Linear.Quaternion:$ccos from linear-1.19.1.3

Percentage Accurate: 100.0% → 100.0%

Time: 8.4s

Alternatives: 12

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

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

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 100.0%

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

   1. clear-num 100.0%

      \[\leadsto \sin x \cdot \color{blue}{\frac{1}{\frac{y}{\sinh y}}} \]

   2. un-div-inv 100.0%

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

3. Applied egg-rr 100.0%

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

4. Final simplification 100.0%

   \[\leadsto \frac{\sin x}{\frac{y}{\sinh y}} \]

Alternative 2: 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. Final simplification 100.0%

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

Alternative 3: 84.5% accurate, 1.8× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 0.38 \lor \neg \left(y \leq 3.6 \cdot 10^{+148}\right):\\ \;\;\;\;\sin x \cdot \left(1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot \frac{\sinh y}{y}\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (if (or (<= y 0.38) (not (<= y 3.6e+148)))
   (* (sin x) (+ 1.0 (* 0.16666666666666666 (* y y))))
   (* x (/ (sinh y) y))))

C

double code(double x, double y) {
	double tmp;
	if ((y <= 0.38) || !(y <= 3.6e+148)) {
		tmp = sin(x) * (1.0 + (0.16666666666666666 * (y * y)));
	} else {
		tmp = x * (sinh(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 <= 0.38d0) .or. (.not. (y <= 3.6d+148))) then
        tmp = sin(x) * (1.0d0 + (0.16666666666666666d0 * (y * y)))
    else
        tmp = x * (sinh(y) / y)
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if ((y <= 0.38) || !(y <= 3.6e+148)) {
		tmp = Math.sin(x) * (1.0 + (0.16666666666666666 * (y * y)));
	} else {
		tmp = x * (Math.sinh(y) / y);
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if (y <= 0.38) or not (y <= 3.6e+148):
		tmp = math.sin(x) * (1.0 + (0.16666666666666666 * (y * y)))
	else:
		tmp = x * (math.sinh(y) / y)
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if ((y <= 0.38) || !(y <= 3.6e+148))
		tmp = Float64(sin(x) * Float64(1.0 + Float64(0.16666666666666666 * Float64(y * y))));
	else
		tmp = Float64(x * Float64(sinh(y) / y));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if ((y <= 0.38) || ~((y <= 3.6e+148)))
		tmp = sin(x) * (1.0 + (0.16666666666666666 * (y * y)));
	else
		tmp = x * (sinh(y) / y);
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[Or[LessEqual[y, 0.38], N[Not[LessEqual[y, 3.6e+148]], $MachinePrecision]], N[(N[Sin[x], $MachinePrecision] * N[(1.0 + N[(0.16666666666666666 * N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(x * N[(N[Sinh[y], $MachinePrecision] / y), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 0.38 or 3.60000000000000006e148 < y

   1. Initial program 100.0%

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

   2. Taylor expanded in y around 0 87.2%

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

      1. unpow2 87.2%

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

   4. Simplified 87.2%

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

   if 0.38 < y < 3.60000000000000006e148

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 74.2%

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

4. Final simplification 85.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 0.38 \lor \neg \left(y \leq 3.6 \cdot 10^{+148}\right):\\ \;\;\;\;\sin x \cdot \left(1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot \frac{\sinh y}{y}\\ \end{array} \]

Alternative 4: 70.1% accurate, 1.8× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 0.192:\\ \;\;\;\;\sin x\\ \mathbf{elif}\;y \leq 2.45 \cdot 10^{+210}:\\ \;\;\;\;x \cdot \frac{\sinh y}{y}\\ \mathbf{else}:\\ \;\;\;\;y \cdot \left(y \cdot \left(\sin x \cdot 0.16666666666666666\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (if (<= y 0.192)
   (sin x)
   (if (<= y 2.45e+210)
     (* x (/ (sinh y) y))
     (* y (* y (* (sin x) 0.16666666666666666))))))

C

double code(double x, double y) {
	double tmp;
	if (y <= 0.192) {
		tmp = sin(x);
	} else if (y <= 2.45e+210) {
		tmp = x * (sinh(y) / y);
	} else {
		tmp = y * (y * (sin(x) * 0.16666666666666666));
	}
	return tmp;
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if (y <= 0.192d0) then
        tmp = sin(x)
    else if (y <= 2.45d+210) then
        tmp = x * (sinh(y) / y)
    else
        tmp = y * (y * (sin(x) * 0.16666666666666666d0))
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (y <= 0.192) {
		tmp = Math.sin(x);
	} else if (y <= 2.45e+210) {
		tmp = x * (Math.sinh(y) / y);
	} else {
		tmp = y * (y * (Math.sin(x) * 0.16666666666666666));
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if y <= 0.192:
		tmp = math.sin(x)
	elif y <= 2.45e+210:
		tmp = x * (math.sinh(y) / y)
	else:
		tmp = y * (y * (math.sin(x) * 0.16666666666666666))
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (y <= 0.192)
		tmp = sin(x);
	elseif (y <= 2.45e+210)
		tmp = Float64(x * Float64(sinh(y) / y));
	else
		tmp = Float64(y * Float64(y * Float64(sin(x) * 0.16666666666666666)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 0.192)
		tmp = sin(x);
	elseif (y <= 2.45e+210)
		tmp = x * (sinh(y) / y);
	else
		tmp = y * (y * (sin(x) * 0.16666666666666666));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[y, 0.192], N[Sin[x], $MachinePrecision], If[LessEqual[y, 2.45e+210], N[(x * N[(N[Sinh[y], $MachinePrecision] / y), $MachinePrecision]), $MachinePrecision], N[(y * N[(y * N[(N[Sin[x], $MachinePrecision] * 0.16666666666666666), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if y < 0.192

   1. Initial program 99.9%

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

   2. Taylor expanded in y around 0 68.8%

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

   if 0.192 < y < 2.45000000000000003e210

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 76.2%

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

   if 2.45000000000000003e210 < y

   1. Initial program 100.0%

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

   2. Taylor expanded in y around 0 100.0%

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

      1. unpow2 100.0%

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

   4. Simplified 100.0%

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

   5. Taylor expanded in y around inf 100.0%

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

      1. unpow2 100.0%

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

      2. *-commutative 100.0%

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

      3. associate-*r* 100.0%

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

      4. *-commutative 100.0%

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

      5. associate-*l* 88.4%

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

      6. *-commutative 88.4%

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

   7. Simplified 88.4%

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

4. Final simplification 71.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 0.192:\\ \;\;\;\;\sin x\\ \mathbf{elif}\;y \leq 2.45 \cdot 10^{+210}:\\ \;\;\;\;x \cdot \frac{\sinh y}{y}\\ \mathbf{else}:\\ \;\;\;\;y \cdot \left(y \cdot \left(\sin x \cdot 0.16666666666666666\right)\right)\\ \end{array} \]

Alternative 5: 62.4% accurate, 1.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 2.3 \cdot 10^{+17}:\\ \;\;\;\;\sin x\\ \mathbf{elif}\;y \leq 2.15 \cdot 10^{+64}:\\ \;\;\;\;{x}^{3} \cdot -0.16666666666666666\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (if (<= y 2.3e+17)
   (sin x)
   (if (<= y 2.15e+64)
     (* (pow x 3.0) -0.16666666666666666)
     (* x (+ 1.0 (* 0.16666666666666666 (* y y)))))))

C

double code(double x, double y) {
	double tmp;
	if (y <= 2.3e+17) {
		tmp = sin(x);
	} else if (y <= 2.15e+64) {
		tmp = pow(x, 3.0) * -0.16666666666666666;
	} else {
		tmp = x * (1.0 + (0.16666666666666666 * (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 <= 2.3d+17) then
        tmp = sin(x)
    else if (y <= 2.15d+64) then
        tmp = (x ** 3.0d0) * (-0.16666666666666666d0)
    else
        tmp = x * (1.0d0 + (0.16666666666666666d0 * (y * y)))
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (y <= 2.3e+17) {
		tmp = Math.sin(x);
	} else if (y <= 2.15e+64) {
		tmp = Math.pow(x, 3.0) * -0.16666666666666666;
	} else {
		tmp = x * (1.0 + (0.16666666666666666 * (y * y)));
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if y <= 2.3e+17:
		tmp = math.sin(x)
	elif y <= 2.15e+64:
		tmp = math.pow(x, 3.0) * -0.16666666666666666
	else:
		tmp = x * (1.0 + (0.16666666666666666 * (y * y)))
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (y <= 2.3e+17)
		tmp = sin(x);
	elseif (y <= 2.15e+64)
		tmp = Float64((x ^ 3.0) * -0.16666666666666666);
	else
		tmp = Float64(x * Float64(1.0 + Float64(0.16666666666666666 * Float64(y * y))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 2.3e+17)
		tmp = sin(x);
	elseif (y <= 2.15e+64)
		tmp = (x ^ 3.0) * -0.16666666666666666;
	else
		tmp = x * (1.0 + (0.16666666666666666 * (y * y)));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[y, 2.3e+17], N[Sin[x], $MachinePrecision], If[LessEqual[y, 2.15e+64], N[(N[Power[x, 3.0], $MachinePrecision] * -0.16666666666666666), $MachinePrecision], N[(x * N[(1.0 + N[(0.16666666666666666 * N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if y < 2.3e17

   1. Initial program 99.9%

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

   2. Taylor expanded in y around 0 68.1%

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

   if 2.3e17 < y < 2.1499999999999999e64

   1. Initial program 100.0%

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

   2. Taylor expanded in y around 0 2.7%

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

   3. Taylor expanded in x around 0 41.2%

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

   4. Taylor expanded in x around inf 41.0%

      \[\leadsto \color{blue}{-0.16666666666666666 \cdot {x}^{3}} \]
   5. Step-by-step derivation

      1. *-commutative 41.0%

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

   6. Simplified 41.0%

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

   if 2.1499999999999999e64 < y

   1. Initial program 100.0%

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

   2. Taylor expanded in y around 0 71.0%

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

      1. unpow2 71.0%

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

   4. Simplified 71.0%

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

   5. Taylor expanded in x around 0 68.1%

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

4. Final simplification 66.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 2.3 \cdot 10^{+17}:\\ \;\;\;\;\sin x\\ \mathbf{elif}\;y \leq 2.15 \cdot 10^{+64}:\\ \;\;\;\;{x}^{3} \cdot -0.16666666666666666\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\right)\\ \end{array} \]

Alternative 6: 69.3% accurate, 1.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 0.162:\\ \;\;\;\;\sin x\\ \mathbf{else}:\\ \;\;\;\;x \cdot \frac{\sinh y}{y}\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (if (<= y 0.162) (sin x) (* x (/ (sinh y) y))))

C

double code(double x, double y) {
	double tmp;
	if (y <= 0.162) {
		tmp = sin(x);
	} else {
		tmp = x * (sinh(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 <= 0.162d0) then
        tmp = sin(x)
    else
        tmp = x * (sinh(y) / y)
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (y <= 0.162) {
		tmp = Math.sin(x);
	} else {
		tmp = x * (Math.sinh(y) / y);
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if y <= 0.162:
		tmp = math.sin(x)
	else:
		tmp = x * (math.sinh(y) / y)
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (y <= 0.162)
		tmp = sin(x);
	else
		tmp = Float64(x * Float64(sinh(y) / y));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 0.162)
		tmp = sin(x);
	else
		tmp = x * (sinh(y) / y);
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[y, 0.162], N[Sin[x], $MachinePrecision], N[(x * N[(N[Sinh[y], $MachinePrecision] / y), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 0.162000000000000005

   1. Initial program 99.9%

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

   2. Taylor expanded in y around 0 68.8%

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

   if 0.162000000000000005 < y

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 78.8%

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

4. Final simplification 71.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 0.162:\\ \;\;\;\;\sin x\\ \mathbf{else}:\\ \;\;\;\;x \cdot \frac{\sinh y}{y}\\ \end{array} \]

Alternative 7: 61.9% accurate, 2.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 6.4 \cdot 10^{+50}:\\ \;\;\;\;\sin x\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (if (<= y 6.4e+50) (sin x) (* x (+ 1.0 (* 0.16666666666666666 (* y y))))))

C

double code(double x, double y) {
	double tmp;
	if (y <= 6.4e+50) {
		tmp = sin(x);
	} else {
		tmp = x * (1.0 + (0.16666666666666666 * (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 <= 6.4d+50) then
        tmp = sin(x)
    else
        tmp = x * (1.0d0 + (0.16666666666666666d0 * (y * y)))
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (y <= 6.4e+50) {
		tmp = Math.sin(x);
	} else {
		tmp = x * (1.0 + (0.16666666666666666 * (y * y)));
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if y <= 6.4e+50:
		tmp = math.sin(x)
	else:
		tmp = x * (1.0 + (0.16666666666666666 * (y * y)))
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (y <= 6.4e+50)
		tmp = sin(x);
	else
		tmp = Float64(x * Float64(1.0 + Float64(0.16666666666666666 * Float64(y * y))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 6.4e+50)
		tmp = sin(x);
	else
		tmp = x * (1.0 + (0.16666666666666666 * (y * y)));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[y, 6.4e+50], N[Sin[x], $MachinePrecision], N[(x * N[(1.0 + N[(0.16666666666666666 * N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 6.39999999999999966e50

   1. Initial program 99.9%

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

   2. Taylor expanded in y around 0 64.5%

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

   if 6.39999999999999966e50 < y

   1. Initial program 100.0%

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

   2. Taylor expanded in y around 0 65.9%

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

      1. unpow2 65.9%

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

   4. Simplified 65.9%

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

   5. Taylor expanded in x around 0 64.9%

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

4. Final simplification 64.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 6.4 \cdot 10^{+50}:\\ \;\;\;\;\sin x\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\right)\\ \end{array} \]

Alternative 8: 37.8% accurate, 22.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 0.013:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;0.16666666666666666 \cdot \left(x \cdot \left(y \cdot y\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (if (<= y 0.013) x (* 0.16666666666666666 (* x (* y y)))))

C

double code(double x, double y) {
	double tmp;
	if (y <= 0.013) {
		tmp = x;
	} else {
		tmp = 0.16666666666666666 * (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 <= 0.013d0) then
        tmp = x
    else
        tmp = 0.16666666666666666d0 * (x * (y * y))
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (y <= 0.013) {
		tmp = x;
	} else {
		tmp = 0.16666666666666666 * (x * (y * y));
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if y <= 0.013:
		tmp = x
	else:
		tmp = 0.16666666666666666 * (x * (y * y))
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (y <= 0.013)
		tmp = x;
	else
		tmp = Float64(0.16666666666666666 * Float64(x * Float64(y * y)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 0.013)
		tmp = x;
	else
		tmp = 0.16666666666666666 * (x * (y * y));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[y, 0.013], x, N[(0.16666666666666666 * N[(x * N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 0.0129999999999999994

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 52.8%

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

   3. Taylor expanded in y around 0 28.9%

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

   if 0.0129999999999999994 < y

   1. Initial program 100.0%

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

   2. Taylor expanded in y around 0 53.4%

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

      1. unpow2 53.4%

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

   4. Simplified 53.4%

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

   5. Taylor expanded in x around 0 51.7%

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

   6. Taylor expanded in y around inf 51.7%

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

      1. associate-*r* 51.7%

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

      2. *-commutative 51.7%

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

      3. associate-*l* 51.7%

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

      4. unpow2 51.7%

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

   8. Simplified 51.7%

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

   9. Taylor expanded in y around 0 51.7%

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

       1. unpow2 51.7%

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

       2. *-commutative 51.7%

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

   11. Simplified 51.7%

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

4. Final simplification 34.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 0.013:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;0.16666666666666666 \cdot \left(x \cdot \left(y \cdot y\right)\right)\\ \end{array} \]

Alternative 9: 48.5% accurate, 22.8× speedup

Math

\[\begin{array}{l} \\ x \cdot \left(1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\right) \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (* x (+ 1.0 (* 0.16666666666666666 (* y y)))))

C

double code(double x, double y) {
	return x * (1.0 + (0.16666666666666666 * (y * y)));
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = x * (1.0d0 + (0.16666666666666666d0 * (y * y)))
end function

Java

public static double code(double x, double y) {
	return x * (1.0 + (0.16666666666666666 * (y * y)));
}

Python

def code(x, y):
	return x * (1.0 + (0.16666666666666666 * (y * y)))

Julia

function code(x, y)
	return Float64(x * Float64(1.0 + Float64(0.16666666666666666 * Float64(y * y))))
end

MATLAB

function tmp = code(x, y)
	tmp = x * (1.0 + (0.16666666666666666 * (y * y)));
end

Wolfram

code[x_, y_] := N[(x * N[(1.0 + N[(0.16666666666666666 * N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 100.0%

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

2. Taylor expanded in y around 0 77.1%

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

   1. unpow2 77.1%

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

4. Simplified 77.1%

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

5. Taylor expanded in x around 0 44.6%

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

6. Final simplification 44.6%

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

Alternative 10: 48.4% accurate, 22.8× speedup

Math

\[\begin{array}{l} \\ x + x \cdot \left(y \cdot \left(y \cdot 0.16666666666666666\right)\right) \end{array} \]

FPCore

(FPCore (x y) :precision binary64 (+ x (* x (* y (* y 0.16666666666666666)))))

C

double code(double x, double y) {
	return x + (x * (y * (y * 0.16666666666666666)));
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = x + (x * (y * (y * 0.16666666666666666d0)))
end function

Java

public static double code(double x, double y) {
	return x + (x * (y * (y * 0.16666666666666666)));
}

Python

def code(x, y):
	return x + (x * (y * (y * 0.16666666666666666)))

Julia

function code(x, y)
	return Float64(x + Float64(x * Float64(y * Float64(y * 0.16666666666666666))))
end

MATLAB

function tmp = code(x, y)
	tmp = x + (x * (y * (y * 0.16666666666666666)));
end

Wolfram

code[x_, y_] := N[(x + N[(x * N[(y * N[(y * 0.16666666666666666), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 100.0%

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

2. Taylor expanded in y around 0 77.1%

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

   1. unpow2 77.1%

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

4. Simplified 77.1%

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

5. Taylor expanded in x around 0 44.6%

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

   1. +-commutative 44.6%

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

   2. distribute-lft-in 44.6%

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

   3. *-commutative 44.6%

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

   4. associate-*l* 44.6%

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

   5. *-rgt-identity 44.6%

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

7. Applied egg-rr 44.6%

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

8. Final simplification 44.6%

   \[\leadsto x + x \cdot \left(y \cdot \left(y \cdot 0.16666666666666666\right)\right) \]

Alternative 11: 30.0% accurate, 29.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 6.4 \cdot 10^{+50}:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot y}{y}\\ \end{array} \end{array} \]

FPCore

(FPCore (x y) :precision binary64 (if (<= y 6.4e+50) x (/ (* x y) y)))

C

double code(double x, double y) {
	double tmp;
	if (y <= 6.4e+50) {
		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 <= 6.4d+50) 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 <= 6.4e+50) {
		tmp = x;
	} else {
		tmp = (x * y) / y;
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if y <= 6.4e+50:
		tmp = x
	else:
		tmp = (x * y) / y
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (y <= 6.4e+50)
		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 <= 6.4e+50)
		tmp = x;
	else
		tmp = (x * y) / y;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[y, 6.4e+50], x, N[(N[(x * y), $MachinePrecision] / y), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 6.39999999999999966e50

   1. Initial program 99.9%

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

   2. Taylor expanded in x around 0 53.1%

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

   3. Taylor expanded in y around 0 27.0%

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

   if 6.39999999999999966e50 < y

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 83.0%

      \[\leadsto \color{blue}{x} \cdot \frac{\sinh y}{y} \]
   3. Step-by-step derivation

      1. *-commutative 83.0%

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

      2. associate-*l/ 83.0%

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

   4. Applied egg-rr 83.0%

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

   5. Taylor expanded in y around 0 18.8%

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

4. Final simplification 25.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 6.4 \cdot 10^{+50}:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot y}{y}\\ \end{array} \]

Alternative 12: 27.1% 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. Taylor expanded in x around 0 59.3%

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

3. Taylor expanded in y around 0 22.0%

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

4. Final simplification 22.0%

   \[\leadsto x \]

Reproduce

herbie shell --seed 2023224 
(FPCore (x y)
  :name "Linear.Quaternion:$ccos from linear-1.19.1.3"
  :precision binary64
  (* (sin x) (/ (sinh y) y)))