^{2} + (r^{2} − (δ_{0}/2)^{2})^{2} + 2r(r^{2} + (δ_{0}/2)^{2})t + r^{2}(δ_{0})^{2}t^{2}.

Lemma 4.1. The value |D_{0}(r, ψ, δ_{0})| is increasing and concave (convex upwards) function from t = cos(ψ),

(13)  min_{−1≤t≤1} |D_{0}(r, ψ, δ_{0})|^{2} = |D_{0}(r, π, δ_{0})|^{2} = |r^{2} − r + (δ_{0}/2)^{2}|.

Proof. According (12),

(14)  |D_{0}(r, ψ, δ_{0})| = \sqrt{at^{2} + bt + c},

where

(15)  a = r^{2}(δ_{0})^{2}, b = 2r(r^{2} + (δ_{0}/2)^{2}), c = r^{2} + (r^{2} − (δ_{0}/2)^{2})^{2}.

Then

\frac{∂}{∂t}|D_{0}(r, ψ, δ_{0})|^{2} = 2at + b = 2r^{2}(δ_{0})^{2}t + 2r(r^{2} + (δ_{0}/2)^{2}) ≥

2r^{2}(δ_{0})^{2}t + 2r|δ_{0}|(r^{2} + (δ_{0}/2)^{2}) ≥ 2r^{2} ≥ 2r|δ_{0}|(r − |δ_{0}|/2)^{2} ≥ 0.

The inequality

(\frac{∂}{∂t})^{2} |D_{0}(r, ψ, δ_{0})| ≤ 0

proved in [53], Lemma 3.1.1. ■

Lemma 4.2. Let −1 < δ_{0} < 1, 0 < r. Then there exist a continuos function R_{0}(ψ) = R_{0}(r, ψ, δ_{0}), defined on ℝ, such that

(16)  R_{0}(r, ψ, δ_{0})^{2} = D_{0}(r, ψ, δ_{0}),

(17)  R_{0}(r, 0, δ_{0}) > 0,

(18)  R_{0}(r, −ψ, δ_{0}) = \overline{R_{0}(r, ψ, δ_{0})},

(19)  R_{0}(r, ψ + 2πk, δ_{0}) = (−1)^{k}R_{0}(r, ψ, δ_{0}),

when 2r ∈ [1 − \sqrt{1 − (δ_{0})^{2}}, 1 + \sqrt{1 − (δ_{0})^{2}}) , ψ ∈ ℝ,

(20)  R_{0}(r, ψ + 2πk, δ_{0}) = R_{0}(r, ψ, δ_{0}), sign(2r − 1)R_{0}(r, π, δ_{0}) > 0,

when 2r ∉ [1 − \sqrt{1 − δ_{0}^{2}}, 1 + \sqrt{1 − (δ_{0})^{2}}) , ψ ∈ ℝ.

Proof. According to (13), if

(21)  2r ≠ 1 ± \sqrt{1 − (δ_{0})^{2}},

then |D_{0}(r, ψ, δ_{0})| > 0 for all ψ ∈ ℝ. Therefore, in view of (9), in this case for any fixed r and δ_{0} there exist a single continuos function

ψ_{1} = ψ_{1}(ψ) = ψ_{1}(r, ψ, δ_{0})

such that

(22)  D_{0}(r, ψ, δ_{0}) = |D_{0}(r, ψ, δ_{0})| exp(−iψ_{1}(r, ψ, δ_{0}))

for each ψ ∈ ℝ and

(23)  ψ_{1}(r, 0, δ_{0}) = 0

(See, for example, [29], supplement to the chapter 9, section 2). Since

(24)  D_{0}(r, −ψ, δ_{0}) = \overline{D_{0}(r, ψ, δ_{0})},

it follows that there exists m_{1} ∈ ℤ with property

ψ_{1}(r, −ψ, δ_{0}) = −ψ_{1}(r, ψ, δ_{0}) + 2m_{1}π.

In view of (23), m = 0 and

(25)  ψ_{1}(r, −ψ, δ_{0}) = −ψ_{1}(r, ψ, δ_{0}).

Since D_{0}(r, ψ, δ_{0}) = D_{0}(r, ψ +2π, δ_{0}) = |D_{0}(r, ψ, δ_{0})| exp(−iψ_{1}(r, ψ +2π, δ_{0})), it follows that

(26)  ψ_{1}(r, ψ + 2π, δ_{0}) = ψ_{1}(r, ψ, δ_{0}) + 2m_{0}π

with constante m_{0} ∈ ℤ. If {ψ^{∨}, ψ^{∧}} ⊂ ℝ and

|ψ_{1}(r, ψ^{∨}, δ_{0}) − ψ_{1}(r, ψ^{∧}, δ_{0})| ≥ π,

then there exist ψ^{∗} between ψ^{∧} and ψ^{∨} such that ψ_{1}(r, ψ^{∗}, δ_{0})/π ∈ ℤ and therefore

(27)  Im(D_{0}(r, ψ^{∗}, δ_{0})) = 0.

But in view of (7), if 2r ≥ (δ_{0})^{2}, then equality (27) holds only for ψ = kπ with k ∈ ℤ; therefore, if {ψ^{∨}, ψ^{∧}} ⊂ (kπ, (k + 1)π) for some k ∈ ℤ, then

(28)  |ψ_{1}(r, ψ^{∨}, δ_{0}) − ψ_{1}(r, ψ^{∧}, δ_{0})| < π,

If

2r ≥ 1 + \sqrt{1 − (δ_{0})^{2}}

then 2r > 1 > (δ_{0})^{2} and, according to (10), (25), (28) and (23),

(29)  ψ_{1}(r, ±π, δ_{0}) = 0,

if 2r > 1 + \sqrt{1 − (δ_{0})^{2}}. Taking now in (26) ψ = −π, we see that, according to (29), m_{0} = 0 and

(30)  ψ_{1}(r, ψ + 2πk, δ_{0}) = ψ_{1}(r, ψ, δ_{0}),

for any k ∈ ℤ and 2r > 1 + \sqrt{1 − (δ_{0})^{2}}. If

(31)  (δ_{0})^{2} ≤ 2r < 1 + \sqrt{1 − (δ_{0})^{2}},

then, according to (10), (23), (25), (28), (31) and to the inequality

(32)  1 − \sqrt{1 − (δ_{0})^{2}} < (δ_{0})^{2}

we have

(33)  ψ_{1}(r, επ, δ_{0}) = επ,

where ε^{2} = 1. Substituting in (26) ψ = −π, we conclude, in view of (23), that m_{0} = 1 and

(34)  ψ_{1}(r, ψ + 2πk, δ_{0}) = ψ_{1}(r, ψ, δ_{0}) + 2πk,

if k ∈ ℤ and (δ_{0})^{2} ≤ 2r < 1 + \sqrt{1 − (δ_{0})^{2}}. If

(35)  2r < (δ_{0})^{2},

and

(ψ^{∨} − arccos(−2r/(δ_{0}) − 2kπ)(ψ^{∧} − arccos(−2r/(δ_{0})^{2}) − 2kπ) > 0,

where {ψ^{∨}, ψ^{∧}} ⊂ (2kπ, 2kπ + π) for some k ∈ ℤ, then

(36)  |ψ_{1}(r, ψ^{∨}, δ_{0}) − ψ_{1}(r, ψ^{∧}, δ_{0})| < π.

According to (8),

(37)  Re(D_{0}(r, arccos(−2r/(δ_{0})^{2}), δ_{0})) = r^{2} − (δ_{0}/2)^{2} <
(δ_{0}^{2}/2)^{2} − (δ_{0}/2)^{2} < 0,

if (35) holds; therefore, in view of (23), (25), (28), (36), if (35) holds, then

(38)  ψ_{1}(r, ε arccos(−2r/(δ_{0})^{2}, δ_{0}) = επ,
(39)  |ψ_{1}(r, π, δ_{0}) − ψ_{1}(r, arccos(−2r/(δ_{0})^{2}), δ_{0})| ≤ π.

If

1 − \sqrt{1 − (δ_{0})^{2}} < 2r < (δ_{0})^{2},

then Re(D_{0}(r, π, δ_{0})) < 0, in view of (39), (38),

ψ_{1}(r, π, δ_{0}) = ψ_{1}(r, arccos(−2r/(δ_{0})^{2}, δ_{0})

and (33) holds. Substituting in (26) ψ = −π, we conclude, in view of (33), that m_{0} = 1 and (34) holds. If 2r < 1−\sqrt{1 − (δ_{0})^{2}}, then (39) holds again, but now Re(D_{0}(r, π, δ_{0})) > 0. In view of (8), the value Re(D_{0}(r, ψ, δ_{0})) decreases together with increasing ψ ∈ (0, arccos(−2r/(δ_{0})^{2})) and it increases with increasing ψ ∈ (arccos(−2r/(δ_{0})^{2}), π). Moreover, in view of (7),

\frac{∂}{∂ψ} (−Im(D_{0}(r, ψ, δ_{0}))) |_{ψ=arccos(−2r/(δ_{0})^{2}))} = 2(r^{2} − (δ_{0}/)^{4}/4)/(δ_{0})^{2} < 0.

Therefore now

(40)  ψ_{1}(r, επ, δ_{0}) = 2επ.

Taking in (26) ψ = −π, we see, that, in view of (40), m_{0} = 2 and

(41)  ψ_{1}(r, ψ + 2πk, δ_{0}) = ψ_{1}(r, ψ, δ_{0}) + 4kπ,

If holds (21), let

(42)  R_{0}(r, ψ, δ_{0}) = \sqrt{|D_{0}(r, ψ, δ_{0})|} exp(−iψ_{1}(r, ψ, δ_{0})/2).

The previous considerations show, that in this case all the assertions of the Lemma are fulfilled for R_{0}(r, ψ, δ_{0}) from (42).

We consider the case

(43)  |2r − 1| = \sqrt{1 − (δ_{0})^{2}}.

now. Then holds the equality

(44)  r^{2} − r + (δ_{0}/2)^{2} = 0.

Substracting (44) from (6) we obtain the equality

(45)  D_{0}(r, ψ, δ_{0}) = (1 + exp(−iψ))(r + (δ_{0}/2)^{2}(exp(−iψ) − 1))

Let

(46)  D_{h}(r, ψ, δ_{0}) = r + (δ_{0}/2)^{2}(exp(−iψ) − 1).

Clearly, the value Im(D_{h}(r, ψ, δ_{0})) = −(δ_{0}/2)^{2} sin(ψ) vanish only for ψ = kπ with k ∈ ℤ; but then

(47)  Re(D_{h}(r, kπ, δ_{0})) = r + (δ_{0}/2)^{2}((−1)^{k} − 1) ≠ 0,

because of

(48)  1 − \sqrt{1 − (δ_{0})^{2}} < (δ_{0})^{2} < 1 + \sqrt{1 − (δ_{0})^{2}}.

Therefore D_{h}(r, ψ, δ_{0}) not vanish for ψ ∈ ℝ and for any fixed r and δ_{0} there exist a single continuos function ψ_{h} = ψ_{h}(ψ) = ψ_{h}(r, ψ, δ_{0}) such that

(49)  D_{h}(r, ψ, δ_{0}) = |D_{h}(r, ψ, δ_{0})| exp(−iψ_{h}(r, ψ, δ_{0}))

for each ψ ∈ ℝ and

(50)  ψ_{h}(r, 0, δ_{0}) = 0.

Reasoning, as before, and using (47), we see that

(51)  ψ_{h}(r, −ψ, δ_{0}) = −ψ_{h}(r, ψ, δ_{0})

(52) ψ_{h} (\frac{1 + \sqrt{1 − (δ_{0})^{2}}}{2} , π, δ_{0}) = 0, ψ_{h} (\frac{1 − \sqrt{1 − (δ_{0})^{2}}}{2} , επ, δ_{0}) = επ,

with ε^{2} = 1,

(53)  ψ_{h}(r, ψ + 2π, δ_{0}) = ψ_{h}(r, ψ, δ_{0}) + 2m_{h}π

where m_{h} = 0, if 2r = 1 + \sqrt{1 − (δ_{0})^{2}}, and m_{h} = 1, if 2r = 1 − \sqrt{1 − (δ_{0})^{2}}. Let

(54)  R_{h}(r, ψ, δ_{0}) = \sqrt{|D_{h}(r, ψ, δ_{0})|} exp(iψ_{h}(r, ψ, δ_{0})/2).

Then R_{h}(r, ψ, δ_{0}) as function of ψ is continuos,

(55)  R_{h}(r, ψ, δ_{0})^{2} = D_{h}(r, ψ, δ_{0}), R_{h}(r, 0, δ_{0}) > 0,

(56)  R_{h}(r, ψ + 2πk, δ_{0}) = (−1)^{k}R_{h}(r, ψ, δ_{0}).

if 2r = 1 − \sqrt{1 − (δ_{0})^{2}}, ψ ∈ ℝ and

(57)  R_{h}(r, ψ + 2πk, δ_{0}) = R_{h}(r, ψ, δ_{0}).

if 2r = 1 + \sqrt{1 − (δ_{0})^{2}}, ψ ∈ ℝ. Clearly, the function

(58)  R^{∨}(ψ) = sign(cos(ψ/2))\sqrt{2| cos(ψ/2)|} exp(−iψ/2)

is continuos 2π−periodical function on ℝ such that

R^{∨}(ψ)^{2} = 1 + exp(−iψ), R^{∨}(−ψ) = \overline{R^{∨}(ψ)}.

Let now

(59)  R_{0}(r, ψ, δ_{0}) = R^{∨}(ψ, δ_{0})R_{h}(r, ψ, δ_{0}),

where r satisfies (43). The previous considerations show that in this case all the assertions of the Lemma are fulfilled for R_{0}(r, ψ, δ_{0}) from (59). ■

We considered in (6) the value D_{0}(r, ψ, δ_{0}) only for |δ_{0}| < 1, but it may be considered also for δ_{0} = 1 and in this case from (6) follows that

(60)  D_{0}(r, ψ, 1) = (r + \frac{exp(−iψ)}{2})^{2} .

If we let

(61)  R_{0}(r, ψ, 1)) = r + \frac{exp(−iψ)}{2},

then we obtain 2π - periodical continuos function R_{0}(r, ψ, 1) = R_{0}(r, ψ) such that

(62) (R_{0}(r, ψ))^{2} = D_{0}(r, ψ, 1), R_{0}(r, 0) > 0, R_{0}(r, −ψ) = \overline{R_{0}(r, ψ)}.

According to (61), for r > 0, |δ_{0}| < 1 and k = 0, 1 the values

(63)  θ_{k}^{∨}(r, ψ, δ_{0}) = R_{0}(r, ψ, 1) + (−1)^{k}R_{0}(r, ψ, δ_{0})

compose the set of all the roots of the polynomial (5), the values

(64)  θ_{k}^{∧}(r, ψ, δ_{0}) = − exp(iψ)(R_{0}(r, ψ, 1) + (−1)^{k}R_{0}(r, ψ, δ_{0}))

compose the set of all the roots of the polynomial (4) and the values

(65) η_{k}^{∧}(r, ψ, δ_{0}) = −(1 + γ_{1})exp(iψ)(R_{0}(r, ψ, 1) + (−1)^{k}R_{0}(r, ψ, δ_{0}))

compose the set of all the roots of the polynomial (1) .

It follows from (65), Lemma 4.2 and (62) that

(66)  η_{k}^{∧}(r, −ψ, δ_{0}) = \overline{η_{k}^{∧}(r, ψ, δ_{0})}

where ψ ∈ ℝ, r > 0, |δ_{0}| < 1, k = 0, 1.

Let ε^{3} = ε,

(67)  R_{ε}^{∗}(r, ψ, δ_{0}) = r + (exp(−iψ))/2 − ε exp(−iψ)(1 + δ_{0})/2;

according to (65), (61),

(68)  η_{k}^{∧}(r, ψ, δ_{0}) + ε = −\frac{2 exp(iψ)}{1 + δ_{0}}(R_{ε}^{∗}(r, ψ, δ_{0}) + (−1)^{k}R_{0}(r, ψ, δ_{0})),

where ε^{3} = ε, k = 0, 1.

Lemma 4.3. The values

|η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2},

where k = 0, 1, 0 < r, ψ ∈ ℝ, ε^{3} = ε, are nonnegative zeros of the polyinomial

(69)  x^{2} − xp_{1+ε}(r, ψ, δ_{0}) + q_{1+ε}(r, ψ, δ_{0}),

where ε^{3} = ε, and

(70)  p_{1+ε}(r, ψ, δ_{0}) = 8\frac{|R_{ε}^{∗}(r, ψ, δ_{0})|^{2} + |R_{0}(r, ψ, δ_{0})|^{2}}{(1 + δ_{0})^{2}} ,

(71)  ∏_{k=0}^{1} |η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2} = q_{1+ε}(r, ψ, δ_{0}) = |D^{∨}(z, −ε)|^{2}.

Proof. The proof may be found in [53], Lemma 3.1.2, or in [47], Lemma 2.

Lemma 4.4. Let ε^{2} = ε. For k = 0, 1 with increasing ψ ∈ (0, π) the value

(−1)^{k}|η_{k}^{∧}(r, ψ, δ_{0})|

monotonically decreases and, if 2r ≠ δ_{0}, then the value

(−1)^{k}|η_{k}^{∧}(r, ψ, δ_{0}) + 1|sign(2r − δ_{0})

also monotonically decreases. Moreover

(72)  |η_{1}^{∧}(r, ψ, δ_{0})| < |η_{0}^{∧}(r, ψ, δ_{0})| ,

if ψ ∈ [0, π) ;

(73)  sign(2r − δ_{0})(|η_{1}^{∧}(r, ψ, δ_{0}) + 1| − |η_{0}^{∧}(r, ψ, δ_{0}) + 1|) < 0 ,

if ψ ∈ [0, π) and 2r ≠ δ_{0} ;

(74)  |η_{1}^{∧}(r, ψ, δ_{0}) + 1| = |η_{0}^{∧}(r, ψ, δ_{0}) + 1| = (2δ_{0}/(1 + δ_{0}))^{1/2} ,

if ψ ∈ [0, π] and 2r = δ_{0} ;

(75)  (2sign(2r − δ_{0}) + 1)^{ε}(|η_{1}^{∧}(r, π, δ_{0}) + ε| − |η_{0}^{∧}(r, π, δ_{0}) + ε|) < 0 ,

if |2r − 1| > \sqrt{1 − (δ_{0})^{2}} ;

(76)  |η_{1}^{∧}(r, π, δ_{0}) + ε| = |η_{0}^{∧}(r, π, δ_{0}) + ε| = \sqrt{q_{1+ε}} ,

if

(77)  |2r − 1| ≤ \sqrt{1 − (δ_{0})^{2}} ;

(78)  |η_{1}^{∧}(r, π, δ_{0}) + ε| ≤ |\frac{1 − δ_{0}}{1 + |δ|} + ε| = \frac{1 − |δ_{0}|}{1 + δ_{0}} + ε ≤

\frac{1 + |δ_{0}|}{1 + δ_{0}} + ε ≤ |η_{0}^{∧}(r, π, δ_{0}) + ε| ,

if r ≥ 1; moreover the first and the last inequaliies (78) are strong, if r > 1, the middle of these inequalities is strong, if δ_{0} ≠ 0 ;

(79)  |η_{1}^{∧}(r, π, δ_{0}) + ε| = |η_{0}^{∧}(r, π, δ_{0}) + ε| = 1 + ε ,

if r = 1 and δ = 0 .

Proof. For r ≥ 1 the proof coincides with the proof of the Lemma 3.1.3 in [53]; we must make only the supplements for the case 0 < r < 1 to this proof. For r = δ_{0}/2 the inequality (3.1.49) in [53] holds and the value (3.1.51) in [53] for r = δ_{0}/2 is equal to 0; therefore (74) holds. It follows from (3.1.52) in [53] that

(80)  |η_{0}^{∧}(r, 0, δ_{0})| = \frac{2r + 1 + 2\sqrt{r^{2} + r + (δ_{0}/2)^{2}}}{1 + δ_{0}} >

|η_{1}^{∧}(r, 0, δ_{0})| = \frac{|2r + 1 − 2\sqrt{r^{2} + r + (δ_{0}/2)^{2}}|}{1 + δ_{0}}

and, if 2r ≠ δ_{0}, then

(81)  sign(2r − δ_{0})|η_{0}^{∧}(r, 0, δ_{0}) + 1| =

sign(2r − δ_{0})\frac{2r − δ_{0} + 2\sqrt{r^{2} + r + (δ_{0}/2)^{2}}}{1 + δ_{0}} >

sign(2r − δ_{0})|η_{1}^{∧}(r, 0, δ_{0}) + 1| =

sign(2r − δ_{0})\frac{|2r − δ_{0} − 2\sqrt{r^{2} + r + (δ_{0}/2)^{2}}|}{1 + δ_{0}}.

Therefore are fulfilled (72) and (73). The equality (3.1.54) in [53] holds now for ε = 0 but for ε = 1 it must be replaced by

(82)  |η_{k}^{∧}(r, ψ, δ_{0}) + 1|^{2} = \frac{p_{2}}{2} + sign(2r − δ_{0})(−1)^{k}\sqrt{(\frac{p_{2}}{2} )^{2} − q_{2}}.

The equality (3.1.54) in [53] holds now for ε = 0; therefore, in view of (3.1.43) in [53], the value

(−1)^{k}|η_{k}^{∧}(r, ψ, δ_{0})|

monotonically increases with increasing t = cos(ψ) ∈ (−1, 1).

In view of (82), if 2r ≠ δ_{0}, then with increasing t = cos(ψ) ∈ (−1, 1) the value

|η_{(1−sign(2r−δ_{0}))/2}^{∧}(r, ψ, δ_{0}) + 1|

monotonically increases and, in view of (3.1.43) in [53], the value

|η_{(1+sign(2r−δ_{0}))/2}^{∧}(r, ψ, δ_{0}) + 1|

monotonically decreases with increasing t = cosψ ∈ (−1, 1).

The equalities (74) follow from (3.1.43) in [53] and (82). For t = −1 each of the inequalities (3.1.46), (3.1.49) in [53] is equivalent to inequality

|2r − 1| ≤ \sqrt{1 − (δ_{0})^{2}}.

Therfore (76) holds.

In view of (3.1.54) with ε = 0 in [53], also (75) with ε = 0 holds. Clearly,

(83)  (1 − \sqrt{1 − (δ_{0})^{2}})/2 ≤ |δ_{0}|/2 ≤ (1 + \sqrt{1 − (δ_{0})^{2}})/2,

and the inequality |2r − 1| > \sqrt{1 − (δ_{0})^{2}} implies the inequality 2r ≠ δ_{0}; therefore (75) holds also for ε = 1. ■

Remark 1. The equality (3.1.55) in [53] for ε^{3} = ε, k^{2} = k, in view of (65), (67), (68) and (20), must be modified as follows:

if |2r − 1| ≥ \sqrt{1 − δ_{0}^{2}}, then

(84)  η_{k}^{∧}(r, π, δ_{0}) + ε =

\frac{2 (r − \frac{1}{2} + ε\frac{1+δ_{0}}{2} + (−1)^{k}sign(2r − 1)\sqrt{r^{2} − r + (\frac{δ_{0}}{2} )^{2}})}{1 + δ_{0}}.

and if |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 and 2r > 1 then from (84) follows the relation

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

Remark 2. Let us to check (75) directly for r < (1 − \sqrt{1 − (δ_{0})^{2}})/2. In view of (83),now sign(2r + δ_{0}) = sign(δ_{0}) and 2r < 1. Therefore (75) in this case follows directly from (65) and second of the inequalities (20). This understrike the importance of the Lemma 4.2.

Corollary 1. The following inequalities hold

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

\sqrt{\frac{1 − δ_{0}}{1 + δ_{0}}} ≤ η_{0}^{∧}(r, π, δ_{0}) ≤ η_{0}^{∧}(r, π, δ_{0}),

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

1 + \sqrt{\frac{1 − δ_{0}}{1 + δ_{0}}} ,

(88)  4r/(1 + δ_{0} + \sqrt{(1 − (δ_{0})^{2}} ≤ |η_{0}^{∧}(r, ψ, δ_{0}) + 1|

and equalities hold only if ψ = π, |2r − 1| ≤ \sqrt{1 − (δ_{0})^{2}}.

Proof. The first and the last inequalities in (86) follow directly from assertions of the Lemma, the second and the third inequalities in (86) follow from the first equality (3.1.43) in [53], from (72), from (75) for ε = 0 and from (76) for ε = 0. The first inequality (87) is obvious, the others directly follow from (86). The inequality (88) directly follow from (87) and second equality (3.1.43) in [53]. ■

Corollary 2. If 2r − 1 > \sqrt{1 − (δ_{0})^{2}}, ψ ∈ (0, π), then

(89)  |η_{0}^{∧}(r, ψ, δ_{0})| > η_{0}^{∧}(r, π, δ_{0}) =

\frac{2(r − 1/2 + \sqrt{r^{2} − r + (δ_{0}/2)^{2}})}{1 + δ_{0}} ≥

\frac{2(r − 1/2 − \sqrt{r^{2} − r + (δ_{0}/2)^{2}})}{1 + δ_{0}} =

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

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

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

\frac{2(r + δ_{0}/2 − \sqrt{r^{2} − r + (δ_{0}/2)^{2}})}{1 + δ_{0}} ≤

\frac{2(r + δ_{0}/2 + \sqrt{r^{2} − r + (δ_{0}/2)^{2}})}{1 + δ_{0}} =

η_{0}^{∧}(r, π, δ_{0}) + 1 = 4r(η_{1}^{∧}(r, π, δ_{0}) + 1)^{−1}/(1 + δ_{0}) <
4r|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{−1}/(1 + δ_{0}) = |η_{0}^{∧}(r, ψ, δ_{0}) + 1|.

Proof. The assertions of the corollary follow from (65), (3.1.43) in [53], the assertions of the Lemma 4.4, Lemma 4.2 and Lemma 4.3. ■

Lemma 4.5. The following equalities hold:

η_{k}^{∧}(r, ψ, δ_{0}) = 1/η_{1−k}^{∧}(r, ψ, −δ_{0}),

where k^{2} = k, |δ_{0}| < 1, ψ ∈ ℝ.

Proof. In view of (6)

D_{0}(r, ψ, δ_{0}) = D_{0}(r, ψ, −δ_{0}).

Therefore, for D_{0}(r, ψ, δ_{0}) from the Lemma 4.2 we have the equality

R_{0}(r, ψ, δ_{0}) = R_{0}(r, ψ, −δ_{0}).

By substitution δ → −δ the value 1 + γ_{1} = 2/(1 + δ_{0}) turns into 2/(1 − δ_{0}). In view of (65),

η_{k}^{∧}(r, ψ, δ_{0}/η_{k}^{∧}(r, ψ, −δ_{0} = γ_{1},

and γ_{1} is the prodoct of both roots of the polynomial (1). ■

Corollary. Let δ_{0} ≥ 0. For k = 0, 1 the value

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

monotonically increases with increasing ψ ∈ (0, π).

Proof. According to the Lemma 4.5, and,if 2r ≠ δ_{0}, then the value

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

because 2r > −δ_{0}. ■

Let

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

(92)  ψ_{0}(r, ψ) = arg(r exp(iψ) + 1) ∈ (−π, π),

where ψ ∈ (−π, π). Clearly, ψ_{0}(r, −ψ) = −ψ_{0}(r, ψ), when ψ ∈ (−π, π).

If 0 < r ≤ 1, then ψ_{0}(r, ψ) increases in (0, arcsin(r)) together with increasing ψ in (0, π/2 + arcsin(r)) and defines bijection of (0, π/2 + arcsin(r)) onto (0, arcsin(r)). If 0 < r < 1, then ψ_{0}(r, ψ) decreases in (0, arcsin(r)) with increasing ψ in (π/2+arcsin(r), π) and defines bijection of (π/2+arcsin(r), π) onto (0, arcsin(r)). If 1 < r, then the function ψ_{0}(r, ψ) increases in (0, π) with increasing ψ in (0, π) and defines bijection of (0, π) onto (0, π). Clearly,

(93)  |r − 1|/2 ≤ s ≤ (r + 1)/2,

for s in (91). It follows from (91) that

(94)  t = cos(ψ) = (4s^{2} − r^{2} − 1)/2r

Lemma 4.6. If at least one of the following conditions holds:

a)

(95)  2r > max(1 + \sqrt{1 − (δ_{0})^{2}}, 2 + δ_{0})

and |r − 1|/2 = s, b)

(96)  2r < 1 − \sqrt{1 − (δ_{0})^{2}}

and |r − 1|/2 = s, c) max(|r − 1|/2, δ_{0}/4) < s ≤ (r + 1)/2, then

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

If at least one of the following conditions holds:

d) 2s = |r − 1|, 2r = 2 + δ_{0},

e) 2s = |r − 1|, |2r − 1| ≤ \sqrt{1 − (δ_{0})^{2}},

f) 2|r − 1| ≤ 4s = δ_{0} ≤ 2|r + 1|,

then

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

If at least one of the following conditions hold g)

(99)  1 + \sqrt{1 − (δ_{0})^{2}} < 2r < 2 + δ_{0}.

and 2s = |r − 1|, h)

(100)  2|r − 1| < 4s < δ_{0},

then

(101)  |η_{0}^{∧}(r, ψ, δ_{0}) − 1| < |η_{1}^{∧}(r, ψ, δ_{0}) − 1|,

Proof. The left part of (3.1.73) in [53] for s = (r + 1)/2 is equal to the positive number r^{2} + r + (δ_{0}/2)^{2}; therefore s = (r + 1)/2 is not a solution of the system (3.1.73) - (3.1.74) in [53]; only s = \frac{1−r}{2} and s = \frac{δ_{0}}{4} may be a solutions of this system. In view of (91) and (93), the equality s = \frac{1−r}{2} is possible only if the inequality r ≤ 1 and equality ψ = π are fulfilled in the same time. In view of (84), if |2r − 1| ≥ \sqrt{1 − (δ_{0})^{2}}, then

(102)  η_{k}^{∧}(r, π, δ_{0}) − 1 =

\frac{2 (r − 1 − \frac{δ_{0}}{2} + (−1)^{k}sign(2r − 1)\sqrt{r^{2} − r + (\frac{δ_{0}}{2} )^{2}})}{1 + δ_{0}};

if |2r − 1| ≥ \sqrt{1 − (δ_{0})^{2}}, then

(103)  η_{k}^{∧}(r, π, δ_{0}) = \overline{η_{1−k}^{∧}(r, π, δ_{0})}

where k = 0, 1. If (95) is fulfiiled, then 2r > 1, in view of (102),

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

and therefore the inequality (97) holds. If (96) holds, then 2r < 1; making use (102) and the inequality (1 − \sqrt{1 − (δ_{0})^{2}})/2 < 1 + δ_{0}/2, we conclude that the inequality (104) holds and therefore in the case b) the inequality (97) holds . In the cases d) and e) the equality (98) directly follows from (103). In the case g) the inequalities (99) hold, therefore 2r > 1 and, according to the equality (102),

(105)  |η_{0}^{∧}(r, π, δ_{0}) − 1| < |η_{1}^{∧}(r, π, δ_{0}) − 1|.

Consequently, (101) holds in the case g).

We consider now s = δ_{0}/4 as possible solution of system, consisting of the inequality (3.1.73) in [53], inequalities (93) and equation (3.1.74) in [53]; according to (93), the equality

(106)  s = δ_{0}/4

is possible only if

(107)  |r − 1| ≤ δ_{0}/2;

If for s = δ_{0}/4 in (107) an equality achieves, then r = 1±δ_{0}/2 with δ_{0} ≥ 0, the case r = 1+δ_{0}/2 coincides with case d) considered above, and if r = 1−δ_{0}/2 with δ_{0} > 0 then |2r − 1| = |1 − δ_{0}| < |1 − δ_{0}^{2}|^{1/2} < (1 + \sqrt{1 − (δ_{0})^{2}})/2 and this case coincides with case e) considered above.

We must consider now the case

(108)  |r − 1|/2 < s = δ_{0}/4;

But (3.1.73) in [53] for s = δ_{0}/4 turns into inequality

δ_{0}(r^{2} − (δ_{0}/2)^{2})/2 > 0,

which follows from (108).

So, if (107) is fulfilled, then s = δ_{0}/4 is the solution of system, consisting of inequality (3.1.73) - (3.1.74) in [53] and inequalities (93). Therefore in case f) (98) holds .

In view of (3.1.52) in [53], the inequality (97) holds in the case c).

It remains to consider the case h).

If 2r ≥ 2 + δ_{0}, then inequality 2|r − 1| < δ_{0} is impossible. If (99) holds, then (103) holds and, consequently, (101) holds.

If 2r ≤ 1 + \sqrt{1 − (δ_{0})^{2}}, then 2 − 2r ≥ 1 − \sqrt{1 − (δ_{0})^{2}} > 0. Since

1 − \sqrt{1 − (δ_{0})^{2}} < 2 − δ_{0} < 1 + \sqrt{1 − (δ_{0})^{2}},

if δ_{0} > 0, it follows that we must consider the case

2 − δ_{0} < 2r ≤ 1 + \sqrt{1 − (δ_{0})^{2}}.

Let us consider the function |η_{0}^{∧}(r, ψ, δ_{0}) − 1| − |η_{1}^{∧}(r, ψ, δ_{0}) − 1| in the open triangle

2 − δ_{0} < 2r < 2 + δ_{0}, 2|r − 1| < 4s < δ_{0},

where, in view of (94), t = cos(ψ) = arccos(4s^{2} − r^{2} − 1)/2r; in this triaqngle it does not vanish and continuos; (101) holds in the part of this triangle, defined by means (99). ■

Lemma 4.7. Let

(109)  f(Δ) = \frac{Δ + 1}{Δ} (\frac{Δ + 1}{Δ − 1})^{Δ−1} ,

where Δ > 1. Then f(2) = 9/2,

(110)  f(Δ) ≥ 16/3

for Δ ∈ 2 + ℕ.

Proof. For Δ = 3 the inequality (114) turns into equality, for Δ = 4 and Δ = 5 it may be obtained directly. Finally,

((\frac{∂}{∂Δ} log(f))) (Δ) = log(1 + 2/(Δ − 1)) − \frac{1}{Δ + 1} − \frac{1}{Δ} ≥

\frac{2}{Δ − 1} − \frac{2}{(Δ − 1)^{2}} − \frac{1}{Δ + 1} − \frac{1}{Δ} = \frac{1}{Δ^{2} − Δ} − \frac{2}{(Δ − 1)^{2}} + \frac{2}{Δ^{2} − 1} =

\frac{1}{Δ^{2} − Δ} − \frac{4}{(Δ^{2} − 1)(Δ − 1)} > 0,

when Δ > 5. ■

Lemma 4.8. 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

(111)  \frac{x_{min}}{x_{max}} ≥ μ^{2}

is equivalent to the inequality

(112)  \frac{p/2}{\sqrt{q}} ≤ \frac{μ^{2} + 1}{2μ}.

Proof. We have

\frac{x_{min}}{x_{max}} = \frac{(x_{min})^{2}}{q} =

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

Therefore (111) is equivalent to the inequality

\frac{p/2}{\sqrt{q}} − μ > \sqrt{(\frac{p/2}{\sqrt{q}})^{2} − 1},

which is equivalent to (112) ■.

It follows from the Lemma 3.1.4 and its proof that

(113)  1 ≤ \frac{|η_{0}^{∧}(r, π, δ_{0})|}{|η_{1}^{∧}(r, π, δ_{0})|} ≤ \frac{|η_{0}^{∧}(r, ψ, δ_{0})|}{|η_{1}^{∧}(r, ψ, δ_{0})|},

and if 2r ≥ δ_{0}, then

(114)  1 ≤ \frac{|η_{0}^{∧}(r, π, δ_{0}) + 1|}{|η_{1}^{∧}(r, π, δ_{0}) + 1|} ≤ \frac{|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 (113) and left inequality in (114) are strong; on the other hand, when r ≥ δ_{0}/2 and ψ ∈ [0, π), then right inequality in (113) and right inequality in (114) are strong.

Let δ_{0} > 0, 2r ≥ δ_{0},

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

(116)  c_{0}(r, δ_{0}) = \frac{(μ_{0}(r, δ_{0}))^{2} + 1}{2μ_{0}(r, δ_{0})}.

According to (3.1.43) in [29] and (113) - (115), if 2r ≥ δ_{0}, then

(117)  0 < μ_{0}(r, δ_{0}) ≤ 1,

(118)  1/μ_{0}(r, δ_{0})) ≤ (\frac{|η_{0}^{∧}(r, ψ, δ_{0})|}{|η_{1}^{∧}(r, ψ, δ_{0})|})^{1/δ_{0}−1} \frac{|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 (117) is strong.

Lemma 4.9. Let δ_{0} ≥ 0 and at least one of the following conditionsis is fulfilled:

A) 2r > max(1 + \sqrt{1 − (δ_{0})^{2}}, 2 + δ_{0}), |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) |r − 1|/2 = s ≤ (r + 1)/2, 1 + \sqrt{1 − (δ_{0})^{2}} < 2r, E) the case h) of the Lemma 4.6 holds and

(119)  s > \frac{δ_{0}((1 + δ_{0}/2)^{2} − r^{2})}{4(c_{0}(r, δ_{0})(1 + δ_{0}) − δ_{0}/2 − r)},

then

(120)  |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. According to Lemma 4.4 and Lemma 4.6, (120) holds in each of of cases A), B), C), D); we must consider the case E). The inequality (120) is equivalent to the inequality

(121)  (\frac{|η_{0}^{∧}(r, ψ, δ_{0})|}{|η_{1}^{∧}(r, ψ, δ_{0})|})^{(1/δ_{0}−1)} ×

\frac{|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 (118), will be fulfilled, if the inequality

(122)  |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, it follows that

x_{max} = |η_{1}^{∧}(r, π, δ_{0}) − 1|^{2}, x_{min} = |η_{0}^{∧}(r, π, δ_{0}) − 1|^{2}

and (122) will be fulfilled, when (112) will be fulfilled with p = p_{0}, q = q_{0}, where p_{0} and q_{0} are taken from (70) and (71) respectively.

The equality (3.1.69) in [53] may be rewritten in the form

(123)  |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

(124)  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 [53] and (123), 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}).

Therefore (112) will be fulfilled, if the inequaliti

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 (100),

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

Therefore (112) will be fulfilled, if the inequality

(125)  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 (119). ■

Corollary. If r = 1, 1/δ_{0} ≥ 3, then (119) may be replaced by the inequality

\frac{1}{51} < s.

Proof. Clearly,

η_{0}^{∧}(1, π, δ_{0}) = 1, η_{0}^{∧}(1, π, δ_{0}) + 1 = 2,

η_{1}^{∧}(1, π, δ_{0}) = \frac{1 − δ_{0}}{1 + δ_{0}} , η_{1}^{∧}(1, π, δ_{0}) + 1 = \frac{2}{1 + δ_{0}} .

According to (115), for Δ = 1/δ ∈ 2 + ℕ holds the equality

μ_{0}(1, δ_{0}) = (\frac{Δ − 1}{Δ + 1})^{Δ−1} \frac{Δ}{Δ + 1}.

L.A.Gutnik, On the dimension of some linear spaces over ℚ 18

According to the Lemma 4.6, μ_{0}(1, δ_{0}) < \frac{3}{16}, and therefore c_{0}(1, δ_{0}) > \frac{265}{96},

4(c_{0}(r, δ_{0})(1 + δ_{0}) − δ_{0}/2 − r) > 4(265 − 16 − 96)/96 = 153/24 = 51/8,

δ_{0}((1 + δ_{/}2)^{2} − r^{2}) ≤ 13/108,

\frac{δ_{0}((1 + δ_{0}/2)^{2} − r^{2})/2)}{4(c_{0}(r, δ_{0})(1 + δ_{0}) − δ_{0}/2 − r)} < \frac{1}{51} \frac{26}{27}. ■

Remark. If r = 1, ψ = −π + 2lπ/m, where

m = 2, , …, 150, l = 1 , …, m − 1],

then the inequality s = cos(ψ/2) > 1/51 will be fulfilled. In fact,

cos(−π/2 + lπ/m) = sin(lπ/m) > sin(π/m) > sin(π/150) >

sin(1/50) > 1/50 − 1/750000 > 1/51.

Since

(126)  |η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2} + |η_{k}^{∧}(r, ψ, δ_{0}) + 1|^{2} = 2|η_{k}^{∧}(r, ψ, δ_{0})|^{2} + 2,

it follows that

(127)  ∑_{k=0}^{1} ∑_{l=0}^{1}(−1)^{k}|η_{k}^{∧}(r, ψ, δ_{0}) + (−1)^{l}|^{2} = 2 ∑_{k=0}^{1}(−1)^{k}|η_{k}^{∧}(r, ψ, δ_{0})|^{2},
(128)  \frac{∂}{∂t}|η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2} = ∑_{ε=0}^{1}(−1)^{ε}2^{1−ε} \frac{∂}{∂t}|η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2}.

According to (3.1.54) in [53] with ε = 0 and (82),

(129)  |η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2} = 2 + \frac{2p_{1} − p_{2}}{2} +

(−1)^{k} (2\sqrt{(\frac{p_{1}}{2} )^{2} − q_{1}} − sign(2r − δ_{0})\sqrt{(\frac{p_{2}}{2} )^{2} − q_{2}}) .

According to the Lemma 4.4, if

ε^{2} = ε, t ∈ (−1, 1], 2r ≠ δ_{0},

then Δ_{1+ε}^{∗}(r, ψ, δ_{0}) ≠ 0; therefore, in view of (3.1.43) in [53],

(130)  \frac{∂}{∂t}(|η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2}) =

(−1)^{k} (\frac{∂}{∂t} \frac{p_{1+ε}}{2} ) \frac{|η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2}}{(sign(2r − δ))^{ε}\sqrt{(\frac{p_{1+ε}}{2} )^{2} − q_{1+ε}}} =

(\frac{∂}{∂t}p_{1+ε}) \frac{(−1)^{k}|η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2}}{∑_{k=0}^{1}(−1)^{k}|η_{k}^{∧}(r, ψ, δ_{0}), ε|^{2}} ,

where

ε^{2} = ε, k^{2} = k, |δ_{0}| < 1, |ψ| < π, 2r ≠ δ_{0}.

According to (128), (129) and (130), if

k^{2} = k, |δ_{0}| < 1, |ψ| < π, 2r ≠ δ_{0},

then

(131)  \frac{∂}{∂t}|η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2} =

∑_{ε=0}^{1}(−1)^{ε}2^{1−ε}(−1)^{k} (\frac{∂}{∂t} \frac{p_{1+ε}}{2} ) \frac{|η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2}}{(sign(2r − δ))^{ε}\sqrt{(\frac{p_{1+ε}}{2} )^{2} − q_{1+ε}}} =

∑_{ε=0}^{1}(−1)^{ε}2^{1−ε}(−1)^{k} (\frac{∂}{∂t}p_{1+ε}) \frac{|η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2}}{∑_{k=0}^{1}(−1)^{k}|η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2}} .

In view of (74), if 2r = δ_{0}, then

\frac{∂}{∂t}|η_{k}^{∧}(r, ψ, δ_{0}) + 1|^{2} = 0

and, according to (128), if 2r = δ_{0}, k = 0, 1 then

\frac{∂}{∂t}|η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2} = 2\frac{∂}{∂t}|η_{k}^{∧}(r, ψ, δ_{0})|^{2};

therefore, if 2r = δ_{0}, k = 0, 1 then the value (−1)^{k}|η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2} monotonically decreases with increasing ψ ∈ (0, π).

According to the proof of the Lemma 3.1.3 in [53],

(132)  \frac{∂}{∂t} \frac{(1 + δ_{0})p_{1}}{8} = r + \frac{∂}{∂t}|D_{0}(r, ψ, δ_{0})| >

−(δ_{0})r + \frac{∂}{∂t}|D_{0}(r, ψ, δ_{0})| = \frac{∂}{∂t} \frac{(1 + δ_{0})p_{2}}{8} > 0

Therefore, in view of (131), if 2r > δ_{0}, then

(133)  (−1)^{k} \frac{∂}{∂t}|η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2} >

− (\frac{∂}{∂t}p_{j}) ∑_{ε=0}^{1}(−2)^{1−ε} \frac{|η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2}}{∑_{k=0}^{1}(−1)^{k}|η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2}} ,

where j = 1, 2 and k = 0, 1.

Lemma 4.10. If 2 ≤ 2r ≤ 2 + δ_{0}, then the value

|η_{0}^{∧}(r, ψ, δ_{0}) − 1|

increases together with increasing s ∈ ((r − 1)/2, δ_{0}/4).

Proof. Since s ∈ ((r − 1)/2, δ_{0}/4), it follows that we have the case h) of the Lemma 4.6. Therefore according to (127),

(134)  ∑_{k=0}^{1}(−1)^{k}|η_{k}^{∧}(r, ψ, δ_{0}) + 1|^{2} >

∑_{k=0}^{1}(−1)^{k} ∑_{l=0}^{1} |η_{k}^{∧}(r, ψ, δ_{0}) + (−1)^{l}|^{2} = 2 ∑_{k=0}^{1}(−1)^{k}|η_{k}^{∧}(r, ψ, δ_{0})|^{2}.

According to (132), (133) and (134),

(135)  (−1)^{k} \frac{∂}{∂t}|η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2} >

− (\frac{∂}{∂t}p_{2}) ∑_{ε=0}^{1}(−4)^{1−ε} \frac{|η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2}}{∑_{k=0}^{1}(−1)^{k}|η_{k}^{∧}(r, ψ, δ_{0}) + 1|^{2}} ,

In view of (3.1.57) in [53],

|η_{0}^{∧}(r, ψ, δ_{0}) + 1|^{2} ≤ 4|η_{0}^{∧}(r, ψ, δ_{0})|^{2}. ■

In view of (1) ,

(136)  D^{∨}(z, η) = (−η + 1)(−η + γ_{1})+
2(1 + γ_{1})(1 + r exp(iψ)) exp(−iπ)(−η) =
(−η + 1)(−η + γ_{1}) + 4(1 + γ_{1})s_{0}(r, ψ) exp(iψ_{0}(r, ψ − π))(−η),

where s_{0}(r, ψ) and ψ_{0}(r, ψ) are defined in (91) and (92) respectively. Therefore, in view of (72)

(137)  η_{k}^{∧}(r, ψ, δ_{0}) = −η_{k}^{∧}(2s_{0}(r, ψ), ψ_{0}(r, ψ) − π, δ_{0}) =
−\overline{η_{k}^{∧}(2s_{0}(r, ψ), π − ψ_{0}(r, ψ)π, δ_{0})},

where k = 0, 1, −1 < δ_{0} < 1, 1 ≤ r, −π < ψ < π. Consequently,

(138)  1 − η_{k}^{∧}(r, ψ, δ_{0}) = 1 + η_{k}^{∧}(2s_{0}(r, ψ), ψ_{0}(r, ψ) − π, δ_{0}) =

Making use (138) and Lemma 4.4, we can give Another proof of the Lemma 4.6. Let

r^{∧} = 2s_{0}(r, ψ), ψ^{∧} = ψ_{0}(r, ψ) − π.

Then

sign(2r^{∧} − δ_{0}) = sign(4s_{0}(r, ψ) − δ_{0})

and, if 0 < ψ < π, then assertion of the Lemma 4.6 directly follow from (138) and Lemma 4.4. If ψ = 0, then (138) implies the equalities

1 − η_{k}^{∧}(r, 0, δ_{0}) = 1 + η_{k}^{∧}(1 + r, −π, δ_{0}), r^{∧} = r + 1

and the equality (76) is impossible. If r > 1 and ψ = π, then (138) implies the equalities

1 − η_{k}^{∧}(r, π, δ_{0}) = 1 + η_{k}^{∧}(r − 1, 0, δ_{0}), r^{∧} = r − 1

and, according to the Lemma 4.4, all depends from the sign(r − 1 − δ_{0}/2) in correspondence with the Lemma 4.6. If r < 1 and ψ = π, then (138) implies the equalities

1 − η_{k}^{∧}(r, π, δ_{0}) = \overline{1 + η_{k}^{∧}(1 − r, π, δ_{0})}, r^{∧} = 1 − r,

the condition of the case e) of the Lemma 4.6 coincides with (77) for r^{∧} instead r, the condition of the case d) of the Lemma 4.6 fulfilled now only for δ_{0} < 0, it may be rewritten in the form

s = |r − 1|/2, 1 − 2r^{∧} = 2r − 1 = 1 − |δ_{0}| ≤ \sqrt{1 − (δ_{0})^{2}}

and we have (77) for r^{∧} intead r; if (95) holds, then

2r^{∧} − 1 = 1 − 2r < −\sqrt{1 − (δ_{0})^{2}}

and, in view of the inequality r^{∧} > 0 > δ_{0}/2, we have the case of (75) for ε = 1 and r^{∧} intead r; if (96) holds, then

2r^{∧} = 2 − 2r > 1 + \sqrt{1 − (δ_{0})^{2}} > δ_{0}

and we again have the case of (75) for ε = 1 and r^{∧} intead r; if (99) holds, then

δ_{0} > 0, 2r^{∧} = 2 − 2r < 1 − \sqrt{1 − (δ_{0})^{2}} < δ_{0}

and we again have the case of (75) for ε = 1 and r^{∧} intead r. In the case, when r = 1, ψ = π, we have case g) of the Lemma 4.6, if δ_{0} > 0, case d) of the Lemma 4.6 if δ_{0} = 0, and case a) of the Lemma 4.6, if δ_{0} < 0; the assertions of the Lemma 4.6 may be directly checcked now by means of (3.1.55) in [53]. ■

Let

(139)  r^{∼} = r^{∼}(r, ψ, δ_{0}) = 4rs_{0}(r, ψ)(1 + γ_{1})^{2}/(1 + γ_{1}^{2}),
(140) δ^{∼} = δ^{∼}(δ_{0}) = (1 − γ_{1}^{2})/(1 + γ_{1}^{2}), ψ^{∼} = ψ_{0}^{∼}(r, ψ) = ψ_{0}(r, ψ) + ψ.

Multiplying the equality

(η_{k}^{∧}(r, ψ, δ_{0}) + 1)(η_{k}^{∧}(r, ψ, δ_{0}) + γ_{1}) = −2r exp(iψ)(1 + γ_{1})η_{k}^{∧}(r, ψ, δ_{0})

with the equality

(−η_{k}^{∧}(r, ψ, δ_{0}) + 1)(−η_{k}^{∧}(r, ψ, δ_{0}) + γ_{1}) =
−4(1 + γ_{1})s_{0}(r, ψ) exp(ψ_{0}(r, ψ))(η_{k}^{∧}(r, ψ, δ_{0})),

we obtain the equality

(141)  (−(η_{k}^{∧}(r, ψ, δ_{0}))^{2} + 1)(−(η_{k}^{∧}(r, ψ, δ_{0}))^{2} + (γ_{1})^{2}) =

−r^{∼}(r, ψ, δ_{0})(1 + (γ_{1})^{2})s_{0}(r, ψ) exp(ψ^{∼}(r, ψ))(−(η_{k}^{∧}(r, ψ, δ_{0}))^{2}).

In view of (72)

(142)  −(η_{k}^{∧}(r, ψ, δ_{0}))^{2}) = η_{k}^{∧}(r^{∼}, ψ^{∼}, δ_{0})).

where k = 0, 1 The equalities (139), (140), (142) and Lemma 4.4 let us to make more preciese assertions about the behavior of value |(η_{k}^{∧}(r, ψ, δ_{0}))^{2}−1|. Lemma 4.11. If

(143)  \frac{|r − 1|}{2} ≤ s = \frac{δ_{0}}{8r} ≤ \frac{r + 1}{2},

or

(144)  r ≤ 1, δ_{0}^{2} ≤ 4r(1 − r), s = (1 − r)/2

then

(145)  ∑_{k=0}^{1}(−1)^{k}|(η_{k}^{∧}(r, ψ, δ_{0}))^{2} − 1| = 0.

If (4r(r + 1) − δ_{0})((4r|r − 1|) − δ_{0}) > 0, then

(146)  sign (∑_{k=0}^{1}(−1)^{k}|(η_{k}^{∧}(r, ψ, δ_{0}))^{2} − 1|) (sign(4r max(r, 1) − δ_{0}).

If 4r|r − 1|) < δ_{0} ≤ 4(r(r + 1) and |r − 1| < 2s ≤ (r + 1), then

(147)  sign (∑_{k=0}^{1}(−1)^{k}|(η_{k}^{∧}(r, ψ, δ_{0}))^{2} − 1|) = (sign(s − δ_{0}/8r)

Proof. In view of (141), we apply the Lemma 4.4, taking r^{∼}, δ_{0}^{∼}, and

−(η_{k}^{∧}(r, ψ, δ_{0}))^{2}, k = 0, 1

respectively in the role of r, δ_{0}, and η_{k}^{∧}(r, ψ, δ_{0}), k = 0, 1. In view of the equalities (139) and (140),

(148)  sign(r^{∼}(r, ψ, δ_{0}) − δ^{∼}(δ_{0})/2) = sign(s_{0}(r, ψ) − δ_{0}/8r)

According to the Lemma 4.4, the equality (145) may be fulfilled only if the equality

(149)  r^{∼}(r, ψ, δ_{0}) = δ^{∼}(δ_{0})/2

is fulfilled or both conditions

(150)  ψ_{0}(r, ψ) + ψ ∈ π + 2πℤ,

and

(151)  |2r^{∼} − 1| ≤ (1 + δ_{0}^{∼})^{2})^{1/2}

are fulfilled. In view of (148) and (93), the condition (149) is equivalent to the condition (143).

In view of (91), (92) and (141), the condition (150) is equivalent to the condition

(152) (1 + r exp(iψ)) exp(iψ) = −2s_{0}(r, ψ) = −(r^{2} + 2rt + 1)^{1/2} ≤ 0,

which is equivalent to the system

(153) (1 + 2rt)) sin(ψ)) = 0, t + r(2t^{2} − 1) = t((1 + 2rt) − r ≤ 0, t = cos(ψ);

the system (153) have only the following solutions:

(154)  t = −1, r ≤ 1

and

(155)  t = −1/2r, r ≥ 1/2.

In view of (139) and (140), the condition (149) is equivalent to the condition

|8rs_{0}(r, ψ)(1 + γ_{1})^{2} − (1 + γ_{1}^{2})| ≤ 2γ_{1},

i.e to the condition

(156)  δ_{0}^{2} ≤ 8rs_{0}(r, ψ) ≤ 1.

The system of conditions (154) - (156) implies the equality 2s = 1 − r and is equivalent to (144). In view of (152), if 2r ≥ 1, then

2s_{0}(r, arccos(−1/2r)) = r

and therefore the system of conditions (155) - (156) takes the form

δ_{0}^{2} ≤ 4r^{2} ≤ 1, 2r ≥ 1

and for r = 1/2 coincides with (144). Other assertions of the Lemma directly follow from Lemma 4.4. ■

Lemma 4.12. Let 2r ≥ 1, 0 ≤ ψ ≤ π. Then

(157)  −sign(arccos(−1/2r) − ψ)Re(η_{0}^{∧}(r, ψ, δ_{0})) ≥

.

−sign(arccos(−1/2r) − ψ)Re(η_{1}^{∧}(r, ψ, δ_{0})) ≥ 0

and

(158) Re(η_{0}^{∧}(r, arccos(−1/2r), δ_{0})) = Re(η_{1}^{∧}(r, arccos(−1/2r), δ_{0})) = 0.

.

Proof. In view of (42),

exp(iψ)R_{0}(r, ψ, δ_{0}) = \sqrt{|D_{0}(r, ψ, δ_{0})|}×
exp(i(ψ − ψ_{1}(r, ψ, δ_{0})/2)).

In view of (6),

(159)  (exp(iψ)R_{0}(r, ψ, δ_{0}))^{2} = r^{2} exp(2iψ) + r exp(iψ) + (δ_{0}/2)^{2} =
= |R_{0}(r, ψ, δ_{0})|^{2}(exp(i(2ψ − ψ_{1}(r, ψ, δ_{0})),
(160)  Re((exp(iψ)R_{0}(r, ψ, δ_{0}))^{2}) = r^{2} cos(2ψ) + r cos(ψ) + δ_{0}^{2}/4

(161)  Im((exp(iψ)R_{0}(r, ψ, δ_{0}))^{2}) =
r^{2} sin(2ψ) + r sin(ψ) = r sin(ψ)(2r cos(ψ) + 1),
(162)  sign(Im((exp(iψ)R_{0}(r, ψ, δ_{0}))^{2})) = sign(2rcos(ψ) + 1),

if 0 < ψ < π; In view of (8) and (9),

−Im(D_{0}(r, ψ, δ_{0}))/Re(D_{0}(r, ψ, δ_{0})) =

\frac{((δ_{0})^{2}/2)t + r)sin(ψ)}{t((δ_{0})^{2}/2)t + r) + r^{2} − (δ_{0}/2)^{2}} ≤ tan(ψ),

if 0 ≤ ψ < π/2, t = cos(ψ) and r ≥ |δ_{0}/2|; consequently,

ψ_{1}(r, ψ, δ_{0}) ≤ ψ

and therefore

(163)  0 ≤ ψ/2 ≤ ψ − ψ_{1}(r, ψ, δ_{0})/2 ≤ π,

if 0 ≤ ψ < π/2 and r ≥ |δ_{0}/2|. In view of (33),

(164)  π − ψ_{1}(r, π, δ_{0})/2 = π/2.

If

Im(exp(iψ)R_{0}(r, ψ, δ_{0})) = 0,

then also

Im((exp(iψ)R_{0}(r, ψ, δ_{0}))^{2}) = 0

and (2r cos(ψ) + 1)r sin(ψ) = 0. Therefore as continuous functions of ψ the values

Im(exp(iψ)R_{0}(r, ψ, δ_{0}))

and

Im((exp(iψ)R_{0}(r, ψ, δ_{0}))^{2})

keep sign in each of the intervals (0, arccos(−1/2r), (arccos(−1/2r), π); therefore ψ −ψ_{1}(r, ψ, δ_{0})/2 and π −(ψ −ψ_{1}(r, ψ, δ_{0})/2) keep sign in each of the intervals (0, arccos(−1/2r), (arccos(−1/2r), π); consequently, in view of (163),

(165)  0 ≤ ψ − ψ_{1}(r, ψ, δ_{0})/2 ≤ π,

for ψ ∈ (0, arccos(−1/2r) and in view of (164), the inequalities (165) hold also for ψ ∈ (arccos(−1/2r), π); so, the inequalities (165) hold for ψ ∈ (0, π); moreover, in view of (162),

sign(2ψ − ψ_{1}(r, ψ, δ_{0}) − π) = sign(ψ − arccos(−1/2r)),

(166) |ψ − ψ_{1}(r, ψ, δ_{0})/2 − π/2| ≤ π/2, sign(ψ − ψ_{1}(r, ψ, δ_{0})/2 − π/2) =
sign(ψ − arccos(−1/2r)).

Let A = Re(exp(iψ)R_{0}(r, ψ, δ_{0}))^{2}), B = Im(exp(iψ)R_{0}(r, ψ, δ_{0}))^{2}). In view of (166),

(167)  exp(iψ)R_{0}(r, ψ, δ_{0}) =

sign (arccos (\frac{−1}{2r}) − ψ) \sqrt{\frac{\sqrt{A^{2} + B^{2}} + A}{2}}+

i\sqrt{\frac{\sqrt{A^{2} + B^{2}} − A}{2}}.

Further we have \frac{∂}{∂x}(\sqrt{x^{2} + B^{2}} + x) = (\sqrt{x^{2} + B^{2}} + x)/\sqrt{x^{2} + B^{2}} > 0, where x ∈ ℝ, B ∈ ℝ⧹{0}. In view of (160),

(168)  Re(exp(iψ)R_{0}(r, ψ, 1))^{2}) − Re(exp(iψ)R_{0}(r, ψ, δ_{0}))^{2}) > 0

In view of (65),

−Reη_{1}^{∧}(r, ψ, δ_{0}) = (1 + γ_{1})(Re(exp(iψ)R_{0}(r, ψ, 1)) − Re(exp(iψ)R_{0}(r, ψ, δ_{0})))

and, in view of (168), (167),

(169)  −signReη_{1}^{∧}(r, ψ, δ_{0}) = sign(arccos(−1/2r) − ψ)

In view of (65),

(170) − ∑_{k=0}^{1}(−1)^{k}Re(η_{0}^{∧}(r, ψ, δ_{0})) = 2(1 + γ_{1})Re(exp(iψ)R_{0}(r, ψ, δ_{0})) =

= 2(1 + γ_{1})sign(arccos(−1/2r) − ψ)\sqrt{\frac{\sqrt{A^{2} + B^{2}} + A}{2}}.

Therefore

(171)  −sign(arccos(−1/2r) − ψ) ∑_{k=0}^{1}(−1)^{k}Re(η_{k}^{∧}(r, ψ, δ_{0})) ≥ 0.

The inequalities (157) follow from (169) and (171). In view of (65) and (164), the equalities (158) hold, if ψ = arccos(−1/2r). ■

Lemma 4.13. If 1 ≤ 2r, then the value of

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

decreases together with increasing ψ ∈ (0, arccos(−1/2r)).

Proof. Clearly,

(−1)^{k} \frac{∂}{∂t}|η_{k}^{∧}(r, ψ, δ_{0})^{2} − 1|^{2} =

(−1)^{k} ∑_{l=0}^{1} |η_{k}^{∧}(r, ψ, δ_{0}) − (−1)^{l}|^{2} \frac{∂}{∂t}|η_{k}^{∧}(r, ψ, δ_{0}) + (−1)^{l}|^{2} =

|η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2}(−1)^{k} \frac{∂}{∂t}|η_{k}^{∧}(r, ψ, δ_{0}) + 1|^{2} + |η_{k}^{∧}(r, ψ, δ_{0}) + 1|^{2}×

∑_{ε=0}^{1}(−1)^{ε}2^{1−ε}(−1)^{k} \frac{∂}{∂t}|η_{k}^{∧}(r, ψ, δ_{0}) + ε|^{2} =

(−1)^{k} (\frac{∂}{∂t}|η_{k}^{∧}(r, ψ, δ_{0}) + 1|^{2}) ×

∑_{l=0}^{1}(−1)^{l}|η_{k}^{∧}(r, ψ, δ_{0}) − (−1)^{l}|^{2}+

2|η_{k}^{∧}(r, ψ, δ_{0}) + 1|^{2}(−1)^{k} \frac{∂}{∂t}|η_{k}^{∧}(r, ψ, δ_{0})|^{2}.

Assertion of the Lemma now directly follows from the Lemma 4.4. and the Lemma 4.12.■

Lemma 4.14. If |r − 1| < 2s < r + 1, then

(172)  (−1)^{k}sign(4s − δ_{0}) =
sign(|η_{0}^{∧}(r, ψ, δ_{0}) − 1|^{2} − 8s/(1 + δ_{0})),

where k = 0, 1.

Proof. In view of (3.1.34) in [53], (1) and (91),

(173)  q_{0}(r, ψ, δ_{0}) = |D^{∨}(z, 1)|^{2} =
4(1 + γ_{1})^{2}(2s)^{2}

and (172) follows now directly from the Lemma 4.6. ■

Lemma 4.15. The value

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

increases with increasing of s ∈ ((1 + δ_{0})/(4 + 2δ_{0}), (r + 1)/2). If r = 1, then the value

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

increases with increasing of s ∈ (max(δ_{0}/4, 0), (1 + δ_{0})/(6 + 2δ_{0}).

If r = 1, 0 ≤ δ_{0}, then the value (174) increases together with increasing of s ∈ (0, δ_{0}/4).

Proof. According to (3.1.31) and (3.1.33) in [53],

|R_{ε}^{∗}(r, ψ, δ_{0})|^{2} = r^{2} + r(2 + δ_{0})t + (1 + δ_{0}/2)^{2},

(176)  p_{0}(r, ψ, δ_{0}) =
8(r^{2} + r(2 + δ_{0})t + (1 + δ_{0}/2)^{2} + |D_{0}(r, ψ, δ_{0})|)/(1 + δ_{0})^{2}.

If r = 1, then for a, b and c in (15), we have

0 ≤ a < 1, 0 < b < 5/2, c > 25/16, 4c > b^{2}

and therefore, according to the Lemma 4.1,

0 ≤ \frac{∂}{∂t}|D_{0}(r, ψ, δ_{0})|,

where t ∈ (−1, 1),

\frac{∂}{∂t}|D_{0}(r, ψ, δ_{0})||_{t=1} = \sqrt{\frac{4a^{2} + 4ab + b^{2}}{4a + 4b + 4c}} < 1;

therefore

(177)  \frac{∂}{∂t}p_{0}(r, ψ, δ_{0})/2 ≥ 4r(2 + δ_{0})/(1 + δ_{0})^{2}

and

\frac{∂}{∂t}p_{0}(1, ψ, δ_{0})/2 ≤ 4(3 + δ_{0})/(1 + δ_{0})^{2}

According to the Lemma 4.6,

Δ_{0}^{∗}(r, ψ, δ_{0}) ≠ 0

if s ∈ (|r − 1|/2, (r + 1)/2), s ≠ δ_{0}/4; therefore, in view of (173),

(178)  \frac{∂}{∂t}(|η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2}) =

(−1)^{k} (\frac{∂}{∂t} \frac{p_{0}}{2} ) \frac{|η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2}}{(sign(4s − δ_{0}))\sqrt{(\frac{p_{0}}{2} )^{2} − q_{0}}} −

\frac{4r(1 + γ_{1})^{2}}{(sign(4s − δ_{0}))\sqrt{(\frac{p_{0}}{2} )^{2} − q_{0}}} .

In view of (172), (178), (176), (173) and Lemma 4.1,

(179)  \sqrt{(\frac{p_{0}}{2} )^{2} − q_{0}} \frac{∂}{∂t}(|η_{0}^{∧}(r, ψ, δ_{0}) − 1|^{2}) ≥
(4r(2 + δ_{0})/(1 + δ_{0})^{2})4(1 + γ_{1})s − 16r/(1 + δ_{0})^{2} =
16r((4 + 2δ_{0})s/(1 + δ_{0}) − 1)/(1 + δ_{0})^{2}.

Since δ_{0}/4 < (1 + δ_{0})/(2 + δ_{0}) the first assertion of the Lemma is proved. Let r = 1 and 4s > δ_{0}. Then

(180)  −\frac{∂}{∂t}(|η_{1}^{∧}(r, ψ, δ_{0}) − 1|^{2}) ≤

(4(3 + δ_{0})/(1 + δ_{0})^{2})4(1 + γ_{1})s − 16/(1 + δ_{0})^{2} =
16((6 + 2δ_{0})s/(1 + δ_{0}) − 1)/(1 + δ_{0})^{2}.

Since

(181)  δ_{0}/4 − (1 + δ_{0})/(6 + 2δ_{0}) = (δ_{0}^{2} + δ_{0} − 2)/(12 + 4δ_{0}) < 0,

it follows that the second assertion of the Lemma is proved. Let r = 1 and 4s < δ_{0}. Then

(182)  −\frac{∂}{∂t}(|η_{0}^{∧}(r, ψ, δ_{0}) − 1|^{2}) ≤
(4(3 + δ_{0})/(1 + δ_{0})^{2})4(1 + γ_{1})s − 16/(1 + δ_{0})^{2} =
16((6 + 2δ_{0})s/(1 + δ_{0}) − 1)/(1 + δ_{0})^{2}.

In view of (181), the third assertion of the Lemma is proved. ■

Remark. The third assertion of the Lemma corresponds

to the Lemma 4.10

Lemma 4.16. If 0 ≤ δ_{0}, then

(−1)^{k} \frac{∂}{∂t} log(|η_{k}^{∧}(r, ψ, δ_{0})|) ≥ (−1)^{k} \frac{∂}{∂t} log(|η_{0}^{∧}(r, ψ, δ_{k})| + 1)

Proof. Assertion of the Lemma follows from the Corollary to the Lemma 4.5. ■

Lemma 4.17. Let a < b. Let the functions p = p(t) and q = q(t) are defined on (a, b) and belong to the class C^{1}. Let further

(183)  q(t) > 0, p(t) > 2\sqrt{q(t)}

(184)  \frac{∂}{∂t}q(t) > 0

for any t ∈ (a, b). Let further

x_{k} = x_{k}(t) = p(t)/2 + (−1)^{k}\sqrt{(p(t)/2)^{2} − q(t)},

where k^{2} = k.

Let k^{2} = k, t ∈ (a, b). Then the condition

(185)  0 < (−1)^{k} ( \frac{\frac{∂}{∂t}p(t)}{\frac{∂}{∂t}q(t)}q(t) − p(t)/2)

is equivalent to the condition

(186)  0 < \frac{∂}{∂t} (x_{k}(t)/x_{1−k}(t))

and imply the following relations:

(187)  0 < \frac{∂}{∂t}x_{k}(t),

(188)  0 < \frac{∂}{∂t}(x_{k}(t) − x_{1−k}(t))

(189)  −sign (\frac{∂}{∂t}x_{1−k}(t)) =

(190)  sign (( \frac{\frac{∂}{∂t}p(t)}{\frac{∂}{∂t}q(t)}q(t) − p(t)) \frac{∂}{∂t}p(t) + \frac{∂}{∂t}q(t)) .

Proof Clearly,

(191)  (−1)^{k} \frac{∂}{∂t}x_{k} = \frac{x_{k} \frac{∂}{∂t}(p/2) − \frac{∂}{∂t}(q/2)}{\sqrt{(p/2)^{2} − q}}.

Therefore

(192)  ∑_{k=0}^{1}(−1)^{k}x_{1−k} \frac{∂}{∂t}x_{k} = \frac{2q \frac{∂}{∂t}(p/2) − p\frac{∂}{∂t}(q/2)}{\sqrt{(p/2)^{2} − q}}.

The condition (185) is equivalent to the the condition

(193)  0 < (−1)^{k} \frac{∂}{∂t} (\frac{p/2}{\sqrt{q}}) ,

and, in view of (192), to (186). According to (186), (183) and (184),

(194)  \frac{∂}{∂t}x_{k} = \frac{∂}{∂t}\sqrt{\frac{x_{k}}{x_{1−k}} q} > 0.

The condition (193) implies the inequality

(195)  0 < (−1)^{k} \frac{∂}{∂t}\sqrt{(\frac{p/2}{\sqrt{q}})^{2} − 1},

which together with (183) and (184) implies the inequality

(196)  \frac{∂}{∂t} (x_{k}(t) − x_{1−k}(t)) = (−1)^{k} \frac{∂}{∂t} (\sqrt{(\frac{p/2}{\sqrt{q}})^{2} − 1} \sqrt{q}) > 0.

In view of (194) and (191),

(197) 0 < \frac{∂}{∂t}x_{k} = (−1)^{k} \frac{x_{k} \frac{∂}{∂t}p − \frac{∂}{∂t}q}{\sqrt{p^{2} − 4q}} = \frac{x_{k} \frac{∂}{∂t}q}{q\sqrt{p^{2} − 4q}} (−1)^{k} (\frac{q \frac{∂}{∂t}p}{\frac{∂}{∂t}q} − x_{1−k}) ,

(198)  −\frac{∂}{∂t}x_{1−k} = \frac{x_{1−k} \frac{∂}{∂t}q}{q\sqrt{p^{2} − 4q}} (−1)^{k} ( \frac{q \frac{∂}{∂t}p}{\frac{∂}{∂t}(q)} − x_{k}) .

Multiplying (197) and (198) we obtain the equality

−sign (\frac{∂}{∂t}x_{1−k}) = sign (\frac{(\frac{∂}{∂t}p)^{2} q^{2}}{(\frac{∂}{∂t}(q))^{2}} − \frac{pq \frac{∂}{∂t}p}{\frac{∂}{∂t}q} + q) =

sign (( \frac{\frac{∂}{∂t}p}{\frac{∂}{∂t}q}q − p) \frac{∂}{∂t}p + \frac{∂}{∂t}q) . ■

Remark. The conditions and assertions of the Lemma are invariant relative to mulyiplying of both x_{k} on a positive constant γ and to substition t = t(τ) with continuous \frac{∂}{∂τ} t > 0. Therefore one cane replace p and q by γp and γ^{2}q.

Lemma 4.18. Let r = 1, δ_{0} > 0,

(199)  p = (1 + 1/δ_{0})^{2}p_{0}/8, q = (1 + 1/δ_{0})^{4}q_{0}/64.

Then

(200)  sign ( \frac{\frac{∂}{∂t}p(t)}{\frac{∂}{∂t}q(t)}q(t) − p(t)/2) = sign(4s − δ_{0}),

where s ∈ (0, 1).

Proof. Let

(201)  τ ∈ (0, 1/δ_{0}), s = \sqrt{δ_{0}τ}, ω = 1/δ_{0},

(202)  ℜ = \sqrt{4τ^{2} + (4ω^{3} − 3ω)τ + 1/16},

(203)  U = \frac{\frac{∂}{∂τ} p}{\frac{∂}{∂τ} q}q − p/2.

In view of (91), t = 2s^{2} −1 = 2δ_{0}τ −1, and (200) is equivalent to the equality

(204)  sign(U) = sign(16ωτ − 1).,

Since

r = 1, δ_{0} > 0, p = (1 + ω)^{2}p_{0}/8, q = (1 + ω)^{4}q_{0}/64,

it follows, according to (3.1.70) and (3.1.71) in [53], that

(205)  p = (4ω + 2)τ + 1/4 + ℜ, q = (1 + ω)^{2}ωτ.

\frac{\frac{∂}{∂t}p}{\frac{∂}{∂t}q}q − p/2 = (2ω + 1)τ − 1/8 + (2τ^{2} − 1/32)ℜ^{−1}.

Therefore for τ ∉ (1/(16ω + 8), 1/8) the equality (200) is fulfilled. Since

(2ω + 1)τ − 1/8 − (2τ^{2} − 1/32)ℜ^{−1} > 0,

if τ ∈ (1/(16ω + 8), 1/8), it follows that

sign ( \frac{\frac{∂}{∂t}p(t)}{\frac{∂}{∂t}q(t)}q(t) − p(t)/2) = sign(Q_{0}(τ)),

where τ ∈ (1/(16ω + 8), 1/8), and

Q_{0}(τ)) = ((2ω + 1)τ − 1/8)^{2}ℜ^{2} − (2τ^{2} − 1/32)^{2} =
(16ω^{2} + 16ω)τ^{4} + (16ω^{5} + 16ω^{4} − 8ω^{3} − 12ω^{2} − 5ω − 1)τ^{3}+
(−2ω^{4} − ω3 + 7ω^{2}/4 + ω + 1/4)τ^{2} + ω^{3}/16 − 5ω/64 − 1/64)τ =
τ(τ − 1/(16ω)ω(ω + 1)×
16τ^{2} + (16ω^{3} − 8ω − 4)τ − ω^{2} + ω + 1/4;

moreover, if τ(16ω + 8) > 1, then

(16τ^{2} + (16ω^{3} − 8ω − 4)τ − ω^{2} + ω + 1/4)/τ >
(16ω^{3} − 8ω − 4) − (ω^{2} − ω − 1/4)(16ω + 8) =
8ω^{2} + 4ω − 2 > 0,

because ω > 1. ■

Lemma 4.19. Let are fulfilled conditions of the lemma 4.18. Then

(206)  ( \frac{\frac{∂}{∂t}p}{\frac{∂}{∂t}q}q − p) \frac{∂}{∂t}p + \frac{∂}{∂t}q ≤ 0,

and an equality reaches here only if s = δ_{0}/4.

Proof. In view of (201) we must proof that

(207)  W = ( \frac{\frac{∂}{∂τ} p}{\frac{∂}{∂τ} q}q − p) \frac{∂}{∂τ} p + \frac{∂}{∂τ} q ≤ 0,

and an equality reaches here only if τ = 1/(16ω). First we prove that for τ = 1/(16ω) equality reaches. Let

V = −\frac{p}{2} \frac{∂}{∂τ} p + \frac{∂}{∂τ} q.

Then W = U \frac{∂}{∂τ} p + V with U in (203). Since, accoding to the Lemma 18, U|_{τ=1/(16ω)} = 0 we must prove that V |_{τ=1/(16ω)} = 0. But in, view of (199),

−V = \frac{∂}{∂τ} ((p/2)^{2} − q) = (1 + 1/δ_{0})^{4} \frac{∂}{∂τ} (\frac{1}{2} ∑_{k=0}^{1}(−1)^{k}|η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2})^{2} =

(1 + 1/δ_{0})^{4} (∑_{k=0}^{1}(−1)^{k}|η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2}) ×

\frac{∂}{∂τ} (\frac{1}{2} ∑_{k=0}^{1}(−1)^{k}|η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2}) ,

according to the Lemma 4.4,

\frac{∂}{∂τ} |η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2} =

\frac{∂}{∂τ} |η_{k}^{∧}(r, ψ, δ_{0})1|^{2} − \frac{∂}{∂τ} (2|η_{k}^{∧}(r, ψ, δ_{0})1| cos(ψ)) ,

exists at the point τ = 1/(16ω) and, according to the Lemma 4.6,

(∑_{k=0}^{1}(−1)^{k}|η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2}) |_{τ=1/(16ω)} = 0.

So, W |_{τ=1/(16ω)} = 0. In view of (202) and (205)

\frac{\frac{∂}{∂t}p}{\frac{∂}{∂t}q}q − p = −1/4 − ℜ/2 + (2τ^{2} − 1/32)ℜ^{−1} =

−(ℜ/4 + (4ω^{3} − 3ω)τ/2 + 1/16)ℜ^{−1},

ℜ^{2} (( \frac{\frac{∂}{∂τ} p}{\frac{∂}{∂τ} q}q − p) + \frac{∂}{∂τ} q) =

−(ℜ/4 + (4ω^{3} − 3ω)τ/2 + 1/16)×

((4ω + 2)ℜ + τ + (4ω^{3} − 3ω)/2)+

(1 + ω)^{2}ωℜ^{2}.

Let

A_{1} = (4ω^{3} − 3ω)τ/2 + 1/16) , B_{1} = 1/4, X_{1} = B_{1}ℜ,
A_{2} = (4ω^{3} − 3ω)/2 + 4τ, B_{2} = 4ω + 2, X_{2} = B_{2}ℜ.

Then

ℜ^{2} (( \frac{\frac{∂}{∂τ} p}{\frac{∂}{∂τ} q}q − p) \frac{∂}{∂τ} p + \frac{∂}{∂τ} q) =

−(A_{1} + X_{1})(A_{2} + X_{2}) + (1 + ω)^{2}ωℜ^{2} = −(A_{3} + X_{3}) + (1 + ω)^{2}ωℜ^{2},

where A_{3} = A_{1}A_{2} + X_{1}X_{2}, X_{3} = A_{1}X_{2} + A_{2}X_{1}. In view of the inequality X_{3} ≥ 0, it is sufficient to prove that

(X_{3})^{2} ≥ ((1 + ω)^{2}ωℜ^{2} − A_{3})^{2}.

and an equality reaches here only if τ = 1/(16ω). Furter we have

A_{1}^{2} − X_{1}^{2} = (ω^{2} − 1)(4ω^{2} − 1)^{2}τ^{2}/4,

A_{2}^{2} = (4ω^{3} − 3ω)^{2}/4 + 4(4ω^{3} − 3ω)τ + 16τ^{2} = 4ℜ^{2} + (ω^{2} − 1)(4ω^{2} − 1)^{2}/4,

X_{2}^{2} − A_{2}^{2} = 16(ω + 1)ωℜ^{2} − (ω^{2} − 1)(4ω^{2} − 1)^{2}/4,
X_{3}^{2} − A_{3}^{2} = −(A_{1}^{2} − X_{1}^{2})(A_{2}^{2} − X_{2}^{2}) = (ω + 1)^{2}(ω − 1)(4ω^{2} − 1)^{2}×
(16ωℜ^{2} − (ω − 1)(4ω^{2} − 1)^{2}/4)τ^{2}/4,
2A_{3} − ω(1 + ω)^{2}ℜ^{2} = (ω − 1)(A_{4}(4τ^{2} + 1/16) + +A_{5}τ,

where A_{4} = 3ω^{2} + ω − 1, A5 = (8ω^{5} − 8ω^{4} − 18ω^{3} + 2ω^{2} + 5ω − 1)/2,

(2A_{3} − ω(1 + ω)^{2}ℜ^{2})ω(1 + ω)^{2}ℜ^{2} = ω(1 + ω)^{2}(ω − 1)×
(4τ^{2} + 1/16 + (4ω^{3} − 3ω)τ)×

L.A.Gutnik, On the dimension of some linear spaces over ℚ 33

(A_{4}(4τ^{2} + 1/16) + A_{5}τ) =

ω(1 + ω)^{2}(ω − 1)×
(A_{4}(4τ^{2} + 1/16)^{2}+
(A_{4}(4ω^{3} − 3ω) + A_{5})(4τ^{2} + 1/16)τ+
(4ω^{3} − 3ω)A_{5}τ^{2})),
X_{3}^{2} − A_{3}^{2} + (2A_{3} − ω(1 + ω)^{2}ℜ^{2})ω(1 + ω)^{2}ℜ^{2} = (ω + 1)^{2}(ω − 1)T_{1},

where T_{1}(ω, τ) = ∑_{k=0}^{4} c_{k}τ^{4−k} and

c_{0} = 16A_{4}ω + 16ω(4ω^{2} − 1)^{2} = 16ω((3ω^{2} + ω − 1) + 16ω^{4} − 8ω^{2} + 1) =

16(16ω^{5} − 5ω^{3} + ω^{2}),

c_{1} = 4ωA_{4}(4ω^{3} − 3ω) + 4ωA_{5} + 4ω(4ω^{2} − 1)^{2}(4ω^{3} − 3ω) =

(12ω^{3} + 4ω^{2} − 4ω)(4ω^{3} − 3ω) + (64ω^{5} − 32ω^{3} + 4ω)(4ω^{3} − 3ω) + 4ωA_{5} =

(64ω^{5} − 20ω^{3} + 4ω^{2})(4ω^{3} − 3ω)+

16ω^{6} − 16ω^{5} − 36ω^{4} + 4ω^{3} + 10ω^{2} − 2ω =
256ω^{8} − 256ω^{6} + 24ω^{4} − 8ω^{3} + 10ω^{2} − 2ω,
c_{2} = (4ω^{2} − 1)^{2}ω/4 − (ω − 1)(4ω^{2} − 1)^{4}/16+
A_{4}ω/2 + A_{5}ω(4ω^{3} − 3ω) =

(4ω^{5} − 2ω^{3} + ω/4) − (ω − 1)(256ω^{8} − 256ω^{6} + 96ω^{4} − 16ω^{2} + 1)/16+ (3ω^{3} + ω^{2} − ω)/2 + (4ω^{4} − 3ω^{2})(8ω^{5} − 8ω^{4} − 18ω^{3} + 2ω^{2} + 5ω − 1)/2 =

−32ω^{7} + 35ω^{5} + ω^{4} − 7ω^{3} + ω^{2} − 5ω/16 + 1/16,

c_{3} = A_{4}(4ω^{3} − 3ω) + A_{5})ω/16 =

(3ω^{2} + ω − 1)(8ω^{4} − 6ω^{2})/32 + (8ω^{6} − 8ω^{5} − 18ω^{4} + 2ω^{3} + 5ω^{2} − ω)/32 = (32ω^{6} − 44ω^{4} − 4ω^{3} + 11ω^{2} − ω)/32,

c_{4} = A_{4}ω/256 = (3ω^{3} + ω^{2} − ω)/256

We apply the Horner’s method now twice and obtain the equality

T_{1} = (τ − 1/(16ω))((256ω^{5} − 80ω^{3} + 16ω^{2})τ^{2}+

(256ω^{8} − 256ω^{6} + 56ω^{4} − 8ω^{3} − 20ω^{2}))τ+
3ω^{5} + ω^{4} − ω^{3};

moreover, since ω ≥ 1, it follows that

256ω^{5} − 80ω^{3} + 16ω^{2} ≥ 192,

256ω^{8} − 256ω^{6} + 56ω^{4} − 8ω^{3} − 20ω^{2} ≥ 28,
3ω^{5} + ω^{4} − ω^{3} ≥ 3

and therefore T_{1} ≥ 3 > 0 for τ ≥ 0 and ω ≥ 1.■.

Lemma 4.20. Let r = 1, δ_{0} > 0, k^{2} = k. Then

(208)  \frac{∂}{∂ψ} |η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2} ≤ 0,

where ψ ∈ (0, π),

(209)  |η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2} ≥ 2δ_{0}/(1 + δ_{0}),

if s ≥ δ_{0}/4.

Proof. If s < δ_{0}/4, then, according to the Lemma 4.6, for k = 0, 1 the role of x_{k} in the Lemma 4.17 plais |η_{1−k}^{∧}(r, ψ, δ_{0}) − 1|^{2}, according to the Lemma 4.18 and to lemma 4.19, all the conditions of the Lemma 4.17 are fulfilled and both |η_{0}^{∧}(r, ψ, δ_{0}) − 1|^{2} and |η_{1}^{∧}(r, ψ, δ_{0}) − 1|^{2} increases together with increasing of s ∈ (0, δ_{0}/4). If s < δ_{0}/4, then, according to the Lemma 4.6, for k = 0, 1 the role of x_{k} in the Lemma 4.17 plais |η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2}, according to the Lemma 4.18 and Lemma 4.19, both |η_{0}^{∧}(r, ψ, δ_{0}) − 1|^{2} and |η_{1}^{∧}(r, ψ, δ_{0}) − 1|^{2} increases together with increasing of s ∈ (δ_{0}/4, 0).

Moreover, \frac{∂}{∂ψ}s < 0. Therefore (208 is proved. It follows from (208 that

|η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2}|_{s≥δ_{0}/4} ≥ |η_{k}^{∧}(r, ψ, δ_{0}) − 1|^{2}|_{s=δ_{0}/4}, ■.

Corollary. Let r = 1, δ_{0} ≥ 0. Then the value

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

decreases with increasing of psi ∈ (0, π).

Proof. According to the Lemma 4.4 and Lemma 4.20, the value

|(η_{0}^{∧}(r, ψ, δ_{0}))^{2} + ε|^{2},

where ε^{3} = ε, decreases together with increasing of psi ∈ (0, π). ■.

Lemma 4.21. Let r = 1, δ_{0} ≥ 0. Then the value

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

increases with increasing of s ∈ (δ_{0}/4, 1).

Proof. For δ_{0} = 0 assertion follows from the Lemma 4.4. Let δ_{0} > 0. According to the lemma 4.16 and (209),

−\frac{∂}{∂t} (log (|η_{1}^{∧}(r, ψ, δ_{0})|^{2(1/δ_{0}−1)}|η_{1}^{∧}(r, ψ, δ_{0})^{2} − 1)) =

−(1/δ_{0} − 1)\frac{∂}{∂t} log (|η_{1}^{∧}(r, ψ, δ_{0})|^{2}) −

\frac{∂}{∂t} log (|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}) −

−∂∂t log (|η_{1}^{∧}(r, ψ, δ_{0}) − 1|^{2}) =

−(1/δ_{0} − 1) \frac{\frac{∂}{∂t}|η_{1}^{∧}(r, ψ, δ_{0})|^{2}}{|η_{1}^{∧}(r, ψ, δ_{0})|^{2}} − 2 \frac{\frac{∂}{∂t}|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}}{|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}} +

\frac{\frac{∂}{∂t}|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}}{|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}} + \frac{\frac{∂}{∂t}|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}}{|η_{1}^{∧}(r, ψ, δ_{0}) − 1|^{2}} −

L.A.Gutnik, On the dimension of some linear spaces over ℚ 35

2 \frac{\frac{∂}{∂t}|η_{1}^{∧}(r, ψ, δ_{0})|^{2}}{|η_{1}^{∧}(r, ψ, δ_{0})|^{2}} \frac{|η_{1}^{∧}(r, ψ, δ_{0})|^{2}}{|η_{1}^{∧}(r, ψ, δ_{0}) − 1|^{2}} ≥

−(1/δ_{0} − 1) \frac{\frac{∂}{∂t}|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}}{|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}} − 2 \frac{\frac{∂}{∂t}|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}}{|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}} +

\frac{\frac{∂}{∂t}|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}}{|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}} + \frac{\frac{∂}{∂t}|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}}{|η_{1}^{∧}(r, ψ, δ_{0}) − 1|^{2}} −

2 \frac{\frac{∂}{∂t}|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}}{|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}} \frac{|η_{1}^{∧}(r, ψ, δ_{0})|^{2}}{|η_{1}^{∧}(r, ψ, δ_{0}) − 1|^{2}} =

−(1/δ_{0} + 1) \frac{\frac{∂}{∂t}|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}}{|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}} +

\frac{\frac{∂}{∂t}|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}}{|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}} \frac{2 + 2|η_{1}^{∧}(r, ψ, δ_{0})|^{2}}{|η_{1}^{∧}(r, ψ, δ_{0}) − 1|^{2}} −

2 \frac{\frac{∂}{∂t}|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}}{|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}} \frac{|η_{1}^{∧}(r, ψ, δ_{0})|^{2}}{|η_{1}^{∧}(r, ψ, δ_{0}) − 1|^{2}} =

− \frac{\frac{∂}{∂t}|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}}{|η_{1}^{∧}(r, ψ, δ_{0}) + 1|^{2}} (\frac{δ_{0} + 1}{δ_{0}} − \frac{2}{|η_{1}^{∧}(r, ψ, δ_{0}) − 1|^{2}} ) ≥ 0. ■

§2. Some corrections to [53].

There are some misprints in the paper [53].

In the equality (3.1.41) instead of

p_{1} = 8(|R_{0}^{∗}(r, ψ, δ_{0})|^{2} + |R_{0}(r, ψ, δ_{0})|^{2})/(1 + δ_{0})^{2} =

8 (r^{2} + rt + \frac{1}{4} + |D_{0}(r, ψ, δ_{0})|)/(1 + δ_{0})^{2}

should be

p_{1} = 8(|R_{0}^{∗}(r, ψ, δ_{0})|^{2} + |R_{0}(r, ψ, δ_{0})|^{2})/(1 + δ_{0})^{2} =

8 (r^{2} + rt + \frac{1}{4} + |D_{0}(r, ψ, δ_{0})|) /(1 + δ_{0})^{2}.

In (3.1.51) instead of

2r(1 + δ_{0}(r − δ/2))(1 + t)

should be

2r(1 + δ_{0}(r − δ_{0}/2))(1 + t).

In (3.1.53) instead of

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

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

should be

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

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

On the page 9, line 7 from the bottom instead of

\frac{∂}{∂t}p_{2} = −rδ_{0} + \frac{∂}{∂t}|D_{0}(r, ψ, δ_{0} >

should be

\frac{∂}{∂t}p_{2} = −rδ_{0} + \frac{∂}{∂t}|D_{0}(r, ψ, δ_{0})| > .

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[9] , 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, 1979, t.288, 965 – 967;

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

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

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

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

[14] A. Dubitskas, An approximation of logarithms of some numbers, Diophantine approximations II,Moscow, 1986, 20 – 34; [15] , On approximation of π/\sqrt{3} by rational fractions,

Vestnik MGU, series 1, 1987, 6, 73 – 76;

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

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

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

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

[20] E.C.Titchmarsh, The Theory of Functions, 1939, Oxford University Press;

[21] E.T.Whittaker and G.N. Watson, A course of modern analysis, 1927, Cambridge University Press;

[22] O.Perron, Über die Poincaresche Differenzengleichumg, Journal für die reine und angewandte mathematik, 1910, 137, 6 – 64;

[23] A.O.Gelfond, Differenzenrechnung (in Russian), 1967, Nauka, Moscow.

[24] A.O.Gelfond and I.M.Kubenskaya, On the theorem of Perron

in the theory of differrence equations (in Russian),

IAN USSR, math. ser., 1953, 17, 2, 83 – 86.

[25] M.A.Evgrafov, New proof of the theorem of Perron (in Russian),IAN USSR, math. ser., 1953, 17, 2, 77 – 82;

[26] G.A.Frejman, On theorems of of Poincaré and Perron (in Russian), UMN, 1957, 12, 3 (75), 243 – 245;

[27] N.E.Nörlund, Differenzenrechnung, Berlin, Springer Verlag, 1924;

[28] I.M.Vinogradov, Foundtions of the Number Theory, (in Russian), 1952, GIT-TL;

[29] , J. Diedonne, Foundations of modern analysis, Institut des Hautes Études Scientifiques, Paris, Academic Press, New York and London, 1960

[30] CH.-J. de la Vallée Poussin, Course d’analyse infinitésimale, Russian translation by G.M.Fikhtengolts, GT-TI, 1933;

[31] L.A.Gutnik, On the decomposition of the difference operators of Poincaré type (in Russian), VINITI, Moscow, 1992, 2468 - 92, 1 – 55;

[32] , On the decomposition of the difference operators of Poincaré type in Banach algebras (in Russian), VINITI, Moscow, 1992, 3443 - 92, 1 – 36;

[33] , On the difference equations of Poincaré type (in Russian), VINITI, Moscow 1993, 443 -B93, 1 – 41;

[34] , On the difference equations of Poincaré type in normed algebras (in Russian), VINITI, Moscow, 1994, 668 -B94, 1 – 44;

[35] , On the decomposition of the difference equations of Poincaré type (in Russian), VINITI, Moscow, 1997, 2062 -B97, 1 – 41;

[36] , The difference equations of Poincaré type with characteristic polynomial having roots equal to zero (in Russian), VINITI, Moscow, 1997, 2418 - 97, 1 – 20;

[37] , On the behavior of solutions of difference equations of Poincaré type (in Russian), VINITI, Moscow, 1997, 3384-B97, 1 - 41;

[38] , On the variability of solutions of difference equations of Poincaré type (in Russian), VINITI, Moscow, 1999, 361-B99, 1 – 9;

[39] , To the question of the variability of solutions of difference equations of Poincaré type (in Russian), VINITI, Moscow, 2000, 2416-B00, 1 – 22;

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

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

[42] , On the Irrationality of Some Quantities Containing ζ(3), Eleven papers translated from the Russian, American Mathematical Society, 1988, 140, 45 - 56;

[43] , Linear independence over ℚ of dilogarithms at rational points (in Russian), UMN, 37 (1982), 179-180; english transl. in Russ. Math. surveys 37 (1982), 176-177;

[44] , On a measure of the irrationality of dilogarithms at rational points (in Russian), VINITI, 1984, 4345-84, 1 – 74;

[45] , To the question of the smallness of some linear forms (in Russian), VINITI, 1993, 2413-B93, 1 – 94;

[46] , About linear forms, whose coefficients are logarithms of algebraic numbers (in Russian), VINITI, 1995, 135-B95, 1 – 149;

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

[48] , About linear forms, whose coefficients are ℚ-proportional to the number log 2, and the values of ζ(s) for integer s (in Russian), VINITI, 1996, 3258-B96, 1 – 70;

[49] , The lower estimate for some linear forms, coefficients of which are proportional to the values of ζ(s) for integer s (in Russian), VINITI, 1997, 3072-B97, 1 – 77;

[50] , On linear forms with coefficients in ℕζ(1 + ℕ)

Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 2000, 3, 1 – 13;

[51] , On linear forms with coefficients in ℕζ(1 + ℕ)

(the detailed version,part 1), Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 2001, 15, 1 – 20;

[52] , On linear forms with coefficients in ℕζ(1 + ℕ)

(the detailed version,part 2), Max-Plank-Institut für Mathematik, Bonn, Preprint Series,2001, 104, 1 – 36;

[53] , On linear forms with coefficients in ℕζ(1 + ℕ)

(the detailed version,part 3), Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 2002, 57, 1 – 33;

[54] , On the rank over ℚ of some real matrices (in Russian), VINITI, 1984, 5736-84; 1 – 29;

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

[56] , On linear forms with coefficients in ℕζ(1 + ℕ)

(the detailed version, part 4), Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 2002, 142, 1 – 27;

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

[58] , On the dimension of some linear spaces over ℚ (part 3), Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 2003, 16, 1 – 45;

[59] , On the difference equations of Poincaré type Max-Plank-Institut für Mathematik, Bonn, Preprint Series, 2003, 52, 1 – 44.