Toniolo and Linder, Equation (3b), real

Percentage Accurate: 93.7% → 99.7%

Time: 19.4s

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: 93.7% 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

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 91.1%

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

   1. remove-double-neg 91.1%

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

   2. sin-neg 91.1%

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

   3. neg-mul-1 91.1%

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

   4. *-commutative 91.1%

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

   5. associate-*l* 91.1%

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

   6. associate-*l/ 88.5%

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

   7. associate-/r/ 88.5%

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

   8. associate-*l/ 91.1%

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

   9. associate-/r/ 91.0%

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

   10. sin-neg 91.0%

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

   11. neg-mul-1 91.0%

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

   12. associate-/r* 91.0%

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

   13. associate-/r/ 91.1%

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

3. Simplified 99.7%

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

4. Final simplification 99.7%

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

Alternative 2: 44.3% accurate, 1.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \sin th \cdot \frac{\sin ky}{\sin kx}\\ \mathbf{if}\;\sin kx \leq -0.01:\\ \;\;\;\;\left|t_1\right|\\ \mathbf{elif}\;\sin kx \leq 5 \cdot 10^{-206}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;t_1\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (let* ((t_1 (* (sin th) (/ (sin ky) (sin kx)))))
   (if (<= (sin kx) -0.01) (fabs t_1) (if (<= (sin kx) 5e-206) (sin th) t_1))))

C

double code(double kx, double ky, double th) {
	double t_1 = sin(th) * (sin(ky) / sin(kx));
	double tmp;
	if (sin(kx) <= -0.01) {
		tmp = fabs(t_1);
	} else if (sin(kx) <= 5e-206) {
		tmp = sin(th);
	} else {
		tmp = t_1;
	}
	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) :: t_1
    real(8) :: tmp
    t_1 = sin(th) * (sin(ky) / sin(kx))
    if (sin(kx) <= (-0.01d0)) then
        tmp = abs(t_1)
    else if (sin(kx) <= 5d-206) then
        tmp = sin(th)
    else
        tmp = t_1
    end if
    code = tmp
end function

Java

public static double code(double kx, double ky, double th) {
	double t_1 = Math.sin(th) * (Math.sin(ky) / Math.sin(kx));
	double tmp;
	if (Math.sin(kx) <= -0.01) {
		tmp = Math.abs(t_1);
	} else if (Math.sin(kx) <= 5e-206) {
		tmp = Math.sin(th);
	} else {
		tmp = t_1;
	}
	return tmp;
}

Python

def code(kx, ky, th):
	t_1 = math.sin(th) * (math.sin(ky) / math.sin(kx))
	tmp = 0
	if math.sin(kx) <= -0.01:
		tmp = math.fabs(t_1)
	elif math.sin(kx) <= 5e-206:
		tmp = math.sin(th)
	else:
		tmp = t_1
	return tmp

Julia

function code(kx, ky, th)
	t_1 = Float64(sin(th) * Float64(sin(ky) / sin(kx)))
	tmp = 0.0
	if (sin(kx) <= -0.01)
		tmp = abs(t_1);
	elseif (sin(kx) <= 5e-206)
		tmp = sin(th);
	else
		tmp = t_1;
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	t_1 = sin(th) * (sin(ky) / sin(kx));
	tmp = 0.0;
	if (sin(kx) <= -0.01)
		tmp = abs(t_1);
	elseif (sin(kx) <= 5e-206)
		tmp = sin(th);
	else
		tmp = t_1;
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := Block[{t$95$1 = N[(N[Sin[th], $MachinePrecision] * N[(N[Sin[ky], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Sin[kx], $MachinePrecision], -0.01], N[Abs[t$95$1], $MachinePrecision], If[LessEqual[N[Sin[kx], $MachinePrecision], 5e-206], N[Sin[th], $MachinePrecision], t$95$1]]]

Derivation

1. Split input into 3 regimes

2. if (sin.f64 kx) < -0.0100000000000000002

   1. Initial program 99.5%

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

   2. Taylor expanded in ky around 0 17.7%

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

      1. add-sqr-sqrt 9.3%

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

      2. sqrt-unprod 26.3%

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

      3. pow2 26.3%

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

   4. Applied egg-rr 26.3%

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

      1. unpow2 26.3%

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

      2. rem-sqrt-square 29.0%

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

   6. Simplified 29.0%

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

      1. add-sqr-sqrt 23.2%

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

      2. sqrt-unprod 31.0%

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

      3. pow2 31.0%

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

      4. *-commutative 31.0%

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

      5. add-sqr-sqrt 17.6%

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

      6. fabs-sqr 17.6%

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

      7. add-sqr-sqrt 31.0%

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

   8. Applied egg-rr 31.0%

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

      1. unpow2 31.0%

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

      2. rem-sqrt-square 40.1%

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

   10. Simplified 40.1%

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

   if -0.0100000000000000002 < (sin.f64 kx) < 5e-206

   1. Initial program 79.3%

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

   2. Taylor expanded in kx around 0 41.5%

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

   if 5e-206 < (sin.f64 kx)

   1. Initial program 97.2%

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

   2. Taylor expanded in ky around 0 45.7%

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

4. Final simplification 42.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin kx \leq -0.01:\\ \;\;\;\;\left|\sin th \cdot \frac{\sin ky}{\sin kx}\right|\\ \mathbf{elif}\;\sin kx \leq 5 \cdot 10^{-206}:\\ \;\;\;\;\sin th\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\sin kx}\\ \end{array} \]

Alternative 3: 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 91.1%

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

   1. expm1-log1p-u 90.9%

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

   2. expm1-udef 38.4%

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

3. Applied egg-rr 41.7%

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

   1. expm1-def 99.6%

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

   2. expm1-log1p 99.7%

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

   3. *-commutative 99.7%

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

   4. associate-*r/ 95.4%

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

   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 91.0%

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

   8. unpow2 91.0%

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

   9. unpow2 91.0%

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

   10. +-commutative 91.0%

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

   11. unpow2 91.0%

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

   12. unpow2 91.0%

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

   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 4: 63.0% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \mathsf{hypot}\left(\sin kx, \sin ky\right)\\ \mathbf{if}\;th \leq 0.0195:\\ \;\;\;\;\frac{\sin ky}{t_1 \cdot \left(\frac{1}{th} + th \cdot 0.16666666666666666\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{ky \cdot \sin th}{t_1}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (let* ((t_1 (hypot (sin kx) (sin ky))))
   (if (<= th 0.0195)
     (/ (sin ky) (* t_1 (+ (/ 1.0 th) (* th 0.16666666666666666))))
     (/ (* ky (sin th)) t_1))))

C

double code(double kx, double ky, double th) {
	double t_1 = hypot(sin(kx), sin(ky));
	double tmp;
	if (th <= 0.0195) {
		tmp = sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	} else {
		tmp = (ky * sin(th)) / t_1;
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double t_1 = Math.hypot(Math.sin(kx), Math.sin(ky));
	double tmp;
	if (th <= 0.0195) {
		tmp = Math.sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	} else {
		tmp = (ky * Math.sin(th)) / t_1;
	}
	return tmp;
}

Python

def code(kx, ky, th):
	t_1 = math.hypot(math.sin(kx), math.sin(ky))
	tmp = 0
	if th <= 0.0195:
		tmp = math.sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)))
	else:
		tmp = (ky * math.sin(th)) / t_1
	return tmp

Julia

function code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky))
	tmp = 0.0
	if (th <= 0.0195)
		tmp = Float64(sin(ky) / Float64(t_1 * Float64(Float64(1.0 / th) + Float64(th * 0.16666666666666666))));
	else
		tmp = Float64(Float64(ky * sin(th)) / t_1);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky));
	tmp = 0.0;
	if (th <= 0.0195)
		tmp = sin(ky) / (t_1 * ((1.0 / th) + (th * 0.16666666666666666)));
	else
		tmp = (ky * sin(th)) / t_1;
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := Block[{t$95$1 = N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision]}, If[LessEqual[th, 0.0195], N[(N[Sin[ky], $MachinePrecision] / N[(t$95$1 * N[(N[(1.0 / th), $MachinePrecision] + N[(th * 0.16666666666666666), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(ky * N[Sin[th], $MachinePrecision]), $MachinePrecision] / t$95$1), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if th < 0.0195

   1. Initial program 93.1%

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

      1. associate-*l/ 89.8%

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

      2. associate-/l* 93.0%

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

      3. unpow2 93.0%

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

      4. unpow2 93.0%

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

      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 68.2%

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

      1. +-commutative 68.2%

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

      2. unpow2 68.2%

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

      3. unpow2 68.2%

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

      4. hypot-def 71.1%

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

      5. associate-*r* 71.1%

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

      6. unpow2 71.1%

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

      7. unpow2 71.1%

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

      8. hypot-def 71.1%

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

      9. distribute-rgt-out 71.1%

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

   6. Simplified 71.1%

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

   if 0.0195 < th

   1. Initial program 84.2%

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

      1. expm1-log1p-u 83.7%

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

      2. expm1-udef 59.7%

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

   3. Applied egg-rr 70.1%

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

      1. expm1-def 99.5%

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

      2. expm1-log1p 99.7%

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

      3. associate-*l/ 99.6%

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

   5. Simplified 99.6%

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

   6. Taylor expanded in ky around 0 52.8%

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

4. Final simplification 66.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 0.0195:\\ \;\;\;\;\frac{\sin ky}{\mathsf{hypot}\left(\sin kx, \sin ky\right) \cdot \left(\frac{1}{th} + th \cdot 0.16666666666666666\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{ky \cdot \sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \end{array} \]

Alternative 5: 55.6% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;th \leq 0.003:\\ \;\;\;\;\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{th}}\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= th 0.003)
   (/ (sin ky) (/ (hypot (sin kx) (sin ky)) th))
   (* (sin th) (fabs (/ (sin ky) (sin kx))))))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (th <= 0.003) {
		tmp = sin(ky) / (hypot(sin(kx), sin(ky)) / th);
	} else {
		tmp = sin(th) * fabs((sin(ky) / sin(kx)));
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (th <= 0.003) {
		tmp = Math.sin(ky) / (Math.hypot(Math.sin(kx), Math.sin(ky)) / th);
	} else {
		tmp = Math.sin(th) * Math.abs((Math.sin(ky) / Math.sin(kx)));
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if th <= 0.003:
		tmp = math.sin(ky) / (math.hypot(math.sin(kx), math.sin(ky)) / th)
	else:
		tmp = math.sin(th) * math.fabs((math.sin(ky) / math.sin(kx)))
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (th <= 0.003)
		tmp = Float64(sin(ky) / Float64(hypot(sin(kx), sin(ky)) / th));
	else
		tmp = Float64(sin(th) * abs(Float64(sin(ky) / sin(kx))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (th <= 0.003)
		tmp = sin(ky) / (hypot(sin(kx), sin(ky)) / th);
	else
		tmp = sin(th) * abs((sin(ky) / sin(kx)));
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[th, 0.003], N[(N[Sin[ky], $MachinePrecision] / N[(N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision] / th), $MachinePrecision]), $MachinePrecision], N[(N[Sin[th], $MachinePrecision] * N[Abs[N[(N[Sin[ky], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if th < 0.0030000000000000001

   1. Initial program 93.1%

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

      1. associate-*l/ 89.8%

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

      2. associate-/l* 93.0%

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

      3. unpow2 93.0%

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

      4. unpow2 93.0%

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

      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 67.6%

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

      1. associate-*l/ 67.7%

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

      2. unpow2 67.7%

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

      3. unpow2 67.7%

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

      4. hypot-def 70.6%

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

      5. *-lft-identity 70.6%

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

   6. Simplified 70.6%

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

   if 0.0030000000000000001 < th

   1. Initial program 84.2%

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

   2. Taylor expanded in ky around 0 18.0%

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

      1. add-sqr-sqrt 9.5%

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

      2. sqrt-unprod 16.6%

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

      3. pow2 16.6%

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

   4. Applied egg-rr 16.6%

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

      1. unpow2 16.6%

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

      2. rem-sqrt-square 19.0%

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

   6. Simplified 19.0%

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

4. Final simplification 58.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 0.003:\\ \;\;\;\;\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{th}}\\ \mathbf{else}:\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \end{array} \]

Alternative 6: 62.5% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \mathsf{hypot}\left(\sin kx, \sin ky\right)\\ \mathbf{if}\;th \leq 0.01:\\ \;\;\;\;\frac{\sin ky}{\frac{t_1}{th}}\\ \mathbf{else}:\\ \;\;\;\;\frac{ky \cdot \sin th}{t_1}\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (let* ((t_1 (hypot (sin kx) (sin ky))))
   (if (<= th 0.01) (/ (sin ky) (/ t_1 th)) (/ (* ky (sin th)) t_1))))

C

double code(double kx, double ky, double th) {
	double t_1 = hypot(sin(kx), sin(ky));
	double tmp;
	if (th <= 0.01) {
		tmp = sin(ky) / (t_1 / th);
	} else {
		tmp = (ky * sin(th)) / t_1;
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double t_1 = Math.hypot(Math.sin(kx), Math.sin(ky));
	double tmp;
	if (th <= 0.01) {
		tmp = Math.sin(ky) / (t_1 / th);
	} else {
		tmp = (ky * Math.sin(th)) / t_1;
	}
	return tmp;
}

Python

def code(kx, ky, th):
	t_1 = math.hypot(math.sin(kx), math.sin(ky))
	tmp = 0
	if th <= 0.01:
		tmp = math.sin(ky) / (t_1 / th)
	else:
		tmp = (ky * math.sin(th)) / t_1
	return tmp

Julia

function code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky))
	tmp = 0.0
	if (th <= 0.01)
		tmp = Float64(sin(ky) / Float64(t_1 / th));
	else
		tmp = Float64(Float64(ky * sin(th)) / t_1);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	t_1 = hypot(sin(kx), sin(ky));
	tmp = 0.0;
	if (th <= 0.01)
		tmp = sin(ky) / (t_1 / th);
	else
		tmp = (ky * sin(th)) / t_1;
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := Block[{t$95$1 = N[Sqrt[N[Sin[kx], $MachinePrecision] ^ 2 + N[Sin[ky], $MachinePrecision] ^ 2], $MachinePrecision]}, If[LessEqual[th, 0.01], N[(N[Sin[ky], $MachinePrecision] / N[(t$95$1 / th), $MachinePrecision]), $MachinePrecision], N[(N[(ky * N[Sin[th], $MachinePrecision]), $MachinePrecision] / t$95$1), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if th < 0.0100000000000000002

   1. Initial program 93.1%

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

      1. associate-*l/ 89.8%

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

      2. associate-/l* 93.0%

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

      3. unpow2 93.0%

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

      4. unpow2 93.0%

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

      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 67.6%

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

      1. associate-*l/ 67.7%

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

      2. unpow2 67.7%

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

      3. unpow2 67.7%

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

      4. hypot-def 70.6%

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

      5. *-lft-identity 70.6%

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

   6. Simplified 70.6%

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

   if 0.0100000000000000002 < th

   1. Initial program 84.2%

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

      1. expm1-log1p-u 83.7%

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

      2. expm1-udef 59.7%

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

   3. Applied egg-rr 70.1%

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

      1. expm1-def 99.5%

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

      2. expm1-log1p 99.7%

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

      3. associate-*l/ 99.6%

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

   5. Simplified 99.6%

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

   6. Taylor expanded in ky around 0 52.8%

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

4. Final simplification 66.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;th \leq 0.01:\\ \;\;\;\;\frac{\sin ky}{\frac{\mathsf{hypot}\left(\sin kx, \sin ky\right)}{th}}\\ \mathbf{else}:\\ \;\;\;\;\frac{ky \cdot \sin th}{\mathsf{hypot}\left(\sin kx, \sin ky\right)}\\ \end{array} \]

Alternative 7: 49.8% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq 2 \cdot 10^{-41}:\\ \;\;\;\;\sin th \cdot \mathsf{expm1}\left(\frac{ky}{\sin kx}\right)\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) 2e-41) (* (sin th) (expm1 (/ ky (sin kx)))) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= 2e-41) {
		tmp = sin(th) * expm1((ky / sin(kx)));
	} else {
		tmp = sin(th);
	}
	return tmp;
}

Java

public static double code(double kx, double ky, double th) {
	double tmp;
	if (Math.sin(ky) <= 2e-41) {
		tmp = Math.sin(th) * Math.expm1((ky / Math.sin(kx)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= 2e-41:
		tmp = math.sin(th) * math.expm1((ky / math.sin(kx)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= 2e-41)
		tmp = Float64(sin(th) * expm1(Float64(ky / sin(kx))));
	else
		tmp = sin(th);
	end
	return tmp
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], 2e-41], N[(N[Sin[th], $MachinePrecision] * N[(Exp[N[(ky / N[Sin[kx], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (sin.f64 ky) < 2.00000000000000001e-41

   1. Initial program 87.4%

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

      1. remove-double-neg 87.4%

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

      2. sin-neg 87.4%

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

      3. neg-mul-1 87.4%

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

      4. *-commutative 87.4%

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

      5. associate-*l* 87.4%

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

      6. associate-*l/ 83.9%

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

      7. associate-/r/ 83.9%

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

      8. associate-*l/ 87.4%

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

      9. associate-/r/ 87.4%

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

      10. sin-neg 87.4%

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

      11. neg-mul-1 87.4%

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

      12. associate-/r* 87.4%

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

      13. associate-/r/ 87.4%

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

   3. Simplified 99.7%

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

      1. expm1-log1p-u 99.7%

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

      2. hypot-udef 87.3%

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

      3. unpow2 87.3%

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

      4. unpow2 87.3%

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

      5. +-commutative 87.3%

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

      6. unpow2 87.3%

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

      7. unpow2 87.3%

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

      8. hypot-def 99.7%

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

   5. Applied egg-rr 99.7%

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

   6. Taylor expanded in ky around 0 44.6%

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

   if 2.00000000000000001e-41 < (sin.f64 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 55.9%

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

4. Final simplification 48.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq 2 \cdot 10^{-41}:\\ \;\;\;\;\sin th \cdot \mathsf{expm1}\left(\frac{ky}{\sin kx}\right)\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 8: 41.5% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq 2 \cdot 10^{-41}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) 2e-41) (* (sin th) (/ (sin ky) (sin kx))) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= 2e-41) {
		tmp = sin(th) * (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) <= 2d-41) then
        tmp = sin(th) * (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) <= 2e-41) {
		tmp = Math.sin(th) * (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) <= 2e-41:
		tmp = math.sin(th) * (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) <= 2e-41)
		tmp = Float64(sin(th) * 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) <= 2e-41)
		tmp = sin(th) * (sin(ky) / sin(kx));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], 2e-41], N[(N[Sin[th], $MachinePrecision] * N[(N[Sin[ky], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (sin.f64 ky) < 2.00000000000000001e-41

   1. Initial program 87.4%

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

   2. Taylor expanded in ky around 0 31.4%

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

   if 2.00000000000000001e-41 < (sin.f64 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 55.9%

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

4. Final simplification 38.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq 2 \cdot 10^{-41}:\\ \;\;\;\;\sin th \cdot \frac{\sin ky}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 9: 41.5% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq 2 \cdot 10^{-41}:\\ \;\;\;\;\sin ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) 2e-41) (* (sin ky) (/ (sin th) (sin kx))) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= 2e-41) {
		tmp = sin(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 (sin(ky) <= 2d-41) then
        tmp = sin(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 (Math.sin(ky) <= 2e-41) {
		tmp = Math.sin(ky) * (Math.sin(th) / Math.sin(kx));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= 2e-41:
		tmp = math.sin(ky) * (math.sin(th) / math.sin(kx))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= 2e-41)
		tmp = Float64(sin(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 (sin(ky) <= 2e-41)
		tmp = sin(ky) * (sin(th) / sin(kx));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], 2e-41], N[(N[Sin[ky], $MachinePrecision] * N[(N[Sin[th], $MachinePrecision] / N[Sin[kx], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (sin.f64 ky) < 2.00000000000000001e-41

   1. Initial program 87.4%

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

   2. Taylor expanded in ky around 0 31.4%

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

   3. Taylor expanded in ky around inf 29.8%

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

      1. *-commutative 29.8%

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

      2. *-rgt-identity 29.8%

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

      3. times-frac 31.4%

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

      4. /-rgt-identity 31.4%

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

   5. Simplified 31.4%

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

   if 2.00000000000000001e-41 < (sin.f64 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 55.9%

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

4. Final simplification 38.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq 2 \cdot 10^{-41}:\\ \;\;\;\;\sin ky \cdot \frac{\sin th}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 10: 33.0% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 3.7 \cdot 10^{-29}:\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 3.7e-29) (* (sin th) (fabs (/ (sin ky) (sin kx)))) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 3.7e-29) {
		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 <= 3.7d-29) 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 <= 3.7e-29) {
		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 <= 3.7e-29:
		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 <= 3.7e-29)
		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 <= 3.7e-29)
		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, 3.7e-29], 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 < 3.6999999999999997e-29

   1. Initial program 87.6%

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

   2. Taylor expanded in ky around 0 31.0%

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

      1. add-sqr-sqrt 14.7%

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

      2. sqrt-unprod 26.6%

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

      3. pow2 26.6%

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

   4. Applied egg-rr 26.6%

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

      1. unpow2 26.6%

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

      2. rem-sqrt-square 29.9%

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

   6. Simplified 29.9%

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

   if 3.6999999999999997e-29 < 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 32.7%

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

4. Final simplification 30.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 3.7 \cdot 10^{-29}:\\ \;\;\;\;\sin th \cdot \left|\frac{\sin ky}{\sin kx}\right|\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 11: 33.5% accurate, 3.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\sin ky \leq 10^{-101}:\\ \;\;\;\;\sin th \cdot \frac{1}{0.16666666666666666 \cdot \left(ky \cdot kx\right) + \frac{kx}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= (sin ky) 1e-101)
   (* (sin th) (/ 1.0 (+ (* 0.16666666666666666 (* ky kx)) (/ kx ky))))
   (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (sin(ky) <= 1e-101) {
		tmp = sin(th) * (1.0 / ((0.16666666666666666 * (ky * kx)) + (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 (sin(ky) <= 1d-101) then
        tmp = sin(th) * (1.0d0 / ((0.16666666666666666d0 * (ky * kx)) + (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 (Math.sin(ky) <= 1e-101) {
		tmp = Math.sin(th) * (1.0 / ((0.16666666666666666 * (ky * kx)) + (kx / ky)));
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if math.sin(ky) <= 1e-101:
		tmp = math.sin(th) * (1.0 / ((0.16666666666666666 * (ky * kx)) + (kx / ky)))
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (sin(ky) <= 1e-101)
		tmp = Float64(sin(th) * Float64(1.0 / Float64(Float64(0.16666666666666666 * Float64(ky * kx)) + Float64(kx / ky))));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (sin(ky) <= 1e-101)
		tmp = sin(th) * (1.0 / ((0.16666666666666666 * (ky * kx)) + (kx / ky)));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[N[Sin[ky], $MachinePrecision], 1e-101], N[(N[Sin[th], $MachinePrecision] * N[(1.0 / N[(N[(0.16666666666666666 * N[(ky * kx), $MachinePrecision]), $MachinePrecision] + N[(kx / ky), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (sin.f64 ky) < 1.00000000000000005e-101

   1. Initial program 86.8%

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

   2. Taylor expanded in ky around 0 30.2%

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

   3. Taylor expanded in kx around 0 22.4%

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

      1. clear-num 22.4%

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

      2. inv-pow 22.4%

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

   5. Applied egg-rr 22.4%

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

      1. unpow-1 22.4%

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

   7. Simplified 22.4%

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

   8. Taylor expanded in ky around 0 22.6%

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

   if 1.00000000000000005e-101 < (sin.f64 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 53.0%

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

4. Final simplification 32.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\sin ky \leq 10^{-101}:\\ \;\;\;\;\sin th \cdot \frac{1}{0.16666666666666666 \cdot \left(ky \cdot kx\right) + \frac{kx}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 12: 33.3% accurate, 3.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 5.8 \cdot 10^{-39}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 5.8e-39) (* (sin th) (/ ky (sin kx))) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 5.8e-39) {
		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 <= 5.8d-39) 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 <= 5.8e-39) {
		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 <= 5.8e-39:
		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 <= 5.8e-39)
		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 <= 5.8e-39)
		tmp = sin(th) * (ky / sin(kx));
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 5.8e-39], 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 < 5.79999999999999975e-39

   1. Initial program 87.5%

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

   2. Taylor expanded in ky around 0 27.4%

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

      1. associate-/l* 29.0%

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

      2. associate-/r/ 29.0%

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

   4. Simplified 29.0%

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

   if 5.79999999999999975e-39 < 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 33.6%

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

4. Final simplification 30.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 5.8 \cdot 10^{-39}:\\ \;\;\;\;\sin th \cdot \frac{ky}{\sin kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 13: 33.4% accurate, 3.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 3 \cdot 10^{-41}:\\ \;\;\;\;\frac{\sin th}{\frac{\sin kx}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 3e-41) (/ (sin th) (/ (sin kx) ky)) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 3e-41) {
		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 <= 3d-41) 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 <= 3e-41) {
		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 <= 3e-41:
		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 <= 3e-41)
		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 <= 3e-41)
		tmp = sin(th) / (sin(kx) / ky);
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 3e-41], 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 < 2.99999999999999989e-41

   1. Initial program 87.5%

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

      1. *-commutative 87.5%

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

      2. clear-num 87.4%

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

      3. un-div-inv 87.5%

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

      4. unpow2 87.5%

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

      5. unpow2 87.5%

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

      6. hypot-def 99.6%

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

   3. Applied egg-rr 99.6%

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

   4. Taylor expanded in ky around 0 29.0%

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

   if 2.99999999999999989e-41 < 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 33.6%

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

4. Final simplification 30.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 3 \cdot 10^{-41}:\\ \;\;\;\;\frac{\sin th}{\frac{\sin kx}{ky}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 14: 26.3% accurate, 6.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 10^{-101}:\\ \;\;\;\;\sin th \cdot \frac{ky}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 1e-101) (* (sin th) (/ ky kx)) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 1e-101) {
		tmp = sin(th) * (ky / 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 <= 1d-101) then
        tmp = sin(th) * (ky / 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 <= 1e-101) {
		tmp = Math.sin(th) * (ky / kx);
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 1e-101:
		tmp = math.sin(th) * (ky / kx)
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= 1e-101)
		tmp = Float64(sin(th) * Float64(ky / kx));
	else
		tmp = sin(th);
	end
	return tmp
end

MATLAB

function tmp_2 = code(kx, ky, th)
	tmp = 0.0;
	if (ky <= 1e-101)
		tmp = sin(th) * (ky / kx);
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 1e-101], N[(N[Sin[th], $MachinePrecision] * N[(ky / kx), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if ky < 1.00000000000000005e-101

   1. Initial program 86.9%

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

   2. Taylor expanded in ky around 0 30.0%

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

   3. Taylor expanded in kx around 0 22.2%

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

   4. Taylor expanded in ky around 0 22.0%

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

   if 1.00000000000000005e-101 < 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 33.0%

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

4. Final simplification 25.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 10^{-101}:\\ \;\;\;\;\sin th \cdot \frac{ky}{kx}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 15: 23.3% accurate, 6.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;ky \leq 2 \cdot 10^{-101}:\\ \;\;\;\;\frac{ky}{\frac{kx}{th}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \end{array} \]

FPCore

(FPCore (kx ky th)
 :precision binary64
 (if (<= ky 2e-101) (/ ky (/ kx th)) (sin th)))

C

double code(double kx, double ky, double th) {
	double tmp;
	if (ky <= 2e-101) {
		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 <= 2d-101) 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 <= 2e-101) {
		tmp = ky / (kx / th);
	} else {
		tmp = Math.sin(th);
	}
	return tmp;
}

Python

def code(kx, ky, th):
	tmp = 0
	if ky <= 2e-101:
		tmp = ky / (kx / th)
	else:
		tmp = math.sin(th)
	return tmp

Julia

function code(kx, ky, th)
	tmp = 0.0
	if (ky <= 2e-101)
		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 <= 2e-101)
		tmp = ky / (kx / th);
	else
		tmp = sin(th);
	end
	tmp_2 = tmp;
end

Wolfram

code[kx_, ky_, th_] := If[LessEqual[ky, 2e-101], N[(ky / N[(kx / th), $MachinePrecision]), $MachinePrecision], N[Sin[th], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if ky < 2.0000000000000001e-101

   1. Initial program 86.9%

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

   2. Taylor expanded in ky around 0 30.0%

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

   3. Taylor expanded in kx around 0 22.2%

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

   4. Taylor expanded in ky around 0 22.0%

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

   5. Taylor expanded in th around 0 15.7%

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

      1. associate-/l* 16.8%

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

   7. Simplified 16.8%

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

   if 2.0000000000000001e-101 < 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 33.0%

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

4. Final simplification 22.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;ky \leq 2 \cdot 10^{-101}:\\ \;\;\;\;\frac{ky}{\frac{kx}{th}}\\ \mathbf{else}:\\ \;\;\;\;\sin th\\ \end{array} \]

Alternative 16: 13.4% 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 91.1%

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

2. Taylor expanded in ky around 0 24.6%

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

3. Taylor expanded in kx around 0 17.4%

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

4. Taylor expanded in ky around 0 17.0%

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

5. Taylor expanded in th around 0 12.0%

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

   1. associate-/l* 13.1%

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

7. Simplified 13.1%

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

8. Final simplification 13.1%

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

Reproduce

herbie shell --seed 2023319 
(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)))