Question

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 \(E_{i+1}=-a{E_i}^2\).

Answer 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

Answer 2

i guess it should be
\[x_{i+1}=x_i-\frac{1-\frac{1}{ax_i}}{-\frac{1}{ax_i^2}}\]

Answer 3

Ya I have proved that part. I need help with the error part

Answer 4

nope i am wrong again
it should be
\[x_{i+1}=x_i-\frac{1-\frac{1}{ax_i}}{\frac{1}{ax_i^2}}\]

Answer 5

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

Answer 6

i think it has something to do with taylor series, but i should really shut up

Answer 7

ohh that's k. Thankk u sooo much for helping me:)

Answer 8

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

Answer 9

on second page for "division"

Answer 10

Thankkkk u sooooo much. It is really verryy useful:)

Answer 11

yw, and good luck

Answer 12

thank u:)