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Please help:) By applying Newton-Raphson method to \(f(x)=1-\large\frac{1}{ax}\) obtain the recurrence formula \(x_{i+1}=x_i(2-ax_i)\) 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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\[x_{i+1}=x_i-\frac{f(x_i)}{f'(x_i)}\] \[x_{i+1}=x_i-\frac{1-\frac{1}{ax}}{-\frac{1}{ax^2}}\]and some algebra should work
i guess it should be \[x_{i+1}=x_i-\frac{1-\frac{1}{ax_i}}{-\frac{1}{ax_i^2}}\]
Ya I have proved that part. I need help with the error part

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Other answers:

nope i am wrong again it should be \[x_{i+1}=x_i-\frac{1-\frac{1}{ax_i}}{\frac{1}{ax_i^2}}\]
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
i think it has something to do with taylor series, but i should really shut up
ohh that's k. Thankk u sooo much for helping me:)
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
1 Attachment
on second page for "division"
Thankkkk u sooooo much. It is really verryy useful:)
yw, and good luck
thank u:)

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