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Where in health am I? : Home / Articles and Reports / Best of Best / Medical Tourism - Have You Decided?
Medical Tourism - Have You Decided?
Jan 31, 2011 | James L. Jones | Best of Best
Jan 31, 2011, James L. Jones, Best of Best
Are you still undecided about getting that medical procedure done outside the country instead of back home? Esmeralda's experience may help you!
image by: Jernej Furman
Well, since this story begins with our driving into Queretaro, Mexico you've probably figured out we went to Mexico. But before we made a decision to participate in medical tourism we checked out the prices back home.
Esmeralda had an Oregon based PPO individual policy that would cover her operation. So we went to a plan gynecologist in Portland and had her evaluated. The doctor ordered an ultrasound at an outside lab, and some lab work, examined her and asked us to come to his office for a consultation. As we were waiting in his consultation room I noticed some of his plaques, certifications, diplomas and licenses. Dr. Plan, as I'll call him, went to medical school in Guadalajara and was a member of the American Society of Gynecologists and Obstetricians.
Dr. Plan agreed Esmeralda needed a hysterectomy, he would be able to do it in about 6 weeks. He thanked us for choosing him and indicated his front desk staff would fill us in on the details. Total cost, hospital and physician, would be $46,000, of which Esmeralda would pay $20,500 out of pocket. The ultrasound and lab we'd had done would be a separate billing item and it would be a two day stay in the hospital.
As we found out in Part I, Esmeralda would have been eligible for two of Mexico's government sponsored health plans, IMSS and Seguro Popular.
But, both had waiting periods of anywhere between 6 months and 3 years and might have required waivers from their coverage of pre-existing conditions. Neither requires a physical exam but IMSS requires you to pass a questionnaire that asks about current and past health problems and can deny applications for prior conditions. As well, she elected not to go to the county hospital.
We had received a referral to a gynecologist we'll call Dr. A. from Esmeralda's cousin who is a nurse at one of the IMSS clincs. "He's very smart, and he's handsome too." She had told us. He also spoke excellent English.
As we were waiting for Dr. A. in his consultation room, I noticed his diplomas and licensures on the wall. Amazing! He went to the same Guadalajara medical school as Dr. Plan and was a member of the same American Society of Gynecology and Obstetrics. I had noticed certificates of membership in a sub-specialty of Reproductive Medicine in both offices also, but we were obviously not interested in that.
Dr. A. listened to us, did a complete history and physical exam, performed his own ultrasound, then sat us down for a talk. He also agreed that Esmeralda needed a hysterectomy and told us the total cost would be $3600 dollars total, all payable by Esmeralda since she would not be using any insurance. His consultation and ultrasound, the hospital, the OR people, the anesthesiologist, the pathologist, the food, and all the medicines were included in the $3600. The hospital part of the $3600 was $1100 and included the two-day stay and all the nursing care.
We agreed. Dr. A. said we could pay him whenever we wanted, the hospital needed their cash up front, upon registration. He opened up his laptop. Asked a few more medical questions, then proceeded to type in his history and physical exam, accessed the hospital by internet, one of the newer private hospitals in the area, and checked the OR schedule.
"How about Thursday, we've been a little busy lately." It was Monday. We were to arrive at 6 AM and Esmeralda would be the first case of the day. He transmitted his medical record so it would be on the chart when we checked into the hospital and entered some pre-op orders.
Esmeralda had a large private room, the TV was included in the hospital fee. And she had an uneventful 2-day stay. The low point was the drive home. Mexico's roads are riddled with potholes and the slightest bump was quite painful to her incision. But we made it and had an uneventful postop visit with Dr. A in about a week.
The only tip I would have for those who choose to have their procedures done in Mexico is to bring your own pain medicine and plan on using it liberally. They're a little stingy on the pain medicine, not because they want to save money, but because they have an older philosophy on pain management, similar to ours in the U.S. 20 years ago before several reports showed physicians were undertreating pain in a lot of patients. Mexico will catch up, but until then bring your own pain medicine.
Good luck with your trip and your procedure and don't drive around or through Mexico. Best to fly and take a cab from the airport, the rental cars are clearly marked as such and sitting targets for whomever. Central Mexico is still safe and once you get there you should have an enjoyable stay and its best not to completely trust those guards who want to back you out of the parking places. But give them a tip anyway.
J. Linder Jones MD MHA practiced emergency medicine for over two decades in Southern California and now writes as a senior correspondent for HealthWorldNet.
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Last Updated : Wednesday, November 16, 2022


Our students have participated in the 8th Hong Kong Geography Olympiad organized by the Hong Kong Geographical Association. Two of them got excellent results and won the Individual Champion and Second runner-up Awards and it is also honourable for us to receive the School First runner-up Award. A total of 696 secondary school students joined the competition.
Besides, Sze Ka Hong (6E) is selected to be one of the representatives of the Hong Kong team to participate in the 15th International Geography Olympiad which will be held in Québec City, Canada from 31st July to 6th August, 2018.


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\section{Introduction and preliminary results}

In studying Prudnikov's  sequence of orthogonal polynomials  (see \cite{YAP})  with  the weight function $x^\alpha \rho_\nu(x)$, where \  $ \rho_{\nu}(x)= 2 x^{\nu/2} K_\nu(2\sqrt x),\ x >0, \nu \ge 0,\ \alpha > -1$ and  $K_\nu(z)$ is  the Macdonald or modified Bessel function  \cite{Bateman}, Vol. II the author interpreted this sequence in terms of the  Laguerre composition orthogonality, involving the differential operator $\theta= xDx$, where $D= d/dx$.  The main aim of this paper is to extend the method, employing classical orthogonal polynomials (Hermite, Laguerre, Jacobi) \cite{Bateman} to find new sequences of orthogonal polynomials with non-classical weights such as, for instance,  the square of Macdonald function and other hypergeometric functions.  In fact,   we will define the composition orthogonality in the following way.

{\bf Definition 1.} {\it Let  $\omega(t), \varphi(t) \ t \in [a,b],\  -\infty \le a < b \le \infty$ be  nonnegative  functions, $\theta = tDt$.    Let $f, g$ be complex-valued functions such that the product of two operators $f(\theta) g(\theta)$ commutes. Then  $f,\ g $ are compositionally orthogonal with respect to the measure $\omega(t)dt$ relatively to the function $\varphi$ if }
$$\int_a^b  f(\theta) g(\theta) \{ \varphi(t)\} \omega(t) dt = 0.\eqno(1.1)$$
We will see in the sequel that this definition suits well with the vector space of polynomials over $\mathbb{R}$, when  the composition orthonormality of the sequence $\{P_n\}_{n\ge 0}$ is defined by

$$\int_a^b  P_n(\theta) P_m(\theta) \{ \varphi(t)\} \omega(t) dt = \delta_{n,m},\eqno(1.2)$$
where $\delta_{n,m},\   n,m\in\mathbb{N}_{0}$ is the Kronecker symbol.   Moreover, for some class of functions $\varphi$ it is possible to transform the left-hand side of the equality (1.2)  to the usual  orthogonality  with respect to a new weight function.   Then  since the $n$-th power of the operator $\theta$ satisfies the Viskov-type identity \cite{Viskov}

$$ \theta^n = \left( xDx\right)^n = x^n D^n x^n,\quad   n \in\mathbb{N}_{0},\eqno(1.3)$$
we can get via integration by parts new properties of the sequence $\{P_n\}_{n\ge 0}$ and its relationship, for instance, with classical orthogonal polynomials.  In fact, let us consider a class of functions $\varphi$ representable by the modified Laplace transform of some nonnegative function $\psi$

$$\varphi(t)= {1\over t} \int_0^\infty e^{-x/t} \psi(x) dx,\ t \in [a,b] \subset (0, \infty).\eqno(1.4)$$
Hence it is easily seen that

$$\theta^k \left\{ t^{-1} e^{-x/t} \right\}  = \left( t D t\right)^k \left\{ t^{-1} e^{-x/t} \right\} = x^k  t^{-1}  e^{-x/t},\quad k \in \mathbb{N}_0,\eqno(1.5)$$
and therefore, differentiating under the integral sign in (1.4), we derive

$$ \theta^k \left\{ \varphi(t)  \right\} = {1\over t} \int_0^\infty e^{-x/t}  x^k \psi(x) dx,\quad  k \in \mathbb{N}_0. \eqno(1.6)$$
It is indeed allowed, for instance,  due to the assumed convergence of the integral

$$ \int_0^\infty e^{-x/b}  x^k \psi(x) dx < \infty, \quad   k \in \mathbb{N}_0.\eqno(1.7)$$
Consequently, returning to (1.2) we write its left-hand side in the form

$$ \int_a^b  P_n(\theta) P_m(\theta) \{ \varphi(t)\} \omega(t) dt  = \int_a^b   \int_0^\infty e^{-x/t} P_n(x) P_m(x) \psi(x)  \omega(t) {dx dt \over t}.\eqno(1.8) $$
The interchange of the order of integration on the right-hand side in (1.8) is permitted by Fubini's theorem under the imposed condition

$$ \int_a^b   \int_0^\infty e^{-x/t}  x^k \psi(x)  \omega(t) {dx dt \over t} < \infty,\quad  k \in \mathbb{N}_0, \eqno(1.9)$$
and, combining with (1.2), we find the equalities

$$\int_a^b  P_n(\theta) P_m(\theta) \{ \varphi(t)\} \omega(t) dt =  \int_0^\infty  P_n(x) P_m(x) \psi(x) \Omega (x) dx = \delta_{n,m},\eqno(1.10)$$
where

$$ \Omega (x) = \int_a^b  e^{-x/t}  \omega(t) {dt \over t},\quad  x >0.\eqno(1.11) $$
Thus we see that the sequence  $\{P_n\}_{n\ge 0}$ is orthonormal over $(0,\infty)$ with respect to the measure $\Omega (x) dx$.  Moreover, one can consider
the left-hand side of the first equality in (1.10) as an  inner product on the vector space of  polynomials over $\mathbb{R}$  (a pre-Hilbert space)

$$\langle p, q \rangle =  \int_a^b  p (\theta) q (\theta) \{ \varphi(t)\} \omega(t) dt,\eqno(1.12)$$
inducing the norm by the equality

$$ ||p||= \sqrt{\langle p, p \rangle} = \left( \int_0^\infty  p^2(x) \psi(x) \Omega (x) dx\right)^{1/2}.\eqno(1.13)$$

\section{The use of classical orthogonal polynomials}

\subsection{Laguerre polynomials}
We begin to consider sequences of orthogonal polynomials which are compositionally orthogonal   with respect to the measure $t^\nu e^{-t} dt$ over $\mathbb{R}_+$ related to   Laguerre polynomials $\{L_n^\nu\}_{n\ge 0},\ \nu > -1$. In fact, letting $\omega(t)=  t^\nu e^{-t},\ t > 0$ in (1.2), we get

$$\int_0^\infty  P_n(\theta) P_m(\theta) \{ \varphi(t)\}  t^\nu e^{-t} dt = \delta_{n,m}.\eqno(2.1)$$
The corresponding integral (1.11) is calculated in \cite{Bateman}, Vol. II, and we obtain

$$\Omega(x)= \int_0^\infty  e^{-x/t -t}  t^{\nu-1} dt = 2 x^{\nu/2} K_\nu\left( 2\sqrt x\right) \equiv \rho_\nu(x),\ x > 0,\eqno(2.2)$$
where $K_\nu(z)$ is the modified Bessel function or Macdonald function \cite{YaL}.  The  function  $\rho_{\nu}$  has  the Mellin-Barnes integral representation in the form (cf. \cite{YAP})
$$
\rho_\nu(x)=  \frac{1}{2\pi i} \int_{\gamma-i\infty}^{\gamma+i\infty} \Gamma(\nu+s) \Gamma (s) x^{-s} ds\ , \quad x, \gamma \in \mathbb{R}_{+},\ \nu  \in \mathbb{R},\eqno(2.3)
$$
where $\Gamma(z)$ is Euler's gamma-function \cite{Bateman}, Vol. I. The asymptotic behavior of the modified Bessel function at infinity and near the origin \cite{Bateman}, Vol. II gives the corresponding values for the  function $\rho_\nu,\ \nu \in \mathbb{R}$.  Precisely, we have
$$\rho_\nu (x)= O\left( x^{(\nu-|\nu|)/2}\right),\  x \to 0,\ \nu\neq 0, \quad  \rho_0(x)= O( \log x),\ x \to 0,\eqno(2.4)$$

$$ \rho_\nu(x)= O\left( x^{\nu/2- 1/4} e^{- 2\sqrt x} \right),\ x \to +\infty.\eqno(2.5)$$
Therefore, if the condition (cf. (1.9))

$$ \int_0^\infty   x^k \rho_\nu(x)  \psi(x) dx   < \infty,\quad  k \in \mathbb{N}_0 \eqno(2.6)$$
holds valid,  we arrive at the following proposition.

{\bf Proposition 1.} {\it Let $\nu > -1, \varphi, \psi$  be  nonnegative functions defined on  $\mathbb{R}_+$ which are related by the modified Laplace transform $(1.4)$.  Then under condition $(2.6)$  the finiteness of the integral

$$ \int_0^\infty   x^k e^{-x/M}  \psi(x) dx   < \infty,\quad  k \in \mathbb{N}_0 \eqno(2.7)$$
for some $M >0$ the sequence $\{P_n\}_{n\ge 0}$ of orthogonal polynomials with respect to the measure $\rho_\nu(x)  \psi(x) dx$ over $\mathbb{R}_+$ is compositionally orthogonal in the sense of Laguerre relatively to the function $\varphi$, i.e.}

$$  \int_0^\infty  P_n(\theta) P_m(\theta) \{ \varphi(t)\}  t^\nu e^{-t} dt= \int_0^\infty  P_n(x) P_m(x)  \rho_\nu(x)  \psi(x) dx = \delta_{n,m}.\eqno(2.8)$$

\begin{proof}  Since  via (2.7) the integral

$$  {1\over t} \int_0^\infty e^{-x/t}  x^k \psi(x) dx,  \quad  k \in \mathbb{N}_0$$
converges uniformly with respect to  $t \in [1/M, M], M >0$, the consecutive differentiation under the integral sign is allowed, and we derive, recalling (1.6)

$$P_n(\theta) P_m(\theta) \{ \varphi(t)\}  =  P_n(\theta) P_m(\theta) \left\{ {1\over t} \int_0^\infty e^{-x/t} \psi(x) dx \right\}$$

$$ =   {1\over t} \int_0^\infty e^{-x/t}  P_n(x) P_m(x) \psi(x) dx.$$
Then

$$ \int_0^\infty  P_n(\theta) P_m(\theta) \{ \varphi(t)\}  t^\nu e^{-t} dt = \lim_{M\to \infty}   \int_{1/M}^M \int_0^\infty e^{-x/t}  P_n(x) P_m(x) \psi(x)   t^{\nu-1}  e^{-t} dx dt $$

$$=  \int_0^\infty  P_n(x) P_m(x)  \rho_\nu(x)  \psi(x) dx = \delta_{n,m},$$
where the latter equality is guaranteed by  condition (2.6) and Fubini's theorem.

\end{proof}

{\bf Remark 1.} {\it Letting $\nu \ge 0, \psi(x)= x^\alpha,\ \alpha > -1$, we find the sequence $\{P^{\nu,\alpha}_n\}_{n\ge 0}$ of Prudnikov's orthogonal polynomials  studied in \cite{YAP}}

$$ \int_0^\infty  P^{\nu,\alpha}_n(x) P^{\nu,\alpha}_m(x)  \rho_\nu(x)  x^\alpha dx =  \Gamma(\alpha +1 ) \int_0^\infty  P_n(\theta) P_m(\theta) \{ t^\alpha\}  t^\nu e^{-t} dt = \delta_{n,m}.\eqno(2.9)$$

\subsection{Hermite polynomials}   Other interesting case is the Hermite orthogonality with respect to the measure $e^{-t^2} dt$ over $\mathbb{R}$. Our method will work on the following even extension  of the modified Laplace transform   (1.4)

$$\varphi( |t |)= {1\over| t|} \int_0^\infty e^{-x/|t| } \psi(x) dx,\ t \in \mathbb{R} \backslash{\{0\}}.\eqno(2.10)$$
Hence when  the Parseval identity for the Mellin transform \cite{Tit} suggests the equality

$$ \int_0^\infty  e^{-x/t -t^2}  {dt\over t} = {1\over 4\pi i} \int_{\gamma-i\infty}^{\gamma+i\infty} \Gamma\left({s\over 2}\right) \Gamma (s)\  x^{-s} ds,\quad   x,\gamma > 0. \eqno(2.11)$$
Invoking the duplication formula for the gamma function \cite{Bateman}, we get the right-hand side of (2.11) in the form

$$  {1\over 4\pi i} \int_{\gamma-i\infty}^{\gamma+i\infty} \Gamma\left({s\over 2}\right) \Gamma (s)\  x^{-s} ds = {1\over 4\pi^{3/2} i} \int_{\gamma/2 -i\infty}^{\gamma /2 +i\infty} \Gamma^2 \left(s\right) \Gamma \left({1\over 2}+ s \right)\ \left( {x\over 2}\right)^{- 2s} ds$$

$$  = {1\over 2\sqrt\pi}\  \rho_{1/2, 2} \left({x^2\over 4}\right),\eqno(2.12)$$
where by $\rho_{\nu, k}(x)$ we denote the ultra-exponential weight function introduced in \cite{YAP}

$$\rho_{\nu, k}(x) =  {1\over 2\pi i} \int_{\gamma-i\infty}^{\gamma+i\infty} \Gamma^k \left(s\right) \Gamma \left(\nu + s \right) x^{-s}ds, \  x,\nu,\gamma >0,  \ k \in \mathbb{N}_0 .\eqno(2.13)$$

{\bf Proposition 2.} {\it Let $t >0,\  x \in \mathbb{R},\  \varphi(t) , \psi (x)$  be  nonnegative functions  which are related by $(2.10)$.  If $\psi$ is even then under conditions $(2.7)$ and

$$ \int_{-\infty}^\infty   |x|^k  \rho_{1/2, 2} \left({x^2\over 4}\right)  \psi(x) dx   < \infty,\quad  k \in \mathbb{N}_0 \eqno(2.14)$$
the sequence $\{P_n\}_{n\ge 0}$ of orthogonal polynomials with respect to the measure $2^{-1}\pi^{-1/2} \rho_{1/2, 2} \left({x^2/4}\right)  \psi(x) dx$ over $\mathbb{R}$ is compositionally orthogonal in the sense of Hermite  relatively to the function $\varphi(|t|)$, i.e.}

$$ \int_{-\infty}^\infty  P_n(\theta) P_m(\theta) \{ \varphi(|t|)\}   e^{-t^2} dt= \int_{-\infty}^\infty  P_n(x) P_m(x)   \rho_{1/2, 2} \left({x^2\over 4}\right)   \psi(x)  { dx\over \sqrt \pi}   = \delta_{n,m}.\eqno(2.15)$$

\begin{proof}  Indeed,  since

$$  \int_{-\infty}^\infty  P_n(\theta) P_m(\theta) \{ \varphi(|t|)\}   e^{-t^2} dt =  \int_{0}^\infty  \left[ P_n(\theta) P_m(\theta) + P_n(- \theta) P_m(- \theta)\right] \{ \varphi(t)\}   e^{-t^2} dt $$
we have due to (1.4), (1.6), (2.7)

$$\left[ P_n(\theta) P_m(\theta) + P_n(- \theta) P_m(- \theta)\right] \{ \varphi(t)\} =  {1\over t} \int_0^\infty e^{-x/t}  \left[  P_n(x) P_m(x)  + P_n(- x) P_m(- x)  \right]  \psi(x) dx.\eqno(2.16)$$
Thus appealing to  (2.11), (2.12),  (2.14) and Fubini's theorem, we obtain finally  from (2.16)

$$   \int_{0}^\infty  \int_0^\infty   e^{-x/t- t^2} \left[  P_n(x) P_m(x)  + P_n(- x) P_m(- x)  \right]  \psi(x)  { dx dt \over t}$$

$$=  \int_0^\infty \left[  P_n(x) P_m(x)  + P_n(- x) P_m(- x)  \right]  \rho_{1/2, 2} \left({x^2\over 4}\right)   \psi(x) {dx\over \sqrt \pi}$$

$$= \int_{-\infty}^\infty  P_n(x) P_m(x)   \rho_{1/2, 2} \left({x^2\over 4}\right)   \psi(x)  { dx\over 2\sqrt \pi}   = \delta_{n,m}.$$

\end{proof}

\subsection{Jacobi  polynomials} The modified sequence $\{P_n^{\alpha,\beta}(2t-1)\}_{n\ge 0}$ of these   classical polynomials is orthogonal with respect to the measure $(1-t)^{\alpha} t^\beta dt,\ \alpha, \beta  > -1$ over the interval $[0,1].$ Therefore  the kernel $\Omega(x),\ x >0$ (1.11) is calculated accordingly, and we obtain

$$\Omega(x) = \int_{0}^1  (1-t)^{\alpha } t^{\beta-1} e^{-x/t} dt =  e^{-x} \int_{0}^\infty   t^{\alpha}\ (1+t)^{-\alpha- \beta-1} e^{-xt} dt= \Gamma(1+\alpha) e^{-x} U(1+\alpha, 1-\beta, x),\eqno(2.17)$$
where $U(a,b,z)$ is the Tricomi function \cite{nist}.  Hence we arrive at

{\bf Proposition 3.} {\it Let $\alpha > -1, \beta  > 0, \varphi, \psi$  be  nonnegative functions defined on  $\mathbb{R}_+$ which are related by the modified Laplace transform $(1.4)$.  Then under the condition

$$\int_0^\infty  x^k e^{-x} \psi(x) dx < \infty, \quad k \in \mathbb{N}_0\eqno(2.18)$$
the sequence $\{P_n\}_{n\ge 0}$ of orthogonal polynomials with respect to the measure $e^{-x} U(1+\alpha, 1-\beta, x) \psi(x) dx$ over $\mathbb{R}_+$ is compositionally orthogonal in the sense of Jacobi relatively to the function $\varphi$, i.e.}

$$  \int_0^1  P_n(\theta) P_m(\theta) \{ \varphi(t)\}  (1-t)^\alpha  t^\beta dt= \Gamma(1+\alpha) \int_0^\infty  P_n(x) P_m(x)  e^{-x} U(1+\alpha, 1-\beta, x) \psi(x) dx = \delta_{n,m}.\eqno(2.19)$$

\begin{proof} Indeed, since

$$ P_n(\theta) P_m(\theta) \{ \varphi(t)\}  = P_n(\theta) P_m(\theta) \left\{ {1\over t} \int_0^\infty e^{-x/t} \psi(x) dx \right\} = {1\over t} \int_0^\infty e^{-x/t}  P_n(x) P_m(x) \psi(x) dx ,$$
where the consecutive differentiation under the integral sign is permitted owing to the estimate

$$ {1\over t} \int_0^\infty e^{-x/t} x^k \psi(x) dx  \le {1\over \delta}  \int_0^\infty e^{-x} x^k \psi(x) dx,\  0 < \delta \le t \le 1,\ k \in \mathbb{N}_0,$$
and the latter integral by $x$ is finite via (2.18).  Hence, taking into account (2.17),

$$  \int_0^1  P_n(\theta) P_m(\theta) \{ \varphi(t)\}  (1-t)^\alpha  t^\beta dt=  \lim_{\delta \to 0+} \int_\delta^1  P_n(\theta) P_m(\theta) \{ \varphi(t)\}  (1-t)^\alpha  t^\beta dt$$

$$= \lim_{\delta \to 0+} \int_\delta^1  \int_0^\infty e^{-x/t}  P_n(x) P_m(x) \psi(x)  (1-t)^\alpha  t^{\beta-1} dx dt $$

$$=   \int_0^1  \int_0^\infty e^{-x/t}  P_n(x) P_m(x) \psi(x)  (1-t)^\alpha  t^{\beta-1} dx dt$$

$$  = \Gamma(1+\alpha) \int_0^\infty  P_n(x) P_m(x)  e^{-x} U(1+\alpha, 1-\beta, x) \psi(x) dx = \delta_{n,m},$$
where the interchange of the order of integration is possible by Fubini's theorem due to (2.18) and an elementary estimate of the Tricomi function

$$\Gamma(1+\alpha) U(1+\alpha, 1-\beta, x) = \int_{0}^\infty   t^{\alpha}\ (1+t)^{-\alpha- \beta-1} e^{-xt} dt \le   \int_{0}^\infty   t^{\alpha}\ (1+t)^{-\alpha- \beta-1} dt =
B(1+\alpha,\beta),$$
where $B(a,b)$ is the Euler beta function \cite{Bateman}, Vol. I.

\end{proof}

\section{Properties of Prudnikov's weight functions and their products}

In this section we will exhibit properties of the weight functions $\rho_{\nu+1}(x)\rho_\nu(x),\ \rho_\nu^2(x),\ x >0$, where $\rho_\nu(x)$ is defined by (2.2), (2.3), in order to involve them in the sequel to investigate the corresponding  orthogonal and multiple orthogonal polynomials.  In particular, we will establish their differential properties, integral representations and differential equations. Concerning the function $\rho_\nu$, we found in \cite{YAP}  the following integral representation in terms of Laguerre polynomials

$${(-1)^n x^n\over n!}\  \rho_\nu(x)=   \int_0^\infty t^{\nu+n -1} e^{-t - x/t}  L_n^\nu(t) dt,\quad    n \in\mathbb{N}_{0}.\eqno(3.1)$$
It has a relationship with the  Riemann-Liouville fractional integral \cite{YaL}

$$ \left( I_{-}^\alpha  f \right) (x) \equiv \left( I_{-}^\alpha  f(x) \right)  = {1\over \Gamma(\alpha)} \int_x^\infty (t-x)^{\alpha-1} f(t) dt,\quad  {\rm Re} \alpha > 0,\eqno(3.2)$$
namely,  we get  the  formula
$$\rho_\nu(x)= \left( I_{-}^\nu \rho_0 \right) (x),\ \nu >0.\eqno(3.3)$$
Further, the index law for fractional integrals immediately implies

$$ \rho_{\nu+\mu} (x)= \left( I_{-}^\nu \rho_\mu \right) (x)=   \left( I_{-}^\mu \rho_\nu \right) (x).\eqno(3.4)$$
The corresponding definition of the fractional derivative presumes the relation $ D^\mu_{-}= - D  I_{-}^{1-\mu}$.   Hence for the ordinary $n$-th derivative of $\rho_\nu$ we find

$$D^n \rho_\nu(x)= (-1)^n \rho_{\nu-n} (x),\quad n \in \mathbb{N}_0.\eqno(3.5)$$
 The  function $\rho_\nu$ possesses  the following recurrence relation (see \cite{YAP})

$$\rho_{\nu+1} (x) =    \nu \rho_\nu(x)+ x \rho_{\nu-1} (x),\quad \nu \in \mathbb{R}.\eqno(3.6)$$
In the operator form it can be written as follows

$$\rho_{\nu+1} (x) = \left( \nu - xD \right) \rho_\nu(x).\eqno(3.7)$$

Now we will derive the Mellin-Barnes representations for the product of functions $\rho_{\nu+1}\rho_\nu$ and for the square $\rho_\nu^2$. In fact, it can be done,  using the related formulas for the product of Macdonald functions.  Thus, appealing to Entries 8.4.23.31, 8.4.23.27 in \cite{PrudnikovMarichev}, Vol. III, we obtain

$$\rho_{\nu+1} (x) \rho_\nu (x) = {4^{-\nu} \over 2 i \sqrt\pi} \int_{\gamma-i\infty}^{\gamma +i\infty} \frac{\Gamma(   s+2\nu+1)  \Gamma(s+\nu) \Gamma(s)}
{\Gamma(s+\nu+1/2)}  (4x)^{-s} ds,\quad  x,\gamma > 0,\eqno(3.8) $$

$$\rho^2_\nu (x) = {4^{-\nu} \over i\sqrt\pi} \int_{\gamma-i\infty}^{\gamma +i\infty} \frac{\Gamma(   s+2\nu)  \Gamma(s+\nu) \Gamma(s)}
{\Gamma(s+\nu+1/2)}  (4x)^{-s} ds,\quad  x,\gamma > 0.\eqno(3.9) $$
Immediate consequences of these formulas are relationships of  the products of Prudnikov's weight functions $\rho_{\nu+1}\rho_\nu,\  \rho_\nu^2$ with $\rho_{2\nu+1},\ \rho_{2\nu}$, correspondingly. Indeed, it can be done via Entry 8.4.2.3   in \cite{PrudnikovMarichev}, Vol. III and the Parseval equality for the Mellin transform. Hence we deduce from (3.8), (3.9), respectively,

$$\rho_{\nu+1} (x) \rho_\nu (x)  = 4^{-\nu} \int_0^1 (1-t)^{-1/2} t^{\nu-1} \rho_{2\nu+1}\left({4x\over t}\right) dt$$

$$=  {x^{\nu} \sqrt\pi\over \Gamma(\nu+1/2)}  \int_0^1 (1-t)^{\nu-1/2} t^{-\nu-1} \rho_{\nu+1}\left({4x\over t}\right) dt,\eqno(3.10)$$

$$ \rho^2_\nu (x)  = 2^{1-2\nu} \int_0^1 (1-t)^{-1/2} t^{\nu-1} \rho_{2\nu}\left({4x\over t}\right) dt$$

$$= {2 x^\nu \sqrt\pi \over \Gamma(\nu+1/2)}  \int_0^1 (1-t)^{\nu-1/2} t^{-\nu-1} \rho_{\nu}\left({4x\over t}\right) dt .\eqno(3.11)$$
An ordinary differential equation for the function $\rho_{\nu+1}\rho_\nu$ is given by

{\bf Proposition 4.} {\it The function $u_\nu = \rho_{\nu+1}\rho_\nu$ satisfies the following third order differential equation}

$$x^2{d^3u_\nu\over dx^3} + x(2- 3\nu) {d^2 u_\nu\over dx^2} + 2 (\nu( \nu-1) -2x) {d u_\nu\over dx} + 2(2\nu-1) u_\nu(x) = 0.\eqno(3.12)$$

\begin{proof}  Differentiating $u_\nu$ with the use of (3.5), we have

$$ {d u_\nu\over dx} = - \rho_\nu^2(x) -   \rho_{\nu+1}(x) \rho_{\nu-1} (x).\eqno(3.13)$$
Multiplying both sides of (3.13) by $x$ and employing the recurrence relation (3.6), we get

$$x {d u_\nu\over dx} = - x \rho_\nu^2(x) + \nu  \rho_{\nu+1}(x) \rho_{\nu} (x) -  \rho^2_{\nu+1}(x).$$
Differentiating both sides of the latter equality, using (3.6), (3.13) and the notation $u_\nu= \rho_{\nu+1}\rho_\nu$, we obtain

$${d\over dx} \left(x {d u_\nu\over dx}\right)  = -  (1+2\nu) \rho_\nu^2(x)  +\nu {d u_\nu\over dx}+ 4u_\nu (x).\eqno(3.14)$$
Differentiating (3.14) and multiplying the result by $x$, we recall (3.6) to find

$$x  {d^2\over dx^2} \left(x {d u_\nu\over dx}\right)  =  - 2\nu  (1+2\nu) \rho^2_\nu(x)   +\nu x {d^2 u_\nu\over dx^2}+ 4x {d u_\nu\over dx} + 2(1+2\nu) u_\nu(x).\eqno(3.15)$$
Hence, expressing $ (1+2\nu) \rho^2_\nu(x)$ from (3.14) and fulfilling the differentiation on the left-hand side of (3.15), we get (3.12).

\end{proof}

Concerning the differential equation for the function $\rho_\nu^2$, we prove

{\bf Proposition 5.} {\it The function $h_\nu = \rho^2_\nu$ satisfies the following third order differential equation}

$$x^2{d^3h_\nu\over dx^3} +  3x(1- \nu) {d^2 h_\nu\over dx^2} +  (2\nu^2+1- 3\nu- 4x) {d h_\nu\over dx} + 2(2\nu-1) h_\nu(x) = 0.\eqno(3.16)$$

\begin{proof}  Since $x \left(\rho^2_\nu(x) \right)^\prime = -2x \rho_{\nu-1}(x) \rho_{\nu} (x) = 2\nu \rho^2_\nu(x) - 2 \rho_{\nu+1}(x) \rho_{\nu} (x) = 2\nu h_\nu(x)  - 2 \rho_{\nu+1}(x) \rho_{\nu} (x)  $  we derive, owing to (3.5), (3.6)

$$x {d\over dx} \left(  x {d h_\nu\over dx} - 2\nu h_\nu(x) \right) = -2 x {d\over dx} \left( \rho_{\nu+1}(x) \rho_{\nu} (x)
\right)$$

$$ = 2x\ h_\nu(x) - 2\nu   \rho_{\nu+1}(x) \rho_{\nu} (x) + 2 \rho_{\nu+1}^2(x)=  2 \left( x - \nu^2 \right)  h_\nu(x)   +\nu x  {d h_\nu\over dx}  + 2 \rho_{\nu+1}^2(x).$$
Hence one more differentiation yields

$$  {d\over dx} \left( x {d\over dx} \left(  x {d h_\nu\over dx} - 2\nu h_\nu(x) \right) \right) =    {d\over dx} \left(  2 \left( x - \nu^2 \right)  h_\nu(x)   +\nu x  {d h_\nu\over dx}\right) - 4  \rho_{\nu+1}(x) \rho_{\nu} (x).$$
But from the beginning of the proof we find

$$2 \rho_{\nu+1}(x) \rho_{\nu} (x) = 2\nu  h_\nu(x) -  x {d h_\nu\over dx}.\eqno(3.17)$$
Therefore, substituting this expression into the previous equality, we fulfil the differentiation to arrive at (3.16) and to complete the proof.

\end{proof}

{\bf Corollary 1}.  {\it The following recurrence relations between functions $u_\nu,\ h_\nu$ hold}

$$ u_\nu=  \nu h_\nu + x u_{\nu-1},\eqno(3.18)$$

$$h_{\nu+1} = \nu^2 h_\nu + 2 x \nu u_{\nu-1} + x^2 h_{\nu-1} = 2\nu u_\nu+ x^2 h_{\nu-1} -  \nu^2 h_\nu .\eqno(3.19)$$

\begin{proof} Equality (3.18) is a direct consequence of (3.17) and (3.5).  Equalities (3.19), in turn,  are obtained, taking squares of both sides of (3.6) and employing (3.18).

\end{proof}

\section{Orthogonal polynomials with  $\rho^2_\nu$ weight function}

 The object of this section is  to  characterize the sequence of orthogonal polynomials $\{P_n\}_{n\ge 0}$, satisfying the orthogonality conditions

$$\int_{0}^{\infty}  P_{m} (x) P_{n}(x)  \rho^2_\nu(x) dx = \delta_{n,m}, \quad \nu > - {1\over 2}.\eqno(4.1)$$
Clearly,  up to a normalization factor conditions (4.1) are equivalent to the equalities

$$\int_{0}^{\infty}  P_{n} (x)  \rho^2_\nu(x)   x^m dx = 0,\quad m=0,1,\dots, n-1,\ \quad n \in \mathbb{N}.\eqno(4.2)$$
The moments of the weight $\rho^2_\nu(x)$ can be obtained immediately from the Mellin-Barnes representation (3.9), treating it as the inverse Mellin transform \cite{Tit}. Hence we get

$$ \int_{0}^{\infty}   \rho^2_\nu(x)   x^\mu dx =  \sqrt\pi \ \frac{\Gamma(   1+\mu+2\nu)  \Gamma(1+\mu+\nu) \Gamma(1+\mu)}
{ 2^{1+2(\nu+\mu)}\  \Gamma(\mu+\nu+3/2)}, \  \mu > \max\{-1, -1-\nu,\ 1-2\nu\}.\eqno(4.3)$$
Furthermore, the sequence $ \left\{P_n\right\}_{n \ge 0}$ satisfies the 3-term recurrence relation in the form

$$x  P_n (x) = A_{n+1}  P_{n+1} (x) + B_n  P_n (x) + A_n  P_{n-1} (x),\eqno(4.4)$$
where $P_{-1}^{\nu} (x)\equiv 0,\  P_n(x)= \sum_{k=0}^n a_{n,k} x^k,\  a_{n,n} \neq 0$ and

$$A_{n+1}= {a_n\over a_{n+1} }, \quad B_{n}= {b_n\over a_n} -  {b_{n+1}\over a_{n+1}},\quad  a_n\equiv a_{n,n},\  b_n\equiv a_{n,n-1}.$$
Basing on properties of the functions $\rho^2_\nu,\ \rho_{\nu+1}\rho_\nu$ and orthonormality (4.1), we establish

{\bf Proposition 6}.  {\it Let $\nu > -1$. The following formulas hold}

$$ \int_{0}^{\infty}   P^2_{n}(x)  \rho_{\nu+1}(x)\rho_\nu(x)  dx =  {1\over 2} + \nu + n,\eqno(4.5)$$

$$ \int_{0}^{\infty}   P^2_{n}(x) \ x \rho_{\nu}(x) \rho_{\nu-1}(x) dx =   {1\over 2}  + n,\eqno(4.6)$$

$$ \int_{0}^{\infty}   P^2_{n}(x) \ x^2 \rho_{\nu}(x) \rho_{\nu-2}(x) dx =  B_n- (\nu-1) \left( {1\over 2}  + n\right),\eqno(4.7)$$

$$ \int_{0}^{\infty}   P^2_{n}(x)  \rho_{\nu+2}(x)\rho_\nu(x)  dx = B_n+ (\nu+1)\left( {1\over 2} + \nu + n\right).\eqno(4.8)$$

\begin{proof} In fact, recurrence relation (3.6) and orthonormality (4.1) implies

$$ \int_{0}^{\infty}   P^2_{n}(x)  \rho_{\nu+1}(x)\rho_\nu(x)  dx  = \nu +  \int_{0}^{\infty}   P^2_{n}(x)\  x \rho_{\nu}(x)\rho_{\nu-1}(x)  dx.\eqno(4.9)$$
 Hence, integrating by parts in the latter integral and eliminating integrated terms by virtue of the asymptotic behavior  (2.4), (2.5) , we find

 $$\int_{0}^{\infty}   P^2_{n}(x) \ x \rho_{\nu}(x)\rho_{\nu-1}(x)  dx = \int_0^\infty  \left( {1\over 2} P^2_{n}(x) +  x P_{n}(x) P_n^\prime (x) \right) \rho^2_{\nu}(x) dx.\eqno(4.10) $$
Therefore, we have from (4.9), (4.10)  and orthogonality (4.1)

$$ \int_{0}^{\infty}   P^2_{n}(x)  \rho_{\nu+1}(x)\rho_\nu(x)  dx = {1\over 2} + \nu + n,$$
which proves (4.5).   Concerning equality (4.6), it is a direct consequence of (3.6), (4.1).  The same idea is to prove (4.7), (4.8), employing (4.3) as well.

\end{proof}

The composition orthogonality which is associated with the sequence  $\{P_n\}_{n\ge 0}$ is given by

{\bf Theorem 1.} {\it Let $\nu > -1/2$. The sequence $\{P_n\}_{n\ge 0}$ is compositionally orthogonal in the sense of Laguerre relatively to the function $t^\nu e^t \Gamma(-\nu,t)$, where $\Gamma(\mu,z)$ is the incomplete gamma function, i.e.}

$$\ \int_{0}^{\infty}   t^{\nu }   e^{-t}  P_n\left(\theta\right)  P_m\left(\theta\right)\left(  t^\nu e^t \Gamma(-\nu, t) \right) dt = { \delta_{n,m}\over \Gamma(1+\nu)}.\eqno(4.11)$$

\begin{proof} Writing $P_n$ explicitly, we appeal to the integral representation (3.1) for $\rho_\nu(x)$ to write the left-hand side of (4.2) in the form

$$ \int_{0}^{\infty}  P_{n} (x)  \rho^2_\nu(x)   x^m dx =  (-1)^m m! \sum_{k=0}^n a_{n,k}  (-1)^k k!  \int_{0}^{\infty} \int_{0}^{\infty}   t^{\nu+k -1} e^{-t - x/t}  L_k^\nu(t) dt$$

$$\times \int_{0}^{\infty}   y^{\nu+m -1} e^{-y - x/y}  L_m^\nu(y) dy dx.\eqno(4.12) $$
The Fubini theorem allows to interchange the order of integration in (4.12) due to the convergence of the following integral for all $r_1, r_2  \in \mathbb{N}_0$

$$\int_{0}^{\infty} \int_{0}^{\infty}   t^{\nu+k+r_1 -1} e^{-t - x/t} dt  \int_{0}^{\infty}   y^{\nu+m +r_2 -1} e^{-y - x/y}  dy\ dx $$

$$=   \int_{0}^{\infty}   \int_{0}^{\infty}   t^{\nu+k+r_1 }   y^{\nu+m +r_2} e^{-t-y } {dt dy\over t+ y} \le {1\over 2} \int_{0}^{\infty}   \int_{0}^{\infty}   t^{\nu+k+r_1 -1/2 }   y^{\nu+m +r_2-1/2 } e^{-t-y } dt dy$$

$$ =   {1\over 2} \Gamma(\nu+k+r_1 +1/2)  \Gamma(\nu+m+r_2 +1/2).$$
Therefore after integration with respect to $x$ we find from (4.12)

$$ \int_{0}^{\infty}  P_{n} (x)  \rho^2_\nu(x)   x^m dx =  (-1)^m m! \sum_{k=0}^n a_{n,k}  (-1)^k k!  \int_{0}^{\infty} \int_{0}^{\infty}   t^{\nu+k}   y^{\nu+m } e^{-t-y}  L_k^\nu(t)  L_m^\nu(y) {dt dy \over t+y}.\eqno(4.13) $$
However the Rodrigues formula for Laguerre polynomials and Viskov-type identity  (1.3) suggest to write the right-hand side of (4.13) as follows

$$  (-1)^m m! \sum_{k=0}^n a_{n,k}  (-1)^k k!  \int_{0}^{\infty} \int_{0}^{\infty}   t^{\nu+k}   y^{\nu+m } e^{-t-y}  L_k^\nu(t)  L_m^\nu(y) {dt dy \over t+y}$$

$$=   (-1)^m \sum_{k=0}^n a_{n,k}  (-1)^k k!  \int_{0}^{\infty}   t^{\nu+k}   e^{-t}   L_k^\nu(t)  \int_{0}^{\infty}   \theta^m \left( y^\nu e^{-y} \right) {dy dt \over t+y} . \eqno(4.14)$$
The inner integral with respect to $y$ on the right-hand side of the latter equality can be treated via integration by parts, and we obtain

$$ \int_{0}^{\infty}   \theta^m \left( y^\nu e^{-y} \right) {dy  \over t+y}  = (-1)^m \int_{0}^{\infty}    y^\nu e^{-y} \theta^m \left( {1\over t+y} \right) dy.\eqno(4.15)$$
Working out the differentiation, we find

$$ \theta^m \left( {1\over t+y} \right) =  y^m {d^m\over dy^m} \left( \sum_{k=0}^m \binom{m}{k} (-1)^{m-k} t^{m-k} (t+y)^{k-1}\right)  $$

$$=  (-1)^m t^m y^m {d^m\over dy^m} \left(  {1\over t+y}\right) = {m! \  t^m y^m \over (t+y)^{m+1} }.$$
Consequently, combining with (4.14), (4.15), equality (4.13) becomes

$$ \int_{0}^{\infty}  P_{n} (x)  \rho^2_\nu(x)   x^m dx  =    m! \sum_{k=0}^n a_{n,k}  (-1)^k k!  \int_{0}^{\infty}   t^{\nu+k+m }   e^{-t}   L_k^\nu(t)  \int_{0}^{\infty}
  {y^{\nu+m}  e^{-y}  \over (t+y)^{m+1} } dy dt.\eqno(4.16) $$
Meanwhile, the integral with respect to $y$ in (4.16) is calculated in  (2.17) in terms of the Tricomi function up to a simple change of variables.  Thus we get

 $$ \int_{0}^{\infty}  P_{n} (x)  \rho^2_\nu(x)   x^m dx  =    m!  \Gamma(\nu+m+1) \sum_{k=0}^n a_{n,k}  (-1)^k k!  \int_{0}^{\infty}   t^{2\nu+k+m }   e^{-t}   L_k^\nu(t)  U(\nu+m+1, 1+\nu, t) dt.\eqno(4.17) $$
But appealing to the differential formula 13.3.24 in \cite{nist} for the Tricomi function, we have

$$  m! \ \Gamma(\nu+m+1)  t^{\nu+m}  U(\nu+m+1, 1+\nu, t) = \Gamma(\nu+1) \theta^m\left( t^\nu U(1+\nu,1+\nu,t)\right)$$

$$ = \Gamma(\nu+1) \theta^m\left( t^\nu e^t \Gamma(-\nu, t) \right),\eqno(4.18)$$
 %
where $\Gamma(a,z)$ is the incomplete gamma function

$$ \Gamma(a,z) = \int_z^\infty e^{-u} u^{a-1} du.\eqno(4.19)$$
Hence,  plugging this result in (4.17), recalling the Rodrigues  formula for Laguerre polynomials and integrating by parts on its right-hand side, we  combine  with (4.2) to  end up with equalities

$$ \int_{0}^{\infty}  P_{n} (x)  \rho^2_\nu(x)   x^m dx = \Gamma(\nu+1) $$

$$\times  \int_{0}^{\infty}   t^{\nu }   e^{-t}  P_n\left(\theta\right) \theta^m \left(  t^\nu e^t \Gamma(-\nu, t) \right) dt = 0,\quad m=0,1,\dots, n-1,\  n \in \mathbb{N},\eqno(4.20)$$
which yield (4.11) and complete  the proof of Theorem 1.

\end{proof}

Further, the right-hand side of the first equality in (4.20) can be rewritten via integration by parts as follows

$$ \int_{0}^{\infty}   t^{\nu+m }   e^{-t}    L_m^\nu(t) P_n\left(\theta\right)  \left(  t^\nu e^t \Gamma(-\nu, t) \right) dt = 0,\quad m=0,1,\dots, n-1,\  n \in \mathbb{N}.\eqno(4.21)$$
Then we expand the function $ F_n(t)= \theta^n \left(  t^\nu e^t \Gamma(-\nu, t) \right)$  in a series of Laguerre polynomials

$$ F_n(t)=   \sum_{r=0}^\infty c_{n,r} L_r^\nu(t),\eqno(4.22)$$
where   the coefficients $c_{n,r}$ are calculated in terms of the ${}_3F_2$-hypergeometric functions at  unity with the aid of Entry 3.31.12.1 in \cite{Brychkov}. Precisely, we get via (4.18)

$$c_{n,r}= {r!\over \Gamma(1+\nu+r)} \int_0^\infty t^\nu e^{-t}  L_r^\nu(t) \theta^n \left(  t^\nu e^t \Gamma(-\nu, t) \right) dt$$

$$= {r! n! (1+\nu)_n \over \Gamma(1+\nu+r)} \int_0^\infty t^{2\nu+n}  e^{-t}  L_r^\nu(t) U(\nu+n+1, 1+\nu, t)  dt$$

$$=    {(1+\nu)_n\over \Gamma(1+\nu)}   \left[ {n! r! \ \Gamma(\nu) \over (1+\nu)_r}  \  {}_3F_2 \left(  n+1, r+1, \nu+n+1; 1-\nu,  1; \ 1 \right) \right.$$

$$\left.  +   \Gamma(2\nu+n+1) \Gamma(-\nu)\  {}_3F_2 \left( \nu+ r+1,  \nu+n+1,  2\nu+n+1; 1+\nu,  1+\nu; \ 1 \right)\right],\eqno(4.23)$$
where $(a)_z$ is the Pochhammer symbol \cite{Bateman}, Vol. I and it is valid for nonnegative integers $\nu$ by continuity.  Hence after substitution the series (4.22) into (4.21) these orthogonality conditions  take the form

$$  \int_{0}^{\infty}   t^{\nu+m }   e^{-t}    L_m^\nu(t) \sum_{k=0}^n a_{n,k} \sum_{r=0}^{2m}  c_{k,r} L_r^\nu(t)   dt = 0,\quad m=0,1,\dots, n-1,\  n \in \mathbb{N}.\eqno(4.24)$$
Calculating the integral in (4.24) via relation (2.19.14.8) in \cite{PrudnikovMarichev}, Vol. II we obtain

 $$d_{m,r} =  \int_{0}^{\infty}   t^{\nu+m }   e^{-t}    L_m^\nu(t)  L_r^\nu(t)   dt = { (-1)^r   \over  r! } \  (1+\nu)_r\  \Gamma(1+\nu+m) $$

 $$\times  {}_3F_2 \left( -r, \nu+ m+1,  m+1; 1+\nu,  1; \ 1 \right).\eqno(4.25)$$
Hence, equalities (4.24) become the linear system of $n$ algebraic equations with $n+1$ unknowns

$$\sum_{k=0}^n a_{n,k} f_{k,m} = 0,\quad m=0,1,\dots, n-1,\  n \in \mathbb{N},\eqno(4.26)$$
where

$$f_{k,m} = \sum_{r=0}^{2m}  c_{k,r} d_{m,r}.\eqno(4.27)$$
Consequently,  explicit values of the coefficients $a_{n,k},\ k =1,2,\dots, n$ can be expressed via Cramer's rule in terms of the free coefficient $a_{n,0}$ as follows

$$  a_{n,k} = -    a_{n,0}\  { D_{n,k}  \over D_{n}},\quad   k = 1,\dots, n,\eqno(4.28)$$
where

$$ D_{n}= \begin{vmatrix}

f_{1, 0}  & f_{2, 0}&  \dots&  \dots&   f_{n, 0} \\

f_{1, 1} &  \dots &   \dots&    \dots&    f_{n, 1} \\

\dots&  \dots &   \dots&   \dots&   \dots \\

 \vdots  &   \ddots  &  \ddots & \ddots&   \vdots\\

 f_{1, n-1} &   \dots&  \dots&   \dots&    f_{n, n-1}\\

 \end{vmatrix},\eqno(4.29)$$

$$ D_{n,k}= \begin{vmatrix}

f_{1, 0}  & \dots & f_{k-1, 0}&  f_{0,0} &   f_{k+1, 0} & \dots&   f_{n, 0}  \\

f_{1, 1} &  \dots & f_{k-1, 1}&  f_{0,1} &   f_{k+1, 1} &  \dots&   f_{n, 1}  \\

 \dots&  \dots &   \dots&   \dots&  \dots&   \dots&  \dots \\

 \vdots  &   \ddots  &  \ddots & \vdots&  \ddots  &  \ddots & \vdots\\

 \vdots  &   \ddots  &  \ddots & \vdots&  \ddots  &  \ddots & \vdots\\

 \vdots  &   \ddots  &  \ddots & \vdots&  \ddots  &  \ddots & \vdots\\

f_{1, n-1} &  \dots & f_{k-1, n-1}&  f_{0,n-1} &   f_{k+1, 1} &  \dots&   f_{n, n-1}  \\

 \end{vmatrix}.\eqno(4.30)$$
The free  coefficient can be determined, in turn,  from the orthogonality conditions (4.1), (4.2), which imply the formula

$$\int_{0}^{\infty}  P_{n} (x)   \rho^2_\nu(x) \  x^n dx = {1\over a_{n,n}}.\eqno(4.31)$$
Therefore from (4.28) and (4.3)  we derive

$$ {1\over a_{n,n}} =  -   { a_{n,0} \sqrt\pi \over D_n  }\  \frac{\Gamma(   1+2\nu)  \Gamma(1+\nu) }
{ 2^{1+2\nu}\  \Gamma(\nu+3/2)} \ \sum_{k=0}^n  D_{n,k} \ (n+k)! \ \frac{(   1+2\nu)_{n+k}   (1+\nu)_{n+k} }
{ 4^{n+k}\  (\nu+3/2)_{n+k}},\eqno(4.32)$$
where $D_{n,0}\equiv - D_n.$  Hence

$$  a_{n,0}  =    \pm \  { D_n  \over [ D_{n,n} ]^{1/2} }\left[ \sqrt\pi \ \frac{\Gamma(   1+2\nu)  \Gamma(1+\nu) }
{ 2^{1+2\nu}\  \Gamma(\nu+3/2)} \ \sum_{k=0}^n  D_{n,k} \ (n+k)! \ \frac{(   1+2\nu)_{n+k}   (1+\nu)_{n+k} }
{ 4^{n+k}\  (\nu+3/2)_{n+k}} \right]^{-1/2},\eqno(4.33)$$
where the sign can be  chosen  accordingly, making positive expressions under the square roots.   Assuming also the positivity of the leading coefficient $a_{n,n}$ we have its value, correspondingly,

$$   a_{n,n}  =    \mp \   [ D_{n,n}]^{1/2} \left[  \sqrt\pi \ \frac{\Gamma(   1+2\nu)  \Gamma(1+\nu) }
{ 2^{1+2\nu}\  \Gamma(\nu+3/2)} \ \sum_{k=0}^n  D_{n,k} \ (n+k)! \ \frac{(   1+2\nu)_{n+k}   (1+\nu)_{n+k} }
{ 4^{n+k}\  (\nu+3/2)_{n+k}}  \right]^{-1/2}.\eqno(4.34)$$

{\bf Theorem 2}.  {\it Let $\nu > -1/2$. The sequence of orthogonal polynomials $\left\{P_n\right\}_{n \ge 0}$  can be expressed explicitly, where the coefficients $a_{n,k},\ k=1,2,\dots, n$ are calculated by relations  $(4.28)$ and the free term $a_{n,0}$ is defined by the equality $(4.33)$. Besides,  it satisfies the  3-term recurrence relation $(4.4)$, where }

$$A_{n+1}= { a_{n,0} \  D_{n+1} \ D_{n,n}   \over a_{n+1,0}\  D_n \ D_{n+1,n+1}},  \quad  B_n= { D_{n,n-1}  \over D_{n,n}} -  { D_{n+1,n}  \over D_{n+1,n+1}}.\eqno(4.35)$$
An analog of the Rodrigues formula for the sequence $ \left\{P_n\right\}_{n \in \mathbb{N}_0}$ can be established, appealing to the   integral representation  of an arbitrary polynomial in terms of the associated polynomial of degree $2n$ (see \cite{YAP}). Hence, employing (2.2), we deduce

$$P_n(x)= {1\over \rho_\nu(x)}   \int_0^\infty t^{\nu-1} e^{-t -x/t  }  q_{2n}(t) dt = {1\over \rho^2_\nu(x)}     \int_0^\infty u^{\nu-1} e^{-u -x/u  }  du \int_0^\infty t^{\nu-1} e^{-t -x/t  }  q_{2n}(t) dt $$

$$ = {(-1)^n \over \rho^2_\nu(x)}    {d^n\over dx^n } \int_0^\infty  \int_0^\infty (u t)^{\nu+n-1} e^{-u -t -x(1/u +1/t)  }  q_{2n}(t) {dt du\over (u+t)^n},\eqno(4.36)$$
where
$$ q_{2n}(x) =  \sum_{k=0}^{n} a_{n,k} (-1)^k k! x^k L_k^\nu(x)\eqno(4.37)$$
and the $k$-th differentiation under the integral sign is permitted due to the estimate

$$\int_0^\infty  \int_0^\infty (u t)^{\nu+k-1} e^{-u -t -x(1/u +1/t)  } \left| q_{2n}(t)\right|  {dt du\over (u+t)^k}$$

$$ \le 2^{-k}  \int_0^\infty   u^{\nu+ k/2 -1}  e^{-u} du  \int_0^\infty   t^{\nu+ k/2 -1}  e^{-t} \left| q_{2n}(t)\right|  dt $$

$$=   2^{-k}  \Gamma\left( \nu+ {k\over 2} \right)  \int_0^\infty   t^{\nu+ k/2 -1}  e^{-t} \left| q_{2n}(t)\right|  dt   < \infty,\quad  k = 1,\dots, n.$$
Then writing

$${(ut)^n\over (u+t)^n} = {1\over (n-1)!} \int_0^\infty e^{ - (1/u+1/t) y} y^{n-1} dy,$$
we get from (4.36)

$$P_n(x)=  {(-1)^n \over (n-1)!  \ \rho^2_\nu(x)}    {d^n\over dx^n } \int_0^\infty  \int_0^\infty   \int_0^\infty  y^{n-1} (u t)^{\nu-1} e^{-u -t -(x+y)/u -(x+y)/t  }  q_{2n}(t) dt du dy$$

$$=  {(-1)^n \over (n-1)!  \ \rho^2_\nu(x)}    {d^n\over dx^n }  \int_0^\infty   \int_0^\infty  y^{n-1} t^{\nu-1} e^{ -t  -(x+y)/t  }  q_{2n}(t) \rho_{\nu}(x+y) dt dy.\eqno(4.38)$$
Then, expressing $q_{2n}$ in terms of the  Laguerre polynomials

$$  q_{2n}(x)= \sum_{k=0}^{2n} h_{2n,k} L_k^\nu(x),\eqno(4.39)$$
where the coefficients $h_{2n,k}$ are calculated by virtue of (4.28),  (4.37) and relation (2.19.14.15) in \cite{PrudnikovMarichev}, Vol. II, namely,

$$ h_{2n,k} =  {k!\over \Gamma(1+\nu+k)} \int_0^\infty t^\nu e^{-t} L_k^\nu(t)  q_{2n}(t) dt $$

$$= -  {k! \ a_{n,0} \over  D_n\ \Gamma(1+\nu+k)} \sum_{r=0}^n D_{n,r} (-1)^r r!  \int_0^\infty t^{\nu+r}  e^{-t} L_k^\nu(t)  L_r^\nu(t) dt $$

$$= -  { a_{n,0} \over  D_n} \sum_{r=0}^n D_{n,r} \  r! \ (1+\nu)_r  \ {}_3F_2 \left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right). $$
So,

$$ h_{2n,k} = -  { a_{n,0} \over  D_n} \sum_{r=0}^n D_{n,r} \  r! \ (1+\nu)_r  \ {}_3F_2 \left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right),\eqno(4.40) $$
and  the values of the generalized hypergeometric function can be simplified via relations (7.4.4; 90,91,92,93) in \cite{PrudnikovMarichev}, Vol. III. Precisely, we get for $k =0,1\dots, n$

$$ {}_3F_2 \left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right) = 0,\quad k > 2r,\eqno(4.41)$$

$$ (1+\nu)_r\  {}_3F_2 \left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right) = {  (2r) ! \over  r!  },\quad k=2r,$$

$$ (1+\nu)_r\  {}_3F_2 \left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right) = -  {  (2r-1) ! (\nu+ 3r ) \over  (r-1)! },  \quad k=2r-1, $$

$$ (1+\nu)_r\  {}_3F_2 \left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right) = {  (2(r-1)) ! \over 2\  r! }
\left( 2 r^2 (2r+\nu-1)(2r-1) \right.$$

$$\left. +  r(r-1)(r+\nu-1)(r+\nu ) \right),\quad k=2(r-1).$$
Therefore, returning to (4.38) and minding (4.39), (4.40), (4.41), (3.1), (3.2) we find

$$P_n(x)= {(-1)^{n+1}  a_{n,0}  \over  D_n\  (n-1)!  \ \rho^2_\nu(x)}    {d^n\over dx^n }   \sum_{k=0}^{2n}   \sum_{r=0}^n   D_{n,r} \  r! \ (1+\nu)_r  \ {}_3F_2 \left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right) $$

$$\times \int_0^\infty   \int_0^\infty  y^{n-1} t^{\nu-1} e^{ -t  -(x+y)/t  } L_k^\nu(t)  \rho_{\nu}(x+y) dt dy$$

$$= {(-1)^{n+1}  a_{n,0}  \over  D_n\  (n-1)!  \ \rho^2_\nu(x)}    {d^n\over dx^n }   \sum_{k=0}^{2n} {1\over k!}   \sum_{r=0}^n   D_{n,r} \  r! \ (1+\nu)_r  \ {}_3F_2 \left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right) $$

$$\times \int_0^\infty    y^{n-1}  {d^k\over dx^k} \left(  (x+y)^k \rho_\nu(x+y) \right) \rho_{\nu}(x+y) dy$$

$$= {(-1)^{n+1}  a_{n,0}  \over  D_n\  (n-1)!  \ \rho^2_\nu(x)}    {d^n\over dx^n }   \sum_{k=0}^{2n} {1\over k!}  \sum_{r=0}^n   D_{n,r} \  r! \ (1+\nu)_r  \ {}_3F_2 \left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right) $$

$$\times  \int_x^\infty  (y-x)^{n-1}  {d^k\over dy^k} \left(y^k \rho_{\nu}(y)\right) \rho_\nu(y)  dy =  - { a_{n,0}  \over  D_n \ \rho_\nu(x)}    \sum_{k=0}^{2n}  {d^k\over dx^k}  \left( x^k \rho_{\nu}(x)\right)  \sum_{r=0}^n   D_{n,r} \  r! $$

$$\times \ {(1+\nu)_r\over k!}   \ {}_3F_2 \left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right)$$

$$=  - { a_{n,0}  \over  D_n \ \rho_\nu(x)}   \sum_{r=0}^n    D_{n,r} \  r!    \ (1+\nu)_r \sum_{k=0}^{2r}    {d^k\over dx^k}  \left( x^k \rho_{\nu}(x)\right)  {1\over k!} \ {}_3F_2 \left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right).$$

Thus it proves

{\bf Theorem 3}. {\it Let $\nu > -1/2,\ n \in \mathbb{N}_0$.  Orthogonal polynomials $P_n$ satisfy  the  Rodrigues-type  formula

$$P_n(x)= - { a_{n,0}  \over  D_n \ \rho_\nu(x)}   \sum_{r=0}^n    D_{n,r}   r! (1+\nu)_r \sum_{k=0}^{2r} {1\over k!}  {d^k\over dx^k}  \left( x^k \rho_{\nu}(x)\right){}_3F_2 \left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right),\eqno(4.42)$$
where  $a_{n,0}$ is defined by  $(4.33)$ and $D_n,\ D_{n,r}$ by $(4.29), (4.30)$, respectively.}

{\bf Corollary 2}. {\it  Orthogonal polynomials  $P_n$  have the following representation

$$P_n(x)= - { a_{n,0}  \over  D_n \ }   \sum_{r=0}^n    D_{n,r}   r! (1+\nu)_r\left[  \sum_{k=0}^{r} {A_{k,k-1} (x) \over (2k)!}   {}_3F_2 \left(-2k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right)\right.$$

$$\left. +  \sum_{k=0}^{r-1} {A_{k,k} (x) \over (2k+1)!}   {}_3F_2 \left(-2k-1,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right) \right], \eqno(4.43)$$
where $ A_{k,k-1},  A_{k,k}$ are the type $1$ multiple orthogonal polynomials of degree $k$,  associated with the vector of weight functions $(\rho_\nu, \rho_{\nu+1})$ }.

\begin{proof} In fact, we write (4.42) in the form

 $$P_n(x)= - { a_{n,0}  \over  D_n \ \rho_\nu(x)}   \sum_{r=0}^n    D_{n,r}   r! (1+\nu)_r \left[ \sum_{k=0}^{r} {1\over (2k)!}  {d^{2k} \over dx^{2k}}  \left( x^{2k} \rho_{\nu}(x)\right){}_3F_2 \left(-2k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right)\right.$$

$$\left. +  \sum_{k=0}^{r-1} {1\over (2k+1)!}  {d^{2k+1} \over dx^{2k+1}}  \left( x^{2k+1} \rho_{\nu}(x)\right){}_3F_2 \left(-2k-1,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right)\right].\eqno(4.44)$$
Meanwhile,  appealing to the Rodrigues formulas for the type $1$  multiple orthogonal polynomials  associated with the vector of weight functions $(\rho_\nu, \rho_{\nu+1})$ (see in \cite{AsscheYakubov2000}), it gives

$$ {d^{2k} \over dx^{2k}}  \left( x^{2k} \rho_{\nu}(x)\right) = A_{k,k-1} (x)\rho_\nu(x) + B_{k,k-1} (x)\rho_{\nu+1}(x),$$

$$ {d^{2k+1} \over dx^{2k+1}}  \left( x^{2k+1} \rho_{\nu}(x)\right) = A_{k,k} (x)\rho_\nu(x) + B_{k,k} (x)\rho_{\nu+1}(x),$$
where $  A_{k,k-1},  B_{k,k-1}$ are polynomials of degree $k, k-1$, respectively, and  $  A_{k,k},  B_{k,k}$ are polynomials of degree $k$.   Therefore, substituting these expressions into (4.44), we obtain

$$P_n(x)= - { a_{n,0}  \over  D_n \ }   \sum_{r=0}^n    D_{n,r}   r! (1+\nu)_r\left[  \sum_{k=0}^{r} {A_{k,k-1} (x) \over (2k)!}   {}_3F_2 \left(-2k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right)\right.$$

$$\left. +  \sum_{k=0}^{r-1} {A_{k,k} (x) \over (2k+1)!}   {}_3F_2 \left(-2k-1,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right) \right] $$

$$  - { a_{n,0} \ \rho_{\nu+1}(x)  \over  D_n \ \rho_\nu(x)}   \sum_{r=0}^n    D_{n,r}   r! (1+\nu)_r \left[   \sum_{k=0}^{r} {B_{k,k-1} (x) \over (2k)!}   {}_3F_2 \left(-2k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right)\right.$$

$$\left. +  \sum_{k=0}^{r-1} {B_{k,k} (x) \over (2k+1)!}   {}_3F_2 \left(-2k-1,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right) \right] .$$
But the existence of  a multiple orthogonal polynomial sequence with respect to the vector of weight functions $(\rho_\nu, \rho_{\nu+1})$ implies the identity

$$ \sum_{r=0}^n    D_{n,r}   r! (1+\nu)_r \left[ \sum_{k=0}^{r} {B_{k,k-1} (x) \over (2k)!}   {}_3F_2 \left(-2k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right)\right.$$

$$\bigg. +  \sum_{k=0}^{r-1} {B_{k,k} (x) \over (2k+1)!}   {}_3F_2 \left(-2k-1,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1 \right) \bigg]\equiv 0,$$
which drives to (4.43) and completes the proof.

\end{proof}

{\bf Remark 2}.  In a similar manner orthogonal polynomials with the weight function $\rho_{\nu+1}(x)\rho_\nu(x)$ can be investigated. We leave this topic to the interested reader.

 Finally, in this section we establish  the generating function for  polynomials $P_n$, which is  defined as usually  by the equality

$$G(x,z) = \sum_{n=0}^\infty P_n (x) {z^n\over n!} ,\quad  x >0,\  z \in \mathbb{C},\eqno(4.45)$$
where $|z| < h_x$ and $h_x >0$ is a  convergence radius of the power series. To do this,  we employ (3.1), (4.36) and (4.39), having  the following equality from (4.45)

$$  G(x,z) =  {1 \over \rho_\nu(x) } \sum_{n=0}^\infty  {z^n\over n!}  \sum_{k=0}^{2n} {h_{2n,k} \over k! }   {d^{k}\over dx^{k}} \left[  x^{k}  \rho_\nu(x) \right]=  {1\over \rho_\nu(x) } \sum_{n=0}^\infty  {z^n\over n!}  \sum_{k=0}^{2n}  h_{2n,k}   \sum_{j=0}^k \binom{k}{j}  {(-1)^j\over (k-j)!} \  x^j \rho_{\nu-j}(x) . $$
Meanwhile, the product  $x^j\rho_{\nu-j}(x)$ is expressed in \cite{Cous}  as follows

$$x^{ j}  \rho_{\nu-j}(x) = x^{ j/2} r_j(2\sqrt x; \nu ) \rho_{\nu}(x) +  x^{ (j-1)/2} r_{j-1} (2\sqrt x; \nu -1) \rho_{\nu+1}(x), \quad  j \in {\mathbb N}_0,$$
where $r_{-1}(z;\nu)=0$,

$$ x^{ j/2} r_j(2\sqrt x; \nu ) = (-1)^j \sum_{i=0}^{[j/2]}   (\nu+i-j+1)_{j-2i} (j-2i+1)_i  {x^i\over i!}.\eqno(4.46)$$
Therefore this leads to the final  expression of  the generating function for the sequence $\left(P_n\right)_{n \in \mathbb{N}_0}$, namely,

$$  G(x,z) =   \sum_{n=0}^\infty  \sum_{k=0}^{2n} \sum_{j=0}^k  \binom{k}{j} { (-1)^j \ h_{2n,k} \over n!\  (k-j)! }   x^{j/2}  r_j(2\sqrt x; \nu ) z^n $$

$$+  {\rho_{\nu+1}(x) \over \rho_\nu(x) } \sum_{n=0}^\infty \sum_{k=0}^{2n} \sum_{j=0}^k  \binom{k}{j}  { (-1)^j \  h_{2n,k} \over n!\  (k-j)! }  x^{(j-1)/2} r_{j-1}(2\sqrt x; \nu-1 ) z^n,\eqno(4.47) $$
where coefficients $h_{2n,k}$ are defined by (4.40).

\section{Note on the multiple orthogonal polynomials}

In this section we will exhibit two types of multiple orthogonal polynomials for the vector of weight functions $(\rho^2_\nu,\  \rho^2_{\nu+1}, \  \rho_\nu\rho_{\nu+1}),\ \nu > -1/2 $ over $\mathbb{R}_+$ with an additional factor $x^\alpha,\ \alpha > -1$. Precisely, we consider the type $1$ polynomials $(A^\alpha_n,\ B^\alpha_{n-1},\ C^\alpha_{n-1}),\ n \in \mathbb{N}$ of degree $n,\ n-1$, respectively, satisfying the orthogonality conditions

$$\int_0^\infty \left[ A^\alpha_n(x) \rho^2_\nu (x) +  B^\alpha_{n-1}(x) \rho^2_{\nu+1} (x)+  C^\alpha_{n-1}(x) \rho_\nu (x) \rho_{\nu+1} (x) \right] x^{\alpha + m} dx = 0, \ m= 0,1,\dots, 3n-1.\eqno(5.1)$$
So, we have $3n$ linear homogeneous equations with $3n+1$ unknown coefficients of polynomials $A^\alpha_n,\ B^\alpha_{n-1},\ C^\alpha_{n-1}$. Therefore we can find type $1$ polynomials up to a multiplicative factor.  Let us denote the function $q^\alpha_{n,n-1,n-1}$ for the convenience

$$ q^\alpha_{n,n-1,n-1}(x) =  A^\alpha_n(x) \rho^2_\nu (x) +  B^\alpha_{n-1}(x) \rho^2_{\nu+1} (x)+  C^\alpha_{n-1}(x) \rho_\nu (x) \rho_{\nu+1} (x) .\eqno(5.2)$$
Type $2$ polynomials $p^\alpha_{n,n-1,n}$ are monic polynomials of degree $3n-1$ which satisfy the multiple orthogonality conditions

$$ \int_0^\infty  p^\alpha_{n,n-1,n}(x) \rho^2_\nu (x) x^{\alpha+m} dx = 0,\quad m= 0,1,\dots, n-1,\eqno(5.3)$$

 $$ \int_0^\infty  p^\alpha_{n,n-1,n}(x) \rho^2_{\nu+1} (x) x^{\alpha+m} dx = 0,\quad m= 0,1,\dots, n-2,\eqno(5.4)$$

$$ \int_0^\infty  p^\alpha_{n,n-1,n}(x) \rho_\nu (x) \rho_{\nu+1}(x) x^{\alpha+m} dx = 0,\quad m= 0,1,\dots, n-1.\eqno(5.5)$$
This gives $3n-1$ linear equations with $3n-1$ unknown coefficients  since the leading coefficient is equal to $1$. Hence it can be uniquely determined.

When $|\nu| < 1/2$ the uniqueness of the representation (5.2)  is validated by the following

{\bf Theorem 4}.  {\it Let $n, m, l  \in \mathbb{N}_0, |\nu| < 1/2, \   f_n,\ g_{m},\ h_l$ be polynomials of  degree at most $n,\ m,\ l$, respectively.  Let

 $$f_n(x) \rho^2_\nu(x)+ g_{m} (x) \rho^2_{\nu+1} (x) +  h_{l} (x) \rho_\nu(x) \rho_{\nu+1} (x) = 0\eqno(5.6)$$
 %
 for all $x >0$. Then $f_n \equiv 0, \ g_{m} \equiv 0,\ h_l \equiv 0.$ }

 \begin{proof} As is known (cf. \cite{YAP})  the quotient $\rho_\nu/ \rho_{\nu+1}$ is represented by the Ismail integral

$$ {\rho_{\nu}(x) \over \rho_{\nu+1}(x) } = {1\over \pi^2} \int_0^\infty { s^{-1} ds \over (x+s)( J_{\nu+1}^2(2\sqrt s) +
Y_{\nu+1}^2(2\sqrt s) )},\eqno(5.7)$$
where $J_\nu, Y_\nu$ are Bessel functions of the first and second kind, respectively \cite{PrudnikovMarichev}, Vol. II.  In fact,   let $r \ge \max\{ n, m+1, l+1\}. $  Hence, dividing (5.6) by $\rho_\nu(x) \rho_{\nu+1} (x)$ and using (3.6), we find

$$f_n(x) {\rho_\nu(x)\over \rho_{\nu+1} (x)}  +  x g_{m} (x) {\rho_{\nu-1} (x)\over \rho_{\nu} (x) } + \nu g_{m} (x) + h_{l} (x) = 0.\eqno(5.8)$$
 %
Then, differentiating $r$ times, it gives

 $${d^{r}\over dx^{r} } \left[ f_n(x) {\rho_\nu(x)\over \rho_{\nu+1} (x)} \right] +  {d^{r}\over dx^{r} } \left[ x g_m(x) {\rho_{\nu-1} (x)\over \rho_{\nu} (x)} \right] = 0.\eqno(5.9)$$
Assuming $f_n(x)= \sum_{k=0}^n f_{n,k}\  x^k$ and employing (5.7), the first term on the left-hand side of (5.9) can be treated as follows

$$ {d^{r}\over dx^{r} } \left[ f_n(x) {\rho_\nu(x)\over \rho_{\nu+1} (x)} \right]  = {1\over \pi^2}  {d^{r}\over dx^{r} }   \sum_{k=0}^n f_{n,k} x^k \int_0^\infty e^{-xy } dy \int_0^\infty { e^{-sy} \ s^{-1} ds \over  J_{\nu+1}^2(2\sqrt s) +
Y_{\nu+1}^2(2\sqrt s)}$$

$$=  {1\over \pi^2}    \sum_{k=0}^n f_{n,k} (-1)^k  {d^{r}\over dx^{r} }  \int_0^\infty  {d^k\over dy^k} \left[ e^{-xy } \right] dy \int_0^\infty { e^{-sy} \ s^{-1} ds \over  J_{\nu+1}^2(2\sqrt s) + Y_{\nu+1}^2(2\sqrt s)}$$

$$=  {1\over \pi^2}    \sum_{k=0}^n f_{n,k} (-1)^k   \int_0^\infty  {\partial^{k+r} \over \partial y^k \partial x^{r} } \left[ e^{-xy } \right] dy \int_0^\infty { e^{-sy} \ s^{-1} ds \over  J_{\nu+1}^2(2\sqrt s) + Y_{\nu+1}^2(2\sqrt s)}$$

 $$=  {1\over \pi^2}    \sum_{k=0}^n f_{n,k} (-1)^{k+r}    \int_0^\infty  {d^k\over dy^k} \left[  y^{r}  e^{-xy } \right] dy \int_0^\infty { e^{-sy} \ s^{-1} ds \over  J_{\nu+1}^2(2\sqrt s) + Y_{\nu+1}^2(2\sqrt s)},$$
where the differentiation  under the integral sign is possible via the absolute and uniform convergence.  Now, we integrate  $k$ times by parts in the outer integral with respect to $y$ on the right-hand side of the latter equality, eliminating the  integrated terms due to the choice of $r$,  and then differentiate  under the  integral sign in the inner integral with respect to $s$ owing to the same arguments, to obtain

$$  {d^{r}\over dx^{r} } \left[ f_n(x) {\rho_\nu(x)\over \rho_{\nu+1} (x)} \right]   =  {1\over \pi^2}     \int_0^\infty   y^{r}  e^{-xy }  \int_0^\infty { e^{-sy} \ s^{-1}  \over  J_{\nu+1}^2(2\sqrt s) + Y_{\nu+1}^2(2\sqrt s)}   \left(    \sum_{k=0}^n f_{n,k} (-1)^{k+r}  s^k \right) ds .\eqno(5.10)$$
In the same fashion the second term in (5.9) is worked out to find  $( g_m(x)=  \sum_{k=0}^m g_{m,k}\  x^k$ )

$$  {d^{r}\over dx^{r} } \left[ x g_m(x) {\rho_{\nu-1} (x)\over \rho_{\nu} (x)} \right]   =  {1\over \pi^2}     \int_0^\infty   y^{r}  e^{-xy }  \int_0^\infty { e^{-sy}  \over  J_{\nu}^2(2\sqrt s) + Y_{\nu}^2(2\sqrt s)}   \left(    \sum_{k=0}^m g_{m,k} (-1)^{k+r+1}  s^k \right) ds.\eqno(5.11)$$
Substituting (5.10), (5.11) into (5.9) and cancelling twice  the Laplace transform  via its injectivity for integrable and continuous functions \cite{Tit},  we arrive at the equality

$$x g_m(-x) \left[  J_{\nu+1}^2(2\sqrt x) + Y_{\nu+1}^2(2\sqrt x) \right] + f_n(-x) \left[   J_{\nu}^2(2\sqrt x) + Y_{\nu}^2(2\sqrt x)\right] = 0,\ x > 0.\eqno(5.12) $$
The sum of squares of Bessel functions in brackets is called the Nicholson kernel, which has the Mellin-Barnes representation by virtue of Entry 8.4.20.35 in \cite{PrudnikovMarichev}, Vol. III

$$x^k  \left[   J_{\nu}^2(2\sqrt x) + Y_{\nu}^2(2\sqrt x)\right]  = {2^{1-2k}  \cos(\pi\nu)\over 2\pi^{7/2} i} \int_{\gamma-i\infty}^{\gamma+i\infty}
\Gamma(s+k) \Gamma(s+k+\nu) \Gamma(s+k-\nu) $$

$$\times \Gamma\left({1\over 2} -s-k\right) (4x)^{-s} ds,\quad |\nu | - k < \gamma < {1\over 2} - k.\eqno(5.13)$$
Then, using on the right-hand side of (5.13) the reflection formula for gamma function  \cite{Bateman}, Vol. I,  it can be written as follows

$$ {2^{1-2k}  \cos(\pi\nu)\over 2\pi^{7/2} i} \int_{\gamma-i\infty}^{\gamma+i\infty} \Gamma(s+k) \Gamma(s+k+\nu) \Gamma(s+k-\nu) \Gamma\left({1\over 2} -s-k\right) (4x)^{-s} ds$$

$$=  {2^{1-2k}  (-1)^k \cos(\pi\nu)\over 2\pi^{7/2} i} \int_{\gamma-i\infty}^{\gamma+i\infty} \Gamma(s+k) \Gamma(s+k+\nu) \Gamma(s+k-\nu) { \Gamma(1/2 -s ) \over (1/2+s)_k } (4x)^{-s} ds.\eqno(5.14)$$
Our goal now is to shift the  contour to the right to make integration along  the straight line with $|\nu| < {\rm Re} s < 1/2$. To do this we should take into account the residues at  $k$ simple poles $s_m= -1/2- m,\  m=0, 1,\dots, k-1$ which have the values

$$\hbox{Res}_{s= s_m}  \left(\frac{\Gamma(s+k) \Gamma(s+k+\nu) \Gamma(s+k-\nu)  \Gamma(1/2 -s )} {(1/2+s) (3/2+s)\dots (s +k-1/2) } (4x)^{-s}\right)$$

$$=  \frac{\Gamma(s_m+k) \Gamma(s_m+k+\nu) \Gamma(s_m+k-\nu)  \Gamma(1/2 -s_m )} {(1/2+s_m) (1/2+s_m+1)\dots (1/2+s_m +m-1) (1/2+s_m +m+1)\dots(1/2+s_m +k-1) } (4x)^{-s_m}$$

$$= { (-1)^m (4x)^{1/2+m} \over (k-m-1)!} \ \Gamma(k-m-1/2) \Gamma(k-m-1/2+\nu) \Gamma(k-m-1/2-\nu) . $$
Therefore we get from (5.14)

$$ {2^{1-2k}  \cos(\pi\nu)\over 2\pi^{7/2} i} \int_{\gamma-i\infty}^{\gamma+i\infty} \Gamma(s+k) \Gamma(s+k+\nu) \Gamma(s+k-\nu) \Gamma\left({1\over 2} -s-k\right) (4x)^{-s} ds$$

$$=  {2^{1-2k}  (-1)^k \cos(\pi\nu)\over 2\pi^{7/2} i} \int_{\mu-i\infty}^{\mu+i\infty} \Gamma(s+k) \Gamma(s+k+\nu) \Gamma(s+k-\nu) { \Gamma(1/2 -s ) \over (1/2+s)_k } (4x)^{-s} ds$$

$$-  {4^{1-k}  \sqrt x \cos(\pi\nu)\over \pi^{5/2}}\sum_{m=0}^{k-1} {(-1)^{k+m} (4x)^m \over (k-m-1)!}  \ \Gamma(k-m-1/2) \Gamma(k-m-1/2+\nu) \Gamma(k-m-1/2-\nu),\eqno(5.15)$$
where $|\nu| < \mu  < 1/2$.  Now, recalling  Parseval's equality for the Mellin transform and Entries 8.4.2.5, 8.4.23.27 in \cite{PrudnikovMarichev}, Vol. III,  we derive from (5.13), (5.15)

$$x^k  \left[   J_{\nu}^2(2\sqrt x) + Y_{\nu}^2(2\sqrt x)\right]  = {2^{3- 2k} (-1)^k  \sqrt x \over \pi^{3}}  \cos(\pi\nu) \int_0^\infty K^2_\nu(\sqrt t) \ {t^{k-1/2}\over 4x +t} dt$$

$$- \ {4^{1-k}  \sqrt x \cos(\pi\nu)\over \pi^{5/2}}\sum_{m=0}^{k-1} {(-1)^{k+m} (4x)^m \over (k-m-1)!}  \ \Gamma(k-m-1/2) \Gamma(k-m-1/2+\nu) \Gamma(k-m-1/2-\nu).\eqno(5.16)$$
Analogously, we find

 $$x^{k+1}  \left[   J_{\nu+1}^2(2\sqrt x) + Y_{\nu+1}^2(2\sqrt x)\right]  = {2^{1- 2k} (-1)^k  \sqrt x \over \pi^{3}}  \cos(\pi\nu) \int_0^\infty K^2_{\nu+1}(\sqrt t) \ {t^{k+1/2}\over 4x +t} dt$$

 $$ - \ {4^{-k}  \sqrt x \cos(\pi\nu)\over \pi^{5/2}}\sum_{m=0}^{k} {(-1)^{k+m+1} (4x)^m \over (k-m)!}  \ \Gamma(k-m+1/2) \Gamma(k-m+1/2+\nu) \Gamma(k-m+1/2-\nu).\eqno(5.17)$$
Substituting  expressions (5.16), (5.17) in (5.12), we derive after straightforward simplifications

$$  \int_0^\infty K^2_{\nu+1}(2 \sqrt t) \  g_m (t)  {\sqrt t \over x +t} dt +    \int_0^\infty K^2_{\nu}(2 \sqrt t) \  f_n(t)  {dt  \over \sqrt t (x +t)}$$

 $$ + \sqrt\pi \sum_{k=0}^m   \sum_{j=0}^{k} { (-x)^j \over  4^{k-j} (k-j)!}  \ \Gamma(k-j+1/2) \Gamma(k-j+1/2+\nu) \Gamma(k-j+1/2-\nu)$$

$$- \sqrt\pi \sum_{k=0}^{n-1}  \sum_{j=0}^{k} { (-x)^j \over 4^{k-j}  (k-j)!}  \ \Gamma(k-j+1/2) \Gamma(k-j+1/2+\nu) \Gamma(k-j+1/2-\nu) = 0,\  x >0.\eqno(5.18)$$
Last two terms in (5.18) are polynomials of degree $m, n-1$, respectively. Hence, differentiating through $r_1 \ge \max\{n, m+1\}$ times by $x$, we obtain

$$ {d^{r_1}\over dx^{r_1}}  \int_0^\infty K^2_{\nu+1}(2 \sqrt t) \  g_m (t)  {\sqrt t \over x +t} dt +     {d^{r_1}\over dx^{r_1}}  \int_0^\infty K^2_{\nu}(2 \sqrt t) \  f_n(t)  {dt  \over \sqrt t (x +t)} = 0.\eqno(5.19)$$
The left-hand side of (5.19) is the $r_1$-th derivative of the sum of two Stieltjes transforms which are, in turn,  two fold Laplace transforms. Consequently, fulfilling the differentiation under the integral sign in (5.19) as above owing to the absolute and uniform convergence by $x \ge x_0 >0 $,  we cancel Laplace transforms of integrable functions via the injectivity.  Then with (2.2)  it yields the equality

$$f_n(x) \rho^2_\nu(x)+ g_{m} (x) \rho^2_{\nu+1} (x) = 0,\quad x >0.\eqno(5.20)$$
Comparing with (5.6), we see that $  h_l \equiv 0.$  Further,  identity (5.20) implies that $f_n, \ g_m$ have the same positive roots, if any.    Let $x >0$ be not a root of $g_m$.  Then dividing (5.20) by $g_m$ and making a differentiation, we get

$$- 2 \rho_\nu (x) \rho_{\nu-1} (x) {f_n(x)\over g_m(x)} +   \rho^2_\nu(x) {f^\prime_n(x) g_m(x)- f_n(x) g_m^\prime(x) \over g_m^2(x)} - 2 \rho_{\nu+1}(x)\rho_\nu(x) =0.$$
Since $\rho_\nu(x) >0$, we divide the previous equation by $\rho_\nu$, multiply by $x,  g_m^2$ and employ  (3.6) to find

$$- 2 g_m(x) \rho_{\nu+1}(x) \left( f_n(x) +x g_m(x) \right) + \rho_\nu(x) \left( 2\nu f_n(x) g_m(x) +   x \left( f^\prime_n(x) g_m(x)- f_n(x) g_m^\prime(x) \right) \right) =0.\eqno(5.21)$$
However, the existence of the type 1 multiple orthogonal polynomials with respect to the vector of weight functions $(\rho_\nu, \rho_{\nu+1})$ suggests the equalities

$$ g_m(x) \left( f_n(x) +x g_m(x) \right) \equiv 0,\quad\quad  2\nu f_n(x) g_m(x) +   x \left( f^\prime_n(x) g_m(x)- f_n(x) g_m^\prime(x)\right)\equiv 0.\eqno(5.22)$$
So, if $g_m \equiv 0$, it proves the theorem. Otherwise  $f_n(x)+ x g_m(x) \equiv 0$,  and with the second equation in (5.22) we have $ x(2\nu+1) g^2_m(x) \equiv 0.$ Thus $g_m \equiv 0$ and $f_n \equiv 0$ from (5.20). Theorem 4 is proved.

  \end{proof}

  {\bf Remark 3}.  The choice of $\nu$ is important. For instance, for $\nu= -1/2$  the theorem  fails.  This can be seen,  taking $f_n(x) \equiv -x,\  g_m(x)\equiv  1, \ h_l(x) \equiv 0$.

{\bf Theorem 5}. {\it Let $\nu \in [0,1/2)$. For every $\alpha > 0$

$${d\over dx} \left[ x^\alpha  q^\alpha_{n,n-1,n-1}(x) \right] = x^{\alpha-1} q^{\alpha-1}_{n,n-1,n} (x)\eqno(5.23)$$
and the following differential recurrence relations hold}

$$ A^{\alpha-1} _n(x) =   (\alpha +2\nu) A^\alpha_n(x) + x [ A^\alpha_n(x) ]^\prime   - x  C^\alpha_{n-1}(x) ,\eqno(5.24)$$

$$ B^{\alpha-1} _{n-1}(x) = \alpha B^\alpha_{n-1}(x)  + x  [B^\alpha_{n-1}(x) ]^\prime  - C^\alpha_{n-1}(x),\eqno(5.25)$$

$$ C^{\alpha-1} _n(x) = (\alpha+\nu)  C^\alpha_{n-1}(x) + x  [C^\alpha_{n-1}(x)]^\prime   - 2  A^\alpha_n(x)  -2x B^\alpha_{n-1}(x).\eqno(5.26)$$

\begin{proof}  From (5.1), (5.2) and integration by parts we get

$$ \int_0^\infty     q^\alpha_{n,n-1,n-1}(x) x^{\alpha+m} dx =  -  {1\over m+1} \int_0^\infty  {d\over dx} \left[  x^\alpha q^\alpha_{n,n-1,n-1}(x) \right] x^{m+1} dx = 0,$$
where the integrated terms vanish for every $\alpha > -1$ via asymptotic behavior  (2.4), (2.5).  Hence (5.1) suggests the equality

$$\int_0^\infty  {d\over dx} \left[  x^\alpha q^\alpha_{n,n-1,n-1}(x) \right] x^{m} dx = 0, \quad    m= 1,\dots, 3n.$$
But,  evidently,

$$\int_0^\infty  {d\over dx} \left[  x^\alpha q^\alpha_{n,n-1,n-1}(x) \right]  dx = 0,\quad \alpha > 0.$$
Therefore

$$\int_0^\infty  {d\over dx} \left[  x^\alpha q^\alpha_{n,n-1,n-1}(x) \right] x^{m} dx = 0, \quad    m= 0,\dots, 3n.\eqno(5.27)$$
Now, working out the differentiation in (5.27), involving (3.5), (3.6), we find

$$ {d\over dx} \left[  x^\alpha q^\alpha_{n,n-1,n-1}(x) \right] = \alpha\  x^{\alpha-1}  q^\alpha_{n,n-1,n-1}(x) + x^{\alpha-1}  \left(  x [ A^\alpha_n(x) ]^\prime  \rho^2_\nu (x) + x  [B^\alpha_{n-1}(x) ]^\prime \rho^2_{\nu+1} (x)+  x  [C^\alpha_{n-1}(x)]^\prime \rho_\nu (x) \rho_{\nu+1} (x) \right. $$

$$\left. + 2  A^\alpha_n(x) \rho_\nu (x) (\nu \rho_\nu(x)- \rho_{\nu+1} (x) )  -2x B^\alpha_{n-1}(x) \rho_{\nu+1} (x)\rho_\nu(x)\right. $$

$$\left. - x  C^\alpha_{n-1}(x) \rho^2_\nu (x) +   C^\alpha_{n-1}(x)  \rho_{\nu+1} (x)  (\nu \rho_\nu(x)- \rho_{\nu+1} (x) ) \right).$$
Thus

$$  {d\over dx} \left[  x^\alpha q^\alpha_{n,n-1,n-1}(x) \right] = x^{\alpha-1} \left[ A^{\alpha-1}_n(x) \rho^2_\nu (x) +  B^{\alpha-1}_{n-1}(x) \rho^2_{\nu+1} (x)+  C^{\alpha-1}_{n}(x) \rho_\nu (x) \rho_{\nu+1} (x) \right],\eqno(5.28)$$
where $A^{\alpha-1}_n(x) ,\   B^{\alpha-1}_{n-1}(x),  \ C^{\alpha-1}_{n}(x) $ are polynomials of degree at most $n, n-1, n$, respectively, being defined by formulas (5.24), (5.25), (5.26).  The linear homogeneous system (5.27) of $3n+1$ equations contains $3n+ 2$ unknowns.  Therefore up to a constant, choosing to be one,  the left-hand side of (5.28) is equal to $ x^{\alpha-1}  q^{\alpha-1}_{n,n-1,n}(x)$, and the representation (5.28) is unique by virtue of Theorem 4.  This proves (5.23) and completes the proof of Theorem 5.

\end{proof}

{\bf Remark 4}. The same analysis for the function $q^\alpha_{n,n-1,n}(x)$ does not work. In fact, in this case we derive analogously

$$\int_0^\infty  {d\over dx} \left[  x^\alpha q^\alpha_{n,n-1,n}(x) \right] x^{m} dx = 0, \quad    m= 0,\dots, 3n+1.$$
However, working out the differentiation, we will get polynomials of degree at most $n+1, n, n$, respectively, which implies $3n+4$ unknowns for $3n+2$ equations (the so-called quasi multiple orthogonal case.)

Finally, we establish the differentiation property for the type $2$ multiple orthogonal polynomials  $p^\alpha_{n,n-1,n}$ (or $3$-orthogonal polynomials).

{\bf Theorem 6.} {\it For every $\nu \ge 0,\ \alpha > -1$

$${d\over dx} \left[ p^\alpha_{n,n-1,n} (x)\right] =   (3n-1) p^{\alpha+1}_{n,n-1,n-1} (x).\eqno(5.29)$$

\begin{proof} Recalling (3.5), (3.6) and asymptotic behavior of the weight functions (2.4), (2.5), we integrate by parts in (5.3), eliminating the integrated terms,  to  deduce

$$ \int_0^\infty {d\over dx} \left[  p^\alpha_{n,n-1,n}(x) \right]  \rho^2_\nu (x) x^{\alpha+1+m} dx = 0,\quad m= 0,1,\dots, n-1.\eqno(5.30)$$
Concerning equalities (5.4), (5.5),  it corresponds the following ones

 $$ \int_0^\infty  {d\over dx} \left[  p^\alpha_{n,n-1,n}(x) \right] \rho^2_{\nu+1} (x) x^{\alpha+1+m} dx = 0,\quad m= 0,1,\dots, n-2,\eqno(5.31)$$

$$ \int_0^\infty  {d\over dx} \left[  p^\alpha_{n,n-1,n}(x) \right] \rho_\nu (x) \rho_{\nu+1}(x) x^{\alpha+1+m} dx = 0,\quad m= 0,1,\dots, n-2.\eqno(5.32)$$
 Now   $\left[  p^\alpha_{n,n-1,n}(x) \right]^\prime$ is a polynomial of degree $3n-2$ with leading coefficient $3n-1$ and by equalities (5.30), (5.31), (5.32) it satisfies orthogonality conditions (5.3), (5.4), (5.5) for the type $2$ multiple orthogonal polynomial $ p^{\alpha+1}_{n,n-1,n-1} (x)$.  Hence we get (5.29) by unicity. Theorem 6 is proved.

\end{proof}

\bibliographystyle{amsplain}

redpajama_set_name: RedPajamaArXiv


text:
The Supreme x Chris Cunningham Night-Vision Collection Has Officially Dropped
Style | Supreme | November 12, 2018
Literally straight after the Nike Air Force 1 collaboration with COMME des GARÇONS Shirt, Supreme has released yet another collaborative capsule , this time with artist Chris Cunningham.
Chris Cunningham is a British artist, best known for his filmmaking, visual artistry, photography and music production. He's no stranger to collaborations, having worked with electronic musician Richard David James aka Aphex Twin, many a times, which began in 1997 with the music video for "Come to Daddy."
A post shared by Supreme (@supremenewyork) on Nov 7, 2018 at 9:00am PST
The Chris Cunningham for Supreme capsule includes a hoodie and two graphic T-shirts featuring Cunningham's character, Rubber Johnny. In 2005, Cunningham first released Rubber Johnny into the world in a six-minute short that was soundtracked by Aphex Twin and shot in night vision, now carried over into the Supreme capsule.
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Cover image via Supreme.

redpajama_set_name: RedPajamaCommonCrawl


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- UDC A fun and educational quiz on Shakespeare' s first lines. The introduction of the Romeo and Juliet essay should.
Wilson Introduce our study guide questions 1, there were many times read more essays, compare contrast the introduction. Each essay is 2, 500 to 5, 000 words in length, and all essays conclude with a list of " Works Cited, " along with endnotes.
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The opening scene of Romeo and Juliet immediately shows us that the theme of the play predominantly will be tragedy. Free Romeo and Juliet Essays and Papers - 123HelpMe.
Leadership and responsibility in the second world. Romeo and Juliet' s fates were intertwined from the moment of their birth, families driven by hate pushing two young lovers together.
Critical Context: Original Introductory Essays. In Romeo and Juliet, the lovers are doomed from the start, and sure.
The concept of fate functions as a central theme in Shakespeare' s Romeo and Juliet. Romeo' s Impulsiveness in Romeo and Juliet Anonymous 11th.
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The Opening Scene in Shakespeare' s Romeo and Juliet Essay. Get free homework help on William Shakespeare' s Romeo and Juliet: play summary, scene summary and analysis and original text, quotes, essays, character analysis, and filmography courtesy of CliffsNotes.
The introduction of the Romeo and Juliet essay should highlight the main idea of the Romeo and Juliet essay. Romeo and Juliet Essay.
Intro Paragraphs & Thesis Statements. At their impressionable age, love can seem.
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The general type of story. Hook and background information.
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An Intro to Intro Paragraphs. English 1: Romeo and Juliet Essay.
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Romeo and Juliet Introduction by Cassie Lowe on Prezi Free Romeo and Juliet papers, essays, and research papers. This line leads many readers to believe that Romeo and Juliet are inescapably destined to fall in love and equally destined.
The essays within this edition of Critical Insights are no exception to the variety of perspectives that Romeo and Juliet inspires. STEP THREE: ( 20 points) Outline traditionally or with a web.
A Review of the Play Romeo and Juliet by William Shakespeare. The press is the light that makes the roaches scatter " charles blow.


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I have supported several stage productions in various ways, from promotion shots to cast profiles and photos of the performance itself.
With my background being in portraiture, I often instinctively pick out individuals in my shots of show performances, so that I can convey the passion and energy of the performer and the event. I also occasionally offer individual portraits of performers either on location or in the photography studio, away from the busy stage environment.


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SSI Celebrates Grand Opening in Manassas Park HUBZone
Manassas Park, VA September 25, 2014 - SSI, a 7 year old firm providing a variety of information technology services to help government leaders make wise technology investments recently opened their new headquarters at 1 Park Place in Manassas Park. Virginia Senator Charles Colgan, Manassas Park Mayor Frank Jones, and Ms. Syd Dorsey from the Office of the Virginia Secretary of Commerce and Trade joined SSI President/CEO Steve Mackie in making remarks and cutting the ribbon to commemorate the opening. The festivities continued with a luncheon and tours of SSI's new facility. SSI's Manassas Park location is in a Historically Underutilized Business Zone (HUBZone) as designated by the U.S. Small Business Administration (SBA). Steve Mackie, President and CEO, was enthusiastic when discussing his decision to locate within the Manassas Park HUBZone calling it "a natural choice" for SSI.
HUBZones promote economic development in historically underutilized business zones through access to federal procurement opportunities. Program benefits for HUBZone certified companies include competitive and sole source contracting and a ten percent price evaluation preference in full and open contract competitions, as well as subcontracting opportunities.
The event attracted a significant crowd of business leaders, supporters, and elected officials to highlight both the achievement of SSI and the business-friendly climate of Manassas Park.
Senator Colgan, the longest serving state senator in Virginia history, told the crowd that he introduced the bill to make Manassas Park an official city during his first term in office.
Mayor Jones stated that SSI is a great fit for the HUBZone location and he and his staff want to do whatever they can to help other businesses interested in locating within the City's HUBZone.
Pastor Ralph Weitz of Immanuel Bible Church gave the Invocation and made brief remarks in which he congratulated the City of Manassas Park for "discovering Steve Mackie." Pastor Weitz had the pleasure of working with Mr. Mackie on a number of community and charitable projects and knew firsthand his ability to get things done.
Syd Dorsey, Advisor for Small Business Equity and Development for the Virginia Secretary of Commerce and Trade, was pleased to help celebrate SSI's opening, particularly given that Information Technology is one of Virginia's key industries. She noted that small business has been the biggest creator of jobs since the 2008-2009 Recession. Ms. Dorsey also applauded SSI's commitment to and investment in the local community, particularly their establishment of a workforce development program through the SSI Career Resource Center.
SSI's Career Resource Center is a new service provided free of charge to the Manassas Park Community as a collaboration between SSI and City of Manassas Park. The Center is the innovative brainstorm of Mr. Mackie and his colleagues who wanted to provide a tangible service to the local community. The mission of the Career Resource Center is to help local residents learn the necessary skills to work in jobs requiring basic Information Technology skills. SSI's friendly, bi-lingual staff can help residents with a number of different activities on the computer, both job related and for personal use including: basic computer use, Microsoft Windows basic use, web browser use, email setup and use, basic internet search skills, basic use of Microsoft Office Suite and resume formatting.
SSI was founded in 2007 by Steve Mackie, a retired Navy Commander, to enable government leaders to make technology investments based on real needs, rather than the sales cycle of an equipment vendor. SSI is a Department of Veterans Affairs verified Service-Disabled Veteran-Owned Small Business (SDVOSB) and certified by the U.S. SBA as a HUBZone Small Business. Page 3 of 3
The company employs over 30 employees, 12 of whom reside in Manassas Park or the surrounding community. Forty percent of SSI's workforce are veterans and six percent are active Reservists. SSI is a Certiport Authorized Testing Center (CATC) providing certification testing services for a number of information technology vendors including Microsoft, Adobe, and CompTIA.
SSI serves a number of customers across a variety of sectors including the Department of Defense, the Intelligence Community, and Civilian Federal Agencies.
To interview Steve Mackie or obtain additional information, please contact SSI at 703-451-1860.


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Polyhymno inermis is a moth of the family Gelechiidae. It was described by Edward Meyrick in 1913. It is found in Mpumalanga, South Africa.

The wingspan is about 8 mm. The forewings are ochreous brownish with a fine white oblique striga from the costa at two-thirds, the costal edge blackish on each side of this, its apex extended as a leaden-grey line to near the termen beneath the apex, then sharply angulated and continued near the termen to near the tornus. There are two wedge-shaped white marks from the costa towards the apex. The hindwings are grey.

References

Endemic moths of South Africa
Moths described in 1913
Polyhymno

redpajama_set_name: RedPajamaWikipedia


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Baillères is a surname. Notable people with the surname include:

Alberto Baillères (1931–2022), Mexican billionaire businessman
Alejandro Baillères (born 1960), Mexican businessman
Raúl Baillères (1895–1967), Mexican businessman