Yet another π formula
|
01-09-2021, 09:22 PM
(This post was last modified: 11-06-2021 04:21 PM by Albert Chan.)
Post: #14
|
|||
|
|||
RE: Yet another π formula
Hi, Gerson W. Barbosa
I don't know how to quantify "optimal" correction. May be you want to proof this instead ? \(\pi = \displaystyle\left( 2 + 4\sum_{k=1}^{n}\frac{(-1)^{k-1}}{4k^{2}-1} \right ) + \lim_{k\rightarrow \infty }\left(\frac{(-1)^{n}} {b\;-\;\frac{3}{\large b+8\;-\; \frac{60}{\frac{\ddots }{b+4k(k-1)\;-\;\frac{k^2 (4k^2-1)}{ b+4k(k+1) }}}}}\right) ,\quad b = 2n(n+1) + {3\over2}\) This is not a proof, but I am convinced that above is true. Nice work First, try n=0, i.e. summation term = 0. Relied only on limit, convergence rate is bad, but it is indeed approach pi. >>> from gmpy2 import * >>> get_context().precision = 1000 # ~ 300 dec. digits >>> pi = const_pi() >>> for i in range(7): print i, format(pi - test_pi(n = 0, cf = 10**i), 'g') ... 0 0.297148 1 0.0675375 2 0.00772836 3 0.000784124 4 7.85271e-05 5 7.85385e-06 6 7.85397e-07 Now, try combination of summation and limit. >>> for i in range(7): print i, format(pi - test_pi(n = 9, cf = 10**i), 'g') ... 0 -5.30332e-21 1 -1.07719e-25 2 -1.47115e-39 3 -8.06555e-58 4 -9.64229e-77 5 -9.81688e-96 6 -9.83452e-115 >>> for i in range(7): print i, format(pi - test_pi(n = 10, cf = 10**i), 'g') ... 0 4.18442e-23 1 5.76716e-28 2 6.21078e-43 3 4.82312e-63 4 5.98619e-84 5 6.11764e-105 6 6.13095e-126 Note: test_pi() from previous post, modified to allow extra cf terms, beyond n. Note: unlike summation term, CF correction convergence is one-sided. |
|||
« Next Oldest | Next Newest »
|
User(s) browsing this thread: 1 Guest(s)