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Math Help - Newton's Method

  1. #1
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    Newton's Method

    approximate the root of h(x)=x^3 -3 using Newton's Method (correct to hundrehts)
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  2. #2
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    Hi there, have a look at this post.

    http://www.mathhelpforum.com/math-he...ngly-easy.html

    make your starting guess x_1 = 1.5
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  3. #3
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    Quote Originally Posted by wisezeta View Post
    approximate the root of h(x)=x^3 -3 using Newton's Method (correct to hundrehts)
    you should have a valid rough estimate of the root , x_0 , to start the iterations.

    note that the cube root of 3 is somewhere between 1 and 2

    x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
    Last edited by skeeter; December 21st 2009 at 03:48 PM. Reason: fix equation
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  4. #4
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    The actual root is 1.442, but the answer I get using 1.5 as my estimation gives me 1.5556. Is this right?
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  5. #5
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    Are you converging on this figure? how many iterations have you done?
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  6. #6
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    sorry for the typo ... the iteration equation should be a minus

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

    try it again.
    Last edited by skeeter; December 21st 2009 at 04:19 PM.
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