Find the Conditions on A for Convergence of f(x) Root

  • Context: MHB 
  • Thread starter Thread starter house2012
  • Start date Start date
  • Tags Tags
    Convergence
Click For Summary
SUMMARY

The discussion focuses on determining the conditions on the parameter A for the convergence of the iterative method defined by $$x_{n+1}=x_n-Af(x_n)$$ to a root of the function f. The key findings indicate that convergence occurs when the initial estimate is sufficiently close to the root, specifically when the condition $$|1-Af'(x_0)|<1$$ is satisfied. Additionally, it is established that A must be greater than zero and that f must cross the x-axis with a positive slope for convergence to be guaranteed. The discussion also outlines specific intervals and conditions for monotonic and oscillating convergence.

PREREQUISITES
  • Understanding of Taylor series expansion
  • Familiarity with iterative methods for root finding
  • Knowledge of convergence criteria in numerical analysis
  • Basic calculus, including derivatives and function behavior
NEXT STEPS
  • Study the implications of the condition $$|1-Af'(x_0)|<1$$ on convergence rates
  • Explore the concept of monotonic vs. oscillating convergence in iterative methods
  • Learn about other root-finding algorithms, such as Newton's method and their convergence properties
  • Investigate the role of the derivative $$f'(x_0)$$ in determining the stability of the iterations
USEFUL FOR

Mathematicians, numerical analysts, and students studying numerical methods for solving equations will benefit from this discussion, particularly those interested in the convergence behavior of iterative root-finding techniques.

house2012
Messages
2
Reaction score
0
Hey guys, I can't get his question dealing with orders of convergence at all so any help would be nice.

Q: Find the conditions on A so that the iteration $$x_{n+1}=x_n-Af(x_n)$$ will converge to a root of f if stared near the root.

I know I should look at the taylor series expansion of f about its root but I am stuck with working out.

Thanks for your help!
 
Physics news on Phys.org
house2012 said:
Hey guys, I can't get his question dealing with orders of convergence at all so any help would be nice.

Q: Find the conditions on A so that the iteration $$x_{n+1}=x_n-Af(x_n)$$ will converge to a root of f if stared near the root.

I know I should look at the taylor series expansion of f about its root but I am stuck with working out.

Thanks for your help!

Write \(x_n=x_0+\varepsilon_n\), where \(x_0\) is the root of \(f(x)\)

Then:

\[
x_{n+1}=x_0+\varepsilon_{n+1}=x_0+\varepsilon_n - A\{f(x_0)+\varepsilon_n f'(x_0)+...\}
\]

Ignoring terms or order 2 and higher in \(\varepsilon_n\) we find:

\[\varepsilon_{n+1}=\varepsilon_n(1-Af'(x_0)) \]

So convergence occurs when the initial estimate is close enough to the solution when:

\[|1-Af'(x_0)|<1\]

CB
 
Last edited:
house2012 said:
Hey guys, I can't get his question dealing with orders of convergence at all so any help would be nice.

Q: Find the conditions on A so that the iteration $$x_{n+1}=x_n-Af(x_n)$$ will converge to a root of f if stared near the root.

I know I should look at the taylor series expansion of f about its root but I am stuck with working out.

Thanks for your help!

In order to avoid confusion we indicate with $x^{*}$ the root of $f(*)$ and with $x_{0}$ the starting point of iterations. Other hypotheses are...

a) $A>0$...

b) $f(*)$ crosses the x axes with positive slope...

If a) and b) are satisfied, then, as explained in...

http://www.mathhelpboards.com/showthread.php?426-Difference-equation-tutorial-draft-of-part-I

... the sequence $x_{n}$ will converge to $x^{*}$ if it exists an interval $a<x<b$ which contains $x^{*}$ and $x_{0}$ ad where for any $x \ne x^{*}$ is...$\displaystyle |A\ f(x)|<2\ |x-x^{*}|$ (1)

More precisely if is...

$\displaystyle |A\ f(x)|\le |x-x^{*}|$ (2)

... the convergence will be 'monotonic' and if is...

$\displaystyle |x-x^{*}|<|A\ f(x)|<2\ |x-x^{*}|$ (3)

... the convergence will be 'oscillating'...

Kind regards

$\chi$ $\sigma$
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
10
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 1 ·
Replies
1
Views
10K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K