A154747 - OEIS

A154747

Decimal expansion of sqrt(sqrt(2) - 1), the radius vector of the point of bisection of the arc of the unit lemniscate (x^2 + y^2)^2 = x^2 - y^2 in the first quadrant.

6, 4, 3, 5, 9, 4, 2, 5, 2, 9, 0, 5, 5, 8, 2, 6, 2, 4, 7, 3, 5, 4, 4, 3, 4, 3, 7, 4, 1, 8, 2, 0, 9, 8, 0, 8, 9, 2, 4, 2, 0, 2, 7, 4, 2, 4, 4, 4, 0, 0, 7, 6, 5, 1, 1, 5, 6, 1, 5, 2, 0, 0, 9, 3, 5, 2, 0, 7, 4, 8, 5, 0, 3, 2, 1, 8, 3, 6, 5, 1, 9, 5, 4, 5, 1, 3, 4, 2, 4, 6, 5, 9, 5

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OFFSET

0,1

COMMENTS

A root of r^4 + 2 r^2 - 1 = 0.

Also real part of sqrt(1 + i)^3, where i is the imaginary unit such that i^2 = -1. - Alonso del Arte, Sep 09 2019

From Bernard Schott, Dec 19 2020: (Start)

Length of the shortest line segment which divides a right isosceles triangle with AB = AC = 1 into two parts of equal area; this is the answer to the 2nd problem proposed during the final round of the 18th British Mathematical Olympiad in 1993 (see link BM0 and Gardiner reference).

The length of this shortest line segment IJ with I on a short side and J on the hypotenuse is sqrt(sqrt(2)-1), and AI = AJ = 1/sqrt(sqrt(2)) = A228497 (see link Figure for B.M.O. 1993, Problem 2). (End)

This algebraic number and its negation equal the real roots of the quartic x^4 + 2*x^2 - 1 (minimal polynomial). The imaginary roots are +A278928*i and -A278928*i. - Wolfdieter Lang, Sep 23 2022

REFERENCES

A. Gardiner, The Mathematical Olympiad Handbook: An Introduction to Problem Solving, Oxford University Press, 1997, reprinted 2011, Problem 2, pages 56 and 104-105 (1993).

C. L. Siegel, Topics in Complex Function Theory, Volume I: Elliptic Functions and Uniformization Theory, Wiley-Interscience, 1969, page 5

LINKS

G. C. Greubel, Table of n, a(n) for n = 0..5000.

British Mathematical Olympiad 1993, Problem 2.

Bernard Schott, Figure for B.M.O. 1993, Problem 2.

Index entries for algebraic numbers, degree 4

Index to sequences related to curves.

Index to sequences related to Olympiads.

FORMULA

From Peter Bala, Jul 01 2024: (Start)

This constant occurs in the evaluation of Integral_{x = 0..Pi/2} sin^2(x)/(1 + sin^4(x)) dx = Pi/4 * sqrt(sqrt(2) - 1).

Equals (1/2)*Sum_{n >= 0} (-1/16)^n * binomial(4*n+2, 2*n+1) (a slowly converging series). (End)