Kahan's exp quotient

Average Error: 40.4 → 0.3

Time: 3.1s

Precision: binary64

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;x \le -0.00016599485325579491:\\ \;\;\;\;\frac{{\left({\left(e^{x}\right)}^{3}\right)}^{3} - {\left({1}^{3}\right)}^{3}}{x \cdot \left(\left({\left(e^{x}\right)}^{2} + 1 \cdot \left(e^{x} + 1\right)\right) \cdot \left({\left(e^{x}\right)}^{6} + \left({1}^{6} + {\left(e^{x}\right)}^{3} \cdot {1}^{3}\right)\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(0.5 + x \cdot 0.16666666666666666\right)\\ \end{array}\]

C

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

↓

double code(double x) {
	double VAR;
	if ((x <= -0.00016599485325579491)) {
		VAR = ((double) (((double) (((double) pow(((double) pow(((double) exp(x)), 3.0)), 3.0)) - ((double) pow(((double) pow(1.0, 3.0)), 3.0)))) / ((double) (x * ((double) (((double) (((double) pow(((double) exp(x)), 2.0)) + ((double) (1.0 * ((double) (((double) exp(x)) + 1.0)))))) * ((double) (((double) pow(((double) exp(x)), 6.0)) + ((double) (((double) pow(1.0, 6.0)) + ((double) (((double) pow(((double) exp(x)), 3.0)) * ((double) pow(1.0, 3.0))))))))))))));
	} else {
		VAR = ((double) (1.0 + ((double) (x * ((double) (0.5 + ((double) (x * 0.16666666666666666))))))));
	}
	return VAR;
}

Error

Bits error versus x

Target

Original	40.4
Target	40.8
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 < -1.65994853255794914e-4

   1. Initial program 0.0

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

   3. Applied flip3-- 0.0

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

   4. Applied associate-/l/ 0.0

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

   5. Simplified 0.0

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

   7. Applied flip3-- 0.0

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

   8. Applied associate-/l/ 0.0

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

   9. Simplified 0.0

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

   if -1.65994853255794914e-4 < x

   1. Initial program 60.3

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

   2. Taylor expanded around 0 0.4

      \[\leadsto \color{blue}{0.16666666666666666 \cdot {x}^{2} + \left(0.5 \cdot x + 1\right)}\]

   3. Simplified 0.4

      \[\leadsto \color{blue}{1 + x \cdot \left(0.5 + x \cdot 0.16666666666666666\right)}\]
3. Recombined 2 regimes into one program.

4. Final simplification 0.3

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -0.00016599485325579491:\\ \;\;\;\;\frac{{\left({\left(e^{x}\right)}^{3}\right)}^{3} - {\left({1}^{3}\right)}^{3}}{x \cdot \left(\left({\left(e^{x}\right)}^{2} + 1 \cdot \left(e^{x} + 1\right)\right) \cdot \left({\left(e^{x}\right)}^{6} + \left({1}^{6} + {\left(e^{x}\right)}^{3} \cdot {1}^{3}\right)\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(0.5 + x \cdot 0.16666666666666666\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020184 
(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))