Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $a<-1/5$.

share|improve this question
1  
Introduce $f(x)=-\exp(x/2)/5$. Then try to prove that $[a,0]$ is stable under $f$, and then bound $|f'|$ there. –  1015 Feb 16 '13 at 23:30
    
Here is a related problem. –  Mhenni Benghorbal Feb 17 '13 at 3:11
add comment

1 Answer

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 $[a,0].$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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