2nthrt (problem 3.4.6)

Average Error: 33.0 → 23.8

Time: 11.0s

Precision: 64

• could not determine a ground truth for program body

  1. x = 2.4317934399292146e+251
  2. n = 5.114661377009554e-83

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \le -0.0170348940873527761:\\ \;\;\;\;e^{\log \left(x + 1\right) \cdot \frac{1}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;\frac{1}{n} \le 3.5103443479232361 \cdot 10^{-5}:\\ \;\;\;\;\frac{\frac{1}{n}}{x} - \left(\log \left(e^{\frac{0.5}{{x}^{2} \cdot n}}\right) - \frac{\log x \cdot 1}{x \cdot {n}^{2}}\right)\\ \mathbf{else}:\\ \;\;\;\;e^{\left(\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\right) \cdot \frac{1}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array}\]

C

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 VAR;
	if (((1.0 / n) <= -0.017034894087352776)) {
		VAR = (exp((log((x + 1.0)) * (1.0 / n))) - pow(x, (1.0 / n)));
	} else {
		double VAR_1;
		if (((1.0 / n) <= 3.510344347923236e-05)) {
			VAR_1 = (((1.0 / n) / x) - (log(exp((0.5 / (pow(x, 2.0) * n)))) - ((log(x) * 1.0) / (x * pow(n, 2.0)))));
		} else {
			VAR_1 = (exp(((((1.0 * x) + log(1.0)) - (0.5 * (pow(x, 2.0) / pow(1.0, 2.0)))) * (1.0 / n))) - pow(x, (1.0 / n)));
		}
		VAR = VAR_1;
	}
	return VAR;
}

Error

Bits error versus x

Bits error versus n

Derivation

1. Split input into 3 regimes

2. if (/ 1.0 n) < -0.017034894087352776

   1. Initial program 0.2

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

   3. Applied add-exp-log 0.2

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

   4. Applied pow-exp 0.2

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

   if -0.017034894087352776 < (/ 1.0 n) < 3.510344347923236e-05

   1. Initial program 45.0

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

   2. Taylor expanded around inf 33.2

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

   3. Simplified 32.6

      \[\leadsto \color{blue}{\frac{\frac{1}{n}}{x} - \left(\frac{\frac{0.5}{n}}{{x}^{2}} - \frac{\log x \cdot 1}{x \cdot {n}^{2}}\right)}\]
   4. Using strategy rm

   5. Applied add-log-exp 32.7

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

   6. Simplified 32.7

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

   if 3.510344347923236e-05 < (/ 1.0 n)

   1. Initial program 5.8

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

   3. Applied add-exp-log 5.8

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

   4. Applied pow-exp 5.8

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

   5. Taylor expanded around 0 2.5

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

4. Final simplification 23.8

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \le -0.0170348940873527761:\\ \;\;\;\;e^{\log \left(x + 1\right) \cdot \frac{1}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;\frac{1}{n} \le 3.5103443479232361 \cdot 10^{-5}:\\ \;\;\;\;\frac{\frac{1}{n}}{x} - \left(\log \left(e^{\frac{0.5}{{x}^{2} \cdot n}}\right) - \frac{\log x \cdot 1}{x \cdot {n}^{2}}\right)\\ \mathbf{else}:\\ \;\;\;\;e^{\left(\left(1 \cdot x + \log 1\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}\right) \cdot \frac{1}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array}\]

Reproduce

herbie shell --seed 2020100 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1) (/ 1 n)) (pow x (/ 1 n))))