Something related to the Bailey-Borwein-Plaff formula.
How to put this method into practice?
Something related to the Bailey-Borwein-Plaff formula.
How to put this method into practice?
The method of Buffon (Monte Carlo) try. More matches threw more accuracy. It may be easier to throw another match than to calculate the next member of the series. You can also use a dirty hack, insert a value already calculated by someone else into memory and request the Nth sign from there, then there will be infinite performance
From the formula and the statement â€śwhat can be calculated without the previous onesâ€ť it turns out that the kth hexadecimal character is calculated like this:
4/(8k+1) - 2/(8k+4) - 1/(8k+5) - 1/(8k+6)
but I donâ€™t really understand why the remnants do not affect the other signs as a result, let's try to calculate, say, the 100th sign:
4/(8*100+1) - 2/(8*100+4)-1/(8*100+5) - 1/(8*100+6) = 0,0012....
The total result affects 103 characters ... here are obviously the previous results you need .... i.e. one fig in order to accurately calculate the kth sign, you must know the results of <k. Or can there be some mega conversion with this formula?
The previous members are not needed, since by adding discharges we can only influence the older, younger digits do not change. But, therefore, we need, in addition to the N terms of the sum, to also count the Log <sub> 16 </ sub> (N) members in order to take into account their influence on the Nth hexadecimal digit. Somewhere on the Internet, an example of code was found, in principle, an algorithm is understood by reference to the code .
Source: https://ru.stackoverflow.com/questions/134088/