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

1. satellite73

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

2. satellite73

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. satellite73

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. satellite73

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. satellite73

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. satellite73

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. satellite73

on second page for "division"

10. ajprincess

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

11. satellite73

yw, and good luck

12. ajprincess

thank u:)