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Math Help - numerical method

  1. #1
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    numerical method

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

    converges, it will converge to  \sqrt{N}
    Usually the questions tell you what  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  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 x_n and x_{n+1} in the formula by x.

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

    IF this sequence converges, then \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  x = \frac{1}{2}(x+ \frac{N}{x})

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

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

     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  x = \frac{1}{2}(x+ \frac{N}{x})

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

     2x = Nx

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

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

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

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

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