approximate the root of h(x)=x^3 -3 using Newton's Method (correct to hundrehts)
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approximate the root of h(x)=x^3 -3 using Newton's Method (correct to hundrehts)
Hi there, have a look at this post.
http://www.mathhelpforum.com/math-he...ngly-easy.html
make your starting guess $\displaystyle x_1 = 1.5$
The actual root is 1.442, but the answer I get using 1.5 as my estimation gives me 1.5556. Is this right?
Are you converging on this figure? how many iterations have you done?
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.