Ralphson-Newton Method And Interval Bisection

  • Thread starter Thread starter Procrastinate
  • Start date Start date
  • Tags Tags
    Interval Method
Click For Summary
SUMMARY

The discussion focuses on determining an optimal starting point for the Ralphson-Newton Method applied to the equation x3 - x - 1 = 0. The suggested starting point is 1.5, derived from the interval bisection method, as the polynomial changes sign between x = 1 and x = 2. An alternative starting point of approximately 1.289 is proposed using the secant method, although it offers minimal improvement in iteration count. Ultimately, starting at x = 1 is also considered a viable option.

PREREQUISITES
  • Understanding of the Ralphson-Newton Method
  • Familiarity with interval bisection technique
  • Basic knowledge of polynomial functions
  • Concept of the secant method
NEXT STEPS
  • Study the implementation of the Ralphson-Newton Method in numerical analysis
  • Explore the interval bisection method for root-finding
  • Learn about the secant method and its applications
  • Investigate convergence criteria for iterative methods
USEFUL FOR

Mathematicians, engineers, and computer scientists involved in numerical methods and root-finding algorithms.

Procrastinate
Messages
155
Reaction score
0
I know quite well how to do these. However, most of the time, the starting points are just given to me i.e. the a and b values to starting iterating.

I was just wondering if you were to find a starting point for x3-x-1 = 0, what would be the best starting point to use the Ralphson-Newton Method. My notes say 1.5 but I would like to know the reason behind that.

Thanks.
 
Physics news on Phys.org
An "educated guess". If x= 1, x^3- x- 1= 1- 1- 1= -2 and if x= 2, x^3- x- 1= 8- 2- 1= 5. Since the polynomial changes sign between 1 and 2, there is a root between 1 and 2. 1.5, half way between 1 and 2, is a good starting point. That's really applying the "interval bisection" to the first step. But, in fact, any number in the vicinity of 1 and 2 will work.

You might even do this: y goes from -2 to 5, a difference of 7. To go to 0 from -2 is just a difference of 2 so perhaps 2/7 of the way from 1 to 2, 1.289, would be better. (That's the "secant" method.) But that only changes the number of iterations to get the same accuracy by, maybe, one or two. Frankly, I would have been inclined to start with x= 1.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
5K
Replies
16
Views
7K
  • · Replies 1 ·
Replies
1
Views
10K