# If $a_{n+1}=\cos(a_n)$ for $n\ge0$ and $a_0 \in [-\pi/2,\pi/2]$, find $\lim_{n \to \infty}a_n$ if it exists

Let $a_{n+1}=\cos(a_n)$ for $n\ge0$ and $a_0 \in [0,\pi/2]$

Find $\lim_{n \to \infty}a_n$ if it exists.

I drew some sketches and it does seem like the limit exists, it's probably $x$ such that $\cos(x)=x$

I have no idea how to go about solving this, hints would really be appreciated.

• I think that is all you need to say: it's $x$ such that $\cos(x) = x$. I don't believe there's an expression of this $x$ in terms of elementary functions. Commented Jul 24, 2014 at 11:39
• Try induction to prove that $a_{2n}$ and $a_{2n+1}$ is monotone. Commented Jul 24, 2014 at 11:40
• Oh, I just realize you need to show the limit exists too. Commented Jul 24, 2014 at 11:40
• Here's a related post. Commented Jul 24, 2014 at 12:23
• If you want to find the limit you can use Newton's method. Commented Jul 24, 2014 at 12:35

Let $f=\cos$. Note that $a_1$ is in $I=[0,1]$ and that $f(I)\subset I$. Furthermore, for every $x$ in $I$, $-c\lt f'(x)\leqslant0$ with $c\lt1$ hence $(a_{2n})_{n\geqslant1}$ and $(a_{2n-1})_{n\geqslant1}$ are monotonous sequences with values in $I$ and they both converge to the unique solution $\ell$ of $\cos\ell=\ell$ in $[0,1]$.
(Numerically, $c=\sin(1)\lt.85$, $|a_{n+1}-\ell|\leqslant c\cdot|a_n-\ell|$ for every $n\geqslant1$, which implies that $|a_n-\ell|\leqslant c^{n-1}$ for every $n\geqslant1$, and $\ell\approx0.739085$.)
The "Fixed Point" method in numerical analysis says that if a function is continuously differentiable in a neighbourhood of a fixed point $x_0$ and $$|f'(x)| \lt 1$$ it will converge to a fixed point
• In this case one has only $\mid f'(x)\mid\leq1$. Commented Jul 24, 2014 at 11:49
• But in a neighbourhood of the answer, $|f'(x)|<1$. Commented Jul 24, 2014 at 12:05