Linear.Quaternion:$ccos from linear-1.19.1.3

Percentage Accurate: 100.0% → 100.0%

Time: 7.6s

Alternatives: 7

Speedup: 1.0×

• 49.2% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

\[\begin{array}{l} \\ \sin x \cdot \frac{\sinh y}{y} \end{array} \]

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Initial Program: 100.0% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \sin x \cdot \frac{\sinh y}{y} \end{array} \]

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Alternative 1: 100.0% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \sin x \cdot \frac{\sinh y}{y} \end{array} \]

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 100.0%

   \[\sin x \cdot \frac{\sinh y}{y} \]
2. Add Preprocessing

3. Final simplification 100.0%

   \[\leadsto \sin x \cdot \frac{\sinh y}{y} \]
4. Add Preprocessing

Alternative 2: 53.8% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 1.36 \cdot 10^{+19}:\\ \;\;\;\;\sin x\\ \mathbf{elif}\;y \leq 3.4 \cdot 10^{+189} \lor \neg \left(y \leq 2.4 \cdot 10^{+290}\right):\\ \;\;\;\;\frac{x \cdot y}{y}\\ \mathbf{else}:\\ \;\;\;\;-0.16666666666666666 \cdot {x}^{3}\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (if (<= y 1.36e+19)
   (sin x)
   (if (or (<= y 3.4e+189) (not (<= y 2.4e+290)))
     (/ (* x y) y)
     (* -0.16666666666666666 (pow x 3.0)))))

C

double code(double x, double y) {
	double tmp;
	if (y <= 1.36e+19) {
		tmp = sin(x);
	} else if ((y <= 3.4e+189) || !(y <= 2.4e+290)) {
		tmp = (x * y) / y;
	} else {
		tmp = -0.16666666666666666 * pow(x, 3.0);
	}
	return tmp;
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if (y <= 1.36d+19) then
        tmp = sin(x)
    else if ((y <= 3.4d+189) .or. (.not. (y <= 2.4d+290))) then
        tmp = (x * y) / y
    else
        tmp = (-0.16666666666666666d0) * (x ** 3.0d0)
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (y <= 1.36e+19) {
		tmp = Math.sin(x);
	} else if ((y <= 3.4e+189) || !(y <= 2.4e+290)) {
		tmp = (x * y) / y;
	} else {
		tmp = -0.16666666666666666 * Math.pow(x, 3.0);
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if y <= 1.36e+19:
		tmp = math.sin(x)
	elif (y <= 3.4e+189) or not (y <= 2.4e+290):
		tmp = (x * y) / y
	else:
		tmp = -0.16666666666666666 * math.pow(x, 3.0)
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (y <= 1.36e+19)
		tmp = sin(x);
	elseif ((y <= 3.4e+189) || !(y <= 2.4e+290))
		tmp = Float64(Float64(x * y) / y);
	else
		tmp = Float64(-0.16666666666666666 * (x ^ 3.0));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 1.36e+19)
		tmp = sin(x);
	elseif ((y <= 3.4e+189) || ~((y <= 2.4e+290)))
		tmp = (x * y) / y;
	else
		tmp = -0.16666666666666666 * (x ^ 3.0);
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[y, 1.36e+19], N[Sin[x], $MachinePrecision], If[Or[LessEqual[y, 3.4e+189], N[Not[LessEqual[y, 2.4e+290]], $MachinePrecision]], N[(N[(x * y), $MachinePrecision] / y), $MachinePrecision], N[(-0.16666666666666666 * N[Power[x, 3.0], $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if y < 1.36e19

   1. Initial program 100.0%

      \[\sin x \cdot \frac{\sinh y}{y} \]
   2. Add Preprocessing

   3. Taylor expanded in y around 0 64.9%

      \[\leadsto \color{blue}{\sin x} \]

   if 1.36e19 < y < 3.39999999999999983e189 or 2.4000000000000001e290 < y

   1. Initial program 100.0%

      \[\sin x \cdot \frac{\sinh y}{y} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. *-commutative 100.0%

         \[\leadsto \color{blue}{\frac{\sinh y}{y} \cdot \sin x} \]

      2. associate-*l/ 100.0%

         \[\leadsto \color{blue}{\frac{\sinh y \cdot \sin x}{y}} \]

   4. Applied egg-rr 100.0%

      \[\leadsto \color{blue}{\frac{\sinh y \cdot \sin x}{y}} \]

   5. Taylor expanded in y around 0 2.6%

      \[\leadsto \frac{\color{blue}{y \cdot \sin x}}{y} \]

   6. Taylor expanded in x around 0 25.8%

      \[\leadsto \frac{\color{blue}{x \cdot y}}{y} \]

   if 3.39999999999999983e189 < y < 2.4000000000000001e290

   1. Initial program 100.0%

      \[\sin x \cdot \frac{\sinh y}{y} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. *-commutative 100.0%

         \[\leadsto \color{blue}{\frac{\sinh y}{y} \cdot \sin x} \]

      2. associate-*l/ 100.0%

         \[\leadsto \color{blue}{\frac{\sinh y \cdot \sin x}{y}} \]

   4. Applied egg-rr 100.0%

      \[\leadsto \color{blue}{\frac{\sinh y \cdot \sin x}{y}} \]

   5. Taylor expanded in y around 0 2.7%

      \[\leadsto \frac{\color{blue}{y \cdot \sin x}}{y} \]

   6. Taylor expanded in x around 0 20.0%

      \[\leadsto \color{blue}{x \cdot \left(1 + -0.16666666666666666 \cdot {x}^{2}\right)} \]
   7. Step-by-step derivation

      1. distribute-rgt-in 20.0%

         \[\leadsto \color{blue}{1 \cdot x + \left(-0.16666666666666666 \cdot {x}^{2}\right) \cdot x} \]

      2. *-lft-identity 20.0%

         \[\leadsto \color{blue}{x} + \left(-0.16666666666666666 \cdot {x}^{2}\right) \cdot x \]

      3. associate-*l* 20.0%

         \[\leadsto x + \color{blue}{-0.16666666666666666 \cdot \left({x}^{2} \cdot x\right)} \]

      4. unpow2 20.0%

         \[\leadsto x + -0.16666666666666666 \cdot \left(\color{blue}{\left(x \cdot x\right)} \cdot x\right) \]

      5. unpow3 20.0%

         \[\leadsto x + -0.16666666666666666 \cdot \color{blue}{{x}^{3}} \]

   8. Simplified 20.0%

      \[\leadsto \color{blue}{x + -0.16666666666666666 \cdot {x}^{3}} \]

   9. Taylor expanded in x around inf 19.7%

      \[\leadsto \color{blue}{-0.16666666666666666 \cdot {x}^{3}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 55.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 1.36 \cdot 10^{+19}:\\ \;\;\;\;\sin x\\ \mathbf{elif}\;y \leq 3.4 \cdot 10^{+189} \lor \neg \left(y \leq 2.4 \cdot 10^{+290}\right):\\ \;\;\;\;\frac{x \cdot y}{y}\\ \mathbf{else}:\\ \;\;\;\;-0.16666666666666666 \cdot {x}^{3}\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 61.8% accurate, 1.8× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 2.65 \cdot 10^{+19}:\\ \;\;\;\;\sin x\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(1 + 0.16666666666666666 \cdot {y}^{2}\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (if (<= y 2.65e+19)
   (sin x)
   (* x (+ 1.0 (* 0.16666666666666666 (pow y 2.0))))))

C

double code(double x, double y) {
	double tmp;
	if (y <= 2.65e+19) {
		tmp = sin(x);
	} else {
		tmp = x * (1.0 + (0.16666666666666666 * pow(y, 2.0)));
	}
	return tmp;
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if (y <= 2.65d+19) then
        tmp = sin(x)
    else
        tmp = x * (1.0d0 + (0.16666666666666666d0 * (y ** 2.0d0)))
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (y <= 2.65e+19) {
		tmp = Math.sin(x);
	} else {
		tmp = x * (1.0 + (0.16666666666666666 * Math.pow(y, 2.0)));
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if y <= 2.65e+19:
		tmp = math.sin(x)
	else:
		tmp = x * (1.0 + (0.16666666666666666 * math.pow(y, 2.0)))
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (y <= 2.65e+19)
		tmp = sin(x);
	else
		tmp = Float64(x * Float64(1.0 + Float64(0.16666666666666666 * (y ^ 2.0))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 2.65e+19)
		tmp = sin(x);
	else
		tmp = x * (1.0 + (0.16666666666666666 * (y ^ 2.0)));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[y, 2.65e+19], N[Sin[x], $MachinePrecision], N[(x * N[(1.0 + N[(0.16666666666666666 * N[Power[y, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 2.65e19

   1. Initial program 100.0%

      \[\sin x \cdot \frac{\sinh y}{y} \]
   2. Add Preprocessing

   3. Taylor expanded in y around 0 64.9%

      \[\leadsto \color{blue}{\sin x} \]

   if 2.65e19 < y

   1. Initial program 100.0%

      \[\sin x \cdot \frac{\sinh y}{y} \]
   2. Add Preprocessing

   3. Taylor expanded in y around 0 61.4%

      \[\leadsto \color{blue}{\sin x + 0.16666666666666666 \cdot \left({y}^{2} \cdot \sin x\right)} \]
   4. Step-by-step derivation

      1. associate-*r* 61.4%

         \[\leadsto \sin x + \color{blue}{\left(0.16666666666666666 \cdot {y}^{2}\right) \cdot \sin x} \]

   5. Simplified 61.4%

      \[\leadsto \color{blue}{\sin x + \left(0.16666666666666666 \cdot {y}^{2}\right) \cdot \sin x} \]

   6. Taylor expanded in x around 0 58.7%

      \[\leadsto \color{blue}{x \cdot \left(1 + 0.16666666666666666 \cdot {y}^{2}\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 63.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 2.65 \cdot 10^{+19}:\\ \;\;\;\;\sin x\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(1 + 0.16666666666666666 \cdot {y}^{2}\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 61.8% accurate, 1.8× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 4.2 \cdot 10^{+15}:\\ \;\;\;\;\sin x\\ \mathbf{else}:\\ \;\;\;\;0.16666666666666666 \cdot \left(x \cdot {y}^{2}\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (if (<= y 4.2e+15) (sin x) (* 0.16666666666666666 (* x (pow y 2.0)))))

C

double code(double x, double y) {
	double tmp;
	if (y <= 4.2e+15) {
		tmp = sin(x);
	} else {
		tmp = 0.16666666666666666 * (x * pow(y, 2.0));
	}
	return tmp;
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if (y <= 4.2d+15) then
        tmp = sin(x)
    else
        tmp = 0.16666666666666666d0 * (x * (y ** 2.0d0))
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (y <= 4.2e+15) {
		tmp = Math.sin(x);
	} else {
		tmp = 0.16666666666666666 * (x * Math.pow(y, 2.0));
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if y <= 4.2e+15:
		tmp = math.sin(x)
	else:
		tmp = 0.16666666666666666 * (x * math.pow(y, 2.0))
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (y <= 4.2e+15)
		tmp = sin(x);
	else
		tmp = Float64(0.16666666666666666 * Float64(x * (y ^ 2.0)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 4.2e+15)
		tmp = sin(x);
	else
		tmp = 0.16666666666666666 * (x * (y ^ 2.0));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[y, 4.2e+15], N[Sin[x], $MachinePrecision], N[(0.16666666666666666 * N[(x * N[Power[y, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 4.2e15

   1. Initial program 100.0%

      \[\sin x \cdot \frac{\sinh y}{y} \]
   2. Add Preprocessing

   3. Taylor expanded in y around 0 64.9%

      \[\leadsto \color{blue}{\sin x} \]

   if 4.2e15 < y

   1. Initial program 100.0%

      \[\sin x \cdot \frac{\sinh y}{y} \]
   2. Add Preprocessing

   3. Taylor expanded in y around 0 61.4%

      \[\leadsto \color{blue}{\sin x + 0.16666666666666666 \cdot \left({y}^{2} \cdot \sin x\right)} \]
   4. Step-by-step derivation

      1. associate-*r* 61.4%

         \[\leadsto \sin x + \color{blue}{\left(0.16666666666666666 \cdot {y}^{2}\right) \cdot \sin x} \]

   5. Simplified 61.4%

      \[\leadsto \color{blue}{\sin x + \left(0.16666666666666666 \cdot {y}^{2}\right) \cdot \sin x} \]

   6. Taylor expanded in x around 0 58.7%

      \[\leadsto \color{blue}{x \cdot \left(1 + 0.16666666666666666 \cdot {y}^{2}\right)} \]

   7. Taylor expanded in y around inf 58.7%

      \[\leadsto \color{blue}{0.16666666666666666 \cdot \left(x \cdot {y}^{2}\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 63.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 4.2 \cdot 10^{+15}:\\ \;\;\;\;\sin x\\ \mathbf{else}:\\ \;\;\;\;0.16666666666666666 \cdot \left(x \cdot {y}^{2}\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 54.0% accurate, 1.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 2.95 \cdot 10^{+17}:\\ \;\;\;\;\sin x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot y}{y}\\ \end{array} \end{array} \]

FPCore

(FPCore (x y) :precision binary64 (if (<= y 2.95e+17) (sin x) (/ (* x y) y)))

C

double code(double x, double y) {
	double tmp;
	if (y <= 2.95e+17) {
		tmp = sin(x);
	} else {
		tmp = (x * y) / y;
	}
	return tmp;
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if (y <= 2.95d+17) then
        tmp = sin(x)
    else
        tmp = (x * y) / y
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (y <= 2.95e+17) {
		tmp = Math.sin(x);
	} else {
		tmp = (x * y) / y;
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if y <= 2.95e+17:
		tmp = math.sin(x)
	else:
		tmp = (x * y) / y
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (y <= 2.95e+17)
		tmp = sin(x);
	else
		tmp = Float64(Float64(x * y) / y);
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 2.95e+17)
		tmp = sin(x);
	else
		tmp = (x * y) / y;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[y, 2.95e+17], N[Sin[x], $MachinePrecision], N[(N[(x * y), $MachinePrecision] / y), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 2.95e17

   1. Initial program 100.0%

      \[\sin x \cdot \frac{\sinh y}{y} \]
   2. Add Preprocessing

   3. Taylor expanded in y around 0 64.9%

      \[\leadsto \color{blue}{\sin x} \]

   if 2.95e17 < y

   1. Initial program 100.0%

      \[\sin x \cdot \frac{\sinh y}{y} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. *-commutative 100.0%

         \[\leadsto \color{blue}{\frac{\sinh y}{y} \cdot \sin x} \]

      2. associate-*l/ 100.0%

         \[\leadsto \color{blue}{\frac{\sinh y \cdot \sin x}{y}} \]

   4. Applied egg-rr 100.0%

      \[\leadsto \color{blue}{\frac{\sinh y \cdot \sin x}{y}} \]

   5. Taylor expanded in y around 0 2.7%

      \[\leadsto \frac{\color{blue}{y \cdot \sin x}}{y} \]

   6. Taylor expanded in x around 0 24.3%

      \[\leadsto \frac{\color{blue}{x \cdot y}}{y} \]
3. Recombined 2 regimes into one program.

4. Final simplification 55.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 2.95 \cdot 10^{+17}:\\ \;\;\;\;\sin x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot y}{y}\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 29.0% accurate, 20.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 2 \cdot 10^{+36}:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot y}{y}\\ \end{array} \end{array} \]

FPCore

(FPCore (x y) :precision binary64 (if (<= y 2e+36) x (/ (* x y) y)))

C

double code(double x, double y) {
	double tmp;
	if (y <= 2e+36) {
		tmp = x;
	} else {
		tmp = (x * y) / y;
	}
	return tmp;
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if (y <= 2d+36) then
        tmp = x
    else
        tmp = (x * y) / y
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (y <= 2e+36) {
		tmp = x;
	} else {
		tmp = (x * y) / y;
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if y <= 2e+36:
		tmp = x
	else:
		tmp = (x * y) / y
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (y <= 2e+36)
		tmp = x;
	else
		tmp = Float64(Float64(x * y) / y);
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 2e+36)
		tmp = x;
	else
		tmp = (x * y) / y;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[y, 2e+36], x, N[(N[(x * y), $MachinePrecision] / y), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 2.00000000000000008e36

   1. Initial program 100.0%

      \[\sin x \cdot \frac{\sinh y}{y} \]
   2. Add Preprocessing

   3. Taylor expanded in y around 0 80.6%

      \[\leadsto \color{blue}{\sin x + 0.16666666666666666 \cdot \left({y}^{2} \cdot \sin x\right)} \]
   4. Step-by-step derivation

      1. associate-*r* 80.6%

         \[\leadsto \sin x + \color{blue}{\left(0.16666666666666666 \cdot {y}^{2}\right) \cdot \sin x} \]

   5. Simplified 80.6%

      \[\leadsto \color{blue}{\sin x + \left(0.16666666666666666 \cdot {y}^{2}\right) \cdot \sin x} \]

   6. Taylor expanded in x around 0 46.0%

      \[\leadsto \color{blue}{x \cdot \left(1 + 0.16666666666666666 \cdot {y}^{2}\right)} \]

   7. Taylor expanded in y around 0 32.4%

      \[\leadsto \color{blue}{x} \]

   if 2.00000000000000008e36 < y

   1. Initial program 100.0%

      \[\sin x \cdot \frac{\sinh y}{y} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. *-commutative 100.0%

         \[\leadsto \color{blue}{\frac{\sinh y}{y} \cdot \sin x} \]

      2. associate-*l/ 100.0%

         \[\leadsto \color{blue}{\frac{\sinh y \cdot \sin x}{y}} \]

   4. Applied egg-rr 100.0%

      \[\leadsto \color{blue}{\frac{\sinh y \cdot \sin x}{y}} \]

   5. Taylor expanded in y around 0 2.7%

      \[\leadsto \frac{\color{blue}{y \cdot \sin x}}{y} \]

   6. Taylor expanded in x around 0 24.6%

      \[\leadsto \frac{\color{blue}{x \cdot y}}{y} \]
3. Recombined 2 regimes into one program.

4. Final simplification 30.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 2 \cdot 10^{+36}:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot y}{y}\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 26.0% accurate, 205.0× speedup

Math

\[\begin{array}{l} \\ x \end{array} \]

FPCore

(FPCore (x y) :precision binary64 x)

C

double code(double x, double y) {
	return x;
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = x
end function

Java

public static double code(double x, double y) {
	return x;
}

Python

def code(x, y):
	return x

Julia

function code(x, y)
	return x
end

MATLAB

function tmp = code(x, y)
	tmp = x;
end

Wolfram

code[x_, y_] := x

Derivation

1. Initial program 100.0%

   \[\sin x \cdot \frac{\sinh y}{y} \]
2. Add Preprocessing

3. Taylor expanded in y around 0 76.6%

   \[\leadsto \color{blue}{\sin x + 0.16666666666666666 \cdot \left({y}^{2} \cdot \sin x\right)} \]
4. Step-by-step derivation

   1. associate-*r* 76.6%

      \[\leadsto \sin x + \color{blue}{\left(0.16666666666666666 \cdot {y}^{2}\right) \cdot \sin x} \]

5. Simplified 76.6%

   \[\leadsto \color{blue}{\sin x + \left(0.16666666666666666 \cdot {y}^{2}\right) \cdot \sin x} \]

6. Taylor expanded in x around 0 49.0%

   \[\leadsto \color{blue}{x \cdot \left(1 + 0.16666666666666666 \cdot {y}^{2}\right)} \]

7. Taylor expanded in y around 0 26.0%

   \[\leadsto \color{blue}{x} \]

8. Final simplification 26.0%

   \[\leadsto x \]
9. Add Preprocessing

Reproduce

herbie shell --seed 2024079 
(FPCore (x y)
  :name "Linear.Quaternion:$ccos from linear-1.19.1.3"
  :precision binary64
  (* (sin x) (/ (sinh y) y)))