# Newtons Method

• Jul 1st 2006, 01:59 AM
nath_quam
Newtons Method
I don't know how to do this question would anyone be able to help thanks nath
• Jul 1st 2006, 04:08 AM
rgep
Newton's Method solves the equation f(x) = 0 as follows. Let x_1 be an approximation to the true root x, say x_1 = x - h. Then 0 = f(x) = f(x_1 + h) = f(x_1) + h.f'(x_1) + (error) and we assume that the error is negligible. Then h is approximately - f(x_1)/f'(x_1) and so another approximation to the root x is x_2 = x_1 + h = x_1 - f(x_1)/f'(x_1).

In your case f(x) = x^{1/3} and f'(x) = (1/3)x^{-2/3}, so h = -3x_1.
• Jul 1st 2006, 03:32 PM
nath_quam
$\displaystyle a1\;=\;a\;-\;\frac{f(a)}{f'(a)}\;where\;x\;=\;a\;is\;close\;t o\;the\;root$

I know i have to use the above but how
• Jul 1st 2006, 06:02 PM
ThePerfectHacker
The point does not need to "be close to" the root. It can be in any interval that satisfies the necessary conditions for this algorithm to work. (I just do not remember the exact conditions).
• Jul 2nd 2006, 04:12 PM
nath_quam
How do i apply this to my question
• Jul 2nd 2006, 09:16 PM
CaptainBlack
Quote:

Originally Posted by nath_quam
$\displaystyle a1\;=\;a\;-\;\frac{f(a)}{f'(a)}\;where\;x\;=\;a\;is\;close\;t o\;the\;root$

I know i have to use the above but how

As rgep indicates, you set

$\displaystyle f(x)=x^{1/3}\$

Then :

$\displaystyle f'(x)=\frac{1}{3}\ x^{-2/3}$

So:

$\displaystyle a_1=a-3\ a$

Note this does not converge for any $\displaystyle a \ne 0$, (as is hinted
at in the question).

RonL