Finding the roots through Newtons method

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SUMMARY

This discussion focuses on applying Newton's method for approximating the zeros of a function, specifically using the iteration formula x_{n+1}=x_n-\displaystyle\frac{f(x_n)}{f'(x_n)}. Participants clarify the stopping criteria for the iteration process, emphasizing that the iterations should continue until the absolute difference between successive approximations, |x_{n+1}-x_n|, is less than or equal to 0.001. Additionally, the discussion touches on the conditions involving the first and second derivatives of the function, which can influence the convergence rate of the method.

PREREQUISITES
  • Understanding of Newton's method for root-finding
  • Knowledge of calculus, specifically derivatives and their properties
  • Familiarity with convergence criteria in numerical methods
  • Basic proficiency in mathematical notation and inequalities
NEXT STEPS
  • Study the convergence properties of Newton's method in detail
  • Learn about alternative root-finding methods such as the Bisection method and Secant method
  • Explore the implications of Lipschitz continuity on derivative functions
  • Practice implementing Newton's method in programming languages like Python or MATLAB
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Students and professionals in mathematics, engineering, and computer science who are interested in numerical analysis and root-finding techniques.

Telemachus
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Homework Statement


In each of the following items approximate the zeros of f using Newton's method. Continue iterating until making two successive approximations differ at most in 0.001.

The Newton iteration: x_{n+1}=x_n-\displaystyle\frac{f(x_n)}{f'(x_n)}

Well, I have a doubt about this. I'm not sure if it's asking me to iterate till |x_{n+1}-x_n|\leq{0.001}|, or if I should apply some of this:

k_1&gt;0, |f&#039;(x)|\geq{k_1} and |f&#039;&#039;(x)|\leq{}k_2 for all x\in{<b>}</b>, then:

|x_{n+1}-r|&lt;\displaystyle\frac{k_2}{2k_1}|x_n-r|^2

If r\in{}[r-\delta,r+\delta]\subset{[a,b]}, and \delta&lt;2(\displaystyle\frac{k_1}{k_2})

|x_{n+1}-r|&lt;\displaystyle\frac{2k_1}{k_2}(\displaystyle\frac{\delta}{\displaystyle\frac{2k_1}{k_2}})^2n
 
Last edited:
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Stop when |x_{n+1} - x_n| &lt; 0.001.
 

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