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Math Help - Approximation of root 2.

  1. #1
    Senior Member slevvio's Avatar
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    Approximation of root 2.

    Hello it's me again and this is the last algebra question for a while!

    Start with  a_1 = b_1 = 1, and define

    a_{n+1} = a_n + 2b_n, \:\ b_{n+1} = a_n + b_n \:\ (n \geq\ 20)

    Work out  \frac{a_n}{b_n} for n = \{1,2,4,5,6,\}. You should find that  \frac{a_n}{b_n} is getting closer to \surd{2} as n increases.

    Prove by induction that  a_n^2 - 2b_n^2 = (-1)^n,
    and deduce that  \frac{a_n}{b_n} \rightarrow \surd{2} as  n \rightarrow \infty.

    I have no problem up to the point of proving the formula by induction but I have no clue as to how to use the formula to make the deduction. Any help would be appreciated.
    Last edited by slevvio; November 5th 2007 at 06:52 AM.
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  2. #2
    Super Member

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    Hello, slevvio!


    Prove by induction that  a_n^2 - 2b_n^2 \:= \:(-1)^n.

    Deduce that  \frac{a_n}{b_n} \rightarrow \sqrt{2} as  n \rightarrow \infty.
    You got the inductive proof . . . Good for you!


    We have: . a_n^2 \;=\;2b_n^2 + (-1)^n

    Divide by b_n^2\!:\;\;\frac{a_n^2}{b_n^2} \;=\;2 + \frac{(-1)^n}{b^n}

    . . Then: . \left(\frac{a_n}{b_n}\right)^2 \;=\;2 + \frac{(-1)^n}{b_n}

    Take limits: . \lim_{n\to\infty}\left(\frac{a_n}{b_n}\right)^2 \;=\;\lim_{n\to\infty}\left[2 + \frac{(-1)^n}{b_n}\right]


    As n\to\infty, \;b_n \to \infty also.
    Hence, that last fraction approaches 0.

    . . and we have: . \lim_{n\to\infty}\left(\frac{a_n}{b_n}\right)^2 \;=\;2 + 0 \;=\;2


    Therefore: . \lim_{n\to\infty}\,\frac{a_n}{b_n} \;=\;\sqrt{2}

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