Re: Pi Day's ramblings Message #31 Posted by Dieter on 15 Mar 2011, 12:58 p.m., in response to message #1 by Valentin Albillo
Here's my approach for the second challenge.
The key is a constant found by Ramanujan who showed that
R = e^{(Pi * sqrt(163))}
is almost an integer. In fact, the exact value for R is
R = 262 537 412 640 768 743,999 999 999 999 250...
and so
R = ~= 262 537 412 640 768 744
In other words:
Pi ~= sqrt(163) * ln R
The value of this expression agrees with Pi in its first 30 decimals.
Now, R also is the intermediate value 2,62537412641 E+17 mentioned in the challenge. At least is looks quite close  the given 12 digits are the same as in R.
So the remaining question is: Why is the real solution of the given cubic equation, raised to the 24th power, so close to R? Or, more precisely, indistinguishable in its first 12 digits.
The analytic solution to the given cubic equation is
x = 2 + (5 + sqrt(489)/9)^{1/3} + (5  sqrt(489)/9)^{1/3}
= 5,318 628 217 750 185 659 109 680 ...
Now, the 24th power of x is
x^{24} = 262 537 412 640 768 767,999 999 999 999 251 ...
which is quite exactly R + 24. Which in turn is close enough to R to that also
R ~= x^{24} ~= e^{(Pi * sqrt(163))}
or finally
Pi ~= ln(x^{24}) * sqrt(163)
which is the value that was calculated in the challenge.
This also explains why the Piapproximation gets much better if the
intermediate is corrected by 24, as Gerson mentioned.
However, there still is one thing left for the perfect solution:
Assume R resp. R^{1/24} is given. Now find a cubic equation with integer coefficents where its real solution is as close as possible to this value.
Dieter
Edited: 15 Mar 2011, 1:22 p.m.
