Average Error: 41.3 → 0.7
Time: 7.3s
Precision: binary64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \leq 1.1857370595327356 \cdot 10^{-16}:\\ \;\;\;\;\frac{e^{x}}{\frac{{\left(e^{x}\right)}^{3} - 1}{1 + e^{x} \cdot \left(e^{x} + 1\right)}}\\ \mathbf{else}:\\ \;\;\;\;x \cdot 0.08333333333333333 + \left(0.5 + \frac{1}{x}\right)\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \leq 1.1857370595327356 \cdot 10^{-16}:\\
\;\;\;\;\frac{e^{x}}{\frac{{\left(e^{x}\right)}^{3} - 1}{1 + e^{x} \cdot \left(e^{x} + 1\right)}}\\

\mathbf{else}:\\
\;\;\;\;x \cdot 0.08333333333333333 + \left(0.5 + \frac{1}{x}\right)\\

\end{array}
(FPCore (x) :precision binary64 (/ (exp x) (- (exp x) 1.0)))
↓
(FPCore (x)
 :precision binary64
 (if (<= (exp x) 1.1857370595327356e-16)
   (/
    (exp x)
    (/ (- (pow (exp x) 3.0) 1.0) (+ 1.0 (* (exp x) (+ (exp x) 1.0)))))
   (+ (* x 0.08333333333333333) (+ 0.5 (/ 1.0 x)))))
double code(double x) {
	return exp(x) / (exp(x) - 1.0);
}
↓
double code(double x) {
	double tmp;
	if (exp(x) <= 1.1857370595327356e-16) {
		tmp = exp(x) / ((pow(exp(x), 3.0) - 1.0) / (1.0 + (exp(x) * (exp(x) + 1.0))));
	} else {
		tmp = (x * 0.08333333333333333) + (0.5 + (1.0 / x));
	}
	return tmp;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original41.3
Target40.9
Herbie0.7
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp.f64 x) < 1.1857370595327356e-16

    1. Initial program 0

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

    3. Applied flip3--_binary64_7550.0

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

    4. Simplified0.0

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

    5. Simplified0.0

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

    if 1.1857370595327356e-16 < (exp.f64 x)

    1. Initial program 61.5

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

    2. Taylor expanded around 0 1.1

      \[\leadsto \color{blue}{0.08333333333333333 \cdot x + \left(\frac{1}{x} + 0.5\right)}\]

    3. Simplified1.1

      \[\leadsto \color{blue}{x \cdot 0.08333333333333333 + \left(0.5 + \frac{1}{x}\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.7

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \leq 1.1857370595327356 \cdot 10^{-16}:\\ \;\;\;\;\frac{e^{x}}{\frac{{\left(e^{x}\right)}^{3} - 1}{1 + e^{x} \cdot \left(e^{x} + 1\right)}}\\ \mathbf{else}:\\ \;\;\;\;x \cdot 0.08333333333333333 + \left(0.5 + \frac{1}{x}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020303 
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

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

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