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Math Help - Can anyone help me in this proving?

  1. #1
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    Can anyone help me in this proving?

    Given

    x_(n+1) = x_n - f(x)/f'(x_n) (1)

    x_(n+1) = { x_n + -f'(x_n) √[f'^2(x_n) - 2f(x_n)f"(y_n)] } / f''(y_n) (2)
    n=1,2,...

    (2) can be write as,
    x_(n+1) = x_n + { [2f(x_n)f'(x_n)] / [2f'^2(x_n) - f''(y_n)f(x_n)] } (3)

    f(x_n) = f'(r) [ e_n + c_2 (e_n)^2 + c_3 (e_n)^3 + ... ] (4)
    f'(x_n) = f'(r) [ 1 + 2c_2 e_n + 3c_3 (e_n)^2 + 4c_4 (e_n)^3 + ...] (5)

    From (1) and (2),

    f(x_n) / f'(x_n) = e_n - c_2 (e_n)^2 + 2[ (c_2)^2 - c_3] (e_n)^3 + ... (6)


    How do I get eqn (6) using (4) and (5)? Please ...

    Thanks !!!

    P/s: _ represent subscript while ^ represent superscript / power
    Last edited by chris_panda85; March 31st 2008 at 06:03 AM. Reason: Missing informations
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by chris_panda85 View Post
    Given

    x_(n+1) = x_n - f(x)/f'(x_n) (1)

    x_(n+1) = { x_n + -f'(x_n) √[f'^2(x_n) - 2f(x_n)f"(y_n)] } / f''(y_n) (2)
    n=1,2,...

    (2) can be write as,
    x_(n+1) = x_n + { [2f(x_n)f'(x_n)] / [2f'^2(x_n) - f''(y_n)f(x_n)] } (3)

    f(x_n) = f'(r) [ e_n + c_2 (e_n)^2 + c_3 (e_n)^3 + ... ] (4)
    f'(x_n) = f'(r) [ 1 + 2c_2 e_n + 3c_3 (e_n)^2 + 4c_4 (e_n)^3 + ...] (5)

    From (1) and (2),

    f(x_n) / f'(x_n) = e_n - c_2 (e_n)^2 + 2[ (c_2)^2 - c_3] (e_n)^3 + ... (6)


    How do I get eqn (6) using (4) and (5)? Please ...

    Thanks !!!

    P/s: _ represent subscript while ^ represent superscript / power
    Assume the errors are small so:

    \frac{1}{f'(x_n)} = \frac{1}{f'(r)} ~\frac{1}{ 1 + [2c_2 e_n + 3c_3 (e_n)^2 + 4c_4 (e_n)^3 + ...]}

    Now expand the second term as a power series:

    \frac{1}{f'(x_n)} =  \frac{1}{f'(r)} [1 - [2c_2 e_n + 3c_3 (e_n)^2 + 4c_4 (e_n)^3 + ...]+ [2c_2 e_n + 3c_3 (e_n)^2 + 4c_4 (e_n)^3 + ...]^2 - ... ]

    etc.

    RonL
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