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Thread: numerical method

  1. #1
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    numerical method

    Show that by putting $\displaystyle x_{n+1} = x_{n} = x $ that if the sequence defined by $\displaystyle x_{n+1} = \frac{1}{2}( x_{n} + \frac{N}{x_{n}}) $

    converges, it will converge to $\displaystyle \sqrt{N} $
    Usually the questions tell you what $\displaystyle x_{0} = $ some number, but not sure how to start with this one, any help appreciated

    thank you
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  2. #2
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    Quote Originally Posted by Tweety View Post
    Usually the questions tell you what $\displaystyle x_{0} = $ some number, but not sure how to start with this one, any help appreciated

    thank you
    Would it shock you if I told you that the way to start is to do what they tell you to do? Replace both $\displaystyle x_n$ and $\displaystyle x_{n+1}$ in the formula by x.

    That gives you $\displaystyle x= \frac{1}{2}\left(x+ \frac{N}{x}\right)$ solve that for x.

    IF this sequence converges, then $\displaystyle \lim x_{n+1}= \frac{1}{2}\left(\lim x_n+ \frac{N}{\lim x_n}\right)$ and each of those limits is of the same sequence and so is equal to x. This is, by the way, a well known iterative method for finding square roots. In fact, it is often known as the "Babylonian method"!
    Methods of computing square roots - Wikipedia, the free encyclopedia
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  3. #3
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    okay thanks,

    When I solve for x I get $\displaystyle x = \frac{1}{2}(x+ \frac{N}{x}) $

    $\displaystyle x = \frac{1}{2}(\frac{Nx^{2}}{x}) $

    $\displaystyle x = \frac{1}{2}(Nx) $

    $\displaystyle 2x = Nx $

    x=0
    Is this correct?
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  4. #4
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    Quote Originally Posted by Tweety View Post
    okay thanks,

    When I solve for x I get $\displaystyle x = \frac{1}{2}(x+ \frac{N}{x}) $

    $\displaystyle x = \frac{1}{2}(\frac{Nx^{2}}{x}) $
    You need practice adding fractions! $\displaystyle x+ \frac{N}{x}= \frac{x^2}{x}+ \frac{N}{x}= \frac{x^2+ N}{x}$

    The simpler thing to do is just multiply the entire equation, $\displaystyle x= \frac{1}{2}(x+ \frac{N}{x})$, by 2x.
    $\displaystyle x = \frac{1}{2}(Nx) $

    $\displaystyle 2x = Nx $

    x=0
    Is this correct?
    Clearly not, since you were told that this sequence was supposed to converge to $\displaystyle \sqrt{N}$!

    By the way, there is a slight inaccuracy in the problem, related to the fact that they did not give you $\displaystyle x_0$.

    For any $\displaystyle x_0> 0$, this recursive sequence converges to $\displaystyle \sqrt{N}$.

    But if $\displaystyle x_0< 0$, it converges to $\displaystyle -\sqrt{N}$.

    (And, of course, if N< 0, the sequence does not converge.)
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