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How can I solve this problem?

Let $$x(n+1)=-\frac{\exp(x(n)/2)}{5}$$ be a given sequence. Prove using the Banach contraction principle that this sequence converges to some fixed point $X$ with $x(0)$ in some interval $[a,0]$ where $a<-1/5$.

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    $\begingroup$ Introduce $f(x)=-\exp(x/2)/5$. Then try to prove that $[a,0]$ is stable under $f$, and then bound $|f'|$ there. $\endgroup$ – Julien Feb 16 '13 at 23:30
  • $\begingroup$ Here is a related problem. $\endgroup$ – Mhenni Benghorbal Feb 17 '13 at 3:11
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As julien says, consider the function $f(x) = -\frac{1}{5} e^{x/2}$ and note that if $x\in [a,0]$ for some $a < -1/5,$ then $a < -\frac{1}{5} \le f(x) \le -1/5 e^{a/2} < 0,$ so $f(x) \in [a,0],$ from which any number of iterations of $f$ applied to $x$ will still be in the interval. Hence, our sequence is contained in $[a,0]$ and $x(n+1) = f(x(n)).$

Consider $f^{\prime}(x) = -1/10 e^{x/2}$ and note that in the interval $[a,0],$ $-1/10 \le f^{\prime}(x) \le -1/10 e^{a/2},$ so $|f^{\prime}(x)| \le 1/10.$ By the Mean Value Inequality, $|f(x) - f(y)| \le 1/10 |x-y|$ in the interval. Using the Banach fixed point theorem gives that the sequence $x(0), f(x(0)) = x(1), \ldots, f(x(n)) = x(n+1), \ldots$ converges to a point in $[a,0].$

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