# Newton's Method

• Dec 21st 2009, 03:14 PM
wisezeta
Newton's Method
approximate the root of h(x)=x^3 -3 using Newton's Method (correct to hundrehts)
• Dec 21st 2009, 03:22 PM
pickslides
Hi there, have a look at this post.

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

make your starting guess $x_1 = 1.5$
• Dec 21st 2009, 03:23 PM
skeeter
Quote:

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

you should have a valid rough estimate of the root , $x_0$ , to start the iterations.

note that the cube root of 3 is somewhere between 1 and 2

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$
• Dec 21st 2009, 03:33 PM
wisezeta
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, 03:47 PM
pickslides
Are you converging on this figure? how many iterations have you done?
• Dec 21st 2009, 03:48 PM
skeeter
sorry for the typo ... the iteration equation should be a minus

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

try it again.