Talk:Bailey–Borwein–Plouffe formula

Backporting?
I translated the article into German (see the Deutsch link). I changed the formulas a bit and think I discovered an error (multiplication by 16 instead of 16^n). --Marc van Woerkom 01:47, 7 Jan 2005 (UTC) A base-￼ BBP-type formula is a convergent series formula of the type ￼	(1)

where ￼and ￼are integer polynomials in ￼(Bailey 2000; Borwein and Bailey 2003, pp. 54 and 128-129). Bailey (2000) and Borwein and Bailey (2003, pp. 128-129) give a collection of such formulas: ￼	￼	￼	(2) ￼	￼	￼	(3) ￼	￼	￼	(4) ￼	￼	￼	(5) ￼	￼	￼	(6) ￼	￼	￼	(7) ￼	￼	￼	(8) ￼	￼	￼	(9) ￼	￼	￼	(10) ￼	￼	￼	(11) ￼	￼	￼	(12) ￼	￼	￼	(13) ￼	￼	￼	(14) ￼	￼	￼	(15) ￼	￼	￼	(16) ￼	￼	￼	(17) ￼	￼	￼	(18) ￼	￼	￼	(19) ￼	￼	￼	(20) ￼	￼	￼	(21) ￼	￼	￼	(22) ￼	￼	￼	(23) ￼	￼	￼	(24) ￼	￼	￼	(25) ￼	￼	￼	(26) ￼	￼	￼	(27) ￼	￼	￼	(28) ￼	￼	￼	(29) ￼	￼	￼	(30) ￼	￼	￼	(31)

where ￼is Catalan's constant, ￼is the hyperbolic volume of the figure eight knot complement, ￼is Clausen's integral, and ￼is also the hyperbolic volume of the knot complement of the figure eight knot. Another example is the Dirichlet L-series ￼	(32)

(Bailey and Borwein 2005; Bailey et al. 2007, pp. 5 and 62). Note that this sort of sum is closely related to the polygamma function since, for example, the above sum can also be written ￼	(33)

Borwein et al. (2004) have recently shown that ￼has no Machin-type BBP arctangent formula that is not binary, although this does not rule out a completely different scheme for digit-extraction algorithms in other bases. A beautiful example of a BBP-type formula in a non-integer base is ￼	(34)

where ￼is the golden ratio, found by B. Cloitre (Cloitre; Borwein and Chamberland 2005; Bailey et al. 2007, p. 277).

Chadman8000 (talk) 02:40, 5 March 2009 (UTC) chad miller


 * All the constant characters in the above render as "Unicode Character 'OBJECT REPLACEMENT CHARACTER' (U+FFFC)" on my system (MacOS X 10.13.4 Beta 17E197a). This is in both the page-view and edit-view. Is there a way to extract the Unicode code point hex from the page, or was this lost when Chadman8000 entered the text to begin with? Jimw338 (talk) 15:20, 30 March 2018 (UTC)
 * They were saved here as U+FFFC. --bdijkstra (talk) 17:35, 30 March 2018 (UTC)

Confusion from the start
Reference seems to made to a formula that, for any positive integer n, computes the n-th binary digit of pi, but no such formula is shown. I expect to see something like

n-th digit of pi = ... some formula involving n ...

Maybe I'm expecting too much? 213.122.8.48 (talk) 00:18, 5 March 2010 (UTC)


 * At first, I was confused, too, but then I followed the link to Spigot algorithm. This article explains how to compute the n-th digit using such a formula. Besides, it is also explained in the section BBP digit-extraction algorithm for π. – Adrianwn (talk) 10:03, 20 July 2010 (UTC)

BBP for arctan (1/3) and arctan (1/7)
$$

\arctan\left(\frac{1}{3}\right) = \sum_{k = 0}^{\infty}\left[ \frac{1}{2^{12k}} \left( \frac{2^{-2}}{8k + 1} + \frac{2^{-3}}{8k + 2} + \frac{2^{-5}}{8k + 3} - \frac{2^{-8}}{8k + 5}- \frac{2^{-9}}{8k + 6} -\frac{2^{-11}}{8k + 7}\right) \right] $$

$$

\arctan\left(\frac{1}{7}\right) = \sum_{k = 0}^{\infty}\left[ \frac{1}{2^{20k}} \left( \frac{2^{-3}}{8k + 1} + \frac{2^{-5}}{8k + 2} + \frac{2^{-8}}{8k + 3} - \frac{2^{-13}}{8k + 5}- \frac{2^{-15}}{8k + 6} -\frac{2^{-18}}{8k + 7}\right) \right] $$

Using the later identity and this arctangent relation $$ \frac{\pi}{4}=2\arctan\left(\frac{1}{2}\right)-\arctan\left(\frac{1}{7}\right)$$ obtain

$$

\pi = \sum_{k = 0}^{\infty}\left[ \frac{1}{2^{20k}} \left( \frac{2^{2}}{20k + 1} - \frac{2^{0}}{20k + 3} - \frac{2^{-4}}{20k + 5} - \frac{2^{-4}}{20k + 7}+ \frac{2^{-6}}{20k + 9}- \frac{2^{-8}}{20k + 11} +\frac{2^{-10}}{20k + 13}+\frac{2^{-14}}{20k + 15}+\frac{2^{-14}}{20k + 17}-\frac{2^{-16}}{20k + 19}-\frac{2^{-1}}{8k + 1}-\frac{2^{-6}}{8k + 3}+\frac{2^{-11}}{8k + 5}+\frac{2^{-16}}{8k + 7}\right) \right] $$

The proof begins by finding the arctangent and Ln components for the expression below for g=(1,2,3,4,5,6,7)and solving various linear equations

$$\sum_{k = 0}^{\infty}\left[ m^{k} \left( \frac{a_g}{8k+g} \right) \right]$$

Source: http://iamned.com/math

Simplification possible
$$ \pi = \sum_{k = 0}^{\infty}\left[ \frac{1}{16^k} \left( \frac{4}{8k + 1} - \frac{2}{8k + 4} - \frac{1}{8k + 5} - \frac{1}{8k + 6} \right) \right]$$

This can be simplified to $$ \pi = \sum_{k = 0}^{\infty}\left[ \frac{1}{16^k} \left( \frac{4}{8k + 1} - \frac{1}{4k + 2} - \frac{1}{8k + 5} - \frac{1}{8k + 6} \right) \right]$$

Pseudocode
I for one would be very grateful if someone could post a piece of pseudocode for how the n:th digit of pi is calculated. Please, do add this. Earl of Warwick 11:03, 28 February 2012 (UTC)

Time complexity
I think the following is important enough that it should be mentioned in the intro:
 * "Though the BBP formula can directly calculate the value of any given digit of π with less computational effort than formulas that must calculate all intervening digits, BBP remains linearithmic whereby successively larger values of n require increasingly more time to calculate; that is, the "further out" a digit is, the longer it takes BBP to calculate it, just like the standard π-computing algorithms."

When I first heard about this it was presented as if you could calculate the n:th digit in constant time which would have been truly amazing. Maybe it is immediately obvious to some that is not the case(?) but i doubt it is to everyone not familiar with the subject. 85.230.137.182 (talk) 15:16, 21 October 2013 (UTC)
 * It should be pretty obvious. From the formula you can already see that you would need at least a $$\sum_{k=n}^{n+C}$$-style calculation to compute the n-th digit. As n increases, the ratio between $1⁄16k$ and the other part increases, which slows down the convergence rate (i.e. causes C to increase). --Bdijkstra (talk) 22:46, 14 March 2015 (UTC)
 * Not at all obvious. I mean, all the pop-lit descriptions of it makes it sound like its constant time. You have to not only see the formula, but also think about it for a while. And even then, its not clear: is it time complexity O(N) or O(Nlog N)? I guess maybe its not O(N^2) but that's not obvious. One is left feeling that maybe there's some trick one didn't think of. 67.198.37.16 (talk) 18:23, 22 December 2020 (UTC)

Most efficient method for parallel computation of pi?
The section "BBP compared to other methods of computing π" states that using bbp to calculate individual digits is 'the fastest way to compute all the digits from 1 to n". This claim seems suspicious to me, because bbp is not the most efficient method for calculating digits of pi on a single (serial) computer. Furthermore, the source cited at the end of the paragraph does not corroborate said statement. Is there an alternative source for it that i missed? 131.174.192.244 (talk) 11:12, 11 July 2016 (UTC)

Formula to calculate the nth digit
The whole article seems to be about finding the nth digit of pi, yet doesn't even once state the formula for it. It has a formula for pi, and an explanation of digit-extraction--but never states what the digit extraction formula actually is, as a whole.

In the "The BBP formula for π" section, it claims: "Once this is done, the four summations are put back into the sum to π". We already have been presented a handful of summations in this section. Which "four summations" is it talking about? Seems confusing. Why not write the whole formula at once? — Preceding unsigned comment added by 173.179.131.54 (talk) 01:49, 20 March 2019 (UTC)
 * The formula was already present as the first equation in the entire article when you wrote this comment, and it is still there now. Double sharp (talk) 04:16, 13 September 2019 (UTC)

Not decimal confusing
Near the top the article says "This does not compute the nth decimal of π (i.e., in base 10).", which I think is true but could be taken as meaning, it is impossible to calculate the nth decimal digit by itself, whereas mathworld says it is. See https://mathworld.wolfram.com/Digit-ExtractionAlgorithm.html

Could the text be rephrased? Or perhaps just delete the sentence? Bill W102102 (talk) 19:18, 6 March 2022 (UTC)

CN
The "citation-needed" that has appeared is slightly idiotic. A citation is hardly possible for a general assumption. — Preceding unsigned comment added by 31.48.26.170 (talk) 13:08, 26 March 2019 (UTC)
 * See the sixth line from the top. — Preceding unsigned comment added by 31.48.26.170 (talk) 13:54, 26 March 2019 (UTC)
 * I added a citation and removed the "citations-needed" (refimprove) message. I've studied this area at least a small amount, and do not think that its a requirement that further citations be added. Yes, of course there are likely more, but I think the current amount is adequate. Sanpitch (talk) 20:54, 11 January 2021 (UTC)

0x
What is the utility of hexadecimal here? kencf0618 (talk) 12:35, 26 May 2024 (UTC)