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jossy04
 one year 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)
jossy04
 one year 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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satellite73
 one year ago
Best ResponseYou've already chosen the best response.1\[x_{n+1}=x_{n}\frac{f(x_n)}{f'(x_n)}\]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1first guess is 1 \(f(1)=5\) so \(1\) is kind of a lousy guess, since you want it to be 0

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1\[f'(x)=3x^2+1\] so \[f'(1)=4\]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1second guess is therefore \(x_2=1\frac{5}{4}=\frac{1}{4}\)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1lather, rinse repeat it gets real messy real fast

jossy04
 one year ago
Best ResponseYou've already chosen the best response.0I'm kind of remembering how to do this

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1you can also do some algebra if you like before you start \[x\frac{f(x)}{f'(x)}=x\frac{x^3+x+3}{3x^2+1}\] \[=\frac{2x^33}{3x^2+1}\] and plug in there, but it amounts to the same thing

jossy04
 one year ago
Best ResponseYou've already chosen the best response.0Thank you very much for your help :)
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