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Math Help - Convergense

  1. #1
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    Convergense

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    \sqrt(b), \sqrt(1+\sqrt(b)), \sqrt(1+\sqrt(1+\sqrt(b)))
    Is convergente and calculate it's value.
    How would I do something like that?
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  2. #2
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    Let I be an interval of the real line and f:I\rightarrow{I} be a

    continuous function from I onto itself. Apply it repeatedly and the process is

    kind of a feedback loop. x results in an output y=f(x).

    After a single period, the new value of x is f(x), after 2 periods it's f(f(x))=f^{2}(x), and on and on.

    The sequence: x, f(x), f^{2}(x), f^{3}(x),.... is called the orbit of x.


    Let's see. Let I be the set of non negative reals and define:

    f(x)=\sqrt{1+x}

    for every x in I. We have the iterations:

    x_{n+1}=f(x_{n}), n=1,2,3,.....


    The orbit of 1:

    1, \sqrt{b}, \sqrt{1+\sqrt{b}}, \sqrt{1+\sqrt{1+\sqrt{b}}}, ......

    This sequence is increasing since every term after the second is gotten from

    the previous one by replacing the final b with 1+\sqrt{b}

    Therefore, the orbit(bounded and

    monotonic), is convergent.

    If we let its limit be L, then:

    x_{n+1}=\sqrt{1+x_{n}}

    shows that, as n\to\infty,

    L=\sqrt{1+L}

    Therefore, L is a fixed point of f. A solution of x=f(x).

    With some basic algebra we see L=\frac{(1+\sqrt{5})}{2}


    I hope I didn't delve into too much. I just wanted to explain best I could.

    I wanted to show more about its convergence, but I'm tired of fighting with this contrary LaTex.
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  3. #3
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    I hope I didn't delve into too much. I just wanted to explain best I could.
    No it's just great, superb answer, I think.
    I haven't 'got' the solution eith, but I am working on it, thanks a lot.
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  4. #4
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    I haven't 'got' the solution eith, but I am working on it, thanks a lot.
    The solution is the "Golden Ratio", or approx. 1.61803398875.
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  5. #5
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    here is another way to prove that the sequence converges.

    we wish to show that a_n<2 this is obviously true for a_1=\sqrt{1}<2 so we have a_{n+1}=\sqrt{1+a_n}<\sqrt{1+2}=\sqrt{3}<2
    so with that induction proof we have shown that there is an upper bound. (the upper bound we are using is 2)

    then we use an argument similar to the one above to show that the sequence is increasing.
    (Then by Bolzano-Weierstrass Theorem it converges).

    in any case the limit is \frac{1+\sqrt{5}}{2}
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  6. #6
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    Have to say it again Galactus, great solution.
    I have now fully (I think) grassped it and it helped me evolve in my studies, great thing!
    And yes I noticed that it was the golden ratio, the answer over all supprised somewhat since I at first thought an exact calculation of the limit was not possible (That is something like maybe \frac{\sqrt(b)+1}{2} would be the result.)
    Over all it was an interesting assigment with an even more interesting answer.

    To you putnam120 I have to say that I don't get the soulution at all...
    It probably has something to with the fact that I have no knowledge what so ever about the Bolzano-Weierstrass Theorem.
    I would sure like to get it thought since induction is the most satisfying parth of math, I will check on it later when the B-W T is avavible to me (if it will becom avavilbe, I think so thought the I have much more material to go trought).

    And again, thank you all!
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