Kahan's exp quotient

Average Error: 39.8 → 0.6

Time: 2.9s

Precision: binary64

Math

\[\frac{e^{x} - 1}{x}\]

↓

\[\begin{array}{l} \mathbf{if}\;\frac{e^{x} - 1}{x} \le -0.0 \lor \neg \left(\frac{e^{x} - 1}{x} \le 8.5931795027175715 \cdot 10^{-7} \lor \neg \left(\frac{e^{x} - 1}{x} \le 1.0122055456987034\right)\right):\\ \;\;\;\;\frac{{\left(\frac{1}{6} \cdot {x}^{2}\right)}^{3} + {\left(\frac{1}{2} \cdot x + 1\right)}^{3}}{\left(\frac{1}{2} \cdot x + 1\right) \cdot \left(\left(\frac{1}{2} \cdot x + 1\right) - \frac{1}{6} \cdot {x}^{2}\right) + \frac{1}{36} \cdot \left({x}^{2} \cdot {x}^{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt[3]{{\left(e^{x} - 1\right)}^{3}}}{x}\\ \end{array}\]

C

double code(double x) {
	return ((double) (((double) (((double) exp(x)) - 1.0)) / x));
}

↓

double code(double x) {
	double VAR;
	if (((((double) (((double) (((double) exp(x)) - 1.0)) / x)) <= -0.0) || !((((double) (((double) (((double) exp(x)) - 1.0)) / x)) <= 8.593179502717572e-07) || !(((double) (((double) (((double) exp(x)) - 1.0)) / x)) <= 1.0122055456987034)))) {
		VAR = ((double) (((double) (((double) pow(((double) (0.16666666666666666 * ((double) pow(x, 2.0)))), 3.0)) + ((double) pow(((double) (((double) (0.5 * x)) + 1.0)), 3.0)))) / ((double) (((double) (((double) (((double) (0.5 * x)) + 1.0)) * ((double) (((double) (((double) (0.5 * x)) + 1.0)) - ((double) (0.16666666666666666 * ((double) pow(x, 2.0)))))))) + ((double) (0.027777777777777776 * ((double) (((double) pow(x, 2.0)) * ((double) pow(x, 2.0))))))))));
	} else {
		VAR = ((double) (((double) cbrt(((double) pow(((double) (((double) exp(x)) - 1.0)), 3.0)))) / x));
	}
	return VAR;
}

Error

Bits error versus x

Target

Original	39.8
Target	40.3
Herbie	0.6

\[\begin{array}{l} \mathbf{if}\;x \lt 1 \land x \gt -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

1. Split input into 2 regimes

2. if (/ (- (exp x) 1.0) x) < -0.0 or 8.5931795027175715e-7 < (/ (- (exp x) 1.0) x) < 1.0122055456987034

   1. Initial program 59.8

      \[\frac{e^{x} - 1}{x}\]

   2. Taylor expanded around 0 0.7

      \[\leadsto \color{blue}{\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)}\]
   3. Using strategy rm

   4. Applied flip3-+ 0.7

      \[\leadsto \color{blue}{\frac{{\left(\frac{1}{6} \cdot {x}^{2}\right)}^{3} + {\left(\frac{1}{2} \cdot x + 1\right)}^{3}}{\left(\frac{1}{6} \cdot {x}^{2}\right) \cdot \left(\frac{1}{6} \cdot {x}^{2}\right) + \left(\left(\frac{1}{2} \cdot x + 1\right) \cdot \left(\frac{1}{2} \cdot x + 1\right) - \left(\frac{1}{6} \cdot {x}^{2}\right) \cdot \left(\frac{1}{2} \cdot x + 1\right)\right)}}\]

   5. Simplified 0.7

      \[\leadsto \frac{{\left(\frac{1}{6} \cdot {x}^{2}\right)}^{3} + {\left(\frac{1}{2} \cdot x + 1\right)}^{3}}{\color{blue}{\left(\frac{1}{2} \cdot x + 1\right) \cdot \left(\left(\frac{1}{2} \cdot x + 1\right) - \frac{1}{6} \cdot {x}^{2}\right) + \frac{1}{36} \cdot \left({x}^{2} \cdot {x}^{2}\right)}}\]

   if -0.0 < (/ (- (exp x) 1.0) x) < 8.5931795027175715e-7 or 1.0122055456987034 < (/ (- (exp x) 1.0) x)

   1. Initial program 0.3

      \[\frac{e^{x} - 1}{x}\]
   2. Using strategy rm

   3. Applied add-cbrt-cube 0.4

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

   4. Simplified 0.4

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

4. Final simplification 0.6

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{e^{x} - 1}{x} \le -0.0 \lor \neg \left(\frac{e^{x} - 1}{x} \le 8.5931795027175715 \cdot 10^{-7} \lor \neg \left(\frac{e^{x} - 1}{x} \le 1.0122055456987034\right)\right):\\ \;\;\;\;\frac{{\left(\frac{1}{6} \cdot {x}^{2}\right)}^{3} + {\left(\frac{1}{2} \cdot x + 1\right)}^{3}}{\left(\frac{1}{2} \cdot x + 1\right) \cdot \left(\left(\frac{1}{2} \cdot x + 1\right) - \frac{1}{6} \cdot {x}^{2}\right) + \frac{1}{36} \cdot \left({x}^{2} \cdot {x}^{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt[3]{{\left(e^{x} - 1\right)}^{3}}}{x}\\ \end{array}\]

Reproduce

herbie shell --seed 2020150 
(FPCore (x)
  :name "Kahan's exp quotient"
  :precision binary64

  :herbie-target
  (if (and (< x 1.0) (> x -1.0)) (/ (- (exp x) 1.0) (log (exp x))) (/ (- (exp x) 1.0) x))

  (/ (- (exp x) 1.0) x))