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Math Help - Newton's Method

  1. #1
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    Newton's Method

    x^3=17x+34

    = x^3 - 17x - 34 = 0

    x^3 - 17x - 34 (that is f(x))

    X0 = 6

    Find X1 and |X1-X0|

    So Xn+1 = Xn - f(Xn) / f'(Xn)

    So Xn+1 = Xn - Xn^3 - 17xn - 34 / 3xn^2 - 17n?

    how do i derive that 17xn? I'm confused at that part basically, because i dont know what to do with 17n (if i did do that right).

    thanks.
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  2. #2
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    f(x)=x^3-17x-34
    f'(x)=3x^2-17

    x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{x^3_n-17x_n-34}{3x_n^2-17}=\frac{3x_n^3-17x_n-x_n^3+17x_n+34}{3x_n^2-17}=\frac{2x_n^3+34}{3x_n^2-17}
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  3. #3
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    Quote Originally Posted by lilwayne View Post
    x^3=17x+34

    = x^3 - 17x - 34 = 0

    x^3 - 17x - 34 (that is f(x))

    X0 = 6

    Find X1 and |X1-X0|

    So Xn+1 = Xn - f(Xn) / f'(Xn)

    So Xn+1 = Xn - Xn^3 - 17xn - 34 / 3xn^2 - 17n?

    how do i derive that 17xn? I'm confused at that part basically, because i dont know what to do with 17n (if i did do that right).

    thanks.
    Unfortunately, you didn't do it right...

    f(x)=x^3-17x-34

    f'(x)=3x^2-17

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

    x_1=x_0-\frac{f\left(x_0\right)}{f'\left(x_0\right)}
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  4. #4
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    Quote Originally Posted by MathoMan View Post
    f(x)=x^3-17x-34
    f'(x)=3x^2-17

    x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{x^3_n-17x_n-34}{3x_n^2-17}=\frac{3x_n^3-17x_n-x_n^3+17x_n+34}{3x_n^2-17}=\frac{2x_n^3+34}{3x_n^2-17}
    Your answer was right but i dont get how you turned Xn - xn^3 ..... to 3Xn^3 .....

    how did you do that :S
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  5. #5
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    x_n-\frac{x^3_n-17x_n-34}{3x_n^2-17}=\frac{x_n\cdot (3x_n^2-17)-(x_n^3-17x_n-34)}{3x_n^2-17}=\frac{3x_n^3-17x_n-x_n^3+17x_n+34}{3x_n^2-17}=\frac{2x_n^3+34}{3x_n^2-17}
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  6. #6
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    Isn't that the same thing as Xn - Xn^3 - 17xn - 34 / 3xn^2 - 17 ?
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  7. #7
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    Quote Originally Posted by lilwayne View Post
    Isn't that the same thing as Xn - Xn^3 - 17xn - 34 / 3xn^2 - 17 ?
    Yes,

    it's a matter of choice.
    You may prefer to substitute the x values into the single fraction (which has only 2 terms in x)
    or into the direct Newton's method formula.
    Either way, done correctly will give the same result for the next approximation to the root.
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