Open AccessArticle

A New Closed-Form Formula of the Gauss Hypergeometric Function at Specific Arguments

Yue-Wu Li

and

Feng Qi

^2,3,*

School of Mathematics and Physics, Hulunbuir University, Hulunbuir 021008, China

School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo 454010, China

Independent Researcher, University Village, Dallas, TX 75252, USA

Author to whom correspondence should be addressed.

Axioms 2024, 13(5), 317; https://doi.org/10.3390/axioms13050317

Submission received: 28 March 2024 / Revised: 21 April 2024 / Accepted: 9 May 2024 / Published: 10 May 2024

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Abstract

In this paper, the authors briefly review some closed-form formulas of the Gauss hypergeometric function at specific arguments, alternatively prove four of these formulas, newly extend a closed-form formula of the Gauss hypergeometric function at some specific arguments, successfully apply a special case of the newly extended closed-form formula to derive an alternative form for the Maclaurin power series expansion of the Wilf function, and discover two novel increasing rational approximations to a quarter of the circular constant.

Keywords:

Gauss hypergeometric function; Euler integral representation; Lerch transcendent; specific argument; closed-form formula; contiguous function; power series expansion; Wilf function; rational approximation; circular constant

MSC:

Primary 33C05; Secondary 11B37, 11B83, 26A09, 33B10, 41A20, 41A58

1. Simple Preliminaries

For

α_{i} \in C

β_{i} \in C ∖ {0, - 1, - 2, \dots}

p, q \in N = {1, 2, \dots}

, and

z \in C

, in terms of the rising factorial, also known as the Pochhammer symbol,

{(z)}_{n} = \prod_{ℓ = 0}^{n - 1} (z + ℓ) = \{\begin{matrix} z (z + 1) \dots (z + n - 1), & n \in N; \\ 1, & n = 0, \end{matrix}

the generalized hypergeometric series is defined in [1] (p. 1020) by

_{p} F_{q} (α_{1}, α_{2}, \dots, α_{p}; β_{1}, β_{2}, \dots, β_{q}; z) = \sum_{n = 0}^{\infty} \frac{{(α_{1})}_{n} {(α_{2})}_{n} \dots {(α_{p})}_{n}}{{(β_{1})}_{n} {(β_{2})}_{n} \dots {(β_{q})}_{n}} \frac{z^{n}}{n!} .

(1)

In particular, when taking

(p, q) = (2, 1)

in (1), the function

_{2} F_{1} (α_{1}, α_{2}; β_{1}; z)

is called the Gauss hypergeometric function. See also [2] (Chapter II) and [3] (Chapter 14).

The classical Euler gamma function

Γ (z)

can be defined [4] (Chapter 3) by

Γ (z) = lim_{n \to \infty} \frac{n! n^{z}}{\prod_{k = 0}^{n} (z + k)}, z \in C ∖ {0, - 1, - 2, \dots} .

The logarithmic derivative

{[ln Γ (z)]}^{'} = \frac{Γ^{'} (z)}{Γ (z)}

is denoted by

ψ (z)

and is called the psi function or the digamma function. The reciprocal

\frac{1}{Γ (z)}

is an entire function possessing simple zeros at the points

1 - k

for

k \in N

(see [5] (p. 255, Entry 6.1.3)). The beta function

B (z, w)

can be defined by

B (z, w) = \frac{Γ (z) Γ (w)}{Γ (z + w)}, z, w \in C ∖ {0, - 1, - 2, \dots} .

(2)

We note that the definition (2) of

B (z, w)

extends the following classical definition:

B (z, w) = \int_{0}^{1} t^{z - 1} {(1 - t)}^{w - 1} d t = \int_{0}^{\infty} \frac{t^{z - 1}}{{(1 + t)}^{z + w}} d t, ℜ (z), ℜ (w) > 0 .

2. A Brief Review

In general, it is not easy to write out elementary, closed-form, explicit expressions of the Gauss hypergeometric function

_{2} F_{1} (a, b; c; z)

at specific arguments

(a, b; c; z)

. See the short and simple review in [6] (Section 4).

In the paper [7], the authors reviewed many results obtained in the papers [8,9,10] and other historical literature about the generalized hypergeometric series

_{p} F_{q}

. In the recently published papers [11,12,13,14], the authors derived more significant conclusions for

_{p} F_{q}

at some specific arguments.

Entry 15.1.21 in [5] (p. 557), Corollary 3.1.2 in [15] (p. 126), Entry 15.4.6 in [16] (p. 387), Theorem 26 in [17] (p. 68), and the first equality in [18] (p. 184, Section 4.13) read that, for

a - b + 1 \neq 0, - 1, - 2, \dots

_{2} F_{1} (a, b; a - b + 1; - 1) = \frac{Γ (a - b + 1) Γ (\frac{a}{2} + 1)}{Γ (a + 1) Γ (\frac{a}{2} - b + 1)} = \frac{\sqrt{π}}{2^{a}} \frac{Γ (a - b + 1)}{Γ (\frac{a}{2} - b + 1) Γ (\frac{a}{2} + \frac{1}{2})} .

(3)

Entry 15.1.22 in [5] (p. 557) states that, for

a - b + 2 \neq 0, - 1, - 2, \dots

_{2} F_{1} (a, b; a - b + 2; - 1) = \frac{\sqrt{π}}{2^{a}} \frac{Γ (a - b + 2)}{b - 1} [\frac{1}{Γ (\frac{a}{2}) Γ (\frac{a}{2} - b + \frac{3}{2})} - \frac{1}{Γ (\frac{a}{2} + \frac{1}{2}) Γ (\frac{a}{2} - b + 1)}] .

(4)

In [19] (pp. 453–496, Section 7.3), the authors included many closed-form expressions of

_{2} F_{1} (a, b; c, z)

for specific values of

(a, b; c, z)

, including the following data: Formulas (3) and (4); many values of

_{2} F_{1} (a, b; c, \pm 1)

for specific

(a, b; c)

; many values of

_{2} F_{1} (a, b; c, \frac{1}{2})

for specific

(a, b; c)

; many values of

_{2} F_{1} (- n, b; c, 2)

for

n \in N

and specific

(b; c)

; and many values of

_{2} F_{1} (- n, b; c, z_{0})

for

z_{0} \neq \pm 1, 2^{\pm 1}

. In [19] (p. 489, Eq. 7.3.6.4), it was given that, for

a - b + 3 \in C ∖ {0, - 1, - 2, \dots}

\begin{matrix} _{2} F_{1} (a, b; a - b + 3; - 1) \\ = {\begin{matrix} \frac{\sqrt{π} Γ (a + 2)}{2^{a} Γ (\frac{a}{2}) Γ (\frac{a}{2} + \frac{1}{2}) Γ (\frac{a}{2} + 1)} [Γ (\frac{a}{2}) - a Γ (\frac{a}{2}) ψ (\frac{a}{2} + 1) + 2 Γ (\frac{a}{2} + 1) ψ (\frac{a}{2} + \frac{1}{2})], b = 1; \\ \frac{a}{2} \{1 + (a - 1) [ψ (\frac{a}{2}) - ψ (\frac{a}{2} + \frac{1}{2})]\}, b = 2; \\ \frac{\sqrt{π} Γ (a - b + 3)}{(b - 1) (b - 2) 2^{a - 1}} [\frac{a - b + 1}{2 Γ (\frac{a}{2} + \frac{1}{2}) Γ (\frac{a}{2} - b + 2)} - \frac{1}{Γ (\frac{a}{2}) Γ (\frac{a}{2} - b + \frac{3}{2})}], b \neq 1, 2 . \end{matrix} \end{matrix}

(5)

Replacing a by

a - 1

in [19] (p. 489, Eq. 7.3.6.1) gives, for

a - b + 1 \in C ∖ {0, - 1, - 2, \dots}

_{2} F_{1} (a + 1, b; a - b + 1; - 1) = \frac{\sqrt{π} Γ (a - b + 1)}{2^{a + 1}} [\frac{1}{Γ (\frac{a}{2} + \frac{1}{2}) Γ (\frac{a}{2} - b + 1)} + \frac{1}{Γ (\frac{a}{2} + 1) Γ (\frac{a}{2} - b + \frac{1}{2})}] .

(6)

On 8 December 2022, Henri Cohen (Université de Bordeaux, France) gave the explicit Formula (6) on the website https://mathoverflow.net/a/436154 (accessed on 8 December 2022) without referring to any references.

In [5] (p. 557), we find the following formulas:

\begin{matrix} _{2} F_{1} (a, b; \frac{a + b + 1}{2}; \frac{1}{2}) & = \sqrt{π} \frac{Γ (\frac{a + b + 1}{2})}{Γ (\frac{a + 1}{2}) Γ (\frac{b + 1}{2})}, \end{matrix}

(7)

\begin{matrix} _{2} F_{1} (a, 1 - a; b; \frac{1}{2}) & = \frac{2^{1 - b} \sqrt{π} Γ (b)}{Γ (\frac{a + b}{2}) Γ (\frac{b - a - 1}{2})}, \end{matrix}

(8)

\begin{matrix} _{2} F_{1} (a, a; a + 1; \frac{1}{2}) & = 2^{a - 1} a [ψ (\frac{a + 1}{2}) - ψ (\frac{a}{2})], \end{matrix}

(9)

\begin{matrix} _{2} F_{1} (a, a + \frac{1}{2}; \frac{3}{2} - 2 a; - \frac{1}{3}) & = {(\frac{9}{8})}^{2 a} \frac{Γ (\frac{4}{3}) Γ (\frac{3}{2} - 2 a)}{Γ (\frac{3}{2}) Γ (\frac{4}{3} - 2 a)}, \end{matrix}

(10)

\begin{matrix} _{2} F_{1} (a, a + \frac{1}{2}; \frac{2}{3} a + \frac{5}{6}; \frac{1}{9}) & = {(\frac{3}{4})}^{a} \sqrt{π} \frac{Γ (\frac{2}{3} a + \frac{5}{6})}{Γ (\frac{a}{3} + \frac{1}{2}) Γ (\frac{a}{3} + \frac{5}{6})}, \end{matrix}

(11)

and

_{2} F_{1} (a, b; \frac{a + b}{2} + 1; \frac{1}{2}) = \frac{2 \sqrt{π}}{a - b} Γ (\frac{a + b}{2} + 1) [\frac{1}{Γ (\frac{a}{2}) Γ (\frac{b + 1}{2})} - \frac{1}{Γ (\frac{a + 1}{2}) Γ (\frac{b}{2})}] .

(12)

Formula (11) corrects a typo appearing in [5] (p. 557, Entry 15.1.30).

In the paper [7], among other things, Rakha and Rathie established several closed-form formulas of the Gauss hypergeometric function

_{2} F_{1} (a, b; c; z)

for

z = - 1, \frac{1}{2}

as follows.

For $j = 0, 1, 2, \dots$ ,

$_{2} F_{1} (a, b; \frac{a + b + j + 1}{2}; \frac{1}{2}) = \frac{Γ (\frac{1}{2}) Γ (\frac{a + b + j + 1}{2})}{Γ (\frac{b}{2}) Γ (\frac{b + 1}{2})} \frac{Γ (\frac{a - b - j + 1}{2})}{Γ (\frac{a - b + j + 1}{2})} \sum_{r = 0}^{j} {(- 1)}^{r} (\binom{j}{r}) \frac{Γ (\frac{b + r}{2})}{Γ (\frac{a - j + r + 1}{2})} .$

(13)
For $j = 0, 1, 2, \dots$ ,

$_{2} F_{1} (a, b; \frac{a + b - j + 1}{2}; \frac{1}{2}) = \frac{Γ (\frac{1}{2}) Γ (\frac{a + b - j + 1}{2})}{Γ (\frac{b}{2}) Γ (\frac{b + 1}{2})} \sum_{r = 0}^{j} (\binom{j}{r}) \frac{Γ (\frac{b + r}{2})}{Γ (\frac{a - j + r + 1}{2})} .$
For $j = 0, 1, 2, \dots$ ,

$_{2} F_{1} (a, b; a - b + j + 1; - 1) = \frac{2^{- a} Γ (\frac{1}{2}) Γ (b - j) Γ (a - b + j + 1)}{Γ (b) Γ (\frac{a - 2 b + j + 1}{2}) Γ (\frac{a - 2 b + j + 2}{2})} \sum_{r = 0}^{j} {(- 1)}^{r} (\binom{j}{r}) \frac{Γ (\frac{a - 2 b + j + r + 1}{2})}{Γ (\frac{a - j + r + 1}{2})} .$

(14)

This equality extends the Equalities (3)–(6) mentioned above.
For $j = 0, 1, 2, \dots$ ,

$_{2} F_{1} (a, b; a - b - j + 1; - 1) = \frac{2^{- a} Γ (\frac{1}{2}) Γ (a - b - j + 1)}{Γ (\frac{a - 2 b - j + 1}{2}) Γ (\frac{a - 2 b - j + 2}{2})} \sum_{r = 0}^{j} (\binom{j}{r}) \frac{Γ (\frac{a - 2 b - j + r + 1}{2})}{Γ (\frac{a - j + r + 1}{2})} .$
For $j = 0, 1, 2, \dots$ ,

$_{2} F_{1} (a, 1 - a + j; c; \frac{1}{2}) = \frac{2^{1 + j - c} Γ (\frac{1}{2}) Γ (c) Γ (a - j)}{Γ (a) Γ (\frac{c - a}{2}) Γ (\frac{c - a + 1}{2})} \sum_{r = 0}^{j} {(- 1)}^{r} (\binom{j}{r}) \frac{Γ (\frac{c - a + r}{2})}{Γ (\frac{c + a + r - 2 j}{2})} .$
For $j = 0, 1, 2, \dots$ ,

$_{2} F_{1} (a, 1 - a - j; c; \frac{1}{2}) = \frac{2^{1 - j - c} Γ (\frac{1}{2}) Γ (c + \frac{j}{2})}{Γ (\frac{c - a}{2}) Γ (\frac{c - a + 1}{2})} \sum_{r = 0}^{j} (\binom{j}{r}) \frac{Γ (\frac{c - a + r}{2})}{Γ (\frac{c + a + r}{2})} .$

These six closed-form formulas generalize Gauss’, Kummer’s, and Bailey’s summation theorems and many of the identities mentioned above.

In the paper [20], as well as [19] (p. 477, Eq. 162) and [21] (Section 6), the closed-form formula

_{2} F_{1} (1, 2; \frac{1}{2}; z) = \frac{z + 2}{2 {(1 - z)}^{2}} + \frac{3}{2} \frac{\sqrt{z}}{{(1 - z)}^{5 / 2}} arcsin \sqrt{z}

was established, discussed, and applied.

In [22] (Lemma 2.6), for

0 \neq | t | < 1

and

n \in N

, Qi successfully discovered and applied the closed-form formula

_{2} F_{1} (\frac{1 - n}{2}, \frac{2 - n}{2}; 1 - n; \frac{1}{t^{2}}) = \frac{t}{2^{n} i \sqrt{1 - t^{2}}} [{(1 + \frac{i \sqrt{1 - t^{2}}}{t})}^{n} - {(1 - \frac{i \sqrt{1 - t^{2}}}{t})}^{n}],

(15)

where

i = \sqrt{- 1}

is the imaginary unit. In [22] (Remark 6.6), Qi conjectured that the range of

n \in N

in (15) can be extended to

n \in R

. This conjecture still remains open at present. See also [23] (Section 3.9).

In [24] (Corollary 4.1), Qi established the closed-form formula

_{2} F_{1} (n + \frac{1}{2}, n + 1; n + \frac{3}{2}; - 1) = \frac{(2 n + 1)!!}{(2 n)!!} \frac{π}{4} + \frac{2 n + 1}{2^{2 n}} \sum_{k = 1}^{n} {(- 1)}^{k} (\binom{2 n - k}{n}) \frac{2^{k / 2}}{k} sin \frac{3 k π}{4}

(16)

for

n \in N_{0} = {0} \cup N

In Section 3 of this paper, we will alternatively compute four Gauss hypergeometric function:

\begin{matrix} _{2} F_{1} (2 α + 1, α + 1; α + 3; - 1), & _{2} F_{1} (a, b; a - b + 3; - 1), \\ _{2} F_{1} (a + 1, b; a - b + 1; - 1), & _{2} F_{1} (2 α + 1, 2; α + 3; \frac{1}{2}) . \end{matrix}

(17)

In Section 4 of this paper, more importantly, we will extend the closed-form Formula (16) by establishing a closed-form expression of the Gauss hypergeometric function

_{2} F_{1} (n + \frac{1}{2}, n + 1; n + \frac{3}{2}; - z^{2})

(18)

for

n \in N_{0}

and

z \in C

. In Section 5, we will apply a special case of the newly extended closed-form formula for the function (18) to derive an alternative form for the Maclaurin power series expansion of the Wilf function

W (z) = \frac{arctan \sqrt{2 e^{- z} - 1}}{\sqrt{2 e^{- z} - 1}},

(19)

which was investigated in the conference paper [25] and the preprints on the site https://arxiv.org/abs/2110.08576 (accessed on 1 May 2022); moreover, we will discover two novel increasing rational approximations to the irrational constant

\frac{π}{4}

. In the final section, Section 6, we will list some more remarks on our main results and related findings.

3. Alternative Proofs of Four Known Results

Now, we set out to alternatively compute the four Gauss hypergeometric function listed in (17).

Theorem 1.

For

α \neq - 3, - 4, \dots

, we have

_{2} F_{1} (2 α + 1, α + 1; α + 3; - 1) = {\begin{matrix} 2 (2 ln 2 - 1), & α = 0; \\ 3 (\frac{3}{2} - 2 ln 2), & α = 1; \\ \frac{1}{2^{2 α}} \frac{(α + 1) (α + 2)}{(α - 1) α} [1 + α - α B (\frac{1}{2}, α)], & α \neq 0, 1, - 3, - 4, \dots \end{matrix}

(20)

and

_{2} F_{1} (2 α + 1, 2; α + 3; \frac{1}{2}) = {\begin{matrix} 4 (2 ln 2 - 1), & α = 0; \\ 24 (\frac{3}{2} - 2 ln 2), & α = 1; \\ \frac{2 (α + 1) (α + 2)}{(α - 1) α} [1 + α - α B (\frac{1}{2}, α)], & α \neq 0, 1, - 3, - 4, \dots . \end{matrix}

(21)

Proof.

The closed-form Formula (20) is a special case of (14) for

a = 2 α + 1

b = α + 1

, and

j = 2

, as well as a special case of the closed-form Formula (5) for

a = 2 α + 1

and

b = α + 1

. Its alternative proof is as follows.

Setting

a = 2 α + 1

and

b = α + 1

in the first equality of (3) results in

_{2} F_{1} (2 α + 1, α + 1; α + 1; - 1) = \frac{1}{\sqrt{π}} \frac{Γ (α + 1) Γ (α + \frac{3}{2})}{Γ (2 α + 2)}, α \neq - 1, - 2, \dots .

(22)

Letting

a = 2 α + 1

and

b = α + 1

in (4) gives

_{2} F_{1} (2 α + 1, α + 1; α + 2; - 1) = {\begin{matrix} ln 2, & α = 0; \\ \frac{\sqrt{π} Γ (α + 2)}{2^{2 α + 1} α} [\frac{1}{Γ (α + \frac{1}{2})} - \frac{1}{\sqrt{π} Γ (α + 1)}], & α \neq 0, - 2, - 3, \dots . \end{matrix}

(23)

Entry 15.5.18 in [16] (p. 388), a relation of contiguous functions, says that

\begin{matrix} c (c - 1) (z - 1)_{2} F_{1} (a, b; c - 1; z) + c [c - 1 - (2 c - a - b - 1) z]_{2} F_{1} (a, b; c; z) \\ + (c - a) (c - b) z_{2} F_{1} (a, b; c + 1; z) = 0 . \end{matrix}

(24)

Taking

a = 2 α + 1

b = α + 1

c = α + 2

, and

z = - 1

in (24) reveals that

\begin{matrix} 2 (α + 2)_{2} F_{1} (2 α + 1, α + 1; α + 2; - 1) - 2 (α + 2) (α + 1)_{2} F_{1} (2 α + 1, α + 1; α + 1; - 1) \\ - (1 - α)_{2} F_{1} (2 α + 1, α + 1; α + 3; - 1) = 0 \end{matrix}

(25)

for

α + 1 \neq - 1, - 2, \dots

Substituting (22) and (23) into (25), and simplifying the result, yields

\begin{matrix} _{2} F_{1} (2 α + 1, α + 1; α + 3; - 1) & = {\begin{matrix} 2 (2 ln 2 - 1), α = 0; \\ 3 (\frac{3}{2} - 2 ln 2), α = 1; \\ \frac{2 Γ (α + 3)}{1 - α} [\frac{1}{2^{2 α + 1} α} \frac{\sqrt{π} Γ (α + 1) - Γ (α + \frac{1}{2})}{Γ (α + 1) Γ (α + \frac{1}{2})} - \frac{Γ (α + \frac{3}{2})}{\sqrt{π} Γ (2 α + 2)}], α \neq 0, 1, - 3, - 4, \dots \end{matrix} \\ = {\begin{matrix} 2 (2 ln 2 - 1), & α = 0; \\ 3 (\frac{3}{2} - 2 ln 2), & α = 1; \\ \frac{1}{2^{2 α}} \frac{(α + 1) (α + 2)}{(α - 1) α} [1 + α - α B (\frac{1}{2}, α)], & α \neq 0, 1, - 3, - 4, \dots \end{matrix} \end{matrix}

Formula (20) is thus alternatively proved.

The closed-form Formula (21) is a special case of (13) for

a = 2 α + 1

b = 2

, and

j = 2

. Its alternative proof is as follows.

By virtue of

\int_{0}^{\infty} x^{λ - 1} {(1 + x)}^{ν} {(1 + a x)}^{μ} d x = B (λ, - μ - ν - λ)_{2} F_{1} (- μ, λ; - μ - ν; 1 - a)

(26)

for

| arg (a) | < π

and

- ℜ (μ + ν) > ℜ (λ) > 0

, which is taken from [1] (p. 320, Entry 5), we obtain

\int_{0}^{\infty} \frac{u {(u + 1)}^{α - 2}}{{(u + 2)}^{2 α + 1}} d u = \frac{B (2, α + 1)}{2^{2 α + 1}}_{2} F_{1} (2 α + 1, 2; α + 3; \frac{1}{2}) .

(27)

By virtue of

\int_{0}^{\infty} x^{λ - 1} {(1 + x)}^{- μ + ν} {(x + β)}^{- ν} d x = B (μ - λ, λ)_{2} F_{1} (ν, μ - λ; μ; 1 - β)

(28)

for

ℜ (μ) > ℜ (λ) > 0

, which is taken from [1] (p. 320, Entry 9), we acquire

\int_{0}^{\infty} \frac{u {(u + 1)}^{α - 2}}{{(u + 2)}^{2 α + 1}} d u = B (2, α + 1)_{2} F_{1} (2 α + 1, α + 1; α + 3; - 1) .

(29)

Therefore, comparing (27) with (29) and making use of Formula (20), we derive

\begin{matrix} _{2} F_{1} (2 α + 1, 2; α + 3; \frac{1}{2}) & = 2^{2 α + 1}_{2} F_{1} (2 α + 1, α + 1; α + 3; - 1) \\ = {\begin{matrix} 4 (2 ln 2 - 1), & α = 0; \\ 24 (\frac{3}{2} - 2 ln 2), & α = 1; \\ \frac{2 (α + 1) (α + 2)}{(α - 1) α} [1 + α - α B (\frac{1}{2}, α)], & α \neq 0, 1, - 3, - 4, \dots . \end{matrix} \end{matrix}

Formula (21) is thus alternatively proved. □

Remark 1.

The identities (20) and (21) were announced as a problem on the site https://mathoverflow.net/q/436124 (accessed on 8 December 2022). On the website https://mathoverflow.net/a/436154 (accessed on 8 December 2022), Henri Cohen immediately sketched out an alternative proof of the identities in (20) and (21).

Theorem 2.

The closed-form Formulas (5) and (6) are valid.

For

μ + ν \in C ∖ {0, - 1, - 2, \dots}

, we have

_{2} F_{1} (μ, λ; μ + ν; 1 - a) = \frac{_{2} F_{1} (μ, μ + ν - λ; μ + ν; 1 - \frac{1}{a})}{a^{μ}} .

(30)

Proof.

The alternative proof of (5) is as follows. Taking

z = - 1

and

c = a - b + 2

in (24) and employing (3) and (4) lead to

\begin{matrix} - 2 (a - b + 2) (a - b + 1)_{2} F_{1} (a, b; a - b + 1; - 1) \\ + 2 (a - b + 2) (a - 2 b + 2)_{2} F_{1} (a, b; a - b + 2; - 1) \\ - (2 - b) (a - 2 b + 2)_{2} F_{1} (a, b; a - b + 3; - 1) = 0, \end{matrix}

which is equivalent to

\begin{matrix} _{2} F_{1} (a, b; a - b + 3; - 1) & = \frac{2 (a - b + 2)}{2 - b}_{2} F_{1} (a, b; a - b + 2; - 1) - \frac{2 (a - b + 2) (a - b + 1)}{(2 - b) (a - 2 b + 2)}_{2} F_{1} (a, b; a - b + 1; - 1) \\ = \frac{\sqrt{π} Γ (a - b + 3)}{(b - 1) (b - 2) 2^{a - 1}} [\frac{a - b + 1}{2 Γ (\frac{a}{2} + \frac{1}{2}) Γ (\frac{a}{2} - b + 2)} - \frac{1}{Γ (\frac{a}{2}) Γ (\frac{a}{2} - b + \frac{3}{2})}] . \end{matrix}

The explicit Formula (5) is thus alternatively proved.

The alternative proof of (6) is as follows. Entry 15.5.14 in [16] (p. 388), a relation of contiguous functions, reads that

c [a + (b - c) z]_{2} F_{1} (a; b; c; z) - a c (1 - z)_{2} F_{1} (a + 1; b; c; z) + (c - a) (c - b) z_{2} F_{1} (a; b; c + 1; z) = 0 .

(31)

Letting

z = - 1

and

c = a - b + 1

in (31) and further substituting (3) and (4) into (31) give

\begin{matrix} _{2} F_{1} (a + 1; b; a - b + 1; - 1) & = \frac{2 a - 2 b + 1}{2 a}_{2} F_{1} (a; b; a - b + 1; - 1) + \frac{(b - 1) (a - 2 b + 1)}{2 a (a - b + 1)}_{2} F_{1} (a; b; a - b + 2; - 1) \\ = \frac{\sqrt{π}}{2^{a + 1} a} [a \frac{Γ (a - b + 1)}{Γ (\frac{a}{2} + \frac{1}{2}) Γ (\frac{a}{2} - b + 1)} + \frac{a - 2 b + 1}{a - b + 1} \frac{Γ (a - b + 2)}{Γ (\frac{a}{2}) Γ (\frac{a}{2} - b + \frac{3}{2})}] \\ = \frac{\sqrt{π} Γ (a - b + 1)}{2^{a + 1}} [\frac{1}{Γ (\frac{a}{2} + \frac{1}{2}) Γ (\frac{a}{2} - b + 1)} + \frac{1}{Γ (\frac{a}{2} + 1) Γ (\frac{a}{2} - b + \frac{1}{2})}] . \end{matrix}

The explicit Formula (6) is thus alternatively proved.

The equality (30) follows from comparing (26) with (28). □

Applying Equation (30), we can derive more explicit formulas of the Gauss hypergeometric function

_{2} F_{1} (a, b; c; z)

at specific arguments, as follows.

Corollary 1.

Under suitable conditions such that

_{2} F_{1} (a, b; c; z)

is defined and convergent, the Gauss hypergeometric function

_{2} F_{1} (a, b; c; z)

has the following explicit formulas:

\begin{matrix} _{2} F_{1} (a, \frac{a - b + 1}{2}; \frac{a + b + 1}{2}; - 1) & = \frac{\sqrt{π}}{2^{a}} \frac{Γ (\frac{a + b + 1}{2})}{Γ (\frac{a + 1}{2}) Γ (\frac{b + 1}{2})}, \end{matrix}

(32)

\begin{matrix} _{2} F_{1} (a, a + b - 1; b; - 1) & = \frac{\sqrt{π}}{2^{a + b - 1}} \frac{Γ (b)}{Γ (\frac{a + b}{2}) Γ (\frac{b - a - 1}{2})}, \end{matrix}

(33)

\begin{matrix} _{2} F_{1} (a, 1; a + 1; - 1) & = \frac{a}{2} [ψ (\frac{a + 1}{2}) - ψ (\frac{a}{2})], \end{matrix}

(34)

\begin{matrix} _{2} F_{1} (a, 1 - 3 a; \frac{3}{2} - 2 a; \frac{1}{4}) & = {(\frac{324}{192})}^{a} \frac{Γ (\frac{4}{3}) Γ (\frac{3}{2} - 2 a)}{Γ (\frac{3}{2}) Γ (\frac{4}{3} - 2 a)}, \end{matrix}

(35)

\begin{matrix} _{2} F_{1} (a, \frac{1 - a}{3}; \frac{2}{3} a + \frac{5}{6}; \frac{1}{8}) & = {(\frac{2}{3})}^{a} \sqrt{π} \frac{Γ (\frac{2}{3} a + \frac{5}{6})}{Γ (\frac{a}{3} + \frac{1}{2}) Γ (\frac{a}{3} + \frac{5}{6})}, \end{matrix}

(36)

and

_{2} F_{1} (a, \frac{a - b}{2} + 1; \frac{a + b}{2} + 1; - 1) = \frac{\sqrt{π}}{(a - b) 2^{a - 1}} Γ (\frac{a + b}{2} + 1) [\frac{1}{Γ (\frac{a}{2}) Γ (\frac{b + 1}{2})} - \frac{1}{Γ (\frac{a + 1}{2}) Γ (\frac{b}{2})}] .

(37)

Proof.

Taking

a = \frac{1}{2}

μ = a

λ = b

, and

ν = \frac{b - a - 1}{2}

in (30) leads to

_{2} F_{1} (a, b; \frac{a + b + 1}{2}; \frac{1}{2}) = 2^{a}_{2} F_{1} (a, \frac{a - b + 1}{2}; \frac{a + b + 1}{2}; - 1) .

Combining this with (7) produces (32).

Taking

a = \frac{1}{2}

μ = a

λ = 1 - a

, and

ν = b - a

in (30), we obtain

_{2} F_{1} (a, 1 - a; b; \frac{1}{2}) = 2^{a}_{2} F_{1} (a, a + b - 1; b; - 1) .

Comparing this with (8) yields (33).

Taking

a = \frac{1}{2}

μ = a

λ = a

, and

ν = 1

in (30) reveals

_{2} F_{1} (a, a; a + 1; \frac{1}{2}) = 2^{a}_{2} F_{1} (a, 1; a + 1; - 1) .

Comparing this with (9) yields (34).

Taking

a = \frac{4}{3}

μ = a

λ = a + \frac{1}{2}

, and

ν = \frac{3}{2} - 3 a

in (30) reveals

_{2} F_{1} (a, a + \frac{1}{2}; \frac{3}{2} - 2 a; - \frac{1}{3}) = {(\frac{3}{4})}^{a}_{2} F_{1} (a, 1 - 3 a; \frac{3}{2} - 2 a; \frac{1}{4}) .

Comparing this with (10) yields (35).

Taking

a = \frac{8}{9}

μ = a

λ = a + \frac{1}{2}

, and

ν = \frac{5}{6} - \frac{a}{3}

in (30) reveals

_{2} F_{1} (a, a + \frac{1}{2}; \frac{2}{3} a + \frac{5}{6}; \frac{1}{9}) = {(\frac{9}{8})}^{a}_{2} F_{1} (a, \frac{1 - a}{3}; \frac{2}{3} a + \frac{5}{6}; \frac{1}{8}) .

Comparing this with (11) yields (36).

Taking

a = \frac{1}{2}

μ = a

λ = b

, and

ν = \frac{b}{2} - \frac{a}{2} + 1

in (30) shows

_{2} F_{1} (a, b; \frac{a + b}{2} + 1; \frac{1}{2}) = 2^{a}_{2} F_{1} (a, \frac{a - b}{2} + 1; \frac{a + b}{2} + 1; - 1) .

Comparing this with (12) yields (37). The proof of Corollary 1 is complete. □

4. A New Closed-Form Formula

In this section, we start off to derive a closed-form formula for the specific Gauss hypergeometric function in (18). This result generalizes the closed-form Formula (16), which was established by Qi in [24] (Corollary 4.1).

Theorem 3.

For

n \in N_{0}

, we have

_{2} F_{1} (n + \frac{1}{2}, n + 1; n + \frac{3}{2}; - z^{2}) = P_{n} (z^{2}) \frac{arctan z}{z} + Q_{n} (z^{2}) \frac{1}{1 + z^{2}},

(38)

where

P_{n} (z) = \frac{(2 n + 1)!!}{(2 n)!!} \frac{1}{z^{n}}, n \in N_{0}

(39)

and

Q_{n} (z) = - \frac{P_{n} (z)}{{(1 + z)}^{n - 1}} \sum_{k = 0}^{n - 1} [\sum_{j = 0}^{k} \frac{{(- 1)}^{j}}{2 j + 1} (\binom{n}{k - j})] z^{k}, n \in N_{0} .

(40)

Proof.

In [4] (p. 109, Example 5.1), it is given that

_{2} F_{1} (\frac{1}{2}, 1; \frac{3}{2}; - z^{2}) = \frac{arctan z}{z} .

(41)

By virtue of Abel’s limit theorem in [26] (p. 245, Theorem 9.31), we can take

z = 1

in (41) and obtain

_{2} F_{1} (\frac{1}{2}, 1; \frac{3}{2}; - 1) = \frac{π}{4} .

This can also be derived from (14) by taking

a = \frac{1}{2}

b = 1

, and

j = 1

In [5] (p. 556, Entry 15.1.8), the formula

_{2} F_{1} (a, b; b; z) = \frac{1}{{(1 - z)}^{a}}

(42)

is formulated. Taking

a = 1

and

b = \frac{3}{2}

in (42) leads to

_{2} F_{1} (1, \frac{3}{2}; \frac{3}{2}; z) =_{2} F_{1} (\frac{3}{2}, 1; \frac{3}{2}; z) = \frac{1}{1 - z}

(43)

and, by virtue of Abel’s limit theorem in [26] (p. 245, Theorem 9.31),

_{2} F_{1} (\frac{3}{2}, 1; \frac{3}{2}; - 1) = \frac{1}{2} .

This can also be derived from (14) by taking

a = \frac{3}{2}

b = 1

, and

j = 0

Theorem 1.1 in the paper [27] reads that, for any integers k, ℓ, and m, there are unique functions

P_{k, ℓ, m} (a, b; c; z)

and

Q_{k, ℓ, m} (a, b; c; z)

, rational in the parameters a, b, c, and z, with

P_{0, 0, 0} (a, b; c; z) = Q_{1, 0, 0} (a, b; c; z) = 1 and P_{1, 0, 0} (a, b; c; z) = Q_{0, 0, 0} (a, b; c; z) = 0,

such that

_{2} F_{1} (a + k, b + ℓ; c + m; z) = P_{k, ℓ, m} (a, b; c; z)_{2} F_{1} (a, b; c; z) + Q_{k, ℓ, m} (a, b; c; z)_{2} F_{1} (a + 1, b; c; z) .

(44)

In particular, letting

k = ℓ = m = n \in N_{0}

, setting

(a, b, c) = (\frac{1}{2}, 1, \frac{3}{2})

, replacing z by

- z^{2}

in (44), making use of Formula (41), and replacing z by

- z^{2}

in (43) all yield the following:

\begin{matrix} _{2} F_{1} (n + \frac{1}{2}, n + 1; n + \frac{3}{2}; - z^{2}) & = P_{n} (z^{2})_{2} F_{1} (\frac{1}{2}, 1; \frac{3}{2}; - z^{2}) + Q_{n} (z^{2})_{2} F_{1} (\frac{3}{2}, 1; \frac{3}{2}; - z^{2}) \\ = P_{n} (z^{2}) \frac{arctan z}{z} + Q_{n} (z^{2}) \frac{1}{1 + z^{2}}, \end{matrix}

(45)

where

P_{n} (z^{2}) = P_{n, n, n} (\frac{1}{2}, 1; \frac{3}{2}; - z^{2}) and Q_{n} (z^{2}) = Q_{n, n, n} (\frac{1}{2}, 1; \frac{3}{2}; - z^{2})

are rational in the parameter

z^{2}

, with

P_{0} (z^{2}) = 1 and Q_{0} (z^{2}) = 0 .

(46)

From [16] (p. 388, Entry 15.5.19), we obtained

\begin{matrix} z (1 - z) (a + 1) (b + 1)_{2} F_{1} (a + 2, b + 2; c + 2; z) \\ + [c - (a + b + 1) z] (c + 1)_{2} F_{1} (a + 1, b + 1; c + 1; z) - c (c + 1)_{2} F_{1} (a, b; c; z) = 0 . \end{matrix}

(47)

Replacing z by

- z^{2}

and letting

(a, b, c) = (n + \frac{1}{2}, n + 1, n + \frac{3}{2})

for

n \in N_{0}

in (47) produce

\begin{matrix} z^{2} (1 + z^{2}) (n + \frac{3}{2}) (n + 2)_{2} F_{1} (n + \frac{5}{2}, n + 3; n + \frac{7}{2}; - z^{2}) \\ - [n + \frac{3}{2} + (2 n + \frac{5}{2}) z^{2}] (n + \frac{5}{2})_{2} F_{1} (n + \frac{3}{2}, n + 2; n + \frac{5}{2}; - z^{2}) \\ + (n + \frac{3}{2}) (n + \frac{5}{2})_{2} F_{1} (n + \frac{1}{2}, n + 1; n + \frac{3}{2}; - z^{2}) = 0 . \end{matrix}

(48)

In [5] (p. 556, Entry 15.1.10), the formula

_{2} F_{1} (a, \frac{1}{2} + a; \frac{3}{2}; z^{2}) = \frac{{(1 + z)}^{1 - 2 a} - {(1 - z)}^{1 - 2 a}}{2 (1 - 2 a) z}

(49)

is obtained. Setting

a = \frac{3}{2}

and replacing z by

z i

in (49) result in

_{2} F_{1} (\frac{3}{2}, 2; \frac{3}{2}; - z^{2}) = \frac{1}{4 z i} [\frac{1}{{(1 - z i)}^{2}} - \frac{1}{{(1 + z i)}^{2}}] = \frac{1}{{(1 + z^{2})}^{2}} .

(50)

In [5] (p. 558, Entry 15.2.20), the formula

c (1 - z)_{2} F_{1} (a, b; c; z) - c_{2} F_{1} (a - 1, b; c; z) + (c - b) z_{2} F_{1} (a, b; c + 1; z) = 0

(51)

is taken. Letting

(a, b, c) = (\frac{3}{2}, 2, \frac{3}{2})

, replacing z by

- z^{2}

in (51), and employing (50) reveal

\begin{matrix} _{2} F_{1} (\frac{3}{2}, 2; \frac{5}{2}; - z^{2}) & = \frac{3}{z^{2}} [_{2} F_{1} (\frac{1}{2}, 2; \frac{3}{2}; - z^{2}) - (1 + z^{2})_{2} F_{1} (\frac{3}{2}, 2; \frac{3}{2}; - z^{2})] \\ = \frac{3}{z^{2}} [_{2} F_{1} (\frac{1}{2}, 2; \frac{3}{2}; - z^{2}) - \frac{1}{1 + z^{2}}] . \end{matrix}

(52)

In [5] (p. 558, Entry 15.2.11), the formula

(c - b)_{2} F_{1} (a, b - 1; c; z) + (2 b - c - b z + a z)_{2} F_{1} (a, b; c; z) + b (z - 1)_{2} F_{1} (a, b + 1; c; z) = 0

(53)

is formulated. Letting

(a, b, c) = (\frac{1}{2}, 1, \frac{3}{2})

, we replace z by

- z^{2}

in (53) and utilize Formula (41), which lead to

\begin{matrix} _{2} F_{1} (\frac{1}{2}, 2; \frac{3}{2}; - z^{2}) & = \frac{1}{2 (z^{2} + 1)}_{2} F_{1} (\frac{1}{2}, 0; \frac{3}{2}; - z^{2}) + \frac{1}{2}_{2} F_{1} (\frac{1}{2}, 1; \frac{3}{2}; - z^{2}) \\ = \frac{1}{2} (\frac{1}{1 + z^{2}} + \frac{arctan z}{z}) . \end{matrix}

(54)

Substituting (54) into (52) and then simplifying the result yield the following:

_{2} F_{1} (\frac{3}{2}, 2; \frac{5}{2}; - z^{2}) = \frac{3}{2 z^{2}} (\frac{arctan z}{z} - \frac{1}{1 + z^{2}}) .

(55)

This means that

P_{1} (z^{2}) = \frac{3}{2 z^{2}} and Q_{1} (z^{2}) = - \frac{3}{2 z^{2}} .

(56)

Taking

n = 0

in (48), utilizing (41) and (55), and then reformulating these formulas allow us to determine

\begin{matrix} _{2} F_{1} (\frac{5}{2}, 3; \frac{7}{2}; - z^{2}) & = \frac{5}{4 z^{2} (1 + z^{2})} [\frac{3 + 5 z^{2}}{3}_{2} F_{1} (\frac{3}{2}, 2; \frac{5}{2}; - z^{2}) -_{2} F_{1} (\frac{1}{2}, 1; \frac{3}{2}; - z^{2})] \\ = \frac{15}{8 z^{4}} [\frac{arctan z}{z} - \frac{3 + 5 z^{2}}{3 {(1 + z^{2})}^{2}}] . \end{matrix}

This means that

P_{2} (z^{2}) = \frac{15}{8 z^{4}} and Q_{2} (z^{2}) = - \frac{15}{8 z^{4}} \frac{3 + 5 z^{2}}{3 (1 + z^{2})} .

(57)

Through similar arguments as those above, taking

n = 1, 2

in (48) and considering the explicit Formulas (56) and (57), we repeatedly derive

\begin{matrix} P_{3} (z^{2}) & = \frac{35}{16 z^{6}}, & Q_{3} (z^{2}) & = - \frac{35}{16 z^{6}} \frac{15 + 40 z^{2} + 33 z^{4}}{15 {(1 + z^{2})}^{2}}, \\ P_{4} (z^{2}) & = \frac{315}{128 z^{8}}, & Q_{4} (z^{2}) & = - \frac{315}{128 z^{8}} \frac{105 + 385 z^{2} + 511 z^{4} + 279 z^{6}}{105 {(1 + z^{2})}^{3}} . \end{matrix}

(58)

Based on the data acquired from (46), (56)–(58), we consider the factor in front of the constant

\frac{π}{4}

in the first term of Formula (16) in addition to being motivated by two sequences displayed on the sites listed below:

https://oeis.org/A001803 (accessed on 18 August 2023);
https://oeis.org/A025547 (accessed on 18 August 2023); and
https://oeis.org/A350670 (accessed on 18 August 2023).

Then, we guess that rational functions

P_{n} (z^{2})

and

Q_{n} (z^{2})

defined in (45) should be (39) and

Q_{n} (z^{2}) = - P_{n} (z^{2}) \frac{\sum_{k = 0}^{n - 1} c_{n, k} z^{2 k}}{c_{n, 0} {(1 + z^{2})}^{n - 1}}, n \in N_{0},

(59)

where we assume

c_{0, 0} = c_{1, 0} = 1

and an empty sum is understood to be zero. We also guess that the numbers

c_{n, k}

for

0 \leq k \leq n - 1

and

n \in N

are positive integers.

We list the first few values of the coefficients

c_{n, k}

for

0 \leq k \leq n - 1

and

1 \leq n \leq 8

in Table 1, which were announced by Qi on the website https://mathoverflow.net/q/436464/ (accessed on 27 March 2024) as a problem.

Substituting (45) into (48) and then simplifying the outcome yield

\begin{matrix} z^{2} (1 + z^{2}) (n + \frac{3}{2}) (n + 2) [P_{n + 2} (z^{2}) \frac{arctan z}{z} + Q_{n + 2} (z^{2}) \frac{1}{1 + z^{2}}] \\ - [n + \frac{3}{2} + (2 n + \frac{5}{2}) z^{2}] (n + \frac{5}{2}) [P_{n + 1} (z^{2}) \frac{arctan z}{z} + Q_{n + 1} (z^{2}) \frac{1}{1 + z^{2}}] \\ + (n + \frac{3}{2}) (n + \frac{5}{2}) [P_{n} (z^{2}) \frac{arctan z}{z} + Q_{n} (z^{2}) \frac{1}{1 + z^{2}}] = 0, \end{matrix}

for

n \in N

. This can be written as two recurrent relations

\begin{matrix} z^{2} (1 + z^{2}) (n + \frac{3}{2}) (n + 2) P_{n + 2} (z^{2}) - [n + \frac{3}{2} + (2 n + \frac{5}{2}) z^{2}] (n + \frac{5}{2}) P_{n + 1} (z^{2}) \\ + (n + \frac{3}{2}) (n + \frac{5}{2}) P_{n} (z^{2}) = 0 \end{matrix}

(60)

and

\begin{matrix} z^{2} (1 + z^{2}) (n + \frac{3}{2}) (n + 2) Q_{n + 2} (z^{2}) - [n + \frac{3}{2} + (2 n + \frac{5}{2}) z^{2}] (n + \frac{5}{2}) Q_{n + 1} (z^{2}) \\ + (n + \frac{3}{2}) (n + \frac{5}{2}) Q_{n} (z^{2}) = 0 \end{matrix}

(61)

for

n \in N

, with initial values in (46), (56)–(58).

Substituting (39) into (60) and then simplifying result in

\frac{(2 n + 5) {(2 n + 3)}^{2}}{8 (n + 1)} \frac{1 + z^{2}}{z^{2}} + (n + \frac{3}{2}) (n + \frac{5}{2}) - [n + \frac{3}{2} + (2 n + \frac{5}{2}) z^{2}] (n + \frac{5}{2}) \frac{2 n + 3}{2 (n + 1)} \frac{1}{z^{2}} = 0

for

n \in N

. This equality can be straightforwardly verified to be true. As a result, Formula (39) is valid for

n \in N_{0}

Substituting (59) into (61) and then simplifying result in

\begin{matrix} - z^{2} (1 + z^{2}) (n + \frac{3}{2}) (n + 2) \frac{(2 n + 5)!!}{(2 n + 4)!!} \frac{1}{z^{2 n + 4}} \frac{\sum_{k = 0}^{n + 1} c_{n + 2, k} z^{2 k}}{c_{n + 2, 0} {(1 + z^{2})}^{n + 1}} \\ + [n + \frac{3}{2} + (2 n + \frac{5}{2}) z^{2}] (n + \frac{5}{2}) \frac{(2 n + 3)!!}{(2 n + 2)!!} \frac{1}{z^{2 n + 2}} \frac{\sum_{k = 0}^{n} c_{n + 1, k} z^{2 k}}{c_{n + 1, 0} {(1 + z^{2})}^{n}} \\ - (n + \frac{3}{2}) (n + \frac{5}{2}) \frac{(2 n + 1)!!}{(2 n)!!} \frac{1}{z^{2 n}} \frac{\sum_{k = 0}^{n - 1} c_{n, k} z^{2 k}}{c_{n, 0} {(1 + z^{2})}^{n - 1}} = 0, \end{matrix}

that is,

(2 n + 3) \sum_{k = 0}^{n + 1} \frac{c_{n + 2, k}}{c_{n + 2, 0}} z^{2 k} - [(2 n + 3) + (4 n + 5) z^{2}] \sum_{k = 0}^{n} \frac{c_{n + 1, k}}{c_{n + 1, 0}} z^{2 k} + 2 (n + 1) z^{2} (1 + z^{2}) \sum_{k = 0}^{n - 1} \frac{c_{n, k}}{c_{n, 0}} z^{2 k} = 0

for

n \in N

. By introducing the notation

C_{n, k} = \frac{c_{n, k}}{c_{n, 0}}, 0 \leq k \leq n - 1, n \in N

(62)

and combining coefficients of the terms

z^{2 k}

for

0 \leq k \leq n + 1

, we deduce

\begin{matrix} (2 n + 3) [C_{n + 2, 0} - C_{n + 1, 0}] \\ + [(2 n + 3) (C_{n + 2, 1} - C_{n + 1, 1}) - (4 n + 5) C_{n + 1, 0} + 2 (n + 1) C_{n, 0}] z^{2} \\ + \sum_{k = 2}^{n} [(2 n + 3) (C_{n + 2, k} - C_{n + 1, k}) - (4 n + 5) C_{n + 1, k - 1} + 2 (n + 1) (C_{n, k - 1} + C_{n, k - 2})] z^{2 k} \\ + [(2 n + 3) C_{n + 2, n + 1} - (4 n + 5) C_{n + 1, n} + 2 (n + 1) C_{n, n - 1}] z^{2 (n + 1)} = 0 . \end{matrix}

Using the fact that

C_{n, 0} = 1, n \in N,

(63)

which is a direct consequence of the definition (62), and equating the coefficients of

z^{2 k}

for

0 \leq k \leq n + 1

derives

\begin{matrix} C_{n + 2, 1} - C_{n + 1, 1} = 1, \end{matrix}

(64)

\begin{matrix} (2 n + 3) (C_{n + 2, n + 1} - C_{n + 1, n}) = 2 (n + 1) (C_{n + 1, n} - C_{n, n - 1}), \end{matrix}

(65)

and

(2 n + 3) (C_{n + 2, k} - C_{n + 1, k} - C_{n + 1, k - 1}) = 2 (n + 1) (C_{n + 1, k - 1} - C_{n, k - 1} - C_{n, k - 2})

(66)

for

2 \leq k \leq n

and

n \in N

The second formula in (57) implies that

C_{2, 1} = \frac{5}{3}

. Applying this to the recurrent relation (64), we obtain

C_{n, 1} = \frac{3 n - 1}{3}, n \geq 2 .

(67)

From the initial values

C_{1, 0} = 1

and

C_{2, 1} = \frac{5}{3}

, and recurring the relation (65), we arrive at

C_{n + 2, n + 1} - C_{n + 1, n} = \frac{(2 n + 2)!!}{(2 n + 3)!!}, n \in N .

Further recurring this relation, we find

C_{n + 1, n} = \frac{2 n + 3}{2} B (\frac{1}{2}, n + 2) - 1 = \frac{(2 n + 2)!!}{(2 n + 1)!!} - 1, n \in N_{0} .

(68)

Letting

k = 2

in (66) and utilizing (63) and (67) lead to

\begin{matrix} C_{n + 2, 2} - C_{n + 1, 2} = n + \frac{2}{3} . \end{matrix}

Taking

n = 2

in (68) gives

C_{3, 2} = \frac{11}{5}

. Using this as a boundary value and recurring the above relation result in

C_{n, 2} = \frac{15 n^{2} - 25 n + 6}{30}, n \geq 3 .

(69)

Letting

k = 3

in (66) and considering (67) and (69) result in

C_{n + 2, 3} - C_{n + 1, 3} = \frac{30 n^{3} + 55 n^{2} + 7 n - 12}{30 (2 n + 3)} .

Using

C_{4, 3} = \frac{93}{35}

, which is deduced by letting

n = 4

in (68), as a boundary value to recur the above relation, demonstrates that

C_{n, 3} = \frac{35 n^{3} - 140 n^{2} + 147 n - 30}{210}, n \geq 4 .

(70)

From (66), consecutively and inductively recurring, considering (67) and (69), we conclude that

C_{n + 2, k} - C_{n + 1, k} - C_{n + 1, k - 1} = (C_{n - k + 4, 2} - C_{n - k + 3, 2} - C_{n - k + 3, 1}) \prod_{j = - 1}^{k - 4} \frac{2 (n - j)}{2 (n - j) + 1} = 0,

that is,

C_{n + 2, k} = C_{n + 1, k} + C_{n + 1, k - 1}, 1 \leq k \leq n .

(71)

Based on the explicit Formulas (67), (69) and (70), Alexander Burstein (Department of Mathematics, Howard University, USA) estimated

C_{n, k} = \sum_{j = 0}^{k} \frac{{(- 1)}^{j}}{2 j + 1} (\binom{n}{k - j}), 0 \leq k \leq n - 1, n \in N;

(72)

see Burstein’s comments on 14 December 2022 on the following site:

https://mathoverflow.net/q/436464/ask-for-a-generating-function-or-an-explicit-expression-of-a-triangle-of-positiv#comment1125097_436464

(accessed on 15 December 2022). By Pascal’s rule

(\binom{n + 2}{k}) = (\binom{n + 1}{k}) + (\binom{n + 1}{k - 1}), k, n \in Z,

(73)

it is easy to inductively verify that Burstein’s guess (72) is true. Consequently, we discover the explicit Formula (40). The proof of Theorem 3 is thus complete. □

Remark 2.

We can regard the recursive relation (71) as a generalization of Pascal’s rule (73). Both the binomial coefficients

(\binom{n}{k})

and the sequence (72) are solutions to the recursive relation (71). Are there any more solutions to the relation (71)? On 24 March 2024, on the site https://mathoverflow.net/a/467616 (accessed on 26 March 2024), Max Alekseyev (George Washington University, USA, https://home.gwu.edu/~maxal/ (accessed on 26 March 2024)) suggested to check paper [28].

Corollary 2.

For

n \in N_{0}

, we have

_{2} F_{1} (n + \frac{1}{2}, n + 1; n + \frac{3}{2}; - 1) = \frac{(2 n + 1)!!}{(2 n)!!} [\frac{π}{4} - \frac{1}{2^{n}} \sum_{j = 0}^{n - 1} \frac{{(- 1)}^{j}}{2 j + 1} \sum_{ℓ = 0}^{n - j - 1} (\binom{n}{ℓ})]

(74)

and

\sum_{j = 0}^{n - 1} \frac{{(- 1)}^{j}}{2 j + 1} \sum_{ℓ = 0}^{n - j - 1} (\binom{n}{ℓ}) = - \frac{1}{(2 n - 1)!!} \sum_{ℓ = 1}^{n} {(- 1)}^{ℓ} \frac{(2 n - ℓ)!}{(n - ℓ)!} \frac{2^{ℓ / 2}}{ℓ} sin \frac{3 ℓ π}{4} .

(75)

Proof.

The closed-form Formula (74) follows from using Abel’s limit theorem stated in [26] (p. 245, Theorem 9.31) and taking

z \to 1

in Equation (38) of Theorem 3.

The identity (75) follows from comparing the closed-form Formulas (16) and (74) and from simplification. □

Remark 3.

Since the expression (72) is an alternating sum, we cannot directly confirm the positivity of the rational sequence

C_{n, k}

from its appearance.

From the explicit Formula (72), we cannot clearly see what the closed-form formula of the sequence

c_{n, k}

for

0 \leq k \leq n - 1

and

n \in N

is, nor can we clearly see whether the numbers

c_{n, k}

for

0 \leq k \leq n - 1

and

n \in N

are positive integers.

5. The Third Problem by Wilf and Rational Approximations

The third problem posed by Herbert S. Wilf (1931–2012) on the site https://www2.math.upenn.edu/~wilf/website/UnsolvedProblems.pdf (accessed on 26 July 2021) states that, if the function

W (z)

defined in (19) has the Maclaurin power series expansion

W (z) = \sum_{n = 0}^{\infty} a_{n} z^{n},

find the first term of the asymptotic behaviour of the

a_{n}

’s.

In the conference paper [25], Ward considered this problem. We now recite the texts of the review https://mathscinet.ams.org/mathscinet-getitem?mr=2735366 (accessed on 1 August 2021) by Tian-Xiao He for the paper [25] as follows.

The coefficient

a_{n}

can be written as

a_{n} = b_{n} π - c_{n},

(76)

where

b_{n}

and

c_{n}

are non-negative rational numbers. In fact,

lim_{n \to \infty} \frac{a_{n}}{b_{n}} = 0,

and the rational numbers of the form

\frac{c_{n}}{b_{n}}

provide approximations to

π

. A complete expansion of the coefficients

a_{n}

is found by the author. It is probably the best that can be performed, given the oscillatory nature of the terms.

Wilf’s comments on the paper [25] on 13 December 2010 is quoted as follows:

“Mark Ward has found a complete expansion of these coefficients. It’s not quite an asymptotic series in the usual sense, but it is probably the best that can be done, given the oscillatory nature of the terms.”

In the preprints on the site https://arxiv.org/abs/2110.08576 (accessed on 1 May 2022), among other findings, Qi discovered Formula (16) and expanded the Wilf function

W (z)

into

W (z) = \frac{π}{4} + \sum_{n = 1}^{\infty} {(- 1)}^{n} [\sum_{k = 1}^{n} {(- 1)}^{k} S (n, k) (2 k - 1)!! (\frac{π}{4} + \frac{1}{(\binom{2 k}{k})} \sum_{ℓ = 1}^{k} {(- 1)}^{ℓ} (\binom{2 k - ℓ}{k}) \frac{2^{ℓ / 2}}{ℓ} sin \frac{3 ℓ π}{4})] \frac{z^{n}}{n!}

(77)

for

| z | < ln 2

, where the Stirling numbers of the second kind

S (n, k)

for

n \geq k \geq 0

can be analytically generated (see [29] (p. 51) and [30]) by

{(\frac{e^{z} - 1}{z})}^{k} = \sum_{n = 0}^{\infty} \frac{S (n + k, k)}{(\binom{n + k}{k})} \frac{z^{n}}{n!}, k \geq 0 .

According to the notations used in (76), the Maclaurin power series expansion (77) can be alternatively expressed as

b_{n} = \frac{1}{4} \frac{{(- 1)}^{n}}{n!} \sum_{k = 0}^{n} {(- 1)}^{k} S (n, k) (2 k - 1)!!

and

c_{n} = \frac{{(- 1)}^{n + 1}}{n!} \sum_{k = 1}^{n} {(- 1)}^{k} S (n, k) \frac{k!}{2^{k}} \sum_{ℓ = 1}^{k} {(- 1)}^{ℓ} (\binom{2 k - ℓ}{k}) \frac{2^{ℓ / 2}}{ℓ} sin \frac{3 ℓ π}{4}

(78)

for

n \in N

In [24] (Theorems 6.2 and 6.3), among other findings, Qi proved the following points:

The sequence $4 n! b_{n}$ for $n \geq 0$ is positive, increasing, and logarithmically convex;
The limits

$lim_{n \to \infty} b_{n} = \infty, lim_{n \to \infty} c_{n} = \infty, lim_{n \to \infty} \frac{c_{n}}{b_{n}} = π$

(79)

are valid.

Making use of the equality (75), we can reformulate the sequence

c_{n}

in (78) as

c_{n} = \frac{{(- 1)}^{n}}{n!} \sum_{k = 1}^{n} {(- 1)}^{k} S (n, k) \frac{(2 k - 1)!!}{2^{k}} \sum_{j = 0}^{k - 1} \frac{{(- 1)}^{j}}{2 j + 1} \sum_{ℓ = 0}^{k - j - 1} (\binom{k}{ℓ}), n \in N .

(80)

Employing (80), we can rewrite the Maclaurin power series expansion (77) as

W (z) = \frac{π}{4} + \sum_{n = 1}^{\infty} {(- 1)}^{n} [\sum_{k = 1}^{n} {(- 1)}^{k} S (n, k) (2 k - 1)!! (\frac{π}{4} - \frac{1}{2^{k}} \sum_{j = 0}^{k - 1} \frac{{(- 1)}^{j}}{2 j + 1} \sum_{ℓ = 0}^{k - j - 1} (\binom{k}{ℓ}))] \frac{z^{n}}{n!}

(81)

for

| z | < ln 2

. As a result, we derive an alternative form (81) for the Maclaurin power series expansion of the Wilf function

W (z)

defined by (19).

The third limit in (79) can be explicitly formulated as

lim_{n \to \infty} \frac{c_{n}}{b_{n}} = 4 lim_{n \to \infty} \frac{\sum_{k = 1}^{n} {(- 1)}^{k} S (n, k) (2 k - 1)!! [\frac{1}{2^{k}} \sum_{j = 0}^{k - 1} \frac{{(- 1)}^{j}}{2 j + 1} \sum_{ℓ = 0}^{k - j - 1} (\binom{k}{ℓ})]}{\sum_{k = 1}^{n} {(- 1)}^{k} S (n, k) (2 k - 1)!!} = π .

(82)

Motivated by the identity (74) and the difference in the parentheses in (81), stimulated by numerical computation, and hinted by the limit (82) and the Stolz–Cesàro theorem for calculating limits, we guess that the sequences

\frac{1}{2^{k}} \sum_{j = 0}^{k - 1} \frac{{(- 1)}^{j}}{2 j + 1} \sum_{ℓ = 0}^{k - j - 1} (\binom{k}{ℓ}) = - \frac{1}{(\binom{2 k}{k})} \sum_{ℓ = 1}^{k} {(- 1)}^{ℓ} (\binom{2 k - ℓ}{k}) \frac{2^{ℓ / 2}}{ℓ} sin \frac{3 ℓ π}{4}

(83)

are increasing in

k \in N

and tend to

\frac{π}{4}

k \to \infty

. This guess was also posted on the site https://math.stackexchange.com/q/4883527 (accessed on 19 March 2024).

Perhaps it is difficult to directly verify the above guess. However, we find out a simple proof of the above guess as follows.

Theorem 4.

The rational sequences in (83) are increasing in

k \in N

and tend to the irrational constant

\frac{π}{4}

k \to \infty

Proof.

The Euler integral representation of the Gauss hypergeometric function

_{2} F_{1}

(see [15] (p. 66, Theorem 2.2.1) and [31] (Theorem 1.1)) reads that, if

ℜ (c) > ℜ (b) > 0

, then

_{2} F_{1} (a, b; c; x) = \frac{Γ (c)}{Γ (b) Γ (c - b)} \int_{0}^{1} t^{b - 1} {(1 - t)}^{c - b - 1} {(1 - x t)}^{- a} d t

(84)

in the x plane cut along the real axis from 1 to ∞, where it is understood that

arg t = arg (1 - t) = 0

and

{(1 - x t)}^{- a}

has its principle value. Setting

(a, b; c; x) = (n + \frac{1}{2}, n + 1; n + \frac{3}{2}; - 1), n \in N_{0}

in (84) and simplifying give

_{2} F_{1} (n + \frac{1}{2}, n + 1; n + \frac{3}{2}; - 1) = \frac{(2 n + 1)!!}{2^{n + 1} n!} \int_{0}^{1} {(\frac{t}{1 + t})}^{n} \frac{1}{\sqrt{1 - t^{2}}} d t

for

n \in N_{0}

. Hence, we obtain

\frac{(2 n)!!}{(2 n + 1)!!}_{2} F_{1} (n + \frac{1}{2}, n + 1; n + \frac{3}{2}; - 1) = \frac{1}{2} \int_{0}^{1} {(\frac{t}{1 + t})}^{n} \frac{1}{\sqrt{1 - t^{2}}} d t

(85)

for

n \in N_{0}

, which is decreasing in

n \in N_{0}

and tends to 0 as

n \to \infty

. Combining the integral representation (85) with Formula (74) reveals

\frac{1}{2^{n}} \sum_{j = 0}^{n - 1} \frac{{(- 1)}^{j}}{2 j + 1} \sum_{ℓ = 0}^{n - j - 1} (\binom{n}{ℓ}) = \frac{π}{4} - \frac{1}{2} \int_{0}^{1} {(\frac{t}{1 + t})}^{n} \frac{1}{\sqrt{1 - t^{2}}} d t, n \in N_{0},

(86)

which is increasing in

n \in N_{0}

and tends to

\frac{π}{4}

. The proof of Theorem 4 is thus complete. □

6. More Remarks

In this section, we list more remarks on our main results and related ones.

Remark 4.

It is known [16] (p. 612, Entry 25.14.5) that the function

Φ (z, s, a) = \sum_{k = 0}^{\infty} \frac{z^{k}}{{(a + k)}^{s}} = \frac{1}{Γ (s)} \int_{0}^{\infty} \frac{x^{s - 1} e^{(1 - a) x}}{e^{x} - z} d x

(87)

for

ℜ (s) > 0

ℜ (a) > 0

, and

z \in C ∖ [1, \infty)

is called the Lerch transcendent. See also [1] (p. 1050, 9.556), the proof of [24] (Theorem 6.3), [32] (Lemma 6), [33] (Theorem 2), and [34] (p. 348).

Combining the formula

Φ (z, 1, v) = \frac{_{2} F_{1} (1, v; 1 + v; z)}{v}

in [1] (p. 1050, Entry 9.559) with the integral representation in (87) results in

_{2} F_{1} (1, v; 1 + v; z) = \frac{1}{1 - z} - z \int_{0}^{\infty} \frac{e^{(1 - v) x}}{{(e^{x} - z)}^{2}} d x

(88)

for

ℜ (v) > 0

and

z \in C ∖ [1, \infty)

. From (88), we can derive, for example,

\begin{matrix} _{2} F_{1} (1, \frac{1}{2}; \frac{3}{2}; z) & = \frac{arctanh \sqrt{z}}{\sqrt{z}}, & _{2} F_{1} (1, \frac{3}{2}; \frac{5}{2}; z) & = \frac{3}{z} (\frac{arctanh \sqrt{z}}{\sqrt{z}} - 1), \\ _{2} F_{1} (1, 1; 2; z) & = - \frac{ln (1 - z)}{z}, & _{2} F_{1} (1, 2; 3; z) & = - \frac{2}{z} [\frac{ln (1 - z)}{z} + 1] . \end{matrix}

The left two results can be found in [1] (p. 61), [4] (p. 109), and [6] (Section 4.2), respectively. All these four formulas can be found in [19] (p. 473, Eq. 83; p. 476, Eq. 148; p. 477, Eq. 157; p. 477, Eq. 165), respectively. Generally, we conclude the following formulas:

_{2} F_{1} (1, \frac{2 k - 1}{2}; \frac{2 k + 1}{2}; z) = \frac{2 k - 1}{z^{k - 1}} (\frac{arctanh \sqrt{z}}{\sqrt{z}} - \sum_{j = 0}^{k - 2} \frac{z^{j}}{2 j + 1})

and

_{2} F_{1} (1, k; 1 + k; z) = - \frac{k}{z^{k - 1}} [\frac{ln (1 - z)}{z} + \sum_{j = 0}^{k - 2} \frac{z^{j}}{j + 1}]

for

k \in N

, where an empty sum is understood to be zero.

From (88), it follows that

{(- 1)}^{n} \frac{d^{n}}{d v^{n}} [\frac{1}{z (1 - z)} - \frac{_{2} F_{1} (1, v; 1 + v; z)}{z}] = \int_{0}^{\infty} \frac{x^{n} e^{(1 - v) x}}{{(e^{x} - z)}^{2}} d x > 0

for

n \in N_{0}

v > 0

, and

z \in (- \infty, 1)

. This means that, for any fixed real number

z \in (- \infty, 1)

, the real function

\frac{1}{z (1 - z)} - \frac{_{2} F_{1} (1, v; 1 + v; z)}{z}

is completely monotonic with respect to the variable

v \in (0, \infty)

. For details about completely monotonic functions, please refer to the review article [35] and closely related references therein.

Remark 5.

On the site https://mathoverflow.net/q/423800 (accessed on 30 March 2023), Qi asked the question: can one find an elementary function

f (t)

such that

_{2} F_{1} (\frac{1}{2}, \frac{1}{2}; 2; t) = f (t), | t | \leq 1 ?

(89)

On the site https://mathoverflow.net/a/423802 (accessed on 6 June 2022), Gerald A. Edgar (Ohio State University, USA) answered this question as follows.

Entry 15.5.16 in [16] (p. 388), a relation of contiguous functions, states that

c_{2} F_{1} (a - 1, b; c; t) + c (t - 1)_{2} F_{1} (a, b; c; t) + (b - c) t_{2} F_{1} (a, b; c + 1; t) = 0 .

(90)

Taking

a = \frac{1}{2}

b = \frac{1}{2}

, and

c = 1

in (90) yields

_{2} F_{1} (- \frac{1}{2}, \frac{1}{2}; 1; t) + (t - 1)_{2} F_{1} (\frac{1}{2}, \frac{1}{2}; 1; t) - \frac{t}{2}_{2} F_{1} (\frac{1}{2}, \frac{1}{2}; 2; t) = 0 .

(91)

In [4] (p. 128), we can find two relations

K (t) = \frac{π}{2}_{2} F_{1} (\frac{1}{2}, \frac{1}{2}; 1; t^{2}) and E (t) = \frac{π}{2}_{2} F_{1} (- \frac{1}{2}, \frac{1}{2}; 1; t^{2})

(92)

for $| t | < 1$ between the Gauss hypergeometric function $_{2} F_{1}$ and the complete elliptic integrals of the first and second kinds $K (t)$ and $E (t)$ . Substituting two formulas in (92) into (91) gives

$\frac{2}{π} E (\sqrt{t}) + (t - 1) \frac{2}{π} K (\sqrt{t}) - \frac{t}{2}_{2} F_{1} (\frac{1}{2}, \frac{1}{2}; 2; t) = 0,$

that is,

_{2} F_{1} (\frac{1}{2}, \frac{1}{2}; 2; t) = \frac{4}{π} [(1 - \frac{1}{t}) K (\sqrt{t}) + \frac{1}{t} E (\sqrt{t})] .

(93)

Formula (93) reveals that the Gauss hypergeometric function $_{2} F_{1} (\frac{1}{2}, \frac{1}{2}; 2; t)$ for $| t | < 1$ should not be an elementary function.

The above question with its motivation and the above answer were mentioned in [6] (Section 4.2).

Remark 6.

As a continuation of the question (89) and the answer by Gerald A. Edgar on the site https://mathoverflow.net/a/423802 (accessed on 2 June 2022), Qi asked an alternative question on https://math.stackexchange.com/q/4669567 (accessed on 30 March 2023) which can be revised and quoted as follows.

Can one write out a closed-form formula for the general term of the coefficients in the Maclaurin power series expansion of the power function

${[_{2} F_{1} (\frac{1}{2}, \frac{1}{2}; 2; t)]}^{m}, m \in N ?$
In other words, is there a closed-form expression for the coefficients $C_{m, n}$ in the power series expansion

${[_{2} F_{1} (\frac{1}{2}, \frac{1}{2}; 2; t)]}^{m} = \sum_{n = 0}^{\infty} C_{m, n} \frac{t^{n}}{n!}, m \in N ?$

The intention of this question is the same one as that stated in [6] (Section 4.3).

As performed in the proof of [6] (Theorem 1), we can derive a recursive relation for the coefficients

C_{m, n}

. However, we are more interested in a possible closed-form formula for the coefficients

C_{m, n}

Remark 7.

On the site https://mathoverflow.net/q/448555 (accessed on 15 June 2023), Qi asked the following two questions:

Is the generalized hypergeometric function $_{1} F_{2} (1; a, a + \frac{1}{2}; - x^{2})$ for $a > - 1$ elementary?
For $a \geq - 1$ , how about the positivity, monotonicity, and convexity of the generalized hypergeometric function $_{1} F_{2} (1; a, a + \frac{1}{2}; - x^{2})$ in x?

These problems originated and proposed from [36] (Remark 15).

On the site https://mathoverflow.net/a/458242 (accessed on 13 November 2023), the expression

_{1} F_{2} (1; a, a + \frac{1}{2}; - x^{2}) = \frac{f (2 x i) + f (- 2 x i)}{2}

was given, where

\begin{matrix} f (t) & = 1 + \frac{t}{2 a} + \frac{t^{2}}{2 a (2 a + 1)} + \frac{t^{3}}{2 a (2 a + 1) (2 a + 2)} + \dots \\ = \frac{2 a - 1}{t^{2 a - 1}} e^{t} [Γ (2 a - 1) - Γ (2 a - 1, t)] \end{matrix}

and the incomplete gamma function

Γ (z, x)

is defined by

Γ (z, x) = \int_{x}^{\infty} e^{- t} t^{z - 1} d t

for

ℜ (z) > 0

and

x \in N_{0}

(see [37] (p. 429)).

On the site https://mathoverflow.net/a/458325 (accessed on 13 November 2023), Gerald A. Edgar (Ohio State University, USA) wrote that, when taking

a = \frac{1}{4}

, the famous software Maple presents

_{1} F_{2} (1; \frac{1}{4}, \frac{3}{4}; - x^{2}) = 1 + 2 \sqrt{π x} [cos (2 x) S (2 \sqrt{\frac{x}{π}}) - sin (2 x) C (2 \sqrt{\frac{x}{π}})],

(94)

where

S (x) = \int_{0}^{x} sin \frac{π t^{2}}{2} d t and C (x) = \int_{0}^{x} cos \frac{π t^{2}}{2} d t

are called the Fresnel integrals [5] (Section 7.3, p. 321). Because

S (x)

and

C (x)

are not elementary, he guessed that the combination (94) is also not elementary. Gerald A. Edgar also simplified and acquired

_{1} F_{2} (1; \frac{1}{4}, \frac{3}{4}; - x^{2}) = 1 + 2 \sqrt{x} \int_{- x}^{0} \frac{sin (2 r)}{\sqrt{r + x}} d r, x > 0 .

He pointed out that the proof of

S (x)

being not elementary may also work for this.

In [36] (p. 16), Qi and his coauthors obtained

\begin{matrix} _{1} F_{2} (1; n + 1, n + \frac{3}{2}; - \frac{x^{2}}{4}) & = {\begin{matrix} {(- 1)}^{n} \frac{(2 n + 1)!}{x^{2 n + 1}} [sin x - \sum_{k = 0}^{n - 1} {(- 1)}^{k} \frac{x^{2 k + 1}}{(2 k + 1)!}], & x \neq 0 \\ 1, & x = 0 \end{matrix} \\ = {SinR}_{n} (x) \end{matrix}

(95)

and

\begin{matrix} _{1} F_{2} (1; n + \frac{1}{2}, n + 1; - \frac{x^{2}}{4}) & = {\begin{matrix} {(- 1)}^{n} \frac{(2 n)!}{x^{2 n}} [cos x - \sum_{k = 0}^{n - 1} {(- 1)}^{k} \frac{x^{2 k}}{(2 k)!}], & x \neq 0 \\ 1, & x = 0 \end{matrix} \\ = {CosR}_{n} (x) \end{matrix}

(96)

for

n \in N

, where the quantities

{SinR}_{n} (x)

and

{CosR}_{n} (x)

are called the normalized tails of the Maclaurin power series expansions of sine and cosine, respectively. On the other hand, it is not difficult to show

_{1} F_{2} (1; 1, \frac{3}{2}; - \frac{x^{2}}{4}) = {\begin{matrix} \frac{4 arcsinh (\frac{x}{2})}{x \sqrt{x^{2} + 4}}, & x \neq 0 \\ 1, & x = 0 \end{matrix}

(97)

and

_{1} F_{2} (1; \frac{1}{2}, 1; - \frac{x^{2}}{4}) = \frac{2}{\sqrt{x^{2} + 4}} .

(98)

Combining (95) and (96) with (97) and (98) reveals that the generalized hypergeometric functions

_{1} F_{2} (1; n, n + \frac{1}{2}; - x^{2}) and_{1} F_{2} (1; n - \frac{1}{2}, n; - x^{2})

for

n \in N

are elementary. Equivalently, the generalized hypergeometric function

_{1} F_{2} (1; \frac{n}{2}, \frac{n + 1}{2}; - x^{2}), n \in N

has a closed-form expression, so it is elementary.

In [36] (Theorems 1 and 2) and [36] (Remarks 3 and 10), among other things, Qi and his coauthors discovered the following:

Both of the normalized tail ${SinR}_{n} (x)$ for $n \in N$ and the normalized tail ${CosR}_{n} (x)$ for $n \geq 2$ are positive and decreasing in $x \in (0, \infty)$ ;
When $n \in N$ , the normalized remainder ${SinR}_{n} (x)$ is concave on $(0, π)$ ;
When $n \geq 2$ , the normalized remainder ${SinR}_{n} (x)$ is concave on $(0, \frac{4 π}{3})$ ;
When $n \geq 3$ , the normalized remainder ${SinR}_{n} (x)$ is concave on $(0, \frac{3 π}{2})$ ;
When $n \geq 4$ , the normalized remainder ${SinR}_{n} (x)$ is concave on $(0, 2 π)$ ;
When $n \geq 2$ , the normalized remainder ${CosR}_{n} (x)$ is concave on $(0, π)$ ;
When $n \geq 3$ , the normalized remainder ${CosR}_{n} (x)$ is concave on $(0, \frac{3 π}{2})$ ;
When $n \geq 4$ , the normalized remainder ${CosR}_{n} (x)$ is concave on $(0, \frac{7 π}{4})$ ;
When $n \geq 5$ , the normalized remainder ${CosR}_{n} (x)$ is concave on $(0, 2 π)$ .

Consequently, by virtue of the relations

{SinR}_{n} (x) =_{1} F_{2} (1; n + 1, n + \frac{3}{2}; - \frac{x^{2}}{4}), n \in N

and

{CosR}_{n} (x) =_{1} F_{2} (1; n + \frac{1}{2}, n + 1; - \frac{x^{2}}{4}), n \in N,

(99)

see (95) and (96), we conclude that the generalized hypergeometric function

_{1} F_{2} (1; \frac{n + 3}{2}, \frac{n + 4}{2}; - \frac{x^{2}}{4})

for

n \in N

is positive and decreasing in

x \in (0, \infty)

, while the following occur:

The generalized hypergeometric function $_{1} F_{2} (1; n + 1, n + \frac{3}{2}; - \frac{x^{2}}{4})$ is concave on the interval

${\begin{matrix} (0, π) & for n = 1; \\ (0, \frac{4 π}{3}) & for n = 2; \\ (0, \frac{3 π}{2}) & for n = 3; \\ (0, 2 π) & for n \geq 4, \end{matrix}$
The generalized hypergeometric function $_{1} F_{2} (1; n + \frac{1}{2}, n + 1; - \frac{x^{2}}{4})$ is concave on the interval

${\begin{matrix} (0, π) & for n = 2; \\ (0, \frac{3 π}{2}) & for n = 3; \\ (0, \frac{7 π}{4}) & for n = 4; \\ (0, 2 π) & for n \geq 5 . \end{matrix}$

Summing up, the generalized hypergeometric function

_{1} F_{2} (1; \frac{n + 3}{2}, \frac{n + 4}{2}; - \frac{x^{2}}{4}), n \in N

is positive and decreasing in

x \in (0, \infty)

, while it is concave in

x \in {\begin{matrix} (0, π) & for n = 1, 2; \\ (0, \frac{4 π}{3}) & for n = 3; \\ (0, \frac{3 π}{2}) & for n = 4, 5; \\ (0, \frac{7 π}{4}) & for n = 6; \\ (0, 2 π), & n \geq 7 . \end{matrix}

Some of these observations were also posted on the site https://mathoverflow.net/a/470042 (accessed on 26 April 2024).

We can also connect the main results in [38] with the hyperbolic function in (99) as follows:

In [38] (Theorem 1), among other findings, the function

$ln {CosR}_{n} (x) = ln [_{1} F_{2} (1; n + \frac{1}{2}, n + 1; - \frac{x^{2}}{4})], n \in N$

(100)

was expanded into a Maclaurin power series at $x = 0$ .
In [38] (Theorem 2), among other findings, the function $ln {CosR}_{n} (x)$ for $n \geq 2$ in (100) was proven to be decreasing and concave on $(0, \frac{π}{2})$ . These results are weaker than the corresponding ones in [36] (Theorem 2), not only because a positive concave function must be a logarithmically concave function (but the converse is not true), but also because we consider the including relations $(0, \frac{π}{2}) \subset (0, \infty)$ and $(0, \frac{π}{2}) \subset (0, π)$ .
In [38] (Theorem 3), the function

$\frac{ln {CosR}_{2} (x)}{ln cos x} = \frac{ln [_{1} F_{2} (1; \frac{5}{2}, 3; - \frac{x^{2}}{4})]}{ln cos x}$

was proven to be decreasing on $(0, \frac{π}{2})$ .

These observations were also announced as a part of an answer on the site https://mathoverflow.net/a/470042 (accessed on 27 April 2024).

Remark 8.

From the identity (83), Henry Ricardo (Westchester Area Math Circle, Purchase, New York, USA) noticed that the identity

\sum_{j = 0}^{\infty} \frac{{(- 1)}^{j}}{2 j + 1} = \frac{π}{4}

(101)

is called the Leibniz formula for the circular constant π; see the site https://en.wikipedia.org/wiki/leibniz_formula_for_%cf%80 (accessed on 23 March 2024). It is the special case

arctan (\pm 1) = \pm \frac{π}{4}

of the power series expansion

arctan x = \sum_{j = 0}^{\infty} \frac{{(- 1)}^{j}}{2 j + 1} x^{2 j + 1}, | x | \leq 1 .

Formula (101) can also be deduced from the general formula

\sum_{k = 1}^{\infty} {(- 1)}^{k - 1} \frac{cos [(2 k - 1) x]}{2 k - 1} = {\begin{matrix} \frac{π}{4}, & - \frac{π}{2} < x < \frac{π}{2}; \\ - \frac{π}{4}, & \frac{π}{2} < x < \frac{3 π}{2}, \end{matrix}

which is taken from [1] (p. 46), by taking

x = 0

On the other hand, since

\frac{{(arctan t)}^{n}}{n!} = \sum_{k = 0}^{\infty} {(- 1)}^{k} (\prod_{m = 1}^{n - 1} \sum_{ℓ_{m} = 0}^{ℓ_{m + 1}} \frac{1}{2 ℓ_{m} + m}) \frac{t^{2 k + n}}{2 k + n}, | t | \leq 1

for

n \in N

(see [37] (Section 6.1)), the Leibniz formula (101) can be generalized as

\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{2 k + n} (\prod_{m = 1}^{n - 1} \sum_{ℓ_{m} = 0}^{ℓ_{m + 1}} \frac{1}{2 ℓ_{m} + m}) = \frac{1}{n!} {(\frac{π}{4})}^{n}, n \in N .

(102)

For example, taking

n = 2, 3, 4

in (102) leads to

\begin{matrix} \begin{matrix} \sum_{k = 0}^{\infty} {(- 1)}^{k} (\sum_{ℓ = 0}^{k} \frac{1}{2 ℓ + 1}) \frac{1}{2 k + 2} & = \frac{1}{2} - (1 + \frac{1}{3}) \frac{1}{4} + (1 + \frac{1}{3} + \frac{1}{5}) \frac{1}{6} - \dots \\ = \frac{1}{2!} {(\frac{π}{4})}^{2}, \end{matrix} \\ \begin{matrix} \sum_{k = 0}^{\infty} {(- 1)}^{k} (\sum_{ℓ_{2} = 0}^{k} \frac{1}{2 ℓ_{2} + 2} \sum_{ℓ_{1} = 0}^{ℓ_{2}} \frac{1}{2 ℓ_{1} + 1}) \frac{1}{2 k + 3} \\ = \frac{1}{2} \cdot \frac{1}{3} - [\frac{1}{2} + \frac{1}{4} (1 + \frac{1}{3})] \frac{1}{5} + [\frac{1}{2} + \frac{1}{4} (1 + \frac{1}{3}) + \frac{1}{6} (1 + \frac{1}{3} + \frac{1}{5})] \frac{1}{7} - \dots \\ = \frac{1}{3!} {(\frac{π}{4})}^{3}, \end{matrix} \end{matrix}

and

\sum_{k = 0}^{\infty} {(- 1)}^{k} (\sum_{ℓ_{3} = 0}^{k} \frac{1}{2 ℓ_{3} + 3} \sum_{ℓ_{2} = 0}^{ℓ_{3}} \frac{1}{2 ℓ_{2} + 2} \sum_{ℓ_{1} = 0}^{ℓ_{2}} \frac{1}{2 ℓ_{1} + 1}) \frac{1}{2 k + 4} = \frac{1}{4!} {(\frac{π}{4})}^{4} .

Remark 9.

Due to Theorem 4, we can regard the sequences in (83) as two increasing rational approximations of the irrational constant

\frac{π}{4}

Remark 10.

The Equation (86) can be reformulated as

\int_{0}^{1} {(\frac{t}{1 + t})}^{n} \frac{1}{\sqrt{1 - t^{2}}} d t = \frac{π}{2} - \frac{1}{2^{n - 1}} \sum_{j = 0}^{n - 1} \frac{{(- 1)}^{j}}{2 j + 1} \sum_{ℓ = 0}^{n - j - 1} (\binom{n}{ℓ})

for

n \in N_{0}

. Further utilizing the identity (83) results in

\int_{0}^{1} {(\frac{t}{1 + t})}^{n} \frac{1}{\sqrt{1 - t^{2}}} d t = \frac{π}{2} + \frac{2}{(\binom{2 n}{n})} \sum_{ℓ = 1}^{n} {(- 1)}^{ℓ} (\binom{2 n - ℓ}{n}) \frac{2^{ℓ / 2}}{ℓ} sin \frac{3 ℓ π}{4}

for

n \in N_{0}

. Generally, combining Theorem 3 with the Euler integral representation (84) reveals that

\int_{0}^{1} \frac{t^{n}}{{(1 - t)}^{1 / 2} {(1 + z^{2} t)}^{n + 1 / 2}} d t = \frac{2^{n + 1} n!}{(2 n + 1)!!} [P_{n} (z^{2}) \frac{arctan z}{z} + Q_{n} (z^{2}) \frac{1}{1 + z^{2}}]

for

n \in N_{0}

and

z^{2} \in C ∖ (- \infty, - 1)

, where the functions

P_{n} (z)

and

Q_{n} (z)

are defined by (39) and (40).

We believe that it is also difficult to directly calculate these improper integrals.

Author Contributions

Writing—original draft, Y.-W.L. and F.Q.; writing—review and editing, Y.-W.L. and F.Q. All authors contributed equally to the manuscript. All authors have read and agreed to the published version of the manuscript.

Funding

The first author, Yue-Wu Li, was supported in part by the Doctors Foundation of Hulunbuir University (grant no. 2018BS12), China.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Data sharing is not applicable to this article as no new data were created or analyzed in this study.

Acknowledgments

The authors appreciate several mathematicians for their careful reading and beneficial comments on the original version of this paper and for their recommendation of the papers [7,8,9,10]. The authors also thank Tibor K. Pogány (University of Rijeka, Croatia) for his several corrections and helpful suggestions to the original version of this paper and for thoroughly reviewing the handbook [19]. The authors are grateful to the anonymous referees for their valuable comments on and careful suggestions to the original version of this paper.

Conflicts of Interest

The authors declare no conflicts of interest.

References

Gradshteyn, I.S.; Ryzhik, I.M. Table of Integrals, Series, and Products, 7th ed.; Translated from the Russian; Zwillinger, D., Moll, V., Eds.; Elsevier: Amsterdam, The Netherlands; Academic Press: Amsterdam, The Netherlands, 2015. [Google Scholar] [CrossRef]
Erdélyi, A.; Magnus, W.; Oberhettinger, F.; Tricomi, F.G. Higher Transcendental Functions; Vol. I. Based on notes left by Harry Bateman. With a preface by Mina Rees. With a foreword by E. C. Watson. Reprint of the 1953 original; Robert E. Krieger Publishing Co., Inc.: Melbourne, FL, USA, 1981. [Google Scholar]
Whittaker, E.T.; Watson, G.N. A Course of Modern Analysis—An Introduction to the General Theory of Infinite Processes and of Analytic Functions with an Account of the Principal Transcendental Functions, 5th ed.; Moll, V.H., Ed.; Foreword by S. J. Patterson; Cambridge University Press: Cambridge, UK, 2021. [Google Scholar]
Temme, N.M. Special Functions: An Introduction to Classical Functions of Mathematical Physics; A Wiley-Interscience Publication; John Wiley & Sons, Inc.: New York, NY, USA, 1996. [Google Scholar] [CrossRef]
Abramowitz, M.; StegunI, A. (Eds.) Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables; National Bureau of Standards, Applied Mathematics Series 55; Reprint of the 1972 edition; Dover Publications, Inc.: New York, NY, USA, 1992. [Google Scholar]
Du, W.-S.; Lim, D.; Qi, F. Several recursive and closed-form formulas for some specific values of partial Bell polynomials. Adv. Theory Nonlinear Anal. Appl. 2022, 6, 528–537. [Google Scholar] [CrossRef]
Rakha, M.A.; Rathie, A.K. Generalizations of classical summation theorems for the series ₂F₁ and ₃F₂ with applications. Integral Transform. Spec. Funct. 2011, 22, 823–840. [Google Scholar] [CrossRef]
Lavoie, J.L.; Grondin, F.; Rathie, A.K. Generalizations of Watson’s theorem on the sum of a ₃F₂. Indian J. Math. 1992, 34, 23–32. [Google Scholar]
Lavoie, J.L.; Grondin, F.; Rathie, A.K. Generalizations of Whipple’s theorem on the sum of a ₃F₂. J. Comput. Appl. Math. 1996, 72, 293–300. [Google Scholar] [CrossRef]
Lavoie, J.L.; Grondin, F.; Rathie, A.K.; Arora, K. Generalizations of Dixon’s theorem on the sum of a ₃F₂. Math. Comp. 1994, 62, 267–276. [Google Scholar] [CrossRef]
Kumar, B.R.S.; Lim, D.; Rathie, A.K. A note on two new closed-form evaluations of the generalized hypergeometric function ₅F₄ with argument $\frac{1}{256}$ . J. Korean Soc. Math. Educ. Ser. B Pure Appl. Math. 2023, 30, 131–138. [Google Scholar] [CrossRef]
Kumar, B.R.S.; Lim, D.; Rathie, A.K. On several new closed-form evaluations for the generalized hypergeometric functions. Commun. Comb. Optim. 2023, 8, 737–749. [Google Scholar] [CrossRef]
Kumar, B.R.S.; Rathie, A.K.; Choi, J. Four families of summation formulas for ₄F₃(1) with applications. Axioms 2024, 13, 164. [Google Scholar] [CrossRef]
Lim, D.; Kulkarni, V.; Vyas, Y.; Rathie, A.K. On a new class of summation formulas involving generalized hypergeometric functions. Proc. Jangjeon Math. Soc. 2023, 26, 325–340. [Google Scholar]
Andrews, G.E.; Askey, R.; Roy, R. Special Functions, Encyclopedia of Mathematics and Its Applications; Cambridge University Press: Cambridge, UK, 1999; p. 71. [Google Scholar] [CrossRef]
Olver, F.W.J.; Lozier, D.W.; Boisvert, R.F.; Clark, C.W. (Eds.) NIST Handbook of Mathematical Functions; Cambridge University Press: New York, NY, USA, 2010. Available online: http://dlmf.nist.gov/ (accessed on 20 December 2022).
Rainville, E.D. Special Functions; Macmillan: New York, NY, USA, 1960. [Google Scholar]
Wang, Z.X.; Guo, D.R. Special Functions; Translated from the Chinese by Guo and X. J. Xia; World Scientific Publishing Co., Inc.: Teaneck, NJ, USA, 1989. [Google Scholar] [CrossRef]
Prudnikov, A.P.; Brychkov, Y.A.; Marichev, O.I. Integrals and Series; More Special Functions; Translated from the Russian by G. G. Gould; Gordon and Breach Science Publishers: New York, NY, USA, 1990; Volume 3. [Google Scholar]
Amdeberhan, T.; Guan, X.; Jiu, L.; Moll, V.H.; Vignat, C. A series involving Catalan numbers: Proofs and demonstrations. Elem. Math. 2016, 71, 109–121. [Google Scholar] [CrossRef]
Qi, F.; Guo, B.-N. Integral representations of the Catalan numbers and their applications. Mathematics 2017, 5, 40. [Google Scholar] [CrossRef]
Qi, F.; Zou, Q.; Guo, B.-N. The inverse of a triangular matrix and several identities of the Catalan numbers. Appl. Anal. Discrete Math. 2019, 13, 518–541. [Google Scholar] [CrossRef]
Qi, F.; Guo, B.-N. Sums of infinite power series whose coefficients involve products of the Catalan–Qi numbers. Montes Taurus J. Pure Appl. Math. 2019, 1, 1–12. [Google Scholar]
Qi, F.; Ward, M.D. Closed-form formulas and properties of coefficients in Maclaurin’s series expansion of Wilf’s function composited by inverse tangent, square root, and exponential functions. arXiv 2022, arXiv:2110.08576v2. [Google Scholar]
Ward, M.D. Asymptotic rational approximation to Pi: Solution of an “unsolved problem” posed by Herbert Wilf. In Proceedings of the 21st International Meeting on Probabilistic, Combinatorial, and Asymptotic Methods in the Analysis of Algorithms (AofA’10), Vienna, Austria, 28 June–2 July 2010; pp. 591–601. Available online: https://hal.inria.fr/hal-01185575 (accessed on 20 December 2022).
Apostol, T.M. Mathematical Analysis, 2nd ed.; Addison-Wesley Publishing Co.: Boston, MA, USA, 1974. [Google Scholar]
Vidūnas, R. Contiguous relations of hypergeometric series, Proceedings of the Sixth International Symposium on Orthogonal Polynomials, Special Functions and Their Applications (Rome, 2001). J. Comput. Appl. Math. 2003, 153, 507–519. [Google Scholar] [CrossRef]
Salas, J.; Sokal, A.D. The Graham–Knuth–Patashnik recurrence: Symmetries and continued fractions. Electron. J. Combin. 2021, 28, 18. [Google Scholar] [CrossRef] [PubMed]
Comtet, L. Advanced Combinatorics: The Art of Finite and Infinite Expansions; Revised and Enlarged Edition; D. Reidel Publishing Co.: Dordrecht, The Netherlands, 1974. [Google Scholar] [CrossRef]
Qi, F. Diagonal recurrence relations, inequalities, and monotonicity related to the Stirling numbers of the second kind. Math. Inequal. Appl. 2016, 19, 313–323. [Google Scholar] [CrossRef]
Driver, K.A.; Johnston, S.J. An integral representation of some hypergeometric functions. Electron. Trans. Numer. Anal. 2006, 25, 115–120. [Google Scholar]
Qi, F.; Guo, B.-N. A diagonal recurrence relation for the Stirling numbers of the first kind. Appl. Anal. Discrete Math. 2018, 12, 153–165. [Google Scholar] [CrossRef]
Qi, F.; Wang, J.-L.; Guo, B.-N. Notes on a family of inhomogeneous linear ordinary differential equations. Adv. Appl. Math. Sci. 2018, 17, 361–368. [Google Scholar]
Sofo, A. Integrals of polylogarithmic functions with negative argument. Acta Univ. Sapientiae Math. 2018, 10, 347–367. [Google Scholar] [CrossRef]
Qi, F.; Agarwal, R.P. Several functions originating from Fisher–Rao geometry of Dirichlet distributions and involving polygamma functions. Mathematics 2024, 12, 44. [Google Scholar] [CrossRef]
Zhang, T.; Yang, Z.-H.; Qi, F.; Du, W.-S. Some properties of normalized tails of Maclaurin power series expansions of sine and cosine. Fractal Fract. 2024, 8, 257. [Google Scholar] [CrossRef]
Guo, B.-N.; Lim, D.; Qi, F. Maclaurin’s series expansions for positive integer powers of inverse (hyperbolic) sine and tangent functions, closed-form formula of specific partial Bell polynomials, and series representation of generalized logsine function. Appl. Anal. Discret. Math. 2022, 16, 427–466. [Google Scholar] [CrossRef]
Wan, A.; Qi, F. Power series expansion, decreasing property, and concavity related to logarithm of normalized tail of power series expansion of cosine. Electron. Res. Arch. 2024, 32, 3130–3144. [Google Scholar] [CrossRef]

Table 1. The coefficients

c_{n, k}

for

0 \leq k \leq n - 1

and

1 \leq n \leq 8

Table 1. The coefficients

c_{n, k}

for

0 \leq k \leq n - 1

and

1 \leq n \leq 8

$c_{n, k}$	$n = 1$	$n = 2$	$n = 3$	$n = 4$	$n = 5$	$n = 6$	$n = 7$	$n = 8$
$k = 0$	1	3	15	105	315	3465	45,045	45,045
$k = 1$		5	40	385	1470	19,635	300,300	345,345
$k = 2$			33	511	2688	45,738	849,849	1,150,149
$k = 3$				279	2370	55,638	1,317,888	2,167,737
$k = 4$					965	36,685	1,200,199	2,518,087
$k = 5$						11,895	631,540	1,831,739
$k = 6$							169,995	801,535
$k = 7$								184,331

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Li, Y.-W.; Qi, F. A New Closed-Form Formula of the Gauss Hypergeometric Function at Specific Arguments. Axioms 2024, 13, 317. https://doi.org/10.3390/axioms13050317

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Li Y-W, Qi F. A New Closed-Form Formula of the Gauss Hypergeometric Function at Specific Arguments. Axioms. 2024; 13(5):317. https://doi.org/10.3390/axioms13050317

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A New Closed-Form Formula of the Gauss Hypergeometric Function at Specific Arguments

Abstract

1. Simple Preliminaries

2. A Brief Review

3. Alternative Proofs of Four Known Results

4. A New Closed-Form Formula

5. The Third Problem by Wilf and Rational Approximations

6. More Remarks

Author Contributions

Funding

Institutional Review Board Statement

Informed Consent Statement

Data Availability Statement

Acknowledgments

Conflicts of Interest

References

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