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Math Help - Find the limit of a recursion formula

  1. #1
    Member pberardi's Avatar
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    Find the limit of a recursion formula

    Hello,
    I am looking at this recursion formula in my Calculus book.
    x_n+1 = 1/2(x_n + 2/x_n)
    It goes on to say that we can find the limit of this sequence by doing this:
    L = 1/2(L + 2/L) fine
    but now it says that this can be rewritten as L^2 = 2 and that the limit is the square root of 2. How?

    Could someone please explain the last two steps? Thank you so much.
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  2. #2
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    Hello, pberardi!

    It's only algebra . . .


    I am looking at this recursion formula in my Calculus book:
    . . x_{n+1} \:=\: \frac{1}{2}\left(x_n + \frac{2}{x_n}\right)

    It goes on to say that we can find the limit of this sequence by doing this:
    . . L \:=\: \frac{1}{2}\left(L + \frac{2}{L}\right) . . . fine

    but now it says that this can be rewritten as L^2 = 2
    and that the limit is \sqrt{2}. .How?

    We have: . L \:=\:\frac{L}{2} + \frac{1}{L}

    Multiply by 2L\!:\quad 2L^2 \:=\:L^2 + 2 \quad\Rightarrow\quad L^2 \:=\:2 \quad\Rightarrow\quad L \:=\:\sqrt{2}

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  3. #3
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    Quote Originally Posted by pberardi View Post
    Hello,
    I am looking at this recursion formula in my Calculus book.
    x_n+1 = 1/2(x_n + 2/x_n)
    It goes on to say that we can find the limit of this sequence by doing this:
    L = 1/2(L + 2/L) fine
    but now it says that this can be rewritten as L^2 = 2 and that the limit is the square root of 2. How?

    Could someone please explain the last two steps? Thank you so much.
    Multiply both sides of that equation by 2L to get rid of the fractions- 2L^2= L^2+ 2 which is the same as L^2= 2. That actually has two roots. In order to determine whether the limit is \sqrt{2} or -\sqrt{2} for a specific sequence you must also be given a "starting" value of x_0. Are you given something like " x_0= 1".

    In fact, it should be clear that if x_n is positive, then x_{n+1}= (1/2)(x_n+ 1/x_n) is also positive while if x_n is negative, then x_{n+1}= (1/2)(x_n+ 1/x_n) is also negative.
    That is, if x_0 is positive, the limit is \sqrt{2} and if x_0 is negative, the limit is -\sqrt{2}.

    By the way, they get that original formula by arguing that if \lim_{x\to \infty} x_n exists then you can call that limit L and use the laws of limits to argue that lim x_{n+1}= (1/2)(lim x_n+ 1/lim x_n) and because " x_{n+1}" is just the same sequence renumbered, each of those limits is L: L= (1/2)(L+ 1/L).

    To be rigorous, you would still have to show the limit exists.

    (Blast! Soroban got in ahead of me again!)
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  4. #4
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    Cool, I'm doing a variation of this question for uni. It's starting to come together for me now. The actual question is:
    x_{n+1} = \frac{1}{2}(x_n + \frac{\alpha}{x_n})
    and priors are \alpha > 0 and x_1 >\sqrt{\alpha}

    My main question is how does the substitution for L work? You substitute L for x_(n+1) and for x_n? I guess that requires the assumption that the sequence is converging, right? If not, why not? If so, how do you prove monotonic decrease and find the limit, given the series is recursive?
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