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Math Help - Newton's method of approximation

  1. #1
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    Newton's method of approximation

    Approximate by applying Newton's Method to the equation
    Round your answer to 4 decimal places.

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  2. #2
    Senior Member x3bnm's Avatar
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    Re: Newton's method of approximation

    Procedure for Newton’s Method

    1. Guess a first approximation to a solution of the equation f(x) = 0. A graph
    of y = f(x) may help.
    2. Use the first approximation to get a second, the second to get a third, and so
    on, using the formula

     x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}, \text{ if } f'(x_n) \neq 0 .....................(1)


    Example:
    So to find a positive root of this equation f(x) = x^2 -2

    we have to guess a value of x_n. It doesn't have to be exact.

    Let x = x_n so  f(x) = x^2 - 2 becomes f(x_n) = (x_n)^2 - 2 if we plugin this into (1) and after simplification we get:

    \begin{aligned}x_{n+1} = & x_n - \frac{(x_n)^2 - 2}{2 * x_n} \\ x_{n+1} = & \frac{x_n}{2} + \frac{1}{x_n} \text{ ..............................(2)}\end{aligned}

    The graph of f(x) = x^2 - 2 is:

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    Figure 1


    Let's guess a approximate value where the graph of f(x) = y = x^2 - 2 =  0 so from the graph(Figure 1) say x_0 = 1 where n = 0 and plug in into equation (2) and solve for x_1 and then plug-in x_1 and find x_2 and this goes on until you get a good approximation up to a particular decimal places:

    \begin{array}{|c|c|c|}\hline \text{ value of x}    & \text{Error} & \text{Number of Correct Digits} \\  \hline  x_0 = 1 & -0.41421 & 1 \\  \hline    x_1 = 1.5 & 0.08579 & 1 \\  \hline   x_2 = 1.41667 & 0.00246 & 3 \\  \hline x_3 = 1.41422 & 0.00001 & 5 \\  \hline \end{array}


    And that's it. You approximated the positive root of f(x) = x^2 - 2 to five decimal places.
    Last edited by x3bnm; August 1st 2012 at 02:57 PM.
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