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Math Help - Newton raphson method

  1. #1
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    Newton raphson method

    i need to solve this equation with newton raphson method

    x^3-3x+1
    thank you
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    By Bolzano's Theorem there is a solution in (0,1) . Choose x_0=0 and construct the corresponding sequence.
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  3. #3
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    Surely you are not expecting people to do this problem for you. Show what you have done. Of course, Newton-Raphson is an iterative method- for each approximation, x_n, to the solution of f(x)= 0, x_{n+1}= x_n- \frac{f(x_n)}{f'(x_n)} should be a better approximation and, hopefully, the sequence will converge to a solution.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    Surely you are not expecting people to do this problem for you. Show what you have done. Of course, Newton-Raphson is an iterative method- for each approximation, x_n, to the solution of f(x)= 0, x_{n+1}= x_n- \frac{f(x_n)}{f'(x_n)} should be a better approximation and, hopefully, the sequence will converge to a solution.

    maybe i should find the location of the root by the method of tabulation ?
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  5. #5
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    Look at FernandoRevilla's post!
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  6. #6
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    for an equation like this ... x^3 -8x-4

    newton raphson iterative scheme is given by

    x_n_+_1 = x_n - f(x_n)/f'(x_n)

    for the given equation f(x)= x^3-8x-4

    first we find the location of the root by the method of tabulation . .

    the table for f(x) is

    x 0 1 2 3 4
    f(x) -4 -13 -12 -1 28

    the positive root is near x = 3

    we take x_0 = 3 in newton raphson iterative scheme

    x_n_+_1 =x_n  - x_n^3-8x_n-4/3x_n^2-8

    we get , x_1 = 3 -     27-24-4/27-8   = 3.0526

    similarly , x_2 = 3.05138

    and

    x_3= 3.05138


    thus , the positive root is 3.0514 , correct to five significant digits
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