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Math Help - Newton's Method of convergence

  1. #1
    Junior Member
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    Newton's Method of convergence

    I'm stuck with this whole question on convergence -

    f(x) = xe^(-x) + 3.
    For what values of xo ("x naught") does Newton's method converge to the root, and for what values of xo will Newton's method fail to find the root?

    Thanks
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  2. #2
    MHF Contributor
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    Hi

    f(x) = xe^{-x} + 3
    f'(x) = (1-x)e^{-x}

    Newton's method starting from x_0 will give

    x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}

    If f'(x_0) = 0 you will not be able to calculate x_1
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  3. #3
    Junior Member
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    Thanks,

    so am i right in saying that Newton's method converges to the root for xo not equal to 1? Or greater than or equal to 1?
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