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%I #34 Nov 06 2016 08:57:51
%S 4,5,4,9,1,1,2,5,5,6,5,0,7,7,4,3,2,3,9,2,0,3,2,2,5,0,3,9,6,9,0,2,9,6,
%T 7,7,7,9,7,7,7,5,1,5,7,1,2,1,2,5,5,3,0,9,7,8,5,2,9,4,1,0,1,2,5,6,2,6,
%U 3,8,4,8,1,7,4,2,5,6,4,3,0,8,4,6,0,0,4,9,4,5,2,0,9,7,4,1,6,9,4,3
%N Decimal approximation of x such that f(x)=4, where f is the Fibonacci function.
%H Eric W. Weisstein, <a href="http://mathworld.wolfram.com/FibonacciNumber.html">MathWorld: Fibonacci Number</a>
%F f(x) = (phi^x - cos(Pi*x) * phi^(-x))/sqrt(5), where phi = (1+sqrt(5))/2 (the golden ratio). The function f, a generalization over the reals of the Binet formula, gives Fibonacci numbers for integer values of x; e.g., f(3) = 2, f(4) = 3, f(5) = 5. [Corrected by _Daniel Forgues_, Oct 05 2016]
%e 4.549112556507743239203225039690296777977751571212553...
%t r = GoldenRatio; s = 1/Sqrt[5];
%t f[x_] := s*(r^x - Cos[Pi*x] * r^(-x));
%t x /. FindRoot[Fibonacci[x] == 4, {x, 5}, WorkingPrecision -> 100]
%t RealDigits[%, 10]
%t (Show[Plot[#1, #2], ListPlot[Table[{x, #1}, #2]]] &)[
%t Fibonacci[x], {x, -7, 7}] (* _Peter J. C. Moses_, Jun 21 2011 *)
%o (PARI) phi = (1+sqrt(5))/2; solve(x=4, 5, (phi^x - cos(Pi*x) * phi^(-x))/sqrt(5) - 4) \\ _Michel Marcus_, Oct 05 2016
%Y Cf. A192039, A192040, A192041, A192042, A192043, A192044 (these correspond to f(x) = 6, 7, 1/2, 3/2, phi, phi^2 respectively); A171909, A172081.
%K nonn,cons
%O 1,1
%A _Clark Kimberling_, Jun 21 2011
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