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 $a<-1/5$.

  • 1
    $\begingroup$ 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

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 $[a,0].$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.