Average Error: 33.0 → 13.1
Time: 14.4s
Precision: binary64
\[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}\]
↓
\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -1.7964466032532866 \cdot 10^{-06}:\\ \;\;\;\;\sqrt[3]{{\left({\left(1 + x\right)}^{\left(\frac{1}{n}\right)}\right)}^{3}} - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;\frac{1}{n} \leq -4.3467324602633687 \cdot 10^{-88}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x} + \frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot x} \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)\\ \mathbf{elif}\;\frac{1}{n} \leq 5.078480559969997 \cdot 10^{-117}:\\ \;\;\;\;\left(\frac{\log \left(1 + x\right)}{n} + 0.5 \cdot \frac{{\log \left(1 + x\right)}^{2}}{n \cdot n}\right) - \left(\frac{\log x}{n} + 0.5 \cdot \frac{{\log x}^{2}}{n \cdot n}\right)\\ \mathbf{elif}\;\frac{1}{n} \leq 4.857132074199983:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x} + \frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot x} \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \end{array}\]
{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}
↓
\begin{array}{l}
\mathbf{if}\;\frac{1}{n} \leq -1.7964466032532866 \cdot 10^{-06}:\\
\;\;\;\;\sqrt[3]{{\left({\left(1 + x\right)}^{\left(\frac{1}{n}\right)}\right)}^{3}} - {x}^{\left(\frac{1}{n}\right)}\\

\mathbf{elif}\;\frac{1}{n} \leq -4.3467324602633687 \cdot 10^{-88}:\\
\;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x} + \frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot x} \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)\\

\mathbf{elif}\;\frac{1}{n} \leq 5.078480559969997 \cdot 10^{-117}:\\
\;\;\;\;\left(\frac{\log \left(1 + x\right)}{n} + 0.5 \cdot \frac{{\log \left(1 + x\right)}^{2}}{n \cdot n}\right) - \left(\frac{\log x}{n} + 0.5 \cdot \frac{{\log x}^{2}}{n \cdot n}\right)\\

\mathbf{elif}\;\frac{1}{n} \leq 4.857132074199983:\\
\;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x} + \frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot x} \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)\\

\mathbf{else}:\\
\;\;\;\;\left(1 + \frac{x}{n}\right) - {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) -1.7964466032532866e-06)
   (- (cbrt (pow (pow (+ 1.0 x) (/ 1.0 n)) 3.0)) (pow x (/ 1.0 n)))
   (if (<= (/ 1.0 n) -4.3467324602633687e-88)
     (+
      (/ (pow x (/ 1.0 n)) (* n x))
      (* (/ (pow x (/ 1.0 n)) (* x x)) (- (/ 0.5 (* n n)) (/ 0.5 n))))
     (if (<= (/ 1.0 n) 5.078480559969997e-117)
       (-
        (+ (/ (log (+ 1.0 x)) n) (* 0.5 (/ (pow (log (+ 1.0 x)) 2.0) (* n n))))
        (+ (/ (log x) n) (* 0.5 (/ (pow (log x) 2.0) (* n n)))))
       (if (<= (/ 1.0 n) 4.857132074199983)
         (+
          (/ (pow x (/ 1.0 n)) (* n x))
          (* (/ (pow x (/ 1.0 n)) (* x x)) (- (/ 0.5 (* n n)) (/ 0.5 n))))
         (- (+ 1.0 (/ 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) <= -1.7964466032532866e-06) {
		tmp = cbrt(pow(pow((1.0 + x), (1.0 / n)), 3.0)) - pow(x, (1.0 / n));
	} else if ((1.0 / n) <= -4.3467324602633687e-88) {
		tmp = (pow(x, (1.0 / n)) / (n * x)) + ((pow(x, (1.0 / n)) / (x * x)) * ((0.5 / (n * n)) - (0.5 / n)));
	} else if ((1.0 / n) <= 5.078480559969997e-117) {
		tmp = ((log(1.0 + x) / n) + (0.5 * (pow(log(1.0 + x), 2.0) / (n * n)))) - ((log(x) / n) + (0.5 * (pow(log(x), 2.0) / (n * n))));
	} else if ((1.0 / n) <= 4.857132074199983) {
		tmp = (pow(x, (1.0 / n)) / (n * x)) + ((pow(x, (1.0 / n)) / (x * x)) * ((0.5 / (n * n)) - (0.5 / n)));
	} else {
		tmp = (1.0 + (x / n)) - pow(x, (1.0 / 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 4 regimes

  2. if (/.f64 1 n) < -1.7964466032532866e-6

    1. Initial program 1.1

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}\]
    2. Using strategy rm

    3. Applied add-cbrt-cube_binary64_1141.2

      \[\leadsto \color{blue}{\sqrt[3]{\left({\left(x + 1\right)}^{\left(\frac{1}{n}\right)} \cdot {\left(x + 1\right)}^{\left(\frac{1}{n}\right)}\right) \cdot {\left(x + 1\right)}^{\left(\frac{1}{n}\right)}}} - {x}^{\left(\frac{1}{n}\right)}\]

    4. Simplified1.2

      \[\leadsto \sqrt[3]{\color{blue}{{\left({\left(x + 1\right)}^{\left(\frac{1}{n}\right)}\right)}^{3}}} - {x}^{\left(\frac{1}{n}\right)}\]

    if -1.7964466032532866e-6 < (/.f64 1 n) < -4.34673246026336867e-88 or 5.0784805599699968e-117 < (/.f64 1 n) < 4.8571320741999831

    1. Initial program 53.0

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

    2. Taylor expanded around inf 31.4

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

    3. Simplified31.4

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

    if -4.34673246026336867e-88 < (/.f64 1 n) < 5.0784805599699968e-117

    1. Initial program 41.4

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

    2. Taylor expanded around inf 10.8

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

    3. Simplified10.8

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

    if 4.8571320741999831 < (/.f64 1 n)

    1. Initial program 3.0

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

    2. Taylor expanded around 0 1.9

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

    3. Simplified1.9

      \[\leadsto \color{blue}{\left(1 + \frac{x}{n}\right) - {x}^{\left(\frac{1}{n}\right)}}\]
  3. Recombined 4 regimes into one program.

  4. Final simplification13.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -1.7964466032532866 \cdot 10^{-06}:\\ \;\;\;\;\sqrt[3]{{\left({\left(1 + x\right)}^{\left(\frac{1}{n}\right)}\right)}^{3}} - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;\frac{1}{n} \leq -4.3467324602633687 \cdot 10^{-88}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x} + \frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot x} \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)\\ \mathbf{elif}\;\frac{1}{n} \leq 5.078480559969997 \cdot 10^{-117}:\\ \;\;\;\;\left(\frac{\log \left(1 + x\right)}{n} + 0.5 \cdot \frac{{\log \left(1 + x\right)}^{2}}{n \cdot n}\right) - \left(\frac{\log x}{n} + 0.5 \cdot \frac{{\log x}^{2}}{n \cdot n}\right)\\ \mathbf{elif}\;\frac{1}{n} \leq 4.857132074199983:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x} + \frac{{x}^{\left(\frac{1}{n}\right)}}{x \cdot x} \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \end{array}\]

Reproduce

herbie shell --seed 2020339 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))