On a Resolution of Another Isolated Case of a Kummer’s Quadratic Transformation for 2 F 1

: It is well-known that the Kummer quadratic transformation formula is valid provided that its parameters fulﬁll some speciﬁc conditions (see Gradshteyn, Ryzhik, Tables of Integrals, Series and Products, 9.130, 9.134.1). Very recently, one of us established a new identity when one of these conditions is not fulﬁlled. In this paper, we aim to discuss another isolated case which completely different from the ﬁrst. Moreover, in the end, we mention two interesting consequences of these two new results


Introduction
Gauss [1] defined his famous infinite series as follows 1 + ab c z 1! + a(a + 1)b(b + 1) c(c + 1) This infinite series (1) is usually denoted by the notation 2 F 1 (a, b; c; z) or simply F and is commonly known as the Gauss's function or the hypergeometric series.Gauss's function or the hypergeometric series is a solution of a second order differential equation.The convergence conditions of 2 F 1 are as follows, (a + b − c) < 0, the series converges (absolutely) throughout the entire unit circle, 3.
(a + b − c) ≥ 1, the series diverges on the entire unit circle.
It is well-known that the Kummer quadratic transformation formula 2 F 1 (α, β; 2α; z) = (1 is valid provided that {2α + 1, α + 3 2 } are not natural numbers and α − β is not an integer (see Gradshteyn, Ryzhik, Tables of Integrals, Series and Products, 9.130, 9.134.1).Very recently, one of us established a new identity for an isolated case where α − β is an integer by letting α to be a negative integer and β to be an even positive integer which we extended here into any even integer not necessarily positive and where we gave explicitly the expressions u (β) α (z) such that [2] 2 F 1 (α, β; 2α; z) = (1 α (z).
In this paper, we aim to discuss another isolated case with α to be a negative integer and β to be an odd integer where we gave explicitly the expressions v The results for u In this paper, we deal with Gauss's function and exactly with its connection with the following interesting and useful Kummer's quadratic transformation [2] for the hypergeometric function 2 F 1 (2).This quadratic transformation is valid for {2α + 1, α + 3 2 } are not natural numbers and α − β is NOT an integer.In this paper, we consider the case when α − β is an integer.
This transformation formula is recorded in several standard texts on special functions and handbooks of mathematics, for example, in the standard text of I.S. Gradshteyn and I.M. Ryzhik [3] (9.134 and 9.134.1)and G. Andrews and al. [4] (3.1.7 page 127 with a slight modification), in the handbook by Abramowitz-Stegun [5] (15.3.20) and in DLMF: NIST Digital Library of Mathematical Functions, https://dlmf.nist.gov/[6], accessed on 15 December 2022, 15.8.13.
either a is itself a positive integer, i.e., a ∈ N, this situation was considered in a previous paper [7] where we proved that (5) remains true for n = 0 but for n = 1 (5) becomes and for n ≥ 2 (5) should be written as which, using the notation we wrote it under the form or a is itself a negative integer, i.e., a ∈ Z − and this situation is solved only by using the Pochhammer symbol in the formula of u (a) n : and ( 8) remains true, • or a is not an integer but half of an integer, i.e., a ∈ { • • • } and, in this case, we prove that (5) remains true for n = 0 but for n = 1 and using the fact that (−1 and which using the notation , ( 11) As an interesting consequence of our new expressions u n and v (a) n , a ∈ Z. (13) Please note the following that this paper is a continuation of [7].Many authors dealt with the quadratic transformation (2) recorded in [8][9][10][11] but always with the restrictions {2α + 1, α + 3 2 } are not natural numbers and α − β is not an integer (see Gradshteyn,Ryzhik,9.130).This paper deals with some isolated cases related to α − β as an integer and will be organized as follows.First we prove promptly that the result published in [7] does not hold for a ∈ { Second, for any a ∈ R we give and prove some relations involving 2 F 1 (a, a + 1 2 ; −n + 3 2 ; z 2 ) and 1 (1±z) 2a 2 F 1 (2a, −n + 1; −2n + 2; ±2z 1±z ).Finally we prove (12) and (13).

No Concordance with Previous Result
), Proof.In fact, with n = 2 and a = 5 2 and taking into account the "+" sign we get 6  .and 2 F 1 ( One Here is another curious counter-example.If we take n = 2, a = 1 2 and z = − 1 2 we find

Relations Involving the First Sum
In the following lemma, we give three relations involving 2 F 1 (a, a + 1 2 ; −n + 3 2 ; z 2 ).
Lemma 1.For any positive integer n and for any a ∈ R we have the following results 2 F 1 (a, a + Proof.The proof of ( 14) is a direct consequence of the following formula [6], 15.5.1, [3-5] Second, let us prove (15).
2 F 1 (a, a + Let us prove (16).It is easy to see that 2 F 1 (a, a +

Relations Involving the Second Sum
In the following lemma we give three relations involving 1 Lemma 2. For any positive integer n and for any a ∈ R we have the following results d dz Proof.First, let us prove (18) and let us prove it for the "+"sign, To prove (19) we begin considering the following change of variable for the + sign whereas for the − sign we assume For the + sign, we should prove that 2 F 1 (2a, −n + 1; −2n + 2; Let us prove (22).The LHS of (22) gives Using the following relation ).
Using the following relation the LHS vanishes.
Lemma 3. In the following lemma, we give these relations involving u u as well as Proof.The proof of (23) is a direct consequence of the combination of ( 8) with ( 16) and (20), whereas the proof of (24) is a direct consequence of the combination of (12) with ( 16) and (20).

Relations between New Added Terms
Using this lemma, (2) and results given in [7] we give the following result Theorem 2. For any a ∈ R we have the following results v as well as u Proof.Using ( 19) and (2) we have ).
Using results of [7] we have and for n ≥ 2 we have If we combine all these quantities together we obtain In the following proposition, we give the simplified expression of v Proof.We have just proved that Using (7) we get . Notation 1.In the sequel, we denote by For any a ∈ Z we have the following result (Fv) (a) because 2a − 1 is an odd integer (among the conditions of the convergence of the series we should add z / ∈ {−1, 0, 1}).The same calculations lead to (Fv) (a) because 2a is an even integer.Using ( 24) and ( 16) we easily get for all n ∈ N and a ∈ Z • • • } and for some values of n ∈ N ∪ {0} we have arrived to write a new identity between a well-defined series with infinitely many terms and a well-defined series with finitely many terms as follows

•
for a ∈ N ∪ {0} and n ∈ N\{1} the series 2 F 1 (a, a + 1 2 ; −n + 3 2 ; z 2 ) is well defined with infinitely many terms is equal to the series which is well-defined with finitely many terms.

•
For a ∈ Z − \{0} and n ∈ N ∪ {0} the series 2 F 1 (a, a + 1 2 ; −n + 3 2 ; z 2 ) is well defined with finitely many terms whereas the series is well defined with finitely many terms provided that 1 − a < n.

Open Problem
By same technique, we are working on other well-known quadratic transformations available in the literature.The work is under investigation and will form a part of the subsequent paper in this direction.
• a hypergeometric series terminates if a or b is equal to a negative integer or zero.For c = −n(n = 0, 1, 2, • • • ), the hypergeometric series is indeterminate if neither a nor b is equal to −m (where m < n and m is a natural number), • if we exclude these values of the parameters a, b, c, a hypergeometric series converges in the unit circle |z| < 1. 2 F 1 then has a branch point at z = 1.Then we have the following conditions for convergence on the unit circle: 1. 0 ≤ (a + b − c) < 1, the series converges throughout the entire unit circle, except at the point z = 1, 2.

α
(z)  are given in the paper.Moreover, in the end, we mention two interesting consequences of our main result involving a symmetric role for u (β) α (z) with an odd β and v (β) α (z) with an even β.

3 =Remark 1 .
0. Then, by recurrence, it is easy to prove that (Fv)(a) n = 0, n ∈ N.The same steps lead to the proof of (Fv) (a) n = 0.The explanation of the consequence is as follows: for some values of a