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Thread: convergence

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

    Prove convergence and find the limit for the sequence $\displaystyle \{a_n\}$, where
    $\displaystyle a_{n+1} = \frac{a_n^2+3}{2(a_n+1)}$ for $\displaystyle n\geq 1$, and $\displaystyle a_1=0.$
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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 $\displaystyle \{a_n\}$, where
    $\displaystyle a_{n+1} = \frac{a_n^2+3}{2(a_n+1)}$ for $\displaystyle n\geq 1$, and $\displaystyle a_1=0.$
    1) Prove that $\displaystyle a_n\geq 1\,\,\,\forall n$

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

    3) Now prove that $\displaystyle \forall n\,,\,\,a_n\geq a_{n+1}$ . Hint: $\displaystyle \displaystyle{a_n\geq a_{n+1}=\frac{a_n^2+3}{2(a_n+1)}\Longleftrightarro w 2a_n^2+2a_n\geq a_n^2+3$ 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
    MHF Contributor
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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...

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

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



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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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