approximate the root of h(x)=x^3 -3 using Newton's Method (correct to hundrehts)
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Hi there, have a look at this post. http://www.mathhelpforum.com/math-he...ngly-easy.html make your starting guess $\displaystyle x_1 = 1.5$
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 , $\displaystyle x_0$ , to start the iterations. note that the cube root of 3 is somewhere between 1 and 2 $\displaystyle x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$
Last edited by skeeter; Dec 21st 2009 at 02:48 PM. Reason: fix equation
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.
Last edited by skeeter; Dec 21st 2009 at 03:19 PM.
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