Average Error: 41.2 → 0.9
Time: 2.8s
Precision: binary64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 1.7331556436317241 \cdot 10^{-227}:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 1.7331556436317241 \cdot 10^{-227}:\\
\;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\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)) <= 1.733155643631724e-227)) {
		VAR = ((double) (((double) (((double) exp(x)) / ((double) (((double) pow(((double) exp(x)), 3.0)) - ((double) pow(1.0, 3.0)))))) * ((double) (((double) (((double) exp(x)) * ((double) exp(x)))) + ((double) (((double) (1.0 * 1.0)) + ((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

Original41.2
Target40.8
Herbie0.9
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 1.7331556436317241e-227

    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. Applied associate-/r/0

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

    if 1.7331556436317241e-227 < (exp x)

    1. Initial program 61.4

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

    2. Taylor expanded around 0 1.3

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

  4. Final simplification0.9

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 1.7331556436317241 \cdot 10^{-227}:\\ \;\;\;\;\frac{e^{x}}{{\left(e^{x}\right)}^{3} - {1}^{3}} \cdot \left(e^{x} \cdot e^{x} + \left(1 \cdot 1 + e^{x} \cdot 1\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} + \left(\frac{1}{12} \cdot x + \frac{1}{x}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020153 
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

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

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