Average Error: 33.1 → 7.2
Time: 33.7s
Precision: binary64
\[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
↓
\[\begin{array}{l} \mathbf{if}\;x \leq 7.4052989362426835:\\ \;\;\;\;\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\log \left(x + 1\right) - \log x}{n}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}\\ \end{array} \]
{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}
↓
\begin{array}{l}
\mathbf{if}\;x \leq 7.4052989362426835:\\
\;\;\;\;\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\log \left(x + 1\right) - \log x}{n}\right)\right)\\

\mathbf{else}:\\
\;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot 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 (<= x 7.4052989362426835)
   (expm1 (log1p (/ (- (log (+ x 1.0)) (log x)) n)))
   (/ (pow x (/ 1.0 n)) (* 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 (x <= 7.4052989362426835) {
		tmp = expm1(log1p((log(x + 1.0) - log(x)) / n));
	} else {
		tmp = pow(x, (1.0 / n)) / (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 2 regimes

  2. if x < 7.4052989362426835

    1. Initial program 47.0

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

    2. Taylor expanded in n around inf 13.9

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

    3. Applied expm1-log1p-u_binary6413.9

      \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\log \left(1 + x\right) - \log x}{n}\right)\right)} \]

    if 7.4052989362426835 < x

    1. Initial program 21.4

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

    2. Taylor expanded in x around inf 1.6

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

    3. Simplified1.6

      \[\leadsto \color{blue}{\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}} \]
  3. Recombined 2 regimes into one program.

  4. Final simplification7.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 7.4052989362426835:\\ \;\;\;\;\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\log \left(x + 1\right) - \log x}{n}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot n}\\ \end{array} \]

Reproduce

herbie shell --seed 2021215 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))