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Math Help - Algebra involving errors and root 2

  1. #1
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    Post Algebra involving errors and root 2

    I can do the first bit, which is just manipulation of the LHS, but I'm not sure how to start on the second and third bit.

    Say we wish to estimate the square rot of 2. We begin with a good first estimate x_1, say  x_1 = 1.5 . In fact, it is a little too big. We note that 2/1.5 is a little too small, so we form a new guess by averaging them,  x_2 = \frac{1}{2}(1.5+2/1.5) . We continue this method to give a sequence:

     x_{n+1} = \frac{1}{2}\bigg(\frac{2}{x_n} + x_n\bigg)

    To ease notation, we set  x_n = \sqrt{2} + \epsilon_n , so,  \epsilon_1 < 1/2^3

    i) Show that

    \bigg(\frac{2}{\sqrt{2} + \epsilon_n} - \big(\sqrt{2} - \epsilon_n\big)\bigg) = \frac{\epsilon_n^2}{\sqrt{2} + \epsilon_n};

    ii) show that

     \epsilon_{n+1} = \frac{1}{2} \frac{\epsilon_n^2}{\sqrt{2}+\epsilon_n} < \frac{1}{2}\epsilon_n^2

    iii) How big do we require n so that x_n is within 2^{-15} of \sqrt{2}. Find the fraction x_n
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  2. #2
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    Errors and Square Roots

    Hello mitch_nufc
    Quote Originally Posted by mitch_nufc View Post
    I can do the first bit, which is just manipulation of the LHS, but I'm not sure how to start on the second and third bit.

    Say we wish to estimate the square rot of 2. We begin with a good first estimate x_1, say  x_1 = 1.5 . In fact, it is a little too big. We note that 2/1.5 is a little too small, so we form a new guess by averaging them,  x_2 = \frac{1}{2}(1.5+2/1.5) . We continue this method to give a sequence:

     x_{n+1} = \frac{1}{2}\bigg(\frac{2}{x_n} + x_n\bigg)

    To ease notation, we set  x_n = \sqrt{2} + \epsilon_n , so,  \epsilon_1 < 1/2^3

    i) Show that

    \bigg(\frac{2}{\sqrt{2} + \epsilon_n} - \big(\sqrt{2} - \epsilon_n\big)\bigg) = \frac{\epsilon_n^2}{\sqrt{2} + \epsilon_n};

    ii) show that

     \epsilon_{n+1} = \frac{1}{2} \frac{\epsilon_n^2}{\sqrt{2}+\epsilon_n} < \frac{1}{2}\epsilon_n^2

    iii) How big do we require n so that x_n is within 2^{-15} of \sqrt{2}. Find the fraction x_n
    As you say, part (i) is straightforward. So:

    (ii) x_{n+1} = \sqrt{2}+ \epsilon_{n+1}

    \Rightarrow \epsilon_{n+1} = x_{n+1} - \sqrt{2}

    = \frac{1}{2}\left(\frac{2}{x_n}+ x_n\right) - \sqrt{2}

    = \frac{1}{2}\left(\frac{2}{\sqrt{2} +\epsilon_n}+ \sqrt{2} +\epsilon_n\right) - \sqrt{2}

    = \frac{1}{2}\left(\frac{2}{\sqrt{2} +\epsilon_n}- (\sqrt{2} -\epsilon_n)\right) - \sqrt{2}

    = \frac{1}{2}\frac{\epsilon_n^2}{\sqrt{2}+\epsilon_n  }

    = \frac{1}{2}\frac{\epsilon_n^2}{x_n}

    < \frac{1}{2}\epsilon_n^2, since x_n > 1

    (iii) \epsilon_1 < 2^{-3} \Rightarrow \epsilon_2 < \tfrac{1}{2}\cdot 2^{-6}=2^{-7}

    \Rightarrow \epsilon_3 < \tfrac{1}{2}\cdot 2^{-14}= 2^{-15}

    x_2 = \frac{1}{2}\left(\frac{3}{2}+ \frac{4}{3}\right) = \frac{17}{12}

    \Rightarrow x_3 = ... = \frac{577}{408}

    Grandad
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  3. #3
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    Correction to part (ii)

    Hello mitch_nufc

    Sorry, I left an extra \sqrt{2} term in part (ii), so here's the corrected version:

    (ii) x_{n+1} = \sqrt{2}+ \epsilon_{n+1}

    \Rightarrow \epsilon_{n+1} = x_{n+1} - \sqrt{2}

    = \frac{1}{2}\left(\frac{2}{x_n}+ x_n\right) - \sqrt{2}

    = \frac{1}{2}\left(\frac{2}{\sqrt{2} +\epsilon_n}+ \sqrt{2} +\epsilon_n\right) - \sqrt{2}

    = \frac{1}{2}\left(\frac{2}{\sqrt{2} +\epsilon_n}- (\sqrt{2} -\epsilon_n)\right)

    = \frac{1}{2}\frac{\epsilon_n^2}{\sqrt{2}+\epsilon_n  }

    = \frac{1}{2}\frac{\epsilon_n^2}{x_n}

    < \frac{1}{2}\epsilon_n^2, since x_n > 1

    (iii) \epsilon_1 < 2^{-3} \Rightarrow \epsilon_2 < \tfrac{1}{2}\cdot 2^{-6}=2^{-7}

    \Rightarrow \epsilon_3 < \tfrac{1}{2}\cdot 2^{-14}= 2^{-15}

    x_2 = \frac{1}{2}\left(\frac{3}{2}+ \frac{4}{3}\right) = \frac{17}{12}

    \Rightarrow x_3 = ... = \frac{577}{408}

    Grandad
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