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jossy04
Group Title
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)
 one year ago
 one year ago
jossy04 Group Title
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)
 one year ago
 one year ago

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

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

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

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

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

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

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

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