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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)
 10 months ago
 10 months 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)
 10 months ago
 10 months ago

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

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

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

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

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

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