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Math Help - Newton's method

  1. #1
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    Newton's method

    Solve e^{-2x}-2x-2 = 0 using Newton's method, giving your answer correct two decimal places.

    f(x) = e^{-2x}-2x-2

    f'(x) = -2e^{-2x}-2

    x_{n+1} = x_{n}-\dfrac{f(x_{n})}{f'(x_{n})}

    I can choose an initial value and what-not, no problem. My problem is, when do we stop? When we have reached a stage where x_{n+k} has no effect on x_{n+k-1}, where the latter is correct to two decimal places?
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  2. #2
    A Plied Mathematician
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    You generally stop when the difference between one iteration and the next is smaller than some pre-defined tolerance, such as 0.0001. The tolerance is problem-dependent.
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