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Math Help - Newtons Method

  1. #1
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    Newtons Method

    I am working on problems using Newton's Method in which i'm given an equation and am told to find a root between 2 numbers.

    For this particular question there is no 'x' in the equation so I'm not quite sure what to do:
    Estimate ^4\sqrt{19} to 2 decimal place accuracy. Use 2 as your initial estimate.
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  2. #2
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    Make f(x) = x^{1\over 4}
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  3. #3
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    could you explain further please
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  4. #4
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    I understand Newton's Method to be

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

    Now find f'(x) and start iterating.
    Last edited by pickslides; January 11th 2010 at 02:21 PM. Reason: bad latex
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  5. #5
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    So I'm using f(x) = x^\frac{1}{4}
    and the derivative f'(x) = \frac{1}{4}x^{-\frac{3}{4}} and iterating until it equals ^4\sqrt{19} ?
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  6. #6
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    No!

    Use f(x)=x^4.

    Now you are looking for the root of x^4-19=0

    f'(x)=4x^3

    Take a first shot at the answer, call it x_n

    Say x_n=4

    Then x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}
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  7. #7
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    I just want to clean this up,
    as i answered in a very general way...

    f(x)=x^4-19

    f'(x)=4x^3

    This is a very steep curve and an initial guess at the solution
    can be within a range of x.
    Simplest is to take the square root of the square root of 16,
    or the fourth root of 16 which is 2.

    x_n=2

    The next approximation to the 4th root of 19 is then found from

    x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=2-\frac{f(2)}{f'(2)}

    x_{n+1}=2-\frac{2^4-19}{(4)2^3}=2-\frac{-3}{32}

    Note... (2+\frac{3}{32})^4=19.2176
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