Approximating $\pi$ using the sine function

Question

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.

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

Answer 2

Your method is an improvement over Newton's method, which would be to look at the sequence $u_n$ defined by $u_0=3$ and $$u_{n+1}=u_n-\dfrac{\sin(u_n)}{\sin'(u_n)}$$ You're using some apriori knowledge about the root, hence getting a quicker convergence. Newton's method is known for doubling the number of accurate digits in each iteration.

Answer 3

Note that $$|a_{n+1}-\pi|=|a_{n}+\sin(a_n)-\pi|leq =|(a_{n}-\pi)-\sin(a_n-\pi)|\leq C\cdot\frac{|a_{n}-\pi|^3}{6}$$ because $\sin(t)=t-\frac{t^3}{6}+O(t^5)$ as $t\to 0$. So the order of convergence is $3$ So the decimal expansion of $a_{n+1}$ should have about three times more zeros than $a_n$.

Answer 4

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.

Answer 5

Denote $c_n = a_n - \pi$, then you ask how fast the sequence approaches zero. Now we have

$$\begin{align}c_{n+1} &= \sin(\pi + c_n) + c_n\\ &= c_n - \sin(c_n)\\ &= \frac{c_n^3}{6} +o(c_n^3)\end{align}$$

So the decimal expansion of $c_{n+1}$ should have roughly three times more zeros than $c_n$, which explains the tripling of accurate digits.

Answer 6

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\mbox{Newton-Raphson:}\ \left\{\begin{array}{rcl} \ds{x_{0}} & \ds{=} & \ds{\color{#f00}{3}} \\ \ds{x_{n}} & \ds{=} & \ds{x_{n - 1} - {\sin\pars{x_{n - 1}} \over \cos\pars{x_{n - 1}}} = x_{n - 1} - \tan\pars{x_{n - 1}}\,,\qquad n \geq 1} \end{array}\right.}$

$\texttt{Mathematica}$:

(*Newton-Raphson*)
Clear[n, x];
Module[{n = 0, x = N[3,50]},
While[n++ < 20, x -= N[Tan[x], 50]];
N[x, 50]]

Result:3.1415926535897932384626433832795028841971693993751

All the digits are 'correct'.