2016
DOI: 10.2298/fil1610711t
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A new iteration scheme for approximating fixed points of nonexpansive mappings

Abstract: In this paper, we introduce a new three-step iteration scheme and establish convergence results for approximation of fixed points of nonexpansive mappings in the framework of Banach space. Further, we show that the new iteration process is faster than a number of existing iteration processes. To support the claim, we consider a numerical example and approximated the fixed point numerically by computer using Matlab.

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Cited by 88 publications
(69 citation statements)
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“…x n+1 = (1 − α n )Ty n + α n Tz n y n = (1 − β n )Tx n + β n Tz n z n = (1 − γ n )x n + γ n Tx n , n ≥ 0, where {α n }, {β n }, {γ n } are real number sequences in (0, 1). In the sequel, we will consider the following iterative process defined by Thakur et al in [7] for numerical reckoning fixed points of nonexpansive mappings; see, also [8]: for an arbitrary chosen element x 0 ∈ C, the sequence {x n } is generated by…”
Section: Introductionmentioning
confidence: 99%
“…x n+1 = (1 − α n )Ty n + α n Tz n y n = (1 − β n )Tx n + β n Tz n z n = (1 − γ n )x n + γ n Tx n , n ≥ 0, where {α n }, {β n }, {γ n } are real number sequences in (0, 1). In the sequel, we will consider the following iterative process defined by Thakur et al in [7] for numerical reckoning fixed points of nonexpansive mappings; see, also [8]: for an arbitrary chosen element x 0 ∈ C, the sequence {x n } is generated by…”
Section: Introductionmentioning
confidence: 99%
“…This was in part due to the fact that Picard's iterative sequence for nonexpansive mappings does not necessarily converge. For more recently introduced iterative schemes, one can see Noor [8], Agrawal et al [9], Abbas and Nazir, [10], Sintunavarat and Pitea [11], Thakur et al [12][13][14], etc.…”
Section: Introductionmentioning
confidence: 99%
“…In 2016, Thakur et al introduced the iterative method for nonlinear mapping in which sequence { x n } is generated by {left leftarrayxn+1=(1αn)Tzn+αnTynarrayyn=(1βn)zn+βnTznarrayzn=(1γn)xn+γnTxn …”
Section: Introductionmentioning
confidence: 99%
“…x nþ1 ¼ ð1 − α n ÞTx n þ α n Ty n y n ¼ ð1 − β n Þx n þ β n Tx n (6) ∀n ≥ 0, where the sequences {α n } and {β n } are in (0,1). They proved some convergence theorems and show that the independent and rate of convergence of method (6) is better than methods (3) and (4), respectively.…”
mentioning
confidence: 99%
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