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A075778
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Decimal expansion of the real root of x^3 + x^2 - 1.
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24
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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
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OFFSET
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0,1
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COMMENTS
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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
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REFERENCES
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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.
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LINKS
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FORMULA
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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).
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EXAMPLE
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0.7548776662466927600495088963585286918946066...
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MAPLE
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1/3-root[3](25/2-3*sqrt(69)/2)/3 -root[3](25/2+3*sqrt(69)/2)/3;
-% ;
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MATHEMATICA
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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 *)
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PROG
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(PARI) solve(x=0, 1, x^3+x^2-1)
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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