# Question regarding Newton's Method

1. Nov 7, 2015

### King_Silver

So I know how to do Newton's Method without any trouble, taking a value for n then taking it away from the function divided by the derivative of the function I get that entirely and have done roughly 22-23 examples of it over the past 3-4 days. However I have come across one question regarding Newton's Method and it just doesn't make sense to me at all.
It is regarding a starting value x = 0.5 and a "tolerance". No examples in my notes or maths book ever have mentioned about tolerance so I'll just ask the question here.
NOTE: Do not do it for me, that defeats the purpose of my question, I want a step-by-step explanation on how the starting value and the tolerance would be introduced to this sort of question.

Question: f(x) = (x-2)3 = 0
Find the derivative of this function, then compute the roots of this equation using Newton's method.
starting value x =0.5
vary the tolerance from 10-3 to 10-13 reducing the tolerance by 10-1 each time

2. Nov 7, 2015

### Ronie Bayron

May be what it means is you have nth Xn approximate, starting from 10-3, 10-4 until 10-13
Tolerance means the extent of the value when the approximate became acceptable and it could be set arbitrarily by the user depending on the significance of the magnitude that is needed enough to impact a solution.

3. Nov 7, 2015

### King_Silver

So in other words, do Newton's Method as it is normally done and then with the value I get, round it up based on the tolerance? :)

4. Nov 7, 2015

### Ronie Bayron

Nope, hence, you have to iterate over and over again, that tolerance will be your condition that satisfies your approximate to stop.

5. Nov 7, 2015

### King_Silver

Ok so, I take a value for Xn i.e. 3 for this example, which would mean 2- function/derivative of function = answer. In that circumstances where would the tolerance come into play? :/

6. Nov 7, 2015

### King_Silver

*Xn = 2 sorry

7. Nov 7, 2015

### Ronie Bayron

You would use your answer as new inputs until your answer magnitude would reach at 10-3 or 1/1000 scale

8. Nov 7, 2015

### King_Silver

That makes perfect sense now :) thank you very much for that, much appreciated.

9. Nov 7, 2015

### Ronie Bayron

10. Nov 9, 2015

### King_Silver

Ok I've attempted the sum and I want to make sure its going right. For initial value X0 =0.5 I plugged it into Newton's Method as follows
0.5 - ((0.5)-2)^3)/(3((0.5-2)^2) and got it = 1
I then used the value "1" and subbed it in for x for the next iteration and got 1.33333.... then 1.5555..... then 1.7037 then 1.802466667 and finally 1.868311111. Am I following the procedure correctly? do I just continue going down this path until the answer is within the scale of 1/1000 or 10-3? thanks in advance

11. Nov 9, 2015

### Staff: Mentor

Your answers are converging very slowly. The equation you're working with is $(x - 2)^3 = 0$, so obviously the solution is x = 2. The reason that convergence is so slow is that the function involved (f(x = (x - 2)3) has a horizontal tangent at x = 2, which makes it difficult for Newton's method to produce results close to the actual solution very rapidly.

12. Nov 9, 2015

### Ronie Bayron

Yes.

Take note that Newton's Method is only an approximation tool out of several known methods available.
It fail somehow at certain conditions. You may see, further explanation on line (pls. click : Pauls online Notes)

Basic programming knowledge on Matlab, Turbo C++ or Java would help to solve your problems on lengthy iterations.

Last edited: Nov 9, 2015