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Today is Pi Approximation Day, at least in countries which use DD/MM date format.

No new outstanding pi approximations, just a minor improvement

on one of Ramanujan's:

\(\sqrt[4]{\frac{2143+\left({6+\sqrt{6-\frac{6}{6^6}}}\right)^{-6}}{22}}\)

on one of mine:

\(\frac{\ln\left({\frac{16\times\ln\left({878}\right)}{\ln\left({16\ln\left({878​}\right)}\right)}}\right)}{1+\left({\frac{5}{94+\sqrt{2}}}\right)^8}\)


Happy Pi Approximation Day!

(Edited to fix a typo)

Impressive and mind-blowing.

Looks like both of you should receive the Nobel Prize!

(07-22-2023 09:00 PM)Gerson W. Barbosa Wrote: [ -> ][image]https://live.staticflickr.com/65535/53041849078_abc8fd75a9_z.jpg[/img]

Today is Pi Approximation Day, at least in countries which use DD/MM date format.

No new outstanding pi approximations, just a minor improvement

on one of Ramanujan's:

\(\sqrt[4]{\frac{2143+\left({6+\sqrt{6-\frac{6}{6^6}}}\right)^{-6}}{22}}\)

on one of mine:

\(\frac{\ln\left({\frac{16\times\ln\left({878}\right)}{\ln\left({16\ln\left({878​}\right)}\right)}}\right)}{1+\left({\frac{5}{94+\sqrt{2}}}\right)^8}\)


Happy Pi Approximation Day!

(Edited to fix a typo)

There's a recent Mathologer video:




Or you can read about it here: The Infinite series of Pi and it’s approximation by Madhava

This gives the following approximation:

\(
\begin{align}
\pi &\approx 4 \cdot \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \frac{37}{149 \cdot 6} \right) \\
&\approx 3.14159427448...
\end{align}
\)

(07-23-2023 11:52 PM)Thomas Klemm Wrote: [ -> ]There's a recent Mathologer video:




Or you can read about it here: The Infinite series of Pi and it’s approximation by Madhava

This gives the following approximation:

\(
\begin{align}
\pi &\approx 4 \cdot \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \frac{37}{149 \cdot 6} \right) \\
&\approx 3.14159427448...
\end{align}
\)

Nice video and article, thanks! I had watched the video already, but the article is new to me.

The correction term is equivalent to the continued fraction

\(\frac{1}{4\times6+\frac{1}{6+\frac{4}{4\times6}}}=\frac{37}{894}\)

For maximum efficiency the continued fraction would need to have four more terms, matching the number of terms of the series, as we have seen in this post from 2014.
Interestingly the video suggests that Madhava was able to derive the correction continued fraction by comparing a few partial sums of the series with the approximation 355/113, which gives only the first six decimal digits of pi, which is quite a remarkable feat. It was not difficult for me to find it, but I had at hand a wp34s set to double precision (34 significant digits).
This post on another subject might give an idea of how these continued fractions can be found numerically.

Gerson.