5 ms·
The formula involves a parameter c=10^10 and initially seems unbelievable because one's initial inclination is to assume that the 10^10 is particularly importan
by xamuel 8y ago
The formula involves a parameter c=10^10 and initially seems unbelievable because one's initial inclination is to assume that the 10^10 is particularly important, like it has to be that particular value in order for the formula to work. The "trick" is that higher values of c, like c=10^19, give even more accuracy. Obviously c=10^10 was chosen to optimize the formula's aesthetic beauty.
- madcaptenor 8y agoEven c = 2 works. > n = seq(-1000, 1000); #assume 1000 is big enough to be infinity > sum(exp(-n^2/2))^2 / 2 [1] 3.141593
- miduil 8y agoWhich programming language is that? Edit: Ah, it's "R".
- madcaptenor 8y agoYes, it's R. Not necessarily the ideal language for this but I had an RStudio window open.