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.
Best Answer
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 ($d_n=3d_{n-1}+0.778$).