2nthrt (problem 3.4.6)

Average Error: 32.9 → 24.6

Time: 12.8s

Precision: 64

• could not determine a ground truth for program body

  1. x = 3.968919822528514e+262
  2. n = 3.869481019488881e-237

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \le -192.91395020987812 \lor \neg \left(\frac{1}{n} \le 1.4546846550253057 \cdot 10^{-14}\right):\\ \;\;\;\;\mathsf{fma}\left({\left(\sqrt[3]{x + 1} \cdot \sqrt[3]{x + 1}\right)}^{\left(\frac{1}{n}\right)}, {\left(\sqrt[3]{x + 1}\right)}^{\left(\frac{1}{n}\right)}, -{x}^{\left(\frac{1}{n}\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(1, \frac{1}{x \cdot n}, -\mathsf{fma}\left(0.5, \frac{1}{{x}^{2} \cdot n}, 1 \cdot \frac{\log \left(\frac{1}{x}\right)}{x \cdot {n}^{2}}\right)\right)\\ \end{array}\]

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) <= -192.91395020987812) || !((1.0 / n) <= 1.4546846550253057e-14))) {
		VAR = fma(pow((cbrt((x + 1.0)) * cbrt((x + 1.0))), (1.0 / n)), pow(cbrt((x + 1.0)), (1.0 / n)), -pow(x, (1.0 / n)));
	} else {
		VAR = fma(1.0, (1.0 / (x * n)), -fma(0.5, (1.0 / (pow(x, 2.0) * n)), (1.0 * (log((1.0 / x)) / (x * pow(n, 2.0))))));
	}
	return VAR;
}

Error

Bits error versus x

Bits error versus n

Derivation

1. Split input into 2 regimes

2. if (/ 1.0 n) < -192.91395020987812 or 1.4546846550253057e-14 < (/ 1.0 n)

   1. Initial program 3.5

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

   3. Applied add-cube-cbrt 3.5

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

   4. Applied unpow-prod-down 3.5

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

   5. Applied fma-neg 3.5

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

   if -192.91395020987812 < (/ 1.0 n) < 1.4546846550253057e-14

   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 33.3

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

   3. Simplified 33.3

      \[\leadsto \color{blue}{\mathsf{fma}\left(1, \frac{1}{x \cdot n}, -\mathsf{fma}\left(0.5, \frac{1}{{x}^{2} \cdot n}, 1 \cdot \frac{\log \left(\frac{1}{x}\right)}{x \cdot {n}^{2}}\right)\right)}\]
3. Recombined 2 regimes into one program.

4. Final simplification 24.6

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \le -192.91395020987812 \lor \neg \left(\frac{1}{n} \le 1.4546846550253057 \cdot 10^{-14}\right):\\ \;\;\;\;\mathsf{fma}\left({\left(\sqrt[3]{x + 1} \cdot \sqrt[3]{x + 1}\right)}^{\left(\frac{1}{n}\right)}, {\left(\sqrt[3]{x + 1}\right)}^{\left(\frac{1}{n}\right)}, -{x}^{\left(\frac{1}{n}\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(1, \frac{1}{x \cdot n}, -\mathsf{fma}\left(0.5, \frac{1}{{x}^{2} \cdot n}, 1 \cdot \frac{\log \left(\frac{1}{x}\right)}{x \cdot {n}^{2}}\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020079 +o rules:numerics
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1) (/ 1 n)) (pow x (/ 1 n))))