OFFSET
1,1
COMMENTS
Let a be an integer and let p be a prime of the form a^2 + 3*a + 9 (see A005471). Shanks introduced a family of cyclic cubic fields generated by the roots of the polynomial x^3 - a*x^2 - (a + 3)*x - 1. The cubic polynomial has discriminant equal to p^2 and has three real roots, one positive and two negative. Here we consider the positive root in the case a = 1 corresponding to the prime p = 13. See A348721 and A348722 for the negative roots.
Let r_0 = 4*cos(2*Pi/13)*cos(3*Pi/13): r_0 is the positive root of the cubic equation x^3 - x^2 - 4*x - 1 = 0. The negative roots are r_1 = - 4*cos(4*Pi/13)*cos(6*Pi/13) = - 0.2738905549... and r_2 = - 4*cos(8*Pi/13)*cos(12*Pi/13) = - 1.3772028539....
The roots of the cubic are permuted by the linear fractional transformation x -> - 1/(1 + x) of order 3:
r_1 = - 1/(1 + r_0); r_2 = - 1/(1 + r_1); r_0 = - 1/(1 + r_2).
The quadratic mapping z -> z^2 - 2*z - 2 also cyclically permutes the roots. The mapping z -> - z^2 + z + 3 gives the inverse cyclic permutation of the roots.
The algebraic number field Q(r_0) is a totally real cubic field with class number 1 and discriminant equal to 13^2. The Galois group of Q(r_0) over Q is a cyclic group of order 3. See Shanks, Table 1, entry corresponding to a = 1.
The real numbers r_0 and 1 + r_0 are two independent fundamental units of the field Q(r_0). See Shanks. In Cusick and Schoenfeld, r_0 and r_1 (denoted there by E_1 and E_2) are taken as a fundamental pair of units (see case 4 in the table).
LINKS
T. W. Cusick and Lowell Schoenfeld, A table of fundamental pairs of units in totally real cubic fields, Math. Comp. 48 (1987), 147-158
D. Shanks, The simplest cubic fields, Math. Comp., 28 (1974), 1137-1152.
FORMULA
r_0 = 2*(cos(Pi/13) + cos(5*Pi/13)).
r_0 = sin(4*Pi/13)*sin(6*Pi/13) / (sin(2*Pi/13)*sin(3*Pi/13)).
r_0 = Product_{n >= 0} (13*n+4)*(13*n+6)*(13*n+7)*(13*n+9)/( (13*n+2)*(13*n+3)*(13*n+10)*(13*n+11) ).
r_1 = 2*(cos(3*Pi/13) - cos(2*Pi/13)).
r_1 = - sin(Pi/13)*sin(5*Pi/13)/(sin(4*Pi/13)*sin(6*Pi/13)).
r_1 = - Product_{n >= 0} (13*n+1)*(13*n+5)*(13*n+8)*(13*n+12)/( (13*n+4)*(13*n+6)*(13*n+7)*(13*n+9) ).
r_2 = 2*(cos(7*Pi/13) - cos(4*Pi/13)).
r_2 = - sin(2*Pi/13)*sin(3*Pi/13)/(sin(Pi/13)*sin(5*Pi/13)).
r_2 = - Product_{n >= 0} (13*n+2)*(13*n+3)*(13*n+10)*(13*n+11)/( (13*n+1)*(13*n+5)*(13*n+8)*(13*n+12) ).
Equivalently, let z = exp(2*Pi*i/13). Then
r_0 = abs( (1 - z^4)*(1 - z^6)/((1 - z^2)*(1 - z^3)) );
r_1 = - abs( (1 - z)*(1 - z^5)/((1 - z^4)*(1 - z^6)) );
r_2 = - abs( (1 - z^2)*(1 - z^3)/((1 - z)*(1 - z^5)) ).
Note: C = {1, 5, 8, 12} is the subgroup of nonzero cubic residues in the finite field Z_13 with cosets 2*C = {2, 3, 10, 11} and 4*C = {4, 6, 7, 9}.
Equals (-1)^(1/13) + (-1)^(5/13) - (-1)^(8/13) - (-1)^(12/13). - Peter Luschny, Nov 08 2021
EXAMPLE
2.651093408937175306253240337787615403132441075705596684018767...
MAPLE
evalf(4*cos(2*Pi/13)*cos(3*Pi/13), 100);
MATHEMATICA
RealDigits[4*Cos[2*Pi/13]*Cos[3*Pi/13], 10, 100][[1]] (* Amiram Eldar, Nov 08 2021 *)
PROG
(PARI) polrootsreal(x^3 - x^2 - 4*x - 1)[3] \\ Charles R Greathouse IV, Oct 30 2023
CROSSREFS
KEYWORD
AUTHOR
Peter Bala, Oct 31 2021
STATUS
approved