Linear.Quaternion:$cexp from linear-1.19.1.3

Percentage Accurate: 99.8% → 99.8%

Time: 5.4s

Alternatives: 9

Speedup: 1.0×

Specification

Math

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

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Initial Program: 99.8% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Alternative 1: 99.8% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 99.8%

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

   1. clear-num 99.7%

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

   2. un-div-inv 99.8%

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

3. Applied egg-rr 99.8%

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

4. Final simplification 99.8%

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

Alternative 2: 99.8% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 99.8%

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

2. Final simplification 99.8%

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

Alternative 3: 57.9% accurate, 9.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 1.85 \cdot 10^{+15}:\\ \;\;\;\;x \cdot \left(1 + \left(y \cdot y\right) \cdot -0.16666666666666666\right)\\ \mathbf{else}:\\ \;\;\;\;6 \cdot \frac{x}{y \cdot y}\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (if (<= y 1.85e+15)
   (* x (+ 1.0 (* (* y y) -0.16666666666666666)))
   (* 6.0 (/ x (* y y)))))

C

double code(double x, double y) {
	double tmp;
	if (y <= 1.85e+15) {
		tmp = x * (1.0 + ((y * y) * -0.16666666666666666));
	} else {
		tmp = 6.0 * (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 <= 1.85d+15) then
        tmp = x * (1.0d0 + ((y * y) * (-0.16666666666666666d0)))
    else
        tmp = 6.0d0 * (x / (y * y))
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (y <= 1.85e+15) {
		tmp = x * (1.0 + ((y * y) * -0.16666666666666666));
	} else {
		tmp = 6.0 * (x / (y * y));
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if y <= 1.85e+15:
		tmp = x * (1.0 + ((y * y) * -0.16666666666666666))
	else:
		tmp = 6.0 * (x / (y * y))
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (y <= 1.85e+15)
		tmp = Float64(x * Float64(1.0 + Float64(Float64(y * y) * -0.16666666666666666)));
	else
		tmp = Float64(6.0 * Float64(x / Float64(y * y)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 1.85e+15)
		tmp = x * (1.0 + ((y * y) * -0.16666666666666666));
	else
		tmp = 6.0 * (x / (y * y));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[y, 1.85e+15], N[(x * N[(1.0 + N[(N[(y * y), $MachinePrecision] * -0.16666666666666666), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(6.0 * N[(x / N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 1.85e15

   1. Initial program 99.8%

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

   2. Taylor expanded in y around 0 69.3%

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

      1. unpow2 69.3%

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

   4. Simplified 69.3%

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

   if 1.85e15 < y

   1. Initial program 99.7%

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

      1. clear-num 99.6%

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

      2. un-div-inv 99.7%

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

   3. Applied egg-rr 99.7%

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

   4. Taylor expanded in y around 0 32.0%

      \[\leadsto \frac{x}{\color{blue}{1 + 0.16666666666666666 \cdot {y}^{2}}} \]
   5. Step-by-step derivation

      1. unpow2 32.0%

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

   6. Simplified 32.0%

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

   7. Taylor expanded in y around inf 32.0%

      \[\leadsto \color{blue}{6 \cdot \frac{x}{{y}^{2}}} \]
   8. Step-by-step derivation

      1. unpow2 32.0%

         \[\leadsto 6 \cdot \frac{x}{\color{blue}{y \cdot y}} \]

   9. Simplified 32.0%

      \[\leadsto \color{blue}{6 \cdot \frac{x}{y \cdot y}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 59.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 1.85 \cdot 10^{+15}:\\ \;\;\;\;x \cdot \left(1 + \left(y \cdot y\right) \cdot -0.16666666666666666\right)\\ \mathbf{else}:\\ \;\;\;\;6 \cdot \frac{x}{y \cdot y}\\ \end{array} \]

Alternative 4: 57.9% accurate, 9.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 1.85 \cdot 10^{+15}:\\ \;\;\;\;x + -0.16666666666666666 \cdot \left(x \cdot \left(y \cdot y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;6 \cdot \frac{x}{y \cdot y}\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (if (<= y 1.85e+15)
   (+ x (* -0.16666666666666666 (* x (* y y))))
   (* 6.0 (/ x (* y y)))))

C

double code(double x, double y) {
	double tmp;
	if (y <= 1.85e+15) {
		tmp = x + (-0.16666666666666666 * (x * (y * y)));
	} else {
		tmp = 6.0 * (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 <= 1.85d+15) then
        tmp = x + ((-0.16666666666666666d0) * (x * (y * y)))
    else
        tmp = 6.0d0 * (x / (y * y))
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (y <= 1.85e+15) {
		tmp = x + (-0.16666666666666666 * (x * (y * y)));
	} else {
		tmp = 6.0 * (x / (y * y));
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if y <= 1.85e+15:
		tmp = x + (-0.16666666666666666 * (x * (y * y)))
	else:
		tmp = 6.0 * (x / (y * y))
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (y <= 1.85e+15)
		tmp = Float64(x + Float64(-0.16666666666666666 * Float64(x * Float64(y * y))));
	else
		tmp = Float64(6.0 * Float64(x / Float64(y * y)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 1.85e+15)
		tmp = x + (-0.16666666666666666 * (x * (y * y)));
	else
		tmp = 6.0 * (x / (y * y));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[y, 1.85e+15], N[(x + N[(-0.16666666666666666 * N[(x * N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(6.0 * N[(x / N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 1.85e15

   1. Initial program 99.8%

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

   2. Taylor expanded in y around 0 69.3%

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

      1. unpow2 69.3%

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

   4. Simplified 69.3%

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

      1. distribute-rgt-in 69.3%

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

      2. *-un-lft-identity 69.3%

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

      3. +-commutative 69.3%

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

      4. associate-*l* 69.3%

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

   6. Applied egg-rr 69.3%

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

   if 1.85e15 < y

   1. Initial program 99.7%

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

      1. clear-num 99.6%

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

      2. un-div-inv 99.7%

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

   3. Applied egg-rr 99.7%

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

   4. Taylor expanded in y around 0 32.0%

      \[\leadsto \frac{x}{\color{blue}{1 + 0.16666666666666666 \cdot {y}^{2}}} \]
   5. Step-by-step derivation

      1. unpow2 32.0%

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

   6. Simplified 32.0%

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

   7. Taylor expanded in y around inf 32.0%

      \[\leadsto \color{blue}{6 \cdot \frac{x}{{y}^{2}}} \]
   8. Step-by-step derivation

      1. unpow2 32.0%

         \[\leadsto 6 \cdot \frac{x}{\color{blue}{y \cdot y}} \]

   9. Simplified 32.0%

      \[\leadsto \color{blue}{6 \cdot \frac{x}{y \cdot y}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 59.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 1.85 \cdot 10^{+15}:\\ \;\;\;\;x + -0.16666666666666666 \cdot \left(x \cdot \left(y \cdot y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;6 \cdot \frac{x}{y \cdot y}\\ \end{array} \]

Alternative 5: 58.2% accurate, 11.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 2.5:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;6 \cdot \frac{x}{y \cdot y}\\ \end{array} \end{array} \]

FPCore

(FPCore (x y) :precision binary64 (if (<= y 2.5) x (* 6.0 (/ x (* y y)))))

C

double code(double x, double y) {
	double tmp;
	if (y <= 2.5) {
		tmp = x;
	} else {
		tmp = 6.0 * (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 <= 2.5d0) then
        tmp = x
    else
        tmp = 6.0d0 * (x / (y * y))
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (y <= 2.5) {
		tmp = x;
	} else {
		tmp = 6.0 * (x / (y * y));
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if y <= 2.5:
		tmp = x
	else:
		tmp = 6.0 * (x / (y * y))
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (y <= 2.5)
		tmp = x;
	else
		tmp = Float64(6.0 * Float64(x / Float64(y * y)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 2.5)
		tmp = x;
	else
		tmp = 6.0 * (x / (y * y));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[y, 2.5], x, N[(6.0 * N[(x / N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 2.5

   1. Initial program 99.8%

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

   2. Taylor expanded in y around 0 69.2%

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

   if 2.5 < y

   1. Initial program 99.6%

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

      1. clear-num 99.6%

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

      2. un-div-inv 99.6%

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

   3. Applied egg-rr 99.6%

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

   4. Taylor expanded in y around 0 31.5%

      \[\leadsto \frac{x}{\color{blue}{1 + 0.16666666666666666 \cdot {y}^{2}}} \]
   5. Step-by-step derivation

      1. unpow2 31.5%

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

   6. Simplified 31.5%

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

   7. Taylor expanded in y around inf 31.5%

      \[\leadsto \color{blue}{6 \cdot \frac{x}{{y}^{2}}} \]
   8. Step-by-step derivation

      1. unpow2 31.5%

         \[\leadsto 6 \cdot \frac{x}{\color{blue}{y \cdot y}} \]

   9. Simplified 31.5%

      \[\leadsto \color{blue}{6 \cdot \frac{x}{y \cdot y}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 59.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 2.5:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;6 \cdot \frac{x}{y \cdot y}\\ \end{array} \]

Alternative 6: 63.9% accurate, 11.7× speedup

Math

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

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 99.8%

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

   1. clear-num 99.7%

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

   2. un-div-inv 99.8%

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

3. Applied egg-rr 99.8%

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

4. Taylor expanded in y around 0 64.8%

   \[\leadsto \frac{x}{\color{blue}{1 + 0.16666666666666666 \cdot {y}^{2}}} \]
5. Step-by-step derivation

   1. unpow2 64.8%

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

6. Simplified 64.8%

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

7. Final simplification 64.8%

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

Alternative 7: 57.6% accurate, 14.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 5 \cdot 10^{-5}:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;y \cdot \frac{x}{y}\\ \end{array} \end{array} \]

FPCore

(FPCore (x y) :precision binary64 (if (<= y 5e-5) x (* y (/ x y))))

C

double code(double x, double y) {
	double tmp;
	if (y <= 5e-5) {
		tmp = x;
	} else {
		tmp = y * (x / 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 <= 5d-5) then
        tmp = x
    else
        tmp = y * (x / y)
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (y <= 5e-5) {
		tmp = x;
	} else {
		tmp = y * (x / y);
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if y <= 5e-5:
		tmp = x
	else:
		tmp = y * (x / y)
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (y <= 5e-5)
		tmp = x;
	else
		tmp = Float64(y * Float64(x / y));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 5e-5)
		tmp = x;
	else
		tmp = y * (x / y);
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[y, 5e-5], x, N[(y * N[(x / y), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 5.00000000000000024e-5

   1. Initial program 99.8%

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

   2. Taylor expanded in y around 0 69.6%

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

   if 5.00000000000000024e-5 < y

   1. Initial program 99.6%

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

   2. Taylor expanded in x around 0 99.6%

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

   3. Taylor expanded in y around 0 5.8%

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

      1. associate-/l* 30.8%

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

      2. div-inv 30.8%

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

      3. clear-num 29.4%

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

   5. Applied egg-rr 29.4%

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

4. Final simplification 58.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 5 \cdot 10^{-5}:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;y \cdot \frac{x}{y}\\ \end{array} \]

Alternative 8: 57.9% accurate, 14.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 0.038:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{y}{\frac{y}{x}}\\ \end{array} \end{array} \]

FPCore

(FPCore (x y) :precision binary64 (if (<= y 0.038) x (/ y (/ y x))))

C

double code(double x, double y) {
	double tmp;
	if (y <= 0.038) {
		tmp = x;
	} else {
		tmp = y / (y / x);
	}
	return tmp;
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if (y <= 0.038d0) then
        tmp = x
    else
        tmp = y / (y / x)
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (y <= 0.038) {
		tmp = x;
	} else {
		tmp = y / (y / x);
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if y <= 0.038:
		tmp = x
	else:
		tmp = y / (y / x)
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (y <= 0.038)
		tmp = x;
	else
		tmp = Float64(y / Float64(y / x));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 0.038)
		tmp = x;
	else
		tmp = y / (y / x);
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[y, 0.038], x, N[(y / N[(y / x), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 0.0379999999999999991

   1. Initial program 99.8%

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

   2. Taylor expanded in y around 0 69.4%

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

   if 0.0379999999999999991 < y

   1. Initial program 99.7%

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

   2. Taylor expanded in x around 0 99.6%

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

   3. Taylor expanded in y around 0 5.2%

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

      1. associate-/l* 30.5%

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

      2. div-inv 30.5%

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

      3. clear-num 29.2%

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

   5. Applied egg-rr 29.2%

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

      1. clear-num 30.5%

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

      2. un-div-inv 30.5%

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

   7. Applied egg-rr 30.5%

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

4. Final simplification 59.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 0.038:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{y}{\frac{y}{x}}\\ \end{array} \]

Alternative 9: 52.2% accurate, 105.0× 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 99.8%

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

2. Taylor expanded in y around 0 52.7%

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

3. Final simplification 52.7%

   \[\leadsto x \]

Reproduce

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