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Decimal expansion of 4*cos(2*Pi/13)*cos(3*Pi/13).
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%I #20 Oct 30 2023 23:53:28

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

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

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

%N Decimal expansion of 4*cos(2*Pi/13)*cos(3*Pi/13).

%C 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.

%C 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....

%C The roots of the cubic are permuted by the linear fractional transformation x -> - 1/(1 + x) of order 3:

%C r_1 = - 1/(1 + r_0); r_2 = - 1/(1 + r_1); r_0 = - 1/(1 + r_2).

%C 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.

%C 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.

%C 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).

%H T. W. Cusick and Lowell Schoenfeld, <a href="https://doi.org/10.1090/S0025-5718-1987-0866105-8">A table of fundamental pairs of units in totally real cubic fields</a>, Math. Comp. 48 (1987), 147-158

%H D. Shanks, <a href="http://dx.doi.org/10.1090/S0025-5718-1974-0352049-8">The simplest cubic fields</a>, Math. Comp., 28 (1974), 1137-1152.

%H <a href="/index/Al#algebraic_03">Index entries for algebraic numbers, degree 3</a>

%F r_0 = 2*(cos(Pi/13) + cos(5*Pi/13)).

%F r_0 = sin(4*Pi/13)*sin(6*Pi/13) / (sin(2*Pi/13)*sin(3*Pi/13)).

%F 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) ).

%F r_1 = 2*(cos(3*Pi/13) - cos(2*Pi/13)).

%F r_1 = - sin(Pi/13)*sin(5*Pi/13)/(sin(4*Pi/13)*sin(6*Pi/13)).

%F 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) ).

%F r_2 = 2*(cos(7*Pi/13) - cos(4*Pi/13)).

%F r_2 = - sin(2*Pi/13)*sin(3*Pi/13)/(sin(Pi/13)*sin(5*Pi/13)).

%F 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) ).

%F Equivalently, let z = exp(2*Pi*i/13). Then

%F r_0 = abs( (1 - z^4)*(1 - z^6)/((1 - z^2)*(1 - z^3)) );

%F r_1 = - abs( (1 - z)*(1 - z^5)/((1 - z^4)*(1 - z^6)) );

%F r_2 = - abs( (1 - z^2)*(1 - z^3)/((1 - z)*(1 - z^5)) ).

%F 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}.

%F Equals (-1)^(1/13) + (-1)^(5/13) - (-1)^(8/13) - (-1)^(12/13). - _Peter Luschny_, Nov 08 2021

%e 2.651093408937175306253240337787615403132441075705596684018767...

%p evalf(4*cos(2*Pi/13)*cos(3*Pi/13), 100);

%t RealDigits[4*Cos[2*Pi/13]*Cos[3*Pi/13], 10, 100][[1]] (* _Amiram Eldar_, Nov 08 2021 *)

%o (PARI) polrootsreal(x^3 - x^2 - 4*x - 1)[3] \\ _Charles R Greathouse IV_, Oct 30 2023

%Y Cf. A160389, A255240, A255241, A255249, A348721-A348729

%K nonn,cons,easy

%O 1,1

%A _Peter Bala_, Oct 31 2021