Talk:Chudnovsky algorithm

Plagiarism

The materials which I saw in this footnote are first wrong and also stolen from another web site. Sunos 6 (talk | contribs) 05:15, 9 April 2008 (UTC)

Usage

How can the reader use this algorithm - from what point he can certainly know the n-th digit after the decimal dot is correct? 79.179.42.44 (talk) 21:16, 16 February 2012 (UTC)

The error will be approximately equal to the next term, so by estimating very roughly the size of the next term, you know up to where the approximation is correct. The factor (6k)!/(3k)!k!^3 grows by a factor 693, 982, 1147, 1252 for the first 5 terms, and ~ 1500 for the next 20 terms. This is to be divided by 262537412640768000, which yields a ratio of ~ 1.5e14 between subsequent terms. If this is not wrong, it should yield roughly 14 more digits at each step. — MFH:Talk 17:50, 14 March 2018 (UTC)

multiple of e ?

The link between 640320 and e^(pi sqrt 163) is given, but is there a simple explanation for 13591409 x 2e-7 = 2.7182818 ~ e? — MFH:Talk 17:32, 14 March 2018 (UTC)

Integer division in the Python code

In the Python section, on the line

   M = (K**3 - 16*K) * M // k**3

Why is this an integer division given that the whole calculation is in arbitrary-precision floats? — Preceding unsigned comment added by Eje211 (talk • contribs) 01:21, 27 May 2018 (UTC)

possible reason

I tried changing '//' to '/' and got an error because '/' is not supported by Decimal objects. The good news is that I checked the output of the longest of the runs, and compared them with the output of a completely different algorithm, and they were identical. The code (based on https://gist.github.com/markhamilton1/9716714) is much simpler, but also pretty fast (they both run in about a second on my hardware).

   q, r, t, k, n, l = 1, 0, 1, 1, 3, 3                                                                                               
   while True:
       if 4*q+r-t < n*t:
           yield n
           nr = 10*(r-n*t)
           n  = ((10*(3*q+r))//t)-10*n
           q  *= 10
           r  = nr
       else:
           nr = (2*q+r)*l
           nn = (q*(7*k)+2+(r*l))//(t*l)
           q  *= k
           t  *= l
           l  += 2
           k += 1
           n  = nn
           r  = nr

kogorman (talk) 00:26, 10 October 2018 (UTC)

How to choose parameters

It's not clear to me what parameters to choose if I want to run this beyond the most precise example. What's the best relationship among maxk, prec and disp?

kogorman (talk) 00:34, 10 October 2018 (UTC)

Incorrect Algorithm

Please correct me if I am wrong, but I believe the formulae for M_q+1, K₀, and K_q+1 are incorrect as written:

${\begin{alignedat}{4}M_{q+1}&=M_{q}\cdot \left({\frac {(12q+2)(12q+6)(12q+10)}{(q+1)^{3}}}\right)\,\,&&{\textrm {where}}\,\,M_{0}&&=1\\[4pt]\end{alignedat}}$

The computation of M_q can be further optimized by introducing an additional term K_q as follows:

${\begin{alignedat}{4}K_{q+1}&=K_{q}+12\,\,&&{\textrm {where}}\,\,K_{0}&&=6\\[4pt]M_{q+1}&=M_{q}\cdot \left({\frac {K_{q}^{3}-16K_{q}}{(q+1)^{3}}}\right)\,\,&&{\textrm {where}}\,\,M_{0}&&=1\\[12pt]\end{alignedat}}$

---

I believe the formulae should instead be:

${\begin{alignedat}{4}M_{q+1}&=M_{q}\cdot \left({\frac {(12q-2)(12q-6)(12q-10)}{q^{3}}}\right)\,\,&&{\textrm {where}}\,\,M_{0}&&=1\\[4pt]\end{alignedat}}$

The computation of M_q can be further optimized by introducing an additional term K_q as follows:

${\begin{alignedat}{4}K_{q+1}&=K_{q}+12\,\,&&{\textrm {where}}\,\,K_{0}&&=-6\\[4pt]M_{q+1}&=M_{q}\cdot \left({\frac {K_{q}^{3}-16K_{q}}{q^{3}}}\right)\,\,&&{\textrm {where}}\,\,M_{0}&&=1\\[12pt]\end{alignedat}}$

By making the above changes, I am able to correctly calculate π in my python program.

ScowlingFlavoredJava (talk) 18:43, 4 June 2021 (UTC)