Kahan's exp quotient

Average Error: 38.8 → 0.3

Time: 3.3s

Precision: 64

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;x \le -1.7487706170283131 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{3}}{1 \cdot \left(1 + e^{x}\right) + {\left(e^{x}\right)}^{2}}}{x} - \frac{\frac{{1}^{3}}{1 \cdot \left(1 + e^{x}\right) + {\left(e^{x}\right)}^{2}}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)\\ \end{array}\]

C

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

↓

double code(double x) {
	double VAR;
	if ((x <= -0.0001748770617028313)) {
		VAR = (((pow(exp(x), 3.0) / ((1.0 * (1.0 + exp(x))) + pow(exp(x), 2.0))) / x) - ((pow(1.0, 3.0) / ((1.0 * (1.0 + exp(x))) + pow(exp(x), 2.0))) / x));
	} else {
		VAR = ((0.16666666666666666 * pow(x, 2.0)) + ((0.5 * x) + 1.0));
	}
	return VAR;
}

Error

Bits error versus x

Target

Original	38.8
Target	39.1
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 < -0.0001748770617028313

   1. Initial program 0.0

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

   3. Applied div-inv 0.0

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

   5. Applied flip3-- 0.0

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

   6. Applied associate-*l/ 0.0

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

   7. Simplified 0.0

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

   9. Applied div-sub 0.0

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

   10. Applied div-sub 0.0

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

   11. Simplified 0.0

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

   12. Simplified 0.0

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

   if -0.0001748770617028313 < 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. Recombined 2 regimes into one program.

4. Final simplification 0.3

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.7487706170283131 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{3}}{1 \cdot \left(1 + e^{x}\right) + {\left(e^{x}\right)}^{2}}}{x} - \frac{\frac{{1}^{3}}{1 \cdot \left(1 + e^{x}\right) + {\left(e^{x}\right)}^{2}}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020102 
(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))