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First of all I don't understand why we need Banach's theorem, as a result I can't make it intuitive for me to understand how it works but I tried to solve an example.

https://en.wikipedia.org/wiki/Banach_fixed-point_theorem#Statement

let $f(x) = x^3$

a) for which fixed points will it converge ?

b) choose an interval $I$ such that all conditions for Banach's theorem are fulfilled (for function $f$)

My try :

a) we have 3 roots $( 0, 1, -1)$ with $0$ being a convergent point. (I think I solved a) but I don't understand what kind of interval are they asking about, and how to compute it)

PS: If some of the text doesn't make sense, please let me know, I translated it from another language.

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You need to find an interval that contains $o$ and in wich the conditions of the theorem are satisfied. You are looking for an interval $[a,b]$ such that

  1. $0\in[a,b]$
  2. $f([a,b])\subset[a,b]$
  3. $|f(x)-f(y)|\le L\,|x-y|$ for some constant $L$, $0\le L<1$ and all $x\in[a,b]$.

Since $|x|^3\le|x|$ for $|x|\le1$, it seems reasonable to try an interval $[-a,a]$ with $0<a\le1$. With this choice, 1. and 2. are satisfied. The easiest way to verify 3. is to impose $|f'(x)|\le L$. Since $|f'(x)|=3\,|x|^2$, the maximum of $|f'(x)|$ on the interval $[-a,a]$ is $3\,|a|^2$. This implies that it is enough to take $a<1/\sqrt3$.

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  • $\begingroup$ $|f(x)-f(y)|\le L\,|x-y|$ you moved $|x-y|$ to the left and got $|f'(x)|$ which is nice, but what statement can I make about this interval ? now that I found it what does it really give me ? $\endgroup$ – Oleg Jul 8 '17 at 17:44
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    $\begingroup$ It gives you an interval in which you know that there is a unique solution of the equation $f(x)=x$, and gives you a procedure to approximate that solution. Start with any $x_0$ in the interval and construct the sequence of iterates $x_{n+1}=f(x_n)$; then $x_n$ converges to the solution. Following the proof of the theorem, you also get estimates on the error of the approximation. $\endgroup$ – Julián Aguirre Jul 8 '17 at 17:52
  • $\begingroup$ thank you Julian ! one last thing , is there any though logic why you took an interval from $[-a, a]$ and not $[0, a]$ or anything else ? $\endgroup$ – Oleg Jul 8 '17 at 17:55
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    $\begingroup$ No, It would have worked also with $[0,a]$. But in a more general setting, you may have to look for an interval in the form $[a,b]$, specially if you do not know a priori the solution to the equation. $\endgroup$ – Julián Aguirre Jul 8 '17 at 18:07
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Hint: You have to choose a closed interval $I$ such that $f(I) \subseteq I$ (i.e., $f$ maps $I$ into itself) and such that $f$ is a contraction on $I$. If $f$ is differentiable (as is in this case), the last condition is satisfied if there exists a constant $L\in [0,1)$ such that $$ |f'(x)| \leq L \qquad\forall x\in I. $$

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