2nthrt (problem 3.4.6)

Percentage Accurate: 53.8% → 85.1%

Time: 40.7s

Alternatives: 17

Speedup: 1.7×

• could not determine a ground truth

  1. x = -2
  2. n = 4.94066e-324

Specification

Math

\[\begin{array}{l} \\ {\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))

C

double code(double x, double n) {
	return pow((x + 1.0), (1.0 / n)) - pow(x, (1.0 / n));
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    code = ((x + 1.0d0) ** (1.0d0 / n)) - (x ** (1.0d0 / n))
end function

Java

public static double code(double x, double n) {
	return Math.pow((x + 1.0), (1.0 / n)) - Math.pow(x, (1.0 / n));
}

Python

def code(x, n):
	return math.pow((x + 1.0), (1.0 / n)) - math.pow(x, (1.0 / n))

Julia

function code(x, n)
	return Float64((Float64(x + 1.0) ^ Float64(1.0 / n)) - (x ^ Float64(1.0 / n)))
end

MATLAB

function tmp = code(x, n)
	tmp = ((x + 1.0) ^ (1.0 / n)) - (x ^ (1.0 / n));
end

Wolfram

code[x_, n_] := N[(N[Power[N[(x + 1.0), $MachinePrecision], N[(1.0 / n), $MachinePrecision]], $MachinePrecision] - N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]

Initial Program: 53.8% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ {\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))

C

double code(double x, double n) {
	return pow((x + 1.0), (1.0 / n)) - pow(x, (1.0 / n));
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    code = ((x + 1.0d0) ** (1.0d0 / n)) - (x ** (1.0d0 / n))
end function

Java

public static double code(double x, double n) {
	return Math.pow((x + 1.0), (1.0 / n)) - Math.pow(x, (1.0 / n));
}

Python

def code(x, n):
	return math.pow((x + 1.0), (1.0 / n)) - math.pow(x, (1.0 / n))

Julia

function code(x, n)
	return Float64((Float64(x + 1.0) ^ Float64(1.0 / n)) - (x ^ Float64(1.0 / n)))
end

MATLAB

function tmp = code(x, n)
	tmp = ((x + 1.0) ^ (1.0 / n)) - (x ^ (1.0 / n));
end

Wolfram

code[x_, n_] := N[(N[Power[N[(x + 1.0), $MachinePrecision], N[(1.0 / n), $MachinePrecision]], $MachinePrecision] - N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]

Alternative 1: 85.1% accurate, 0.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{\sqrt[3]{{x}^{\left(\frac{3}{n}\right)}}}{n \cdot x}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-5}:\\ \;\;\;\;t\_0\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-153}:\\ \;\;\;\;\frac{\left(\mathsf{log1p}\left(x\right) + \frac{0.5 \cdot \left({\left(\mathsf{log1p}\left(x\right)\right)}^{2} - {\log x}^{2}\right) + 0.16666666666666666 \cdot \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{3} - {\log x}^{3}}{n}}{n}\right) - \log x}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5000:\\ \;\;\;\;t\_0\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{x}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (/ (cbrt (pow x (/ 3.0 n))) (* n x))))
   (if (<= (/ 1.0 n) -2e-5)
     t_0
     (if (<= (/ 1.0 n) 5e-153)
       (/
        (-
         (+
          (log1p x)
          (/
           (+
            (* 0.5 (- (pow (log1p x) 2.0) (pow (log x) 2.0)))
            (*
             0.16666666666666666
             (/ (- (pow (log1p x) 3.0) (pow (log x) 3.0)) n)))
           n))
         (log x))
        n)
       (if (<= (/ 1.0 n) 5000.0) t_0 (- (exp (/ x n)) (pow x (/ 1.0 n))))))))

C

double code(double x, double n) {
	double t_0 = cbrt(pow(x, (3.0 / n))) / (n * x);
	double tmp;
	if ((1.0 / n) <= -2e-5) {
		tmp = t_0;
	} else if ((1.0 / n) <= 5e-153) {
		tmp = ((log1p(x) + (((0.5 * (pow(log1p(x), 2.0) - pow(log(x), 2.0))) + (0.16666666666666666 * ((pow(log1p(x), 3.0) - pow(log(x), 3.0)) / n))) / n)) - log(x)) / n;
	} else if ((1.0 / n) <= 5000.0) {
		tmp = t_0;
	} else {
		tmp = exp((x / n)) - pow(x, (1.0 / n));
	}
	return tmp;
}

Java

public static double code(double x, double n) {
	double t_0 = Math.cbrt(Math.pow(x, (3.0 / n))) / (n * x);
	double tmp;
	if ((1.0 / n) <= -2e-5) {
		tmp = t_0;
	} else if ((1.0 / n) <= 5e-153) {
		tmp = ((Math.log1p(x) + (((0.5 * (Math.pow(Math.log1p(x), 2.0) - Math.pow(Math.log(x), 2.0))) + (0.16666666666666666 * ((Math.pow(Math.log1p(x), 3.0) - Math.pow(Math.log(x), 3.0)) / n))) / n)) - Math.log(x)) / n;
	} else if ((1.0 / n) <= 5000.0) {
		tmp = t_0;
	} else {
		tmp = Math.exp((x / n)) - Math.pow(x, (1.0 / n));
	}
	return tmp;
}

Julia

function code(x, n)
	t_0 = Float64(cbrt((x ^ Float64(3.0 / n))) / Float64(n * x))
	tmp = 0.0
	if (Float64(1.0 / n) <= -2e-5)
		tmp = t_0;
	elseif (Float64(1.0 / n) <= 5e-153)
		tmp = Float64(Float64(Float64(log1p(x) + Float64(Float64(Float64(0.5 * Float64((log1p(x) ^ 2.0) - (log(x) ^ 2.0))) + Float64(0.16666666666666666 * Float64(Float64((log1p(x) ^ 3.0) - (log(x) ^ 3.0)) / n))) / n)) - log(x)) / n);
	elseif (Float64(1.0 / n) <= 5000.0)
		tmp = t_0;
	else
		tmp = Float64(exp(Float64(x / n)) - (x ^ Float64(1.0 / n)));
	end
	return tmp
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[(N[Power[N[Power[x, N[(3.0 / n), $MachinePrecision]], $MachinePrecision], 1/3], $MachinePrecision] / N[(n * x), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-5], t$95$0, If[LessEqual[N[(1.0 / n), $MachinePrecision], 5e-153], N[(N[(N[(N[Log[1 + x], $MachinePrecision] + N[(N[(N[(0.5 * N[(N[Power[N[Log[1 + x], $MachinePrecision], 2.0], $MachinePrecision] - N[Power[N[Log[x], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(0.16666666666666666 * N[(N[(N[Power[N[Log[1 + x], $MachinePrecision], 3.0], $MachinePrecision] - N[Power[N[Log[x], $MachinePrecision], 3.0], $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision]), $MachinePrecision] - N[Log[x], $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 5000.0], t$95$0, N[(N[Exp[N[(x / n), $MachinePrecision]], $MachinePrecision] - N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 #s(literal 1 binary64) n) < -2.00000000000000016e-5 or 5.00000000000000033e-153 < (/.f64 #s(literal 1 binary64) n) < 5e3

   1. Initial program 82.2%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in x around inf 94.3%

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
   4. Step-by-step derivation

      1. mul-1-neg 94.3%

         \[\leadsto \frac{e^{\color{blue}{-\frac{\log \left(\frac{1}{x}\right)}{n}}}}{n \cdot x} \]

      2. log-rec 94.3%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]

      3. mul-1-neg 94.3%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]

      4. distribute-neg-frac 94.3%

         \[\leadsto \frac{e^{\color{blue}{\frac{--1 \cdot \log x}{n}}}}{n \cdot x} \]

      5. mul-1-neg 94.3%

         \[\leadsto \frac{e^{\frac{-\color{blue}{\left(-\log x\right)}}{n}}}{n \cdot x} \]

      6. remove-double-neg 94.3%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]

      7. *-rgt-identity 94.3%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x \cdot 1}}{n}}}{n \cdot x} \]

      8. associate-/l* 94.3%

         \[\leadsto \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{n \cdot x} \]

      9. exp-to-pow 94.3%

         \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]

      10. *-commutative 94.3%

          \[\leadsto \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{x \cdot n}} \]

   5. Simplified 94.3%

      \[\leadsto \color{blue}{\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}} \]
   6. Step-by-step derivation

      1. add-cbrt-cube 94.3%

         \[\leadsto \frac{\color{blue}{\sqrt[3]{\left({x}^{\left(\frac{1}{n}\right)} \cdot {x}^{\left(\frac{1}{n}\right)}\right) \cdot {x}^{\left(\frac{1}{n}\right)}}}}{x \cdot n} \]

      2. pow3 94.3%

         \[\leadsto \frac{\sqrt[3]{\color{blue}{{\left({x}^{\left(\frac{1}{n}\right)}\right)}^{3}}}}{x \cdot n} \]

      3. pow-pow 94.3%

         \[\leadsto \frac{\sqrt[3]{\color{blue}{{x}^{\left(\frac{1}{n} \cdot 3\right)}}}}{x \cdot n} \]

   7. Applied egg-rr 94.3%

      \[\leadsto \frac{\color{blue}{\sqrt[3]{{x}^{\left(\frac{1}{n} \cdot 3\right)}}}}{x \cdot n} \]
   8. Step-by-step derivation

      1. associate-*l/ 94.3%

         \[\leadsto \frac{\sqrt[3]{{x}^{\color{blue}{\left(\frac{1 \cdot 3}{n}\right)}}}}{x \cdot n} \]

      2. metadata-eval 94.3%

         \[\leadsto \frac{\sqrt[3]{{x}^{\left(\frac{\color{blue}{3}}{n}\right)}}}{x \cdot n} \]

   9. Simplified 94.3%

      \[\leadsto \frac{\color{blue}{\sqrt[3]{{x}^{\left(\frac{3}{n}\right)}}}}{x \cdot n} \]

   if -2.00000000000000016e-5 < (/.f64 #s(literal 1 binary64) n) < 5.00000000000000033e-153

   1. Initial program 39.0%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around -inf 86.6%

      \[\leadsto \color{blue}{-1 \cdot \frac{\left(-1 \cdot \log \left(1 + x\right) + -1 \cdot \frac{\left(-1 \cdot \frac{-0.16666666666666666 \cdot {\log \left(1 + x\right)}^{3} - -0.16666666666666666 \cdot {\log x}^{3}}{n} + 0.5 \cdot {\log \left(1 + x\right)}^{2}\right) - 0.5 \cdot {\log x}^{2}}{n}\right) - -1 \cdot \log x}{n}} \]

   5. Simplified 86.6%

      \[\leadsto \color{blue}{\frac{\log x - \left(\mathsf{log1p}\left(x\right) + \frac{0.5 \cdot \left({\left(\mathsf{log1p}\left(x\right)\right)}^{2} - {\log x}^{2}\right) + 0.16666666666666666 \cdot \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{3} - {\log x}^{3}}{n}}{n}\right)}{-n}} \]

   if 5e3 < (/.f64 #s(literal 1 binary64) n)

   1. Initial program 58.4%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around 0 58.4%

      \[\leadsto \color{blue}{e^{\frac{\log \left(1 + x\right)}{n}} - e^{\frac{\log x}{n}}} \]
   4. Step-by-step derivation

      1. log1p-define 100.0%

         \[\leadsto e^{\frac{\color{blue}{\mathsf{log1p}\left(x\right)}}{n}} - e^{\frac{\log x}{n}} \]

      2. *-rgt-identity 100.0%

         \[\leadsto e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - e^{\frac{\color{blue}{\log x \cdot 1}}{n}} \]

      3. associate-/l* 100.0%

         \[\leadsto e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - e^{\color{blue}{\log x \cdot \frac{1}{n}}} \]

      4. exp-to-pow 100.0%

         \[\leadsto e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - \color{blue}{{x}^{\left(\frac{1}{n}\right)}} \]

   5. Simplified 100.0%

      \[\leadsto \color{blue}{e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}} \]

   6. Taylor expanded in x around 0 100.0%

      \[\leadsto e^{\color{blue}{\frac{x}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
3. Recombined 3 regimes into one program.

4. Final simplification 92.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-5}:\\ \;\;\;\;\frac{\sqrt[3]{{x}^{\left(\frac{3}{n}\right)}}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-153}:\\ \;\;\;\;\frac{\left(\mathsf{log1p}\left(x\right) + \frac{0.5 \cdot \left({\left(\mathsf{log1p}\left(x\right)\right)}^{2} - {\log x}^{2}\right) + 0.16666666666666666 \cdot \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{3} - {\log x}^{3}}{n}}{n}\right) - \log x}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5000:\\ \;\;\;\;\frac{\sqrt[3]{{x}^{\left(\frac{3}{n}\right)}}}{n \cdot x}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{x}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 85.0% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{\sqrt[3]{{x}^{\left(\frac{3}{n}\right)}}}{n \cdot x}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-5}:\\ \;\;\;\;t\_0\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-153}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(x\right) + \left(0.5 \cdot \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2} - {\log x}^{2}}{n} - \log x\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5000:\\ \;\;\;\;t\_0\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{x}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (/ (cbrt (pow x (/ 3.0 n))) (* n x))))
   (if (<= (/ 1.0 n) -2e-5)
     t_0
     (if (<= (/ 1.0 n) 5e-153)
       (/
        (+
         (log1p x)
         (- (* 0.5 (/ (- (pow (log1p x) 2.0) (pow (log x) 2.0)) n)) (log x)))
        n)
       (if (<= (/ 1.0 n) 5000.0) t_0 (- (exp (/ x n)) (pow x (/ 1.0 n))))))))

C

double code(double x, double n) {
	double t_0 = cbrt(pow(x, (3.0 / n))) / (n * x);
	double tmp;
	if ((1.0 / n) <= -2e-5) {
		tmp = t_0;
	} else if ((1.0 / n) <= 5e-153) {
		tmp = (log1p(x) + ((0.5 * ((pow(log1p(x), 2.0) - pow(log(x), 2.0)) / n)) - log(x))) / n;
	} else if ((1.0 / n) <= 5000.0) {
		tmp = t_0;
	} else {
		tmp = exp((x / n)) - pow(x, (1.0 / n));
	}
	return tmp;
}

Java

public static double code(double x, double n) {
	double t_0 = Math.cbrt(Math.pow(x, (3.0 / n))) / (n * x);
	double tmp;
	if ((1.0 / n) <= -2e-5) {
		tmp = t_0;
	} else if ((1.0 / n) <= 5e-153) {
		tmp = (Math.log1p(x) + ((0.5 * ((Math.pow(Math.log1p(x), 2.0) - Math.pow(Math.log(x), 2.0)) / n)) - Math.log(x))) / n;
	} else if ((1.0 / n) <= 5000.0) {
		tmp = t_0;
	} else {
		tmp = Math.exp((x / n)) - Math.pow(x, (1.0 / n));
	}
	return tmp;
}

Julia

function code(x, n)
	t_0 = Float64(cbrt((x ^ Float64(3.0 / n))) / Float64(n * x))
	tmp = 0.0
	if (Float64(1.0 / n) <= -2e-5)
		tmp = t_0;
	elseif (Float64(1.0 / n) <= 5e-153)
		tmp = Float64(Float64(log1p(x) + Float64(Float64(0.5 * Float64(Float64((log1p(x) ^ 2.0) - (log(x) ^ 2.0)) / n)) - log(x))) / n);
	elseif (Float64(1.0 / n) <= 5000.0)
		tmp = t_0;
	else
		tmp = Float64(exp(Float64(x / n)) - (x ^ Float64(1.0 / n)));
	end
	return tmp
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[(N[Power[N[Power[x, N[(3.0 / n), $MachinePrecision]], $MachinePrecision], 1/3], $MachinePrecision] / N[(n * x), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-5], t$95$0, If[LessEqual[N[(1.0 / n), $MachinePrecision], 5e-153], N[(N[(N[Log[1 + x], $MachinePrecision] + N[(N[(0.5 * N[(N[(N[Power[N[Log[1 + x], $MachinePrecision], 2.0], $MachinePrecision] - N[Power[N[Log[x], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision]), $MachinePrecision] - N[Log[x], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 5000.0], t$95$0, N[(N[Exp[N[(x / n), $MachinePrecision]], $MachinePrecision] - N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 #s(literal 1 binary64) n) < -2.00000000000000016e-5 or 5.00000000000000033e-153 < (/.f64 #s(literal 1 binary64) n) < 5e3

   1. Initial program 82.2%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in x around inf 94.3%

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
   4. Step-by-step derivation

      1. mul-1-neg 94.3%

         \[\leadsto \frac{e^{\color{blue}{-\frac{\log \left(\frac{1}{x}\right)}{n}}}}{n \cdot x} \]

      2. log-rec 94.3%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]

      3. mul-1-neg 94.3%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]

      4. distribute-neg-frac 94.3%

         \[\leadsto \frac{e^{\color{blue}{\frac{--1 \cdot \log x}{n}}}}{n \cdot x} \]

      5. mul-1-neg 94.3%

         \[\leadsto \frac{e^{\frac{-\color{blue}{\left(-\log x\right)}}{n}}}{n \cdot x} \]

      6. remove-double-neg 94.3%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]

      7. *-rgt-identity 94.3%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x \cdot 1}}{n}}}{n \cdot x} \]

      8. associate-/l* 94.3%

         \[\leadsto \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{n \cdot x} \]

      9. exp-to-pow 94.3%

         \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]

      10. *-commutative 94.3%

          \[\leadsto \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{x \cdot n}} \]

   5. Simplified 94.3%

      \[\leadsto \color{blue}{\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}} \]
   6. Step-by-step derivation

      1. add-cbrt-cube 94.3%

         \[\leadsto \frac{\color{blue}{\sqrt[3]{\left({x}^{\left(\frac{1}{n}\right)} \cdot {x}^{\left(\frac{1}{n}\right)}\right) \cdot {x}^{\left(\frac{1}{n}\right)}}}}{x \cdot n} \]

      2. pow3 94.3%

         \[\leadsto \frac{\sqrt[3]{\color{blue}{{\left({x}^{\left(\frac{1}{n}\right)}\right)}^{3}}}}{x \cdot n} \]

      3. pow-pow 94.3%

         \[\leadsto \frac{\sqrt[3]{\color{blue}{{x}^{\left(\frac{1}{n} \cdot 3\right)}}}}{x \cdot n} \]

   7. Applied egg-rr 94.3%

      \[\leadsto \frac{\color{blue}{\sqrt[3]{{x}^{\left(\frac{1}{n} \cdot 3\right)}}}}{x \cdot n} \]
   8. Step-by-step derivation

      1. associate-*l/ 94.3%

         \[\leadsto \frac{\sqrt[3]{{x}^{\color{blue}{\left(\frac{1 \cdot 3}{n}\right)}}}}{x \cdot n} \]

      2. metadata-eval 94.3%

         \[\leadsto \frac{\sqrt[3]{{x}^{\left(\frac{\color{blue}{3}}{n}\right)}}}{x \cdot n} \]

   9. Simplified 94.3%

      \[\leadsto \frac{\color{blue}{\sqrt[3]{{x}^{\left(\frac{3}{n}\right)}}}}{x \cdot n} \]

   if -2.00000000000000016e-5 < (/.f64 #s(literal 1 binary64) n) < 5.00000000000000033e-153

   1. Initial program 39.0%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 86.6%

      \[\leadsto \color{blue}{\frac{\left(\log \left(1 + x\right) + 0.5 \cdot \frac{{\log \left(1 + x\right)}^{2}}{n}\right) - \left(\log x + 0.5 \cdot \frac{{\log x}^{2}}{n}\right)}{n}} \]
   4. Step-by-step derivation

   5. Simplified 86.6%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) + \left(0.5 \cdot \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2} - {\log x}^{2}}{n} - \log x\right)}{n}} \]

   if 5e3 < (/.f64 #s(literal 1 binary64) n)

   1. Initial program 58.4%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around 0 58.4%

      \[\leadsto \color{blue}{e^{\frac{\log \left(1 + x\right)}{n}} - e^{\frac{\log x}{n}}} \]
   4. Step-by-step derivation

      1. log1p-define 100.0%

         \[\leadsto e^{\frac{\color{blue}{\mathsf{log1p}\left(x\right)}}{n}} - e^{\frac{\log x}{n}} \]

      2. *-rgt-identity 100.0%

         \[\leadsto e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - e^{\frac{\color{blue}{\log x \cdot 1}}{n}} \]

      3. associate-/l* 100.0%

         \[\leadsto e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - e^{\color{blue}{\log x \cdot \frac{1}{n}}} \]

      4. exp-to-pow 100.0%

         \[\leadsto e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - \color{blue}{{x}^{\left(\frac{1}{n}\right)}} \]

   5. Simplified 100.0%

      \[\leadsto \color{blue}{e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}} \]

   6. Taylor expanded in x around 0 100.0%

      \[\leadsto e^{\color{blue}{\frac{x}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
3. Recombined 3 regimes into one program.

4. Final simplification 92.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-5}:\\ \;\;\;\;\frac{\sqrt[3]{{x}^{\left(\frac{3}{n}\right)}}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-153}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(x\right) + \left(0.5 \cdot \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2} - {\log x}^{2}}{n} - \log x\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5000:\\ \;\;\;\;\frac{\sqrt[3]{{x}^{\left(\frac{3}{n}\right)}}}{n \cdot x}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{x}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 84.9% accurate, 0.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{\sqrt[3]{{x}^{\left(\frac{3}{n}\right)}}}{n \cdot x}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-5}:\\ \;\;\;\;t\_0\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-153}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5000:\\ \;\;\;\;t\_0\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{x}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (/ (cbrt (pow x (/ 3.0 n))) (* n x))))
   (if (<= (/ 1.0 n) -2e-5)
     t_0
     (if (<= (/ 1.0 n) 5e-153)
       (/ (log (/ (+ 1.0 x) x)) n)
       (if (<= (/ 1.0 n) 5000.0) t_0 (- (exp (/ x n)) (pow x (/ 1.0 n))))))))

C

double code(double x, double n) {
	double t_0 = cbrt(pow(x, (3.0 / n))) / (n * x);
	double tmp;
	if ((1.0 / n) <= -2e-5) {
		tmp = t_0;
	} else if ((1.0 / n) <= 5e-153) {
		tmp = log(((1.0 + x) / x)) / n;
	} else if ((1.0 / n) <= 5000.0) {
		tmp = t_0;
	} else {
		tmp = exp((x / n)) - pow(x, (1.0 / n));
	}
	return tmp;
}

Java

public static double code(double x, double n) {
	double t_0 = Math.cbrt(Math.pow(x, (3.0 / n))) / (n * x);
	double tmp;
	if ((1.0 / n) <= -2e-5) {
		tmp = t_0;
	} else if ((1.0 / n) <= 5e-153) {
		tmp = Math.log(((1.0 + x) / x)) / n;
	} else if ((1.0 / n) <= 5000.0) {
		tmp = t_0;
	} else {
		tmp = Math.exp((x / n)) - Math.pow(x, (1.0 / n));
	}
	return tmp;
}

Julia

function code(x, n)
	t_0 = Float64(cbrt((x ^ Float64(3.0 / n))) / Float64(n * x))
	tmp = 0.0
	if (Float64(1.0 / n) <= -2e-5)
		tmp = t_0;
	elseif (Float64(1.0 / n) <= 5e-153)
		tmp = Float64(log(Float64(Float64(1.0 + x) / x)) / n);
	elseif (Float64(1.0 / n) <= 5000.0)
		tmp = t_0;
	else
		tmp = Float64(exp(Float64(x / n)) - (x ^ Float64(1.0 / n)));
	end
	return tmp
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[(N[Power[N[Power[x, N[(3.0 / n), $MachinePrecision]], $MachinePrecision], 1/3], $MachinePrecision] / N[(n * x), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-5], t$95$0, If[LessEqual[N[(1.0 / n), $MachinePrecision], 5e-153], N[(N[Log[N[(N[(1.0 + x), $MachinePrecision] / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 5000.0], t$95$0, N[(N[Exp[N[(x / n), $MachinePrecision]], $MachinePrecision] - N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 #s(literal 1 binary64) n) < -2.00000000000000016e-5 or 5.00000000000000033e-153 < (/.f64 #s(literal 1 binary64) n) < 5e3

   1. Initial program 82.2%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in x around inf 94.3%

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
   4. Step-by-step derivation

      1. mul-1-neg 94.3%

         \[\leadsto \frac{e^{\color{blue}{-\frac{\log \left(\frac{1}{x}\right)}{n}}}}{n \cdot x} \]

      2. log-rec 94.3%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]

      3. mul-1-neg 94.3%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]

      4. distribute-neg-frac 94.3%

         \[\leadsto \frac{e^{\color{blue}{\frac{--1 \cdot \log x}{n}}}}{n \cdot x} \]

      5. mul-1-neg 94.3%

         \[\leadsto \frac{e^{\frac{-\color{blue}{\left(-\log x\right)}}{n}}}{n \cdot x} \]

      6. remove-double-neg 94.3%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]

      7. *-rgt-identity 94.3%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x \cdot 1}}{n}}}{n \cdot x} \]

      8. associate-/l* 94.3%

         \[\leadsto \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{n \cdot x} \]

      9. exp-to-pow 94.3%

         \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]

      10. *-commutative 94.3%

          \[\leadsto \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{x \cdot n}} \]

   5. Simplified 94.3%

      \[\leadsto \color{blue}{\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}} \]
   6. Step-by-step derivation

      1. add-cbrt-cube 94.3%

         \[\leadsto \frac{\color{blue}{\sqrt[3]{\left({x}^{\left(\frac{1}{n}\right)} \cdot {x}^{\left(\frac{1}{n}\right)}\right) \cdot {x}^{\left(\frac{1}{n}\right)}}}}{x \cdot n} \]

      2. pow3 94.3%

         \[\leadsto \frac{\sqrt[3]{\color{blue}{{\left({x}^{\left(\frac{1}{n}\right)}\right)}^{3}}}}{x \cdot n} \]

      3. pow-pow 94.3%

         \[\leadsto \frac{\sqrt[3]{\color{blue}{{x}^{\left(\frac{1}{n} \cdot 3\right)}}}}{x \cdot n} \]

   7. Applied egg-rr 94.3%

      \[\leadsto \frac{\color{blue}{\sqrt[3]{{x}^{\left(\frac{1}{n} \cdot 3\right)}}}}{x \cdot n} \]
   8. Step-by-step derivation

      1. associate-*l/ 94.3%

         \[\leadsto \frac{\sqrt[3]{{x}^{\color{blue}{\left(\frac{1 \cdot 3}{n}\right)}}}}{x \cdot n} \]

      2. metadata-eval 94.3%

         \[\leadsto \frac{\sqrt[3]{{x}^{\left(\frac{\color{blue}{3}}{n}\right)}}}{x \cdot n} \]

   9. Simplified 94.3%

      \[\leadsto \frac{\color{blue}{\sqrt[3]{{x}^{\left(\frac{3}{n}\right)}}}}{x \cdot n} \]

   if -2.00000000000000016e-5 < (/.f64 #s(literal 1 binary64) n) < 5.00000000000000033e-153

   1. Initial program 39.0%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 85.3%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   4. Step-by-step derivation

      1. log1p-define 85.3%

         \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   5. Simplified 85.3%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   6. Step-by-step derivation

      1. log1p-undefine 85.3%

         \[\leadsto \frac{\color{blue}{\log \left(1 + x\right)} - \log x}{n} \]

      2. diff-log 85.6%

         \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]

   7. Applied egg-rr 85.6%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]
   8. Step-by-step derivation

      1. +-commutative 85.6%

         \[\leadsto \frac{\log \left(\frac{\color{blue}{x + 1}}{x}\right)}{n} \]

   9. Simplified 85.6%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{x + 1}{x}\right)}}{n} \]

   if 5e3 < (/.f64 #s(literal 1 binary64) n)

   1. Initial program 58.4%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around 0 58.4%

      \[\leadsto \color{blue}{e^{\frac{\log \left(1 + x\right)}{n}} - e^{\frac{\log x}{n}}} \]
   4. Step-by-step derivation

      1. log1p-define 100.0%

         \[\leadsto e^{\frac{\color{blue}{\mathsf{log1p}\left(x\right)}}{n}} - e^{\frac{\log x}{n}} \]

      2. *-rgt-identity 100.0%

         \[\leadsto e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - e^{\frac{\color{blue}{\log x \cdot 1}}{n}} \]

      3. associate-/l* 100.0%

         \[\leadsto e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - e^{\color{blue}{\log x \cdot \frac{1}{n}}} \]

      4. exp-to-pow 100.0%

         \[\leadsto e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - \color{blue}{{x}^{\left(\frac{1}{n}\right)}} \]

   5. Simplified 100.0%

      \[\leadsto \color{blue}{e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}} \]

   6. Taylor expanded in x around 0 100.0%

      \[\leadsto e^{\color{blue}{\frac{x}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
3. Recombined 3 regimes into one program.

4. Final simplification 92.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-5}:\\ \;\;\;\;\frac{\sqrt[3]{{x}^{\left(\frac{3}{n}\right)}}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-153}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5000:\\ \;\;\;\;\frac{\sqrt[3]{{x}^{\left(\frac{3}{n}\right)}}}{n \cdot x}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{x}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 85.0% accurate, 0.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-5}:\\ \;\;\;\;\frac{t\_0}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-153}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5000:\\ \;\;\;\;\frac{e^{\frac{\log x}{n}}}{n \cdot x}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{x}{n}} - t\_0\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))))
   (if (<= (/ 1.0 n) -2e-5)
     (/ t_0 (* n x))
     (if (<= (/ 1.0 n) 5e-153)
       (/ (log (/ (+ 1.0 x) x)) n)
       (if (<= (/ 1.0 n) 5000.0)
         (/ (exp (/ (log x) n)) (* n x))
         (- (exp (/ x n)) t_0))))))

C

double code(double x, double n) {
	double t_0 = pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-5) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 5e-153) {
		tmp = log(((1.0 + x) / x)) / n;
	} else if ((1.0 / n) <= 5000.0) {
		tmp = exp((log(x) / n)) / (n * x);
	} else {
		tmp = exp((x / n)) - t_0;
	}
	return tmp;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    real(8) :: t_0
    real(8) :: tmp
    t_0 = x ** (1.0d0 / n)
    if ((1.0d0 / n) <= (-2d-5)) then
        tmp = t_0 / (n * x)
    else if ((1.0d0 / n) <= 5d-153) then
        tmp = log(((1.0d0 + x) / x)) / n
    else if ((1.0d0 / n) <= 5000.0d0) then
        tmp = exp((log(x) / n)) / (n * x)
    else
        tmp = exp((x / n)) - t_0
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double t_0 = Math.pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-5) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 5e-153) {
		tmp = Math.log(((1.0 + x) / x)) / n;
	} else if ((1.0 / n) <= 5000.0) {
		tmp = Math.exp((Math.log(x) / n)) / (n * x);
	} else {
		tmp = Math.exp((x / n)) - t_0;
	}
	return tmp;
}

Python

def code(x, n):
	t_0 = math.pow(x, (1.0 / n))
	tmp = 0
	if (1.0 / n) <= -2e-5:
		tmp = t_0 / (n * x)
	elif (1.0 / n) <= 5e-153:
		tmp = math.log(((1.0 + x) / x)) / n
	elif (1.0 / n) <= 5000.0:
		tmp = math.exp((math.log(x) / n)) / (n * x)
	else:
		tmp = math.exp((x / n)) - t_0
	return tmp

Julia

function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	tmp = 0.0
	if (Float64(1.0 / n) <= -2e-5)
		tmp = Float64(t_0 / Float64(n * x));
	elseif (Float64(1.0 / n) <= 5e-153)
		tmp = Float64(log(Float64(Float64(1.0 + x) / x)) / n);
	elseif (Float64(1.0 / n) <= 5000.0)
		tmp = Float64(exp(Float64(log(x) / n)) / Float64(n * x));
	else
		tmp = Float64(exp(Float64(x / n)) - t_0);
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	t_0 = x ^ (1.0 / n);
	tmp = 0.0;
	if ((1.0 / n) <= -2e-5)
		tmp = t_0 / (n * x);
	elseif ((1.0 / n) <= 5e-153)
		tmp = log(((1.0 + x) / x)) / n;
	elseif ((1.0 / n) <= 5000.0)
		tmp = exp((log(x) / n)) / (n * x);
	else
		tmp = exp((x / n)) - t_0;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-5], N[(t$95$0 / N[(n * x), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 5e-153], N[(N[Log[N[(N[(1.0 + x), $MachinePrecision] / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 5000.0], N[(N[Exp[N[(N[Log[x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision] / N[(n * x), $MachinePrecision]), $MachinePrecision], N[(N[Exp[N[(x / n), $MachinePrecision]], $MachinePrecision] - t$95$0), $MachinePrecision]]]]]

Derivation

1. Split input into 4 regimes

2. if (/.f64 #s(literal 1 binary64) n) < -2.00000000000000016e-5

   1. Initial program 97.9%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in x around inf 99.0%

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
   4. Step-by-step derivation

      1. mul-1-neg 99.0%

         \[\leadsto \frac{e^{\color{blue}{-\frac{\log \left(\frac{1}{x}\right)}{n}}}}{n \cdot x} \]

      2. log-rec 99.0%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]

      3. mul-1-neg 99.0%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]

      4. distribute-neg-frac 99.0%

         \[\leadsto \frac{e^{\color{blue}{\frac{--1 \cdot \log x}{n}}}}{n \cdot x} \]

      5. mul-1-neg 99.0%

         \[\leadsto \frac{e^{\frac{-\color{blue}{\left(-\log x\right)}}{n}}}{n \cdot x} \]

      6. remove-double-neg 99.0%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]

      7. *-rgt-identity 99.0%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x \cdot 1}}{n}}}{n \cdot x} \]

      8. associate-/l* 99.0%

         \[\leadsto \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{n \cdot x} \]

      9. exp-to-pow 99.0%

         \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]

      10. *-commutative 99.0%

          \[\leadsto \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{x \cdot n}} \]

   5. Simplified 99.0%

      \[\leadsto \color{blue}{\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}} \]

   if -2.00000000000000016e-5 < (/.f64 #s(literal 1 binary64) n) < 5.00000000000000033e-153

   1. Initial program 39.0%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 85.3%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   4. Step-by-step derivation

      1. log1p-define 85.3%

         \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   5. Simplified 85.3%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   6. Step-by-step derivation

      1. log1p-undefine 85.3%

         \[\leadsto \frac{\color{blue}{\log \left(1 + x\right)} - \log x}{n} \]

      2. diff-log 85.6%

         \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]

   7. Applied egg-rr 85.6%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]
   8. Step-by-step derivation

      1. +-commutative 85.6%

         \[\leadsto \frac{\log \left(\frac{\color{blue}{x + 1}}{x}\right)}{n} \]

   9. Simplified 85.6%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{x + 1}{x}\right)}}{n} \]

   if 5.00000000000000033e-153 < (/.f64 #s(literal 1 binary64) n) < 5e3

   1. Initial program 21.1%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in x around inf 76.1%

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
   4. Step-by-step derivation

      1. mul-1-neg 76.1%

         \[\leadsto \frac{e^{\color{blue}{-\frac{\log \left(\frac{1}{x}\right)}{n}}}}{n \cdot x} \]

      2. log-rec 76.2%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]

      3. mul-1-neg 76.2%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]

      4. distribute-neg-frac 76.2%

         \[\leadsto \frac{e^{\color{blue}{\frac{--1 \cdot \log x}{n}}}}{n \cdot x} \]

      5. mul-1-neg 76.2%

         \[\leadsto \frac{e^{\frac{-\color{blue}{\left(-\log x\right)}}{n}}}{n \cdot x} \]

      6. remove-double-neg 76.2%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]

      7. *-commutative 76.2%

         \[\leadsto \frac{e^{\frac{\log x}{n}}}{\color{blue}{x \cdot n}} \]

   5. Simplified 76.2%

      \[\leadsto \color{blue}{\frac{e^{\frac{\log x}{n}}}{x \cdot n}} \]

   if 5e3 < (/.f64 #s(literal 1 binary64) n)

   1. Initial program 58.4%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around 0 58.4%

      \[\leadsto \color{blue}{e^{\frac{\log \left(1 + x\right)}{n}} - e^{\frac{\log x}{n}}} \]
   4. Step-by-step derivation

      1. log1p-define 100.0%

         \[\leadsto e^{\frac{\color{blue}{\mathsf{log1p}\left(x\right)}}{n}} - e^{\frac{\log x}{n}} \]

      2. *-rgt-identity 100.0%

         \[\leadsto e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - e^{\frac{\color{blue}{\log x \cdot 1}}{n}} \]

      3. associate-/l* 100.0%

         \[\leadsto e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - e^{\color{blue}{\log x \cdot \frac{1}{n}}} \]

      4. exp-to-pow 100.0%

         \[\leadsto e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - \color{blue}{{x}^{\left(\frac{1}{n}\right)}} \]

   5. Simplified 100.0%

      \[\leadsto \color{blue}{e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}} \]

   6. Taylor expanded in x around 0 100.0%

      \[\leadsto e^{\color{blue}{\frac{x}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
3. Recombined 4 regimes into one program.

4. Final simplification 92.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-5}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-153}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5000:\\ \;\;\;\;\frac{e^{\frac{\log x}{n}}}{n \cdot x}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{x}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 81.7% accurate, 0.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-5}:\\ \;\;\;\;\frac{t\_0}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-153}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5000:\\ \;\;\;\;\frac{e^{\frac{\log x}{n}}}{n \cdot x}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + x \cdot \frac{1 + 0.5 \cdot \frac{x}{n}}{n}\right) - t\_0\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))))
   (if (<= (/ 1.0 n) -2e-5)
     (/ t_0 (* n x))
     (if (<= (/ 1.0 n) 5e-153)
       (/ (log (/ (+ 1.0 x) x)) n)
       (if (<= (/ 1.0 n) 5000.0)
         (/ (exp (/ (log x) n)) (* n x))
         (- (+ 1.0 (* x (/ (+ 1.0 (* 0.5 (/ x n))) n))) t_0))))))

C

double code(double x, double n) {
	double t_0 = pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-5) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 5e-153) {
		tmp = log(((1.0 + x) / x)) / n;
	} else if ((1.0 / n) <= 5000.0) {
		tmp = exp((log(x) / n)) / (n * x);
	} else {
		tmp = (1.0 + (x * ((1.0 + (0.5 * (x / n))) / n))) - t_0;
	}
	return tmp;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    real(8) :: t_0
    real(8) :: tmp
    t_0 = x ** (1.0d0 / n)
    if ((1.0d0 / n) <= (-2d-5)) then
        tmp = t_0 / (n * x)
    else if ((1.0d0 / n) <= 5d-153) then
        tmp = log(((1.0d0 + x) / x)) / n
    else if ((1.0d0 / n) <= 5000.0d0) then
        tmp = exp((log(x) / n)) / (n * x)
    else
        tmp = (1.0d0 + (x * ((1.0d0 + (0.5d0 * (x / n))) / n))) - t_0
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double t_0 = Math.pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-5) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 5e-153) {
		tmp = Math.log(((1.0 + x) / x)) / n;
	} else if ((1.0 / n) <= 5000.0) {
		tmp = Math.exp((Math.log(x) / n)) / (n * x);
	} else {
		tmp = (1.0 + (x * ((1.0 + (0.5 * (x / n))) / n))) - t_0;
	}
	return tmp;
}

Python

def code(x, n):
	t_0 = math.pow(x, (1.0 / n))
	tmp = 0
	if (1.0 / n) <= -2e-5:
		tmp = t_0 / (n * x)
	elif (1.0 / n) <= 5e-153:
		tmp = math.log(((1.0 + x) / x)) / n
	elif (1.0 / n) <= 5000.0:
		tmp = math.exp((math.log(x) / n)) / (n * x)
	else:
		tmp = (1.0 + (x * ((1.0 + (0.5 * (x / n))) / n))) - t_0
	return tmp

Julia

function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	tmp = 0.0
	if (Float64(1.0 / n) <= -2e-5)
		tmp = Float64(t_0 / Float64(n * x));
	elseif (Float64(1.0 / n) <= 5e-153)
		tmp = Float64(log(Float64(Float64(1.0 + x) / x)) / n);
	elseif (Float64(1.0 / n) <= 5000.0)
		tmp = Float64(exp(Float64(log(x) / n)) / Float64(n * x));
	else
		tmp = Float64(Float64(1.0 + Float64(x * Float64(Float64(1.0 + Float64(0.5 * Float64(x / n))) / n))) - t_0);
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	t_0 = x ^ (1.0 / n);
	tmp = 0.0;
	if ((1.0 / n) <= -2e-5)
		tmp = t_0 / (n * x);
	elseif ((1.0 / n) <= 5e-153)
		tmp = log(((1.0 + x) / x)) / n;
	elseif ((1.0 / n) <= 5000.0)
		tmp = exp((log(x) / n)) / (n * x);
	else
		tmp = (1.0 + (x * ((1.0 + (0.5 * (x / n))) / n))) - t_0;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-5], N[(t$95$0 / N[(n * x), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 5e-153], N[(N[Log[N[(N[(1.0 + x), $MachinePrecision] / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 5000.0], N[(N[Exp[N[(N[Log[x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision] / N[(n * x), $MachinePrecision]), $MachinePrecision], N[(N[(1.0 + N[(x * N[(N[(1.0 + N[(0.5 * N[(x / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - t$95$0), $MachinePrecision]]]]]

Derivation

1. Split input into 4 regimes

2. if (/.f64 #s(literal 1 binary64) n) < -2.00000000000000016e-5

   1. Initial program 97.9%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in x around inf 99.0%

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
   4. Step-by-step derivation

      1. mul-1-neg 99.0%

         \[\leadsto \frac{e^{\color{blue}{-\frac{\log \left(\frac{1}{x}\right)}{n}}}}{n \cdot x} \]

      2. log-rec 99.0%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]

      3. mul-1-neg 99.0%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]

      4. distribute-neg-frac 99.0%

         \[\leadsto \frac{e^{\color{blue}{\frac{--1 \cdot \log x}{n}}}}{n \cdot x} \]

      5. mul-1-neg 99.0%

         \[\leadsto \frac{e^{\frac{-\color{blue}{\left(-\log x\right)}}{n}}}{n \cdot x} \]

      6. remove-double-neg 99.0%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]

      7. *-rgt-identity 99.0%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x \cdot 1}}{n}}}{n \cdot x} \]

      8. associate-/l* 99.0%

         \[\leadsto \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{n \cdot x} \]

      9. exp-to-pow 99.0%

         \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]

      10. *-commutative 99.0%

          \[\leadsto \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{x \cdot n}} \]

   5. Simplified 99.0%

      \[\leadsto \color{blue}{\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}} \]

   if -2.00000000000000016e-5 < (/.f64 #s(literal 1 binary64) n) < 5.00000000000000033e-153

   1. Initial program 39.0%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 85.3%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   4. Step-by-step derivation

      1. log1p-define 85.3%

         \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   5. Simplified 85.3%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   6. Step-by-step derivation

      1. log1p-undefine 85.3%

         \[\leadsto \frac{\color{blue}{\log \left(1 + x\right)} - \log x}{n} \]

      2. diff-log 85.6%

         \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]

   7. Applied egg-rr 85.6%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]
   8. Step-by-step derivation

      1. +-commutative 85.6%

         \[\leadsto \frac{\log \left(\frac{\color{blue}{x + 1}}{x}\right)}{n} \]

   9. Simplified 85.6%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{x + 1}{x}\right)}}{n} \]

   if 5.00000000000000033e-153 < (/.f64 #s(literal 1 binary64) n) < 5e3

   1. Initial program 21.1%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in x around inf 76.1%

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
   4. Step-by-step derivation

      1. mul-1-neg 76.1%

         \[\leadsto \frac{e^{\color{blue}{-\frac{\log \left(\frac{1}{x}\right)}{n}}}}{n \cdot x} \]

      2. log-rec 76.2%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]

      3. mul-1-neg 76.2%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]

      4. distribute-neg-frac 76.2%

         \[\leadsto \frac{e^{\color{blue}{\frac{--1 \cdot \log x}{n}}}}{n \cdot x} \]

      5. mul-1-neg 76.2%

         \[\leadsto \frac{e^{\frac{-\color{blue}{\left(-\log x\right)}}{n}}}{n \cdot x} \]

      6. remove-double-neg 76.2%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]

      7. *-commutative 76.2%

         \[\leadsto \frac{e^{\frac{\log x}{n}}}{\color{blue}{x \cdot n}} \]

   5. Simplified 76.2%

      \[\leadsto \color{blue}{\frac{e^{\frac{\log x}{n}}}{x \cdot n}} \]

   if 5e3 < (/.f64 #s(literal 1 binary64) n)

   1. Initial program 58.4%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in x around 0 63.3%

      \[\leadsto \color{blue}{\left(1 + x \cdot \left(x \cdot \left(0.5 \cdot \frac{1}{{n}^{2}} - 0.5 \cdot \frac{1}{n}\right) + \frac{1}{n}\right)\right)} - {x}^{\left(\frac{1}{n}\right)} \]

   4. Taylor expanded in n around inf 80.9%

      \[\leadsto \left(1 + x \cdot \color{blue}{\frac{1 + \left(-0.5 \cdot x + 0.5 \cdot \frac{x}{n}\right)}{n}}\right) - {x}^{\left(\frac{1}{n}\right)} \]

   5. Taylor expanded in n around 0 80.9%

      \[\leadsto \left(1 + x \cdot \frac{1 + \color{blue}{0.5 \cdot \frac{x}{n}}}{n}\right) - {x}^{\left(\frac{1}{n}\right)} \]
   6. Step-by-step derivation

      1. *-commutative 80.9%

         \[\leadsto \left(1 + x \cdot \frac{1 + \color{blue}{\frac{x}{n} \cdot 0.5}}{n}\right) - {x}^{\left(\frac{1}{n}\right)} \]

   7. Simplified 80.9%

      \[\leadsto \left(1 + x \cdot \frac{1 + \color{blue}{\frac{x}{n} \cdot 0.5}}{n}\right) - {x}^{\left(\frac{1}{n}\right)} \]
3. Recombined 4 regimes into one program.

4. Final simplification 88.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-5}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-153}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5000:\\ \;\;\;\;\frac{e^{\frac{\log x}{n}}}{n \cdot x}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + x \cdot \frac{1 + 0.5 \cdot \frac{x}{n}}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 81.7% accurate, 1.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ t_1 := \frac{t\_0}{n \cdot x}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-5}:\\ \;\;\;\;t\_1\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-153}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5000:\\ \;\;\;\;t\_1\\ \mathbf{else}:\\ \;\;\;\;\left(1 + x \cdot \frac{1 + 0.5 \cdot \frac{x}{n}}{n}\right) - t\_0\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))) (t_1 (/ t_0 (* n x))))
   (if (<= (/ 1.0 n) -2e-5)
     t_1
     (if (<= (/ 1.0 n) 5e-153)
       (/ (log (/ (+ 1.0 x) x)) n)
       (if (<= (/ 1.0 n) 5000.0)
         t_1
         (- (+ 1.0 (* x (/ (+ 1.0 (* 0.5 (/ x n))) n))) t_0))))))

C

double code(double x, double n) {
	double t_0 = pow(x, (1.0 / n));
	double t_1 = t_0 / (n * x);
	double tmp;
	if ((1.0 / n) <= -2e-5) {
		tmp = t_1;
	} else if ((1.0 / n) <= 5e-153) {
		tmp = log(((1.0 + x) / x)) / n;
	} else if ((1.0 / n) <= 5000.0) {
		tmp = t_1;
	} else {
		tmp = (1.0 + (x * ((1.0 + (0.5 * (x / n))) / n))) - t_0;
	}
	return tmp;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    real(8) :: t_0
    real(8) :: t_1
    real(8) :: tmp
    t_0 = x ** (1.0d0 / n)
    t_1 = t_0 / (n * x)
    if ((1.0d0 / n) <= (-2d-5)) then
        tmp = t_1
    else if ((1.0d0 / n) <= 5d-153) then
        tmp = log(((1.0d0 + x) / x)) / n
    else if ((1.0d0 / n) <= 5000.0d0) then
        tmp = t_1
    else
        tmp = (1.0d0 + (x * ((1.0d0 + (0.5d0 * (x / n))) / n))) - t_0
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double t_0 = Math.pow(x, (1.0 / n));
	double t_1 = t_0 / (n * x);
	double tmp;
	if ((1.0 / n) <= -2e-5) {
		tmp = t_1;
	} else if ((1.0 / n) <= 5e-153) {
		tmp = Math.log(((1.0 + x) / x)) / n;
	} else if ((1.0 / n) <= 5000.0) {
		tmp = t_1;
	} else {
		tmp = (1.0 + (x * ((1.0 + (0.5 * (x / n))) / n))) - t_0;
	}
	return tmp;
}

Python

def code(x, n):
	t_0 = math.pow(x, (1.0 / n))
	t_1 = t_0 / (n * x)
	tmp = 0
	if (1.0 / n) <= -2e-5:
		tmp = t_1
	elif (1.0 / n) <= 5e-153:
		tmp = math.log(((1.0 + x) / x)) / n
	elif (1.0 / n) <= 5000.0:
		tmp = t_1
	else:
		tmp = (1.0 + (x * ((1.0 + (0.5 * (x / n))) / n))) - t_0
	return tmp

Julia

function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	t_1 = Float64(t_0 / Float64(n * x))
	tmp = 0.0
	if (Float64(1.0 / n) <= -2e-5)
		tmp = t_1;
	elseif (Float64(1.0 / n) <= 5e-153)
		tmp = Float64(log(Float64(Float64(1.0 + x) / x)) / n);
	elseif (Float64(1.0 / n) <= 5000.0)
		tmp = t_1;
	else
		tmp = Float64(Float64(1.0 + Float64(x * Float64(Float64(1.0 + Float64(0.5 * Float64(x / n))) / n))) - t_0);
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	t_0 = x ^ (1.0 / n);
	t_1 = t_0 / (n * x);
	tmp = 0.0;
	if ((1.0 / n) <= -2e-5)
		tmp = t_1;
	elseif ((1.0 / n) <= 5e-153)
		tmp = log(((1.0 + x) / x)) / n;
	elseif ((1.0 / n) <= 5000.0)
		tmp = t_1;
	else
		tmp = (1.0 + (x * ((1.0 + (0.5 * (x / n))) / n))) - t_0;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 / N[(n * x), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-5], t$95$1, If[LessEqual[N[(1.0 / n), $MachinePrecision], 5e-153], N[(N[Log[N[(N[(1.0 + x), $MachinePrecision] / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 5000.0], t$95$1, N[(N[(1.0 + N[(x * N[(N[(1.0 + N[(0.5 * N[(x / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - t$95$0), $MachinePrecision]]]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 #s(literal 1 binary64) n) < -2.00000000000000016e-5 or 5.00000000000000033e-153 < (/.f64 #s(literal 1 binary64) n) < 5e3

   1. Initial program 82.2%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in x around inf 94.3%

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
   4. Step-by-step derivation

      1. mul-1-neg 94.3%

         \[\leadsto \frac{e^{\color{blue}{-\frac{\log \left(\frac{1}{x}\right)}{n}}}}{n \cdot x} \]

      2. log-rec 94.3%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]

      3. mul-1-neg 94.3%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]

      4. distribute-neg-frac 94.3%

         \[\leadsto \frac{e^{\color{blue}{\frac{--1 \cdot \log x}{n}}}}{n \cdot x} \]

      5. mul-1-neg 94.3%

         \[\leadsto \frac{e^{\frac{-\color{blue}{\left(-\log x\right)}}{n}}}{n \cdot x} \]

      6. remove-double-neg 94.3%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]

      7. *-rgt-identity 94.3%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x \cdot 1}}{n}}}{n \cdot x} \]

      8. associate-/l* 94.3%

         \[\leadsto \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{n \cdot x} \]

      9. exp-to-pow 94.3%

         \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]

      10. *-commutative 94.3%

          \[\leadsto \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{x \cdot n}} \]

   5. Simplified 94.3%

      \[\leadsto \color{blue}{\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}} \]

   if -2.00000000000000016e-5 < (/.f64 #s(literal 1 binary64) n) < 5.00000000000000033e-153

   1. Initial program 39.0%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 85.3%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   4. Step-by-step derivation

      1. log1p-define 85.3%

         \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   5. Simplified 85.3%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   6. Step-by-step derivation

      1. log1p-undefine 85.3%

         \[\leadsto \frac{\color{blue}{\log \left(1 + x\right)} - \log x}{n} \]

      2. diff-log 85.6%

         \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]

   7. Applied egg-rr 85.6%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]
   8. Step-by-step derivation

      1. +-commutative 85.6%

         \[\leadsto \frac{\log \left(\frac{\color{blue}{x + 1}}{x}\right)}{n} \]

   9. Simplified 85.6%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{x + 1}{x}\right)}}{n} \]

   if 5e3 < (/.f64 #s(literal 1 binary64) n)

   1. Initial program 58.4%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in x around 0 63.3%

      \[\leadsto \color{blue}{\left(1 + x \cdot \left(x \cdot \left(0.5 \cdot \frac{1}{{n}^{2}} - 0.5 \cdot \frac{1}{n}\right) + \frac{1}{n}\right)\right)} - {x}^{\left(\frac{1}{n}\right)} \]

   4. Taylor expanded in n around inf 80.9%

      \[\leadsto \left(1 + x \cdot \color{blue}{\frac{1 + \left(-0.5 \cdot x + 0.5 \cdot \frac{x}{n}\right)}{n}}\right) - {x}^{\left(\frac{1}{n}\right)} \]

   5. Taylor expanded in n around 0 80.9%

      \[\leadsto \left(1 + x \cdot \frac{1 + \color{blue}{0.5 \cdot \frac{x}{n}}}{n}\right) - {x}^{\left(\frac{1}{n}\right)} \]
   6. Step-by-step derivation

      1. *-commutative 80.9%

         \[\leadsto \left(1 + x \cdot \frac{1 + \color{blue}{\frac{x}{n} \cdot 0.5}}{n}\right) - {x}^{\left(\frac{1}{n}\right)} \]

   7. Simplified 80.9%

      \[\leadsto \left(1 + x \cdot \frac{1 + \color{blue}{\frac{x}{n} \cdot 0.5}}{n}\right) - {x}^{\left(\frac{1}{n}\right)} \]
3. Recombined 3 regimes into one program.

4. Final simplification 88.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-5}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-153}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5000:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + x \cdot \frac{1 + 0.5 \cdot \frac{x}{n}}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 79.9% accurate, 1.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ t_1 := \frac{t\_0}{n \cdot x}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-5}:\\ \;\;\;\;t\_1\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-153}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5000:\\ \;\;\;\;t\_1\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{+251}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - t\_0\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{\frac{0.5 + \frac{-0.3333333333333333}{x}}{x} + -1}{-x}}{n}\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))) (t_1 (/ t_0 (* n x))))
   (if (<= (/ 1.0 n) -2e-5)
     t_1
     (if (<= (/ 1.0 n) 5e-153)
       (/ (log (/ (+ 1.0 x) x)) n)
       (if (<= (/ 1.0 n) 5000.0)
         t_1
         (if (<= (/ 1.0 n) 1e+251)
           (- (+ 1.0 (/ x n)) t_0)
           (/
            (/ (+ (/ (+ 0.5 (/ -0.3333333333333333 x)) x) -1.0) (- x))
            n)))))))

C

double code(double x, double n) {
	double t_0 = pow(x, (1.0 / n));
	double t_1 = t_0 / (n * x);
	double tmp;
	if ((1.0 / n) <= -2e-5) {
		tmp = t_1;
	} else if ((1.0 / n) <= 5e-153) {
		tmp = log(((1.0 + x) / x)) / n;
	} else if ((1.0 / n) <= 5000.0) {
		tmp = t_1;
	} else if ((1.0 / n) <= 1e+251) {
		tmp = (1.0 + (x / n)) - t_0;
	} else {
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n;
	}
	return tmp;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    real(8) :: t_0
    real(8) :: t_1
    real(8) :: tmp
    t_0 = x ** (1.0d0 / n)
    t_1 = t_0 / (n * x)
    if ((1.0d0 / n) <= (-2d-5)) then
        tmp = t_1
    else if ((1.0d0 / n) <= 5d-153) then
        tmp = log(((1.0d0 + x) / x)) / n
    else if ((1.0d0 / n) <= 5000.0d0) then
        tmp = t_1
    else if ((1.0d0 / n) <= 1d+251) then
        tmp = (1.0d0 + (x / n)) - t_0
    else
        tmp = ((((0.5d0 + ((-0.3333333333333333d0) / x)) / x) + (-1.0d0)) / -x) / n
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double t_0 = Math.pow(x, (1.0 / n));
	double t_1 = t_0 / (n * x);
	double tmp;
	if ((1.0 / n) <= -2e-5) {
		tmp = t_1;
	} else if ((1.0 / n) <= 5e-153) {
		tmp = Math.log(((1.0 + x) / x)) / n;
	} else if ((1.0 / n) <= 5000.0) {
		tmp = t_1;
	} else if ((1.0 / n) <= 1e+251) {
		tmp = (1.0 + (x / n)) - t_0;
	} else {
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n;
	}
	return tmp;
}

Python

def code(x, n):
	t_0 = math.pow(x, (1.0 / n))
	t_1 = t_0 / (n * x)
	tmp = 0
	if (1.0 / n) <= -2e-5:
		tmp = t_1
	elif (1.0 / n) <= 5e-153:
		tmp = math.log(((1.0 + x) / x)) / n
	elif (1.0 / n) <= 5000.0:
		tmp = t_1
	elif (1.0 / n) <= 1e+251:
		tmp = (1.0 + (x / n)) - t_0
	else:
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n
	return tmp

Julia

function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	t_1 = Float64(t_0 / Float64(n * x))
	tmp = 0.0
	if (Float64(1.0 / n) <= -2e-5)
		tmp = t_1;
	elseif (Float64(1.0 / n) <= 5e-153)
		tmp = Float64(log(Float64(Float64(1.0 + x) / x)) / n);
	elseif (Float64(1.0 / n) <= 5000.0)
		tmp = t_1;
	elseif (Float64(1.0 / n) <= 1e+251)
		tmp = Float64(Float64(1.0 + Float64(x / n)) - t_0);
	else
		tmp = Float64(Float64(Float64(Float64(Float64(0.5 + Float64(-0.3333333333333333 / x)) / x) + -1.0) / Float64(-x)) / n);
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	t_0 = x ^ (1.0 / n);
	t_1 = t_0 / (n * x);
	tmp = 0.0;
	if ((1.0 / n) <= -2e-5)
		tmp = t_1;
	elseif ((1.0 / n) <= 5e-153)
		tmp = log(((1.0 + x) / x)) / n;
	elseif ((1.0 / n) <= 5000.0)
		tmp = t_1;
	elseif ((1.0 / n) <= 1e+251)
		tmp = (1.0 + (x / n)) - t_0;
	else
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 / N[(n * x), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-5], t$95$1, If[LessEqual[N[(1.0 / n), $MachinePrecision], 5e-153], N[(N[Log[N[(N[(1.0 + x), $MachinePrecision] / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 5000.0], t$95$1, If[LessEqual[N[(1.0 / n), $MachinePrecision], 1e+251], N[(N[(1.0 + N[(x / n), $MachinePrecision]), $MachinePrecision] - t$95$0), $MachinePrecision], N[(N[(N[(N[(N[(0.5 + N[(-0.3333333333333333 / x), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision] + -1.0), $MachinePrecision] / (-x)), $MachinePrecision] / n), $MachinePrecision]]]]]]]

Derivation

1. Split input into 4 regimes

2. if (/.f64 #s(literal 1 binary64) n) < -2.00000000000000016e-5 or 5.00000000000000033e-153 < (/.f64 #s(literal 1 binary64) n) < 5e3

   1. Initial program 82.2%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in x around inf 94.3%

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
   4. Step-by-step derivation

      1. mul-1-neg 94.3%

         \[\leadsto \frac{e^{\color{blue}{-\frac{\log \left(\frac{1}{x}\right)}{n}}}}{n \cdot x} \]

      2. log-rec 94.3%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]

      3. mul-1-neg 94.3%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]

      4. distribute-neg-frac 94.3%

         \[\leadsto \frac{e^{\color{blue}{\frac{--1 \cdot \log x}{n}}}}{n \cdot x} \]

      5. mul-1-neg 94.3%

         \[\leadsto \frac{e^{\frac{-\color{blue}{\left(-\log x\right)}}{n}}}{n \cdot x} \]

      6. remove-double-neg 94.3%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]

      7. *-rgt-identity 94.3%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x \cdot 1}}{n}}}{n \cdot x} \]

      8. associate-/l* 94.3%

         \[\leadsto \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{n \cdot x} \]

      9. exp-to-pow 94.3%

         \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]

      10. *-commutative 94.3%

          \[\leadsto \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{x \cdot n}} \]

   5. Simplified 94.3%

      \[\leadsto \color{blue}{\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}} \]

   if -2.00000000000000016e-5 < (/.f64 #s(literal 1 binary64) n) < 5.00000000000000033e-153

   1. Initial program 39.0%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 85.3%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   4. Step-by-step derivation

      1. log1p-define 85.3%

         \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   5. Simplified 85.3%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   6. Step-by-step derivation

      1. log1p-undefine 85.3%

         \[\leadsto \frac{\color{blue}{\log \left(1 + x\right)} - \log x}{n} \]

      2. diff-log 85.6%

         \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]

   7. Applied egg-rr 85.6%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]
   8. Step-by-step derivation

      1. +-commutative 85.6%

         \[\leadsto \frac{\log \left(\frac{\color{blue}{x + 1}}{x}\right)}{n} \]

   9. Simplified 85.6%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{x + 1}{x}\right)}}{n} \]

   if 5e3 < (/.f64 #s(literal 1 binary64) n) < 1e251

   1. Initial program 73.4%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in x around 0 71.7%

      \[\leadsto \color{blue}{\left(1 + \frac{x}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]

   if 1e251 < (/.f64 #s(literal 1 binary64) n)

   1. Initial program 12.0%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 8.1%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   4. Step-by-step derivation

      1. log1p-define 8.1%

         \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   5. Simplified 8.1%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   6. Step-by-step derivation

      1. expm1-log1p-u 8.1%

         \[\leadsto \frac{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\mathsf{log1p}\left(x\right) - \log x\right)\right)}}{n} \]

   7. Applied egg-rr 8.1%

      \[\leadsto \frac{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\mathsf{log1p}\left(x\right) - \log x\right)\right)}}{n} \]

   8. Taylor expanded in x around -inf 82.4%

      \[\leadsto \frac{\color{blue}{-1 \cdot \frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} - 1}{x}}}{n} \]
   9. Step-by-step derivation

      1. mul-1-neg 82.4%

         \[\leadsto \frac{\color{blue}{-\frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} - 1}{x}}}{n} \]

      2. distribute-neg-frac2 82.4%

         \[\leadsto \frac{\color{blue}{\frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} - 1}{-x}}}{n} \]

      3. sub-neg 82.4%

         \[\leadsto \frac{\frac{\color{blue}{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} + \left(-1\right)}}{-x}}{n} \]

      4. associate-*r/ 82.4%

         \[\leadsto \frac{\frac{\color{blue}{\frac{-1 \cdot \left(0.3333333333333333 \cdot \frac{1}{x} - 0.5\right)}{x}} + \left(-1\right)}{-x}}{n} \]

      5. sub-neg 82.4%

         \[\leadsto \frac{\frac{\frac{-1 \cdot \color{blue}{\left(0.3333333333333333 \cdot \frac{1}{x} + \left(-0.5\right)\right)}}{x} + \left(-1\right)}{-x}}{n} \]

      6. metadata-eval 82.4%

         \[\leadsto \frac{\frac{\frac{-1 \cdot \left(0.3333333333333333 \cdot \frac{1}{x} + \color{blue}{-0.5}\right)}{x} + \left(-1\right)}{-x}}{n} \]

      7. distribute-lft-in 82.4%

         \[\leadsto \frac{\frac{\frac{\color{blue}{-1 \cdot \left(0.3333333333333333 \cdot \frac{1}{x}\right) + -1 \cdot -0.5}}{x} + \left(-1\right)}{-x}}{n} \]

      8. neg-mul-1 82.4%

         \[\leadsto \frac{\frac{\frac{\color{blue}{\left(-0.3333333333333333 \cdot \frac{1}{x}\right)} + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      9. associate-*r/ 82.4%

         \[\leadsto \frac{\frac{\frac{\left(-\color{blue}{\frac{0.3333333333333333 \cdot 1}{x}}\right) + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      10. metadata-eval 82.4%

          \[\leadsto \frac{\frac{\frac{\left(-\frac{\color{blue}{0.3333333333333333}}{x}\right) + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      11. distribute-neg-frac 82.4%

          \[\leadsto \frac{\frac{\frac{\color{blue}{\frac{-0.3333333333333333}{x}} + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      12. metadata-eval 82.4%

          \[\leadsto \frac{\frac{\frac{\frac{\color{blue}{-0.3333333333333333}}{x} + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      13. metadata-eval 82.4%

          \[\leadsto \frac{\frac{\frac{\frac{-0.3333333333333333}{x} + \color{blue}{0.5}}{x} + \left(-1\right)}{-x}}{n} \]

      14. metadata-eval 82.4%

          \[\leadsto \frac{\frac{\frac{\frac{-0.3333333333333333}{x} + 0.5}{x} + \color{blue}{-1}}{-x}}{n} \]

   10. Simplified 82.4%

       \[\leadsto \frac{\color{blue}{\frac{\frac{\frac{-0.3333333333333333}{x} + 0.5}{x} + -1}{-x}}}{n} \]
3. Recombined 4 regimes into one program.

4. Final simplification 87.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-5}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-153}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5000:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{+251}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{\frac{0.5 + \frac{-0.3333333333333333}{x}}{x} + -1}{-x}}{n}\\ \end{array} \]
5. Add Preprocessing

Alternative 8: 79.8% accurate, 1.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ t_1 := \frac{t\_0}{n \cdot x}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-5}:\\ \;\;\;\;t\_1\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-153}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5000:\\ \;\;\;\;t\_1\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{+251}:\\ \;\;\;\;1 - t\_0\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{\frac{0.5 + \frac{-0.3333333333333333}{x}}{x} + -1}{-x}}{n}\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))) (t_1 (/ t_0 (* n x))))
   (if (<= (/ 1.0 n) -2e-5)
     t_1
     (if (<= (/ 1.0 n) 5e-153)
       (/ (log (/ (+ 1.0 x) x)) n)
       (if (<= (/ 1.0 n) 5000.0)
         t_1
         (if (<= (/ 1.0 n) 1e+251)
           (- 1.0 t_0)
           (/
            (/ (+ (/ (+ 0.5 (/ -0.3333333333333333 x)) x) -1.0) (- x))
            n)))))))

C

double code(double x, double n) {
	double t_0 = pow(x, (1.0 / n));
	double t_1 = t_0 / (n * x);
	double tmp;
	if ((1.0 / n) <= -2e-5) {
		tmp = t_1;
	} else if ((1.0 / n) <= 5e-153) {
		tmp = log(((1.0 + x) / x)) / n;
	} else if ((1.0 / n) <= 5000.0) {
		tmp = t_1;
	} else if ((1.0 / n) <= 1e+251) {
		tmp = 1.0 - t_0;
	} else {
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n;
	}
	return tmp;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    real(8) :: t_0
    real(8) :: t_1
    real(8) :: tmp
    t_0 = x ** (1.0d0 / n)
    t_1 = t_0 / (n * x)
    if ((1.0d0 / n) <= (-2d-5)) then
        tmp = t_1
    else if ((1.0d0 / n) <= 5d-153) then
        tmp = log(((1.0d0 + x) / x)) / n
    else if ((1.0d0 / n) <= 5000.0d0) then
        tmp = t_1
    else if ((1.0d0 / n) <= 1d+251) then
        tmp = 1.0d0 - t_0
    else
        tmp = ((((0.5d0 + ((-0.3333333333333333d0) / x)) / x) + (-1.0d0)) / -x) / n
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double t_0 = Math.pow(x, (1.0 / n));
	double t_1 = t_0 / (n * x);
	double tmp;
	if ((1.0 / n) <= -2e-5) {
		tmp = t_1;
	} else if ((1.0 / n) <= 5e-153) {
		tmp = Math.log(((1.0 + x) / x)) / n;
	} else if ((1.0 / n) <= 5000.0) {
		tmp = t_1;
	} else if ((1.0 / n) <= 1e+251) {
		tmp = 1.0 - t_0;
	} else {
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n;
	}
	return tmp;
}

Python

def code(x, n):
	t_0 = math.pow(x, (1.0 / n))
	t_1 = t_0 / (n * x)
	tmp = 0
	if (1.0 / n) <= -2e-5:
		tmp = t_1
	elif (1.0 / n) <= 5e-153:
		tmp = math.log(((1.0 + x) / x)) / n
	elif (1.0 / n) <= 5000.0:
		tmp = t_1
	elif (1.0 / n) <= 1e+251:
		tmp = 1.0 - t_0
	else:
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n
	return tmp

Julia

function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	t_1 = Float64(t_0 / Float64(n * x))
	tmp = 0.0
	if (Float64(1.0 / n) <= -2e-5)
		tmp = t_1;
	elseif (Float64(1.0 / n) <= 5e-153)
		tmp = Float64(log(Float64(Float64(1.0 + x) / x)) / n);
	elseif (Float64(1.0 / n) <= 5000.0)
		tmp = t_1;
	elseif (Float64(1.0 / n) <= 1e+251)
		tmp = Float64(1.0 - t_0);
	else
		tmp = Float64(Float64(Float64(Float64(Float64(0.5 + Float64(-0.3333333333333333 / x)) / x) + -1.0) / Float64(-x)) / n);
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	t_0 = x ^ (1.0 / n);
	t_1 = t_0 / (n * x);
	tmp = 0.0;
	if ((1.0 / n) <= -2e-5)
		tmp = t_1;
	elseif ((1.0 / n) <= 5e-153)
		tmp = log(((1.0 + x) / x)) / n;
	elseif ((1.0 / n) <= 5000.0)
		tmp = t_1;
	elseif ((1.0 / n) <= 1e+251)
		tmp = 1.0 - t_0;
	else
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 / N[(n * x), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-5], t$95$1, If[LessEqual[N[(1.0 / n), $MachinePrecision], 5e-153], N[(N[Log[N[(N[(1.0 + x), $MachinePrecision] / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 5000.0], t$95$1, If[LessEqual[N[(1.0 / n), $MachinePrecision], 1e+251], N[(1.0 - t$95$0), $MachinePrecision], N[(N[(N[(N[(N[(0.5 + N[(-0.3333333333333333 / x), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision] + -1.0), $MachinePrecision] / (-x)), $MachinePrecision] / n), $MachinePrecision]]]]]]]

Derivation

1. Split input into 4 regimes

2. if (/.f64 #s(literal 1 binary64) n) < -2.00000000000000016e-5 or 5.00000000000000033e-153 < (/.f64 #s(literal 1 binary64) n) < 5e3

   1. Initial program 82.2%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in x around inf 94.3%

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
   4. Step-by-step derivation

      1. mul-1-neg 94.3%

         \[\leadsto \frac{e^{\color{blue}{-\frac{\log \left(\frac{1}{x}\right)}{n}}}}{n \cdot x} \]

      2. log-rec 94.3%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]

      3. mul-1-neg 94.3%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]

      4. distribute-neg-frac 94.3%

         \[\leadsto \frac{e^{\color{blue}{\frac{--1 \cdot \log x}{n}}}}{n \cdot x} \]

      5. mul-1-neg 94.3%

         \[\leadsto \frac{e^{\frac{-\color{blue}{\left(-\log x\right)}}{n}}}{n \cdot x} \]

      6. remove-double-neg 94.3%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]

      7. *-rgt-identity 94.3%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x \cdot 1}}{n}}}{n \cdot x} \]

      8. associate-/l* 94.3%

         \[\leadsto \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{n \cdot x} \]

      9. exp-to-pow 94.3%

         \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]

      10. *-commutative 94.3%

          \[\leadsto \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{x \cdot n}} \]

   5. Simplified 94.3%

      \[\leadsto \color{blue}{\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}} \]

   if -2.00000000000000016e-5 < (/.f64 #s(literal 1 binary64) n) < 5.00000000000000033e-153

   1. Initial program 39.0%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 85.3%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   4. Step-by-step derivation

      1. log1p-define 85.3%

         \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   5. Simplified 85.3%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   6. Step-by-step derivation

      1. log1p-undefine 85.3%

         \[\leadsto \frac{\color{blue}{\log \left(1 + x\right)} - \log x}{n} \]

      2. diff-log 85.6%

         \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]

   7. Applied egg-rr 85.6%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]
   8. Step-by-step derivation

      1. +-commutative 85.6%

         \[\leadsto \frac{\log \left(\frac{\color{blue}{x + 1}}{x}\right)}{n} \]

   9. Simplified 85.6%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{x + 1}{x}\right)}}{n} \]

   if 5e3 < (/.f64 #s(literal 1 binary64) n) < 1e251

   1. Initial program 73.4%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in x around 0 70.5%

      \[\leadsto \color{blue}{1 - e^{\frac{\log x}{n}}} \]
   4. Step-by-step derivation

      1. *-rgt-identity 70.5%

         \[\leadsto 1 - e^{\frac{\color{blue}{\log x \cdot 1}}{n}} \]

      2. associate-/l* 70.5%

         \[\leadsto 1 - e^{\color{blue}{\log x \cdot \frac{1}{n}}} \]

      3. exp-to-pow 70.5%

         \[\leadsto 1 - \color{blue}{{x}^{\left(\frac{1}{n}\right)}} \]

   5. Simplified 70.5%

      \[\leadsto \color{blue}{1 - {x}^{\left(\frac{1}{n}\right)}} \]

   if 1e251 < (/.f64 #s(literal 1 binary64) n)

   1. Initial program 12.0%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 8.1%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   4. Step-by-step derivation

      1. log1p-define 8.1%

         \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   5. Simplified 8.1%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   6. Step-by-step derivation

      1. expm1-log1p-u 8.1%

         \[\leadsto \frac{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\mathsf{log1p}\left(x\right) - \log x\right)\right)}}{n} \]

   7. Applied egg-rr 8.1%

      \[\leadsto \frac{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\mathsf{log1p}\left(x\right) - \log x\right)\right)}}{n} \]

   8. Taylor expanded in x around -inf 82.4%

      \[\leadsto \frac{\color{blue}{-1 \cdot \frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} - 1}{x}}}{n} \]
   9. Step-by-step derivation

      1. mul-1-neg 82.4%

         \[\leadsto \frac{\color{blue}{-\frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} - 1}{x}}}{n} \]

      2. distribute-neg-frac2 82.4%

         \[\leadsto \frac{\color{blue}{\frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} - 1}{-x}}}{n} \]

      3. sub-neg 82.4%

         \[\leadsto \frac{\frac{\color{blue}{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} + \left(-1\right)}}{-x}}{n} \]

      4. associate-*r/ 82.4%

         \[\leadsto \frac{\frac{\color{blue}{\frac{-1 \cdot \left(0.3333333333333333 \cdot \frac{1}{x} - 0.5\right)}{x}} + \left(-1\right)}{-x}}{n} \]

      5. sub-neg 82.4%

         \[\leadsto \frac{\frac{\frac{-1 \cdot \color{blue}{\left(0.3333333333333333 \cdot \frac{1}{x} + \left(-0.5\right)\right)}}{x} + \left(-1\right)}{-x}}{n} \]

      6. metadata-eval 82.4%

         \[\leadsto \frac{\frac{\frac{-1 \cdot \left(0.3333333333333333 \cdot \frac{1}{x} + \color{blue}{-0.5}\right)}{x} + \left(-1\right)}{-x}}{n} \]

      7. distribute-lft-in 82.4%

         \[\leadsto \frac{\frac{\frac{\color{blue}{-1 \cdot \left(0.3333333333333333 \cdot \frac{1}{x}\right) + -1 \cdot -0.5}}{x} + \left(-1\right)}{-x}}{n} \]

      8. neg-mul-1 82.4%

         \[\leadsto \frac{\frac{\frac{\color{blue}{\left(-0.3333333333333333 \cdot \frac{1}{x}\right)} + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      9. associate-*r/ 82.4%

         \[\leadsto \frac{\frac{\frac{\left(-\color{blue}{\frac{0.3333333333333333 \cdot 1}{x}}\right) + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      10. metadata-eval 82.4%

          \[\leadsto \frac{\frac{\frac{\left(-\frac{\color{blue}{0.3333333333333333}}{x}\right) + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      11. distribute-neg-frac 82.4%

          \[\leadsto \frac{\frac{\frac{\color{blue}{\frac{-0.3333333333333333}{x}} + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      12. metadata-eval 82.4%

          \[\leadsto \frac{\frac{\frac{\frac{\color{blue}{-0.3333333333333333}}{x} + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      13. metadata-eval 82.4%

          \[\leadsto \frac{\frac{\frac{\frac{-0.3333333333333333}{x} + \color{blue}{0.5}}{x} + \left(-1\right)}{-x}}{n} \]

      14. metadata-eval 82.4%

          \[\leadsto \frac{\frac{\frac{\frac{-0.3333333333333333}{x} + 0.5}{x} + \color{blue}{-1}}{-x}}{n} \]

   10. Simplified 82.4%

       \[\leadsto \frac{\color{blue}{\frac{\frac{\frac{-0.3333333333333333}{x} + 0.5}{x} + -1}{-x}}}{n} \]
3. Recombined 4 regimes into one program.

4. Final simplification 87.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-5}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-153}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5000:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{+251}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{\frac{0.5 + \frac{-0.3333333333333333}{x}}{x} + -1}{-x}}{n}\\ \end{array} \]
5. Add Preprocessing

Alternative 9: 64.9% accurate, 1.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{+166}:\\ \;\;\;\;t\_0\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-153}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5000:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{+251}:\\ \;\;\;\;t\_0\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{\frac{0.5 + \frac{-0.3333333333333333}{x}}{x} + -1}{-x}}{n}\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (- 1.0 (pow x (/ 1.0 n)))))
   (if (<= (/ 1.0 n) -1e+166)
     t_0
     (if (<= (/ 1.0 n) 5e-153)
       (/ (log (/ (+ 1.0 x) x)) n)
       (if (<= (/ 1.0 n) 5000.0)
         (/ (/ 1.0 x) n)
         (if (<= (/ 1.0 n) 1e+251)
           t_0
           (/
            (/ (+ (/ (+ 0.5 (/ -0.3333333333333333 x)) x) -1.0) (- x))
            n)))))))

C

double code(double x, double n) {
	double t_0 = 1.0 - pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -1e+166) {
		tmp = t_0;
	} else if ((1.0 / n) <= 5e-153) {
		tmp = log(((1.0 + x) / x)) / n;
	} else if ((1.0 / n) <= 5000.0) {
		tmp = (1.0 / x) / n;
	} else if ((1.0 / n) <= 1e+251) {
		tmp = t_0;
	} else {
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n;
	}
	return tmp;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    real(8) :: t_0
    real(8) :: tmp
    t_0 = 1.0d0 - (x ** (1.0d0 / n))
    if ((1.0d0 / n) <= (-1d+166)) then
        tmp = t_0
    else if ((1.0d0 / n) <= 5d-153) then
        tmp = log(((1.0d0 + x) / x)) / n
    else if ((1.0d0 / n) <= 5000.0d0) then
        tmp = (1.0d0 / x) / n
    else if ((1.0d0 / n) <= 1d+251) then
        tmp = t_0
    else
        tmp = ((((0.5d0 + ((-0.3333333333333333d0) / x)) / x) + (-1.0d0)) / -x) / n
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double t_0 = 1.0 - Math.pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -1e+166) {
		tmp = t_0;
	} else if ((1.0 / n) <= 5e-153) {
		tmp = Math.log(((1.0 + x) / x)) / n;
	} else if ((1.0 / n) <= 5000.0) {
		tmp = (1.0 / x) / n;
	} else if ((1.0 / n) <= 1e+251) {
		tmp = t_0;
	} else {
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n;
	}
	return tmp;
}

Python

def code(x, n):
	t_0 = 1.0 - math.pow(x, (1.0 / n))
	tmp = 0
	if (1.0 / n) <= -1e+166:
		tmp = t_0
	elif (1.0 / n) <= 5e-153:
		tmp = math.log(((1.0 + x) / x)) / n
	elif (1.0 / n) <= 5000.0:
		tmp = (1.0 / x) / n
	elif (1.0 / n) <= 1e+251:
		tmp = t_0
	else:
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n
	return tmp

Julia

function code(x, n)
	t_0 = Float64(1.0 - (x ^ Float64(1.0 / n)))
	tmp = 0.0
	if (Float64(1.0 / n) <= -1e+166)
		tmp = t_0;
	elseif (Float64(1.0 / n) <= 5e-153)
		tmp = Float64(log(Float64(Float64(1.0 + x) / x)) / n);
	elseif (Float64(1.0 / n) <= 5000.0)
		tmp = Float64(Float64(1.0 / x) / n);
	elseif (Float64(1.0 / n) <= 1e+251)
		tmp = t_0;
	else
		tmp = Float64(Float64(Float64(Float64(Float64(0.5 + Float64(-0.3333333333333333 / x)) / x) + -1.0) / Float64(-x)) / n);
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	t_0 = 1.0 - (x ^ (1.0 / n));
	tmp = 0.0;
	if ((1.0 / n) <= -1e+166)
		tmp = t_0;
	elseif ((1.0 / n) <= 5e-153)
		tmp = log(((1.0 + x) / x)) / n;
	elseif ((1.0 / n) <= 5000.0)
		tmp = (1.0 / x) / n;
	elseif ((1.0 / n) <= 1e+251)
		tmp = t_0;
	else
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[(1.0 - N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -1e+166], t$95$0, If[LessEqual[N[(1.0 / n), $MachinePrecision], 5e-153], N[(N[Log[N[(N[(1.0 + x), $MachinePrecision] / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 5000.0], N[(N[(1.0 / x), $MachinePrecision] / n), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 1e+251], t$95$0, N[(N[(N[(N[(N[(0.5 + N[(-0.3333333333333333 / x), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision] + -1.0), $MachinePrecision] / (-x)), $MachinePrecision] / n), $MachinePrecision]]]]]]

Derivation

1. Split input into 4 regimes

2. if (/.f64 #s(literal 1 binary64) n) < -9.9999999999999994e165 or 5e3 < (/.f64 #s(literal 1 binary64) n) < 1e251

   1. Initial program 88.2%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in x around 0 68.1%

      \[\leadsto \color{blue}{1 - e^{\frac{\log x}{n}}} \]
   4. Step-by-step derivation

      1. *-rgt-identity 68.1%

         \[\leadsto 1 - e^{\frac{\color{blue}{\log x \cdot 1}}{n}} \]

      2. associate-/l* 68.1%

         \[\leadsto 1 - e^{\color{blue}{\log x \cdot \frac{1}{n}}} \]

      3. exp-to-pow 68.1%

         \[\leadsto 1 - \color{blue}{{x}^{\left(\frac{1}{n}\right)}} \]

   5. Simplified 68.1%

      \[\leadsto \color{blue}{1 - {x}^{\left(\frac{1}{n}\right)}} \]

   if -9.9999999999999994e165 < (/.f64 #s(literal 1 binary64) n) < 5.00000000000000033e-153

   1. Initial program 58.9%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 72.5%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   4. Step-by-step derivation

      1. log1p-define 72.5%

         \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   5. Simplified 72.5%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   6. Step-by-step derivation

      1. log1p-undefine 72.5%

         \[\leadsto \frac{\color{blue}{\log \left(1 + x\right)} - \log x}{n} \]

      2. diff-log 72.6%

         \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]

   7. Applied egg-rr 72.6%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]
   8. Step-by-step derivation

      1. +-commutative 72.6%

         \[\leadsto \frac{\log \left(\frac{\color{blue}{x + 1}}{x}\right)}{n} \]

   9. Simplified 72.6%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{x + 1}{x}\right)}}{n} \]

   if 5.00000000000000033e-153 < (/.f64 #s(literal 1 binary64) n) < 5e3

   1. Initial program 21.1%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 45.1%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   4. Step-by-step derivation

      1. log1p-define 45.1%

         \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   5. Simplified 45.1%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]

   6. Taylor expanded in x around inf 68.7%

      \[\leadsto \frac{\color{blue}{\frac{1}{x}}}{n} \]

   if 1e251 < (/.f64 #s(literal 1 binary64) n)

   1. Initial program 12.0%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 8.1%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   4. Step-by-step derivation

      1. log1p-define 8.1%

         \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   5. Simplified 8.1%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   6. Step-by-step derivation

      1. expm1-log1p-u 8.1%

         \[\leadsto \frac{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\mathsf{log1p}\left(x\right) - \log x\right)\right)}}{n} \]

   7. Applied egg-rr 8.1%

      \[\leadsto \frac{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\mathsf{log1p}\left(x\right) - \log x\right)\right)}}{n} \]

   8. Taylor expanded in x around -inf 82.4%

      \[\leadsto \frac{\color{blue}{-1 \cdot \frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} - 1}{x}}}{n} \]
   9. Step-by-step derivation

      1. mul-1-neg 82.4%

         \[\leadsto \frac{\color{blue}{-\frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} - 1}{x}}}{n} \]

      2. distribute-neg-frac2 82.4%

         \[\leadsto \frac{\color{blue}{\frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} - 1}{-x}}}{n} \]

      3. sub-neg 82.4%

         \[\leadsto \frac{\frac{\color{blue}{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} + \left(-1\right)}}{-x}}{n} \]

      4. associate-*r/ 82.4%

         \[\leadsto \frac{\frac{\color{blue}{\frac{-1 \cdot \left(0.3333333333333333 \cdot \frac{1}{x} - 0.5\right)}{x}} + \left(-1\right)}{-x}}{n} \]

      5. sub-neg 82.4%

         \[\leadsto \frac{\frac{\frac{-1 \cdot \color{blue}{\left(0.3333333333333333 \cdot \frac{1}{x} + \left(-0.5\right)\right)}}{x} + \left(-1\right)}{-x}}{n} \]

      6. metadata-eval 82.4%

         \[\leadsto \frac{\frac{\frac{-1 \cdot \left(0.3333333333333333 \cdot \frac{1}{x} + \color{blue}{-0.5}\right)}{x} + \left(-1\right)}{-x}}{n} \]

      7. distribute-lft-in 82.4%

         \[\leadsto \frac{\frac{\frac{\color{blue}{-1 \cdot \left(0.3333333333333333 \cdot \frac{1}{x}\right) + -1 \cdot -0.5}}{x} + \left(-1\right)}{-x}}{n} \]

      8. neg-mul-1 82.4%

         \[\leadsto \frac{\frac{\frac{\color{blue}{\left(-0.3333333333333333 \cdot \frac{1}{x}\right)} + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      9. associate-*r/ 82.4%

         \[\leadsto \frac{\frac{\frac{\left(-\color{blue}{\frac{0.3333333333333333 \cdot 1}{x}}\right) + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      10. metadata-eval 82.4%

          \[\leadsto \frac{\frac{\frac{\left(-\frac{\color{blue}{0.3333333333333333}}{x}\right) + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      11. distribute-neg-frac 82.4%

          \[\leadsto \frac{\frac{\frac{\color{blue}{\frac{-0.3333333333333333}{x}} + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      12. metadata-eval 82.4%

          \[\leadsto \frac{\frac{\frac{\frac{\color{blue}{-0.3333333333333333}}{x} + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      13. metadata-eval 82.4%

          \[\leadsto \frac{\frac{\frac{\frac{-0.3333333333333333}{x} + \color{blue}{0.5}}{x} + \left(-1\right)}{-x}}{n} \]

      14. metadata-eval 82.4%

          \[\leadsto \frac{\frac{\frac{\frac{-0.3333333333333333}{x} + 0.5}{x} + \color{blue}{-1}}{-x}}{n} \]

   10. Simplified 82.4%

       \[\leadsto \frac{\color{blue}{\frac{\frac{\frac{-0.3333333333333333}{x} + 0.5}{x} + -1}{-x}}}{n} \]
3. Recombined 4 regimes into one program.

4. Final simplification 71.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{+166}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-153}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5000:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{+251}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{\frac{0.5 + \frac{-0.3333333333333333}{x}}{x} + -1}{-x}}{n}\\ \end{array} \]
5. Add Preprocessing

Alternative 10: 55.6% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;x \leq 1.18 \cdot 10^{-79}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;x \leq 3.2 \cdot 10^{-34}:\\ \;\;\;\;\frac{\log x}{-n}\\ \mathbf{elif}\;x \leq 3.1 \cdot 10^{+97}:\\ \;\;\;\;\frac{\frac{\frac{0.5 + \frac{-0.3333333333333333}{x}}{x} + -1}{-x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{-0.25}{n \cdot {x}^{4}}\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (if (<= x 1.18e-79)
   (- 1.0 (pow x (/ 1.0 n)))
   (if (<= x 3.2e-34)
     (/ (log x) (- n))
     (if (<= x 3.1e+97)
       (/ (/ (+ (/ (+ 0.5 (/ -0.3333333333333333 x)) x) -1.0) (- x)) n)
       (/ -0.25 (* n (pow x 4.0)))))))

C

double code(double x, double n) {
	double tmp;
	if (x <= 1.18e-79) {
		tmp = 1.0 - pow(x, (1.0 / n));
	} else if (x <= 3.2e-34) {
		tmp = log(x) / -n;
	} else if (x <= 3.1e+97) {
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n;
	} else {
		tmp = -0.25 / (n * pow(x, 4.0));
	}
	return tmp;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    real(8) :: tmp
    if (x <= 1.18d-79) then
        tmp = 1.0d0 - (x ** (1.0d0 / n))
    else if (x <= 3.2d-34) then
        tmp = log(x) / -n
    else if (x <= 3.1d+97) then
        tmp = ((((0.5d0 + ((-0.3333333333333333d0) / x)) / x) + (-1.0d0)) / -x) / n
    else
        tmp = (-0.25d0) / (n * (x ** 4.0d0))
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double tmp;
	if (x <= 1.18e-79) {
		tmp = 1.0 - Math.pow(x, (1.0 / n));
	} else if (x <= 3.2e-34) {
		tmp = Math.log(x) / -n;
	} else if (x <= 3.1e+97) {
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n;
	} else {
		tmp = -0.25 / (n * Math.pow(x, 4.0));
	}
	return tmp;
}

Python

def code(x, n):
	tmp = 0
	if x <= 1.18e-79:
		tmp = 1.0 - math.pow(x, (1.0 / n))
	elif x <= 3.2e-34:
		tmp = math.log(x) / -n
	elif x <= 3.1e+97:
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n
	else:
		tmp = -0.25 / (n * math.pow(x, 4.0))
	return tmp

Julia

function code(x, n)
	tmp = 0.0
	if (x <= 1.18e-79)
		tmp = Float64(1.0 - (x ^ Float64(1.0 / n)));
	elseif (x <= 3.2e-34)
		tmp = Float64(log(x) / Float64(-n));
	elseif (x <= 3.1e+97)
		tmp = Float64(Float64(Float64(Float64(Float64(0.5 + Float64(-0.3333333333333333 / x)) / x) + -1.0) / Float64(-x)) / n);
	else
		tmp = Float64(-0.25 / Float64(n * (x ^ 4.0)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	tmp = 0.0;
	if (x <= 1.18e-79)
		tmp = 1.0 - (x ^ (1.0 / n));
	elseif (x <= 3.2e-34)
		tmp = log(x) / -n;
	elseif (x <= 3.1e+97)
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n;
	else
		tmp = -0.25 / (n * (x ^ 4.0));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, n_] := If[LessEqual[x, 1.18e-79], N[(1.0 - N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[x, 3.2e-34], N[(N[Log[x], $MachinePrecision] / (-n)), $MachinePrecision], If[LessEqual[x, 3.1e+97], N[(N[(N[(N[(N[(0.5 + N[(-0.3333333333333333 / x), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision] + -1.0), $MachinePrecision] / (-x)), $MachinePrecision] / n), $MachinePrecision], N[(-0.25 / N[(n * N[Power[x, 4.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 4 regimes

2. if x < 1.18e-79

   1. Initial program 59.9%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in x around 0 59.9%

      \[\leadsto \color{blue}{1 - e^{\frac{\log x}{n}}} \]
   4. Step-by-step derivation

      1. *-rgt-identity 59.9%

         \[\leadsto 1 - e^{\frac{\color{blue}{\log x \cdot 1}}{n}} \]

      2. associate-/l* 59.9%

         \[\leadsto 1 - e^{\color{blue}{\log x \cdot \frac{1}{n}}} \]

      3. exp-to-pow 59.9%

         \[\leadsto 1 - \color{blue}{{x}^{\left(\frac{1}{n}\right)}} \]

   5. Simplified 59.9%

      \[\leadsto \color{blue}{1 - {x}^{\left(\frac{1}{n}\right)}} \]

   if 1.18e-79 < x < 3.20000000000000003e-34

   1. Initial program 26.7%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 62.9%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   4. Step-by-step derivation

      1. log1p-define 62.9%

         \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   5. Simplified 62.9%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]

   6. Taylor expanded in x around 0 62.9%

      \[\leadsto \frac{\color{blue}{-1 \cdot \log x}}{n} \]
   7. Step-by-step derivation

      1. neg-mul-1 62.9%

         \[\leadsto \frac{\color{blue}{-\log x}}{n} \]

   8. Simplified 62.9%

      \[\leadsto \frac{\color{blue}{-\log x}}{n} \]

   if 3.20000000000000003e-34 < x < 3.09999999999999981e97

   1. Initial program 56.3%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 41.7%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   4. Step-by-step derivation

      1. log1p-define 41.7%

         \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   5. Simplified 41.7%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   6. Step-by-step derivation

      1. expm1-log1p-u 41.7%

         \[\leadsto \frac{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\mathsf{log1p}\left(x\right) - \log x\right)\right)}}{n} \]

   7. Applied egg-rr 41.7%

      \[\leadsto \frac{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\mathsf{log1p}\left(x\right) - \log x\right)\right)}}{n} \]

   8. Taylor expanded in x around -inf 45.9%

      \[\leadsto \frac{\color{blue}{-1 \cdot \frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} - 1}{x}}}{n} \]
   9. Step-by-step derivation

      1. mul-1-neg 45.9%

         \[\leadsto \frac{\color{blue}{-\frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} - 1}{x}}}{n} \]

      2. distribute-neg-frac2 45.9%

         \[\leadsto \frac{\color{blue}{\frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} - 1}{-x}}}{n} \]

      3. sub-neg 45.9%

         \[\leadsto \frac{\frac{\color{blue}{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} + \left(-1\right)}}{-x}}{n} \]

      4. associate-*r/ 45.9%

         \[\leadsto \frac{\frac{\color{blue}{\frac{-1 \cdot \left(0.3333333333333333 \cdot \frac{1}{x} - 0.5\right)}{x}} + \left(-1\right)}{-x}}{n} \]

      5. sub-neg 45.9%

         \[\leadsto \frac{\frac{\frac{-1 \cdot \color{blue}{\left(0.3333333333333333 \cdot \frac{1}{x} + \left(-0.5\right)\right)}}{x} + \left(-1\right)}{-x}}{n} \]

      6. metadata-eval 45.9%

         \[\leadsto \frac{\frac{\frac{-1 \cdot \left(0.3333333333333333 \cdot \frac{1}{x} + \color{blue}{-0.5}\right)}{x} + \left(-1\right)}{-x}}{n} \]

      7. distribute-lft-in 45.9%

         \[\leadsto \frac{\frac{\frac{\color{blue}{-1 \cdot \left(0.3333333333333333 \cdot \frac{1}{x}\right) + -1 \cdot -0.5}}{x} + \left(-1\right)}{-x}}{n} \]

      8. neg-mul-1 45.9%

         \[\leadsto \frac{\frac{\frac{\color{blue}{\left(-0.3333333333333333 \cdot \frac{1}{x}\right)} + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      9. associate-*r/ 45.9%

         \[\leadsto \frac{\frac{\frac{\left(-\color{blue}{\frac{0.3333333333333333 \cdot 1}{x}}\right) + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      10. metadata-eval 45.9%

          \[\leadsto \frac{\frac{\frac{\left(-\frac{\color{blue}{0.3333333333333333}}{x}\right) + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      11. distribute-neg-frac 45.9%

          \[\leadsto \frac{\frac{\frac{\color{blue}{\frac{-0.3333333333333333}{x}} + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      12. metadata-eval 45.9%

          \[\leadsto \frac{\frac{\frac{\frac{\color{blue}{-0.3333333333333333}}{x} + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      13. metadata-eval 45.9%

          \[\leadsto \frac{\frac{\frac{\frac{-0.3333333333333333}{x} + \color{blue}{0.5}}{x} + \left(-1\right)}{-x}}{n} \]

      14. metadata-eval 45.9%

          \[\leadsto \frac{\frac{\frac{\frac{-0.3333333333333333}{x} + 0.5}{x} + \color{blue}{-1}}{-x}}{n} \]

   10. Simplified 45.9%

       \[\leadsto \frac{\color{blue}{\frac{\frac{\frac{-0.3333333333333333}{x} + 0.5}{x} + -1}{-x}}}{n} \]

   if 3.09999999999999981e97 < x

   1. Initial program 79.6%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 79.6%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   4. Step-by-step derivation

      1. log1p-define 79.6%

         \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   5. Simplified 79.6%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]

   6. Taylor expanded in x around -inf 62.5%

      \[\leadsto \frac{\color{blue}{-1 \cdot \frac{-1 \cdot \frac{-1 \cdot \frac{0.25 \cdot \frac{1}{x} - 0.3333333333333333}{x} - 0.5}{x} - 1}{x}}}{n} \]

   7. Taylor expanded in x around 0 79.6%

      \[\leadsto \color{blue}{\frac{-0.25}{n \cdot {x}^{4}}} \]
3. Recombined 4 regimes into one program.

4. Final simplification 63.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 1.18 \cdot 10^{-79}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;x \leq 3.2 \cdot 10^{-34}:\\ \;\;\;\;\frac{\log x}{-n}\\ \mathbf{elif}\;x \leq 3.1 \cdot 10^{+97}:\\ \;\;\;\;\frac{\frac{\frac{0.5 + \frac{-0.3333333333333333}{x}}{x} + -1}{-x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{-0.25}{n \cdot {x}^{4}}\\ \end{array} \]
5. Add Preprocessing

Alternative 11: 52.4% accurate, 1.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;x \leq 1.6 \cdot 10^{-79}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;x \leq 3.1 \cdot 10^{-34}:\\ \;\;\;\;\frac{\log x}{-n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{\frac{0.5 + \frac{-0.3333333333333333}{x}}{x} + -1}{-x}}{n}\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (if (<= x 1.6e-79)
   (- 1.0 (pow x (/ 1.0 n)))
   (if (<= x 3.1e-34)
     (/ (log x) (- n))
     (/ (/ (+ (/ (+ 0.5 (/ -0.3333333333333333 x)) x) -1.0) (- x)) n))))

C

double code(double x, double n) {
	double tmp;
	if (x <= 1.6e-79) {
		tmp = 1.0 - pow(x, (1.0 / n));
	} else if (x <= 3.1e-34) {
		tmp = log(x) / -n;
	} else {
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n;
	}
	return tmp;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    real(8) :: tmp
    if (x <= 1.6d-79) then
        tmp = 1.0d0 - (x ** (1.0d0 / n))
    else if (x <= 3.1d-34) then
        tmp = log(x) / -n
    else
        tmp = ((((0.5d0 + ((-0.3333333333333333d0) / x)) / x) + (-1.0d0)) / -x) / n
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double tmp;
	if (x <= 1.6e-79) {
		tmp = 1.0 - Math.pow(x, (1.0 / n));
	} else if (x <= 3.1e-34) {
		tmp = Math.log(x) / -n;
	} else {
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n;
	}
	return tmp;
}

Python

def code(x, n):
	tmp = 0
	if x <= 1.6e-79:
		tmp = 1.0 - math.pow(x, (1.0 / n))
	elif x <= 3.1e-34:
		tmp = math.log(x) / -n
	else:
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n
	return tmp

Julia

function code(x, n)
	tmp = 0.0
	if (x <= 1.6e-79)
		tmp = Float64(1.0 - (x ^ Float64(1.0 / n)));
	elseif (x <= 3.1e-34)
		tmp = Float64(log(x) / Float64(-n));
	else
		tmp = Float64(Float64(Float64(Float64(Float64(0.5 + Float64(-0.3333333333333333 / x)) / x) + -1.0) / Float64(-x)) / n);
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	tmp = 0.0;
	if (x <= 1.6e-79)
		tmp = 1.0 - (x ^ (1.0 / n));
	elseif (x <= 3.1e-34)
		tmp = log(x) / -n;
	else
		tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, n_] := If[LessEqual[x, 1.6e-79], N[(1.0 - N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[x, 3.1e-34], N[(N[Log[x], $MachinePrecision] / (-n)), $MachinePrecision], N[(N[(N[(N[(N[(0.5 + N[(-0.3333333333333333 / x), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision] + -1.0), $MachinePrecision] / (-x)), $MachinePrecision] / n), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if x < 1.59999999999999994e-79

   1. Initial program 59.9%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in x around 0 59.9%

      \[\leadsto \color{blue}{1 - e^{\frac{\log x}{n}}} \]
   4. Step-by-step derivation

      1. *-rgt-identity 59.9%

         \[\leadsto 1 - e^{\frac{\color{blue}{\log x \cdot 1}}{n}} \]

      2. associate-/l* 59.9%

         \[\leadsto 1 - e^{\color{blue}{\log x \cdot \frac{1}{n}}} \]

      3. exp-to-pow 59.9%

         \[\leadsto 1 - \color{blue}{{x}^{\left(\frac{1}{n}\right)}} \]

   5. Simplified 59.9%

      \[\leadsto \color{blue}{1 - {x}^{\left(\frac{1}{n}\right)}} \]

   if 1.59999999999999994e-79 < x < 3.0999999999999998e-34

   1. Initial program 26.7%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 62.9%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   4. Step-by-step derivation

      1. log1p-define 62.9%

         \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   5. Simplified 62.9%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]

   6. Taylor expanded in x around 0 62.9%

      \[\leadsto \frac{\color{blue}{-1 \cdot \log x}}{n} \]
   7. Step-by-step derivation

      1. neg-mul-1 62.9%

         \[\leadsto \frac{\color{blue}{-\log x}}{n} \]

   8. Simplified 62.9%

      \[\leadsto \frac{\color{blue}{-\log x}}{n} \]

   if 3.0999999999999998e-34 < x

   1. Initial program 71.2%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 65.9%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   4. Step-by-step derivation

      1. log1p-define 65.9%

         \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   5. Simplified 65.9%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   6. Step-by-step derivation

      1. expm1-log1p-u 65.9%

         \[\leadsto \frac{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\mathsf{log1p}\left(x\right) - \log x\right)\right)}}{n} \]

   7. Applied egg-rr 65.9%

      \[\leadsto \frac{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\mathsf{log1p}\left(x\right) - \log x\right)\right)}}{n} \]

   8. Taylor expanded in x around -inf 56.5%

      \[\leadsto \frac{\color{blue}{-1 \cdot \frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} - 1}{x}}}{n} \]
   9. Step-by-step derivation

      1. mul-1-neg 56.5%

         \[\leadsto \frac{\color{blue}{-\frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} - 1}{x}}}{n} \]

      2. distribute-neg-frac2 56.5%

         \[\leadsto \frac{\color{blue}{\frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} - 1}{-x}}}{n} \]

      3. sub-neg 56.5%

         \[\leadsto \frac{\frac{\color{blue}{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} + \left(-1\right)}}{-x}}{n} \]

      4. associate-*r/ 56.5%

         \[\leadsto \frac{\frac{\color{blue}{\frac{-1 \cdot \left(0.3333333333333333 \cdot \frac{1}{x} - 0.5\right)}{x}} + \left(-1\right)}{-x}}{n} \]

      5. sub-neg 56.5%

         \[\leadsto \frac{\frac{\frac{-1 \cdot \color{blue}{\left(0.3333333333333333 \cdot \frac{1}{x} + \left(-0.5\right)\right)}}{x} + \left(-1\right)}{-x}}{n} \]

      6. metadata-eval 56.5%

         \[\leadsto \frac{\frac{\frac{-1 \cdot \left(0.3333333333333333 \cdot \frac{1}{x} + \color{blue}{-0.5}\right)}{x} + \left(-1\right)}{-x}}{n} \]

      7. distribute-lft-in 56.5%

         \[\leadsto \frac{\frac{\frac{\color{blue}{-1 \cdot \left(0.3333333333333333 \cdot \frac{1}{x}\right) + -1 \cdot -0.5}}{x} + \left(-1\right)}{-x}}{n} \]

      8. neg-mul-1 56.5%

         \[\leadsto \frac{\frac{\frac{\color{blue}{\left(-0.3333333333333333 \cdot \frac{1}{x}\right)} + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      9. associate-*r/ 56.5%

         \[\leadsto \frac{\frac{\frac{\left(-\color{blue}{\frac{0.3333333333333333 \cdot 1}{x}}\right) + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      10. metadata-eval 56.5%

          \[\leadsto \frac{\frac{\frac{\left(-\frac{\color{blue}{0.3333333333333333}}{x}\right) + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      11. distribute-neg-frac 56.5%

          \[\leadsto \frac{\frac{\frac{\color{blue}{\frac{-0.3333333333333333}{x}} + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      12. metadata-eval 56.5%

          \[\leadsto \frac{\frac{\frac{\frac{\color{blue}{-0.3333333333333333}}{x} + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

      13. metadata-eval 56.5%

          \[\leadsto \frac{\frac{\frac{\frac{-0.3333333333333333}{x} + \color{blue}{0.5}}{x} + \left(-1\right)}{-x}}{n} \]

      14. metadata-eval 56.5%

          \[\leadsto \frac{\frac{\frac{\frac{-0.3333333333333333}{x} + 0.5}{x} + \color{blue}{-1}}{-x}}{n} \]

   10. Simplified 56.5%

       \[\leadsto \frac{\color{blue}{\frac{\frac{\frac{-0.3333333333333333}{x} + 0.5}{x} + -1}{-x}}}{n} \]
3. Recombined 3 regimes into one program.

4. Final simplification 58.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 1.6 \cdot 10^{-79}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;x \leq 3.1 \cdot 10^{-34}:\\ \;\;\;\;\frac{\log x}{-n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{\frac{0.5 + \frac{-0.3333333333333333}{x}}{x} + -1}{-x}}{n}\\ \end{array} \]
5. Add Preprocessing

Alternative 12: 46.8% accurate, 1.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq 1.1 \cdot 10^{+164}:\\ \;\;\;\;\frac{\frac{1}{n} + \frac{0.3333333333333333 \cdot \frac{1}{n \cdot x} + 0.5 \cdot \frac{-1}{n}}{x}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{\log x}{-n}\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (if (<= n 1.1e+164)
   (/
    (+
     (/ 1.0 n)
     (/ (+ (* 0.3333333333333333 (/ 1.0 (* n x))) (* 0.5 (/ -1.0 n))) x))
    x)
   (/ (log x) (- n))))

C

double code(double x, double n) {
	double tmp;
	if (n <= 1.1e+164) {
		tmp = ((1.0 / n) + (((0.3333333333333333 * (1.0 / (n * x))) + (0.5 * (-1.0 / n))) / x)) / x;
	} else {
		tmp = log(x) / -n;
	}
	return tmp;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    real(8) :: tmp
    if (n <= 1.1d+164) then
        tmp = ((1.0d0 / n) + (((0.3333333333333333d0 * (1.0d0 / (n * x))) + (0.5d0 * ((-1.0d0) / n))) / x)) / x
    else
        tmp = log(x) / -n
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double tmp;
	if (n <= 1.1e+164) {
		tmp = ((1.0 / n) + (((0.3333333333333333 * (1.0 / (n * x))) + (0.5 * (-1.0 / n))) / x)) / x;
	} else {
		tmp = Math.log(x) / -n;
	}
	return tmp;
}

Python

def code(x, n):
	tmp = 0
	if n <= 1.1e+164:
		tmp = ((1.0 / n) + (((0.3333333333333333 * (1.0 / (n * x))) + (0.5 * (-1.0 / n))) / x)) / x
	else:
		tmp = math.log(x) / -n
	return tmp

Julia

function code(x, n)
	tmp = 0.0
	if (n <= 1.1e+164)
		tmp = Float64(Float64(Float64(1.0 / n) + Float64(Float64(Float64(0.3333333333333333 * Float64(1.0 / Float64(n * x))) + Float64(0.5 * Float64(-1.0 / n))) / x)) / x);
	else
		tmp = Float64(log(x) / Float64(-n));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	tmp = 0.0;
	if (n <= 1.1e+164)
		tmp = ((1.0 / n) + (((0.3333333333333333 * (1.0 / (n * x))) + (0.5 * (-1.0 / n))) / x)) / x;
	else
		tmp = log(x) / -n;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, n_] := If[LessEqual[n, 1.1e+164], N[(N[(N[(1.0 / n), $MachinePrecision] + N[(N[(N[(0.3333333333333333 * N[(1.0 / N[(n * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(0.5 * N[(-1.0 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision], N[(N[Log[x], $MachinePrecision] / (-n)), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < 1.10000000000000003e164

   1. Initial program 65.3%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 46.3%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   4. Step-by-step derivation

      1. log1p-define 46.3%

         \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   5. Simplified 46.3%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]

   6. Taylor expanded in x around -inf 49.6%

      \[\leadsto \color{blue}{-1 \cdot \frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{n \cdot x} - 0.5 \cdot \frac{1}{n}}{x} - \frac{1}{n}}{x}} \]

   if 1.10000000000000003e164 < n

   1. Initial program 39.6%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 96.4%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   4. Step-by-step derivation

      1. log1p-define 96.4%

         \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   5. Simplified 96.4%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]

   6. Taylor expanded in x around 0 60.7%

      \[\leadsto \frac{\color{blue}{-1 \cdot \log x}}{n} \]
   7. Step-by-step derivation

      1. neg-mul-1 60.7%

         \[\leadsto \frac{\color{blue}{-\log x}}{n} \]

   8. Simplified 60.7%

      \[\leadsto \frac{\color{blue}{-\log x}}{n} \]
3. Recombined 2 regimes into one program.

4. Final simplification 51.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq 1.1 \cdot 10^{+164}:\\ \;\;\;\;\frac{\frac{1}{n} + \frac{0.3333333333333333 \cdot \frac{1}{n \cdot x} + 0.5 \cdot \frac{-1}{n}}{x}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{\log x}{-n}\\ \end{array} \]
5. Add Preprocessing

Alternative 13: 46.5% accurate, 10.0× speedup

Math

\[\begin{array}{l} \\ \frac{\frac{1}{n} + \frac{0.3333333333333333 \cdot \frac{1}{n \cdot x} + 0.5 \cdot \frac{-1}{n}}{x}}{x} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (/
  (+
   (/ 1.0 n)
   (/ (+ (* 0.3333333333333333 (/ 1.0 (* n x))) (* 0.5 (/ -1.0 n))) x))
  x))

C

double code(double x, double n) {
	return ((1.0 / n) + (((0.3333333333333333 * (1.0 / (n * x))) + (0.5 * (-1.0 / n))) / x)) / x;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    code = ((1.0d0 / n) + (((0.3333333333333333d0 * (1.0d0 / (n * x))) + (0.5d0 * ((-1.0d0) / n))) / x)) / x
end function

Java

public static double code(double x, double n) {
	return ((1.0 / n) + (((0.3333333333333333 * (1.0 / (n * x))) + (0.5 * (-1.0 / n))) / x)) / x;
}

Python

def code(x, n):
	return ((1.0 / n) + (((0.3333333333333333 * (1.0 / (n * x))) + (0.5 * (-1.0 / n))) / x)) / x

Julia

function code(x, n)
	return Float64(Float64(Float64(1.0 / n) + Float64(Float64(Float64(0.3333333333333333 * Float64(1.0 / Float64(n * x))) + Float64(0.5 * Float64(-1.0 / n))) / x)) / x)
end

MATLAB

function tmp = code(x, n)
	tmp = ((1.0 / n) + (((0.3333333333333333 * (1.0 / (n * x))) + (0.5 * (-1.0 / n))) / x)) / x;
end

Wolfram

code[x_, n_] := N[(N[(N[(1.0 / n), $MachinePrecision] + N[(N[(N[(0.3333333333333333 * N[(1.0 / N[(n * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(0.5 * N[(-1.0 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision]

Derivation

1. Initial program 62.2%

   \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
2. Add Preprocessing

3. Taylor expanded in n around inf 52.4%

   \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
4. Step-by-step derivation

   1. log1p-define 52.4%

      \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

5. Simplified 52.4%

   \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]

6. Taylor expanded in x around -inf 48.8%

   \[\leadsto \color{blue}{-1 \cdot \frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{n \cdot x} - 0.5 \cdot \frac{1}{n}}{x} - \frac{1}{n}}{x}} \]

7. Final simplification 48.8%

   \[\leadsto \frac{\frac{1}{n} + \frac{0.3333333333333333 \cdot \frac{1}{n \cdot x} + 0.5 \cdot \frac{-1}{n}}{x}}{x} \]
8. Add Preprocessing

Alternative 14: 46.5% accurate, 15.1× speedup

Math

\[\begin{array}{l} \\ \frac{\frac{\frac{0.5 + \frac{-0.3333333333333333}{x}}{x} + -1}{-x}}{n} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (/ (/ (+ (/ (+ 0.5 (/ -0.3333333333333333 x)) x) -1.0) (- x)) n))

C

double code(double x, double n) {
	return ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    code = ((((0.5d0 + ((-0.3333333333333333d0) / x)) / x) + (-1.0d0)) / -x) / n
end function

Java

public static double code(double x, double n) {
	return ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n;
}

Python

def code(x, n):
	return ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n

Julia

function code(x, n)
	return Float64(Float64(Float64(Float64(Float64(0.5 + Float64(-0.3333333333333333 / x)) / x) + -1.0) / Float64(-x)) / n)
end

MATLAB

function tmp = code(x, n)
	tmp = ((((0.5 + (-0.3333333333333333 / x)) / x) + -1.0) / -x) / n;
end

Wolfram

code[x_, n_] := N[(N[(N[(N[(N[(0.5 + N[(-0.3333333333333333 / x), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision] + -1.0), $MachinePrecision] / (-x)), $MachinePrecision] / n), $MachinePrecision]

Derivation

1. Initial program 62.2%

   \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
2. Add Preprocessing

3. Taylor expanded in n around inf 52.4%

   \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
4. Step-by-step derivation

   1. log1p-define 52.4%

      \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

5. Simplified 52.4%

   \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
6. Step-by-step derivation

   1. expm1-log1p-u 52.1%

      \[\leadsto \frac{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\mathsf{log1p}\left(x\right) - \log x\right)\right)}}{n} \]

7. Applied egg-rr 52.1%

   \[\leadsto \frac{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\mathsf{log1p}\left(x\right) - \log x\right)\right)}}{n} \]

8. Taylor expanded in x around -inf 48.8%

   \[\leadsto \frac{\color{blue}{-1 \cdot \frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} - 1}{x}}}{n} \]
9. Step-by-step derivation

   1. mul-1-neg 48.8%

      \[\leadsto \frac{\color{blue}{-\frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} - 1}{x}}}{n} \]

   2. distribute-neg-frac2 48.8%

      \[\leadsto \frac{\color{blue}{\frac{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} - 1}{-x}}}{n} \]

   3. sub-neg 48.8%

      \[\leadsto \frac{\frac{\color{blue}{-1 \cdot \frac{0.3333333333333333 \cdot \frac{1}{x} - 0.5}{x} + \left(-1\right)}}{-x}}{n} \]

   4. associate-*r/ 48.8%

      \[\leadsto \frac{\frac{\color{blue}{\frac{-1 \cdot \left(0.3333333333333333 \cdot \frac{1}{x} - 0.5\right)}{x}} + \left(-1\right)}{-x}}{n} \]

   5. sub-neg 48.8%

      \[\leadsto \frac{\frac{\frac{-1 \cdot \color{blue}{\left(0.3333333333333333 \cdot \frac{1}{x} + \left(-0.5\right)\right)}}{x} + \left(-1\right)}{-x}}{n} \]

   6. metadata-eval 48.8%

      \[\leadsto \frac{\frac{\frac{-1 \cdot \left(0.3333333333333333 \cdot \frac{1}{x} + \color{blue}{-0.5}\right)}{x} + \left(-1\right)}{-x}}{n} \]

   7. distribute-lft-in 48.8%

      \[\leadsto \frac{\frac{\frac{\color{blue}{-1 \cdot \left(0.3333333333333333 \cdot \frac{1}{x}\right) + -1 \cdot -0.5}}{x} + \left(-1\right)}{-x}}{n} \]

   8. neg-mul-1 48.8%

      \[\leadsto \frac{\frac{\frac{\color{blue}{\left(-0.3333333333333333 \cdot \frac{1}{x}\right)} + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

   9. associate-*r/ 48.8%

      \[\leadsto \frac{\frac{\frac{\left(-\color{blue}{\frac{0.3333333333333333 \cdot 1}{x}}\right) + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

   10. metadata-eval 48.8%

       \[\leadsto \frac{\frac{\frac{\left(-\frac{\color{blue}{0.3333333333333333}}{x}\right) + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

   11. distribute-neg-frac 48.8%

       \[\leadsto \frac{\frac{\frac{\color{blue}{\frac{-0.3333333333333333}{x}} + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

   12. metadata-eval 48.8%

       \[\leadsto \frac{\frac{\frac{\frac{\color{blue}{-0.3333333333333333}}{x} + -1 \cdot -0.5}{x} + \left(-1\right)}{-x}}{n} \]

   13. metadata-eval 48.8%

       \[\leadsto \frac{\frac{\frac{\frac{-0.3333333333333333}{x} + \color{blue}{0.5}}{x} + \left(-1\right)}{-x}}{n} \]

   14. metadata-eval 48.8%

       \[\leadsto \frac{\frac{\frac{\frac{-0.3333333333333333}{x} + 0.5}{x} + \color{blue}{-1}}{-x}}{n} \]

10. Simplified 48.8%

    \[\leadsto \frac{\color{blue}{\frac{\frac{\frac{-0.3333333333333333}{x} + 0.5}{x} + -1}{-x}}}{n} \]

11. Final simplification 48.8%

    \[\leadsto \frac{\frac{\frac{0.5 + \frac{-0.3333333333333333}{x}}{x} + -1}{-x}}{n} \]
12. Add Preprocessing

Alternative 15: 41.0% accurate, 42.2× speedup

Math

\[\begin{array}{l} \\ \frac{\frac{1}{n}}{x} \end{array} \]

FPCore

(FPCore (x n) :precision binary64 (/ (/ 1.0 n) x))

C

double code(double x, double n) {
	return (1.0 / n) / x;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    code = (1.0d0 / n) / x
end function

Java

public static double code(double x, double n) {
	return (1.0 / n) / x;
}

Python

def code(x, n):
	return (1.0 / n) / x

Julia

function code(x, n)
	return Float64(Float64(1.0 / n) / x)
end

MATLAB

function tmp = code(x, n)
	tmp = (1.0 / n) / x;
end

Wolfram

code[x_, n_] := N[(N[(1.0 / n), $MachinePrecision] / x), $MachinePrecision]

Derivation

1. Initial program 62.2%

   \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
2. Add Preprocessing

3. Taylor expanded in n around inf 52.4%

   \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
4. Step-by-step derivation

   1. log1p-define 52.4%

      \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

5. Simplified 52.4%

   \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]

6. Taylor expanded in x around inf 39.7%

   \[\leadsto \color{blue}{\frac{1}{n \cdot x}} \]
7. Step-by-step derivation

   1. associate-/r* 40.2%

      \[\leadsto \color{blue}{\frac{\frac{1}{n}}{x}} \]

8. Simplified 40.2%

   \[\leadsto \color{blue}{\frac{\frac{1}{n}}{x}} \]
9. Add Preprocessing

Alternative 16: 40.6% accurate, 42.2× speedup

Math

\[\begin{array}{l} \\ \frac{1}{n \cdot x} \end{array} \]

FPCore

(FPCore (x n) :precision binary64 (/ 1.0 (* n x)))

C

double code(double x, double n) {
	return 1.0 / (n * x);
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    code = 1.0d0 / (n * x)
end function

Java

public static double code(double x, double n) {
	return 1.0 / (n * x);
}

Python

def code(x, n):
	return 1.0 / (n * x)

Julia

function code(x, n)
	return Float64(1.0 / Float64(n * x))
end

MATLAB

function tmp = code(x, n)
	tmp = 1.0 / (n * x);
end

Wolfram

code[x_, n_] := N[(1.0 / N[(n * x), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 62.2%

   \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
2. Add Preprocessing

3. Taylor expanded in n around inf 52.4%

   \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
4. Step-by-step derivation

   1. log1p-define 52.4%

      \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

5. Simplified 52.4%

   \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]

6. Taylor expanded in x around inf 39.7%

   \[\leadsto \color{blue}{\frac{1}{n \cdot x}} \]
7. Add Preprocessing

Alternative 17: 4.5% accurate, 70.3× speedup

Math

\[\begin{array}{l} \\ \frac{x}{n} \end{array} \]

FPCore

(FPCore (x n) :precision binary64 (/ x n))

C

double code(double x, double n) {
	return x / n;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    code = x / n
end function

Java

public static double code(double x, double n) {
	return x / n;
}

Python

def code(x, n):
	return x / n

Julia

function code(x, n)
	return Float64(x / n)
end

MATLAB

function tmp = code(x, n)
	tmp = x / n;
end

Wolfram

code[x_, n_] := N[(x / n), $MachinePrecision]

Derivation

1. Initial program 62.2%

   \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
2. Add Preprocessing

3. Taylor expanded in x around 0 38.4%

   \[\leadsto \color{blue}{\left(1 + \frac{x}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]

4. Taylor expanded in x around inf 4.5%

   \[\leadsto \color{blue}{\frac{x}{n}} \]
5. Add Preprocessing

Reproduce

herbie shell --seed 2024118 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))