Average Error: 40.5 → 0.3
Time: 2.5s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.8041726414346648 \cdot 10^{-4}:\\ \;\;\;\;\frac{e^{x} \cdot e^{x} - 1 \cdot 1}{x \cdot \left(e^{x} + 1\right)}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.8041726414346648 \cdot 10^{-4}:\\
\;\;\;\;\frac{e^{x} \cdot e^{x} - 1 \cdot 1}{x \cdot \left(e^{x} + 1\right)}\\

\mathbf{else}:\\
\;\;\;\;\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)\\

\end{array}
double code(double x) {
	return ((exp(x) - 1.0) / x);
}
↓
double code(double x) {
	double temp;
	if ((x <= -0.00018041726414346648)) {
		temp = (((exp(x) * exp(x)) - (1.0 * 1.0)) / (x * (exp(x) + 1.0)));
	} else {
		temp = fma(0.16666666666666666, pow(x, 2.0), fma(0.5, x, 1.0));
	}
	return temp;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original40.5
Target40.8
Herbie0.3
\[\begin{array}{l} \mathbf{if}\;x \lt 1 \land x \gt -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -0.00018041726414346648

    1. Initial program 0.0

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

    3. Applied flip--0.0

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

    4. Applied associate-/l/0.0

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

    if -0.00018041726414346648 < x

    1. Initial program 60.0

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

    2. Taylor expanded around 0 0.5

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

    3. Simplified0.5

      \[\leadsto \color{blue}{\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.8041726414346648 \cdot 10^{-4}:\\ \;\;\;\;\frac{e^{x} \cdot e^{x} - 1 \cdot 1}{x \cdot \left(e^{x} + 1\right)}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020066 +o rules:numerics
(FPCore (x)
  :name "Kahan's exp quotient"
  :precision binary64

  :herbie-target
  (if (and (< x 1) (> x -1)) (/ (- (exp x) 1) (log (exp x))) (/ (- (exp x) 1) x))

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