Approximating $\pi$ using the sine function

If we have some approximation $x$ for $\pi$, it is possible to improve that approximation by calculating $\sin(x) + x$ if $x$ is sufficiently close to $\pi$. The reason why this works is that for $x \approx \pi$, $\sin(x) \approx \pi - x$ (note that $\sin'(\pi) = -1$), so $x + \sin(x) \approx x + \pi - x = \pi$.

I am interested in the number of good digits when approximating $\pi$ by iteratively applying this technique iteratively starting with the number $3$. In other words, I am interested in the following sequences:

$$ a_0=3; a_{n+1}=\sin(a_n)+a_n\\ b_n=\text{The number of digits of accuracy of }a_n $$

The first few elements of $b$ are $\{0, 3, 10, 32, 99, 300, 902, 2702\}$. I did not find this sequence in OEIS. Interestingly, the number of correct digits seems to almost triple with every step.

Why does this method of approximating $\pi$ triple the number of accurate digits? If this approximation or sequence has been studied before, any pointers are welcome as well.

Solution 1:

The Taylor Series for $\sin(x)$ for $x$ near $\pi$ says $$ \sin(x)=\sin(\pi-x)=(\pi-x)-\frac{(\pi-x)^3}6+O\!\left((\pi-x)^5\right) $$ Thus $$ x+\sin(x)-\pi=\frac{(x-\pi)^3}6+O\!\left((\pi-x)^5\right) $$ That is, $$ x_{n+1}-\pi\sim\frac{(x_n-\pi)^3}6 $$ which means the number of correct digits more than triples with each iteration ($d_n=3d_{n-1}+0.778$).

Solution 2:

The Taylor expansion at $x=\pi$ is $$\sin(x)= \pi-x + \frac{1}{6}(x-\pi)^3- O((x-\pi)^4)$$ $$\sin(x) +x = \pi + \frac{1}{6}(x-\pi)^3- O((x-\pi)^4)$$ Therefore $$a_{n+1}-\pi = \sin(a_n)+a_n-\pi = \frac{1}{6}(a_n-\pi)^3- O((a_n-\pi)^4)$$

This means that the correct digits triple with each step, after convergence has set-it.