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Question

What is the most effective way to approximate $\pi$. There are many a way that we can approximate $\pi$. The way I am looking for the most effective approximation of $\pi$ is via the equations in forms of products or sums. This means that there will be an exclusion of the complex realm as you would have to approximate $e$ first as well as the $cos(x)$ and $sin(x)$ functions due to the involvement of $e$.

So far I am currently aware of 2 methods of the approximation of $\pi$ being. $${{\pi}\over{2}} = \prod_{n=1}^{\infty}{{4n^2}\over{4n^2-1}}$$

$${\pi^2\over 6}={\zeta (2)}=\sum_{n=1}^{\infty}{1\over n^2}$$

I have been able to calculate that the first product method is more accurate as when you apply a variable to the $\infty$ of both you are able to see that the product method approaches $\pi$ at a faster rate.

I am more so interested in what other methods there are of the approximation of forumlae of $\pi$ and I do not require anyone to make the calculations of which is more accurate. If anyone has any ideas, please leave them down below. Thank you.

Morgan
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2 Answers2

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The most accurate way to approximate a real number by a fraction with small denominator is its continued fraction decomposition.

Given a real number $\alpha \in \mathbb R$, we compute $a_0 := \lfloor \alpha \rfloor$ and $x_0 = \alpha$, and once $a_0, \ldots, a_k$ and $x_0, \ldots, x_k$ are computed, we set $$ x_{k+1} := \frac{1}{x_k - a_k}, ~~~~~~~~~~~~~~~~~~ a_{k+1} := \lfloor x_{k+1} \rfloor $$ so that we obtain inductively $$ x = a_0 + \frac{1}{a_1 + \frac{1}{{a_2 + \ddots}\atop{~~~~~~~~~~~~~~~~~~~a_k +\frac{1}{x_{k+1}}}}}, $$ that is, $$ x = a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \frac{1}{a_3 + \ddots}}}. $$ Each truncated so-called partial fraction is closest to the given real number among all fractions that have denominator less than or equal to it. By expanding the partial fraction, one obtains a series, which converges rapidly to the number one wants to approximate.

Unfortunately, the partial fraction decomposition of $\pi$ doesn't have an obvious pattern. If there were one, one would obtain a method of rapidly approximating $\pi$.

Cloudscape
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This is to answer the question "what is the quickest and most effective way of approximating $\pi$ on your calculator?" as you have not made the case that you need greater accuracy than this brings (10 to 12 decimal places).

To enter $\pi$ on your calculator then $4\arctan(1)$ is probably the quickest and most effective, but that hides how slow to converge the well known infinite series approximation for $\arctan(1)$ really is $\left( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{2n-1} \right)$.

If you have a summation function along with the factorial or nCr (Binomial combinations function) on your calculator then this series is probably the quickest and easiest to remember

$$\pi=\lim_\limits{m\to\infty}2\sum_{n=2}^m\frac{(n!)^22^n}{(2n)!}=\lim_\limits{m\to\infty}2\sum_{n=2}^m\frac{2^n}{nCr(2n,n)}=\lim_\limits{m\to\infty}2\sum_{n=2}^m\frac{2^n}{\binom{2n}{n}}$$

where $1+2\arcsin \left( \frac{1}{\sqrt{2}}\right)=\sum_{n=1}^{\infty}\frac{(n!)^2 2^n}{(2n)!}$

Directly entering the function $4\arcsin \left(\frac{1}{\sqrt{2}} \right)$ is reasonably fast on the calculator, but the standard infinite series form only involves irrational terms

$$\pi=\lim_\limits{m\to\infty} 4\sum_{n=0}^m \frac{\binom{2n}{n}}{2^{2n}(2n+1)(\sqrt{2})^{2n+1}}$$

James Arathoon
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