Let $a_{1}=a$ and $a_{n+1}=\cos(a_{n})\;\forall \;n\;\in \mathbb{N}.$

Then $\lim\limits_{n\rightarrow \infty}(a_{n+2}-a_{n})$ is

Try: $a_{n+2}=\cos(a_{n+1})=\cos(\cos(a_{n}))=\cos(\cos(\cos (a_{n-1})))=\cdots \cdots \cos(\cos\cos\cos(\cdots \cdots \cos(a)))))))$

Did not know how can i solve it, could some help me , Thanks

  • 2
    $\begingroup$ Try to show that $\lim\limits_{n\rightarrow\infty}a_n$ exists, first (e.g. here). Then the limit of the difference is the difference of the limits. Or the sequence is Cauchy. $\endgroup$ – rtybase Feb 26 at 13:45

Hint: Apply Banach's fixed point theorem to $\varphi(x) = \cos(\cos(x))$.

  • $\begingroup$ Did not understand , Can you explain me. $\endgroup$ – DXT Feb 26 at 13:52
  • $\begingroup$ @DXT What exactly did you not understand? $\endgroup$ – Klaus Feb 26 at 13:53
  • $\begingroup$ I mean Branch Fixed point Theorem $\endgroup$ – DXT Feb 26 at 13:57
  • 1
    $\begingroup$ @DXT If you were given this exercise, I am sure you had Banach's fixed point theorem in class. You can read it up here for example: en.wikipedia.org/wiki/Banach_fixed-point_theorem ($T$ is $\varphi$ and $X$ is any sufficiently large interval here) $\endgroup$ – Klaus Feb 26 at 14:00


  • $\cos: [0,1] \rightarrow [0,1]$
  • $|\cos'(x)| = |\sin x| \leq \sin 1 < \frac{9}{10}\Rightarrow \cos$ is contractive on $[0,1]$
  • $\Rightarrow$ $a_{n+1} = \cos a_n$ converges to the only fixpoint in $[0,1]$.

Now, reason why irrespective of the starting value $a$ the sequence $a_n$ will have to fall into $[0,1]$ "earlier or later".

  • $\begingroup$ $\cos{x}$ is $[-1,1]\rightarrow[-1,1]$. $\endgroup$ – rtybase Feb 26 at 13:55
  • $\begingroup$ @rtybase : $\cos (-1) = \cos 1 \in [0,1]$. So, my one is a bit "narrower". $\endgroup$ – trancelocation Feb 26 at 13:59
  • $\begingroup$ Say initial condition is $a_0=3$, then $\cos(a_0)=-0.9899924...$. It's still fine to consider $\cos$ as a contraction mapping on $[-1,1]$, it has one fixed point anyway. $\endgroup$ – rtybase Feb 26 at 14:03
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    $\begingroup$ @rtybase In any case, for a sufficiently large $n$ (actually quite small), you have that $a_n\in[0,1]$. $\endgroup$ – egreg Feb 26 at 14:05
  • $\begingroup$ @egreg I know. It's not me who should provide these details. And it's really easy to update the answer and close this little gap. $\endgroup$ – rtybase Feb 26 at 14:06

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