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let $f:[a,b]\to [a,b]$ be Continuous function,Assmue that sequence $\{x_{n}\}(n\ge 0)$ such $$x_{0}=x,x_{1}=f(x_{0}),x_{2}=f(x_{1}),\cdots,x_{n+1}=f(x_{n}),\forall n\in N^{+}$$ and $$\lim_{n\to\infty}(x_{n+1}-x_{n})=0$$ show that: $$\lim_{n\to\infty}x_{n}$$ is exsit

My idea: first we note $$x_{n+1}-x_{n}=f(x_{n})-x_{n}$$ so we have $$\lim_{n\to\infty}(f(x_{n})-x_{n})=0$$ then I can't prove this limits is exsit?I fell can't easy prove it.?

math110
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  • I suspect a typo, as in the "show that" part of the problem, $f$ isn't mentioned at all. – GFauxPas Nov 28 '14 at 16:43
  • @GFauxPas It is, as $x_n=f(\ldots f(x_0)\ldots)$. – user2345215 Nov 28 '14 at 16:43
  • I see that your question related to the Newten-Raphson algorithm, and it's a special case from it , The algorithm is always converge .at the moment see :http://en.wikipedia.org/wiki/Newton%27s_method – zeraoulia rafik Nov 28 '14 at 17:40
  • We can easily show that the sequence $x_n$ is cauchy and so it converges. Let $x_n$ converge to $x$. Then, by continuity of $f$, $x$ is a fixed point of $f$. We know that for the given function, a fixed point will exist.I feel the uniqueness of the fixed point can be shown by Banach fixed-point theorem. – Srinivas K Nov 28 '14 at 18:00

1 Answers1

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WLOG $f:[0,1]\to[0,1]$. Let $a=\underline\lim x_n$, $b=\overline\lim x_n$. Assume $a<b$, clearly $f$ can't be identically $x$ on any subinterval of $[a,b]$, as the sequence would stop due to the limit condition. So there's some $c\in (a,b)$ so that WLOG $f(c)>c$.

$f$ is continuous in $c$, so let $\varepsilon>0$ be so small that $f(x)>c+\varepsilon$ for all $x\in(c-\varepsilon,c+\varepsilon)$ and $c-\varepsilon>a$, $c+\varepsilon<b$. Together with the limit condition this guarantees that $x_n$ can never cross the $(c-\varepsilon,c+\varepsilon)$ interval on a path from $b$ to $a$ for sufficiently large $n$, which is a contradiction.

user2345215
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