# Thread:

1. ## Newtons Method

I don't know how to do this question would anyone be able to help thanks nath
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2. 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.

3. $<br /> a1\;=\;a\;-\;\frac{f(a)}{f'(a)}\;where\;x\;=\;a\;is\;close\;t o\;the\;root<br />$

I know i have to use the above but how

4. 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).

5. How do i apply this to my question

6. Originally Posted by nath_quam
$<br /> a1\;=\;a\;-\;\frac{f(a)}{f'(a)}\;where\;x\;=\;a\;is\;close\;t o\;the\;root<br />$

I know i have to use the above but how
As rgep indicates, you set

$<br /> f(x)=x^{1/3}\$

Then :

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

So:

$<br /> a_1=a-3\ a<br />$

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

RonL
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