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Math Help - Newtonian Method, just need algebra help

  1. #1
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    Newtonian Method, just need algebra help

    f(x) = 4x^4 - 4x^2
    x0 = sqrt(21)/7
    x1 = -sqrt(21)/7

    i need to show, by hand, that you can calculate x1 from x0. i tried, but the answers didn't match up. exact square roots (no decimals) are to be used.
    the equation i used was x1 = x0 - f(x0)/f '(x0), is that correct?
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  2. #2
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    Quote Originally Posted by mistykz View Post
    f(x) = 4x^4 - 4x^2
    x0 = sqrt(21)/7
    x1 = -sqrt(21)/7

    i need to show, by hand, that you can calculate x1 from x0. i tried, but the answers didn't match up. exact square roots (no decimals) are to be used.
    the equation i used was x1 = x0 - f(x0)/f '(x0), is that correct?
    Yes, you have the correct equation.
    f(x) = 4x^4 - 4x^2

    Thus
    f^{\prime}(x) = 16x^3 - 8x

    Thus
    x_{n + 1} = x_n - \frac{4x_n^4 - 4x_n^2}{16x_n^3 - 8x_n}

    So with x_0 = \frac{\sqrt{21}}{7} we get
    x_1 = \frac{\sqrt{21}}{7} - \frac{4 \left ( \frac{\sqrt{21}}{7} \right )^4 - 4 \left ( \frac{\sqrt{21}}{7} \right ) ^2}{16 \left ( \frac{\sqrt{21}}{7} \right ) ^3 - 8 \left ( \frac{\sqrt{21}}{7} \right )}

    = \frac{\sqrt{21}}{7} - \frac{4 \left ( \frac{21^2}{7^4} \right ) - 4 \left ( \frac{21}{7^2} \right ) }{16 \left ( \frac{21 \sqrt{21}}{7^3} \right )  - 8 \left ( \frac{\sqrt{21}}{7} \right )}

    = \frac{\sqrt{21}}{7} - \frac{4 \left ( \frac{21^2}{7^4} \right ) - 4 \left ( \frac{21}{7^2} \right ) }{16 \left ( \frac{21 \sqrt{21}}{7^3} \right ) - 8 \left ( \frac{\sqrt{21}}{7} \right )} \cdot \frac{ \frac{ \sqrt{21}}{7^4} }{ \frac{ \sqrt{21}}{7^4} }

    = \frac{\sqrt{21}}{7} - \frac{4 \cdot 21^2 \sqrt{21} - 4 \cdot 21 \sqrt{21} \cdot 7^2 }{16 \cdot 21^2 \cdot 7 - 8 \cdot 21 \cdot 7^3}

    I'm going to factor that last fraction a bit...
    = \frac{\sqrt{21}}{7} - \frac{21 - 7^2}{2 \cdot 21 - 7^2} \cdot \frac{4 \cdot 21 \sqrt{21}}{8 \cdot 21 \cdot 7}

    = \frac{\sqrt{21}}{7} - \frac{-28}{-7} \cdot \frac{\sqrt{21}}{2 \cdot 7}

    = \frac{\sqrt{21}}{7} - 2 \cdot \frac{\sqrt{21}}{7}

    = -\frac{\sqrt{21}}{7}

    (whew! )

    -Dan
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  3. #3
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    wow, i was nowhere near close. thanks a ton!
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