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Math Help - sequence related question...

  1. #1
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    sequence related question...

    hello all,

    I have a tough past exam question that I can not seem to fully understand. If we use newtons method to approx \sqrt(a) by looking for the 0 of the function f defined by f(x) = x^2 - a, the sequence \{x_n\} as n=1 to \infty is generated, where x_1 = 1 and x_{n+1} = \frac{1}{2}(x_n+\frac{a}{x_n}).

    The question states that show that if:
    0<x_m<\sqrt(a), then x_{m+1}>\sqrt(a).

    • Hint: add x_{m+1}-\sqrt(a), and complete the square.

    Not sure how to approach this one, but using the hint would I do something like..
    \frac{1}{2}(x_m+\frac{a}{m_n})-\sqrt(a) = \frac{x_m}{2}+\frac{a}{m_n} - \sqrt(a) ..


    Thanks..
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: sequence related question...

    Quote Originally Posted by Oiler View Post
    hello all,

    I have a tough past exam question that I can not seem to fully understand. If we use newtons method to approx \sqrt(a) by looking for the 0 of the function f defined by f(x) = x^2 - a, the sequence \{x_n\} as n=1 to \infty is generated, where x_1 = 1 and x_{n+1} = \frac{1}{2}(x_n+\frac{a}{x_n}).

    The question states that show that if:

    0<x_m<\sqrt(a), then x_{m+1}>\sqrt(a).

    • Hint: add x_{m+1}-\sqrt(a), and complete the square.
    Not sure how to approach this one, but using the hint would I do something like..
    \frac{1}{2}(x_m+\frac{a}{m_n})-\sqrt(a) = \frac{x_m}{2}+\frac{a}{m_n} - \sqrt(a) ..


    Thanks..
    The procedure to be adopted in this type of problem is illustrated in...

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

    The difference equation can be written in the form...

    \Delta_{n}= x_{n+1}-x_{n}= -\frac{1}{2}\ (x_{n}-\frac{a}{x_{n}})= f(x_{n}) (1)

    If we limit our attention to x>0, is easy to see that f(x) has only one 'attractive fixed point' [a point x_{0} where is f(x_{0})=0 and f^{'}(x_{0})<0...] in x_{0}=\sqrt{a} and, considering that is...

    f(x)= -\frac{1}{2}\ (x-\frac{a}{x}) \implies  f^{'}(x)= -\frac{1}{2}\ (1+\frac{a}{x^{2}}) (2)

    ... we can conclude that...

    a) for all the x_{1} >0 the sequence defined by (1) will converge to \sqrt{a} no matter which is a...

    b) if x_{1}>\sqrt{a} then the condition |f(x)|<|x-x_{0}| is verified and the convergence is 'monotonic'...

    c)if x_{1}<\sqrt{a} then the condition |f(x)|<|x-x_{0}| isn't verified and the convergence is 'oscillating'...

    Kind regards

    \chi \sigma
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