Results & Lemmas (10)
Each result is auto-extracted from the paper via OCR - the math is shown verbatim and may carry occasional transcription artifacts. Click to expand the full statement.
Lemma 2.1.
Lemma 2.1. [40] Let p ∈P be given by (1.7), then pk
Lemma 2.1. [40] Let p ∈P be given by (1.7), then pk
Lemma 2.2.
Lemma 2.2. [19,39] Let p ∈P be given by (1.7), for complex number µ, we have cn+k −µcnck ≤2 max n 1, 2µ −1
Lemma 2.2. [19,39] Let p ∈P be given by (1.7), for complex number µ, we have cn+k −µcnck ≤2 max n 1, 2µ −1
Lemma 2.3.
Lemma 2.3. [40] Let p ∈P be given by (1.7), then ξp3 1 −ϖp1p2 + σp3 ≤2 (|ξ| + |ϖ −2ξ| + |ξ −ϖ + σ|) (2.7)
Lemma 2.3. [40] Let p ∈P be given by (1.7), then ξp3 1 −ϖp1p2 + σp3 ≤2 (|ξ| + |ϖ −2ξ| + |ξ −ϖ + σ|) (2.7)
Lemma 2.4.
Lemma 2.4. [41] Let α, β, r and b satisfy the inequalities 0 < α < 1, 0 < b < 1 and 8b(1 −b) h (αβ −2r)2 + (α(b + α) −β)2i + α(1 −α)(β…
Lemma 2.4. [41] Let α, β, r and b satisfy the inequalities 0 < α < 1 , 0 < b < 1 and 8b(1 −b) h (αβ −2r)2 + (α(b + α) −β)2i + α(1 −α)(β −2bα)2 ≤4α2(1 −α)2b(1 −b) . (2.8) If p ∈P, be given by (1.7), then rp4 1 + bp2 2 + 2αp1p3 −3 2βp2 1p2 −p4 ≤2. (2.9)
Lemma 2.5.
Lemma 2.5. [42] If p ∈P be given by (1.7), then p5 1 + 3p1p2 2 + 3p2 1p3 −4p3 1p2 −2p1p4 −2p2p3 + p5 ≤2. (2.10) p6 1 + 6p2 1p2 2 + 4p3 1p3…
Lemma 2.5. [42] If p ∈P be given by (1.7), then p5 1 + 3p1p2 2 + 3p2 1p3 −4p3 1p2 −2p1p4 −2p2p3 + p5 ≤2. (2.10) p6 1 + 6p2 1p2 2 + 4p3 1p3 + 2p1p5 + 2p2p4 + p2 3 −p3 2 −5p4
Lemma 2.6.
Lemma 2.6. [43] Let p ∈P be given by (1.7), then for some complex valued x with |x| ≤1, some complex valued ϱ with |ϱ| ≤1 and some complex…
Lemma 2.6. [43] Let p ∈P be given by (1.7), then for some complex valued x with |x| ≤1, some complex valued ϱ with |ϱ| ≤1 and some complex valued ψ with |ψ| ≤1, we have 2p2 = p2 1 + x 4 −p2 1 , (2.12) 4p3 = p3
Theorem 3.1.
Theorem 3.1. If f ∈A(δ); (0 ≤δ ≤1), then we have the sharp bounds
Theorem 3.1. If f ∈A(δ); (0 ≤δ ≤1), then we have the sharp bounds
Theorem 3.2.
Theorem 3.2. For any α ∈C, where C is the set of complex numbers, if we consider the function f defined by (1.4) to belong to the class…
Theorem 3.2. For any α ∈C, where C is the set of complex numbers, if we consider the function f defined by (1.4) to belong to the class A(δ); (0 ≤δ ≤1), then s1 −αs2 2 ≤ 1 2 + δ max ( 1,
Corollary 3.1.
Corollary 3.1. Let f given by (1.1), be in the class A(δ); (0 ≤δ ≤1). Then for any α ∈C, we have s1 −αs2 2 ≤1 2 max 1, |2α −1|. α = 1, we…
Corollary 3.1. Let f given by (1.1), be in the class A(δ); (0 ≤δ ≤1). Then for any α ∈C, we have s1 −αs2 2 ≤1 2 max {1, |2α −1|} . α = 1, we get the following result in form of corollary
Corollary 3.2.
Corollary 3.2. Let f given by (1.1), be in the class A(δ); (0 ≤δ ≤1). Then for any α ∈C, we have s1 −s2 2 ≤ 1 2 + δ max ( 1,
Corollary 3.2. Let f given by (1.1), be in the class A(δ); (0 ≤δ ≤1). Then for any α ∈C, we have s1 −s2 2 ≤ 1 2 + δ max ( 1,
Function classes studied:
Related Papers