2nthrt (problem 3.4.6)

Percentage Accurate: 53.5% → 86.6%
Time: 25.7s
Alternatives: 21
Speedup: 3.8×

Specification

?
\[\begin{array}{l} \\ {\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \end{array} \]
(FPCore (x n)
 :precision binary64
 (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))
double code(double x, double n) {
	return pow((x + 1.0), (1.0 / n)) - pow(x, (1.0 / n));
}
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
public static double code(double x, double n) {
	return Math.pow((x + 1.0), (1.0 / n)) - Math.pow(x, (1.0 / n));
}
def code(x, n):
	return math.pow((x + 1.0), (1.0 / n)) - math.pow(x, (1.0 / n))
function code(x, n)
	return Float64((Float64(x + 1.0) ^ Float64(1.0 / n)) - (x ^ Float64(1.0 / n)))
end
function tmp = code(x, n)
	tmp = ((x + 1.0) ^ (1.0 / n)) - (x ^ (1.0 / n));
end
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]
\begin{array}{l}

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

Sampling outcomes in binary64 precision:

Local Percentage Accuracy vs ?

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

Accuracy vs Speed?

Herbie found 21 alternatives:

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

Initial Program: 53.5% accurate, 1.0× speedup?

\[\begin{array}{l} \\ {\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \end{array} \]
(FPCore (x n)
 :precision binary64
 (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))
double code(double x, double n) {
	return pow((x + 1.0), (1.0 / n)) - pow(x, (1.0 / n));
}
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
public static double code(double x, double n) {
	return Math.pow((x + 1.0), (1.0 / n)) - Math.pow(x, (1.0 / n));
}
def code(x, n):
	return math.pow((x + 1.0), (1.0 / n)) - math.pow(x, (1.0 / n))
function code(x, n)
	return Float64((Float64(x + 1.0) ^ Float64(1.0 / n)) - (x ^ Float64(1.0 / n)))
end
function tmp = code(x, n)
	tmp = ((x + 1.0) ^ (1.0 / n)) - (x ^ (1.0 / n));
end
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]
\begin{array}{l}

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

Alternative 1: 86.6% accurate, 0.2× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\ \;\;\;\;\frac{t\_0}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-7}:\\ \;\;\;\;\frac{\frac{\mathsf{fma}\left({\left(\mathsf{log1p}\left(x\right)\right)}^{2} - {\log x}^{2}, 0.5, \frac{\mathsf{fma}\left({\left(\mathsf{log1p}\left(x\right)\right)}^{3} - {\log x}^{3}, -0.16666666666666666, \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{4} - {\log x}^{4}}{n} \cdot -0.041666666666666664\right)}{-n}\right)}{n} - \left(\log x - \mathsf{log1p}\left(x\right)\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t\_0\\ \end{array} \end{array} \]
(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))))
   (if (<= (/ 1.0 n) -2e-6)
     (/ t_0 (* x n))
     (if (<= (/ 1.0 n) 5e-7)
       (/
        (-
         (/
          (fma
           (- (pow (log1p x) 2.0) (pow (log x) 2.0))
           0.5
           (/
            (fma
             (- (pow (log1p x) 3.0) (pow (log x) 3.0))
             -0.16666666666666666
             (*
              (/ (- (pow (log1p x) 4.0) (pow (log x) 4.0)) n)
              -0.041666666666666664))
            (- n)))
          n)
         (- (log x) (log1p x)))
        n)
       (- (exp (/ (log1p x) n)) t_0)))))
double code(double x, double n) {
	double t_0 = pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-6) {
		tmp = t_0 / (x * n);
	} else if ((1.0 / n) <= 5e-7) {
		tmp = ((fma((pow(log1p(x), 2.0) - pow(log(x), 2.0)), 0.5, (fma((pow(log1p(x), 3.0) - pow(log(x), 3.0)), -0.16666666666666666, (((pow(log1p(x), 4.0) - pow(log(x), 4.0)) / n) * -0.041666666666666664)) / -n)) / n) - (log(x) - log1p(x))) / n;
	} else {
		tmp = exp((log1p(x) / n)) - t_0;
	}
	return tmp;
}
function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	tmp = 0.0
	if (Float64(1.0 / n) <= -2e-6)
		tmp = Float64(t_0 / Float64(x * n));
	elseif (Float64(1.0 / n) <= 5e-7)
		tmp = Float64(Float64(Float64(fma(Float64((log1p(x) ^ 2.0) - (log(x) ^ 2.0)), 0.5, Float64(fma(Float64((log1p(x) ^ 3.0) - (log(x) ^ 3.0)), -0.16666666666666666, Float64(Float64(Float64((log1p(x) ^ 4.0) - (log(x) ^ 4.0)) / n) * -0.041666666666666664)) / Float64(-n))) / n) - Float64(log(x) - log1p(x))) / n);
	else
		tmp = Float64(exp(Float64(log1p(x) / n)) - t_0);
	end
	return tmp
end
code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-6], N[(t$95$0 / N[(x * n), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 5e-7], N[(N[(N[(N[(N[(N[Power[N[Log[1 + x], $MachinePrecision], 2.0], $MachinePrecision] - N[Power[N[Log[x], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] * 0.5 + N[(N[(N[(N[Power[N[Log[1 + x], $MachinePrecision], 3.0], $MachinePrecision] - N[Power[N[Log[x], $MachinePrecision], 3.0], $MachinePrecision]), $MachinePrecision] * -0.16666666666666666 + N[(N[(N[(N[Power[N[Log[1 + x], $MachinePrecision], 4.0], $MachinePrecision] - N[Power[N[Log[x], $MachinePrecision], 4.0], $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision] * -0.041666666666666664), $MachinePrecision]), $MachinePrecision] / (-n)), $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision] - N[(N[Log[x], $MachinePrecision] - N[Log[1 + x], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision], N[(N[Exp[N[(N[Log[1 + x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision] - t$95$0), $MachinePrecision]]]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := {x}^{\left(\frac{1}{n}\right)}\\
\mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\
\;\;\;\;\frac{t\_0}{x \cdot n}\\

\mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-7}:\\
\;\;\;\;\frac{\frac{\mathsf{fma}\left({\left(\mathsf{log1p}\left(x\right)\right)}^{2} - {\log x}^{2}, 0.5, \frac{\mathsf{fma}\left({\left(\mathsf{log1p}\left(x\right)\right)}^{3} - {\log x}^{3}, -0.16666666666666666, \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{4} - {\log x}^{4}}{n} \cdot -0.041666666666666664\right)}{-n}\right)}{n} - \left(\log x - \mathsf{log1p}\left(x\right)\right)}{n}\\

\mathbf{else}:\\
\;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t\_0\\


\end{array}
\end{array}
Derivation
  1. Split input into 3 regimes
  2. if (/.f64 #s(literal 1 binary64) n) < -1.99999999999999991e-6

    1. Initial program 98.8%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
    2. Add Preprocessing
    3. Taylor expanded in x around inf

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
    4. Step-by-step derivation
      1. lower-/.f64N/A

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

        \[\leadsto \frac{e^{-1 \cdot \frac{\color{blue}{\mathsf{neg}\left(\log x\right)}}{n}}}{n \cdot x} \]
      3. mul-1-negN/A

        \[\leadsto \frac{e^{-1 \cdot \frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]
      4. associate-*r/N/A

        \[\leadsto \frac{e^{\color{blue}{\frac{-1 \cdot \left(-1 \cdot \log x\right)}{n}}}}{n \cdot x} \]
      5. associate-*r*N/A

        \[\leadsto \frac{e^{\frac{\color{blue}{\left(-1 \cdot -1\right) \cdot \log x}}{n}}}{n \cdot x} \]
      6. metadata-evalN/A

        \[\leadsto \frac{e^{\frac{\color{blue}{1} \cdot \log x}{n}}}{n \cdot x} \]
      7. *-commutativeN/A

        \[\leadsto \frac{e^{\frac{\color{blue}{\log x \cdot 1}}{n}}}{n \cdot x} \]
      8. associate-/l*N/A

        \[\leadsto \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{n \cdot x} \]
      9. exp-to-powN/A

        \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
      10. lower-pow.f64N/A

        \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
      11. lower-/.f64N/A

        \[\leadsto \frac{{x}^{\color{blue}{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
      12. lower-*.f6498.9

        \[\leadsto \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{n \cdot x}} \]
    5. Applied rewrites98.9%

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

    if -1.99999999999999991e-6 < (/.f64 #s(literal 1 binary64) n) < 4.99999999999999977e-7

    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

      \[\leadsto \color{blue}{-1 \cdot \frac{\left(-1 \cdot \log \left(1 + x\right) + -1 \cdot \frac{\left(-1 \cdot \frac{\left(-1 \cdot \frac{\frac{1}{24} \cdot {\log \left(1 + x\right)}^{4} - \frac{1}{24} \cdot {\log x}^{4}}{n} + \frac{-1}{6} \cdot {\log \left(1 + x\right)}^{3}\right) - \frac{-1}{6} \cdot {\log x}^{3}}{n} + \frac{1}{2} \cdot {\log \left(1 + x\right)}^{2}\right) - \frac{1}{2} \cdot {\log x}^{2}}{n}\right) - -1 \cdot \log x}{n}} \]
    4. Applied rewrites81.3%

      \[\leadsto \color{blue}{\frac{\left(\log x - \mathsf{log1p}\left(x\right)\right) - \frac{\mathsf{fma}\left({\left(\mathsf{log1p}\left(x\right)\right)}^{2} - {\log x}^{2}, 0.5, \frac{\mathsf{fma}\left({\left(\mathsf{log1p}\left(x\right)\right)}^{3} - {\log x}^{3}, -0.16666666666666666, -0.041666666666666664 \cdot \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{4} - {\log x}^{4}}{n}\right)}{-n}\right)}{n}}{-n}} \]

    if 4.99999999999999977e-7 < (/.f64 #s(literal 1 binary64) n)

    1. Initial program 52.9%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
    2. Add Preprocessing
    3. Step-by-step derivation
      1. lift-pow.f64N/A

        \[\leadsto \color{blue}{{\left(x + 1\right)}^{\left(\frac{1}{n}\right)}} - {x}^{\left(\frac{1}{n}\right)} \]
      2. pow-to-expN/A

        \[\leadsto \color{blue}{e^{\log \left(x + 1\right) \cdot \frac{1}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
      3. lower-exp.f64N/A

        \[\leadsto \color{blue}{e^{\log \left(x + 1\right) \cdot \frac{1}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
      4. lift-/.f64N/A

        \[\leadsto e^{\log \left(x + 1\right) \cdot \color{blue}{\frac{1}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
      5. un-div-invN/A

        \[\leadsto e^{\color{blue}{\frac{\log \left(x + 1\right)}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
      6. lower-/.f64N/A

        \[\leadsto e^{\color{blue}{\frac{\log \left(x + 1\right)}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
      7. lift-+.f64N/A

        \[\leadsto e^{\frac{\log \color{blue}{\left(x + 1\right)}}{n}} - {x}^{\left(\frac{1}{n}\right)} \]
      8. +-commutativeN/A

        \[\leadsto e^{\frac{\log \color{blue}{\left(1 + x\right)}}{n}} - {x}^{\left(\frac{1}{n}\right)} \]
      9. lower-log1p.f6497.3

        \[\leadsto e^{\frac{\color{blue}{\mathsf{log1p}\left(x\right)}}{n}} - {x}^{\left(\frac{1}{n}\right)} \]
    4. Applied rewrites97.3%

      \[\leadsto \color{blue}{e^{\frac{\mathsf{log1p}\left(x\right)}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification89.7%

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-7}:\\ \;\;\;\;\frac{\frac{\mathsf{fma}\left({\left(\mathsf{log1p}\left(x\right)\right)}^{2} - {\log x}^{2}, 0.5, \frac{\mathsf{fma}\left({\left(\mathsf{log1p}\left(x\right)\right)}^{3} - {\log x}^{3}, -0.16666666666666666, \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{4} - {\log x}^{4}}{n} \cdot -0.041666666666666664\right)}{-n}\right)}{n} - \left(\log x - \mathsf{log1p}\left(x\right)\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
  5. Add Preprocessing

Alternative 2: 86.5% accurate, 0.2× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\ \;\;\;\;\frac{t\_0}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-10}:\\ \;\;\;\;\frac{\frac{\mathsf{fma}\left({\left(\mathsf{log1p}\left(x\right)\right)}^{2} - {\log x}^{2}, 0.5, \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{3} - {\log x}^{3}}{n} \cdot 0.16666666666666666\right)}{n} - \left(\log x - \mathsf{log1p}\left(x\right)\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t\_0\\ \end{array} \end{array} \]
(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))))
   (if (<= (/ 1.0 n) -2e-6)
     (/ t_0 (* x n))
     (if (<= (/ 1.0 n) 2e-10)
       (/
        (-
         (/
          (fma
           (- (pow (log1p x) 2.0) (pow (log x) 2.0))
           0.5
           (*
            (/ (- (pow (log1p x) 3.0) (pow (log x) 3.0)) n)
            0.16666666666666666))
          n)
         (- (log x) (log1p x)))
        n)
       (- (exp (/ (log1p x) n)) t_0)))))
double code(double x, double n) {
	double t_0 = pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-6) {
		tmp = t_0 / (x * n);
	} else if ((1.0 / n) <= 2e-10) {
		tmp = ((fma((pow(log1p(x), 2.0) - pow(log(x), 2.0)), 0.5, (((pow(log1p(x), 3.0) - pow(log(x), 3.0)) / n) * 0.16666666666666666)) / n) - (log(x) - log1p(x))) / n;
	} else {
		tmp = exp((log1p(x) / n)) - t_0;
	}
	return tmp;
}
function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	tmp = 0.0
	if (Float64(1.0 / n) <= -2e-6)
		tmp = Float64(t_0 / Float64(x * n));
	elseif (Float64(1.0 / n) <= 2e-10)
		tmp = Float64(Float64(Float64(fma(Float64((log1p(x) ^ 2.0) - (log(x) ^ 2.0)), 0.5, Float64(Float64(Float64((log1p(x) ^ 3.0) - (log(x) ^ 3.0)) / n) * 0.16666666666666666)) / n) - Float64(log(x) - log1p(x))) / n);
	else
		tmp = Float64(exp(Float64(log1p(x) / n)) - t_0);
	end
	return tmp
end
code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-6], N[(t$95$0 / N[(x * n), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e-10], N[(N[(N[(N[(N[(N[Power[N[Log[1 + x], $MachinePrecision], 2.0], $MachinePrecision] - N[Power[N[Log[x], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] * 0.5 + N[(N[(N[(N[Power[N[Log[1 + x], $MachinePrecision], 3.0], $MachinePrecision] - N[Power[N[Log[x], $MachinePrecision], 3.0], $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision] * 0.16666666666666666), $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision] - N[(N[Log[x], $MachinePrecision] - N[Log[1 + x], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision], N[(N[Exp[N[(N[Log[1 + x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision] - t$95$0), $MachinePrecision]]]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := {x}^{\left(\frac{1}{n}\right)}\\
\mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\
\;\;\;\;\frac{t\_0}{x \cdot n}\\

\mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-10}:\\
\;\;\;\;\frac{\frac{\mathsf{fma}\left({\left(\mathsf{log1p}\left(x\right)\right)}^{2} - {\log x}^{2}, 0.5, \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{3} - {\log x}^{3}}{n} \cdot 0.16666666666666666\right)}{n} - \left(\log x - \mathsf{log1p}\left(x\right)\right)}{n}\\

\mathbf{else}:\\
\;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t\_0\\


\end{array}
\end{array}
Derivation
  1. Split input into 3 regimes
  2. if (/.f64 #s(literal 1 binary64) n) < -1.99999999999999991e-6

    1. Initial program 98.8%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
    2. Add Preprocessing
    3. Taylor expanded in x around inf

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
    4. Step-by-step derivation
      1. lower-/.f64N/A

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

        \[\leadsto \frac{e^{-1 \cdot \frac{\color{blue}{\mathsf{neg}\left(\log x\right)}}{n}}}{n \cdot x} \]
      3. mul-1-negN/A

        \[\leadsto \frac{e^{-1 \cdot \frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]
      4. associate-*r/N/A

        \[\leadsto \frac{e^{\color{blue}{\frac{-1 \cdot \left(-1 \cdot \log x\right)}{n}}}}{n \cdot x} \]
      5. associate-*r*N/A

        \[\leadsto \frac{e^{\frac{\color{blue}{\left(-1 \cdot -1\right) \cdot \log x}}{n}}}{n \cdot x} \]
      6. metadata-evalN/A

        \[\leadsto \frac{e^{\frac{\color{blue}{1} \cdot \log x}{n}}}{n \cdot x} \]
      7. *-commutativeN/A

        \[\leadsto \frac{e^{\frac{\color{blue}{\log x \cdot 1}}{n}}}{n \cdot x} \]
      8. associate-/l*N/A

        \[\leadsto \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{n \cdot x} \]
      9. exp-to-powN/A

        \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
      10. lower-pow.f64N/A

        \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
      11. lower-/.f64N/A

        \[\leadsto \frac{{x}^{\color{blue}{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
      12. lower-*.f6498.9

        \[\leadsto \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{n \cdot x}} \]
    5. Applied rewrites98.9%

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

    if -1.99999999999999991e-6 < (/.f64 #s(literal 1 binary64) n) < 2.00000000000000007e-10

    1. Initial program 26.2%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
    2. Add Preprocessing
    3. Taylor expanded in n around -inf

      \[\leadsto \color{blue}{-1 \cdot \frac{\left(-1 \cdot \log \left(1 + x\right) + -1 \cdot \frac{\left(-1 \cdot \frac{\frac{-1}{6} \cdot {\log \left(1 + x\right)}^{3} - \frac{-1}{6} \cdot {\log x}^{3}}{n} + \frac{1}{2} \cdot {\log \left(1 + x\right)}^{2}\right) - \frac{1}{2} \cdot {\log x}^{2}}{n}\right) - -1 \cdot \log x}{n}} \]
    4. Applied rewrites81.1%

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

    if 2.00000000000000007e-10 < (/.f64 #s(literal 1 binary64) n)

    1. Initial program 53.7%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
    2. Add Preprocessing
    3. Step-by-step derivation
      1. lift-pow.f64N/A

        \[\leadsto \color{blue}{{\left(x + 1\right)}^{\left(\frac{1}{n}\right)}} - {x}^{\left(\frac{1}{n}\right)} \]
      2. pow-to-expN/A

        \[\leadsto \color{blue}{e^{\log \left(x + 1\right) \cdot \frac{1}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
      3. lower-exp.f64N/A

        \[\leadsto \color{blue}{e^{\log \left(x + 1\right) \cdot \frac{1}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
      4. lift-/.f64N/A

        \[\leadsto e^{\log \left(x + 1\right) \cdot \color{blue}{\frac{1}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
      5. un-div-invN/A

        \[\leadsto e^{\color{blue}{\frac{\log \left(x + 1\right)}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
      6. lower-/.f64N/A

        \[\leadsto e^{\color{blue}{\frac{\log \left(x + 1\right)}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
      7. lift-+.f64N/A

        \[\leadsto e^{\frac{\log \color{blue}{\left(x + 1\right)}}{n}} - {x}^{\left(\frac{1}{n}\right)} \]
      8. +-commutativeN/A

        \[\leadsto e^{\frac{\log \color{blue}{\left(1 + x\right)}}{n}} - {x}^{\left(\frac{1}{n}\right)} \]
      9. lower-log1p.f6497.0

        \[\leadsto e^{\frac{\color{blue}{\mathsf{log1p}\left(x\right)}}{n}} - {x}^{\left(\frac{1}{n}\right)} \]
    4. Applied rewrites97.0%

      \[\leadsto \color{blue}{e^{\frac{\mathsf{log1p}\left(x\right)}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification89.7%

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-10}:\\ \;\;\;\;\frac{\frac{\mathsf{fma}\left({\left(\mathsf{log1p}\left(x\right)\right)}^{2} - {\log x}^{2}, 0.5, \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{3} - {\log x}^{3}}{n} \cdot 0.16666666666666666\right)}{n} - \left(\log x - \mathsf{log1p}\left(x\right)\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
  5. Add Preprocessing

Alternative 3: 86.5% accurate, 0.3× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\ \;\;\;\;\frac{t\_0}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-10}:\\ \;\;\;\;\frac{\mathsf{fma}\left(\frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2} - {\log x}^{2}}{n}, 0.5, \mathsf{log1p}\left(x\right) - \log x\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t\_0\\ \end{array} \end{array} \]
(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))))
   (if (<= (/ 1.0 n) -2e-6)
     (/ t_0 (* x n))
     (if (<= (/ 1.0 n) 2e-10)
       (/
        (fma
         (/ (- (pow (log1p x) 2.0) (pow (log x) 2.0)) n)
         0.5
         (- (log1p x) (log x)))
        n)
       (- (exp (/ (log1p x) n)) t_0)))))
double code(double x, double n) {
	double t_0 = pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-6) {
		tmp = t_0 / (x * n);
	} else if ((1.0 / n) <= 2e-10) {
		tmp = fma(((pow(log1p(x), 2.0) - pow(log(x), 2.0)) / n), 0.5, (log1p(x) - log(x))) / n;
	} else {
		tmp = exp((log1p(x) / n)) - t_0;
	}
	return tmp;
}
function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	tmp = 0.0
	if (Float64(1.0 / n) <= -2e-6)
		tmp = Float64(t_0 / Float64(x * n));
	elseif (Float64(1.0 / n) <= 2e-10)
		tmp = Float64(fma(Float64(Float64((log1p(x) ^ 2.0) - (log(x) ^ 2.0)) / n), 0.5, Float64(log1p(x) - log(x))) / n);
	else
		tmp = Float64(exp(Float64(log1p(x) / n)) - t_0);
	end
	return tmp
end
code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-6], N[(t$95$0 / N[(x * n), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e-10], N[(N[(N[(N[(N[Power[N[Log[1 + x], $MachinePrecision], 2.0], $MachinePrecision] - N[Power[N[Log[x], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision] * 0.5 + N[(N[Log[1 + x], $MachinePrecision] - N[Log[x], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision], N[(N[Exp[N[(N[Log[1 + x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision] - t$95$0), $MachinePrecision]]]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := {x}^{\left(\frac{1}{n}\right)}\\
\mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\
\;\;\;\;\frac{t\_0}{x \cdot n}\\

\mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-10}:\\
\;\;\;\;\frac{\mathsf{fma}\left(\frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2} - {\log x}^{2}}{n}, 0.5, \mathsf{log1p}\left(x\right) - \log x\right)}{n}\\

\mathbf{else}:\\
\;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t\_0\\


\end{array}
\end{array}
Derivation
  1. Split input into 3 regimes
  2. if (/.f64 #s(literal 1 binary64) n) < -1.99999999999999991e-6

    1. Initial program 98.8%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
    2. Add Preprocessing
    3. Taylor expanded in x around inf

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
    4. Step-by-step derivation
      1. lower-/.f64N/A

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

        \[\leadsto \frac{e^{-1 \cdot \frac{\color{blue}{\mathsf{neg}\left(\log x\right)}}{n}}}{n \cdot x} \]
      3. mul-1-negN/A

        \[\leadsto \frac{e^{-1 \cdot \frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]
      4. associate-*r/N/A

        \[\leadsto \frac{e^{\color{blue}{\frac{-1 \cdot \left(-1 \cdot \log x\right)}{n}}}}{n \cdot x} \]
      5. associate-*r*N/A

        \[\leadsto \frac{e^{\frac{\color{blue}{\left(-1 \cdot -1\right) \cdot \log x}}{n}}}{n \cdot x} \]
      6. metadata-evalN/A

        \[\leadsto \frac{e^{\frac{\color{blue}{1} \cdot \log x}{n}}}{n \cdot x} \]
      7. *-commutativeN/A

        \[\leadsto \frac{e^{\frac{\color{blue}{\log x \cdot 1}}{n}}}{n \cdot x} \]
      8. associate-/l*N/A

        \[\leadsto \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{n \cdot x} \]
      9. exp-to-powN/A

        \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
      10. lower-pow.f64N/A

        \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
      11. lower-/.f64N/A

        \[\leadsto \frac{{x}^{\color{blue}{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
      12. lower-*.f6498.9

        \[\leadsto \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{n \cdot x}} \]
    5. Applied rewrites98.9%

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

    if -1.99999999999999991e-6 < (/.f64 #s(literal 1 binary64) n) < 2.00000000000000007e-10

    1. Initial program 26.2%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
    2. Add Preprocessing
    3. Taylor expanded in n around inf

      \[\leadsto \color{blue}{\frac{\left(\log \left(1 + x\right) + \frac{1}{2} \cdot \frac{{\log \left(1 + x\right)}^{2}}{n}\right) - \left(\log x + \frac{1}{2} \cdot \frac{{\log x}^{2}}{n}\right)}{n}} \]
    4. Step-by-step derivation
      1. lower-/.f64N/A

        \[\leadsto \color{blue}{\frac{\left(\log \left(1 + x\right) + \frac{1}{2} \cdot \frac{{\log \left(1 + x\right)}^{2}}{n}\right) - \left(\log x + \frac{1}{2} \cdot \frac{{\log x}^{2}}{n}\right)}{n}} \]
    5. Applied rewrites81.1%

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

    if 2.00000000000000007e-10 < (/.f64 #s(literal 1 binary64) n)

    1. Initial program 53.7%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
    2. Add Preprocessing
    3. Step-by-step derivation
      1. lift-pow.f64N/A

        \[\leadsto \color{blue}{{\left(x + 1\right)}^{\left(\frac{1}{n}\right)}} - {x}^{\left(\frac{1}{n}\right)} \]
      2. pow-to-expN/A

        \[\leadsto \color{blue}{e^{\log \left(x + 1\right) \cdot \frac{1}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
      3. lower-exp.f64N/A

        \[\leadsto \color{blue}{e^{\log \left(x + 1\right) \cdot \frac{1}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
      4. lift-/.f64N/A

        \[\leadsto e^{\log \left(x + 1\right) \cdot \color{blue}{\frac{1}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
      5. un-div-invN/A

        \[\leadsto e^{\color{blue}{\frac{\log \left(x + 1\right)}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
      6. lower-/.f64N/A

        \[\leadsto e^{\color{blue}{\frac{\log \left(x + 1\right)}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
      7. lift-+.f64N/A

        \[\leadsto e^{\frac{\log \color{blue}{\left(x + 1\right)}}{n}} - {x}^{\left(\frac{1}{n}\right)} \]
      8. +-commutativeN/A

        \[\leadsto e^{\frac{\log \color{blue}{\left(1 + x\right)}}{n}} - {x}^{\left(\frac{1}{n}\right)} \]
      9. lower-log1p.f6497.0

        \[\leadsto e^{\frac{\color{blue}{\mathsf{log1p}\left(x\right)}}{n}} - {x}^{\left(\frac{1}{n}\right)} \]
    4. Applied rewrites97.0%

      \[\leadsto \color{blue}{e^{\frac{\mathsf{log1p}\left(x\right)}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification89.6%

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-10}:\\ \;\;\;\;\frac{\mathsf{fma}\left(\frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2} - {\log x}^{2}}{n}, 0.5, \mathsf{log1p}\left(x\right) - \log x\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
  5. Add Preprocessing

Alternative 4: 78.6% accurate, 0.4× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ t_1 := {\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - t\_0\\ t_2 := 1 - t\_0\\ \mathbf{if}\;t\_1 \leq -0.05:\\ \;\;\;\;t\_2\\ \mathbf{elif}\;t\_1 \leq 5 \cdot 10^{-12}:\\ \;\;\;\;\frac{\log \left(\frac{x}{x + 1}\right)}{-n}\\ \mathbf{else}:\\ \;\;\;\;t\_2\\ \end{array} \end{array} \]
(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n)))
        (t_1 (- (pow (+ x 1.0) (/ 1.0 n)) t_0))
        (t_2 (- 1.0 t_0)))
   (if (<= t_1 -0.05)
     t_2
     (if (<= t_1 5e-12) (/ (log (/ x (+ x 1.0))) (- n)) t_2))))
double code(double x, double n) {
	double t_0 = pow(x, (1.0 / n));
	double t_1 = pow((x + 1.0), (1.0 / n)) - t_0;
	double t_2 = 1.0 - t_0;
	double tmp;
	if (t_1 <= -0.05) {
		tmp = t_2;
	} else if (t_1 <= 5e-12) {
		tmp = log((x / (x + 1.0))) / -n;
	} else {
		tmp = t_2;
	}
	return tmp;
}
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) :: t_2
    real(8) :: tmp
    t_0 = x ** (1.0d0 / n)
    t_1 = ((x + 1.0d0) ** (1.0d0 / n)) - t_0
    t_2 = 1.0d0 - t_0
    if (t_1 <= (-0.05d0)) then
        tmp = t_2
    else if (t_1 <= 5d-12) then
        tmp = log((x / (x + 1.0d0))) / -n
    else
        tmp = t_2
    end if
    code = tmp
end function
public static double code(double x, double n) {
	double t_0 = Math.pow(x, (1.0 / n));
	double t_1 = Math.pow((x + 1.0), (1.0 / n)) - t_0;
	double t_2 = 1.0 - t_0;
	double tmp;
	if (t_1 <= -0.05) {
		tmp = t_2;
	} else if (t_1 <= 5e-12) {
		tmp = Math.log((x / (x + 1.0))) / -n;
	} else {
		tmp = t_2;
	}
	return tmp;
}
def code(x, n):
	t_0 = math.pow(x, (1.0 / n))
	t_1 = math.pow((x + 1.0), (1.0 / n)) - t_0
	t_2 = 1.0 - t_0
	tmp = 0
	if t_1 <= -0.05:
		tmp = t_2
	elif t_1 <= 5e-12:
		tmp = math.log((x / (x + 1.0))) / -n
	else:
		tmp = t_2
	return tmp
function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	t_1 = Float64((Float64(x + 1.0) ^ Float64(1.0 / n)) - t_0)
	t_2 = Float64(1.0 - t_0)
	tmp = 0.0
	if (t_1 <= -0.05)
		tmp = t_2;
	elseif (t_1 <= 5e-12)
		tmp = Float64(log(Float64(x / Float64(x + 1.0))) / Float64(-n));
	else
		tmp = t_2;
	end
	return tmp
end
function tmp_2 = code(x, n)
	t_0 = x ^ (1.0 / n);
	t_1 = ((x + 1.0) ^ (1.0 / n)) - t_0;
	t_2 = 1.0 - t_0;
	tmp = 0.0;
	if (t_1 <= -0.05)
		tmp = t_2;
	elseif (t_1 <= 5e-12)
		tmp = log((x / (x + 1.0))) / -n;
	else
		tmp = t_2;
	end
	tmp_2 = tmp;
end
code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, Block[{t$95$1 = N[(N[Power[N[(x + 1.0), $MachinePrecision], N[(1.0 / n), $MachinePrecision]], $MachinePrecision] - t$95$0), $MachinePrecision]}, Block[{t$95$2 = N[(1.0 - t$95$0), $MachinePrecision]}, If[LessEqual[t$95$1, -0.05], t$95$2, If[LessEqual[t$95$1, 5e-12], N[(N[Log[N[(x / N[(x + 1.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-n)), $MachinePrecision], t$95$2]]]]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := {x}^{\left(\frac{1}{n}\right)}\\
t_1 := {\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - t\_0\\
t_2 := 1 - t\_0\\
\mathbf{if}\;t\_1 \leq -0.05:\\
\;\;\;\;t\_2\\

\mathbf{elif}\;t\_1 \leq 5 \cdot 10^{-12}:\\
\;\;\;\;\frac{\log \left(\frac{x}{x + 1}\right)}{-n}\\

\mathbf{else}:\\
\;\;\;\;t\_2\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (-.f64 (pow.f64 (+.f64 x #s(literal 1 binary64)) (/.f64 #s(literal 1 binary64) n)) (pow.f64 x (/.f64 #s(literal 1 binary64) n))) < -0.050000000000000003 or 4.9999999999999997e-12 < (-.f64 (pow.f64 (+.f64 x #s(literal 1 binary64)) (/.f64 #s(literal 1 binary64) n)) (pow.f64 x (/.f64 #s(literal 1 binary64) n)))

    1. Initial program 78.1%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
    2. Add Preprocessing
    3. Taylor expanded in x around 0

      \[\leadsto \color{blue}{1} - {x}^{\left(\frac{1}{n}\right)} \]
    4. Step-by-step derivation
      1. Applied rewrites75.6%

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

      if -0.050000000000000003 < (-.f64 (pow.f64 (+.f64 x #s(literal 1 binary64)) (/.f64 #s(literal 1 binary64) n)) (pow.f64 x (/.f64 #s(literal 1 binary64) n))) < 4.9999999999999997e-12

      1. Initial program 45.6%

        \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
      2. Add Preprocessing
      3. Taylor expanded in n around inf

        \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
      4. Step-by-step derivation
        1. lower-/.f64N/A

          \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
        2. lower--.f64N/A

          \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
        3. lower-log1p.f64N/A

          \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
        4. lower-log.f6483.3

          \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
      5. Applied rewrites83.3%

        \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
      6. Step-by-step derivation
        1. Applied rewrites82.9%

          \[\leadsto \frac{-\log \left(\frac{x}{1 + x}\right)}{n} \]
      7. Recombined 2 regimes into one program.
      8. Final simplification80.7%

        \[\leadsto \begin{array}{l} \mathbf{if}\;{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \leq -0.05:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \leq 5 \cdot 10^{-12}:\\ \;\;\;\;\frac{\log \left(\frac{x}{x + 1}\right)}{-n}\\ \mathbf{else}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
      9. Add Preprocessing

      Alternative 5: 78.5% accurate, 0.4× speedup?

      \[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ t_1 := {\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - t\_0\\ t_2 := 1 - t\_0\\ \mathbf{if}\;t\_1 \leq -0.05:\\ \;\;\;\;t\_2\\ \mathbf{elif}\;t\_1 \leq 5 \cdot 10^{-12}:\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;t\_2\\ \end{array} \end{array} \]
      (FPCore (x n)
       :precision binary64
       (let* ((t_0 (pow x (/ 1.0 n)))
              (t_1 (- (pow (+ x 1.0) (/ 1.0 n)) t_0))
              (t_2 (- 1.0 t_0)))
         (if (<= t_1 -0.05)
           t_2
           (if (<= t_1 5e-12) (/ (log (/ (+ x 1.0) x)) n) t_2))))
      double code(double x, double n) {
      	double t_0 = pow(x, (1.0 / n));
      	double t_1 = pow((x + 1.0), (1.0 / n)) - t_0;
      	double t_2 = 1.0 - t_0;
      	double tmp;
      	if (t_1 <= -0.05) {
      		tmp = t_2;
      	} else if (t_1 <= 5e-12) {
      		tmp = log(((x + 1.0) / x)) / n;
      	} else {
      		tmp = t_2;
      	}
      	return tmp;
      }
      
      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) :: t_2
          real(8) :: tmp
          t_0 = x ** (1.0d0 / n)
          t_1 = ((x + 1.0d0) ** (1.0d0 / n)) - t_0
          t_2 = 1.0d0 - t_0
          if (t_1 <= (-0.05d0)) then
              tmp = t_2
          else if (t_1 <= 5d-12) then
              tmp = log(((x + 1.0d0) / x)) / n
          else
              tmp = t_2
          end if
          code = tmp
      end function
      
      public static double code(double x, double n) {
      	double t_0 = Math.pow(x, (1.0 / n));
      	double t_1 = Math.pow((x + 1.0), (1.0 / n)) - t_0;
      	double t_2 = 1.0 - t_0;
      	double tmp;
      	if (t_1 <= -0.05) {
      		tmp = t_2;
      	} else if (t_1 <= 5e-12) {
      		tmp = Math.log(((x + 1.0) / x)) / n;
      	} else {
      		tmp = t_2;
      	}
      	return tmp;
      }
      
      def code(x, n):
      	t_0 = math.pow(x, (1.0 / n))
      	t_1 = math.pow((x + 1.0), (1.0 / n)) - t_0
      	t_2 = 1.0 - t_0
      	tmp = 0
      	if t_1 <= -0.05:
      		tmp = t_2
      	elif t_1 <= 5e-12:
      		tmp = math.log(((x + 1.0) / x)) / n
      	else:
      		tmp = t_2
      	return tmp
      
      function code(x, n)
      	t_0 = x ^ Float64(1.0 / n)
      	t_1 = Float64((Float64(x + 1.0) ^ Float64(1.0 / n)) - t_0)
      	t_2 = Float64(1.0 - t_0)
      	tmp = 0.0
      	if (t_1 <= -0.05)
      		tmp = t_2;
      	elseif (t_1 <= 5e-12)
      		tmp = Float64(log(Float64(Float64(x + 1.0) / x)) / n);
      	else
      		tmp = t_2;
      	end
      	return tmp
      end
      
      function tmp_2 = code(x, n)
      	t_0 = x ^ (1.0 / n);
      	t_1 = ((x + 1.0) ^ (1.0 / n)) - t_0;
      	t_2 = 1.0 - t_0;
      	tmp = 0.0;
      	if (t_1 <= -0.05)
      		tmp = t_2;
      	elseif (t_1 <= 5e-12)
      		tmp = log(((x + 1.0) / x)) / n;
      	else
      		tmp = t_2;
      	end
      	tmp_2 = tmp;
      end
      
      code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, Block[{t$95$1 = N[(N[Power[N[(x + 1.0), $MachinePrecision], N[(1.0 / n), $MachinePrecision]], $MachinePrecision] - t$95$0), $MachinePrecision]}, Block[{t$95$2 = N[(1.0 - t$95$0), $MachinePrecision]}, If[LessEqual[t$95$1, -0.05], t$95$2, If[LessEqual[t$95$1, 5e-12], N[(N[Log[N[(N[(x + 1.0), $MachinePrecision] / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision], t$95$2]]]]]
      
      \begin{array}{l}
      
      \\
      \begin{array}{l}
      t_0 := {x}^{\left(\frac{1}{n}\right)}\\
      t_1 := {\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - t\_0\\
      t_2 := 1 - t\_0\\
      \mathbf{if}\;t\_1 \leq -0.05:\\
      \;\;\;\;t\_2\\
      
      \mathbf{elif}\;t\_1 \leq 5 \cdot 10^{-12}:\\
      \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\
      
      \mathbf{else}:\\
      \;\;\;\;t\_2\\
      
      
      \end{array}
      \end{array}
      
      Derivation
      1. Split input into 2 regimes
      2. if (-.f64 (pow.f64 (+.f64 x #s(literal 1 binary64)) (/.f64 #s(literal 1 binary64) n)) (pow.f64 x (/.f64 #s(literal 1 binary64) n))) < -0.050000000000000003 or 4.9999999999999997e-12 < (-.f64 (pow.f64 (+.f64 x #s(literal 1 binary64)) (/.f64 #s(literal 1 binary64) n)) (pow.f64 x (/.f64 #s(literal 1 binary64) n)))

        1. Initial program 78.1%

          \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
        2. Add Preprocessing
        3. Taylor expanded in x around 0

          \[\leadsto \color{blue}{1} - {x}^{\left(\frac{1}{n}\right)} \]
        4. Step-by-step derivation
          1. Applied rewrites75.6%

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

          if -0.050000000000000003 < (-.f64 (pow.f64 (+.f64 x #s(literal 1 binary64)) (/.f64 #s(literal 1 binary64) n)) (pow.f64 x (/.f64 #s(literal 1 binary64) n))) < 4.9999999999999997e-12

          1. Initial program 45.6%

            \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
          2. Add Preprocessing
          3. Taylor expanded in n around inf

            \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
          4. Step-by-step derivation
            1. lower-/.f64N/A

              \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
            2. lower--.f64N/A

              \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
            3. lower-log1p.f64N/A

              \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
            4. lower-log.f6483.3

              \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
          5. Applied rewrites83.3%

            \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
          6. Step-by-step derivation
            1. Applied rewrites82.9%

              \[\leadsto \color{blue}{\frac{\log \left(\frac{1 + x}{x}\right)}{n}} \]
          7. Recombined 2 regimes into one program.
          8. Final simplification80.7%

            \[\leadsto \begin{array}{l} \mathbf{if}\;{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \leq -0.05:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \leq 5 \cdot 10^{-12}:\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
          9. Add Preprocessing

          Alternative 6: 86.4% accurate, 0.6× speedup?

          \[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\ \;\;\;\;\frac{t\_0}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{\log \left(\frac{x}{x + 1}\right)}{-n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t\_0\\ \end{array} \end{array} \]
          (FPCore (x n)
           :precision binary64
           (let* ((t_0 (pow x (/ 1.0 n))))
             (if (<= (/ 1.0 n) -2e-6)
               (/ t_0 (* x n))
               (if (<= (/ 1.0 n) 2e-14)
                 (/ (log (/ x (+ x 1.0))) (- n))
                 (- (exp (/ (log1p x) n)) t_0)))))
          double code(double x, double n) {
          	double t_0 = pow(x, (1.0 / n));
          	double tmp;
          	if ((1.0 / n) <= -2e-6) {
          		tmp = t_0 / (x * n);
          	} else if ((1.0 / n) <= 2e-14) {
          		tmp = log((x / (x + 1.0))) / -n;
          	} else {
          		tmp = exp((log1p(x) / n)) - t_0;
          	}
          	return tmp;
          }
          
          public static double code(double x, double n) {
          	double t_0 = Math.pow(x, (1.0 / n));
          	double tmp;
          	if ((1.0 / n) <= -2e-6) {
          		tmp = t_0 / (x * n);
          	} else if ((1.0 / n) <= 2e-14) {
          		tmp = Math.log((x / (x + 1.0))) / -n;
          	} else {
          		tmp = Math.exp((Math.log1p(x) / n)) - t_0;
          	}
          	return tmp;
          }
          
          def code(x, n):
          	t_0 = math.pow(x, (1.0 / n))
          	tmp = 0
          	if (1.0 / n) <= -2e-6:
          		tmp = t_0 / (x * n)
          	elif (1.0 / n) <= 2e-14:
          		tmp = math.log((x / (x + 1.0))) / -n
          	else:
          		tmp = math.exp((math.log1p(x) / n)) - t_0
          	return tmp
          
          function code(x, n)
          	t_0 = x ^ Float64(1.0 / n)
          	tmp = 0.0
          	if (Float64(1.0 / n) <= -2e-6)
          		tmp = Float64(t_0 / Float64(x * n));
          	elseif (Float64(1.0 / n) <= 2e-14)
          		tmp = Float64(log(Float64(x / Float64(x + 1.0))) / Float64(-n));
          	else
          		tmp = Float64(exp(Float64(log1p(x) / n)) - t_0);
          	end
          	return tmp
          end
          
          code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-6], N[(t$95$0 / N[(x * n), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e-14], N[(N[Log[N[(x / N[(x + 1.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-n)), $MachinePrecision], N[(N[Exp[N[(N[Log[1 + x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision] - t$95$0), $MachinePrecision]]]]
          
          \begin{array}{l}
          
          \\
          \begin{array}{l}
          t_0 := {x}^{\left(\frac{1}{n}\right)}\\
          \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\
          \;\;\;\;\frac{t\_0}{x \cdot n}\\
          
          \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\
          \;\;\;\;\frac{\log \left(\frac{x}{x + 1}\right)}{-n}\\
          
          \mathbf{else}:\\
          \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t\_0\\
          
          
          \end{array}
          \end{array}
          
          Derivation
          1. Split input into 3 regimes
          2. if (/.f64 #s(literal 1 binary64) n) < -1.99999999999999991e-6

            1. Initial program 98.8%

              \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
            2. Add Preprocessing
            3. Taylor expanded in x around inf

              \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
            4. Step-by-step derivation
              1. lower-/.f64N/A

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

                \[\leadsto \frac{e^{-1 \cdot \frac{\color{blue}{\mathsf{neg}\left(\log x\right)}}{n}}}{n \cdot x} \]
              3. mul-1-negN/A

                \[\leadsto \frac{e^{-1 \cdot \frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]
              4. associate-*r/N/A

                \[\leadsto \frac{e^{\color{blue}{\frac{-1 \cdot \left(-1 \cdot \log x\right)}{n}}}}{n \cdot x} \]
              5. associate-*r*N/A

                \[\leadsto \frac{e^{\frac{\color{blue}{\left(-1 \cdot -1\right) \cdot \log x}}{n}}}{n \cdot x} \]
              6. metadata-evalN/A

                \[\leadsto \frac{e^{\frac{\color{blue}{1} \cdot \log x}{n}}}{n \cdot x} \]
              7. *-commutativeN/A

                \[\leadsto \frac{e^{\frac{\color{blue}{\log x \cdot 1}}{n}}}{n \cdot x} \]
              8. associate-/l*N/A

                \[\leadsto \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{n \cdot x} \]
              9. exp-to-powN/A

                \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
              10. lower-pow.f64N/A

                \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
              11. lower-/.f64N/A

                \[\leadsto \frac{{x}^{\color{blue}{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
              12. lower-*.f6498.9

                \[\leadsto \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{n \cdot x}} \]
            5. Applied rewrites98.9%

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

            if -1.99999999999999991e-6 < (/.f64 #s(literal 1 binary64) n) < 2e-14

            1. Initial program 25.9%

              \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
            2. Add Preprocessing
            3. Taylor expanded in n around inf

              \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
            4. Step-by-step derivation
              1. lower-/.f64N/A

                \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
              2. lower--.f64N/A

                \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
              3. lower-log1p.f64N/A

                \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
              4. lower-log.f6480.5

                \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
            5. Applied rewrites80.5%

              \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
            6. Step-by-step derivation
              1. Applied rewrites80.8%

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

              if 2e-14 < (/.f64 #s(literal 1 binary64) n)

              1. Initial program 54.0%

                \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
              2. Add Preprocessing
              3. Step-by-step derivation
                1. lift-pow.f64N/A

                  \[\leadsto \color{blue}{{\left(x + 1\right)}^{\left(\frac{1}{n}\right)}} - {x}^{\left(\frac{1}{n}\right)} \]
                2. pow-to-expN/A

                  \[\leadsto \color{blue}{e^{\log \left(x + 1\right) \cdot \frac{1}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
                3. lower-exp.f64N/A

                  \[\leadsto \color{blue}{e^{\log \left(x + 1\right) \cdot \frac{1}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
                4. lift-/.f64N/A

                  \[\leadsto e^{\log \left(x + 1\right) \cdot \color{blue}{\frac{1}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
                5. un-div-invN/A

                  \[\leadsto e^{\color{blue}{\frac{\log \left(x + 1\right)}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
                6. lower-/.f64N/A

                  \[\leadsto e^{\color{blue}{\frac{\log \left(x + 1\right)}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
                7. lift-+.f64N/A

                  \[\leadsto e^{\frac{\log \color{blue}{\left(x + 1\right)}}{n}} - {x}^{\left(\frac{1}{n}\right)} \]
                8. +-commutativeN/A

                  \[\leadsto e^{\frac{\log \color{blue}{\left(1 + x\right)}}{n}} - {x}^{\left(\frac{1}{n}\right)} \]
                9. lower-log1p.f6496.2

                  \[\leadsto e^{\frac{\color{blue}{\mathsf{log1p}\left(x\right)}}{n}} - {x}^{\left(\frac{1}{n}\right)} \]
              4. Applied rewrites96.2%

                \[\leadsto \color{blue}{e^{\frac{\mathsf{log1p}\left(x\right)}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
            7. Recombined 3 regimes into one program.
            8. Final simplification89.4%

              \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{\log \left(\frac{x}{x + 1}\right)}{-n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
            9. Add Preprocessing

            Alternative 7: 82.8% accurate, 1.1× speedup?

            \[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\ \;\;\;\;\frac{t\_0}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{\log \left(\frac{x}{x + 1}\right)}{-n}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\mathsf{fma}\left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}, x, \frac{1}{n}\right), x, 1 - t\_0\right)\\ \end{array} \end{array} \]
            (FPCore (x n)
             :precision binary64
             (let* ((t_0 (pow x (/ 1.0 n))))
               (if (<= (/ 1.0 n) -2e-6)
                 (/ t_0 (* x n))
                 (if (<= (/ 1.0 n) 2e-14)
                   (/ (log (/ x (+ x 1.0))) (- n))
                   (fma (fma (- (/ 0.5 (* n n)) (/ 0.5 n)) x (/ 1.0 n)) x (- 1.0 t_0))))))
            double code(double x, double n) {
            	double t_0 = pow(x, (1.0 / n));
            	double tmp;
            	if ((1.0 / n) <= -2e-6) {
            		tmp = t_0 / (x * n);
            	} else if ((1.0 / n) <= 2e-14) {
            		tmp = log((x / (x + 1.0))) / -n;
            	} else {
            		tmp = fma(fma(((0.5 / (n * n)) - (0.5 / n)), x, (1.0 / n)), x, (1.0 - t_0));
            	}
            	return tmp;
            }
            
            function code(x, n)
            	t_0 = x ^ Float64(1.0 / n)
            	tmp = 0.0
            	if (Float64(1.0 / n) <= -2e-6)
            		tmp = Float64(t_0 / Float64(x * n));
            	elseif (Float64(1.0 / n) <= 2e-14)
            		tmp = Float64(log(Float64(x / Float64(x + 1.0))) / Float64(-n));
            	else
            		tmp = fma(fma(Float64(Float64(0.5 / Float64(n * n)) - Float64(0.5 / n)), x, Float64(1.0 / n)), x, Float64(1.0 - t_0));
            	end
            	return tmp
            end
            
            code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-6], N[(t$95$0 / N[(x * n), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e-14], N[(N[Log[N[(x / N[(x + 1.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-n)), $MachinePrecision], N[(N[(N[(N[(0.5 / N[(n * n), $MachinePrecision]), $MachinePrecision] - N[(0.5 / n), $MachinePrecision]), $MachinePrecision] * x + N[(1.0 / n), $MachinePrecision]), $MachinePrecision] * x + N[(1.0 - t$95$0), $MachinePrecision]), $MachinePrecision]]]]
            
            \begin{array}{l}
            
            \\
            \begin{array}{l}
            t_0 := {x}^{\left(\frac{1}{n}\right)}\\
            \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\
            \;\;\;\;\frac{t\_0}{x \cdot n}\\
            
            \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\
            \;\;\;\;\frac{\log \left(\frac{x}{x + 1}\right)}{-n}\\
            
            \mathbf{else}:\\
            \;\;\;\;\mathsf{fma}\left(\mathsf{fma}\left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}, x, \frac{1}{n}\right), x, 1 - t\_0\right)\\
            
            
            \end{array}
            \end{array}
            
            Derivation
            1. Split input into 3 regimes
            2. if (/.f64 #s(literal 1 binary64) n) < -1.99999999999999991e-6

              1. Initial program 98.8%

                \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
              2. Add Preprocessing
              3. Taylor expanded in x around inf

                \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
              4. Step-by-step derivation
                1. lower-/.f64N/A

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

                  \[\leadsto \frac{e^{-1 \cdot \frac{\color{blue}{\mathsf{neg}\left(\log x\right)}}{n}}}{n \cdot x} \]
                3. mul-1-negN/A

                  \[\leadsto \frac{e^{-1 \cdot \frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]
                4. associate-*r/N/A

                  \[\leadsto \frac{e^{\color{blue}{\frac{-1 \cdot \left(-1 \cdot \log x\right)}{n}}}}{n \cdot x} \]
                5. associate-*r*N/A

                  \[\leadsto \frac{e^{\frac{\color{blue}{\left(-1 \cdot -1\right) \cdot \log x}}{n}}}{n \cdot x} \]
                6. metadata-evalN/A

                  \[\leadsto \frac{e^{\frac{\color{blue}{1} \cdot \log x}{n}}}{n \cdot x} \]
                7. *-commutativeN/A

                  \[\leadsto \frac{e^{\frac{\color{blue}{\log x \cdot 1}}{n}}}{n \cdot x} \]
                8. associate-/l*N/A

                  \[\leadsto \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{n \cdot x} \]
                9. exp-to-powN/A

                  \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
                10. lower-pow.f64N/A

                  \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
                11. lower-/.f64N/A

                  \[\leadsto \frac{{x}^{\color{blue}{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
                12. lower-*.f6498.9

                  \[\leadsto \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{n \cdot x}} \]
              5. Applied rewrites98.9%

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

              if -1.99999999999999991e-6 < (/.f64 #s(literal 1 binary64) n) < 2e-14

              1. Initial program 25.9%

                \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
              2. Add Preprocessing
              3. Taylor expanded in n around inf

                \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
              4. Step-by-step derivation
                1. lower-/.f64N/A

                  \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                2. lower--.f64N/A

                  \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                3. lower-log1p.f64N/A

                  \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                4. lower-log.f6480.5

                  \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
              5. Applied rewrites80.5%

                \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
              6. Step-by-step derivation
                1. Applied rewrites80.8%

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

                if 2e-14 < (/.f64 #s(literal 1 binary64) n)

                1. Initial program 54.0%

                  \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                2. Add Preprocessing
                3. Taylor expanded in x around 0

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

                    \[\leadsto \color{blue}{\left(x \cdot \left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) + 1\right)} - e^{\frac{\log x}{n}} \]
                  2. associate--l+N/A

                    \[\leadsto \color{blue}{x \cdot \left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) + \left(1 - e^{\frac{\log x}{n}}\right)} \]
                  3. *-commutativeN/A

                    \[\leadsto \color{blue}{\left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) \cdot x} + \left(1 - e^{\frac{\log x}{n}}\right) \]
                  4. lower-fma.f64N/A

                    \[\leadsto \color{blue}{\mathsf{fma}\left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}, x, 1 - e^{\frac{\log x}{n}}\right)} \]
                5. Applied rewrites86.3%

                  \[\leadsto \color{blue}{\mathsf{fma}\left(\mathsf{fma}\left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}, x, \frac{1}{n}\right), x, 1 - {x}^{\left(\frac{1}{n}\right)}\right)} \]
              7. Recombined 3 regimes into one program.
              8. Final simplification87.9%

                \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{\log \left(\frac{x}{x + 1}\right)}{-n}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\mathsf{fma}\left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}, x, \frac{1}{n}\right), x, 1 - {x}^{\left(\frac{1}{n}\right)}\right)\\ \end{array} \]
              9. Add Preprocessing

              Alternative 8: 82.5% accurate, 1.3× speedup?

              \[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\ \;\;\;\;\frac{t\_0}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{\log \left(\frac{x}{x + 1}\right)}{-n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{+152}:\\ \;\;\;\;\left(\frac{x}{n} + 1\right) - t\_0\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{x}{n \cdot n} \cdot x\right) \cdot 0.5\\ \end{array} \end{array} \]
              (FPCore (x n)
               :precision binary64
               (let* ((t_0 (pow x (/ 1.0 n))))
                 (if (<= (/ 1.0 n) -2e-6)
                   (/ t_0 (* x n))
                   (if (<= (/ 1.0 n) 2e-14)
                     (/ (log (/ x (+ x 1.0))) (- n))
                     (if (<= (/ 1.0 n) 2e+152)
                       (- (+ (/ x n) 1.0) t_0)
                       (* (* (/ x (* n n)) x) 0.5))))))
              double code(double x, double n) {
              	double t_0 = pow(x, (1.0 / n));
              	double tmp;
              	if ((1.0 / n) <= -2e-6) {
              		tmp = t_0 / (x * n);
              	} else if ((1.0 / n) <= 2e-14) {
              		tmp = log((x / (x + 1.0))) / -n;
              	} else if ((1.0 / n) <= 2e+152) {
              		tmp = ((x / n) + 1.0) - t_0;
              	} else {
              		tmp = ((x / (n * n)) * x) * 0.5;
              	}
              	return tmp;
              }
              
              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-6)) then
                      tmp = t_0 / (x * n)
                  else if ((1.0d0 / n) <= 2d-14) then
                      tmp = log((x / (x + 1.0d0))) / -n
                  else if ((1.0d0 / n) <= 2d+152) then
                      tmp = ((x / n) + 1.0d0) - t_0
                  else
                      tmp = ((x / (n * n)) * x) * 0.5d0
                  end if
                  code = tmp
              end function
              
              public static double code(double x, double n) {
              	double t_0 = Math.pow(x, (1.0 / n));
              	double tmp;
              	if ((1.0 / n) <= -2e-6) {
              		tmp = t_0 / (x * n);
              	} else if ((1.0 / n) <= 2e-14) {
              		tmp = Math.log((x / (x + 1.0))) / -n;
              	} else if ((1.0 / n) <= 2e+152) {
              		tmp = ((x / n) + 1.0) - t_0;
              	} else {
              		tmp = ((x / (n * n)) * x) * 0.5;
              	}
              	return tmp;
              }
              
              def code(x, n):
              	t_0 = math.pow(x, (1.0 / n))
              	tmp = 0
              	if (1.0 / n) <= -2e-6:
              		tmp = t_0 / (x * n)
              	elif (1.0 / n) <= 2e-14:
              		tmp = math.log((x / (x + 1.0))) / -n
              	elif (1.0 / n) <= 2e+152:
              		tmp = ((x / n) + 1.0) - t_0
              	else:
              		tmp = ((x / (n * n)) * x) * 0.5
              	return tmp
              
              function code(x, n)
              	t_0 = x ^ Float64(1.0 / n)
              	tmp = 0.0
              	if (Float64(1.0 / n) <= -2e-6)
              		tmp = Float64(t_0 / Float64(x * n));
              	elseif (Float64(1.0 / n) <= 2e-14)
              		tmp = Float64(log(Float64(x / Float64(x + 1.0))) / Float64(-n));
              	elseif (Float64(1.0 / n) <= 2e+152)
              		tmp = Float64(Float64(Float64(x / n) + 1.0) - t_0);
              	else
              		tmp = Float64(Float64(Float64(x / Float64(n * n)) * x) * 0.5);
              	end
              	return tmp
              end
              
              function tmp_2 = code(x, n)
              	t_0 = x ^ (1.0 / n);
              	tmp = 0.0;
              	if ((1.0 / n) <= -2e-6)
              		tmp = t_0 / (x * n);
              	elseif ((1.0 / n) <= 2e-14)
              		tmp = log((x / (x + 1.0))) / -n;
              	elseif ((1.0 / n) <= 2e+152)
              		tmp = ((x / n) + 1.0) - t_0;
              	else
              		tmp = ((x / (n * n)) * x) * 0.5;
              	end
              	tmp_2 = tmp;
              end
              
              code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-6], N[(t$95$0 / N[(x * n), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e-14], N[(N[Log[N[(x / N[(x + 1.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-n)), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e+152], N[(N[(N[(x / n), $MachinePrecision] + 1.0), $MachinePrecision] - t$95$0), $MachinePrecision], N[(N[(N[(x / N[(n * n), $MachinePrecision]), $MachinePrecision] * x), $MachinePrecision] * 0.5), $MachinePrecision]]]]]
              
              \begin{array}{l}
              
              \\
              \begin{array}{l}
              t_0 := {x}^{\left(\frac{1}{n}\right)}\\
              \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\
              \;\;\;\;\frac{t\_0}{x \cdot n}\\
              
              \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\
              \;\;\;\;\frac{\log \left(\frac{x}{x + 1}\right)}{-n}\\
              
              \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{+152}:\\
              \;\;\;\;\left(\frac{x}{n} + 1\right) - t\_0\\
              
              \mathbf{else}:\\
              \;\;\;\;\left(\frac{x}{n \cdot n} \cdot x\right) \cdot 0.5\\
              
              
              \end{array}
              \end{array}
              
              Derivation
              1. Split input into 4 regimes
              2. if (/.f64 #s(literal 1 binary64) n) < -1.99999999999999991e-6

                1. Initial program 98.8%

                  \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                2. Add Preprocessing
                3. Taylor expanded in x around inf

                  \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
                4. Step-by-step derivation
                  1. lower-/.f64N/A

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

                    \[\leadsto \frac{e^{-1 \cdot \frac{\color{blue}{\mathsf{neg}\left(\log x\right)}}{n}}}{n \cdot x} \]
                  3. mul-1-negN/A

                    \[\leadsto \frac{e^{-1 \cdot \frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]
                  4. associate-*r/N/A

                    \[\leadsto \frac{e^{\color{blue}{\frac{-1 \cdot \left(-1 \cdot \log x\right)}{n}}}}{n \cdot x} \]
                  5. associate-*r*N/A

                    \[\leadsto \frac{e^{\frac{\color{blue}{\left(-1 \cdot -1\right) \cdot \log x}}{n}}}{n \cdot x} \]
                  6. metadata-evalN/A

                    \[\leadsto \frac{e^{\frac{\color{blue}{1} \cdot \log x}{n}}}{n \cdot x} \]
                  7. *-commutativeN/A

                    \[\leadsto \frac{e^{\frac{\color{blue}{\log x \cdot 1}}{n}}}{n \cdot x} \]
                  8. associate-/l*N/A

                    \[\leadsto \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{n \cdot x} \]
                  9. exp-to-powN/A

                    \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
                  10. lower-pow.f64N/A

                    \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
                  11. lower-/.f64N/A

                    \[\leadsto \frac{{x}^{\color{blue}{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
                  12. lower-*.f6498.9

                    \[\leadsto \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{n \cdot x}} \]
                5. Applied rewrites98.9%

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

                if -1.99999999999999991e-6 < (/.f64 #s(literal 1 binary64) n) < 2e-14

                1. Initial program 25.9%

                  \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                2. Add Preprocessing
                3. Taylor expanded in n around inf

                  \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                4. Step-by-step derivation
                  1. lower-/.f64N/A

                    \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                  2. lower--.f64N/A

                    \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                  3. lower-log1p.f64N/A

                    \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                  4. lower-log.f6480.5

                    \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                5. Applied rewrites80.5%

                  \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                6. Step-by-step derivation
                  1. Applied rewrites80.8%

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

                  if 2e-14 < (/.f64 #s(literal 1 binary64) n) < 2.0000000000000001e152

                  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

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

                      \[\leadsto \color{blue}{\left(\frac{x}{n} + 1\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                    2. *-rgt-identityN/A

                      \[\leadsto \left(\frac{\color{blue}{x \cdot 1}}{n} + 1\right) - {x}^{\left(\frac{1}{n}\right)} \]
                    3. associate-*r/N/A

                      \[\leadsto \left(\color{blue}{x \cdot \frac{1}{n}} + 1\right) - {x}^{\left(\frac{1}{n}\right)} \]
                    4. lower-+.f64N/A

                      \[\leadsto \color{blue}{\left(x \cdot \frac{1}{n} + 1\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                    5. associate-*r/N/A

                      \[\leadsto \left(\color{blue}{\frac{x \cdot 1}{n}} + 1\right) - {x}^{\left(\frac{1}{n}\right)} \]
                    6. *-rgt-identityN/A

                      \[\leadsto \left(\frac{\color{blue}{x}}{n} + 1\right) - {x}^{\left(\frac{1}{n}\right)} \]
                    7. lower-/.f6483.8

                      \[\leadsto \left(\color{blue}{\frac{x}{n}} + 1\right) - {x}^{\left(\frac{1}{n}\right)} \]
                  5. Applied rewrites83.8%

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

                  if 2.0000000000000001e152 < (/.f64 #s(literal 1 binary64) n)

                  1. Initial program 14.2%

                    \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                  2. Add Preprocessing
                  3. Taylor expanded in x around 0

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

                      \[\leadsto \color{blue}{\left(x \cdot \left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) + 1\right)} - e^{\frac{\log x}{n}} \]
                    2. associate--l+N/A

                      \[\leadsto \color{blue}{x \cdot \left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) + \left(1 - e^{\frac{\log x}{n}}\right)} \]
                    3. *-commutativeN/A

                      \[\leadsto \color{blue}{\left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) \cdot x} + \left(1 - e^{\frac{\log x}{n}}\right) \]
                    4. lower-fma.f64N/A

                      \[\leadsto \color{blue}{\mathsf{fma}\left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}, x, 1 - e^{\frac{\log x}{n}}\right)} \]
                  5. Applied rewrites89.2%

                    \[\leadsto \color{blue}{\mathsf{fma}\left(\mathsf{fma}\left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}, x, \frac{1}{n}\right), x, 1 - {x}^{\left(\frac{1}{n}\right)}\right)} \]
                  6. Taylor expanded in n around -inf

                    \[\leadsto -1 \cdot \color{blue}{\frac{x \cdot \left(\frac{1}{2} \cdot x - 1\right) - -1 \cdot \log x}{n}} \]
                  7. Step-by-step derivation
                    1. Applied rewrites17.1%

                      \[\leadsto \frac{\mathsf{fma}\left(\mathsf{fma}\left(0.5, x, -1\right), x, \log x\right)}{\color{blue}{-n}} \]
                    2. Taylor expanded in x around inf

                      \[\leadsto \frac{{x}^{2} \cdot \left(\left(\frac{1}{2} + -1 \cdot \frac{\log \left(\frac{1}{x}\right)}{{x}^{2}}\right) - \frac{1}{x}\right)}{\mathsf{neg}\left(n\right)} \]
                    3. Step-by-step derivation
                      1. Applied rewrites25.5%

                        \[\leadsto \frac{\left(\left(\frac{\log x}{x \cdot x} + 0.5\right) - \frac{1}{x}\right) \cdot \left(x \cdot x\right)}{-n} \]
                      2. Taylor expanded in n around 0

                        \[\leadsto \frac{1}{2} \cdot \color{blue}{\frac{{x}^{2}}{{n}^{2}}} \]
                      3. Step-by-step derivation
                        1. Applied rewrites89.2%

                          \[\leadsto \left(x \cdot \frac{x}{n \cdot n}\right) \cdot \color{blue}{0.5} \]
                      4. Recombined 4 regimes into one program.
                      5. Final simplification87.9%

                        \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{\log \left(\frac{x}{x + 1}\right)}{-n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{+152}:\\ \;\;\;\;\left(\frac{x}{n} + 1\right) - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{x}{n \cdot n} \cdot x\right) \cdot 0.5\\ \end{array} \]
                      6. Add Preprocessing

                      Alternative 9: 82.4% accurate, 1.4× speedup?

                      \[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\ \;\;\;\;\frac{t\_0}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{\log \left(\frac{x}{x + 1}\right)}{-n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{+152}:\\ \;\;\;\;1 - t\_0\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{x}{n \cdot n} \cdot x\right) \cdot 0.5\\ \end{array} \end{array} \]
                      (FPCore (x n)
                       :precision binary64
                       (let* ((t_0 (pow x (/ 1.0 n))))
                         (if (<= (/ 1.0 n) -2e-6)
                           (/ t_0 (* x n))
                           (if (<= (/ 1.0 n) 2e-14)
                             (/ (log (/ x (+ x 1.0))) (- n))
                             (if (<= (/ 1.0 n) 2e+152) (- 1.0 t_0) (* (* (/ x (* n n)) x) 0.5))))))
                      double code(double x, double n) {
                      	double t_0 = pow(x, (1.0 / n));
                      	double tmp;
                      	if ((1.0 / n) <= -2e-6) {
                      		tmp = t_0 / (x * n);
                      	} else if ((1.0 / n) <= 2e-14) {
                      		tmp = log((x / (x + 1.0))) / -n;
                      	} else if ((1.0 / n) <= 2e+152) {
                      		tmp = 1.0 - t_0;
                      	} else {
                      		tmp = ((x / (n * n)) * x) * 0.5;
                      	}
                      	return tmp;
                      }
                      
                      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-6)) then
                              tmp = t_0 / (x * n)
                          else if ((1.0d0 / n) <= 2d-14) then
                              tmp = log((x / (x + 1.0d0))) / -n
                          else if ((1.0d0 / n) <= 2d+152) then
                              tmp = 1.0d0 - t_0
                          else
                              tmp = ((x / (n * n)) * x) * 0.5d0
                          end if
                          code = tmp
                      end function
                      
                      public static double code(double x, double n) {
                      	double t_0 = Math.pow(x, (1.0 / n));
                      	double tmp;
                      	if ((1.0 / n) <= -2e-6) {
                      		tmp = t_0 / (x * n);
                      	} else if ((1.0 / n) <= 2e-14) {
                      		tmp = Math.log((x / (x + 1.0))) / -n;
                      	} else if ((1.0 / n) <= 2e+152) {
                      		tmp = 1.0 - t_0;
                      	} else {
                      		tmp = ((x / (n * n)) * x) * 0.5;
                      	}
                      	return tmp;
                      }
                      
                      def code(x, n):
                      	t_0 = math.pow(x, (1.0 / n))
                      	tmp = 0
                      	if (1.0 / n) <= -2e-6:
                      		tmp = t_0 / (x * n)
                      	elif (1.0 / n) <= 2e-14:
                      		tmp = math.log((x / (x + 1.0))) / -n
                      	elif (1.0 / n) <= 2e+152:
                      		tmp = 1.0 - t_0
                      	else:
                      		tmp = ((x / (n * n)) * x) * 0.5
                      	return tmp
                      
                      function code(x, n)
                      	t_0 = x ^ Float64(1.0 / n)
                      	tmp = 0.0
                      	if (Float64(1.0 / n) <= -2e-6)
                      		tmp = Float64(t_0 / Float64(x * n));
                      	elseif (Float64(1.0 / n) <= 2e-14)
                      		tmp = Float64(log(Float64(x / Float64(x + 1.0))) / Float64(-n));
                      	elseif (Float64(1.0 / n) <= 2e+152)
                      		tmp = Float64(1.0 - t_0);
                      	else
                      		tmp = Float64(Float64(Float64(x / Float64(n * n)) * x) * 0.5);
                      	end
                      	return tmp
                      end
                      
                      function tmp_2 = code(x, n)
                      	t_0 = x ^ (1.0 / n);
                      	tmp = 0.0;
                      	if ((1.0 / n) <= -2e-6)
                      		tmp = t_0 / (x * n);
                      	elseif ((1.0 / n) <= 2e-14)
                      		tmp = log((x / (x + 1.0))) / -n;
                      	elseif ((1.0 / n) <= 2e+152)
                      		tmp = 1.0 - t_0;
                      	else
                      		tmp = ((x / (n * n)) * x) * 0.5;
                      	end
                      	tmp_2 = tmp;
                      end
                      
                      code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-6], N[(t$95$0 / N[(x * n), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e-14], N[(N[Log[N[(x / N[(x + 1.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-n)), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e+152], N[(1.0 - t$95$0), $MachinePrecision], N[(N[(N[(x / N[(n * n), $MachinePrecision]), $MachinePrecision] * x), $MachinePrecision] * 0.5), $MachinePrecision]]]]]
                      
                      \begin{array}{l}
                      
                      \\
                      \begin{array}{l}
                      t_0 := {x}^{\left(\frac{1}{n}\right)}\\
                      \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\
                      \;\;\;\;\frac{t\_0}{x \cdot n}\\
                      
                      \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\
                      \;\;\;\;\frac{\log \left(\frac{x}{x + 1}\right)}{-n}\\
                      
                      \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{+152}:\\
                      \;\;\;\;1 - t\_0\\
                      
                      \mathbf{else}:\\
                      \;\;\;\;\left(\frac{x}{n \cdot n} \cdot x\right) \cdot 0.5\\
                      
                      
                      \end{array}
                      \end{array}
                      
                      Derivation
                      1. Split input into 4 regimes
                      2. if (/.f64 #s(literal 1 binary64) n) < -1.99999999999999991e-6

                        1. Initial program 98.8%

                          \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                        2. Add Preprocessing
                        3. Taylor expanded in x around inf

                          \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
                        4. Step-by-step derivation
                          1. lower-/.f64N/A

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

                            \[\leadsto \frac{e^{-1 \cdot \frac{\color{blue}{\mathsf{neg}\left(\log x\right)}}{n}}}{n \cdot x} \]
                          3. mul-1-negN/A

                            \[\leadsto \frac{e^{-1 \cdot \frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]
                          4. associate-*r/N/A

                            \[\leadsto \frac{e^{\color{blue}{\frac{-1 \cdot \left(-1 \cdot \log x\right)}{n}}}}{n \cdot x} \]
                          5. associate-*r*N/A

                            \[\leadsto \frac{e^{\frac{\color{blue}{\left(-1 \cdot -1\right) \cdot \log x}}{n}}}{n \cdot x} \]
                          6. metadata-evalN/A

                            \[\leadsto \frac{e^{\frac{\color{blue}{1} \cdot \log x}{n}}}{n \cdot x} \]
                          7. *-commutativeN/A

                            \[\leadsto \frac{e^{\frac{\color{blue}{\log x \cdot 1}}{n}}}{n \cdot x} \]
                          8. associate-/l*N/A

                            \[\leadsto \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{n \cdot x} \]
                          9. exp-to-powN/A

                            \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
                          10. lower-pow.f64N/A

                            \[\leadsto \frac{\color{blue}{{x}^{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
                          11. lower-/.f64N/A

                            \[\leadsto \frac{{x}^{\color{blue}{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
                          12. lower-*.f6498.9

                            \[\leadsto \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{n \cdot x}} \]
                        5. Applied rewrites98.9%

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

                        if -1.99999999999999991e-6 < (/.f64 #s(literal 1 binary64) n) < 2e-14

                        1. Initial program 25.9%

                          \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                        2. Add Preprocessing
                        3. Taylor expanded in n around inf

                          \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                        4. Step-by-step derivation
                          1. lower-/.f64N/A

                            \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                          2. lower--.f64N/A

                            \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                          3. lower-log1p.f64N/A

                            \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                          4. lower-log.f6480.5

                            \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                        5. Applied rewrites80.5%

                          \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                        6. Step-by-step derivation
                          1. Applied rewrites80.8%

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

                          if 2e-14 < (/.f64 #s(literal 1 binary64) n) < 2.0000000000000001e152

                          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

                            \[\leadsto \color{blue}{1} - {x}^{\left(\frac{1}{n}\right)} \]
                          4. Step-by-step derivation
                            1. Applied rewrites83.3%

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

                            if 2.0000000000000001e152 < (/.f64 #s(literal 1 binary64) n)

                            1. Initial program 14.2%

                              \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                            2. Add Preprocessing
                            3. Taylor expanded in x around 0

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

                                \[\leadsto \color{blue}{\left(x \cdot \left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) + 1\right)} - e^{\frac{\log x}{n}} \]
                              2. associate--l+N/A

                                \[\leadsto \color{blue}{x \cdot \left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) + \left(1 - e^{\frac{\log x}{n}}\right)} \]
                              3. *-commutativeN/A

                                \[\leadsto \color{blue}{\left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) \cdot x} + \left(1 - e^{\frac{\log x}{n}}\right) \]
                              4. lower-fma.f64N/A

                                \[\leadsto \color{blue}{\mathsf{fma}\left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}, x, 1 - e^{\frac{\log x}{n}}\right)} \]
                            5. Applied rewrites89.2%

                              \[\leadsto \color{blue}{\mathsf{fma}\left(\mathsf{fma}\left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}, x, \frac{1}{n}\right), x, 1 - {x}^{\left(\frac{1}{n}\right)}\right)} \]
                            6. Taylor expanded in n around -inf

                              \[\leadsto -1 \cdot \color{blue}{\frac{x \cdot \left(\frac{1}{2} \cdot x - 1\right) - -1 \cdot \log x}{n}} \]
                            7. Step-by-step derivation
                              1. Applied rewrites17.1%

                                \[\leadsto \frac{\mathsf{fma}\left(\mathsf{fma}\left(0.5, x, -1\right), x, \log x\right)}{\color{blue}{-n}} \]
                              2. Taylor expanded in x around inf

                                \[\leadsto \frac{{x}^{2} \cdot \left(\left(\frac{1}{2} + -1 \cdot \frac{\log \left(\frac{1}{x}\right)}{{x}^{2}}\right) - \frac{1}{x}\right)}{\mathsf{neg}\left(n\right)} \]
                              3. Step-by-step derivation
                                1. Applied rewrites25.5%

                                  \[\leadsto \frac{\left(\left(\frac{\log x}{x \cdot x} + 0.5\right) - \frac{1}{x}\right) \cdot \left(x \cdot x\right)}{-n} \]
                                2. Taylor expanded in n around 0

                                  \[\leadsto \frac{1}{2} \cdot \color{blue}{\frac{{x}^{2}}{{n}^{2}}} \]
                                3. Step-by-step derivation
                                  1. Applied rewrites89.2%

                                    \[\leadsto \left(x \cdot \frac{x}{n \cdot n}\right) \cdot \color{blue}{0.5} \]
                                4. Recombined 4 regimes into one program.
                                5. Final simplification87.9%

                                  \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-6}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{\log \left(\frac{x}{x + 1}\right)}{-n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{+152}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{x}{n \cdot n} \cdot x\right) \cdot 0.5\\ \end{array} \]
                                6. Add Preprocessing

                                Alternative 10: 59.9% accurate, 1.4× speedup?

                                \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5000000:\\ \;\;\;\;\frac{0.3333333333333333}{\left(\left(x \cdot x\right) \cdot x\right) \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{+152}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{x}{n \cdot n} \cdot x\right) \cdot 0.5\\ \end{array} \end{array} \]
                                (FPCore (x n)
                                 :precision binary64
                                 (if (<= (/ 1.0 n) -5000000.0)
                                   (/ 0.3333333333333333 (* (* (* x x) x) n))
                                   (if (<= (/ 1.0 n) 2e-14)
                                     (/ (- x (log x)) n)
                                     (if (<= (/ 1.0 n) 2e+152)
                                       (- 1.0 (pow x (/ 1.0 n)))
                                       (* (* (/ x (* n n)) x) 0.5)))))
                                double code(double x, double n) {
                                	double tmp;
                                	if ((1.0 / n) <= -5000000.0) {
                                		tmp = 0.3333333333333333 / (((x * x) * x) * n);
                                	} else if ((1.0 / n) <= 2e-14) {
                                		tmp = (x - log(x)) / n;
                                	} else if ((1.0 / n) <= 2e+152) {
                                		tmp = 1.0 - pow(x, (1.0 / n));
                                	} else {
                                		tmp = ((x / (n * n)) * x) * 0.5;
                                	}
                                	return tmp;
                                }
                                
                                real(8) function code(x, n)
                                    real(8), intent (in) :: x
                                    real(8), intent (in) :: n
                                    real(8) :: tmp
                                    if ((1.0d0 / n) <= (-5000000.0d0)) then
                                        tmp = 0.3333333333333333d0 / (((x * x) * x) * n)
                                    else if ((1.0d0 / n) <= 2d-14) then
                                        tmp = (x - log(x)) / n
                                    else if ((1.0d0 / n) <= 2d+152) then
                                        tmp = 1.0d0 - (x ** (1.0d0 / n))
                                    else
                                        tmp = ((x / (n * n)) * x) * 0.5d0
                                    end if
                                    code = tmp
                                end function
                                
                                public static double code(double x, double n) {
                                	double tmp;
                                	if ((1.0 / n) <= -5000000.0) {
                                		tmp = 0.3333333333333333 / (((x * x) * x) * n);
                                	} else if ((1.0 / n) <= 2e-14) {
                                		tmp = (x - Math.log(x)) / n;
                                	} else if ((1.0 / n) <= 2e+152) {
                                		tmp = 1.0 - Math.pow(x, (1.0 / n));
                                	} else {
                                		tmp = ((x / (n * n)) * x) * 0.5;
                                	}
                                	return tmp;
                                }
                                
                                def code(x, n):
                                	tmp = 0
                                	if (1.0 / n) <= -5000000.0:
                                		tmp = 0.3333333333333333 / (((x * x) * x) * n)
                                	elif (1.0 / n) <= 2e-14:
                                		tmp = (x - math.log(x)) / n
                                	elif (1.0 / n) <= 2e+152:
                                		tmp = 1.0 - math.pow(x, (1.0 / n))
                                	else:
                                		tmp = ((x / (n * n)) * x) * 0.5
                                	return tmp
                                
                                function code(x, n)
                                	tmp = 0.0
                                	if (Float64(1.0 / n) <= -5000000.0)
                                		tmp = Float64(0.3333333333333333 / Float64(Float64(Float64(x * x) * x) * n));
                                	elseif (Float64(1.0 / n) <= 2e-14)
                                		tmp = Float64(Float64(x - log(x)) / n);
                                	elseif (Float64(1.0 / n) <= 2e+152)
                                		tmp = Float64(1.0 - (x ^ Float64(1.0 / n)));
                                	else
                                		tmp = Float64(Float64(Float64(x / Float64(n * n)) * x) * 0.5);
                                	end
                                	return tmp
                                end
                                
                                function tmp_2 = code(x, n)
                                	tmp = 0.0;
                                	if ((1.0 / n) <= -5000000.0)
                                		tmp = 0.3333333333333333 / (((x * x) * x) * n);
                                	elseif ((1.0 / n) <= 2e-14)
                                		tmp = (x - log(x)) / n;
                                	elseif ((1.0 / n) <= 2e+152)
                                		tmp = 1.0 - (x ^ (1.0 / n));
                                	else
                                		tmp = ((x / (n * n)) * x) * 0.5;
                                	end
                                	tmp_2 = tmp;
                                end
                                
                                code[x_, n_] := If[LessEqual[N[(1.0 / n), $MachinePrecision], -5000000.0], N[(0.3333333333333333 / N[(N[(N[(x * x), $MachinePrecision] * x), $MachinePrecision] * n), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e-14], N[(N[(x - N[Log[x], $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e+152], N[(1.0 - N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[(N[(x / N[(n * n), $MachinePrecision]), $MachinePrecision] * x), $MachinePrecision] * 0.5), $MachinePrecision]]]]
                                
                                \begin{array}{l}
                                
                                \\
                                \begin{array}{l}
                                \mathbf{if}\;\frac{1}{n} \leq -5000000:\\
                                \;\;\;\;\frac{0.3333333333333333}{\left(\left(x \cdot x\right) \cdot x\right) \cdot n}\\
                                
                                \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\
                                \;\;\;\;\frac{x - \log x}{n}\\
                                
                                \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{+152}:\\
                                \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\
                                
                                \mathbf{else}:\\
                                \;\;\;\;\left(\frac{x}{n \cdot n} \cdot x\right) \cdot 0.5\\
                                
                                
                                \end{array}
                                \end{array}
                                
                                Derivation
                                1. Split input into 4 regimes
                                2. if (/.f64 #s(literal 1 binary64) n) < -5e6

                                  1. Initial program 100.0%

                                    \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                  2. Add Preprocessing
                                  3. Taylor expanded in n around inf

                                    \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                  4. Step-by-step derivation
                                    1. lower-/.f64N/A

                                      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                    2. lower--.f64N/A

                                      \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                                    3. lower-log1p.f64N/A

                                      \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                                    4. lower-log.f6454.1

                                      \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                                  5. Applied rewrites54.1%

                                    \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                                  6. Taylor expanded in x around inf

                                    \[\leadsto \frac{\left(\frac{\frac{1}{3}}{n \cdot {x}^{2}} + \frac{1}{n}\right) - \frac{\frac{1}{2}}{n \cdot x}}{\color{blue}{x}} \]
                                  7. Step-by-step derivation
                                    1. Applied rewrites18.5%

                                      \[\leadsto \frac{\left(\frac{0.3333333333333333}{\left(x \cdot x\right) \cdot n} + \frac{1}{n}\right) - \frac{0.5}{n \cdot x}}{\color{blue}{x}} \]
                                    2. Taylor expanded in x around 0

                                      \[\leadsto \frac{\frac{1}{3}}{n \cdot \color{blue}{{x}^{3}}} \]
                                    3. Step-by-step derivation
                                      1. Applied rewrites75.9%

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

                                      if -5e6 < (/.f64 #s(literal 1 binary64) n) < 2e-14

                                      1. Initial program 26.2%

                                        \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                      2. Add Preprocessing
                                      3. Taylor expanded in n around inf

                                        \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                      4. Step-by-step derivation
                                        1. lower-/.f64N/A

                                          \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                        2. lower--.f64N/A

                                          \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                                        3. lower-log1p.f64N/A

                                          \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                                        4. lower-log.f6479.5

                                          \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                                      5. Applied rewrites79.5%

                                        \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                                      6. Taylor expanded in x around 0

                                        \[\leadsto \frac{x - \log x}{n} \]
                                      7. Step-by-step derivation
                                        1. Applied rewrites58.9%

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

                                        if 2e-14 < (/.f64 #s(literal 1 binary64) n) < 2.0000000000000001e152

                                        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

                                          \[\leadsto \color{blue}{1} - {x}^{\left(\frac{1}{n}\right)} \]
                                        4. Step-by-step derivation
                                          1. Applied rewrites83.3%

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

                                          if 2.0000000000000001e152 < (/.f64 #s(literal 1 binary64) n)

                                          1. Initial program 14.2%

                                            \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                          2. Add Preprocessing
                                          3. Taylor expanded in x around 0

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

                                              \[\leadsto \color{blue}{\left(x \cdot \left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) + 1\right)} - e^{\frac{\log x}{n}} \]
                                            2. associate--l+N/A

                                              \[\leadsto \color{blue}{x \cdot \left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) + \left(1 - e^{\frac{\log x}{n}}\right)} \]
                                            3. *-commutativeN/A

                                              \[\leadsto \color{blue}{\left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) \cdot x} + \left(1 - e^{\frac{\log x}{n}}\right) \]
                                            4. lower-fma.f64N/A

                                              \[\leadsto \color{blue}{\mathsf{fma}\left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}, x, 1 - e^{\frac{\log x}{n}}\right)} \]
                                          5. Applied rewrites89.2%

                                            \[\leadsto \color{blue}{\mathsf{fma}\left(\mathsf{fma}\left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}, x, \frac{1}{n}\right), x, 1 - {x}^{\left(\frac{1}{n}\right)}\right)} \]
                                          6. Taylor expanded in n around -inf

                                            \[\leadsto -1 \cdot \color{blue}{\frac{x \cdot \left(\frac{1}{2} \cdot x - 1\right) - -1 \cdot \log x}{n}} \]
                                          7. Step-by-step derivation
                                            1. Applied rewrites17.1%

                                              \[\leadsto \frac{\mathsf{fma}\left(\mathsf{fma}\left(0.5, x, -1\right), x, \log x\right)}{\color{blue}{-n}} \]
                                            2. Taylor expanded in x around inf

                                              \[\leadsto \frac{{x}^{2} \cdot \left(\left(\frac{1}{2} + -1 \cdot \frac{\log \left(\frac{1}{x}\right)}{{x}^{2}}\right) - \frac{1}{x}\right)}{\mathsf{neg}\left(n\right)} \]
                                            3. Step-by-step derivation
                                              1. Applied rewrites25.5%

                                                \[\leadsto \frac{\left(\left(\frac{\log x}{x \cdot x} + 0.5\right) - \frac{1}{x}\right) \cdot \left(x \cdot x\right)}{-n} \]
                                              2. Taylor expanded in n around 0

                                                \[\leadsto \frac{1}{2} \cdot \color{blue}{\frac{{x}^{2}}{{n}^{2}}} \]
                                              3. Step-by-step derivation
                                                1. Applied rewrites89.2%

                                                  \[\leadsto \left(x \cdot \frac{x}{n \cdot n}\right) \cdot \color{blue}{0.5} \]
                                              4. Recombined 4 regimes into one program.
                                              5. Final simplification68.8%

                                                \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5000000:\\ \;\;\;\;\frac{0.3333333333333333}{\left(\left(x \cdot x\right) \cdot x\right) \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{+152}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{x}{n \cdot n} \cdot x\right) \cdot 0.5\\ \end{array} \]
                                              6. Add Preprocessing

                                              Alternative 11: 60.9% accurate, 1.9× speedup?

                                              \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;x \leq 0.9:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;x \leq 3.2 \cdot 10^{+109}:\\ \;\;\;\;\frac{\frac{1 - \frac{0.5 - \frac{0.3333333333333333 - \frac{0.25}{x}}{x}}{x}}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;1 - 1\\ \end{array} \end{array} \]
                                              (FPCore (x n)
                                               :precision binary64
                                               (if (<= x 0.9)
                                                 (/ (- x (log x)) n)
                                                 (if (<= x 3.2e+109)
                                                   (/ (/ (- 1.0 (/ (- 0.5 (/ (- 0.3333333333333333 (/ 0.25 x)) x)) x)) x) n)
                                                   (- 1.0 1.0))))
                                              double code(double x, double n) {
                                              	double tmp;
                                              	if (x <= 0.9) {
                                              		tmp = (x - log(x)) / n;
                                              	} else if (x <= 3.2e+109) {
                                              		tmp = ((1.0 - ((0.5 - ((0.3333333333333333 - (0.25 / x)) / x)) / x)) / x) / n;
                                              	} else {
                                              		tmp = 1.0 - 1.0;
                                              	}
                                              	return tmp;
                                              }
                                              
                                              real(8) function code(x, n)
                                                  real(8), intent (in) :: x
                                                  real(8), intent (in) :: n
                                                  real(8) :: tmp
                                                  if (x <= 0.9d0) then
                                                      tmp = (x - log(x)) / n
                                                  else if (x <= 3.2d+109) then
                                                      tmp = ((1.0d0 - ((0.5d0 - ((0.3333333333333333d0 - (0.25d0 / x)) / x)) / x)) / x) / n
                                                  else
                                                      tmp = 1.0d0 - 1.0d0
                                                  end if
                                                  code = tmp
                                              end function
                                              
                                              public static double code(double x, double n) {
                                              	double tmp;
                                              	if (x <= 0.9) {
                                              		tmp = (x - Math.log(x)) / n;
                                              	} else if (x <= 3.2e+109) {
                                              		tmp = ((1.0 - ((0.5 - ((0.3333333333333333 - (0.25 / x)) / x)) / x)) / x) / n;
                                              	} else {
                                              		tmp = 1.0 - 1.0;
                                              	}
                                              	return tmp;
                                              }
                                              
                                              def code(x, n):
                                              	tmp = 0
                                              	if x <= 0.9:
                                              		tmp = (x - math.log(x)) / n
                                              	elif x <= 3.2e+109:
                                              		tmp = ((1.0 - ((0.5 - ((0.3333333333333333 - (0.25 / x)) / x)) / x)) / x) / n
                                              	else:
                                              		tmp = 1.0 - 1.0
                                              	return tmp
                                              
                                              function code(x, n)
                                              	tmp = 0.0
                                              	if (x <= 0.9)
                                              		tmp = Float64(Float64(x - log(x)) / n);
                                              	elseif (x <= 3.2e+109)
                                              		tmp = Float64(Float64(Float64(1.0 - Float64(Float64(0.5 - Float64(Float64(0.3333333333333333 - Float64(0.25 / x)) / x)) / x)) / x) / n);
                                              	else
                                              		tmp = Float64(1.0 - 1.0);
                                              	end
                                              	return tmp
                                              end
                                              
                                              function tmp_2 = code(x, n)
                                              	tmp = 0.0;
                                              	if (x <= 0.9)
                                              		tmp = (x - log(x)) / n;
                                              	elseif (x <= 3.2e+109)
                                              		tmp = ((1.0 - ((0.5 - ((0.3333333333333333 - (0.25 / x)) / x)) / x)) / x) / n;
                                              	else
                                              		tmp = 1.0 - 1.0;
                                              	end
                                              	tmp_2 = tmp;
                                              end
                                              
                                              code[x_, n_] := If[LessEqual[x, 0.9], N[(N[(x - N[Log[x], $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision], If[LessEqual[x, 3.2e+109], N[(N[(N[(1.0 - N[(N[(0.5 - N[(N[(0.3333333333333333 - N[(0.25 / x), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision] / n), $MachinePrecision], N[(1.0 - 1.0), $MachinePrecision]]]
                                              
                                              \begin{array}{l}
                                              
                                              \\
                                              \begin{array}{l}
                                              \mathbf{if}\;x \leq 0.9:\\
                                              \;\;\;\;\frac{x - \log x}{n}\\
                                              
                                              \mathbf{elif}\;x \leq 3.2 \cdot 10^{+109}:\\
                                              \;\;\;\;\frac{\frac{1 - \frac{0.5 - \frac{0.3333333333333333 - \frac{0.25}{x}}{x}}{x}}{x}}{n}\\
                                              
                                              \mathbf{else}:\\
                                              \;\;\;\;1 - 1\\
                                              
                                              
                                              \end{array}
                                              \end{array}
                                              
                                              Derivation
                                              1. Split input into 3 regimes
                                              2. if x < 0.900000000000000022

                                                1. Initial program 43.4%

                                                  \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                2. Add Preprocessing
                                                3. Taylor expanded in n around inf

                                                  \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                4. Step-by-step derivation
                                                  1. lower-/.f64N/A

                                                    \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                  2. lower--.f64N/A

                                                    \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                                                  3. lower-log1p.f64N/A

                                                    \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                                                  4. lower-log.f6453.4

                                                    \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                                                5. Applied rewrites53.4%

                                                  \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                                                6. Taylor expanded in x around 0

                                                  \[\leadsto \frac{x - \log x}{n} \]
                                                7. Step-by-step derivation
                                                  1. Applied rewrites53.4%

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

                                                  if 0.900000000000000022 < x < 3.2000000000000001e109

                                                  1. Initial program 47.7%

                                                    \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                  2. Add Preprocessing
                                                  3. Taylor expanded in n around inf

                                                    \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                  4. Step-by-step derivation
                                                    1. lower-/.f64N/A

                                                      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                    2. lower--.f64N/A

                                                      \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                                                    3. lower-log1p.f64N/A

                                                      \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                                                    4. lower-log.f6443.3

                                                      \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                                                  5. Applied rewrites43.3%

                                                    \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                                                  6. Step-by-step derivation
                                                    1. Applied rewrites41.5%

                                                      \[\leadsto \frac{-\log \left(\frac{x}{1 + x}\right)}{n} \]
                                                    2. Taylor expanded in x around -inf

                                                      \[\leadsto \frac{-1 \cdot \frac{-1 \cdot \frac{-1 \cdot \frac{\frac{1}{4} \cdot \frac{1}{x} - \frac{1}{3}}{x} - \frac{1}{2}}{x} - 1}{x}}{n} \]
                                                    3. Applied rewrites61.3%

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

                                                    if 3.2000000000000001e109 < x

                                                    1. Initial program 87.4%

                                                      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                    2. Add Preprocessing
                                                    3. Taylor expanded in x around 0

                                                      \[\leadsto \color{blue}{1} - {x}^{\left(\frac{1}{n}\right)} \]
                                                    4. Step-by-step derivation
                                                      1. Applied rewrites36.7%

                                                        \[\leadsto \color{blue}{1} - {x}^{\left(\frac{1}{n}\right)} \]
                                                      2. Taylor expanded in n around inf

                                                        \[\leadsto 1 - \color{blue}{1} \]
                                                      3. Step-by-step derivation
                                                        1. Applied rewrites87.4%

                                                          \[\leadsto 1 - \color{blue}{1} \]
                                                      4. Recombined 3 regimes into one program.
                                                      5. Add Preprocessing

                                                      Alternative 12: 60.7% accurate, 1.9× speedup?

                                                      \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;x \leq 0.7:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 3.2 \cdot 10^{+109}:\\ \;\;\;\;\frac{\frac{1 - \frac{0.5 - \frac{0.3333333333333333 - \frac{0.25}{x}}{x}}{x}}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;1 - 1\\ \end{array} \end{array} \]
                                                      (FPCore (x n)
                                                       :precision binary64
                                                       (if (<= x 0.7)
                                                         (/ (- (log x)) n)
                                                         (if (<= x 3.2e+109)
                                                           (/ (/ (- 1.0 (/ (- 0.5 (/ (- 0.3333333333333333 (/ 0.25 x)) x)) x)) x) n)
                                                           (- 1.0 1.0))))
                                                      double code(double x, double n) {
                                                      	double tmp;
                                                      	if (x <= 0.7) {
                                                      		tmp = -log(x) / n;
                                                      	} else if (x <= 3.2e+109) {
                                                      		tmp = ((1.0 - ((0.5 - ((0.3333333333333333 - (0.25 / x)) / x)) / x)) / x) / n;
                                                      	} else {
                                                      		tmp = 1.0 - 1.0;
                                                      	}
                                                      	return tmp;
                                                      }
                                                      
                                                      real(8) function code(x, n)
                                                          real(8), intent (in) :: x
                                                          real(8), intent (in) :: n
                                                          real(8) :: tmp
                                                          if (x <= 0.7d0) then
                                                              tmp = -log(x) / n
                                                          else if (x <= 3.2d+109) then
                                                              tmp = ((1.0d0 - ((0.5d0 - ((0.3333333333333333d0 - (0.25d0 / x)) / x)) / x)) / x) / n
                                                          else
                                                              tmp = 1.0d0 - 1.0d0
                                                          end if
                                                          code = tmp
                                                      end function
                                                      
                                                      public static double code(double x, double n) {
                                                      	double tmp;
                                                      	if (x <= 0.7) {
                                                      		tmp = -Math.log(x) / n;
                                                      	} else if (x <= 3.2e+109) {
                                                      		tmp = ((1.0 - ((0.5 - ((0.3333333333333333 - (0.25 / x)) / x)) / x)) / x) / n;
                                                      	} else {
                                                      		tmp = 1.0 - 1.0;
                                                      	}
                                                      	return tmp;
                                                      }
                                                      
                                                      def code(x, n):
                                                      	tmp = 0
                                                      	if x <= 0.7:
                                                      		tmp = -math.log(x) / n
                                                      	elif x <= 3.2e+109:
                                                      		tmp = ((1.0 - ((0.5 - ((0.3333333333333333 - (0.25 / x)) / x)) / x)) / x) / n
                                                      	else:
                                                      		tmp = 1.0 - 1.0
                                                      	return tmp
                                                      
                                                      function code(x, n)
                                                      	tmp = 0.0
                                                      	if (x <= 0.7)
                                                      		tmp = Float64(Float64(-log(x)) / n);
                                                      	elseif (x <= 3.2e+109)
                                                      		tmp = Float64(Float64(Float64(1.0 - Float64(Float64(0.5 - Float64(Float64(0.3333333333333333 - Float64(0.25 / x)) / x)) / x)) / x) / n);
                                                      	else
                                                      		tmp = Float64(1.0 - 1.0);
                                                      	end
                                                      	return tmp
                                                      end
                                                      
                                                      function tmp_2 = code(x, n)
                                                      	tmp = 0.0;
                                                      	if (x <= 0.7)
                                                      		tmp = -log(x) / n;
                                                      	elseif (x <= 3.2e+109)
                                                      		tmp = ((1.0 - ((0.5 - ((0.3333333333333333 - (0.25 / x)) / x)) / x)) / x) / n;
                                                      	else
                                                      		tmp = 1.0 - 1.0;
                                                      	end
                                                      	tmp_2 = tmp;
                                                      end
                                                      
                                                      code[x_, n_] := If[LessEqual[x, 0.7], N[((-N[Log[x], $MachinePrecision]) / n), $MachinePrecision], If[LessEqual[x, 3.2e+109], N[(N[(N[(1.0 - N[(N[(0.5 - N[(N[(0.3333333333333333 - N[(0.25 / x), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision] / n), $MachinePrecision], N[(1.0 - 1.0), $MachinePrecision]]]
                                                      
                                                      \begin{array}{l}
                                                      
                                                      \\
                                                      \begin{array}{l}
                                                      \mathbf{if}\;x \leq 0.7:\\
                                                      \;\;\;\;\frac{-\log x}{n}\\
                                                      
                                                      \mathbf{elif}\;x \leq 3.2 \cdot 10^{+109}:\\
                                                      \;\;\;\;\frac{\frac{1 - \frac{0.5 - \frac{0.3333333333333333 - \frac{0.25}{x}}{x}}{x}}{x}}{n}\\
                                                      
                                                      \mathbf{else}:\\
                                                      \;\;\;\;1 - 1\\
                                                      
                                                      
                                                      \end{array}
                                                      \end{array}
                                                      
                                                      Derivation
                                                      1. Split input into 3 regimes
                                                      2. if x < 0.69999999999999996

                                                        1. Initial program 43.4%

                                                          \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                        2. Add Preprocessing
                                                        3. Taylor expanded in n around inf

                                                          \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                        4. Step-by-step derivation
                                                          1. lower-/.f64N/A

                                                            \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                          2. lower--.f64N/A

                                                            \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                                                          3. lower-log1p.f64N/A

                                                            \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                                                          4. lower-log.f6453.4

                                                            \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                                                        5. Applied rewrites53.4%

                                                          \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                                                        6. Taylor expanded in x around 0

                                                          \[\leadsto \frac{-1 \cdot \log x}{n} \]
                                                        7. Step-by-step derivation
                                                          1. Applied rewrites53.0%

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

                                                          if 0.69999999999999996 < x < 3.2000000000000001e109

                                                          1. Initial program 47.7%

                                                            \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                          2. Add Preprocessing
                                                          3. Taylor expanded in n around inf

                                                            \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                          4. Step-by-step derivation
                                                            1. lower-/.f64N/A

                                                              \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                            2. lower--.f64N/A

                                                              \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                                                            3. lower-log1p.f64N/A

                                                              \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                                                            4. lower-log.f6443.3

                                                              \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                                                          5. Applied rewrites43.3%

                                                            \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                                                          6. Step-by-step derivation
                                                            1. Applied rewrites41.5%

                                                              \[\leadsto \frac{-\log \left(\frac{x}{1 + x}\right)}{n} \]
                                                            2. Taylor expanded in x around -inf

                                                              \[\leadsto \frac{-1 \cdot \frac{-1 \cdot \frac{-1 \cdot \frac{\frac{1}{4} \cdot \frac{1}{x} - \frac{1}{3}}{x} - \frac{1}{2}}{x} - 1}{x}}{n} \]
                                                            3. Applied rewrites61.3%

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

                                                            if 3.2000000000000001e109 < x

                                                            1. Initial program 87.4%

                                                              \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                            2. Add Preprocessing
                                                            3. Taylor expanded in x around 0

                                                              \[\leadsto \color{blue}{1} - {x}^{\left(\frac{1}{n}\right)} \]
                                                            4. Step-by-step derivation
                                                              1. Applied rewrites36.7%

                                                                \[\leadsto \color{blue}{1} - {x}^{\left(\frac{1}{n}\right)} \]
                                                              2. Taylor expanded in n around inf

                                                                \[\leadsto 1 - \color{blue}{1} \]
                                                              3. Step-by-step derivation
                                                                1. Applied rewrites87.4%

                                                                  \[\leadsto 1 - \color{blue}{1} \]
                                                              4. Recombined 3 regimes into one program.
                                                              5. Add Preprocessing

                                                              Alternative 13: 55.9% accurate, 2.7× speedup?

                                                              \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{-5}:\\ \;\;\;\;\frac{0.3333333333333333}{\left(\left(x \cdot x\right) \cdot x\right) \cdot n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{n} - \frac{\frac{0.5}{n} - \frac{0.3333333333333333}{x \cdot n}}{x}}{x}\\ \end{array} \end{array} \]
                                                              (FPCore (x n)
                                                               :precision binary64
                                                               (if (<= (/ 1.0 n) -5e-5)
                                                                 (/ 0.3333333333333333 (* (* (* x x) x) n))
                                                                 (/ (- (/ 1.0 n) (/ (- (/ 0.5 n) (/ 0.3333333333333333 (* x n))) x)) x)))
                                                              double code(double x, double n) {
                                                              	double tmp;
                                                              	if ((1.0 / n) <= -5e-5) {
                                                              		tmp = 0.3333333333333333 / (((x * x) * x) * n);
                                                              	} else {
                                                              		tmp = ((1.0 / n) - (((0.5 / n) - (0.3333333333333333 / (x * n))) / x)) / x;
                                                              	}
                                                              	return tmp;
                                                              }
                                                              
                                                              real(8) function code(x, n)
                                                                  real(8), intent (in) :: x
                                                                  real(8), intent (in) :: n
                                                                  real(8) :: tmp
                                                                  if ((1.0d0 / n) <= (-5d-5)) then
                                                                      tmp = 0.3333333333333333d0 / (((x * x) * x) * n)
                                                                  else
                                                                      tmp = ((1.0d0 / n) - (((0.5d0 / n) - (0.3333333333333333d0 / (x * n))) / x)) / x
                                                                  end if
                                                                  code = tmp
                                                              end function
                                                              
                                                              public static double code(double x, double n) {
                                                              	double tmp;
                                                              	if ((1.0 / n) <= -5e-5) {
                                                              		tmp = 0.3333333333333333 / (((x * x) * x) * n);
                                                              	} else {
                                                              		tmp = ((1.0 / n) - (((0.5 / n) - (0.3333333333333333 / (x * n))) / x)) / x;
                                                              	}
                                                              	return tmp;
                                                              }
                                                              
                                                              def code(x, n):
                                                              	tmp = 0
                                                              	if (1.0 / n) <= -5e-5:
                                                              		tmp = 0.3333333333333333 / (((x * x) * x) * n)
                                                              	else:
                                                              		tmp = ((1.0 / n) - (((0.5 / n) - (0.3333333333333333 / (x * n))) / x)) / x
                                                              	return tmp
                                                              
                                                              function code(x, n)
                                                              	tmp = 0.0
                                                              	if (Float64(1.0 / n) <= -5e-5)
                                                              		tmp = Float64(0.3333333333333333 / Float64(Float64(Float64(x * x) * x) * n));
                                                              	else
                                                              		tmp = Float64(Float64(Float64(1.0 / n) - Float64(Float64(Float64(0.5 / n) - Float64(0.3333333333333333 / Float64(x * n))) / x)) / x);
                                                              	end
                                                              	return tmp
                                                              end
                                                              
                                                              function tmp_2 = code(x, n)
                                                              	tmp = 0.0;
                                                              	if ((1.0 / n) <= -5e-5)
                                                              		tmp = 0.3333333333333333 / (((x * x) * x) * n);
                                                              	else
                                                              		tmp = ((1.0 / n) - (((0.5 / n) - (0.3333333333333333 / (x * n))) / x)) / x;
                                                              	end
                                                              	tmp_2 = tmp;
                                                              end
                                                              
                                                              code[x_, n_] := If[LessEqual[N[(1.0 / n), $MachinePrecision], -5e-5], N[(0.3333333333333333 / N[(N[(N[(x * x), $MachinePrecision] * x), $MachinePrecision] * n), $MachinePrecision]), $MachinePrecision], N[(N[(N[(1.0 / n), $MachinePrecision] - N[(N[(N[(0.5 / n), $MachinePrecision] - N[(0.3333333333333333 / N[(x * n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision]]
                                                              
                                                              \begin{array}{l}
                                                              
                                                              \\
                                                              \begin{array}{l}
                                                              \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{-5}:\\
                                                              \;\;\;\;\frac{0.3333333333333333}{\left(\left(x \cdot x\right) \cdot x\right) \cdot n}\\
                                                              
                                                              \mathbf{else}:\\
                                                              \;\;\;\;\frac{\frac{1}{n} - \frac{\frac{0.5}{n} - \frac{0.3333333333333333}{x \cdot n}}{x}}{x}\\
                                                              
                                                              
                                                              \end{array}
                                                              \end{array}
                                                              
                                                              Derivation
                                                              1. Split input into 2 regimes
                                                              2. if (/.f64 #s(literal 1 binary64) n) < -5.00000000000000024e-5

                                                                1. Initial program 99.9%

                                                                  \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                2. Add Preprocessing
                                                                3. Taylor expanded in n around inf

                                                                  \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                4. Step-by-step derivation
                                                                  1. lower-/.f64N/A

                                                                    \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                  2. lower--.f64N/A

                                                                    \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                                                                  3. lower-log1p.f64N/A

                                                                    \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                                                                  4. lower-log.f6453.8

                                                                    \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                                                                5. Applied rewrites53.8%

                                                                  \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                                                                6. Taylor expanded in x around inf

                                                                  \[\leadsto \frac{\left(\frac{\frac{1}{3}}{n \cdot {x}^{2}} + \frac{1}{n}\right) - \frac{\frac{1}{2}}{n \cdot x}}{\color{blue}{x}} \]
                                                                7. Step-by-step derivation
                                                                  1. Applied rewrites18.3%

                                                                    \[\leadsto \frac{\left(\frac{0.3333333333333333}{\left(x \cdot x\right) \cdot n} + \frac{1}{n}\right) - \frac{0.5}{n \cdot x}}{\color{blue}{x}} \]
                                                                  2. Taylor expanded in x around 0

                                                                    \[\leadsto \frac{\frac{1}{3}}{n \cdot \color{blue}{{x}^{3}}} \]
                                                                  3. Step-by-step derivation
                                                                    1. Applied rewrites75.0%

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

                                                                    if -5.00000000000000024e-5 < (/.f64 #s(literal 1 binary64) n)

                                                                    1. Initial program 32.3%

                                                                      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                    2. Add Preprocessing
                                                                    3. Taylor expanded in n around inf

                                                                      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                    4. Step-by-step derivation
                                                                      1. lower-/.f64N/A

                                                                        \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                      2. lower--.f64N/A

                                                                        \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                                                                      3. lower-log1p.f64N/A

                                                                        \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                                                                      4. lower-log.f6464.6

                                                                        \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                                                                    5. Applied rewrites64.6%

                                                                      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                                                                    6. Taylor expanded in x around -inf

                                                                      \[\leadsto -1 \cdot \color{blue}{\frac{-1 \cdot \frac{\frac{1}{3} \cdot \frac{1}{n \cdot x} - \frac{1}{2} \cdot \frac{1}{n}}{x} - \frac{1}{n}}{x}} \]
                                                                    7. Step-by-step derivation
                                                                      1. Applied rewrites42.8%

                                                                        \[\leadsto \frac{\frac{\frac{0.3333333333333333}{n \cdot x} - \frac{0.5}{n}}{-x} - \frac{1}{n}}{\color{blue}{-x}} \]
                                                                    8. Recombined 2 regimes into one program.
                                                                    9. Final simplification53.9%

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

                                                                    Alternative 14: 55.9% accurate, 3.4× speedup?

                                                                    \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{-5}:\\ \;\;\;\;\frac{0.3333333333333333}{\left(\left(x \cdot x\right) \cdot x\right) \cdot n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1 - \frac{0.5 - \frac{0.3333333333333333}{x}}{x}}{x}}{n}\\ \end{array} \end{array} \]
                                                                    (FPCore (x n)
                                                                     :precision binary64
                                                                     (if (<= (/ 1.0 n) -5e-5)
                                                                       (/ 0.3333333333333333 (* (* (* x x) x) n))
                                                                       (/ (/ (- 1.0 (/ (- 0.5 (/ 0.3333333333333333 x)) x)) x) n)))
                                                                    double code(double x, double n) {
                                                                    	double tmp;
                                                                    	if ((1.0 / n) <= -5e-5) {
                                                                    		tmp = 0.3333333333333333 / (((x * x) * x) * n);
                                                                    	} else {
                                                                    		tmp = ((1.0 - ((0.5 - (0.3333333333333333 / x)) / x)) / x) / n;
                                                                    	}
                                                                    	return tmp;
                                                                    }
                                                                    
                                                                    real(8) function code(x, n)
                                                                        real(8), intent (in) :: x
                                                                        real(8), intent (in) :: n
                                                                        real(8) :: tmp
                                                                        if ((1.0d0 / n) <= (-5d-5)) then
                                                                            tmp = 0.3333333333333333d0 / (((x * x) * x) * n)
                                                                        else
                                                                            tmp = ((1.0d0 - ((0.5d0 - (0.3333333333333333d0 / x)) / x)) / x) / n
                                                                        end if
                                                                        code = tmp
                                                                    end function
                                                                    
                                                                    public static double code(double x, double n) {
                                                                    	double tmp;
                                                                    	if ((1.0 / n) <= -5e-5) {
                                                                    		tmp = 0.3333333333333333 / (((x * x) * x) * n);
                                                                    	} else {
                                                                    		tmp = ((1.0 - ((0.5 - (0.3333333333333333 / x)) / x)) / x) / n;
                                                                    	}
                                                                    	return tmp;
                                                                    }
                                                                    
                                                                    def code(x, n):
                                                                    	tmp = 0
                                                                    	if (1.0 / n) <= -5e-5:
                                                                    		tmp = 0.3333333333333333 / (((x * x) * x) * n)
                                                                    	else:
                                                                    		tmp = ((1.0 - ((0.5 - (0.3333333333333333 / x)) / x)) / x) / n
                                                                    	return tmp
                                                                    
                                                                    function code(x, n)
                                                                    	tmp = 0.0
                                                                    	if (Float64(1.0 / n) <= -5e-5)
                                                                    		tmp = Float64(0.3333333333333333 / Float64(Float64(Float64(x * x) * x) * n));
                                                                    	else
                                                                    		tmp = Float64(Float64(Float64(1.0 - Float64(Float64(0.5 - Float64(0.3333333333333333 / x)) / x)) / x) / n);
                                                                    	end
                                                                    	return tmp
                                                                    end
                                                                    
                                                                    function tmp_2 = code(x, n)
                                                                    	tmp = 0.0;
                                                                    	if ((1.0 / n) <= -5e-5)
                                                                    		tmp = 0.3333333333333333 / (((x * x) * x) * n);
                                                                    	else
                                                                    		tmp = ((1.0 - ((0.5 - (0.3333333333333333 / x)) / x)) / x) / n;
                                                                    	end
                                                                    	tmp_2 = tmp;
                                                                    end
                                                                    
                                                                    code[x_, n_] := If[LessEqual[N[(1.0 / n), $MachinePrecision], -5e-5], N[(0.3333333333333333 / N[(N[(N[(x * x), $MachinePrecision] * x), $MachinePrecision] * n), $MachinePrecision]), $MachinePrecision], N[(N[(N[(1.0 - N[(N[(0.5 - N[(0.3333333333333333 / x), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision] / n), $MachinePrecision]]
                                                                    
                                                                    \begin{array}{l}
                                                                    
                                                                    \\
                                                                    \begin{array}{l}
                                                                    \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{-5}:\\
                                                                    \;\;\;\;\frac{0.3333333333333333}{\left(\left(x \cdot x\right) \cdot x\right) \cdot n}\\
                                                                    
                                                                    \mathbf{else}:\\
                                                                    \;\;\;\;\frac{\frac{1 - \frac{0.5 - \frac{0.3333333333333333}{x}}{x}}{x}}{n}\\
                                                                    
                                                                    
                                                                    \end{array}
                                                                    \end{array}
                                                                    
                                                                    Derivation
                                                                    1. Split input into 2 regimes
                                                                    2. if (/.f64 #s(literal 1 binary64) n) < -5.00000000000000024e-5

                                                                      1. Initial program 99.9%

                                                                        \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                      2. Add Preprocessing
                                                                      3. Taylor expanded in n around inf

                                                                        \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                      4. Step-by-step derivation
                                                                        1. lower-/.f64N/A

                                                                          \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                        2. lower--.f64N/A

                                                                          \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                                                                        3. lower-log1p.f64N/A

                                                                          \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                                                                        4. lower-log.f6453.8

                                                                          \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                                                                      5. Applied rewrites53.8%

                                                                        \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                                                                      6. Taylor expanded in x around inf

                                                                        \[\leadsto \frac{\left(\frac{\frac{1}{3}}{n \cdot {x}^{2}} + \frac{1}{n}\right) - \frac{\frac{1}{2}}{n \cdot x}}{\color{blue}{x}} \]
                                                                      7. Step-by-step derivation
                                                                        1. Applied rewrites18.3%

                                                                          \[\leadsto \frac{\left(\frac{0.3333333333333333}{\left(x \cdot x\right) \cdot n} + \frac{1}{n}\right) - \frac{0.5}{n \cdot x}}{\color{blue}{x}} \]
                                                                        2. Taylor expanded in x around 0

                                                                          \[\leadsto \frac{\frac{1}{3}}{n \cdot \color{blue}{{x}^{3}}} \]
                                                                        3. Step-by-step derivation
                                                                          1. Applied rewrites75.0%

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

                                                                          if -5.00000000000000024e-5 < (/.f64 #s(literal 1 binary64) n)

                                                                          1. Initial program 32.3%

                                                                            \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                          2. Add Preprocessing
                                                                          3. Taylor expanded in n around inf

                                                                            \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                          4. Step-by-step derivation
                                                                            1. lower-/.f64N/A

                                                                              \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                            2. lower--.f64N/A

                                                                              \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                                                                            3. lower-log1p.f64N/A

                                                                              \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                                                                            4. lower-log.f6464.6

                                                                              \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                                                                          5. Applied rewrites64.6%

                                                                            \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                                                                          6. Step-by-step derivation
                                                                            1. Applied rewrites64.8%

                                                                              \[\leadsto \frac{-\log \left(\frac{x}{1 + x}\right)}{n} \]
                                                                            2. Taylor expanded in x around -inf

                                                                              \[\leadsto \frac{-1 \cdot \frac{-1 \cdot \frac{\frac{1}{3} \cdot \frac{1}{x} - \frac{1}{2}}{x} - 1}{x}}{n} \]
                                                                            3. Step-by-step derivation
                                                                              1. Applied rewrites42.8%

                                                                                \[\leadsto \frac{\frac{1 - \frac{0.5 - \frac{0.3333333333333333}{x}}{x}}{x}}{n} \]
                                                                            4. Recombined 2 regimes into one program.
                                                                            5. Add Preprocessing

                                                                            Alternative 15: 55.6% accurate, 3.7× speedup?

                                                                            \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5000000:\\ \;\;\;\;\frac{0.3333333333333333}{\left(\left(x \cdot x\right) \cdot x\right) \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{+40}:\\ \;\;\;\;\frac{-1}{x} \cdot \frac{-1}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{x}{n \cdot n} \cdot x\right) \cdot 0.5\\ \end{array} \end{array} \]
                                                                            (FPCore (x n)
                                                                             :precision binary64
                                                                             (if (<= (/ 1.0 n) -5000000.0)
                                                                               (/ 0.3333333333333333 (* (* (* x x) x) n))
                                                                               (if (<= (/ 1.0 n) 1e+40)
                                                                                 (* (/ -1.0 x) (/ -1.0 n))
                                                                                 (* (* (/ x (* n n)) x) 0.5))))
                                                                            double code(double x, double n) {
                                                                            	double tmp;
                                                                            	if ((1.0 / n) <= -5000000.0) {
                                                                            		tmp = 0.3333333333333333 / (((x * x) * x) * n);
                                                                            	} else if ((1.0 / n) <= 1e+40) {
                                                                            		tmp = (-1.0 / x) * (-1.0 / n);
                                                                            	} else {
                                                                            		tmp = ((x / (n * n)) * x) * 0.5;
                                                                            	}
                                                                            	return tmp;
                                                                            }
                                                                            
                                                                            real(8) function code(x, n)
                                                                                real(8), intent (in) :: x
                                                                                real(8), intent (in) :: n
                                                                                real(8) :: tmp
                                                                                if ((1.0d0 / n) <= (-5000000.0d0)) then
                                                                                    tmp = 0.3333333333333333d0 / (((x * x) * x) * n)
                                                                                else if ((1.0d0 / n) <= 1d+40) then
                                                                                    tmp = ((-1.0d0) / x) * ((-1.0d0) / n)
                                                                                else
                                                                                    tmp = ((x / (n * n)) * x) * 0.5d0
                                                                                end if
                                                                                code = tmp
                                                                            end function
                                                                            
                                                                            public static double code(double x, double n) {
                                                                            	double tmp;
                                                                            	if ((1.0 / n) <= -5000000.0) {
                                                                            		tmp = 0.3333333333333333 / (((x * x) * x) * n);
                                                                            	} else if ((1.0 / n) <= 1e+40) {
                                                                            		tmp = (-1.0 / x) * (-1.0 / n);
                                                                            	} else {
                                                                            		tmp = ((x / (n * n)) * x) * 0.5;
                                                                            	}
                                                                            	return tmp;
                                                                            }
                                                                            
                                                                            def code(x, n):
                                                                            	tmp = 0
                                                                            	if (1.0 / n) <= -5000000.0:
                                                                            		tmp = 0.3333333333333333 / (((x * x) * x) * n)
                                                                            	elif (1.0 / n) <= 1e+40:
                                                                            		tmp = (-1.0 / x) * (-1.0 / n)
                                                                            	else:
                                                                            		tmp = ((x / (n * n)) * x) * 0.5
                                                                            	return tmp
                                                                            
                                                                            function code(x, n)
                                                                            	tmp = 0.0
                                                                            	if (Float64(1.0 / n) <= -5000000.0)
                                                                            		tmp = Float64(0.3333333333333333 / Float64(Float64(Float64(x * x) * x) * n));
                                                                            	elseif (Float64(1.0 / n) <= 1e+40)
                                                                            		tmp = Float64(Float64(-1.0 / x) * Float64(-1.0 / n));
                                                                            	else
                                                                            		tmp = Float64(Float64(Float64(x / Float64(n * n)) * x) * 0.5);
                                                                            	end
                                                                            	return tmp
                                                                            end
                                                                            
                                                                            function tmp_2 = code(x, n)
                                                                            	tmp = 0.0;
                                                                            	if ((1.0 / n) <= -5000000.0)
                                                                            		tmp = 0.3333333333333333 / (((x * x) * x) * n);
                                                                            	elseif ((1.0 / n) <= 1e+40)
                                                                            		tmp = (-1.0 / x) * (-1.0 / n);
                                                                            	else
                                                                            		tmp = ((x / (n * n)) * x) * 0.5;
                                                                            	end
                                                                            	tmp_2 = tmp;
                                                                            end
                                                                            
                                                                            code[x_, n_] := If[LessEqual[N[(1.0 / n), $MachinePrecision], -5000000.0], N[(0.3333333333333333 / N[(N[(N[(x * x), $MachinePrecision] * x), $MachinePrecision] * n), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 1e+40], N[(N[(-1.0 / x), $MachinePrecision] * N[(-1.0 / n), $MachinePrecision]), $MachinePrecision], N[(N[(N[(x / N[(n * n), $MachinePrecision]), $MachinePrecision] * x), $MachinePrecision] * 0.5), $MachinePrecision]]]
                                                                            
                                                                            \begin{array}{l}
                                                                            
                                                                            \\
                                                                            \begin{array}{l}
                                                                            \mathbf{if}\;\frac{1}{n} \leq -5000000:\\
                                                                            \;\;\;\;\frac{0.3333333333333333}{\left(\left(x \cdot x\right) \cdot x\right) \cdot n}\\
                                                                            
                                                                            \mathbf{elif}\;\frac{1}{n} \leq 10^{+40}:\\
                                                                            \;\;\;\;\frac{-1}{x} \cdot \frac{-1}{n}\\
                                                                            
                                                                            \mathbf{else}:\\
                                                                            \;\;\;\;\left(\frac{x}{n \cdot n} \cdot x\right) \cdot 0.5\\
                                                                            
                                                                            
                                                                            \end{array}
                                                                            \end{array}
                                                                            
                                                                            Derivation
                                                                            1. Split input into 3 regimes
                                                                            2. if (/.f64 #s(literal 1 binary64) n) < -5e6

                                                                              1. Initial program 100.0%

                                                                                \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                              2. Add Preprocessing
                                                                              3. Taylor expanded in n around inf

                                                                                \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                              4. Step-by-step derivation
                                                                                1. lower-/.f64N/A

                                                                                  \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                                2. lower--.f64N/A

                                                                                  \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                                                                                3. lower-log1p.f64N/A

                                                                                  \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                                                                                4. lower-log.f6454.1

                                                                                  \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                                                                              5. Applied rewrites54.1%

                                                                                \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                                                                              6. Taylor expanded in x around inf

                                                                                \[\leadsto \frac{\left(\frac{\frac{1}{3}}{n \cdot {x}^{2}} + \frac{1}{n}\right) - \frac{\frac{1}{2}}{n \cdot x}}{\color{blue}{x}} \]
                                                                              7. Step-by-step derivation
                                                                                1. Applied rewrites18.5%

                                                                                  \[\leadsto \frac{\left(\frac{0.3333333333333333}{\left(x \cdot x\right) \cdot n} + \frac{1}{n}\right) - \frac{0.5}{n \cdot x}}{\color{blue}{x}} \]
                                                                                2. Taylor expanded in x around 0

                                                                                  \[\leadsto \frac{\frac{1}{3}}{n \cdot \color{blue}{{x}^{3}}} \]
                                                                                3. Step-by-step derivation
                                                                                  1. Applied rewrites75.9%

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

                                                                                  if -5e6 < (/.f64 #s(literal 1 binary64) n) < 1.00000000000000003e40

                                                                                  1. Initial program 30.9%

                                                                                    \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                  2. Add Preprocessing
                                                                                  3. Taylor expanded in n around inf

                                                                                    \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                                  4. Step-by-step derivation
                                                                                    1. lower-/.f64N/A

                                                                                      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                                    2. lower--.f64N/A

                                                                                      \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                                                                                    3. lower-log1p.f64N/A

                                                                                      \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                                                                                    4. lower-log.f6474.6

                                                                                      \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                                                                                  5. Applied rewrites74.6%

                                                                                    \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                                                                                  6. Taylor expanded in x around inf

                                                                                    \[\leadsto \frac{1}{\color{blue}{n \cdot x}} \]
                                                                                  7. Step-by-step derivation
                                                                                    1. Applied rewrites38.1%

                                                                                      \[\leadsto \frac{1}{\color{blue}{n \cdot x}} \]
                                                                                    2. Step-by-step derivation
                                                                                      1. Applied rewrites40.2%

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

                                                                                      if 1.00000000000000003e40 < (/.f64 #s(literal 1 binary64) n)

                                                                                      1. Initial program 41.4%

                                                                                        \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                      2. Add Preprocessing
                                                                                      3. Taylor expanded in x around 0

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

                                                                                          \[\leadsto \color{blue}{\left(x \cdot \left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) + 1\right)} - e^{\frac{\log x}{n}} \]
                                                                                        2. associate--l+N/A

                                                                                          \[\leadsto \color{blue}{x \cdot \left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) + \left(1 - e^{\frac{\log x}{n}}\right)} \]
                                                                                        3. *-commutativeN/A

                                                                                          \[\leadsto \color{blue}{\left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) \cdot x} + \left(1 - e^{\frac{\log x}{n}}\right) \]
                                                                                        4. lower-fma.f64N/A

                                                                                          \[\leadsto \color{blue}{\mathsf{fma}\left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}, x, 1 - e^{\frac{\log x}{n}}\right)} \]
                                                                                      5. Applied rewrites89.6%

                                                                                        \[\leadsto \color{blue}{\mathsf{fma}\left(\mathsf{fma}\left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}, x, \frac{1}{n}\right), x, 1 - {x}^{\left(\frac{1}{n}\right)}\right)} \]
                                                                                      6. Taylor expanded in n around -inf

                                                                                        \[\leadsto -1 \cdot \color{blue}{\frac{x \cdot \left(\frac{1}{2} \cdot x - 1\right) - -1 \cdot \log x}{n}} \]
                                                                                      7. Step-by-step derivation
                                                                                        1. Applied rewrites12.9%

                                                                                          \[\leadsto \frac{\mathsf{fma}\left(\mathsf{fma}\left(0.5, x, -1\right), x, \log x\right)}{\color{blue}{-n}} \]
                                                                                        2. Taylor expanded in x around inf

                                                                                          \[\leadsto \frac{{x}^{2} \cdot \left(\left(\frac{1}{2} + -1 \cdot \frac{\log \left(\frac{1}{x}\right)}{{x}^{2}}\right) - \frac{1}{x}\right)}{\mathsf{neg}\left(n\right)} \]
                                                                                        3. Step-by-step derivation
                                                                                          1. Applied rewrites17.2%

                                                                                            \[\leadsto \frac{\left(\left(\frac{\log x}{x \cdot x} + 0.5\right) - \frac{1}{x}\right) \cdot \left(x \cdot x\right)}{-n} \]
                                                                                          2. Taylor expanded in n around 0

                                                                                            \[\leadsto \frac{1}{2} \cdot \color{blue}{\frac{{x}^{2}}{{n}^{2}}} \]
                                                                                          3. Step-by-step derivation
                                                                                            1. Applied rewrites59.1%

                                                                                              \[\leadsto \left(x \cdot \frac{x}{n \cdot n}\right) \cdot \color{blue}{0.5} \]
                                                                                          4. Recombined 3 regimes into one program.
                                                                                          5. Final simplification54.4%

                                                                                            \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5000000:\\ \;\;\;\;\frac{0.3333333333333333}{\left(\left(x \cdot x\right) \cdot x\right) \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{+40}:\\ \;\;\;\;\frac{-1}{x} \cdot \frac{-1}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{x}{n \cdot n} \cdot x\right) \cdot 0.5\\ \end{array} \]
                                                                                          6. Add Preprocessing

                                                                                          Alternative 16: 55.6% accurate, 3.8× speedup?

                                                                                          \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5000000:\\ \;\;\;\;\frac{0.3333333333333333}{\left(\left(x \cdot x\right) \cdot x\right) \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{+40}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{x}{n \cdot n} \cdot x\right) \cdot 0.5\\ \end{array} \end{array} \]
                                                                                          (FPCore (x n)
                                                                                           :precision binary64
                                                                                           (if (<= (/ 1.0 n) -5000000.0)
                                                                                             (/ 0.3333333333333333 (* (* (* x x) x) n))
                                                                                             (if (<= (/ 1.0 n) 1e+40) (/ (/ 1.0 x) n) (* (* (/ x (* n n)) x) 0.5))))
                                                                                          double code(double x, double n) {
                                                                                          	double tmp;
                                                                                          	if ((1.0 / n) <= -5000000.0) {
                                                                                          		tmp = 0.3333333333333333 / (((x * x) * x) * n);
                                                                                          	} else if ((1.0 / n) <= 1e+40) {
                                                                                          		tmp = (1.0 / x) / n;
                                                                                          	} else {
                                                                                          		tmp = ((x / (n * n)) * x) * 0.5;
                                                                                          	}
                                                                                          	return tmp;
                                                                                          }
                                                                                          
                                                                                          real(8) function code(x, n)
                                                                                              real(8), intent (in) :: x
                                                                                              real(8), intent (in) :: n
                                                                                              real(8) :: tmp
                                                                                              if ((1.0d0 / n) <= (-5000000.0d0)) then
                                                                                                  tmp = 0.3333333333333333d0 / (((x * x) * x) * n)
                                                                                              else if ((1.0d0 / n) <= 1d+40) then
                                                                                                  tmp = (1.0d0 / x) / n
                                                                                              else
                                                                                                  tmp = ((x / (n * n)) * x) * 0.5d0
                                                                                              end if
                                                                                              code = tmp
                                                                                          end function
                                                                                          
                                                                                          public static double code(double x, double n) {
                                                                                          	double tmp;
                                                                                          	if ((1.0 / n) <= -5000000.0) {
                                                                                          		tmp = 0.3333333333333333 / (((x * x) * x) * n);
                                                                                          	} else if ((1.0 / n) <= 1e+40) {
                                                                                          		tmp = (1.0 / x) / n;
                                                                                          	} else {
                                                                                          		tmp = ((x / (n * n)) * x) * 0.5;
                                                                                          	}
                                                                                          	return tmp;
                                                                                          }
                                                                                          
                                                                                          def code(x, n):
                                                                                          	tmp = 0
                                                                                          	if (1.0 / n) <= -5000000.0:
                                                                                          		tmp = 0.3333333333333333 / (((x * x) * x) * n)
                                                                                          	elif (1.0 / n) <= 1e+40:
                                                                                          		tmp = (1.0 / x) / n
                                                                                          	else:
                                                                                          		tmp = ((x / (n * n)) * x) * 0.5
                                                                                          	return tmp
                                                                                          
                                                                                          function code(x, n)
                                                                                          	tmp = 0.0
                                                                                          	if (Float64(1.0 / n) <= -5000000.0)
                                                                                          		tmp = Float64(0.3333333333333333 / Float64(Float64(Float64(x * x) * x) * n));
                                                                                          	elseif (Float64(1.0 / n) <= 1e+40)
                                                                                          		tmp = Float64(Float64(1.0 / x) / n);
                                                                                          	else
                                                                                          		tmp = Float64(Float64(Float64(x / Float64(n * n)) * x) * 0.5);
                                                                                          	end
                                                                                          	return tmp
                                                                                          end
                                                                                          
                                                                                          function tmp_2 = code(x, n)
                                                                                          	tmp = 0.0;
                                                                                          	if ((1.0 / n) <= -5000000.0)
                                                                                          		tmp = 0.3333333333333333 / (((x * x) * x) * n);
                                                                                          	elseif ((1.0 / n) <= 1e+40)
                                                                                          		tmp = (1.0 / x) / n;
                                                                                          	else
                                                                                          		tmp = ((x / (n * n)) * x) * 0.5;
                                                                                          	end
                                                                                          	tmp_2 = tmp;
                                                                                          end
                                                                                          
                                                                                          code[x_, n_] := If[LessEqual[N[(1.0 / n), $MachinePrecision], -5000000.0], N[(0.3333333333333333 / N[(N[(N[(x * x), $MachinePrecision] * x), $MachinePrecision] * n), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 1e+40], N[(N[(1.0 / x), $MachinePrecision] / n), $MachinePrecision], N[(N[(N[(x / N[(n * n), $MachinePrecision]), $MachinePrecision] * x), $MachinePrecision] * 0.5), $MachinePrecision]]]
                                                                                          
                                                                                          \begin{array}{l}
                                                                                          
                                                                                          \\
                                                                                          \begin{array}{l}
                                                                                          \mathbf{if}\;\frac{1}{n} \leq -5000000:\\
                                                                                          \;\;\;\;\frac{0.3333333333333333}{\left(\left(x \cdot x\right) \cdot x\right) \cdot n}\\
                                                                                          
                                                                                          \mathbf{elif}\;\frac{1}{n} \leq 10^{+40}:\\
                                                                                          \;\;\;\;\frac{\frac{1}{x}}{n}\\
                                                                                          
                                                                                          \mathbf{else}:\\
                                                                                          \;\;\;\;\left(\frac{x}{n \cdot n} \cdot x\right) \cdot 0.5\\
                                                                                          
                                                                                          
                                                                                          \end{array}
                                                                                          \end{array}
                                                                                          
                                                                                          Derivation
                                                                                          1. Split input into 3 regimes
                                                                                          2. if (/.f64 #s(literal 1 binary64) n) < -5e6

                                                                                            1. Initial program 100.0%

                                                                                              \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                            2. Add Preprocessing
                                                                                            3. Taylor expanded in n around inf

                                                                                              \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                                            4. Step-by-step derivation
                                                                                              1. lower-/.f64N/A

                                                                                                \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                                              2. lower--.f64N/A

                                                                                                \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                                                                                              3. lower-log1p.f64N/A

                                                                                                \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                                                                                              4. lower-log.f6454.1

                                                                                                \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                                                                                            5. Applied rewrites54.1%

                                                                                              \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                                                                                            6. Taylor expanded in x around inf

                                                                                              \[\leadsto \frac{\left(\frac{\frac{1}{3}}{n \cdot {x}^{2}} + \frac{1}{n}\right) - \frac{\frac{1}{2}}{n \cdot x}}{\color{blue}{x}} \]
                                                                                            7. Step-by-step derivation
                                                                                              1. Applied rewrites18.5%

                                                                                                \[\leadsto \frac{\left(\frac{0.3333333333333333}{\left(x \cdot x\right) \cdot n} + \frac{1}{n}\right) - \frac{0.5}{n \cdot x}}{\color{blue}{x}} \]
                                                                                              2. Taylor expanded in x around 0

                                                                                                \[\leadsto \frac{\frac{1}{3}}{n \cdot \color{blue}{{x}^{3}}} \]
                                                                                              3. Step-by-step derivation
                                                                                                1. Applied rewrites75.9%

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

                                                                                                if -5e6 < (/.f64 #s(literal 1 binary64) n) < 1.00000000000000003e40

                                                                                                1. Initial program 30.9%

                                                                                                  \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                2. Add Preprocessing
                                                                                                3. Taylor expanded in n around inf

                                                                                                  \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                                                4. Step-by-step derivation
                                                                                                  1. lower-/.f64N/A

                                                                                                    \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                                                  2. lower--.f64N/A

                                                                                                    \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                                                                                                  3. lower-log1p.f64N/A

                                                                                                    \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                                                                                                  4. lower-log.f6474.6

                                                                                                    \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                                                                                                5. Applied rewrites74.6%

                                                                                                  \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                                                                                                6. Taylor expanded in x around inf

                                                                                                  \[\leadsto \frac{\frac{1}{x}}{n} \]
                                                                                                7. Step-by-step derivation
                                                                                                  1. Applied rewrites40.2%

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

                                                                                                  if 1.00000000000000003e40 < (/.f64 #s(literal 1 binary64) n)

                                                                                                  1. Initial program 41.4%

                                                                                                    \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                  2. Add Preprocessing
                                                                                                  3. Taylor expanded in x around 0

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

                                                                                                      \[\leadsto \color{blue}{\left(x \cdot \left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) + 1\right)} - e^{\frac{\log x}{n}} \]
                                                                                                    2. associate--l+N/A

                                                                                                      \[\leadsto \color{blue}{x \cdot \left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) + \left(1 - e^{\frac{\log x}{n}}\right)} \]
                                                                                                    3. *-commutativeN/A

                                                                                                      \[\leadsto \color{blue}{\left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) \cdot x} + \left(1 - e^{\frac{\log x}{n}}\right) \]
                                                                                                    4. lower-fma.f64N/A

                                                                                                      \[\leadsto \color{blue}{\mathsf{fma}\left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}, x, 1 - e^{\frac{\log x}{n}}\right)} \]
                                                                                                  5. Applied rewrites89.6%

                                                                                                    \[\leadsto \color{blue}{\mathsf{fma}\left(\mathsf{fma}\left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}, x, \frac{1}{n}\right), x, 1 - {x}^{\left(\frac{1}{n}\right)}\right)} \]
                                                                                                  6. Taylor expanded in n around -inf

                                                                                                    \[\leadsto -1 \cdot \color{blue}{\frac{x \cdot \left(\frac{1}{2} \cdot x - 1\right) - -1 \cdot \log x}{n}} \]
                                                                                                  7. Step-by-step derivation
                                                                                                    1. Applied rewrites12.9%

                                                                                                      \[\leadsto \frac{\mathsf{fma}\left(\mathsf{fma}\left(0.5, x, -1\right), x, \log x\right)}{\color{blue}{-n}} \]
                                                                                                    2. Taylor expanded in x around inf

                                                                                                      \[\leadsto \frac{{x}^{2} \cdot \left(\left(\frac{1}{2} + -1 \cdot \frac{\log \left(\frac{1}{x}\right)}{{x}^{2}}\right) - \frac{1}{x}\right)}{\mathsf{neg}\left(n\right)} \]
                                                                                                    3. Step-by-step derivation
                                                                                                      1. Applied rewrites17.2%

                                                                                                        \[\leadsto \frac{\left(\left(\frac{\log x}{x \cdot x} + 0.5\right) - \frac{1}{x}\right) \cdot \left(x \cdot x\right)}{-n} \]
                                                                                                      2. Taylor expanded in n around 0

                                                                                                        \[\leadsto \frac{1}{2} \cdot \color{blue}{\frac{{x}^{2}}{{n}^{2}}} \]
                                                                                                      3. Step-by-step derivation
                                                                                                        1. Applied rewrites59.1%

                                                                                                          \[\leadsto \left(x \cdot \frac{x}{n \cdot n}\right) \cdot \color{blue}{0.5} \]
                                                                                                      4. Recombined 3 regimes into one program.
                                                                                                      5. Final simplification54.4%

                                                                                                        \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5000000:\\ \;\;\;\;\frac{0.3333333333333333}{\left(\left(x \cdot x\right) \cdot x\right) \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{+40}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{x}{n \cdot n} \cdot x\right) \cdot 0.5\\ \end{array} \]
                                                                                                      6. Add Preprocessing

                                                                                                      Alternative 17: 48.3% accurate, 3.8× speedup?

                                                                                                      \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{+18}:\\ \;\;\;\;1 - 1\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{+40}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{x}{n \cdot n} \cdot x\right) \cdot 0.5\\ \end{array} \end{array} \]
                                                                                                      (FPCore (x n)
                                                                                                       :precision binary64
                                                                                                       (if (<= (/ 1.0 n) -5e+18)
                                                                                                         (- 1.0 1.0)
                                                                                                         (if (<= (/ 1.0 n) 1e+40) (/ (/ 1.0 x) n) (* (* (/ x (* n n)) x) 0.5))))
                                                                                                      double code(double x, double n) {
                                                                                                      	double tmp;
                                                                                                      	if ((1.0 / n) <= -5e+18) {
                                                                                                      		tmp = 1.0 - 1.0;
                                                                                                      	} else if ((1.0 / n) <= 1e+40) {
                                                                                                      		tmp = (1.0 / x) / n;
                                                                                                      	} else {
                                                                                                      		tmp = ((x / (n * n)) * x) * 0.5;
                                                                                                      	}
                                                                                                      	return tmp;
                                                                                                      }
                                                                                                      
                                                                                                      real(8) function code(x, n)
                                                                                                          real(8), intent (in) :: x
                                                                                                          real(8), intent (in) :: n
                                                                                                          real(8) :: tmp
                                                                                                          if ((1.0d0 / n) <= (-5d+18)) then
                                                                                                              tmp = 1.0d0 - 1.0d0
                                                                                                          else if ((1.0d0 / n) <= 1d+40) then
                                                                                                              tmp = (1.0d0 / x) / n
                                                                                                          else
                                                                                                              tmp = ((x / (n * n)) * x) * 0.5d0
                                                                                                          end if
                                                                                                          code = tmp
                                                                                                      end function
                                                                                                      
                                                                                                      public static double code(double x, double n) {
                                                                                                      	double tmp;
                                                                                                      	if ((1.0 / n) <= -5e+18) {
                                                                                                      		tmp = 1.0 - 1.0;
                                                                                                      	} else if ((1.0 / n) <= 1e+40) {
                                                                                                      		tmp = (1.0 / x) / n;
                                                                                                      	} else {
                                                                                                      		tmp = ((x / (n * n)) * x) * 0.5;
                                                                                                      	}
                                                                                                      	return tmp;
                                                                                                      }
                                                                                                      
                                                                                                      def code(x, n):
                                                                                                      	tmp = 0
                                                                                                      	if (1.0 / n) <= -5e+18:
                                                                                                      		tmp = 1.0 - 1.0
                                                                                                      	elif (1.0 / n) <= 1e+40:
                                                                                                      		tmp = (1.0 / x) / n
                                                                                                      	else:
                                                                                                      		tmp = ((x / (n * n)) * x) * 0.5
                                                                                                      	return tmp
                                                                                                      
                                                                                                      function code(x, n)
                                                                                                      	tmp = 0.0
                                                                                                      	if (Float64(1.0 / n) <= -5e+18)
                                                                                                      		tmp = Float64(1.0 - 1.0);
                                                                                                      	elseif (Float64(1.0 / n) <= 1e+40)
                                                                                                      		tmp = Float64(Float64(1.0 / x) / n);
                                                                                                      	else
                                                                                                      		tmp = Float64(Float64(Float64(x / Float64(n * n)) * x) * 0.5);
                                                                                                      	end
                                                                                                      	return tmp
                                                                                                      end
                                                                                                      
                                                                                                      function tmp_2 = code(x, n)
                                                                                                      	tmp = 0.0;
                                                                                                      	if ((1.0 / n) <= -5e+18)
                                                                                                      		tmp = 1.0 - 1.0;
                                                                                                      	elseif ((1.0 / n) <= 1e+40)
                                                                                                      		tmp = (1.0 / x) / n;
                                                                                                      	else
                                                                                                      		tmp = ((x / (n * n)) * x) * 0.5;
                                                                                                      	end
                                                                                                      	tmp_2 = tmp;
                                                                                                      end
                                                                                                      
                                                                                                      code[x_, n_] := If[LessEqual[N[(1.0 / n), $MachinePrecision], -5e+18], N[(1.0 - 1.0), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 1e+40], N[(N[(1.0 / x), $MachinePrecision] / n), $MachinePrecision], N[(N[(N[(x / N[(n * n), $MachinePrecision]), $MachinePrecision] * x), $MachinePrecision] * 0.5), $MachinePrecision]]]
                                                                                                      
                                                                                                      \begin{array}{l}
                                                                                                      
                                                                                                      \\
                                                                                                      \begin{array}{l}
                                                                                                      \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{+18}:\\
                                                                                                      \;\;\;\;1 - 1\\
                                                                                                      
                                                                                                      \mathbf{elif}\;\frac{1}{n} \leq 10^{+40}:\\
                                                                                                      \;\;\;\;\frac{\frac{1}{x}}{n}\\
                                                                                                      
                                                                                                      \mathbf{else}:\\
                                                                                                      \;\;\;\;\left(\frac{x}{n \cdot n} \cdot x\right) \cdot 0.5\\
                                                                                                      
                                                                                                      
                                                                                                      \end{array}
                                                                                                      \end{array}
                                                                                                      
                                                                                                      Derivation
                                                                                                      1. Split input into 3 regimes
                                                                                                      2. if (/.f64 #s(literal 1 binary64) n) < -5e18

                                                                                                        1. Initial program 100.0%

                                                                                                          \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                        2. Add Preprocessing
                                                                                                        3. Taylor expanded in x around 0

                                                                                                          \[\leadsto \color{blue}{1} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                        4. Step-by-step derivation
                                                                                                          1. Applied rewrites44.6%

                                                                                                            \[\leadsto \color{blue}{1} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                          2. Taylor expanded in n around inf

                                                                                                            \[\leadsto 1 - \color{blue}{1} \]
                                                                                                          3. Step-by-step derivation
                                                                                                            1. Applied rewrites57.8%

                                                                                                              \[\leadsto 1 - \color{blue}{1} \]

                                                                                                            if -5e18 < (/.f64 #s(literal 1 binary64) n) < 1.00000000000000003e40

                                                                                                            1. Initial program 32.4%

                                                                                                              \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                            2. Add Preprocessing
                                                                                                            3. Taylor expanded in n around inf

                                                                                                              \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                                                            4. Step-by-step derivation
                                                                                                              1. lower-/.f64N/A

                                                                                                                \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                                                              2. lower--.f64N/A

                                                                                                                \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                                                                                                              3. lower-log1p.f64N/A

                                                                                                                \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                                                                                                              4. lower-log.f6473.1

                                                                                                                \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                                                                                                            5. Applied rewrites73.1%

                                                                                                              \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                                                                                                            6. Taylor expanded in x around inf

                                                                                                              \[\leadsto \frac{\frac{1}{x}}{n} \]
                                                                                                            7. Step-by-step derivation
                                                                                                              1. Applied rewrites39.5%

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

                                                                                                              if 1.00000000000000003e40 < (/.f64 #s(literal 1 binary64) n)

                                                                                                              1. Initial program 41.4%

                                                                                                                \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                              2. Add Preprocessing
                                                                                                              3. Taylor expanded in x around 0

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

                                                                                                                  \[\leadsto \color{blue}{\left(x \cdot \left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) + 1\right)} - e^{\frac{\log x}{n}} \]
                                                                                                                2. associate--l+N/A

                                                                                                                  \[\leadsto \color{blue}{x \cdot \left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) + \left(1 - e^{\frac{\log x}{n}}\right)} \]
                                                                                                                3. *-commutativeN/A

                                                                                                                  \[\leadsto \color{blue}{\left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}\right) \cdot x} + \left(1 - e^{\frac{\log x}{n}}\right) \]
                                                                                                                4. lower-fma.f64N/A

                                                                                                                  \[\leadsto \color{blue}{\mathsf{fma}\left(x \cdot \left(\frac{1}{2} \cdot \frac{1}{{n}^{2}} - \frac{1}{2} \cdot \frac{1}{n}\right) + \frac{1}{n}, x, 1 - e^{\frac{\log x}{n}}\right)} \]
                                                                                                              5. Applied rewrites89.6%

                                                                                                                \[\leadsto \color{blue}{\mathsf{fma}\left(\mathsf{fma}\left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}, x, \frac{1}{n}\right), x, 1 - {x}^{\left(\frac{1}{n}\right)}\right)} \]
                                                                                                              6. Taylor expanded in n around -inf

                                                                                                                \[\leadsto -1 \cdot \color{blue}{\frac{x \cdot \left(\frac{1}{2} \cdot x - 1\right) - -1 \cdot \log x}{n}} \]
                                                                                                              7. Step-by-step derivation
                                                                                                                1. Applied rewrites12.9%

                                                                                                                  \[\leadsto \frac{\mathsf{fma}\left(\mathsf{fma}\left(0.5, x, -1\right), x, \log x\right)}{\color{blue}{-n}} \]
                                                                                                                2. Taylor expanded in x around inf

                                                                                                                  \[\leadsto \frac{{x}^{2} \cdot \left(\left(\frac{1}{2} + -1 \cdot \frac{\log \left(\frac{1}{x}\right)}{{x}^{2}}\right) - \frac{1}{x}\right)}{\mathsf{neg}\left(n\right)} \]
                                                                                                                3. Step-by-step derivation
                                                                                                                  1. Applied rewrites17.2%

                                                                                                                    \[\leadsto \frac{\left(\left(\frac{\log x}{x \cdot x} + 0.5\right) - \frac{1}{x}\right) \cdot \left(x \cdot x\right)}{-n} \]
                                                                                                                  2. Taylor expanded in n around 0

                                                                                                                    \[\leadsto \frac{1}{2} \cdot \color{blue}{\frac{{x}^{2}}{{n}^{2}}} \]
                                                                                                                  3. Step-by-step derivation
                                                                                                                    1. Applied rewrites59.1%

                                                                                                                      \[\leadsto \left(x \cdot \frac{x}{n \cdot n}\right) \cdot \color{blue}{0.5} \]
                                                                                                                  4. Recombined 3 regimes into one program.
                                                                                                                  5. Final simplification47.7%

                                                                                                                    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{+18}:\\ \;\;\;\;1 - 1\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{+40}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{x}{n \cdot n} \cdot x\right) \cdot 0.5\\ \end{array} \]
                                                                                                                  6. Add Preprocessing

                                                                                                                  Alternative 18: 46.8% accurate, 5.8× speedup?

                                                                                                                  \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{+18}:\\ \;\;\;\;1 - 1\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \end{array} \end{array} \]
                                                                                                                  (FPCore (x n)
                                                                                                                   :precision binary64
                                                                                                                   (if (<= (/ 1.0 n) -5e+18) (- 1.0 1.0) (/ (/ 1.0 x) n)))
                                                                                                                  double code(double x, double n) {
                                                                                                                  	double tmp;
                                                                                                                  	if ((1.0 / n) <= -5e+18) {
                                                                                                                  		tmp = 1.0 - 1.0;
                                                                                                                  	} else {
                                                                                                                  		tmp = (1.0 / x) / n;
                                                                                                                  	}
                                                                                                                  	return tmp;
                                                                                                                  }
                                                                                                                  
                                                                                                                  real(8) function code(x, n)
                                                                                                                      real(8), intent (in) :: x
                                                                                                                      real(8), intent (in) :: n
                                                                                                                      real(8) :: tmp
                                                                                                                      if ((1.0d0 / n) <= (-5d+18)) then
                                                                                                                          tmp = 1.0d0 - 1.0d0
                                                                                                                      else
                                                                                                                          tmp = (1.0d0 / x) / n
                                                                                                                      end if
                                                                                                                      code = tmp
                                                                                                                  end function
                                                                                                                  
                                                                                                                  public static double code(double x, double n) {
                                                                                                                  	double tmp;
                                                                                                                  	if ((1.0 / n) <= -5e+18) {
                                                                                                                  		tmp = 1.0 - 1.0;
                                                                                                                  	} else {
                                                                                                                  		tmp = (1.0 / x) / n;
                                                                                                                  	}
                                                                                                                  	return tmp;
                                                                                                                  }
                                                                                                                  
                                                                                                                  def code(x, n):
                                                                                                                  	tmp = 0
                                                                                                                  	if (1.0 / n) <= -5e+18:
                                                                                                                  		tmp = 1.0 - 1.0
                                                                                                                  	else:
                                                                                                                  		tmp = (1.0 / x) / n
                                                                                                                  	return tmp
                                                                                                                  
                                                                                                                  function code(x, n)
                                                                                                                  	tmp = 0.0
                                                                                                                  	if (Float64(1.0 / n) <= -5e+18)
                                                                                                                  		tmp = Float64(1.0 - 1.0);
                                                                                                                  	else
                                                                                                                  		tmp = Float64(Float64(1.0 / x) / n);
                                                                                                                  	end
                                                                                                                  	return tmp
                                                                                                                  end
                                                                                                                  
                                                                                                                  function tmp_2 = code(x, n)
                                                                                                                  	tmp = 0.0;
                                                                                                                  	if ((1.0 / n) <= -5e+18)
                                                                                                                  		tmp = 1.0 - 1.0;
                                                                                                                  	else
                                                                                                                  		tmp = (1.0 / x) / n;
                                                                                                                  	end
                                                                                                                  	tmp_2 = tmp;
                                                                                                                  end
                                                                                                                  
                                                                                                                  code[x_, n_] := If[LessEqual[N[(1.0 / n), $MachinePrecision], -5e+18], N[(1.0 - 1.0), $MachinePrecision], N[(N[(1.0 / x), $MachinePrecision] / n), $MachinePrecision]]
                                                                                                                  
                                                                                                                  \begin{array}{l}
                                                                                                                  
                                                                                                                  \\
                                                                                                                  \begin{array}{l}
                                                                                                                  \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{+18}:\\
                                                                                                                  \;\;\;\;1 - 1\\
                                                                                                                  
                                                                                                                  \mathbf{else}:\\
                                                                                                                  \;\;\;\;\frac{\frac{1}{x}}{n}\\
                                                                                                                  
                                                                                                                  
                                                                                                                  \end{array}
                                                                                                                  \end{array}
                                                                                                                  
                                                                                                                  Derivation
                                                                                                                  1. Split input into 2 regimes
                                                                                                                  2. if (/.f64 #s(literal 1 binary64) n) < -5e18

                                                                                                                    1. Initial program 100.0%

                                                                                                                      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                                    2. Add Preprocessing
                                                                                                                    3. Taylor expanded in x around 0

                                                                                                                      \[\leadsto \color{blue}{1} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                                    4. Step-by-step derivation
                                                                                                                      1. Applied rewrites44.6%

                                                                                                                        \[\leadsto \color{blue}{1} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                                      2. Taylor expanded in n around inf

                                                                                                                        \[\leadsto 1 - \color{blue}{1} \]
                                                                                                                      3. Step-by-step derivation
                                                                                                                        1. Applied rewrites57.8%

                                                                                                                          \[\leadsto 1 - \color{blue}{1} \]

                                                                                                                        if -5e18 < (/.f64 #s(literal 1 binary64) n)

                                                                                                                        1. Initial program 33.8%

                                                                                                                          \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                                        2. Add Preprocessing
                                                                                                                        3. Taylor expanded in n around inf

                                                                                                                          \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                                                                        4. Step-by-step derivation
                                                                                                                          1. lower-/.f64N/A

                                                                                                                            \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                                                                          2. lower--.f64N/A

                                                                                                                            \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                                                                                                                          3. lower-log1p.f64N/A

                                                                                                                            \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                                                                                                                          4. lower-log.f6463.3

                                                                                                                            \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                                                                                                                        5. Applied rewrites63.3%

                                                                                                                          \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                                                                                                                        6. Taylor expanded in x around inf

                                                                                                                          \[\leadsto \frac{\frac{1}{x}}{n} \]
                                                                                                                        7. Step-by-step derivation
                                                                                                                          1. Applied rewrites40.0%

                                                                                                                            \[\leadsto \frac{\frac{1}{x}}{n} \]
                                                                                                                        8. Recombined 2 regimes into one program.
                                                                                                                        9. Add Preprocessing

                                                                                                                        Alternative 19: 46.8% accurate, 5.8× speedup?

                                                                                                                        \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{+18}:\\ \;\;\;\;1 - 1\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \end{array} \end{array} \]
                                                                                                                        (FPCore (x n)
                                                                                                                         :precision binary64
                                                                                                                         (if (<= (/ 1.0 n) -5e+18) (- 1.0 1.0) (/ (/ 1.0 n) x)))
                                                                                                                        double code(double x, double n) {
                                                                                                                        	double tmp;
                                                                                                                        	if ((1.0 / n) <= -5e+18) {
                                                                                                                        		tmp = 1.0 - 1.0;
                                                                                                                        	} else {
                                                                                                                        		tmp = (1.0 / n) / x;
                                                                                                                        	}
                                                                                                                        	return tmp;
                                                                                                                        }
                                                                                                                        
                                                                                                                        real(8) function code(x, n)
                                                                                                                            real(8), intent (in) :: x
                                                                                                                            real(8), intent (in) :: n
                                                                                                                            real(8) :: tmp
                                                                                                                            if ((1.0d0 / n) <= (-5d+18)) then
                                                                                                                                tmp = 1.0d0 - 1.0d0
                                                                                                                            else
                                                                                                                                tmp = (1.0d0 / n) / x
                                                                                                                            end if
                                                                                                                            code = tmp
                                                                                                                        end function
                                                                                                                        
                                                                                                                        public static double code(double x, double n) {
                                                                                                                        	double tmp;
                                                                                                                        	if ((1.0 / n) <= -5e+18) {
                                                                                                                        		tmp = 1.0 - 1.0;
                                                                                                                        	} else {
                                                                                                                        		tmp = (1.0 / n) / x;
                                                                                                                        	}
                                                                                                                        	return tmp;
                                                                                                                        }
                                                                                                                        
                                                                                                                        def code(x, n):
                                                                                                                        	tmp = 0
                                                                                                                        	if (1.0 / n) <= -5e+18:
                                                                                                                        		tmp = 1.0 - 1.0
                                                                                                                        	else:
                                                                                                                        		tmp = (1.0 / n) / x
                                                                                                                        	return tmp
                                                                                                                        
                                                                                                                        function code(x, n)
                                                                                                                        	tmp = 0.0
                                                                                                                        	if (Float64(1.0 / n) <= -5e+18)
                                                                                                                        		tmp = Float64(1.0 - 1.0);
                                                                                                                        	else
                                                                                                                        		tmp = Float64(Float64(1.0 / n) / x);
                                                                                                                        	end
                                                                                                                        	return tmp
                                                                                                                        end
                                                                                                                        
                                                                                                                        function tmp_2 = code(x, n)
                                                                                                                        	tmp = 0.0;
                                                                                                                        	if ((1.0 / n) <= -5e+18)
                                                                                                                        		tmp = 1.0 - 1.0;
                                                                                                                        	else
                                                                                                                        		tmp = (1.0 / n) / x;
                                                                                                                        	end
                                                                                                                        	tmp_2 = tmp;
                                                                                                                        end
                                                                                                                        
                                                                                                                        code[x_, n_] := If[LessEqual[N[(1.0 / n), $MachinePrecision], -5e+18], N[(1.0 - 1.0), $MachinePrecision], N[(N[(1.0 / n), $MachinePrecision] / x), $MachinePrecision]]
                                                                                                                        
                                                                                                                        \begin{array}{l}
                                                                                                                        
                                                                                                                        \\
                                                                                                                        \begin{array}{l}
                                                                                                                        \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{+18}:\\
                                                                                                                        \;\;\;\;1 - 1\\
                                                                                                                        
                                                                                                                        \mathbf{else}:\\
                                                                                                                        \;\;\;\;\frac{\frac{1}{n}}{x}\\
                                                                                                                        
                                                                                                                        
                                                                                                                        \end{array}
                                                                                                                        \end{array}
                                                                                                                        
                                                                                                                        Derivation
                                                                                                                        1. Split input into 2 regimes
                                                                                                                        2. if (/.f64 #s(literal 1 binary64) n) < -5e18

                                                                                                                          1. Initial program 100.0%

                                                                                                                            \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                                          2. Add Preprocessing
                                                                                                                          3. Taylor expanded in x around 0

                                                                                                                            \[\leadsto \color{blue}{1} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                                          4. Step-by-step derivation
                                                                                                                            1. Applied rewrites44.6%

                                                                                                                              \[\leadsto \color{blue}{1} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                                            2. Taylor expanded in n around inf

                                                                                                                              \[\leadsto 1 - \color{blue}{1} \]
                                                                                                                            3. Step-by-step derivation
                                                                                                                              1. Applied rewrites57.8%

                                                                                                                                \[\leadsto 1 - \color{blue}{1} \]

                                                                                                                              if -5e18 < (/.f64 #s(literal 1 binary64) n)

                                                                                                                              1. Initial program 33.8%

                                                                                                                                \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                                              2. Add Preprocessing
                                                                                                                              3. Taylor expanded in n around inf

                                                                                                                                \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                                                                              4. Step-by-step derivation
                                                                                                                                1. lower-/.f64N/A

                                                                                                                                  \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                                                                                2. lower--.f64N/A

                                                                                                                                  \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                                                                                                                                3. lower-log1p.f64N/A

                                                                                                                                  \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                                                                                                                                4. lower-log.f6463.3

                                                                                                                                  \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                                                                                                                              5. Applied rewrites63.3%

                                                                                                                                \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                                                                                                                              6. Taylor expanded in x around inf

                                                                                                                                \[\leadsto \frac{1}{\color{blue}{n \cdot x}} \]
                                                                                                                              7. Step-by-step derivation
                                                                                                                                1. Applied rewrites38.3%

                                                                                                                                  \[\leadsto \frac{1}{\color{blue}{n \cdot x}} \]
                                                                                                                                2. Step-by-step derivation
                                                                                                                                  1. Applied rewrites40.0%

                                                                                                                                    \[\leadsto \frac{\frac{1}{n}}{x} \]
                                                                                                                                3. Recombined 2 regimes into one program.
                                                                                                                                4. Add Preprocessing

                                                                                                                                Alternative 20: 46.3% accurate, 6.8× speedup?

                                                                                                                                \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{+18}:\\ \;\;\;\;1 - 1\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{x \cdot n}\\ \end{array} \end{array} \]
                                                                                                                                (FPCore (x n)
                                                                                                                                 :precision binary64
                                                                                                                                 (if (<= (/ 1.0 n) -5e+18) (- 1.0 1.0) (/ 1.0 (* x n))))
                                                                                                                                double code(double x, double n) {
                                                                                                                                	double tmp;
                                                                                                                                	if ((1.0 / n) <= -5e+18) {
                                                                                                                                		tmp = 1.0 - 1.0;
                                                                                                                                	} else {
                                                                                                                                		tmp = 1.0 / (x * n);
                                                                                                                                	}
                                                                                                                                	return tmp;
                                                                                                                                }
                                                                                                                                
                                                                                                                                real(8) function code(x, n)
                                                                                                                                    real(8), intent (in) :: x
                                                                                                                                    real(8), intent (in) :: n
                                                                                                                                    real(8) :: tmp
                                                                                                                                    if ((1.0d0 / n) <= (-5d+18)) then
                                                                                                                                        tmp = 1.0d0 - 1.0d0
                                                                                                                                    else
                                                                                                                                        tmp = 1.0d0 / (x * n)
                                                                                                                                    end if
                                                                                                                                    code = tmp
                                                                                                                                end function
                                                                                                                                
                                                                                                                                public static double code(double x, double n) {
                                                                                                                                	double tmp;
                                                                                                                                	if ((1.0 / n) <= -5e+18) {
                                                                                                                                		tmp = 1.0 - 1.0;
                                                                                                                                	} else {
                                                                                                                                		tmp = 1.0 / (x * n);
                                                                                                                                	}
                                                                                                                                	return tmp;
                                                                                                                                }
                                                                                                                                
                                                                                                                                def code(x, n):
                                                                                                                                	tmp = 0
                                                                                                                                	if (1.0 / n) <= -5e+18:
                                                                                                                                		tmp = 1.0 - 1.0
                                                                                                                                	else:
                                                                                                                                		tmp = 1.0 / (x * n)
                                                                                                                                	return tmp
                                                                                                                                
                                                                                                                                function code(x, n)
                                                                                                                                	tmp = 0.0
                                                                                                                                	if (Float64(1.0 / n) <= -5e+18)
                                                                                                                                		tmp = Float64(1.0 - 1.0);
                                                                                                                                	else
                                                                                                                                		tmp = Float64(1.0 / Float64(x * n));
                                                                                                                                	end
                                                                                                                                	return tmp
                                                                                                                                end
                                                                                                                                
                                                                                                                                function tmp_2 = code(x, n)
                                                                                                                                	tmp = 0.0;
                                                                                                                                	if ((1.0 / n) <= -5e+18)
                                                                                                                                		tmp = 1.0 - 1.0;
                                                                                                                                	else
                                                                                                                                		tmp = 1.0 / (x * n);
                                                                                                                                	end
                                                                                                                                	tmp_2 = tmp;
                                                                                                                                end
                                                                                                                                
                                                                                                                                code[x_, n_] := If[LessEqual[N[(1.0 / n), $MachinePrecision], -5e+18], N[(1.0 - 1.0), $MachinePrecision], N[(1.0 / N[(x * n), $MachinePrecision]), $MachinePrecision]]
                                                                                                                                
                                                                                                                                \begin{array}{l}
                                                                                                                                
                                                                                                                                \\
                                                                                                                                \begin{array}{l}
                                                                                                                                \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{+18}:\\
                                                                                                                                \;\;\;\;1 - 1\\
                                                                                                                                
                                                                                                                                \mathbf{else}:\\
                                                                                                                                \;\;\;\;\frac{1}{x \cdot n}\\
                                                                                                                                
                                                                                                                                
                                                                                                                                \end{array}
                                                                                                                                \end{array}
                                                                                                                                
                                                                                                                                Derivation
                                                                                                                                1. Split input into 2 regimes
                                                                                                                                2. if (/.f64 #s(literal 1 binary64) n) < -5e18

                                                                                                                                  1. Initial program 100.0%

                                                                                                                                    \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                                                  2. Add Preprocessing
                                                                                                                                  3. Taylor expanded in x around 0

                                                                                                                                    \[\leadsto \color{blue}{1} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                                                  4. Step-by-step derivation
                                                                                                                                    1. Applied rewrites44.6%

                                                                                                                                      \[\leadsto \color{blue}{1} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                                                    2. Taylor expanded in n around inf

                                                                                                                                      \[\leadsto 1 - \color{blue}{1} \]
                                                                                                                                    3. Step-by-step derivation
                                                                                                                                      1. Applied rewrites57.8%

                                                                                                                                        \[\leadsto 1 - \color{blue}{1} \]

                                                                                                                                      if -5e18 < (/.f64 #s(literal 1 binary64) n)

                                                                                                                                      1. Initial program 33.8%

                                                                                                                                        \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                                                      2. Add Preprocessing
                                                                                                                                      3. Taylor expanded in n around inf

                                                                                                                                        \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                                                                                      4. Step-by-step derivation
                                                                                                                                        1. lower-/.f64N/A

                                                                                                                                          \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
                                                                                                                                        2. lower--.f64N/A

                                                                                                                                          \[\leadsto \frac{\color{blue}{\log \left(1 + x\right) - \log x}}{n} \]
                                                                                                                                        3. lower-log1p.f64N/A

                                                                                                                                          \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]
                                                                                                                                        4. lower-log.f6463.3

                                                                                                                                          \[\leadsto \frac{\mathsf{log1p}\left(x\right) - \color{blue}{\log x}}{n} \]
                                                                                                                                      5. Applied rewrites63.3%

                                                                                                                                        \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
                                                                                                                                      6. Taylor expanded in x around inf

                                                                                                                                        \[\leadsto \frac{1}{\color{blue}{n \cdot x}} \]
                                                                                                                                      7. Step-by-step derivation
                                                                                                                                        1. Applied rewrites38.3%

                                                                                                                                          \[\leadsto \frac{1}{\color{blue}{n \cdot x}} \]
                                                                                                                                      8. Recombined 2 regimes into one program.
                                                                                                                                      9. Final simplification44.7%

                                                                                                                                        \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{+18}:\\ \;\;\;\;1 - 1\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{x \cdot n}\\ \end{array} \]
                                                                                                                                      10. Add Preprocessing

                                                                                                                                      Alternative 21: 31.3% accurate, 57.8× speedup?

                                                                                                                                      \[\begin{array}{l} \\ 1 - 1 \end{array} \]
                                                                                                                                      (FPCore (x n) :precision binary64 (- 1.0 1.0))
                                                                                                                                      double code(double x, double n) {
                                                                                                                                      	return 1.0 - 1.0;
                                                                                                                                      }
                                                                                                                                      
                                                                                                                                      real(8) function code(x, n)
                                                                                                                                          real(8), intent (in) :: x
                                                                                                                                          real(8), intent (in) :: n
                                                                                                                                          code = 1.0d0 - 1.0d0
                                                                                                                                      end function
                                                                                                                                      
                                                                                                                                      public static double code(double x, double n) {
                                                                                                                                      	return 1.0 - 1.0;
                                                                                                                                      }
                                                                                                                                      
                                                                                                                                      def code(x, n):
                                                                                                                                      	return 1.0 - 1.0
                                                                                                                                      
                                                                                                                                      function code(x, n)
                                                                                                                                      	return Float64(1.0 - 1.0)
                                                                                                                                      end
                                                                                                                                      
                                                                                                                                      function tmp = code(x, n)
                                                                                                                                      	tmp = 1.0 - 1.0;
                                                                                                                                      end
                                                                                                                                      
                                                                                                                                      code[x_, n_] := N[(1.0 - 1.0), $MachinePrecision]
                                                                                                                                      
                                                                                                                                      \begin{array}{l}
                                                                                                                                      
                                                                                                                                      \\
                                                                                                                                      1 - 1
                                                                                                                                      \end{array}
                                                                                                                                      
                                                                                                                                      Derivation
                                                                                                                                      1. Initial program 55.5%

                                                                                                                                        \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                                                      2. Add Preprocessing
                                                                                                                                      3. Taylor expanded in x around 0

                                                                                                                                        \[\leadsto \color{blue}{1} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                                                      4. Step-by-step derivation
                                                                                                                                        1. Applied rewrites36.6%

                                                                                                                                          \[\leadsto \color{blue}{1} - {x}^{\left(\frac{1}{n}\right)} \]
                                                                                                                                        2. Taylor expanded in n around inf

                                                                                                                                          \[\leadsto 1 - \color{blue}{1} \]
                                                                                                                                        3. Step-by-step derivation
                                                                                                                                          1. Applied rewrites31.7%

                                                                                                                                            \[\leadsto 1 - \color{blue}{1} \]
                                                                                                                                          2. Add Preprocessing

                                                                                                                                          Reproduce

                                                                                                                                          ?
                                                                                                                                          herbie shell --seed 2024235 
                                                                                                                                          (FPCore (x n)
                                                                                                                                            :name "2nthrt (problem 3.4.6)"
                                                                                                                                            :precision binary64
                                                                                                                                            (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))