Prove that the sequence (xn)n≥1 is (δ-)contractive for some δ ∈ (0, 1)
Question
Prove that the sequence (xn)n≥1 is (δ-)contractive for some δ ∈ (0, 1)
Solution
To prove that a sequence (xn)n≥1 is (δ-)contractive for some δ ∈ (0, 1), we need to show that there exists a δ such that for all n, the distance between two successive terms of the sequence is less than δ times the distance between the previous terms.
Here are the steps to prove it:
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Assume that (xn)n≥1 is a sequence in a metric space (X, d).
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We say that the sequence is δ-contractive if there exists a δ ∈ (0, 1) such that for all n ≥ 1, we have d(xn+1, xn) ≤ δ * d(xn, xn-1).
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To prove this, we need to show that such a δ exists. This usually involves manipulating the definition of the sequence or using properties of the metric space.
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For example, if the sequence is defined by a recurrence relation xn+1 = f(xn) where f is a contraction mapping, then we can use the contraction mapping theorem to show that such a δ exists.
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If the sequence is defined in some other way, we may need to use other properties of the sequence or the metric space. For example, if the sequence is monotone and bounded, then it is convergent, and we can use the properties of convergent sequences to find such a δ.
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Once we have found such a δ, we can conclude that the sequence is δ-contractive.
Remember, the exact details of the proof will depend on the specific sequence and metric space in question.
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