The following result is given in [18 , p. 441, formula 43] (see also [6 ]).
Corollary.
For any natural number m we have
$${}_{p}F_{q} \left({\left. {\begin{array}{*{20}c} {a_{1},\,\,a_{2},\,\, \cdots} & {a_{p}} \\ {b_{1},\,\,b_{2},\,\, \ldots} & {b_{q}} \\ \end{array}} \right|\,z} \right) = \sum\limits_{k = 0}^{m - 1} {\frac{{(a_{1})_{k} (a_{2})_{k} \ldots (a_{p})_{k}}}{{(b_{1})_{k} (b_{2})_{k} \cdots (b_{q})_{k}}}\,\frac{{z^{k}}}{k!}\,{}_{(mp + 1)}F_{(mq + m)} \left({\left. {\begin{array}{*{20}c} {\vec{A}_{1,k},\,\,\vec{A}_{2,k},\,\, \ldots} & {\vec{A}_{p,k},\,\,1} \\ {\vec{B}_{1,k},\,\,\vec{B}_{2,k},\,\, \ldots} & {\vec{B}_{q,k},\vec{I}_{1,k}} \\ \end{array}} \right|\,m^{(p - q - 1)m} z^{m}} \right)} \,,$$
(2)
where
$$\begin{aligned} \vec{A}_{j,k} = \left( {\frac{{a_{j} + k}}{m},\frac{{a_{j} + 1 + k}}{m},\, \ldots \,,\,\frac{{a_{j} + m - 1 + k}}{m}} \right)\,\,\,\,\,(j = 1,\;2,\, \ldots ,\,\;p), \hfill \\ \vec{B}_{j,k} = \left( {\frac{{b_{j} + k}}{m},\frac{{b_{j} + 1 + k}}{m},\, \ldots \,,\,\frac{{b_{j} + m - 1 + k}}{m}} \right)\,\,\,\,\,\,\,(j = 1,\;2,\, \ldots \,,\,\;q), \hfill \\ \end{aligned}$$
and
$$\vec{I}_{1,k} = \left( {\frac{1 + k}{m},\frac{2 + k}{m},\, \ldots \,,\,\frac{m + k}{m}} \right).$$
Application 1.
Identities (2 ) can be interpreted as a decomposition formula for many elementary special functions whose hypergeometric representations are known.
Example 1.
Take \((p,q) = (0,0)\) and \(m = 2,3,4\) in (2 ) to, respectively, obtain
$$e^{z} = {}_{0}F_{1} \left( {\left. {\begin{array}{*{20}c} - \\ {1/2} \\ \end{array} \,} \right|\,\,\frac{1}{4}z^{2} } \right) + z\,{}_{0}F_{1} \left( {\left. {\begin{array}{*{20}c} - \\ {3/2} \\ \end{array} \,} \right|\,\,\frac{1}{4}z^{2} } \right) = \cosh z + \sinh z,$$
$$e^{z} = {}_{0}F_{2} \left( {\left. {\begin{array}{*{20}c} - \\ {1/3,\,\,\,2/3} \\ \end{array} \,} \right|\,\,\frac{1}{27}z^{3} } \right) + z{}_{0}F_{2} \left( {\left. {\begin{array}{*{20}c} - \\ {2/3,\,\,\,4/3} \\ \end{array} \,} \right|\,\,\frac{1}{27}z^{3} } \right) + \frac{{z^{2} }}{2}{}_{0}F_{2} \left( {\left. {\begin{array}{*{20}c} - \\ {4/3,\,\,\,5/3} \\ \end{array} \,} \right|\,\,\frac{1}{27}z^{3} } \right),$$
and
$$\begin{aligned} e^{z} = {}_{0}F_{3} \left( {\left. {\begin{array}{*{20}c} - \\ {1/4,\,\,\,1/2,\,\,\,3/4} \\ \end{array} \,} \right|\,\,\frac{1}{256}z^{4} } \right) + z\,{}_{0}F_{3} \left( {\left. {\begin{array}{*{20}c} - \\ {1/2,\,\,\,3/4,\,\,\,5/4} \\ \end{array} \,} \right|\,\,\frac{1}{256}z^{4} } \right) \hfill \\ \;\;\; +\quad \frac{{z^{2} }}{2}{}_{0}F_{3} \left( {\left. {\begin{array}{*{20}c} - \\ {3/4,\,\,\,5/4,\,\,\,3/2} \\ \end{array} \,} \right|\,\,\frac{1}{256}z^{4} } \right) + \frac{{z^{3} }}{6}{}_{0}F_{3} \left( {\left. {\begin{array}{*{20}c} - \\ {5/4,\,\,\,3/2,\,\,\,7/4} \\ \end{array} \,} \right|\,\,\frac{1}{256}z^{4} } \right)\,. \hfill \\ \end{aligned}$$
Note that the first representation corresponds to the decomposition of the exponential function in the even and odd parts. Moreover, the general case of the three above relations is given as
$$e^{z} = \sum\limits_{k = 0}^{m - 1} {\,\frac{{z^{k} }}{k!}\,{}_{1}F_{m} \left( {\left. {\begin{array}{*{20}c} 1 \\ {\frac{1 + k}{m},\frac{2 + k}{m},\, \ldots \,,\,\frac{m + k}{m}} \\ \end{array} \,} \right|\,\,\frac{1}{{m^{m} }}z^{m} } \right)} \,.$$
Example 2.
Since \({}_{1}F_{0} \left( {\left. {\begin{array}{*{20}c} a \\ - \\ \end{array} } \right|\,z} \right) = (1 - z)^{ - a}\) , identity (2 ) for \(m = 2,3,4\) , respectively, reads as
$$(1 - z)^{ - a} = {}_{2}F_{1} \left( {\left. {\begin{array}{*{20}c} {a/2,\,\,\,(a + 1)/2} \\ {1/2} \\ \end{array} \,} \right|\,z^{2} } \right) + az{}_{2}F_{1} \left( {\left. {\begin{array}{*{20}c} {(a + 1)/2,\,\,\,(a + 2)/2} \\ {3/2} \\ \end{array} \,} \right|\,z^{2} } \right),$$
$$\begin{aligned} (1 - z)^{ - a} = {}_{3}F_{2} \left( {\left. {\begin{array}{*{20}c} {a/3,\,\,\,(a + 1)/3,\,\,\,(a + 2)/3} \\ {1/3,\,\,\,\,2/3} \\ \end{array} \,} \right|\,z^{3} } \right) + a\frac{z}{1!}{}_{3}F_{2} \left( {\left. {\begin{array}{*{20}c} {(a + 1)/3,\,\,\,(a + 2)/3,\,\,\,(a + 3)/3} \\ {2/3,\,\,\,\,4/3} \\ \end{array} \,} \right|\,z^{3} } \right) \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + a(a + 1)\frac{{z^{2} }}{2}{}_{3}F_{2} \left( {\left. {\begin{array}{*{20}c} {(a + 2)/3,\,\,\,(a + 3)/3,\,\,\,(a + 4)/3} \\ {4/3,\,\,\,\,5/3} \\ \end{array} \,} \right|\,z^{3} } \right)\,, \hfill \\ \end{aligned}$$
and
$$\begin{aligned} (1 - z)^{ - a} = {}_{4}F_{3} \left( {\left. {\begin{array}{*{20}c} {a/4,\,\,\,(a + 1)/4,\,\,\,(a + 2)/4,\,\,(a + 3)/4} \\ {1/4,\,\,\,\,1/2,\,\,\,\,3/4} \\ \end{array} \,} \right|\,z^{4} } \right) \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\; + az{}_{4}F_{3} \left( {\left. {\begin{array}{*{20}c} {(a + 1)/4,\,\,\,(a + 2)/4,\,\,(a + 3)/4,\,\,(a + 4)/4} \\ {1/2,\,\,\,\,3/4,\,\,\,\,5/4} \\ \end{array} \,} \right|\,z^{4} } \right) \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\; + a(a + 1)\frac{{z^{2} }}{2}{}_{4}F_{3} \left( {\left. {\begin{array}{*{20}c} {(a + 2)/4,\,\,\,(a + 3)/4,\,\,(a + 4)/4,\,\,(a + 5)/4} \\ {3/4,\,\,\,\,5/4,\,\,\,\,3/2} \\ \end{array} \,} \right|\,z^{4} } \right)\, \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\; + a(a + 1)(a + 2)\frac{{z^{3} }}{6}{}_{4}F_{3} \left( {\left. {\begin{array}{*{20}c} {(a + 3)/4,\,\,\,(a + 4)/4,\,\,(a + 5)/4,\,\,(a + 6)/4} \\ {5/4,\,\,\,\,3/2,\,\,\,\,7/4} \\ \end{array} \,} \right|\,z^{4} } \right)\,. \hfill \\ \end{aligned}$$
Example 3.
First, replacing \((p,q) = (0,1)\) in (2 ) for \(m = 2\) yields
$${}_{0}F_{1} \left( {\left. {\begin{array}{*{20}c} - \\ b \\ \end{array} \,} \right|\,\,z} \right) = {}_{0}F_{3} \left( {\left. {\begin{array}{*{20}c} - \\ {b/2,\,\,\,(b + 1)/2,\,\,\,1/2} \\ \end{array} \,} \right|\,\,\frac{1}{16}z^{2} } \right) + \frac{z}{b}{}_{0}F_{3} \left( {\left. {\begin{array}{*{20}c} - \\ {(b + 1)/2,\,\,\,(b + 2)/2,\,\,\,3/2} \\ \end{array} \,} \right|\,\,\frac{1}{16}z^{2} } \right).$$
(3)
Now since the cosine and sine functions can be written as
$$\cos z = {}_{0}F_{1} \left( {\left. {\begin{array}{*{20}c} - \\ {1/2} \\ \end{array} \,} \right|\, - \frac{1}{4}z^{2} } \right)\;{\text{and }}\sin z = z\,{}_{0}F_{1} \left( {\left. {\begin{array}{*{20}c} - \\ {3/2} \\ \end{array} \,} \right|\, - \frac{1}{4}z^{2} } \right),\;$$
relation (3 ), respectively, yields
$$\cos z = {}_{0}F_{3} \left( {\left. {\begin{array}{*{20}c} - \\ {1/4,\,\,1/2,\,\,\,3/4} \\ \end{array} \,} \right|\,\,\frac{1}{256}z^{4} } \right) - \frac{{z^{2} }}{2}{}_{0}F_{3} \left( {\left. {\begin{array}{*{20}c} - \\ {3/4,\,\,\,5/4,\,\,3/2} \\ \end{array} \,} \right|\,\,\frac{1}{256}z^{4} } \right),$$
and
$$\sin z = z{}_{0}F_{3} \left( {\left. {\begin{array}{*{20}c} - \\ {1/2,\,\,3/4,\,\,5/4} \\ \end{array} \,} \right|\,\,\frac{1}{256}z^{4} } \right) - \frac{{z^{3} }}{6}{}_{0}F_{3} \left( {\left. {\begin{array}{*{20}c} - \\ {5/4,\,\,\,3/2,\,\,7/4} \\ \end{array} \,} \right|\,\,\frac{1}{256}z^{4} } \right).$$
Application 2.
A generalization of Euler’s identity.
If \(m = 2\) is replaced in (2 ), then we have
$$\begin{aligned} {}_{p}F_{q} \left( {\left. {\begin{array}{*{20}c} {a_{1} ,\,\,a_{2} ,\,\; \ldots } & {a_{p} } \\ {b_{1} ,\,\,b_{2} ,\,\, \ldots } & {b_{q} } \\ \end{array} } \right|\,z} \right) = {}_{2p}F_{2q + 1} \left( {\left. {\begin{array}{*{20}c} {a_{1} /2,\,\,(1 + a_{1} )/2,\,\, \ldots } & {a_{p} /2,\,(1 + a_{p} )/2} \\ {b_{1} /2,\,\,(1 + b_{1} )/2,\,\, \ldots } & {b_{q} /2,\,(1 + b_{q} )/2\,,\,1/2} \\ \end{array} } \right|\,4^{p - q - 1} z^{2} } \right) \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,\, + \,\frac{{a_{1} a_{2} \ldots \,a_{p} }}{{b_{1} b_{2} \ldots \,b_{q} }}z\,{}_{2p}F_{2q + 1} \left( {\left. {\begin{array}{*{20}c} {(a_{1} + 1)/2,\,\,(a_{1} + 2)/2,\,\, \ldots } & {(a_{p} + 1)/2,\,(a_{p} + 2)/2} \\ {(b_{1} + 1)/2,\,\,(b_{1} + 2)/2,\,\, \ldots } & {(b_{q} + 1)/2,\,(b_{q} + 2)/2\,,\,3/2} \\ \end{array} } \right|\,4^{p - q - 1} z^{2} } \right)\,. \hfill \\ \end{aligned}$$
(4)
By noting that \(z\) may be a complex variable, \(z = i\,x\) in (4 ) gives
$$\begin{aligned} {}_{p}F_{q} \left( {\left. {\begin{array}{*{20}c} {a_{1} ,\,\,a_{2} ,\,\, \ldots } & {a_{p} } \\ {b_{1} ,\,\,b_{2} ,\,\, \ldots } & {b_{q} } \\ \end{array} } \right|\,ix} \right) = {}_{2p}F_{2q + 1} \left( {\left. {\begin{array}{*{20}c} {a_{1} /2,\,\,(1 + a_{1} )/2,\,\, \ldots } & {a_{p} /2,\,(1 + a_{p} )/2} \\ {b_{1} /2,\,\,(1 + b_{1} )/2,\,\, \ldots } & {b_{q} /2,\,(1 + b_{q} )/2\,,\,1/2} \\ \end{array} } \right|\, - 4^{p - q - 1} x^{2} } \right) \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, + \,i\,\frac{{a_{1} a_{2} \ldots \,a_{p} }}{{b_{1} b_{2} \ldots b_{q} }}x\,{}_{2p}F_{2q + 1} \left( {\left. {\begin{array}{*{20}c} {(a_{1} + 1)/2,\,\,(a_{1} + 2)/2,\,\, \ldots } & {(a_{p} + 1)/2,\,(a_{p} + 2)/2} \\ {(b_{1} + 1)/2,\,\,(b_{1} + 2)/2,\,\, \ldots } & {(b_{q} + 1)/2,\,(b_{q} + 2)/2\,,\,3/2} \\ \end{array} } \right|\, - 4^{p - q - 1} x^{2} } \right)\,, \hfill \\ \end{aligned}$$
(5)
which is a generalization of Euler’s identity for \(p = q = 0\) , because
$${}_{0}F_{0} \left( {\left. {\begin{array}{*{20}c} - \\ - \\ \end{array} \,} \right|\,ix} \right) = {}_{0}F_{1} \left( {\left. {\begin{array}{*{20}c} - \\ {1/2} \\ \end{array} } \right|\, - x^{2} /4} \right) + \,i\,x\,{}_{0}F_{1} \left( {\left. {\begin{array}{*{20}c} - \\ {3/2} \\ \end{array} } \right|\, - x^{2} /4} \right) \Leftrightarrow e^{ix} = \cos x\, + \,i\,\sin x.$$
Subsequently, the well-known hyperbolic functions [2 , 16 ] can be generalized via (4 ) and (5 ) and be defined, respectively, as
$$\begin{aligned} {}_{p}Ch\,_{q} \,\left( {\left. {\begin{array}{*{20}c} {a_{1} ,\,\,a_{2} ,\,\, \ldots } & {a_{p} } \\ {b_{1} ,\,\,b_{2} ,\,\, \ldots } & {b_{q} } \\ \end{array} } \right|\,x} \right) = \frac{1}{2}\left( {{}_{p}F_{q} \left( {\left. {\begin{array}{*{20}c} {a_{1} ,\,\,a_{2} ,\,\, \ldots } & {a_{p} } \\ {b_{1} ,\,\,b_{2} ,\,\, \ldots } & {b_{q} } \\ \end{array} } \right|\,x} \right) + {}_{p}F_{q} \left( {\left. {\begin{array}{*{20}c} {a_{1} ,\,\,a_{2} ,\,\, \ldots } & {a_{p} } \\ {b_{1} ,\,\,b_{2} ,\,\, \ldots } & {b_{q} } \\ \end{array} } \right|\, - x} \right)} \right) \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {}_{2p}F_{2q + 1} \left( {\left. {\begin{array}{*{20}c} {a_{1} /2,\,\,(1 + a_{1} )/2,\,\, \ldots } & {a_{p} /2,\,(1 + a_{p} )/2} \\ {b_{1} /2,\,\,(1 + b_{1} )/2,\,\, \ldots } & {b_{q} /2,\,(1 + b_{q} )/2\,,\,1/2} \\ \end{array} } \right|\,4^{p - q - 1} x^{2} } \right), \hfill \\ \end{aligned}$$
(6)
and
$$\begin{aligned} {}_{p}Sh\,_{q} \,\left( {\left. {\begin{array}{*{20}c} {a_{1} ,\,\,a_{2} ,\,\, \ldots } & {a_{p} } \\ {b_{1} ,\,\,b_{2} ,\,\, \ldots } & {b_{q} } \\ \end{array} } \right|\,x} \right) = \frac{1}{2}\left( {{}_{p}F_{q} \left( {\left. {\begin{array}{*{20}c} {a_{1} ,\,\,a_{2} ,\,\, \ldots } & {a_{p} } \\ {b_{1} ,\,\,b_{2} ,\,\, \ldots } & {b_{q} } \\ \end{array} } \right|\,x} \right) - {}_{p}F_{q} \left( {\left. {\begin{array}{*{20}c} {a_{1} ,\,\,a_{2} ,\,\, \ldots } & {a_{p} } \\ {b_{1} ,\,\,b_{2} ,\,\, \ldots } & {b_{q} } \\ \end{array} } \right|\, - x} \right)} \right) \hfill \\ = \frac{{a_{1} a_{2} \ldots a_{p} }}{{b_{1} b_{2} \ldots b_{q} }}x\,{}_{2p}F_{2q + 1} \left( {\left. {\begin{array}{*{20}c} {(a_{1} + 1)/2,\,\,(a_{1} + 2)/2,\,\, \ldots } & {(a_{p} + 1)/2,\,(a_{p} + 2)/2} \\ {(b_{1} + 1)/2,\,\,(b_{1} + 2)/2,\,\, \ldots } & {(b_{p} + 1)/2,\,(b_{p} + 1)/2\,,\,3/2} \\ \end{array} } \right|\,4^{p - q - 1} x^{2} } \right). \hfill \\ \end{aligned}$$
(7)
It is clear that for \((p,q) = (0,0)\) , relations (6 ) and (7 ) reduce to
$${}_{0}Ch_{0} \,\left( {\left. {\begin{array}{*{20}c} - \\ - \\ \end{array} } \right|\,x} \right) = \cosh x = \frac{1}{2}(e^{x} + e^{ - x} )\;{\text{and}}\;{}_{0}Sh_{0} \,\left( {\left. {\begin{array}{*{20}c} - \\ - \\ \end{array} } \right|\,x} \right) = \sinh x = \frac{1}{2}(e^{x} - e^{ - x} ).$$
Application 3.
A decomposition formula for classical hypergeometric orthogonal polynomials: there are ten sequences of hypergeometric polynomials [7 , 12 –14 ] that are orthogonal with respect to the Pearson distribution family
$$W\,\left( {\left. {\begin{array}{*{20}c} {\begin{array}{*{20}c} {d,} & e \\ \end{array} } \\ {a,\,\,\,b,\,\,\,c} \\ \end{array} } \right|\,x} \right) = \exp \left( {\int {\frac{dx + e}{{ax^{2} + bx + c}}dx} } \right)\,\,\,\,\,\,\,\,\,\,\,(a,b,c,d,e \in {\mathbf{\mathbb{R}}}),$$
and its symmetric analog [11 ]
$$W^{*} \left( {\left. {\begin{array}{*{20}c} {r,} & s \\ {p,} & q \\ \end{array} \,} \right|\,x} \right) = \exp \left( {\int {\frac{{r\,x^{2} + s}}{{x(px^{2} + q)}}} \,dx} \right)\,\,\,\,\,\,\,\,\,\,(p,q,r,s \in {\mathbf{\mathbb{R}}}).$$
Five of them are infinitely orthogonal with respect to special cases of the two above-mentioned weight functions and five other ones are finitely orthogonal [12 –14 ] which are limited to some parametric constraints. The following Table 1 shows their main characteristics.
Table 1 Characteristics of ten sequences of orthogonal polynomials
where the sequence
$$\varPhi_{n} (x) = S_{n} \left( {\left. {\begin{array}{*{20}c} {r,} & s \\ {p,} & q \\ \end{array} \,} \right|\,x} \right) = \sum\limits_{k = 0}^{[n/2]} {\,\left( {\begin{array}{*{20}c} {[n/2]} \\ k \\ \end{array} } \right)\,\,\left( {\prod\limits_{i = 0}^{[n/2] - (k + 1)} {\frac{{\left( {2i + ( - 1)^{n + 1} + 2\,[n/2]} \right)\,p + r}}{{\left( {2i + ( - 1)^{n + 1} + 2} \right)\,\,q + s}}} } \right)\,\,x^{n - 2k} } ,$$
is a basic class of symmetric orthogonal polynomials [12 ] satisfying the equation
$$x^{2} (px^{2} + q)\,\varPhi_{n}^{\prime \prime } (x) + x(rx^{2} + s)\,\varPhi_{n}^{\prime } (x) - \left( {n(r + (n - 1)p)x^{2} + (1 - ( - 1)^{n} )\,s/2} \right)\,\varPhi_{n} (x) = 0.$$
For example, consider the shifted Jacobi polynomials [20 ]
$$P_{n, + }^{(\alpha ,\beta )} (x) = {}_{2}F_{1} \left( {\left. {\begin{array}{*{20}c} {\begin{array}{*{20}c} { - n,} & {n + \alpha + \beta + 1} \\ \end{array} } \\ {\alpha + 1} \\ \end{array} \,} \right|\,x} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\alpha ,\,\beta > - 1),$$
(8)
which are orthogonal with respect to the shifted beta distribution on \([0,1]\) as
$$\int_{\,0}^{\,1} {x^{\alpha } (1 - x)^{\beta } P_{n, + }^{(\alpha ,\beta )} (x)P_{m, + }^{(\alpha ,\beta )} (x)dx} = \frac{{n!\,\,\varGamma^{2} (\alpha + 1)\varGamma (n + \beta + 1)}}{(2n + \alpha + \beta + 1)\varGamma (n + \alpha + \beta + 1)\varGamma (n + \alpha + 1)}\,\,\delta_{n,m} .$$
If (8 ) is replaced in (2 ) for, e.g., \(m = 2\) , then one gets (see also [10 ] in this regard)
$$\begin{aligned} P_{n, + }^{(\alpha ,\beta )} (x) = \,{}_{4}F_{3} \left( {\left. {\begin{array}{*{20}c} {\begin{array}{*{20}c} { - n/2,} & {(1 - n)/2,} & {(n + \alpha + \beta + 1)/2,} & {(n + \alpha + \beta + 2)/2} \\ \end{array} } \\ {\begin{array}{*{20}c} {(\alpha + 1)/2,} & {(\alpha + 2)/2,} & {1/2} \\ \end{array} } \\ \end{array} \,} \right|\,x^{2} } \right)\, \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\; - \frac{n(n + \alpha + \beta + 1)}{\alpha + 1}\,x{}_{4}F_{3} \left( {\left. {\begin{array}{*{20}c} {\begin{array}{*{20}c} {(1 - n)/2,} & {(2 - n)/2,} & {(n + \alpha + \beta + 2)/2,} & {(n + \alpha + \beta + 3)/2} \\ \end{array} } \\ {\begin{array}{*{20}c} {(\alpha + 2)/2,} & {(\alpha + 3)/2,} & {3/2} \\ \end{array} } \\ \end{array} \,} \right|\,x^{2} } \right)\,. \hfill \\ \end{aligned}$$
(9)
Hence, two sequences of polynomials can be defined by (9 ) as
$$\frac{{P_{n, + }^{(\alpha ,\beta )} (\sqrt x ) + P_{n, + }^{(\alpha ,\beta )} ( - \sqrt x )}}{2} = {}_{4}F_{3} \left( {\left. {\begin{array}{*{20}c} {\begin{array}{*{20}c} { - n/2,} & {(1 - n)/2,} & {(n + \alpha + \beta + 1)/2,} & {(n + \alpha + \beta + 2)/2} \\ \end{array} } \\ {\begin{array}{*{20}c} {(\alpha + 1)/2,} & {(\alpha + 2)/2,} & {1/2} \\ \end{array} } \\ \end{array} \,} \right|\,x} \right),$$
and
$$\begin{aligned} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{P_{n, + }^{(\alpha ,\beta )} (\sqrt x ) - P_{n, + }^{(\alpha ,\beta )} ( - \sqrt x )}}{2\sqrt x } = - \frac{n(n + \alpha + \beta + 1)}{\alpha + 1}\,\,\, \hfill \\ \times {}_{4}F_{3} \left( {\left. {\begin{array}{*{20}c} {\begin{array}{*{20}c} {(1 - n)/2,} & {(2 - n)/2,} & {(n + \alpha + \beta + 2)/2,} & {(n + \alpha + \beta + 3)/2} \\ \end{array} } \\ {\begin{array}{*{20}c} {(\alpha + 2)/2,} & {(\alpha + 3)/2,} & {3/2} \\ \end{array} } \\ \end{array} \,} \right|\,x} \right)\,. \hfill \\ \end{aligned}$$
Another classical example are the Laguerre polynomials
$$y = \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{L}_{n}^{(\alpha )} (x) = {}_{1}F_{1} \left( {\left. {\begin{array}{*{20}c} { - n} \\ {\alpha + 1} \\ \end{array} \,} \right|\,x} \right)\,\,\,\,\,\,\,\,\,(\alpha > - 1),$$
(10)
that satisfy the differential equation [7 ]
$$x\,y^{\prime \prime } + (\alpha + 1 - x)\,y^{\prime } + n\,y = 0,$$
and the orthogonality relation [7 ]
$$\int_{0}^{\infty} {x^{\alpha} e^{- x} \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{L}_{n}^{(\alpha)} (x)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{L}_{m}^{(\alpha)} (x)\,{\text{d}}x} = \frac{n!\varGamma (\alpha + 1)}{{(\alpha + 1)_{n}}}\delta_{n,m}.$$
If (10 ) is replaced in (2 ) for e.g. \(m = 3\) , then one gets
$$\begin{aligned} \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{L}_{n}^{(\alpha)} (x) = \,{}_{3}F_{5} \left({\left. {\begin{array}{*{20}c} {- n/3,\,\,\,(1 - n)/3,\,\,\,\,(2 - n)/3} \\ {(\alpha + 1)/3,\,\,\,(\alpha + 2)/3,\,\,\,(\alpha + 3)/3,\,\,\,1/3,\,\,\,2/3} \\ \end{array} \,} \right|\,\frac{{x^{3}}}{27}} \right)\, \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \frac{n}{\alpha + 1}\,x\,{}_{3}F_{5} \left({\left. {\begin{array}{*{20}c} {(1 - n)/3,\,\,\,(2 - n)/3,\,\,\,\,(3 - n)/3} \\ {(\alpha + 2)/3,\,\,\,(\alpha + 3)/3,\,\,\,(\alpha + 4)/3,\,\,\,2/3,\,\,\,4/3} \\ \end{array} \,} \right|\,\frac{{x^{3}}}{27}} \right) \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \frac{n(n - 1)}{2(\alpha + 1)(\alpha + 2)}\,x^{2} {}_{3}F_{5} \left({\left. {\begin{array}{*{20}c} {(2 - n)/3,\,\,\,(3 - n)/3,\,\,\,\,(4 - n)/3} \\ {(\alpha + 3)/3,\,\,\,(\alpha + 4)/3,\,\,\,(\alpha + 5)/3,\,\,\,4/3,\,\,\,5/3} \\ \end{array} \,} \right|\,\frac{{x^{3}}}{27}} \right)\,. \hfill \\ \end{aligned}$$
See also [9 ] in the sense of incomplete symmetric orthogonal polynomials of Laguerre type. Some other works related to classical hypergeometric orthogonal polynomials can be found in [3 , 4 , 19 ].
Application 4.
The classical summation theorems of hypergeometric series (such as the Gauss, Kummer and Bailey theorems for \(_{2} F_{1}\) and the Watson and Dixon theorems for \(_{3} F_{2}\) ) play an important role in evaluating hypergeometric series at specific points [18 ]. In this section, by expressing the aforesaid theorems we employ identity (2 ) for them.
Gauss’s theorem:
$$_{2} F_{1} \left( {\left. {\begin{array}{*{20}c} {\begin{array}{*{20}c} {a,} & b \\ \end{array} } \\ c \\ \end{array} } \right|\,\,1} \right) = \frac{\varGamma (c)\varGamma (c - a - b)}{\varGamma (c - a)\varGamma (c - b)},$$
provided that \(\text{Re} (c - a - b) > 0.\)
Gauss’s second theorem:
$$_{2} F_{1} \left( {\left. {\begin{array}{*{20}c} {\begin{array}{*{20}c} {a,} & b \\ \end{array} } \\ {(a + b + 1)/2\,} \\ \end{array} } \right|\,\,\frac{1}{2}} \right) = \frac{{\sqrt \pi \,\varGamma \left( {\frac{a + b + 1}{2}} \right)}}{{\varGamma \left( {\frac{a + 1}{2}} \right)\varGamma \left( {\frac{b + 1}{2}} \right)}},$$
provided that \(\text{Re} (a) > - 1\) and \(\text{Re} (b) > - 1\) .
Kummer’s theorem:
$$_{2} F_{1} \left( {\left. {\begin{array}{*{20}c} {\begin{array}{*{20}c} {a,} & b \\ \end{array} } \\ {1 + a - b} \\ \end{array} \,} \right|\, - \,1} \right) = \frac{{\varGamma (1 + a - b)\varGamma \left( {1 + \frac{a}{2}} \right)}}{{\varGamma (1 + a)\,\varGamma \left( {1 + \frac{a}{2} - b} \right)}}.$$
provided that \(\text{Re} (a) > - 1,\,\,\text{Re} (b) < 1\) and \(\text{Re} (a - b) > - 1\) .
Bailey’s theorem [18 ]:
$$_{2} F_{1} \left( {\left. {\begin{array}{*{20}c} {\begin{array}{*{20}c} {a,} & {1 - a} \\ \end{array} } \\ c \\ \end{array} \,} \right|\,\frac{1}{2}} \right) = \frac{{\varGamma \left( {\frac{c}{2}} \right)\varGamma \left( {\frac{c + 1}{2}} \right)}}{{\varGamma \left( {\frac{c + a}{2}} \right)\varGamma \left( {\frac{c - a + 1}{2}} \right)}}.$$
provided that \(\text{Re} (c) > \,\text{Re} (a) > 0\) .
Watson’s theorem [18 ]:
$$_{3} F_{2} \left( {\left. {\begin{array}{*{20}c} {a,\,\,\,\,\,\,\,\,\,\,\,\,b,\,\,\,\,\,\,\,\,\,\,\,\,c} \\ {(a + b + 1)/2,\,\,2c} \\ \end{array} \,} \right|\,\,1} \right) = \frac{{\sqrt \pi \,\varGamma \left( {c + \frac{1}{2}} \right)\varGamma \left( {\frac{a + b + 1}{2}} \right)\,\varGamma \left( {c - \frac{a + b - 1}{2}} \right)}}{{\varGamma \left( {\frac{a + 1}{2}} \right)\,\varGamma \left( {\frac{b + 1}{2}} \right)\,\varGamma \left( {c - \frac{a - 1}{2}} \right)\,\varGamma \left( {c - \frac{b - 1}{2}} \right)}}.$$
provided that \(\text{Re} (2c - a - b) > - 1.\)
Dixon’s theorem [18 ]:
$$_{3} F_{2} \left( {\left. {\begin{array}{*{20}c} {a,\,\,\,\,\,\,\,\,\,\,\,\,b,\,\,\,\,\,\,\,\,\,\,\,\,c} \\ {1 + a - b,\,\,1 + a - c} \\ \end{array} \,} \right|\,\,1} \right) = \frac{{\varGamma (1 + \frac{a}{2})\varGamma (1 + a - b)\,\varGamma (1 + a - c)\varGamma (1 + \frac{a}{2} - b - c)}}{{\varGamma (1 + a)\,\varGamma (1 + \frac{a}{2} - b)\,\varGamma (1 + \frac{a}{2} - c)\varGamma (1 + a - b - c)}}.$$
provided that \(\text{Re} (a - 2b - 2c) > - 2.\)
Now, if we substitute Gauss’s theorem into (8 ) for e.g. \(m = 2,3\) , then we, respectively, obtain
$$\begin{aligned} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{}_{4}F_{3} \left( {\left. {\begin{array}{*{20}c} {a/2,\,\,(a + 1)/2\,,\,\,b/2,\,\,(b + 1)/2} \\ {c/2,\,\,\,\,(c + 1)/2,\,\,\,\,\,1/2} \\ \end{array} \,} \right|\,\,1} \right) \hfill \\ + \frac{ab}{c}{}_{4}F_{3} \left( {\left. {\begin{array}{*{20}c} {(a + 1)/2,\,\,(a + 2)/2\,,\,\,(b + 1)/2,\,\,(b + 2)/2} \\ {(c + 1)/2,\,\,\,\,(c + 2)/2,\,\,\,\,3/2} \\ \end{array} \,} \right|\,\,1} \right) = \frac{\varGamma (c)\varGamma (c - a - b)}{\varGamma (c - a)\varGamma (c - b)}\,, \hfill \\ \end{aligned}$$
and
$$\begin{aligned} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{}_{6}F_{5} \left( {\left. {\begin{array}{*{20}c} {a/3,\,\,(a + 1)/3\,,\,\,(a + 2)/3,\,\,\,b/3,\,\,(b + 1)/3,\,\,(b + 2)/3} \\ {c/3,\,\,(c + 1)/3\,,\,\,(c + 2)/3,\,\,\,1/3,\,\,2/3} \\ \end{array} \,} \right|\,\,1} \right) \hfill \\ \,\,\,\,\,\,\,\,\, + \frac{ab}{c}{}_{6}F_{5} \left( {\left. {\begin{array}{*{20}c} {(a + 1)/3,\,\,(a + 2)/3\,,\,\,(a + 3)/3,\,\,\,(b + 1)/3,\,\,(b + 2)/3,\,\,(b + 3)/3} \\ {(c + 1)/3,\,\,(c + 2)/3\,,\,\,(c + 3)/3,\,\,\,2/3,\,\,4/3} \\ \end{array} \,} \right|\,\,1} \right) \hfill \\ + \frac{a(a + 1)b(b + 1)}{2c(c + 1)}{}_{6}F_{5} \left( {\left. {\begin{array}{*{20}c} {(a + 2)/3,\,\,(a + 3)/3\,,\,\,(a + 4)/3,\,\,\,(b + 2)/3,\,\,(b + 3)/3,\,\,(b + 4)/3} \\ {(c + 2)/3,\,\,(c + 3)/3\,,\,\,(c + 4)/3,\,\,\,4/3,\,\,5/3} \\ \end{array} \,} \right|\,\,1} \right) \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{\varGamma (c)\varGamma (c - a - b)}{\varGamma (c - a)\varGamma (c - b)}\,.\, \hfill \\ \end{aligned}$$
The general case of the two above identities for any natural m is as
$$\sum\limits_{k = 0}^{m - 1} {\frac{{(a)_{k} (b)_{k} }}{{(c)_{k} k!}}\,{}_{2m + 1}F_{2m} \left( {\left. {\begin{array}{*{20}c} {\frac{a + k}{m},\, \ldots \,,\,\frac{a + m - 1 + k}{m},\,\,\frac{b + k}{m},\, \ldots ,\,\frac{b + m - 1 + k}{m},} & 1 \\ {\frac{c + k}{m},\, \ldots \,,\,\frac{c + m - 1 + k}{m},\,\frac{1 + k}{m},\, \ldots \,,\,\frac{m + k}{m}} & {} \\ \end{array} \,} \right|\,\,1} \right)} \, = \frac{\varGamma (c)\varGamma (c - a - b)}{\varGamma (c - a)\varGamma (c - b)}.$$
And the general case for Gauss’s second theorem takes the form
$$\begin{aligned} \sum\limits_{k = 0}^{m - 1} {\frac{{(a)_{k} (b)_{k} \,2^{ - k} }}{{\left( {\frac{a + b + 1}{2}} \right)_{k} k!}}\,{}_{2m + 1}F_{2m} \left( {\left. {\begin{array}{*{20}c} {\frac{a + k}{m},\, \ldots \,,\,\frac{a + m - 1 + k}{m},\,\,\frac{b + k}{m},\, \ldots \,,\,\frac{b + m - 1 + k}{m},\,\,\,1} \\ {\frac{k + (a + b + 1)/2}{m},\, \ldots \,,\,\frac{m - 1 + k + (a + b + 1)/2}{m},\,\frac{1 + k}{m},\, \ldots ,\,\frac{m + k}{m}} \\ \end{array} \,} \right|\,\,2^{ - m} } \right)} \, \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\sqrt \pi \,\varGamma \left( {\frac{a + b + 1}{2}} \right)}}{{\varGamma \left( {\frac{a + 1}{2}} \right)\varGamma \left( {\frac{b + 1}{2}} \right)}}, \hfill \\ \end{aligned}$$
and for Kummer theorem takes the form
$$\begin{aligned} \sum\limits_{k = 0}^{m - 1} {\frac{{(a)_{k} (b)_{k} ( - 1)^{k} }}{{(1 + a - b)_{k} k!}}\,{}_{2m + 1}F_{2m} \left( {\left. {\begin{array}{*{20}c} {\frac{a + k}{m},\, \ldots ,\,\frac{a + m - 1 + k}{m},\,\,\frac{b + k}{m},\, \ldots \,,\,\frac{b + m - 1 + k}{m},\,\,\,1} \\ {\frac{1 + a - b + k}{m},\, \ldots ,\,\frac{a - b + m + k}{m},\,\frac{1 + k}{m},\, \ldots ,\,\frac{m + k}{m}} \\ \end{array} \,} \right|\,( - 1)^{m} } \right)} \, \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\varGamma (1 + a - b)\varGamma \left( {1 + \frac{a}{2}} \right)}}{{\varGamma (1 + a)\,\varGamma \left( {1 + \frac{a}{2} - b} \right)}}, \hfill \\ \end{aligned}$$
and for Bailey theorem takes the form
$$\begin{aligned} \sum\limits_{k = 0}^{m - 1} {\frac{{(a)_{k} (1 - a)_{k} (2)^{ - k} }}{{(c)_{k} k!}}\,{}_{2m + 1}F_{2m} \left( {\left. {\begin{array}{*{20}c} {\frac{a + k}{m},\, \ldots \,,\,\frac{a + m - 1 + k}{m},\,\,\frac{1 - a + k}{m},\, \ldots \,,\,\frac{ - a + m + k}{m},\,\,\,1} \\ {\frac{c + k}{m},\, \ldots ,\,\frac{c + m - 1 + k}{m},\,\frac{1 + k}{m},\, \ldots \,,\,\frac{m + k}{m}} \\ \end{array} \,} \right|\,2^{ - m} } \right)} \, \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\varGamma \left( {\frac{c}{2}} \right)\varGamma \left( {\frac{c + 1}{2}} \right)}}{{\varGamma \left( {\frac{c + a}{2}} \right)\varGamma \left( {\frac{c - a + 1}{2}} \right)}}, \hfill \\ \end{aligned}$$
and for Watson theorem takes the form
$$\begin{aligned} \sum\limits_{k = 0}^{m - 1} {\frac{{(a)_{k} (b)_{k} (c)_{k} }}{{((a + b + 1)/2)_{k} (2c)_{k} k!}}\,} \times \hfill \\ {}_{3m + 1}F_{3m} \left( {\left. {\begin{array}{*{20}c} {\frac{a + k}{m}, \ldots ,\,\frac{a + m - 1 + k}{m},\,\,\frac{b + k}{m}, \ldots ,\,\frac{b + m - 1 + k}{m},\,\frac{c + k}{m}, \ldots ,\,\frac{c + m - 1 + k}{m},\,\,1} \\ {\frac{k + (a + b + 1)/2}{m}, \ldots ,\,\frac{m - 1 + k + (a + b + 1)/2}{m},\,\frac{2c + k}{m}, \ldots ,\,\frac{2c + m - 1 + k}{m},\frac{1 + k}{m}, \ldots ,\,\frac{m + k}{m}} \\ \end{array} \,} \right|\,1} \right) \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\sqrt \pi \,\varGamma (c + \frac{1}{2})\varGamma (\frac{a + b + 1}{2})\,\varGamma (c - \frac{a + b - 1}{2})}}{{\varGamma (\frac{a + 1}{2})\,\varGamma (\frac{b + 1}{2})\,\varGamma (c - \frac{a - 1}{2})\,\varGamma (c - \frac{b - 1}{2})}}\,, \hfill \\ \end{aligned}$$
and finally for Dixon’s theorem takes the form
$$\begin{aligned} \sum\limits_{k = 0}^{m - 1} {\frac{{(a)_{k} (b)_{k} (c)_{k} }}{{(1 + a - b)_{k} (1 + a - c)_{k} k!}}\,} \times \hfill \\ {}_{3m + 1}F_{3m} \left( {\left. {\begin{array}{*{20}c} {\frac{a + k}{m}, \ldots ,\,\frac{a + m - 1 + k}{m},\,\,\frac{b + k}{m}, \ldots ,\,\frac{b + m - 1 + k}{m},\,\frac{c + k}{m}, \ldots ,\,\frac{c + m - 1 + k}{m},\,\,1} \\ {\frac{1 + a - b + k}{m}, \ldots ,\,\frac{a - b + m + k}{m},\,\frac{1 + a - c + k}{m}, \ldots ,\,\frac{a - c + m + k}{m},\frac{1 + k}{m}, \ldots ,\,\frac{m + k}{m}} \\ \end{array} \,} \right|\,1} \right) \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\varGamma (1 + \frac{a}{2})\varGamma (1 + a - b)\,\varGamma (1 + a - c)\varGamma (1 + \frac{a}{2} - b - c)}}{{\varGamma (1 + a)\,\varGamma (1 + \frac{a}{2} - b)\,\varGamma (1 + \frac{a}{2} - c)\varGamma (1 + a - b - c)}}. \hfill \\ \end{aligned}$$