Average Error: 39.8 → 0.3
Time: 6.5s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.2153975327080830811602579766272924644 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{e^{x + x} - 1 \cdot 1}{e^{x} + 1}}{x}\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(\frac{1}{2} + \frac{1}{6} \cdot x\right) + 1\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.2153975327080830811602579766272924644 \cdot 10^{-4}:\\
\;\;\;\;\frac{\frac{e^{x + x} - 1 \cdot 1}{e^{x} + 1}}{x}\\

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

\end{array}
double f(double x) {
        double r85338 = x;
        double r85339 = exp(r85338);
        double r85340 = 1.0;
        double r85341 = r85339 - r85340;
        double r85342 = r85341 / r85338;
        return r85342;
}
↓
double f(double x) {
        double r85343 = x;
        double r85344 = -0.00012153975327080831;
        bool r85345 = r85343 <= r85344;
        double r85346 = r85343 + r85343;
        double r85347 = exp(r85346);
        double r85348 = 1.0;
        double r85349 = r85348 * r85348;
        double r85350 = r85347 - r85349;
        double r85351 = exp(r85343);
        double r85352 = r85351 + r85348;
        double r85353 = r85350 / r85352;
        double r85354 = r85353 / r85343;
        double r85355 = 0.5;
        double r85356 = 0.16666666666666666;
        double r85357 = r85356 * r85343;
        double r85358 = r85355 + r85357;
        double r85359 = r85343 * r85358;
        double r85360 = 1.0;
        double r85361 = r85359 + r85360;
        double r85362 = r85345 ? r85354 : r85361;
        return r85362;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.8
Target40.2
Herbie0.3
\[\begin{array}{l} \mathbf{if}\;x \lt 1 \land x \gt -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -0.00012153975327080831

    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

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

    if -0.00012153975327080831 < 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

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.2153975327080830811602579766272924644 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{e^{x + x} - 1 \cdot 1}{e^{x} + 1}}{x}\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(\frac{1}{2} + \frac{1}{6} \cdot x\right) + 1\\ \end{array}\]

Reproduce

herbie shell --seed 2019351 
(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))