Average Error: 40.8 → 0.3
Time: 3.0s
Precision: 64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 1.00006043012207457:\\ \;\;\;\;\frac{e^{x}}{{x}^{2} \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right) + x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{1 - \frac{1}{e^{x}}}\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 1.00006043012207457:\\
\;\;\;\;\frac{e^{x}}{{x}^{2} \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right) + x}\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{1 - \frac{1}{e^{x}}}\\

\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)) <= 1.0000604301220746)) {
		VAR = ((double) (((double) exp(x)) / ((double) (((double) (((double) pow(x, 2.0)) * ((double) (((double) (x * 0.16666666666666666)) + 0.5)))) + x))));
	} else {
		VAR = ((double) (1.0 / ((double) (1.0 - ((double) (1.0 / ((double) exp(x))))))));
	}
	return VAR;
}

Error

Bits error versus x

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Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original40.8
Target40.4
Herbie0.3
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 1.0000604301220746

    1. Initial program 41.0

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

    2. Taylor expanded around 0 11.1

      \[\leadsto \frac{e^{x}}{\color{blue}{\frac{1}{2} \cdot {x}^{2} + \left(\frac{1}{6} \cdot {x}^{3} + x\right)}}\]

    3. Simplified0.2

      \[\leadsto \frac{e^{x}}{\color{blue}{{x}^{2} \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right) + x}}\]

    if 1.0000604301220746 < (exp x)

    1. Initial program 29.6

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

    3. Applied clear-num29.7

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

    4. Simplified1.5

      \[\leadsto \frac{1}{\color{blue}{1 - \frac{1}{e^{x}}}}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 1.00006043012207457:\\ \;\;\;\;\frac{e^{x}}{{x}^{2} \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right) + x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{1 - \frac{1}{e^{x}}}\\ \end{array}\]

Reproduce

herbie shell --seed 2020122 
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

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

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