Linear.Quaternion:$csin from linear-1.19.1.3

Percentage Accurate: 100.0% → 100.0%

Time: 8.4s

Precision: binary64

Cost: 13120

• 48.3% of points produce a very large (infinite) output. You may want to add a precondition.

Math

\[\cos x \cdot \frac{\sinh y}{y} \]

↓

\[\cos x \cdot \frac{\sinh y}{y} \]

FPCore

(FPCore (x y) :precision binary64 (* (cos x) (/ (sinh y) y)))

↓

(FPCore (x y) :precision binary64 (* (cos x) (/ (sinh y) y)))

C

double code(double x, double y) {
	return cos(x) * (sinh(y) / y);
}

↓

double code(double x, double y) {
	return cos(x) * (sinh(y) / y);
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = cos(x) * (sinh(y) / y)
end function

↓

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = cos(x) * (sinh(y) / y)
end function

Java

public static double code(double x, double y) {
	return Math.cos(x) * (Math.sinh(y) / y);
}

↓

public static double code(double x, double y) {
	return Math.cos(x) * (Math.sinh(y) / y);
}

Python

def code(x, y):
	return math.cos(x) * (math.sinh(y) / y)

↓

def code(x, y):
	return math.cos(x) * (math.sinh(y) / y)

Julia

function code(x, y)
	return Float64(cos(x) * Float64(sinh(y) / y))
end

↓

function code(x, y)
	return Float64(cos(x) * Float64(sinh(y) / y))
end

MATLAB

function tmp = code(x, y)
	tmp = cos(x) * (sinh(y) / y);
end

↓

function tmp = code(x, y)
	tmp = cos(x) * (sinh(y) / y);
end

Wolfram

code[x_, y_] := N[(N[Cos[x], $MachinePrecision] * N[(N[Sinh[y], $MachinePrecision] / y), $MachinePrecision]), $MachinePrecision]

↓

code[x_, y_] := N[(N[Cos[x], $MachinePrecision] * N[(N[Sinh[y], $MachinePrecision] / y), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 100.0%

   \[\cos x \cdot \frac{\sinh y}{y} \]

2. Final simplification 100.0%

   \[\leadsto \cos x \cdot \frac{\sinh y}{y} \]

Alternatives

Alternative 1
Accuracy	100.0%
Cost	13120

\[\cos x \cdot \frac{\sinh y}{y} \]

Alternative 2
Accuracy	84.9%
Cost	7376

\[\begin{array}{l} t_0 := y \cdot \left(0.16666666666666666 \cdot \left(\cos x \cdot y\right)\right)\\ t_1 := \frac{1}{\frac{1 - 0.16666666666666666 \cdot \left(y \cdot y\right)}{1 - \left(y \cdot y\right) \cdot \left(\left(y \cdot y\right) \cdot 0.027777777777777776\right)}}\\ \mathbf{if}\;y \leq -1.32 \cdot 10^{+154}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;y \leq -1.1 \cdot 10^{+44}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;y \leq 1.8 \cdot 10^{+68}:\\ \;\;\;\;\cos x\\ \mathbf{elif}\;y \leq 1.35 \cdot 10^{+154}:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 3
Accuracy	85.2%
Cost	7376

\[\begin{array}{l} t_0 := 0.16666666666666666 \cdot \left(y \cdot y\right)\\ t_1 := y \cdot \left(0.16666666666666666 \cdot \left(\cos x \cdot y\right)\right)\\ t_2 := \frac{1}{\frac{1 - t_0}{1 - \left(y \cdot y\right) \cdot \left(\left(y \cdot y\right) \cdot 0.027777777777777776\right)}}\\ \mathbf{if}\;y \leq -1.32 \cdot 10^{+154}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;y \leq -1.15 \cdot 10^{+44}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;y \leq 7 \cdot 10^{+73}:\\ \;\;\;\;\cos x \cdot \left(1 + t_0\right)\\ \mathbf{elif}\;y \leq 1.35 \cdot 10^{+154}:\\ \;\;\;\;t_2\\ \mathbf{else}:\\ \;\;\;\;t_1\\ \end{array} \]

Alternative 4
Accuracy	78.9%
Cost	6860

\[\begin{array}{l} t_0 := 0.16666666666666666 \cdot \left(y \cdot y\right)\\ t_1 := \frac{1}{\frac{1 - t_0}{1 - \left(y \cdot y\right) \cdot \left(\left(y \cdot y\right) \cdot 0.027777777777777776\right)}}\\ \mathbf{if}\;y \leq -1.32 \cdot 10^{+154}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;y \leq -1.12 \cdot 10^{+44}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;y \leq 8.6 \cdot 10^{+66}:\\ \;\;\;\;\cos x\\ \mathbf{elif}\;y \leq 1.35 \cdot 10^{+154}:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;1 + t_0\\ \end{array} \]

Alternative 5
Accuracy	56.2%
Cost	1608

\[\begin{array}{l} t_0 := 0.16666666666666666 \cdot \left(y \cdot y\right)\\ \mathbf{if}\;y \leq -1.32 \cdot 10^{+154}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;y \leq 1.35 \cdot 10^{+154}:\\ \;\;\;\;\frac{1}{\frac{1 - t_0}{1 - \left(y \cdot y\right) \cdot \left(\left(y \cdot y\right) \cdot 0.027777777777777776\right)}}\\ \mathbf{else}:\\ \;\;\;\;1 + t_0\\ \end{array} \]

Alternative 6
Accuracy	47.1%
Cost	585

\[\begin{array}{l} \mathbf{if}\;y \leq -2.4 \lor \neg \left(y \leq 2.5\right):\\ \;\;\;\;0.16666666666666666 \cdot \left(y \cdot y\right)\\ \mathbf{else}:\\ \;\;\;\;1\\ \end{array} \]

Alternative 7
Accuracy	47.2%
Cost	448

\[1 + 0.16666666666666666 \cdot \left(y \cdot y\right) \]

Alternative 8
Accuracy	29.2%
Cost	64

\[1 \]

Reproduce

herbie shell --seed 2023272 
(FPCore (x y)
  :name "Linear.Quaternion:$csin from linear-1.19.1.3"
  :precision binary64
  (* (cos x) (/ (sinh y) y)))