I don't know how to do this question would anyone be able to help thanks nath
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.
As rgep indicates, you setOriginally Posted by nath_quam
$\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