2nthrt (problem 3.4.6)

?

Percentage Accurate: 48.8% → 82.9%
Time: 56.8s
Precision: binary64
Cost: 79368

?

\[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-30}:\\ \;\;\;\;\frac{\frac{e^{\frac{\log x}{n}}}{n}}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-7}:\\ \;\;\;\;\left(\mathsf{fma}\left(0.5, \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2}}{n \cdot n}, \frac{\mathsf{log1p}\left(x\right) - \log x}{n}\right) + \frac{0.16666666666666666 \cdot \left({\left(\mathsf{log1p}\left(x\right)\right)}^{3} - {\log x}^{3}\right)}{{n}^{3}}\right) + \frac{{\log x}^{2}}{n \cdot n} \cdot -0.5\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
(FPCore (x n)
 :precision binary64
 (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))
(FPCore (x n)
 :precision binary64
 (if (<= (/ 1.0 n) -1e-30)
   (/ (/ (exp (/ (log x) n)) n) x)
   (if (<= (/ 1.0 n) 1e-7)
     (+
      (+
       (fma 0.5 (/ (pow (log1p x) 2.0) (* n n)) (/ (- (log1p x) (log x)) n))
       (/
        (* 0.16666666666666666 (- (pow (log1p x) 3.0) (pow (log x) 3.0)))
        (pow n 3.0)))
      (* (/ (pow (log x) 2.0) (* n n)) -0.5))
     (- (exp (/ (log1p x) 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));
}
double code(double x, double n) {
	double tmp;
	if ((1.0 / n) <= -1e-30) {
		tmp = (exp((log(x) / n)) / n) / x;
	} else if ((1.0 / n) <= 1e-7) {
		tmp = (fma(0.5, (pow(log1p(x), 2.0) / (n * n)), ((log1p(x) - log(x)) / n)) + ((0.16666666666666666 * (pow(log1p(x), 3.0) - pow(log(x), 3.0))) / pow(n, 3.0))) + ((pow(log(x), 2.0) / (n * n)) * -0.5);
	} else {
		tmp = exp((log1p(x) / n)) - pow(x, (1.0 / n));
	}
	return tmp;
}
function code(x, n)
	return Float64((Float64(x + 1.0) ^ Float64(1.0 / n)) - (x ^ Float64(1.0 / n)))
end
function code(x, n)
	tmp = 0.0
	if (Float64(1.0 / n) <= -1e-30)
		tmp = Float64(Float64(exp(Float64(log(x) / n)) / n) / x);
	elseif (Float64(1.0 / n) <= 1e-7)
		tmp = Float64(Float64(fma(0.5, Float64((log1p(x) ^ 2.0) / Float64(n * n)), Float64(Float64(log1p(x) - log(x)) / n)) + Float64(Float64(0.16666666666666666 * Float64((log1p(x) ^ 3.0) - (log(x) ^ 3.0))) / (n ^ 3.0))) + Float64(Float64((log(x) ^ 2.0) / Float64(n * n)) * -0.5));
	else
		tmp = Float64(exp(Float64(log1p(x) / n)) - (x ^ Float64(1.0 / n)));
	end
	return tmp
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]
code[x_, n_] := If[LessEqual[N[(1.0 / n), $MachinePrecision], -1e-30], N[(N[(N[Exp[N[(N[Log[x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision] / x), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 1e-7], N[(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[(0.16666666666666666 * N[(N[Power[N[Log[1 + x], $MachinePrecision], 3.0], $MachinePrecision] - N[Power[N[Log[x], $MachinePrecision], 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[Power[n, 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(N[(N[Power[N[Log[x], $MachinePrecision], 2.0], $MachinePrecision] / N[(n * n), $MachinePrecision]), $MachinePrecision] * -0.5), $MachinePrecision]), $MachinePrecision], N[(N[Exp[N[(N[Log[1 + x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision] - N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]
{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}
\begin{array}{l}
\mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-30}:\\
\;\;\;\;\frac{\frac{e^{\frac{\log x}{n}}}{n}}{x}\\

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

\mathbf{else}:\\
\;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\\


\end{array}

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.

Herbie found 15 alternatives:

AlternativeAccuracySpeedup

Accuracy vs Speed

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.

Bogosity?

Bogosity

Derivation?

  1. Split input into 3 regimes
  2. if (/.f64 1 n) < -1e-30

    1. Initial program 87.4%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
    2. Taylor expanded in x around inf 98.1%

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
    3. Simplified98.1%

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

      [Start]98.1%

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

      mul-1-neg [=>]98.1%

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

      log-rec [=>]98.1%

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

      *-commutative [=>]98.1%

      \[ \frac{e^{-\frac{-\log x}{n}}}{\color{blue}{x \cdot n}} \]
    4. Taylor expanded in x around 0 98.1%

      \[\leadsto \color{blue}{\frac{e^{--1 \cdot \frac{\log x}{n}}}{n \cdot x}} \]
    5. Simplified98.2%

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

      [Start]98.1%

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

      associate-/r* [=>]98.2%

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

      neg-mul-1 [<=]98.2%

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

      remove-double-neg [=>]98.2%

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

    if -1e-30 < (/.f64 1 n) < 9.9999999999999995e-8

    1. Initial program 32.1%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
    2. Taylor expanded in n around -inf 86.9%

      \[\leadsto \color{blue}{\left(0.5 \cdot \frac{{\log \left(1 + x\right)}^{2}}{{n}^{2}} + \left(-1 \cdot \frac{-0.16666666666666666 \cdot {\log \left(1 + x\right)}^{3} - -0.16666666666666666 \cdot {\log x}^{3}}{{n}^{3}} + -1 \cdot \frac{-1 \cdot \log \left(1 + x\right) - -1 \cdot \log x}{n}\right)\right) - 0.5 \cdot \frac{{\log x}^{2}}{{n}^{2}}} \]
    3. Simplified86.9%

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

      [Start]86.9%

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

      sub-neg [=>]86.9%

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

    if 9.9999999999999995e-8 < (/.f64 1 n)

    1. Initial program 90.4%

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
    2. Taylor expanded in n around 0 90.4%

      \[\leadsto \color{blue}{e^{\frac{\log \left(1 + x\right)}{n}} - e^{\frac{\log x}{n}}} \]
    3. Simplified100.0%

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

      [Start]90.4%

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

      log1p-def [=>]100.0%

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

      *-rgt-identity [<=]100.0%

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

      associate-*r/ [<=]100.0%

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

      unpow-1 [<=]100.0%

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

      exp-to-pow [=>]100.0%

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

      /-rgt-identity [<=]100.0%

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

      metadata-eval [<=]100.0%

      \[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{{n}^{-1}}{\color{blue}{\frac{2}{2}}}\right)} \]

      associate-/l* [<=]100.0%

      \[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\color{blue}{\left(\frac{{n}^{-1} \cdot 2}{2}\right)}} \]

      *-commutative [<=]100.0%

      \[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{\color{blue}{2 \cdot {n}^{-1}}}{2}\right)} \]

      *-commutative [=>]100.0%

      \[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{\color{blue}{{n}^{-1} \cdot 2}}{2}\right)} \]

      associate-/l* [=>]100.0%

      \[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\color{blue}{\left(\frac{{n}^{-1}}{\frac{2}{2}}\right)}} \]

      metadata-eval [=>]100.0%

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

      /-rgt-identity [=>]100.0%

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

      unpow-1 [=>]100.0%

      \[ e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\color{blue}{\left(\frac{1}{n}\right)}} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification91.2%

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-30}:\\ \;\;\;\;\frac{\frac{e^{\frac{\log x}{n}}}{n}}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-7}:\\ \;\;\;\;\left(\mathsf{fma}\left(0.5, \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2}}{n \cdot n}, \frac{\mathsf{log1p}\left(x\right) - \log x}{n}\right) + \frac{0.16666666666666666 \cdot \left({\left(\mathsf{log1p}\left(x\right)\right)}^{3} - {\log x}^{3}\right)}{{n}^{3}}\right) + \frac{{\log x}^{2}}{n \cdot n} \cdot -0.5\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy82.9%
Cost79368
\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-30}:\\ \;\;\;\;\frac{\frac{e^{\frac{\log x}{n}}}{n}}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-7}:\\ \;\;\;\;\left(\mathsf{fma}\left(0.5, \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2}}{n \cdot n}, \frac{\mathsf{log1p}\left(x\right) - \log x}{n}\right) + \frac{0.16666666666666666 \cdot \left({\left(\mathsf{log1p}\left(x\right)\right)}^{3} - {\log x}^{3}\right)}{{n}^{3}}\right) + \frac{{\log x}^{2}}{n \cdot n} \cdot -0.5\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
Alternative 2
Accuracy82.9%
Cost46664
\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-30}:\\ \;\;\;\;\frac{\frac{e^{\frac{\log x}{n}}}{n}}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-7}:\\ \;\;\;\;\mathsf{fma}\left(0.5, \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2}}{n \cdot n}, \frac{\mathsf{log1p}\left(x\right) - \log x}{n}\right) + \frac{{\log x}^{2}}{n \cdot n} \cdot -0.5\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
Alternative 3
Accuracy82.8%
Cost20232
\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-30}:\\ \;\;\;\;\frac{\frac{e^{\frac{\log x}{n}}}{n}}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-12}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(x\right) - \log x}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
Alternative 4
Accuracy82.0%
Cost13640
\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-30}:\\ \;\;\;\;\frac{\frac{t_0}{n}}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-12}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(x\right) - \log x}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - t_0\\ \end{array} \]
Alternative 5
Accuracy82.0%
Cost13640
\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-30}:\\ \;\;\;\;\frac{\frac{e^{\frac{\log x}{n}}}{n}}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-12}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(x\right) - \log x}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
Alternative 6
Accuracy80.9%
Cost7628
\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -0.1:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq -1 \cdot 10^{-30}:\\ \;\;\;\;\frac{1}{n} \cdot \frac{1}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-12}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]
Alternative 7
Accuracy82.1%
Cost7560
\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-30}:\\ \;\;\;\;\frac{\frac{t_0}{n}}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-12}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - t_0\\ \end{array} \]
Alternative 8
Accuracy81.9%
Cost7368
\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-30}:\\ \;\;\;\;\frac{t_0}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-12}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;1 - t_0\\ \end{array} \]
Alternative 9
Accuracy81.9%
Cost7368
\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-30}:\\ \;\;\;\;\frac{\frac{t_0}{n}}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{-12}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;1 - t_0\\ \end{array} \]
Alternative 10
Accuracy72.0%
Cost6852
\[\begin{array}{l} \mathbf{if}\;x \leq 1.35:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;x \leq 2 \cdot 10^{+70}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;x \leq 3.4 \cdot 10^{+89}:\\ \;\;\;\;\frac{1}{n \cdot x}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]
Alternative 11
Accuracy71.8%
Cost6788
\[\begin{array}{l} \mathbf{if}\;x \leq 1:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 2 \cdot 10^{+70}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;x \leq 3.4 \cdot 10^{+89}:\\ \;\;\;\;\frac{1}{n \cdot x}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]
Alternative 12
Accuracy54.7%
Cost580
\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -0.1:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \end{array} \]
Alternative 13
Accuracy36.7%
Cost320
\[\frac{1}{n \cdot x} \]
Alternative 14
Accuracy37.2%
Cost320
\[\frac{\frac{1}{x}}{n} \]
Alternative 15
Accuracy4.7%
Cost192
\[\frac{x}{n} \]

Reproduce?

herbie shell --seed 2023263 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))