Get Billions of Correct Digits of Pi from a Wrong Formula (1999) [pdf]
academics.rowan.edu
academics.rowan.edu
Use the fact that the approximation in the paper is equal to the inner product of e^{-x^2} with a Dirac comb [1]. The amazing thing about the Gaussian function and the Dirac comb is that they're both preserved by the Fourier transform. So apply the Fourier transform to both of them, observe that the Fourier transform is unitary (essentially a rotation), and therefore doesn't affect the inner product; then expand the inner product of the Fourier transforms.
Essentially, it's the same proof, but not in the language of Theta functions. In fact, I'd argue it's a better proof, because it generalises.
Generalisation: This technique applies to all Riemann sums, as long as you can compute the Fourier transform of the function. The thinner the tails of the function's Fourier transform, the faster its Riemann sums will converge.
A better generalisation: if a function is thin-tailed, and its Fourier transform is thin-tailed, then it's Riemann sums converge to its integral at an exponential rate. In particular, this applies to the Gaussian function.
I'll check out what you said.
> n = seq(-1000, 1000); #assume 1000 is big enough to be infinity
> sum(exp(-n^2/2))^2 / 2
[1] 3.141593Edit: Ah, it's "R".
Other than that, it's just a Riemann sum presented in an aesthetically pleasing manner.
But the point is not to drop small terms at the extremes - it is still an infinite sum. The point is to approximate sections of the function by rectangles, and those rectangles are a bad approximation around 0 too.
Generally I'd expect a numerical integration scheme to have an error like h^k for some constant k - for example for Riemann sums the error goes like h^2 and for Simpson's rule like h^5. h is the distance between the sampling points and so is analogous to 1/c. So this doesn't just follow from Riemann summation.
(1 + 1/n)^n = e^(log(1 + 1/n) * n)
log(1 + t) = t - t^2/2 + t^3/3 - t^4/4 + ...
e^(log(1 + 1/n) * n)
= e^((1/n - 1/(2 n^2) + 1/(3 n^3) - ...) n)
= e^(1 - 1/(2n) + 1/(3n^2) - ...)
e - e/(2n) + O(1/n^2)
Edit: I'm an idiot, and blind. It was 'n', not pi.
https://en.wikipedia.org/wiki/Normal_number
Quoting:
> It is widely believed that the (computable) numbers √2, π, and e are normal, but a proof remains elusive.
Summary: It's "seems" quite random, but has yet to proven as such.
Pi is not a random number even if it digit sequence looks as pseudo-random sequence.
Basically there's an awful lot of room between zero and mathematical infinity, and normal large numbers that we experience or even try to reason about in everyday life don't and can't begin to stretch even a little way towards the ultimate mathematical limits.
Quoting:
> It is widely believed that the (computable) numbers √2, π, and e are normal, but a proof remains elusive.
Not all sequences are uniformly distributed, thus it very unlikely that we will be able to find long pieces of non-uniformly distributed sequences in uniformly distributed sequence.
Not as accurate, either, given that the title of the paper is "Get Billions and Billions of Correct Digits of pi from a Wrong Formula," and it was published before "click-bait" was a thing.