Average Error: 41.2 → 0.3
Time: 3.6s
Precision: binary64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \leq 1.0000173072590348:\\ \;\;\;\;\frac{e^{x}}{x + x \cdot \left(x \cdot \left(0.5 + x \cdot 0.16666666666666666\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{1 - e^{-x}}\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \leq 1.0000173072590348:\\
\;\;\;\;\frac{e^{x}}{x + x \cdot \left(x \cdot \left(0.5 + x \cdot 0.16666666666666666\right)\right)}\\

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

\end{array}
(FPCore (x) :precision binary64 (/ (exp x) (- (exp x) 1.0)))
↓
(FPCore (x)
 :precision binary64
 (if (<= (exp x) 1.0000173072590348)
   (/ (exp x) (+ x (* x (* x (+ 0.5 (* x 0.16666666666666666))))))
   (/ 1.0 (- 1.0 (exp (- x))))))
double code(double x) {
	return exp(x) / (exp(x) - 1.0);
}
↓
double code(double x) {
	double tmp;
	if (exp(x) <= 1.0000173072590348) {
		tmp = exp(x) / (x + (x * (x * (0.5 + (x * 0.16666666666666666)))));
	} else {
		tmp = 1.0 / (1.0 - exp(-x));
	}
	return tmp;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original41.2
Target40.8
Herbie0.3
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp.f64 x) < 1.0000173072590348

    1. Initial program 41.4

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

    2. Taylor expanded around 0 10.8

      \[\leadsto \frac{e^{x}}{\color{blue}{x + \left(0.5 \cdot {x}^{2} + 0.16666666666666666 \cdot {x}^{3}\right)}}\]

    3. Simplified0.3

      \[\leadsto \frac{e^{x}}{\color{blue}{x + x \cdot \left(x \cdot \left(0.5 + x \cdot 0.16666666666666666\right)\right)}}\]

    if 1.0000173072590348 < (exp.f64 x)

    1. Initial program 27.3

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

    3. Applied clear-num_binary64_110027.4

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

    4. Simplified1.9

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \leq 1.0000173072590348:\\ \;\;\;\;\frac{e^{x}}{x + x \cdot \left(x \cdot \left(0.5 + x \cdot 0.16666666666666666\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{1 - e^{-x}}\\ \end{array}\]

Reproduce

herbie shell --seed 2020354 
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

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

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