Average Error: 39.6 → 0.4
Time: 17.8s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -0.00013964996942021534:\\ \;\;\;\;\frac{e^{\left(x + x\right) + x} + -1}{x \cdot \left(\left(e^{x} + 1\right) + e^{x} \cdot e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(\frac{1}{2} + x \cdot \frac{1}{6}\right) + 1\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -0.00013964996942021534:\\
\;\;\;\;\frac{e^{\left(x + x\right) + x} + -1}{x \cdot \left(\left(e^{x} + 1\right) + e^{x} \cdot e^{x}\right)}\\

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

\end{array}
double f(double x) {
        double r11924385 = x;
        double r11924386 = exp(r11924385);
        double r11924387 = 1.0;
        double r11924388 = r11924386 - r11924387;
        double r11924389 = r11924388 / r11924385;
        return r11924389;
}
↓
double f(double x) {
        double r11924390 = x;
        double r11924391 = -0.00013964996942021534;
        bool r11924392 = r11924390 <= r11924391;
        double r11924393 = r11924390 + r11924390;
        double r11924394 = r11924393 + r11924390;
        double r11924395 = exp(r11924394);
        double r11924396 = -1.0;
        double r11924397 = r11924395 + r11924396;
        double r11924398 = exp(r11924390);
        double r11924399 = 1.0;
        double r11924400 = r11924398 + r11924399;
        double r11924401 = r11924398 * r11924398;
        double r11924402 = r11924400 + r11924401;
        double r11924403 = r11924390 * r11924402;
        double r11924404 = r11924397 / r11924403;
        double r11924405 = 0.5;
        double r11924406 = 0.16666666666666666;
        double r11924407 = r11924390 * r11924406;
        double r11924408 = r11924405 + r11924407;
        double r11924409 = r11924390 * r11924408;
        double r11924410 = r11924409 + r11924399;
        double r11924411 = r11924392 ? r11924404 : r11924410;
        return r11924411;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.6
Target38.8
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 < -0.00013964996942021534

    1. Initial program 0.0

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

    3. Applied flip3--0.0

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

    4. Applied associate-/l/0.0

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

    5. Simplified0.0

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

    if -0.00013964996942021534 < x

    1. Initial program 60.0

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

    2. Taylor expanded around 0 0.6

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

    3. Simplified0.6

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -0.00013964996942021534:\\ \;\;\;\;\frac{e^{\left(x + x\right) + x} + -1}{x \cdot \left(\left(e^{x} + 1\right) + e^{x} \cdot e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(\frac{1}{2} + x \cdot \frac{1}{6}\right) + 1\\ \end{array}\]

Reproduce

herbie shell --seed 2019107 
(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))