Approximative properties of Fourier-Meixner sums Текст научной статьи по специальности «Математика»

Probl. Anal. Issues Anal. Vol. 7(25), No. 1, 2018, pp. 23-40

DOI: 10.15393/j3.art.2018.4390

23

UDC 517.521

R. M. GADZHIMIRZAEV

APPROXIMATIVE PROPERTIES OF FOURIER -MEIXNER SUMS

Abstract. We consider the problem of approximation of discrete functions f = f (x) defined on the set Qg = {0, 5, 25, ...}, where 5 = N, N > 0, using the Fourier sums in the modified Meixner polynomials Ma,N (x) = (Nx) (n = 0,1,...), which for a > —1 constitute an orthogonal system on the grid Qg with

the weight function w(x) = e-x r(NX + a +1) • We study the

11 x I 1)

approximative properties of partial sums of Fourier series in polynomials M^n (x), with particular attention paid to estimating their Lebesgue function.

Key words: Meixner polynomials, Fourier series, Lebesgue function

2010 Mathematical Subject Classification: 41A10

1. Introduction. Suppose Us = {0, 5, 25,...}, where 5 = N, N > 0. Following [5] we denote by M^ N (x) = M^ (Nx) (n = 0,1,...) the modified Meixner polynomials that constitute for a > —1 an orthogonal system

t ^-ii • i r • / \ -r r(Nx + a +1)

on discrete set with the weight function w(x) = e —wm--\—,

r(Nx + 1)

that is,

M°N(x)Mk>(x)w(x) = (1 - e-5)-(X-1KnSnk, a > — 1.

Here ha N = (n+a)en5r(a + 1), and the corresponding orthonormal polynomials with the weight function pN(x) = (1 — e-5)a+1 w(x) are denoted by m<a N (x) = (ha n)"1/2Ma N(x) (n = 0, ]_,... ). These polynomials

(Petrozavodsk State University, 2018

find applications in various problems of mathematical statistics, quantum physics, mathematical physics, computational mathematics and signal processing by spectral methods. In turn, these applications lead to the study of the approximative properties of Fourier - Meixner sums SaN(f, x), where f is the function given on the grid fi. We present the main result of this paper in Theorem 1. It holds an upper estimate for the Lebesgue function X^ N(x) of the partial sums SaN(f, x) for x E [0, ^¡t], where 6n = 4n + 2a + 2. In the case a = — \ this problem was solved in [2].

2. Some properties of Meixner polynomials. To study the approximative properties of the partial sums of Fourier - Meixner series we need several properties of Meixner polynomials Ma N (x). For q = 0 and an arbitrary a E R, the classical Meixner polynomials [1, 4, 5] can be defined by

Ma (x) = Ma(x,q) = (n+a) ± ^ (i — 1 )

where x[k = x(x — 1)... (x — k + 1), (a)k = a(a +1)... (a + k — 1). It is well known [1, 4, 5] that for a > —1 and 0 < q < 1 Meixner polynomials Ma(x) constitute an orthogonal system on the grid {0,1,...} with the

r(x + a + 1)

weight function p(x) = p(x,a,q) = q —^^—+1)—, that is,

Y,Ma(x)Ma(x)p(x) = (1 - q)-a-1ha (q)5nk, 0 <q< 1,a> -1,

x=0

where h£(q) = (n+a)q-nr(a + 1).

Suppose N> 0, 5 = 1/N, q = e-5, = {0, 25,...}. If a > -1, then the polynomials N (x) = Ma(Nx,e-5) and m^ N (x) = ma (Nx,e-5) =

_i/o

= {ha(e-5)} MaN(x) constitute orthogonal and orthonormal on systems, respectively, with the weight function pN(x) = (1 — e-5)a+1w(x). Further, the following Christoffel-Darboux formula

Kn,N (t,x) = ^ mk,N (t)mk,N (x) = (e/2 —e-s/2)(x — t) x

5y/ (n + 1)(n + a + 1)

k=0

X [ma+1 ,N (t)ma N (x) — m<aN (t)ma+i,N (x)] (1)

plays an important role for the estimate of the Lebesgue function. The formula (1) can be written [3] as

KAt, x) -

an a /,\ a f \ anan— 1

;mn,N (t)mn,N (x) +

5

1

(an+an-i)

a

(an + an-i) (e2-e-2) (x-1)

x [m<aN (x) (c+1,N (t) - mn-1,N (t)) -

m

,N (t) (mn+1 ,N (x) — mn-1 ,N (

(x)

(2)

where an = y/(n + 1)(n + a + 1), m"1N(x) =0. For 0 <5 < 1, N = ■, A > 0, 1 < n < XN, a > —1,0 < x < œ, 9n = 4n + 2a + 2 the following estimates [5, 6] hold:

e 2 № N(x)l < c(a,A)0n 2 Aa(x),

(3)

ûa

0n ,

Aaa (x) =

0.

0n(0n + \x - 0n|)

e 4.

0 < x < TT,

it <x < %,

Tn < x 31 n 2 < x < 2 ,

31n < x < OO,

(4)

e 2 Imn+1,N (x) - mZ-1,N (x)| <

0 < x < 1-,

in 1

< c(a, A) <

——1 ü2 1

0n ,

^ — 3 a I 1

0n 4 x- 2 + 4 ,

a ^ — 3

x-a0n4

e 4,

on + \x - 0n\

it <x < %,

Tn < x 3in

2 < x < 2 ,

31n < x < OO,

(5)

where hereinafter c, c(a), c(a, A) are positive numbers depending only on the indicated parameters.

3. Formulation of the main result. We denote by C(Us) the space of discrete functions f : Us ^ R, such that

(6)

lim \f(x)\e-x/2 =0

The norm in this space we define as follows:

c(QS) = sup e"x/2 \f(x)

x

4 x 2 4

The following lemma holds.

Lemma 1. Suppose that a > —1, p > 1 and lpN is the space of functions defined on fis with

(E If (x)\PPN (x))

VxeOs J

1/p

IPN = (E If(x)rPN(x)| < ^ (7)

xE^s

Then C(fis) C lpN for 1 <p< 2.

Proof. The proof of the lemma follows immediately from (6) and (7). □

It follows from lemma 1 that for an arbitrary function f E C(fis) we can define Fourier - Meixner coefficients

fa = E f (t)ma,N (t)PN (t) (8)

tens

and Fourier - Meixner series

f (x) -E famak,N(x). (9)

k=0

We denote by Sa N (f,x) the partial sum of the series (9):

n

S<a, N (f, x) = E famk, N (x), k=0

which in view of (1) and (8) can be represented as

Sa,N (f,x) = E f (t)Ka,N (t,x)e-t r(NN+ a +)1) (1 — e-S )a+1. (10)

tens ( + )

Let En(f,S) be the best approximation of function f E C(fis) in the metric of the space C(fis) by algebraic polynomials of degree n, that is,

En(f,S)= inf t \\f — pn\\o(ns)'

where Hn is the subspace of algebraic polynomials pn (x) of degree less than or equal to n. Further, let pn(f) = pn(f,x) be a polynomial of the

best approximation f in C(fis) for which En(f,5) = \\f — pn(f)\\c(ns). Then, since Sa N (pn) = pn for pn E Hn, we get

|f (x) — S^ N (f, x) | = |f (x) — Pn (f,x) + Pn (f,x) — SZt N (f, x) | < < \f (x) — Pn (f,x)\ + ^ n (Pn — f,x)f . By the last inequality and (10) we have e-f ff(x) — Sa,n(f,x)f < e-f \f(x) —pn(f,x)\ + e-f fSa,n(Pn — f,x)f <

< En(f, 5)(1 + Aa , n(x)),

where

Aa N (x) = y e-^ r(Nt + a +1) (1 - e-5)a+i K

tens

r(Nt + 1)

n, N

(11)

(t,x)\ . (12)

The inequality (11) needs estimating on [0, to) the Lebesgue function n (x) defined in (12). In this article we analyse this problem only for the cases x E G1 = [0, jf] and x E G2 = [ jf, jr]. The case when x E (jf, to) will be discussed in another paper. We note that this problem without proof of the main result was presented in [3]. In this subsection we give, with full proof, the result announced in paper [3]. Namely, the following theorem holds.

Theorem 1. Suppose that a> —1,9n = 4n + 2a + 2, A > 0, 0 <5 < 1, 1 < n < AN. We have the following: 1) if x E G1, then

1,

a E (-1, -2),

An,n(x) < c(a,A) { log(n + 1), a = -2,

a > - 2 ;

(13)

na+ 2 ,

2) if x E G2, then

( log(nx + 1), Aa , N (x) < c(a,A) \ log(n + 1),

,log(n + 1) + ( n )'

a E (-1, - 2) a = - 2,

a > - 2.

(14)

4. Proof of Theorem 1. Suppose that x E G1, then

an , n (x) = a + ,

2 1 4

where

h < c(a)5 £ e^"?1 (t + 5)

tens,

n (t)ma n (x)

k=Q

/2 =

E

_ t+x r(Nt + a + 1) e 2 r(Nt + 1)

(1 — e_s )a+1 \K%t n (t,x)\.

Let us estimate /1. From (3) and (4) we obtain

/1 < c{a)5 Y, (t + 5)a£ \e_2mak,N(x)||e_2m\(t)|

<

tens,

0<t<

k=q

< c(a,A)5 Y (t + 5)a£ < c(a,A)5 Y (t + 5)a^+1 <

tens,

0<t< -4

Vn

< c(a, A)&.

a+1 n

k=Q

T +s

5a+1 + J (t + 5)adt

Q

tens, 0<t< tt-

< c(a, A).

(16)

Now we proceed to estimating /2. Taking (2) into account, we can write

/2 < /21 + /22 + /23, (17)

where

/21 =

an

an + an 1

e 2 \mann(x)\x

x

tens, i4- <t<&

e_ 2 r(Nt + a + 1) r(Nt + 1)

(1 — e_s)a+1 \ma, n(t)\ ,

/22 =

an an_1 5 _ x \ a / \ a / \ \

, s!2 e 2 \mn+1 ,N(x) — mn_1 ,N(x)\ x

an + an-1 e6'2 — e 6/2 1 ' ' 1

x

e _ 2 r(Nt + a + 1)

^ r(Nt + 1)(t — x)

i4- <t<&

(1 — e _6)a+1 \ma, n(t)\ ,

a

t

n

4

<t<

V

n

n

n

n

anan-1 5 -x \ a , , \

123 = -;--5^--J/2 e 2 \mn, N(x)\ x

X

an + an-1 e5/2 - e

E CNN+f +■) (1 - e-5 )°+1 lmS+1, n (t) - mn-1,n (t)|

e,

' - <t<x>

r(Nt+1)(t - x)

Let us estimate I21. From (3) and (4) we have

- 2 r(Nt + a I a

I21 < c(a, A)0n E We put

tens,

r(Nt + 1)

-(1 - e-5)a+1 \ma,N(t)\. (18)

W = y ^2+ a)+1) (1 - e-5)a+1 K ,N(t)\ = W1 + W2, (19)

tens

r(Nt + 1)

where

W1 = E "2r(Nt+ a)+1)(1 - e~5)a+1 |m°N(t)\,

tens, 0<t<

r(Nt + 1)

W2 = E " 2 IN + "+J) (1 - e-5)a + 1 \maN (t)\

3en 2

tens,

<t<x>

r(Nt + 1)

By the Cauchy-Bunyakovsky inequality

V

W1 < J2 (1 - e-5)

1/2

-5^+1 r(Nt + a + 1)

r(Nt + 1)

x

x

V 0

V

0

\

1/2

tenô, 3e

< c(a)

E (1 - e-5)

/ +5

V

-5^+1 e-tr(Nt + a + 1)

r(Nt + 1)

1/2

{ma,N (t)f

<

\

5a+1 + J (t + 5)adt

0

< c{a)Q,

rn + 1 n2

(20)

)

te a

_a w—^ t _a

W2 < c(a, A)0_ 2 5 J2 e_ t (t + 5)a < c(a,A)0_ 2 e_n. (21)

tens, 32n<t<&

From (19)-(21) we conclude that

a+1

W < c(a, A)6n2 . (22)

From (18), (19) and the last inequality we have

/21 < c(a,A)ea+1. (23)

Now we proceed to estimating /22. From (5) and (3) we have

/22 < c(a,A)n0a_ 10_a5 V = 4 + /22 + /32,

tx

tens,

Vn

where

taK(t) i

tx

P22 = c(a, A)n0_^ £ ^ , i = 1, 2, 3,

teBt

B1 = (4/0n, On/2] n Us, B2 = (On/2,30n/2] n Us, B3 = (30n/2, rc) n Us.

Turning to inequality (4), we obtain

a-a-1

a _ 1 x-- 1,1, 2 4 a _ 1 x-- a 5

/1 < c(a,A)0n 45 Y -< c(a,A)0n 45 ^ 12_i <

teBi x teBi

1, a e (—1, — 2),

< c(a, A) { log(n + 1), a = ±, (24)

0^ _ 2, a > 1.

r2 ^ f ta[01 + \t — 0n\]_4 . , _i

/22 < c(a,A)0n 5 —--— < c(a,A)0n x

teB2 x

3V

n 2

i 1 1 a —5 3 a_1

x [0n + \t — 0n\]_idt < c(a,A)0a 40n4 < c(a,A)0a 2, (25)

ta e-4 r

I232 < c(a, A)5 E -—- < c(a, A) tae-4dt < c(a, A)e-n. (26)

teB3 3en__ 5

2

Combining estimates (24)-(26), we obtain

a E (—1, — 2),

I22 < c(a, A) ^log(n +1), a = 1, (27)

C-2, a>1.

Proceed to estimating I23 for x E G1. Using (3) and (4), we can write

a|

~ 1 a (4-\

a x—^ t t I I

I23 < c(a,X)n6n 5 ^T^2 \™a+1 , N (t) - ma-1, N (t)\ =

teas,

4 <t< rv,

= I213 + I223 + I233, (28)

where

I23 = c(a,A)nen 2 \mn+1 ,n(t) - <-1 ,n(t)\ , i = 1, 2, 3.

teBi

a I 1

On 4 ta t-a+1 , , +ta+4

1 a X-> O^ t t 2 ■ 4 a _L 1 X-> t 2 ■ 4

I13 < c(a, A)nOni 5 ^ ^- < c(a, A)0n 4 5 ^ — <

_ c(a,..,_,„ t x t x

teBi teBi

< c(a,A)0S+4 2 - 4 <

teBi

1, a E (-1, -2),

< c(a, A) { log(n +1), a = -2, (29)

ea+ 2, a > - 2,

I2

I23 <

c(a, A)nOn 5

E

teB2

On 4 tat- 2 [0| + \t - 0n |] 4 tx

3

< c(a,A)ea-4 x

x

[on + \t - en\]-4dt < c(a,A)o:

a-3 ^

a 4 r\ A

n

on = c(a,A)0a+1, (30)

^ — 5

2 5

3

/33 < c(a, A)n0ni 5

a <- v^ tae 4

tx

teB3

<

a_1

< c(a,A)0n 1

a_i

ta_1 e_idt < c(a,A)0n 1 e_n. (31)

s

The inequality (28) and estimates (29)-(31) yield

(1,

a e (—1, — 2),

/23 < c(a

(a, A) < log(n + 1), a = — |,

a > — 2,

(n

Qa+ 2 0n 2 ,

(32)

From inequalities (17), (23), (27) and (32) we obtain

, a e (—1, — 2)

/2 < c(a, A) I log(n + 1), a = — 2,

n

a+1

na + 2,

a > — 2.

(33)

In turn, from (15), (16) and (33) we have

1,

a e (—1, — 2),

A^n(x) < c(a, A) { log(n + 1), a = — 2,

na+ 2,

1

2 ■

a > — 2.

Thus, (13) is proved.

Now let us prove (14). Assume that x e G2. We introduce the notation: D1 = [0, x — yjx/0n] n Us, D2 = (x — y/x/0n, x + y/x/0n] n Us, D3 = (x + y/x/0n, to) n Us. Then

Aa n (x) = J1 + J2 + J3,

where

Ji = e" X V e _ 2 r(Nt + a + 1) (1 — e _s )a(1 n (t,x)\,i = 1, 2, 3.

teDi

r(Nt + 1)

2

Let us estimate J2. To this end, observe that the Cauchy-Bunyakovsky inequality yields

\Kn,N (t,x)\ < \Kn,N (t,t)\ 1 \Kn,N (x,x)\ 1 .

Further, if -3 < x < , then x — \ j~ji~ > in, moreover, for t e D2, we have c1x < t < c2x. Then

J2 < c(a) \ e_rKaN(x,x)\1/2 5 ^ ta \e_tK<a, N(t,t)\1/2 .

teD2

Let us now estimate \e_tKa N(t,t)\. Using (1), (3), (4) and (5), repeating almost verbatim the arguments of the proof of lemma 7.1 [7], we can prove the following assertion.

Lemma 2. If a> —1, 0n = 4n + 2a + 2, A > 0, 1 < n < AN, t > 3/0n then

\e_tKlN(t,t)\ < c(a,A)t_a_ 1/2n1/2. By lemma 2 we have

J2 < c(a, A)x_a_4n15 ^ (t + 5)at_a_4n1 = c(a, A)x _a_1 n25x

teD2

x £ (t + 5)a_1 < c(a,A)x_2n2 Y 5 < c(a,A). (34)

teD2 teD2

Let us estimate J1. Using (2) we can write

J1 < J11 + J12 + J13,

where

J11 = c(a)e 2 2 (t + 5)a \m% , n (x)m£ , n (t)| ,

teDi

J12 = c(a)ne - x \mZ+1 , N (x) - mu -1 , N (x)\ 5 £ e |t^l+f) \mn , N (t)\

\ t - x\ teDi 1 1

J13 = c(a)ne x \ma, N (x)\ 5 e \ma+1 ' N (t) - ma -1 , N (t)\

te D

To estimate J11 we have

j11 < c(a)(jh

¡1),

(35)

where

a _ 1 a 1 V--

Jn < c(a,A)en 4x-2-45 (t + 5)a < c(a,A)a

a~1 _a_i

n 4 x 2 4 X

tens, 0<t< -1

(

X

i- +s

\

5a+1 + (t + 5)adt

a 1 a_1 1

< c(a,A)x-2 - 4 en 4 e-a-1 =

V

)

= c(a, A)(xen)-2- 4 e-1 < c(a, A)en 2,

(36)

J121 < c(a,A)x-2 - 4 en 2 5 J] (t + 5)at-2 - 4 <

tens,

~a <t<f \ / -p

< c(a,A)x2 en 2. From the inequalities (35), (36) and (37) we have

(37)

J11 < c(a, A)

+ e 2

en

In order to estimate J12 we represent it as

where

J12 = J12 + JJ\_2,

J\2 < c(a,A)ne-3x-a+1 eaa5 £ <

tens, \<-1^ 1

xt

a i 1 a i 1 1 x—-V

< c(a,A)en2 +4 x-2+4 -5 (t + 5)a <

x tens, 0<t'

ilt + 4™-a~ 3 a-a-1

< c(a, A)e„ 4 x-2- 4 e-a-1 = c(a, A) (xen)

_ a _ 3

2 4

(38)

(39)

(40)

-

2

-

1 \ a 1 1 A 2-4

5

( it )

J22 < c(a,A)n0n 4 x -a+4On 4 5

(

< c(a, A)x-a+1 ,

V

1-

< c(a, A)

E

tsfij ,

<t<x

(t + 5)at-a - 4

xt

<

x - ( it )

+

a _ 1

t 2 4

xt

dt

1

9n

<

y

a _ 1

y2 4

1 - y

dy < c(a, A) y2 4 dy+

< c(a, A)

1 2 !-

--dy < c(a, A)(1 + log -VxOn).

1 - y 3

From (39)-(41) it follows

2

j12 < c(a,A)(1 + log^v/^). Using the same arguments that led to (40)-(42), we obtain

J13 < c(a, A)(1 + log y/xk). From (38), (42) and (43) we have

J1 < c(a, A)(1 + log ^fxfn). Let us estimate J3. By (2) we have

J3 < (J31 + j32 + J33),

(41)

(42)

(43)

(44)

(45)

where

J31 = e-x \mZ, n(x)\ £ e-2 ! +)1) (1 - e-)a+1 \m%, n(t)\ ,

teD3 ( + )

J32 = c(a)ne-x \ma+1;N (x) - mn-1,N (x)\ 5 £ e ( +x5) \maN (t)\

teD3

tx

x

9

n

n

j33 = c(a)ne 2 im^n(x)| 5 j2 e 2(-+x5 |mn+1,N(t) — mz-1,n(t)|

teDs

To estimate J31, we note that the inequality (19) yields

J31 < e 2 m^N (x)| w,

hence, from (22) we obtain

a+1 _1 , / e

J31 < c(a,A)en 2 en 4x-a- 4 = c(a, A) —

(46)

To estimate J32, we represent it as

J32 = J32 ' 72

32

3

32,

(47)

where

J12 < c(a,A)nen 4 x-2 + 4 5

E

tens,

f+Vif<t< +v 0

taen 41-a - 4

. . _a j_ 1 c

< c(a,A)x 2 +4 5

E

tens,

f+J it<t< % W 0-

,a_ 1

12 4

tx

tx

<

Let us examine three cases:

1) If a = 2, then (

J32 < c(A)

0—

2

(5

1

V

+

tx

dt

< c(A) log

)

f - x

+ 1 I . (48)

V VJ

2) If —1 < a < 2, then

a _i_1 2 + 4

1

e

x

J32 < c(a,A)x 2 +1

(x + a/e—

+

dt

y

2

5

e

x

2x

< c(a, A)

dt

tx

+ c(a,A)x 2 +4

t 2-4 dt <

2x

< c(a, A) log \f0~x.

(49)

3) If a > 2 then

9

a

2

x

2x

9n I I x 2 V 9n

J12 < c(a,A) [ + c(a,A)x-a+4 I a - 4

tx

t2 - 4 dt <

(

< c(a, A)

log +

2x

a- 1 \

2 4 \

V

V

x

y

Consequently, from (48) - (50) we deduce the estimate:

log

( \T- ) '

a = 2,

J32 < c(a, A) <

log \Jx0n,

__/ 9n I / x \

log vonx + ( 2 x9n )

a_ 1

2 4

a > 2 .

Further

J322 < c(a,A)nx-a+40- 4 5x

£ (t + 5)a 0-

tens,

X

tx

0n + \t - 0n

<

(50)

a G (-1, 1 ), (51)

< c(a,A)x 2 +4 02

= c(a,A)x 2 +4 on

(on + \t - On\)

- 4 dt

tx

9n I / x _^

2 V 9n

V V 3 39n

Vn-Vn 2

[ + f

9n I / x_r 1

2 +V Qn 0 in-1,3

0 + \t - 0n\)'

dt

tx

<

x

9

2 +A/ 9

2

n

2

< c(a,A)( 0f) T - 4 log e^H^' (52)

v xJ ¥ + J et — x

J32 < c(a,A)nx-a+4en 4 5

E

tens,

i4-<t<

e-44 (t + 5)a 6, tx

aft 2

n

<

a I 1 _ a +1 X-> t 1 30 —

< c(a, A)x-2+4 en 2 +4 5 J] e-4ta-1 < c(a,A)e^^. (53)

tens,

2

From (47) and (51)-(53) we obtain the estimate

f log(nx + 1), J32 < c(a, A) < (e—

a E (—1, 2),

T_i log

enx

0— 2

a > 2.

(54)

Now we estimate J33 using the same scheme as for J32. We have

J33 = J33 + J33 + J33,

(55)

where

J33 < c(a,A)nen 4 x 2 4 5

< c(a,A)x 2 4 5

E

tens,

0-<t< 0.

ta en 41-2+4

tx

<

E

tens, <t<

,a +1

t 2 + 4

tx

log

( ) ,

<

a = — 2 ,

< c(a, A)

log y/x6.

a E (—1, — 1 ), (56)

_ / 0— + / f \

lo^v/e^ + ( 2 /0— J

, a> —2.

J23 < c(a,A)n6n 4 x 2 4 5

E

tens, 0— + 1 f 'i 30— 2 + V 0— <t< 2

taen 4 (el/3 + \t — 6n\)1/4 tx

<

a 1 — -J- —

< c(a,A)x-2 - 4 02 4 5

£ f 0n/3 + 0n -1 \

V t /

1/4

tens

a 1 a _L 1 X->

+ c(a,A)x-2 - 4 02 +4 5 J]

tsfij ,

0n1/

3+t-0n

t5/4

1/4

<

tx

< c(a, Am —

„ a I 1 /

0 2 + 4

1 + log

0nx

V

in + X__x

2 + V in x

(57)

J333 < c(a,A)n0n 4 x-2- 4 5

< c(a,A)n3/4x 2 4 5

E

<t<

tsfij ,

(t + 5)a e-t/4

tx

<

ta-1 e-t/4 < c(a,A)e-3n/2. (58)

From (55) - (58) we obtain

J33 < c(a, A) <

log(nx + 1) , log(n + 1),

log vxön + ft)a+1 (1 + log ^+n-i_x) ,

K \ 2 +V 9n xj

a G (-1, -2),

a = - 2,

a > - 2.

(59)

In turn, from (45), (46), (54) and (59) we deduce the estimate

J3 < c(a, A)

log(nx + 1), log(n + 1),

__a +1 /

log vxn + (-T) 2+4 ^1 + log ÖT^-X

a G (-1, -2),

x,

a = - 2,

a > - 2.

1

9n<t<^2n

39n<t<,x>

2

Combining (34), (44) and (60), we see that

(log(nx + 1), a e (-1, -1 ),

\1n(x) < c(a,AW log(n +1),

n +1), a = -1

2 '

a I 1

I2 , a > -1.

Therefore, (14) is proved. This completes the proof of the theorem.

References

[1] Bateman H, Erdeyi A. Higher transcendental functions. Vol. 2. McGraw-Hill, New York-Toronto-London, 1953.

[2] Gadzhieva Z. D., Esetov F. E., Yuzbekova M. N. Approximation properties of Fourier - Meixner sums on [0, œ). Proceedings of Dagestan State Pedagogical University. Natural and exact sciences, 2015. 3 (32), pp. 6-8. (in Russian)

[3] Gadzhimirzaev R. M. Approximation of functions defined, on the grid {0, 5, 25,...} by Fourier - Meixner sums. Daghestan electronic mathematical reports, 2017, iss. 7, pp. 61-65. (in Russian)

[4] Nikiforov A. F, Uvarov V. B., Suslov S. K. Classical orthogonal polynomials of a discrete variable. Springer-Verlag Berlin Heidelberg, 1991.

[5] Sharapudinov I. I. Polynomials orthogonal on the grid. Theory and Applications. Makhachkala: DSU publishing, 1997. (in Russian)

[6] Sharapudinov I. I. Asymptotics and weighted estimates of Meixner polynomials orthogonal on the gird {0,5, 25,...}. Math. Notes (1997) 62:501, pp. 501-512. DOI: 10.1007/BF02358983.

[7] Sharapudinov I. I. Special series in Laguerre polynomials and their approximation properties. Siberian Mathematical Journal, 2017, vol. 58, no. 2, pp. 338-362. DOI: 10.1134/S0037446617020173.

Received February 05, 2018.

In revised form, April 13, 2018.

Accepted April 16, 2018.

Published online May 11, 2018.

>g(n +1)+ 2+4, a > -1.

Dagestan Scientific Center RAS

45, M. Gadzhieva st., Makhachkala, 367025, Russia

E-mail: [email protected]