Return to Article Details Asymptotic approximation with Stancu Beta operators

REVUE D'ANALYSE NUMÉRIQUE ET DE THÉORIE DE L'APPROXIMATION

Rev. Anal. Numér. Théor. Approx., vol. 27 (1998) no. 1, pp. 5-13
ictp.acad.ro/jnaat

ASYMPTOTIC APPROXIMATION
WITH STANCU BETA OPERATORS

ULRICH ABELDedicated to Prof. Dr. D. D. Stancu on the occasion of his 70th birthday

Abstract

The concern of this paper is a beta type operator $L_{n}$ recently introduced by D. D. Stancu. We present the complete asymptotic expansion for $L_{n}$ as $n$ tends to infinity. All coefficients of $n^{- k} (k = 1, 2, \dots)$ are calculated explicitly in terms of Stirling numbers of the first and second kind. Moreover, we give an asymptotic expansion for $L_{n}$ into a series of reciprocal factorials.

MSC 2000. 41A36.

1. INTRODUCTION

In his recent paper [17] D. D. Stancu introduced a new beta operator of second kind

L_{n}

associating, for

n \in N

, with each

f \in M [0, \infty)

\begin{aligned} L_{n} (f; x) & \equiv L_{n} (f (\cdot); x) \\ (1) & = \frac{1}{B (n x, n + 1)} \int_{0}^{\infty} f (t) \frac{t^{n x - 1}}{(1 + t)^{n x + n + 1}} d t (x \in (0, \infty)) \end{aligned}

where

M [0, \infty)

denotes the space of bounded and locally integrable functions on

[0, + \infty)

. Actually,

L_{n}

is applicable to functions of polynomial growth if we restrict ourselves to

n \geq n_{0}

with

n_{0} \in N

sufficiently great.

The operators (1) are constructed, roughly speaking, by applying a beta second kind transform

T_{p, q}

to the Baskakov operators

\begin{matrix} (2) & B_{n} (f; x) = \sum_{k = 0}^{\infty} (\binom{n + k - 1}{k}) x^{k} (1 + x)^{n + k} f (\frac{k}{n}) (x \in (0, \infty)) \end{matrix}

and then choosing

p = n x

and

q = n + 1

.
The

L_{n}

are positive linear operators reproducing linear functions. It is mentioned in [17] that

L_{n}

is distinct from the other beta type operators considered earlier in the papers

[14, 13, 18, 11, 6]

Furthermore, D. D. Stancu gave estimations for the degree of approximation by using the moduli of continuity of first and second orders. He established also an asymptotic formula of Voronovskaja-type

\begin{matrix} (3) & lim_{n \to \infty} n ((L_{n} f) (x) - f (x)) = \frac{x (1 + x)}{2} f^{''} (x) \end{matrix}

for all

f \in M [0, \infty)

admitting a derivative of second order at

x (x > 0)

. The first result of this type was given by Voronovskaja [19] for the classical Bernstein polynomials and then generalized by Bernstein [8]. It should be worth noting that the right-hand term of Eq. (3) is the same as for the Baskakov operators (2).

The purpose of this paper is to present the complete asymptotic expansion for the operators (1) in the form

\begin{matrix} (4) & (L_{n} f) (x) \sim f (x) + \sum_{k = 1}^{\infty} a_{k} (f; x) n^{- k} (n \to \infty), \end{matrix}

provided

f

possesses derivatives of sufficiently high order at

x (x > 0)

. Formula (4) means that

(L_{n} f) (x) = f (x) + \sum_{k = 1}^{q} a_{k} (f; x) n^{- k} + o (n^{- q}) (n \to \infty)

for all

q \in N

. The Voronovskaja-type result (3) is the special case

q = 1

with

a_{1} (f; x) = (1 / 2) x (1 + x) f^{''} (x)

. We give explicit expressions for all coefficients

a_{k} (f; x) (k = 1, 2, \dots)

. The Stirling numbers of the first, resp. second, kind play an important role in our investigations.

It turns out that the representation of the moments

L_{n} e_{r}

with

e_{r} (x) = x^{r} (r = 0, 1, 2, \dots)

requires an infinite series. In order to avoid this drawback we, finally, derive a closed expression for the moments by means of a finite series of reciprocal factorials. This leads to a further asymptotic expansion of the form

\begin{matrix} (5) & (L_{n} f) (x) \sim f (x) + \sum_{k = 1}^{\infty} b_{k} (f; x) \frac{1}{n^{\underset{―}{k}}} (n \to \infty), \end{matrix}

where

n^{\underset{―}{k}}

denotes the falling factorial

n^{\underset{―}{k}} = n (n - 1) \dots (n - k + 1), n^{\underset{―}{0}} = 1

.
We remark that in

1, 3, 2, 4

the author gave the analogous results for the operators of Meyer-König and Zeller, for the operators of Bleimann, Butzer and Hahn and the Stancu operators, respectively.

2. THE MOMENTS FOR THE STANCU BETA OPERATOR

A change of variable replacing

t

t / (1 - t)

in the definition (1) yields

\begin{matrix} (6) & (L_{n} f) (x) = \frac{1}{B (n x, n + 1)} \int_{0}^{1} f (\frac{t}{1 - t}) t^{n x - 1} (1 - t)^{n} d t . \end{matrix}

For the monomials

e_{r}

we get

\begin{matrix} (7) & (L_{n} e_{r}) (x) = \frac{B (n x + r, n + 1 - r)}{B (n x, n + 1)} = \frac{(n x)^{\bar{r}}}{n^{\underset{―}{r}}} (n > r - 1), \end{matrix}

where

n^{\bar{k}} = n (n + 1) \dots (n + k - 1), n^{\overset{―}{0}} = 1

is the rising factorial. Note that, in particular,

L_{n} e_{r} = e_{r} (r = 0, 1)

. For

r \geq 2

, we take advantage of the
well-known identities

(n x)^{\bar{r}} = \sum_{j = 0}^{r} (- 1)^{r - j} S_{r}^{j} (n x)^{j}

and

{[(1 - \frac{1}{n}) (1 - \frac{2}{n}) \dots (1 - \frac{r - 1}{n})]}^{- 1} = \sum_{i = 0}^{\infty} σ_{r - 1 + i}^{r - 1} n^{- i} (n > r - 1)

(see, e.g., [10, §60, Eq. (2), p.175]) which imply, by Eq. (7),

\begin{aligned} (L_{n} e_{r}) (x) & = n^{- r} \sum_{j = 0}^{r} (- 1)^{j} S_{r}^{r - j} (n x)^{r - j} \sum_{i = 0}^{\infty} σ_{r - 1 + i}^{r - 1} n^{- i} \\ (8) & = x^{r} + \sum_{k = 1}^{\infty} n^{- k} \sum_{j = 0}^{min {k, r}} (- 1)^{j} S_{r}^{r - j} σ_{r - 1 + k - j}^{r - 1} x^{r - j} (n > r - 1) \end{aligned}

Recall that the Stirling numbers

S_{j}^{i}

and

σ_{j}^{i}

of the first, resp. second, kind are defined by

x^{\underset{―}{j}} = \sum_{i = 0}^{j} S_{j}^{i} x^{i} and x^{j} = \sum_{i = 0}^{j} σ_{j}^{i} x^{\underset{―}{i}} (j \in N_{0}) .

Now we study the central moments

T_{n, s} (x) = L_{n} ((\cdot - x)^{s}; x) (s = 0, 1, 2, \dots)

Since

L_{n}

reproduces linear functions we have

T_{n, 0} (x) = 1

and

T_{n, 1} (x) = 0

. Application of the binomial formula yields

\begin{matrix} (9) & T_{n, s} (x) = \sum_{r = 0}^{s} (\binom{s}{r}) (- x)^{s - r} (L_{n} e_{r}) (x) \end{matrix}

and we get, by Eq. (8),
Proposition 1. For each

x > 0

and

s = 1, 2, \dots

the central moments

T_{n, s}

of the operators

L_{n}

possess, for

n \geq s

, the representation

T_{n, s} (x) = \sum_{k = 1}^{\infty} n^{- k} \sum_{j = 0}^{min {s, k}} x^{s - j} \sum_{r = max {j, 1}}^{s} (- 1)^{s - r + j} (\binom{s}{r}) S_{r}^{r - j} σ_{r - 1 + k - j}^{r - 1} .

In order to derive as our main result the complete asymptotic expansion of the Stancu beta operators we use a general approximation theorem for positive linear operators due to Sikkema [16, Theorems 1 and 2].

As in 15 let

K^{[q]} (x)

be the class of all functions

f \in M [0, \infty)

which are

q

times differentiable at

x (x > 0)

Theorem A. For

q \in N

and fixed

x > 0

let

A_{n} : K^{[2 q]} (x) \to C (0, \infty)

be a sequence of positive linear operators. If

\begin{matrix} (10) & A_{n} ((\cdot - x)^{s}; x) = O (n^{- ⌊ (s + 1) / 2 ⌋}) (n \to \infty) (s = 0, 1, \dots, 2 q + 2) \end{matrix}

then we have for each

f \in K^{[2 q]} (x)

\begin{matrix} (11) & A_{n} (f (\cdot); x) = \sum_{s = 0}^{2 q} \frac{f^{(s)} (x)}{s!} A_{n} ((\cdot - x)^{s}; x) + o (n^{- q}) (n \to \infty) \end{matrix}

Furthermore, if

f \in K^{[2 q + 2]} (x)

, the term

o (n^{- q})

in (11) can be replaced by

O (n^{- (q + 1)})

In order to apply Theorem A we have to show that assumption (10) is valid for the operators

L_{n}

. For this reason we prove

Proposition 2. For each

x > 0

and

s = 0, 1, 2, \dots

the central moments

T_{n, s}

of the operators

L_{n}

satisfy

\begin{matrix} (12) & T_{n, s} (x) = O (n^{- ⌊ (s + 1) / 2 ⌋}) (n \to \infty) \end{matrix}

Proof. The case

s = 0

is obvious. Let now

s \geq 1

and put

\begin{matrix} (13) & Q (k, s, j) = \sum_{r = max {j, 1}}^{s} (- 1)^{r} (\binom{s}{r}) S_{r}^{r - j} σ_{r - 1 + k - j}^{r - 1} \end{matrix}

By Proposition 1, it is sufficient to show that

Q (k, s, j) = 0 (j = 0, \dots, min {s, k})

for all

k \in N

with

k < ⌊ (s + 1) / 2 ⌋

. The Stirling numbers of the first, resp. second, kind occurring in Eq. (13) possess the representations

\begin{matrix} (14) & S_{r}^{r - j} = \sum_{α = 0}^{j} C_{j, j - α} (\binom{r}{j + α}) (j = 0, \dots, r) \end{matrix}

resp.

\begin{matrix} (15) & σ_{r - 1 + ℓ}^{r - 1} = \sum_{β = 0}^{ℓ} {\bar{C}}_{ℓ, ℓ - β} (\binom{r - 1 + ℓ}{ℓ + β}) (r \geq 1, ℓ = 0, 1, 2, \dots) \end{matrix}

(see [10, p. 151, Eq. (5)], resp. [10, p. 171, Eq. (7)]). The coefficients

C_{j, i}

and

{\bar{C}}_{j, i}

are independent on

r

and satisfy certain partial difference equations whose general solutions are unknown ([10, p. 150]). Some closed expressions for

C_{j, i}

and

{\bar{C}}_{j, i}

can be found in [3].

In the case

j = 0

we have

Q (k, s, j = 0) = \sum_{r = 1}^{s} (- 1)^{r} (\binom{s}{r}) σ_{r - 1 + k}^{r - 1}

By Eq. (15),

σ_{r - 1 + k}^{r - 1}

is a polynomial in

r

of degree

\leq 2 k

without constant summand if

k \geq 1

. Therefore, we have

Q (k, s, j = 0) = 0

0 < 2 k < s

On the other hand, if

j \geq 1

we insert (14) into Eq. (13) and get

\begin{aligned} Q (k, s, j) & = \sum_{α = 0}^{j} C_{j, j - α} \sum_{r = j}^{s} (- 1)^{r} (\binom{s}{r}) (\binom{r}{j + α}) σ_{r - 1 + k - j}^{r - 1} \\ = \sum_{α = 0}^{j} C_{j, j - α} \frac{s^{\underset{―}{j}}}{(j + α)!} \sum_{r = 0}^{s - j} (- 1)^{r + j} (\binom{s - j}{r}) r^{\underset{―}{α}} σ_{r - 1 + k}^{r - 1 + j} . \end{aligned}

Since

r^{\underset{―}{α}} σ_{r - 1 + k}^{r - 1 + j}

is a polynomial in

r

of degree

\leq j + 2 (k - j)

we have

Q (k, s, j) =

0 if

j + 2 (k - j) < s - j

, i.e., if

2 k < s

. Note that we only have to consider the case

s - j > 0

. This completes the proof of Proposition 2.

3. THE COMPLETE ASYMPTOTIC EXPANSION

Combining Proposition 1, Theorem A and Proposition 2 we get our first main result.

Theorem 1. Let

x > 0, q \in N

and

f \in K^{[2 q]} (x)

. Then, the Stancu beta operators possess the asymptotic expansion

\begin{matrix} (16) & (L_{n} f) (x) = f (x) + \sum_{k = 1}^{q} a_{k} (f; x) n^{- k} + o (n^{- q}) (n \to \infty) \end{matrix}

where the coefficients

a_{k} (f; x)

are given by

\begin{matrix} (17) & a_{k} (f; x) = \sum_{s = 2}^{2 k} \frac{f^{(s)} (x)}{s!} \sum_{j = 0}^{min {k, s}} x^{s - j} \sum_{r = max {j, 1}}^{s} (- 1)^{s - r + j} (\binom{s}{r}) S_{r}^{r - j} σ_{r - 1 + k - j}^{r - 1} . \end{matrix}

For the convenience of the reader we calculate the initial coefficients. As usual, we put

X = x (1 + x)

(cf. [9, Theorem 1.1, p. 303]) and

X^{'} = 1 + 2 x

a_{1} (f; x) = \frac{1}{2} X f^{''} (x)

\begin{aligned} a_{2} (f; x) = & \frac{1}{2} X f^{(2)} (x) + \frac{1}{3} X X^{'} f^{(3)} (x) + \frac{1}{8} X^{2} f^{(4)} (x), \\ a_{3} (f; x) = & \frac{1}{2} X f^{(2)} (x) + X X^{'} f^{(3)} (x) \\ + \frac{1}{4} X (6 X + 1) f^{(4)} (x) + \frac{1}{6} X^{2} X^{'} f^{(5)} (x) + \frac{1}{48} X^{3} f^{(6)} (x), \\ a_{4} (f; x) = & \frac{1}{2} X f^{(2)} (x) + \frac{7}{3} X X^{'} f^{(3)} (x) + \frac{1}{8} X (61 X + 12) f^{(4)} (x) \\ + \frac{1}{15} X X^{'} (31 X + 3) f^{(5)} (x) + \frac{1}{144} X^{2} (131 X + 26) f^{(6)} (x) \\ + \frac{1}{24} X^{3} X^{'} f^{(7)} (x) + \frac{1}{384} X^{4} f^{(8)} (x) . \end{aligned}

We mention that, in particular, the coefficient

a_{3} (f; x)

possesses the concise form

a_{3} (f; x) = \frac{1}{48} X {(\frac{d}{d x})}^{(4)} (X^{2} f^{(2)} (x))

For the sake of comparison we give without proof the initial terms of the asymptotic expansion of the Baskakov operators for

x > 0

and

f \in K^{[6]} (x)

\begin{aligned} B_{n} (f; x) = & f (x) + \frac{1}{2 n} X f^{''} (x) + \frac{1}{24 n^{2}} (4 X X^{'} f^{(3)} (x) + 3 X^{2} f^{(4)} (x)) \\ + \frac{1}{48 n^{3}} (2 X (6 X + 1) f^{(4)} (x) + 4 X^{2} X^{'} f^{(5)} (x) + X^{3} f^{(6)} (x)) \\ + o (n^{- 3}) (n \to \infty) \end{aligned}

4. THE COMPLETE ASYMPTOTIC EXPANSION INTO A SERIES OF RECIPROCAL FACTORIALS

Direct computation of the moments

(L_{n} e_{r}) (x) = (n x)^{\bar{r}} / n^{\underset{―}{r}} (n > r - 1)

as given in Eq. (7) yields

\begin{aligned} (L_{n} e_{1}) (x) = x \\ (L_{n} e_{2}) (x) = x^{2} + \frac{x^{2} + x}{n - 1} \\ (L_{n} e_{3}) (x) = x^{3} + 3 \frac{x^{3} + x^{2}}{n - 1} + 2 \frac{2 x^{3} + 3 x^{2} + x}{(n - 1) (n - 2)} \end{aligned}

These expressions suggest to study the expansion of the operators

L_{n}

into a series of reciprocal factorials. Proposition 3 exhibits the general expression for the moments.

Proposition 3. For

r = 0, 1, 2, \dots

the moments

L_{n} e_{r}

of the operators

L_{n}

possess the representation

\begin{matrix} (18) & (L_{n + 1} e_{r}) (x) = x^{r} + \sum_{k = 1}^{r - 1} \frac{(- 1)^{k}}{n^{\underset{―}{k}}} \sum_{j = 0}^{k} x^{r - j} \sum_{i = 0}^{k - j} (- 1)^{i} (\binom{r - 1 - j}{i}) S_{r}^{r - j} σ_{r - 1 - j - i}^{r - 1 - k} (r - 1)^{i}, \end{matrix}

n \geq r

, where the sum

\sum_{k = 1}^{r - 1}

is to be read as 0 in the case

r \leq 1

.
Proof. The case

r = 0

is obvious. Let

r \in N

and

n \geq r

. We take advantage of the identities

\begin{matrix} (n x)^{\bar{r}} = \sum_{j = 1}^{r} (- 1)^{r - j} S_{r}^{j} (n x)^{j} \\ n^{j} = n (n - r + 1 + r - 1)^{j - 1} = n \sum_{i = 0}^{j - 1} (- 1)^{i} (\binom{j - 1}{i}) (r - 1)^{j - 1 - i} [- (n - r + 1)]^{i} \\ [- (n - r + 1)]^{i} = \sum_{k = 0}^{i} σ_{i}^{k} [- (n - r + 1)]^{\underset{―}{k}} = \sum_{k = 0}^{i} σ_{i}^{k} (- 1)^{k} (n - r + k)^{\underset{―}{k}}, \end{matrix}

and therefore

\frac{n^{j}}{n^{\underset{―}{r}}} = n \sum_{i = 0}^{j - 1} \sum_{k = 0}^{i} (- 1)^{i + k} (\binom{j - 1}{i}) σ_{i}^{k} (r - 1)^{j - 1 - i} \frac{1}{n^{\underset{―}{r - k}}}

Combining these formulae with Eq. (7) we receive

\begin{aligned} (L_{n} e_{r}) (x) & = \frac{(n x)^{\bar{r}}}{n \underset{―}{\underset{―}{r}}} \\ = \sum_{j = 0}^{r - 1} (- 1)^{r - j - 1} S_{r}^{j + 1} x^{j + 1} \sum_{i = 0}^{j} \sum_{k = 0}^{i} (- 1)^{i + k} (\binom{j}{i}) σ_{i}^{k} (r - 1)^{j - i} \frac{n}{n \underset{―}{r - k}} \\ = \sum_{k = 0}^{r - 1} \frac{(- 1)^{k}}{(n - 1)^{\underset{―}{r - 1 - k}}} \sum_{j = k}^{r - 1} x^{j + 1} \sum_{i = k}^{j} (- 1)^{r + i - j - 1} (\binom{j}{i}) S_{r}^{j + 1} σ_{i}^{k} (r - 1)^{j - i} \end{aligned}

which yields Eq. (18) after some manipulations.

By Eq. (9) and Proposition 3, we get for the central moments

T_{n, s}

of the operators

L_{n}

, for

n + 1 \geq s \geq 1

\begin{aligned} T_{n + 1, s} (x) = \\ = \sum_{k = 1}^{s - 1} \frac{(- 1)^{k}}{n^{\underset{―}{k}}} \sum_{j = 0}^{k} x^{s - j} \sum_{r = k + 1}^{s} (- 1)^{s - r} (\binom{s}{r}) \sum_{i = 0}^{k - j} (- 1)^{i} (\binom{r - 1 - j}{i}) S_{r}^{r - j} σ_{r - 1 - j - i}^{r - 1 - k} (r - 1)^{i} \end{aligned}

Thus, we receive as our second main result, by Theorem A and Proposition 2
Theorem 2. Let

x > 0, q \in N

and

f \in K^{[2 q]} (x)

. Then, the Stancu beta operators possess an asymptotic expansion into a reciprocal factorial series

\begin{matrix} (19) & (L_{n} f) (x) = f (x) + \sum_{k = 1}^{q} \frac{b_{k} (f; x)}{(n - 1)^{\underset{―}{k}}} + o (n^{- q}) (n \to \infty) \end{matrix}

where the coefficients

b_{k} (f; x)

are given by

\begin{aligned} (20) & b_{k} (f; x) = \\ = (- 1)^{k} \sum_{s = k + 1}^{2 k} \frac{f^{(s)} (x)}{s!} \sum_{j = 0}^{k} x^{s - j} \sum_{r = k + 1}^{s} (- 1)^{s - r} (\binom{s}{r}) \sum_{i = 0}^{k - j} (- 1)^{i} (\binom{r - 1 - j}{i}) S_{r}^{r - j} σ_{r - 1 - j - i}^{r - 1 - k} (r - 1)^{i} \end{aligned}

For the convenience of the reader we calculate the initial coefficients:

\begin{aligned} b_{1} (f; x) = & \frac{1}{2} X f^{''} (x) \\ b_{2} (f; x) = & \frac{1}{3} X X^{'} f^{(3)} (x) + \frac{1}{8} X^{2} f^{(4)} (x) \\ b_{3} (f; x) = & \frac{1}{8} X (9 X + 2) f^{(4)} (x) + \frac{1}{6} X^{2} X^{'} f^{(5)} (x) + \frac{1}{48} X^{3} f^{(6)} (x) \\ b_{4} (f; x) = & \frac{1}{15} X X^{'} (16 X + 3) f^{(5)} (x) + \frac{1}{144} X^{2} (113 X + 26) f^{(6)} (x) \\ + \frac{1}{24} X^{3} X^{'} f^{(7)} (x) + \frac{1}{384} X^{4} f^{(8)} (x) \\ b_{5} (f; x) = & (\frac{1}{6} X + \frac{125}{144} X^{2} (5 X + 2)) f^{(6)} (x) + \frac{1}{120} X^{2} X^{'} (109 X + 22) f^{(7)} (x) \\ + \frac{1}{576} X^{3} (145 X + 34) f^{(8)} (x) + \frac{1}{144} X^{4} X^{'} f^{(9)} (x) + \frac{1}{3840} X^{5} f^{(10)} (x) \end{aligned}

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Received: May 30, 1997.

Ulrich Abel Lerchenweg 2, D-35435 Wettenberg, Germany,

e-mail: Ulrich.Abel@mnd.fh-friedberg.de