3 ms·
Even 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
by madcaptenor 8y ago
Even 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.