Simultaneous approximation by Gauss–Weierstrass–Wachnicki operators

Ulrich Abel Fachbereich MND,
Technische Hochschule Mittelhessen,
Wilhelm-Leuschner-Straße 13,
61169 Friedberg, Germany ulrich.abel@mnd.thm.de
ORCID: 0000-0003-1889-4850 and Octavian Agratini Babeş-Bolyai University,
Faculty of Mathematics and Computer Science,
Str. Kogălniceanu, 1, 400084 Cluj-Napoca, Romania
and
Tiberiu Popoviciu Institute of Numerical Analysis,
Romanian Academy,
Str. Fântânele, 57, 400320 Cluj-Napoca, Romania agratini@math.ubbcluj.ro
ORCID: 0000-0002-2406-4274

Abstract.

In this note we spotlight a generalization of Weierstrass integral operators introduced by Eugeniusz Wachnicki in the year 2000. The operators are correlated with diffusion equation. Our main result consists in the obtaining of the asymptotic expansion of derivatives of any order of Wachnicki’s operators. All coefficients are explicitly calculated and distinct expressions are provided for analytical functions.

Mathematics Subject Classification: 41A36, 41A60, 26A24.

Keywords and phrases: Gauss–Weierstrass operator, Bessel function, Gamma function, hypergeometric function, asymptotic expansion.

1. Introduction

The starting point of this note is the following generalized Gauss-Weierstrass transform

	$\displaystyle W(f;x,t)$	$\displaystyle=\displaystyle\frac{1}{2\sqrt{\pi t}}\int_{\mathbb{R}}\exp\left(-% \displaystyle\frac{(x-y)^{2}}{4t}\right)f(y)dy$
		$\displaystyle=\displaystyle\frac{1}{2\sqrt{\pi t}}\int_{\mathbb{R}}f(x-y)\exp% \left(-\displaystyle\frac{y^{2}}{4t}\right)dy,$		(1)

where $t>0$ is a parameter, $x\in\mathbb{R}$ and $f:\mathbb{R}\to\mathbb{R}$ is chosen such that the integral exists and is finite. Actually, (1) represents the convolution of $f$ with the density of the normal distribution (also called Gaussian distribution) having the expectation null and the variance $2t$ . $W(f;\cdot,t)$ is a smoothed out version of $f$ and physically $W(\cdot;\cdot,t)\equiv W_{t}$ is correlated with a heat or diffusion equation for $t$ times units. It is additive

W_{t_{1}}\circ W_{t_{2}}=W_{t_{1}+t_{2}},\ (t_{1},t_{2})\in(0,\infty)\times(0,% \times),

this being read as follows: diffusion for $t_{1}$ time units and then $t_{2}$ time units, is equivalent to diffusion for $t_{1}+t_{2}$ times units.

Examining relation (1) shows that $W_{t}$ is translation invariant, meaning that the transform of $f(x+a)$ is $W(f;x+a,t)$ .

We mention that $W_{t}$ transform one can extend to $t=0$ by setting $W_{0}$ to be the convolution with the Dirac delta function. This case does not come to our attention.

The special case $t=1$ can be interpreted this way. Taking in view the following formula

e^{u^{2}}=\displaystyle\frac{1}{2\sqrt{\pi}}\int_{\mathbb{R}}e^{-uy}e^{-y^{2}/% 4}dy,\ u\in\mathbb{R},

if we replace $u$ with the formal differential operator $D=\displaystyle\frac{d}{dx}$ and we utilize the Lagrange shift operator

e^{-yD}f(x)=f(x-y),\ x\in\mathbb{R},

then we get

e^{D^{2}}f(x)=\displaystyle\frac{1}{2\sqrt{\pi}}\int_{\mathbb{R}}e^{-yD}f(x)e^% {-y^{2}/4}dy=W(f;x,1)

which allows us to get the following formal expression for this particular transform

W_{1}=e^{D^{2}}.

The operator $e^{D^{2}}$ is to be understood as acting on the signal $f$ as

e^{D^{2}}f(x)=\sum_{k=0}^{\infty}\displaystyle\frac{D^{2k}f(x)}{k!},\ x\in% \mathbb{R}.

For more documentation on $W_{t}$ transform, see Zayed’s book [5, Chapter 18: The Weierstrass Transform].

In 2000 Eugeniusz Wachnicki [6] defined and studied an integral operator representing a generalization of $W_{t}$ which involves modified Bessel function. Our paper focuses bringing to light new properties of Weierstrass–Wachnicki integral operators. In the next section we present these operators pointing out some already established properties based on which we will highlight noteworthy features of them. The main results are set out in Section 3.

2. The operators

At the beginning we recall the modified Bessel function of the first kind and fractional order $\alpha>-1$ , see [2, Chapter 10] or [8]. Using the traditional notation $I_{\alpha}$ , it is described by the series

I_{\alpha}(z)=\sum_{k=0}^{\infty}\displaystyle\frac{1}{k!\Gamma(k+\alpha+1)}% \left(\displaystyle\frac{z}{2}\right)^{2k+\alpha},

(2)

where $\Gamma$ is the Gamma function. $I_{\alpha}$ forms a class of particular solutions of the following ordinary linear differential equation

z^{2}w^{\prime\prime}(z)+zw^{\prime}(z)-(\alpha^{2}+z^{2})w(z)=0.

In the motivation of our results we will also use the low-order differentiation with respect to $z$ described as follows

\displaystyle\frac{d}{dz}(z^{-\alpha}I_{\alpha}(z))=z^{-\alpha}I_{\alpha+1}(z),

(3)

see [8, Formula 03.02.20.0009.01].

Set $\mathbb{R}_{+}=[0,\infty)$ . For a fixed constant $K\geq 0$ , we consider the space

	$\displaystyle E_{K}=\{$	$\displaystyle f:\mathbb{R}_{+}\to\mathbb{R}\mid f\mbox{ is locally integrable % and exists }M_{f}\geq 0,$
		$\displaystyle\|f(s)\|\leq M_{f}\exp(Ks^{2}),\ s>0\}.$

The space can be endowed with the norm $\|\cdot\|_{K}$ as such

\|f\|_{K}=\sup_{s\in\mathbb{R}_{+}}|f(s)|\exp(-Ks^{2}).

We consider the operator $W_{\alpha}$ defined on $E_{K}$ by the following relation

W_{\alpha}(f;r,t)=\displaystyle\frac{1}{2t}\int_{0}^{\infty}r^{-\alpha}s^{% \alpha+1}\exp\left(-\displaystyle\frac{r^{2}+s^{2}}{4t}\right)I_{\alpha}\left(% \displaystyle\frac{rs}{2t}\right)f(s)ds,

(4)

where $\alpha\geq-1/2$ , $(r,t)\in(0,\infty)\times(0,\times)$ and $I_{\alpha}$ is given at (2).

This operator was introduced in [6, Eq. (1)] with a minor modification of the domain $E_{K}$ in which the author inserted $f\in C(\mathbb{R}_{+})$ , the space of all real-valued continuous functions defined on $\mathbb{R}_{+}$ . Since $W_{\alpha}f$ , $f\in E_{K}$ , is well defined for any $K>0$ , we can consider the domain of $W_{\alpha}$ as

E=\displaystyle\bigcup_{K>0}E_{K}.

Wachnicki [6, Theorem 4] showed the convergence

\lim\limits_{t\to 0^{+}}W_{\alpha}(f;r,t)=f(r),\ f\in E\cap C(\mathbb{R}_{+}),

uniformly on compact subintervals of $(0,\infty)$ .

Also, in [6] the author specified that for $\alpha=-1/2$ the operator defined by (4) turns out to be the authentic Weierstrass operators. Because this statement was not accompanied by a proof, we insert it as a detail in our paper. More precisely we prove

W_{-1/2}f=W\widehat{f},\ f\in E_{K},

(5)

see (1), where $\widehat{f}(s)=f(({\rm sgn}\,s)s)$ , $s\in\mathbb{R}$ .

By using the hyperbolic cosine, the identity

I_{-1/2}(z)=\sqrt{\displaystyle\frac{2}{\pi z}}\cosh(z)

takes place, see [2, page 443]. Further, for any $(r,t)\in(0,\infty)\times(0,\infty)$ we can write successively

	$\displaystyle\quad\ W_{-1/2}(f;r,t)$
	$\displaystyle=\displaystyle\frac{1}{2t}\sqrt{\displaystyle\frac{4t}{\pi}}\int_% {0}^{\infty}\exp\left(-\displaystyle\frac{r^{2}+s^{2}}{4t}\right)\cosh\left(% \displaystyle\frac{rs}{2t}\right)f(s)ds$
	$\displaystyle=\displaystyle\frac{1}{2\sqrt{\pi t}}\int_{0}^{\infty}\exp\left(-% \displaystyle\frac{r^{2}+s^{2}}{4t}\right)\left(\exp\left(\displaystyle\frac{% rs}{2t}\right)+\exp\left(-\displaystyle\frac{rs}{2t}\right)\right)f(s)ds$
	$\displaystyle=\displaystyle\frac{1}{2\sqrt{\pi t}}\int_{0}^{\infty}\left(\exp% \left(-\displaystyle\frac{(r-s)^{2}}{4t}\right)+\exp\left(-\displaystyle\frac{% (r+s)^{2}}{4t}\right)\right)f(s)ds$
	$\displaystyle=\displaystyle\frac{1}{2\sqrt{\pi t}}\left(\int_{0}^{\infty}\exp% \left(-\displaystyle\frac{(r-s)^{2}}{4t}\right)f(s)ds+\int_{-\infty}^{0}\exp% \left(-\displaystyle\frac{(r-s)^{2}}{4t}\right)\right)f(-s)ds$
	$\displaystyle=W(\widehat{f};r,t)$

and the statement (5) is completed.

$W_{\alpha}f$ is intimately connected to a generalized heat equation having the following expression, see [3, Eq. (1.3)]

\displaystyle\frac{\partial u(r,t)}{\partial t}=\Delta_{\mu}\cdot u(r,t)

(6)

where $\mu=2(\alpha+1)$ , $\alpha>-1/2$ , and the operator

\Delta_{\mu}=\displaystyle\frac{\partial^{2}}{\partial r^{2}}+\displaystyle% \frac{\mu-1}{r}\,\displaystyle\frac{\partial}{\partial r}

is the Laplacian in radial coordinates when $\mu=n\in\mathbb{N}$ . It is usual to refer to $t$ as time. If $f\in E_{K}\cap C(\mathbb{R})$ , then $W_{\alpha}f$ with $\alpha=n/2-1$ is a caloric function, this means it is a solution of equation (6) on a certain domain $D$ ,

D=\left\{(r,t):\ r>0,\ 0<t<\displaystyle\frac{1}{4K}\right\}.

For the detailed proof see [3, pages 254-255].

Recently, these operators have come back to the attention of some authors. For example in [4] an extension of $W_{\alpha}f$ was achieved for continuous functions defined on the domain $(0,\infty)\times\mathbb{R}$ and bounded by certain two-dimensional exponential functions. In [1], the authors obtained the asymptotic expansion of the operator $W_{\alpha}(f;r,t)$ as $t\to 0^{+}$ , for functions $f\in E$ being sufficiently smooth at a point $r>0$ .

In order to obtain an autonomous exposure, in this preliminary section we recall some notions which will be used in establishing our results.

The factorial powers (falling and rising factorial, respectively) are marked as follows

u^{\underline{j}}=\prod_{l=0}^{j-1}(u-l),\quad u^{\overline{j}}=\prod_{l=0}^{j% -1}(u+l),\quad j\in\mathbb{N}.

An empty product $(j=0)$ is taken to be 1.

For $|z|<1$ and generic parameters $a, b, c$ , the Gauss hypergeometric function ${}_{2}F_{1}$ is defined by

_{2}F_{1}(a,b;c;z)=\sum_{j\geq 0}\displaystyle\frac{a^{\overline{j}}b^{% \overline{j}}}{c^{\overline{j}}}\,\displaystyle\frac{z^{j}}{j!},

(7)

this series being convergent, see [2, Chapter 15]. Outside the disk with unit radius the function is defined as the analytic continuation with respect to $z$ of this sum, with the parameters $a, b, c$ held fixed. For particular case $z=1$ , the following identity

_{2}F_{1}(a,b;c;1)=\displaystyle\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)% \Gamma(c-b)},\ {\rm Re}(c-a-b)>0,

(8)

$c\neq 0,-1,-2,\ldots$ , takes place.

3. Results

In the first stage we establish some technical formulas gathered in a few lemmas. Set $\mathbb{N}_{0}=\{0\}\cup\mathbb{N}$ .

The first result is an explicit representation of $\left(\displaystyle\frac{\partial}{\partial r}\right)^{m}W_{\alpha}(f;r,t)$ in terms of $W_{\alpha+j}(f;r,t)$ , $j\in\mathbb{N}_{0}$ . We mention the identity that we will state contains a finite sum.

Lemma 1.

Let $f$ belong to $E_{K}$ and let $W_{\alpha}f$ be defined by (1). For any $m\in\mathbb{N}_{0}$ ,

\left(\displaystyle\frac{\partial}{\partial r}\right)^{m}W_{\alpha}(f;r,t)=% \sum_{j\geq 0}\displaystyle\frac{(2j)!}{2^{j}j!}\binom{m}{2j}r^{m-2j}\left(% \displaystyle\frac{1}{2t}\Delta\right)^{m-j}W_{\alpha}(f;r,t)

(9)

holds, where $\Delta$ denoted the forward difference of step one with respect to $\alpha$ .

Proof.

If $f\in E_{K}$ , for $s>0$ we have

\displaystyle\frac{\partial}{\partial r}W_{\alpha}(f;r,t)=\displaystyle\frac{1% }{2t}\int_{0}^{\infty}s^{\alpha+1}\displaystyle\frac{\partial}{\partial r}% \left(\exp\left(-\displaystyle\frac{r^{2}+s^{2}}{4t}\right)r^{-\alpha}I_{% \alpha}\left(\displaystyle\frac{rs}{2t}\right)\right)f(s)ds.

Formula (3) yields the relation

\displaystyle\frac{\partial}{\partial r}W_{\alpha}(f;r,t)=\displaystyle\frac{r% }{2t}(W_{\alpha+1}(f;r,t)-W_{\alpha}(f;r,t))=\displaystyle\frac{r}{2t}\Delta W% _{\alpha}(f;r,t).

(10)

We will realize the proof of identity (10) by mathematical induction. Obviously, the assertion is valid for $m=0$ . Assuming that it is true for an arbitrary $m$ , we show that it takes place for $m+1$ . Relations (9) and (10) imply

	$\displaystyle\quad\ \left(\displaystyle\frac{\partial}{\partial r}\right)^{m+1% }W_{\alpha}(f;r,t)=\sum_{j\geq 0}\displaystyle\frac{(2j)!}{2^{j}j!}% \displaystyle\binom{m}{2j}$
	$\displaystyle\times\left[(m-2j)r^{m-2j-1}\left(\displaystyle\frac{1}{2t}\Delta% \right)^{m-j}+r^{m-2j+1}\left(\displaystyle\frac{1}{2t}\Delta\right)^{m-j+1}% \right]W_{\alpha}(f;r,t)$
	$\displaystyle=r^{m+1}\left(\displaystyle\frac{1}{2t}\Delta\right)^{m+1}W_{% \alpha}(f;r,t)$
	$\displaystyle+\sum_{j\geq 1}\left[\displaystyle\frac{(2j-2)!}{2^{j-1}(j-1)!}% \displaystyle\binom{m}{2j-2}(m-2j+2)+\displaystyle\frac{(2j)!}{2^{j}j!}% \displaystyle\binom{m}{2j}\right]$
	$\displaystyle\times r^{m-2j+1}\left(\displaystyle\frac{1}{2t}\Delta\right)^{m+% 1-j}W_{\alpha}(f;r,t)$
	$\displaystyle=r^{m+1}\left(\displaystyle\frac{1}{2t}\Delta\right)^{m+1}W_{% \alpha}(f;r,t)$
	$\displaystyle+\sum_{j\geq 1}\displaystyle\frac{(2j)!}{2^{j}j!}\left[% \displaystyle\binom{m}{2j-1}+\displaystyle\binom{m}{2j}\right]r^{m+1-2j}\left(% \displaystyle\frac{1}{2t}\Delta\right)^{m+1-j}W_{\alpha}(f;r,t).$

Considering the elementary identity

\displaystyle\binom{m}{2j-1}+\displaystyle\binom{m}{2j}=\displaystyle\binom{m+% 1}{2j},

we obtain that (9) is true for $m+1$ and the induction is completed. ∎

From [1, Theorem 3] we extract the following relation.

If $f$ belonging to $E$ is a real analytic function, then

W_{\alpha}(f;r,t)\sim\sum_{n=0}^{\infty}c_{n}(\alpha,f,r)t^{n}\quad(t\to 0^{+}),

where

c_{n}(\alpha,f,r)=\displaystyle\frac{4^{n}}{n!r^{2n}}\left(\displaystyle\frac{% \partial}{\partial w}\right)^{n}\left[w^{n+\alpha}\left(\displaystyle\frac{% \partial}{\partial w}\right)^{n}f\left(r\sqrt{w}\right)\right]\Big{|}_{w=1}.

(11)

In the following $\Delta[f]$ stands for forward difference with the step $h=1$ of a function $f$ .

Lemma 2.

Let $f$ belong to $E_{K}$ and let $W_{\alpha}f$ be defined by (1). For any $m\in\mathbb{N}_{0}$ ,

\Delta^{m}W_{\alpha}(f;r,t)\sim\sum_{n=m}^{\infty}\Delta^{m}c_{n}(\alpha,f,r)t% ^{n}\quad(t\to 0^{+}),

where the quantities $c_{n}(\alpha,f,r)$ are described by (11). For $m\leq n$ ,

\Delta^{m}c_{n}(\alpha,f,r)=\displaystyle\frac{4^{n}}{(n-m)!r^{2n}}\left(% \displaystyle\frac{\partial}{\partial w}\right)^{n-m}\left[w^{n+\alpha}\left(% \displaystyle\frac{\partial}{\partial w}\right)^{n}f\left(r\sqrt{w}\right)% \right]\Big{|}_{w=1}

(12)

and for $m<n$ , $\Delta^{n}c_{n}(\alpha,f,r)=0$ .

Proof.

Based on the $m$ -th order forward difference with the step $h=1$ , we get

\Delta^{m}w^{m+\alpha}=\sum_{k=0}^{m}(-1)^{m-k}\displaystyle\binom{m}{k}w^{n+% \alpha+k}=w^{n+\alpha}(w-1)^{m}.

Consequently, taking into accounr (11), we can write

\Delta^{m}c_{n}(\alpha,f,r)=\displaystyle\frac{4^{n}}{n!r^{2n}}\left(% \displaystyle\frac{\partial}{\partial w}\right)^{n}\left[w^{n+\alpha}(w-1)^{m}% \left(\displaystyle\frac{\partial}{\partial w}\right)^{n}f\left(f\sqrt{w}% \right)\right]\Big{|}_{w=1}.

Since

\left(\displaystyle\frac{\partial}{\partial w}\right)^{k}(w-1)^{m}\Big{|}_{w=1% }=0

for $k\neq m$ , we deduce that $\Delta^{m}c_{n}(\alpha,f,r)$ is null for $m>n$ . Otherwise, if $m\leq n$ , application of the Leibniz rule for differentiation yields

\Delta^{m}c_{n}(\alpha,f,r)=\displaystyle\frac{4^{n}}{n!r^{2n}}\displaystyle% \binom{n}{m}m!\left(\displaystyle\frac{\partial}{\partial w}\right)^{n-m}\left% [w^{n+\alpha}\left(\displaystyle\frac{\partial}{\partial w}\right)^{n}f\left(r% \sqrt{w}\right)\right]\Big{|}_{w=1}

which leads us at (12) and the proof of our lemma is completed. ∎

Our first main result can be read as follows.

Theorem 3.

Let $f$ belong to $E_{K}$ and let $W_{\alpha}f$ be defined by (1). For any $m\in\mathbb{N}_{0}$ the following relation

\left(\displaystyle\frac{\partial}{\partial r}\right)^{m}W_{\alpha}(f;r,t)\sim% \sum_{n=0}^{\infty}c_{n}^{[m]}(\alpha,f,r)t^{n}\quad(t\to 0^{+}),

(13)

holds, where

	$\displaystyle c_{n}^{[m]}(\alpha,f,r)$	$\displaystyle=\displaystyle\frac{1}{n!}\left(\displaystyle\frac{2}{r}\right)^{% 2n+m}\sum_{j\geq 0}\displaystyle\frac{(2j)!}{4^{j}j!}\displaystyle\binom{m}{2j% }\left(\displaystyle\frac{\partial}{\partial w}\right)^{n}$
		$\displaystyle\times\left[w^{n+m-j+\alpha}\left(\displaystyle\frac{\partial}{% \partial w}\right)^{n+m-j}f\left(r\sqrt{w}\right)\right]\Big{\|}_{w=1}.$		(14)

Proof.

Let $m\in\mathbb{N}_{0}$ be fixed. Concatenating the conclusions of Lemmas 1 and 2, we can write

		$\displaystyle\quad\ \left(\displaystyle\frac{\partial}{\partial r}\right)^{m}W% _{\alpha}(f;r,t)$
		$\displaystyle\sim\displaystyle\frac{1}{2^{m}}\sum_{j\geq 0}\displaystyle\frac{% (2j)!}{j!}\displaystyle\binom{m}{2j}r^{m-2j}\sum_{n=m-j}^{\infty}\Delta^{m-j}c% _{n}(\alpha,f,r)t^{n-m+j}$
		$\displaystyle=\displaystyle\frac{1}{2^{m}}\sum_{j\geq 0}\displaystyle\frac{(2j% )!}{j!}\displaystyle\binom{m}{2j}r^{m-2j}\sum_{n=0}^{\infty}\Delta^{m-j}c_{n+m% -j}(\alpha,f,r)t^{n}\quad(t\to 0^{+}).$		(15)

Further, for each $n\in\mathbb{N}_{0}$ we process the coefficient of $t^{n}$ with the help of the relation (12) applied for $\Delta^{m-j}c_{n+m-j}(\alpha,f,r)$ .

	$\displaystyle\quad\ \displaystyle\frac{1}{2^{m}}\sum_{j\geq 0}\displaystyle% \frac{(2j)!}{j!}\displaystyle\binom{m}{2j}r^{m-2j}\Delta^{m-j}c_{n+m-j}(\alpha% ,f,r)$
	$\displaystyle=\displaystyle\frac{1}{2^{m}}\sum_{j\geq 0}\displaystyle\frac{(2j% )!}{j!}\displaystyle\binom{m}{2j}r^{m-2j}\displaystyle\frac{4^{n+m-j}}{n!r^{2(% n+m-j)}}$
	$\displaystyle\times\left(\displaystyle\frac{\partial}{\partial w}\right)^{n}% \left[w^{n+m-j+\alpha}\left(\displaystyle\frac{\partial}{\partial w}\right)^{n% +m-j}f\left(r\sqrt{w}\right)\right]\Big{\|}_{w=1}$
	$\displaystyle:=c_{n}^{[m]}(\alpha,f,r),$

see (3) Returning at (3), the statement (13) is proven. ∎

Remark. Choosing in (3) $m=0$ , the sum is reduced to a single term $(j=0)$ and $c_{n}^{[0]}(\alpha,f,r)$ coincides with $c_{n}(\alpha,f,r)$ defined at (11).

Next we focus on a special case concerning the function $f$ , namely we consider that it is an analytic function on an open set $D\subset[0,\infty)$ . Consequently, for any $x_{0}\in D$ one can write

f(x)=\sum_{n=0}^{\infty}a_{n}(x-x_{0})^{n}

for $x$ in a neighborhood of $x_{0}$ .

We mention that the above sum can be finite, see the case where $f$ is a polynomial.

Since $f$ is a linear combination of the monomials $e_{i}$ , $e_{i}(r)=r^{i}$ $(i\in\mathbb{N}_{0})$ , and $W_{\alpha}(f;\cdot,\cdot)$ is a linear operator with respect to $f$ , it is sufficient to consider the behavior of the operator on the functions $e_{i}$ .

To present our result, we first need to establish some identities that involve hypergeometric functions defined by (7) and (8).

Lemma 4.

Let $m\in\mathbb{N}_{0}$ and $x>m-1$ , $x\not\in\mathbb{Z}$ , arbitrarily chosen.

(i) For any $z$ , $|z|<1/4$ , the identity

\sum_{j\geq 0}\displaystyle\binom{2j}{j}\!\displaystyle\binom{x}{m-j}\!\!\left% (\!\displaystyle\binom{m}{2j}/\displaystyle\binom{m}{j}\!\right)z^{j}\!=\!% \displaystyle\binom{x}{m}\ _{2}F_{1}\left(\displaystyle\frac{-m}{2},% \displaystyle\frac{-m+1}{2};x\!-\!m\!+\!1;4z\!\right),

(16)

holds, where ${}_{2}F_{1}$ is given by (7).

(ii) For $z=1/4$ , the identity

\sum_{j\geq 0}\displaystyle\frac{(2j)!}{4^{j}j!}\displaystyle\binom{m}{2j}x^{% \underline{m-j}}=2^{-m}(2x)^{\underline{m}},

(17)

holds.

Proof.

(i) The sum of the left-hand side of (16) is finite and has terms only for integer values of $j$ satisfying $0\leq j\leq\lfloor m/2\rfloor$ , where $\lfloor\cdot\rfloor$ stands for the floor function. Using well-known basic formulas

w^{\underline{p}}=\displaystyle\frac{\Gamma(w+1)}{\Gamma(w-p+1)}=\displaystyle% \frac{w!}{(w-p)!},\quad w^{\overline{p}}=(-1)^{p}(-w)^{\underline{p}}=(w+p-1)^% {\underline{p}},

valid for any real or complex $w$ , we can write the next set of identities

	$\displaystyle\quad\ \displaystyle\binom{2j}{j}\displaystyle\binom{x}{m-j}\left% (\displaystyle\binom{m}{2j}/\displaystyle\binom{m}{j}\right)$
	$\displaystyle=\displaystyle\frac{m^{\underline{2j}}}{j!m!}x^{\underline{m-j}}=% \displaystyle\frac{(-m)^{\overline{2j}}}{j!m!}\,\displaystyle\frac{x^{% \underline{m}}}{(x-m+j)^{\underline{j}}}$
	$\displaystyle=\displaystyle\frac{x^{\underline{m}}}{m!}\,\displaystyle\frac{2^% {j}\left(-\displaystyle\frac{m}{2}\right)^{\overline{j}}2^{j}\left(-% \displaystyle\frac{m-1}{2}\right)^{\overline{j}}}{(x-m+1)^{\overline{j}}j!}.$

Since

\displaystyle\frac{x^{\underline{m}}}{m!}=\displaystyle\binom{x}{m},

from the above relations we obtain

\sum_{j\geq 0}\displaystyle\binom{2j}{j}\displaystyle\binom{x}{m-j}\left(% \displaystyle\binom{m}{2j}/\displaystyle\binom{m}{j}\right)z^{j}=\displaystyle% \binom{x}{m}\sum_{j\geq 0}\displaystyle\frac{\left(-\displaystyle\frac{m}{2}^{% \overline{j}}\right)\left(-\displaystyle\frac{m-1}{2}\right)^{\overline{j}}}{(% x-m+1)^{\overline{j}}}\,\displaystyle\frac{(4z)^{j}}{j!}.

Considering (7), the relation (16) is proved.

(ii) Choosing $z=1/4$ in (16) and using (8), this identity can thus be rewritten

\sum_{j\geq 0}\displaystyle\frac{(2j)!}{4^{j}j!}\displaystyle\binom{m}{2j}x^{% \underline{m-j}}=x^{\underline{m}}\,\displaystyle\frac{\Gamma(x-m+1)\Gamma% \left(x+\displaystyle\frac{1}{2}\right)}{\Gamma\left(x-\displaystyle\frac{m}{2% }+1\right)\Gamma\left(x-\displaystyle\frac{m}{2}+\displaystyle\frac{1}{2}% \right)}.

(18)

The Legendre duplication formula for the Gamma function

\Gamma(2w)=\displaystyle\frac{1}{\sqrt{\pi}}\,2^{2w-1}\Gamma(w)\Gamma\left(w+% \displaystyle\frac{1}{2}\right),\ 2w\neq 0,-1,-2,\ldots,

see, e.g., [7, Eq. 5.5.5], allows us to write

	$\displaystyle\displaystyle\frac{\Gamma(x-m+1)\Gamma\left(x+\displaystyle\frac{% 1}{2}\right)}{\Gamma\left(x-\displaystyle\frac{m}{2}+1\right)\Gamma\left(x-% \displaystyle\frac{m}{2}+\displaystyle\frac{1}{2}\right)}$	$\displaystyle=\displaystyle\frac{\Gamma(x-m+1)\Gamma(2x+1)2^{-2x}}{\Gamma(x+1)% \Gamma(2x-m+1)2^{-2x+m}}$
		$\displaystyle=2^{-m}\,\displaystyle\frac{(2x)^{\underline{m}}}{x^{\underline{m% }}}.$

Returning at (18), identity (17) is obtained and our lemma is completely motivated. ∎

In the light of those discussed before, exploring the case when $f$ is an analytic function, we present

Theorem 5.

For any $m\in\mathbb{N}_{0}$ and $n\in\mathbb{N}_{0}$ , the following identity

c_{n}^{[m]}(\alpha,e_{i},r)=\left(\displaystyle\frac{\partial}{\partial r}% \right)^{m}c_{n}(\alpha,e_{i},r)

(19)

holds, where $c_{n}^{[m]}(\alpha,\cdot,r)$ and $c_{n}(\alpha,\cdot,r)$ are respectively defined by (3) and (11).

Proof.

For $m=0$ we have highlighted this identity, see Remark. Further we consider $m\in\mathbb{N}$ .

		$\displaystyle\quad\ c_{n}^{[m]}(\alpha,e_{i},r)$
		$\displaystyle=\displaystyle\frac{1}{n!}\left(\displaystyle\frac{2}{r}\right)^{% 2n+m}\sum_{j\geq 0}\displaystyle\frac{(2j)!}{4^{j}j!}\displaystyle\binom{m}{2j% }\left(\displaystyle\frac{\partial}{\partial w}\right)^{n}\left[w^{n+m-j+% \alpha}\left(\displaystyle\frac{\partial}{\partial w}\right)^{n+m-j}r^{i}w^{i/% 2}\right]\Big{\|}_{w=1}$
		$\displaystyle=\displaystyle\frac{r^{i}}{n!}\left(\displaystyle\frac{2}{r}% \right)^{2n+m}\sum_{j\geq 0}\displaystyle\frac{(2j)!}{4^{j}j!}\displaystyle% \binom{m}{2j}\left(\displaystyle\frac{\partial}{\partial w}\right)^{n}\left[% \left(\displaystyle\frac{i}{2}\right)^{\underline{n+m-j}}w^{i/2+\alpha}\right]% \Big{\|}_{w=1}$
		$\displaystyle=\displaystyle\frac{r^{i}}{n!}\left(\displaystyle\frac{2}{r}% \right)^{2n+m}\sum_{j\geq 0}\displaystyle\frac{(2j)!}{4^{j}j!}\displaystyle% \binom{m}{2j}\left(\displaystyle\frac{i}{2}\right)^{\underline{n+m-j}}\left(% \displaystyle\frac{i}{2}+\alpha\right)^{\underline{n}}.$		(20)

On the other hand,

		$\displaystyle\quad\ \left(\displaystyle\frac{\partial}{\partial r}\right)^{m}c% _{n}(\alpha,e_{i},r)$
		$\displaystyle=\displaystyle\frac{4^{n}}{n!}\left[\left(\displaystyle\frac{% \partial}{\partial r}\right)^{m}\,\displaystyle\frac{r^{i}}{r^{2n}}\right]% \left(\displaystyle\frac{\partial}{\partial w}\right)^{n}\left[w^{n+\alpha}% \left(\displaystyle\frac{\partial}{\partial w}\right)^{n}w^{i/2}\right]\Big{\|}% _{w=1}$
		$\displaystyle=\displaystyle\frac{4^{n}}{n!}(i-2n)^{\underline{m}}r^{i-2n-m}% \left(\displaystyle\frac{\partial}{\partial w}\right)^{n}\left[w^{n+\alpha}% \left(\displaystyle\frac{i}{2}\right)^{\underline{n}}w^{i/2-n}\right]\Big{\|}_{% w=1}$
		$\displaystyle=\displaystyle\frac{r^{i}}{n!}\,\displaystyle\frac{2^{2n}}{i^{2n+% m}}(i-2n)^{\underline{m}}\left(\displaystyle\frac{i}{2}\right)^{\underline{n}}% \left(\displaystyle\frac{i}{2}+\alpha\right)^{\underline{n}}.$		(21)

To match the expressions (3) and (3) it remains to be shown that

2^{m}\sum_{j\geq 0}\displaystyle\frac{(2j)!}{4^{j}j!}\displaystyle\binom{m}{2j% }\left(\displaystyle\frac{i}{2}-n\right)^{\underline{m-j}}=(i-2n)^{\underline{% m}}.

(22)

We have taken into account that

\left(\displaystyle\frac{i}{2}\right)^{\underline{n+m-j}}/\left(\displaystyle% \frac{i}{2}\right)^{\underline{n}}=\left(\displaystyle\frac{i}{2}-n\right)^{% \underline{m-j}}.

Now we turn to Lemma 4. By choosing in the identity (17) $x:=i/2-n$ , we obtain exactly (22). Thus relation (19) is proven. ∎

References

[1] Abel, U., Agratini, O., On Wachnicki’s generalization of the Gauss-Weierstrass integral, In: Recent Advances in Mathematical Analysis (Eds. Anna Maria Candela, Mirella Cappelletti Montano, Elisabeta Mangino), Springer, 2022.
[2] Abramowitz, M., Stegun, I.A. (Eds.), Handbook of Mathematical Functions with Formulas, Graphs and Mathematical Tables, National Bureau of Standards Applied Mathematics Series 55, Issued June 1964, Tenth Printing, December 1972, with corrections.
[3] Bragg, L.R., The radial heat polynomials and related functions, Trans. Amer. Math. Soc., 119(1965), 270–290.
[4] Krech, G., Krech, I., On some bivariate Gauss-Weierstrass operators, Constr. Math. Anal., 2(2019), No. 2, 57–63.
[5] Zayed, A.I., Handbook of Function and Generalized Function Transformations, $1^{st}$ Edition, CRC Press, London, 1996.
[6] Wachnicki, E., On a Gauss–Weierstrass generalized integral, Rocznik Naukowo–Dydaktyczny Akademii Pedagogicznej W Krakowie, Prace Matematyczne, 17(2000), 251–263.
[7] Digital Library of Mathematical Functions, https://dlmf.nist.gov
[8] https://functions.wolfram.com/Bessel-TypeFunctions/BesselI

Simultaneous approximation by Gauss–Weierstrass–Wachnicki operators

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Simultaneous approximation by Gauss–Weierstrass–Wachnicki operators

Abstract.

1. Introduction

2. The operators

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Lemma 1.

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Lemma 2.

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Theorem 3.

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Lemma 4.

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Theorem 5.

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