Convergence Interval for Newton's Method

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SUMMARY

The discussion centers on determining the convergence interval for Newton's Method applied to the function f(x) = e^(-1/x^2) with the root at x = 0. The formula for the iterative method is given as xn+1 = xn - f(xn) / f'(xn). A critical finding is that the exponential function does not yield a real root, thus raising the question of convergence towards infinity rather than a finite root. The suggested interval for initial guesses, x0, is [-1, 1], but this leads to the conclusion that there are no real roots for the function.

PREREQUISITES
  • Understanding of Newton's Method for root-finding
  • Familiarity with the exponential function and its properties
  • Knowledge of derivatives, specifically f'(x) and f''(x)
  • Basic concepts of convergence in numerical methods
NEXT STEPS
  • Investigate the behavior of f(x) = e^(-1/x^2) as x approaches 0 and infinity
  • Learn about convergence criteria for Newton's Method in non-linear functions
  • Explore alternative root-finding methods for functions without real roots
  • Examine the implications of convergence intervals in numerical analysis
USEFUL FOR

Mathematicians, numerical analysts, and students studying numerical methods who are interested in the convergence properties of iterative algorithms like Newton's Method.

Scootertaj
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1. The problem statement:

In what region can we choose x0 and get convergence to the root x = 0 for f(x) = e-1/x^2

Homework Equations


xn+1 = xn - f(xn) / f'(xn)


The Attempt at a Solution


The only thing I've come across is a formula that says |root - initial point| < 1/M where M = max|f''(x)|/(2min|f'(x)| where x belongs to a "sufficiently small interval"

My thought: [-1,1]
 
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Scootertaj said:
1. The problem statement:

In what region can we choose x0 and get convergence to the root x = 0 for f(x) = e-1/x^2

Homework Equations


xn+1 = xn - f(xn) / f'(xn)

The Attempt at a Solution


The only thing I've come across is a formula that says |root - initial point| < 1/M where M = max|f''(x)|/(2min|f'(x)| where x belongs to a "sufficiently small interval"

My thought: [-1,1]

The exponential function does not equal zero for any real argument, so there is NO root. (I suppose you could regard x = +-infinity as "roots", but things like |root - x_n| are then not real numbers, either.) If you regard the question as: "for what x_0 does x_n --> + infinity (or -infinity)?", then you might have a sensible question.

RGV
 
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