Average Error: 39.8 → 0.4
Time: 2.6s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.69715010585821388 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{\mathsf{fma}\left(-1, 1, e^{x + x}\right)}{e^{x} + 1}}{x}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{expm1}\left(\mathsf{fma}\left(\frac{5}{96}, {x}^{2}, \mathsf{fma}\left(\frac{1}{4}, x, \log 2\right)\right)\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.69715010585821388 \cdot 10^{-4}:\\
\;\;\;\;\frac{\frac{\mathsf{fma}\left(-1, 1, e^{x + x}\right)}{e^{x} + 1}}{x}\\

\mathbf{else}:\\
\;\;\;\;\mathsf{expm1}\left(\mathsf{fma}\left(\frac{5}{96}, {x}^{2}, \mathsf{fma}\left(\frac{1}{4}, x, \log 2\right)\right)\right)\\

\end{array}
double code(double x) {
	return ((exp(x) - 1.0) / x);
}
↓
double code(double x) {
	double temp;
	if ((x <= -0.0001697150105858214)) {
		temp = ((fma(-1.0, 1.0, exp((x + x))) / (exp(x) + 1.0)) / x);
	} else {
		temp = expm1(fma(0.052083333333333336, pow(x, 2.0), fma(0.25, x, log(2.0))));
	}
	return temp;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.8
Target40.2
Herbie0.4
\[\begin{array}{l} \mathbf{if}\;x \lt 1 \land x \gt -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -0.0001697150105858214

    1. Initial program 0.1

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

    3. Applied flip--0.1

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

    4. Simplified0.0

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

    if -0.0001697150105858214 < 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

      \[\leadsto \color{blue}{\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)}\]
    4. Using strategy rm

    5. Applied expm1-log1p-u0.5

      \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)\right)\right)}\]

    6. Taylor expanded around 0 0.5

      \[\leadsto \mathsf{expm1}\left(\color{blue}{\frac{5}{96} \cdot {x}^{2} + \left(\frac{1}{4} \cdot x + \log 2\right)}\right)\]

    7. Simplified0.5

      \[\leadsto \mathsf{expm1}\left(\color{blue}{\mathsf{fma}\left(\frac{5}{96}, {x}^{2}, \mathsf{fma}\left(\frac{1}{4}, x, \log 2\right)\right)}\right)\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.69715010585821388 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{\mathsf{fma}\left(-1, 1, e^{x + x}\right)}{e^{x} + 1}}{x}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{expm1}\left(\mathsf{fma}\left(\frac{5}{96}, {x}^{2}, \mathsf{fma}\left(\frac{1}{4}, x, \log 2\right)\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020057 +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))