Average Error: 32.4 → 10.9
Time: 15.2s
Precision: binary64
\[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2.2779687884622282 \cdot 10^{-19}:\\ \;\;\;\;\frac{e^{-\frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2.8270861523315105 \cdot 10^{-23}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - e^{\frac{\log x}{n}}\\ \end{array} \]
{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}
\begin{array}{l}
\mathbf{if}\;\frac{1}{n} \leq -2.2779687884622282 \cdot 10^{-19}:\\
\;\;\;\;\frac{e^{-\frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}\\

\mathbf{elif}\;\frac{1}{n} \leq 2.8270861523315105 \cdot 10^{-23}:\\
\;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\

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


\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) -2.2779687884622282e-19)
   (/ (exp (- (/ (log (/ 1.0 x)) n))) (* n x))
   (if (<= (/ 1.0 n) 2.8270861523315105e-23)
     (/ (log (/ (+ 1.0 x) x)) n)
     (- (exp (/ (log1p x) n)) (exp (/ (log x) 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) <= -2.2779687884622282e-19) {
		tmp = exp(-(log(1.0 / x) / n)) / (n * x);
	} else if ((1.0 / n) <= 2.8270861523315105e-23) {
		tmp = log((1.0 + x) / x) / n;
	} else {
		tmp = exp(log1p(x) / n) - exp(log(x) / n);
	}
	return tmp;
}

Error

Bits error versus x

Bits error versus n

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Split input into 3 regimes
  2. if (/.f64 1 n) < -2.2779687884622282e-19

    1. Initial program 4.0

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
    2. Taylor expanded in x around inf 3.3

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

    if -2.2779687884622282e-19 < (/.f64 1 n) < 2.8270861523315105e-23

    1. Initial program 44.1

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
    2. Taylor expanded in n around inf 13.4

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
    3. Simplified13.4

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
    4. Applied log1p-udef_binary6413.4

      \[\leadsto \frac{\color{blue}{\log \left(1 + x\right)} - \log x}{n} \]
    5. Applied diff-log_binary6413.3

      \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]
    6. Simplified13.3

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

    if 2.8270861523315105e-23 < (/.f64 1 n)

    1. Initial program 12.4

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
    2. Taylor expanded in n around 0 12.4

      \[\leadsto \color{blue}{e^{\frac{\log \left(1 + x\right)}{n}} - e^{\frac{\log x}{n}}} \]
    3. Simplified9.6

      \[\leadsto \color{blue}{e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - e^{\frac{\log x}{n}}} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification10.9

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2.2779687884622282 \cdot 10^{-19}:\\ \;\;\;\;\frac{e^{-\frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2.8270861523315105 \cdot 10^{-23}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - e^{\frac{\log x}{n}}\\ \end{array} \]

Reproduce

herbie shell --seed 2022068 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))