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hendelsherif · Aug 30, 2011, 07:20 PM

This is used in calculus and mathematics.

Unknown008 · Aug 31, 2011, 01:23 AM

I did a wrong assumption that the method I used was called the newton method. Sorry for that. See the post of ebaines for the description of the Newton method :)

I think that you are referring to the iteration method which I think was also called Newton method.

Take that for example:

$x^2 - x - 6 = 0$

The usual way around is to factor then solve. The Newton method converts it into another form where x is the subject of formula, and you start using a value of x which you think a near point:

$x = \sqrt{x + 6}$

Say you don't know the solution. You know however that 2.9 is close to the solution. (this is often given, or you are asked to show that the root lies between two numbers and are asked to start with 2.9). You proceed with the first iteration:

$x_0 = 2.9$

$x_1 = \sqrt{(2.9) + 6} = 2.9832867780352594$

Now you take that result and put it into another iteration:

$x_2 = \sqrt{(2.9832867780352594) + 6} = 2.997213168601002$

And you go on an on:

until you get a consistent answer:

$x_5 = 2.999987097$

$x_6 = 2.999997849$

Which gives you an answer of 3.0000 rounded to 4 decimal places.

Of course, this method is used when the equation cannot be easily solved. :)

Sometimes even, this method can have tweaks, in that you might get values getting large too quickly, which means you are not converging towards the root, but diverging away. So, the way you write down your iteration sometimes matters.

ebaines · Aug 31, 2011, 11:35 AM

I'm going to disagree with Unk on this one. Newton's method is indeed an iterative method, but it's for solving an equation of the form f(x)=0, and it involves using the slope of the function to make a better estimate of the next guess. Mathematically it's this:

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

To use Unk's example: solving for f(x)=x^2-x-6=0, the slope of this function is f'(x)=2x-1. So if you make a first estimate of x = 2.9, then f(x) = -0.49, and f'(x) = 4.8, and the next estimate would be 2.9-(-0.49)/4.8 = 3.002083. Contining on, the next estimate is 3.000000867. You can see this technique converges on the answer of 3 very quickly.

Unknown008 · Aug 31, 2011, 11:38 AM

Ah, okay, I actually didn't do this one. The Newton's method seems longer than the typical iteration method :rolleyes:

So, I did a wrong assumption in that the iteration that I used to do was called the Newton method, good to know :)