Toniolo and Linder, Equation (3b), real

Percentage Accurate: 94.1% → 99.7%

Time: 17.6s

Alternatives: 16

Speedup: 1.4×

Specification

Math

\[\begin{array}{l} \\ \frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (* (/ (sin ky) (sqrt (+ (pow (sin kx) 2.0) (pow (sin ky) 2.0)))) (sin th)))

C

double code(double kx, double ky, double th) {
	return (sin(ky) / sqrt((pow(sin(kx), 2.0) + pow(sin(ky), 2.0)))) * sin(th);
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    code = (sin(ky) / sqrt(((sin(kx) ** 2.0d0) + (sin(ky) ** 2.0d0)))) * sin(th)
end function

Java

public static double code(double kx, double ky, double th) {
	return (Math.sin(ky) / Math.sqrt((Math.pow(Math.sin(kx), 2.0) + Math.pow(Math.sin(ky), 2.0)))) * Math.sin(th);
}

Python

def code(kx, ky, th):
	return (math.sin(ky) / math.sqrt((math.pow(math.sin(kx), 2.0) + math.pow(math.sin(ky), 2.0)))) * math.sin(th)

Julia

function code(kx, ky, th)
	return Float64(Float64(sin(ky) / sqrt(Float64((sin(kx) ^ 2.0) + (sin(ky) ^ 2.0)))) * sin(th))
end

MATLAB

function tmp = code(kx, ky, th)
	tmp = (sin(ky) / sqrt(((sin(kx) ^ 2.0) + (sin(ky) ^ 2.0)))) * sin(th);
end

Wolfram

code[kx_, ky_, th_] := N[(N[(N[Sin[ky], $MachinePrecision] / N[Sqrt[N[(N[Power[N[Sin[kx], $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[Sin[ky], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Sin[th], $MachinePrecision]), $MachinePrecision]

Initial Program: 94.1% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (* (/ (sin ky) (sqrt (+ (pow (sin kx) 2.0) (pow (sin ky) 2.0)))) (sin th)))

C

double code(double kx, double ky, double th) {
	return (sin(ky) / sqrt((pow(sin(kx), 2.0) + pow(sin(ky), 2.0)))) * sin(th);
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    code = (sin(ky) / sqrt(((sin(kx) ** 2.0d0) + (sin(ky) ** 2.0d0)))) * sin(th)
end function

Java

public static double code(double kx, double ky, double th) {
	return (Math.sin(ky) / Math.sqrt((Math.pow(Math.sin(kx), 2.0) + Math.pow(Math.sin(ky), 2.0)))) * Math.sin(th);
}

Python

def code(kx, ky, th):
	return (math.sin(ky) / math.sqrt((math.pow(math.sin(kx), 2.0) + math.pow(math.sin(ky), 2.0)))) * math.sin(th)

Julia

function code(kx, ky, th)
	return Float64(Float64(sin(ky) / sqrt(Float64((sin(kx) ^ 2.0) + (sin(ky) ^ 2.0)))) * sin(th))
end

MATLAB

function tmp = code(kx, ky, th)
	tmp = (sin(ky) / sqrt(((sin(kx) ^ 2.0) + (sin(ky) ^ 2.0)))) * sin(th);
end

Wolfram

code[kx_, ky_, th_] := N[(N[(N[Sin[ky], $MachinePrecision] / N[Sqrt[N[(N[Power[N[Sin[kx], $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[Sin[ky], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[Sin[th], $MachinePrecision]), $MachinePrecision]

Alternative 1: 99.7% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \frac{\sin th}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin ky}} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (/ (sin th) (/ (hypot (sin kx) (sin ky)) (sin ky))))

C

double code(double kx, double ky, double th) {
	return sin(th) / (hypot(sin(kx), sin(ky)) / sin(ky));
}

Java

public static double code(double kx, double ky, double th) {
	return Math.sin(th) / (Math.hypot(Math.sin(kx), Math.sin(ky)) / Math.sin(ky));
}

Python

def code(kx, ky, th):
	return math.sin(th) / (math.hypot(math.sin(kx), math.sin(ky)) / math.sin(ky))

Julia

function code(kx, ky, th)
	return Float64(sin(th) / Float64(hypot(sin(kx), sin(ky)) / sin(ky)))
end

MATLAB

function tmp = code(kx, ky, th)
	tmp = sin(th) / (hypot(sin(kx), sin(ky)) / sin(ky));
end

Wolfram

code[kx_, ky_, th_] := N[(N[Sin[th], $MachinePrecision] / N[(N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision] / N[Sin[ky], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 92.8%

   \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
2. Step-by-step derivation

   1. *-commutative 92.8%

      \[\leadsto \color{blue}{\sin th \cdot \frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

   2. clear-num 92.7%

      \[\leadsto \sin th \cdot \color{blue}{\frac{1}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin ky}}} \]

   3. un-div-inv 92.8%

      \[\leadsto \color{blue}{\frac{\sin th}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin ky}}} \]

   4. unpow2 92.8%

      \[\leadsto \frac{\sin th}{\frac{\sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}}}{\sin ky}} \]

   5. unpow2 92.8%

      \[\leadsto \frac{\sin th}{\frac{\sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}}}{\sin ky}} \]

   6. hypot-def 99.7%

      \[\leadsto \frac{\sin th}{\frac{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}}{\sin ky}} \]

3. Applied egg-rr 99.7%

   \[\leadsto \color{blue}{\frac{\sin th}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin ky}}} \]

4. Final simplification 99.7%

   \[\leadsto \frac{\sin th}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin ky}} \]

Alternative 2: 65.3% accurate, 1.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin th \leq -0.005:\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{elif}\;\sin th \leq 0.002:\\ \;\;\;\;\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{th}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sin ky}{\left|\frac{\sin kx}{\sin th}\right|}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin th) -0.005)
   (* (sin th) (fabs (/ (sin ky) (sin kx))))
   (if (<= (sin th) 0.002)
     (/ (sin ky) (/ (hypot (sin ky) (sin kx)) th))
     (/ (sin ky) (fabs (/ (sin kx) (sin th)))))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(th) <= -0.005) {
		tmp = sin(th) * fabs((sin(ky) / sin(kx)));
	} else if (sin(th) <= 0.002) {
		tmp = sin(ky) / (hypot(sin(ky), sin(kx)) / th);
	} else {
		tmp = sin(ky) / fabs((sin(kx) / sin(th)));
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(th) <= -0.005) {
		tmp = Math.sin(th) * Math.abs((Math.sin(ky) / Math.sin(kx)));
	} else if (Math.sin(th) <= 0.002) {
		tmp = Math.sin(ky) / (Math.hypot(Math.sin(ky), Math.sin(kx)) / th);
	} else {
		tmp = Math.sin(ky) / Math.abs((Math.sin(kx) / Math.sin(th)));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(th) <= -0.005:
		tmp = math.sin(th) * math.fabs((math.sin(ky) / math.sin(kx)))
	elif math.sin(th) <= 0.002:
		tmp = math.sin(ky) / (math.hypot(math.sin(ky), math.sin(kx)) / th)
	else:
		tmp = math.sin(ky) / math.fabs((math.sin(kx) / math.sin(th)))
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(th) <= -0.005)
		tmp = Float64(sin(th) * abs(Float64(sin(ky) / sin(kx))));
	elseif (sin(th) <= 0.002)
		tmp = Float64(sin(ky) / Float64(hypot(sin(ky), sin(kx)) / th));
	else
		tmp = Float64(sin(ky) / abs(Float64(sin(kx) / sin(th))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(th) <= -0.005)
		tmp = sin(th) * abs((sin(ky) / sin(kx)));
	elseif (sin(th) <= 0.002)
		tmp = sin(ky) / (hypot(sin(ky), sin(kx)) / th);
	else
		tmp = sin(ky) / abs((sin(kx) / sin(th)));
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[th], $MachinePrecision], -0.005], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(N[Sin[ky], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[th], $MachinePrecision], 0.002], N[(N[Sin[ky], $MachinePrecision] / N[(N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision] / th), $MachinePrecision]), $MachinePrecision], N[(N[Sin[ky], $MachinePrecision] / N[Abs[N[(N[Sin[kx], $MachinePrecision] / N[Sin[th], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 th) < -0.0050000000000000001

   1. Initial program 91.5%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in ky around 0 22.3%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sin kx}} \cdot \sin th \]
   3. Step-by-step derivation

      1. add-sqr-sqrt 12.4%

         \[\leadsto \color{blue}{\left(\sqrt{\frac{\sin ky}{\sin kx}} \cdot \sqrt{\frac{\sin ky}{\sin kx}}\right)} \cdot \sin th \]

      2. sqrt-unprod 19.5%

         \[\leadsto \color{blue}{\sqrt{\frac{\sin ky}{\sin kx} \cdot \frac{\sin ky}{\sin kx}}} \cdot \sin th \]

      3. pow2 19.5%

         \[\leadsto \sqrt{\color{blue}{{\left(\frac{\sin ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]

   4. Applied egg-rr 19.5%

      \[\leadsto \color{blue}{\sqrt{{\left(\frac{\sin ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]
   5. Step-by-step derivation

      1. unpow2 19.5%

         \[\leadsto \sqrt{\color{blue}{\frac{\sin ky}{\sin kx} \cdot \frac{\sin ky}{\sin kx}}} \cdot \sin th \]

      2. rem-sqrt-square 25.5%

         \[\leadsto \color{blue}{\left|\frac{\sin ky}{\sin kx}\right|} \cdot \sin th \]

   6. Simplified 25.5%

      \[\leadsto \color{blue}{\left|\frac{\sin ky}{\sin kx}\right|} \cdot \sin th \]

   if -0.0050000000000000001 < (sin.f64 th) < 2e-3

   1. Initial program 93.2%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. associate-*l/ 90.8%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

      2. associate-/l* 93.1%

         \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin th}}} \]

      3. unpow2 93.1%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}}}{\sin th}} \]

      4. unpow2 93.1%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}}}{\sin th}} \]

      5. hypot-def 99.8%

         \[\leadsto \frac{\sin ky}{\frac{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}}{\sin th}} \]

   3. Applied egg-rr 99.8%

      \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin th}}} \]

   4. Taylor expanded in th around 0 91.8%

      \[\leadsto \frac{\sin ky}{\color{blue}{\frac{1}{th} \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]
   5. Step-by-step derivation

      1. associate-*l/ 92.0%

         \[\leadsto \frac{\sin ky}{\color{blue}{\frac{1 \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{th}}} \]

      2. unpow2 92.0%

         \[\leadsto \frac{\sin ky}{\frac{1 \cdot \sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}}}{th}} \]

      3. unpow2 92.0%

         \[\leadsto \frac{\sin ky}{\frac{1 \cdot \sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}}}{th}} \]

      4. hypot-def 98.5%

         \[\leadsto \frac{\sin ky}{\frac{1 \cdot \color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}}{th}} \]

      5. *-lft-identity 98.5%

         \[\leadsto \frac{\sin ky}{\frac{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}}{th}} \]

      6. hypot-def 92.0%

         \[\leadsto \frac{\sin ky}{\frac{\color{blue}{\sqrt{\sin kx \cdot \sin kx + \sin ky \cdot \sin ky}}}{th}} \]

      7. unpow2 92.0%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\color{blue}{{\sin kx}^{2}} + \sin ky \cdot \sin ky}}{th}} \]

      8. unpow2 92.0%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{{\sin kx}^{2} + \color{blue}{{\sin ky}^{2}}}}{th}} \]

      9. +-commutative 92.0%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}}{th}} \]

      10. unpow2 92.0%

          \[\leadsto \frac{\sin ky}{\frac{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}}{th}} \]

      11. unpow2 92.0%

          \[\leadsto \frac{\sin ky}{\frac{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}}{th}} \]

      12. hypot-def 98.5%

          \[\leadsto \frac{\sin ky}{\frac{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}}{th}} \]

   6. Simplified 98.5%

      \[\leadsto \frac{\sin ky}{\color{blue}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{th}}} \]

   if 2e-3 < (sin.f64 th)

   1. Initial program 93.3%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in ky around 0 25.2%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sin kx}} \cdot \sin th \]
   3. Step-by-step derivation

      1. associate-*l/ 25.2%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sin kx}} \]

      2. associate-/l* 25.2%

         \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\sin kx}{\sin th}}} \]

   4. Applied egg-rr 25.2%

      \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\sin kx}{\sin th}}} \]
   5. Step-by-step derivation

      1. add-sqr-sqrt 24.2%

         \[\leadsto \frac{\sin ky}{\color{blue}{\sqrt{\frac{\sin kx}{\sin th}} \cdot \sqrt{\frac{\sin kx}{\sin th}}}} \]

      2. sqrt-unprod 37.5%

         \[\leadsto \frac{\sin ky}{\color{blue}{\sqrt{\frac{\sin kx}{\sin th} \cdot \frac{\sin kx}{\sin th}}}} \]

      3. pow2 37.5%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{{\left(\frac{\sin kx}{\sin th}\right)}^{2}}}} \]

   6. Applied egg-rr 37.5%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sqrt{{\left(\frac{\sin kx}{\sin th}\right)}^{2}}}} \]
   7. Step-by-step derivation

      1. unpow2 37.5%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{\frac{\sin kx}{\sin th} \cdot \frac{\sin kx}{\sin th}}}} \]

      2. rem-sqrt-square 42.9%

         \[\leadsto \frac{\sin ky}{\color{blue}{\left|\frac{\sin kx}{\sin th}\right|}} \]

   8. Simplified 42.9%

      \[\leadsto \frac{\sin ky}{\color{blue}{\left|\frac{\sin kx}{\sin th}\right|}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 64.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin th \leq -0.005:\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{elif}\;\sin th \leq 0.002:\\ \;\;\;\;\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{th}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sin ky}{\left|\frac{\sin kx}{\sin th}\right|}\\ \end{array} \]

Alternative 3: 39.9% accurate, 1.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.04:\\ \;\;\;\;\frac{\sin ky}{\left|\frac{\sin kx}{\sin th}\right|}\\ \mathbf{elif}\;\sin ky \leq 5 \cdot 10^{-7}:\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) -0.04)
   (/ (sin ky) (fabs (/ (sin kx) (sin th))))
   (if (<= (sin ky) 5e-7) (* (sin th) (fabs (/ (sin ky) (sin kx)))) (sin th))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= -0.04) {
		tmp = sin(ky) / fabs((sin(kx) / sin(th)));
	} else if (sin(ky) <= 5e-7) {
		tmp = sin(th) * fabs((sin(ky) / sin(kx)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (sin(ky) <= (-0.04d0)) then
        tmp = sin(ky) / abs((sin(kx) / sin(th)))
    else if (sin(ky) <= 5d-7) then
        tmp = sin(th) * abs((sin(ky) / sin(kx)))
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= -0.04) {
		tmp = Math.sin(ky) / Math.abs((Math.sin(kx) / Math.sin(th)));
	} else if (Math.sin(ky) <= 5e-7) {
		tmp = Math.sin(th) * Math.abs((Math.sin(ky) / Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= -0.04:
		tmp = math.sin(ky) / math.fabs((math.sin(kx) / math.sin(th)))
	elif math.sin(ky) <= 5e-7:
		tmp = math.sin(th) * math.fabs((math.sin(ky) / math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= -0.04)
		tmp = Float64(sin(ky) / abs(Float64(sin(kx) / sin(th))));
	elseif (sin(ky) <= 5e-7)
		tmp = Float64(sin(th) * abs(Float64(sin(ky) / sin(kx))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= -0.04)
		tmp = sin(ky) / abs((sin(kx) / sin(th)));
	elseif (sin(ky) <= 5e-7)
		tmp = sin(th) * abs((sin(ky) / sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], -0.04], N[(N[Sin[ky], $MachinePrecision] / N[Abs[N[(N[Sin[kx], $MachinePrecision] / N[Sin[th], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Sin[ky], $MachinePrecision], 5e-7], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(N[Sin[ky], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 ky) < -0.0400000000000000008

   1. Initial program 99.8%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in ky around 0 6.3%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sin kx}} \cdot \sin th \]
   3. Step-by-step derivation

      1. associate-*l/ 6.3%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sin kx}} \]

      2. associate-/l* 6.3%

         \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\sin kx}{\sin th}}} \]

   4. Applied egg-rr 6.3%

      \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\sin kx}{\sin th}}} \]
   5. Step-by-step derivation

      1. add-sqr-sqrt 3.6%

         \[\leadsto \frac{\sin ky}{\color{blue}{\sqrt{\frac{\sin kx}{\sin th}} \cdot \sqrt{\frac{\sin kx}{\sin th}}}} \]

      2. sqrt-unprod 6.9%

         \[\leadsto \frac{\sin ky}{\color{blue}{\sqrt{\frac{\sin kx}{\sin th} \cdot \frac{\sin kx}{\sin th}}}} \]

      3. pow2 6.9%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{{\left(\frac{\sin kx}{\sin th}\right)}^{2}}}} \]

   6. Applied egg-rr 6.9%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sqrt{{\left(\frac{\sin kx}{\sin th}\right)}^{2}}}} \]
   7. Step-by-step derivation

      1. unpow2 6.9%

         \[\leadsto \frac{\sin ky}{\sqrt{\color{blue}{\frac{\sin kx}{\sin th} \cdot \frac{\sin kx}{\sin th}}}} \]

      2. rem-sqrt-square 7.2%

         \[\leadsto \frac{\sin ky}{\color{blue}{\left|\frac{\sin kx}{\sin th}\right|}} \]

   8. Simplified 7.2%

      \[\leadsto \frac{\sin ky}{\color{blue}{\left|\frac{\sin kx}{\sin th}\right|}} \]

   if -0.0400000000000000008 < (sin.f64 ky) < 4.99999999999999977e-7

   1. Initial program 85.5%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in ky around 0 45.5%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sin kx}} \cdot \sin th \]
   3. Step-by-step derivation

      1. add-sqr-sqrt 20.7%

         \[\leadsto \color{blue}{\left(\sqrt{\frac{\sin ky}{\sin kx}} \cdot \sqrt{\frac{\sin ky}{\sin kx}}\right)} \cdot \sin th \]

      2. sqrt-unprod 41.4%

         \[\leadsto \color{blue}{\sqrt{\frac{\sin ky}{\sin kx} \cdot \frac{\sin ky}{\sin kx}}} \cdot \sin th \]

      3. pow2 41.4%

         \[\leadsto \sqrt{\color{blue}{{\left(\frac{\sin ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]

   4. Applied egg-rr 41.4%

      \[\leadsto \color{blue}{\sqrt{{\left(\frac{\sin ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]
   5. Step-by-step derivation

      1. unpow2 41.4%

         \[\leadsto \sqrt{\color{blue}{\frac{\sin ky}{\sin kx} \cdot \frac{\sin ky}{\sin kx}}} \cdot \sin th \]

      2. rem-sqrt-square 50.3%

         \[\leadsto \color{blue}{\left|\frac{\sin ky}{\sin kx}\right|} \cdot \sin th \]

   6. Simplified 50.3%

      \[\leadsto \color{blue}{\left|\frac{\sin ky}{\sin kx}\right|} \cdot \sin th \]

   if 4.99999999999999977e-7 < (sin.f64 ky)

   1. Initial program 99.6%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 58.9%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 3 regimes into one program.

4. Final simplification 42.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq -0.04:\\ \;\;\;\;\frac{\sin ky}{\left|\frac{\sin kx}{\sin th}\right|}\\ \mathbf{elif}\;\sin ky \leq 5 \cdot 10^{-7}:\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 4: 99.6% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \sin ky \cdot \frac{\sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (* (sin ky) (/ (sin th) (hypot (sin ky) (sin kx)))))

C

double code(double kx, double ky, double th) {
	return sin(ky) * (sin(th) / hypot(sin(ky), sin(kx)));
}

Java

public static double code(double kx, double ky, double th) {
	return Math.sin(ky) * (Math.sin(th) / Math.hypot(Math.sin(ky), Math.sin(kx)));
}

Python

def code(kx, ky, th):
	return math.sin(ky) * (math.sin(th) / math.hypot(math.sin(ky), math.sin(kx)))

Julia

function code(kx, ky, th)
	return Float64(sin(ky) * Float64(sin(th) / hypot(sin(ky), sin(kx))))
end

MATLAB

function tmp = code(kx, ky, th)
	tmp = sin(ky) * (sin(th) / hypot(sin(ky), sin(kx)));
end

Wolfram

code[kx_, ky_, th_] := N[(N[Sin[ky], $MachinePrecision] * N[(N[Sin[th], $MachinePrecision] / N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 92.8%

   \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
2. Step-by-step derivation

   1. expm1-log1p-u 92.7%

      \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th\right)\right)} \]

   2. expm1-udef 42.2%

      \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th\right)} - 1} \]

3. Applied egg-rr 44.1%

   \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin th\right)} - 1} \]
4. Step-by-step derivation

   1. expm1-def 99.6%

      \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin th\right)\right)} \]

   2. expm1-log1p 99.6%

      \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin th} \]

   3. *-commutative 99.6%

      \[\leadsto \color{blue}{\sin th \cdot \frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

   4. associate-*r/ 96.4%

      \[\leadsto \color{blue}{\frac{\sin th \cdot \sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

   5. associate-*l/ 99.6%

      \[\leadsto \color{blue}{\frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin ky} \]

   6. *-commutative 99.6%

      \[\leadsto \color{blue}{\sin ky \cdot \frac{\sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

   7. hypot-def 92.8%

      \[\leadsto \sin ky \cdot \frac{\sin th}{\color{blue}{\sqrt{\sin kx \cdot \sin kx + \sin ky \cdot \sin ky}}} \]

   8. unpow2 92.8%

      \[\leadsto \sin ky \cdot \frac{\sin th}{\sqrt{\color{blue}{{\sin kx}^{2}} + \sin ky \cdot \sin ky}} \]

   9. unpow2 92.8%

      \[\leadsto \sin ky \cdot \frac{\sin th}{\sqrt{{\sin kx}^{2} + \color{blue}{{\sin ky}^{2}}}} \]

   10. +-commutative 92.8%

       \[\leadsto \sin ky \cdot \frac{\sin th}{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}} \]

   11. unpow2 92.8%

       \[\leadsto \sin ky \cdot \frac{\sin th}{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}} \]

   12. unpow2 92.8%

       \[\leadsto \sin ky \cdot \frac{\sin th}{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}} \]

   13. hypot-def 99.6%

       \[\leadsto \sin ky \cdot \frac{\sin th}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

5. Simplified 99.6%

   \[\leadsto \color{blue}{\sin ky \cdot \frac{\sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}} \]

6. Final simplification 99.6%

   \[\leadsto \sin ky \cdot \frac{\sin th}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]

Alternative 5: 99.7% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \sin th \cdot \frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (* (sin th) (/ (sin ky) (hypot (sin ky) (sin kx)))))

C

double code(double kx, double ky, double th) {
	return sin(th) * (sin(ky) / hypot(sin(ky), sin(kx)));
}

Java

public static double code(double kx, double ky, double th) {
	return Math.sin(th) * (Math.sin(ky) / Math.hypot(Math.sin(ky), Math.sin(kx)));
}

Python

def code(kx, ky, th):
	return math.sin(th) * (math.sin(ky) / math.hypot(math.sin(ky), math.sin(kx)))

Julia

function code(kx, ky, th)
	return Float64(sin(th) * Float64(sin(ky) / hypot(sin(ky), sin(kx))))
end

MATLAB

function tmp = code(kx, ky, th)
	tmp = sin(th) * (sin(ky) / hypot(sin(ky), sin(kx)));
end

Wolfram

code[kx_, ky_, th_] := N[(N[Sin[th], $MachinePrecision] * N[(N[Sin[ky], $MachinePrecision] / N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 92.8%

   \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
2. Step-by-step derivation

   1. remove-double-neg 92.8%

      \[\leadsto \frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \color{blue}{\left(-\left(-\sin th\right)\right)} \]

   2. sin-neg 92.8%

      \[\leadsto \frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \left(-\color{blue}{\sin \left(-th\right)}\right) \]

   3. neg-mul-1 92.8%

      \[\leadsto \frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \color{blue}{\left(-1 \cdot \sin \left(-th\right)\right)} \]

   4. *-commutative 92.8%

      \[\leadsto \frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \color{blue}{\left(\sin \left(-th\right) \cdot -1\right)} \]

   5. associate-*l* 92.8%

      \[\leadsto \color{blue}{\left(\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin \left(-th\right)\right) \cdot -1} \]

   6. associate-*l/ 91.6%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin \left(-th\right)}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \cdot -1 \]

   7. associate-/r/ 91.6%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin \left(-th\right)}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{-1}}} \]

   8. associate-*l/ 92.8%

      \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{-1}} \cdot \sin \left(-th\right)} \]

   9. associate-/r/ 92.8%

      \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{-1}}{\sin \left(-th\right)}}} \]

   10. sin-neg 92.8%

       \[\leadsto \frac{\sin ky}{\frac{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{-1}}{\color{blue}{-\sin th}}} \]

   11. neg-mul-1 92.8%

       \[\leadsto \frac{\sin ky}{\frac{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{-1}}{\color{blue}{-1 \cdot \sin th}}} \]

   12. associate-/r* 92.8%

       \[\leadsto \frac{\sin ky}{\color{blue}{\frac{\frac{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{-1}}{-1}}{\sin th}}} \]

   13. associate-/r/ 92.8%

       \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{-1}}{-1}} \cdot \sin th} \]

3. Simplified 99.6%

   \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \cdot \sin th} \]

4. Final simplification 99.6%

   \[\leadsto \sin th \cdot \frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right)} \]

Alternative 6: 62.7% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;th \leq 0.22:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right) \cdot \left(\frac{1}{th} + th \cdot 0.16666666666666666\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sin th \cdot ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= th 0.22)
   (/
    (sin ky)
    (* (hypot (sin ky) (sin kx)) (+ (/ 1.0 th) (* th 0.16666666666666666))))
   (/ (* (sin th) ky) (hypot (sin kx) (sin ky)))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (th <= 0.22) {
		tmp = sin(ky) / (hypot(sin(ky), sin(kx)) * ((1.0 / th) + (th * 0.16666666666666666)));
	} else {
		tmp = (sin(th) * ky) / hypot(sin(kx), sin(ky));
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (th <= 0.22) {
		tmp = Math.sin(ky) / (Math.hypot(Math.sin(ky), Math.sin(kx)) * ((1.0 / th) + (th * 0.16666666666666666)));
	} else {
		tmp = (Math.sin(th) * ky) / Math.hypot(Math.sin(kx), Math.sin(ky));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if th <= 0.22:
		tmp = math.sin(ky) / (math.hypot(math.sin(ky), math.sin(kx)) * ((1.0 / th) + (th * 0.16666666666666666)))
	else:
		tmp = (math.sin(th) * ky) / math.hypot(math.sin(kx), math.sin(ky))
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (th <= 0.22)
		tmp = Float64(sin(ky) / Float64(hypot(sin(ky), sin(kx)) * Float64(Float64(1.0 / th) + Float64(th * 0.16666666666666666))));
	else
		tmp = Float64(Float64(sin(th) * ky) / hypot(sin(kx), sin(ky)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (th <= 0.22)
		tmp = sin(ky) / (hypot(sin(ky), sin(kx)) * ((1.0 / th) + (th * 0.16666666666666666)));
	else
		tmp = (sin(th) * ky) / hypot(sin(kx), sin(ky));
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[th, 0.22], N[(N[Sin[ky], $MachinePrecision] / N[(N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision] * N[(N[(1.0 / th), $MachinePrecision] + N[(th * 0.16666666666666666), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Sin[th], $MachinePrecision] * ky), $MachinePrecision] / N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if th < 0.220000000000000001

   1. Initial program 92.8%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. associate-*l/ 91.2%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

      2. associate-/l* 92.7%

         \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin th}}} \]

      3. unpow2 92.7%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}}}{\sin th}} \]

      4. unpow2 92.7%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}}}{\sin th}} \]

      5. hypot-def 99.6%

         \[\leadsto \frac{\sin ky}{\frac{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}}{\sin th}} \]

   3. Applied egg-rr 99.6%

      \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin th}}} \]

   4. Taylor expanded in th around 0 61.3%

      \[\leadsto \frac{\sin ky}{\color{blue}{0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right) + \frac{1}{th} \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]
   5. Step-by-step derivation

      1. +-commutative 61.3%

         \[\leadsto \frac{\sin ky}{\color{blue}{\frac{1}{th} \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}} + 0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right)}} \]

      2. unpow2 61.3%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}} + 0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right)} \]

      3. unpow2 61.3%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}} + 0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right)} \]

      4. hypot-def 65.5%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} + 0.16666666666666666 \cdot \left(th \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}\right)} \]

      5. associate-*r* 65.5%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right) + \color{blue}{\left(0.16666666666666666 \cdot th\right) \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

      6. unpow2 65.5%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right) + \left(0.16666666666666666 \cdot th\right) \cdot \sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}}} \]

      7. unpow2 65.5%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right) + \left(0.16666666666666666 \cdot th\right) \cdot \sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}}} \]

      8. hypot-def 65.6%

         \[\leadsto \frac{\sin ky}{\frac{1}{th} \cdot \mathsf{hypot}\left(\sin kx, \sin ky\right) + \left(0.16666666666666666 \cdot th\right) \cdot \color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

      9. distribute-rgt-out 65.6%

         \[\leadsto \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \left(\frac{1}{th} + 0.16666666666666666 \cdot th\right)}} \]

   6. Simplified 65.6%

      \[\leadsto \frac{\sin ky}{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right) \cdot \left(\frac{1}{th} + 0.16666666666666666 \cdot th\right)}} \]

   if 0.220000000000000001 < th

   1. Initial program 92.9%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. expm1-log1p-u 92.6%

         \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th\right)\right)} \]

      2. expm1-udef 65.9%

         \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th\right)} - 1} \]

   3. Applied egg-rr 68.7%

      \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin th\right)} - 1} \]
   4. Step-by-step derivation

      1. expm1-def 99.4%

         \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin th\right)\right)} \]

      2. expm1-log1p 99.5%

         \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin th} \]

      3. associate-*l/ 99.5%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

   5. Simplified 99.5%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

   6. Taylor expanded in ky around 0 45.4%

      \[\leadsto \frac{\color{blue}{ky \cdot \sin th}}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 59.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 0.22:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin ky, \sin kx\right) \cdot \left(\frac{1}{th} + th \cdot 0.16666666666666666\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sin th \cdot ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \end{array} \]

Alternative 7: 62.3% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;th \leq 0.05:\\ \;\;\;\;\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{th}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sin th \cdot ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= th 0.05)
   (/ (sin ky) (/ (hypot (sin ky) (sin kx)) th))
   (/ (* (sin th) ky) (hypot (sin kx) (sin ky)))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (th <= 0.05) {
		tmp = sin(ky) / (hypot(sin(ky), sin(kx)) / th);
	} else {
		tmp = (sin(th) * ky) / hypot(sin(kx), sin(ky));
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (th <= 0.05) {
		tmp = Math.sin(ky) / (Math.hypot(Math.sin(ky), Math.sin(kx)) / th);
	} else {
		tmp = (Math.sin(th) * ky) / Math.hypot(Math.sin(kx), Math.sin(ky));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if th <= 0.05:
		tmp = math.sin(ky) / (math.hypot(math.sin(ky), math.sin(kx)) / th)
	else:
		tmp = (math.sin(th) * ky) / math.hypot(math.sin(kx), math.sin(ky))
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (th <= 0.05)
		tmp = Float64(sin(ky) / Float64(hypot(sin(ky), sin(kx)) / th));
	else
		tmp = Float64(Float64(sin(th) * ky) / hypot(sin(kx), sin(ky)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (th <= 0.05)
		tmp = sin(ky) / (hypot(sin(ky), sin(kx)) / th);
	else
		tmp = (sin(th) * ky) / hypot(sin(kx), sin(ky));
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[th, 0.05], N[(N[Sin[ky], $MachinePrecision] / N[(N[Sqrt[N[Sin[ky], $MachinePrecision] ^ 2 + N[Sin[kx], $MachinePrecision] ^ 2], $MachinePrecision] / th), $MachinePrecision]), $MachinePrecision], N[(N[(N[Sin[th], $MachinePrecision] * ky), $MachinePrecision] / N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if th < 0.050000000000000003

   1. Initial program 92.8%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. associate-*l/ 91.2%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

      2. associate-/l* 92.7%

         \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin th}}} \]

      3. unpow2 92.7%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}}}{\sin th}} \]

      4. unpow2 92.7%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}}}{\sin th}} \]

      5. hypot-def 99.6%

         \[\leadsto \frac{\sin ky}{\frac{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}}{\sin th}} \]

   3. Applied egg-rr 99.6%

      \[\leadsto \color{blue}{\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin th}}} \]

   4. Taylor expanded in th around 0 60.6%

      \[\leadsto \frac{\sin ky}{\color{blue}{\frac{1}{th} \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]
   5. Step-by-step derivation

      1. associate-*l/ 60.8%

         \[\leadsto \frac{\sin ky}{\color{blue}{\frac{1 \cdot \sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{th}}} \]

      2. unpow2 60.8%

         \[\leadsto \frac{\sin ky}{\frac{1 \cdot \sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}}}{th}} \]

      3. unpow2 60.8%

         \[\leadsto \frac{\sin ky}{\frac{1 \cdot \sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}}}{th}} \]

      4. hypot-def 65.0%

         \[\leadsto \frac{\sin ky}{\frac{1 \cdot \color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}}{th}} \]

      5. *-lft-identity 65.0%

         \[\leadsto \frac{\sin ky}{\frac{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}}{th}} \]

      6. hypot-def 60.8%

         \[\leadsto \frac{\sin ky}{\frac{\color{blue}{\sqrt{\sin kx \cdot \sin kx + \sin ky \cdot \sin ky}}}{th}} \]

      7. unpow2 60.8%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\color{blue}{{\sin kx}^{2}} + \sin ky \cdot \sin ky}}{th}} \]

      8. unpow2 60.8%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{{\sin kx}^{2} + \color{blue}{{\sin ky}^{2}}}}{th}} \]

      9. +-commutative 60.8%

         \[\leadsto \frac{\sin ky}{\frac{\sqrt{\color{blue}{{\sin ky}^{2} + {\sin kx}^{2}}}}{th}} \]

      10. unpow2 60.8%

          \[\leadsto \frac{\sin ky}{\frac{\sqrt{\color{blue}{\sin ky \cdot \sin ky} + {\sin kx}^{2}}}{th}} \]

      11. unpow2 60.8%

          \[\leadsto \frac{\sin ky}{\frac{\sqrt{\sin ky \cdot \sin ky + \color{blue}{\sin kx \cdot \sin kx}}}{th}} \]

      12. hypot-def 65.0%

          \[\leadsto \frac{\sin ky}{\frac{\color{blue}{\mathsf{hypot}\left(\sin ky, \sin kx\right)}}{th}} \]

   6. Simplified 65.0%

      \[\leadsto \frac{\sin ky}{\color{blue}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{th}}} \]

   if 0.050000000000000003 < th

   1. Initial program 92.9%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. expm1-log1p-u 92.6%

         \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th\right)\right)} \]

      2. expm1-udef 65.9%

         \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th\right)} - 1} \]

   3. Applied egg-rr 68.7%

      \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin th\right)} - 1} \]
   4. Step-by-step derivation

      1. expm1-def 99.4%

         \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin th\right)\right)} \]

      2. expm1-log1p 99.5%

         \[\leadsto \color{blue}{\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \cdot \sin th} \]

      3. associate-*l/ 99.5%

         \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

   5. Simplified 99.5%

      \[\leadsto \color{blue}{\frac{\sin ky \cdot \sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}} \]

   6. Taylor expanded in ky around 0 45.4%

      \[\leadsto \frac{\color{blue}{ky \cdot \sin th}}{\mathsf{hypot}\left(\sin kx, \sin ky\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 59.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 0.05:\\ \;\;\;\;\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin ky, \sin kx\right)}{th}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sin th \cdot ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \end{array} \]

Alternative 8: 32.7% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 4.5 \cdot 10^{-7}:\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 4.5e-7) (* (sin th) (fabs (/ (sin ky) (sin kx)))) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 4.5e-7) {
		tmp = sin(th) * fabs((sin(ky) / sin(kx)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (ky <= 4.5d-7) then
        tmp = sin(th) * abs((sin(ky) / sin(kx)))
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 4.5e-7) {
		tmp = Math.sin(th) * Math.abs((Math.sin(ky) / Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 4.5e-7:
		tmp = math.sin(th) * math.fabs((math.sin(ky) / math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= 4.5e-7)
		tmp = Float64(sin(th) * abs(Float64(sin(ky) / sin(kx))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (ky <= 4.5e-7)
		tmp = sin(th) * abs((sin(ky) / sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 4.5e-7], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(N[Sin[ky], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if ky < 4.4999999999999998e-7

   1. Initial program 90.4%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in ky around 0 32.1%

      \[\leadsto \frac{\sin ky}{\color{blue}{\sin kx}} \cdot \sin th \]
   3. Step-by-step derivation

      1. add-sqr-sqrt 15.0%

         \[\leadsto \color{blue}{\left(\sqrt{\frac{\sin ky}{\sin kx}} \cdot \sqrt{\frac{\sin ky}{\sin kx}}\right)} \cdot \sin th \]

      2. sqrt-unprod 29.9%

         \[\leadsto \color{blue}{\sqrt{\frac{\sin ky}{\sin kx} \cdot \frac{\sin ky}{\sin kx}}} \cdot \sin th \]

      3. pow2 29.9%

         \[\leadsto \sqrt{\color{blue}{{\left(\frac{\sin ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]

   4. Applied egg-rr 29.9%

      \[\leadsto \color{blue}{\sqrt{{\left(\frac{\sin ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]
   5. Step-by-step derivation

      1. unpow2 29.9%

         \[\leadsto \sqrt{\color{blue}{\frac{\sin ky}{\sin kx} \cdot \frac{\sin ky}{\sin kx}}} \cdot \sin th \]

      2. rem-sqrt-square 35.9%

         \[\leadsto \color{blue}{\left|\frac{\sin ky}{\sin kx}\right|} \cdot \sin th \]

   6. Simplified 35.9%

      \[\leadsto \color{blue}{\left|\frac{\sin ky}{\sin kx}\right|} \cdot \sin th \]

   if 4.4999999999999998e-7 < ky

   1. Initial program 99.8%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 29.0%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 2 regimes into one program.

4. Final simplification 34.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 4.5 \cdot 10^{-7}:\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 9: 31.4% accurate, 2.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 2.5 \cdot 10^{-6}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 2.5e-6) (* (sin th) (fabs (/ ky (sin kx)))) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 2.5e-6) {
		tmp = sin(th) * fabs((ky / sin(kx)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (ky <= 2.5d-6) then
        tmp = sin(th) * abs((ky / sin(kx)))
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 2.5e-6) {
		tmp = Math.sin(th) * Math.abs((ky / Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 2.5e-6:
		tmp = math.sin(th) * math.fabs((ky / math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= 2.5e-6)
		tmp = Float64(sin(th) * abs(Float64(ky / sin(kx))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (ky <= 2.5e-6)
		tmp = sin(th) * abs((ky / sin(kx)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 2.5e-6], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if ky < 2.5000000000000002e-6

   1. Initial program 90.4%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in ky around 0 28.2%

      \[\leadsto \color{blue}{\frac{ky \cdot \sin th}{\sin kx}} \]
   3. Step-by-step derivation

      1. associate-/l* 30.6%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{\sin th}}} \]

      2. associate-/r/ 30.6%

         \[\leadsto \color{blue}{\frac{ky}{\sin kx} \cdot \sin th} \]

   4. Simplified 30.6%

      \[\leadsto \color{blue}{\frac{ky}{\sin kx} \cdot \sin th} \]
   5. Step-by-step derivation

      1. add-sqr-sqrt 14.1%

         \[\leadsto \color{blue}{\left(\sqrt{\frac{ky}{\sin kx}} \cdot \sqrt{\frac{ky}{\sin kx}}\right)} \cdot \sin th \]

      2. sqrt-unprod 28.0%

         \[\leadsto \color{blue}{\sqrt{\frac{ky}{\sin kx} \cdot \frac{ky}{\sin kx}}} \cdot \sin th \]

      3. pow2 28.0%

         \[\leadsto \sqrt{\color{blue}{{\left(\frac{ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]

   6. Applied egg-rr 28.0%

      \[\leadsto \color{blue}{\sqrt{{\left(\frac{ky}{\sin kx}\right)}^{2}}} \cdot \sin th \]
   7. Step-by-step derivation

      1. unpow2 28.0%

         \[\leadsto \sqrt{\color{blue}{\frac{ky}{\sin kx} \cdot \frac{ky}{\sin kx}}} \cdot \sin th \]

      2. rem-sqrt-square 34.0%

         \[\leadsto \color{blue}{\left|\frac{ky}{\sin kx}\right|} \cdot \sin th \]

   8. Simplified 34.0%

      \[\leadsto \color{blue}{\left|\frac{ky}{\sin kx}\right|} \cdot \sin th \]

   if 2.5000000000000002e-6 < ky

   1. Initial program 99.8%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 29.0%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 2 regimes into one program.

4. Final simplification 32.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 2.5 \cdot 10^{-6}:\\ \;\;\;\;\sin th \cdot \left|\frac{ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 10: 32.0% accurate, 3.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 4.9 \cdot 10^{-106}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 4.9e-106) (* (sin th) (/ ky (sin kx))) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 4.9e-106) {
		tmp = sin(th) * (ky / sin(kx));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (ky <= 4.9d-106) then
        tmp = sin(th) * (ky / sin(kx))
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 4.9e-106) {
		tmp = Math.sin(th) * (ky / Math.sin(kx));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 4.9e-106:
		tmp = math.sin(th) * (ky / math.sin(kx))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= 4.9e-106)
		tmp = Float64(sin(th) * Float64(ky / sin(kx)));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (ky <= 4.9e-106)
		tmp = sin(th) * (ky / sin(kx));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 4.9e-106], N[(N[Sin[th], $MachinePrecision] * N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if ky < 4.90000000000000016e-106

   1. Initial program 89.3%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in ky around 0 28.3%

      \[\leadsto \color{blue}{\frac{ky \cdot \sin th}{\sin kx}} \]
   3. Step-by-step derivation

      1. associate-/l* 31.0%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{\sin th}}} \]

      2. associate-/r/ 31.0%

         \[\leadsto \color{blue}{\frac{ky}{\sin kx} \cdot \sin th} \]

   4. Simplified 31.0%

      \[\leadsto \color{blue}{\frac{ky}{\sin kx} \cdot \sin th} \]

   if 4.90000000000000016e-106 < ky

   1. Initial program 99.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 31.5%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 2 regimes into one program.

4. Final simplification 31.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 4.9 \cdot 10^{-106}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 11: 32.0% accurate, 3.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 6.6 \cdot 10^{-107}:\\ \;\;\;\;ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 6.6e-107) (* ky (/ (sin th) (sin kx))) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 6.6e-107) {
		tmp = ky * (sin(th) / sin(kx));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (ky <= 6.6d-107) then
        tmp = ky * (sin(th) / sin(kx))
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 6.6e-107) {
		tmp = ky * (Math.sin(th) / Math.sin(kx));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 6.6e-107:
		tmp = ky * (math.sin(th) / math.sin(kx))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= 6.6e-107)
		tmp = Float64(ky * Float64(sin(th) / sin(kx)));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (ky <= 6.6e-107)
		tmp = ky * (sin(th) / sin(kx));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 6.6e-107], N[(ky * N[(N[Sin[th], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if ky < 6.60000000000000007e-107

   1. Initial program 89.3%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in ky around 0 28.3%

      \[\leadsto \color{blue}{\frac{ky \cdot \sin th}{\sin kx}} \]
   3. Step-by-step derivation

      1. associate-/l* 31.0%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{\sin th}}} \]

      2. associate-/r/ 31.0%

         \[\leadsto \color{blue}{\frac{ky}{\sin kx} \cdot \sin th} \]

   4. Simplified 31.0%

      \[\leadsto \color{blue}{\frac{ky}{\sin kx} \cdot \sin th} \]
   5. Step-by-step derivation

      1. *-commutative 31.0%

         \[\leadsto \color{blue}{\sin th \cdot \frac{ky}{\sin kx}} \]

      2. clear-num 31.0%

         \[\leadsto \sin th \cdot \color{blue}{\frac{1}{\frac{\sin kx}{ky}}} \]

      3. un-div-inv 31.0%

         \[\leadsto \color{blue}{\frac{\sin th}{\frac{\sin kx}{ky}}} \]

   6. Applied egg-rr 31.0%

      \[\leadsto \color{blue}{\frac{\sin th}{\frac{\sin kx}{ky}}} \]
   7. Step-by-step derivation

      1. associate-/r/ 31.0%

         \[\leadsto \color{blue}{\frac{\sin th}{\sin kx} \cdot ky} \]

   8. Simplified 31.0%

      \[\leadsto \color{blue}{\frac{\sin th}{\sin kx} \cdot ky} \]

   if 6.60000000000000007e-107 < ky

   1. Initial program 99.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 31.5%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 2 regimes into one program.

4. Final simplification 31.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 6.6 \cdot 10^{-107}:\\ \;\;\;\;ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 12: 32.0% accurate, 3.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 6.2 \cdot 10^{-106}:\\ \;\;\;\;\frac{\sin th}{\frac{\sin kx}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 6.2e-106) (/ (sin th) (/ (sin kx) ky)) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 6.2e-106) {
		tmp = sin(th) / (sin(kx) / ky);
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (ky <= 6.2d-106) then
        tmp = sin(th) / (sin(kx) / ky)
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 6.2e-106) {
		tmp = Math.sin(th) / (Math.sin(kx) / ky);
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 6.2e-106:
		tmp = math.sin(th) / (math.sin(kx) / ky)
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= 6.2e-106)
		tmp = Float64(sin(th) / Float64(sin(kx) / ky));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (ky <= 6.2e-106)
		tmp = sin(th) / (sin(kx) / ky);
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 6.2e-106], N[(N[Sin[th], $MachinePrecision] / N[(N[Sin[kx], $MachinePrecision] / ky), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if ky < 6.19999999999999971e-106

   1. Initial program 89.3%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]
   2. Step-by-step derivation

      1. *-commutative 89.3%

         \[\leadsto \color{blue}{\sin th \cdot \frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}} \]

      2. clear-num 89.2%

         \[\leadsto \sin th \cdot \color{blue}{\frac{1}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin ky}}} \]

      3. un-div-inv 89.3%

         \[\leadsto \color{blue}{\frac{\sin th}{\frac{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}}{\sin ky}}} \]

      4. unpow2 89.3%

         \[\leadsto \frac{\sin th}{\frac{\sqrt{\color{blue}{\sin kx \cdot \sin kx} + {\sin ky}^{2}}}{\sin ky}} \]

      5. unpow2 89.3%

         \[\leadsto \frac{\sin th}{\frac{\sqrt{\sin kx \cdot \sin kx + \color{blue}{\sin ky \cdot \sin ky}}}{\sin ky}} \]

      6. hypot-def 99.7%

         \[\leadsto \frac{\sin th}{\frac{\color{blue}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}}{\sin ky}} \]

   3. Applied egg-rr 99.7%

      \[\leadsto \color{blue}{\frac{\sin th}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{\sin ky}}} \]

   4. Taylor expanded in ky around 0 31.0%

      \[\leadsto \frac{\sin th}{\color{blue}{\frac{\sin kx}{ky}}} \]

   if 6.19999999999999971e-106 < ky

   1. Initial program 99.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 31.5%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 2 regimes into one program.

4. Final simplification 31.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 6.2 \cdot 10^{-106}:\\ \;\;\;\;\frac{\sin th}{\frac{\sin kx}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 13: 25.6% accurate, 6.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 1.52 \cdot 10^{-105}:\\ \;\;\;\;ky \cdot \frac{\sin th}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 1.52e-105) (* ky (/ (sin th) kx)) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 1.52e-105) {
		tmp = ky * (sin(th) / kx);
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (ky <= 1.52d-105) then
        tmp = ky * (sin(th) / kx)
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 1.52e-105) {
		tmp = ky * (Math.sin(th) / kx);
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 1.52e-105:
		tmp = ky * (math.sin(th) / kx)
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= 1.52e-105)
		tmp = Float64(ky * Float64(sin(th) / kx));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (ky <= 1.52e-105)
		tmp = ky * (sin(th) / kx);
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 1.52e-105], N[(ky * N[(N[Sin[th], $MachinePrecision] / kx), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if ky < 1.52e-105

   1. Initial program 89.3%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in ky around 0 28.3%

      \[\leadsto \color{blue}{\frac{ky \cdot \sin th}{\sin kx}} \]
   3. Step-by-step derivation

      1. associate-/l* 31.0%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{\sin th}}} \]

      2. associate-/r/ 31.0%

         \[\leadsto \color{blue}{\frac{ky}{\sin kx} \cdot \sin th} \]

   4. Simplified 31.0%

      \[\leadsto \color{blue}{\frac{ky}{\sin kx} \cdot \sin th} \]

   5. Taylor expanded in kx around 0 21.6%

      \[\leadsto \color{blue}{\frac{ky}{kx}} \cdot \sin th \]

   6. Taylor expanded in ky around 0 18.8%

      \[\leadsto \color{blue}{\frac{ky \cdot \sin th}{kx}} \]
   7. Step-by-step derivation

      1. associate-*r/ 21.6%

         \[\leadsto \color{blue}{ky \cdot \frac{\sin th}{kx}} \]

   8. Simplified 21.6%

      \[\leadsto \color{blue}{ky \cdot \frac{\sin th}{kx}} \]

   if 1.52e-105 < ky

   1. Initial program 99.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 31.5%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 2 regimes into one program.

4. Final simplification 25.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 1.52 \cdot 10^{-105}:\\ \;\;\;\;ky \cdot \frac{\sin th}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 14: 25.6% accurate, 6.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 1.2 \cdot 10^{-105}:\\ \;\;\;\;\frac{\sin th}{\frac{kx}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 1.2e-105) (/ (sin th) (/ kx ky)) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 1.2e-105) {
		tmp = sin(th) / (kx / ky);
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (ky <= 1.2d-105) then
        tmp = sin(th) / (kx / ky)
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 1.2e-105) {
		tmp = Math.sin(th) / (kx / ky);
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 1.2e-105:
		tmp = math.sin(th) / (kx / ky)
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= 1.2e-105)
		tmp = Float64(sin(th) / Float64(kx / ky));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (ky <= 1.2e-105)
		tmp = sin(th) / (kx / ky);
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 1.2e-105], N[(N[Sin[th], $MachinePrecision] / N[(kx / ky), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if ky < 1.20000000000000007e-105

   1. Initial program 89.3%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in ky around 0 28.3%

      \[\leadsto \color{blue}{\frac{ky \cdot \sin th}{\sin kx}} \]
   3. Step-by-step derivation

      1. associate-/l* 31.0%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{\sin th}}} \]

      2. associate-/r/ 31.0%

         \[\leadsto \color{blue}{\frac{ky}{\sin kx} \cdot \sin th} \]

   4. Simplified 31.0%

      \[\leadsto \color{blue}{\frac{ky}{\sin kx} \cdot \sin th} \]

   5. Taylor expanded in kx around 0 21.6%

      \[\leadsto \color{blue}{\frac{ky}{kx}} \cdot \sin th \]
   6. Step-by-step derivation

      1. *-commutative 21.6%

         \[\leadsto \color{blue}{\sin th \cdot \frac{ky}{kx}} \]

      2. clear-num 21.6%

         \[\leadsto \sin th \cdot \color{blue}{\frac{1}{\frac{kx}{ky}}} \]

      3. un-div-inv 21.6%

         \[\leadsto \color{blue}{\frac{\sin th}{\frac{kx}{ky}}} \]

   7. Applied egg-rr 21.6%

      \[\leadsto \color{blue}{\frac{\sin th}{\frac{kx}{ky}}} \]

   if 1.20000000000000007e-105 < ky

   1. Initial program 99.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 31.5%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 2 regimes into one program.

4. Final simplification 25.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 1.2 \cdot 10^{-105}:\\ \;\;\;\;\frac{\sin th}{\frac{kx}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 15: 23.1% accurate, 6.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 3.1 \cdot 10^{-107}:\\ \;\;\;\;\frac{ky}{\frac{kx}{th}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 3.1e-107) (/ ky (/ kx th)) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 3.1e-107) {
		tmp = ky / (kx / th);
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    real(8) :: tmp
    if (ky <= 3.1d-107) then
        tmp = ky / (kx / th)
    else
        tmp = sin(th)
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 3.1e-107) {
		tmp = ky / (kx / th);
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 3.1e-107:
		tmp = ky / (kx / th)
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= 3.1e-107)
		tmp = Float64(ky / Float64(kx / th));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (ky <= 3.1e-107)
		tmp = ky / (kx / th);
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 3.1e-107], N[(ky / N[(kx / th), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if ky < 3.10000000000000022e-107

   1. Initial program 89.3%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in ky around 0 28.3%

      \[\leadsto \color{blue}{\frac{ky \cdot \sin th}{\sin kx}} \]
   3. Step-by-step derivation

      1. associate-/l* 31.0%

         \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{\sin th}}} \]

      2. associate-/r/ 31.0%

         \[\leadsto \color{blue}{\frac{ky}{\sin kx} \cdot \sin th} \]

   4. Simplified 31.0%

      \[\leadsto \color{blue}{\frac{ky}{\sin kx} \cdot \sin th} \]

   5. Taylor expanded in kx around 0 21.6%

      \[\leadsto \color{blue}{\frac{ky}{kx}} \cdot \sin th \]

   6. Taylor expanded in th around 0 14.9%

      \[\leadsto \color{blue}{\frac{ky \cdot th}{kx}} \]
   7. Step-by-step derivation

      1. associate-/l* 17.7%

         \[\leadsto \color{blue}{\frac{ky}{\frac{kx}{th}}} \]

   8. Simplified 17.7%

      \[\leadsto \color{blue}{\frac{ky}{\frac{kx}{th}}} \]

   if 3.10000000000000022e-107 < ky

   1. Initial program 99.7%

      \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

   2. Taylor expanded in kx around 0 31.5%

      \[\leadsto \color{blue}{\sin th} \]
3. Recombined 2 regimes into one program.

4. Final simplification 22.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 3.1 \cdot 10^{-107}:\\ \;\;\;\;\frac{ky}{\frac{kx}{th}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 16: 13.6% accurate, 141.8× speedup

Math

\[\begin{array}{l} \\ \frac{ky}{\frac{kx}{th}} \end{array} \]

FPCore

(FPCore (kx ky th) :precision binary64 (/ ky (/ kx th)))

C

double code(double kx, double ky, double th) {
	return ky / (kx / th);
}

Fortran

real(8) function code(kx, ky, th)
    real(8), intent (in) :: kx
    real(8), intent (in) :: ky
    real(8), intent (in) :: th
    code = ky / (kx / th)
end function

Java

public static double code(double kx, double ky, double th) {
	return ky / (kx / th);
}

Python

def code(kx, ky, th):
	return ky / (kx / th)

Julia

function code(kx, ky, th)
	return Float64(ky / Float64(kx / th))
end

MATLAB

function tmp = code(kx, ky, th)
	tmp = ky / (kx / th);
end

Wolfram

code[kx_, ky_, th_] := N[(ky / N[(kx / th), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 92.8%

   \[\frac{\sin ky}{\sqrt{{\sin kx}^{2} + {\sin ky}^{2}}} \cdot \sin th \]

2. Taylor expanded in ky around 0 21.9%

   \[\leadsto \color{blue}{\frac{ky \cdot \sin th}{\sin kx}} \]
3. Step-by-step derivation

   1. associate-/l* 23.7%

      \[\leadsto \color{blue}{\frac{ky}{\frac{\sin kx}{\sin th}}} \]

   2. associate-/r/ 23.7%

      \[\leadsto \color{blue}{\frac{ky}{\sin kx} \cdot \sin th} \]

4. Simplified 23.7%

   \[\leadsto \color{blue}{\frac{ky}{\sin kx} \cdot \sin th} \]

5. Taylor expanded in kx around 0 15.8%

   \[\leadsto \color{blue}{\frac{ky}{kx}} \cdot \sin th \]

6. Taylor expanded in th around 0 11.1%

   \[\leadsto \color{blue}{\frac{ky \cdot th}{kx}} \]
7. Step-by-step derivation

   1. associate-/l* 12.9%

      \[\leadsto \color{blue}{\frac{ky}{\frac{kx}{th}}} \]

8. Simplified 12.9%

   \[\leadsto \color{blue}{\frac{ky}{\frac{kx}{th}}} \]

9. Final simplification 12.9%

   \[\leadsto \frac{ky}{\frac{kx}{th}} \]

Reproduce

herbie shell --seed 2023337 
(FPCore (kx ky th)
  :name "Toniolo and Linder, Equation (3b), real"
  :precision binary64
  (* (/ (sin ky) (sqrt (+ (pow (sin kx) 2.0) (pow (sin ky) 2.0)))) (sin th)))