I tried to approximate the solution to $x=f(x)$ for some given $f$, by guessing $x=a$, then I observed that $x=f(a)$ was an even better approximation, and $x=f(f(a))$ and so on was even better, so why does this method work and for which f, is it sufficient that f is continuous?
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I would also appreciate a different phrasing of this question. You know, for those of us who are easily confused... – The Chaz 2.0 Oct 27 '13 at 21:43
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1It doesn't always; it depends on the slope of $f(x)$ at the point of solution. For instance, if $f(x)=x^2$, then this method will give you the fixed point at $x=0$, but not the one at $x=1$. – mjqxxxx Oct 27 '13 at 21:43
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4See here for the standard condition that guarantees this behavior. – Micah Oct 27 '13 at 21:45
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1This will happen if $f$ is a contraction (and in other special cases). The iteration can take you away from a fixed point as well as towards one, and it won't find a fixed point if there isn't one. – Mark Bennet Oct 27 '13 at 21:45
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This method certainly works if the function is Lipschitz continuous with Lipschitz constant $L<1$. In general, it is not true.
For example let $f(x)=x+1$, then for any starting value $x_0$, the sequence $x_n=f(x_{n-1})$ ($n\geq1$) diverges.
Daniel Robert-Nicoud
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1Since there is no reaction for my comment, then -1. ...And who is upvoting this answer? – Artem Oct 27 '13 at 22:19
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@Artem Sorry, I forgot to specify that you must have Lipschitz constant (strictly) less than $1$. Then (the proof of) Banach's fixed point theorem tells you that the sequence of $x_n$'s converges to the (unique) fixed point of the function. – Daniel Robert-Nicoud Oct 27 '13 at 23:08