Extension of Chu–Vandermonde Identity and Quadratic Transformation Conditions

Abstract

In 1812, Gauss stated the following identity: ${\displaystyle {}_{2}F_1(a,b;c;1)=\frac{\mathrm{\Gamma }\left(c\right)\mathrm{\Gamma }(c-a-b)}{\mathrm{\Gamma }(c-a)\mathrm{\Gamma }(c-b)}}$, where, in the real case, $c-a-b>0$ and as an immediat consequence the Chu–Vandermonde identity: ${\displaystyle {}_{2}F_1(a,-n;c;1)=\frac{{(c-a)}_n}{{\left(c\right)}_n}}$ for any positive integer n. In this paper, we investigate the case when $c-a-b<0$ by taking $c=2b=-2n$, n and a are positive integers ($c-a-b=-n-a<0$). We give two significant applications stemming from these findings. The second part of the paper will be devoted to Kummer’s conditions concerning hypergeometric quadratic transformations, particularly focusing on the distinctions between the conditions provided by Gradshteyn and Ryzhik (GR) and those by Erdélyi, Magnus, Oberhettinger, and Tricomi (EMOI) are outlined. We establish that the conditions given by GR differ from those of EMOI, and we explore the methodologies employed by both groups in deriving their results. This leads us to conclude that the search for exact and unified conditions remains an open problem.

1. Introduction and Preliminaries

Researchers in the fields of classical orthogonal polynomials, special functions, and related areas have undoubtedly utilized Gauss’s hypergeometric function:

$${}_{2}F_1\left(\begin{array}{c}a,b\\ c\end{array};z\right)=\sum _{n\ge 0}{\displaystyle \frac{{\left(a\right)}_n{\left(b\right)}_n}{{\left(c\right)}_n}}{\displaystyle \frac{z^n}{n!}},$$

(1)

where ${\left(a\right)}_n$ is the Pochhammer symbol defined for a a complex number $a\ne 0$ by

$${\left(a\right)}_n=\left\{\begin{array}{cc}a(a+1)\cdots (a+n-1),\hfill & n\in \mathbb{N},\hfill \\ 1,\hfill & n=0,\hfill \end{array}\right.$$

(2)

which in terms of the well-known Gamma function, ${\left(a\right)}_n$, is represented by

$${\left(a\right)}_n={\displaystyle \frac{\mathrm{\Gamma }(a+n)}{\mathrm{\Gamma }\left(a\right)}}.$$

Consider selecting three parameters $a,b,$ and c, typically rational numbers, and examining (1) in terms of the complex variable z. This expression is well defined provided that c is not a negative integer. It can be shown that its radius of convergence is 1, assuming that a and b are not negative integers. If either a or b is a negative integer, the infinite series becomes a polynomial. For simplicity, we will denote (1) by $f\left(z\right)$ and let $\theta$ represent the differential operator ${\displaystyle z\frac{d}{dz}}$. A straightforward calculation, leveraging the factorial structure of the coefficients in $f\left(z\right)$, demonstrates that

$$z(\theta +a)(\theta +b)f=\theta (\theta +c-1)f.$$

(3)

By replacing $\theta$ with ${\displaystyle z\frac{d}{dz}}$, the expansion of (4) gives the following second-order differential equation:

$$z(z-1)f^{″}+((a+b+1)z-c)f^{\prime }+abf=0.$$

(4)

The differential Equation (4) appears across numerous branches of mathematics, mathematical physics, and applied sciences. Later, Gauss studied hypergeometric functions not only as values of Euler’s hypergeometric series but also as solutions to the hypergeometric equation throughout the complex plane, an approach that was entirely new at that time. In this way Gauss very soon became aware of the problem of their multivaluedness, known nowadays as the monodromy problem. He published a part of his findings in 1812 [1], p. 123, with further results on complex-plane behavior discovered in his nachlass [1], p. 207. Due to his contributions, these functions, defined by Euler’s hypergeometric series, are often referred to as Gauss hypergeometric functions.

The next significant advancement came from Riemann. In his 1857 article [2], he provided a comprehensive description of the monodromy group associated with Gauss’s hypergeometric function. The monodromy group of a linear differential equation in the complex plane characterizes the behavior of analytic continuation for its solutions. This approach established hypergeometric functions as a crucial testing ground for Riemann’s groundbreaking ideas on analytic continuation. His work was later extended by H. A. Schwarz [3], Felix Klein, and others, leading to applications in the early development of algebraic geometry and modular forms. In his 1893 lectures, Klein [4] offered an in-depth exposition of Riemann’s ideas and their implications.

By the late nineteenth century, various generalizations of Gauss hypergeometric functions had been introduced, expanding the number of parameters, variables, or both. With few exceptions, the subsequent studies were largely descriptive, requiring the determination of (partial) differential equations, the identification of a basis of solutions around specific points, and the analysis of their domains of convergence. For further insights into the special nature of hypergeometric functions, see [5].

Researchers in the fields of classical orthogonal polynomials, special functions, and related disciplines have undoubtedly relied on foundational results such as the Gauss identity, the Chu–Vandermonde identity, and the quadratic transformation formulas for (1). These identities and transformation formulas serve as essential tools for both students and researchers. Their origins trace back to the pioneering work of Gauss [6,7] and Kummer [8]. Subsequently, Whipple [9,10,11] and Bailey [12,13,14,15], among others, expanded these results to encompass higher-order hypergeometric functions, which have since found broad applications.

However, none of the former researchers have considered isolated and/or non-defined cases, including but not limited to

$${}_{2}F_1\left(\begin{array}{c}a,-n\\ -2n\end{array};1\right),$$

where n is a positive integer and

$${\displaystyle {(1-\frac{z}{2})}^u{}_{2}F_1\left(\begin{array}{c}u,v\\ 2v\end{array};z\right),}$$

where v is a negative integer, which forms the central focus of this paper.

Let us begin with some preliminaries. Gauss’s Hypergeometric theorem says that if $c-a-b>0$, we have [16,17,18,19]

$${}_{2}F_1\left(\begin{array}{c}a,b\\ c\end{array};1\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(a\right)}_n{\left(b\right)}_n}{n!{\left(c\right)}_n}}={\displaystyle \frac{\mathrm{\Gamma }\left(c\right)\mathrm{\Gamma }(c-a-b)}{\mathrm{\Gamma }(c-a)\mathrm{\Gamma }(c-b)}}.$$

(5)

An immediate consequence of this theorem is the Chu–Vandermonde identity, where $b=-n$, $n\in {\mathbb{Z}}_{+}$ and $c-a+n>0$ and where we obtain [16,18,19]

$${}_{2}F_1\left(\begin{array}{c}a,-n\\ c\end{array};1\right)={\displaystyle \frac{{(c-a)}_n}{{\left(c\right)}_n}}.$$

(6)

Taking $c=-2n$ and $a\in {\mathbb{Z}}_{+}$, then $c-a+n=-n-a$ is not positive and then one wonders if the result remains true. We prove that this identity remains true and we give two applications for it.

In Section 3, we consider the Kummer quadratic transformation [18]

$${\displaystyle {\left(1-\frac{z}{2}\right)}^u{}_{2}F_1\left(\begin{array}{c}u,v\\ 2v\end{array};z\right)={}_{2}F_1\left(\begin{array}{c}\frac{u}{2},\frac{u+1}{2}\\ v+\frac{1}{2}\end{array};{\left(\frac{z}{2-z}\right)}^2\right),}$$

(7)

and we take $u=2$ and $v=-1$. The left-hand side (LHS) of (7) becomes

$${\displaystyle {\left(1-\frac{z}{2}\right)}^u{}_{2}F_1\left(\begin{array}{c}u,v\\ 2v\end{array};z\right)=\frac{1}{4}{(z-2)}^2(z+1),}$$

whereas the right-hand side (RHS) of (21) gives

$${}_{2}F_1\left(\begin{array}{c}{\displaystyle \frac{u}{2}},{\displaystyle \frac{u+1}{2}}\\ v+{\displaystyle \frac{1}{2}}\end{array};{\left({\displaystyle \frac{z}{2-z}}\right)}^2\right)={\displaystyle \frac{(z^4-2z^3+4z-2)}{8{(z-2)}^2{(z-1)}^3}}.$$

Kummer transformation (7) is not valid for every value of u and v. The main two books that gave the conditions opf Kummer’s quadratic transformation are presented by Gradshteyn and Ryzhik (GR) and by Erdelyi, Magnus, Oberhettinger, and Tricomi (EMOI):

• In [19], on page 1008 paragraph 9.13, Gradshteyn and Ryzhik state that the series (1) will, in the cut plane, yield a single-valued analytic continuation, which we can obtain by means of the formulas below provided $c+1$ is not a natural number, $a-b$ and $c-a-b$ are not integers;
• In [20], on page 64 paragraph 2.1.5, Magnus and his coauthors state that there exists a quadratic transformation if and only if the numbers

  $$\pm (1-c),\pm (a-b),\pm (a+b-c)$$

  have the property that one of them equals ${\displaystyle \frac{1}{2}}$ or that two of them are equal.

The second part of this paper will be devoted to, starting from Section 3, examine Kummer’s quadratic transformation conditions as presented by either Gradshteyn and Ryzhik (GR) or by Erdelyi, Magnus, Oberhettinger, and Tricomi (EMOI). We demonstrate that the conditions proposed by GR differ from those of EMOI and provide an explanation of how each group derived their conditions. This discussion highlights that the quest for precise conditions remains open.

Let us now start with some preliminaries. Gauss [21] defined his famous infinite series as follows:

$$1+{\displaystyle \frac{ab}{c}}{\displaystyle \frac{z}{1!}}+{\displaystyle \frac{a(a+1)b(b+1)}{c(c+1)}}{\displaystyle \frac{z^2}{1!}}+\dots$$

(8)

which is usually denoted by (1), and alternatively denoted by F. This function is widely recognized as Gauss’s function or the hypergeometric series. Gauss’s function, or the hypergeometric series, serves as a solution to a second-order differential equation. In this function, there are two parameters in the numerator, a and b, and one parameter in the denominator, c, which can be real or complex numbers, with the exception that c must not be zero or a negative integer. The variable z represents the argument of the series. Notably, when $a=1$ and $b=c$ (or equivalently $b=1$ and $a=c$), the infinite series (8) simplifies to the well-known “geometric series”. This characteristic is what gives the series (8) the name “hypergeometric series”.

By the ratio test, it is not difficult to verify that the infinite series (1) has the following properties:

• Is convergent for all values of z provided that $\mid z\mid < 1$ and divergent when $\mid z\mid > 1$;
• Is convergent for $z=1$ provided that Re$(c-a-b)>0$ and divergent when Re$(c-a-b)\le 0$;
• Is absolutely convergent for $z=-1$ provided that Re$(c-a-b)>0$, convergent, but not absolutely, for $-1<$ Re$(c-a-b)\le 0$, and divergent for Re$(c-a-b)<-1$.

Remark in passing that almost all elementary functions of mathematics and mathematical physics are special cases or limiting cases of Gauss’s hypergeometric function. For more details about Gauss’s hypergeometric function, we refer to the standard text of Rainville [22].

2. Isolated Cases of Gauss Theorem for Hypergeometric Functions

In [18], on page 66 theorem 2.2.2, we have the following: (Gauss [1812]) for Re$(c-a-b)>0$, we have

$${}_{2}F_1\left(\begin{array}{c}a,b\\ c\end{array};1\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(a\right)}_n{\left(b\right)}_n}{n!{\left(c\right)}_n}}={\displaystyle \frac{\mathrm{\Gamma }\left(c\right)\mathrm{\Gamma }(c-a-b)}{\mathrm{\Gamma }(c-a)\mathrm{\Gamma }(c-b)}}.$$

(9)

Remark 1.

In the sequel of the paper we consider only the case when $a,b$, and c are real (not complex).

Let us see first prove which is based on Euler’s integral representation.

2.1. Euler’s Integral Representation

The following theorem shows an important integral representation of the ${}_{2}F_1$ function based on Euler’s work.

Proposition 1.

If $c>0$ and $b>0$ then

$${}_{2}F_1\left(\begin{array}{c}a,b\\ c\end{array};z\right)={\displaystyle \frac{\mathrm{\Gamma }\left(c\right)}{\mathrm{\Gamma }\left(b\right)\mathrm{\Gamma }(c-b)}}{\int }_0^1{t}^{b-1}{(1-t)}^{c-b-1}{(1-zt)}^{-a}dt$$

(10)

in the z plane cut along the real axis from 1 to ∞, where Arg$\left(t\right)=$ Arg$(1-t)=0$, and $(1-zt)$ has its principal value.

Remark 2.

We recall here the proof of this integral representation in order to point out the importance of the condition $c>b>0$ whereas, in the sequel, we will be interested by the case $c<b<0$.

Proof of Proposition 1.

If we assume $\mid z\mid <1$ and if we expand ${(1-zt)}^{-a}$ the right-hand side of (10), it becomes

$${\displaystyle \frac{\mathrm{\Gamma }\left(c\right)}{\mathrm{\Gamma }\left(b\right)\mathrm{\Gamma }(c-b)}}\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(a\right)}_n}{n!}}{x}^n{\int }_0^1{t}^{n+b-1}{(1-t)}^{c-b-1}dt.$$

The integral portion is the beta function equal to ${\displaystyle \frac{\mathrm{\Gamma }(n+b)\mathrm{\Gamma }(c-b)}{\mathrm{\Gamma }(n+c)}}$. The whole expression becomes

$${\displaystyle \frac{\mathrm{\Gamma }\left(c\right)}{\mathrm{\Gamma }\left(b\right)}}\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(a\right)}_n}{n!}}{\displaystyle \frac{\mathrm{\Gamma }(n+b)}{\mathrm{\Gamma }(n+c)}}{x}^n={}_{2}F_1\left(\begin{array}{c}a,b\\ c\end{array};z\right).$$

This proves the result for $\mid z\mid <1$. Since the integral is analytic in the cut plane, the theorem holds for z in this region as well.

By applying Euler’s integral representation, we can prove Gauss’s theorem for the hypergeometric series [18] (page 66). □

Proposition 2.

For $(c-a-b)>0$, we have

$${}_{2}F_1\left(\begin{array}{c}a,b\\ c\end{array};1\right)={\displaystyle \frac{\mathrm{\Gamma }\left(c\right)\mathrm{\Gamma }(c-a-b)}{\mathrm{\Gamma }\left(b\right)\mathrm{\Gamma }(c-b)}}.$$

(11)

Proof. 

Let $x\to 1^{-}$ be Euler’s integral for ${}_{2}F_1$. By Abel’s continuity theorem, we can write for $c>b>0$ and $c-a-b>0$,

$${}_{2}F_1\left(\begin{array}{c}a,b\\ c\end{array};1\right)={\displaystyle \frac{\mathrm{\Gamma }\left(c\right)}{\mathrm{\Gamma }\left(b\right)\mathrm{\Gamma }(c-b)}}{\int }_0^1{t}^{b-1}{(1-t)}^{c-a-b-1}dt={\displaystyle \frac{\mathrm{\Gamma }\left(c\right)\mathrm{\Gamma }(c-a-b)}{\mathrm{\Gamma }(c-a)\mathrm{\Gamma }(c-b)}},$$

The condition $c>b>0$ can be removed by continuation. □

As a consequence of Gauss’s theorem, we have the Chu–Vandermonde identity.

For $n\in {\mathbb{Z}}^{+}$ and for $a\in {\mathbb{Z}}^{+}$$(c-a+n)>0$, we have

$${}_{2}F_1\left(\begin{array}{c}a,-n\\ c\end{array};1\right)={\displaystyle \frac{{(c-a)}_n}{{\left(c\right)}_n}}.$$

(12)

We are interested by the case when $c=-2n\in {\mathbb{Z}}^{-}$ and $-n-a\in {\mathbb{Z}}^{-}$. We begin by proving, by induction on n, that the Chu–Vandrmonde identity remains valid with $-n-a\in {\mathbb{Z}}^{-}$.

Proposition 3.

For $n\in {\mathbb{Z}}_{+}$, we have the following proposition ${\mathfrak{P}}_n^a$ depending on n and a:

$${\mathfrak{P}}_n^a:{}_{2}F_1\left(\begin{array}{c}a,-n\\ -2n\end{array};1\right)={\displaystyle \frac{{(-2n-a)}_n}{{(-2n)}_n}}.$$

(13)

Proof. 

The proof will be carried out by induction on n. For $n=1$, the left-hand side of (12) becomes

$${}_{2}F_1(\begin{array}{c}a,-n\\ -2n\end{array};1)=\sum _{k=0}^1{\displaystyle \frac{{\left(a\right)}_k{(-1)}_k}{{(-2)}_{k}k!}}=1+{\displaystyle \frac{-a}{-2}}={\displaystyle \frac{a}{2}}+1$$

and the right-hand side gives

$${\displaystyle \frac{{(-2-a)}_1}{{(-2)}_1}}={\displaystyle \frac{-2-a}{-2}}={\displaystyle \frac{a}{2}}+1.$$

Suppose in this inductive step (IS) that the property is true for any k such that $1\le k\le n$, and we prove that it is still true for $n+1$.

Let us start with the right-hand side. It is easy to see that

$${\displaystyle \frac{{(-2(n+1)-a)}_{n+1}}{{(-2(n+1))}_{n+1}}}={\displaystyle \frac{(a+2n+1)(a+2n+2)}{2(2n+1)(a+n+1)}}{\displaystyle \frac{{(-2n-a)}_n}{{(-2n)}_n}}.$$

(14)

Let us prove the left-hand side. We take into account

$$\begin{array}{ccc}& & {}_{2}F_1\left(\begin{array}{c}a,-(n+1)\\ -2(n+1)\end{array};1\right)-{}_{2}F_1\left(\begin{array}{c}a,-(n+1)\\ -2n-1\end{array};1\right)=-{\displaystyle \frac{a}{2(2n+1)}}{}_{2}F_1\left(\begin{array}{c}a+1,-n\\ -2n\end{array};1\right)\hfill \\ & & {}_{2}F_1\left(\begin{array}{c}a,-n-1\\ -2n-1\end{array};1\right)-{}_{2}F_1\left(\begin{array}{c}a,-n\\ -2n\end{array};1\right)={\displaystyle \frac{a}{2(2n+1)}}{}_{2}F_1\left(\begin{array}{c}a+1,-n\\ -2n+1\end{array};1\right),\hfill \end{array}$$

the left-hand side becomes

$$\begin{array}{cc}\hfill {}_{2}F_1\left(\begin{array}{c}a,-(n+1)\\ -2(n+1)\end{array};1\right)=& {\displaystyle \frac{a}{2(2n+1)}}{}_{2}F_1\left(\begin{array}{c}a+1,-n\\ -2n+1\end{array};1\right)-{\displaystyle \frac{a}{2(2n+1)}}{}_{2}F_1\left(\begin{array}{c}a+1,-n\\ -2n\end{array};1\right)\hfill \\ \hfill =& {\displaystyle \frac{a}{2(2n+1)}}{}_{2}F_1\left(\begin{array}{c}a+1,-n\\ -2n+1\end{array};1\right)-{\displaystyle \frac{a}{2(2n+1)}}{}_{2}F_1\left(\begin{array}{c}a+1,-n\\ -2n\end{array};1\right)\hfill \\ \hfill =& {\displaystyle \frac{a}{2(2n+1)}}\left({}_{2}F_1\left(\begin{array}{c}a+1,-n\\ -2n+1\end{array};1\right)-{}_{2}F_1\left(\begin{array}{c}a+1,-n\\ -2n\end{array};1\right)\right)\hfill \\ \hfill =& {\displaystyle \frac{a(a+1)}{4(4n^2-1)}}{}_{2}F_1\left(\begin{array}{c}a+2,-n+1\\ -2n+2\end{array};1\right)={\displaystyle \frac{a(a+1)}{4(4n^2-1)}}{\mathfrak{P}}_{n-1}^{a+2}.\hfill \end{array}$$

Using IS, we can express

$$\begin{array}{ccc}\hfill {\mathfrak{P}}_{n+1}^a& & ={\mathfrak{P}}_n^a+{\displaystyle \frac{a(a+1)}{4(4n^2-1)}}{\mathfrak{P}}_{n-1}^{a+2}\hfill \\ & & ={\displaystyle \frac{{(-2n-a)}_n}{{(-2n)}_n}}+{\displaystyle \frac{a(a+1)}{4(4n^2-1)}}{\displaystyle \frac{{(-2n+2-a-2)}_{n-1}}{{(-2n+2)}_{n-1}}}\hfill \\ & & ={\displaystyle \frac{{(-2n-a)}_n}{{(-2n)}_n}}+{\displaystyle \frac{a(a+1)}{4(4n^2-1)}}{\displaystyle \frac{{(-2n-a)}_{n-1}}{{(-2n+2)}_{n-2}}}\hfill \\ & & ={\displaystyle \frac{{(-2n-a)}_n}{{(-2n)}_n}}+{\displaystyle \frac{a(a+1)}{4(4n^2-1)}}{\displaystyle \frac{{(-2n-a)}_n}{(-n-a-1){(-2n+2)}_{n-2}}}\hfill \\ & & ={\displaystyle \frac{{(-2n-a)}_n}{{(-2n)}_n}}+{\displaystyle \frac{a(a+1)}{4(4n^2-1)}}{\displaystyle \frac{(-2n)(-2n+1){(-2n-a)}_n}{(-n)(-n-a-1){(-2n)}_n}}\hfill \\ & & ={\displaystyle \frac{{(-2n-a)}_n}{{(-2n)}_n}}\left(1+{\displaystyle \frac{a(a+1)}{4(4n^2-1)}}{\displaystyle \frac{2(-2n+1)}{(-n-a-1)}}\right)\hfill \\ & & ={\displaystyle \frac{{(-2n-a)}_n}{{(-2n)}_n}}\left(1+{\displaystyle \frac{a(a+1)}{4(2n-1)}}{\displaystyle \frac{2(2n-1)}{(n+a+1)}}\right)\hfill \end{array}$$

and this is exactly (14). □

2.2. Applications

• First application
  In the introduction, we stated that Kummer’s quadratic transformation (21) given by

  $${\displaystyle {\left(1-\frac{z}{2}\right)}^u{}_{2}F_1\left(\begin{array}{c}u,v\\ 2v\end{array};z\right)={}_{2}F_1\left(\begin{array}{c}\frac{u}{2},\frac{u+1}{2}\\ v+\frac{1}{2}\end{array};{\left(\frac{z}{2-z}\right)}^2\right).}$$

  is not fulfilled for $u=2$ and $v=-1$. In this regard, if we consider the case when $v\in {\mathbb{Z}}^{-}$ and $u\in \mathbb{Z}$, very recently, Atia [23] and Atia with Al-Mohaimeed [24] established the following result (please see the Appendix A below):

  $${\displaystyle {}_{2}F_1\left(\begin{array}{c}u,v\\ 2v\end{array};z\right)={\left(1-\frac{z}{2}\right)}^{-u}{}_{2}F_1\left(\begin{array}{c}\frac{u}{2},\frac{u+1}{2}\\ v+\frac{1}{2}\end{array};{\left(\frac{z}{2-z}\right)}^2\right)-\frac{1+{(-1)}^u}{2}{b}_u^v\left(z\right)}$$

  $$-{\displaystyle \frac{1-{(-1)}^u}{2}}{c}_u^v\left(z\right),$$

  (15)

  where $b_u^v$ is defined for $v\in {\mathbb{Z}}^{-}$ and $u\in 2\mathbb{Z}$, where $2\mathbb{Z}$ is the set of even numbers, by

  $$b_v^{\left(u\right)}\left(z\right)={\displaystyle \frac{\sqrt{\pi }(2-z){(z-1)}^{v-u}{\left(\frac{z}{2}\right)}^{u-2v-1}{\left(\frac{u}{2}\right)}_{-v+1}{}_{2}F_1\left(\begin{array}{c}1-\frac{u}{2},\frac{1}{2}+v-\frac{u}{2}\\ \frac{3}{2}\end{array};{\left(\frac{2}{z}-1\right)}^2\right)}{\mathrm{\Gamma }(-v+\frac{1}{2})}},$$

  (16)

  and $c_u^v$ is defined for $v\in {\mathbb{Z}}^{-}$ and $u\in 1+2\mathbb{Z}$ (the set of odd numbers) by

  $$c_v^{\left(u\right)}\left(z\right)=-{\displaystyle \frac{\sqrt{\pi }{(z-1)}^{v-u}{\left(\frac{z}{2}\right)}^{u-2v}{\left(\frac{u+1}{2}\right)}_{-v}{}_{2}F_1\left(\begin{array}{c}\frac{1-u}{2},v-\frac{u}{2}\\ \frac{1}{2}\end{array};{\left(\frac{2}{z}-1\right)}^2\right)}{\mathrm{\Gamma }(-v+\frac{1}{2})}}.$$

  (17)
  Using this result, we have the following new result in the case of the Chu–Vandermonde identity.
  Theorem 1.

  For any positive integers n and u, we have

  $$\begin{array}{ccc}& & {\displaystyle {\left(\frac{1}{2}\right)}^{-u}\underset{z\to 1}{lim}{}_{2}F_1\left(\begin{array}{c}\frac{u}{2},\frac{u+1}{2}\\ -n+\frac{1}{2}\end{array};{\left(\frac{z}{2-z}\right)}^2\right)}\hfill \\ & & ={\displaystyle \frac{1+{(-1)}^u}{2}}\underset{z\to 1}{lim}{b}_u^{-n}\left(z\right)+{\displaystyle \frac{1-{(-1)}^u}{2}}\underset{z\to 1}{lim}{c}_u^{-n}\left(z\right)+{\displaystyle \frac{{(-2n-u)}_n}{{(-2n)}_n}}.\hfill \end{array}$$
  Remark 3.

  This theorem is a direct consequence of a recent result given in [23], which is expanded into a more general case with Al-Mohaimeed [24,25] by taking into account (22). We prove now that both

  $$\underset{z\to 1}{lim}{}_{2}F_1\left(\begin{array}{c}{\displaystyle \frac{u}{2}},{\displaystyle \frac{u+1}{2}}\\ -n+{\displaystyle \frac{1}{2}}\end{array};{\left({\displaystyle \frac{z}{2-z}}\right)}^2\right),$$

  and

  $${\displaystyle \frac{1+{(-1)}^u}{2}}\underset{z\to 1}{lim}{b}_u^{-n}\left(z\right)+{\displaystyle \frac{1-{(-1)}^u}{2}}\underset{z\to 1}{lim}{c}_u^{-n}\left(z\right),$$

  have $z=1$ as a pole of order $u+n$.
  To prove that

  $${}_{2}F_1\left(\begin{array}{c}{\displaystyle \frac{u}{2}},{\displaystyle \frac{u+1}{2}}\\ -n+{\displaystyle \frac{1}{2}}\end{array};{\left({\displaystyle \frac{z}{2-z}}\right)}^2\right)$$

  is a rational fraction with $z=1$ as a pole, it suffices to prove the following:

  –

  $${\displaystyle \frac{1+{(-1)}^u}{2}}\underset{z\to 1}{lim}{b}_u^{-n}\left(z\right)+{\displaystyle \frac{1-{(-1)}^u}{2}}\underset{z\to 1}{lim}{c}_u^{-n}\left(z\right),$$

  has $z=1$ as a pole of order $u+n$.

  –

  Also, ${}_{2}F_1\left(\begin{array}{c}u,-n\\ -2n\end{array};z\right)$ is a finite sum without any singularity at $z=1$.

  The expression

  $${\displaystyle \frac{1+{(-1)}^u}{2}}\underset{z\to 1}{lim}{b}_u^{-n}\left(z\right)+{\displaystyle \frac{1-{(-1)}^u}{2}}\underset{z\to 1}{lim}{c}_u^{-n}\left(z\right),$$

  has $z=1$ as a pole of order $u+n$ because of the following:
  • $b_u^{-n}$ is defined for $n\in {\mathbb{Z}}^{+}$ and $u\in 2\mathbb{N}$ by

    $$\begin{array}{ccc}\hfill b_{-n}^{\left(u\right)}\left(z\right)=& & {\displaystyle \frac{\sqrt{\pi }(2-z){(z-1)}^{-n-u}{\left(\frac{z}{2}\right)}^{u+2n-1}{\left(\frac{u}{2}\right)}_{n+1}}{\mathrm{\Gamma }(n+\frac{1}{2})}}\hfill \\ & & \times {}_{2}F_1\left(\begin{array}{c}1-{\displaystyle \frac{u}{2}},{\displaystyle \frac{1}{2}}-n-{\displaystyle \frac{u}{2}}\\ {\displaystyle \frac{3}{2}}\end{array};{({\displaystyle \frac{2}{z}}-1)}^2\right)\hfill \end{array}$$

    which has ${(z-1)}^{-n-u}$ with ${}_{2}F_1$ as a sum with a finite number of terms with variable ${\displaystyle {\left(\frac{2}{z}-1\right)}^2.}$
  • $c_u^v$ is defined for $n\in {\mathbb{Z}}^{+}$ and $u\in 1+2\mathbb{N}$ by

    $$c_{-n}^{\left(u\right)}\left(z\right)=-{\displaystyle \frac{\sqrt{\pi }{(z-1)}^{-n-u}{\left(\frac{z}{2}\right)}^{u+2n}{\left(\frac{u+1}{2}\right)}_n{}_{2}F_1\left(\begin{array}{c}\frac{1-u}{2},-n-\frac{u}{2}\\ \frac{1}{2}\end{array};{\left(\frac{2}{z}-1\right)}^2\right)}{\mathrm{\Gamma }(n+\frac{1}{2})}},$$

    which has ${(z-1)}^{-n-u}$ with ${}_{2}F_1$ as a sum with a finite number of terms with variable ${\left({\displaystyle \frac{2}{z}}-1\right)}^2$.
  The expression ${}_{2}F_1\left(\begin{array}{c}u,-n\\ -2n\end{array};z\right)$ is a sum with a finite number of terms without any singularity at $z=1$.
  Remark 4.

  Some computation may assure readers that the expression

  $${}_{2}F_1\left(\begin{array}{c}{\displaystyle \frac{u}{2}},{\displaystyle \frac{u+1}{2}}\\ -n+{\displaystyle \frac{1}{2}}\end{array};{\left({\displaystyle \frac{z}{2-z}}\right)}^2\right),$$

  is a rational fraction with $z=1$ as a pole. The first easiest case is when $u=1$ and $n=0$, where we find ${\displaystyle \sum _{k=0}^{\infty }{\left(\frac{z}{2-z}\right)}^{2k}}$, which, for $\mid z\mid <1$, gives ${\displaystyle \frac{1}{1-{\left(\frac{z}{2-z}\right)}^2}=\frac{{(2-z)}^2}{-4(z-1)}}$.

  The second easiest case is when $u=1$ and $n=1$, where we find ${\displaystyle \sum _{k=0}^{\infty }(1-2k){\left(\frac{z}{2-z}\right)}^{2k}}$, which, for $\mid z\mid <1$, gives ${\displaystyle \frac{{(2-z)}^2(z^2+2z-2)}{-8{(z-1)}^2}}$.
• Second application on Bailey’s balanced ${}_{4}F_3$ summation
  The formula of Bailey (see [13] p. 512 and [26] p. 245, (IIL20)) is given by

  $${}_{4}F_3\left(\begin{array}{cccc}{\displaystyle \frac{a}{2}},& {\displaystyle \frac{a+1}{2}},& c+n,& -n\\ {\displaystyle \frac{c}{2}},& {\displaystyle \frac{c+1}{2}},& a+1& \end{array};1\right)={\displaystyle \frac{{(c-a)}_n}{{\left(c\right)}_n}}.$$

  (18)
  We are interested by the case $c=-2n$, where the summation ${}_{4}F_3$ reduces to the ${}_{3}F_2$ summation, and we obtain

  $$\begin{array}{ccc}\hfill {}_{4}F_3\left(\begin{array}{cccc}{\displaystyle \frac{a}{2}},& {\displaystyle \frac{a+1}{2}},& -n,& -n\\ -n,& -n+{\displaystyle \frac{1}{2}},& a+1& \end{array};1\right)& & ={}_{3}F_2\left(\begin{array}{ccc}{\displaystyle \frac{a}{2}},& {\displaystyle \frac{a+1}{2}},& -n\\ -n+{\displaystyle \frac{1}{2}},& a+1& \end{array};1\right)\hfill \\ & & ={\displaystyle \frac{{(-2n-a)}_n}{{(-2n)}_n}}.\hfill \end{array}$$

  (19)
  Using (15), we obtain the following new result in the case of the Chu–Vandermonde identity.
  Theorem 2.

  For any positive integers n and u, we have

  $$\begin{array}{ccc}\hfill & \hfill & {\displaystyle {}_{3}F_2\left(\begin{array}{ccc}\frac{u}{2},& \frac{u+1}{2},& -n\\ -n+\frac{1}{2},& u+1& \end{array};1\right)={\left(\frac{1}{2}\right)}^{-u}\underset{z\to 1}{lim}{}_{2}F_1\left(\begin{array}{c}\frac{u}{2},\frac{u+1}{2}\\ -n+\frac{1}{2}\end{array};{\left(\frac{z}{2-z}\right)}^2\right)}\hfill \\ \hfill & \hfill & -{\displaystyle \frac{1+{(-1)}^u}{2}}\underset{z\to 1}{lim}{b}_u^{-n}\left(z\right)-{\displaystyle \frac{1-{(-1)}^u}{2}}\underset{z\to 1}{lim}{c}_u^{-n}\left(z\right)={\displaystyle \frac{{(-2n-u)}_n}{{(-2n)}_n}}.\hfill \end{array}$$

  (20)

3. Kummer’s Quadratic Transformation Conditions by CR and EMOI

In this paragraph, we are focused on the following Kummer quadratic transformation [8]:

$${\displaystyle {\left(1-\frac{z}{2}\right)}^u{}_{2}F_1\left(\begin{array}{c}u,v\\ 2v\end{array};z\right)={}_{2}F_1\left(\begin{array}{c}\frac{u}{2},\frac{u+1}{2}\\ v+\frac{1}{2}\end{array};{\left(\frac{z}{2-z}\right)}^2\right).}$$

(21)

For $u=2$ and $v=-1$, the left-hand side (LHS) of (21) becomes

$${\displaystyle {\left(1-\frac{z}{2}\right)}^u{}_{2}F_1\left(\begin{array}{c}u,v\\ 2v\end{array};z\right)=\frac{1}{4}{(z-2)}^2(z+1),}$$

whereas the right-hand side (RHS) of (21) gives

$${}_{2}F_1\left(\begin{array}{c}{\displaystyle \frac{u}{2}},{\displaystyle \frac{u+1}{2}}\\ v+{\displaystyle \frac{1}{2}}\end{array};{\left({\displaystyle \frac{z}{2-z}}\right)}^2\right)={\displaystyle \frac{(z^4-2z^3+4z-2)}{8{(z-2)}^2{(z-1)}^3}}.$$

This proves that (21) is not valid for every value of u and v. So, let us see the conditions on u and v such that (21) holds true:

–

In [19], on page 1008 paragraph 9.13, Gradshteyn and Ryzhik state that the series

$${}_{2}F_1\left(\begin{array}{c}a,b\\ c\end{array};z\right)$$

will, in the cut plane, yield a single-valued analytic continuation, which we can obtain by means of the formulas below provided $c+1$ is not a natural number, $a-b$ and $c-a-b$ are not integers.

–

In [20], on page 64 paragraph 2.1.5, Magnus and his co-authors state that there exists a quadratic transformation if and only if the numbers

$$\pm (1-c),\pm (a-b),\pm (a+b-c),$$

have the property that one of them equals ${\displaystyle \frac{1}{2}}$ or that two of them are equal.

It is clear that the conditions given in [27] are different from those given in [19].

3.1. Understanding Kummer’S Quadratic Transformation Conditions

3.1.1. The Conditions Given by Gradshteyn and Ryzhik

–

The first transformation: In Askey’s book, page 79, one can find the following proposition, noted as corollary 2.3.3 [18].

Proposition 4.

$$\begin{array}{ccc}& & {}_{2}F_1\left(\begin{array}{c}a,b\\ c\end{array};z\right)={\displaystyle \frac{\mathrm{\Gamma }\left(c\right)\mathrm{\Gamma }(c-a-b)}{\mathrm{\Gamma }(c-a)\mathrm{\Gamma }(c-b)}}{}_{2}F_1\left(\begin{array}{c}a,b\\ a+b+1-c\end{array};1-z\right)\hfill \\ & & +{\displaystyle \frac{\mathrm{\Gamma }\left(c\right)\mathrm{\Gamma }(a+b-c)}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}}{(1-z)}^{c-a-b}{}_{2}F_1\left(\begin{array}{c}c-a,c-b\\ c+1-a-b\end{array};1-z\right).\hfill \end{array}$$

This proposition comes as a corollary of a theorem (denoted by Theorem 2.3.2) given on page 78 of Askey’s book [18].

At this stage, we can understand from where condition 3 comes from. Indeed, by applying Proposition 4, we can express that $a+b+1-c$ and $c-a-b+1$ should not be negative integers. This gives two cases:

*

$a+b-c$ and $c-a-b$ should not be in $\{-2,-3,-4,\dots \}$, and this means that $c-a-b$ should not be an integer (which is exactly condition 3);

*

Either $c=a+b$ or $c=a+b-1$, for which $\mathrm{\Gamma }(c-a-b)$ is not defined.

–

A second transformation: In Magnus’ book, one can find the following proposition [27] on page 48.

Proposition 5.

$$\begin{array}{ccc}& & {}_{2}F_1\left(\begin{array}{c}a,b\\ c\end{array};z\right)={\displaystyle \frac{\mathrm{\Gamma }\left(c\right)\mathrm{\Gamma }(b-a)}{\mathrm{\Gamma }\left(b\right)\mathrm{\Gamma }(c-a)}}{(1-z)}^{-a}{}_{2}F_1\left(\begin{array}{c}a,c-b\\ a+1-b\end{array};{\displaystyle \frac{1}{1-z}}\right)\hfill \\ & & +{\displaystyle \frac{\mathrm{\Gamma }\left(c\right)\mathrm{\Gamma }(a-b)}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }(c-b)}}{(1-z)}^{-b}{}_{2}F_1\left(\begin{array}{c}b,c-a\\ b+1-a\end{array};{\displaystyle \frac{1}{1-z}}\right).\hfill \end{array}$$

At this point, we can understand from where condition 2 comes. In fact, using Proposition 5, we shall write that $a+1-b$ and $b+1-a$ should not be negative integers. This gives two cases:

*

$a-b$ and $b-a$ should not be in $\{-2,-3,-4,\dots \}$, and this means that $a-b$ should not be an integer (which is exactly condition 2);

*

Either $a=b$ or $a-b=-1$, for which $\mathrm{\Gamma }(a-b)$ is not defined.

–

A third transformation: In Magnus’ book, one can find the following proposition [27] on page 48, first line.

Proposition 6.

$$\begin{array}{ccc}& & {}_{2}F_1\left(\begin{array}{c}a,b\\ c\end{array};z\right)={\displaystyle \frac{\mathrm{\Gamma }\left(c\right)\mathrm{\Gamma }(b-a)}{\mathrm{\Gamma }\left(b\right)\mathrm{\Gamma }(c-a)}}{(-z)}^{-a}{}_{2}F_1\left(\begin{array}{c}a,a+1-c\\ a+1-b\end{array};{\displaystyle \frac{1}{z}}\right)\hfill \\ & & +{\displaystyle \frac{\mathrm{\Gamma }\left(c\right)\mathrm{\Gamma }(a-b)}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }(c-b)}}{(-z)}^{-b}{}_{2}F_1\left(\begin{array}{c}b,b+1-c\\ b+1-a\end{array};{\displaystyle \frac{1}{z}}\right).\hfill \end{array}$$

At this stage, we can understand from where condition 1 comes. In fact, if we have ${}_{2}F_1(a,b;c;z)$, then using Proposition 5, we shall write that $a-b$ should not be an integer, and this means that applying this condition to the RHS of Proposition 6 with $a\to a,b\to a+1-c$ reveals that $c-1$ should not be an integer. This gives two cases:

*

$c+1$ should not be in $\{1,2,3,4,5,\dots \}$;

*

$c+1$ should not be in $\{\dots ,-3,-2,-1,0\}$ because either the definition of the hypergeometric series or $\mathrm{\Gamma }\left(c\right)$ is not defined.

This leads to the condition that $c+1$ should not be a natural number.

3.1.2. The Conditions Given by ErdéLyi, Magnus, Oberhettinger, and Tricomi

In [20], on page 64 paragraph 2.1.5, it is state that “If and only if the numbers

$$\pm (1-c),\pm (a-b),\pm (a+b-c)$$

have the property that one of them equals ${\displaystyle \frac{1}{2}}$ or that two of them are equal, then there exists a quadratic transformation”. Let us examine each of these six conditions individually:

–

$\pm (1-c)$ has the property where ${\displaystyle \frac{1}{2}}$ means $c={\displaystyle \frac{1}{2}}$ or $c={\displaystyle \frac{3}{2}}$, and this comes from the transformation given in [28] in group 4 in Table 15.8.1

$$\begin{array}{cccc}\mathrm{GROUP}1& \mathrm{GROUP}2& \mathrm{GROUP}3& \mathrm{GROUP}4\\ & c=a-b+1& a=b+{\displaystyle \frac{1}{2}}& \\ c=2a& c=b-a+1& b=a+{\displaystyle \frac{1}{2}}& c={\displaystyle \frac{1}{2}}\\ c=2b& c={\displaystyle \frac{1}{2}}(a+b+1)& c=a+b+{\displaystyle \frac{1}{2}}& c={\displaystyle \frac{3}{2}}\\ & a+b=1& c=a+b-{\displaystyle \frac{1}{2}}\end{array}$$

and in ([28], 15.8.24, 15.8.25, 15.8.26, 15.8.27 and 15.8.28).

–

$\pm (a-b)$ and $\pm (a+b-c)$ have the property where ${\displaystyle \frac{1}{2}}$ comes from the transformation given in [28] in group 3 in Table 15.8.1.

3.1.3. Conclusions

–

We presented two contradictory references, and we do not know which one is correct and which one is incorrect.

–

Are there any conditions that have not been stated? For example, based on Table 15.8.1 given in [28] and based on the same methodology of Erdélyi, Magnus, Oberhettinger, and Tricomi, can we say that $\pm (a-b)$ should have the property that one of them equals $1-c$?

–

What are the exact conditions?

–

We investigated specific cases of Gauss’ theorem in the context of hypergeometric functions and examined isolated instances of the Chu–Vandermonde identity, which emerges as a consequence of Gauss’s theorem. We discussed two significant applications stemming from these findings. The question is how about other specific cases, i.e., can we extend the following result ${\mathfrak{P}}_n^a$ depending on n and a, where

$${\mathfrak{P}}_n^a:{}_{2}F_1\left(\begin{array}{c}a,-n\\ -2n\end{array};1\right)={\displaystyle \frac{{(-2n-a)}_n}{{(-2n)}_n}}$$

(22)

to any a in ${\mathbb{R}}^{+}$?