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anonymous
 3 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)
anonymous
 3 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[x_{n+1}=x_{n}\frac{f(x_n)}{f'(x_n)}\]

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[f'(x)=3x^2+1\] so \[f'(1)=4\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0second guess is therefore \(x_2=1\frac{5}{4}=\frac{1}{4}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lather, rinse repeat it gets real messy real fast

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm kind of remembering how to do this

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you very much for your help :)
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