Average Error: 32.8 → 24.3
Time: 13.2s
Precision: binary64
\[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}\]
↓
\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -1.9573488753029086 \cdot 10^{-11}:\\ \;\;\;\;\log \left(e^{{\left(1 + x\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}}\right)\\ \mathbf{elif}\;\frac{1}{n} \leq 4.184674277867971 \cdot 10^{-15}:\\ \;\;\;\;\left(\frac{1}{n \cdot x} - \frac{0.5}{x \cdot \left(n \cdot x\right)}\right) + \frac{\log x}{x \cdot \left(n \cdot n\right)}\\ \mathbf{else}:\\ \;\;\;\;{\left(1 + x\right)}^{\left(\frac{1}{n}\right)} - {\left(x \cdot x\right)}^{\left(\frac{0.5}{n}\right)}\\ \end{array}\]
{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}
↓
\begin{array}{l}
\mathbf{if}\;\frac{1}{n} \leq -1.9573488753029086 \cdot 10^{-11}:\\
\;\;\;\;\log \left(e^{{\left(1 + x\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}}\right)\\

\mathbf{elif}\;\frac{1}{n} \leq 4.184674277867971 \cdot 10^{-15}:\\
\;\;\;\;\left(\frac{1}{n \cdot x} - \frac{0.5}{x \cdot \left(n \cdot x\right)}\right) + \frac{\log x}{x \cdot \left(n \cdot n\right)}\\

\mathbf{else}:\\
\;\;\;\;{\left(1 + x\right)}^{\left(\frac{1}{n}\right)} - {\left(x \cdot x\right)}^{\left(\frac{0.5}{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) -1.9573488753029086e-11)
   (log (exp (- (pow (+ 1.0 x) (/ 1.0 n)) (pow x (/ 1.0 n)))))
   (if (<= (/ 1.0 n) 4.184674277867971e-15)
     (+ (- (/ 1.0 (* n x)) (/ 0.5 (* x (* n x)))) (/ (log x) (* x (* n n))))
     (- (pow (+ 1.0 x) (/ 1.0 n)) (pow (* x x) (/ 0.5 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) <= -1.9573488753029086e-11) {
		tmp = log(exp(pow((1.0 + x), (1.0 / n)) - pow(x, (1.0 / n))));
	} else if ((1.0 / n) <= 4.184674277867971e-15) {
		tmp = ((1.0 / (n * x)) - (0.5 / (x * (n * x)))) + (log(x) / (x * (n * n)));
	} else {
		tmp = pow((1.0 + x), (1.0 / n)) - pow((x * x), (0.5 / 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) < -1.9573488753029086e-11

    1. Initial program 2.1

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

    3. Applied add-log-exp_binary642.7

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

    4. Applied add-log-exp_binary642.6

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

    5. Applied diff-log_binary642.6

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

    6. Simplified2.6

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

    if -1.9573488753029086e-11 < (/.f64 1 n) < 4.1846742778679711e-15

    1. Initial program 45.2

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

    2. Taylor expanded around inf 32.8

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

    3. Simplified32.6

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

    if 4.1846742778679711e-15 < (/.f64 1 n)

    1. Initial program 8.8

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

    3. Applied sqr-pow_binary648.9

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

    4. Simplified8.9

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

    5. Simplified8.9

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

    7. Applied pow-prod-down_binary6410.3

      \[\leadsto {\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - \color{blue}{{\left(x \cdot x\right)}^{\left(\frac{0.5}{n}\right)}}\]
  3. Recombined 3 regimes into one program.

  4. Final simplification24.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -1.9573488753029086 \cdot 10^{-11}:\\ \;\;\;\;\log \left(e^{{\left(1 + x\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}}\right)\\ \mathbf{elif}\;\frac{1}{n} \leq 4.184674277867971 \cdot 10^{-15}:\\ \;\;\;\;\left(\frac{1}{n \cdot x} - \frac{0.5}{x \cdot \left(n \cdot x\right)}\right) + \frac{\log x}{x \cdot \left(n \cdot n\right)}\\ \mathbf{else}:\\ \;\;\;\;{\left(1 + x\right)}^{\left(\frac{1}{n}\right)} - {\left(x \cdot x\right)}^{\left(\frac{0.5}{n}\right)}\\ \end{array}\]

Reproduce

herbie shell --seed 2020260 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))