

A192408


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


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
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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 independant of excentricity.
In the case of a unit radius circle, total cut length is 4*sqrt(1t^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.
JeanFrançois Alcover, L. A. Graham, Ingenious Mathematical Problems and Methods, Dover, 1959, p. 6.
Eric Weisstein, Ellipse, MathWorld


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/6x*sqrt(1x^2))x) \\ Charles R Greathouse IV, Jun 30 2011


CROSSREFS

Sequence in context: A324651 A342808 A201895 * A074208 A333775 A334205
Adjacent sequences: A192405 A192406 A192407 * A192409 A192410 A192411


KEYWORD

nonn,cons


AUTHOR

JeanFrançois Alcover, Jun 30 2011


STATUS

approved



