Use the --timeout flag to change the timeout.
(FPCore (x y z t a b c)
:precision binary64
(/
x
(+
x
(*
y
(exp
(*
2.0
(-
(/ (* z (sqrt (+ t a))) t)
(* (- b c) (- (+ a (/ 5.0 6.0)) (/ 2.0 (* t 3.0)))))))))))double code(double x, double y, double z, double t, double a, double b, double c) {
return x / (x + (y * exp((2.0 * (((z * sqrt((t + a))) / t) - ((b - c) * ((a + (5.0 / 6.0)) - (2.0 / (t * 3.0)))))))));
}
real(8) function code(x, y, z, t, a, b, c)
real(8), intent (in) :: x
real(8), intent (in) :: y
real(8), intent (in) :: z
real(8), intent (in) :: t
real(8), intent (in) :: a
real(8), intent (in) :: b
real(8), intent (in) :: c
code = x / (x + (y * exp((2.0d0 * (((z * sqrt((t + a))) / t) - ((b - c) * ((a + (5.0d0 / 6.0d0)) - (2.0d0 / (t * 3.0d0)))))))))
end function
public static double code(double x, double y, double z, double t, double a, double b, double c) {
return x / (x + (y * Math.exp((2.0 * (((z * Math.sqrt((t + a))) / t) - ((b - c) * ((a + (5.0 / 6.0)) - (2.0 / (t * 3.0)))))))));
}
def code(x, y, z, t, a, b, c): return x / (x + (y * math.exp((2.0 * (((z * math.sqrt((t + a))) / t) - ((b - c) * ((a + (5.0 / 6.0)) - (2.0 / (t * 3.0)))))))))
function code(x, y, z, t, a, b, c) return Float64(x / Float64(x + Float64(y * exp(Float64(2.0 * Float64(Float64(Float64(z * sqrt(Float64(t + a))) / t) - Float64(Float64(b - c) * Float64(Float64(a + Float64(5.0 / 6.0)) - Float64(2.0 / Float64(t * 3.0)))))))))) end
function tmp = code(x, y, z, t, a, b, c) tmp = x / (x + (y * exp((2.0 * (((z * sqrt((t + a))) / t) - ((b - c) * ((a + (5.0 / 6.0)) - (2.0 / (t * 3.0))))))))); end
code[x_, y_, z_, t_, a_, b_, c_] := N[(x / N[(x + N[(y * N[Exp[N[(2.0 * N[(N[(N[(z * N[Sqrt[N[(t + a), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / t), $MachinePrecision] - N[(N[(b - c), $MachinePrecision] * N[(N[(a + N[(5.0 / 6.0), $MachinePrecision]), $MachinePrecision] - N[(2.0 / N[(t * 3.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\frac{x}{x + y \cdot e^{2 \cdot \left(\frac{z \cdot \sqrt{t + a}}{t} - \left(b - c\right) \cdot \left(\left(a + \frac{5}{6}\right) - \frac{2}{t \cdot 3}\right)\right)}}