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mcastillo356
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- TL;DR Summary
- Want to understand which kind of functions I can be sure that the method of fixed-point iteration is suitable
Hi PF
Not every function works when we try to compute the root with this method
![20220129_185054.jpg 20220129_185054.jpg](https://www.physicsforums.com/data/attachments/279/279452-ddc11a234720b67d3bb5e6f955d19bc6.jpg)
The following theorem guarantees that the method of fixed point iteration will work for a particular class of functions
A fixed point theorem
Suppose that ##f## is defined on an interval ##I=[a,b]## and satisfies the following two conditions
(i) ##f(x)## belongs to ##I## whenever ##x## belongs to ##I##; and
(ii) there exists a constant ##K## with ##0<K<1## such that for every ##u## and ##v## in ##I##
$$|f(u)-f(v)|\leq K|u-v|$$
Then ##f## has a unique fixed point ##r## in ##I##, that is, ##f(r)=r##, and starting with any number ##x_0## in ##I##, the iterates
$$x_1=f(x_0),\qquad{x_2=f(x_1),...}$$
converge to ##r##
Hints for the proof
1- Condition (ii) of theorem implies that ##f## is continuous on ##I=[a,b]##. Use condition (i) to show that ##f## has a unique fixed point ##r## on ##I##. Apply the Intermediate-Value Theorem to ##g(x)=f(x)-x## on ##[a,b]##.
2- Use condition (ii) of theorem and mathematical induction to show that ##|x_n-r|\leq K^{n}|x_0-r|##. Since ##0<K<1##, we know that ##K^{n}\rightarrow{0}## as ##n\rightarrow{\infty}##. This shows that ##\lim_{n\rightarrow{infty}}{x_n=r}##
First: ¿Why bounded Newton quotient means continuity on the interval?.
Thanks
Not every function works when we try to compute the root with this method
![20220129_185054.jpg 20220129_185054.jpg](https://www.physicsforums.com/data/attachments/279/279452-ddc11a234720b67d3bb5e6f955d19bc6.jpg)
The following theorem guarantees that the method of fixed point iteration will work for a particular class of functions
A fixed point theorem
Suppose that ##f## is defined on an interval ##I=[a,b]## and satisfies the following two conditions
(i) ##f(x)## belongs to ##I## whenever ##x## belongs to ##I##; and
(ii) there exists a constant ##K## with ##0<K<1## such that for every ##u## and ##v## in ##I##
$$|f(u)-f(v)|\leq K|u-v|$$
Then ##f## has a unique fixed point ##r## in ##I##, that is, ##f(r)=r##, and starting with any number ##x_0## in ##I##, the iterates
$$x_1=f(x_0),\qquad{x_2=f(x_1),...}$$
converge to ##r##
Hints for the proof
1- Condition (ii) of theorem implies that ##f## is continuous on ##I=[a,b]##. Use condition (i) to show that ##f## has a unique fixed point ##r## on ##I##. Apply the Intermediate-Value Theorem to ##g(x)=f(x)-x## on ##[a,b]##.
2- Use condition (ii) of theorem and mathematical induction to show that ##|x_n-r|\leq K^{n}|x_0-r|##. Since ##0<K<1##, we know that ##K^{n}\rightarrow{0}## as ##n\rightarrow{\infty}##. This shows that ##\lim_{n\rightarrow{infty}}{x_n=r}##
First: ¿Why bounded Newton quotient means continuity on the interval?.
Thanks