A075778 Decimal expansion of the real root of x^3 + x^2 - 1.

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

Offset: 0

Author

Cino Hilliard, Oct 09 2002

Keywords

• nonn
• cons

Comments

Also decimal expansion of the root of x^(1/sqrt(x+1)) = (1/sqrt(x+1))^x. The root of (1/x)^(1/sqrt(x+1)) = (1/sqrt(x+1))^(1/x) is the golden ratio. - Michel Lagneau, Apr 17 2012

The following decomposition holds true: X^3 + X^2 - 1 = (X - r)*(X + i * e^(-i*a) * r^(-1/2))*(X - i * e^(i*a) * r^(-1/2)), where a = arcsin(1/(2*r^(3/2))), see A218197 for the decimal expansion of a and the paper of Witula et al. for details. - Roman Witula, Oct 22 2012

Examples

			0.7548776662466927600495088963585286918946066...
		

References

• Roman Witula, E. Hetmaniok, D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, submitted to Proceedings of the 15th International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012.

Links

• H. R. P. Ferguson, On a Generalization of the Fibonacci Numbers Useful in Memory Allocation Schema or All About the Zeroes of Z^k - Z^{k - 1} - 1, k > 0, Fibonacci Quarterly, Volume 14, Number 3, October, 1976 (see Table 2 p. 238).
• Index entries for algebraic numbers, degree 3

Crossrefs

Cf. A060006 (inverse), A210462, A210463.

Programs

Maple

A075778 := proc()
        1/3-root[3](25/2-3*sqrt(69)/2)/3 -root[3](25/2+3*sqrt(69)/2)/3;
          -% ;
end proc: # R. J. Mathar, Jan 22 2013

Mathematica

RealDigits[N[Solve[x^3 + x^2 - 1 == 0, x] [[1]] [[1, 2]], 111]] [[1]]
RealDigits[x /. FindRoot[x^3 + x^2 == 1, {x, 1}, WorkingPrecision -> 120]][[1]] (* Harvey P. Dale, Nov 23 2012 *)

PARI

solve(x=0, 1, x^3+x^2-1)

PARI

polrootsreal(x^3 + x^2 - 1)[1] \\ Charles R Greathouse IV, Jul 23 2020

Formula

Let 0 < a < 1 be any real number. Then a is the lesser and 1 is the greater and a^2/1 = 1/(a+1) and a^3 + a^2 - 1 = 0. Solving this using PARI we have 0.7548776662466927600495088964... . The general cubic can also be solved in radicals.

Equals -(1/3) + (1/3)*(25/2 - (3*sqrt(69))/2)^(1/3) + (1/3)*((1/2)*(25 + 3*sqrt(69)))^(1/3).

Extensions

More terms from Robert G. Wilson v, Oct 10 2002