i want to verify that when newton's method is used to compute sqrt(R) (by solving the equation x^2=R), the sequence o iterates is defined by

x_(n+1)=.5(x_(n) + (R/x_(n))

f(x) = Xn^2 - R

f'(x) = 2Xn
Xn+1 = Xn - f(Xn)/f'(Xn)
Xn+1 = Xn - (Xn^2-R)/2Xn
Xn+1 = 2Xn^2/2Xn - Xn^2/2Xn + R/2Xn
Xn+1 = Xn/2 +R/2Xn

thank you so much,

i have alot of questions but i don't always get answered on this site. Is it ok if I can get your contact info.

THANKS AGAIN.

I am not around much. Do not want email or anything. I am happy to help if I am free to do so but am usually building boats outside.

To verify that the sequence of iterates in Newton's method for computing the square root of a number R is defined by x_(n+1) = 0.5(x_(n) + (R/x_(n))), you can follow these steps:

1. Start with an initial guess x_0 for the square root of R.

2. Compute x_(n+1) using the formula x_(n+1) = 0.5(x_(n) + (R/x_(n))).

3. Repeat step 2 until the sequence converges to the desired accuracy.

Here's an example to illustrate the verification process using R = 25 and an initial guess x_0 = 5:

Iteration 1:
x_1 = 0.5(5 + (25/5)) = 0.5(5 + 5) = 0.5(10) = 5

Iteration 2:
x_2 = 0.5(5 + (25/5)) = 0.5(5 + 5) = 0.5(10) = 5

Iteration 3:
x_3 = 0.5(5 + (25/5)) = 0.5(5 + 5) = 0.5(10) = 5

As you can see, the sequence of iterates remains constant at 5 after the first iteration. This indicates that the sequence has converged to the square root of R, which in this case is 5. Therefore, the formula x_(n+1) = 0.5(x_(n) + (R/x_(n))) is verified.

Please note that the convergence of Newton's method depends on the initial guess and the function being solved. In some cases, the method may not converge or may converge slowly, so it is important to choose a reasonable initial guess and monitor the convergence behavior.