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Math Help - analysis

  1. #1
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    analysis

    Consider the sequence
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  2. #2
    ynj
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    Quote Originally Posted by flower3 View Post
    Consider the sequence
    x_{n+1}=\frac{1}{4}(1+x_n)\Rightarrow x_{n+1}-\frac{1}{3}=\frac{1}{4}(x_n-\frac{1}{3})
    So x_n-\frac{1}{3}=\frac{1}{4^{n-1}}(x_1-\frac{1}{3})=\frac{1}{4^{n-1}}(\alpha-\frac{1}{3})
    So x_n=\frac{1}{3}+\frac{1}{4^{n-1}}(\alpha-\frac{1}{3})
    so \{x_n\}converges for every \alpha
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