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Math Help - Num. analysis #2

  1. #1
    Member javax's Avatar
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    Num. analysis #2

    Hello. I'm having problem with these:

    a). Prove that the Bisection algorithm gives a sequence that has the bound of error which converges linearly to zero.

    b). The sequence F_n described as F_0, F_1, and F_{n+2}=F_n+F_{n+1}, n\geq0, is called Fibonacci sequence and x_n=\frac{F_{n+1}}{F_n}, \lim_{n\to\infty}{x_n}=x exists, prove that that limit is x=\frac{(1+\sqrt5)}{2}. (This number is called golden ratio)

    Thank you for your time.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by javax View Post
    Hello. I'm having problem with these:

    a). Prove that the Bisection algorithm gives a sequence that has the bound of error which converges linearly to zero.
    If the initial interval length is L, the interval length after n itterations is

    L_n=\frac{L}{2^n}

    So:

    L_{n+1}=L_n/2

    and as L_k is a bound on the error the convergence is linear.

    CB
    Last edited by CaptainBlack; December 21st 2008 at 01:21 AM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by javax View Post
    b). The sequence F_n described as F_0, F_1, and F_{n+2}=F_n+F_{n+1}, n\geq0, is called Fibonacci sequence and x_n=\frac{F_{n+1}}{F_n}, \lim_{n\to\infty}{x_n}=x exists, prove that that limit is x=\frac{(1+\sqrt5)}{2}. (This number is called golden ratio)

    Thank you for your time.
     <br />
x_n=\frac{F_{n+1}}{F_n}=\frac{F_n+F_{n-1}}{F_n}=1+\frac{1}{x_{n-1}}<br />

    Now we are told that the limit as x \to \infty exists, so the limit must satisfy the equation:

     <br />
x=1+\frac{1}{x}<br />

    which may be simplified to:

    x^2-x-1=0

    which has roots:

     <br />
x=\frac{1\pm\sqrt{5}}{2}<br />

    Of these roots we may discard the one with the negative sign as x must be greater than 0 (in fact greater than 1), so the required root is:

     <br />
x=\frac{1+\sqrt{5}}{2}<br />

    CB
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