Average Error: 32.3 → 7.2
Time: 18.1s
Precision: binary64
\[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
↓
\[\begin{array}{l} \mathbf{if}\;x \leq 318382.55505162105:\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{-\frac{\log \left(\frac{1}{x}\right)}{n}}}{x \cdot n}\\ \end{array} \]
{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}
↓
\begin{array}{l}
\mathbf{if}\;x \leq 318382.55505162105:\\
\;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\

\mathbf{else}:\\
\;\;\;\;\frac{e^{-\frac{\log \left(\frac{1}{x}\right)}{n}}}{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 318382.55505162105)
   (/ (log (/ (+ x 1.0) x)) n)
   (/ (exp (- (/ (log (/ 1.0 x)) 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 <= 318382.55505162105) {
		tmp = log((x + 1.0) / x) / n;
	} else {
		tmp = exp(-(log(1.0 / x) / 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 < 318382.555051621

    1. Initial program 46.8

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

    2. Taylor expanded in n around inf 14.1

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

    3. Simplified14.1

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

    4. Applied log1p-udef_binary6414.1

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

    5. Applied diff-log_binary6414.1

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

    6. Simplified14.1

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

    if 318382.555051621 < x

    1. Initial program 19.7

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

    2. Taylor expanded in x around inf 1.3

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

  4. Final simplification7.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 318382.55505162105:\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{-\frac{\log \left(\frac{1}{x}\right)}{n}}}{x \cdot n}\\ \end{array} \]

Reproduce

herbie shell --seed 2021333 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))