Average Error: 41.0 → 0.5
Time: 2.4s
Precision: binary64
\[\frac{e^{x}}{e^{x} - 1}\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \leq 0.787664626492611:\\ \;\;\;\;\frac{e^{x}}{\frac{{\left(e^{x}\right)}^{2} + -1}{e^{x} + 1}}\\ \mathbf{else}:\\ \;\;\;\;0.5 + \left(x \cdot 0.08333333333333333 + \frac{1}{x}\right)\\ \end{array}\]
\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \leq 0.787664626492611:\\
\;\;\;\;\frac{e^{x}}{\frac{{\left(e^{x}\right)}^{2} + -1}{e^{x} + 1}}\\

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

\end{array}
(FPCore (x) :precision binary64 (/ (exp x) (- (exp x) 1.0)))
↓
(FPCore (x)
 :precision binary64
 (if (<= (exp x) 0.787664626492611)
   (/ (exp x) (/ (+ (pow (exp x) 2.0) -1.0) (+ (exp x) 1.0)))
   (+ 0.5 (+ (* x 0.08333333333333333) (/ 1.0 x)))))
double code(double x) {
	return (((double) exp(x)) / ((double) (((double) exp(x)) - 1.0)));
}
↓
double code(double x) {
	double tmp;
	if ((((double) exp(x)) <= 0.787664626492611)) {
		tmp = (((double) exp(x)) / (((double) (((double) pow(((double) exp(x)), 2.0)) + -1.0)) / ((double) (((double) exp(x)) + 1.0))));
	} else {
		tmp = ((double) (0.5 + ((double) (((double) (x * 0.08333333333333333)) + (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

Original41.0
Target40.7
Herbie0.5
\[\frac{1}{1 - e^{-x}}\]

Derivation

  1. Split input into 2 regimes

  2. if (exp.f64 x) < 0.78766462649261104

    1. Initial program 0.0

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

    3. Applied flip--_binary640.0

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

    4. Simplified0.0

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

    if 0.78766462649261104 < (exp.f64 x)

    1. Initial program 61.7

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

    2. Taylor expanded around 0 0.8

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

    3. Simplified0.8

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

  4. Final simplification0.5

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \leq 0.787664626492611:\\ \;\;\;\;\frac{e^{x}}{\frac{{\left(e^{x}\right)}^{2} + -1}{e^{x} + 1}}\\ \mathbf{else}:\\ \;\;\;\;0.5 + \left(x \cdot 0.08333333333333333 + \frac{1}{x}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020219 
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

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

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