Another Ramanujan Trick

[Valentin Albillo]

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Hi again, Gerson:

(09-30-2019 03:10 AM)Gerson W. Barbosa Wrote:  My second approximation uses 10 digits and produces 17 significant digits, but it requires four functions, twice as much compared to the first approximation you mention.

Funny, it seems that my counting methods and yours do differ. In your second approximation I count 13 digits (4,2,1,4,3,6,6,1,6,-6,-6,2,2) and 9 operations (3 additions, 2 divisions, 2 roots and 2 powers).

Quote:They have prompted me to search for an 82-digit one belonging to the same group. I’ve finally found it in Tito Pieza’s blog here

I did follow the link you mention but I was less than impressed. It may produce 82 correct digits but it's far too complex to be regarded as a remarkable approximation, the number of digits and operations needed is on the same ballpark as the number of correct digits produced.

Compare it with the two truly remarkable approximations I gave, or even the simple 355/133, where 6 digits and 1 division gets you almost 8 digits (save for 2 or 3 units in the last place, depending on rounding).

Best regards.
V.
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