Question regarding Newton's Method

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    Method Newton's method
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Discussion Overview

The discussion revolves around the application of Newton's Method for finding roots of the function f(x) = (x-2)³ = 0, specifically focusing on the introduction of a starting value and varying tolerance levels. Participants explore how to implement these concepts in the context of numerical methods.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses familiarity with Newton's Method but seeks clarification on the role of tolerance in the method.
  • Another participant suggests that tolerance indicates the acceptable extent of approximation and can be set arbitrarily based on significance.
  • There is a discussion about iterating values until the result's magnitude reaches a specified tolerance level, such as 10^-3.
  • A participant shares their iterative results starting from x = 0.5 and questions whether they are following the procedure correctly, noting the slow convergence of their results.
  • Another participant points out that the function has a horizontal tangent at the solution, which contributes to the slow convergence observed in the iterations.
  • There is mention of the limitations of Newton's Method and the suggestion that programming knowledge could assist in handling lengthy iterations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the specifics of how to implement tolerance in Newton's Method, and there are varying interpretations of its application. The discussion includes both agreement on the iterative process and differing views on the implications of the results.

Contextual Notes

Some participants note the slow convergence of the method due to the characteristics of the function involved, highlighting the need for caution in applying Newton's Method under certain conditions.

King_Silver
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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
 
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King_Silver said:
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
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.
 
Ronie Bayron said:
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.

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? :)
 
King_Silver said:
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? :)
eq0010MP.gif
Nope, hence, you have to iterate over and over again, that tolerance will be your condition that satisfies your approximate to stop.
 
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Ronie Bayron said:
eq0010MP.gif
Nope, hence, you have to iterate over and over again, that tolerance will be your condition that satisfies your approximate to stop.

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? :/
 
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King_Silver said:
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? :/
*Xn = 2 sorry
 
King_Silver said:
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? :/
You would use your answer as new inputs until your answer magnitude would reach at 10-3 or 1/1000 scale
 
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Ronie Bayron said:
You would use your answer as new inputs until your answer magnitude would reach at 10-3 or 1/1000 scale
That makes perfect sense now :) thank you very much for that, much appreciated.
 
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Your welcome:smile:
 
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  • #10
Ronie Bayron said:
Your welcome:smile:
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
King_Silver said:
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
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.
 
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  • #12
King_Silver said:
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
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.
 
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