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Math Help - convergence

  1. #1
    Senior Member Sambit's Avatar
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    Question convergence

    Prove convergence and find the limit for the sequence \{a_n\}, where
    a_{n+1} = \frac{a_n^2+3}{2(a_n+1)} for n\geq 1, and a_1=0.
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  2. #2
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    Quote Originally Posted by Sambit View Post
    Prove convergence and find the limit for the sequence \{a_n\}, where
    a_{n+1} = \frac{a_n^2+3}{2(a_n+1)} for n\geq 1, and a_1=0.
    1) Prove that a_n\geq 1\,\,\,\forall n

    2) Now prove that a_n\leq 2\,\,\,\forall n

    3) Now prove that \forall n\,,\,\,a_n\geq a_{n+1} . Hint: \displaystyle{a_n\geq a_{n+1}=\frac{a_n^2+3}{2(a_n+1)}\Longleftrightarro  w 2a_n^2+2a_n\geq a_n^2+3 and etc. Use (1) above.

    4) Now prove the seq. converges and using arithmetic of functions prove its limit is 1.

    Tonio
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  3. #3
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    Just practicing creating diagrams...

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  4. #4
    MHF Contributor chisigma's Avatar
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    The difference equation can be written in the form...

    \displaystyle \Delta_{n} = a_{n+1} - a_{n} = \frac{1}{2}\ \frac{3 - 2\ a_{n} - a^{2}_{n}}{1+a_{n}} = f(a_{n}) (1)

    The function f(x) is represented here...



    There is only one 'attractive fixed point' in x_{0}=1 and any a_{1}> -1 will generate a sequence converging at x_{0}. If -1< a_{1} < 1 will be a_{2}>1 and the successive sequence will be monotonically decreasing...

    Kind regards

    \chi \sigma
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