A192408 Decimal expansion of the solution to x = sin( Pi/6 - x*sqrt(1 - x^2) ).

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

Offset

0,1

Comments

Trisecting an ellipse area.

Given the ellipse x^2/a^2 + y^2/b^2 = 1, one way to trisect its area is to use the symmetric lines x = s and x = -s, s being the unique real solution to s = a*sin(Pi/6 - (s*sqrt(a^2 - s^2))/a^2).

Setting s = a * t, the equation in t becomes t = sin( Pi/6 - t*sqrt(1 - t^2) ), which is noticeably independent of eccentricity.

In the case of a unit radius circle, total cut length is 4*sqrt(1-t^2) = 3.857068297..., which is quite larger than cutting along 3 radii.

This constant is also the solution to an elementary problem involving two overlapping circles, known as "Mrs. Miniver's problem" (cf. S. R. Finch, p. 487). The distance between the centers of the two circles is 2*x = 0.5298641692...

References

Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 487.

Links

Table of n, a(n) for n=0..99.

L. A. Graham, Mrs Miniver's problem, Ingenious Mathematical Problems and Methods, Dover, 1959, p. 6.

Eric Weisstein's World of Mathematics, Ellipse.

Example

0.26493208460277686243411649476257106865019006604136445287874489329209025087...

Mathematica

RealDigits[ x /. FindRoot[x == Sin[Pi/6 - x*Sqrt[1 - x^2]], {x, 1/4}, WorkingPrecision -> 100]][[1]]

Prog

(PARI) solve(x=.2, .3, sin(Pi/6-x*sqrt(1-x^2))-x) \\ Charles R Greathouse IV, Jun 30 2011
(PARI) sin(solve(x=0, 1, sin(x)+x-Pi/3)/2) \\ Gleb Koloskov, Aug 25 2021

Crossrefs

Sequence in context: A324651 A342808 A201895 * A074208 A362229 A333775

Adjacent sequences: A192405 A192406 A192407 * A192409 A192410 A192411

Keyword

nonn,cons

Author

Jean-François Alcover, Jun 30 2011

Status

approved