Asymptotic expansions for certain exponential-type operators connected with \(2x^{3/2}\)

Abstract

In the present paper, we consider the complete asymptotic expansion of certain exponential-type operators connected with \(2x^{3/2}\). Also, a modification of such exponential-type operators is provided, which preserve the function \(\mathrm{e}^{Ax}\).

Introduction

Ismail and May [5, (3.16)] defined the operators

$$\begin{aligned} (T_nf)(x)&= \int _0^\infty k_n(x,t)f(t)\,\mathrm{d}t \nonumber \\&= \mathrm{e}^{-n\sqrt{x}}\left\{ n\int _{0}^\infty \mathrm{e}^{-nt/\sqrt{x}}t^{-1/2} I_1(2n \sqrt{t})f(t)\,\mathrm{d}t+f(0)\right\} , \end{aligned}$$
(1)

where \(x\in (0,\infty )\). The kernel is given by

$$\begin{aligned} k_n(x,t)=\mathrm{e}^{-n\sqrt{x}}\left( n \mathrm{e}^{-nt/\sqrt{x}}t^{-1/2} I_1(2n\sqrt{t})+\delta (t)\right) , \end{aligned}$$
(2)

where \(\delta (t)\) is the Dirac delta function and \(I_1\) denotes the modified Bessel function of first kind defined by

$$\begin{aligned} I_1(z)=\sum _{k=0}^\infty \frac{\left( \frac{z}{2}\right) ^{1+2k}}{k!\varGamma (k+2)}. \end{aligned}$$

The operators (1) are exponential-type operators as the kernels \(k_n(x,t)\) of the operators \(T_n\) satisfy the partial differential equation

$$\begin{aligned} \frac{\partial }{\partial x}[k_n(x,t)]=\frac{n(t-x)}{p(x)}k_n(x,t), \end{aligned}$$
(3)

with \(p(x)=2x^{3/2}.\) The operators (1) reproduce constant and linear functions.

Note that the operators (1) can alternatively be written in the form

$$\begin{aligned} \left( T_{n}f\right) \left( x\right)= & {} \frac{n}{\sqrt{x}}\sum _{k=1}^{\infty }s_{k}\left( n\sqrt{x}\right) \int _{0}^{\infty }s_{k-1}\left( \frac{nt}{ \sqrt{x}}\right) f\left( t\right) \,\mathrm{d}t\nonumber \\&+\,\mathrm{e}^{-n\sqrt{x}}f\left( 0\right) , \end{aligned}$$
(4)

where \(x\in \left( 0,\infty \right)\) and

$$\begin{aligned} s_{k}\left( x\right) =\mathrm{e}^{-x}\frac{x^{k}}{k!}. \end{aligned}$$

We observe that the operators \(T_{n}\) are closely related to the well-known Phillips operators given by

$$\begin{aligned} \left( P_{n}f\right) \left( x\right)= & {} n\sum _{k=0}^{\infty }s_{k}\left( nx\right) \int _{0}^{\infty }s_{k-1}\left( nt\right) f\left( t\right) \,\mathrm{d}t\nonumber \\&+\,\mathrm{e}^{-nx}f\left( 0\right) . \end{aligned}$$
(5)

Obviously, we have \(\left( P_{n/\sqrt{x}}f\right) \left( x\right) =\left( T_{n}f\right) \left( x\right)\), but unlike the operators \(T_n\), Phillips operators are not exponential type operators.

The operators \(T_n\) individually have not been studied by researchers much, due to its complicated behaviour. Very recently, Abel and Gupta [1] studied the rate of convergence of these operators on functions of bounded variation. Also, Gupta [2] and Gupta et al. [3] resp. gave some direct results in ordinary and simultaneous approximation. In the present article, we establish the complete asymptotic expansion of the operators \(T_n\) as n tends to infinity.

Asymptotic expansion

In the following, we use the falling factorial given by \(z^{\underline{0}}=1\) , and \(z^{\underline{m}}=z\left( z-1\right) \left( z-2\right) \cdots \left( z-m+1\right)\), for \(m\in \mathbb {N}\).

Firstly, we derive a concise representation of the r-th order moment \(\left( T_{n} e_{r}\right) (x)\) where \(e_{r}\left( t\right) =t^{r},\) \(r\in \mathbb {N}_{0}\), denote the monomials. Obviously, we have \(\left( T_{n} e_{0}\right) (x) =1\).

Lemma 1

For \(r\in \mathbb {N}\), the r-th order moment of the Ismail–May operators \(T_{n}\) has the representation

$$\begin{aligned} \left( T_{n} e_{r}\right) (x) =\sum _{k=0}^{r-1}\frac{1}{n^{k}}{\left( {\begin{array}{c}{r-1}\\ {k }\end{array}}\right) }r^{\underline{k}}x^{r-\frac{k}{2}}. \end{aligned}$$

Proof

For \(r\in \mathbb {N}\), we have

$$\begin{aligned}&\left( T_{n} e_{r}\right) (x)\\&\quad =\mathrm{e}^{-n\sqrt{x}}\left\{ n\int _{0}^{\infty }\mathrm{e}^{-nt/\sqrt{x}}t^{-1/2}I_{1}\left( 2n\sqrt{t}\right) e_{r}\left( t\right) \,\mathrm{d}t +e_{r}(0)\right\} \\&\quad =\mathrm{e}^{-n\sqrt{x}}\left\{ n\sum _{v=0}^{\infty }\frac{n^{2v+1}}{v!(v+1)!} \int _{0}^{\infty }\mathrm{e}^{-nt/\sqrt{x}}t^{v+r}\,\mathrm{d}t+\delta _{r,0}\right\} \\&\quad =\mathrm{e}^{-n\sqrt{x}}\sum _{v=0}^{\infty }\frac{n^{2v+2}\left( \frac{\sqrt{x}}{n} \right) ^{v+r+1}}{v!}\frac{(v+r)!}{(v+1)!} \\&\quad =\mathrm{e}^{-n\sqrt{x}}n^{2}\left( \frac{\sqrt{x}}{n}\right) ^{r+1}\left[ \left( \frac{\partial }{\partial z}\right) ^{r-1}z^{r}\exp \left( n^{2}\frac{\sqrt{x }}{n}z\right) \right] _{z=1}. \end{aligned}$$

Application of the Leibniz rule for differentiation yields

$$\begin{aligned}&\left( T_{n} e_{r}\right) (x)\\&\quad =\mathrm{e}^{-n\sqrt{x}}\frac{x^{(r+1)/2}}{n^{r-1}} \sum _{k=0}^{r-1}{\left( {\begin{array}{c}{r-1}\\ {k}\end{array}}\right) }r^{\underline{j}}z^{r-k}\\&\qquad \cdot \, \left( n \sqrt{x}\right) ^{r-1-k}\mathrm{e}^{n\sqrt{x}z}\biggr |_{z=1} \\&\quad =r!\sum _{k=0}^{r-1}{\left( {\begin{array}{c}{r-1}\\ {k}\end{array}}\right) }\frac{x^{r-\frac{k}{2}}}{n^{k}\left( r-k\right) !} \end{aligned}$$

which is the desired formula. \(\square\)

Now we turn to the central moments \(\left( T_{n} \psi _{x}^{s}\right) (x)\). For real x, we define \(\psi _{x}\left( t\right) =t-x\).

Lemma 2

For \(s=1,2,\ldots\), the central moment of the Ismail-May operator \(T_{n}\) possesses the representation

$$\begin{aligned} \left( T_{n} \psi _{x}^{s}\right) (x) =\sum _{k=\left\lfloor \left( s+1\right) /2\right\rfloor }^{s-1}n^{-k}{\left( {\begin{array}{c}{s}\\ {k}\end{array}}\right) }k^{\underline{s-k}}\left( k-1\right) ^{\underline{2k-s}}x^{s-\frac{k}{2}}. \end{aligned}$$

Proof

For \(s\in \mathbb {N}\), we have

$$\begin{aligned} \left( T_{n}\psi _{x}^{s}\right) (x)= & {} \sum _{r=0}^{s}{\left( {\begin{array}{c}{s}\\ {r}\end{array}}\right) } \left( -x\right) ^{s-r}\sum _{k=0}^{r-1}\frac{1}{n^{k}}{\left( {\begin{array}{c}{r-1}\\ {k}\end{array}}\right) }r^{ \underline{k}}x^{r-\frac{k}{2}} \\= & {} \sum _{k=0}^{s-1}n^{-k}\sum _{r=k+1}^{s}\left( -x\right) ^{s-r}{\left( {\begin{array}{c}{s}\\ {r}\end{array}}\right) }{\left( {\begin{array}{c}{r}\\ {k}\end{array}}\right) }(r-1)^{\underline{k}}x^{r-\frac{k}{2}} . \end{aligned}$$

Using the binomial identity \({\left( {\begin{array}{c}{s}\\ {r}\end{array}}\right) }{\left( {\begin{array}{c}{r}\\ {k}\end{array}}\right) }={\left( {\begin{array}{c}{s} \\ {k}\end{array}}\right) }{\left( {\begin{array}{c}{s-k}\\ {r-k}\end{array}}\right) }\) we obtain

$$\begin{aligned}&\left( T_{n}\psi _{x}^{s}\right) (x)\\&\quad =\sum _{k=0}^{s-1}n^{-k}{\left( {\begin{array}{c}{s}\\ {k }\end{array}}\right) }x^{s-\frac{k}{2}}\sum _{r=k+1}^{s}\left( -1\right) ^{s-r}{\left( {\begin{array}{c}{s-k}\\ { r-k}\end{array}}\right) }(r-1)^{\underline{k}} \\&\quad =\sum _{k=0}^{s-1}n^{-k}{\left( {\begin{array}{c}{s}\\ {k}\end{array}}\right) }x^{s-\frac{k}{2} }\sum _{r=1}^{s-k-r}\left( -1\right) ^{s-k}{\left( {\begin{array}{c}{s-k}\\ {r}\end{array}}\right) }\\&\qquad \left[ \left( \frac{\partial }{\partial z}\right) ^{k}\left( z^{r+k-1}\right) \right] \biggr |_{z=1} \\&\quad =\sum _{k=0}^{s-1}n^{-k}{\left( {\begin{array}{c}{s}\\ {k}\end{array}}\right) }x^{s-\frac{k}{2}}\left[ \left( \frac{\partial }{\partial z}\right) ^{k}z^{k-1}\left( z-1\right) ^{s-k} \right] \biggr |_{z=1} \end{aligned}$$

and the desired formula follows by an application of the Leibniz rule for differentiation. \(\square\)

Lemma 3

The central moments of the operators \(T_n\) satisfy

$$\begin{aligned} \left( T_{n}\psi _{x}^{s}\right) \left( x\right) =O\left( n^{-\left\lfloor \left( s+1\right) /2\right\rfloor }\right) \qquad \left( n\rightarrow \infty \right) . \end{aligned}$$

Let \(C_{\gamma }\left[ 0,\infty \right)\) be the class of continuous functions f on \(\left[ 0,\infty \right)\) satisfying the exponential growth condition \({f}\left( t\right) =O\left( \mathrm{e}^{\gamma t}\right)\) as \(t\rightarrow \infty\), for some \(\gamma >0\).

The next theorem is the main result of this section.

Theorem 1

Let \(q\in \mathbb {N}\) and \(x\in (0,\infty )\). For each function \(f\in C_{\gamma }\left[ 0,\infty \right)\), which has a derivative of order 2q at the point x, the operators \(T_{n}\) possess the asymptotic expansion

$$\begin{aligned} \left( T_{n}f\right) (x)&= f(x)\\&+\,\sum _{k=1}^{q}\frac{1}{k!n^{k}}\sum _{s=1}^{k}{ \left( {\begin{array}{c}{k}\\ {s}\end{array}}\right) }(k-1)^{\underline{k-s}}f^{(k+s)}(x)x^{s+\frac{k}{2} }+o(n^{-q}) \end{aligned}$$

as \(n\rightarrow \infty\).

Proof

We apply a general result by Sikkema [7]. In view of the localization result (Proposition 1), we can assume that the function f has (at most) polynomial growth as the variable tends to infinity. Using Lemma 2, we obtain

$$\begin{aligned} \left( T_{n}f\right) (x)&= f(x)+\sum _{s=1}^{2q}\frac{f^{(s)}(x)}{s!} \sum _{k=\left\lfloor \left( s+1\right) /2\right\rfloor }^{s-1}n^{-k}{ \left( {\begin{array}{c}{s}\\ {k}\end{array}}\right) }k^{\underline{s-k}}\\&\quad \left( k-1\right) ^{\underline{2k-s} }x^{s-\frac{k}{2}}+o(n^{-q}) \\&= f(x)+\sum _{k=1}^{q}n^{-k}\sum _{s=k+1}^{2k}\frac{f^{(s)}(x)}{s!}{\left( {\begin{array}{c}{s} \\ {k}\end{array}}\right) }k^{\underline{s-k}}\\&\quad (k-1)^{\underline{2k-s}}x^{s-\frac{k}{2}}+o(n^{-q})\\&= f(x)+\sum _{k=1}^{q}n^{-k}\sum _{s=1}^{k}\frac{f^{(k+s)}(x)}{k!s!}k^{ \underline{s}}\\&\quad (k-1)^{\underline{k-s}}x^{s+\frac{k}{2}}+o(n^{-q}) \\&= f(x)+\sum _{k=1}^{q}\frac{1}{k!n^{k}}\sum _{s=1}^{k}{\left( {\begin{array}{c}{k}\\ {s}\end{array}}\right) }\\&\quad (k-1)^{\underline{k-s}}f^{(k+s)}(x)x^{s+\frac{k}{2}}+o\left( n^{-q}\right) \end{aligned}$$

as \(n\rightarrow \infty\). \(\square\)

More explicitly, we have

$$\begin{aligned} \left( T_{n}f\right) \left( x\right)&= f\left( x\right) +\frac{x^{3/2}}{n} f^{\left( 2\right) }\left( x\right) \\&\quad+\,\frac{x^{2}}{n^2} \left( f^{\left( 3 \right) }\left( x\right) + \frac{x}{2} f^{\left( 4\right) }\left( x\right) \right) \\&\quad+\,\frac{x^{5/2}}{n^3} \left( f^{\left( 4\right) }\left( x\right) +x f^{\left( 5\right) }\left( x\right) +\frac{x^{2}}{6} f^{\left( 6\right) }\left( x\right) \right) \\&\quad+\,o\left( n^{-3}\right) \end{aligned}$$

as \(n\rightarrow \infty\).

A direct corollary is the following Voronovskaja-type formula.

Corollary 1

Let \(x\in \mathbb {R}\). For each function \(f\in C_{\gamma }\left[ 0,\infty \right)\), which has a second derivative at the point x, we have the asymptotic relation

$$\begin{aligned} \lim _{n\rightarrow \infty }n\left( \left( T_{n}f\right) \left( x\right) -f\left( x\right) \right) =x^{3/2}f^{\prime \prime }\left( x\right) . \end{aligned}$$

Now we prove a localization result which is of use in the proof of Theorem  1. It is of interest in its own. In what follows let \(\exp\) denote the exponential function. For real \(\gamma\), we write \(\exp _{\gamma }\left( t\right) :=\mathrm{e}^{\gamma t}\).

Proposition 1

(Localisation theorem) Let \(\gamma ,\delta >0\) and fix \(x>0\). If \(f\in C_{\gamma }\left[ 0,\infty \right)\) vanishes on the interval \(\left( x-\delta ,x+\delta \right) \cap \left( 0,\infty \right)\) then

$$\begin{aligned} \left( T_{n}f\right) \left( x\right) =O\left( n^{-m}\right) \qquad \left( n\rightarrow \infty \right) , \end{aligned}$$

for arbitrary large \(m>0\).

Proof

Let \(m\in {N}\). For a certain positive constant C, we have

$$\begin{aligned} \left| \left( T_{n}f\right) \left( x\right) \right| &\le \int _{|t-x|\ge \delta }k_{n}(x,t)\mathrm{e}^{\gamma t}\,\mathrm{d}t \\& \le C\delta ^{-m}\int _{0}^{\infty }k_{n}(x,t)\left| t-x\right| ^{m}\mathrm{e}^{\gamma t}\,\mathrm{d}t. \end{aligned}$$

Application of the Schwarz inequality yields

$$\begin{aligned} \left| \left( T_{n}f\right) \left( x\right) \right| \le C\delta ^{-m} \sqrt{\left( T_{n} \exp _{2\gamma }\right) \left( x\right) } \sqrt{\left( T_{n}\psi _{x}^{2m}\right) \left( x\right) }. \end{aligned}$$

The first root is finite, by [5, Prop. 2.5], if n is sufficiently large. The second root satisfies

$$\begin{aligned} \sqrt{\left( T_{n}\psi _{x}^{2m}\right) \left( x\right) }=O\left( n^{-m}\right) \qquad \left( n\rightarrow \infty \right) , \end{aligned}$$

by Lemma 3. This completes the proof. \(\square\)

Preservation of the exponential function \(\mathrm{e}^{Ax}\)

King [6] considered a modification of the Bernstein polynomials so as to preserve the function \(e_{2}\). Also the moments of several operators have been discussed in [4]. We start with the following modified form of operators (1) with \(\left( \widetilde{T} _{n}f\right) (x)=\left( T_{n}f\right) (a_{n}(x))\), i.e.,

$$\begin{aligned} (\widetilde{T}_{n}f)(x)&= \int _{0}^{\infty }k_{n}(a_{n}(x),t)f(t)\,\mathrm{d}t \nonumber \\&= \mathrm{e}^{-n\sqrt{a_{n}(x)}}\left\{ n\int _{0}^{\infty }\mathrm{e}^{-nt/\sqrt{a_{n}(x)} }t^{-1/2}I_{1}(2n\sqrt{t})f(t)\,\mathrm{d}t+f(0)\right\} , \end{aligned}$$
(6)

such that

$$\begin{aligned} \mathrm{e}^{Ax}=\left( \widetilde{T}_{n}\exp _{A}\right) (x)=\int _{0}^{\infty }k_{n}(a_{n}(x),t)\mathrm{e}^{At}\,\mathrm{d}t, \end{aligned}$$

which is the case if

$$\begin{aligned} a_{n}(x)=\frac{x(2n^{2}+A^{2}x)+Ax\sqrt{A^{2}x^{2}+4xn^{2}}}{2n^{2}}. \end{aligned}$$

This modified form of operators preserves constants as well as the exponential function \(\exp _{A}\), but loose to preserve the linear functions. We have the following limits:

$$\begin{aligned} \lim _{n\rightarrow \infty }a_{n}(x)=x=\lim _{A\rightarrow 0}a_{n}(x). \end{aligned}$$

Also, we have

$$\begin{aligned} \lim _{n\rightarrow \infty }a_{n}^{\prime }(x)=1. \end{aligned}$$

After simple computations, we derive the following representation.

Lemma 4

For real numbers C, it holds

$$\begin{aligned} \left( \widetilde{T}_{n}\exp _{C}\right) (x)=\exp \left( \frac{nCa_{n}(x)}{ n-C\sqrt{a_{n}(x)}}\right) . \end{aligned}$$

We may observe here that the r-th order moment \(\widetilde{T}_{n}e_{r}\) is given by

$$\begin{aligned} \left( \widetilde{T}_{n}e_{r}\right) (x)=\left[ \left( \frac{\partial }{ \partial C}\right) ^{r}\exp \left( \frac{nCa_{n}(x)}{n-C\sqrt{a_{n}(x)}} \right) \right] _{C=0}. \end{aligned}$$

In particular, we have

$$\begin{aligned}&\left( \widetilde{T}_{n}e_{0}\right) \left( x\right) =1,\quad \left( \widetilde{T}_{n}e_{1}\right) \left( x\right) =a_{n}(x),\\&\left( \widetilde{T}_{n}e_{2}\right) \left( x\right) =[a_{n}\left( x\right) ]^{2}+ \frac{2[a_{n}\left( x\right) ]^{3/2}}{n}, \\&\left( \widetilde{T}_{n}e_{3}\right) \left( x\right) =[a_{n}(x)]^{3}+ \frac{6[a_{n}(x)]^{5/2}}{n}+\frac{6[a_{n}(x)]^{2}}{n^{2}}, \\&\left( \widetilde{T}_{n}e_{4}\right) \left( x\right) =[a_{n}(x)]^{4}+ \frac{12[a_{n}(x)]^{7/2}}{n}\\&\qquad +\,\frac{36[a_{n}(x)]^{3}}{n^{2}}+\frac{ 24[a_{n}(x)]^{5/2}}{n^{3}}. \end{aligned}$$

Remark 1

The central moments of the operators \(\widetilde{T}_{n}\) have the representation

$$\begin{aligned} \left( \widetilde{T}_{n}\psi _{x}^{s}\right) \left( x\right) =\left[ \left( \frac{\partial }{\partial C}\right) ^{s}\exp \left( -Cx+\frac{nCa_{n}(x)}{n-C \sqrt{a_{n}(x)}}\right) \right] _{C=0}. \end{aligned}$$

Direct computation yields

$$\begin{aligned}&\lim \limits _{n\rightarrow \infty }n\left( \widetilde{T}_{n}\psi _{x}\right) \left( x\right) =Ax^{3/2}, \\&\lim \limits _{n\rightarrow \infty }n\left( \widetilde{T}_{n}\psi _{x}^{2}\right) \left( x\right) =2x^{3/2}, \\&\lim \limits _{n\rightarrow \infty }n^{2}\left( \widetilde{T}_{n}\psi _{x}^{3}\right) \left( x\right) =6x^{2}(1+Ax), \\&\lim _{n\rightarrow \infty }n^{2}\left( \widetilde{T}_{n}\psi _{x}^{4}\right) \left( x\right) =12x^{3}. \end{aligned}$$

Hence, for \(s=0,1,2,3,4\), the central moments satisfy

$$\begin{aligned} \left( \widetilde{T}_{n}\psi _{x}^{s}\right) \left( x\right) =O_{x}(n^{-[(s+1)/2]}),\ \ n\rightarrow \infty . \end{aligned}$$

The proof of the next lemma uses standard techniques.

Corollary 2

Let \(\gamma\) and \(\delta\) be any two positive real numbers and \([a,b]\subset (0,\infty )\) be any bounded interval. Then, for any \(m>0\) there exists a constant M depending on m only such that

$$\begin{aligned} \left( \int _{|t-x|\ge \delta }k_{n}(a_{n}(x),t)\mathrm{e}^{\gamma t}\,\mathrm{d}t\right) \le Mn^{-m}, \end{aligned}$$

where \(\Vert .\Vert\) is the sup-norm over [ab].

By Remark 1 and Corollary 2, we have the following asymptotic formula for the modified operators \(\widetilde{T}_{n}\).

Theorem 2

Let \(f\in C_{\gamma }(0,\infty )\) for some \(\gamma >0.\) If \(f^{\prime \prime }\) exists at a point \(x\in (0,\infty )\), then we have

$$\begin{aligned} \lim _{n\rightarrow \infty }n\left( \left( \widetilde{T}_{n}f\right) (x)-f(x)\right) =Ax^{3/2}f^{\prime }(x)+x^{3/2}f^{\prime \prime }(x). \end{aligned}$$

Without a proof, we state the following formula for the first derivative.

Theorem 3

Let \(f\in C_{\gamma }(0,\infty )\) admitting the derivative of 3 -rd order at a fixed point \(x\in (0,\infty )\), then we have

$$\begin{aligned} \lim _{n\rightarrow \infty }n\left( \left( \widetilde{T}_{n}f\right) ^{\prime }(x)-f^{\prime }(x)\right)= & {} \frac{3Ax^{1/2}}{2}f^{(1)}(x)\\&+\,\left( \frac{3}{ 2}x^{1/2}+Ax^{3/2}\right) f^{(2)}(x) \\&+\,x^{3/2}f^{(3)}(x). \end{aligned}$$

The above result can be derived by using the fact that the kernel of the operator \(\left( \widetilde{T}_nf\right) (x)\) satisfies the partial differential equation

$$\begin{aligned} \frac{\partial }{\partial x} k_n(a_n(x),t)= & {} \frac{n(t-a_n(x))}{ 2[a_n(x)]^{3/2}}k_n(a_n(x),t)\\&\quad \left( 1+\frac{A^2x}{n^2} +\frac{A^3 x^2+3n^2 A x }{n^2 \sqrt{A^2x^2+4n^2x}}\right) . \end{aligned}$$

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Abel, U., Gupta, V. & Kushnirevych, V. Asymptotic expansions for certain exponential-type operators connected with \(2x^{3/2}\). Math Sci (2021). https://doi.org/10.1007/s40096-021-00382-9

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Keywords

  • Exponential-type operators
  • Asymptotic expansion
  • Modified Bessel function
  • Degree of approximation

Mathematics Subject Classification

  • 41A36
  • 41A25