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Math Help - correct? (sequences)

  1. #1
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    correct? (sequences)

    exercise:

    A sequence is recursively defined by:
    a(0) = 5
    a(n+1) = 3 - 2/a(n)

    Show that the sequence converges and calculate the limit.

    my soution:
    Imageshack - img0001swm.jpg

    would you say it is correct?
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  2. #2
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    Re: correct? (sequences)

    Quote Originally Posted by nappysnake View Post
    exercise:
    A sequence is recursively defined by:
    a(0) = 5
    a(n+1) = 3 - 2/a(n)
    Show that the sequence converges and calculate the limit.
    my soution:
    Imageshack - img0001swm.jpg
    would you say it is correct?
    Your work is sound.
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  3. #3
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    Re: correct? (sequences)

    Great!
    Now, second exercise:
    A sequence is recursively defined by a(0) = 0 and a(n+1) = a(n)^2 + 1/4
    Show that the sequence converges and calculate the limit.

    I tried applying a similar solution to the above, but although this sequence is monotone increasing it has only a lower bound = 0.
    Is there any theorem i can apply or any help from you guys on how to show that the sequence converges, hence prove that the limit can be calculated?
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: correct? (sequences)

    Quote Originally Posted by nappysnake View Post
    Great!
    Now, second exercise:
    A sequence is recursively defined by a(0) = 0 and a(n+1) = a(n)^2 + 1/4
    Show that the sequence converges and calculate the limit.

    I tried applying a similar solution to the above, but although this sequence is monotone increasing it has only a lower bound = 0.
    Is there any theorem i can apply or any help from you guys on how to show that the sequence converges, hence prove that the limit can be calculated?
    The difference equation can be written as...

    \Delta_{n}= a_{n+1}-a_{n}= a^{2}_{n} - a_{n} + \frac{1}{4}= f(a_{n}),\ a_{0}=0 (1)

    In...

    http://www.mathhelpforum.com/math-he...-i-188482.html

    ... it has been defined the condition for which e sequence produced by (1) converges or not and the procedure for find the limit if exist. In particular the sequence a_{n} converges to an x_{0} only if is f(x_{0})=0 and in that case we call x_{0} 'fixed point'. If in x=x_{0} f(x) has positive slope, then x_{0} is a 'repulsive fixed point' and only the initial value a_{0}=x_{0} [and a very restricted set of points] produces a sequence converging to it. If in x=x_{0} f(x) has negative slope, then x_{0} is an 'attractive fixed point' and in most cases there is convergence if a_{0} belongs to a set of points around x_{0}. In Your case f(x)=x^{2}-x+\frac{1}{4} has only one zero [even if with multelicity two...] in x_{0}= \frac{1}{2}. In this case f(x) has in x_{0} 'zero slope' so that we can define x_{0}= \frac{1}{2} 'half- attracting fixed point' and that means that only initial values a_{0} \le x_{0} can produce a sequence converging at x_{0}. According to the Theorem 4.1 of the post I indicated the values of a_{0} that satisfy the condition |f(x)| \le |x_{0}-x| will produce a sequence converging at x_{0} and it is easy to verify that any value -\frac{1}{2} \le a_{0} \le \frac{1}{2} will produce a sequence converging at \frac{1}{2} and any other value of a_{0} will produce a diverging sequence...

    Kind regards

    \chi \sigma
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