A188582 Decimal expansion of sqrt(2) - 1.

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, 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, 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

Offset: 0

Author

Robert G. Wilson v, Apr 04 2011

Keywords

• cons
• nonn
• changed

Comments

"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]

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

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

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

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

Examples

			0.414213562373095048801688724209698078569671875376948073...
		

References

• B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, p. 140, Entry 25.
• 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.
• Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, pp. 396 and 486.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..10000
• S. M. Abrarov and B. M. Quine, A Viète-like formula for pi based on infinite sum of the arctangent functions with nested radicals, figshare, 4509014, (2017).
• Steven R. Finch, Errata and Addenda to Mathematical Constants, p. 62.
• Olympiade Mathématique Belge, OMB 2008, Finale Maxi, Question 1.
• Index entries for algebraic numbers, degree 2
• Index to sequences related to Olympiads.

Crossrefs

Cf. A002193, A014176, A020807, A120731, A182168 (sin(Pi/8)), A144981 (cos(Pi/8)).

Programs

Magma

Sqrt(2) - 1; // G. C. Greubel, Jan 31 2018

Mathematica

RealDigits[ Sqrt[2] - 1, 10, 111][[1]]

PARI

sqrt(2) - 1 \\ G. C. Greubel, Jan 31 2018

Formula

Equals exp(asinh(cos(Pi))) = exp(asinh(-1)). - Geoffrey Caveney, Apr 23 2014

Equals tan(Pi/8) = A182168 / A144981 = 1 / A014176. - Bernard Schott, Apr 12 2022

From Antonio Graciá Llorente, Mar 15 2024: (Start)

Equals Product_{k >= 0} ((8*k - 1)*(8*k + 9))/((8*k - 5)*(8*k + 13)).

Equals Product_{k >= 1} A047554(k)/A047447(k). (End)

From Peter Bala, Mar 24 2024: (Start)

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.

For example, taking k = 0 and k = 1 yields

Equals 1/(2 + (1*3)/(4 + (5*7)/(4 + (9*11)/(4 + (13*15)/(4 + ... + (4*n + 1)*(4*n + 3)/(4 + ...)))))) and

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)

Tan(arctan(c) + arctan(c^3)) = 1/2. - Gary W. Adamson, Apr 04 2024