Average Error: 41.3 → 0.6
Time: 3.0s
Precision: 64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 0.989466018642316869:\\ \;\;\;\;\frac{e^{x}}{\frac{\log \left(e^{\mathsf{fma}\left(-1, 1, e^{x + x}\right)}\right)}{e^{x} + 1}}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{12}, x, \frac{1}{x}\right) + \frac{1}{2}\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 0.989466018642316869:\\
\;\;\;\;\frac{e^{x}}{\frac{\log \left(e^{\mathsf{fma}\left(-1, 1, e^{x + x}\right)}\right)}{e^{x} + 1}}\\

\mathbf{else}:\\
\;\;\;\;\mathsf{fma}\left(\frac{1}{12}, x, \frac{1}{x}\right) + \frac{1}{2}\\

\end{array}
double code(double x) {
	return ((double) (((double) exp(x)) / ((double) (((double) exp(x)) - 1.0))));
}
↓
double code(double x) {
	double VAR;
	if ((((double) exp(x)) <= 0.9894660186423169)) {
		VAR = ((double) (((double) exp(x)) / ((double) (((double) log(((double) exp(((double) fma(((double) -(1.0)), 1.0, ((double) exp(((double) (x + x)))))))))) / ((double) (((double) exp(x)) + 1.0))))));
	} else {
		VAR = ((double) (((double) fma(0.08333333333333333, x, ((double) (1.0 / x)))) + 0.5));
	}
	return VAR;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original41.3
Target40.9
Herbie0.6
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 0.9894660186423169

    1. Initial program 0.0

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

    3. Applied flip--0.0

      \[\leadsto \frac{e^{x}}{\color{blue}{\frac{e^{x} \cdot e^{x} - 1 \cdot 1}{e^{x} + 1}}}\]

    4. Simplified0.0

      \[\leadsto \frac{e^{x}}{\frac{\color{blue}{\mathsf{fma}\left(-1, 1, e^{x + x}\right)}}{e^{x} + 1}}\]
    5. Using strategy rm

    6. Applied add-log-exp0.0

      \[\leadsto \frac{e^{x}}{\frac{\color{blue}{\log \left(e^{\mathsf{fma}\left(-1, 1, e^{x + x}\right)}\right)}}{e^{x} + 1}}\]

    if 0.9894660186423169 < (exp x)

    1. Initial program 61.7

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

    2. Taylor expanded around 0 0.9

      \[\leadsto \color{blue}{\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)}\]

    3. Simplified0.9

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

  4. Final simplification0.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 0.989466018642316869:\\ \;\;\;\;\frac{e^{x}}{\frac{\log \left(e^{\mathsf{fma}\left(-1, 1, e^{x + x}\right)}\right)}{e^{x} + 1}}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{12}, x, \frac{1}{x}\right) + \frac{1}{2}\\ \end{array}\]

Reproduce

herbie shell --seed 2020114 +o rules:numerics
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

  :herbie-target
  (/ 1 (- 1 (exp (- x))))

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