The first absolute moment for some operators
Ovidiu T. Pop\(^\S \) Petru I. Braica\(^\ast \)
December 15, 2009.
\(^\S \)National College “Mihai Eminescu", 5 Mihai Eminescu Street, 440014 Satu Mare, Romania, e-mail: ovidiutiberiu@yahoo.com
\(^\ast \) Secondary School “Grigore Moisil", 1 Mileniului Street, 440037 Satu Mare, Romania, e-mail: petrubr@yahoo.com
In this paper we will determinate the first absolute moments for Bernstein, Szász-Mirakjan, Bleimann-Butzer-Hahn, Meyer-König and Zeller operators. For the Szász-Mirakjan operators we give some properties with the absolute moment of high order.
MSC. 41A10, 41A10, 41A36
Keywords. Bernstein’s polynomials, Szász-Mirakjan operators, Bleimann-Butzer-Hahn operators, Meyer-König and Zeller operators, absolute moment of high order
1 Preliminaries
Let \(\mathbb N\) be the set of positive integers and \(\mathbb N_{0}=\mathbb N\cup \{ 0\} \). For any \(m \in \mathbb N\), let \(B_m:C([0,1])\to C([0,1])\) be the Bernstein operators, defined by
\begin{equation} \label{1.1} (B_m f)(x)=\sum \limits _{k=0}^{m}p_{m,k}(x)f\left(\tfrac {k}{m}\right), \end{equation}
1.1
where \(p_{m,k}(x)\) are the fundamental Bernstein’s polynomials, given by
\begin{equation} \label{1.2} p_{m,k}(x)=\tbinom {m}{k}x^k(1-x)^{m-k}, \end{equation}
1.2
for any \(x\in [0,1]\) and any \(k\in \{ 0,1,\ldots ,m\} \) (see [3] or [11]).
For any \(m\in \mathbb N\), let \(S_m:C_2([0,+\infty ))\to C([0,+\infty ))\) be the Szász-Mirakjan operators (see [4], [7], [11] or [12]), defined by
\begin{equation} \label{1.3} (S_m f)(x)={\rm e}^{-mx}\sum \limits _{k=0}^{\infty }\tfrac {(mx)^k}{k!}f\left(\tfrac {k}{m}\right), \end{equation}
1.3
for any \(x\in [0,+\infty )\).
The operators \(Z_m:B([0,1])\to C([0,1])\) defined by
\begin{equation} \label{1.4} (Z_m f)(x)=\left(1-x\right)^{m+1} \sum \limits _{k=0}^{\infty }\tbinom {m+k}{k}x^{k}f\left(\tfrac {k}{m+k}\right), \end{equation}
1.4
for any \(x\in [0,1)\) and \((Z_m f)(1)=f(1)\), \(m \in \mathbb N\), are called Meyer-König and Zeller operators (see [6] or [11]).
For \(m \in \mathbb N\), let the operators \(L_{m}:C_{B}([0,+\infty ))\to C_{B}([0,+\infty ))\), defined by
\begin{equation} \label{1.5} (L_{m}f)(x)=\left(1+x\right)^{-m}\sum _{k=0}^{m} \tbinom {m}{k}x^{k}f\left(\tfrac {k}{m+1-k}\right), \end{equation}
1.5
for any \(x\in [0,\infty )\). These operators are called Bleimann-Butzer-Hahn operators (see [2] or [11]).
In what follows, let \(I\subset \mathbb R\) be an interval. We recall that the functions \(\varphi _{x},\psi _{x}\! :\! I\! \to \! \mathbb R\) are defined by
\[ \varphi _{x}(t)=\left|t-x\right|,\ \ \ \psi _{x}(t)=t-x, \]
for any \((x,t)\in I\times I\). For a sequence of operators \((L_m)_{m\geq 1}\), define \(T_{m,i}\) by
\begin{equation} \label{1.6} (T_{m,i} L_m)(x)=m^i\left(L_m\psi ^i_x\right)(x), \end{equation}
1.6
for any \(x\in I\), any \(m\in \mathbb N\) and any \(i\in \mathbb N_0\).
2 Main Results
In [5] is proved the following result contained in Theorem 2.1.
Theorem
2.1
Let \(m\in \mathbb N\) and \(0\leq x\leq 1\). If \(i=\left[mx\right]\), then
\begin{equation} \label{2.1} \left(B_m \varphi _x\right)\left(x\right)=2x(1-x)p_{m-1,i}\left(x\right). \end{equation}
2.7
Theorem
2.2
Let \(m\in \mathbb N\) and \(0\leq x\). If \(i=\left[mx\right]\), then
\begin{equation} \label{2.2} \left(S_m\varphi _x\right)\left(x\right)=2x{\rm e}^{-mx}\tfrac {\left(mx\right)^i}{i!}. \end{equation}
2.8
From \(i=[mx]\) it results \(i\leq mx{\lt}i+1\), equivalent with \(\cfrac {i}{m}\leq x{\lt}\cfrac {i+1}{m}\). We get
\begin{gather*} (S_m \varphi _x)(x)={\rm e}^{-mx}\sum _{k=0}^{\infty }\tfrac {\left(mx\right)^k}{k!}\left| \tfrac {k}{m}-x\right|\\ =2{\rm e}^{-mx}\sum ^{i}_{k=0} \tfrac {\left(mx\right)^k}{k!}\left(x-\tfrac {k}{m} \right)+{\rm e}^{-mx}\sum ^{\infty }_{k=0}\tfrac {\left(mx\right)^k}{k!} \left(\tfrac {k}{m}-x\right)\\ =2{\rm e}^{-mx}\left[x\sum ^{i}_{k=0} \tfrac {\left(mx\right)^{k}}{k!}-x\sum ^{i}_{k=1} \tfrac {\left(mx\right)^{k-1}}{\left(k-1\right)!}\right]+\left(S_m e_{1}\right)\left(x\right)-x\left(S_{m} e_{0}\right)(x)\\ =2x {\rm e}^{-mx}\tfrac {\left(mx\right)^i}{i!}, \end{gather*}
so (2.2) is obtained.
Theorem
2.3
Let \(m\in \mathbb N\) and \(0\leq x{\lt}m\). If \(i=\left[\tfrac {(m+1)x}{x+1}\right]\), then
\begin{equation} \label{2.3} (L_{m}\varphi _{x})(x)=x(1+x)^{-m}\left(2\tbinom {m}{i}x^{i}-x^{m}\right). \end{equation}
2.9
In the case when \(m\in \mathbb N\) and \(x\geq m\), then
\begin{equation} \label{2.4} (L_{m}\varphi _{x})(x)=x^{m+1}(1+x)^{-m}. \end{equation}
2.10
From \(i=\left[\tfrac {\left(m+1\right)x}{x+1}\right]\) it results \(i\leq \tfrac {\left(m+1\right)x}{x+1}{\lt}i+1\), equivalent with \(xi+i\leq mx+x{\lt} xi+x+i+1\), from where \(\tfrac {i}{m-i+1}\leq x{\lt}\tfrac {i+1}{m-i}\). Taking that \(x{\lt}m\) into account, one obtains \(i=\left[\tfrac {\left(m+1\right)x}{x+1}\right]\leq \tfrac {\left(m+1\right)x}{x+1}{\lt}m\).
We get
\begin{align*} & (L_{m}\varphi _{x})(x)=(1+x)^{-m}\sum \limits _{k=0}^{m}\tbinom {m}{k}x^{k} \left|\tfrac {k}{m-k+1}-x\right|\\ & =2\left(1+x\right)^{-m} \sum \limits _{k=0}^{i}\tbinom {m}{k}x^{k}\left(x-\tfrac {k}{m-k+1}\right)\\ & \quad +(1+x)^{-m}\sum \limits ^{m}_{k=0}\tbinom {m}{k}x^{k}\left(\tfrac {k}{m-k+1}-x\right)\\ & =2\left(1\! +\! x\right)^{-m}\left(\! x\sum ^{i}_{k=1}\tbinom {m}{k}x^{k}-\sum ^{i}_{k=1} \tbinom {m}{k}x^{k}\tfrac {k}{m\! -\! k\! +\! 1}\! \right)\! +\! \left(L_{m}e_{1}\right)(x) \! -\! x\left(L_{m}e_{0}\right)(x)\\ & =2\left(1+x\right)^{-m}\left(x\sum ^{i}_{k=1} \tbinom {m}{k}x^{k}-x\sum ^{i}_{k=1}\tbinom {m}{k-1}x^{k-1} \right)-x\left(\tfrac {x}{1+x}\right)^{m}\\ & =2x(1+x)^{-m}\tbinom {m}{i}x^{i}-x\left(\tfrac {x}{1+x}\right)^{m} = x\left(1+x\right)^{-m}\left(2\tbinom {m}{i}x^{i}-x^{m}\right), \end{align*}
so (2.3) is obtained. If \(x\geq m\), we have that \(\tfrac {k}{m-k+1}\leq x\), for any \(k\in \{ 0,1,\ldots ,m\} \) and then
\begin{gather*} (L_{m}\varphi _{x})(x)=(1+x)^{-m}\sum \limits _{k=0}^{m}\tbinom {m}{k}x^{k} \left(x-\tfrac {k}{m-k+1}\right)\\ =x\left(L_{m}e_{0}\right)(x)-\left(L_{m}e_{1}\right)(x) =x^{m+1}(1+x)^{-m}, \end{gather*}
so (2.4) holds.
Theorem
2.4
Let \(m\in \mathbb N\) and \(0\leq x{\lt}1\). If \(i=\left[\tfrac {mx}{1-x}\right]\), then
\begin{equation} \label{2.5} (Z_{m}\varphi _{x})(x)=2x^{i+1}(1-x)^{m+1}\tbinom {m+i}{i}. \end{equation}
2.11
From \(i=\left[\tfrac {mx}{1-x}\right]\) it results \(i\leq \tfrac {mx}{1-x}{\lt}i+1\), equivalent with \(\tfrac {i}{m+i}\leq x{\lt}\tfrac {i+1}{m+i+1}\). We get
\begin{align*} (Z_m\varphi _{x})(x)& =(1-x)^{m+1}\sum \limits _{k=0}^{i} \tbinom {m+k}{k}x^{k}\left|x-\tfrac {k}{m+k}\right|\\ & =2(1-x)^{m+1}\sum \limits _{k=0}^{i}\tbinom {m+k}{k}x^{k}\left(x-\tfrac {k}{m+k}\right)\\ & \quad +(1-x)^{m+1}\sum \limits _{k=0}^{\infty }\tbinom {m+k}{k}x^{k} \left(\tfrac {k}{m+k}-x\right)\\ & =2(1-x)^{m+1} \left(x\sum \limits ^{i}_{k=0}\tbinom {m+k}{k}x^{k}-\sum ^{i}_{k=1} \tbinom {m+k}{k}x^{k}\tfrac {k}{m+k}\right)+\\ & \quad +(Z_{m}e_{1})(x)-x(Z_{m}e_{0})(x)\\ & = 2x(1-x)^{m+1}\left(1+\sum \limits ^{i}_{k=1} \left(\tbinom {m+k}{k}x^{k}-\tbinom {m+k-1}{k-1}x^{k-1}\right)\right)\\ & =2x^{i+1}(1-x)^{m+1}\tbinom {m+i}{i}, \end{align*}
so (2.5) is obtained.
It is known the result contained in Lemma 2.5.
Lemma
2.5
If \(m,n\in \mathbb N_{0}, m\neq 0\), then \(\left(B_{m}\psi _{x}^{n}\right)(x)\), with \(x\in [0,1]\) is a polynomial in variable \(x\).
Lemma
2.6
If \(m,n\in \mathbb N\) and \(n\) is even, then
\begin{equation} \label{2.6} \left(B_{m}\varphi _{x}^{n}\right)(x)=x(1-x)q_{m,n}(x), \end{equation}
2.12
for any \(x\in [0,1]\), where \(q_{m,n}(x)\) is a polynomial in variable \(x\) and \(q_{m,n}(x){\gt}0\), for any \(x\in [0,1]\).
For \(m=1\), we get
\begin{align*} \left(B_{1}\varphi _{x}^{n}\right)(x)& = \sum \limits _{k=0}^{1}(k-x)^{n}p_{1,k}(x)\\ & =x^{n}p_{1,0}(x)+(1-x)^{n}p_{1,1}(x)=x(1-x) q_{1,n}(x), \end{align*}
where \(q_{1,n}(x)=x^{n-1}+(1-x)^{n-1}\).
For \(m\geq 2\), we get
\begin{align*} \left(B_{m}\varphi _{x}^{n}\right)(x)& =\sum \limits _{k=0}^{m} \left(\tfrac {k}{m}-x\right)^{n}p_{m,k}(x)\\ & =x^{n}(1-x)^{m}+\sum \limits _{k=1}^{m-1} \left(\tfrac {k}{m}-x\right)^{n}p_{m,k}(x)+(1-x)^{n}x^{m}\\ & =x(1-x)q_{m,n}, \end{align*}
where
\begin{align*} q_{m,n}(x)& =x^{n-1}(1-x)^{m-1}\\ & \quad +\sum \limits _{k=1}^{m-1} \left(\tfrac {k}{m}-x\right)^{n} \tbinom {m}{k}x^{k-1}(1-x)^{m-k-1} +(1-x)^{n-1}x^{m-1}. \end{align*}
For any \(x\in [0,1]\), every summand from \(q_{m,n}(x)\) is positive. But for \(k=1\) and \(k=m-1\), which means \(\left(\tfrac {1}{m}-x\right)^n\tbinom {m}{1}(1-x)^{m-2}\), respectively
\(\left(\tfrac {m-1}{m}-x\right)^n\tbinom {m}{m-1}x^{m-2}\) cannot be simultaneously null.
Taking the above remark into account, it results
\[ \left(B_{m}\varphi _{x}^{n}\right)(x)=x(1-x)q_{m,n}(x) \]
is a polynomial in variable \(x\) and \(q_{m,n}(x){\gt}0\), for any \(x\in [0,1]\).
Corollary
2.7
If \(m,n\in \mathbb N\), with \(n\) even, then
\begin{equation} \label{2.7} \left(T_{m,n}B_{m}\right)(x)=m^{n}\left(B_{m}\psi _{x}^{n}\right)(x)=m^{n} \left( B_{m}\varphi _{x}^{n}\right)(x)=m^{n}x(1-x) q_{m,n}(x). \end{equation}
2.13
For the Szász-Mirakjan operators, we get
\begin{equation} \label{2.8} \left(T_{m,s}S_{m}\right)(x)=m^{s}\left(S_{m}\psi _{x}^{s}\right)(x)= m^{s}e^{-mx}\sum \limits _{k=0}^{\infty }\tfrac {(mx)^{k}}{k!}\left(\tfrac {k}{m}-x\right)^{s}, \end{equation}
2.14
where \(m\in \mathbb N\), \(s\in \mathbb N_{0}\) and \(x\in [0,+\infty )\).
Lemma
2.8
If \(m,s\in N\), with \(s\geq 2\), then the following
\begin{equation} \label{2.9} \left((T_{m,s}S_{m})(x)\right)^{\prime }=m\sum _{i=0}^{s-2} \tbinom {s}{i}\left(T_{m,i}S_{m}\right)(x) \end{equation}
2.15
holds, for any \(x\in [0,+\infty )\).
From (2.8), we obtain
\begin{align*} & \left((T_{m,s}S_{m})(x)\right)^{\prime }=\left(m^{s}{\rm e}^{-mx} \sum \limits _{k=0}^{\infty }\tfrac {(mx)^{k}}{k!} \left(\tfrac {k}{m}-x\right)^{s}\right)^{\prime }\\ & =-m^{s+1}{\rm e}^{-mx}\sum \limits _{k=0}^{\infty }\tfrac {(mx)^{k}}{k!} \left(\tfrac {k}{m}-x\right)^{s}+m^{s+1}{\rm e}^{-mx}\sum \limits _{k=0}^{\infty } \tfrac {(mx)^{k-1}}{(k-1)!}\left(\tfrac {k}{m}-x\right)^{s}-\\ & \quad -sm^{s}{\rm e}^{-mx}\sum \limits _{k=0}^{\infty }\tfrac {(mx)^{k}}{k!} \left(\tfrac {k}{m}-x\right)^{s-1}\\ & = -m\left(T_{m,s}S_{m}\right)(x)-ms \left(T_{m,s-1}S_{m}\right)(x)+m^{s+1}{\rm e}^{-mx}\sum \limits _{k=0}^{\infty } \tfrac {(mx)^{k}}{k!}\left(\tfrac {k+1}{m}-x\right)^{s}\\ & =-m\left(T_{m,s}S_{m}\right)(x)\! -ms\left(T_{m,s-1}S_{m}\right) (x)\! +\! m^{s+1}{\rm e}^{-mx}\sum _{k=0}^{\infty }\tfrac {(mx)^{k}}{k!} \left(\! \left(\! \tfrac {k}{m}-x\! \right)\! +\! \tfrac {1}{m}\right)^{s}\\ & =-m\left(T_{m,s}S_{m}\right)(x)-ms\left(T_{m,s-1}S_{m}\right) (x)\\ & \quad +m^{s+1}{\rm e}^{-mx}\sum _{k=0}^{\infty }\tfrac {(mx)^{k}}{k!}\sum _{i=0}^{s} \tbinom {s}{i}\left(\tfrac {1}{m}\right)^{s-i}\left(\tfrac {k}{m}-x\right)^{i}\\ & =-m\left(T_{m,s}S_{m}\right)(x)\! -ms\left(T_{m,s-1}S_{m}\right)(x)\! +\! \sum _{i=0}^{s} \tbinom {s}{i}m^{i+1}{\rm e}^{-mx}\sum _{k=0}^{\infty }\tfrac {(mx)^{k}}{k!} \left(\! \tfrac {k}{m}\! -x\! \right)^{i}\\ & =-m\left(T_{m,s}S_{m}\right)(x)-ms\left(T_{m,s-1}S_{m}\right)(x)+m \sum _{i=0}^{s}\tbinom {s}{i}\left(T_{m,i}S_{m}\right)(x) \end{align*}
from where, the relation (2.9) follows.
Lemma
2.9
If \(m,s\in \mathbb N\), with \(s\geq 2\), then
\begin{equation} \label{2.10} (T_{m,s}S_{m})(x)=m\sum _{i=0}^{s-2}\tbinom {s}{i}\int _{0}^{x}(T_{m,i}S_{m})(t){\rm d}t \end{equation}
2.16
holds, for any \(x\in [0,\infty )\).
It yields immediately, taking that \((T_{m,s}S_{m})(0)=0\) into account and integrate the relation
2.15.
For any \(m\in \mathbb N\) and any \(x\in [0,\infty )\), it is well known that \((T_{m,0}S_{m})(x)=1\), \((T_{m,1}S_{m})(x)=0\), then the following holds:
Corollary
2.10
If \(m,s\in N\), with \(s\geq 4\), then
\begin{equation} \label{2.11} (T_{m,s}S_{m})(x)=mx+m\sum _{i=2}^{s-2}\tbinom {s}{i}\int _{0}^{x}(T_{m,i}S_{m})(t){\rm d}t \end{equation}
2.17
for any \(x\in [0,\infty )\).
Let \(m\in \mathbb N\) and \(x\in [0,\infty )\). We obtain
\begin{align*} (T_{m,0}S_{m})(x)=& (S_{m}e_{0})(x)=1,\\ (T_{m,1}S_{m})(x)=& m(S_{m}\psi _{x})(x)=m((S_{m}e_{1})(x)-x(S_{m}e_{0})(x))=0,\\ (T_{m,2}S_{m})(x)=& m\tbinom {2}{0}\int _{0}^{x}(T_{m,0}S_{m})(t){\rm d}t=mx,\\ (T_{m,3}S_{m})(x)=& m\left(\tbinom {3}{0}\int _{0}^{x} (T_{m,0}S_{m})(t){\rm d}t+\tbinom {3}{1}\int _{0}^{x}(T_{m,1}S_{m})(t){\rm d}t\right)=mx,\\ (T_{m,4}S_{m})(x)=& mx+m\tbinom {4}{2}\int _{0}^{x}(T_{m,2}S_{m})(t){\rm d}t=mx+3m^{2}x^{2}. \end{align*}
Lemma
2.11
If \(m,s\in \mathbb N\), with \(s{\gt}3\), then exist \(a_{2}^{(s)},a_{3}^{(s)},\ldots , a_{\left[{s/2}\right]}^{(s)}\geq 0\), \(a_{\left[{s/2}\right]}^{(s)}\neq 0\), such that
\begin{equation} \label{2.12} (T_{m,s}S_{m})(x)=mx+m\sum \limits _{k=2}^{\left[{s/2}\right]}a_{k}^{(s)}x^{k}, \end{equation}
2.18
for any \(x\in [0,\infty )\), where \(a_{k}^{(s)}\) depends of \(m\) and \(k\in \left\{ 2,3,\ldots ,\left[\tfrac {s}{2}\right]\right\} \).
We prove by mathematical induction.
For \(s=4\) we get \((T_{m,4}S_{m})(x)=mx+3m^{2}x^{2}\), so that \(a_{2}^{(4)}=3m{\gt}0.\)
We assume that
\[ (T_{m,j}S_{m})(x)=mx+m\sum \limits _{k=2}^{\left[{j/2}\right]}a_{k}^{(j)}x^{k}, \]
for \(a_{2}^{(j)},a_{3}^{j},\ldots ,a_{\left[{j/2}\right]}^{(j)}\geq 0\), \(a_{\left[{j/2}\right]}^{(j)}\neq 0\), for any \(j\in \{ 4,5,\ldots ,s\} \). Taking relation 2.17 into account, we obtain
\begin{align*} & (T_{m,s+1}S_{m})(x)=\\ & =mx+m\Bigg(\tbinom {s+1}{2}\int _{0}^{x}(T_{m,2}S_{m})(t){\rm d}t +\tbinom {s+1}{3}\int _{0}^{x}(T_{m,3}S_{m})(t){\rm d}t\\ & \quad + \sum \limits _{i=4}^{s-1}\tbinom {s+1}{i} \int _{0}^{x}\bigg(mt+m\sum \limits _{k=2}^{\left[{i/2}\right]}a_{k}^{(i)}t^{k}\bigg){\rm d}t\Bigg)\\ & =mx+m\bigg(\tbinom {s+1}{2}\tfrac {mx^2}{2}+\tbinom {s+1}{3}\tfrac {mx^2}{2}+\sum \limits _{i=4}^{s-1}\tbinom {s+1}{i} \bigg(\tfrac {mx^2}{2}+m\sum \limits _{k=2}^{\left[{i/2}\right]} \tfrac {a_{k}^{(i)}}{k+1}x^{k+1}\bigg)\bigg)\\ & =mx+\tfrac {1}{2}\left(m\tbinom {s+1}{2}+m\tbinom {s+1}{3}+\ldots +m\tbinom {s+1}{s-1}\right) mx^{2}\\ & \quad +\left(m\tbinom {s+1}{4}\tfrac {a_{2}^{(4)}}{3}+m \tbinom {s+1}{5}\tfrac {a_{2}^{(5)}}{3}+\ldots +m \tbinom {s+1}{s-1}\tfrac {a_{2}^{(s-1)}}{3}\right)mx^{3}+\end{align*}
\begin{align*} & \quad +\ldots +m\cdot a_{\left[\tfrac {s+1}{2}\right]}^{(s+1)}x^{\left[\tfrac {s+1}{2}\right]} \end{align*}
where:
then for \(k\in \left\{ 2,3,\ldots ,\left[\tfrac {s-3}{2} \right]\right\} \) we get \(k+1{\lt}\left[\tfrac {s-2}{2}\right]+1= \left[\tfrac {s-1}{2}\right]+1=\left[\tfrac {s+1}{2}\right]\) and
\[ a_{\left[\tfrac {s+1}{2}\right]}^{(s+1)}=m \tbinom {s+1}{s-2}\tfrac {a_{\left[\tfrac {s-2}{2}\right]}^{(s-2)}} {\left[\tfrac {s-2}{2}\right]+1}+m\tbinom {s+1}{s-1} \tfrac {a_{\left[\tfrac {s-1}{2}\right]}^{(s-1)}}{\left[\tfrac {s-1}{2}\right]+1}{\gt}0; \]
then for \(k\in \{ 2,3\ldots ,\left[\tfrac {s-2}{2}\right]\} \) we get \(k+1{\lt}\left[\tfrac {s-1}{2}\right]+1=\left[\tfrac {s+1}{2}\right]\) and
\[ a_{\left[\tfrac {s+1}{2}\right]}^{(s+1)}=m\tbinom {s+1}{s-1} \tfrac {a_{\left[\tfrac {s-1}{2}\right]}^{(s-1)}}{\left[\tfrac {s-1}{2}\right]+1}{\gt}0. \]
And now, the proof is done.
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