# Newton-Raphson Method for derivation of iteration formula

1. Dec 22, 2004

### KataKoniK

Hi,

Prolly a dumb question, but the general equation for finding Xn+1 is

Xn+1 = Xn - f(Xn) / f(Xn)

What will it be for Xn-1?

2. Dec 22, 2004

### Muzza

X(n - 1) = X(n - 2) - f(X(n - 2)) / f'(X(n - 2)) (obviously only valid for n > 2 or n >= 2).

3. Dec 22, 2004

### ahrkron

Staff Emeritus
Or, if you want to "walk backwards" from xm to x(m-1), you can substitute n=m-1 in your equation and solve for x(m-1):

x((m-1)+1) = x(m-1) - f(x(m-1))/f(x(m-1))
x(m) = x(m-1) - f(x(m-1))/f(x(m-1))
x(m-1) = x(m) + f(x(m-1))/f(x(m-1))

From here I'm not quite sure, since you don't have f(x(m-1)) and its derivative,... maybe we can approximate them using the value at m, like so:

x(m-1) = x(m) + f(x(m))/f`(x(m))

But I'm not certain that's a valid approx... is this reasonable, guys?

4. Dec 22, 2004

### KataKoniK

Thanks for the help.

The question I am doing is the following.

f(x) = x^2 - a

Show that for x1 > 0 is any initial estimate for sqrt(a) then the Newton-Raphson method gives the iteration

Xn-1 = (1/2)(Xn + a/Xn) , n >= 1