text:
Tag: Saturn Kilometric Radiation
Posted on May 3, 2013 December 23, 2015
How Saturn's Magnetic Activity Could Help us Pinpoint Time on the Ringed Planet
He's not even finished his first university degree yet, but Tim Kennelly is already part of a team that is altering our perception of time on Saturn.
The University of Iowa undergrad — in junior year, yet — led a paper describing activity in Saturn's magnetosphere, where charged particles collect and sometimes form auroras. The process changes with the Saturnian seasons and could, the university stated, help scientists better understand how long a Saturn day lasts.
The researchers used information from NASA's Cassini spacecraft, which has been orbiting the planet and its moons since 2004. The research challenge: Saturn is a gas giant full of layers that each have their own rotational speed. That makes it hard to figure out how long Saturn's day is. (It's about 10 hours, but varies by latitude.)
Kennelly made direct observations of seasonal changes in a phenomenon known as Saturn kilometric radiation (SKR). This robust radio signal was first discovered several decades ago and is being examined more closely by Cassini.
"UI space physicist Donald Gurnett and other scientists showed that the north and south poles have their own SKR 'days' that vary over periods of weeks and years," the university stated. "How these different periods arise and are driven through the magnetosphere has become a central question of the Cassini mission, according to NASA officials."
Kennelly observed, from looking at data collected between 2004 and 2011, that SKAs are linked with "flux tubes" that are made up of plasma, or superhot gas. These tubes happen around the same time of instances of SKAs in the northern and southern hemisphere, which changes seasonally.
It's possible that this understanding could be carried over to other planets, the university stated, including our own.
"This finding may alter how scientists look at the Earth's magnetosphere and the Van Allen radiation belts that affect a variety of activities at Earth ranging from space flight safety to satellite and cell phone communications," it added.
This won't be Kennelly's only degree. He is about to apply to graduate schools, and he has aims to earn a doctorate in plasma physics.
"I'm pleased to have contributed to our understanding of Saturn's magnetosphere so early in my career," stated Kennelly. "I hope this trend continues."
The research is described in the American Geophysical Union's Journal of Geophysical Research.
Source: University of Iowa
Dusty Plasma From Enceladus Might Affect Saturn's Magnetosphere
[/caption]
Discovered by the Cassini mission, Saturn Kilometric Radiation (SKR) has been something of an enigma to astronomers. According to the radio and plasma wave instruments, variations occur in sync with the planet's rotation. However, there are periodic "bursts" of radiation which are in line with Saturn's magnetosphere. What makes this odd? The rate isn't quite the same.
Thanks to investigations of Enceladus by Cassini in 2008, new information about the plasma environment surrounding Saturn's satellite could show a marked impact on the magnetosphere. This image and video show a changing pattern of radio waves from Saturn known as Saturn Kilometric Radiation, as detected by NASA's Cassini spacecraft. The colors indicate the emitted power of the radio waves, with red as the strongest.
How is it being affected? Thanks to Enceladus' "spraying" nature, the huge plume of water vapor and ice from its southern pole provides a hefty source of plasma to feed Saturn's magnetosphere and E-Ring. These negatively charged particles are again impacting the motion of the localized plasma.
"These signatures result from half or more of the electrons being attached to dust grains and by the interaction between the surrounding cold plasma and the predominantly negatively charged submicrometer-sized dust grains." says M. W. Morooka (et al). "The dust and plasma properties estimated from the observations clearly show that the dust-plasma interaction is collective."
According to the AGU Journal, this dust-plasma interaction impacts the dynamics of Saturn's magnetosphere, possibly influencing the rate of SKR emissions.
Original Story Source: American Geophysical Union.


text:
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https://www.sfgate.com/bayarea/article/Fish-and-Wildlife-spokesman-Mike-Taugher-dies-4694671.php
Fish and Wildlife spokesman Mike Taugher dies
Ellen Huet
Updated 4:55 pm PDT, Tuesday, July 30, 2013
A longtime environmental reporter for the Contra Costa Times and spokesman for the California Department of Fish and Wildlife died Sunday while snorkeling with his family in Maui, officials said.
Mike Taugher, 50, covered complex state issues with depth, persistence and fairness, and was one of the most knowledgeable reporters on the politics and intricacies of the Sacramento-San Joaquin River Delta, colleagues said Monday.
Mr. Taugher was snorkeling with his family Sunday morning near Black Rock in Kaanapali, Maui police said. His family members came back to shore but were unable to find him immediately.
After a short search, he was found unconscious in the waters fronting Black Rock and was pulled to shore, where emergency crews tried unsuccessfully to revive him. Foul play is not suspected.
Mr. Taugher, who lived in Benicia with his wife, Kim, and two children, Anna, 12, and Zachary, 15, won numerous awards for his reporting over the course of his career, which began at papers in Lake Tahoe and moved him through Greeley, Colo., and Albuquerque before he arrived at the Contra Costa Times in 2000.
There, he reported on environmental issues and public lands, doggedly pursuing stories other reporters didn't have the patience or knowledge to sort through, said Paul Rogers, a co-worker and environmental reporter at the San Jose Mercury News.
"Water is one of the most complicated topics in California, but it's really, really important," Rogers said. "There's billions of dollars riding on the decisions that are made. The health of rivers and lakes and salmon and other species depend on the kind of decisions that are made.
"There are very few reporters that take the time to really understand it and report in a really fair and thoughtful way," Rogers added. "Mike knew more about water than nearly any journalist in California."
Mr. Taugher left the Times a year ago when he was appointed assistant deputy director of communications for the state Department of Fish and Wildlife. Politicians who knew him through his reporting praised his work.
"Mike was unquestionably fair, knowledgeable and deeply insightful about the issues he covered," Rep. George Miller, D-Martinez, said in a statement.
His reporting didn't shy away from stories involving power players. In 2009, he waded through complicated documents to find that Stewart Resnick, a Southern California billionaire, was pulling a large profit as his companies re-sold water.
But in person, Mr. Taugher was "soft-spoken and thoughtful" and just as likely to talk about his family as his work, Rogers said.
"He liked talking about his son's Little League games as much as he liked talking about the complexities of water in the West," Rogers said. "I was e-mailing him on Friday, and he said he was in Maui and I thought to myself, 'Wow, he worked so hard, it's really great that he and his family can get away for a nice vacation like that.' It's so devastating to see it end up like this."
His brother, David Taugher, remembered him as a deeply caring brother and father.
"We're all very proud of him and what he accomplished," he said. "Mike loved his family very dearly."
A memorial is pending.
Ellen Huet is a San Francisco Chronicle staff writer. E-mail: ehuet@sfchronicle.com
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redpajama_set_name: RedPajamaCommonCrawl


Repost our server and get Coin of Lick.
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text:
import Ember from 'ember';
import Util from 'ui/utils/util';
import CattleTransitioningController from 'ui/mixins/cattle-transitioning-controller';

export default Ember.Controller.extend(CattleTransitioningController, {
  needs: ['application'],
  endpoint: Ember.inject.service(),

  actions: {
    activateServices: function() {
      return this.doAction('activateservices');
    },

    deactivateServices: function() {
      return this.doAction('deactivateservices');
    },

    addService: function() {
      this.get('controllers.application').transitionToRoute('service.new', {
        queryParams: {
          environmentId: this.get('model.id'),
        },
      });
    },

    addBalancer: function() {
      this.get('controllers.application').transitionToRoute('service.new-balancer', {
        queryParams: {
          environmentId: this.get('model.id'),
        },
      });
    },

    edit: function() {
      this.get('controllers.application').setProperties({
        editEnvironment: true,
        originalModel: this.get('model'),
      });
    },

    exportConfig: function() {
      var url = this.get('endpoint').addAuthParams(this.get('model').linkFor('composeConfig'));
      Util.download(url);
    },

    viewCode: function() {
      this.get('controllers.application').transitionToRoute('environment.code', this.get('model.id'));
    },

    viewGraph: function() {
      this.get('controllers.application').transitionToRoute('environment.graph', this.get('model.id'));
    },

    delete: function() {
      return this._super().then(() => {
        if ( this.get('controllers.application.currentRouteName') === 'environment.index' )
        {
          this.transitionToRoute('environments');
        }
      });
    },
  },

  availableActions: function() {
    var a = this.get('model.actions');

    var out = [
      { label: 'Start Services',  icon: 'icon icon-play',             action: 'activateServices',    enabled: this.get('model.canActivate') },
      { label: 'Stop Services',   icon: 'icon icon-pause',            action: 'deactivateServices',  enabled: this.get('model.canDeactivate') },
      { label: 'View Graph',      icon: 'icon icon-share',            action: 'viewGraph',            enabled: true },
      { label: 'View Config',     icon: 'icon icon-files',            action: 'viewCode',            enabled: true },
      { label: 'Export Config',   icon: 'icon icon-download',         action: 'exportConfig',        enabled: !!a.exportconfig },
      { divider: true },
      { label: 'Delete',          icon: 'icon icon-trash',            action: 'promptDelete',        enabled: !!a.remove, altAction: 'delete', color: 'text-warning' },
      { label: 'View in API',     icon: 'icon icon-externallink',     action: 'goToApi',             enabled: true },
      { divider: true },
      { label: 'Edit',            icon: 'icon icon-edit',             action: 'edit',                enabled: true },
    ];

    return out;
  }.property('model.actions.{remove,purge,exportconfig}','model.{canActivate,canDeactivate}'),

  state: Ember.computed.alias('model.combinedState'),
});

redpajama_set_name: RedPajamaGithub


I LOVE pinterest. I spend way too much time on it, and find some pretty amazing ideas. I thought I would share some of the ideas I plan on completing this summer. Even if I don't get to them, they are great for inspiration!
These projects may be a bit ambitious for me to tackle all of these (and other projects I want to do this summer) with a baby coming sometime in the next 5 weeks (give or take). I only get a short 6 weeks of maternity leave, and then I have to quickly adjust to being a working mom of two. Feel free to follow me on pinterest, and let me know so I can follow you too. What projects have you found on pinterest?


The Lexus IS was introduced to the market as a 2013 model year vehicle. Since then, an athletic sedan sold to a great level beyond the expectations of Lexus. At present, the current Lexus IS model is a few years old, and the automaker took it upon itself to provide its athletic small entry-level car a refresh. Even though there is not a whole bunch that changed, the headlight modules of the vehicle will take on a special shape, and the front part will be more athletic than that of the outgoing model.
The 2017 Lexus IS sedan will also come with a new set of tail lamps, innovative wheel designs, and even a couple of new colors. On the indoor part of the new vehicle, the cabin comes equipped with a redesigned infotainment display, and there are some other notable refinements inside the cabin. The safety equipment even got a tad love with this redesign. The sedan will be launched with three models, such as the strongest 2017 Lexus IS 350. IS 300 and the base model, IS 200t.
For the most exterior part, the overall look has remained identical, but there are a few extremely attractive changes. Among the external parts, the most eminent exterior change that can be observed in the new vehicle is the front. It comes outfitted with a continuous, absolute-height grille in both standard trim and in the F-Sport trim, in conjunction with a fresh bumper, which is accommodating gigantic corner air channels, and revamped headlamps, at the same time as maintaining the discordant L-shaped Nike symbol-shaped daytime running lamps. However, the F Sport, which is the performance-focused model of the 2017 IS comes outfitted with a grille that comes with a three-dimensional F-net outline with jet-black metallic casing and handy brake ducts.
At the rear, the new-fangled IS comes outfitted with a fresh set of tail lamp furnishings with more articulated L-shaped LED brilliance guides and rectangular chrome exhaust tilts. The new Lexus IS 2017 comes outfitted with a set of freshly styled 10-spoke 17-inch diameter alloy wheels as standard. However, there are also wheels that come with more than a few novel designs with luminous machine finishes as an optional feature. Other than the accessible body colors, the new Lexus sedan will be provided with two innovative body colors that comprise Graphite Black Glass Flak and Deep Blue Mica.
Bizarrely, for a sedan making its first appearance in China, it stays behind as a single wheelbase edition. Other compact lavishness sedans, such as the Volvo S60, Cadillac ATS, and the Audi A4 are all built with an extensive wheelbase to house the favorite of the market for driving the sedan with a foreign driver.
As far as the interior is concerned, it is prepared with a comfortable cabin, which is organized with a bigger infotainment screen of size 10.3 inches to handle utilities in the course of the Remote Touch Interface system of Lexus, the console installed controller, similar to a mouse. It now consists of spare Enter keys, one on every side, in a plan to perk up the simplicity of use.
The auto manufacturer has also improved the cabin beautification of the new Lexus IS 2017 sedan, by providing it with a leather-creased palm rest at the back of the controls of the Remote Touch Interface setup, fresh cup holders, and flush-installed middle-console controls. The minor alterations of the internal part of the novel sedan are completed by a revamped instrument cluster with a sewed cover, fresh clock markings, and by a few other details.
 Lane-departure warning with steering assist.
Within the cabin, there are innovative colors for some markets, a gently modified lower center console to suit completely within the knee bolsters, and a novel sewing on top of the meter wrap, as well.
The 2017 IS comes equipped with the same power plants, similar to those of the 2016 version Lexus. As a result, the 2017 Lexus IS200t, which is the base trim, is going to be powered by a four-cylinder turbocharged power plant with the capacity of 2.0 liters. This power plant delivers the base model the necessary power of 241 horses and the necessary torque of 258 lbs per foot.
Next, the IS300 and IS350 trims will be powered by the V6 power plant train with the same capacity of 3.5 liters. However, the power plant will be tuned in two diverse levels to offer dissimilar power output to the two trims. As a result, the first tuned power plant offers the IS300 trim the necessary power of 255 horses and the necessary torque of 236 lbs per foot, whereas the IS350 trim of the new sedan acquires the necessary power of 306 horses and the necessary torque of 277 lbs per foot.
All power plants concerned trim accelerate from 0-60 mph velocity from the static mode respectively: 6.9 seconds and assisting them to reach the top velocity of 143 mph for IS 200t, 6.1 seconds with a top speed 131 mph for IS 300 and 5.7 seconds with top speed of 131 mph for IS 350. About mileage, these power plants make the 2017 version IS to offer a mileage of 19 mpg within the city, the mileage of 28 mpg on the highways and 21 mpg as the combined mileage. The new version comes equipped with the fuel tank with the highest capacity of 17.4 gallons.
All rear-wheel drive versions will be provided with an eight-rate mechanical transmission, and as the Lexus IS300 and IS350 trims, which are the all-wheel drive vehicles, they will be provided with a six-rate mechanical transmission.
The 2017 Lexus IS release date is anticipated to be the final quarter of 2016. However, the base cost of the new-fangled IS sedan is predictable to be $ 38,500.


text:
var express = require('express');
var app = express();

app.get('/', function (req, res) {
  res.send('Hello World');
});

app.listen(3000, function () {
  console.log('app is listening at port 3000');
});


text:
\section{Introduction}
This work is at confluence of two mathematical research areas namely linear approximation processes and statistical
convergence. To reveal what novelties this paper brings, we briefly present both a background and some historical comments. The main problem of approximation theory consists in finding for a complicated function a close-by simple function.\\
\indent At 70 years old, Karl Wilhelm Theodor Weierstrass (1815--1897) proved the density of the algebraic polynomials in the space $C([a,b])$ and of the trigonometric polynomial in $\tilde{C}([a,b]).$
Weierstrass's approximation theorem stating that every continuous function on a bounded interval can be approximated to arbitrary accuracy by polynomials is such an important example for this process and has been played
the significant role in the development of analysis. By using probability theory Bernstein \cite{B} proved the
Weierstrass's theorem and defined approximate polynomials known as Bernstein polynomials in the literature.
 In 1987, Lupa\c{s} \cite{L} introduced the first $q$-analogue
of Bernstein operator and investigated its approximating and shape-preserving properties.
Another $q$-generalization of the classical Bernstein polynomials is due to Phillips
\cite{PH}. After that many generalizations of well-known positive linear operators, based
on $q$-integers were introduced and studied by several authors. In \cite{OA} Agratini introduced
a new class of $q$-Bernstein-type operators which fix certain polynomials and
studied the limit of iterates of Lupa\c{s} $q$-analogue of the Bernstein operators.
In 1974, Khan \cite{HH} studied approximation of functions in various classes using different
types of operators.

In 1950, Sz\'{a}sz defined and studied the approximation properties of the following operators
\begin{equation}
S_n(f;x)=e^{-n x}\sum_{j=0}^{\infty} \frac{(n x)^j}{j!}f\bigg(\frac{j}{n}\bigg),
\end{equation}
whenever $f$ satisfies exponential-type growth condition \cite{OS}. In 1962, Schurer \cite{FS} introduced and studied the approximation properties of linear positive operators. An extension in $q$-Calculus of Sz\'{a}sz-Mirakyan operators was constructed by Aral \cite{AA} who formulated also a Voronovskaya theorem related to $q$- derivatives for these operators. In 1976, May \cite{M} showed that the Baskakov operators can reduce to the Sz\'{a}sz-Mirakyan operators.  After that several other researchers have studied in this direction and obtained different approximation properties of many operators \cite{KJ}-\cite{MA}.

Motivated by the operators due to Y\"{u}ksel \cite{IY}, in the year 2013, we considered the following
operators.
\newline
\indent Let $p, k, n \in \mathbb{N}$ and $f$ be a real valued continuous function on the interval $[0,\infty)$. We define the Baskakov-Schurer-Sz\'{a}sz type linear positive operators as

\begin{equation}\label{eq1}
\mathcal{L}_{n,p}(f,x)= (n+p) \sum_{k=0}^{\infty} b^k_{n,p}(x)\int_{0}^{\infty} f(t) s^k_{n,p}(t) d t \\
\end{equation}

where  $\displaystyle b^{k}_{n,p}(x)=\binom{n+p+k-1}{k}\frac{x^k}{(1+x)^{n+p+k}}$\,\,
and\,\, $\displaystyle s^k_{n,p}(t)= e^{-(n+p)t}\frac{{((n+p)t)}^k}{k!}.$\\
\newline
Now we give an auxiliary lemma for the Korovkin test functions.
\begin{lemma}
Let $e_m(t) =t^m$, $m = 0,1,2.$ we have
\begin{itemize}
\item[(i)] $\mathcal{L}_{n,p}(e_0;x)=1,$
\item[(ii)] $\mathcal{L}_{n,p}(e_1;x)=x+\frac{1}{(n+p)},$
\item[(iii)] $\mathcal{L}_{n,p}(e_2;x)=\big(1+\frac{1}{n+p}\big)x^2+\frac{4}{(n+p)}x+\frac{2}{(n+p)^2}.$
\end{itemize}
\end{lemma}

\begin{lemma}\label{l1}
Let $p\in \mathbb{N}$. Then following hold:\\
$\alpha_n(x):= \mathcal{L}_{n,p}((t-x);x)=\frac{1}{n+p}.$\\
$\delta_n(x):=\mathcal{L}_{n,p}((t-x)^2;x)=\frac{1}{n+p} x^2+\frac{2}{n+p} x+ \frac{2}{(n+p)^2}.$
\end{lemma}

\section{Some auxiliary results}
Let the space $C_B[0,\infty)$ of all continuous and bounded
functions $f$ on $[0,\infty)$, be endowed with the norm $\|f\|=sup\{ \mid f(x)\mid: x\in[0,\infty)\}.$ Further let us consider the Peetre's K-functional which is defined by
 \begin{equation} K_2(f,\delta)= \inf_{g\in W^2}\{\|f-g\|+\delta \|g''\|\},\end{equation}
where $\delta >0$ and
$W^2_{\infty}=\{g\in C_B[0,\infty):g', g'' \in C_B[0,\infty)\}.$ By the method
as given (\cite{DL} p.177, Theorem 2.4), there exists an absolute constant $C>0$ such that
\begin{equation}\label{1}
K_2(f,\delta) \leq C \omega_2(f,\delta),\end{equation}
 where

\begin{equation}\label{2}
\omega_2(f,{\delta})=\sup_{0<h \leq \delta}\sup_{x\in[0,\infty)}\mid f(x+2h)-2f(x+h)+f(x)\mid \end{equation}
is the second order modulus of smoothness of $f\in C_B[0,\infty).$
Also we set

\begin{equation}\label{3}
\omega(f,{\delta})= \sup_{0<h \leq
\delta}\sup_{x\in[0,\infty)}\mid f(x+h)-f(x)\mid.
\end{equation}
We denote the usual modulus of continuity of $f\in C_B[0,\infty)$.

\begin{theorem}\label{the1}
Let $f \in C_B[0,\infty)$, then for all $x\in
[0,\infty)$, there exists an absolute constant $ C >0 $ such that
\begin{equation}|\mathcal{L}_{n,p}(f,x)-f(x)| \leq C \omega_2\left(f,\sqrt{\delta_n(x) +(\alpha_n(x))^2}\right)+
\omega(f,\alpha_n(x)).
\end{equation}
\end{theorem}
\textbf{Proof.}
Let $g\in W_\infty^2$ and $x,t\in [0,\infty).$ By Taylor's expansion, we have
\begin{equation} \label{ac1}
g(t)=g(x)+g'(x)(t-x)+\int_x^t (t-u)g''(u)du.
\end{equation}
Define \begin{equation}
\mathcal{\widetilde{L}}_{n,p}(f,x) = \mathcal{L}_{n,p}(f,x) + f(x)-f\left(x +\frac{1}{n+p}\right).
\end{equation}
Now, we have $\mathcal{\widetilde{L}}_{n,p}(t-x,x)=0,$ $t\in[0,\infty).$\\
Applying $\mathcal{\widetilde{L}}_{n,p}$ on both sides of (\ref{ac1}), we get
\begin{eqnarray*}
\mathcal{\widetilde{L}}_{n,p}(g,x)-g(x) & = & g'(x)\mathcal{\tilde{L}}_{n,p}((t-x),x)+\mathcal{\tilde{L}}_{n,p} \left(\int_x^t(t-u)g''(u)du,x\right)\\
&=& \mathcal{L}_{n,p}\left(\int_x^t(t-u)g''(u) du, x\right)+\int_x^{x +\frac{1}{n+p}} \left(x +\frac{1}{n+p}-u\right)g''(u)du,
\end{eqnarray*}
on the other hand $\displaystyle \bigg| \int_x^t (x-u)g''(u)du\bigg| \leq \|g''\|(t-x)^2$ and
\begin{eqnarray*}
\int_x^{x +\frac{1}{n+p}} \left(x +\frac{1}{n+p}-u\right)g''(u)du \leq \left(x +\frac{1}{n+p}-x\right)^2\|g''\|&=& \|g''\|\left(\mathcal{L}_{n,p}(t-x,x)\right)^2\\
&=&\|g''\|\left(\alpha_n(x)\right)^2.
\end{eqnarray*}
Thus, one can do this
\begin{eqnarray*}
\bigg|\mathcal{\tilde{L}}_{n,p}(g,x)-g(x)\bigg| &\leq & \bigg|\mathcal{L}_{n,p} \left(\int_x^t(t-u)g''(u)du,x\right)\bigg| +\bigg|\int_x^{x +\frac{1}{n+p}}\left(x +\frac{1}{n+p} -u\right)g''(u)du\bigg|\\
&\leq& \|g''\|\mathcal{L}_{n,p} \left((t-x)^2,x\right)+\|g''\|\left(\alpha_n(x)\right)^2\\
&\leq& \|g''\| \bigg[\delta_n(x) +\left(\alpha_n(x)\right)^2\bigg].
\end{eqnarray*}
We observe that,
\begin{eqnarray*}
\bigg|\mathcal{\widetilde{L}}_{n,p}(f,x)-f(x)\bigg| &\leq & \bigg|\mathcal{\tilde{L}}_{n,p}(f-g,x)-(f-g)(x)\bigg|\\
&+&  \bigg|\mathcal{\tilde{L}}_{n,p} (g,x)-g(x)\bigg|+\bigg|f(x)-f\left(x +\frac{1}{n+p}\right)\bigg|\\
 &\leq & \|f-g\|+\|g''\| \bigg[\delta_n(x)+\left(\alpha_n(x)\right)^2\bigg]+\omega(f,\alpha_n(x)).
\end{eqnarray*}
Now, taking infimum on the right-hand side over all $g\in C_B^2[0,\infty)$  and from (\ref{1}), we get
\begin{eqnarray*}
\bigg|\mathcal{L}_{n,p}(f,x)-f(x)\bigg| &\leq & K_2\left(f,\delta_{n}(x)+\left(\alpha_{n}(x)\right)^2\right)+ \omega(f,\alpha_n(x))\\
&\leq& C \omega_2\left(f,\sqrt{\delta_{n}(x)+\left(\alpha_{n}(x)\right)^2}\right)+ \omega(f,\alpha_{n}(x)),
\end{eqnarray*}
which proves the theorem.\\

\section{Weighted approximation}
In this section, we obtain the Korovkin type weighted approximation by the operators defined in (\ref{eq1}). The weighted Korovkin-type theorems were proved by Gadzhiev \cite{AD}.
A real function $\rho = 1+x^2$ is called a weight function if it is continuous on $%
\mathbb{R}$ and $\lim\limits_{\mid x\mid \rightarrow \infty }\rho (x)=\infty
,~\rho (x)\geq 1$ for all $x\in \mathbb{R}$.

Let $B_{\rho }(\mathbb{R})$  denote the weighted space of real-valued
functions $f$ defined on $\mathbb{R}$ with the property $\mid f(x)\mid \leq
M_{f}~\rho (x)$ for all $x\in \mathbb{R}$, where $M_{f}$ is a constant
depending on the function $f$. We also consider the weighted subspace $%
C_{\rho }(\mathbb{R})$ of $B_{\rho }(\mathbb{R})$ given by $C_{\rho }(%
\mathbb{R})=\{f\in B_{\rho }(\mathbb{R}){:}$ $f$ is continuous on $\mathbb{R} $\} and $C_{\rho}^{*}[0,\infty)$ denotes the subspace of all functions
 $f\in C_{\rho}[0,\infty)$ for which $\lim\limits_{|x|\rightarrow\infty} \frac{ f(x)}{\rho(x)}$ exists finitely.

\begin{theorem}(See \cite{AD} and \cite{AD1})
\begin{itemize}
\item[(i)] There exists a sequence of linear positive operators $A_n(C_{\rho}\rightarrow B_{\rho})$ such that
\begin{equation}\label{wa1}
\lim_{n\rightarrow \infty}\|A_n(\phi^\nu)- \phi^\nu\|_{\rho}=0,~~\nu=0,1,2
\end{equation}
and a function $f^{*}\in C_{\rho} \backslash C^{*}_{\rho}$  with $\lim\limits_{n\rightarrow \infty} \| A_n(f^{*})- f^{*}\|_{\rho}\geq 1.$
\item[(ii)] If a sequence of linear positive operators $A_n(C_{\rho}\rightarrow B_{\rho})$ satisfies conditions (\ref{wa1}) then
\begin{equation}
\lim_{n\rightarrow \infty}\|A_n(f)- f\|_{\rho}=0, \text{  for every } f\in C^{*}_{\rho}.
\end{equation}
\end{itemize}
\end{theorem}
Throughout this paper we take the growth condition as $\rho(x) = 1 + x^2$ and $\rho_{\gamma}(x) = 1 + x^{2+\gamma},~ x\in [0,\infty), \gamma > 0.$
Now we are ready to prove our next result as follows:

\begin{theorem}
For each $f \in C_{\rho}^{*}[0,\infty)$, we have
\begin{equation*}\lim _{n\rightarrow \infty} \| \mathcal{L}_{n,p}(f)-f \|_{\rho} = 0.\end{equation*}
\end{theorem}
\textbf{Proof.} Using the theorem in \cite{AD} we see that it is sufficient to verify the following three conditions
\begin{equation}\label{w1}
\lim_{n\rightarrow \infty} \|
\mathcal{L}_{n,p}(t^r,x)-x^r\|_{\rho}=0, \text{  } r=0,1,2.
\end{equation}
Since, $\mathcal{L}_{n,p}(1,x)=1$, the first condition of (\ref{w1}) is satisfied for $r=0$. Now,
\begin{eqnarray*}
\|\mathcal{L}_{n,p}(t,x)-x\|_{\rho}&=&\sup_{x\in [0,\infty)} \frac{\mid \mathcal{L}_{n,p}(t,x)-x \mid}{1+x^2}\\
&\leq&  \frac{1}{n+p} \sup_{x\in [0,\infty)}\frac{1}{1+x^2}\\
&=& o(1).
\end{eqnarray*}
Finally,
\begin{eqnarray*}
\|\mathcal{L}_{n,p}(t^2,x)-x^2\|_{\rho} & = & \sup_{x\in [0,\infty)} \frac{\mid \mathcal{L}_{n,p}(t^2,x)-x^2 \mid}{1+x^2}\\
&\leq&\left|\frac{1}{(n+p)}\right|\sup_{x\in [0,\infty)} \frac{x^2}{1+x^2}+
\left|\frac{4}{(n+p)}\right|\sup_{x\in [0,\infty)}\frac{x}{1+x^2}+\left| \frac{2}{(n+p)^2}\right| \\
&=& o(1).
\end{eqnarray*}
Thus, from Gadziv's Theorem in \cite{AD} we obtain the desired result of theorem.
\qed

We give the following theorem to approximate all functions in $C_{x^2}[0,\infty)$.
\begin{theorem}
For each $f\in C_{x^2}[0,\infty)$ and $\alpha
>0$, we have \\
$$\lim\limits_{n \rightarrow \infty} \sup_{x\in[0,\infty)} \frac{\mid\mathcal{L}_{n,p}(f,x)-f(x)
\mid}{(1+x^2)^{1+\alpha}}=0.$$
\end{theorem}
\textbf{Proof.} For any fixed $x_0>0$,
\begin{eqnarray*}
\sup_{x\in[0,\infty)}\frac{\mid \mathcal{L}_{n,p}(f,x)-f(x) \mid}{(1+x^2)^{1+\alpha}}&\leq & \sup_{x \leq x_0}\frac{\mid \mathcal{L}_{n,p}(f,x)-f(x) \mid}{(1+x^2)^{1+\alpha}} + \sup_{x \geq x_0}\frac{\mid \mathcal{L}_{n,p}(f,x)-f(x) \mid}{(1+x^2)^{1+\alpha}}\\
&\leq&  \|\mathcal{L}_{n,p}(f)-f\|_{C[0,x_0]} + \|f\|_{x^2}\sup_{x \geq x_0}\frac{\mid \mathcal{L}_{n,p}(1+t^2,x)\mid}{(1+x^2)^{1+\alpha}}\\&&+\sup_{x \geq x_0}\frac{\mid f(x) \mid}{(1+x^2)^{1+\alpha}}.
\end{eqnarray*}
The first term of the above inequality tends to zero from Theorem \ref{t2}. By Lemma \ref{l1}(ii), for any fixed $x_0>0$ it is easily seen that $ \sup_{x\geq x_0} \frac{\mid
\mathcal{L}_{n,p}(1+t^2,x)\mid}{(1+x^2)^{1+\alpha}}$
tends to zero as $n \rightarrow \infty$. We can choose $x_0>0$ so
large that the last part of the above inequality can be made small
enough. Thus the proof is completed.
\qed

\section{Voronovskaja type theorem}
In this section we establish a Voronovskaja type asymptotic formula for the operators $\mathcal{L}_{n,p}.$
\begin{lemma} \label{vt1}
For every $x \in [0,\infty)$, we have
\begin{equation}
\lim_{n \rightarrow \infty}n\mathcal{L}_{n,p}((t-x),x)= 1,
\end{equation}
\begin{equation} \label{vt2}
\lim_{n \rightarrow \infty}n\mathcal{L}_{n,p}((t-x)^2,x)= x(2+x).
\end{equation}
\end{lemma}

\begin{theorem}\label{vt3}
If any $f\in C_{x^2}^*[0,\infty)$ such that $f',f''\in C_{x^2}^*[0,\infty)$ and $x \in[0,\infty)$ then, we have
\begin{eqnarray*}
\lim\limits_{n\rightarrow\infty}n\bigg[\mathcal{L}_{n,p}(f,x)-f(x)\bigg] & = & f'(x)+ \big\{(x^2+ 2x)/2\big\}f''(x).\\
\end{eqnarray*}
\end{theorem}
\textbf{Proof.}
Using Taylor's expansion to prove this identity
\begin{eqnarray*}
f(t)-f(x)=(t-x)f'(x)+\frac{1}{2!} f''(x)(t-x)^2+ r(t,x)(t-x)^2,
\end{eqnarray*}
where $r(t,x)$ is Peano form of the remainder, $r(t,x)\in C_{B}[0,\infty)$ and $ \lim_{t
\rightarrow x} r(t,x)=0$. Applying $\mathcal{L}_{n,p}$ to above, we obtain
\begin{equation*}
n[\mathcal{L}_{n,p}(f,x)-f(x)]=f'(x)n\mathcal{L}_{n,p}(t-x,x) +\frac{n}{2!}f''(x)\mathcal{L}_{n,p}((t-x)^2,x)+n \mathcal{L}_{n,p}(r(t,x)(t-x)^2,x).
\end{equation*}
By Cauchy-Schwarz inequality, we have
\newline
\begin{equation} \label{et2}
\mathcal{L}_{n,p}(r(t,x)(t-x)^2,x)\leq
\sqrt{\mathcal{L}_{n,p}(r(t,x)^2,x)}
\sqrt{\mathcal{L}_{n,p}((t-x)^4,x)}.
\end{equation}We observe that $r^2(x,x)=0$ and $r^2(t,x) \in C_{x^2}[0,\infty)$. Then it follows that
\begin{equation} \label{et3}
\lim_{n\rightarrow \infty}n\mathcal{L}_{n,p}(r(t,x)^2,x)=r^2(x,x)=0,
\end{equation}
uniformly with respect to $x\in [0,A]$, where $A>0$. Now from (\ref{et2}), (\ref{et3}) and Lemma \ref{vt1}, we
obtain
\begin{equation*}
\lim_{n\rightarrow \infty}n\mathcal{L}_{n,p}(r(t,x)(t-x)^2,x)=0.
\end{equation*}
Hence,
\begin{eqnarray*}
&&\lim_{n \rightarrow \infty}n[\mathcal{L}_{n,p}(f,x)-f(x)]\\
&& =\lim_{n \rightarrow \infty}\left(f'(x) n\mathcal{L}_{n,p}(t-x,x)+\frac{n}{2}f''(x)\mathcal{L}_{n,p}((t-x)^2,x)
+n \mathcal{L}_{n,p}(r(t,x)(t-x)^2,x)\right)\\
&& = f'(x)+\big\{x(x+2)/2\big\} f''(x),
\end{eqnarray*}
which completes the proof.
\qed

\section{Error Estimation}
The usual modulus of continuity of $f$ on the closed interval $[0, b]$ is defined
by
$$\omega_b(f,\delta) =\sup_{|t-x|\leq\delta,\, x,t\in[0,b]}|f(t)-f(x)|,\,\,  b>0.$$
It is well known that, for a function $f\in E$, $$ \lim_{\delta\rightarrow 0^+}\omega_b(f,\delta)=0,$$
where\\
$$E:=\left\{f\in C[0,\infty):\lim_{x\rightarrow\infty}\frac{f(x)}{1+x^2}\,\, is\,\, finite \right\}.$$
The next theorem gives the rate of convergence of the operators $\mathcal{L}_{n,p}(f,x)$ to  $f(x),$ for all $f \in E.$

\begin{theorem}  \label{t2}
Let $f\in E$ and $\omega_{b+1}(f,\delta)$ be its modulus of continuity on the
finite interval $[0,b+1]\subset[0,\infty)$, where $a>0$. Then we have
\begin{equation*}
\| \mathcal{L}_{n,p}(f,x)-f \|_{C[0,b]} \leq
M_f(1+b^2)\delta_n(b)+2\omega_{b+1}\left(f,\sqrt{\delta_n(b)}\right).
\end{equation*}
\end{theorem}
\textbf{Proof.}
The proof is based on the following inequality
\begin{equation}\label{t3}
\|\mathcal{L}_{n,p}(f,x)-f \| \leq
M_f(1+b^2)\mathcal{L}_{n,p}((t-x)^2,x)+
\left(1+\frac{\mathcal{L}_{n,p}(|t-x|,x)}{\delta}\right)\omega_{b+1}(f,\delta).
\end{equation}
For all $(x,t)\in [0,b]\times[0,\infty):= S.$
To prove (\ref{t3}), we write\\
$$S=S_1\cup S_2:=\{(x,t):0\leq x\leq b,\, 0\leq t \leq b+1\}\cup\{(x,t):0\leq x\leq b,\,  t> b+1\}.$$
If $(x, t)\in S_1,$ we can write
\begin{equation}\label{o}
|f(t)-f(x)|\leq \omega_{b+1}(f,|t-x|)\leq\left(1+\frac{|t-x|}{\delta}\right)\omega_{b+1}(f,\delta)
\end{equation}
where $\delta > 0.$ On the other hand, if $(x, t)\in S_2,$ using the fact that $t-x > 1$,
we have
\begin{eqnarray}\label{o1}
|f(t)-f(x)|&\leq& M_f(1+x^2+t^2)\\
&\leq& M_f(1+3x^2+2(t-x)^2)\nonumber\\
&\leq& N_f(1+b^2)(t-x)^2\nonumber
\end{eqnarray}
where $N_f = 6M_f.$ Combining (\ref{o}) and (\ref{o1}), we get (\ref{t3}).
Now from (\ref{t3}) it follows that
\begin{eqnarray*}
|\mathcal{L}_{n,p}(f,x)-f(x)|&\leq & N_f(1+b^2)\mathcal{L}_{n,p}((t-x)^2,x)+\left(1+\frac{\mathcal{L}_{n,p}(|t-x|,x)}{\delta}\right)\omega_{b+1}(f,\delta)\\
&\leq& N_f(1+b^2)\mathcal{L}_{n,p}((t-x)^2,x)+\left(1+\frac{{[\mathcal{L}_{n,p}((t-x)^2,x)]}^{1/2}}{\delta}\right)\omega_{b+1}(f,\delta).\\
\end{eqnarray*}
By Lemma \ref{l1},  we have
$$\mathcal{L}_{n,p}(t-x)^2\leq\delta_n(b).$$
\begin{equation*}
\| \mathcal{L}_{n,p}(f,x)-f \| \leq
N_f(1+b^2)\delta_n(b)+\left(1+\frac{\sqrt{\delta_n(b)}}{\delta}\right)\omega_{b+1}(f,\delta).
\end{equation*}
Choosing $\delta =\sqrt{\delta_n(b)},$ we get the desired estimation.
\qed

\section{Stancu approach}
In \cite{s} Stancu introduced the following generalization of Bernstein polynomials
\begin{equation}
S_n^\alpha (f,x)=\sum_{k=0}^n f\left(\frac{k}{n}\right)P_{n,\alpha}^k(x),0\leq x \leq 1,
\end{equation}
where $\displaystyle P_{n,\alpha}^k(x)= \binom{n}{k}\frac{\prod_{s=0}^{k-1}(x+\alpha s)\prod_{s=0}^{n-k-1}(1-x+\alpha s)}{\prod_{s=0}^{n-1}(1+\alpha s)}.$\\
\indent We get the classical Bernstein polynomials by putting $\alpha=0$.
Starting with two parameter $\alpha,\beta$ satisfying the
condition $0\leq \alpha \leq \beta$. In 1983, the other
generalization of Stancu opreators was given in \cite{s1} and
studied the linear positive operators $S_n^{\alpha, \beta}:
C[0,1]\rightarrow C[0,1]$ defined for any $f\in C[0,1]$ as
follows:

\begin{equation}
S_n^{\alpha, \beta}(f,x)=\sum_{k=0}^n
p'_{n,k}(x)f\left(\frac{k+\alpha}{n+\beta}\right), 0\leq x \leq 1,
\end{equation}
where $ \displaystyle p'_{n,k}(x)=\binom{n}{k}x^k(1-x)^{n-k}$ is the Bernstein basis
function(cf. \cite{B}).\\

\indent Recently, Ibrahim \cite{i} introduced Stancu-Chlodowsky polynomial and investigated convergence and approximation properties of these operators. Motivated by such types of operators, we introduce a Stancu type generalization of the Baskakov-Schurer-Sz\'{a}sz  type operators (\ref{eq1}) as follows:

\begin{equation}\label{eq2}
\mathcal{L}_{n,p}^{(\alpha,\beta)}(f,x) = (n+p)\sum_{k=0}^\infty b_{n,p}^k(x) \int_0^{\infty} s_{n,p}^{k}(t)f\left(\frac{nt+\alpha}{n+\beta}\right) dt,
\end{equation}
where $b_{n,p}^k(x)$ and $s_{n,p}^{k}(t)$ defined as same in (\ref{eq1}). The operators $\mathcal{L}_{n,p}^{(\alpha,\beta)}(f,x)$ in (\ref{eq2}) are called Baskakov-Schurer-Sz\'{a}sz-Stancu operators. For $\alpha=0, \,\beta=0$ the operators (\ref{eq2}) reduce to the operators (\ref{eq1}).\\
\newline
We know that $$\sum_{k=0}^{\infty} b_{n,p}^{k}(t)=1 ,\text{   } \int_0^{\infty} b_{n,p}^k(t)=\frac{1}{n + p-1},$$  and $$\sum_{k=0}^{\infty} s_{n,p}^k(x)=1 ,\text{   } \int_0^{\infty}
s_{n,p}^k(x)=\frac{1}{n+p}.$$
Before proceeding further, we need some lemmas for proving our next results.

\begin{lemma} \label{sl1} Let $e_m(t) =t^m$, $m = 0,1,2.$ we have
\begin{eqnarray*}
\mathcal{L}_{n,p}^{(\alpha,\beta)}(e_0,x)&&=1,\\
\mathcal{L}_{n,p}^{(\alpha,\beta)}(e_1,x)&& = \frac{(n x+ \alpha)(n+p) + n}{(n+p)(n+\beta)},\\
\mathcal{L}_{n,p}^{(\alpha,\beta)}(e_2,x)&&= \bigg(\frac{n^2 (n+p+1)}{(n+p)(n+\beta)^2}\bigg) x^2 + \left(\frac{4n^2 + 2 n \alpha (n+p)}{(n+p) (n+\beta)^2} \right) x +\bigg[ \frac{2 n^2+ 2 n\alpha (n+p) + \alpha^2 (n+p)^2 }{(n+p)^2(n+\beta)^2}\bigg].
\end{eqnarray*}
\end{lemma}
\textbf{Proof.}
We observe that, $\mathcal{L}_{n,p}^{(\alpha,\beta)}(e_0,x)=\mathcal{L}_{n,p}(e_0,x)=1,$
\begin{eqnarray*}
\mathcal{L}_{n,p}^{(\alpha,\beta)}(e_1,x)&=&\frac{n}{n+\beta}\mathcal{L}_{n,p}(e_1,x)+\frac{\alpha}{n+\beta}\mathcal{L}_{n,p}(e_1,x)\\
&=& \frac{n}{n+\beta}\bigg[x+\frac{1}{n+p}\bigg] +\frac{\alpha}{n+\beta} = \frac{(n x+ \alpha)(n+p) + n}{(n+p)(n+\beta)}.
\end{eqnarray*}
Finally

\begin{eqnarray*}
\mathcal{L}_{n,p}^{(\alpha,\beta)}(e_2,x)&=&\frac{n^2}{(n+\beta)^2}\mathcal{L}_{n,p}(e_2,x)+
\frac{2n\alpha}{(n+\beta)^2}\mathcal{L}_{n,p}(e_1,x)+\frac{\alpha^2}{(n+\beta)^2}\mathcal{L}_{n,p}(e_0,x)\\
&=&\frac{n^2}{(n+\beta)^2}\left[\bigg(1+\frac{1}{n+p}\bigg) x^2+ \frac{4 x}{(n+p)} +\frac{2}{(n+p)^2}\right]+\frac{2n\alpha}{(n+\beta)^2}\left(x+\frac{1}{n+p}\right)+ \frac{\alpha^2}{(n+\beta)^2}\\
&=& \bigg(\frac{n^2 (n+p+1)}{(n+p)(n+\beta)^2}\bigg) x^2 + \left(\frac{4n^2 + 2 n \alpha (n+p)}{(n+p) (n+\beta)^2} \right) x + \bigg[\frac{2 n^2+ 2 n\alpha (n+p) + \alpha^2 (n+p)^2 }{(n+p)^2(n+\beta)^2}\bigg].
\end{eqnarray*}
\qed

\begin{lemma}
\begin{eqnarray*}
(i) \mathcal{L}_{n,p}^{(\alpha,\beta)}((t-x);x):&=&\mu_{n,p}^1=
\frac{n(1+\alpha)+\alpha p-\beta x (n+p)}{(n+p)(n+\beta)}.\\
(ii) \mathcal{L}_{n,p}^{(\alpha,\beta)}((t-x)^2;x):&=&\mu_{n,p}^2=
\left[\frac{n^2 + \beta^2(n+ p)}{(n+p)(n+\beta)^2}\right] x^2+ \left[\frac{2 n^2 - 2 n \beta (1+\alpha)- 2\alpha p \beta}{(n+p)(n+\beta)^2}\right] x \\&&~~~~+ \left[\frac{ 2 n^2+ 2 n \alpha (n+p) + \alpha^2 (n+p)^2}{(n+\beta)^2 (n+p)^2}\right].
\end{eqnarray*}
\end{lemma}

\begin{theorem}
For each $f \in C_{\rho}^{*}[0,\infty)$, we have
\begin{equation*}\lim _{n\rightarrow \infty} \| \mathcal{L}_{n,p}^{(\alpha,\beta)} (f)-f \|_{\rho} = 0.\end{equation*}
\end{theorem}
\textbf{Proof.}
Using theorem in \cite{AD} we see that it is sufficient to verify the following three conditions
\begin{equation}\label{l6}
\lim_{n\rightarrow \infty} \|
\mathcal{L}_{n,p}^{(\alpha,\beta)}(t^r,x)-x^r\|_{\rho}=0, \text{  } r=0,1,2.
\end{equation}
Since, $\mathcal{L}_{n,p}^{(\alpha,\beta)}(1,x)=1$, the first condition of (\ref{l6}) is satisfied for $r=0$. Now,

\begin{eqnarray*}
\|\mathcal{L}_{n,p}^{(\alpha,\beta)}(t,x)-x\|_{\rho}&=&\sup_{x\in [0,\infty)} \frac{\mid \mathcal{L}_{n,p}^{(\alpha,\beta)}(t,x)-x \mid}{1+x^2}\\
&\leq & \frac{\beta}{(n+\beta)}\sup_{x\in [0,\infty)}\frac{ x}{1+x^2} + \bigg(\frac{n+ \alpha(n+p)}{(n+\beta)(n+p)}\bigg)\\
&\leq & \frac{\beta}{(n+\beta)} + \bigg(\frac{n+ \alpha(n+p)}{(n+\beta)(n+p)}\bigg)
\end{eqnarray*}
\hspace{3.9cm}$\rightarrow 0$ as $n\rightarrow \infty$.\\
Condition of (\ref{l6}) hold for r = 1.
Similarly, we can write

\begin{eqnarray*}
\|\mathcal{L}_{n,p}^{(\alpha,\beta)}(t^2,x)-x^2\|_{\rho} & = & \sup_{x\in [0,\infty)} \frac{\mid \mathcal{L}_{n,p}^{(\alpha,\beta)}(t^2,x)-x^2 \mid}{1+x^2}\\
&\leq & \left[\frac{n^2(n+p+1)}{(n+p)(n+\beta)^2}-1 \right] \sup_{x\in [0,\infty)} \frac{x^2}{1+x^2}+  \left(\frac{4n^2 + 2 n \alpha (n+p)}{(n+p) (n+\beta)^2} \right) \sup_{x\in [0,\infty)} \frac{x}{1+x^2} \\&&+ \bigg[ \frac{2 n^2+ 2 n\alpha (n+p) + \alpha^2 (n+p)^2 }{(n+p)^2(n+\beta)^2}\bigg].
\end{eqnarray*}
which implies that \\
$$\|\mathcal{L}_{n,p}^{(\alpha,\beta)} (t^2;x)- x^2 \|_{\rho} = 0$$ as $n\rightarrow \infty $. Thus, from Gadzhiev's Theorem in \cite{AD} we obtain the desired result of theorem.
\qed

\begin{theorem}
Let $f \in C_{\rho}[0,\infty)$ and $\omega_{[0,a+1]}(f,\delta)$ be its modulus of continuity on the
finite interval $[0,a+1]\subset[0,\infty)$, where $a>0$. Then, we have
\begin{eqnarray*}
\| \mathcal{L}_{n,p}^{(\alpha,\beta)}(f)-f \|_{[0,a]} \leq C\bigg[(1+a^2)\delta_n(x)+2~ \omega_{[0,a+1]}\left(f,\sqrt{\delta_n(x)}\right)\bigg], ~~~ C > 0.
\end{eqnarray*}
\end{theorem}
\textbf{Proof.} The proof is similar to Theorem \ref{t2} with appropriate modifications. We
omit the details. \qed

\section{Rate of convergence}

We can give some estimations of the errors
$|\mathcal{L}_{n,p}^{(\alpha,\beta)}(f)-f |,$ $n\in \mathbb{N}$ for unbounded functions
by using a weighted modulus of smoothness associated to the space $B_{\rho_{\gamma}}{(\mathbb{R}_+)}$.
The weighed modulus of continuity $\Omega_{\rho_{\gamma}}(f;\delta)$ was defined by L\'{o}pez--Moreno in \cite{LM}.
We consider
\begin{equation}\label{r1}
\Omega_{\rho_{\gamma}}(f;\delta) =\sup_{x\geq 0, 0\leq h\leq \delta} \frac{|f(x+h)-f(x)|}{1+(x+h)^{2+\gamma}},\,\,  \delta>0, \gamma\geq 0.
\end{equation}
It is evident that for each $f\in B_{\rho_{\gamma}}{(\mathbb{R}_+)}, \  \Omega_{\rho_{\gamma}}(f; \cdot)$ is well defined and

$$\Omega_{\rho_{\gamma}}(f;\delta)\leq 2\| f \|_{\rho_{\gamma}}.$$
The weighted modulus of smoothness $\Omega_{\rho_{\gamma}}(f; \cdot)$ possesses
the following properties.
\begin{align} \label{p1}
&(i) \Omega_{\rho_{\gamma}}(f; \lambda \delta)\leq (\lambda+1) \Omega_{\rho_{\gamma}}(f;\delta),\delta>0, \lambda>0 \\
&(ii) \Omega_{\rho_{\gamma}}(f; n \delta)\leq n\Omega_{\rho_{\gamma}}(f; \delta), ~~n\in\mathbb{N}\nonumber\\
&(iii) \lim\limits_{\delta \rightarrow  0}\Omega_{\rho_{\gamma}}(f; \delta)= 0.\nonumber
\end{align}
Now, we are ready to prove our next theorem by using above properties.

\begin{theorem}
For all non-decreasing $f \in B_{\rho_{\gamma}}{(\mathbb{R}_+)}$, we have
\begin{equation*}
| \mathcal{L}_{n,p}^{\alpha,\beta} (f,x)-f| \leq \sqrt{\mathcal{L}_{n,p}^{(\alpha,\beta)}(\nu^2_{x,\gamma};x)}\left( 1 + \frac{1}{\delta}\sqrt{\mathcal{L}_{n,p}^{(\alpha,\beta)}(\Psi_{x}^2;x)}\right)\Omega_{\rho_{\gamma}}(f; \delta),
\end{equation*}
$x\geq 0,~ \delta> 0, ~ n\in \mathbb{N},$ where $$ \nu_{x,\gamma}(t):=1+(x+|t-x|)^{2+\gamma}, ~~ \Psi_{x}(t):=|t-x|, ~ t\geq 0.$$
\end{theorem}
\textbf{Proof.}
Let $n\in\mathbb{N}$ and $f \in B_{\rho_\gamma }(\mathbb{R_+}).$ From (\ref{r1}) and (\ref{p1}), we can write

\begin{eqnarray*}
|f(t)-f(x)| &\leq& \bigg(1+(x + |t-x|)^{2+\gamma}\bigg)\bigg(1+\frac{1}{\delta} |t-x|\bigg)\Omega_{\rho_{\gamma}}(f; \delta)\\
&=&\nu_{x,\gamma}(t)\bigg(1+\frac{1}{\delta}\Psi_x(t)\bigg)\Omega_{\rho_{\gamma}}(f; \delta).
\end{eqnarray*}
Now, applying operator $\mathcal{L}_{n,p}^{(\alpha,\beta)}$ on above inequality, we get

\begin{eqnarray*}
| \mathcal{L}_{n,p}^{(\alpha,\beta)} (f,x)-f(x)|
&\leq & \Omega_{\rho_{\gamma}}(f; \delta)\mathcal{L}_{n,p}^{(\alpha,\beta)}\bigg(\nu_{x,\gamma}\bigg(1+\frac{1}{\delta}\Psi_x\bigg);x\bigg)
\end{eqnarray*}
\begin{eqnarray} \label{p2}
~~~~~~~~~~~~~~~~~~~~~~~~~\hspace{2cm}&\leq &\Omega_{\rho_{\gamma}}(f; \delta) \bigg(\mathcal{L}_{n,p}^{(\alpha,\beta)}(\nu_{x,\gamma};x)+\mathcal{L}_{n,p}^{(\alpha,\beta)}\bigg(\frac{\nu_{x,\gamma} \Psi_x}{\delta};x\bigg)\bigg).
\end{eqnarray}
By using the Cauchy-Schwartz inequality, we obtain
\begin{eqnarray*}
\mathcal{L}_{n,p}^{(\alpha,\beta)}\bigg(\frac{\nu_{x,\gamma} \Psi_{x}}{\delta};x\bigg) &\leq & \bigg\{\mathcal{L}_{n,p}^{(\alpha,\beta)}\big(({\nu_{x,\gamma})^2;x)}\bigg\}^{1/2} \bigg\{\mathcal{L}_{n,p}^{(\alpha,\beta)}\bigg(\bigg(\frac{\Psi_x}{\delta}\bigg)^2;x\bigg)\bigg\}^{1/2}\\
&=&\frac{1}{\delta}\bigg\{\mathcal{L}_{n,p}^{(\alpha,\beta)}({\nu^2_{x,\gamma};x)}\bigg\}^{1/2} \bigg\{\mathcal{L}_{n,p}^{(\alpha,\beta)}({\Psi_x}^2;x)\bigg\}^{1/2}.
\end{eqnarray*}
Now, by (\ref{p2}), we get
\begin{equation*}
| \mathcal{L}_{n,p}^{(\alpha,\beta)} (f,x)-f| \leq \sqrt{\mathcal{L}_{n,p}^{(\alpha,\beta)}(\nu^2_{x,\gamma};x)}\left( 1 + \frac{1}{\delta}\sqrt{\mathcal{L}_{n,p}^{(\alpha,\beta)}(\Psi_{x}^2;x)}\right)\Omega_{\rho_{\gamma}}(f; \delta).
\end{equation*}

\section{Better approximation}
It is well know that the operators preserve constant as well as linear functions. To make the convergence faster, King \cite{kin} proposed an approach  to modify the classical Bernstein polynomials, so that this sequence preserves two test functions $e_0$ and $e_1$.
 After this several researchers have studied that many approximating operators $L$, possess these properties i.e.
 $ L(e_i,x)=e_i(x)$ where $e_i(x)=x^i(i=0,1)$, for examples Bernstein, Baskakov and Baskakov-Durrmeyer-Stancu operators.\\
\indent As the operators $\mathcal{L}_{n,p}^{(\alpha,\beta)}$  introduced in (\ref{eq2}) preserve only the constant functions so further modification of these operators is proposed to be made so that the modified operators preserve the constant as well as linear functions, for this purpose the modification of $\mathcal{L}_{n,p}^{(\alpha,\beta)}$ as follows:

\begin{equation*}
\mathcal{L}_{n,p}^{*(\alpha,\beta)}(f,x) = (n+p)\sum_{k=0}^\infty b_{n,p}^k (r_n(x)) \int_0^{\infty/A} s_{n,p}^{k}(t)f\left(\frac{n t+\alpha}{n+\beta}\right) dt
\end{equation*}
where $r_n(x)=\bigg(\frac{(n+\beta)x-\alpha}{n}-\frac{1}{n+p}\bigg)$ and $x\in I_n=\bigg[\frac{n+\alpha(n+p)}{(n+\beta)(n+p)},\infty \bigg)$.
\begin{lemma}
For each $x\in I_n$, we have
$$\mathcal{L^{*}}_{n,p}^{(\alpha,\beta)}(1,x)=1 ,\text{ } \mathcal{L^{*}}_{n,p}^{(\alpha,\beta)}(t,x) = x,$$
\begin{eqnarray*}
\mathcal{L^{*}}_{n,p}^{(\alpha,\beta)}(t^2,x)&=& \bigg(\frac{ n^2(2(n+p)+1)+ (n+p)(n+\beta)^2}{(n+p)(n+\beta)^2} \bigg) x^2 + \left(\frac{6 n^2 + 4 n \alpha (n+p)}{(n+p) (n + \beta)^2} \right)x \\ &&+
\left(\frac{3 n^2 + 4 n \alpha (n+p) + 2 \alpha^2(n+p)^2}{(n+p)^2 (n+\beta)^2}\right).
\end{eqnarray*}
\end{lemma}

\begin{lemma}
For $x\in I_n$, the following holds,
$$\widetilde{\mu}_{n,p}^{1}(x)=\mathcal{L^{*}}_{n,p}^{(\alpha,\beta)}(t-x,x)=0,$$
\begin{eqnarray*}
\widetilde{\mu}_{n,p}^{2}(x) = \mathcal{L^{*}}_{n,p}^{(\alpha,\beta)}((t-x)^2,x)&=& \bigg(\frac{ n^2(2(n+p)+1)}{(n+p)(n+\beta)^2} \bigg) x^2 + \left(\frac{6 n^2 + 4 n \alpha (n+p)}{(n+p) (n + \beta)^2} \right)x \\ &&+
\left(\frac{3 n^2 + 4 n \alpha (n+p) + 2 \alpha^2(n+p)^2}{(n+p)^2 (n+\beta)^2}\right).
\end{eqnarray*}
\end{lemma}

\begin{theorem}
Let $f\in C_B(I_n)$ and $x\in I_n$. Then, there exist an absolute constant $C>0$ such that
$$\bigg|  \mathcal{L^{*}}_{n,p}^{(\alpha,\beta)}(f,x)-f(x)\bigg| \leq C\omega_2\left(f,\sqrt{\widetilde{\mu}_{n,p}^{2}(x)} ~\right).$$
\end{theorem}
\textbf{Proof.}
Let $g \in C_B(I_n)$ and $x,t \in I_n$. By Taylor's expansion we have
\begin{equation}
g(t)=g(x)+ (t-x)g'(x)+\int_x^t(t-u)g''(u)du.
\end{equation}
Applying $\mathcal{L^{*}}_{n,p}^{(\alpha,\beta)}$, we get
\begin{equation*}
\mathcal{L^{*}}_{n,p}^{(\alpha,\beta)}(g,x)-g(x) = g'(x)\mathcal{L^{*}}_{n,p}^{(\alpha,\beta)}((t-x),x)+\mathcal{L^{*}}_{n,p}^{(\alpha,\beta)}\left(\int_x^t(t-u)g''(u)du,x\right).
\end{equation*}
Now, we have $\displaystyle \bigg|\int_x^t(t-x)g''(u)du\bigg| \leq (t-x)^2\|g''\|$,
$$ \bigg| \mathcal{L^{*}}_{n,p}^{(\alpha,\beta)}(g,x)-g(x)\bigg| \leq \mathcal{L^{*}}_{n,p}^{(\alpha,\beta)}((t-x)^2,x)\|g''\|=\widetilde{\mu}_{n,p}^{2}\|g''\|.$$
Since $\bigg|\mathcal{L^{*}}_{n,p}^{(\alpha,\beta)}(f,x)\bigg|\leq \|f\|$,
\begin{eqnarray*}
\bigg| \mathcal{L^{*}}_{n,p}^{(\alpha,\beta)}(f,x)-f(x)\bigg| &\leq& \bigg| \mathcal{L^{*}}_{n,p}^{(\alpha,\beta)} (f-g,x)-(f-g)(x)\bigg| +\bigg|\mathcal{L^{*}}_{n,p}^{(\alpha,\beta)}(g,x)-g(x)\bigg|\\
&\leq& 2\|f-g\|+\widetilde{\mu}_{n,p}^{2}\|g''\|.
\end{eqnarray*}
Taking infimum overall $g\in C^2(I_n)$, we obtain
$$\bigg|\mathcal{L^{*}}_{n,p}^{(\alpha,\beta)}(f,x)-f(x)\bigg|\leq K_2(f,\widetilde{\mu}_{n,p}^{2}).$$
In view of (\ref{1}), we have
$$\bigg|\mathcal{L^{*}}_{n,p}^{(\alpha,\beta)}(f,x)-f(x)\bigg| \leq C \omega_2\left(f,\sqrt{\widetilde{\mu}_{n,p}^{2}}\right),$$
which proves the theorem.

\section{$q$-Baskakov-Schurer-Sz\'{a}sz operator}
Let $p, k, n\in\mathbb{N}, A > 0$ and $f$ be a real valued continuous function on the interval $[0,\infty)$. In \cite{IY} Y\"{u}ksel defined the $q$-Baskakov-Schurer-Sz\'{a}sz type linear positive operators as

\begin{equation}\label{qb}
\mathcal{L}^{q}_{n,p}(f,x)= [n+p]_q \sum_{k=0}^{\infty} b^k_{n,p}(x;q)\int_{0}^{\frac{\infty}{A(1-q)}} f(t) s^k_{n,p}(t;q) d_q t
\end{equation}
where
$$\displaystyle b^k_{n,p}(x;q)=\left[\begin{array}{c} n+p+k-1\\k \end{array}\right]_q q^{k^2}\frac{x^k}{(1+x)_q^{n+k+p}},$$
and\,\, $\displaystyle s^k_{n,p}(t;q)= e^{-[n+p]_qt}\frac{{([n+p]_qt)}^k}{k!}.$\\
\newline
If $q = 1$ then the operators $\mathcal{L}^q_{n,p}(f,x)$
are reduced to Baskakov-Schurer-Sz\'{a}sz type operators defined in (\ref{eq1}).

\begin{lemma}\cite{IY}\label{ql1}
Let $e_m(t) =t^m$, $m = 0,1,2.$ we have
\begin{itemize}
\item[(i)] $\mathcal{L}^q_{n,p}(e_0;x)=1,$
\item[(ii)] $\mathcal{L}^q_{n,p}(e_1;x)=\frac{x}{q^2}+\frac{1}{q[n+p]_q},$
\item[(iii)] $\mathcal{L}^q_{n,p}(e_2;x)=\frac{[n+p+1]_q}{q^{6}[n+p]_q}x^2+\frac{1+2q+q^2}{q^5[n+p]_q}x+\frac{1+q}{q^3[n+p]^2_q}.$
\end{itemize}
\end{lemma}

\begin{lemma}\cite{IY}\label{al1}
Let $ q\in (0,1)$ and  $p\in \mathbb{N}$. Then following hold:\\
$\delta_{n,p}(x;q):=\mathcal{L}^q_{n,p}((t-x)^2;x)\leq \frac{9}{q^6}\left(1-q^3+\frac{1}{[n+p]_q}\right)(x+1)^2.$
\end{lemma}

\begin{remark}\label{rem1}
$\alpha_{n,p}(x;q):=\mathcal{L}^q_{n,p}((t-x);x)=\bigg(1-\frac{1}{q^2}\bigg) x+\frac{1}{q[n+p]_q}.$
\end{remark}
Now, we establish a Voronovskaja type asymptotic formula for the operators $\mathcal{L}^q_{n,p}.$

\begin{theorem}\label{vothm2}
Let $f$  be bounded and integrable on the interval $[0,\infty).$ First and second derivatives of $f$ exists at a fixed point $x\in [0,\infty)$  and $q=q_n\in (0,1)$ such that $q=q_n \rightarrow 1$ as $n\rightarrow \infty$, then
\begin{eqnarray*}
\lim\limits_{n\rightarrow\infty} [n]_q\big[\mathcal{L}^q_{n,p}(f,x)-f(x)\big] & = & f'(x) +  x f''(x).
\end{eqnarray*}
\end{theorem}
\textbf{Proof.}
In order to prove this identity we use Taylor's expansion
\begin{eqnarray*}
f(t)-f(x)=(t-x)f'(x) + (t-x)^2\bigg( \frac{1}{2!} f''(x) + \xi(t-x)\bigg),
\end{eqnarray*}
where $\xi$ is bounded and  $ \lim_{t\rightarrow 0} \xi(t)=0$. Applying $\mathcal{L}^q_{n,p}$ to above, we obtain
\begin{eqnarray*}
[n]_q[\mathcal{L}^q_{n,p}(f,x)-f(x)]&=& f'(x)[n]_q\mathcal{L}^q_{n,p}(t-x,x) +\frac{[n]_q}{2!}f''(x)\mathcal{L}^q_{n,p}((t-x)^2,x)+ [n]_q \mathcal{L}^q_{n,p}(\xi(t-x)(t-x)^2,x)\\
&&= [n]_q f'(x)\alpha_n(q_n,x)+ [n]_q\frac{1}{2} f''(x)\delta_n(q_n,x)+ [n]_q \mathcal{L}^q_{n,p}(\xi(t-x)(t-x)^2,x)
\end{eqnarray*}
where $\alpha_n(q_n,x)$ and $\delta_n (q, x)$ defined as in remark \ref{rem1} and  lemma \ref{al1}.
Using Cauchy-Schwarz inequality, we have

\begin{equation} \label{e2}
[n]_q \mathcal{L}^q_{n,p}(\xi(t-x)(t-x)^2,x)\leq
\sqrt{\mathcal{L}^q_{n,p}(\xi(t-x)^2,x)}
\sqrt{ [n]_q^2 \mathcal{L}^q_{n,p}((t-x)^4,x)}.
\end{equation}
We observe that $\lim_{t\rightarrow x}\xi^2(t,x)=0$ it follows that $\lim_{[n]_q\rightarrow \infty}[n]_q\mathcal{L}^q_{n,p}(\xi(t-x)(t-x)^2,x)=0$

Hence,
\begin{eqnarray*}
&&\lim_{[n]_q \rightarrow \infty} [n]_q [\mathcal{L}^q_{n,p}(f,x)-f(x)]\\
&& =\lim_{n \rightarrow \infty}\left(f'(x) [n]_q\mathcal{L}^q_{n,p}(t-x,x)+\frac{[n]_q}{2}f''(x)\mathcal{L}^q_{n,p}((t-x)^2,x)
+[n]_q \mathcal{L}^q_{n,p}(\xi(t-x)(t-x)^2,x)\right)\\
&& = f'(x)+ x f''(x),
\end{eqnarray*}
which completes the proof.
\qed

\begin{theorem}
For $q_n\in (0, 1),$ the sequence $\{\mathcal{L}^q_{n,p}\}$
 converges to $f$ uniformly on $[0,A],~ A>0$ for each $f \in C^{*}_{x^2} [0,\infty)$ if and only if $\lim_{n\rightarrow\infty} q_n = 1.$
\end{theorem}
\textbf{Proof.}  The proof is similar to the Theorem 2 in \cite{VG}.

\section{Weighted statistical approximation}
Our next concern is the study of statistical convergence of the sequence of the $q$-Baskakov-Schurer-Sz\'{a}sz operators.
Now, we recall the concept of statistical convergence for
sequences of real numbers which was introduced by Fast \cite{HF} and further studied by many others.

Let $K\subseteq \mathbb{N}$ and $K_{n}=\left\{ j\leq n:j\in
K\right\}.$ Then the $natural~density$ of $K$ is defined by $\delta
(K)={\lim\limits_{n}}~ n^{-1}|K_{n}|$ if the limit exists, where $|K_{n}|$ denotes the
cardinality of the set $K_{n}$.

\parindent=8mm A sequence $x=(x_{j})_{j\geq1}$ of real numbers is said to be $%
statistically$ $convergent$ to $L$ provided that for every $\epsilon >0$ the
set $\{j\in \mathbb{N}:|x_{j}-L|\geq \epsilon \}$ has natural density zero,
i.e. for each $\epsilon >0$,
\begin{equation*}
\lim\limits_{n}\frac{1}{n}|\{j\leq n:|x_{j}-L|\geq \epsilon \}|=0.
\end{equation*}
It is denoted by $st-\lim\limits_{n}x_{n}=L$.\\
\newline
We consider a sequence $q=(q_{n}),$ $q_{n}\in $ $(0,1),$ such that
\begin{equation}\label{s1}
\lim\limits_{n\rightarrow \infty }q_{n}=1.
\end{equation}
The condition (\ref{s1}) guarantees that $[n]_{q_{n}}\rightarrow
\infty $ as $n\rightarrow \infty .$\\
\indent In \cite{DO} Doğru and Kanat defined the Kantorovich-type modification of Lupa\c{s} operators as follows:
\begin{equation}\label{dk}
\tilde{R}_n(f;q;x)=[n+1]\sum_{k=0}^{n}\bigg(\int_{\frac{[k]}{[n+1]}}^{\frac{[k+1]}{[n+1]}} ~ f(t) d_{q}t\bigg)\left(\begin{array}{c}n \\k \end{array}\right) \frac{q^{-k}q^{k(k-1)/2} x^k (1-x)^{(n-k)}}{(1-x+qx)\cdots(1-x+q^{n-1} x)}.
\end{equation}
Doğru and Kanat \cite{DO} proved the following statistical Korovkin-type approximation
theorem for operators (\ref{dk}).

\begin{theorem}
Let $q:=(q_n),~ 0 < q < 1$, be a sequence satisfying the following conditions:\\
\begin{equation}\label{thm10}
 st-\lim_n q_n = 1,~ st - \lim_n q_n^n = a ~( a < 1)~ and~ st - \lim_n \frac{1}{[n]_q} = 0,
\end{equation}
then if $f$ is any monotone increasing function defined on $[0, 1]$, for the positive linear operator $\tilde{R}_n(f;q;x)$, then
$$ st- \lim_n {\Vert \tilde{R}_n(f;q;\cdot) - f \Vert}_{C[0, 1]} = 0$$ holds.
\end{theorem}
In \cite{do26} Do\u{g}ru gave some examples so that $(q_{n})$ is statistically convergent to $1$ but it may not convergent to $1$ in the ordinary case.

\begin{theorem}
Let ${{\mathcal{L}^{q}_{n,p}}}$ be the sequence of the operators (\ref{qb}) and the sequence
$q = (q_n)$ satisfies (\ref{thm10}). Then for any function $f \in C[0,\nu] \subset C[0,\infty),~ \nu > 0 ,$ we have
\begin{equation}
st- \lim_n \Vert \mathcal{L}^q_{n,p}(f; \cdot) - f \Vert = 0,
\end{equation}
where $C[0,\nu]$ denotes the space of all real bounded functions $f$ which are continuous in
$[0,\nu].$
\end{theorem}
\textbf{Proof.} The proof is similar to Theorem 2 \cite{KJ} with appropriate modifications. So, we
omit the details.\\

\begin{theorem}\label{thm3}
Let $\mathcal{L}^q_{n,p}$ be the sequence of the operators (\ref{qb}) and the sequence $q=(q_{n})$ satisfies (\ref{thm10}). Then for any function $f\in C_{B}[0,\infty ),$ we have

\begin{equation*}
st-\lim_{n\rightarrow \infty }{\Vert }\mathcal{L}^q_{n,p}(f;.)-f{\Vert }%
_{\rho}=0.
\end{equation*}
\end{theorem}

\hspace{-0.9cm}
\textbf{Proof.} By Lemma (\ref{ql1})(iii), we have  $\mathcal{L}^q_{n,p}(t^2; x) \leq Cx^2,$ where $C$ is a positive constant, $\mathcal{L}^q_{n,p}(f; x)$ is a sequence of positive linear operator acting from $C_{\rho}[0,\infty)$ to $B_{\rho}[0,\infty)$. \\
Using $\mathcal{L}^q_{n,p}(1;x) = 1,$ it is clear that\\

$$st-\lim_{n}\|\mathcal{L}^q_{n,p}(1;x)-1\|_{\rho}=0.$$
Now, by Lemma (\ref{ql1})(ii), we have

$$\|\mathcal{L}^q_{n,p}(t; x)-x\|_{\rho}= \sup_{x\in [0,\infty)} \frac{|\mathcal{L}^q_{n,p}(t;x)-x|}{1+x^2} \leq \bigg( \frac{1-q^2}{q^2}\bigg) \sup_{x\in [0,\infty)} \frac{x}{1+x^2} + \frac{1}{q [n+p]_q}.$$
Using (\ref{thm10}), we get

$$st-\lim_{n}\left( \bigg( \frac{1-q^2}{q^2}\bigg)  + \frac{1}{q [n+p]_q }\right)=0,$$
then $$st-\lim_{n}\|\mathcal{L}^q_{n,p}(t;x)-x\|_{\rho}=0.$$
Finally, by Lemma(\ref{ql1})(iii), we have

\begin{align*}
\Vert\mathcal{L}^q_{n,p}(f,x)(t^2;q_{n};x)-x^2\Vert_{\rho}
&\leq \left({\frac{[n+p+1]_q}{q^6 [n+p]_q}-1} \right)\sup_{x\in[0,\infty)}\frac{x^2}{1+x^2}\\
&~~~+\left(\frac{1+2q+q^2}{q^5[n+p]_q} \right)\sup_{x\in[0,\infty)}\frac{x}{1+x^2} + \frac{1+q}{q^3[n+p]^2_q}.\\
& \leq \left({\frac{[n+p+1]_q}{q^6 [n+p]_q}-1} \right) +\left(\frac{1+2q+q^2}{q^5[n+p]_q} \right)+ \frac{1+q}{q^3[n+p]^2_q}.\\
\end{align*}
Now, If we choose

\begin{equation*}
\alpha _{n}=  \left({\frac{[n+p+1]_q}{q^6 [n+p]_q}-1} \right),
\end{equation*}%
\begin{equation*}
\beta _{n}=\left(\frac{1+2q+q^2}{q^5[n+p]_q} \right),
\end{equation*}%
\begin{equation*}
\gamma _{n}= \frac{1+q}{q^3[n+p]^2_q},
\end{equation*}%
then by $(\ref{thm10}),$ we can write

\begin{equation}\label{3.1}
st-\lim_{n\rightarrow \infty }\alpha _{n}=0=st-\lim_{n\rightarrow \infty
}\beta _{n}=st-\lim_{n\rightarrow \infty }\gamma _{n}.
\end{equation}
Now for given $\epsilon >0$, we define the following four sets:

\begin{equation*}
{S}=\{k: \Vert \mathcal{L}^q_{n,p}(t^{2};q_{k};x)-x^{2}\Vert_{\rho} \geq \epsilon \},
\end{equation*}%
\begin{equation*}
S_{1}=\lbrace k:\alpha _{k}\geq \frac{\epsilon}{3}\rbrace,
\end{equation*}%
\begin{equation*}
S_{2}=\{k:\beta _{k}\geq \frac{\epsilon }{3}\},
\end{equation*}%
\begin{equation*}
S_{3}=\{k:\gamma _{k}\geq \frac{\epsilon }{3}\}.
\end{equation*}%
It is obvious that $S  \subseteq {S_{1}\cup S_{2}\cup S_{3}}$. Then, we obtain%
\begin{eqnarray*}
\delta \big(\{k &\leq &n:\Vert \mathcal{L}^q_{n,p}(t^{2};x)-x^{2}\Vert_{\rho} \geq
\epsilon \}\big) \\
&&~~~~~~\leq \delta\big(\{k\leq n:\alpha _{k}\geq \frac{\epsilon }{3}\}\big)+\delta
\big(\{k \leq n:\beta _{k}\geq \frac{\epsilon }{3}\}\big)+\delta\big(\{k\leq n:\gamma
_{k}\geq \frac{\epsilon }{3}\}\big).
\end{eqnarray*}%
\newline
Using (\ref{3.1}), we get%
\begin{equation*}
st-\lim_{n\rightarrow \infty }\Vert \mathcal{L}^q_{n,p}(t^{2};x)-x^{2}\Vert_{\rho} =0.
\end{equation*}
Since
\begin{eqnarray*}
\Vert \mathcal{L}^q_{n,p}(f;x)-f\Vert_{\rho} \leq \Vert \mathcal{L}^q_{n,p}(t^{2};x)-x^{2}\Vert_{\rho} +\Vert  \mathcal{L}^q_{n,p}(t;x)-x\Vert_{\rho}+{\Vert \mathcal{L}^q_{n,p}(1;x)-1\Vert}_{\rho} ,\newline
\end{eqnarray*}
we get
\begin{eqnarray*}
st-\lim_{n\rightarrow \infty }\Vert\mathcal{L}^q_{n,p}(f;x)-f\Vert_{\rho}
&\leq &st-\lim_{n\rightarrow \infty }\Vert\mathcal{L}^q_{n,p}(t^{2};x)-x^{2}\Vert_{\rho}\\&& +~st-\lim_{n\rightarrow \infty }\Vert \mathcal{L}^q_{n,p}(t;x)-x\Vert_{\rho}\\
&& + ~st-\lim_{n\rightarrow \infty }\Vert \mathcal{L}^q_{n,p}(1;x)-1\Vert_{\rho} ,
\end{eqnarray*}%
which implies that
\begin{equation*}
st-\lim_{n\rightarrow \infty }\Vert \mathcal{L}^q_{n,p}(f;x)-f\Vert_{\rho} =0.
\end{equation*}
This completes the proof of the theorem.\qed

\vspace{0.6cm}
\hspace{-0.9cm}\textbf{Acknowledgment}\\
The authors are thankful to the referees for making valuable suggestions.

\vspace{0.9cm}
\hspace{-0.9cm}
\textbf{References: }

redpajama_set_name: RedPajamaArXiv


text:
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In contrast to most universities in the Czech Republic, the TUL offers students the possibility to stay on campus, fully equipped with modern student halls and residences.


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Looking at Dwayne Johnson's life from only one angle where he sits at the top of a hill alongside luminaries like Jackie Chan, Tom Cruise, and Robert Downey falls short of the whole story. It is similar to a finishing move in wrestling that excites the fans, not the time taken to perfect it. His journey from a teenager to a football player, to a wrestler and finally a Hollywood star has been one that's hard to fathom. There were a number of instances that would have led him to a path of incarceration and limited him to the lifestyle of wrestling that has claimed countless lives, but they did not.
Dwayne Johnson is the son of the legendary pro wrestler, Rocky Johnson. Being the son of a pro wrestler, most people would think that that meant that he had a comfortable childhood, but that was not the case. His family was always in financial despair to the point that they were evicted from their home. This turned the Rock to stealing so as to fend for himself which often got him in police custody.
At this time of despair, the only positive thing Dwayne did with his life was body building. In an interview with Esquire, Johnson talked about how his dad would make him wake up every morning at 6 AM to go to the gym with him. Working out and lifting weights started at a young age. He looked up to those people he saw in the gym every morning. This was because the only successful people he knew were built. He was also lucky to continue going to school. After bouncing around several schools, Dwayne made a final stop at Freedom High School in Bethlehem. It was at this school he started playing football thanks to his massive body frame and stubbornness.
Dwayne impressed at football. His success on the pitch also reflected on his grades, and he was able to get a scholarship to play football at the University of Miami. Football changed his personality, and he quit living a carefree life opting for a goal-oriented lifestyle.
Unfortunately for him, he did not receive as much game time in the University of Miami football team due to the presence of better players. After graduating from the University, no football team expressed interest and Dwayne fell into depression. He went back to body building and after a lot of thought he decided to become a wrestler like his father.
In 1996, Dwayne made his WWF debut. His career as a wrestler went on to blossom winning some significant matches. The Rock dominated the storylines for close to a decade winning the WWE/WWF championship eight times. After a while, he felt that he needed more and in his mind, Hollywood was his dream destination.
His dreams finally came true when he made his Hollywood debut in the movie the Mummy Returns. This is back in 2001. Despite his minor role, Dwayne impressed the producers. He went on to be among the cast in movies such as Be Cool, Journey to the Center of the Earth 2 and The Tooth Fairy.
His breakthrough came in 2013 when Dwayne was accorded the title of #1 international box office star for his role in San Andréas. He got the same accolade in 2015 for his role in Fast and Furious 7. He lives by the code that no matter what one achieves in life, they should always aim for more.


Car cover for garage kept Z06??
Can anyone offer advice on whether a care cover is beneficial for a new Z06 that will be garage kept and is a weekend driver? I think its a good idea to protect from elements even inside. Any thoughts on best type of cover and where to get it?
I have a satin stretch cover. Don't use it during driving months but put it on for storage. The GM cover is also good.
The toy shed is large building that has a lot of skylights and seems to create its own dust. I have a soft custom cover that keeps the light, dust, bugs, etc. off. I don't remember the name of the cover, I bought it for the C7 stingray in 2013, but it works fine for the Z06. I will head into the toy shed tomorrow and post the brand after I read the tag.
Yes I cover mine inside to keep off the dust. Cover craft form fit.
Got a nice indoor cover for my new ZO6 from Zip Corvette for $159 and free shipping. It's fitted for my Stage 3 Aero, has soft inside and comes with a storage bag. I had a similar cover from Zip for my C6 Grand Sport and it served the purpose well.
You can actually cause more paint issues putting these on and taking them off. When I go out to my garage I like to look at cars not covers.
I found that even when you put on a cover then go to take it off there is still some kind of lint from the cover left on the car. I wound up having to dust off the car anyway. If you want to protect the car from someone rubbing up against it then i would say yes but otherwise i would not bother. I have my car for 18 months with not putting a cover on and no problems. Also remember you have to put the cover on and off properly as there is dust built up on the outside of the cover.
You will need to wash the car every time before you put the cover on as the elements on the car from driving can scratch the paint as you drag the cover on and off.
Start with a California Dust Buster and gently wipe off the dust between weekend drives. I bought a cover when I lived in Chicago and used it during the winter hibernation, however after moving to Florida, I haven't taken it out of the storage bag.
I use one in the winter when it's not driven for a month or 2 at a time.
Thanks for all the great advice. I am going to compromise and get one just for winter hibernation and will make sure the car is washed and polished prior to covering. I just had a detached garage built for the car and my kids car but its not insulated or sheetrocked. I am curious if anyone would have any concerns?
Ordered my first vette a coupe in 2005. When it came in the dealer called and said we are celebrating the 50th anniversary of our dealership and would like to display your car "behind ropes" and if you allow that for the weekend, we will compensate you. There were several compensation offers. I took the car cover and have used it on four Vette's including my two Z06s. I cover only in non driving months.
Quick Reply: Car cover for garage kept Z06??


Premier Moves has a team of builders that are fast, honest and professional. We provide our building services at a fair price which is very competitive without compromising on quality. Our expert builders work together extremely hard to provide the highest standard of workmanship. We specialise in design-build and build-only projects for new builds as well as extensions and full refurbishments.
We are innovative in design and specification, and successfully integrate and latest environmentally-friendly trends into our building service, whilst continuing to seek the latest developments and technologies in a fast-moving industry. We are experienced in refurbishment work and have completed many high-end specialist repair and refurbishment contracts on historical and listed buildings.
We continue to build our reputation for quality on personal contact and service by continually developing our communications and expertise and always aim to complete projects that exceed our client's expectations!
General Property Maintenance and a lot more!


Enjoy desk space in one of the most prestigious office buildings in Belfast. Enjoy desk space in one of the most prestigious office buildings in Belfast.
If you're looking to rent office space in Northern Ireland, the shopping mecca of Lisburn must be on your list of potential locations. Welcome to the Office Genie guide to this city with a great location on the Belfast-Dublin corridor and a wealth of office space to choose from.
Why rent offices in Lisburn?
Lying less than 10 miles south west of Belfast city centre, Lisburn is part of the Belfast Metropolitan Area, but gained city status in 2002 as part of Queen Elizabeth II's Golden Jubilee celebrations. Did we say shopping? It's one of Northern Ireland's main retail destinations, with Bow Street Mall, a pedestrianized city centre packed with independent and chain stores alike, and the outdoor Lisburn Square shopping area. There are two more large retail parks just two minutes from the centre, all of which draws consumers from as far afield as Dublin.
Lisburn's location is a big factor in this. Belfast is just 20 minutes' drive away down the M1, which also connects Lisburn to Lurgan, 15 minutes' drive away, and Dungannon (40 minutes). Newry is 40 minutes' drive down the A1, while Dublin is an hour and three-quarters by car.
Rail services are excellent too, with journey times to Belfast just 10-15 minutes and trains to Bangor taking just over an hour. Newry is a 50-minute rail journey, while three trains every hour make the two-hour trip to Dublin.
There's a clear business feel to the city. Lisburn City Council sponsors the Lisburn City Business Awards, which has been going for more than a decade. Business networking, advice and training are offered through the Lisburn Chamber of Commerce and the Lisburn Entrepreneurs Network, while the Lisburn Enterprise Organisation offers information and advice to new and established businesses in the city.
There's a wealth of office rental options in Lisburn. Starting out and need a cheap desk rental? We've got shared offices that could be perfect. Or if you're after a flexible, all-inclusive, executive suite, take a look at our stock of serviced offices. And when you're looking to make the move to a more permanent home, check out our stock of traditional offices for rent on a long-term lease.
If for any reason you can't find what you need straight away, speak to an office expert by dialling the number you'll find at on this page, or why not widen your search to take in nearby Belfast, Newry or Lurgan? And don't forget to check back often, as our stock is always being updated.
Get a FREE customised report on Lisburn office space. Includes full details about availability and prices.


But based on His promise, we wait for the new heavens and a new earth, where righteousness will dwell.
9 thoughts on "Grace Day"
Hi Victoria, think about this "waiting" time as the real part of your life-God is using this time for His kingdom work in your life and in others. He has His own plan for you that is not the same as for anybody else! Live life for Him in the fullest right now! Don't wait!
Thank you,SRT, for the prayers for me! I have lost 92% of the use of the balance nerve in my right ear. I feel a little better each day. I am learning to compensate and am in therapy to regain my balance. I am hopeful for complete healing but, am peaceful that whatever God decides is best. Glory to God!
Our God is steadfast, faithful and true; therein lies my hope.. Even so Lord Jesus come.
His promises. His faithfulness. I am holding on.
This is when righteousness will prevail- with the new heavens and the new earth . It will not come on November 8. As Rick Warren has said, "We do not put our faith in the politics of man but in the promises of God." Come Lord Jesus, come.


PRINCETON, NJ -- President Barack Obama signed the new $787 billion stimulus bill into law today in Denver with the general support of a majority of the American public, albeit a public sharply divided along party lines -- just as was the case in Congress where the bill was deliberated and voted on over the last several weeks. The bill becomes law at a time when almost 8 in 10 Americans believe the U.S. economy is getting worse, and fewer than 1 in 10 say it's a good time to find a quality job.
Gallup's Feb. 11 update on the public's view of the stimulus bill showed 59% in favor of "a new stimulus package of at least $800 billion". This demonstrates a marginal increase in the level of support found in an earlier February poll, as well as one in late January (52% in both cases). Opposition over the last month has been running between 37% in late January to 33% in Feb. 10 polling.
Polling from Feb. 4 shows support for a stimulus bill described as "reduced by up to $200 billion" received increased support from those who do not approve of the initial bill, described as "at least $800 billion". In fact, the final bill, which passed in both houses of Congress and was signed into law by Obama today, came in slightly below the $800 billion figure used in Gallup's survey question. It is not clear if the small reduction in cost of the final bill to $785 billion would change the level of public support from what Gallup measured when the bill was described as "at least $800 billion".
It has been well documented that public support for a stimulus bill is highly partisan, echoing what occurred in Congress. In Gallup's Feb. 11 update, 82% of Democrats supported the bill compared to just 28% of Republicans.
The final bill includes a long list of ways in which the government money is to be spent in the months and years to come. Gallup's recent polling of a short list of things that could be included in the bill showed public support was highest for money to be spent on education (the final bill includes up to $100 billion for education) and for tax cuts for individuals and families. Americans considered the least important aspects of the bill to be money for state and local governments and tax cuts for businesses.
A review of Gallup polling suggests a majority of Americans supported President Obama when he signed the massive economic stimulus bill into law today, but this support is far from universal. About a third of Americans, mostly Republicans, oppose the bill. Americans believe that some aspects of the spending bill -- including education and tax cuts -- are very important, but are much less likely to believe that other items, such as money for state and local governments, are important.
The bill was signed into law at a time when Americans are reporting extremely negative views of the economy and the job market, and it is clear that this bill is expected to address what is overwhelmingly considered by the public to be the number one problem facing the country.
Two-thirds of Americans say Barack Obama has made a sincere effort to work with the Republicans in Congress on the major issues facing the country. Americans are less likely to say the Republicans (38%) or the Democrats (44%) in Congress are acting in a bipartisan manner.
With the April 15 tax deadline looming, just under half of Americans (48%) say the amount of federal income tax they pay is "about right." Since 1956, Gallup has found a higher percentage expressing this view only once -- in 2003. Additionally, 61% regard the income tax they are paying this year as fair.
Congress' job approval rating increased again this month to 39% -- its highest in four years -- due largely to higher ratings from Democrats.


text:
Edmund Bałuka (* 4. Juni 1933 in Machnówka bei Krosno, Polen; † 8. Januar 2015 in Warschau) war ein polnischer Dissident und Gewerkschafter.

Leben
Edmund Bałuka, der 1933 bei Krosno im Karpatenvorland zur Welt kam, versuchte bereits in den 1950er-Jahren in den Westen zu fliehen. Wegen des Fluchtversuchs wurde er zur Arbeit in einem Steinbruch verurteilt. Später arbeitete er als Schlosser und Kranführer in der Stettiner Werft. Nach den blutig niedergeschlagenen Streiks vom Dezember 1970 stand Bałuka an der Spitze der Arbeiterproteste im Januar 1971. Bekannt wurde er auch im Westen durch die 9-stündigen Gespräche, die er für die Streikleitung mit dem PZPR-Vorsitzenden Edward Gierek, dem Premierminister Piotr Jaroszewicz, dem Innenminister Franciszek Szlachcic und dem General und Armeechef Wojciech Jaruzelski führte. 1973 floh er erneut in den Westen. Wie später bekannt wurde, war die Flucht vom Sicherheitsdienst des Innenministeriums vorbereitet worden, um den unbequemen Bałuka loszuwerden. Nach den Augustabkommen von 1980, infolge derer die oppositionelle Gewerkschaft Solidarność legalisiert wurde, kehrte Bałuka nach Polen zurück. Bei Verhängung des Kriegszustands wurde Edmund Bałuka wieder interniert und spielte danach in der Oppositionsbewegung keine entscheidende Rolle mehr. Er kehrte 1985 nach Frankreich zurück und kam erst nach 1989 nach Polen zurück, wo er 2015 in Warschau verstarb. Edward Bałuka war neben Marian Jurczyk eine der bedeutendsten Personen der Stettiner antikommunistischen Arbeiterbewegung.

Weblinks
 Nie żyje Edmund Bałuka, przywódca strajku w styczniu 1971 r.  Nachruf der Gazeta Wyborcza vom 8. Januar 2015 (polnisch)

Solidarność
Opfer der Diktatur der Volksrepublik Polen
Pole
Geboren 1933
Gestorben 2015
Mann

redpajama_set_name: RedPajamaWikipedia


text:
Q: How to typesafe reduce a Collection of Either to only Right Maybe a stupid question but I just don't get it.
I have a Set<Either<Failure, Success>> and want to output a Set<Success> with Arrow-kt.

A: You can map the set like this for right:
val successes = originalSet.mapNotNull { it.orNull() }.toSet()

or if you want the lefts:
val failures = originalSet.mapNotNull { it.swap().orNull() }.toSet()

The final toSet() is optional if you want to keep it as a Set as mapNotNull is an extension function on Iterable and always returns a List
PS: No stupid questions :)
Update:
It can be done avoiding nullables:
val successes = originalSet
  .map { it.toOption() }
  .filter { it is Some }
  .toSet()

We could potentially add Iterable<Option<A>>.filterSome and Iterable<Either<A, B>.mapAsOptions functions.
Update 2:
That last example returns a Set<Option<Success>>. If you want to unwrap the results without using null then one thing you can try is to fold the Set:
val successes = originalSet
  .fold(emptySet<Success>()) { acc, item ->
    item.fold({ acc }, { acc + it })
  }

This last option (unintended pun) doesn't require the use of Option.

redpajama_set_name: RedPajamaStackExchange


a small group discussion in central London (near Victoria Station)?
If you have had a stillborn baby, would you be willing to share your opinions of the care you have received in a small group discussion? If so, you could help us with our research project which is based at the Universities of Birmingham and Bristol.
We are looking for participants over 18 years of age, living in or near London, who have experienced stillbirth between 1st October 1992 and 1st of April 2017.
Our research project is funded by the Economic and Social Sciences Research Council (ESRC) and its full title is: 'Death Before Birth: Understanding, informing and supporting the choices made by people who have experienced miscarriage, termination, and stillbirth.' Our project partners include: SANDS, Miscarriage Association, ARC and the Human Tissue Authority.
This focus group will take place from 12:00 till 1:45 pm on Tuesday the 8th of May 2018 in London (near Victoria station). Light lunch/refreshments will be provided.
If you would like to ask questions about taking part in the research, please contact Dr Karolina Kuberska (k.kuberska@bham.ac.uk). We can refund travel costs up to £20 pp (receipts will be required). If you have mobility/travel needs please discuss this with us since we may be able to help with any additional travel costs entailed.
f you would like to read a little more about the project, please visit our website www.deathbeforebirthproject.org.


A substance use and mental health disorder treatment industry professional since 2002, Curt Maddon has worked in numerous areas of rehabilitation services including all levels of direct client care as a line director, an executive director and over eight years as an officer of two long-term dual diagnosis treatment facilities. In personal recovery from addiction and mental health disorders since October of 2000, Curt and his family are on the same journey he helps clients and families embark upon, and his message of hope coupled with his passion for recovery has helped many begin the process in an effective and lasting way.
Like many of his colleagues, Curt subscribes to the belief that if everyone in the family system learns how to support the recovery of their loved one by practicing the principles of recovery themselves, successful outcomes are far greater. Without this understanding, shift and action, outcomes are diminished.
Also of paramount importance to the process is the continuum of care provided for all involved – spanning intervention into recovery, and the critical progress milestones of early recovery in between.
Studies have shown that more comprehensive, well-designed plans of action that involve the tincture of time and appropriate connected levels of care and accountability produce better outcomes. If one reviews the basic physician's model of addiction treatment, that is, care for doctors with addiction problems, it is revealed that the basic model is often up to five years from the day of the intervention to the day that the patient is fully clear of all obligations of the recovery continuum of care plan he or she agreed to in order to keep their medical license. The outcomes for these types of programs are astounding – recovery rates in the upper 80th to lower 90th percentile. This is, of course, a point of leverage that pertains to a small percentage of the overall the population of people seeking recovery, however, proper assembly of emotional leverage, particularly in shutting down the disease of enabling, has a powerfully positive effect on the process.
Therefore, Curt predominantly refers to programs who have a solid continuum of care philosophy. This usually means inpatient treatment, followed by a comprehensive aftercare plan involving outpatient treatment and/or clinical services, supportive living, accountability and monitoring, surrounded by a Recovery-Oriented System of Care (ROSC).
Curt has noticed over the years that a person who enters into what is ultimately successful recovery is surrounded by numerous key people. Of course, there are family members – those who know them best and love them the most. Generally speaking however, the family tends to know far less about recovery than other key people. This is, of course, normal and the new approach seeks to engage the family in not just an educational process, but a personal and experiential one as well. The other side of the coin is the other key people involved. This could be the interventionist, the counselor in treatment, sponsor in a 12-step group, etc. Let's say there are 10 key non-family people involved. All are a necessary cog in the machine of the person's early recovery. All play an important role. But Curt's observation is that out of those 10, the recovering person will only truly "connect" with about two or three. And one can not always predict who those connected key people will be.
Therefore, Curt's services focus on these key people along with the other support mechanisms we install for the arduous early recovery process. It is not out of the ordinary for Curt to actually NOT be one of the key connected people, but you better believe that he does all he can to maintain strong relationships with the key connected people during the process. People in early recovery are quite impressionable. This approach yields better results.
The holistic approach to family recovery coupled with a long term continuum of care model and attention to village dynamics for the person seeking recovery is the primary framework of Curt's approach to recovery. Whether providing consulting, intervention or marketing services, this model is fully applied to his work. For more information, rates, and availability, please contact Curt at (213) 500-0696 or click on the secure and confidential contact link below.
NOTE: Curt Maddon does not provide therapy, treatment, or medical services. Needs for any of these services are referred to individuals and/or facilities specializing in an appropriate level of care. Curt Maddon, and associated brands, offers consulting services and therefore any recommendation given does not replace the advice of therapists, treatment providers or medical professionals.


Self Defense is a fantastic team building tool which generates trust, mitigates conflict, encourages communication, and increases collaboration. This also sends a very clear message that you, the employer, care about your employee's safety.
Our military relies heavily on team building and due to its significance have created manuals specific to team building.
The team-building approach addresses the interaction and relationships among team members essential to collaboration and critical to high performance. It accelerates collaboration by allowing the team members to further develop qualities of shared vision, trust, competence, and confidence among team members. It is a streamlined approach designed to break through barriers and boundaries that stagnate team development and, ultimately, performance. Compliments of Globalsecurity.org/military.
A Columbia University study shows that the likelihood of job turnover at an organization with rich company culture is a mere 13.9 percent, whereas the probability of job turnover in poor company cultures is 48.4 percent. Companies with a very successful company culture rely heavily on team building. Austin Self Defense would like to be a partner in building your companies culture!


Nanometrics material characterization systems have been built for maximum flexibility and modularity to ensure a wide range metrology solutions in various process application. Nanometrics' Materials Characterization Group has continuously designed systems that specializes in delivering information about critical material parameters which allow users to develop and control processes in many areas. By supplying optical and electrical properties of layers along with defect, thickness, uniformity and bow information, system users can easily control the deposition process and enhance process equipment productivity and yield.


5cm wide x 1.8m long, this luggage strap has the name embroidered on, providing a quick way to identify a suitcase or bag at the carousel. A TSA approved lock means the strap is secure in transit and helps prevent theft. Produced and despatched within 2-3 working days.


"My inspiration for these products is nothing new. I comb through old museum catalogs, I sneak into old churches, and smell the spines of old books. I guess it's that old stuff that gets me. Those pieces from a time when everything was curly, floral, and covered in gold. The world moves too fast to live like that anymore. That decadence from the turn of the century has died, and I can't bring it back no matter how many episodes of Boardwalk Empire I watch.
What I can do, is try my darnedest to bring some of that aesthetic back through these products in my shop. I'm interested in making stuff that stands out from the expected rectangle of screen printed posters. Stuff that doesn't necessarily need a frame. Something with a unique outline- a silhouette that feels vaguely nostalgic, and reminds you of that time when Gramps brought out his box of old stuff from the war and got all twinkly-eyed with his stories".


First floor office/retail space available at the brand new Centric in University Circle. Tons of glass, high ceilings and flexible floorplan options ready for build-out. Attached parking garage for tenants.


text:
Erardo Cóccaro, vollständiger Name Erardo José Cóccaro, (* 20. Januar 1961 in Montevideo oder Dolores) ist ein ehemaliger uruguayischer Fußballspieler und Trainer.

Karriere

Spielerlaufbahn

Verein
Mittelfeldakteur Cóccaro, dessen Nachname auch in der Schreibweise Cócaro geführt wird, stand von 1981 bis 1988 im Kader des uruguayischen Erstligisten Club Atlético Progreso. In der Spielzeit 1988/89 absolvierte er sechs Ligaspiele (kein Tor) für den argentinischen Verein Gimnasia y Esgrima La Plata. Auch soll er in den Jahren 1988 bis 1989 beim Club Atlético Peñarol unter Vertrag gestanden haben. Allerdings wird er weder 1988 noch 1989 im Kader der "Aurinegros" geführt. 1990 folgte ein Engagement bei Bella Vista. Dort wurde er mit der Mannschaft in jenem Jahr Uruguayischer Meister. Für das Jahr 1991 kehrte er nochmals zu Progreso zurück.

Nationalmannschaft
Cóccaro war Mitglied der A-Nationalmannschaft Uruguays. Mit dieser nahm er an der Copa América 1987 teil und gewann den Titel. Im Verlaufe des Turniers wurde er jedoch nicht eingesetzt.

Erfolge
 Copa América: 1987
 Uruguayischer Meister: 1990

Trainertätigkeit

Junioren- / Reservemannschaften
Nach der aktiven Karriere war er in diversen uruguayischen Vereinen als Jugendtrainer tätig. Als solcher wirkte er 1995, 2003 bis 2004, 2007 bis Mitte 2008 und erneut 2009 bei Progreso. Überdies war er in den Jahren 2000 und 2006 in dieser Funktion bei Peñarol aktiv. 2005 trainierte er die Jugend des Club Atlético Rentistas, im ersten Halbjahr 2009 diejenige des Liverpool FC. Beim Liverpool FC hatte er am 7. Januar 2009 in Nachfolge des nach Honduras gewechselten Gustavo Faral das Team der "Cuarta División" übernommen. Mindestens im Jahr 2014 war er Trainer der U-16 Bella Vistas.

Profimannschaften
Auch im Seniorenbereich trat er als Trainer in Erscheinung. Von 1996 bis Mitte 1997 coachte er die Mannschaft Progresos. Später betreute er – eigentlich im Jahr 2000 für die "Formativas" des Klubs verantwortlich – im 1. Meisterschaftsfinalspiel der Saison 2000 Peñarol bei der 0:1-Niederlage gegen Nacional Montevideo am 7. Dezember 2000, während im Rückspiel mit Julio César Ribas wieder der etatmäßige Trainer der Jahre 1999 bis 2001 dort auf der Bank saß. Bei Progreso hatte er erneut von Mitte 2003 bis 2004 sowie ab Juli 2007 die Cheftrainerrolle inne.

Am 8. August 2013 wurde vermeldet, dass er das Traineramt beim Zweitligisten Canadian Soccer Club übernommen habe. Das Engagement endete Anfang November jenen Jahres mit seinem Rücktritt nach einer Niederlage gegen Central Español.

Einzelnachweise

Fußballnationalspieler (Uruguay)
Fußballspieler (Club Atlético Progreso)
Fußballspieler (Gimnasia y Esgrima La Plata)
Fußballspieler (Club Atlético Bella Vista)
Fußballtrainer (Club Atlético Progreso)
Fußballtrainer (Club Atlético Peñarol)
Fußballtrainer (Canadian Soccer Club)
Südamerikameister (Fußball)
Uruguayischer Meister (Fußball)
Geboren 1961
Uruguayer
Mann