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Thread: Recurrence relation

  1. #1
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    Recurrence relation

    A sequence of real numbers $\displaystyle {U_n}$ satisfies the recurrence relation

    $\displaystyle U_{n+1}=\frac{1}{2}U_n+1$ for all positive integer $\displaystyle n$.

    a) Given that $\displaystyle {U_n}$ converges to $\displaystyle l$, find the value of $\displaystyle l$

    b) If $\displaystyle U_n < l$, show that $\displaystyle U_n<U_{n+1}<l$






    I was able to find that $\displaystyle l=2$ for part a.

    For (b), what I have done is:

    $\displaystyle 2U_{n+1}-2=U_n$

    $\displaystyle 2U_{n+1}-2<2$

    $\displaystyle U_{n+1}<0$

    How do I continue? Or am I going in the wrong direction?
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  2. #2
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    Re: Recurrence relation

    Quote Originally Posted by Punch View Post
    $\displaystyle 2U_{n+1}-2<2$

    $\displaystyle U_{n+1}<0$
    The last line should be $\displaystyle 2U_{n+1}<4$.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: Recurrence relation

    The recurrence relation can be written as...

    $\displaystyle \Delta_{n}=u_{n+1}-u_{n}= 1-\frac{u_{n}}{2}= f(u_{n})$ (1)

    As explained in...

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

    ... has only one 'atrractive fixed point' in $\displaystyle x=2$ and because for all x is $\displaystyle |f(x)|\le |x_{0}-x|$ all initial value $\displaystyle u_{0}$ will produce a sequence monotonically convergent to 2...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  4. #4
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    Re: Recurrence relation

    Quote Originally Posted by emakarov View Post
    The last line should be $\displaystyle 2U_{n+1}<4$.
    Then $\displaystyle U_{n+1}<2$, how do I continue to show that $\displaystyle U_n<U_{n+1}<l$?
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  5. #5
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    Re: Recurrence relation

    $\displaystyle U_n<2\implies U_n+U_n<2+U_n\implies U_n<1+U_n/2=U_{n+1}$
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  6. #6
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    Re: Recurrence relation

    Quote Originally Posted by emakarov View Post
    $\displaystyle U_n<2\implies U_n+U_n<2+U_n\implies U_n<1+U_n/2=U_{n+1}$
    Thank you, how did you think of that?
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  7. #7
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    Re: Recurrence relation

    Good question. I just solved the inequality $\displaystyle U_n<U_n/2+1$ and found that it is equivalent to $\displaystyle U_n<2$, so in particular the latter implies the former.

    I have a feeling that solving two inequalities to prove $\displaystyle U_{n+1}<2$ and $\displaystyle U_n<U_{n+1}$ is repeating some work, but to understand it better I probably need to read chisigma's tutorial...
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