Talk:Computing $π$

WTF??

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)[reply]

It seems a little pointless now, but I did put in the serif font:

π

. Kauffner (talk) 18:03, 19 April 2011 (UTC)[reply]

Projects

What Projects need to be researched besides Pi Hex and Background Pi? And does anyone know where good information about Pi Hex still exists since there website seems to have disappeared?

Citations?

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)[reply]

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)[reply]

Error

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).[reply]

I just checked with Mathematica 5.2 and it sais it's pi. Could you explain your reasoning? Deathbob 17:06, 24 February 2007 (UTC)[reply]

π

The following equation is TRUE!

\pi =({\frac {2(2\pi +3)+4}{2}}-5)({\frac {1}{2\pi }})({\frac {\pi ^{n}}{\pi ^{n-1}}})\,\!

It is, Briguychau. But that's just a complicated, unuseful way of expressing

\pi

. Why would you want to do that and why put it into an encyclopedia? You realize that

({\frac {1}{2\pi }})({\frac {\pi ^{n}}{\pi ^{n-1}}})={\frac {1}{2}}

, don't you? Ocolon 12:28, 1 March 2007 (UTC)[reply]

I just put this equation here to make people calculate it to the result

\pi =\pi \,\!

. Briguychau 2:39, 2 March 2007 (UTC)

I have never seen a number being calculated starting from its value.

I agree that is a perfectly useless equation. --Kiam (talk) 01:36, 23 February 2009 (UTC)[reply]

Continued fractions

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)[reply]

Large section removal

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)[reply]

Pi as the limit of a trigonometric expression

Hi there,
(forgive me if the english is not so clear)
one day I came up with a method of calculating the surface of a circle or any perfect polygon. I don't have background in mathematics so I am sure this method was thought of before and is probably taught in schools as well. (if there is an error in this method tell me)

I wanted to add it to the article, because I think it is easier to understand:
we calculate the surface based on the radius and the number of polygon sides
so Pi is not absolute but a function of number_of_sides.
in perfect polygons all the angles are the same.
let's start by thinking of a simple square
in the square, we draw a straight line to the center of the side. it will be 90degrees.
we'll call this the radius. we draw another line to the corner.
then we calculate the surface by using triangles.
(radius*half_of_side_length)/2 .. but we need the other half as well. so *2. meaning, one side is radius*half_of_side_length.
the whole surface is radius*half_of_side_length*number_of_sides.
good. now, we realize that the longer the radius becomes in such perfect polygons, the longer the side length will become.
so side length is a function of radius.
in the small triangle, we see that tan(angle)=radius/half_side_length
half_side_length=radius/tan(angle)
.. good.
now about the angles.
there is a way to calculate the sum of angles in polygons.
it's 180*(n-2) (triangle is 180, square is 360 ..etc.)
since it is a perfect polygon, the line to the corner we use in the little triangle "splits" the angle to two exact angles.
Surface was radius*half_of_side_length*number_of_sides
and half_side_length=radius/tan(angle)
Surface=radius*number_of_sides*radius/tan(angle)
angle is a function of number of sides, meaning: half_of(180*(n-2)/n)
S=(r*n*r)/tan(180*(n-2)/2n) = r^2*(n/tan(90-180/n))
we've reached the end.
if we say S=pi*r^2
let's check pi as n/tan(90-180/n)
pi(4)=4/tan(90-180/4)=4/tan(45)=4/1=4 ;; pi(4)=4
pi(100)=100/tan(90-180/100)=100/tan(88.2)=3.142626604.. ;; pi(100)=3.142626604
pi(1000000000)=3.14159265358979... which is closer to the perfect pi.
a circle has infinite sides, so pi of infinite is the pi that is normally talked about.
again, I'd like to say that I don't have background in math so if there's something obvious that doesn't work forgive me. I just thought of archimedes a little time ago, and how he liked triangles so much, and then I understood how to calculate this with triangles.
I liked that I understood it myself, and was happy to find it
since I understood it easily, I think the readers can also follow, especially with images and basic equations.

Thanks, kobi.
k_lurie@gbrener.org.il

Thanks for your effort. Unfortunately there are two separate reasons why your method is not appropriate for the article.

First, since you developed the formula yourself, it counts as "original research" by Wikipedia's rules. This is not allowed in a Wikipedia article; we report on new discoveries when they have been published in a respectable non-Wikipedia medium such as a scientific journal or a widely-used textbook. You may read more details about this at Wikipedia:No original research.

Second, your method is not a practical way to compute pi, because it requires that you alreay have a way to compute tangents when the angle is given in degrees. I assume you used a calculator to get the above numbers. What the calculator does when you ask for trigonometric functions in degrees is first to convert your argument to radians, and then compute the tangent of the radian value. The conversion from detrees to radians is by multiplying by π/180, and your method will not yield a pi that is better than the approximation that the calculator uses for the radian conversion. In other words, your method only allows calculation of pi if you already know it.

By the way, you're not the first to propose formulas of this kind; here is a discussion of an earlier, similar proposal. –Henning Makholm 17:16, 3 July 2007 (UTC)[reply]

ah ok, I understand now. I thought there must be something obvious that I missed. :)

so if I understand you correctly, for this method to work, we'd need to compute the radius and side ratio without using the angles in the triangle.

and so (according to your explanation) the current method is actually
pi=n/radtan((pi/2)-(pi/n))

if I had no mistake..

anyway, thank you for the clarification.

(I still think it may be easier to understand for those who read and want to understand what pi means, and its connection to a circle - not the formula, just the drawing and conception)

Software for calculating π

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)[reply]

Well that's 1 in favour... 0 against, so I'll do it. - Jimmi Hugh (talk) 08:11, 17 October 2008 (UTC)[reply]

A bit disheartened...

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)[reply]

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)[reply]

Standard methods/As an infinite-sided polygon

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 $π = lim n \to\infty n tan (180 ° / n)$ . In order to evaluate $tan (180 ° / n)$ one has to convert $180 °$ 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)[reply]

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)[reply]