[Math] How are the known digits of $\pi$ guaranteed

decimal-expansionpi

When discussing with my son a few of the many methods to calculate the digits of $\pi$ (15 yo school level), I realized that the methods I know more or less (geometric approximation, Monte Carlo and basic series) are all convergent but none of them explicitly states that the $n$-th digit calculated at some point is indeed a true digit (that it will not change in further calculations).

To take an example, the Gregory–Leibniz series gives us, for each step:

$$
\begin{align}
\frac{4}{1} & = 4\\
\frac{4}{1}-\frac{4}{3} & = 2.666666667…\\
\frac{4}{1}-\frac{4}{3}+\frac{4}{5} & = 3.466666667…\\
\frac{4}{1}-\frac{4}{3}+\frac{4}{5}-\frac{4}{7} & = 2.895238095…
\end{align}
$$

The integer part has changed four times in four steps. Why would we know that $3$ is the correct first digit?

Similarly in Monte Carlo: the larger the sample, the better the result but do we mathematically know that "now that we tried [that many times], we are mathematically sure that $\pi$ starts with $3$".

In other words:

  • does each of the techniques to calculate $\pi$ (or at least the major ones) have a proof that a given digit is now correct?
  • if not, what are examples of the ones which do and do not have this proof?

Note: The great answers so far (thank you!) mention a proof on a specific technique, and/or a proof that a specific digit is indeed the correct one. I was more interested to understand if this applies to all of the (major) techniques (= whether they all certify that this digit is guaranteed correct).

Or that we have some which do (the ones in the two first answers for instance) and others do not (the further we go, the more precise the number but we do not know if something will not jump in at some step and change a previously stable digit. When typing this in and thinking on the fly, I wonder if this would not be a very bad technique in itself, due to that lack of stability)

Best Answer

Note that $\pi=6\arcsin\left(\frac12\right)$. So, since$$\arcsin(x)=\sum_{n=0}^\infty \frac1{2^{2n}}\binom{2n}n\frac{ x^{2n+1}}{2n+1},$$you have$$\pi=\sum_{n=0}^\infty\frac6{2^{4n+1}(2n+1)}\binom{2n}n.$$Now, for each $N\in\mathbb{Z}^+$, let$$S_N=\sum_{n=0}^N\frac6{2^{4n+1}(2n+1)}\binom{2n}n\text{ and let }R_N=\sum_{n=N+1}^\infty\frac6{2^{4n+1}(2n+1)}\binom{2n}n.$$Then:

  • $(\forall N\in\mathbb{Z}^+):\pi=S_N+R_N$;
  • the sequence $(S_N)_{N\in\mathbb{Z}_+}$ is strictly increasing and $\lim_{N\to\infty}S_N=\pi$. In particular, each $S_N$ is a better approximation of $\pi$ than the previous one.

Since$$(\forall n\in\mathbb N):\binom{2n}n<4^n=2^{2n},$$you have$$R_N<\sum_{n=N+1}^\infty\frac6{2^{2n+1}}=\frac1{4^N}.$$So, taking $N=0$, you get that $\pi=S_0+R_0$. But $S_0=3$ and $R_0<1$. So, the first digit of $\pi$ is $3$. If you take $N=3$, then $\pi=S_3+R_3$. But $S_3\approx3.14116$ and $R_3<0.015625$. So, the second digit is $1$. And so on…

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