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Results & Lemmas (7)

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Lemma 1.1 Lemma 1.1 [12] If the function h ∈P is given by the series (1.2), then |hk| ≤2 (k ∈N) (1.8) and h2 – h2 1 2  ≤2 – |h2|2 2. (1.9)
Lemma 1.1 [12] If the function h ∈P is given by the series (1.2), then |hk| ≤2 (k ∈N) (1.8) and h2 – h2 1 2  ≤2 – |h2|2 2 . (1.9)
Lemma 1.2 Lemma 1.2 [14] If the function h ∈P is given by the series (1.2), then 2h2 = h2 1 + x  4 – h2 1  (1.10) and 4h3 = h3 1 + 2  4 – h2 1 
Lemma 1.2 [14] If the function h ∈P is given by the series (1.2), then 2h2 = h2 1 + x  4 – h2 1  (1.10) and 4h3 = h3 1 + 2  4 – h2 1 
Theorem 2.1 Theorem 2.1 Let f ∈m(λ,γ;β) be given by (1.5). Then, am+1a3m+1 – a2 2m+1  ≤ ⎧ ⎨ ⎩ 4(1–β)2 (γ +1)2(mλ+1)[ (m+1)2(1–β)2 (γ +1)2(mλ+1)3…
Theorem 2.1 Let f ∈m(λ,γ ;β) be given by (1.5). Then, am+1a3m+1 – a2 2m+1  ≤ ⎧ ⎨ ⎩ 4(1–β)2 (γ +1)2(mλ+1)[ (m+1)2(1–β)2 (γ +1)2(mλ+1)3 + 6 (γ +2)(γ +3)(3mλ+1)], β ∈[0,τ], 4(1–β)2
Corollary 2.1 Corollary 2.1 [3] Let f ∈m(β) (0 ≤β < 1) be given by (1.5). Then, am+1a3m+1 – a2 2m+1  ≤ ⎧ ⎨ ⎩ 4(1–β)2 m+1 [ (1–β)2 m+1 + 1 3m+1], β…
Corollary 2.1 [3] Let f ∈m(β) (0 ≤β < 1) be given by (1.5). Then, am+1a3m+1 – a2 2m+1  ≤ ⎧ ⎨ ⎩ 4(1–β)2 m+1 [ (1–β)2 m+1 + 1 3m+1], β ∈[0,v], (1–β)2 (2m+1)2 [4 –
Corollary 2.2 Corollary 2.2 Let f ∈(λ,γ;β) (λ ≥1,γ ∈N0,0 ≤β < 1) be given by (1.1). Then, a2a4 – a2 3  ≤ ⎧ ⎨ ⎩ 8(1–β)2 (γ +1)2(λ+1)[ 2(1–β)2 (γ…
Corollary 2.2 Let f ∈(λ,γ ;β) (λ ≥1,γ ∈N0,0 ≤β < 1) be given by (1.1). Then, a2a4 – a2 3  ≤ ⎧ ⎨ ⎩ 8(1–β)2 (γ +1)2(λ+1)[ 2(1–β)2 (γ +1)2(λ+1)3 + 3 (γ +2)(γ +3)(3λ+1)], β ∈[0,ξ], 4(1–β)2
Corollary 2.3 Corollary 2.3 Let f ∈(λ;β) (λ ≥1, 0 ≤β < 1) be given by (1.1). Then, a2a4 – a2 3  ≤ ⎧ ⎨ ⎩ 8(1–β)2 λ+1 [ 2(1–β)2 (λ+1)3 + 1 2(3λ+1)],…
Corollary 2.3 Let f ∈(λ;β) (λ ≥1, 0 ≤β < 1) be given by (1.1). Then, a2a4 – a2 3  ≤ ⎧ ⎨ ⎩ 8(1–β)2 λ+1 [ 2(1–β)2 (λ+1)3 + 1 2(3λ+1)], β ∈[0,ϵ], (1–β)2 (2λ+1)2 [4 –
Corollary 2.4 Corollary 2.4 [10] Let f ∈(β) (0 ≤β < 1) be given by (1.1). Then, a2a4 – a2 3  ≤ ⎧ ⎨ ⎩ (1 – β)2[(1 – β)2 + 1 2], β ∈[0, 11– √ 37 12…
Corollary 2.4 [10] Let f ∈(β) (0 ≤β < 1) be given by (1.1). Then, a2a4 – a2 3  ≤ ⎧ ⎨ ⎩ (1 – β)2[(1 – β)2 + 1 2], β ∈[0, 11– √ 37 12 ], (1–β)2
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