Average Error: 40.9 → 0.7
Time: 6.2s
Precision: binary64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 4.83372930294818627 \cdot 10^{-10}:\\ \;\;\;\;\frac{e^{x}}{\log \left(e^{e^{x} - 1}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 4.83372930294818627 \cdot 10^{-10}:\\
\;\;\;\;\frac{e^{x}}{\log \left(e^{e^{x} - 1}\right)}\\

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

\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)) <= 4.833729302948186e-10)) {
		VAR = ((double) (((double) exp(x)) / ((double) log(((double) exp(((double) (((double) exp(x)) - 1.0))))))));
	} else {
		VAR = ((double) (0.5 + ((double) (((double) (0.08333333333333333 * x)) + ((double) (1.0 / x))))));
	}
	return VAR;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original40.9
Target40.5
Herbie0.7
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 4.83372930294818627e-10

    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 4.83372930294818627e-10 < (exp x)

    1. Initial program 61.5

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

    2. Taylor expanded around 0 1.1

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

  4. Final simplification0.7

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 4.83372930294818627 \cdot 10^{-10}:\\ \;\;\;\;\frac{e^{x}}{\log \left(e^{e^{x} - 1}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020163 
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

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

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