On orthogonal polynomials and quadrature rules related to the second kind of beta distribution

Abstract

We consider a finite class of weighted quadratures with the weight function $x^{-2a}{(1+x^2)}^{-b}$ on $(-\mathrm{\infty },\mathrm{\infty })$, which is valid only for finite values of n (the number of nodes). This means that classical Gauss-Jacobi quadrature rules cannot be considered for this class, because some restrictions such as $\{maxn\}\le a+b-1/2$, $a<1/2$, $b>0$ and ${(-1)}^{2a}=1$ must be satisfied for its orthogonality relation. Some analytic examples are given in this sense.

MSC:41A55, 65D30, 65D32.

1 Introduction

The differential equation

(1)

was introduced in [1], and it was established that the symmetric polynomials

$$\begin{array}{rcl}{\mathrm{\Phi }}_n(x)& =& S_n\left(\begin{array}{cc}r& s\\ p& q\end{array}|x\right)\\ =& \sum _{k=0}^{[n/2]}\left(\begin{array}{c}k\end{array}\right)\left(\prod _{i=0}^{[n/2]-(k+1)}\frac{(2i+{(-1)}^{n+1}+2[n/2])p+r}{(2i+{(-1)}^{n+1}+2)q+s}\right)x^{n-2k}\end{array}$$

(2)

are a basis solution of it. If this equation is written in a self-adjoint form, then the first-order equation

$$x\frac{d}{dx}((p{x}^2+q)W(x))=(r{x}^2+s)W(x)$$

(3)

would appear. The solution of equation (3) is known as an analogue of Pearson distributions family and can be indicated as

$$W\left(\begin{array}{cc}r& s\\ p& q\end{array}|x\right)=exp(\int \frac{(r-2p)x^2+s}{x(p{x}^2+q)}dx).$$

(4)

There are four main sub-classes of distributions family (4) (and consequently, sub-solutions of equation (3)) whose explicit probability density functions are, respectively, as follows:

(5)

(6)

(7)

(8)

where $K_i$; $i=1,2,3,4$ play the normalizing constant role.

Clearly, the value of distribution vanishes at $x=0$ in each of the above mentioned four cases, i.e., $W(p,q,r,s;0)=0$ for $s\ne 0$.

As a special case of (4), let us consider the values $p=1$, $q=1$, $r=-2a-2b+2$ and $s=-2a$ corresponding to distribution (7) and replace them in equation (1) to get

(9)

By solving equation (9), the polynomial solution of monic type is derived

(10)

According to [1], these polynomials are finitely orthogonal with respect to the second kind of beta weight function $x^{-2a}{(1+x^2)}^{-b}$ on $(-\mathrm{\infty },\mathrm{\infty })$ if and only if $\{maxn\}\le a+b-1/2$, i.e., we have

(11)

if $m,n=0,1,\dots$ , $N\le a+b-1/2$, where $N=max\{m,n\}$, ${\delta }_{n,m}=\{\begin{array}{cc}0\hfill & (n\ne m),\hfill \\ 1\hfill & (n=m),\hfill \end{array}$ $a<1/2$, $b>0$ and ${(-1)}^{2a}=1$. Moreover, they satisfy a three-term recurrence relation

(12)

The orthogonality property (11) shows that the polynomials ${\overline{S}}_n(1,1,-2a-2b+2,-2a;x)$ are a suitable tool to finitely approximate the functions that satisfy the Dirichlet conditions [2–5].

For example, if $N=\{maxn\}=3$, $a+b\ge 7/2$, $a<1/2$, $b>0$ and ${(-1)}^{2a}=1$ in (10), then the arbitrary function $f(x)$ can be approximated as

$$f(x)\cong \sum _{m=0}^3{B}_m{\overline{S}}_m(1,1,-2a-2b+2,-2a;x),$$

(13)

where

$$\begin{array}{rcl}{B}_m& =& \left({(-1)}^m\prod _{i=1}^m\frac{(i-(1-{(-1)}^i)a)(i-(1-{(-1)}^i)a-2b)}{(2i-2a-2b+1)(2i-2a-2b-1)}\right)\frac{\mathrm{\Gamma }(b+a-1/2)\mathrm{\Gamma }(-a+1/2)}{\mathrm{\Gamma }(b)}\\ \times {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x^{-2a}}{{(1+x^2)}^b}{\overline{S}}_m\left(\begin{array}{cc}-2a-2b+2& -2a\\ 1& 1\end{array}|x\right)f(x)dx.\end{array}$$

(14)

This means that the finite set ${\{{\overline{S}}_i(1,1,-2a-2b+2,-2a;x)\}}_{i=0}^3$ is a basis space for all polynomials of degree at most three, i.e., for $f(x)=a_3{x}^3+a_2{x}^2+a_{1}x+a_0$, the approximation (13) is exact. This matter helps us to state the application of symmetric orthogonal polynomials (10) in weighted quadrature rules [6–9].

2 Application of ${\overline{S}}_n(1,1,-2a-2b+2,-2a;x)$ in quadrature rules

Consider the general form of a weighted quadrature

$${\int }_{\alpha }^{\beta }w(x)f(x)dx=\sum _{i=1}^n{w}_{i}f(x_i)+\sum _{k=1}^m{v}_{k}f(z_k)+R_{n,m}[f],$$

(15)

where $w(x)$ is a positive function on $[\alpha ,\beta ]$; ${\{w_i\}}_{i=1}^n$, ${\{v_k\}}_{k=1}^m$ are unknown coefficients; ${\{x_i\}}_{i=1}^n$ are unknown nodes; ${\{z_k\}}_{k=1}^m$ are pre-determined nodes [7, 8]; and finally, the residue $R_{n,m}[f]$ is determined (see, e.g., [8]) by

$$R_{n,m}[f]=\frac{f^{(2n+m)}(\xi )}{(2n+m)!}{\int }_{\alpha }^{\beta }w(x)\prod _{k=1}^m(x-z_k)\prod _{i=1}^n{(x-x_i)}^{2}dx;\alpha <\xi <\beta .$$

(16)

It can be shown in (15) that $R_{n,m}[f]=0$ for any linear combination of the sequence $\{1,x,\dots ,x^{2n+m+1}\}$ if and only if ${\{x_i\}}_{i=1}^n$ are the roots of orthogonal polynomials of degree n with respect to the weight function $w(x)$, and ${\{z_k\}}_{k=1}^m$ belong to $[\alpha ,\beta ]$; see [7] for more details. Also, it is proved that to derive ${\{w_i\}}_{i=1}^n$ in (15), when $m=0$, it is not required to solve the following linear system of order $n\times n$:

$$\sum _{i=1}^n{w}_i{x}_i^j={\int }_{\alpha }^{\beta }w(x)x^{j}dx\text{for }j=0,1,\dots ,2n-1.$$

(17)

Rather, one can directly use the relation

$$\frac{1}{w_i}={\hat{P}}_0^2(x_i)+{\hat{P}}_1^2(x_i)+\cdots +{\hat{P}}_{n-1}^2(x_i)\text{for }i=1,2,\dots ,n,$$

(18)

in which ${\hat{P}}_i(x)$ is the orthonormal polynomial of $P_i(x)$, i.e.,

$${\hat{P}}_i(x)={({\int }_{\alpha }^{\beta }w(x)P_i^2(x)dx)}^{-1/2}{P}_i(x).$$

(19)

Now, by noting that the symmetric polynomials (10) are finitely orthogonal with respect to the weight function $W(x,a,b)=x^{-2a}{(1+x^2)}^{-b}$ on the real line, we consider the following finite class of quadrature rules:

(20)

where $x_j$ are the roots of polynomials ${\overline{S}}_n(1,1,-2a-2b+2,-2a;x)$ and $w_j$ are calculated by

$$\frac{1}{w_j}=\sum _{i=0}^{n-1}{({\overline{S}}_i^{\ast }(1,1,-2a-2b+2,-2a;x_j))}^2\text{for }j=0,1,2,\dots ,n.$$

(21)

2.1 An important remark

The change of variable $x=t^{-1/2}{(1-t)}^{1/2}$ in the left-hand side of (20) first changes the interval $(-\mathrm{\infty },\mathrm{\infty })$ to $[0,1]$ such that we have

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x^{-2a}}{{(1+x^2)}^b}f(x)dx={\int }_0^1{t}^{a+b-\frac{3}{2}}{(1-t)}^{-a-\frac{1}{2}}f\left(\sqrt{\frac{1}{t}-1}\right)dt.$$

(22)

As the right-hand integral of (22) shows, the shifted Jacobi weight function ${(1-x)}^u{x}^v$ has appeared for $u=-a-1/2$ and $v=a+b-3/2$. Hence, the shifted Gauss-Jacobi quadrature rule [6, 9] with the special parameters $u=-a-1/2$ and $v=a+b-3/2$ can also be applied for estimating (22). This procedure eventually changes (20) into the form

(23)

where $x_j^{(-a-1/2,a+b-3/2)}$ are the zeros of shifted Jacobi polynomials $P_{n,+}^{(-a-1/2,a+b-3/2)}(x)$ on $[0,1]$. But, there is the main problem for the formula (23). From (16), it is generally known that the residue of quadrature rules depends on $f^{(2n)}(\xi )$; $\alpha <\xi <\beta$. Therefore, by noting (23), we should have

$$\frac{d^{2n}f(\sqrt{x^{-1}-1})}{d{x}^{2n}}=\sum _{i=0}^{2n}{\phi }_i(x)f^{(i)}\left(\sqrt{x^{-1}-1}\right),$$

(24)

where ${\phi }_i$ are real functions to be computed and $f^{(i)}$, $i=0,1,2,\dots ,2n$ are the successive derivatives of the function f. On the other hand, the function f cannot be in the form of an arbitrary polynomial in order that the right-hand side of (24) becomes zero. In other words, the formula (23) cannot be exact for all elements of the basis $f(x)=x^j$; $j=0,1,2,\dots ,2n-1$. This is the main disadvantage of using (23), which shows the importance of the polynomials (10) in estimating a class of weighted quadrature rules [10]. The following examples clarify this remark.

Example 1

Consider the two-point quadrature formula

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^{-2a}{(1+x^2)}^{-b}f(x)dx\cong w_{1}f(x_1)+w_{2}f(x_2),$$

(25)

in which $a+b\ge 5/2$, $a<1/2$, $b>0$ and ${(-1)}^{2a}=1$. According to the explained comments, (25) must be exact for all elements of the basis $f(x)=\{1,x,x^2,x^3\}$ if and only if $x_1$, $x_2$ are two roots of ${\overline{S}}_2(1,1,-2a-2b+2,-2a;x)$. As a particular sample, let us take $a=0$ and $b=3$. Then (25) is reduced to

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{{(1+x^2)}^3}f(x)dx\cong w_{1}f\left(\frac{\sqrt{3}}{3}\right)+w_{2}f(-\frac{\sqrt{3}}{3}),$$

(26)

in which $\sqrt{3}/3$ and $-\sqrt{3}/3$ are zeros of ${\overline{S}}_2(1,1,-4,0;x)$ and $w_1$, $w_2$ are computed by solving the linear system

$$w_1+w_2={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{(1+x^2)}^{-3}dx=\frac{3}{8}\pi ,\frac{\sqrt{3}}{3}(w_1-w_2)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x{(1+x^2)}^{-3}dx=0.$$

(27)

After deriving $w_1$, $w_2$ in (27), the complete form of (26) would be

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{{(1+x^2)}^3}f(x)dx=\frac{3\pi }{16}(f\left(\frac{\sqrt{3}}{3}\right)+f(-\frac{\sqrt{3}}{3}))+R_2[f],$$

(28)

where

$$R_2[f]=\frac{f^{(4)}(\xi )}{4!}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{{(1+x^2)}^3}{\left({\overline{S}}_2\left(\begin{array}{cc}-4& 0\\ 1& 1\end{array}|x\right)\right)}^{2}dx=\frac{\pi }{72}{f}^{(4)}(\xi ),\xi \in \mathbf{R}.$$

(29)

Relation (28) shows that it is exact for any arbitrary polynomial of degree at most three.

Example 2 To have a three-point formula of type (20), first we should note that the conditions $a+b\ge 7/2$, $a<1/2$, $b>0$ and ${(-1)}^{2a}=1$ must be satisfied. For instance, if $a=-1$ and $b=5$, then after some computations, the related formula takes the form

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x^2}{{(1+x^2)}^5}f(x)dx=\frac{\pi }{1,280}(9f\left(\sqrt{\frac{5}{3}}\right)+32f(0)+9f(-\sqrt{\frac{5}{3}}))+R_3[f],$$

(30)

where

$$R_3[f]=\frac{f^{(6)}(\xi )}{6!}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x^2}{{(1+x^2)}^5}{\left({\overline{S}}_3\left(\begin{array}{cc}-6& 2\\ 1& 1\end{array}|x\right)\right)}^{2}dx=\frac{5\pi }{3,456}{f}^{(6)}(\xi ),\xi \in \mathbf{R},$$

(31)

and $x_1=\sqrt{5/3}$, $x_2=0$, $x_3=-\sqrt{5/3}$ are the roots of ${\overline{S}}_3(1,1,-6,2;x)=x^3-(5/3)x$.

Example 3 To derive a four-point formula of type (20), first the conditions $a+b\ge 9/2$, $a<1/2$, $b>0$ and ${(-1)}^{2a}=1$ must be satisfied. For example, if $a=0$ and $b=6$, then

(32)

where

$$\begin{array}{rcl}{R}_4[f]& =& \frac{f^{(8)}(\xi )}{8!}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{{(1+x^2)}^6}{\left({\overline{S}}_4\left(\begin{array}{cc}-10& 0\\ 1& 1\end{array}|x\right)\right)}^{2}dx\\ =& \frac{\pi }{2,822,400}{f}^{(8)}(\xi ),\xi \in \mathbf{R}.\end{array}$$

(33)

This formula is exact for all elements of the basis $f(x)=x^j$; $j=0,1,2,\dots ,7$ and its nodes are the roots of ${\overline{S}}_4(1,0,-8,2;x)=x^4-(6/5)x^2+3/35$.

Tables 1-3 show some numerical examples related to three given examples.

Table 1 Numerical results for two-point formula ( 28 )

Table 2 Numerical results for three-point formula ( 30 )

Table 3 Numerical results for four-point formula ( 32 )