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Math Help - Divergence of sequence

  1. #1
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    Divergence of sequence

    Consider f: \mathbb{C} \rightarrow \mathbb{C}, f(z) = z^2 - 2.
    Show that if |z| > 2, then the sequence z, f(z), f(f(z), ... diverges, i.e. is unbounded.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by EinStone View Post
    Consider f: \mathbb{C} \rightarrow \mathbb{C}, f(z) = z^2 - 2.
    Show that if |z| > 2, then the sequence z, f(z), f(f(z), ... diverges, i.e. is unbounded.
    Isn't this solved almost immediately by the reverse triangle inequality?
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  3. #3
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    Yes good point! Now consider |z| \leq 2 and Im(z) \neq 0 and show that the sequence is unbounded, i.e. leaves the disk of radius 2.
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by EinStone View Post
    Consider f: \mathbb{C} \rightarrow \mathbb{C}, f(z) = z^2 - 2.
    Show that if |z| > 2, then the sequence z, f(z), f(f(z), ... diverges, i.e. is unbounded.
    The sequence is alternatively defined by the recursive relation...

    a_{n+1} = a^{2}_{n}-2\ , \ a_{0}=z (1)

    The procedure for solving (1) has been described a lot of times, the last in...

    http://www.mathhelpforum.com/math-he...ce-176658.html

    Following the procedure you find that...

    a) if |z|=|a_{0}|< 2 the sequence converges to -1

    b) if |z|=|a_{0}|= 2 the sequence converges to 2

    c) if |z|=|a_{0}|> 2 the sequence diverges...

    Kind regards

    \chi \sigma
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  5. #5
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    Ok, but I think this is only for real a_0, now I want to consider the case that a_0 is complex and the imaginary part is non-zero.
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