Average Error: 41.1 → 0.6
Time: 3.7min
Precision: binary64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 0.319271880665959629:\\ \;\;\;\;\frac{e^{x}}{\log \left(e^{e^{x} - 1}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{12} \cdot x + \left(\frac{1}{x} + \frac{1}{2}\right)\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 0.319271880665959629:\\
\;\;\;\;\frac{e^{x}}{\log \left(e^{e^{x} - 1}\right)}\\

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

\end{array}
double code(double x) {
	return (((double) exp(x)) / ((double) (((double) exp(x)) - 1.0)));
}
↓
double code(double x) {
	double VAR;
	if ((((double) exp(x)) <= 0.31927188066595963)) {
		VAR = (((double) exp(x)) / ((double) log(((double) exp(((double) (((double) exp(x)) - 1.0)))))));
	} else {
		VAR = ((double) (((double) (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

Original41.1
Target40.8
Herbie0.6
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 0.319271880665959629

    1. Initial program 0.0

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

    3. Applied add-log-exp0.0

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

    4. Applied add-log-exp0.0

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

    5. Applied diff-log0.0

      \[\leadsto \frac{e^{x}}{\color{blue}{\log \left(\frac{e^{e^{x}}}{e^{1}}\right)}}\]

    6. Simplified0.0

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

    if 0.319271880665959629 < (exp x)

    1. Initial program 61.6

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

    2. Taylor expanded around 0 0.9

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

  4. Final simplification0.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 0.319271880665959629:\\ \;\;\;\;\frac{e^{x}}{\log \left(e^{e^{x} - 1}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{12} \cdot x + \left(\frac{1}{x} + \frac{1}{2}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020181 
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

  :herbie-target
  (/ 1.0 (- 1.0 (exp (neg x))))

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