Results & Lemmas (11)
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Theorem 1.2.
Theorem 1.2. For a, b ∈(0, 1), c = a + b and C = const. > 1, with c ≤1, define the functions f and g on (0, 1) by f(r) = (1 + √r)F(a, b; c;…
Theorem 1.2. For a, b ∈(0, 1), c = a + b and C = const. > 1, with c ≤1, define the functions f and g on (0, 1) by f(r) = (1 + √r)F(a, b; c; r) −F(a, b; c; 4√r/(1 + √r)2), g(r) = CF(a, b; c; r) −F(a, b; c; 4√r/(1 + √r)2), respectively. Then we have: (1) For any a, b ∈(0, 1) with c ≤1 (c < 1, respectively), f is (strictly, respectively) increasing from (0, 1) onto (0, (R−log 16)/B), where B = B(a, b) and R = R(a, b) are defined by (1.4) and (1.6), respectively. In particular, for a, b ∈(0, 1) with c
Theorem 1.4.
Theorem 1.4. For a, b ∈(0, ∞), c = a + b, let a1 = 1 −ab, a2 = 2ab −a −b, a3 = |a1| + |a2|, and define the function f on (0, 1) by f(r) =…
Theorem 1.4. For a, b ∈(0, ∞), c = a + b, let a1 = 1 −ab, a2 = 2ab −a −b, a3 = |a1| + |a2|, and define the function f on (0, 1) by f(r) = BF(a, b; c; r) + 1 r log(1 −r), where B = B(a, b). Then we have: https://doi.org/10.1017/S0027763000025290 Published online by Cambridge University Press
Theorem 1.25
Theorem 1.25] to prove that the function f(r) ≡[K(r) −log(4/r′)][(r′/r)2 log(1/r′)]−1 is strictly decreasing from (0, 1) onto (1/4,π −log…
Theorem 1.25] to prove that the function f(r) ≡[K(r) −log(4/r′)][(r′/r)2 log(1/r′)]−1 is strictly decreasing from (0, 1) onto (1/4,π −log 16), (see Corollary 2.1). Here and in the sequel, we let r′ = √ 1 −r2 for r ∈(0, 1). Consider the function g(r) ≡B(a, b)F(a, b; a + b; r) + log(1 −r) −R(a, b) [(1 −r)/r] log(1/(1 −r)) . Then for a = b = 1/2 we have g(r2) = f(r). Therefore, it is natural to ask if the function g(r) is monotone on (0, 1) for a, b ∈(0, ∞). Our next result answers this question. h
Theorem 1.5.
Theorem 1.5. For a, b ∈(0, ∞), let A1 = A1(a, b) = a + b + ab −3, A2 = A2(a, b) = a + b −3ab + 1, A = |A1| + |A2|, and define the function f…
Theorem 1.5. For a, b ∈(0, ∞), let A1 = A1(a, b) = a + b + ab −3, A2 = A2(a, b) = a + b −3ab + 1, A = |A1| + |A2|, and define the function f on (0, 1) by f(r) = BF(a, b; a + b; r) + log(1 −r) −R [(1 −r)/r] log[1/(1 −r)] , where B = B(a, b) and R = R(a, b) are defined by (1.4) and (1.6), respec- tively. Then we have the following conclusions: (1) If A = 0, then f(r) ≡1. (2) If A ̸= 0 and A1 ≤min{0, A2}, then f is strictly decreasing from (0, 1) onto (ab, B −R). In particular, with this condition, f
Theorem 1.5
Theorem 1.5 gives twosided estimates for the O-term in Ramanujan’s asymptotic formula (1.4). §2. Proofs of Theorems In this section, we…
Theorem 1.5 gives twosided estimates for the O-term in Ramanujan’s asymptotic formula (1.4). §2. Proofs of Theorems In this section, we prove our main theorems stated in Section 1. First of all, let us recall the following formulas, which will later be frequently used, see 6.3.2, 6.3.5, 15.3.10 and 15.3.12 in [AS]: For a, b ∈(0, ∞) and r ∈(0, 1), B(a, b)F(a, b; a+b; r) = ∞ X n=0 (a, n)(b, n) (n!)2 [Rn(a, b)−log(1−r)](1−r)n, (2.1) where
Theorem 1.25
Theorem 1.25]. The double inequality (1.12) is clear. For part (3), we also investigate two cases. Case (3) (i). A1 ≥A2 ≥0 with A ̸= 0. If…
Theorem 1.25]. The double inequality (1.12) is clear. For part (3), we also investigate two cases. Case (3) (i). A1 ≥A2 ≥0 with A ̸= 0. If A2 = 0, then A1 > 0 so that an > 0 for all n ≥1. If A2 > 0, then an ≥A2(n −1) ≥0 for all n ≥1, and an = 0 if and only if A1 = A2 and n = 1. Case (3) (ii). A1 ≥0 ≥A2 with A ̸= 0. If A1 = 0, then A2 < 0 and an = −A2 > 0 for all n ≥0. If A1 > 0, then it is clear that an > 0 for all n ≥1. Consequently, under the condition of part (3), f ′ 5 is strictly increasing
Corollary 2.1.
Corollary 2.1. The function f(r) ≡K(r) −log(4/r′) (r′/r)2 log(1/r′) is strictly decreasing from (0, 1) onto (1/4,π −log 16). We shall next…
Corollary 2.1. The function f(r) ≡K(r) −log(4/r′) (r′/r)2 log(1/r′) is strictly decreasing from (0, 1) onto (1/4,π −log 16). We shall next compare Corollary 2.1 to some earlier bounds for the function K(r). For this purpose we need the following proposition.
Proposition 2.2.
Proposition 2.2. The function g(r) ≡(log(1/r′))/(r2 log(4/r′)) is strictly increasing and convex from (0, 1) onto (1/ log 16, 1). In…
Proposition 2.2. The function g(r) ≡(log(1/r′))/(r2 log(4/r′)) is strictly increasing and convex from (0, 1) onto (1/ log 16, 1). In particular, for all r ∈(0, 1), 1/ log 16 < g(r) < (1/ log 16) + [1 −(1/ log 16)]r. https://doi.org/10.1017/S0027763000025290 Published online by Cambridge University Press
Theorem 1.4
Theorem 1.4 be clearer by detailed computation. In fact, if we let D4 = (a, b) | a, b ∈(0, ∞), ab ≤1, (a, b) ̸= (1, 1), D5 = (a, b) | a, b…
Theorem 1.4 be clearer by detailed computation. In fact, if we let D4 = {(a, b) | a, b ∈(0, ∞), ab ≤1, (a, b) ̸= (1, 1)}, D5 = {(a, b) | a, b ∈(0, ∞), 0 < (a + 1)/(3a −1) ≤b, (a, b) ̸= (1, 1)} https://doi.org/10.1017/S0027763000025290 Published online by Cambridge University Press
Corollary 2.4.
Corollary 2.4. Let D1 and D2 be as in Remark 2.1, D4 and D5 as in Remark 2.2, B = B(a, b) and R = R(a, b). Then for all r ∈(0, 1), max …
Corollary 2.4. Let D1 and D2 be as in Remark 2.1, D4 and D5 as in Remark 2.2, B = B(a, b) and R = R(a, b). Then for all r ∈(0, 1), max ab1 −r r log 1 1 −r, B −R −1 + 1 −r r log 1 1 −r
Theorem 2.5.
Theorem 2.5. For a, b, c ∈(0, ∞) with c < a + b, let d = a + b −c, and define the function f on (0, 1) by f(r) = (1 −r)dF(a, b; c; r). Then…
Theorem 2.5. For a, b, c ∈(0, ∞) with c < a + b, let d = a + b −c, and define the function f on (0, 1) by f(r) = (1 −r)dF(a, b; c; r). Then f ′(r) = 1 c (c −a)(c −b)(1 −r)d−1F(a, b; c + 1; r), (2.41) and we have: https://doi.org/10.1017/S0027763000025290 Published online by Cambridge University Press