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Let $x>0$, $x_1=\cos(x),$ and $x_{n+1}=\cos(x_n), \forall n\geq1.$ Then the sequence $(x_n)_{n\geq 1}$ is

  1. bounded but not monotone,
  2. bounded but not Cauchy,
  3. Cauchy,
  4. convergent.

It is a multiple select question , so it may have more than one correct option.

The above sequence is bounded, it's trivial. Also if the sequence converges then it will converge to a real root $\ \ l\ $ of $\ \ \ l-\cos(l)=0$.

I tried by taking a particular value of $x$. Suppose $x=\pi/2$, then the sequence will become $\{0,1,\cos(1),\cos(\cos(1)),\cos(\cos(\cos(1))),.....\}$. Without using calculator, I think we can't say whether this sequence is monotone or not.

Again while showing 'Cauchy' I considered $|x_m-x_n|\leq |x_{m-1}-x_{n-1}|\leq|x_{m-2}-x_{n-2}|\leq...$

I don't know how can we show whether the sequence Cauchy or not from the above relation.

Any help is appreciated. Thank you.

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  • $\begingroup$ On a figure, try to represent $\cos(x)$ and $f(x)=x$ and use them to graphically represent what happens. It will not be a formal proof but help you understanding what to prove $\endgroup$ – Damien Dec 14 '18 at 6:27
  • $\begingroup$ en.wikipedia.org/wiki/Dottie_number $\endgroup$ – Claude Leibovici Dec 14 '18 at 8:13
  • $\begingroup$ "Without using a calculator..." I strongly advise you to use a calculator as a tool of discovery. What you learn might be surprising and insightful in regards to formulating a strategy to answer your questions rigorously. $\endgroup$ – Lee Mosher Dec 14 '18 at 15:30
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Hint 1. Note that if $0\leq x_n\leq 1$, and $x_n\leq x_{n+1}=\cos(x_n)$ then $x_n\in[0,a]$ and $x_{n+1}\in [a,1]$ where $a\simeq 0.73908$ is the unique solution of the equation $\cos(x)=x$ . Therefore $$x_{n+2}=\cos(x_{n+1})\leq x_{n+1}.$$ So the sequence has an alternating behaviour.

Hint2. If you show that the function $x\to \cos(\cos x)$ is strictly increasing in $[0,1]$ then it follows that the subsequences $\{x_{2n}\}_n$ and $\{x_{2n+1}\}_n$ are strictly monotone and bounded. What may we conclude?

P.S. You may also note that $x\to \cos(\cos x)$ is a contraction in $[0,1]$ and Banach fixed-point theorem can be applied.

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There's only one real root $y$ such that $\cos y=y$ (it's about $0.74$). In the neighborhood of $y$, $$ x_n=y+\varepsilon\implies x_{n+1}=\cos(y+\varepsilon)=\cos y - \varepsilon \sin y + O(\varepsilon^2)=y-\varepsilon\sqrt{1-y^2}+O(\varepsilon^2)... $$ since $\vert{\sin y} \vert< 1$, this leads to exponential (and non-monotonic) convergence to $y$ if $x_n$ ever lands close enough to it. So it seems like you'll sometimes have 1., 3., and 4. all true. However, it's not obvious without further analysis that 3. and 4. have to hold... it might be that you can converge to some other stable limit cycle, or even if there were an unstable cycle, you might stay on it for some initial conditions. (Update: per other answers, in this case there are no other limit cycles and the fixed point is universally attracting.)

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  • $\begingroup$ Actually, you may not. $\endgroup$ – Did Dec 14 '18 at 7:08

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