Linear.Quaternion:$csin from linear-1.19.1.3

Percentage Accurate: 100.0% → 100.0%

Time: 6.5s

Alternatives: 12

Speedup: 1.0×

• 50.3% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

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

FPCore

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

C

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

Java

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)

Julia

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

Wolfram

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

Initial Program: 100.0% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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)

Julia

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

Wolfram

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

Alternative 1: 100.0% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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)

Julia

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

Wolfram

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%

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

Alternative 2: 86.9% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{\sinh y}{y}\\ \mathbf{if}\;t_0 \leq 1.00000000002:\\ \;\;\;\;\cos x\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (let* ((t_0 (/ (sinh y) y))) (if (<= t_0 1.00000000002) (cos x) t_0)))

C

double code(double x, double y) {
	double t_0 = sinh(y) / y;
	double tmp;
	if (t_0 <= 1.00000000002) {
		tmp = cos(x);
	} else {
		tmp = t_0;
	}
	return tmp;
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: t_0
    real(8) :: tmp
    t_0 = sinh(y) / y
    if (t_0 <= 1.00000000002d0) then
        tmp = cos(x)
    else
        tmp = t_0
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double t_0 = Math.sinh(y) / y;
	double tmp;
	if (t_0 <= 1.00000000002) {
		tmp = Math.cos(x);
	} else {
		tmp = t_0;
	}
	return tmp;
}

Python

def code(x, y):
	t_0 = math.sinh(y) / y
	tmp = 0
	if t_0 <= 1.00000000002:
		tmp = math.cos(x)
	else:
		tmp = t_0
	return tmp

Julia

function code(x, y)
	t_0 = Float64(sinh(y) / y)
	tmp = 0.0
	if (t_0 <= 1.00000000002)
		tmp = cos(x);
	else
		tmp = t_0;
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	t_0 = sinh(y) / y;
	tmp = 0.0;
	if (t_0 <= 1.00000000002)
		tmp = cos(x);
	else
		tmp = t_0;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := Block[{t$95$0 = N[(N[Sinh[y], $MachinePrecision] / y), $MachinePrecision]}, If[LessEqual[t$95$0, 1.00000000002], N[Cos[x], $MachinePrecision], t$95$0]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 (sinh.f64 y) y) < 1.00000000002

   1. Initial program 100.0%

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

   2. Taylor expanded in y around 0 100.0%

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

      1. unpow2 100.0%

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

   4. Simplified 100.0%

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

   5. Taylor expanded in y around 0 99.9%

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

   if 1.00000000002 < (/.f64 (sinh.f64 y) y)

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 73.9%

      \[\leadsto \color{blue}{1} \cdot \frac{\sinh y}{y} \]
3. Recombined 2 regimes into one program.

4. Final simplification 86.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{\sinh y}{y} \leq 1.00000000002:\\ \;\;\;\;\cos x\\ \mathbf{else}:\\ \;\;\;\;\frac{\sinh y}{y}\\ \end{array} \]

Alternative 3: 84.8% accurate, 1.8× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 0.051 \lor \neg \left(y \leq 1.35 \cdot 10^{+154}\right):\\ \;\;\;\;\cos x \cdot \left(1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\sinh y}{y} \cdot \left(1 + -0.5 \cdot \left(x \cdot x\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (if (or (<= y 0.051) (not (<= y 1.35e+154)))
   (* (cos x) (+ 1.0 (* 0.16666666666666666 (* y y))))
   (* (/ (sinh y) y) (+ 1.0 (* -0.5 (* x x))))))

C

double code(double x, double y) {
	double tmp;
	if ((y <= 0.051) || !(y <= 1.35e+154)) {
		tmp = cos(x) * (1.0 + (0.16666666666666666 * (y * y)));
	} else {
		tmp = (sinh(y) / y) * (1.0 + (-0.5 * (x * 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 <= 0.051d0) .or. (.not. (y <= 1.35d+154))) then
        tmp = cos(x) * (1.0d0 + (0.16666666666666666d0 * (y * y)))
    else
        tmp = (sinh(y) / y) * (1.0d0 + ((-0.5d0) * (x * x)))
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if ((y <= 0.051) || !(y <= 1.35e+154)) {
		tmp = Math.cos(x) * (1.0 + (0.16666666666666666 * (y * y)));
	} else {
		tmp = (Math.sinh(y) / y) * (1.0 + (-0.5 * (x * x)));
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if (y <= 0.051) or not (y <= 1.35e+154):
		tmp = math.cos(x) * (1.0 + (0.16666666666666666 * (y * y)))
	else:
		tmp = (math.sinh(y) / y) * (1.0 + (-0.5 * (x * x)))
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if ((y <= 0.051) || !(y <= 1.35e+154))
		tmp = Float64(cos(x) * Float64(1.0 + Float64(0.16666666666666666 * Float64(y * y))));
	else
		tmp = Float64(Float64(sinh(y) / y) * Float64(1.0 + Float64(-0.5 * Float64(x * x))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if ((y <= 0.051) || ~((y <= 1.35e+154)))
		tmp = cos(x) * (1.0 + (0.16666666666666666 * (y * y)));
	else
		tmp = (sinh(y) / y) * (1.0 + (-0.5 * (x * x)));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[Or[LessEqual[y, 0.051], N[Not[LessEqual[y, 1.35e+154]], $MachinePrecision]], N[(N[Cos[x], $MachinePrecision] * N[(1.0 + N[(0.16666666666666666 * N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Sinh[y], $MachinePrecision] / y), $MachinePrecision] * N[(1.0 + N[(-0.5 * N[(x * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 0.0509999999999999967 or 1.35000000000000003e154 < y

   1. Initial program 100.0%

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

   2. Taylor expanded in y around 0 85.6%

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

      1. unpow2 85.6%

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

   4. Simplified 85.6%

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

   if 0.0509999999999999967 < y < 1.35000000000000003e154

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 81.9%

      \[\leadsto \color{blue}{\left(1 + -0.5 \cdot {x}^{2}\right)} \cdot \frac{\sinh y}{y} \]
   3. Step-by-step derivation

      1. unpow2 81.9%

         \[\leadsto \left(1 + -0.5 \cdot \color{blue}{\left(x \cdot x\right)}\right) \cdot \frac{\sinh y}{y} \]

   4. Simplified 81.9%

      \[\leadsto \color{blue}{\left(1 + -0.5 \cdot \left(x \cdot x\right)\right)} \cdot \frac{\sinh y}{y} \]
3. Recombined 2 regimes into one program.

4. Final simplification 85.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 0.051 \lor \neg \left(y \leq 1.35 \cdot 10^{+154}\right):\\ \;\;\;\;\cos x \cdot \left(1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\sinh y}{y} \cdot \left(1 + -0.5 \cdot \left(x \cdot x\right)\right)\\ \end{array} \]

Alternative 4: 84.4% accurate, 1.8× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 0.038 \lor \neg \left(y \leq 1.35 \cdot 10^{+154}\right):\\ \;\;\;\;\cos x \cdot \left(1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\sinh y}{y}\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (if (or (<= y 0.038) (not (<= y 1.35e+154)))
   (* (cos x) (+ 1.0 (* 0.16666666666666666 (* y y))))
   (/ (sinh y) y)))

C

double code(double x, double y) {
	double tmp;
	if ((y <= 0.038) || !(y <= 1.35e+154)) {
		tmp = cos(x) * (1.0 + (0.16666666666666666 * (y * y)));
	} else {
		tmp = 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 <= 0.038d0) .or. (.not. (y <= 1.35d+154))) then
        tmp = cos(x) * (1.0d0 + (0.16666666666666666d0 * (y * y)))
    else
        tmp = sinh(y) / y
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if ((y <= 0.038) || !(y <= 1.35e+154)) {
		tmp = Math.cos(x) * (1.0 + (0.16666666666666666 * (y * y)));
	} else {
		tmp = Math.sinh(y) / y;
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if (y <= 0.038) or not (y <= 1.35e+154):
		tmp = math.cos(x) * (1.0 + (0.16666666666666666 * (y * y)))
	else:
		tmp = math.sinh(y) / y
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if ((y <= 0.038) || !(y <= 1.35e+154))
		tmp = Float64(cos(x) * Float64(1.0 + Float64(0.16666666666666666 * Float64(y * y))));
	else
		tmp = Float64(sinh(y) / y);
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if ((y <= 0.038) || ~((y <= 1.35e+154)))
		tmp = cos(x) * (1.0 + (0.16666666666666666 * (y * y)));
	else
		tmp = sinh(y) / y;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[Or[LessEqual[y, 0.038], N[Not[LessEqual[y, 1.35e+154]], $MachinePrecision]], N[(N[Cos[x], $MachinePrecision] * N[(1.0 + N[(0.16666666666666666 * N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sinh[y], $MachinePrecision] / y), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 0.0379999999999999991 or 1.35000000000000003e154 < y

   1. Initial program 100.0%

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

   2. Taylor expanded in y around 0 85.6%

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

      1. unpow2 85.6%

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

   4. Simplified 85.6%

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

   if 0.0379999999999999991 < y < 1.35000000000000003e154

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 66.7%

      \[\leadsto \color{blue}{1} \cdot \frac{\sinh y}{y} \]
3. Recombined 2 regimes into one program.

4. Final simplification 83.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 0.038 \lor \neg \left(y \leq 1.35 \cdot 10^{+154}\right):\\ \;\;\;\;\cos x \cdot \left(1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\sinh y}{y}\\ \end{array} \]

Alternative 5: 61.8% accurate, 2.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\\ \mathbf{if}\;y \leq 1.6 \cdot 10^{-5}:\\ \;\;\;\;\cos x\\ \mathbf{elif}\;y \leq 8.114 \cdot 10^{+146} \lor \neg \left(y \leq 10^{+183} \lor \neg \left(y \leq 4.8 \cdot 10^{+243}\right) \land y \leq 1.8 \cdot 10^{+292}\right):\\ \;\;\;\;\left(1 + -0.5 \cdot \left(x \cdot x\right)\right) \cdot t_0\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (let* ((t_0 (+ 1.0 (* 0.16666666666666666 (* y y)))))
   (if (<= y 1.6e-5)
     (cos x)
     (if (or (<= y 8.114e+146)
             (not
              (or (<= y 1e+183) (and (not (<= y 4.8e+243)) (<= y 1.8e+292)))))
       (* (+ 1.0 (* -0.5 (* x x))) t_0)
       t_0))))

C

double code(double x, double y) {
	double t_0 = 1.0 + (0.16666666666666666 * (y * y));
	double tmp;
	if (y <= 1.6e-5) {
		tmp = cos(x);
	} else if ((y <= 8.114e+146) || !((y <= 1e+183) || (!(y <= 4.8e+243) && (y <= 1.8e+292)))) {
		tmp = (1.0 + (-0.5 * (x * x))) * t_0;
	} else {
		tmp = t_0;
	}
	return tmp;
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: t_0
    real(8) :: tmp
    t_0 = 1.0d0 + (0.16666666666666666d0 * (y * y))
    if (y <= 1.6d-5) then
        tmp = cos(x)
    else if ((y <= 8.114d+146) .or. (.not. (y <= 1d+183) .or. (.not. (y <= 4.8d+243)) .and. (y <= 1.8d+292))) then
        tmp = (1.0d0 + ((-0.5d0) * (x * x))) * t_0
    else
        tmp = t_0
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double t_0 = 1.0 + (0.16666666666666666 * (y * y));
	double tmp;
	if (y <= 1.6e-5) {
		tmp = Math.cos(x);
	} else if ((y <= 8.114e+146) || !((y <= 1e+183) || (!(y <= 4.8e+243) && (y <= 1.8e+292)))) {
		tmp = (1.0 + (-0.5 * (x * x))) * t_0;
	} else {
		tmp = t_0;
	}
	return tmp;
}

Python

def code(x, y):
	t_0 = 1.0 + (0.16666666666666666 * (y * y))
	tmp = 0
	if y <= 1.6e-5:
		tmp = math.cos(x)
	elif (y <= 8.114e+146) or not ((y <= 1e+183) or (not (y <= 4.8e+243) and (y <= 1.8e+292))):
		tmp = (1.0 + (-0.5 * (x * x))) * t_0
	else:
		tmp = t_0
	return tmp

Julia

function code(x, y)
	t_0 = Float64(1.0 + Float64(0.16666666666666666 * Float64(y * y)))
	tmp = 0.0
	if (y <= 1.6e-5)
		tmp = cos(x);
	elseif ((y <= 8.114e+146) || !((y <= 1e+183) || (!(y <= 4.8e+243) && (y <= 1.8e+292))))
		tmp = Float64(Float64(1.0 + Float64(-0.5 * Float64(x * x))) * t_0);
	else
		tmp = t_0;
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	t_0 = 1.0 + (0.16666666666666666 * (y * y));
	tmp = 0.0;
	if (y <= 1.6e-5)
		tmp = cos(x);
	elseif ((y <= 8.114e+146) || ~(((y <= 1e+183) || (~((y <= 4.8e+243)) && (y <= 1.8e+292)))))
		tmp = (1.0 + (-0.5 * (x * x))) * t_0;
	else
		tmp = t_0;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := Block[{t$95$0 = N[(1.0 + N[(0.16666666666666666 * N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[y, 1.6e-5], N[Cos[x], $MachinePrecision], If[Or[LessEqual[y, 8.114e+146], N[Not[Or[LessEqual[y, 1e+183], And[N[Not[LessEqual[y, 4.8e+243]], $MachinePrecision], LessEqual[y, 1.8e+292]]]], $MachinePrecision]], N[(N[(1.0 + N[(-0.5 * N[(x * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * t$95$0), $MachinePrecision], t$95$0]]]

Derivation

1. Split input into 3 regimes

2. if y < 1.59999999999999993e-5

   1. Initial program 100.0%

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

   2. Taylor expanded in y around 0 83.1%

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

      1. unpow2 83.1%

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

   4. Simplified 83.1%

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

   5. Taylor expanded in y around 0 66.9%

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

   if 1.59999999999999993e-5 < y < 8.1140000000000002e146 or 9.99999999999999947e182 < y < 4.8000000000000001e243 or 1.79999999999999998e292 < y

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 85.8%

      \[\leadsto \color{blue}{\left(1 + -0.5 \cdot {x}^{2}\right)} \cdot \frac{\sinh y}{y} \]
   3. Step-by-step derivation

      1. unpow2 85.8%

         \[\leadsto \left(1 + -0.5 \cdot \color{blue}{\left(x \cdot x\right)}\right) \cdot \frac{\sinh y}{y} \]

   4. Simplified 85.8%

      \[\leadsto \color{blue}{\left(1 + -0.5 \cdot \left(x \cdot x\right)\right)} \cdot \frac{\sinh y}{y} \]

   5. Taylor expanded in y around 0 49.2%

      \[\leadsto \left(1 + -0.5 \cdot \left(x \cdot x\right)\right) \cdot \color{blue}{\left(1 + 0.16666666666666666 \cdot {y}^{2}\right)} \]
   6. Step-by-step derivation

      1. unpow2 40.5%

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

   7. Simplified 49.2%

      \[\leadsto \left(1 + -0.5 \cdot \left(x \cdot x\right)\right) \cdot \color{blue}{\left(1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\right)} \]

   if 8.1140000000000002e146 < y < 9.99999999999999947e182 or 4.8000000000000001e243 < y < 1.79999999999999998e292

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 88.2%

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

   3. Taylor expanded in y around 0 77.7%

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

      1. unpow2 89.5%

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

   5. Simplified 77.7%

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

4. Final simplification 64.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 1.6 \cdot 10^{-5}:\\ \;\;\;\;\cos x\\ \mathbf{elif}\;y \leq 8.114 \cdot 10^{+146} \lor \neg \left(y \leq 10^{+183} \lor \neg \left(y \leq 4.8 \cdot 10^{+243}\right) \land y \leq 1.8 \cdot 10^{+292}\right):\\ \;\;\;\;\left(1 + -0.5 \cdot \left(x \cdot x\right)\right) \cdot \left(1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\right)\\ \mathbf{else}:\\ \;\;\;\;1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\\ \end{array} \]

Alternative 6: 47.8% accurate, 12.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\\ \mathbf{if}\;x \leq 2.55 \cdot 10^{+76}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\left(1 + -0.5 \cdot \left(x \cdot x\right)\right) \cdot t_0\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (let* ((t_0 (+ 1.0 (* 0.16666666666666666 (* y y)))))
   (if (<= x 2.55e+76) t_0 (* (+ 1.0 (* -0.5 (* x x))) t_0))))

C

double code(double x, double y) {
	double t_0 = 1.0 + (0.16666666666666666 * (y * y));
	double tmp;
	if (x <= 2.55e+76) {
		tmp = t_0;
	} else {
		tmp = (1.0 + (-0.5 * (x * x))) * t_0;
	}
	return tmp;
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: t_0
    real(8) :: tmp
    t_0 = 1.0d0 + (0.16666666666666666d0 * (y * y))
    if (x <= 2.55d+76) then
        tmp = t_0
    else
        tmp = (1.0d0 + ((-0.5d0) * (x * x))) * t_0
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double t_0 = 1.0 + (0.16666666666666666 * (y * y));
	double tmp;
	if (x <= 2.55e+76) {
		tmp = t_0;
	} else {
		tmp = (1.0 + (-0.5 * (x * x))) * t_0;
	}
	return tmp;
}

Python

def code(x, y):
	t_0 = 1.0 + (0.16666666666666666 * (y * y))
	tmp = 0
	if x <= 2.55e+76:
		tmp = t_0
	else:
		tmp = (1.0 + (-0.5 * (x * x))) * t_0
	return tmp

Julia

function code(x, y)
	t_0 = Float64(1.0 + Float64(0.16666666666666666 * Float64(y * y)))
	tmp = 0.0
	if (x <= 2.55e+76)
		tmp = t_0;
	else
		tmp = Float64(Float64(1.0 + Float64(-0.5 * Float64(x * x))) * t_0);
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	t_0 = 1.0 + (0.16666666666666666 * (y * y));
	tmp = 0.0;
	if (x <= 2.55e+76)
		tmp = t_0;
	else
		tmp = (1.0 + (-0.5 * (x * x))) * t_0;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := Block[{t$95$0 = N[(1.0 + N[(0.16666666666666666 * N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[x, 2.55e+76], t$95$0, N[(N[(1.0 + N[(-0.5 * N[(x * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * t$95$0), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if x < 2.5500000000000001e76

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 70.3%

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

   3. Taylor expanded in y around 0 50.9%

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

      1. unpow2 74.6%

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

   5. Simplified 50.9%

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

   if 2.5500000000000001e76 < x

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 29.8%

      \[\leadsto \color{blue}{\left(1 + -0.5 \cdot {x}^{2}\right)} \cdot \frac{\sinh y}{y} \]
   3. Step-by-step derivation

      1. unpow2 29.8%

         \[\leadsto \left(1 + -0.5 \cdot \color{blue}{\left(x \cdot x\right)}\right) \cdot \frac{\sinh y}{y} \]

   4. Simplified 29.8%

      \[\leadsto \color{blue}{\left(1 + -0.5 \cdot \left(x \cdot x\right)\right)} \cdot \frac{\sinh y}{y} \]

   5. Taylor expanded in y around 0 27.8%

      \[\leadsto \left(1 + -0.5 \cdot \left(x \cdot x\right)\right) \cdot \color{blue}{\left(1 + 0.16666666666666666 \cdot {y}^{2}\right)} \]
   6. Step-by-step derivation

      1. unpow2 78.9%

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

   7. Simplified 27.8%

      \[\leadsto \left(1 + -0.5 \cdot \left(x \cdot x\right)\right) \cdot \color{blue}{\left(1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 46.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 2.55 \cdot 10^{+76}:\\ \;\;\;\;1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\\ \mathbf{else}:\\ \;\;\;\;\left(1 + -0.5 \cdot \left(x \cdot x\right)\right) \cdot \left(1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\right)\\ \end{array} \]

Alternative 7: 47.8% accurate, 15.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;x \leq 2.55 \cdot 10^{+76}:\\ \;\;\;\;1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\\ \mathbf{else}:\\ \;\;\;\;\left(x \cdot x\right) \cdot \left(-0.5 + \left(y \cdot y\right) \cdot -0.08333333333333333\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (if (<= x 2.55e+76)
   (+ 1.0 (* 0.16666666666666666 (* y y)))
   (* (* x x) (+ -0.5 (* (* y y) -0.08333333333333333)))))

C

double code(double x, double y) {
	double tmp;
	if (x <= 2.55e+76) {
		tmp = 1.0 + (0.16666666666666666 * (y * y));
	} else {
		tmp = (x * x) * (-0.5 + ((y * y) * -0.08333333333333333));
	}
	return tmp;
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if (x <= 2.55d+76) then
        tmp = 1.0d0 + (0.16666666666666666d0 * (y * y))
    else
        tmp = (x * x) * ((-0.5d0) + ((y * y) * (-0.08333333333333333d0)))
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (x <= 2.55e+76) {
		tmp = 1.0 + (0.16666666666666666 * (y * y));
	} else {
		tmp = (x * x) * (-0.5 + ((y * y) * -0.08333333333333333));
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if x <= 2.55e+76:
		tmp = 1.0 + (0.16666666666666666 * (y * y))
	else:
		tmp = (x * x) * (-0.5 + ((y * y) * -0.08333333333333333))
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (x <= 2.55e+76)
		tmp = Float64(1.0 + Float64(0.16666666666666666 * Float64(y * y)));
	else
		tmp = Float64(Float64(x * x) * Float64(-0.5 + Float64(Float64(y * y) * -0.08333333333333333)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (x <= 2.55e+76)
		tmp = 1.0 + (0.16666666666666666 * (y * y));
	else
		tmp = (x * x) * (-0.5 + ((y * y) * -0.08333333333333333));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[x, 2.55e+76], N[(1.0 + N[(0.16666666666666666 * N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(x * x), $MachinePrecision] * N[(-0.5 + N[(N[(y * y), $MachinePrecision] * -0.08333333333333333), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < 2.5500000000000001e76

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 70.3%

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

   3. Taylor expanded in y around 0 50.9%

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

      1. unpow2 74.6%

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

   5. Simplified 50.9%

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

   if 2.5500000000000001e76 < x

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 29.8%

      \[\leadsto \color{blue}{\left(1 + -0.5 \cdot {x}^{2}\right)} \cdot \frac{\sinh y}{y} \]
   3. Step-by-step derivation

      1. unpow2 29.8%

         \[\leadsto \left(1 + -0.5 \cdot \color{blue}{\left(x \cdot x\right)}\right) \cdot \frac{\sinh y}{y} \]

   4. Simplified 29.8%

      \[\leadsto \color{blue}{\left(1 + -0.5 \cdot \left(x \cdot x\right)\right)} \cdot \frac{\sinh y}{y} \]

   5. Taylor expanded in x around inf 29.8%

      \[\leadsto \color{blue}{\left(-0.5 \cdot {x}^{2}\right)} \cdot \frac{\sinh y}{y} \]
   6. Step-by-step derivation

      1. unpow2 29.8%

         \[\leadsto \left(-0.5 \cdot \color{blue}{\left(x \cdot x\right)}\right) \cdot \frac{\sinh y}{y} \]

   7. Simplified 29.8%

      \[\leadsto \color{blue}{\left(-0.5 \cdot \left(x \cdot x\right)\right)} \cdot \frac{\sinh y}{y} \]

   8. Taylor expanded in y around 0 27.5%

      \[\leadsto \color{blue}{-0.5 \cdot {x}^{2} + -0.08333333333333333 \cdot \left({x}^{2} \cdot {y}^{2}\right)} \]
   9. Step-by-step derivation

      1. unpow2 27.5%

         \[\leadsto -0.5 \cdot {x}^{2} + -0.08333333333333333 \cdot \left({x}^{2} \cdot \color{blue}{\left(y \cdot y\right)}\right) \]

      2. *-commutative 27.5%

         \[\leadsto -0.5 \cdot {x}^{2} + -0.08333333333333333 \cdot \color{blue}{\left(\left(y \cdot y\right) \cdot {x}^{2}\right)} \]

      3. associate-*r* 27.5%

         \[\leadsto -0.5 \cdot {x}^{2} + \color{blue}{\left(-0.08333333333333333 \cdot \left(y \cdot y\right)\right) \cdot {x}^{2}} \]

      4. distribute-rgt-out 27.8%

         \[\leadsto \color{blue}{{x}^{2} \cdot \left(-0.5 + -0.08333333333333333 \cdot \left(y \cdot y\right)\right)} \]

      5. unpow2 27.8%

         \[\leadsto \color{blue}{\left(x \cdot x\right)} \cdot \left(-0.5 + -0.08333333333333333 \cdot \left(y \cdot y\right)\right) \]

   10. Simplified 27.8%

       \[\leadsto \color{blue}{\left(x \cdot x\right) \cdot \left(-0.5 + -0.08333333333333333 \cdot \left(y \cdot y\right)\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 46.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 2.55 \cdot 10^{+76}:\\ \;\;\;\;1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\\ \mathbf{else}:\\ \;\;\;\;\left(x \cdot x\right) \cdot \left(-0.5 + \left(y \cdot y\right) \cdot -0.08333333333333333\right)\\ \end{array} \]

Alternative 8: 32.9% accurate, 18.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 290000000:\\ \;\;\;\;1\\ \mathbf{else}:\\ \;\;\;\;-0.08333333333333333 \cdot \left(\left(y \cdot y\right) \cdot \left(x \cdot x\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (if (<= y 290000000.0) 1.0 (* -0.08333333333333333 (* (* y y) (* x x)))))

C

double code(double x, double y) {
	double tmp;
	if (y <= 290000000.0) {
		tmp = 1.0;
	} else {
		tmp = -0.08333333333333333 * ((y * y) * (x * 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 <= 290000000.0d0) then
        tmp = 1.0d0
    else
        tmp = (-0.08333333333333333d0) * ((y * y) * (x * x))
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (y <= 290000000.0) {
		tmp = 1.0;
	} else {
		tmp = -0.08333333333333333 * ((y * y) * (x * x));
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if y <= 290000000.0:
		tmp = 1.0
	else:
		tmp = -0.08333333333333333 * ((y * y) * (x * x))
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (y <= 290000000.0)
		tmp = 1.0;
	else
		tmp = Float64(-0.08333333333333333 * Float64(Float64(y * y) * Float64(x * x)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 290000000.0)
		tmp = 1.0;
	else
		tmp = -0.08333333333333333 * ((y * y) * (x * x));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[y, 290000000.0], 1.0, N[(-0.08333333333333333 * N[(N[(y * y), $MachinePrecision] * N[(x * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 2.9e8

   1. Initial program 100.0%

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

   2. Taylor expanded in y around 0 81.6%

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

      1. unpow2 81.6%

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

   4. Simplified 81.6%

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

   5. Taylor expanded in y around 0 65.7%

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

   6. Taylor expanded in x around 0 35.1%

      \[\leadsto \color{blue}{1} \]

   if 2.9e8 < y

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 77.4%

      \[\leadsto \color{blue}{\left(1 + -0.5 \cdot {x}^{2}\right)} \cdot \frac{\sinh y}{y} \]
   3. Step-by-step derivation

      1. unpow2 77.4%

         \[\leadsto \left(1 + -0.5 \cdot \color{blue}{\left(x \cdot x\right)}\right) \cdot \frac{\sinh y}{y} \]

   4. Simplified 77.4%

      \[\leadsto \color{blue}{\left(1 + -0.5 \cdot \left(x \cdot x\right)\right)} \cdot \frac{\sinh y}{y} \]

   5. Taylor expanded in x around inf 29.0%

      \[\leadsto \color{blue}{\left(-0.5 \cdot {x}^{2}\right)} \cdot \frac{\sinh y}{y} \]
   6. Step-by-step derivation

      1. unpow2 29.0%

         \[\leadsto \left(-0.5 \cdot \color{blue}{\left(x \cdot x\right)}\right) \cdot \frac{\sinh y}{y} \]

   7. Simplified 29.0%

      \[\leadsto \color{blue}{\left(-0.5 \cdot \left(x \cdot x\right)\right)} \cdot \frac{\sinh y}{y} \]

   8. Taylor expanded in y around 0 26.3%

      \[\leadsto \color{blue}{-0.5 \cdot {x}^{2} + -0.08333333333333333 \cdot \left({x}^{2} \cdot {y}^{2}\right)} \]
   9. Step-by-step derivation

      1. unpow2 26.3%

         \[\leadsto -0.5 \cdot {x}^{2} + -0.08333333333333333 \cdot \left({x}^{2} \cdot \color{blue}{\left(y \cdot y\right)}\right) \]

      2. *-commutative 26.3%

         \[\leadsto -0.5 \cdot {x}^{2} + -0.08333333333333333 \cdot \color{blue}{\left(\left(y \cdot y\right) \cdot {x}^{2}\right)} \]

      3. associate-*r* 26.3%

         \[\leadsto -0.5 \cdot {x}^{2} + \color{blue}{\left(-0.08333333333333333 \cdot \left(y \cdot y\right)\right) \cdot {x}^{2}} \]

      4. distribute-rgt-out 26.3%

         \[\leadsto \color{blue}{{x}^{2} \cdot \left(-0.5 + -0.08333333333333333 \cdot \left(y \cdot y\right)\right)} \]

      5. unpow2 26.3%

         \[\leadsto \color{blue}{\left(x \cdot x\right)} \cdot \left(-0.5 + -0.08333333333333333 \cdot \left(y \cdot y\right)\right) \]

   10. Simplified 26.3%

       \[\leadsto \color{blue}{\left(x \cdot x\right) \cdot \left(-0.5 + -0.08333333333333333 \cdot \left(y \cdot y\right)\right)} \]

   11. Taylor expanded in y around inf 26.3%

       \[\leadsto \color{blue}{-0.08333333333333333 \cdot \left({x}^{2} \cdot {y}^{2}\right)} \]
   12. Step-by-step derivation

       1. unpow2 26.3%

          \[\leadsto -0.08333333333333333 \cdot \left(\color{blue}{\left(x \cdot x\right)} \cdot {y}^{2}\right) \]

       2. unpow2 26.3%

          \[\leadsto -0.08333333333333333 \cdot \left(\left(x \cdot x\right) \cdot \color{blue}{\left(y \cdot y\right)}\right) \]

   13. Simplified 26.3%

       \[\leadsto \color{blue}{-0.08333333333333333 \cdot \left(\left(x \cdot x\right) \cdot \left(y \cdot y\right)\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 33.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 290000000:\\ \;\;\;\;1\\ \mathbf{else}:\\ \;\;\;\;-0.08333333333333333 \cdot \left(\left(y \cdot y\right) \cdot \left(x \cdot x\right)\right)\\ \end{array} \]

Alternative 9: 47.8% accurate, 18.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;x \leq 2.55 \cdot 10^{+76}:\\ \;\;\;\;1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\\ \mathbf{else}:\\ \;\;\;\;-0.08333333333333333 \cdot \left(\left(y \cdot y\right) \cdot \left(x \cdot x\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (if (<= x 2.55e+76)
   (+ 1.0 (* 0.16666666666666666 (* y y)))
   (* -0.08333333333333333 (* (* y y) (* x x)))))

C

double code(double x, double y) {
	double tmp;
	if (x <= 2.55e+76) {
		tmp = 1.0 + (0.16666666666666666 * (y * y));
	} else {
		tmp = -0.08333333333333333 * ((y * y) * (x * x));
	}
	return tmp;
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if (x <= 2.55d+76) then
        tmp = 1.0d0 + (0.16666666666666666d0 * (y * y))
    else
        tmp = (-0.08333333333333333d0) * ((y * y) * (x * x))
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (x <= 2.55e+76) {
		tmp = 1.0 + (0.16666666666666666 * (y * y));
	} else {
		tmp = -0.08333333333333333 * ((y * y) * (x * x));
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if x <= 2.55e+76:
		tmp = 1.0 + (0.16666666666666666 * (y * y))
	else:
		tmp = -0.08333333333333333 * ((y * y) * (x * x))
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (x <= 2.55e+76)
		tmp = Float64(1.0 + Float64(0.16666666666666666 * Float64(y * y)));
	else
		tmp = Float64(-0.08333333333333333 * Float64(Float64(y * y) * Float64(x * x)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (x <= 2.55e+76)
		tmp = 1.0 + (0.16666666666666666 * (y * y));
	else
		tmp = -0.08333333333333333 * ((y * y) * (x * x));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[x, 2.55e+76], N[(1.0 + N[(0.16666666666666666 * N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(-0.08333333333333333 * N[(N[(y * y), $MachinePrecision] * N[(x * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < 2.5500000000000001e76

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 70.3%

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

   3. Taylor expanded in y around 0 50.9%

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

      1. unpow2 74.6%

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

   5. Simplified 50.9%

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

   if 2.5500000000000001e76 < x

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 29.8%

      \[\leadsto \color{blue}{\left(1 + -0.5 \cdot {x}^{2}\right)} \cdot \frac{\sinh y}{y} \]
   3. Step-by-step derivation

      1. unpow2 29.8%

         \[\leadsto \left(1 + -0.5 \cdot \color{blue}{\left(x \cdot x\right)}\right) \cdot \frac{\sinh y}{y} \]

   4. Simplified 29.8%

      \[\leadsto \color{blue}{\left(1 + -0.5 \cdot \left(x \cdot x\right)\right)} \cdot \frac{\sinh y}{y} \]

   5. Taylor expanded in x around inf 29.8%

      \[\leadsto \color{blue}{\left(-0.5 \cdot {x}^{2}\right)} \cdot \frac{\sinh y}{y} \]
   6. Step-by-step derivation

      1. unpow2 29.8%

         \[\leadsto \left(-0.5 \cdot \color{blue}{\left(x \cdot x\right)}\right) \cdot \frac{\sinh y}{y} \]

   7. Simplified 29.8%

      \[\leadsto \color{blue}{\left(-0.5 \cdot \left(x \cdot x\right)\right)} \cdot \frac{\sinh y}{y} \]

   8. Taylor expanded in y around 0 27.5%

      \[\leadsto \color{blue}{-0.5 \cdot {x}^{2} + -0.08333333333333333 \cdot \left({x}^{2} \cdot {y}^{2}\right)} \]
   9. Step-by-step derivation

      1. unpow2 27.5%

         \[\leadsto -0.5 \cdot {x}^{2} + -0.08333333333333333 \cdot \left({x}^{2} \cdot \color{blue}{\left(y \cdot y\right)}\right) \]

      2. *-commutative 27.5%

         \[\leadsto -0.5 \cdot {x}^{2} + -0.08333333333333333 \cdot \color{blue}{\left(\left(y \cdot y\right) \cdot {x}^{2}\right)} \]

      3. associate-*r* 27.5%

         \[\leadsto -0.5 \cdot {x}^{2} + \color{blue}{\left(-0.08333333333333333 \cdot \left(y \cdot y\right)\right) \cdot {x}^{2}} \]

      4. distribute-rgt-out 27.8%

         \[\leadsto \color{blue}{{x}^{2} \cdot \left(-0.5 + -0.08333333333333333 \cdot \left(y \cdot y\right)\right)} \]

      5. unpow2 27.8%

         \[\leadsto \color{blue}{\left(x \cdot x\right)} \cdot \left(-0.5 + -0.08333333333333333 \cdot \left(y \cdot y\right)\right) \]

   10. Simplified 27.8%

       \[\leadsto \color{blue}{\left(x \cdot x\right) \cdot \left(-0.5 + -0.08333333333333333 \cdot \left(y \cdot y\right)\right)} \]

   11. Taylor expanded in y around inf 27.7%

       \[\leadsto \color{blue}{-0.08333333333333333 \cdot \left({x}^{2} \cdot {y}^{2}\right)} \]
   12. Step-by-step derivation

       1. unpow2 27.7%

          \[\leadsto -0.08333333333333333 \cdot \left(\color{blue}{\left(x \cdot x\right)} \cdot {y}^{2}\right) \]

       2. unpow2 27.7%

          \[\leadsto -0.08333333333333333 \cdot \left(\left(x \cdot x\right) \cdot \color{blue}{\left(y \cdot y\right)}\right) \]

   13. Simplified 27.7%

       \[\leadsto \color{blue}{-0.08333333333333333 \cdot \left(\left(x \cdot x\right) \cdot \left(y \cdot y\right)\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 46.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 2.55 \cdot 10^{+76}:\\ \;\;\;\;1 + 0.16666666666666666 \cdot \left(y \cdot y\right)\\ \mathbf{else}:\\ \;\;\;\;-0.08333333333333333 \cdot \left(\left(y \cdot y\right) \cdot \left(x \cdot x\right)\right)\\ \end{array} \]

Alternative 10: 31.0% accurate, 22.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 3250:\\ \;\;\;\;1\\ \mathbf{else}:\\ \;\;\;\;1 + -0.5 \cdot \left(x \cdot x\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (x y)
 :precision binary64
 (if (<= y 3250.0) 1.0 (+ 1.0 (* -0.5 (* x x)))))

C

double code(double x, double y) {
	double tmp;
	if (y <= 3250.0) {
		tmp = 1.0;
	} else {
		tmp = 1.0 + (-0.5 * (x * 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 <= 3250.0d0) then
        tmp = 1.0d0
    else
        tmp = 1.0d0 + ((-0.5d0) * (x * x))
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (y <= 3250.0) {
		tmp = 1.0;
	} else {
		tmp = 1.0 + (-0.5 * (x * x));
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if y <= 3250.0:
		tmp = 1.0
	else:
		tmp = 1.0 + (-0.5 * (x * x))
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (y <= 3250.0)
		tmp = 1.0;
	else
		tmp = Float64(1.0 + Float64(-0.5 * Float64(x * x)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 3250.0)
		tmp = 1.0;
	else
		tmp = 1.0 + (-0.5 * (x * x));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[y, 3250.0], 1.0, N[(1.0 + N[(-0.5 * N[(x * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 3250

   1. Initial program 100.0%

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

   2. Taylor expanded in y around 0 82.4%

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

      1. unpow2 82.4%

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

   4. Simplified 82.4%

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

   5. Taylor expanded in y around 0 66.3%

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

   6. Taylor expanded in x around 0 35.5%

      \[\leadsto \color{blue}{1} \]

   if 3250 < y

   1. Initial program 100.0%

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

   2. Taylor expanded in y around 0 54.4%

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

      1. unpow2 54.4%

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

   4. Simplified 54.4%

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

   5. Taylor expanded in y around 0 3.1%

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

   6. Taylor expanded in x around 0 15.0%

      \[\leadsto \color{blue}{1 + -0.5 \cdot {x}^{2}} \]
   7. Step-by-step derivation

      1. unpow2 78.1%

         \[\leadsto \left(1 + -0.5 \cdot \color{blue}{\left(x \cdot x\right)}\right) \cdot \frac{\sinh y}{y} \]

   8. Simplified 15.0%

      \[\leadsto \color{blue}{1 + -0.5 \cdot \left(x \cdot x\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 30.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 3250:\\ \;\;\;\;1\\ \mathbf{else}:\\ \;\;\;\;1 + -0.5 \cdot \left(x \cdot x\right)\\ \end{array} \]

Alternative 11: 30.8% accurate, 29.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;y \leq 170000000:\\ \;\;\;\;1\\ \mathbf{else}:\\ \;\;\;\;-0.5 \cdot \left(x \cdot x\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (x y) :precision binary64 (if (<= y 170000000.0) 1.0 (* -0.5 (* x x))))

C

double code(double x, double y) {
	double tmp;
	if (y <= 170000000.0) {
		tmp = 1.0;
	} else {
		tmp = -0.5 * (x * 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 <= 170000000.0d0) then
        tmp = 1.0d0
    else
        tmp = (-0.5d0) * (x * x)
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	double tmp;
	if (y <= 170000000.0) {
		tmp = 1.0;
	} else {
		tmp = -0.5 * (x * x);
	}
	return tmp;
}

Python

def code(x, y):
	tmp = 0
	if y <= 170000000.0:
		tmp = 1.0
	else:
		tmp = -0.5 * (x * x)
	return tmp

Julia

function code(x, y)
	tmp = 0.0
	if (y <= 170000000.0)
		tmp = 1.0;
	else
		tmp = Float64(-0.5 * Float64(x * x));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 170000000.0)
		tmp = 1.0;
	else
		tmp = -0.5 * (x * x);
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := If[LessEqual[y, 170000000.0], 1.0, N[(-0.5 * N[(x * x), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 1.7e8

   1. Initial program 100.0%

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

   2. Taylor expanded in y around 0 81.6%

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

      1. unpow2 81.6%

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

   4. Simplified 81.6%

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

   5. Taylor expanded in y around 0 65.7%

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

   6. Taylor expanded in x around 0 35.1%

      \[\leadsto \color{blue}{1} \]

   if 1.7e8 < y

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 77.4%

      \[\leadsto \color{blue}{\left(1 + -0.5 \cdot {x}^{2}\right)} \cdot \frac{\sinh y}{y} \]
   3. Step-by-step derivation

      1. unpow2 77.4%

         \[\leadsto \left(1 + -0.5 \cdot \color{blue}{\left(x \cdot x\right)}\right) \cdot \frac{\sinh y}{y} \]

   4. Simplified 77.4%

      \[\leadsto \color{blue}{\left(1 + -0.5 \cdot \left(x \cdot x\right)\right)} \cdot \frac{\sinh y}{y} \]

   5. Taylor expanded in x around inf 29.0%

      \[\leadsto \color{blue}{\left(-0.5 \cdot {x}^{2}\right)} \cdot \frac{\sinh y}{y} \]
   6. Step-by-step derivation

      1. unpow2 29.0%

         \[\leadsto \left(-0.5 \cdot \color{blue}{\left(x \cdot x\right)}\right) \cdot \frac{\sinh y}{y} \]

   7. Simplified 29.0%

      \[\leadsto \color{blue}{\left(-0.5 \cdot \left(x \cdot x\right)\right)} \cdot \frac{\sinh y}{y} \]

   8. Taylor expanded in y around 0 14.5%

      \[\leadsto \color{blue}{-0.5 \cdot {x}^{2}} \]
   9. Step-by-step derivation

      1. *-commutative 14.5%

         \[\leadsto \color{blue}{{x}^{2} \cdot -0.5} \]

      2. unpow2 14.5%

         \[\leadsto \color{blue}{\left(x \cdot x\right)} \cdot -0.5 \]

   10. Simplified 14.5%

       \[\leadsto \color{blue}{\left(x \cdot x\right) \cdot -0.5} \]
3. Recombined 2 regimes into one program.

4. Final simplification 30.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 170000000:\\ \;\;\;\;1\\ \mathbf{else}:\\ \;\;\;\;-0.5 \cdot \left(x \cdot x\right)\\ \end{array} \]

Alternative 12: 27.8% accurate, 205.0× speedup

Math

\[\begin{array}{l} \\ 1 \end{array} \]

FPCore

(FPCore (x y) :precision binary64 1.0)

C

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

Fortran

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

Java

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

Python

def code(x, y):
	return 1.0

Julia

function code(x, y)
	return 1.0
end

MATLAB

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

Wolfram

code[x_, y_] := 1.0

Derivation

1. Initial program 100.0%

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

2. Taylor expanded in y around 0 75.4%

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

   1. unpow2 75.4%

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

4. Simplified 75.4%

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

5. Taylor expanded in y around 0 50.5%

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

6. Taylor expanded in x around 0 27.2%

   \[\leadsto \color{blue}{1} \]

7. Final simplification 27.2%

   \[\leadsto 1 \]

Reproduce

herbie shell --seed 2023290 
(FPCore (x y)
  :name "Linear.Quaternion:$csin from linear-1.19.1.3"
  :precision binary64
  (* (cos x) (/ (sinh y) y)))