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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 \(E_{i+1}=a{E_i}^2\).
 one year ago
 one year 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 \(E_{i+1}=a{E_i}^2\).
 one year ago
 one year ago

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satellite73Best ResponseYou've already chosen the best response.1
\[x_{i+1}=x_i\frac{f(x_i)}{f'(x_i)}\] \[x_{i+1}=x_i\frac{1\frac{1}{ax}}{\frac{1}{ax^2}}\]and some algebra should work
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
i guess it should be \[x_{i+1}=x_i\frac{1\frac{1}{ax_i}}{\frac{1}{ax_i^2}}\]
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
Ya I have proved that part. I need help with the error part
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
nope i am wrong again it should be \[x_{i+1}=x_i\frac{1\frac{1}{ax_i}}{\frac{1}{ax_i^2}}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
damn 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
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
i think it has something to do with taylor series, but i should really shut up
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
ohh that's k. Thankk u sooo much for helping me:)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
on 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
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
on second page for "division"
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
Thankkkk u sooooo much. It is really verryy useful:)
 one year ago
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