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jossy04

  • 2 years ago

How does newton's method work ? This is all supposed to be a review, but I'm having trouble remembering how it works... f(x)=x^3 + x + 3 (Initial Guess = 1)

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  1. anonymous
    • 2 years ago
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    \[x_{n+1}=x_{n}-\frac{f(x_n)}{f'(x_n)}\]

  2. anonymous
    • 2 years ago
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    first guess is 1 \(f(1)=5\) so \(1\) is kind of a lousy guess, since you want it to be 0

  3. anonymous
    • 2 years ago
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    \[f'(x)=3x^2+1\] so \[f'(1)=4\]

  4. anonymous
    • 2 years ago
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    second guess is therefore \(x_2=1-\frac{5}{4}=-\frac{1}{4}\)

  5. anonymous
    • 2 years ago
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    lather, rinse repeat it gets real messy real fast

  6. jossy04
    • 2 years ago
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    I'm kind of remembering how to do this

  7. anonymous
    • 2 years ago
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    you can also do some algebra if you like before you start \[x-\frac{f(x)}{f'(x)}=x-\frac{x^3+x+3}{3x^2+1}\] \[=\frac{2x^3-3}{3x^2+1}\] and plug in there, but it amounts to the same thing

  8. jossy04
    • 2 years ago
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    Thank you very much for your help :)

  9. anonymous
    • 2 years ago
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    yw

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