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Math Help - sequence problems

  1. #1
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    sequence problems

    #1
    Prove the sequence \{x_n\} defined in the following is increasing and bounded.
    x_n=1+\frac{x_{n-1}}{1+x_{n-1}}

    I suppose to use MI to prove and let S(n) by the statement
    x_{n+1}\geq x_{n} and 1\leq x_n<2
    but i cant prove the sequence is increasing for the case n=k+1, am i starting with wrong statement ?

    #2
    Set a_1=1 ,and for n\geq2, a_{n+1}=\frac{1}{2+a_n}.Show that the sequence \{a_n\} is convergent.

    I am confusing how to let the statement.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: sequence problems

    Quote Originally Posted by maoro View Post
    #1
    Prove the sequence \{x_n\} defined in the following is increasing and bounded...

    x_n=1+\frac{x_{n-1}}{1+x_{n-1}}
    The 'attack strategy' for this type of problems is illustrated in...

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

    The recurrence relation defining the x_{n} can be written as...

    \Delta_{n}= x_{n+1}-x_{n} = \frac{1+x_{n} - x_{n}^{2}} {1+x_{n}} = f(x_{n}) (1)

    The function f(x) has one 'attractive fixed point' in x_{+}= \frac{1+\sqrt{5}}{2} and one 'repulsive fixed point' in x_{-}= \frac{1-\sqrt{5}}{2} and because the inequality |f(x)|< |x_{+}-x| all the initial values x_{0}> x_{-} will produce a sequence converging to x_{+}. All the initial values x_{0}< x_{-} however will produce a diverging sequence...



    Marry Christmas from Serbia

    \chi \sigma

    P.S. An important detail: all the initial values x_{-}<x_{0}<x_{+} will produce an increasing sequence and all the initial values x_{0}>x_{+} will produce a decreasing sequence...
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: sequence problems

    Quote Originally Posted by maoro View Post
    #2
    Set a_1=1 ,and for n\geq2, a_{n+1}=\frac{1}{2+a_n}.Show that the sequence \{a_n\} is convergent.

    I am confusing how to let the statement.
    The 'strategy' is the same illustrated in my previous post. The difference equation generating the a_{n} can be written as...

    \Delta_{n}= a_{n+1}-a_{n}= \frac{1}{2+a_{n}}-a_{n}=f(a_{n}) (1)

    Here f(x) has two 'attractive fixed points' in x_{-}=-1-\sqrt{2} and x_{+}=-1+\sqrt{2}. The attractive point x_{-} however is 'pratically unarrivable' so that we indagate on x_{+}. For this 'attractive fixed point' the conditions of convergence are satisfied for any 'initial value' a_{0}>-2. For -2<a_{0}<x_{+} the convergence is 'oscillatory' and for a_{0}>x_{+} the convergence will be 'monotonic'...



    Marry Christmas from Serbia

    \chi \sigma
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  4. #4
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    Re: sequence problems

    sorry, but are there easier alternatives to approach them ?
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