Newton's Method

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Dec 21st 2009, 02:14 PM
wisezeta

Newton's Method

approximate the root of h(x)=x^3 -3 using Newton's Method (correct to hundrehts)
Dec 21st 2009, 02:22 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, 02:23 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, 02:33 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, 02:47 PM
pickslides

Are you converging on this figure? how many iterations have you done?
Dec 21st 2009, 02:48 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.

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