value of $\lim\limits_{n\rightarrow \infty}\left(a_{n+2}-a_{n}\right)$ is

Question

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

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. — rtybase, Feb 26 '19 at 13:45

score 3 · Answer 1 · answered Feb 26 '19 at 13:47

3

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

answered Feb 26 '19 at 13:47

Klaus

Did not understand , Can you explain me. – DXT Feb 26 '19 at 13:52
@DXT What exactly did you not understand? – Klaus Feb 26 '19 at 13:53
I mean Branch Fixed point Theorem – DXT Feb 26 '19 at 13:57
1

@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: https://en.wikipedia.org/wiki/Banach_fixed-point_theorem ($T$ is $\varphi$ and $X$ is any sufficiently large interval here) – Klaus Feb 26 '19 at 14:00

score 2 · Answer 2 · answered Feb 26 '19 at 13:52

2

Hints:

$\cos: [0,1] \rightarrow [0,1]$
$|\cos'(x)| = |\sin x| \leq \sin 1 < \frac{9}{10}\Rightarrow \cos$ 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".

answered Feb 26 '19 at 13:52

trancelocation

$\cos{x}$ is $[-1,1]\rightarrow[-1,1]$. – rtybase Feb 26 '19 at 13:55
@rtybase : $\cos (-1) = \cos 1 \in [0,1]$. So, my one is a bit "narrower". – trancelocation Feb 26 '19 at 13:59
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. – rtybase Feb 26 '19 at 14:03
1

@rtybase In any case, for a sufficiently large $n$ (actually quite small), you have that $a_n\in[0,1]$. – egreg Feb 26 '19 at 14:05
@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. – rtybase Feb 26 '19 at 14:06
@rtybase : Maybe you didn't read the last sentence in my answer? There I write exactly this detail :-) – trancelocation Feb 26 '19 at 14:07

2 Answers2