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Thread: newton raphson iterative method

  1. #1
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    newton raphson iterative method

    okay so i got a question conserning newton-raphson method to find a root to 4 d.p.

    the equation is $\displaystyle e^x-3x = 0$ now i've got the equation

    $\displaystyle x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n}}$

    i have been given the information to start with $\displaystyle x_{0} = 2.$

    my question is how do i use the equation. I have no idea. I know what im looking for but i have no idea how to use the equation.

    Cheers guys!
    Last edited by mxmadman_44; Aug 19th 2007 at 12:59 PM. Reason: typo
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  2. #2
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    Quote Originally Posted by mxmadman_44 View Post
    okay so i got a question conserning newton-raphson method to find a root to 4 d.p.

    the equation is $\displaystyle e^x-3x = 0$ now i've got the equation

    $\displaystyle x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}$

    i have been given the information to start with $\displaystyle x_{0} = 2.$

    my question is how do i use the equation. I have no idea. I know what im looking for but i have no idea how to use the equation.

    Cheers guys!
    Ummm... Plug it into the formula?

    $\displaystyle f(x) = e^x - 3x$

    $\displaystyle f^{\prime} = e^x - 3$

    So
    $\displaystyle x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{e^{x_n} - 3x_n}{e^{x_n} - 3}$


    With [tex]x_0 = 2 we have
    $\displaystyle x_1 = 2 - \frac{e^2 - 3 \cdot 2}{e^2 - 3} \approx 1.6835182627834$

    Then do it again with $\displaystyle x_1 = 1.6835182627834$ to get $\displaystyle x_2$, etc.

    I got an answer after about 5 iterations.

    -Dan
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