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ajprincess
 4 years ago
Please help:)
By applying NewtonRaphson method to \(f(x)=1\large\frac{1}{ax}\) obtain the recurrence formula \(x_{i+1}=x_i(2ax_i)\) for the iterative determination of the reciprocal of a. Show that if \(E_i\) denotes the error in the \(x_i\), there follows \(E_{i+1}=a{E_i}^2\).
ajprincess
 4 years ago
Please help:) By applying NewtonRaphson method to \(f(x)=1\large\frac{1}{ax}\) obtain the recurrence formula \(x_{i+1}=x_i(2ax_i)\) for the iterative determination of the reciprocal of a. Show that if \(E_i\) denotes the error in the \(x_i\), there follows \(E_{i+1}=a{E_i}^2\).

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[x_{i+1}=x_i\frac{f(x_i)}{f'(x_i)}\] \[x_{i+1}=x_i\frac{1\frac{1}{ax}}{\frac{1}{ax^2}}\]and some algebra should work

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i guess it should be \[x_{i+1}=x_i\frac{1\frac{1}{ax_i}}{\frac{1}{ax_i^2}}\]

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.0Ya I have proved that part. I need help with the error part

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0nope i am wrong again it should be \[x_{i+1}=x_i\frac{1\frac{1}{ax_i}}{\frac{1}{ax_i^2}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0damn i wish i could help, but i don't know an expression for the error. newton ralphson is always error squaring, i know that much but i am not sure how to prove it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think it has something to do with taylor series, but i should really shut up

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.0ohh that's k. Thankk u sooo much for helping me:)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0on the other hand i am pretty good at googling. try looking at "error analysis" in the attached pdf, seems like exactly what you are looking for

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0on second page for "division"

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.0Thankkkk u sooooo much. It is really verryy useful:)
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