Why does $x_0 + \sin(x_0) \approx \pi$, when computing it multiple times? Why does $x_0 + \sin(x_0) \approx \pi$ when computing this multiple times on the calculator? 
So for any value of up to $\approx 6.25$, doing the following operation ($x_0 + \sin(x_0)$, then the newly obtained value (let's call it $x_1$) is replaced instead of $x_0$: $x_1 + \sin(x_1)$) many times leads to a value, which is very close to $\pi$. 
However, when going above that value, this will result in roughly $3\pi$, $5\pi$, etc.
Could anyone please try to explain why the simple first calculation works?
EDIT: Thank you so much for your answers guys - through a combination of the many, I believe to now understand why!
 A: Have a look at this related post  where it is shown that the sequence 
$$
\begin{cases}
  x_{0}  = c & \\ 
  x_{n}  = x_{n - 1}  + \sin (x_{n - 1} ) &
\end{cases}
$$
graphically translates into the following picture
 
Have a look there to see the conclusion about the attractor points being $\pi +2k \pi$
A: Observing that $\sin x$ is bounded from above by the line $\pi - x$ over $x\le\pi$ and, they eventually meet at the bound $x=\pi$ . Therefore, 
$$x_n = x_{n-1} + \sin x_{n-1}  \le x_{n-1} + (\pi - x_{n-1}) = \pi$$
i.e. 
$$ x_n \le \pi$$
It is also true that,
$$x_n - x_{n-1} = \sin x_{n-1} > 0$$
i.e. $x_n$ already increases with every iteration.
Thus, starting with $x_0 < \pi$ and with each iteration, the successive $x_n$ keeps increasing by $\sin x_{n-1}$, until it reaches the its upper bound $\pi$.
A: Consider $u_n=x_n-\pi$, then as $\sin(u+k\pi)=(-1)^k\sin(u)$,
$$
u_{n+1}=u_n-\sin(u_n)=\frac{u_n^3}{6}-\frac{u_n^5}{120}+...
$$
This shows cubic convergence, this iteration will rapidly converge to $0$ after the iteration reaches a neighborhood of $0$. By the first term of the series expansion this is true for $|u_0|<\sqrt6$, $\pi-\sqrt6<x_0<\pi+\sqrt6$. By the higher degree terms this can be true for a larger interval, the "graphical" properties of the sine give the convergence for $|u_0|<\pi$.
Note that for a valid method to compute $\pi$ you can not use $\pi$ in the computation of $\sin(x_n)$. One can use the usual reduction tricks to accelerate convergence for exponential values, for instance
$$
\cos(x_n)+i\sin(x_n)=(\cos(2^{-k}x_n)+i\sin(2^{-k}x_n))^{2^k}
$$
for some moderately large $k$.
A: You start with $x_0 \in \mathbb R$ and define recursively for $n \ge 0$
$$x_{n+1} = x_n + \sin x_n .$$
What can we say about the convergence of the sequence $(x_n)$?
If it converges to some $x \in \mathbb R$, then necessarily
$$x = \lim x_{n+1} = \lim (x_n + \sin x_n) = \lim x_n + \lim \sin x_n = \lim x_n + \sin (\lim x_n) = x + \sin x .$$
This means $\sin x  = 0$, i.e. $x = k\pi$ for some $k \in \mathbb Z$.
Case 1. $x_0 = k\pi$.
Then all $x_n = x_0$. Hence $(x_n)$ is a constant sequence which trivially converges to $k\pi$.
Case 2. $0 < x_0 < \pi$.
We claim that $(x_n)$ is strictly monotonically increasing such that all $x_n < \pi$. Hence it converges to some $x \in [x_0,\pi]$ and our above limit consideration shows $x =  \pi$. The claim is easily proved by induction using the following facts:


*

*For $0 < \xi < \pi$ we have $\sin \xi > 0$ which implies $\xi < \xi + \sin \xi$.

*For $0 < \eta$ we have $\sin \eta < \eta$. For $\xi < \pi$ we therefore get $\xi + \sin \xi = \xi + \sin (\pi -\xi) < \xi + \pi - \xi = \pi$.
Case 3. $2r\pi < x_0 < (2r+1)\pi$.
Then $(x_n)$ is strictly monotonically increasing such that all $x_n < (2r+1)\pi$ and $\lim x_n = (2r+1)\pi$. This follows from Case 2 by considering the sequence $x'_n = x_n - 2r\pi$. Note that
$$x'_{n+1} = x_{n+1} - 2r\pi  = x_n + \sin x_n - 2r\pi  = x_n - 2r\pi + \sin (x_n - 2r\pi)  = x'_n + \sin x'_n .$$
Case 4. $(2r-1)\pi < x_0 < 2r\pi$.
Then $(x_n)$ is strictly monotonically decreasing such that all $x_n > (2r-1)\pi$ and $\lim x_n = (2r-1)\pi$. This follows from Case 3 by considering the sequence $x'_n = -x_n$. Note that that $2(-r)\pi = -2r\pi < x'_0 = -x_0  < -(2r-1)\pi = (2(-r)+1)\pi$ and
$$x'_{n+1} = -x_{n+1} = -x_n - \sin x_n = -x_n +\sin(-x_n) = x'_n +\sin x'_n . $$
