Kahan's exp quotient

Average Error: 39.2 → 0.3

Time: 2.7s

Precision: 64

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;x \le -1.4964858386511397 \cdot 10^{-4}:\\ \;\;\;\;\frac{\mathsf{expm1}\left(\mathsf{log1p}\left(e^{x} - 1\right)\right)}{x}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{12}, {x}^{2}, \mathsf{fma}\left(\frac{1}{3}, x, 1\right)\right) \cdot \mathsf{fma}\left(\frac{1}{36}, {x}^{2}, \mathsf{fma}\left(\frac{1}{6}, x, 1\right)\right)\\ \end{array}\]

C

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

↓

double code(double x) {
	double temp;
	if ((x <= -0.00014964858386511397)) {
		temp = (expm1(log1p((exp(x) - 1.0))) / x);
	} else {
		temp = (fma(0.08333333333333333, pow(x, 2.0), fma(0.3333333333333333, x, 1.0)) * fma(0.027777777777777776, pow(x, 2.0), fma(0.16666666666666666, x, 1.0)));
	}
	return temp;
}

Error

Bits error versus x

Target

Original	39.2
Target	39.6
Herbie	0.3

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

Derivation

1. Split input into 2 regimes

2. if x < -0.00014964858386511397

   1. Initial program 0.0

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

   3. Applied expm1-log1p-u 0.0

      \[\leadsto \frac{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(e^{x} - 1\right)\right)}}{x}\]

   if -0.00014964858386511397 < x

   1. Initial program 59.8

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

   2. Taylor expanded around 0 0.5

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

   3. Simplified 0.5

      \[\leadsto \color{blue}{\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)}\]
   4. Using strategy rm

   5. Applied add-cube-cbrt 0.5

      \[\leadsto \color{blue}{\left(\sqrt[3]{\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)} \cdot \sqrt[3]{\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)}\right) \cdot \sqrt[3]{\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)}}\]

   6. Taylor expanded around 0 0.5

      \[\leadsto \color{blue}{\left(\frac{1}{12} \cdot {x}^{2} + \left(\frac{1}{3} \cdot x + 1\right)\right)} \cdot \sqrt[3]{\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)}\]

   7. Simplified 0.5

      \[\leadsto \color{blue}{\mathsf{fma}\left(\frac{1}{12}, {x}^{2}, \mathsf{fma}\left(\frac{1}{3}, x, 1\right)\right)} \cdot \sqrt[3]{\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)}\]

   8. Taylor expanded around 0 0.5

      \[\leadsto \mathsf{fma}\left(\frac{1}{12}, {x}^{2}, \mathsf{fma}\left(\frac{1}{3}, x, 1\right)\right) \cdot \color{blue}{\left(\frac{1}{36} \cdot {x}^{2} + \left(\frac{1}{6} \cdot x + 1\right)\right)}\]

   9. Simplified 0.5

      \[\leadsto \mathsf{fma}\left(\frac{1}{12}, {x}^{2}, \mathsf{fma}\left(\frac{1}{3}, x, 1\right)\right) \cdot \color{blue}{\mathsf{fma}\left(\frac{1}{36}, {x}^{2}, \mathsf{fma}\left(\frac{1}{6}, x, 1\right)\right)}\]
3. Recombined 2 regimes into one program.

4. Final simplification 0.3

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.4964858386511397 \cdot 10^{-4}:\\ \;\;\;\;\frac{\mathsf{expm1}\left(\mathsf{log1p}\left(e^{x} - 1\right)\right)}{x}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{12}, {x}^{2}, \mathsf{fma}\left(\frac{1}{3}, x, 1\right)\right) \cdot \mathsf{fma}\left(\frac{1}{36}, {x}^{2}, \mathsf{fma}\left(\frac{1}{6}, x, 1\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020049 +o rules:numerics
(FPCore (x)
  :name "Kahan's exp quotient"
  :precision binary64

  :herbie-target
  (if (and (< x 1) (> x -1)) (/ (- (exp x) 1) (log (exp x))) (/ (- (exp x) 1) x))

  (/ (- (exp x) 1) x))