approximate the root of h(x)=x^3 -3 using Newton's Method (correct to hundrehts)

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- Dec 21st 2009, 02:14 PMwisezetaNewton's Method
approximate the root of h(x)=x^3 -3 using Newton's Method (correct to hundrehts)

- Dec 21st 2009, 02:22 PMpickslides
Hi there, have a look at this post.

http://www.mathhelpforum.com/math-he...ngly-easy.html

make your starting guess $\displaystyle x_1 = 1.5$ - Dec 21st 2009, 02:23 PMskeeter
- Dec 21st 2009, 02:33 PMwisezeta
The actual root is 1.442, but the answer I get using 1.5 as my estimation gives me 1.5556. Is this right?

- Dec 21st 2009, 02:47 PMpickslides
Are you converging on this figure? how many iterations have you done?

- Dec 21st 2009, 02:48 PMskeeter
sorry for the typo ... the iteration equation should be a minus

$\displaystyle x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

try it again.