Average Error: 40.4 → 0.2
Time: 9.5s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -0.00018534463155854764:\\ \;\;\;\;\frac{{\left(e^{x}\right)}^{3} - 1}{x + \left(x + e^{x} \cdot x\right) \cdot e^{x}}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -0.00018534463155854764:\\
\;\;\;\;\frac{{\left(e^{x}\right)}^{3} - 1}{x + \left(x + e^{x} \cdot x\right) \cdot e^{x}}\\

\mathbf{else}:\\
\;\;\;\;1 + x \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right)\\

\end{array}
double f(double x) {
        double r1506517 = x;
        double r1506518 = exp(r1506517);
        double r1506519 = 1.0;
        double r1506520 = r1506518 - r1506519;
        double r1506521 = r1506520 / r1506517;
        return r1506521;
}
↓
double f(double x) {
        double r1506522 = x;
        double r1506523 = -0.00018534463155854764;
        bool r1506524 = r1506522 <= r1506523;
        double r1506525 = exp(r1506522);
        double r1506526 = 3.0;
        double r1506527 = pow(r1506525, r1506526);
        double r1506528 = 1.0;
        double r1506529 = r1506527 - r1506528;
        double r1506530 = r1506525 * r1506522;
        double r1506531 = r1506522 + r1506530;
        double r1506532 = r1506531 * r1506525;
        double r1506533 = r1506522 + r1506532;
        double r1506534 = r1506529 / r1506533;
        double r1506535 = 0.16666666666666666;
        double r1506536 = r1506522 * r1506535;
        double r1506537 = 0.5;
        double r1506538 = r1506536 + r1506537;
        double r1506539 = r1506522 * r1506538;
        double r1506540 = r1506528 + r1506539;
        double r1506541 = r1506524 ? r1506534 : r1506540;
        return r1506541;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original40.4
Target39.5
Herbie0.2
\[\begin{array}{l} \mathbf{if}\;x \lt 1 \land x \gt -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -0.00018534463155854764

    1. Initial program 0.1

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

    3. Applied flip3--0.1

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

    4. Applied associate-/l/0.1

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

    5. Simplified0.1

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

    if -0.00018534463155854764 < x

    1. Initial program 60.2

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

    2. Taylor expanded around 0 0.3

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

    3. Simplified0.3

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -0.00018534463155854764:\\ \;\;\;\;\frac{{\left(e^{x}\right)}^{3} - 1}{x + \left(x + e^{x} \cdot x\right) \cdot e^{x}}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019154 
(FPCore (x)
  :name "Kahan's exp quotient"

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

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