Migdal et al, Equation (64)

Percentage Accurate: 99.5% → 99.1%
Time: 8.9s
Alternatives: 8
Speedup: N/A×

Specification

?
\[\begin{array}{l} \\ \begin{array}{l} t_1 := \frac{\cos th}{\sqrt{2}}\\ t_1 \cdot \left(a1 \cdot a1\right) + t_1 \cdot \left(a2 \cdot a2\right) \end{array} \end{array} \]
(FPCore (a1 a2 th)
 :precision binary64
 (let* ((t_1 (/ (cos th) (sqrt 2.0))))
   (+ (* t_1 (* a1 a1)) (* t_1 (* a2 a2)))))
double code(double a1, double a2, double th) {
	double t_1 = cos(th) / sqrt(2.0);
	return (t_1 * (a1 * a1)) + (t_1 * (a2 * a2));
}
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: t_1
    t_1 = cos(th) / sqrt(2.0d0)
    code = (t_1 * (a1 * a1)) + (t_1 * (a2 * a2))
end function
public static double code(double a1, double a2, double th) {
	double t_1 = Math.cos(th) / Math.sqrt(2.0);
	return (t_1 * (a1 * a1)) + (t_1 * (a2 * a2));
}
def code(a1, a2, th):
	t_1 = math.cos(th) / math.sqrt(2.0)
	return (t_1 * (a1 * a1)) + (t_1 * (a2 * a2))
function code(a1, a2, th)
	t_1 = Float64(cos(th) / sqrt(2.0))
	return Float64(Float64(t_1 * Float64(a1 * a1)) + Float64(t_1 * Float64(a2 * a2)))
end
function tmp = code(a1, a2, th)
	t_1 = cos(th) / sqrt(2.0);
	tmp = (t_1 * (a1 * a1)) + (t_1 * (a2 * a2));
end
code[a1_, a2_, th_] := Block[{t$95$1 = N[(N[Cos[th], $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]}, N[(N[(t$95$1 * N[(a1 * a1), $MachinePrecision]), $MachinePrecision] + N[(t$95$1 * N[(a2 * a2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}

\\
\begin{array}{l}
t_1 := \frac{\cos th}{\sqrt{2}}\\
t_1 \cdot \left(a1 \cdot a1\right) + t_1 \cdot \left(a2 \cdot a2\right)
\end{array}
\end{array}

Sampling outcomes in binary64 precision:

Local Percentage Accuracy vs ?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Accuracy vs Speed?

Herbie found 8 alternatives:

AlternativeAccuracySpeedup
The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

Initial Program: 99.5% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_1 := \frac{\cos th}{\sqrt{2}}\\ t_1 \cdot \left(a1 \cdot a1\right) + t_1 \cdot \left(a2 \cdot a2\right) \end{array} \end{array} \]
(FPCore (a1 a2 th)
 :precision binary64
 (let* ((t_1 (/ (cos th) (sqrt 2.0))))
   (+ (* t_1 (* a1 a1)) (* t_1 (* a2 a2)))))
double code(double a1, double a2, double th) {
	double t_1 = cos(th) / sqrt(2.0);
	return (t_1 * (a1 * a1)) + (t_1 * (a2 * a2));
}
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: t_1
    t_1 = cos(th) / sqrt(2.0d0)
    code = (t_1 * (a1 * a1)) + (t_1 * (a2 * a2))
end function
public static double code(double a1, double a2, double th) {
	double t_1 = Math.cos(th) / Math.sqrt(2.0);
	return (t_1 * (a1 * a1)) + (t_1 * (a2 * a2));
}
def code(a1, a2, th):
	t_1 = math.cos(th) / math.sqrt(2.0)
	return (t_1 * (a1 * a1)) + (t_1 * (a2 * a2))
function code(a1, a2, th)
	t_1 = Float64(cos(th) / sqrt(2.0))
	return Float64(Float64(t_1 * Float64(a1 * a1)) + Float64(t_1 * Float64(a2 * a2)))
end
function tmp = code(a1, a2, th)
	t_1 = cos(th) / sqrt(2.0);
	tmp = (t_1 * (a1 * a1)) + (t_1 * (a2 * a2));
end
code[a1_, a2_, th_] := Block[{t$95$1 = N[(N[Cos[th], $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]}, N[(N[(t$95$1 * N[(a1 * a1), $MachinePrecision]), $MachinePrecision] + N[(t$95$1 * N[(a2 * a2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}

\\
\begin{array}{l}
t_1 := \frac{\cos th}{\sqrt{2}}\\
t_1 \cdot \left(a1 \cdot a1\right) + t_1 \cdot \left(a2 \cdot a2\right)
\end{array}
\end{array}

Alternative 1: 99.1% accurate, 2.0× speedup?

\[\begin{array}{l} a1 = |a1|\\ a2 = |a2|\\ [a1, a2] = \mathsf{sort}([a1, a2])\\ \\ \frac{a2 \cdot \left(a2 \cdot \cos th\right)}{\sqrt{2}} \end{array} \]
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th) :precision binary64 (/ (* a2 (* a2 (cos th))) (sqrt 2.0)))
a1 = abs(a1);
a2 = abs(a2);
assert(a1 < a2);
double code(double a1, double a2, double th) {
	return (a2 * (a2 * cos(th))) / sqrt(2.0);
}
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    code = (a2 * (a2 * cos(th))) / sqrt(2.0d0)
end function
a1 = Math.abs(a1);
a2 = Math.abs(a2);
assert a1 < a2;
public static double code(double a1, double a2, double th) {
	return (a2 * (a2 * Math.cos(th))) / Math.sqrt(2.0);
}
a1 = abs(a1)
a2 = abs(a2)
[a1, a2] = sort([a1, a2])
def code(a1, a2, th):
	return (a2 * (a2 * math.cos(th))) / math.sqrt(2.0)
a1 = abs(a1)
a2 = abs(a2)
a1, a2 = sort([a1, a2])
function code(a1, a2, th)
	return Float64(Float64(a2 * Float64(a2 * cos(th))) / sqrt(2.0))
end
a1 = abs(a1)
a2 = abs(a2)
a1, a2 = num2cell(sort([a1, a2])){:}
function tmp = code(a1, a2, th)
	tmp = (a2 * (a2 * cos(th))) / sqrt(2.0);
end
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := N[(N[(a2 * N[(a2 * N[Cos[th], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
a1 = |a1|\\
a2 = |a2|\\
[a1, a2] = \mathsf{sort}([a1, a2])\\
\\
\frac{a2 \cdot \left(a2 \cdot \cos th\right)}{\sqrt{2}}
\end{array}
Derivation
  1. Initial program 99.6%

    \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
  2. Step-by-step derivation
    1. distribute-lft-out99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
    2. associate-*l/99.7%

      \[\leadsto \color{blue}{\frac{\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)}{\sqrt{2}}} \]
    3. associate-*r/99.6%

      \[\leadsto \color{blue}{\cos th \cdot \frac{a1 \cdot a1 + a2 \cdot a2}{\sqrt{2}}} \]
    4. fma-def99.6%

      \[\leadsto \cos th \cdot \frac{\color{blue}{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}}{\sqrt{2}} \]
  3. Simplified99.6%

    \[\leadsto \color{blue}{\cos th \cdot \frac{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}{\sqrt{2}}} \]
  4. Taylor expanded in a1 around 0 58.0%

    \[\leadsto \color{blue}{\frac{{a2}^{2} \cdot \cos th}{\sqrt{2}}} \]
  5. Taylor expanded in a2 around 0 58.0%

    \[\leadsto \frac{\color{blue}{\cos th \cdot {a2}^{2}}}{\sqrt{2}} \]
  6. Step-by-step derivation
    1. unpow258.0%

      \[\leadsto \frac{\cos th \cdot \color{blue}{\left(a2 \cdot a2\right)}}{\sqrt{2}} \]
    2. *-commutative58.0%

      \[\leadsto \frac{\color{blue}{\left(a2 \cdot a2\right) \cdot \cos th}}{\sqrt{2}} \]
    3. associate-*r*58.0%

      \[\leadsto \frac{\color{blue}{a2 \cdot \left(a2 \cdot \cos th\right)}}{\sqrt{2}} \]
  7. Simplified58.0%

    \[\leadsto \frac{\color{blue}{a2 \cdot \left(a2 \cdot \cos th\right)}}{\sqrt{2}} \]
  8. Final simplification58.0%

    \[\leadsto \frac{a2 \cdot \left(a2 \cdot \cos th\right)}{\sqrt{2}} \]

Alternative 2: 99.1% accurate, 2.0× speedup?

\[\begin{array}{l} a1 = |a1|\\ a2 = |a2|\\ [a1, a2] = \mathsf{sort}([a1, a2])\\ \\ a2 \cdot \left(a2 \cdot \left(\cos th \cdot \sqrt{0.5}\right)\right) \end{array} \]
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th) :precision binary64 (* a2 (* a2 (* (cos th) (sqrt 0.5)))))
a1 = abs(a1);
a2 = abs(a2);
assert(a1 < a2);
double code(double a1, double a2, double th) {
	return a2 * (a2 * (cos(th) * sqrt(0.5)));
}
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    code = a2 * (a2 * (cos(th) * sqrt(0.5d0)))
end function
a1 = Math.abs(a1);
a2 = Math.abs(a2);
assert a1 < a2;
public static double code(double a1, double a2, double th) {
	return a2 * (a2 * (Math.cos(th) * Math.sqrt(0.5)));
}
a1 = abs(a1)
a2 = abs(a2)
[a1, a2] = sort([a1, a2])
def code(a1, a2, th):
	return a2 * (a2 * (math.cos(th) * math.sqrt(0.5)))
a1 = abs(a1)
a2 = abs(a2)
a1, a2 = sort([a1, a2])
function code(a1, a2, th)
	return Float64(a2 * Float64(a2 * Float64(cos(th) * sqrt(0.5))))
end
a1 = abs(a1)
a2 = abs(a2)
a1, a2 = num2cell(sort([a1, a2])){:}
function tmp = code(a1, a2, th)
	tmp = a2 * (a2 * (cos(th) * sqrt(0.5)));
end
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := N[(a2 * N[(a2 * N[(N[Cos[th], $MachinePrecision] * N[Sqrt[0.5], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
a1 = |a1|\\
a2 = |a2|\\
[a1, a2] = \mathsf{sort}([a1, a2])\\
\\
a2 \cdot \left(a2 \cdot \left(\cos th \cdot \sqrt{0.5}\right)\right)
\end{array}
Derivation
  1. Initial program 99.6%

    \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
  2. Step-by-step derivation
    1. distribute-lft-out99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
    2. associate-*l/99.7%

      \[\leadsto \color{blue}{\frac{\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)}{\sqrt{2}}} \]
    3. associate-*r/99.6%

      \[\leadsto \color{blue}{\cos th \cdot \frac{a1 \cdot a1 + a2 \cdot a2}{\sqrt{2}}} \]
    4. fma-def99.6%

      \[\leadsto \cos th \cdot \frac{\color{blue}{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}}{\sqrt{2}} \]
  3. Simplified99.6%

    \[\leadsto \color{blue}{\cos th \cdot \frac{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}{\sqrt{2}}} \]
  4. Taylor expanded in a1 around 0 58.0%

    \[\leadsto \color{blue}{\frac{{a2}^{2} \cdot \cos th}{\sqrt{2}}} \]
  5. Taylor expanded in a2 around 0 58.0%

    \[\leadsto \frac{\color{blue}{\cos th \cdot {a2}^{2}}}{\sqrt{2}} \]
  6. Step-by-step derivation
    1. unpow258.0%

      \[\leadsto \frac{\cos th \cdot \color{blue}{\left(a2 \cdot a2\right)}}{\sqrt{2}} \]
  7. Simplified58.0%

    \[\leadsto \frac{\color{blue}{\cos th \cdot \left(a2 \cdot a2\right)}}{\sqrt{2}} \]
  8. Step-by-step derivation
    1. div-inv57.9%

      \[\leadsto \color{blue}{\left(\cos th \cdot \left(a2 \cdot a2\right)\right) \cdot \frac{1}{\sqrt{2}}} \]
    2. *-commutative57.9%

      \[\leadsto \color{blue}{\left(\left(a2 \cdot a2\right) \cdot \cos th\right)} \cdot \frac{1}{\sqrt{2}} \]
    3. associate-*l*57.9%

      \[\leadsto \color{blue}{\left(a2 \cdot a2\right) \cdot \left(\cos th \cdot \frac{1}{\sqrt{2}}\right)} \]
    4. add-sqr-sqrt57.9%

      \[\leadsto \left(a2 \cdot a2\right) \cdot \left(\cos th \cdot \color{blue}{\left(\sqrt{\frac{1}{\sqrt{2}}} \cdot \sqrt{\frac{1}{\sqrt{2}}}\right)}\right) \]
    5. sqrt-unprod57.9%

      \[\leadsto \left(a2 \cdot a2\right) \cdot \left(\cos th \cdot \color{blue}{\sqrt{\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}}}\right) \]
    6. frac-times57.9%

      \[\leadsto \left(a2 \cdot a2\right) \cdot \left(\cos th \cdot \sqrt{\color{blue}{\frac{1 \cdot 1}{\sqrt{2} \cdot \sqrt{2}}}}\right) \]
    7. metadata-eval57.9%

      \[\leadsto \left(a2 \cdot a2\right) \cdot \left(\cos th \cdot \sqrt{\frac{\color{blue}{1}}{\sqrt{2} \cdot \sqrt{2}}}\right) \]
    8. add-sqr-sqrt57.9%

      \[\leadsto \left(a2 \cdot a2\right) \cdot \left(\cos th \cdot \sqrt{\frac{1}{\color{blue}{2}}}\right) \]
    9. metadata-eval57.9%

      \[\leadsto \left(a2 \cdot a2\right) \cdot \left(\cos th \cdot \sqrt{\color{blue}{0.5}}\right) \]
  9. Applied egg-rr57.9%

    \[\leadsto \color{blue}{\left(a2 \cdot a2\right) \cdot \left(\cos th \cdot \sqrt{0.5}\right)} \]
  10. Step-by-step derivation
    1. associate-*l*57.9%

      \[\leadsto \color{blue}{a2 \cdot \left(a2 \cdot \left(\cos th \cdot \sqrt{0.5}\right)\right)} \]
  11. Simplified57.9%

    \[\leadsto \color{blue}{a2 \cdot \left(a2 \cdot \left(\cos th \cdot \sqrt{0.5}\right)\right)} \]
  12. Final simplification57.9%

    \[\leadsto a2 \cdot \left(a2 \cdot \left(\cos th \cdot \sqrt{0.5}\right)\right) \]

Alternative 3: 99.0% accurate, 2.0× speedup?

\[\begin{array}{l} a1 = |a1|\\ a2 = |a2|\\ [a1, a2] = \mathsf{sort}([a1, a2])\\ \\ \cos th \cdot \frac{a2 \cdot a2}{\sqrt{2}} \end{array} \]
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th) :precision binary64 (* (cos th) (/ (* a2 a2) (sqrt 2.0))))
a1 = abs(a1);
a2 = abs(a2);
assert(a1 < a2);
double code(double a1, double a2, double th) {
	return cos(th) * ((a2 * a2) / sqrt(2.0));
}
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    code = cos(th) * ((a2 * a2) / sqrt(2.0d0))
end function
a1 = Math.abs(a1);
a2 = Math.abs(a2);
assert a1 < a2;
public static double code(double a1, double a2, double th) {
	return Math.cos(th) * ((a2 * a2) / Math.sqrt(2.0));
}
a1 = abs(a1)
a2 = abs(a2)
[a1, a2] = sort([a1, a2])
def code(a1, a2, th):
	return math.cos(th) * ((a2 * a2) / math.sqrt(2.0))
a1 = abs(a1)
a2 = abs(a2)
a1, a2 = sort([a1, a2])
function code(a1, a2, th)
	return Float64(cos(th) * Float64(Float64(a2 * a2) / sqrt(2.0)))
end
a1 = abs(a1)
a2 = abs(a2)
a1, a2 = num2cell(sort([a1, a2])){:}
function tmp = code(a1, a2, th)
	tmp = cos(th) * ((a2 * a2) / sqrt(2.0));
end
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := N[(N[Cos[th], $MachinePrecision] * N[(N[(a2 * a2), $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
a1 = |a1|\\
a2 = |a2|\\
[a1, a2] = \mathsf{sort}([a1, a2])\\
\\
\cos th \cdot \frac{a2 \cdot a2}{\sqrt{2}}
\end{array}
Derivation
  1. Initial program 99.6%

    \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
  2. Step-by-step derivation
    1. distribute-lft-out99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
    2. associate-*l/99.7%

      \[\leadsto \color{blue}{\frac{\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)}{\sqrt{2}}} \]
    3. associate-*r/99.6%

      \[\leadsto \color{blue}{\cos th \cdot \frac{a1 \cdot a1 + a2 \cdot a2}{\sqrt{2}}} \]
    4. fma-def99.6%

      \[\leadsto \cos th \cdot \frac{\color{blue}{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}}{\sqrt{2}} \]
  3. Simplified99.6%

    \[\leadsto \color{blue}{\cos th \cdot \frac{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}{\sqrt{2}}} \]
  4. Taylor expanded in a1 around 0 57.9%

    \[\leadsto \cos th \cdot \color{blue}{\frac{{a2}^{2}}{\sqrt{2}}} \]
  5. Step-by-step derivation
    1. unpow257.9%

      \[\leadsto \cos th \cdot \frac{\color{blue}{a2 \cdot a2}}{\sqrt{2}} \]
  6. Simplified57.9%

    \[\leadsto \cos th \cdot \color{blue}{\frac{a2 \cdot a2}{\sqrt{2}}} \]
  7. Final simplification57.9%

    \[\leadsto \cos th \cdot \frac{a2 \cdot a2}{\sqrt{2}} \]

Alternative 4: 65.3% accurate, 3.5× speedup?

\[\begin{array}{l} a1 = |a1|\\ a2 = |a2|\\ [a1, a2] = \mathsf{sort}([a1, a2])\\ \\ \begin{array}{l} \mathbf{if}\;a1 \cdot a1 \leq 2.4 \cdot 10^{-79}:\\ \;\;\;\;a2 \cdot \sqrt{a2 \cdot \frac{a2}{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\left(a2 \cdot a2\right) \cdot \left(-0.5 \cdot \left(th \cdot th\right) + 1\right)}{\sqrt{2}}\\ \end{array} \end{array} \]
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th)
 :precision binary64
 (if (<= (* a1 a1) 2.4e-79)
   (* a2 (sqrt (* a2 (/ a2 2.0))))
   (/ (* (* a2 a2) (+ (* -0.5 (* th th)) 1.0)) (sqrt 2.0))))
a1 = abs(a1);
a2 = abs(a2);
assert(a1 < a2);
double code(double a1, double a2, double th) {
	double tmp;
	if ((a1 * a1) <= 2.4e-79) {
		tmp = a2 * sqrt((a2 * (a2 / 2.0)));
	} else {
		tmp = ((a2 * a2) * ((-0.5 * (th * th)) + 1.0)) / sqrt(2.0);
	}
	return tmp;
}
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    real(8) :: tmp
    if ((a1 * a1) <= 2.4d-79) then
        tmp = a2 * sqrt((a2 * (a2 / 2.0d0)))
    else
        tmp = ((a2 * a2) * (((-0.5d0) * (th * th)) + 1.0d0)) / sqrt(2.0d0)
    end if
    code = tmp
end function
a1 = Math.abs(a1);
a2 = Math.abs(a2);
assert a1 < a2;
public static double code(double a1, double a2, double th) {
	double tmp;
	if ((a1 * a1) <= 2.4e-79) {
		tmp = a2 * Math.sqrt((a2 * (a2 / 2.0)));
	} else {
		tmp = ((a2 * a2) * ((-0.5 * (th * th)) + 1.0)) / Math.sqrt(2.0);
	}
	return tmp;
}
a1 = abs(a1)
a2 = abs(a2)
[a1, a2] = sort([a1, a2])
def code(a1, a2, th):
	tmp = 0
	if (a1 * a1) <= 2.4e-79:
		tmp = a2 * math.sqrt((a2 * (a2 / 2.0)))
	else:
		tmp = ((a2 * a2) * ((-0.5 * (th * th)) + 1.0)) / math.sqrt(2.0)
	return tmp
a1 = abs(a1)
a2 = abs(a2)
a1, a2 = sort([a1, a2])
function code(a1, a2, th)
	tmp = 0.0
	if (Float64(a1 * a1) <= 2.4e-79)
		tmp = Float64(a2 * sqrt(Float64(a2 * Float64(a2 / 2.0))));
	else
		tmp = Float64(Float64(Float64(a2 * a2) * Float64(Float64(-0.5 * Float64(th * th)) + 1.0)) / sqrt(2.0));
	end
	return tmp
end
a1 = abs(a1)
a2 = abs(a2)
a1, a2 = num2cell(sort([a1, a2])){:}
function tmp_2 = code(a1, a2, th)
	tmp = 0.0;
	if ((a1 * a1) <= 2.4e-79)
		tmp = a2 * sqrt((a2 * (a2 / 2.0)));
	else
		tmp = ((a2 * a2) * ((-0.5 * (th * th)) + 1.0)) / sqrt(2.0);
	end
	tmp_2 = tmp;
end
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := If[LessEqual[N[(a1 * a1), $MachinePrecision], 2.4e-79], N[(a2 * N[Sqrt[N[(a2 * N[(a2 / 2.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[(N[(a2 * a2), $MachinePrecision] * N[(N[(-0.5 * N[(th * th), $MachinePrecision]), $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}
a1 = |a1|\\
a2 = |a2|\\
[a1, a2] = \mathsf{sort}([a1, a2])\\
\\
\begin{array}{l}
\mathbf{if}\;a1 \cdot a1 \leq 2.4 \cdot 10^{-79}:\\
\;\;\;\;a2 \cdot \sqrt{a2 \cdot \frac{a2}{2}}\\

\mathbf{else}:\\
\;\;\;\;\frac{\left(a2 \cdot a2\right) \cdot \left(-0.5 \cdot \left(th \cdot th\right) + 1\right)}{\sqrt{2}}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (*.f64 a1 a1) < 2.40000000000000006e-79

    1. Initial program 99.5%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
    2. Step-by-step derivation
      1. distribute-lft-out99.5%

        \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
      2. associate-*l/99.6%

        \[\leadsto \color{blue}{\frac{\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)}{\sqrt{2}}} \]
      3. associate-*r/99.5%

        \[\leadsto \color{blue}{\cos th \cdot \frac{a1 \cdot a1 + a2 \cdot a2}{\sqrt{2}}} \]
      4. fma-def99.5%

        \[\leadsto \cos th \cdot \frac{\color{blue}{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}}{\sqrt{2}} \]
    3. Simplified99.5%

      \[\leadsto \color{blue}{\cos th \cdot \frac{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}{\sqrt{2}}} \]
    4. Taylor expanded in a1 around 0 81.9%

      \[\leadsto \color{blue}{\frac{{a2}^{2} \cdot \cos th}{\sqrt{2}}} \]
    5. Step-by-step derivation
      1. unpow281.9%

        \[\leadsto \frac{\color{blue}{\left(a2 \cdot a2\right)} \cdot \cos th}{\sqrt{2}} \]
      2. associate-*l*81.9%

        \[\leadsto \frac{\color{blue}{a2 \cdot \left(a2 \cdot \cos th\right)}}{\sqrt{2}} \]
      3. associate-*r/81.9%

        \[\leadsto \color{blue}{a2 \cdot \frac{a2 \cdot \cos th}{\sqrt{2}}} \]
      4. associate-/l*81.9%

        \[\leadsto a2 \cdot \color{blue}{\frac{a2}{\frac{\sqrt{2}}{\cos th}}} \]
    6. Simplified81.9%

      \[\leadsto \color{blue}{a2 \cdot \frac{a2}{\frac{\sqrt{2}}{\cos th}}} \]
    7. Taylor expanded in th around 0 60.7%

      \[\leadsto a2 \cdot \color{blue}{\frac{a2}{\sqrt{2}}} \]
    8. Step-by-step derivation
      1. add-sqr-sqrt32.5%

        \[\leadsto a2 \cdot \color{blue}{\left(\sqrt{\frac{a2}{\sqrt{2}}} \cdot \sqrt{\frac{a2}{\sqrt{2}}}\right)} \]
      2. sqrt-unprod39.8%

        \[\leadsto a2 \cdot \color{blue}{\sqrt{\frac{a2}{\sqrt{2}} \cdot \frac{a2}{\sqrt{2}}}} \]
      3. frac-times39.8%

        \[\leadsto a2 \cdot \sqrt{\color{blue}{\frac{a2 \cdot a2}{\sqrt{2} \cdot \sqrt{2}}}} \]
      4. add-sqr-sqrt39.9%

        \[\leadsto a2 \cdot \sqrt{\frac{a2 \cdot a2}{\color{blue}{2}}} \]
    9. Applied egg-rr39.9%

      \[\leadsto a2 \cdot \color{blue}{\sqrt{\frac{a2 \cdot a2}{2}}} \]
    10. Step-by-step derivation
      1. associate-/l*39.9%

        \[\leadsto a2 \cdot \sqrt{\color{blue}{\frac{a2}{\frac{2}{a2}}}} \]
      2. associate-/r/39.9%

        \[\leadsto a2 \cdot \sqrt{\color{blue}{\frac{a2}{2} \cdot a2}} \]
    11. Simplified39.9%

      \[\leadsto a2 \cdot \color{blue}{\sqrt{\frac{a2}{2} \cdot a2}} \]

    if 2.40000000000000006e-79 < (*.f64 a1 a1)

    1. Initial program 99.7%

      \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
    2. Step-by-step derivation
      1. distribute-lft-out99.7%

        \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
      2. associate-*l/99.7%

        \[\leadsto \color{blue}{\frac{\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)}{\sqrt{2}}} \]
      3. associate-*r/99.7%

        \[\leadsto \color{blue}{\cos th \cdot \frac{a1 \cdot a1 + a2 \cdot a2}{\sqrt{2}}} \]
      4. fma-def99.7%

        \[\leadsto \cos th \cdot \frac{\color{blue}{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}}{\sqrt{2}} \]
    3. Simplified99.7%

      \[\leadsto \color{blue}{\cos th \cdot \frac{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}{\sqrt{2}}} \]
    4. Taylor expanded in a1 around 0 35.1%

      \[\leadsto \color{blue}{\frac{{a2}^{2} \cdot \cos th}{\sqrt{2}}} \]
    5. Taylor expanded in th around 0 10.9%

      \[\leadsto \frac{\color{blue}{{a2}^{2} + -0.5 \cdot \left({th}^{2} \cdot {a2}^{2}\right)}}{\sqrt{2}} \]
    6. Step-by-step derivation
      1. unpow210.9%

        \[\leadsto \frac{\color{blue}{a2 \cdot a2} + -0.5 \cdot \left({th}^{2} \cdot {a2}^{2}\right)}{\sqrt{2}} \]
      2. unpow210.9%

        \[\leadsto \frac{a2 \cdot a2 + -0.5 \cdot \left({th}^{2} \cdot \color{blue}{\left(a2 \cdot a2\right)}\right)}{\sqrt{2}} \]
      3. associate-*r*10.9%

        \[\leadsto \frac{a2 \cdot a2 + \color{blue}{\left(-0.5 \cdot {th}^{2}\right) \cdot \left(a2 \cdot a2\right)}}{\sqrt{2}} \]
      4. distribute-rgt1-in30.0%

        \[\leadsto \frac{\color{blue}{\left(-0.5 \cdot {th}^{2} + 1\right) \cdot \left(a2 \cdot a2\right)}}{\sqrt{2}} \]
      5. unpow230.0%

        \[\leadsto \frac{\left(-0.5 \cdot \color{blue}{\left(th \cdot th\right)} + 1\right) \cdot \left(a2 \cdot a2\right)}{\sqrt{2}} \]
    7. Simplified30.0%

      \[\leadsto \frac{\color{blue}{\left(-0.5 \cdot \left(th \cdot th\right) + 1\right) \cdot \left(a2 \cdot a2\right)}}{\sqrt{2}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification34.8%

    \[\leadsto \begin{array}{l} \mathbf{if}\;a1 \cdot a1 \leq 2.4 \cdot 10^{-79}:\\ \;\;\;\;a2 \cdot \sqrt{a2 \cdot \frac{a2}{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\left(a2 \cdot a2\right) \cdot \left(-0.5 \cdot \left(th \cdot th\right) + 1\right)}{\sqrt{2}}\\ \end{array} \]

Alternative 5: 66.2% accurate, 3.9× speedup?

\[\begin{array}{l} a1 = |a1|\\ a2 = |a2|\\ [a1, a2] = \mathsf{sort}([a1, a2])\\ \\ a2 \cdot \sqrt{a2 \cdot \frac{a2}{2}} \end{array} \]
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th) :precision binary64 (* a2 (sqrt (* a2 (/ a2 2.0)))))
a1 = abs(a1);
a2 = abs(a2);
assert(a1 < a2);
double code(double a1, double a2, double th) {
	return a2 * sqrt((a2 * (a2 / 2.0)));
}
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    code = a2 * sqrt((a2 * (a2 / 2.0d0)))
end function
a1 = Math.abs(a1);
a2 = Math.abs(a2);
assert a1 < a2;
public static double code(double a1, double a2, double th) {
	return a2 * Math.sqrt((a2 * (a2 / 2.0)));
}
a1 = abs(a1)
a2 = abs(a2)
[a1, a2] = sort([a1, a2])
def code(a1, a2, th):
	return a2 * math.sqrt((a2 * (a2 / 2.0)))
a1 = abs(a1)
a2 = abs(a2)
a1, a2 = sort([a1, a2])
function code(a1, a2, th)
	return Float64(a2 * sqrt(Float64(a2 * Float64(a2 / 2.0))))
end
a1 = abs(a1)
a2 = abs(a2)
a1, a2 = num2cell(sort([a1, a2])){:}
function tmp = code(a1, a2, th)
	tmp = a2 * sqrt((a2 * (a2 / 2.0)));
end
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := N[(a2 * N[Sqrt[N[(a2 * N[(a2 / 2.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
a1 = |a1|\\
a2 = |a2|\\
[a1, a2] = \mathsf{sort}([a1, a2])\\
\\
a2 \cdot \sqrt{a2 \cdot \frac{a2}{2}}
\end{array}
Derivation
  1. Initial program 99.6%

    \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
  2. Step-by-step derivation
    1. distribute-lft-out99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
    2. associate-*l/99.7%

      \[\leadsto \color{blue}{\frac{\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)}{\sqrt{2}}} \]
    3. associate-*r/99.6%

      \[\leadsto \color{blue}{\cos th \cdot \frac{a1 \cdot a1 + a2 \cdot a2}{\sqrt{2}}} \]
    4. fma-def99.6%

      \[\leadsto \cos th \cdot \frac{\color{blue}{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}}{\sqrt{2}} \]
  3. Simplified99.6%

    \[\leadsto \color{blue}{\cos th \cdot \frac{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}{\sqrt{2}}} \]
  4. Taylor expanded in a1 around 0 58.0%

    \[\leadsto \color{blue}{\frac{{a2}^{2} \cdot \cos th}{\sqrt{2}}} \]
  5. Step-by-step derivation
    1. unpow258.0%

      \[\leadsto \frac{\color{blue}{\left(a2 \cdot a2\right)} \cdot \cos th}{\sqrt{2}} \]
    2. associate-*l*58.0%

      \[\leadsto \frac{\color{blue}{a2 \cdot \left(a2 \cdot \cos th\right)}}{\sqrt{2}} \]
    3. associate-*r/57.9%

      \[\leadsto \color{blue}{a2 \cdot \frac{a2 \cdot \cos th}{\sqrt{2}}} \]
    4. associate-/l*57.9%

      \[\leadsto a2 \cdot \color{blue}{\frac{a2}{\frac{\sqrt{2}}{\cos th}}} \]
  6. Simplified57.9%

    \[\leadsto \color{blue}{a2 \cdot \frac{a2}{\frac{\sqrt{2}}{\cos th}}} \]
  7. Taylor expanded in th around 0 40.9%

    \[\leadsto a2 \cdot \color{blue}{\frac{a2}{\sqrt{2}}} \]
  8. Step-by-step derivation
    1. add-sqr-sqrt20.0%

      \[\leadsto a2 \cdot \color{blue}{\left(\sqrt{\frac{a2}{\sqrt{2}}} \cdot \sqrt{\frac{a2}{\sqrt{2}}}\right)} \]
    2. sqrt-unprod26.1%

      \[\leadsto a2 \cdot \color{blue}{\sqrt{\frac{a2}{\sqrt{2}} \cdot \frac{a2}{\sqrt{2}}}} \]
    3. frac-times26.0%

      \[\leadsto a2 \cdot \sqrt{\color{blue}{\frac{a2 \cdot a2}{\sqrt{2} \cdot \sqrt{2}}}} \]
    4. add-sqr-sqrt26.1%

      \[\leadsto a2 \cdot \sqrt{\frac{a2 \cdot a2}{\color{blue}{2}}} \]
  9. Applied egg-rr26.1%

    \[\leadsto a2 \cdot \color{blue}{\sqrt{\frac{a2 \cdot a2}{2}}} \]
  10. Step-by-step derivation
    1. associate-/l*26.1%

      \[\leadsto a2 \cdot \sqrt{\color{blue}{\frac{a2}{\frac{2}{a2}}}} \]
    2. associate-/r/26.1%

      \[\leadsto a2 \cdot \sqrt{\color{blue}{\frac{a2}{2} \cdot a2}} \]
  11. Simplified26.1%

    \[\leadsto a2 \cdot \color{blue}{\sqrt{\frac{a2}{2} \cdot a2}} \]
  12. Final simplification26.1%

    \[\leadsto a2 \cdot \sqrt{a2 \cdot \frac{a2}{2}} \]

Alternative 6: 66.1% accurate, 4.0× speedup?

\[\begin{array}{l} a1 = |a1|\\ a2 = |a2|\\ [a1, a2] = \mathsf{sort}([a1, a2])\\ \\ a2 \cdot \frac{a2}{\sqrt{2}} \end{array} \]
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th) :precision binary64 (* a2 (/ a2 (sqrt 2.0))))
a1 = abs(a1);
a2 = abs(a2);
assert(a1 < a2);
double code(double a1, double a2, double th) {
	return a2 * (a2 / sqrt(2.0));
}
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    code = a2 * (a2 / sqrt(2.0d0))
end function
a1 = Math.abs(a1);
a2 = Math.abs(a2);
assert a1 < a2;
public static double code(double a1, double a2, double th) {
	return a2 * (a2 / Math.sqrt(2.0));
}
a1 = abs(a1)
a2 = abs(a2)
[a1, a2] = sort([a1, a2])
def code(a1, a2, th):
	return a2 * (a2 / math.sqrt(2.0))
a1 = abs(a1)
a2 = abs(a2)
a1, a2 = sort([a1, a2])
function code(a1, a2, th)
	return Float64(a2 * Float64(a2 / sqrt(2.0)))
end
a1 = abs(a1)
a2 = abs(a2)
a1, a2 = num2cell(sort([a1, a2])){:}
function tmp = code(a1, a2, th)
	tmp = a2 * (a2 / sqrt(2.0));
end
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := N[(a2 * N[(a2 / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
a1 = |a1|\\
a2 = |a2|\\
[a1, a2] = \mathsf{sort}([a1, a2])\\
\\
a2 \cdot \frac{a2}{\sqrt{2}}
\end{array}
Derivation
  1. Initial program 99.6%

    \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
  2. Step-by-step derivation
    1. distribute-lft-out99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
    2. associate-*l/99.7%

      \[\leadsto \color{blue}{\frac{\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)}{\sqrt{2}}} \]
    3. associate-*r/99.6%

      \[\leadsto \color{blue}{\cos th \cdot \frac{a1 \cdot a1 + a2 \cdot a2}{\sqrt{2}}} \]
    4. fma-def99.6%

      \[\leadsto \cos th \cdot \frac{\color{blue}{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}}{\sqrt{2}} \]
  3. Simplified99.6%

    \[\leadsto \color{blue}{\cos th \cdot \frac{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}{\sqrt{2}}} \]
  4. Taylor expanded in a1 around 0 58.0%

    \[\leadsto \color{blue}{\frac{{a2}^{2} \cdot \cos th}{\sqrt{2}}} \]
  5. Step-by-step derivation
    1. unpow258.0%

      \[\leadsto \frac{\color{blue}{\left(a2 \cdot a2\right)} \cdot \cos th}{\sqrt{2}} \]
    2. associate-*l*58.0%

      \[\leadsto \frac{\color{blue}{a2 \cdot \left(a2 \cdot \cos th\right)}}{\sqrt{2}} \]
    3. associate-*r/57.9%

      \[\leadsto \color{blue}{a2 \cdot \frac{a2 \cdot \cos th}{\sqrt{2}}} \]
    4. associate-/l*57.9%

      \[\leadsto a2 \cdot \color{blue}{\frac{a2}{\frac{\sqrt{2}}{\cos th}}} \]
  6. Simplified57.9%

    \[\leadsto \color{blue}{a2 \cdot \frac{a2}{\frac{\sqrt{2}}{\cos th}}} \]
  7. Taylor expanded in th around 0 40.9%

    \[\leadsto a2 \cdot \color{blue}{\frac{a2}{\sqrt{2}}} \]
  8. Final simplification40.9%

    \[\leadsto a2 \cdot \frac{a2}{\sqrt{2}} \]

Alternative 7: 66.1% accurate, 4.0× speedup?

\[\begin{array}{l} a1 = |a1|\\ a2 = |a2|\\ [a1, a2] = \mathsf{sort}([a1, a2])\\ \\ \frac{a2}{\frac{\sqrt{2}}{a2}} \end{array} \]
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th) :precision binary64 (/ a2 (/ (sqrt 2.0) a2)))
a1 = abs(a1);
a2 = abs(a2);
assert(a1 < a2);
double code(double a1, double a2, double th) {
	return a2 / (sqrt(2.0) / a2);
}
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    code = a2 / (sqrt(2.0d0) / a2)
end function
a1 = Math.abs(a1);
a2 = Math.abs(a2);
assert a1 < a2;
public static double code(double a1, double a2, double th) {
	return a2 / (Math.sqrt(2.0) / a2);
}
a1 = abs(a1)
a2 = abs(a2)
[a1, a2] = sort([a1, a2])
def code(a1, a2, th):
	return a2 / (math.sqrt(2.0) / a2)
a1 = abs(a1)
a2 = abs(a2)
a1, a2 = sort([a1, a2])
function code(a1, a2, th)
	return Float64(a2 / Float64(sqrt(2.0) / a2))
end
a1 = abs(a1)
a2 = abs(a2)
a1, a2 = num2cell(sort([a1, a2])){:}
function tmp = code(a1, a2, th)
	tmp = a2 / (sqrt(2.0) / a2);
end
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := N[(a2 / N[(N[Sqrt[2.0], $MachinePrecision] / a2), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
a1 = |a1|\\
a2 = |a2|\\
[a1, a2] = \mathsf{sort}([a1, a2])\\
\\
\frac{a2}{\frac{\sqrt{2}}{a2}}
\end{array}
Derivation
  1. Initial program 99.6%

    \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
  2. Step-by-step derivation
    1. distribute-lft-out99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
    2. associate-*l/99.7%

      \[\leadsto \color{blue}{\frac{\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)}{\sqrt{2}}} \]
    3. associate-*r/99.6%

      \[\leadsto \color{blue}{\cos th \cdot \frac{a1 \cdot a1 + a2 \cdot a2}{\sqrt{2}}} \]
    4. fma-def99.6%

      \[\leadsto \cos th \cdot \frac{\color{blue}{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}}{\sqrt{2}} \]
  3. Simplified99.6%

    \[\leadsto \color{blue}{\cos th \cdot \frac{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}{\sqrt{2}}} \]
  4. Taylor expanded in a1 around 0 58.0%

    \[\leadsto \color{blue}{\frac{{a2}^{2} \cdot \cos th}{\sqrt{2}}} \]
  5. Taylor expanded in th around 0 40.9%

    \[\leadsto \color{blue}{\frac{{a2}^{2}}{\sqrt{2}}} \]
  6. Step-by-step derivation
    1. unpow240.9%

      \[\leadsto \frac{\color{blue}{a2 \cdot a2}}{\sqrt{2}} \]
    2. associate-/l*40.9%

      \[\leadsto \color{blue}{\frac{a2}{\frac{\sqrt{2}}{a2}}} \]
  7. Simplified40.9%

    \[\leadsto \color{blue}{\frac{a2}{\frac{\sqrt{2}}{a2}}} \]
  8. Final simplification40.9%

    \[\leadsto \frac{a2}{\frac{\sqrt{2}}{a2}} \]

Alternative 8: 66.1% accurate, 4.0× speedup?

\[\begin{array}{l} a1 = |a1|\\ a2 = |a2|\\ [a1, a2] = \mathsf{sort}([a1, a2])\\ \\ \frac{a2 \cdot a2}{\sqrt{2}} \end{array} \]
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
(FPCore (a1 a2 th) :precision binary64 (/ (* a2 a2) (sqrt 2.0)))
a1 = abs(a1);
a2 = abs(a2);
assert(a1 < a2);
double code(double a1, double a2, double th) {
	return (a2 * a2) / sqrt(2.0);
}
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
real(8) function code(a1, a2, th)
    real(8), intent (in) :: a1
    real(8), intent (in) :: a2
    real(8), intent (in) :: th
    code = (a2 * a2) / sqrt(2.0d0)
end function
a1 = Math.abs(a1);
a2 = Math.abs(a2);
assert a1 < a2;
public static double code(double a1, double a2, double th) {
	return (a2 * a2) / Math.sqrt(2.0);
}
a1 = abs(a1)
a2 = abs(a2)
[a1, a2] = sort([a1, a2])
def code(a1, a2, th):
	return (a2 * a2) / math.sqrt(2.0)
a1 = abs(a1)
a2 = abs(a2)
a1, a2 = sort([a1, a2])
function code(a1, a2, th)
	return Float64(Float64(a2 * a2) / sqrt(2.0))
end
a1 = abs(a1)
a2 = abs(a2)
a1, a2 = num2cell(sort([a1, a2])){:}
function tmp = code(a1, a2, th)
	tmp = (a2 * a2) / sqrt(2.0);
end
NOTE: a1 should be positive before calling this function
NOTE: a2 should be positive before calling this function
NOTE: a1 and a2 should be sorted in increasing order before calling this function.
code[a1_, a2_, th_] := N[(N[(a2 * a2), $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
a1 = |a1|\\
a2 = |a2|\\
[a1, a2] = \mathsf{sort}([a1, a2])\\
\\
\frac{a2 \cdot a2}{\sqrt{2}}
\end{array}
Derivation
  1. Initial program 99.6%

    \[\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1\right) + \frac{\cos th}{\sqrt{2}} \cdot \left(a2 \cdot a2\right) \]
  2. Step-by-step derivation
    1. distribute-lft-out99.6%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
    2. associate-*l/99.7%

      \[\leadsto \color{blue}{\frac{\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)}{\sqrt{2}}} \]
    3. associate-*r/99.6%

      \[\leadsto \color{blue}{\cos th \cdot \frac{a1 \cdot a1 + a2 \cdot a2}{\sqrt{2}}} \]
    4. fma-def99.6%

      \[\leadsto \cos th \cdot \frac{\color{blue}{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}}{\sqrt{2}} \]
  3. Simplified99.6%

    \[\leadsto \color{blue}{\cos th \cdot \frac{\mathsf{fma}\left(a1, a1, a2 \cdot a2\right)}{\sqrt{2}}} \]
  4. Taylor expanded in a1 around 0 58.0%

    \[\leadsto \color{blue}{\frac{{a2}^{2} \cdot \cos th}{\sqrt{2}}} \]
  5. Taylor expanded in th around 0 40.9%

    \[\leadsto \color{blue}{\frac{{a2}^{2}}{\sqrt{2}}} \]
  6. Step-by-step derivation
    1. unpow240.9%

      \[\leadsto \frac{\color{blue}{a2 \cdot a2}}{\sqrt{2}} \]
  7. Simplified40.9%

    \[\leadsto \color{blue}{\frac{a2 \cdot a2}{\sqrt{2}}} \]
  8. Final simplification40.9%

    \[\leadsto \frac{a2 \cdot a2}{\sqrt{2}} \]

Reproduce

?
herbie shell --seed 2023213 
(FPCore (a1 a2 th)
  :name "Migdal et al, Equation (64)"
  :precision binary64
  (+ (* (/ (cos th) (sqrt 2.0)) (* a1 a1)) (* (/ (cos th) (sqrt 2.0)) (* a2 a2))))