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%I #120 Apr 05 2024 00:40:38
%S 4,1,4,2,1,3,5,6,2,3,7,3,0,9,5,0,4,8,8,0,1,6,8,8,7,2,4,2,0,9,6,9,8,0,
%T 7,8,5,6,9,6,7,1,8,7,5,3,7,6,9,4,8,0,7,3,1,7,6,6,7,9,7,3,7,9,9,0,7,3,
%U 2,4,7,8,4,6,2,1,0,7,0,3,8,8,5,0,3,8,7,5,3,4,3,2,7,6,4,1,5,7,2,7,3,5,0,1
%N Decimal expansion of sqrt(2) - 1.
%C "In his Book 'The Theory of Poker,' David Sklansky coined the phrase 'Fundamental Theorem of Poker,' a tongue-in-cheek reference to the Fundamental Theorem of Algebra and Fundamental Theorem of Calculus from introductory texts on those two subjects. The constant [sqrt(2) - 1] appears so often in poker analysis that we will in the same vein go so far as to call it 'the golden mean of poker,' and we call it 'r' for short. We will see this value in a number of important results throughout this book." [Chen and Ankenman]
%C If a triangle has sides whose lengths form a harmonic progression in the ratio 1/(1 - d) : 1 : 1/(1 + d) then the triangle inequality condition requires that d be in the range 1 - sqrt(2) < d < sqrt(2) - 1. - _Frank M Jackson_, Oct 01 2013
%C This constant is the 6th smallest radius r < 1 for which a compact packing of the plane exists, with disks of radius 1 and r. - _Jean-François Alcover_, Sep 02 2014, after Steven Finch
%C This constant is also the largest argument of the arctangent function in the Viète-like formula for Pi given by Pi/2^(k+1) = arctan(sqrt(2 - a_(k-1))/a_k), where the index k >= 2 and the nested radicals are defined by recurrence using the relations a_k = sqrt(2 + a_(k-1)), a_1 = sqrt(2). When k = 2 the argument of the arctangent function sqrt(2 - a_1)/a_2 = sqrt(2 - sqrt(2))/sqrt(2 + sqrt(2)) = sqrt(2) - 1 is largest. Consequently, at k = 2 the Viète-like formula for Pi can be written as Pi/8 = arctan(sqrt(2 - sqrt(2))/sqrt(2 + sqrt(2))) = arctan(sqrt(2) - 1) (after Abrarov-Quine, see the article). - _Sanjar Abrarov_, Jan 07 2017
%C If r and R are respectively the inradius and the circumradius of a triangle, then the ratio r/R <= 1/2 (Euler inequality), and this maximum value 1/2 is obtained when the triangle is equilateral. Now, for a right triangle, the ratio r/R <= this constant = sqrt(2) - 1, and this maximum value sqrt(2) - 1 is obtained when the right triangle is isosceles. This is the answer to the question 1 of the Olympiade Mathématique Belge Maxi in 2008. - _Bernard Schott_, Sep 07 2022
%D B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, p. 140, Entry 25.
%D Bill Chen and Jerrod Ankenman, The Mathematics of Poker, Chpt 14 - You Don't Have To Guess: No-Limit Bet Sizing, p. 153, ConJelCo, LLC, Pittsburgh PA 2006.
%D Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, pp. 396 and 486.
%H G. C. Greubel, <a href="/A188582/b188582.txt">Table of n, a(n) for n = 0..10000</a>
%H S. M. Abrarov and B. M. Quine, <a href="https://dx.doi.org/10.6084/m9.figshare.4509014">A Viète-like formula for pi based on infinite sum of the arctangent functions with nested radicals</a>, figshare, 4509014, (2017).
%H Steven R. Finch, <a href="http://arxiv.org/abs/2001.00578">Errata and Addenda to Mathematical Constants</a>, p. 62.
%H Olympiade Mathématique Belge, <a href="http://omb.sbpm.be/modules/finale/article.php?storyid=33">OMB 2008, Finale Maxi, Question 1</a>.
%H <a href="/index/Al#algebraic_02">Index entries for algebraic numbers, degree 2</a>
%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.
%F Equals exp(asinh(cos(Pi))) = exp(asinh(-1)). - _Geoffrey Caveney_, Apr 23 2014
%F Equals tan(Pi/8) = A182168 / A144981 = 1 / A014176. - _Bernard Schott_, Apr 12 2022
%F From _Antonio Graciá Llorente_, Mar 15 2024: (Start)
%F Equals Product_{k >= 0} ((8*k - 1)*(8*k + 9))/((8*k - 5)*(8*k + 13)).
%F Equals Product_{k >= 1} A047554(k)/A047447(k). (End)
%F From _Peter Bala_, Mar 24 2024: (Start)
%F An infinite family of continued fraction expansions for this constant can be obtained from Berndt, Entry 25, by setting n = 1/2 and x = 8*k + 2 for k >= 0.
%F For example, taking k = 0 and k = 1 yields
%F Equals 1/(2 + (1*3)/(4 + (5*7)/(4 + (9*11)/(4 + (13*15)/(4 + ... + (4*n + 1)*(4*n + 3)/(4 + ...)))))) and
%F Equals (21/5) * 1/(10 + (1*3)/(20 + (5*7)/(20 + (9*11)/(20 + (13*15)/(20 + ... + (4*n + 1)*(4*n + 3)/(20 + ...)))))). (End)
%F Tan(arctan(c) + arctan(c^3)) = 1/2. - _Gary W. Adamson_, Apr 04 2024
%e 0.414213562373095048801688724209698078569671875376948073...
%t RealDigits[ Sqrt[2] - 1, 10, 111][[1]]
%o (PARI) sqrt(2) - 1 \\ _G. C. Greubel_, Jan 31 2018
%o (Magma) Sqrt(2) - 1; // _G. C. Greubel_, Jan 31 2018
%Y Cf. A002193, A014176, A020807, A120731, A182168 (sin(Pi/8)), A144981 (cos(Pi/8)).
%K cons,nonn
%O 0,1
%A _Robert G. Wilson v_, Apr 04 2011
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