ON LINEAR FORMS WITH COEFFICIENTS IN ℕ ζ(1 + ℕ)

The detailed version, Part 5

L. A. Gutnik

In memory of Professor Alexander Buchstab

Content

§0. Foreword. §1. Ballad. §2. An application of the Lemma 5.5.

§0. Foreword

I continue the presentation of the details of the proof of the Theorem announced in [35] and proved in [31], [32]. In Section 1, I prove a Theorem (see Lemma 5.5), which gives sufficient conditions for the coefficient a_{0}(ν) in the linear difference equation

∑_{k=0}^{n} a_{k}(ν)y(ν + k) = 0

to be equal to 0. I have decided to select that Theorem in a separate part because of its general character. In Section 2, I apply the general Theorem proved in Section 1 to the situation investigated in this series.

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

§1. Ballad.

Lemma 5.1. Let N ∈ ℕ − 1, m ∈ ℕ. Let

(1)  R(z) = F (z) + ∑_{i=1}^{∞} ∑_{k=0}^{N} β_{i,k}(z + k)^{−i},

Typeset by AMS-TEX 2 L. A. GUTNIK

where F (z) ∈ ℂ[z], β_{i,k} ∈ ℂ for i ∈ ℕ and k = 0, ..., N,

β_{i,k} = 0

for i ∈ ℕ + m and k = 0, ..., N. Let

β_{i}^{∨}(z) = ∑_{k=0}^{N} β_{i,k}z^{k},

where i ∈ ℕ. Let further s ∈ ℕ − 1 and

(2)  z^{s}R(z) = F_{s}(z) + ∑_{i=1}^{∞} ∑_{k=0}^{N} β_{s,i,k}^{∗}(z + k)^{−i},

where F_{s}(z) ∈ ℂ[z], β_{s,i,k} ∈ ℂ for i ∈ ℕ and k = 0, ..., N,

β_{s,i,k}^{∗} = 0

for i ∈ ℕ + m and k = 0, ..., N. Let

β_{s,i}^{∗}(z) = ∑_{k=0}^{N} β_{s,i,k}^{∗}z^{k}.

Then

(3)  β_{s,i}^{∗}(z) = ∑_{l=0}^{s} \binom{s}{l}((−δ)^{l}β_{s+i−l}^{∨})(z),

where δ denotes the operator z \frac{∂}{∂z} .

Proof. Clearly, the equality (3) is true for s = 0. In view of (2),

z^{s+1}R(z) = zF_{s}(z) + ∑_{i=1}^{∞} ∑_{k=0}^{N} β_{s,i,k}^{∗}((z + k)^{−i+1} − k(z + k)^{−i}).

Therefore

F_{s+1}(z) = zF_{s}(z) + β_{s,1}^{∗}(1),
(4)  β_{s+1,i}^{∗}(z) = −δβ_{s,i}^{∗}(z) + β_{s,i+1}^{∗}(z).

ON LINEAR FORMS WITH COEFFICIENTS IN ℕζ(1 + ℕ) 3

Clearly, β_{0,i}^{∗}(z) = β_{i}^{∨}(z), where i ∈ ℕ; therefore, taking in (4) s = 0 we obtain (3) for s = 1. Let the equality (3) is true for some s ∈ ℕ − 1. In view of (4),

β_{s+1,i}^{∗}(z) = −δβ_{s,i}^{∗}(z) + β_{s,i+1}^{∗}(z) =

(∑_{l=0}^{s} \binom{s}{l}(−δ)^{l+1}β_{s+i−l}^{∨}(z)) + ∑_{l=0}^{s} \binom{s}{l}(−δ)^{l}β_{s+i+1−l}^{∨}(z) =

(∑_{l=1}^{s+1} \binom{s}{l − 1}((−δ)^{l}β_{s+i+1−l}^{∨}(z)) + ∑_{l=0}^{s} \binom{s}{l}(−δ)^{l}β_{s+i+1−l}^{∨}(z) =

∑_{l=0}^{s+1} \binom{s + 1}{l}(−δ)^{l}β_{s+1+i−l}^{∨}(z). ■

Lemma 5.2. Let s ∈ ℕ − 1

(x)_{s} = ∏_{i=0}^{s−1}(x + i),

where x is a independent variable. Then the following equality holds in ℂ[x, y] :

(6)  (x + y)_{s} = ∑_{l=0}^{s} \binom{s}{l}(x)_{l}(y)_{s−l}.

Proof. For s = 0 and for s = 1 the equality (6) is true. Let (6) is true for some s ∈ ℕ − 1. Then

(x + y)_{s+1} = (x + y + s)(x + y)_{s} = (x + y + s) ∑_{l=0}^{s} \binom{s}{l}(x)_{l}(y)_{s−l} =

∑_{l=0}^{s} \binom{s}{l}(x)_{l}(y)_{s−l}((x + l) + (y + s − l)) =

(∑_{l=0}^{s} \binom{s}{l}(x)_{l+1}(y)_{s−l}) + ∑_{l=0}^{s} \binom{s}{l}(x)_{l}(y)_{s+1−l} =

(∑_{l=1}^{s+1} \binom{s}{l − 1}l(x)_{l}(y)_{s+1−l}) + ∑_{l=0}^{s} \binom{s}{l}(x)_{l}(y)_{s+1−l} =

∑_{l=0}^{s+1} \binom{s}{l}(x)_{l}(y)_{s+1−l}. ■

Let 𝔈 = (ℂ[z])^{ℕ} be a space of all the sequences of elements contained in ℂ[z]. On this space act two commuting operators: the operator δ = z \frac{∂}{∂z} of the coordinatewise differentiation, transforming each sequence

(7)  (p_{1}(z), p_{2}(z), ...)

into the sequence

(8)  (δp_{1}(z), δp_{2}(z), ...)

and the shift operator ▽^{∗} transforming each sequence (7) into the sequence

(9)  (p_{2}(z), p_{3}(z), ...).

On the space 𝔈 acts also the operator z^{l}(\frac{∂}{∂z})^{l} transforming each sequense (7) into the sequences

(10)  (z^{l} ((\frac{∂}{∂z} )^{l} p_{1}) (z), z^{l} ((\frac{∂}{∂z} )^{l} p_{2}) (z), ...)

I we take in (6) x = −δ, y = ▽^{∗}, then we obtain the equality

(11)  (▽^{∗} − δ)_{s} = ∑_{l=0}^{s} \binom{s}{l}(−δ)_{l}(▽^{∗})_{s−l}.

Lemma 5.3. Let l ∈ ℕ − 1. Then

(12)  (−δ)_{l} = (−z)^{l} (\frac{∂}{∂z} )^{l} .

Proof. For l = 0 and l = 1 assertion of the Lemma is evident. Let l ∈ ℕ + 1 and assrtion of the Lemma is true for l − 1. Then

(−δ)_{l+1} = (−δ + l)(−z)^{l} (\frac{∂}{∂z} )^{l} =

−(−1)^{l}l(z)^{l} (\frac{∂}{∂z} )^{l} + l((−z)^{l} (\frac{∂}{∂z} )^{l} + (−z)(−z)^{l} (\frac{∂}{∂z} )^{l+1} =

(−z)^{l+1} (\frac{∂}{∂z} )^{l+1} . ■

In view of (12),

((−δ)_{l}p)|_{z=1} = ((−1)^{l}(z)^{l} (\frac{∂}{∂z} )^{l} p) |_{z=1},

for any p ∈ ℚ[z]. Let

f(z) = ∑_{k=0}^{∞} a_{k}(z − 1)^{k} ∈ ℂ[[z]]

Let us consider the operator ∂^{∗}, for which

(∂^{∗}f)(z) = ∑_{k=0}^{∞} a_{k+1}(z − 1)^{k} ∈ ℂ[[z]].

Clearly,

(13)  ((∂^{∗})^{n}f)|_{z=1} = ((z)^{n} (\frac{∂}{∂z} )^{n} f) |_{z=1}/n!

for any f ∈ ℂ[[z]] and any n ∈ ℕ − 1. The operator ∂^{∗} acts coordinate-wise on the space 𝔈 of all the sequences (7), transforming each sequence (7) into the sequence

(14)  (∂^{∗}p_{1}(z), ∂^{∗}p_{2}(z), ...)

and commutes with the above operator ▽^{∗}.

Lemma 5.4. Let the conditions of the Lemma 1 is fulfilled and

R(t) = P (t)/Q(t),

where

P (t) ∈ ℂ(t), Q(t) ∈ ℂ(t), deg(Q) − deg(P) = r ≥ 1

and the quotient of the leading coefficients of the polynomials P and Q is equal to the number γ. Then

β_{r−1,1}(1) = γ, β_{s,1}(1) = 0

for 0 ≤ s ≤ r − 2.

Proof. The assertion of the Lemma follows from the equality

β_{s,1}(1) = Res_{z=∞}(z^{s}R),

where s = 0, ..., r − 1.

Lemma 5.5 ([32], Lemma 9). Let δ_{1} > 0, ε_{1} > 0, n ∈ ℕ and for any ν ∈ [1/ε_{1}] + ℕ are given

P (z : ν) ∈ C[z]⧹{0}, Q(z : ν) ∈ C[z]⧹{0}, R(z; ν) = P (z; ν)/Q(z; ν),
N(ν) ∈ ℕ − 1, m(ν) ∈ ℕ, a_{k}(z : ν) ∈ ℂ[z],

where k = 0, … n. Let further γ_{ν} is equal to the result of division of the leading coefficient of the polynomial P (z; ν) by the leading coefficient of the polynomial Q(z; ν),

r(ν) = deg(Q(z; ν)) − deg(P (z; ν)) > 1, r(ν) < r(ν + 1)

where ν ∈ [1/ε_{1}] + ℕ,

z^{s}R(z; ν) = F_{s}(z; ν) + ∑_{i=1}^{∞} ∑_{k=0}^{N(ν)} β_{s,i,k}^{∗}(ν)(z + k)^{−i},

where s ∈ ℕ − 1 and the following conditions are fulfilled:

F_{s}(z; ν) ∈ ℂ[z], β_{s,i,k}^{∗}(ν) = 0,

where s ∈ ℕ − 1, i ∈ m(ν) + ℕ, k = 0, …, N(ν), Let

β_{s,i}^{∗}(z; ν) = ∑_{k=0}^{N(ν)} β_{s,i,k}^{∗}(ν)z^{k}

where s ∈ ℕ − 1, i ∈ ℕ − 1, ν ∈ [1/ε_{1}] + ℕ, and let the following equality holds:

(15)  ∑_{k=0}^{n} a_{k}(z : ν)β_{0,i}^{∗}(z; ν + k) = 0,

where i ∈ ℕ − 1 and ν ∈ [1/ε_{1}] + ℕ, i.e. for each, for each i ∈ ℕ − 1 te polynomials β_{0,i}^{∗}(z : ν) satisfy to the equality

(16)  ∑_{k=0}^{n} a_{k}(z : ν)y(ν + k) = 0,

where ν ∈ [1/ε_{1}] + ℕ. Then

a_{0}(1; ν) = 0,

for ν ∈ [1/ε_{1}] + ℕ.

Proof. Let us consider the ℕ-dimensional vector

(17)  𝔟(z; ν) = (β_{0,1}^{∗}(z; ν), β_{0,2}^{∗}(z; ν), β_{0,3}^{∗}(z; ν), …),

i − th coordinate of which is equal to β_{0,i}^{∗}(z; ν). In view of (15),

(18)  ∑_{k=0}^{n} a_{k}(z : ν)𝔟(z; ν + k) = 0,

Let

(19)  𝔶_{s}(z; ν) = ∑_{k=0}^{s} \binom{s}{k}(−1)^{k}((∂^{∗})^{k}(▽^{∗})_{s−k}𝔟)(z; ν)k!,

where s ∈ ℕ − 1, ν ∈ [1/ε_{1}] + ℕ, and let y_{s}^{∗}(z; ν) be the first coordoinate of the vector 𝔶_{s}^{∗}(z; ν). Let m ∈ [1/ε_{1}] + ℕ. Let us to prove the following

Preparatory Assertion. The equality

(20)  ∑_{k=0}^{n} a_{k}(z : ν)y_{s}^{∗}(ν + k) = 0

holds for any s ∈ [0, r(m) − 1] ∩ ℤ; moreover

y_{r(m)−1}^{∗}(1; m) ≠ 0,
y_{r(m)−1}^{∗}(1; ν) = 0

for ν ∈ ℕ + m, and

y_{s}^{∗}(1; ν) = 0

for s ∈ [0, r(m) − 2] ∩ ℤ, ν ∈ ℕ + m − 1.

To prove the Preparatory Assertion we use induction on s. Since y_{0}^{∗}(1, ν) = β_{0,1}^{∗}(1, ν) = 0 for ν ∈ ℕ − 1 and 0 < r(m) − 1, it follows that for s = 0 the Preparatory Assertion is true. Let s ∈ [1, r(m) − 1] ∩ ℤ and the Preparatory Assertion is true for s−1. Since s−1 < r(m)−1, it follows, according the induction assumption, that y_{s−1}^{∗}(1, ν) = 0 for ν ∈ ℕ + m − 1; therefore

y_{s−1}^{∗}(z, ν)/(z − 1) = (z − 1)^{−1}(y_{s−1}^{∗}(z, ν) − y_{s−1}^{∗}(1, ν)) = ∂^{∗}y_{s−1}^{∗}(z; ν)

satisfies to the equation (16) for ν ∈ ℕ + m − 1, and, according to (19) with s − 1 instead s, coincides with the first coordinate of the vector

(z − 1)^{−1}(𝔶_{s−1}(z, ν) − 𝔶_{s−1}(1, ν)) = ∂^{∗}𝔶_{s−1}(z; ν) =

∑_{k=0}^{s−1} \binom{s − 1}{k}(−1)^{k}((∂^{∗})^{k+1}(▽^{∗})_{s−1−k}𝔟)(z; ν)k! =

∑_{k=1}^{s} \binom{s − 1}{k − 1}(−1)^{k−1}((∂^{∗})^{k}(▽^{∗})_{s−k}𝔟)(z; ν)(k − 1)! =

∑_{k=1}^{s} \binom{s − 1}{k − 1}(−1)^{k−1}(k!(∂^{∗})^{k}(▽^{∗})_{s−k}𝔟)(z; ν)/k =

s^{−1} ∑_{k=1}^{s} \binom{s}{k}(−1)^{k−1}(k!(∂^{∗})^{k}(▽^{∗})_{s−k}𝔟)(z; ν) =

−s^{−1} ∑_{k=0}^{s} \binom{s}{k}(−1)^{k}(k!(∂^{∗})^{k}(▽^{∗})_{s−k}𝔟)(z; ν)+

s^{−1}((▽^{∗})_{s}𝔟)(z; ν),

where, in view of (19), the first coordinate of the sum

∑_{k=0}^{s} \binom{s}{k}(−1)^{k}(k!(∂^{∗})^{k}(▽^{∗})_{s−k}𝔟)(z; ν)

is equal to y_{s}^{∗}(z; ν) and, in view of (17), the first coordinate of the vector

((▽^{∗})_{s}𝔟)(z; ν)

is equal to β_{0,s}^{∗}(z; ν). Therefore

∂^{∗}y_{s−1}^{∗}(z; ν) = −s^{−1}y_{s}^{∗}(z; ν) + s^{−1}β_{0,s}^{∗}(z; ν),

(21)  y_{s}^{∗}(z; ν) = −s∂^{∗}y_{s−1}^{∗}(z; ν) + sβ_{0,s}^{∗}(z; ν).

Since both −s∂^{∗}y_{s−1}^{∗}(z; ν) and sβ_{0,s}^{∗}(z; ν) satisfy to the equation (16) for ν ∈ ℕ + m − 1, it follows, in view of (20), that y_{s}^{∗}(z; ν) also satisfies to the equation (16) for ν ∈ ℕ + m − 1. The number y_{s}^{∗}(1; ν) coincides with the first coordinate of the vector (19) at the point z = 1, i.e. with the first coordinate of the vector

(22)  ∑_{k=0}^{s} \binom{s}{k}(−1)^{k}((∂^{∗})^{k}(▽^{∗})_{s−k}𝔟)(1; ν)k!.

But (▽^{∗} − δ)_{s} is a linear combinaion of the operators (▽^{∗} − δ)^{k}, k = 0, …, s with entire positve coefficients and the coefficient equal to 1 at (▽^{∗} − δ)^{s}. According to the Lemma 5.1, the first coordinate of the vector ((▽^{∗}−δ)^{k}𝔟)(1, ν) for k = 0, …, s is equal to β_{k,1}^{∗}(1; ν). According to the Lemma 5.4, β_{k,1}^{∗}(1; ν) = 0, if k < r(ν) − 1, and β_{k,1}^{∗}(1; ν) = γ_{ν}, if k = r(ν) − 1; in particular, if s < r(m) − 1, ν = m or if s = r(m) − 1, ν > m all the k = 0, …, s are less than r(ν) − 1. Therefore in these cases y_{s}^{∗}(1, ν) = 0. Let now ν = m, s = r(m) − 1. Then k < r(ν) − 1, if k < s and k = r(m) − 1 = r(ν) − 1, if ν = m, s = r(m) − 1, k = s. Therefore y_{r(m)−1}^{∗}(1, m) = γ_{m} ≠ 0. So, the Preparatory Assertion is proved.

Let m ∈ [1/ε_{1}] + ℕ. Then

y_{r(m)−1}^{∗}(1, m) = γ_{m} ≠ 0, y_{r(m)−1}^{∗}(1, ν) = 0,

if ν ∈ ℕ + m, and y_{r(m)−1}^{∗}(1, ν) is a solution of the equation (16) for ν ∈ ℕ + m − 1. Therevore

a_{0}(1; m)γ(m) = ∑_{k=0}^{n} a_{k}(1; m)y_{r(m)−1}^{∗}(m + k) = 0. ■

§2. An application of the Lemma 5.5.

Let us consider the difference equation (3.1.106) in [37], i.e. the following difference equation:

(23)  ∑_{j=0}^{6} q_{j}(z; ν^{−1})x_{ν+j} = 0,

where q_{j}(z; ν^{−1}) ∈ ℚ[z; ν^{−1}] for j = 0 , …, 6 have no common divisors with exeption of different from zero constants. The function (R_{0}(t; ν))^{3}, where R_{0}(t; ν) is defined in (2.6) of the paper [35], plays the role of the function R(z; ν) = P (z; ν)/Q(z; ν) of the Lemma 5.5 now; so we can take

P (z; ν) = ((νΔ)!)^{3} ∏_{κ=ν+1}^{νΔ}(z − κ)^{3},

Q(z; ν) = ((ν(Δ − 1))!)^{3} ∏_{κ=0}^{νΔ}(z − κ)^{3},

in the Lemma 5.5. Then ((νΔ)!)^{3}/((ν(Δ−1))!)^{3}, plays the role of γ_{ν} in the Lemma 5.5,

1 < r(ν) = deg(Q(z; ν)) − deg(P (z; ν)) = 3ν + 3 < r(ν + 1) = 3ν + 6,

where ν ∈ ℕ − 1. The value νΔ plays the role of N(ν) in the Lemma 5.5. now, the number 3 plays the role of m(ν) in the Lemma 5.5. now, the number ε_{2}(F) defined in [37] plays the role of ε_{1} in the Lemma 5.5. now. The defined in (2.21)-(2.23) of [35] polynomials α(z; ν)/(−z)^{ν}, β(z; ν)/(−z)^{ν} and γ(z; ν)/(−z)^{ν} play the role respectively of β_{0,1}^{∗}(z; ν), β_{0,2}^{∗}(z; ν) and β_{0,3}^{∗}(z; ν), …) in the Lemma 5.5. It was proved in the end of the paper [37] that

(24) x_{ν} = α_{0}^{∗}(z; ν) = α^{∗}(z; ν), x_{ν} = α_{1}^{∗}(z; ν) = β^{∗}(z; ν), x_{ν} = α_{2}^{∗}(z; ν) = γ^{∗}(z; ν),

and defined in (2.24), (2.26) and (2.28) of the paper [35] polynomials

(25) x_{ν} = α_{3}^{∗}(z; ν) = ϕ^{∗}(z; ν), x_{ν} = α_{4}^{∗}(z; ν) = ψ^{∗}(z; ν), x_{ν} = α_{5}^{∗}(z; ν) = ξ^{∗}(z; ν)

with ν ∈ ℕ + [1/ε_{2}(F)], |z| > 1 are solutions of the equation (22); the uniqueness theorem shows that x_{ν} in (24) and (25) are solutions of the equation (23) for ν ∈ ℕ + [1/ε_{2}(F)], z ∈ ℂ. Therefore y_{ν} = β_{0,i}^{∗}(z; ν) with i = 1, 2, 3, ν ∈ ℕ + [1/ε_{2}(F)], z ∈ ℂ are solutions of the equation

∑_{j=0}^{6}(−z)^{6−i}q_{j}(z; ν^{−1})y_{ν+j} = 0,

where q_{j}(z; ν^{−1}) with j = 0, …, 6 are coefficients of the equation (23). Applying the Lemma 5.5, we see that

(26)  q_{0}(1; ν^{−1}) = 0

for ν ∈ ℕ + [1/ε_{2}(F)]. Let

q_{j}^{∨}(ν^{−1}) = q_{j+1}((1, ν − 1)^{−1}),

where j = 0, … 5 and ν ∈ ℕ + [1/ε_{2}(F)] + 1. In view of (26), (24) and (25)

(27)  x_{ν} = α_{k}^{∗}(1; ν),

where k = 0, 1, 3, 4, 5 and ν ∈ ℕ + [1/ε_{2}(F)] + 1, is a solution of the difference equation

(28)  ∑_{j=0}^{5} q_{j}^{∨}(ν^{−1})x_{ν+j} = 0.

According to the Lemma 4.1.3 in [39], q_{0}^{∨}(ν^{−1})q_{5}^{∨}(ν^{−1}) ≠ 0_{ℚ[ν^{−1}]}. Therefore there exists ε_{5} > 0 such that

q_{0}^{∨}(ν^{−1})q_{5}^{∨}(ν^{−1}) ≠ 0

for ν ∈ [1/ε_{5}] + ℕ. Let

q_{j}^{∧∗}(ν^{−1}) = q_{j}^{∨}(ν^{−1})/q_{5}^{∨}(ν^{−1}),

where j = 0, …, 5 and ν ∈ [1/ε_{5}] + ℕ. Then polynomials (27) compose a basis of the space of the solutions of the equation

(29)  ∑_{j=0}^{4} q_{j}^{∧∗}(ν^{−1})y(ν + j) + y(ν + 5) = 0,

where y(ν) = x_{ν}, ν ∈ [1/ε_{5}] + ℕ, and q_{0}^{∧∗}(ν^{−1}) ≠ 0 for ν ∈ [1/ε_{5}] + ℕ.

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

[35] , 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.

[36] , 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.

[37] , On some 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.

[38] , 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.

[39] , 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.

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

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

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