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  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 $\displaystyle \sqrt(a)$ by looking for the 0 of the function $\displaystyle f$ defined by $\displaystyle f(x) = x^2 - a$, the sequence $\displaystyle \{x_n\}$ as $\displaystyle n=1$ to $\displaystyle \infty$ is generated, where $\displaystyle x_1$ = 1 and $\displaystyle x_{n+1}$ = $\displaystyle \frac{1}{2}(x_n+\frac{a}{x_n})$.

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

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

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


    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 $\displaystyle \sqrt(a)$ by looking for the 0 of the function $\displaystyle f$ defined by $\displaystyle f(x) = x^2 - a$, the sequence $\displaystyle \{x_n\}$ as $\displaystyle n=1$ to $\displaystyle \infty$ is generated, where $\displaystyle x_1$ = 1 and $\displaystyle x_{n+1}$ = $\displaystyle \frac{1}{2}(x_n+\frac{a}{x_n})$.

    The question states that show that if:

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

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


    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...

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

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

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

    ... we can conclude that...

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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