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Results & Lemmas (6)
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Theorem 2.1.
Theorem 2.1. For 0 ≦α < 1 and γ ≧0, let f ∈Rq (Σ, α, γ). If am = 0 (2 ≦m ≦n −1), then |an| ≦ |1 −α + q(1 −α)| [n]q + γ [n]q [n −1]q (n ≧3).…
Theorem 2.1. For 0 ≦α < 1 and γ ≧0, let f ∈Rq (Σ, α, γ) . If am = 0 (2 ≦m ≦n −1), then |an| ≦ |1 −α + q(1 −α)| [n]q + γ [n]q [n −1]q (n ≧3). (2.3)
Corollary 2.2.
Corollary 2.2. (see [26]) Let f given by (1.1) be in the class Rα,γ Σ (0 ≦α < 1; γ ≧0). If am = 0 (2 ≦m ≦n −1), then |an| ≦ 2 (1 −α) n [1 +…
Corollary 2.2. (see [26]) Let f given by (1.1) be in the class Rα,γ Σ (0 ≦α < 1; γ ≧0). If am = 0 (2 ≦m ≦n −1), then |an| ≦ 2 (1 −α) n [1 + γ(n −1)] (n ∈N \ {1, 2}).
Theorem 2.3.
Theorem 2.3. For 0 ≦α < 1 and 0 ≦γ, let f ∈Rq (Σ, α, γ). Then |a2| ≦min |1 −α + q(1 −α)| [2]q + γ [2]q [1]q, v u u t2(1 + q) |1…
Theorem 2.3. For 0 ≦α < 1 and 0 ≦γ, let f ∈Rq (Σ, α, γ). Then |a2| ≦min |1 −α + q(1 −α)| [2]q + γ [2]q [1]q , v u u t2(1 + q) |1 −α + q(1 −α)| [2]q
Corollary 2.4.
Corollary 2.4. (see [26]) Let f given by (1.1) be in the class Rα,γ Σ (0 ≦α < 1; γ ≧0). Then a2 ≦
Corollary 2.4. (see [26]) Let f given by (1.1) be in the class Rα,γ Σ (0 ≦α < 1; γ ≧0). Then a2 ≦
Theorem 2.5.
Theorem 2.5. For 0 ≦α < 1 and 0 ≦γ, let f ∈Rq (Σ, α, γ, λ). If am = 0 (2 ≦m ≦n −1), then |an| ≦|1 −α + q(1 −α)| [λ + 1]q,n−1 [n]q + γ…
Theorem 2.5. For 0 ≦α < 1 and 0 ≦γ, let f ∈Rq (Σ, α, γ, λ). If am = 0 (2 ≦m ≦n −1), then |an| ≦|1 −α + q(1 −α)| [λ + 1]q,n−1 [n]q + γ [n]q [n −1]q [n]q! (n ≧3). (2.25)
Theorem 2.6.
Theorem 2.6. For 0 ≦α < 1 and γ ≧0, let f ∈Rq (Σ, α, γ, λ). Then |a2| ≦min |1 −α + q(1 −α)| [λ + 1]q,1 [2]q + γ [2]q [1]q [2]q!,…
Theorem 2.6. For 0 ≦α < 1 and γ ≧0, let f ∈Rq (Σ, α, γ, λ). Then |a2| ≦min |1 −α + q(1 −α)| [λ + 1]q,1 [2]q + γ [2]q [1]q [2]q! , v u u t2(1 + q) |1 −α + q(1 −α)| [λ + 1]q,2
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