An Elementary Proof of Wallis’ Product Formula for Pi (2005)

One can actually derive the Euler product for sine perfectly rigorously quite quickly, in just the manner Euler did!

(Euler is often said to have glibly moved from "sin(x)/x has roots at nonzero integer multiples of π" to "sin(x)/x = the product of (1 - x/(kπ)) for each nonzero integer k" in solving the Basel problem, simply ignoring the problem of the fact that many other functions have precisely the same zeros.

But Euler was not, in fact, always so glib in presenting this argument! For example, in Volume 1, Chapter 9 of his Introductio in Analysin Infinitorum (translation by Ian Bruce available at http://www.17centurymaths.com/contents/introductiontoanalysi...), we see that Euler argues for the product formula by noting (what would in modern notation be) that sin(x) = the imaginary part of e^(ix) = the imaginary part of (1 + ix/N)^N for infinitely large N, so that a factorization of sin(x) can be extracted from a factorization of the polynomial Im[(1 + ix/N)^N] for large N.

I'll write out in modern style the extraction of that factorization, but all the ideas are already present in the Euler:)

Let us denote Im[(1 + ix/N)^N] by f_N(x). Note that f_N is a polynomial of degree either N - 1 (if N is even) or N (if N is odd), whose degree 1 term is 1. Furthermore, its roots occur where x/N is the tangent of a multiple of π/N. Putting these together, we get that f_N(x)/x = the product of 1 - x/(N tan(kπ/N)) over the nonzero integers k in (-N/2, N/2).

As N approaches infinity, N tan(kπ/N) approaches kπ. Thus, we have that sin(x)/x = the product of 1 - x/(kπ) over the nonzero integers k.

[This last step is slightly glib, in that we've commuted limits without justification. We can rectify that by bundling together the factors where k differs only in sign, saying f_N(x)/x = the product of 1 - (x/(N tan(kπ/N)))^2 over k in (0, N/2). Now we note that the movement of the factors toward their limit is monotonic in N (considering the k-th factor to be 1 when N/2 <= k), which is sufficient justification for commuting the limits.]

If speed of convergence is what you seek, i've yet to discover a more impressive case than a surprisingly simple method attributed to Gauss using the geometric and arithmetic means of two numbers. Three times around the loop and it's already correct to five decimal places: http://pastebin.com/674YJ9VM

For another famous, but also very slow series converging to pi, try

pi/4 = 1 - 1/3 + 1/5 - 1/7 + ...

Is there a acceleration ? May be it converge much faster than rest of the method after say a billion precision point ?

Are slow convergences useful for testing purposes?

What is being tested with 100million function calls?

maybe I'm missing something (haven't tried to debug it), but that just results in 0.

edit: Oh that's interesting. Python 2 gave me 0, Python 3 worked fine.

I don't think anyone would call it trivial; it might be that the submitter erroneously translates elementary as trivial. Here elementary means that it doesn't use higher math (calculus, analysis, etc), it doesn't say anything about ease or complication, and indeed the elementary proof of the prime number theorem is hella complicated.

In fact, there is a negative correlation between a proof being elementary and a proof being trivial. The reason people like abstraction is that it makes difficult proofs trivial. The reason people don't like abstraction is that it makes elementary proofs non-elementary.