OFFSET
1,2
COMMENTS
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.
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).
REFERENCES
Friedrich Katscher, Die Kubischen Gleichungen bei Nicolo Tartaglia, Verlag der Ă–sterreichischen Akademie der Wissenschaften, 2001, Wien, Aufgabe XXV, pp. 13-16.
LINKS
MacTutor History of Mathematics, Nicolo Tartaglia.
FORMULA
The real solution x1 to x^3 + 4*x - 13 = 0 is
x1 = (1/6)*((1404 + 12*sqrt(14457))^(1/3) - (-1404 + 12*sqrt(14457))^(1/3)).
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
y1 = 2.926590945638182088730632869966915335446... and
z1 = 5.809338022398154156043352227759054154638...
EXAMPLE
x1 = 1.797665494400461460989161943060236461340...
MATHEMATICA
RealDigits[ Solve[x^3 + 4*x - 13 == 0, x][[1, 1, 2]], 10, 111][[1]] (* Robert G. Wilson v, May 22 2015 *)
PROG
(PARI) polrootsreal(x^3+4*x-13)[1] \\ Charles R Greathouse IV, May 21 2015
CROSSREFS
KEYWORD
AUTHOR
Wolfdieter Lang, May 21 2015
STATUS
approved