Migdal et al, Equation (64)

Percentage Accurate: 99.6% → 99.2%

Time: 7.4s

Alternatives: 8

Speedup: N/A×

• 44.5% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

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

FPCore

(FPCore (a1 a2 th)
 :precision binary64
 (let* ((t_1 (/ (cos th) (sqrt 2.0))))
   (+ (* t_1 (* a1 a1)) (* t_1 (* a2 a2)))))

C

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));
}

Fortran

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

Java

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));
}

Python

def code(a1, a2, th):
	t_1 = math.cos(th) / math.sqrt(2.0)
	return (t_1 * (a1 * a1)) + (t_1 * (a2 * a2))

Julia

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

MATLAB

function tmp = code(a1, a2, th)
	t_1 = cos(th) / sqrt(2.0);
	tmp = (t_1 * (a1 * a1)) + (t_1 * (a2 * a2));
end

Wolfram

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]]

Initial Program: 99.6% accurate, 1.0× speedup

Math

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

FPCore

(FPCore (a1 a2 th)
 :precision binary64
 (let* ((t_1 (/ (cos th) (sqrt 2.0))))
   (+ (* t_1 (* a1 a1)) (* t_1 (* a2 a2)))))

C

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));
}

Fortran

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

Java

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));
}

Python

def code(a1, a2, th):
	t_1 = math.cos(th) / math.sqrt(2.0)
	return (t_1 * (a1 * a1)) + (t_1 * (a2 * a2))

Julia

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

MATLAB

function tmp = code(a1, a2, th)
	t_1 = cos(th) / sqrt(2.0);
	tmp = (t_1 * (a1 * a1)) + (t_1 * (a2 * a2));
end

Wolfram

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]]

Alternative 1: 99.2% accurate, 2.0× speedup

Math

\[\begin{array}{l} a1 = |a1|\\ a2 = |a2|\\ [a1, a2] = \mathsf{sort}([a1, a2])\\ \\ a2 \cdot \frac{a2 \cdot \cos th}{\sqrt{2}} \end{array} \]

FPCore

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))))

C

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));
}

Fortran

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

Java

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));
}

Python

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))

Julia

a1 = abs(a1)
a2 = abs(a2)
a1, a2 = sort([a1, a2])
function code(a1, a2, th)
	return Float64(a2 * Float64(Float64(a2 * cos(th)) / sqrt(2.0)))
end

MATLAB

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

Wolfram

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[(a2 * N[Cos[th], $MachinePrecision]), $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.3%

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

   1. distribute-lft-out 99.3%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   2. associate-*l/ 99.3%

      \[\leadsto \color{blue}{\frac{\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)}{\sqrt{2}}} \]

   3. associate-*r/ 99.3%

      \[\leadsto \color{blue}{\cos th \cdot \frac{a1 \cdot a1 + a2 \cdot a2}{\sqrt{2}}} \]

   4. fma-def 99.3%

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

3. Simplified 99.3%

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

4. Taylor expanded in a1 around 0 65.1%

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

   1. unpow2 65.1%

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

   2. associate-*r/ 65.1%

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

   3. associate-*l* 65.1%

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

6. Simplified 65.1%

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

7. Taylor expanded in th around inf 65.1%

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

8. Final simplification 65.1%

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

Alternative 2: 99.2% accurate, 2.0× speedup

Math

\[\begin{array}{l} a1 = |a1|\\ a2 = |a2|\\ [a1, a2] = \mathsf{sort}([a1, a2])\\ \\ a2 \cdot \left(a2 \cdot \left(\cos th \cdot \sqrt{0.5}\right)\right) \end{array} \]

FPCore

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)))))

C

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)));
}

Fortran

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

Java

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)));
}

Python

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)))

Julia

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

MATLAB

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

Wolfram

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]

Derivation

1. Initial program 99.3%

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

   1. distribute-lft-out 99.3%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   2. associate-*l/ 99.3%

      \[\leadsto \color{blue}{\frac{\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)}{\sqrt{2}}} \]

   3. associate-*r/ 99.3%

      \[\leadsto \color{blue}{\cos th \cdot \frac{a1 \cdot a1 + a2 \cdot a2}{\sqrt{2}}} \]

   4. fma-def 99.3%

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

3. Simplified 99.3%

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

4. Taylor expanded in a1 around 0 65.1%

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

   1. unpow2 65.1%

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

   2. associate-*r/ 65.1%

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

   3. associate-*l* 65.1%

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

6. Simplified 65.1%

   \[\leadsto \color{blue}{a2 \cdot \left(a2 \cdot \frac{\cos th}{\sqrt{2}}\right)} \]
7. Step-by-step derivation

   1. expm1-log1p-u 41.6%

      \[\leadsto a2 \cdot \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(a2 \cdot \frac{\cos th}{\sqrt{2}}\right)\right)} \]

   2. expm1-udef 29.4%

      \[\leadsto a2 \cdot \color{blue}{\left(e^{\mathsf{log1p}\left(a2 \cdot \frac{\cos th}{\sqrt{2}}\right)} - 1\right)} \]

   3. div-inv 29.4%

      \[\leadsto a2 \cdot \left(e^{\mathsf{log1p}\left(a2 \cdot \color{blue}{\left(\cos th \cdot \frac{1}{\sqrt{2}}\right)}\right)} - 1\right) \]

   4. add-sqr-sqrt 29.4%

      \[\leadsto a2 \cdot \left(e^{\mathsf{log1p}\left(a2 \cdot \left(\cos th \cdot \color{blue}{\left(\sqrt{\frac{1}{\sqrt{2}}} \cdot \sqrt{\frac{1}{\sqrt{2}}}\right)}\right)\right)} - 1\right) \]

   5. sqrt-unprod 29.4%

      \[\leadsto a2 \cdot \left(e^{\mathsf{log1p}\left(a2 \cdot \left(\cos th \cdot \color{blue}{\sqrt{\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}}}\right)\right)} - 1\right) \]

   6. frac-times 29.4%

      \[\leadsto a2 \cdot \left(e^{\mathsf{log1p}\left(a2 \cdot \left(\cos th \cdot \sqrt{\color{blue}{\frac{1 \cdot 1}{\sqrt{2} \cdot \sqrt{2}}}}\right)\right)} - 1\right) \]

   7. metadata-eval 29.4%

      \[\leadsto a2 \cdot \left(e^{\mathsf{log1p}\left(a2 \cdot \left(\cos th \cdot \sqrt{\frac{\color{blue}{1}}{\sqrt{2} \cdot \sqrt{2}}}\right)\right)} - 1\right) \]

   8. add-sqr-sqrt 29.4%

      \[\leadsto a2 \cdot \left(e^{\mathsf{log1p}\left(a2 \cdot \left(\cos th \cdot \sqrt{\frac{1}{\color{blue}{2}}}\right)\right)} - 1\right) \]

   9. metadata-eval 29.4%

      \[\leadsto a2 \cdot \left(e^{\mathsf{log1p}\left(a2 \cdot \left(\cos th \cdot \sqrt{\color{blue}{0.5}}\right)\right)} - 1\right) \]

8. Applied egg-rr 29.4%

   \[\leadsto a2 \cdot \color{blue}{\left(e^{\mathsf{log1p}\left(a2 \cdot \left(\cos th \cdot \sqrt{0.5}\right)\right)} - 1\right)} \]
9. Step-by-step derivation

   1. expm1-def 41.6%

      \[\leadsto a2 \cdot \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(a2 \cdot \left(\cos th \cdot \sqrt{0.5}\right)\right)\right)} \]

   2. expm1-log1p 65.1%

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

10. Simplified 65.1%

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

11. Final simplification 65.1%

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

Alternative 3: 99.2% accurate, 2.0× speedup

Math

\[\begin{array}{l} a1 = |a1|\\ a2 = |a2|\\ [a1, a2] = \mathsf{sort}([a1, a2])\\ \\ a2 \cdot \left(a2 \cdot \frac{\cos th}{\sqrt{2}}\right) \end{array} \]

FPCore

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)))))

C

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)));
}

Fortran

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

Java

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)));
}

Python

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)))

Julia

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(2.0))))
end

MATLAB

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

Wolfram

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[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.3%

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

   1. distribute-lft-out 99.3%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   2. associate-*l/ 99.3%

      \[\leadsto \color{blue}{\frac{\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)}{\sqrt{2}}} \]

   3. associate-*r/ 99.3%

      \[\leadsto \color{blue}{\cos th \cdot \frac{a1 \cdot a1 + a2 \cdot a2}{\sqrt{2}}} \]

   4. fma-def 99.3%

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

3. Simplified 99.3%

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

4. Taylor expanded in a1 around 0 65.1%

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

   1. unpow2 65.1%

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

   2. associate-*r/ 65.1%

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

   3. associate-*l* 65.1%

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

6. Simplified 65.1%

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

7. Final simplification 65.1%

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

Alternative 4: 67.1% accurate, 3.7× speedup

Math

\[\begin{array}{l} a1 = |a1|\\ a2 = |a2|\\ [a1, a2] = \mathsf{sort}([a1, a2])\\ \\ \frac{1}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \end{array} \]

FPCore

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
 (* (/ 1.0 (sqrt 2.0)) (+ (* a1 a1) (* a2 a2))))

C

a1 = abs(a1);
a2 = abs(a2);
assert(a1 < a2);
double code(double a1, double a2, double th) {
	return (1.0 / sqrt(2.0)) * ((a1 * a1) + (a2 * a2));
}

Fortran

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 = (1.0d0 / sqrt(2.0d0)) * ((a1 * a1) + (a2 * a2))
end function

Java

a1 = Math.abs(a1);
a2 = Math.abs(a2);
assert a1 < a2;
public static double code(double a1, double a2, double th) {
	return (1.0 / Math.sqrt(2.0)) * ((a1 * a1) + (a2 * a2));
}

Python

a1 = abs(a1)
a2 = abs(a2)
[a1, a2] = sort([a1, a2])
def code(a1, a2, th):
	return (1.0 / math.sqrt(2.0)) * ((a1 * a1) + (a2 * a2))

Julia

a1 = abs(a1)
a2 = abs(a2)
a1, a2 = sort([a1, a2])
function code(a1, a2, th)
	return Float64(Float64(1.0 / sqrt(2.0)) * Float64(Float64(a1 * a1) + Float64(a2 * a2)))
end

MATLAB

a1 = abs(a1)
a2 = abs(a2)
a1, a2 = num2cell(sort([a1, a2])){:}
function tmp = code(a1, a2, th)
	tmp = (1.0 / sqrt(2.0)) * ((a1 * a1) + (a2 * a2));
end

Wolfram

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[(1.0 / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision] * N[(N[(a1 * a1), $MachinePrecision] + N[(a2 * a2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.3%

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

   1. distribute-lft-out 99.3%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

3. Simplified 99.3%

   \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

4. Taylor expanded in th around 0 61.6%

   \[\leadsto \color{blue}{\frac{1}{\sqrt{2}}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

5. Final simplification 61.6%

   \[\leadsto \frac{1}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

Alternative 5: 67.1% accurate, 3.8× speedup

Math

\[\begin{array}{l} a1 = |a1|\\ a2 = |a2|\\ [a1, a2] = \mathsf{sort}([a1, a2])\\ \\ \sqrt{0.5} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \end{array} \]

FPCore

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 (* (sqrt 0.5) (+ (* a1 a1) (* a2 a2))))

C

a1 = abs(a1);
a2 = abs(a2);
assert(a1 < a2);
double code(double a1, double a2, double th) {
	return sqrt(0.5) * ((a1 * a1) + (a2 * a2));
}

Fortran

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 = sqrt(0.5d0) * ((a1 * a1) + (a2 * a2))
end function

Java

a1 = Math.abs(a1);
a2 = Math.abs(a2);
assert a1 < a2;
public static double code(double a1, double a2, double th) {
	return Math.sqrt(0.5) * ((a1 * a1) + (a2 * a2));
}

Python

a1 = abs(a1)
a2 = abs(a2)
[a1, a2] = sort([a1, a2])
def code(a1, a2, th):
	return math.sqrt(0.5) * ((a1 * a1) + (a2 * a2))

Julia

a1 = abs(a1)
a2 = abs(a2)
a1, a2 = sort([a1, a2])
function code(a1, a2, th)
	return Float64(sqrt(0.5) * Float64(Float64(a1 * a1) + Float64(a2 * a2)))
end

MATLAB

a1 = abs(a1)
a2 = abs(a2)
a1, a2 = num2cell(sort([a1, a2])){:}
function tmp = code(a1, a2, th)
	tmp = sqrt(0.5) * ((a1 * a1) + (a2 * a2));
end

Wolfram

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[Sqrt[0.5], $MachinePrecision] * N[(N[(a1 * a1), $MachinePrecision] + N[(a2 * a2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.3%

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

   1. distribute-lft-out 99.3%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

3. Simplified 99.3%

   \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
4. Step-by-step derivation

   1. clear-num 99.3%

      \[\leadsto \color{blue}{\frac{1}{\frac{\sqrt{2}}{\cos th}}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   2. associate-/r/ 99.3%

      \[\leadsto \color{blue}{\left(\frac{1}{\sqrt{2}} \cdot \cos th\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   3. pow1/2 99.3%

      \[\leadsto \left(\frac{1}{\color{blue}{{2}^{0.5}}} \cdot \cos th\right) \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   4. pow-flip 99.2%

      \[\leadsto \left(\color{blue}{{2}^{\left(-0.5\right)}} \cdot \cos th\right) \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   5. metadata-eval 99.2%

      \[\leadsto \left({2}^{\color{blue}{-0.5}} \cdot \cos th\right) \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

5. Applied egg-rr 99.2%

   \[\leadsto \color{blue}{\left({2}^{-0.5} \cdot \cos th\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

6. Taylor expanded in th around 0 61.6%

   \[\leadsto \color{blue}{\sqrt{0.5} \cdot \left({a2}^{2} + {a1}^{2}\right)} \]
7. Step-by-step derivation

   1. unpow2 61.6%

      \[\leadsto \sqrt{0.5} \cdot \left(\color{blue}{a2 \cdot a2} + {a1}^{2}\right) \]

   2. unpow2 61.6%

      \[\leadsto \sqrt{0.5} \cdot \left(a2 \cdot a2 + \color{blue}{a1 \cdot a1}\right) \]

8. Simplified 61.6%

   \[\leadsto \color{blue}{\sqrt{0.5} \cdot \left(a2 \cdot a2 + a1 \cdot a1\right)} \]

9. Final simplification 61.6%

   \[\leadsto \sqrt{0.5} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

Alternative 6: 66.9% accurate, 3.9× speedup

Math

\[\begin{array}{l} a1 = |a1|\\ a2 = |a2|\\ [a1, a2] = \mathsf{sort}([a1, a2])\\ \\ \frac{\frac{a2}{\sqrt{2}}}{\frac{1}{a2}} \end{array} \]

FPCore

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)) (/ 1.0 a2)))

C

a1 = abs(a1);
a2 = abs(a2);
assert(a1 < a2);
double code(double a1, double a2, double th) {
	return (a2 / sqrt(2.0)) / (1.0 / a2);
}

Fortran

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)) / (1.0d0 / a2)
end function

Java

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)) / (1.0 / a2);
}

Python

a1 = abs(a1)
a2 = abs(a2)
[a1, a2] = sort([a1, a2])
def code(a1, a2, th):
	return (a2 / math.sqrt(2.0)) / (1.0 / a2)

Julia

a1 = abs(a1)
a2 = abs(a2)
a1, a2 = sort([a1, a2])
function code(a1, a2, th)
	return Float64(Float64(a2 / sqrt(2.0)) / Float64(1.0 / a2))
end

MATLAB

a1 = abs(a1)
a2 = abs(a2)
a1, a2 = num2cell(sort([a1, a2])){:}
function tmp = code(a1, a2, th)
	tmp = (a2 / sqrt(2.0)) / (1.0 / a2);
end

Wolfram

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[Sqrt[2.0], $MachinePrecision]), $MachinePrecision] / N[(1.0 / a2), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.3%

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

   1. distribute-lft-out 99.3%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

3. Simplified 99.3%

   \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]
4. Step-by-step derivation

   1. clear-num 99.3%

      \[\leadsto \color{blue}{\frac{1}{\frac{\sqrt{2}}{\cos th}}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   2. associate-/r/ 99.3%

      \[\leadsto \color{blue}{\left(\frac{1}{\sqrt{2}} \cdot \cos th\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   3. pow1/2 99.3%

      \[\leadsto \left(\frac{1}{\color{blue}{{2}^{0.5}}} \cdot \cos th\right) \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   4. pow-flip 99.2%

      \[\leadsto \left(\color{blue}{{2}^{\left(-0.5\right)}} \cdot \cos th\right) \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

   5. metadata-eval 99.2%

      \[\leadsto \left({2}^{\color{blue}{-0.5}} \cdot \cos th\right) \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

5. Applied egg-rr 99.2%

   \[\leadsto \color{blue}{\left({2}^{-0.5} \cdot \cos th\right)} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right) \]

6. Taylor expanded in th around 0 61.6%

   \[\leadsto \color{blue}{\sqrt{0.5} \cdot \left({a2}^{2} + {a1}^{2}\right)} \]
7. Step-by-step derivation

   1. unpow2 61.6%

      \[\leadsto \sqrt{0.5} \cdot \left(\color{blue}{a2 \cdot a2} + {a1}^{2}\right) \]

   2. unpow2 61.6%

      \[\leadsto \sqrt{0.5} \cdot \left(a2 \cdot a2 + \color{blue}{a1 \cdot a1}\right) \]

8. Simplified 61.6%

   \[\leadsto \color{blue}{\sqrt{0.5} \cdot \left(a2 \cdot a2 + a1 \cdot a1\right)} \]

9. Taylor expanded in a2 around inf 41.6%

   \[\leadsto \color{blue}{\sqrt{0.5} \cdot {a2}^{2}} \]
10. Step-by-step derivation

    1. unpow2 41.6%

       \[\leadsto \sqrt{0.5} \cdot \color{blue}{\left(a2 \cdot a2\right)} \]

11. Simplified 41.6%

    \[\leadsto \color{blue}{\sqrt{0.5} \cdot \left(a2 \cdot a2\right)} \]
12. Step-by-step derivation

    1. *-commutative 41.6%

       \[\leadsto \color{blue}{\left(a2 \cdot a2\right) \cdot \sqrt{0.5}} \]

    2. pow2 41.6%

       \[\leadsto \color{blue}{{a2}^{2}} \cdot \sqrt{0.5} \]

    3. metadata-eval 41.6%

       \[\leadsto {a2}^{\color{blue}{\left(\frac{4}{2}\right)}} \cdot \sqrt{0.5} \]

    4. sqrt-pow1 35.0%

       \[\leadsto \color{blue}{\sqrt{{a2}^{4}}} \cdot \sqrt{0.5} \]

    5. sqrt-prod 35.1%

       \[\leadsto \color{blue}{\sqrt{{a2}^{4} \cdot 0.5}} \]

    6. metadata-eval 35.1%

       \[\leadsto \sqrt{{a2}^{4} \cdot \color{blue}{\frac{1}{2}}} \]

    7. div-inv 35.1%

       \[\leadsto \sqrt{\color{blue}{\frac{{a2}^{4}}{2}}} \]

    8. sqrt-div 35.0%

       \[\leadsto \color{blue}{\frac{\sqrt{{a2}^{4}}}{\sqrt{2}}} \]

    9. sqrt-pow1 41.6%

       \[\leadsto \frac{\color{blue}{{a2}^{\left(\frac{4}{2}\right)}}}{\sqrt{2}} \]

    10. metadata-eval 41.6%

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

    11. pow2 41.6%

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

    12. associate-/l* 41.6%

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

    13. div-inv 41.6%

        \[\leadsto \frac{a2}{\color{blue}{\sqrt{2} \cdot \frac{1}{a2}}} \]

    14. associate-/r* 41.6%

        \[\leadsto \color{blue}{\frac{\frac{a2}{\sqrt{2}}}{\frac{1}{a2}}} \]

13. Applied egg-rr 41.6%

    \[\leadsto \color{blue}{\frac{\frac{a2}{\sqrt{2}}}{\frac{1}{a2}}} \]

14. Final simplification 41.6%

    \[\leadsto \frac{\frac{a2}{\sqrt{2}}}{\frac{1}{a2}} \]

Alternative 7: 66.9% accurate, 4.0× speedup

Math

\[\begin{array}{l} a1 = |a1|\\ a2 = |a2|\\ [a1, a2] = \mathsf{sort}([a1, a2])\\ \\ a2 \cdot \frac{a2}{\sqrt{2}} \end{array} \]

FPCore

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))))

C

a1 = abs(a1);
a2 = abs(a2);
assert(a1 < a2);
double code(double a1, double a2, double th) {
	return a2 * (a2 / sqrt(2.0));
}

Fortran

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

Java

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));
}

Python

a1 = abs(a1)
a2 = abs(a2)
[a1, a2] = sort([a1, a2])
def code(a1, a2, th):
	return a2 * (a2 / math.sqrt(2.0))

Julia

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

MATLAB

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

Wolfram

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]

Derivation

1. Initial program 99.3%

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

   1. distribute-lft-out 99.3%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   2. associate-*l/ 99.3%

      \[\leadsto \color{blue}{\frac{\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)}{\sqrt{2}}} \]

   3. associate-*r/ 99.3%

      \[\leadsto \color{blue}{\cos th \cdot \frac{a1 \cdot a1 + a2 \cdot a2}{\sqrt{2}}} \]

   4. fma-def 99.3%

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

3. Simplified 99.3%

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

4. Taylor expanded in a1 around 0 65.1%

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

   1. unpow2 65.1%

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

   2. associate-*r/ 65.1%

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

   3. associate-*l* 65.1%

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

6. Simplified 65.1%

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

7. Taylor expanded in th around 0 41.6%

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

8. Final simplification 41.6%

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

Alternative 8: 66.9% accurate, 4.0× speedup

Math

\[\begin{array}{l} a1 = |a1|\\ a2 = |a2|\\ [a1, a2] = \mathsf{sort}([a1, a2])\\ \\ \frac{a2 \cdot a2}{\sqrt{2}} \end{array} \]

FPCore

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)))

C

a1 = abs(a1);
a2 = abs(a2);
assert(a1 < a2);
double code(double a1, double a2, double th) {
	return (a2 * a2) / sqrt(2.0);
}

Fortran

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

Java

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);
}

Python

a1 = abs(a1)
a2 = abs(a2)
[a1, a2] = sort([a1, a2])
def code(a1, a2, th):
	return (a2 * a2) / math.sqrt(2.0)

Julia

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

MATLAB

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

Wolfram

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]

Derivation

1. Initial program 99.3%

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

   1. distribute-lft-out 99.3%

      \[\leadsto \color{blue}{\frac{\cos th}{\sqrt{2}} \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)} \]

   2. associate-*l/ 99.3%

      \[\leadsto \color{blue}{\frac{\cos th \cdot \left(a1 \cdot a1 + a2 \cdot a2\right)}{\sqrt{2}}} \]

   3. associate-*r/ 99.3%

      \[\leadsto \color{blue}{\cos th \cdot \frac{a1 \cdot a1 + a2 \cdot a2}{\sqrt{2}}} \]

   4. fma-def 99.3%

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

3. Simplified 99.3%

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

4. Taylor expanded in a1 around 0 65.1%

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

   1. unpow2 65.1%

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

   2. associate-*r/ 65.1%

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

   3. associate-*l* 65.1%

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

6. Simplified 65.1%

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

7. Taylor expanded in th around 0 41.6%

   \[\leadsto \color{blue}{\frac{{a2}^{2}}{\sqrt{2}}} \]
8. Step-by-step derivation

   1. unpow2 41.6%

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

9. Simplified 41.6%

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

10. Final simplification 41.6%

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

Reproduce

herbie shell --seed 2023215 
(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))))