Average Error: 41.5 → 0.7
Time: 2.9s
Precision: binary64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 5.8144600494016733 \cdot 10^{-27}:\\ \;\;\;\;\frac{e^{x}}{\frac{{\left(e^{x}\right)}^{3} \cdot {\left(e^{x}\right)}^{3} - {1}^{3} \cdot {1}^{3}}{\left({\left(e^{x}\right)}^{2} + 1 \cdot \left(e^{x} + 1\right)\right) \cdot \left({\left(e^{x}\right)}^{3} + {1}^{3}\right)}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(x \cdot \frac{1}{12} + \frac{1}{x}\right)\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 5.8144600494016733 \cdot 10^{-27}:\\
\;\;\;\;\frac{e^{x}}{\frac{{\left(e^{x}\right)}^{3} \cdot {\left(e^{x}\right)}^{3} - {1}^{3} \cdot {1}^{3}}{\left({\left(e^{x}\right)}^{2} + 1 \cdot \left(e^{x} + 1\right)\right) \cdot \left({\left(e^{x}\right)}^{3} + {1}^{3}\right)}}\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{2} + \left(x \cdot \frac{1}{12} + \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)) <= 5.814460049401673e-27)) {
		VAR = ((double) (((double) exp(x)) / ((double) (((double) (((double) (((double) pow(((double) exp(x)), 3.0)) * ((double) pow(((double) exp(x)), 3.0)))) - ((double) (((double) pow(1.0, 3.0)) * ((double) pow(1.0, 3.0)))))) / ((double) (((double) (((double) pow(((double) exp(x)), 2.0)) + ((double) (1.0 * ((double) (((double) exp(x)) + 1.0)))))) * ((double) (((double) pow(((double) exp(x)), 3.0)) + ((double) pow(1.0, 3.0))))))))));
	} else {
		VAR = ((double) (0.5 + ((double) (((double) (x * 0.08333333333333333)) + ((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

Original41.5
Target41.1
Herbie0.7
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 5.8144600494016733e-27

    1. Initial program 0

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

    3. Applied flip3--0

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

    4. Simplified0

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

    6. Applied flip--0

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

    7. Applied associate-/l/0

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

    if 5.8144600494016733e-27 < (exp x)

    1. Initial program 61.8

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

    2. Taylor expanded around 0 1.1

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

    3. Simplified1.1

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

  4. Final simplification0.7

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 5.8144600494016733 \cdot 10^{-27}:\\ \;\;\;\;\frac{e^{x}}{\frac{{\left(e^{x}\right)}^{3} \cdot {\left(e^{x}\right)}^{3} - {1}^{3} \cdot {1}^{3}}{\left({\left(e^{x}\right)}^{2} + 1 \cdot \left(e^{x} + 1\right)\right) \cdot \left({\left(e^{x}\right)}^{3} + {1}^{3}\right)}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(x \cdot \frac{1}{12} + \frac{1}{x}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020179 
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

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

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