Talk:Computing π

message: The symbol that wikipedia uses for pi is a disgrace.... it has NO, i repeat, NO curve to it. It looks like two small upper case 't's.... this is sad.... We need to replace it with a REAL pi symbol. One that doesn't suck!!! 24.166.154.108 (talk) 22:44, 27 May 2009 (UTC)
replies: text: It seems a little pointless now, but I did put in the serif font: . Kauffner (talk) 18:03, 19 April 2011 (UTC)
thread_title: WTF??

message: I removed the citations-needed template, since I can't find anything obvious that looks like it needs a citation. Lunkwill 23:33, 18 January 2007 (UTC)
replies: text: Just thought it might be a good idea to have some referances to sources that are respected as being fact. Deathbob 18:39, 19 January 2007 (UTC)
thread_title:  Citations? 

message: http://upload.wikimedia.org/math/3/a/8/3a87d816e9c1cfa57c6c4bf688a17f06.png
Does not evaluate to pi. ~~ —The preceding unsigned comment was added by 71.111.49.18 (talk) 05:06, 21 February 2007 (UTC)
thread_title:  Error 

message: I modified the second continued fraction from one for 4/π to one for π itself. Since the orignal article has been re-edited, I'll restate these continued fractions here.
\pi = 3 + \frac{1^2}{6+} \frac{3^2}{6+} \frac{5^2}{6+} \frac{7^2}{6+} \frac{9^2}{6+} \frac{11^2}{6+} \frac{13^2}{6+} ... = 3 - \sum_{n=1}^\infty \frac{(-1)^n}{n (n+1)(2n+1)} = 3 + \frac{1}{6} - \frac{1}{30} + \frac{1}{84} - \frac{1}{180} + \frac{1}{330} - \frac{1}{546} ... 
Its convergence slows rapidly as additional terms are computed. Not so the second. It is more difficult to compute:
\pi = \frac{4}{1+} \frac{1^2}{3+} \frac{2^2}{5+} \frac{3^2}{7+} \frac{4^2}{9+}  \frac{5^2}{11+} \frac{6^2}{13+} \frac{7^2}{15+} \frac{8^2}{17+} \frac{9^2}{19+} \frac{10^2}{21+} \frac{11^2}{23+} \frac{12^2}{25+} \frac{13^2}{27+} ... 
= \sum_{n=0}^\infty \frac{(-2)^n}{(n+1) \pi_n \pi_n+_1}, where 
\pi_0,\pi_1=0.5, \pi_2=2, \pi_3=4, \pi_4=17, \pi_5=37, \pi_6=164, \pi_7=368, \pi_8=1,667, \pi_9=3,803, \pi_1~_0
=17,452, \pi_1~_1=40,232, \pi_1~_2=186,218, \pi_1~_3=432,386, \pi_1~_4=2,013,368, \pi_1~_5=4,700,096, ...
Its convergence, however, is remarkably stable. For example, dividing the 10th term by the 11th gives -(11/20) x (40,232/3,803) = -5.8184..., and dividing the 14th by the 15th gives -(15/28) x (4,700,096/432,386) = -5.82328.... These quotients appear to converge toward -(3+2\sqrt{2}) or -5.828427..., and I needed to compute many thousands of terms to determine that this is indeed (at least, to 8 decimals) the case. Since log10(5.828) is about 0.765, this continued fraction adds better than 3/4 of a decimal digit of precision per term (more precisely, 13 digits per 17 terms).  Glenn L (talk) 08:21, 30 May 2009 (UTC)
thread_title:  Continued fractions 

message: I removed a large section that appears to be a copy/paste from somewhere, but that is almost useless and very confusing because it misses de formulae. Still it had some information that looks interesting at first glance, so I leave the edit diff here so that it may be used later. - Nabla 18:56, 19 June 2007 (UTC)
thread_title:  Large section removal 

message: I suggested this merge, following the AfD, Wikipedia:Articles for deletion/Software for calculating π, in which despite the Admin's hurried decision, it seemed that the consensus was very much to keep and merge here. The software mentioned is definetly non-notable, but the process, while only a small amount of content is worth saving here. - Jimmi Hugh (talk) 12:56, 12 September 2008 (UTC)
replies: text: Well that's 1 in favour... 0 against, so I'll do it. - Jimmi Hugh (talk) 08:11, 17 October 2008 (UTC)
thread_title:  Software for calculating π 

message: One of my favorite formulas for pi - due to its relative simplicity and the annoying fact of its slow speed - has always been neglected in this and related articles. Since Pi Day is this weekend, I feel I should share this simple, yet un-elegant solution:
:\pi=\sqrt{6 \sum_{k=1}^\infty \frac{1}{k^2}}\!
This short little root of a summation takes 7 iterations to reach 3 and (precisely) 600 to reach 3.14!
Its source has received some slight recognition, though I'm failing to find it here at the moment:
:\frac{\pi^2}{6}=\sum_{k=1}^\infty \frac{1}{k^2} = 1+\frac{1}{4}+\frac{1}{9}+\frac{1}{2}+\frac{1}{16}+\cdots\!
173.88.213.120 (talk) 20:19, 10 March 2009 (UTC)
replies: text: Your series is the Basel problem; that article have a bunch of references. I don't think it is relevant to this article unless the series has (notably) been used "in anger" to compute pi. It gets a bit of spotlight in the main Pi article, though.  –Henning Makholm (talk) 00:50, 11 March 2009 (UTC)
thread_title:  A bit disheartened... 

message: I can understand the intention of whoever created this subsection to avoid defining \pi in terms of itself as seen in the use of degrees instead of radians. But that is unsuccessful given the final result of . In order to evaluate  one has to convert  into radians, and that results in exactly \pi.
I think that we can replace the content of subsection with Liu Hui's π algorithm. Shall we copy some details here or just leave a reference to that article?
Kxx (talk) 09:08, 20 March 2009 (UTC)
replies: text: Pointing to Liu Hui would probably be a good idea. I have removed the circular method, which shows up here in some guise or another with some regularity (see "Pi as the limit of a trigonometric expression" above -- rats, you look away from the article for a year or two, and this thing happens). –Henning Makholm (talk) 12:11, 20 March 2009 (UTC)
thread_title:  Standard methods/As an infinite-sided polygon