Average Error: 39.6 → 0.3
Time: 8.4s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.0015840142054043 \cdot 10^{-4}:\\ \;\;\;\;\frac{e^{2 \cdot x}}{\left(e^{x} + 1\right) \cdot x} - \frac{\frac{1 \cdot 1}{x}}{e^{x} + 1}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.0015840142054043 \cdot 10^{-4}:\\
\;\;\;\;\frac{e^{2 \cdot x}}{\left(e^{x} + 1\right) \cdot x} - \frac{\frac{1 \cdot 1}{x}}{e^{x} + 1}\\

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

\end{array}
double f(double x) {
        double r80351 = x;
        double r80352 = exp(r80351);
        double r80353 = 1.0;
        double r80354 = r80352 - r80353;
        double r80355 = r80354 / r80351;
        return r80355;
}
↓
double f(double x) {
        double r80356 = x;
        double r80357 = -0.00010015840142054043;
        bool r80358 = r80356 <= r80357;
        double r80359 = 2.0;
        double r80360 = r80359 * r80356;
        double r80361 = exp(r80360);
        double r80362 = exp(r80356);
        double r80363 = 1.0;
        double r80364 = r80362 + r80363;
        double r80365 = r80364 * r80356;
        double r80366 = r80361 / r80365;
        double r80367 = r80363 * r80363;
        double r80368 = r80367 / r80356;
        double r80369 = r80368 / r80364;
        double r80370 = r80366 - r80369;
        double r80371 = 1.0;
        double r80372 = 0.16666666666666666;
        double r80373 = r80356 * r80372;
        double r80374 = 0.5;
        double r80375 = r80373 + r80374;
        double r80376 = r80356 * r80375;
        double r80377 = r80371 + r80376;
        double r80378 = r80358 ? r80370 : r80377;
        return r80378;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.6
Target40.0
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 < -0.00010015840142054043

    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

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

    6. Applied div-sub0.1

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

    7. Simplified0.1

      \[\leadsto \color{blue}{\frac{e^{2 \cdot x}}{\left(e^{x} + 1\right) \cdot x}} - \frac{1 \cdot 1}{x \cdot \left(e^{x} + 1\right)}\]
    8. Using strategy rm

    9. Applied associate-/r*0.1

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

    if -0.00010015840142054043 < x

    1. Initial program 60.1

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

    2. Taylor expanded around 0 0.4

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

    3. Simplified0.4

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.0015840142054043 \cdot 10^{-4}:\\ \;\;\;\;\frac{e^{2 \cdot x}}{\left(e^{x} + 1\right) \cdot x} - \frac{\frac{1 \cdot 1}{x}}{e^{x} + 1}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020042 
(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))