ajprincess 3 years ago 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 $$E_{i+1}=-a{E_i}^2$$.

1. anonymous

$x_{i+1}=x_i-\frac{f(x_i)}{f'(x_i)}$ $x_{i+1}=x_i-\frac{1-\frac{1}{ax}}{-\frac{1}{ax^2}}$and some algebra should work

2. anonymous

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

3. ajprincess

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

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. ajprincess

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

8. anonymous

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

9. anonymous

on second page for "division"

10. ajprincess

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

11. anonymous

yw, and good luck

12. ajprincess

thank u:)