Falkner and Boettcher, Appendix A

Percentage Accurate: 90.1% → 99.1%

Time: 12.5s

Alternatives: 10

Speedup: 1.0×

• 23.3% of points produce a very large (infinite) output. You may want to add a precondition.

• could not determine a ground truth

  1. a = -1.9999999999999999e-154
  2. k = 0.0
  3. m = -2.00000000000000007e154

Specification

Math

\[\begin{array}{l} \\ \frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))

C

double code(double a, double k, double m) {
	return (a * pow(k, m)) / ((1.0 + (10.0 * k)) + (k * k));
}

Fortran

real(8) function code(a, k, m)
    real(8), intent (in) :: a
    real(8), intent (in) :: k
    real(8), intent (in) :: m
    code = (a * (k ** m)) / ((1.0d0 + (10.0d0 * k)) + (k * k))
end function

Java

public static double code(double a, double k, double m) {
	return (a * Math.pow(k, m)) / ((1.0 + (10.0 * k)) + (k * k));
}

Python

def code(a, k, m):
	return (a * math.pow(k, m)) / ((1.0 + (10.0 * k)) + (k * k))

Julia

function code(a, k, m)
	return Float64(Float64(a * (k ^ m)) / Float64(Float64(1.0 + Float64(10.0 * k)) + Float64(k * k)))
end

MATLAB

function tmp = code(a, k, m)
	tmp = (a * (k ^ m)) / ((1.0 + (10.0 * k)) + (k * k));
end

Wolfram

code[a_, k_, m_] := N[(N[(a * N[Power[k, m], $MachinePrecision]), $MachinePrecision] / N[(N[(1.0 + N[(10.0 * k), $MachinePrecision]), $MachinePrecision] + N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Initial Program: 90.1% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))

C

double code(double a, double k, double m) {
	return (a * pow(k, m)) / ((1.0 + (10.0 * k)) + (k * k));
}

Fortran

real(8) function code(a, k, m)
    real(8), intent (in) :: a
    real(8), intent (in) :: k
    real(8), intent (in) :: m
    code = (a * (k ** m)) / ((1.0d0 + (10.0d0 * k)) + (k * k))
end function

Java

public static double code(double a, double k, double m) {
	return (a * Math.pow(k, m)) / ((1.0 + (10.0 * k)) + (k * k));
}

Python

def code(a, k, m):
	return (a * math.pow(k, m)) / ((1.0 + (10.0 * k)) + (k * k))

Julia

function code(a, k, m)
	return Float64(Float64(a * (k ^ m)) / Float64(Float64(1.0 + Float64(10.0 * k)) + Float64(k * k)))
end

MATLAB

function tmp = code(a, k, m)
	tmp = (a * (k ^ m)) / ((1.0 + (10.0 * k)) + (k * k));
end

Wolfram

code[a_, k_, m_] := N[(N[(a * N[Power[k, m], $MachinePrecision]), $MachinePrecision] / N[(N[(1.0 + N[(10.0 * k), $MachinePrecision]), $MachinePrecision] + N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Alternative 1: 99.1% accurate, 0.4× speedup

Math

\[\begin{array}{l} \\ \frac{\frac{a \cdot {k}^{m}}{\mathsf{hypot}\left(1, k\right)}}{\mathsf{hypot}\left(1, k\right)} \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (/ (/ (* a (pow k m)) (hypot 1.0 k)) (hypot 1.0 k)))

C

double code(double a, double k, double m) {
	return ((a * pow(k, m)) / hypot(1.0, k)) / hypot(1.0, k);
}

Java

public static double code(double a, double k, double m) {
	return ((a * Math.pow(k, m)) / Math.hypot(1.0, k)) / Math.hypot(1.0, k);
}

Python

def code(a, k, m):
	return ((a * math.pow(k, m)) / math.hypot(1.0, k)) / math.hypot(1.0, k)

Julia

function code(a, k, m)
	return Float64(Float64(Float64(a * (k ^ m)) / hypot(1.0, k)) / hypot(1.0, k))
end

MATLAB

function tmp = code(a, k, m)
	tmp = ((a * (k ^ m)) / hypot(1.0, k)) / hypot(1.0, k);
end

Wolfram

code[a_, k_, m_] := N[(N[(N[(a * N[Power[k, m], $MachinePrecision]), $MachinePrecision] / N[Sqrt[1.0 ^ 2 + k ^ 2], $MachinePrecision]), $MachinePrecision] / N[Sqrt[1.0 ^ 2 + k ^ 2], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 89.8%

   \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
2. Step-by-step derivation

   1. *-commutative 89.8%

      \[\leadsto \frac{a \cdot {k}^{m}}{\left(1 + \color{blue}{k \cdot 10}\right) + k \cdot k} \]

3. Simplified 89.8%

   \[\leadsto \color{blue}{\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}} \]
4. Add Preprocessing

5. Taylor expanded in k around 0 89.2%

   \[\leadsto \frac{a \cdot {k}^{m}}{\color{blue}{1} + k \cdot k} \]
6. Step-by-step derivation

   1. add-sqr-sqrt 64.3%

      \[\leadsto \color{blue}{\sqrt{\frac{a \cdot {k}^{m}}{1 + k \cdot k}} \cdot \sqrt{\frac{a \cdot {k}^{m}}{1 + k \cdot k}}} \]

   2. sqrt-div 58.8%

      \[\leadsto \color{blue}{\frac{\sqrt{a \cdot {k}^{m}}}{\sqrt{1 + k \cdot k}}} \cdot \sqrt{\frac{a \cdot {k}^{m}}{1 + k \cdot k}} \]

   3. hypot-1-def 58.8%

      \[\leadsto \frac{\sqrt{a \cdot {k}^{m}}}{\color{blue}{\mathsf{hypot}\left(1, k\right)}} \cdot \sqrt{\frac{a \cdot {k}^{m}}{1 + k \cdot k}} \]

   4. sqrt-div 58.8%

      \[\leadsto \frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)} \cdot \color{blue}{\frac{\sqrt{a \cdot {k}^{m}}}{\sqrt{1 + k \cdot k}}} \]

   5. hypot-1-def 63.5%

      \[\leadsto \frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)} \cdot \frac{\sqrt{a \cdot {k}^{m}}}{\color{blue}{\mathsf{hypot}\left(1, k\right)}} \]

7. Applied egg-rr 63.5%

   \[\leadsto \color{blue}{\frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)} \cdot \frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)}} \]
8. Step-by-step derivation

   1. unpow2 63.5%

      \[\leadsto \color{blue}{{\left(\frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)}\right)}^{2}} \]

9. Simplified 63.5%

   \[\leadsto \color{blue}{{\left(\frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)}\right)}^{2}} \]
10. Step-by-step derivation

    1. unpow2 63.5%

       \[\leadsto \color{blue}{\frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)} \cdot \frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)}} \]

    2. frac-times 58.8%

       \[\leadsto \color{blue}{\frac{\sqrt{a \cdot {k}^{m}} \cdot \sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right) \cdot \mathsf{hypot}\left(1, k\right)}} \]

    3. add-sqr-sqrt 89.2%

       \[\leadsto \frac{\color{blue}{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right) \cdot \mathsf{hypot}\left(1, k\right)} \]

    4. hypot-udef 89.2%

       \[\leadsto \frac{a \cdot {k}^{m}}{\color{blue}{\sqrt{1 \cdot 1 + k \cdot k}} \cdot \mathsf{hypot}\left(1, k\right)} \]

    5. hypot-udef 89.2%

       \[\leadsto \frac{a \cdot {k}^{m}}{\sqrt{1 \cdot 1 + k \cdot k} \cdot \color{blue}{\sqrt{1 \cdot 1 + k \cdot k}}} \]

    6. rem-square-sqrt 89.2%

       \[\leadsto \frac{a \cdot {k}^{m}}{\color{blue}{1 \cdot 1 + k \cdot k}} \]

    7. remove-double-div 89.2%

       \[\leadsto \frac{a \cdot {k}^{m}}{1 \cdot 1 + k \cdot \color{blue}{\frac{1}{\frac{1}{k}}}} \]

    8. reciprocal-undefine 86.2%

       \[\leadsto \frac{a \cdot {k}^{m}}{1 \cdot 1 + k \cdot \frac{1}{\color{blue}{\mathsf{reciprocal}\left(k\right)}}} \]

    9. *-un-lft-identity 86.2%

       \[\leadsto \frac{a \cdot {k}^{m}}{1 \cdot 1 + \color{blue}{1 \cdot \left(k \cdot \frac{1}{\mathsf{reciprocal}\left(k\right)}\right)}} \]

    10. div-inv 86.2%

        \[\leadsto \frac{a \cdot {k}^{m}}{1 \cdot 1 + 1 \cdot \color{blue}{\frac{k}{\mathsf{reciprocal}\left(k\right)}}} \]

    11. distribute-lft-in 86.2%

        \[\leadsto \frac{a \cdot {k}^{m}}{\color{blue}{1 \cdot \left(1 + \frac{k}{\mathsf{reciprocal}\left(k\right)}\right)}} \]

    12. *-un-lft-identity 86.2%

        \[\leadsto \frac{a \cdot {k}^{m}}{\color{blue}{1 + \frac{k}{\mathsf{reciprocal}\left(k\right)}}} \]

    13. add-sqr-sqrt 86.2%

        \[\leadsto \frac{a \cdot {k}^{m}}{\color{blue}{\sqrt{1 + \frac{k}{\mathsf{reciprocal}\left(k\right)}} \cdot \sqrt{1 + \frac{k}{\mathsf{reciprocal}\left(k\right)}}}} \]

    14. associate-/r* 86.2%

        \[\leadsto \color{blue}{\frac{\frac{a \cdot {k}^{m}}{\sqrt{1 + \frac{k}{\mathsf{reciprocal}\left(k\right)}}}}{\sqrt{1 + \frac{k}{\mathsf{reciprocal}\left(k\right)}}}} \]

11. Applied egg-rr 99.4%

    \[\leadsto \color{blue}{\frac{\frac{a \cdot {k}^{m}}{\mathsf{hypot}\left(1, k\right)}}{\mathsf{hypot}\left(1, k\right)}} \]

12. Final simplification 99.4%

    \[\leadsto \frac{\frac{a \cdot {k}^{m}}{\mathsf{hypot}\left(1, k\right)}}{\mathsf{hypot}\left(1, k\right)} \]
13. Add Preprocessing

Alternative 2: 98.5% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := a \cdot {k}^{m}\\ \mathbf{if}\;\frac{t_0}{\left(1 + k \cdot 10\right) + k \cdot k} \leq 5 \cdot 10^{+178}:\\ \;\;\;\;\frac{{k}^{m}}{\mathsf{hypot}\left(1, k\right)} \cdot \frac{a}{\mathsf{hypot}\left(1, k\right)}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (let* ((t_0 (* a (pow k m))))
   (if (<= (/ t_0 (+ (+ 1.0 (* k 10.0)) (* k k))) 5e+178)
     (* (/ (pow k m) (hypot 1.0 k)) (/ a (hypot 1.0 k)))
     t_0)))

C

double code(double a, double k, double m) {
	double t_0 = a * pow(k, m);
	double tmp;
	if ((t_0 / ((1.0 + (k * 10.0)) + (k * k))) <= 5e+178) {
		tmp = (pow(k, m) / hypot(1.0, k)) * (a / hypot(1.0, k));
	} else {
		tmp = t_0;
	}
	return tmp;
}

Java

public static double code(double a, double k, double m) {
	double t_0 = a * Math.pow(k, m);
	double tmp;
	if ((t_0 / ((1.0 + (k * 10.0)) + (k * k))) <= 5e+178) {
		tmp = (Math.pow(k, m) / Math.hypot(1.0, k)) * (a / Math.hypot(1.0, k));
	} else {
		tmp = t_0;
	}
	return tmp;
}

Python

def code(a, k, m):
	t_0 = a * math.pow(k, m)
	tmp = 0
	if (t_0 / ((1.0 + (k * 10.0)) + (k * k))) <= 5e+178:
		tmp = (math.pow(k, m) / math.hypot(1.0, k)) * (a / math.hypot(1.0, k))
	else:
		tmp = t_0
	return tmp

Julia

function code(a, k, m)
	t_0 = Float64(a * (k ^ m))
	tmp = 0.0
	if (Float64(t_0 / Float64(Float64(1.0 + Float64(k * 10.0)) + Float64(k * k))) <= 5e+178)
		tmp = Float64(Float64((k ^ m) / hypot(1.0, k)) * Float64(a / hypot(1.0, k)));
	else
		tmp = t_0;
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, k, m)
	t_0 = a * (k ^ m);
	tmp = 0.0;
	if ((t_0 / ((1.0 + (k * 10.0)) + (k * k))) <= 5e+178)
		tmp = ((k ^ m) / hypot(1.0, k)) * (a / hypot(1.0, k));
	else
		tmp = t_0;
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, k_, m_] := Block[{t$95$0 = N[(a * N[Power[k, m], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(t$95$0 / N[(N[(1.0 + N[(k * 10.0), $MachinePrecision]), $MachinePrecision] + N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 5e+178], N[(N[(N[Power[k, m], $MachinePrecision] / N[Sqrt[1.0 ^ 2 + k ^ 2], $MachinePrecision]), $MachinePrecision] * N[(a / N[Sqrt[1.0 ^ 2 + k ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], t$95$0]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 (*.f64 a (pow.f64 k m)) (+.f64 (+.f64 1 (*.f64 10 k)) (*.f64 k k))) < 4.9999999999999999e178

   1. Initial program 97.1%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. *-commutative 97.1%

         \[\leadsto \frac{a \cdot {k}^{m}}{\left(1 + \color{blue}{k \cdot 10}\right) + k \cdot k} \]

   3. Simplified 97.1%

      \[\leadsto \color{blue}{\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}} \]
   4. Add Preprocessing

   5. Taylor expanded in k around 0 96.4%

      \[\leadsto \frac{a \cdot {k}^{m}}{\color{blue}{1} + k \cdot k} \]
   6. Step-by-step derivation

      1. *-commutative 96.4%

         \[\leadsto \frac{\color{blue}{{k}^{m} \cdot a}}{1 + k \cdot k} \]

      2. add-sqr-sqrt 96.4%

         \[\leadsto \frac{{k}^{m} \cdot a}{\color{blue}{\sqrt{1 + k \cdot k} \cdot \sqrt{1 + k \cdot k}}} \]

      3. times-frac 95.9%

         \[\leadsto \color{blue}{\frac{{k}^{m}}{\sqrt{1 + k \cdot k}} \cdot \frac{a}{\sqrt{1 + k \cdot k}}} \]

      4. hypot-1-def 95.9%

         \[\leadsto \frac{{k}^{m}}{\color{blue}{\mathsf{hypot}\left(1, k\right)}} \cdot \frac{a}{\sqrt{1 + k \cdot k}} \]

      5. hypot-1-def 98.8%

         \[\leadsto \frac{{k}^{m}}{\mathsf{hypot}\left(1, k\right)} \cdot \frac{a}{\color{blue}{\mathsf{hypot}\left(1, k\right)}} \]

   7. Applied egg-rr 98.8%

      \[\leadsto \color{blue}{\frac{{k}^{m}}{\mathsf{hypot}\left(1, k\right)} \cdot \frac{a}{\mathsf{hypot}\left(1, k\right)}} \]

   if 4.9999999999999999e178 < (/.f64 (*.f64 a (pow.f64 k m)) (+.f64 (+.f64 1 (*.f64 10 k)) (*.f64 k k)))

   1. Initial program 55.6%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 44.4%

         \[\leadsto \color{blue}{\frac{a}{\left(1 + 10 \cdot k\right) + k \cdot k} \cdot {k}^{m}} \]

      2. sqr-neg 44.4%

         \[\leadsto \frac{a}{\left(1 + 10 \cdot k\right) + \color{blue}{\left(-k\right) \cdot \left(-k\right)}} \cdot {k}^{m} \]

      3. associate-+l+ 44.4%

         \[\leadsto \frac{a}{\color{blue}{1 + \left(10 \cdot k + \left(-k\right) \cdot \left(-k\right)\right)}} \cdot {k}^{m} \]

      4. sqr-neg 44.4%

         \[\leadsto \frac{a}{1 + \left(10 \cdot k + \color{blue}{k \cdot k}\right)} \cdot {k}^{m} \]

      5. distribute-rgt-out 44.4%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot \left(10 + k\right)}} \cdot {k}^{m} \]

   3. Simplified 44.4%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)} \cdot {k}^{m}} \]
   4. Add Preprocessing

   5. Taylor expanded in k around 0 100.0%

      \[\leadsto \color{blue}{a \cdot {k}^{m}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 99.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k} \leq 5 \cdot 10^{+178}:\\ \;\;\;\;\frac{{k}^{m}}{\mathsf{hypot}\left(1, k\right)} \cdot \frac{a}{\mathsf{hypot}\left(1, k\right)}\\ \mathbf{else}:\\ \;\;\;\;a \cdot {k}^{m}\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 86.0% accurate, 0.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := a \cdot {k}^{m}\\ \mathbf{if}\;k \leq 0.000455:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;{\left(\frac{\sqrt{t_0}}{k}\right)}^{2}\\ \end{array} \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (let* ((t_0 (* a (pow k m))))
   (if (<= k 0.000455) t_0 (pow (/ (sqrt t_0) k) 2.0))))

C

double code(double a, double k, double m) {
	double t_0 = a * pow(k, m);
	double tmp;
	if (k <= 0.000455) {
		tmp = t_0;
	} else {
		tmp = pow((sqrt(t_0) / k), 2.0);
	}
	return tmp;
}

Fortran

real(8) function code(a, k, m)
    real(8), intent (in) :: a
    real(8), intent (in) :: k
    real(8), intent (in) :: m
    real(8) :: t_0
    real(8) :: tmp
    t_0 = a * (k ** m)
    if (k <= 0.000455d0) then
        tmp = t_0
    else
        tmp = (sqrt(t_0) / k) ** 2.0d0
    end if
    code = tmp
end function

Java

public static double code(double a, double k, double m) {
	double t_0 = a * Math.pow(k, m);
	double tmp;
	if (k <= 0.000455) {
		tmp = t_0;
	} else {
		tmp = Math.pow((Math.sqrt(t_0) / k), 2.0);
	}
	return tmp;
}

Python

def code(a, k, m):
	t_0 = a * math.pow(k, m)
	tmp = 0
	if k <= 0.000455:
		tmp = t_0
	else:
		tmp = math.pow((math.sqrt(t_0) / k), 2.0)
	return tmp

Julia

function code(a, k, m)
	t_0 = Float64(a * (k ^ m))
	tmp = 0.0
	if (k <= 0.000455)
		tmp = t_0;
	else
		tmp = Float64(sqrt(t_0) / k) ^ 2.0;
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, k, m)
	t_0 = a * (k ^ m);
	tmp = 0.0;
	if (k <= 0.000455)
		tmp = t_0;
	else
		tmp = (sqrt(t_0) / k) ^ 2.0;
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, k_, m_] := Block[{t$95$0 = N[(a * N[Power[k, m], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[k, 0.000455], t$95$0, N[Power[N[(N[Sqrt[t$95$0], $MachinePrecision] / k), $MachinePrecision], 2.0], $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if k < 4.55e-4

   1. Initial program 92.8%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 87.3%

         \[\leadsto \color{blue}{\frac{a}{\left(1 + 10 \cdot k\right) + k \cdot k} \cdot {k}^{m}} \]

      2. sqr-neg 87.3%

         \[\leadsto \frac{a}{\left(1 + 10 \cdot k\right) + \color{blue}{\left(-k\right) \cdot \left(-k\right)}} \cdot {k}^{m} \]

      3. associate-+l+ 87.3%

         \[\leadsto \frac{a}{\color{blue}{1 + \left(10 \cdot k + \left(-k\right) \cdot \left(-k\right)\right)}} \cdot {k}^{m} \]

      4. sqr-neg 87.3%

         \[\leadsto \frac{a}{1 + \left(10 \cdot k + \color{blue}{k \cdot k}\right)} \cdot {k}^{m} \]

      5. distribute-rgt-out 87.3%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot \left(10 + k\right)}} \cdot {k}^{m} \]

   3. Simplified 87.3%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)} \cdot {k}^{m}} \]
   4. Add Preprocessing

   5. Taylor expanded in k around 0 99.7%

      \[\leadsto \color{blue}{a \cdot {k}^{m}} \]

   if 4.55e-4 < k

   1. Initial program 84.4%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. *-commutative 84.4%

         \[\leadsto \frac{a \cdot {k}^{m}}{\left(1 + \color{blue}{k \cdot 10}\right) + k \cdot k} \]

   3. Simplified 84.4%

      \[\leadsto \color{blue}{\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k}} \]
   4. Add Preprocessing

   5. Taylor expanded in k around 0 83.2%

      \[\leadsto \frac{a \cdot {k}^{m}}{\color{blue}{1} + k \cdot k} \]
   6. Step-by-step derivation

      1. add-sqr-sqrt 72.1%

         \[\leadsto \color{blue}{\sqrt{\frac{a \cdot {k}^{m}}{1 + k \cdot k}} \cdot \sqrt{\frac{a \cdot {k}^{m}}{1 + k \cdot k}}} \]

      2. sqrt-div 56.4%

         \[\leadsto \color{blue}{\frac{\sqrt{a \cdot {k}^{m}}}{\sqrt{1 + k \cdot k}}} \cdot \sqrt{\frac{a \cdot {k}^{m}}{1 + k \cdot k}} \]

      3. hypot-1-def 56.4%

         \[\leadsto \frac{\sqrt{a \cdot {k}^{m}}}{\color{blue}{\mathsf{hypot}\left(1, k\right)}} \cdot \sqrt{\frac{a \cdot {k}^{m}}{1 + k \cdot k}} \]

      4. sqrt-div 56.4%

         \[\leadsto \frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)} \cdot \color{blue}{\frac{\sqrt{a \cdot {k}^{m}}}{\sqrt{1 + k \cdot k}}} \]

      5. hypot-1-def 64.3%

         \[\leadsto \frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)} \cdot \frac{\sqrt{a \cdot {k}^{m}}}{\color{blue}{\mathsf{hypot}\left(1, k\right)}} \]

   7. Applied egg-rr 64.3%

      \[\leadsto \color{blue}{\frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)} \cdot \frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)}} \]
   8. Step-by-step derivation

      1. unpow2 64.3%

         \[\leadsto \color{blue}{{\left(\frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)}\right)}^{2}} \]

   9. Simplified 64.3%

      \[\leadsto \color{blue}{{\left(\frac{\sqrt{a \cdot {k}^{m}}}{\mathsf{hypot}\left(1, k\right)}\right)}^{2}} \]

   10. Taylor expanded in k around inf 64.2%

       \[\leadsto {\color{blue}{\left(\sqrt{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}} \cdot \frac{1}{k}\right)}}^{2} \]
   11. Step-by-step derivation

       1. associate-*r/ 64.3%

          \[\leadsto {\color{blue}{\left(\frac{\sqrt{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}} \cdot 1}{k}\right)}}^{2} \]

       2. *-rgt-identity 64.3%

          \[\leadsto {\left(\frac{\color{blue}{\sqrt{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}}}{k}\right)}^{2} \]

   12. Simplified 64.3%

       \[\leadsto {\color{blue}{\left(\frac{\sqrt{a \cdot {k}^{m}}}{k}\right)}}^{2} \]
3. Recombined 2 regimes into one program.

4. Final simplification 87.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 0.000455:\\ \;\;\;\;a \cdot {k}^{m}\\ \mathbf{else}:\\ \;\;\;\;{\left(\frac{\sqrt{a \cdot {k}^{m}}}{k}\right)}^{2}\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 96.5% accurate, 0.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := a \cdot {k}^{m}\\ \mathbf{if}\;\frac{t_0}{\left(1 + k \cdot 10\right) + k \cdot k} \leq 5 \cdot 10^{+178}:\\ \;\;\;\;{k}^{m} \cdot \frac{a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (let* ((t_0 (* a (pow k m))))
   (if (<= (/ t_0 (+ (+ 1.0 (* k 10.0)) (* k k))) 5e+178)
     (* (pow k m) (/ a (+ 1.0 (* k (+ k 10.0)))))
     t_0)))

C

double code(double a, double k, double m) {
	double t_0 = a * pow(k, m);
	double tmp;
	if ((t_0 / ((1.0 + (k * 10.0)) + (k * k))) <= 5e+178) {
		tmp = pow(k, m) * (a / (1.0 + (k * (k + 10.0))));
	} else {
		tmp = t_0;
	}
	return tmp;
}

Fortran

real(8) function code(a, k, m)
    real(8), intent (in) :: a
    real(8), intent (in) :: k
    real(8), intent (in) :: m
    real(8) :: t_0
    real(8) :: tmp
    t_0 = a * (k ** m)
    if ((t_0 / ((1.0d0 + (k * 10.0d0)) + (k * k))) <= 5d+178) then
        tmp = (k ** m) * (a / (1.0d0 + (k * (k + 10.0d0))))
    else
        tmp = t_0
    end if
    code = tmp
end function

Java

public static double code(double a, double k, double m) {
	double t_0 = a * Math.pow(k, m);
	double tmp;
	if ((t_0 / ((1.0 + (k * 10.0)) + (k * k))) <= 5e+178) {
		tmp = Math.pow(k, m) * (a / (1.0 + (k * (k + 10.0))));
	} else {
		tmp = t_0;
	}
	return tmp;
}

Python

def code(a, k, m):
	t_0 = a * math.pow(k, m)
	tmp = 0
	if (t_0 / ((1.0 + (k * 10.0)) + (k * k))) <= 5e+178:
		tmp = math.pow(k, m) * (a / (1.0 + (k * (k + 10.0))))
	else:
		tmp = t_0
	return tmp

Julia

function code(a, k, m)
	t_0 = Float64(a * (k ^ m))
	tmp = 0.0
	if (Float64(t_0 / Float64(Float64(1.0 + Float64(k * 10.0)) + Float64(k * k))) <= 5e+178)
		tmp = Float64((k ^ m) * Float64(a / Float64(1.0 + Float64(k * Float64(k + 10.0)))));
	else
		tmp = t_0;
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, k, m)
	t_0 = a * (k ^ m);
	tmp = 0.0;
	if ((t_0 / ((1.0 + (k * 10.0)) + (k * k))) <= 5e+178)
		tmp = (k ^ m) * (a / (1.0 + (k * (k + 10.0))));
	else
		tmp = t_0;
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, k_, m_] := Block[{t$95$0 = N[(a * N[Power[k, m], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(t$95$0 / N[(N[(1.0 + N[(k * 10.0), $MachinePrecision]), $MachinePrecision] + N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 5e+178], N[(N[Power[k, m], $MachinePrecision] * N[(a / N[(1.0 + N[(k * N[(k + 10.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], t$95$0]]

Derivation

1. Split input into 2 regimes

2. if (/.f64 (*.f64 a (pow.f64 k m)) (+.f64 (+.f64 1 (*.f64 10 k)) (*.f64 k k))) < 4.9999999999999999e178

   1. Initial program 97.1%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 94.3%

         \[\leadsto \color{blue}{\frac{a}{\left(1 + 10 \cdot k\right) + k \cdot k} \cdot {k}^{m}} \]

      2. sqr-neg 94.3%

         \[\leadsto \frac{a}{\left(1 + 10 \cdot k\right) + \color{blue}{\left(-k\right) \cdot \left(-k\right)}} \cdot {k}^{m} \]

      3. associate-+l+ 94.3%

         \[\leadsto \frac{a}{\color{blue}{1 + \left(10 \cdot k + \left(-k\right) \cdot \left(-k\right)\right)}} \cdot {k}^{m} \]

      4. sqr-neg 94.3%

         \[\leadsto \frac{a}{1 + \left(10 \cdot k + \color{blue}{k \cdot k}\right)} \cdot {k}^{m} \]

      5. distribute-rgt-out 94.3%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot \left(10 + k\right)}} \cdot {k}^{m} \]

   3. Simplified 94.3%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)} \cdot {k}^{m}} \]
   4. Add Preprocessing

   if 4.9999999999999999e178 < (/.f64 (*.f64 a (pow.f64 k m)) (+.f64 (+.f64 1 (*.f64 10 k)) (*.f64 k k)))

   1. Initial program 55.6%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 44.4%

         \[\leadsto \color{blue}{\frac{a}{\left(1 + 10 \cdot k\right) + k \cdot k} \cdot {k}^{m}} \]

      2. sqr-neg 44.4%

         \[\leadsto \frac{a}{\left(1 + 10 \cdot k\right) + \color{blue}{\left(-k\right) \cdot \left(-k\right)}} \cdot {k}^{m} \]

      3. associate-+l+ 44.4%

         \[\leadsto \frac{a}{\color{blue}{1 + \left(10 \cdot k + \left(-k\right) \cdot \left(-k\right)\right)}} \cdot {k}^{m} \]

      4. sqr-neg 44.4%

         \[\leadsto \frac{a}{1 + \left(10 \cdot k + \color{blue}{k \cdot k}\right)} \cdot {k}^{m} \]

      5. distribute-rgt-out 44.4%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot \left(10 + k\right)}} \cdot {k}^{m} \]

   3. Simplified 44.4%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)} \cdot {k}^{m}} \]
   4. Add Preprocessing

   5. Taylor expanded in k around 0 100.0%

      \[\leadsto \color{blue}{a \cdot {k}^{m}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 95.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{a \cdot {k}^{m}}{\left(1 + k \cdot 10\right) + k \cdot k} \leq 5 \cdot 10^{+178}:\\ \;\;\;\;{k}^{m} \cdot \frac{a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;a \cdot {k}^{m}\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 97.0% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;m \leq 3.3:\\ \;\;\;\;{k}^{m} \cdot \frac{a}{1 + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;a \cdot {k}^{m}\\ \end{array} \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (if (<= m 3.3) (* (pow k m) (/ a (+ 1.0 (* k k)))) (* a (pow k m))))

C

double code(double a, double k, double m) {
	double tmp;
	if (m <= 3.3) {
		tmp = pow(k, m) * (a / (1.0 + (k * k)));
	} else {
		tmp = a * pow(k, m);
	}
	return tmp;
}

Fortran

real(8) function code(a, k, m)
    real(8), intent (in) :: a
    real(8), intent (in) :: k
    real(8), intent (in) :: m
    real(8) :: tmp
    if (m <= 3.3d0) then
        tmp = (k ** m) * (a / (1.0d0 + (k * k)))
    else
        tmp = a * (k ** m)
    end if
    code = tmp
end function

Java

public static double code(double a, double k, double m) {
	double tmp;
	if (m <= 3.3) {
		tmp = Math.pow(k, m) * (a / (1.0 + (k * k)));
	} else {
		tmp = a * Math.pow(k, m);
	}
	return tmp;
}

Python

def code(a, k, m):
	tmp = 0
	if m <= 3.3:
		tmp = math.pow(k, m) * (a / (1.0 + (k * k)))
	else:
		tmp = a * math.pow(k, m)
	return tmp

Julia

function code(a, k, m)
	tmp = 0.0
	if (m <= 3.3)
		tmp = Float64((k ^ m) * Float64(a / Float64(1.0 + Float64(k * k))));
	else
		tmp = Float64(a * (k ^ m));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, k, m)
	tmp = 0.0;
	if (m <= 3.3)
		tmp = (k ^ m) * (a / (1.0 + (k * k)));
	else
		tmp = a * (k ^ m);
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, k_, m_] := If[LessEqual[m, 3.3], N[(N[Power[k, m], $MachinePrecision] * N[(a / N[(1.0 + N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(a * N[Power[k, m], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if m < 3.2999999999999998

   1. Initial program 96.4%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 96.4%

         \[\leadsto \color{blue}{\frac{a}{\left(1 + 10 \cdot k\right) + k \cdot k} \cdot {k}^{m}} \]

      2. sqr-neg 96.4%

         \[\leadsto \frac{a}{\left(1 + 10 \cdot k\right) + \color{blue}{\left(-k\right) \cdot \left(-k\right)}} \cdot {k}^{m} \]

      3. associate-+l+ 96.4%

         \[\leadsto \frac{a}{\color{blue}{1 + \left(10 \cdot k + \left(-k\right) \cdot \left(-k\right)\right)}} \cdot {k}^{m} \]

      4. sqr-neg 96.4%

         \[\leadsto \frac{a}{1 + \left(10 \cdot k + \color{blue}{k \cdot k}\right)} \cdot {k}^{m} \]

      5. distribute-rgt-out 96.4%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot \left(10 + k\right)}} \cdot {k}^{m} \]

   3. Simplified 96.4%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)} \cdot {k}^{m}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. flip3-+ 70.8%

         \[\leadsto \frac{a}{1 + k \cdot \color{blue}{\frac{{10}^{3} + {k}^{3}}{10 \cdot 10 + \left(k \cdot k - 10 \cdot k\right)}}} \cdot {k}^{m} \]

      2. associate-*r/ 68.6%

         \[\leadsto \frac{a}{1 + \color{blue}{\frac{k \cdot \left({10}^{3} + {k}^{3}\right)}{10 \cdot 10 + \left(k \cdot k - 10 \cdot k\right)}}} \cdot {k}^{m} \]

      3. pow3 68.6%

         \[\leadsto \frac{a}{1 + \frac{k \cdot \left({10}^{3} + \color{blue}{\left(k \cdot k\right) \cdot k}\right)}{10 \cdot 10 + \left(k \cdot k - 10 \cdot k\right)}} \cdot {k}^{m} \]

      4. +-commutative 68.6%

         \[\leadsto \frac{a}{1 + \frac{k \cdot \color{blue}{\left(\left(k \cdot k\right) \cdot k + {10}^{3}\right)}}{10 \cdot 10 + \left(k \cdot k - 10 \cdot k\right)}} \cdot {k}^{m} \]

      5. pow3 68.6%

         \[\leadsto \frac{a}{1 + \frac{k \cdot \left(\color{blue}{{k}^{3}} + {10}^{3}\right)}{10 \cdot 10 + \left(k \cdot k - 10 \cdot k\right)}} \cdot {k}^{m} \]

      6. metadata-eval 68.6%

         \[\leadsto \frac{a}{1 + \frac{k \cdot \left({k}^{3} + \color{blue}{1000}\right)}{10 \cdot 10 + \left(k \cdot k - 10 \cdot k\right)}} \cdot {k}^{m} \]

      7. metadata-eval 68.6%

         \[\leadsto \frac{a}{1 + \frac{k \cdot \left({k}^{3} + 1000\right)}{\color{blue}{100} + \left(k \cdot k - 10 \cdot k\right)}} \cdot {k}^{m} \]

      8. distribute-rgt-out-- 68.6%

         \[\leadsto \frac{a}{1 + \frac{k \cdot \left({k}^{3} + 1000\right)}{100 + \color{blue}{k \cdot \left(k - 10\right)}}} \cdot {k}^{m} \]

   6. Applied egg-rr 68.6%

      \[\leadsto \frac{a}{1 + \color{blue}{\frac{k \cdot \left({k}^{3} + 1000\right)}{100 + k \cdot \left(k - 10\right)}}} \cdot {k}^{m} \]
   7. Step-by-step derivation

      1. associate-/l* 70.8%

         \[\leadsto \frac{a}{1 + \color{blue}{\frac{k}{\frac{100 + k \cdot \left(k - 10\right)}{{k}^{3} + 1000}}}} \cdot {k}^{m} \]

      2. sub-neg 70.8%

         \[\leadsto \frac{a}{1 + \frac{k}{\frac{100 + k \cdot \color{blue}{\left(k + \left(-10\right)\right)}}{{k}^{3} + 1000}}} \cdot {k}^{m} \]

      3. metadata-eval 70.8%

         \[\leadsto \frac{a}{1 + \frac{k}{\frac{100 + k \cdot \left(k + \color{blue}{-10}\right)}{{k}^{3} + 1000}}} \cdot {k}^{m} \]

   8. Simplified 70.8%

      \[\leadsto \frac{a}{1 + \color{blue}{\frac{k}{\frac{100 + k \cdot \left(k + -10\right)}{{k}^{3} + 1000}}}} \cdot {k}^{m} \]

   9. Taylor expanded in k around inf 95.6%

      \[\leadsto \frac{a}{1 + \frac{k}{\color{blue}{\frac{1}{k}}}} \cdot {k}^{m} \]
   10. Step-by-step derivation

       1. reciprocal-define 91.0%

          \[\leadsto \frac{a}{1 + \frac{k}{\color{blue}{\mathsf{reciprocal}\left(k\right)}}} \cdot {k}^{m} \]

   11. Simplified 91.0%

       \[\leadsto \frac{a}{1 + \frac{k}{\color{blue}{\mathsf{reciprocal}\left(k\right)}}} \cdot {k}^{m} \]
   12. Step-by-step derivation

       1. div-inv 91.0%

          \[\leadsto \frac{a}{1 + \color{blue}{k \cdot \frac{1}{\mathsf{reciprocal}\left(k\right)}}} \cdot {k}^{m} \]

       2. reciprocal-undefine 95.6%

          \[\leadsto \frac{a}{1 + k \cdot \frac{1}{\color{blue}{\frac{1}{k}}}} \cdot {k}^{m} \]

       3. remove-double-div 95.6%

          \[\leadsto \frac{a}{1 + k \cdot \color{blue}{k}} \cdot {k}^{m} \]

   13. Applied egg-rr 95.6%

       \[\leadsto \frac{a}{1 + \color{blue}{k \cdot k}} \cdot {k}^{m} \]

   if 3.2999999999999998 < m

   1. Initial program 76.5%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 63.5%

         \[\leadsto \color{blue}{\frac{a}{\left(1 + 10 \cdot k\right) + k \cdot k} \cdot {k}^{m}} \]

      2. sqr-neg 63.5%

         \[\leadsto \frac{a}{\left(1 + 10 \cdot k\right) + \color{blue}{\left(-k\right) \cdot \left(-k\right)}} \cdot {k}^{m} \]

      3. associate-+l+ 63.5%

         \[\leadsto \frac{a}{\color{blue}{1 + \left(10 \cdot k + \left(-k\right) \cdot \left(-k\right)\right)}} \cdot {k}^{m} \]

      4. sqr-neg 63.5%

         \[\leadsto \frac{a}{1 + \left(10 \cdot k + \color{blue}{k \cdot k}\right)} \cdot {k}^{m} \]

      5. distribute-rgt-out 63.5%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot \left(10 + k\right)}} \cdot {k}^{m} \]

   3. Simplified 63.5%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)} \cdot {k}^{m}} \]
   4. Add Preprocessing

   5. Taylor expanded in k around 0 100.0%

      \[\leadsto \color{blue}{a \cdot {k}^{m}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 97.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;m \leq 3.3:\\ \;\;\;\;{k}^{m} \cdot \frac{a}{1 + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;a \cdot {k}^{m}\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 97.1% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;m \leq -8.6 \cdot 10^{-18} \lor \neg \left(m \leq 1.95 \cdot 10^{-6}\right):\\ \;\;\;\;a \cdot {k}^{m}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \end{array} \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (if (or (<= m -8.6e-18) (not (<= m 1.95e-6)))
   (* a (pow k m))
   (/ a (+ 1.0 (* k (+ k 10.0))))))

C

double code(double a, double k, double m) {
	double tmp;
	if ((m <= -8.6e-18) || !(m <= 1.95e-6)) {
		tmp = a * pow(k, m);
	} else {
		tmp = a / (1.0 + (k * (k + 10.0)));
	}
	return tmp;
}

Fortran

real(8) function code(a, k, m)
    real(8), intent (in) :: a
    real(8), intent (in) :: k
    real(8), intent (in) :: m
    real(8) :: tmp
    if ((m <= (-8.6d-18)) .or. (.not. (m <= 1.95d-6))) then
        tmp = a * (k ** m)
    else
        tmp = a / (1.0d0 + (k * (k + 10.0d0)))
    end if
    code = tmp
end function

Java

public static double code(double a, double k, double m) {
	double tmp;
	if ((m <= -8.6e-18) || !(m <= 1.95e-6)) {
		tmp = a * Math.pow(k, m);
	} else {
		tmp = a / (1.0 + (k * (k + 10.0)));
	}
	return tmp;
}

Python

def code(a, k, m):
	tmp = 0
	if (m <= -8.6e-18) or not (m <= 1.95e-6):
		tmp = a * math.pow(k, m)
	else:
		tmp = a / (1.0 + (k * (k + 10.0)))
	return tmp

Julia

function code(a, k, m)
	tmp = 0.0
	if ((m <= -8.6e-18) || !(m <= 1.95e-6))
		tmp = Float64(a * (k ^ m));
	else
		tmp = Float64(a / Float64(1.0 + Float64(k * Float64(k + 10.0))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, k, m)
	tmp = 0.0;
	if ((m <= -8.6e-18) || ~((m <= 1.95e-6)))
		tmp = a * (k ^ m);
	else
		tmp = a / (1.0 + (k * (k + 10.0)));
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, k_, m_] := If[Or[LessEqual[m, -8.6e-18], N[Not[LessEqual[m, 1.95e-6]], $MachinePrecision]], N[(a * N[Power[k, m], $MachinePrecision]), $MachinePrecision], N[(a / N[(1.0 + N[(k * N[(k + 10.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if m < -8.6000000000000005e-18 or 1.95e-6 < m

   1. Initial program 87.9%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 81.2%

         \[\leadsto \color{blue}{\frac{a}{\left(1 + 10 \cdot k\right) + k \cdot k} \cdot {k}^{m}} \]

      2. sqr-neg 81.2%

         \[\leadsto \frac{a}{\left(1 + 10 \cdot k\right) + \color{blue}{\left(-k\right) \cdot \left(-k\right)}} \cdot {k}^{m} \]

      3. associate-+l+ 81.2%

         \[\leadsto \frac{a}{\color{blue}{1 + \left(10 \cdot k + \left(-k\right) \cdot \left(-k\right)\right)}} \cdot {k}^{m} \]

      4. sqr-neg 81.2%

         \[\leadsto \frac{a}{1 + \left(10 \cdot k + \color{blue}{k \cdot k}\right)} \cdot {k}^{m} \]

      5. distribute-rgt-out 81.2%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot \left(10 + k\right)}} \cdot {k}^{m} \]

   3. Simplified 81.2%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)} \cdot {k}^{m}} \]
   4. Add Preprocessing

   5. Taylor expanded in k around 0 99.4%

      \[\leadsto \color{blue}{a \cdot {k}^{m}} \]

   if -8.6000000000000005e-18 < m < 1.95e-6

   1. Initial program 93.3%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 93.3%

         \[\leadsto \color{blue}{\frac{a}{\left(1 + 10 \cdot k\right) + k \cdot k} \cdot {k}^{m}} \]

      2. sqr-neg 93.3%

         \[\leadsto \frac{a}{\left(1 + 10 \cdot k\right) + \color{blue}{\left(-k\right) \cdot \left(-k\right)}} \cdot {k}^{m} \]

      3. associate-+l+ 93.3%

         \[\leadsto \frac{a}{\color{blue}{1 + \left(10 \cdot k + \left(-k\right) \cdot \left(-k\right)\right)}} \cdot {k}^{m} \]

      4. sqr-neg 93.3%

         \[\leadsto \frac{a}{1 + \left(10 \cdot k + \color{blue}{k \cdot k}\right)} \cdot {k}^{m} \]

      5. distribute-rgt-out 93.3%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot \left(10 + k\right)}} \cdot {k}^{m} \]

   3. Simplified 93.3%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)} \cdot {k}^{m}} \]
   4. Add Preprocessing

   5. Taylor expanded in m around 0 93.0%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 97.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;m \leq -8.6 \cdot 10^{-18} \lor \neg \left(m \leq 1.95 \cdot 10^{-6}\right):\\ \;\;\;\;a \cdot {k}^{m}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 50.0% accurate, 8.1× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;m \leq 750000:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (if (<= m 750000.0) (/ a (+ 1.0 (* k (+ k 10.0)))) (* -10.0 (* a k))))

C

double code(double a, double k, double m) {
	double tmp;
	if (m <= 750000.0) {
		tmp = a / (1.0 + (k * (k + 10.0)));
	} else {
		tmp = -10.0 * (a * k);
	}
	return tmp;
}

Fortran

real(8) function code(a, k, m)
    real(8), intent (in) :: a
    real(8), intent (in) :: k
    real(8), intent (in) :: m
    real(8) :: tmp
    if (m <= 750000.0d0) then
        tmp = a / (1.0d0 + (k * (k + 10.0d0)))
    else
        tmp = (-10.0d0) * (a * k)
    end if
    code = tmp
end function

Java

public static double code(double a, double k, double m) {
	double tmp;
	if (m <= 750000.0) {
		tmp = a / (1.0 + (k * (k + 10.0)));
	} else {
		tmp = -10.0 * (a * k);
	}
	return tmp;
}

Python

def code(a, k, m):
	tmp = 0
	if m <= 750000.0:
		tmp = a / (1.0 + (k * (k + 10.0)))
	else:
		tmp = -10.0 * (a * k)
	return tmp

Julia

function code(a, k, m)
	tmp = 0.0
	if (m <= 750000.0)
		tmp = Float64(a / Float64(1.0 + Float64(k * Float64(k + 10.0))));
	else
		tmp = Float64(-10.0 * Float64(a * k));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, k, m)
	tmp = 0.0;
	if (m <= 750000.0)
		tmp = a / (1.0 + (k * (k + 10.0)));
	else
		tmp = -10.0 * (a * k);
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, k_, m_] := If[LessEqual[m, 750000.0], N[(a / N[(1.0 + N[(k * N[(k + 10.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(-10.0 * N[(a * k), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if m < 7.5e5

   1. Initial program 95.9%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 95.9%

         \[\leadsto \color{blue}{\frac{a}{\left(1 + 10 \cdot k\right) + k \cdot k} \cdot {k}^{m}} \]

      2. sqr-neg 95.9%

         \[\leadsto \frac{a}{\left(1 + 10 \cdot k\right) + \color{blue}{\left(-k\right) \cdot \left(-k\right)}} \cdot {k}^{m} \]

      3. associate-+l+ 95.9%

         \[\leadsto \frac{a}{\color{blue}{1 + \left(10 \cdot k + \left(-k\right) \cdot \left(-k\right)\right)}} \cdot {k}^{m} \]

      4. sqr-neg 95.9%

         \[\leadsto \frac{a}{1 + \left(10 \cdot k + \color{blue}{k \cdot k}\right)} \cdot {k}^{m} \]

      5. distribute-rgt-out 95.9%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot \left(10 + k\right)}} \cdot {k}^{m} \]

   3. Simplified 95.9%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)} \cdot {k}^{m}} \]
   4. Add Preprocessing

   5. Taylor expanded in m around 0 67.9%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)}} \]

   if 7.5e5 < m

   1. Initial program 77.4%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 64.3%

         \[\leadsto \color{blue}{\frac{a}{\left(1 + 10 \cdot k\right) + k \cdot k} \cdot {k}^{m}} \]

      2. sqr-neg 64.3%

         \[\leadsto \frac{a}{\left(1 + 10 \cdot k\right) + \color{blue}{\left(-k\right) \cdot \left(-k\right)}} \cdot {k}^{m} \]

      3. associate-+l+ 64.3%

         \[\leadsto \frac{a}{\color{blue}{1 + \left(10 \cdot k + \left(-k\right) \cdot \left(-k\right)\right)}} \cdot {k}^{m} \]

      4. sqr-neg 64.3%

         \[\leadsto \frac{a}{1 + \left(10 \cdot k + \color{blue}{k \cdot k}\right)} \cdot {k}^{m} \]

      5. distribute-rgt-out 64.3%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot \left(10 + k\right)}} \cdot {k}^{m} \]

   3. Simplified 64.3%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)} \cdot {k}^{m}} \]
   4. Add Preprocessing

   5. Taylor expanded in m around 0 2.8%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)}} \]

   6. Taylor expanded in k around 0 7.8%

      \[\leadsto \color{blue}{a + -10 \cdot \left(a \cdot k\right)} \]

   7. Taylor expanded in k around inf 18.6%

      \[\leadsto \color{blue}{-10 \cdot \left(a \cdot k\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 51.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;m \leq 750000:\\ \;\;\;\;\frac{a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 8: 33.4% accurate, 9.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;m \leq 880000:\\ \;\;\;\;\frac{a}{1 + k \cdot 10}\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (a k m)
 :precision binary64
 (if (<= m 880000.0) (/ a (+ 1.0 (* k 10.0))) (* -10.0 (* a k))))

C

double code(double a, double k, double m) {
	double tmp;
	if (m <= 880000.0) {
		tmp = a / (1.0 + (k * 10.0));
	} else {
		tmp = -10.0 * (a * k);
	}
	return tmp;
}

Fortran

real(8) function code(a, k, m)
    real(8), intent (in) :: a
    real(8), intent (in) :: k
    real(8), intent (in) :: m
    real(8) :: tmp
    if (m <= 880000.0d0) then
        tmp = a / (1.0d0 + (k * 10.0d0))
    else
        tmp = (-10.0d0) * (a * k)
    end if
    code = tmp
end function

Java

public static double code(double a, double k, double m) {
	double tmp;
	if (m <= 880000.0) {
		tmp = a / (1.0 + (k * 10.0));
	} else {
		tmp = -10.0 * (a * k);
	}
	return tmp;
}

Python

def code(a, k, m):
	tmp = 0
	if m <= 880000.0:
		tmp = a / (1.0 + (k * 10.0))
	else:
		tmp = -10.0 * (a * k)
	return tmp

Julia

function code(a, k, m)
	tmp = 0.0
	if (m <= 880000.0)
		tmp = Float64(a / Float64(1.0 + Float64(k * 10.0)));
	else
		tmp = Float64(-10.0 * Float64(a * k));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, k, m)
	tmp = 0.0;
	if (m <= 880000.0)
		tmp = a / (1.0 + (k * 10.0));
	else
		tmp = -10.0 * (a * k);
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, k_, m_] := If[LessEqual[m, 880000.0], N[(a / N[(1.0 + N[(k * 10.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(-10.0 * N[(a * k), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if m < 8.8e5

   1. Initial program 95.9%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 95.9%

         \[\leadsto \color{blue}{\frac{a}{\left(1 + 10 \cdot k\right) + k \cdot k} \cdot {k}^{m}} \]

      2. sqr-neg 95.9%

         \[\leadsto \frac{a}{\left(1 + 10 \cdot k\right) + \color{blue}{\left(-k\right) \cdot \left(-k\right)}} \cdot {k}^{m} \]

      3. associate-+l+ 95.9%

         \[\leadsto \frac{a}{\color{blue}{1 + \left(10 \cdot k + \left(-k\right) \cdot \left(-k\right)\right)}} \cdot {k}^{m} \]

      4. sqr-neg 95.9%

         \[\leadsto \frac{a}{1 + \left(10 \cdot k + \color{blue}{k \cdot k}\right)} \cdot {k}^{m} \]

      5. distribute-rgt-out 95.9%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot \left(10 + k\right)}} \cdot {k}^{m} \]

   3. Simplified 95.9%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)} \cdot {k}^{m}} \]
   4. Add Preprocessing

   5. Taylor expanded in m around 0 67.9%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)}} \]

   6. Taylor expanded in k around 0 43.8%

      \[\leadsto \frac{a}{1 + \color{blue}{10 \cdot k}} \]
   7. Step-by-step derivation

      1. *-commutative 43.8%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot 10}} \]

   8. Simplified 43.8%

      \[\leadsto \frac{a}{1 + \color{blue}{k \cdot 10}} \]

   if 8.8e5 < m

   1. Initial program 77.4%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 64.3%

         \[\leadsto \color{blue}{\frac{a}{\left(1 + 10 \cdot k\right) + k \cdot k} \cdot {k}^{m}} \]

      2. sqr-neg 64.3%

         \[\leadsto \frac{a}{\left(1 + 10 \cdot k\right) + \color{blue}{\left(-k\right) \cdot \left(-k\right)}} \cdot {k}^{m} \]

      3. associate-+l+ 64.3%

         \[\leadsto \frac{a}{\color{blue}{1 + \left(10 \cdot k + \left(-k\right) \cdot \left(-k\right)\right)}} \cdot {k}^{m} \]

      4. sqr-neg 64.3%

         \[\leadsto \frac{a}{1 + \left(10 \cdot k + \color{blue}{k \cdot k}\right)} \cdot {k}^{m} \]

      5. distribute-rgt-out 64.3%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot \left(10 + k\right)}} \cdot {k}^{m} \]

   3. Simplified 64.3%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)} \cdot {k}^{m}} \]
   4. Add Preprocessing

   5. Taylor expanded in m around 0 2.8%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)}} \]

   6. Taylor expanded in k around 0 7.8%

      \[\leadsto \color{blue}{a + -10 \cdot \left(a \cdot k\right)} \]

   7. Taylor expanded in k around inf 18.6%

      \[\leadsto \color{blue}{-10 \cdot \left(a \cdot k\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 35.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;m \leq 880000:\\ \;\;\;\;\frac{a}{1 + k \cdot 10}\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 9: 24.5% accurate, 11.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;m \leq 6.8 \cdot 10^{+40}:\\ \;\;\;\;a\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (a k m) :precision binary64 (if (<= m 6.8e+40) a (* -10.0 (* a k))))

C

double code(double a, double k, double m) {
	double tmp;
	if (m <= 6.8e+40) {
		tmp = a;
	} else {
		tmp = -10.0 * (a * k);
	}
	return tmp;
}

Fortran

real(8) function code(a, k, m)
    real(8), intent (in) :: a
    real(8), intent (in) :: k
    real(8), intent (in) :: m
    real(8) :: tmp
    if (m <= 6.8d+40) then
        tmp = a
    else
        tmp = (-10.0d0) * (a * k)
    end if
    code = tmp
end function

Java

public static double code(double a, double k, double m) {
	double tmp;
	if (m <= 6.8e+40) {
		tmp = a;
	} else {
		tmp = -10.0 * (a * k);
	}
	return tmp;
}

Python

def code(a, k, m):
	tmp = 0
	if m <= 6.8e+40:
		tmp = a
	else:
		tmp = -10.0 * (a * k)
	return tmp

Julia

function code(a, k, m)
	tmp = 0.0
	if (m <= 6.8e+40)
		tmp = a;
	else
		tmp = Float64(-10.0 * Float64(a * k));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, k, m)
	tmp = 0.0;
	if (m <= 6.8e+40)
		tmp = a;
	else
		tmp = -10.0 * (a * k);
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, k_, m_] := If[LessEqual[m, 6.8e+40], a, N[(-10.0 * N[(a * k), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if m < 6.79999999999999977e40

   1. Initial program 95.0%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 94.5%

         \[\leadsto \color{blue}{\frac{a}{\left(1 + 10 \cdot k\right) + k \cdot k} \cdot {k}^{m}} \]

      2. sqr-neg 94.5%

         \[\leadsto \frac{a}{\left(1 + 10 \cdot k\right) + \color{blue}{\left(-k\right) \cdot \left(-k\right)}} \cdot {k}^{m} \]

      3. associate-+l+ 94.5%

         \[\leadsto \frac{a}{\color{blue}{1 + \left(10 \cdot k + \left(-k\right) \cdot \left(-k\right)\right)}} \cdot {k}^{m} \]

      4. sqr-neg 94.5%

         \[\leadsto \frac{a}{1 + \left(10 \cdot k + \color{blue}{k \cdot k}\right)} \cdot {k}^{m} \]

      5. distribute-rgt-out 94.5%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot \left(10 + k\right)}} \cdot {k}^{m} \]

   3. Simplified 94.5%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)} \cdot {k}^{m}} \]
   4. Add Preprocessing

   5. Taylor expanded in m around 0 64.3%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)}} \]

   6. Taylor expanded in k around 0 29.7%

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

   if 6.79999999999999977e40 < m

   1. Initial program 77.0%

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
   2. Step-by-step derivation

      1. associate-*l/ 63.5%

         \[\leadsto \color{blue}{\frac{a}{\left(1 + 10 \cdot k\right) + k \cdot k} \cdot {k}^{m}} \]

      2. sqr-neg 63.5%

         \[\leadsto \frac{a}{\left(1 + 10 \cdot k\right) + \color{blue}{\left(-k\right) \cdot \left(-k\right)}} \cdot {k}^{m} \]

      3. associate-+l+ 63.5%

         \[\leadsto \frac{a}{\color{blue}{1 + \left(10 \cdot k + \left(-k\right) \cdot \left(-k\right)\right)}} \cdot {k}^{m} \]

      4. sqr-neg 63.5%

         \[\leadsto \frac{a}{1 + \left(10 \cdot k + \color{blue}{k \cdot k}\right)} \cdot {k}^{m} \]

      5. distribute-rgt-out 63.5%

         \[\leadsto \frac{a}{1 + \color{blue}{k \cdot \left(10 + k\right)}} \cdot {k}^{m} \]

   3. Simplified 63.5%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)} \cdot {k}^{m}} \]
   4. Add Preprocessing

   5. Taylor expanded in m around 0 2.9%

      \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)}} \]

   6. Taylor expanded in k around 0 8.6%

      \[\leadsto \color{blue}{a + -10 \cdot \left(a \cdot k\right)} \]

   7. Taylor expanded in k around inf 20.7%

      \[\leadsto \color{blue}{-10 \cdot \left(a \cdot k\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 27.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;m \leq 6.8 \cdot 10^{+40}:\\ \;\;\;\;a\\ \mathbf{else}:\\ \;\;\;\;-10 \cdot \left(a \cdot k\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 10: 19.7% accurate, 114.0× speedup

Math

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

FPCore

(FPCore (a k m) :precision binary64 a)

C

double code(double a, double k, double m) {
	return a;
}

Fortran

real(8) function code(a, k, m)
    real(8), intent (in) :: a
    real(8), intent (in) :: k
    real(8), intent (in) :: m
    code = a
end function

Java

public static double code(double a, double k, double m) {
	return a;
}

Python

def code(a, k, m):
	return a

Julia

function code(a, k, m)
	return a
end

MATLAB

function tmp = code(a, k, m)
	tmp = a;
end

Wolfram

code[a_, k_, m_] := a

Derivation

1. Initial program 89.8%

   \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k} \]
2. Step-by-step derivation

   1. associate-*l/ 85.5%

      \[\leadsto \color{blue}{\frac{a}{\left(1 + 10 \cdot k\right) + k \cdot k} \cdot {k}^{m}} \]

   2. sqr-neg 85.5%

      \[\leadsto \frac{a}{\left(1 + 10 \cdot k\right) + \color{blue}{\left(-k\right) \cdot \left(-k\right)}} \cdot {k}^{m} \]

   3. associate-+l+ 85.5%

      \[\leadsto \frac{a}{\color{blue}{1 + \left(10 \cdot k + \left(-k\right) \cdot \left(-k\right)\right)}} \cdot {k}^{m} \]

   4. sqr-neg 85.5%

      \[\leadsto \frac{a}{1 + \left(10 \cdot k + \color{blue}{k \cdot k}\right)} \cdot {k}^{m} \]

   5. distribute-rgt-out 85.5%

      \[\leadsto \frac{a}{1 + \color{blue}{k \cdot \left(10 + k\right)}} \cdot {k}^{m} \]

3. Simplified 85.5%

   \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)} \cdot {k}^{m}} \]
4. Add Preprocessing

5. Taylor expanded in m around 0 46.5%

   \[\leadsto \color{blue}{\frac{a}{1 + k \cdot \left(10 + k\right)}} \]

6. Taylor expanded in k around 0 22.1%

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

7. Final simplification 22.1%

   \[\leadsto a \]
8. Add Preprocessing

Reproduce

herbie shell --seed 2024024 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))