2nthrt (problem 3.4.6)

Percentage Accurate: 37.3% → 90.4%

Time: 26.0s

Precision: binary64

Cost: 73288

• could not determine a ground truth

  1. x = 6.14389980537994e+99
  2. n = -5.0757214593988915e+297

Math

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

↓

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ t_1 := {\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - t_0\\ \mathbf{if}\;t_1 \leq -\infty:\\ \;\;\;\;1 - t_0\\ \mathbf{elif}\;t_1 \leq 0:\\ \;\;\;\;\mathsf{fma}\left(0.5, \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2}}{n \cdot n}, \frac{\mathsf{log1p}\left(x\right) - \log x}{n}\right) + \frac{{\log x}^{2}}{n \cdot n} \cdot -0.5\\ \mathbf{elif}\;t_1 \leq \infty:\\ \;\;\;\;e^{\frac{x}{n}} - t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{t_0}{x \cdot n}\\ \end{array} \]

FPCore

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

↓

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))) (t_1 (- (pow (+ x 1.0) (/ 1.0 n)) t_0)))
   (if (<= t_1 (- INFINITY))
     (- 1.0 t_0)
     (if (<= t_1 0.0)
       (+
        (fma 0.5 (/ (pow (log1p x) 2.0) (* n n)) (/ (- (log1p x) (log x)) n))
        (* (/ (pow (log x) 2.0) (* n n)) -0.5))
       (if (<= t_1 INFINITY) (- (exp (/ x n)) t_0) (/ t_0 (* x n)))))))

C

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

↓

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 tmp;
	if (t_1 <= -((double) INFINITY)) {
		tmp = 1.0 - t_0;
	} else if (t_1 <= 0.0) {
		tmp = fma(0.5, (pow(log1p(x), 2.0) / (n * n)), ((log1p(x) - log(x)) / n)) + ((pow(log(x), 2.0) / (n * n)) * -0.5);
	} else if (t_1 <= ((double) INFINITY)) {
		tmp = exp((x / n)) - t_0;
	} else {
		tmp = t_0 / (x * n);
	}
	return tmp;
}

Julia

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

↓

function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	t_1 = Float64((Float64(x + 1.0) ^ Float64(1.0 / n)) - t_0)
	tmp = 0.0
	if (t_1 <= Float64(-Inf))
		tmp = Float64(1.0 - t_0);
	elseif (t_1 <= 0.0)
		tmp = Float64(fma(0.5, Float64((log1p(x) ^ 2.0) / Float64(n * n)), Float64(Float64(log1p(x) - log(x)) / n)) + Float64(Float64((log(x) ^ 2.0) / Float64(n * n)) * -0.5));
	elseif (t_1 <= Inf)
		tmp = Float64(exp(Float64(x / n)) - t_0);
	else
		tmp = Float64(t_0 / Float64(x * n));
	end
	return tmp
end

Wolfram

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

↓

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]}, If[LessEqual[t$95$1, (-Infinity)], N[(1.0 - t$95$0), $MachinePrecision], If[LessEqual[t$95$1, 0.0], N[(N[(0.5 * N[(N[Power[N[Log[1 + x], $MachinePrecision], 2.0], $MachinePrecision] / N[(n * n), $MachinePrecision]), $MachinePrecision] + N[(N[(N[Log[1 + x], $MachinePrecision] - N[Log[x], $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision]), $MachinePrecision] + N[(N[(N[Power[N[Log[x], $MachinePrecision], 2.0], $MachinePrecision] / N[(n * n), $MachinePrecision]), $MachinePrecision] * -0.5), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, Infinity], N[(N[Exp[N[(x / n), $MachinePrecision]], $MachinePrecision] - t$95$0), $MachinePrecision], N[(t$95$0 / N[(x * n), $MachinePrecision]), $MachinePrecision]]]]]]

Derivation

1. Split input into 4 regimes

2. if (-.f64 (pow.f64 (+.f64 x 1) (/.f64 1 n)) (pow.f64 x (/.f64 1 n))) < -inf.0

   1. Initial program 100.0%

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

   2. Taylor expanded in x around 0 100.0%

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

   3. Simplified 100.0%

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

      [Start] 100.0%	\[ 1 - e^{\frac{\log x}{n}} \]
      *-rgt-identity [<=] 100.0%	\[ 1 - e^{\frac{\color{blue}{\log x \cdot 1}}{n}} \]
      associate-*r/ [<=] 100.0%	\[ 1 - e^{\color{blue}{\log x \cdot \frac{1}{n}}} \]
      unpow-1 [<=] 100.0%	\[ 1 - e^{\log x \cdot \color{blue}{{n}^{-1}}} \]
      exp-to-pow [=>] 100.0%	\[ 1 - \color{blue}{{x}^{\left({n}^{-1}\right)}} \]
      unpow-1 [=>] 100.0%	\[ 1 - {x}^{\color{blue}{\left(\frac{1}{n}\right)}} \]

   if -inf.0 < (-.f64 (pow.f64 (+.f64 x 1) (/.f64 1 n)) (pow.f64 x (/.f64 1 n))) < 0.0

   1. Initial program 28.3%

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

   2. Taylor expanded in n around inf 82.4%

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

   3. Simplified 82.4%

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

      [Start] 82.4%	\[ \left(0.5 \cdot \frac{{\log \left(1 + x\right)}^{2}}{{n}^{2}} + \frac{\log \left(1 + x\right)}{n}\right) - \left(\frac{\log x}{n} + 0.5 \cdot \frac{{\log x}^{2}}{{n}^{2}}\right) \]
      associate--r+ [=>] 74.5%	\[ \color{blue}{\left(\left(0.5 \cdot \frac{{\log \left(1 + x\right)}^{2}}{{n}^{2}} + \frac{\log \left(1 + x\right)}{n}\right) - \frac{\log x}{n}\right) - 0.5 \cdot \frac{{\log x}^{2}}{{n}^{2}}} \]
      sub-neg [=>] 74.5%	\[ \color{blue}{\left(\left(0.5 \cdot \frac{{\log \left(1 + x\right)}^{2}}{{n}^{2}} + \frac{\log \left(1 + x\right)}{n}\right) - \frac{\log x}{n}\right) + \left(-0.5 \cdot \frac{{\log x}^{2}}{{n}^{2}}\right)} \]

   if 0.0 < (-.f64 (pow.f64 (+.f64 x 1) (/.f64 1 n)) (pow.f64 x (/.f64 1 n))) < +inf.0

   1. Initial program 56.1%

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

   2. Taylor expanded in n around 0 56.1%

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

   3. Simplified 99.9%

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

      [Start] 56.1%	\[ e^{\frac{\log \left(1 + x\right)}{n}} - e^{\frac{\log x}{n}} \]
      log1p-def [=>] 99.9%	\[ e^{\frac{\color{blue}{\mathsf{log1p}\left(x\right)}}{n}} - e^{\frac{\log x}{n}} \]
      *-rgt-identity [<=] 99.9%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - e^{\frac{\color{blue}{\log x \cdot 1}}{n}} \]
      associate-*r/ [<=] 99.9%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - e^{\color{blue}{\log x \cdot \frac{1}{n}}} \]
      unpow-1 [<=] 99.9%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - e^{\log x \cdot \color{blue}{{n}^{-1}}} \]
      exp-to-pow [=>] 99.9%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - \color{blue}{{x}^{\left({n}^{-1}\right)}} \]
      /-rgt-identity [<=] 99.9%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\color{blue}{\left(\frac{{n}^{-1}}{1}\right)}} \]
      metadata-eval [<=] 99.9%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{{n}^{-1}}{\color{blue}{\frac{2}{2}}}\right)} \]
      associate-/l* [<=] 99.9%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\color{blue}{\left(\frac{{n}^{-1} \cdot 2}{2}\right)}} \]
      *-commutative [<=] 99.9%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{\color{blue}{2 \cdot {n}^{-1}}}{2}\right)} \]
      *-commutative [=>] 99.9%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{\color{blue}{{n}^{-1} \cdot 2}}{2}\right)} \]
      associate-/l* [=>] 99.9%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\color{blue}{\left(\frac{{n}^{-1}}{\frac{2}{2}}\right)}} \]
      metadata-eval [=>] 99.9%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{{n}^{-1}}{\color{blue}{1}}\right)} \]
      /-rgt-identity [=>] 99.9%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\color{blue}{\left({n}^{-1}\right)}} \]
      unpow-1 [=>] 99.9%	\[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\color{blue}{\left(\frac{1}{n}\right)}} \]

   4. Taylor expanded in x around 0 99.9%

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

   if +inf.0 < (-.f64 (pow.f64 (+.f64 x 1) (/.f64 1 n)) (pow.f64 x (/.f64 1 n)))

   1. Initial program 0.0%

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

   2. Taylor expanded in x around inf 100.0%

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

   3. Simplified 100.0%

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

      [Start] 100.0%	\[ \frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x} \]
      log-rec [=>] 100.0%	\[ \frac{e^{-1 \cdot \frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]
      mul-1-neg [<=] 100.0%	\[ \frac{e^{-1 \cdot \frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]
      mul-1-neg [=>] 100.0%	\[ \frac{e^{-1 \cdot \frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]
      distribute-frac-neg [=>] 100.0%	\[ \frac{e^{-1 \cdot \color{blue}{\left(-\frac{\log x}{n}\right)}}}{n \cdot x} \]
      neg-mul-1 [<=] 100.0%	\[ \frac{e^{\color{blue}{-\left(-\frac{\log x}{n}\right)}}}{n \cdot x} \]
      remove-double-neg [=>] 100.0%	\[ \frac{e^{\color{blue}{\frac{\log x}{n}}}}{n \cdot x} \]
      *-rgt-identity [<=] 100.0%	\[ \frac{e^{\frac{\color{blue}{\log x \cdot 1}}{n}}}{n \cdot x} \]
      associate-*r/ [<=] 100.0%	\[ \frac{e^{\color{blue}{\log x \cdot \frac{1}{n}}}}{n \cdot x} \]
      unpow-1 [<=] 100.0%	\[ \frac{e^{\log x \cdot \color{blue}{{n}^{-1}}}}{n \cdot x} \]
      exp-to-pow [=>] 100.0%	\[ \frac{\color{blue}{{x}^{\left({n}^{-1}\right)}}}{n \cdot x} \]
      unpow-1 [=>] 100.0%	\[ \frac{{x}^{\color{blue}{\left(\frac{1}{n}\right)}}}{n \cdot x} \]
      *-commutative [=>] 100.0%	\[ \frac{{x}^{\left(\frac{1}{n}\right)}}{\color{blue}{x \cdot n}} \]

3. Recombined 4 regimes into one program.

4. Final simplification 90.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \leq -\infty:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \leq 0:\\ \;\;\;\;\mathsf{fma}\left(0.5, \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2}}{n \cdot n}, \frac{\mathsf{log1p}\left(x\right) - \log x}{n}\right) + \frac{{\log x}^{2}}{n \cdot n} \cdot -0.5\\ \mathbf{elif}\;{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \leq \infty:\\ \;\;\;\;e^{\frac{x}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	90.4%
Cost	73288

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ t_1 := {\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - t_0\\ \mathbf{if}\;t_1 \leq -\infty:\\ \;\;\;\;1 - t_0\\ \mathbf{elif}\;t_1 \leq 0:\\ \;\;\;\;\mathsf{fma}\left(0.5, \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2}}{n \cdot n}, \frac{\mathsf{log1p}\left(x\right) - \log x}{n}\right) + \frac{{\log x}^{2}}{n \cdot n} \cdot -0.5\\ \mathbf{elif}\;t_1 \leq \infty:\\ \;\;\;\;e^{\frac{x}{n}} - t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{t_0}{x \cdot n}\\ \end{array} \]

Alternative 2
Accuracy	90.3%
Cost	54028

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ t_1 := {\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - t_0\\ \mathbf{if}\;t_1 \leq -\infty:\\ \;\;\;\;1 - t_0\\ \mathbf{elif}\;t_1 \leq 0:\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{elif}\;t_1 \leq \infty:\\ \;\;\;\;e^{\frac{x}{n}} - t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{t_0}{x \cdot n}\\ \end{array} \]

Alternative 3
Accuracy	77.5%
Cost	8528

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ t_1 := \frac{1}{x \cdot n}\\ \mathbf{if}\;\frac{1}{n} \leq -10:\\ \;\;\;\;\frac{t_0}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-19}:\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{+101}:\\ \;\;\;\;\frac{1}{n} \cdot \frac{t_0}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{+148}:\\ \;\;\;\;\sqrt[3]{t_1 \cdot \left(t_1 \cdot t_1\right)}\\ \mathbf{else}:\\ \;\;\;\;0.5 \cdot \frac{x \cdot x}{n \cdot n}\\ \end{array} \]

Alternative 4
Accuracy	78.4%
Cost	7820

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -10:\\ \;\;\;\;\frac{t_0}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-19}:\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{+65}:\\ \;\;\;\;\frac{1}{n} \cdot \frac{t_0}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{x}{n} + \left(x \cdot x\right) \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)\\ \end{array} \]

Alternative 5
Accuracy	78.4%
Cost	7820

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -10:\\ \;\;\;\;\frac{t_0}{x \cdot n}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-19}:\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{+65}:\\ \;\;\;\;\frac{1}{\frac{x \cdot n}{t_0}}\\ \mathbf{else}:\\ \;\;\;\;\frac{x}{n} + \left(x \cdot x\right) \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)\\ \end{array} \]

Alternative 6
Accuracy	58.8%
Cost	7696

\[\begin{array}{l} t_0 := \frac{1}{x \cdot n}\\ t_1 := -\frac{\log x}{n}\\ \mathbf{if}\;\frac{1}{n} \leq -10:\\ \;\;\;\;t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-298}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-80}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-48}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 0.05:\\ \;\;\;\;\frac{\frac{1}{x} - \frac{0.5}{x \cdot x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{x}{n} + \left(x \cdot x\right) \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)\\ \end{array} \]

Alternative 7
Accuracy	78.4%
Cost	7692

\[\begin{array}{l} t_0 := \frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}\\ \mathbf{if}\;\frac{1}{n} \leq -10:\\ \;\;\;\;t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-19}:\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{+65}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{x}{n} + \left(x \cdot x\right) \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)\\ \end{array} \]

Alternative 8
Accuracy	75.9%
Cost	7368

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -10:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;\frac{1}{n} \leq 20000000:\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{x}{n} + \left(x \cdot x\right) \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)\\ \end{array} \]

Alternative 9
Accuracy	67.3%
Cost	7048

\[\begin{array}{l} \mathbf{if}\;n \leq -24000000000:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;n \leq 2 \cdot 10^{-309}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;n \leq 0.73:\\ \;\;\;\;\frac{x}{n} + \left(x \cdot x\right) \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{x \cdot n}\\ \end{array} \]

Alternative 10
Accuracy	67.3%
Cost	7048

\[\begin{array}{l} \mathbf{if}\;n \leq -21000000000:\\ \;\;\;\;\frac{1}{n} \cdot \left(x - \log x\right)\\ \mathbf{elif}\;n \leq 2 \cdot 10^{-309}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;n \leq 0.73:\\ \;\;\;\;\frac{x}{n} + \left(x \cdot x\right) \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{x \cdot n}\\ \end{array} \]

Alternative 11
Accuracy	59.3%
Cost	6852

\[\begin{array}{l} \mathbf{if}\;n \leq -2:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;n \leq 2 \cdot 10^{-309} \lor \neg \left(n \leq 0.8\right):\\ \;\;\;\;\frac{1}{x \cdot n}\\ \mathbf{else}:\\ \;\;\;\;\frac{x}{n} + \left(x \cdot x\right) \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)\\ \end{array} \]

Alternative 12
Accuracy	45.2%
Cost	1348

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq 0.05:\\ \;\;\;\;\frac{1}{x \cdot n}\\ \mathbf{else}:\\ \;\;\;\;\frac{x}{n} + \left(x \cdot x\right) \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)\\ \end{array} \]

Alternative 13
Accuracy	44.8%
Cost	836

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq 20000000:\\ \;\;\;\;\frac{1}{x \cdot n}\\ \mathbf{else}:\\ \;\;\;\;0.5 \cdot \frac{x \cdot x}{n \cdot n}\\ \end{array} \]

Alternative 14
Accuracy	35.4%
Cost	585

\[\begin{array}{l} \mathbf{if}\;n \leq 2.5 \cdot 10^{-281} \lor \neg \left(n \leq 1\right):\\ \;\;\;\;\frac{1}{x \cdot n}\\ \mathbf{else}:\\ \;\;\;\;\frac{x}{n}\\ \end{array} \]

Alternative 15
Accuracy	20.1%
Cost	324

\[\begin{array}{l} \mathbf{if}\;n \leq 7.2 \cdot 10^{+34}:\\ \;\;\;\;\frac{x}{n}\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \]

Alternative 16
Accuracy	11.4%
Cost	64

\[0 \]

Reproduce

herbie shell --seed 2023272 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))