A257239 - OEIS

%I #20 Oct 27 2023 10:32:15

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

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

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

%N Decimal expansion of the real root of x^3 + 4*x - 13.

%C This is related to the third of thirty problems posed by Niccolò Tartaglia to Antonio Maria Fiore in the year 1535 (in Venice it was still 1534). See the Katscher reference [in German] pp. 14, 15.

%C The problem is: find me a number which when added to 4 times its cube root gives 13. That is z + z^(1/3) = 13, or, with z = x^3, x^3 + 4*x = 13, with real solution x1. The solution to the problem is then z1 = x1^3 = 13 - 4*x1 (see the formula and example section).

%D Friedrich Katscher, Die Kubischen Gleichungen bei Nicolo Tartaglia, Verlag der Österreichischen Akademie der Wissenschaften, 2001, Wien, Aufgabe XXV, pp. 13-16.

%H MacTutor History of Mathematics, <a href="http://www-history.mcs.st-andrews.ac.uk/Biographies/Tartaglia.html">Nicolo Tartaglia</a>.

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

%F The real solution x1 to x^3 + 4*x - 13 = 0 is

%F x1 = (1/6)*((1404 + 12*sqrt(14457))^(1/3) - (-1404 + 12*sqrt(14457))^(1/3)).

%F The two complex solutions are a + b*i and a - b*i, with a = -x1/2 and b = sqrt(3)*y1/2 where y1 = (1/6)*((1404+12*sqrt(14457))^(1/3) + (-1404 + 12*sqrt(14457))^(1/3)) with

%F y1 = 2.926590945638182088730632869966915335446... and

%F z1 = 5.809338022398154156043352227759054154638...

%e x1 = 1.797665494400461460989161943060236461340...

%t RealDigits[ Solve[x^3 + 4*x - 13 == 0, x][[1, 1, 2]], 10, 111][[1]] (* _Robert G. Wilson v_, May 22 2015 *)

%o (PARI) polrootsreal(x^3+4*x-13)[1] \\ _Charles R Greathouse IV_, May 21 2015

%Y Cf. A257235, A257236, A257237.

%K nonn,easy,cons

%O 1,2

%A _Wolfdieter Lang_, May 21 2015