# Thread: simple newtons method question

1. ## simple newtons method question

so I have a cubic equation here with 3 roots I need to find using newtons method. Basically I wanted to know how to determine the starting point. I picked an x value close to the root from what I can tell on the graph but newtons method keeps giving me larger numbers and then smaller numbers on each recursion.

2. ## Re: simple newtons method question

This method should converge to the nearest root if a close enough starting point is choosen.

Maybe your derivative is wrong? Maybe your calculations are out? Are you doing them by hand or using technology?

3. ## Re: simple newtons method question

That is quite possible. Pick a different one. Without the actual function, it's a bit difficult to say where one might have failed at the process.

Starting closer is one possible solution. Given some bound on the solution, perhaps given by the failed Newton, ry a bisection or two and then go back to Newton, since the exercise seems to demand it.

4. ## Re: simple newtons method question

I already know the roots, for example one of them is ~-1.7, I picked -2 as a starting point but the recursion isn't getting me anywhere.

so basically I started with

-2 - f(-2)/f'(-2)

and then use that value as my new x

5. ## Re: simple newtons method question

is there some reason for not posting the cubic equation?

6. ## Re: simple newtons method question

Originally Posted by skeeter
is there some reason for not posting the cubic equation?
oh sorry lol here it is.

x^3 -2x^2 -4x +4

7. ## Re: simple newtons method question

Originally Posted by Kuma
oh sorry lol here it is.

x^3 -2x^2 -4x +4
$f(x) = x^3 - 2x^2 - 4x + 4$

$f(0) = 4$ , $f(1) = -1$

by the intermediate value theorem, there exists a root between $x = 0$ and $x = 1$

let $x_0 = 0.5$

$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} \approx 0.8095...$

$x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} \approx 0.806062468...$

$x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} \approx 0.8060634335...$

$f(-2) = -4$ , $f(-1) = 5$ ... another root between $x = -2$ and $x = -1$

I get it converging to x = -1.709275359

$f(2) = -4$ , $f(3) = 1$ ... another root between $x = 2$ and $x = 3$

keep going.