# Finding the roots through Newtons method

• Aug 14th 2010, 07:51 AM
Ulysses
Finding the roots through Newtons method
In each of the following items approximate the zeros of $f$ using Newton's method. Continue iterating until making two successive approximations differ at most in 0.001

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

Well, I have a doubt about this. I'm not sure if it's asking me to iterate till $|x_{n+1}-x_n|\leq{0.001}|$, or if I should apply some of this:

$k_1>0$, $|f'(x)|\geq{k_1}$ and $|f''(x)|\leq{}k_2$ for all $x\in{[b]}$, then:

$|x_{n+1}-r|<\displaystyle\frac{k_2}{2k_1}|x_n-r|^2$

If $r\in{}[r-\delta,r+\delta]\subset{[a,b]}$, and $\delta<2(\displaystyle\frac{k_1}{k_2})$

$|x_{n+1}-r|<\displaystyle\frac{2k_1}{k_2}(\displaystyle\fra c{\delta}{\displaystyle\frac{2k_1}{k_2}})^2n$
• Aug 14th 2010, 08:25 AM
Failure
Quote:

Originally Posted by Ulysses
In each of the following items approximate the zeros of $f$ using Newton's method. Continue iterating until making two successive approximations differ at most in 0.001

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

Well, I have a doubt about this. I'm not sure if it's asking me to iterate till $|x_{n+1}-x_n|\leq{0.001}$,

Well, yes, I think this is very clearly what you are asked to do.

Quote:

or if I should apply some of this: <more complicated stuff>
I know that in order to show that Newton's method converges "quadratically" (that is, that the number of valid digits doubles with each iteration), if it converges at all, one does have to do some more fancy guesswork like this. But I think it is obvious that the text of your exercise does not require you to handle such technicalities.