Average Error: 41.8 → 0.7
Time: 2.8s
Precision: binary64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 0.497914740142360979:\\ \;\;\;\;\frac{1}{\sqrt[3]{e^{x} - 1} \cdot \sqrt[3]{e^{x} - 1}} \cdot \frac{e^{x}}{\sqrt[3]{e^{x} - 1}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(x \cdot \frac{1}{12} + \frac{1}{x}\right)\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 0.497914740142360979:\\
\;\;\;\;\frac{1}{\sqrt[3]{e^{x} - 1} \cdot \sqrt[3]{e^{x} - 1}} \cdot \frac{e^{x}}{\sqrt[3]{e^{x} - 1}}\\

\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)) <= 0.497914740142361)) {
		VAR = ((double) (((double) (1.0 / ((double) (((double) cbrt(((double) (((double) exp(x)) - 1.0)))) * ((double) cbrt(((double) (((double) exp(x)) - 1.0)))))))) * ((double) (((double) exp(x)) / ((double) cbrt(((double) (((double) exp(x)) - 1.0))))))));
	} 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

Original41.8
Target41.4
Herbie0.7
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 0.497914740142360979

    1. Initial program 0.0

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

    3. Applied add-cube-cbrt0.0

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

    4. Applied *-un-lft-identity0.0

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

    5. Applied times-frac0.0

      \[\leadsto \color{blue}{\frac{1}{\sqrt[3]{e^{x} - 1} \cdot \sqrt[3]{e^{x} - 1}} \cdot \frac{e^{x}}{\sqrt[3]{e^{x} - 1}}}\]

    if 0.497914740142360979 < (exp x)

    1. Initial program 61.9

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

    2. Taylor expanded around 0 1.0

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

    3. Simplified1.0

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

  4. Final simplification0.7

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 0.497914740142360979:\\ \;\;\;\;\frac{1}{\sqrt[3]{e^{x} - 1} \cdot \sqrt[3]{e^{x} - 1}} \cdot \frac{e^{x}}{\sqrt[3]{e^{x} - 1}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(x \cdot \frac{1}{12} + \frac{1}{x}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020184 
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

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

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