Talk:Chudnovsky algorithm: Difference between revisions
Appearance
Content deleted Content added
→Usage: Rougly 15 digits per k ? |
No edit summary |
||
Line 12: | Line 12: | ||
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? — [[User:MFH|MFH]]:[[User talk:MFH|Talk]] 17:32, 14 March 2018 (UTC) |
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? — [[User:MFH|MFH]]:[[User talk: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? |
Revision as of 01:21, 27 May 2018
Mathematics Stub‑class Low‑priority | ||||||||||
|
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?