[VA] SRC #010 - Pi Day 2022 Special

[Albert Chan]

(03-14-2022 10:29 PM)EdS2 Wrote:  I'm also interested in this process of intuiting the correction terms.
How sure can we be that what seem to be correct terms are in fact correct?

We can get correction term symbolically, to be "sure"

Correction term = product(e*(1-1/k^2)^(k^2), k=n+1 .. inf)
Instead of doing products, we sum the log's instead.

We estimate the size of correction using Euler-Maclaurin formula

XCAS> f := 1 + ln(1-1/x^2)*x^2
XCAS> c := int(f) - f/2 + f'/12 - f'''/720 :;

f = -x^-2/2 - x^-4/3 - x^-6/4 + ... ⇒ c(x = inf) = 0. No need to eval upper limit.

XCAS> C := exp(c)(x=n+1) :;       // PN/C ≈ pi
XCAS> series(C, n=inf, 7)

\(\displaystyle
1 + {1/2 \over n} - {1/8 \over n^2}
+ {13/144 \over n^3} - {77/1152 \over n^4}
+ {547/11520 \over n^5} - {13529/414720 \over n^6} + O\left({1\over n^7}\right)
\)

Continued fraction with taylor series (n=inf) that matches C coefs: (N = 2n+1)

\(\Large C =
1 + \frac{1}{(N - {1 \over 2}) \;
- \frac{1}{{36 \over 17}N \;
+ \frac{1}{{1445 \over 419}N \;+\;...}}}\)

Or, based from continued fraction approximation of little c: (again, N = 2n+1)

\(\Large \ln(C) =
\frac{1}{N \;
- \frac{1}{{9 \over 5}N \;
+ \frac{1}{{125 \over 8}N \;+\;...}}}\)