Average Error: 39.6 → 0.4
Time: 14.5s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -0.00019562550875617412:\\ \;\;\;\;\frac{1}{\frac{\left(1 + e^{x}\right) \cdot x}{e^{x} \cdot e^{x} - 1}}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{6} \cdot x + \frac{1}{2}\right) \cdot x + 1\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -0.00019562550875617412:\\
\;\;\;\;\frac{1}{\frac{\left(1 + e^{x}\right) \cdot x}{e^{x} \cdot e^{x} - 1}}\\

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

\end{array}
double f(double x) {
        double r3725503 = x;
        double r3725504 = exp(r3725503);
        double r3725505 = 1.0;
        double r3725506 = r3725504 - r3725505;
        double r3725507 = r3725506 / r3725503;
        return r3725507;
}
↓
double f(double x) {
        double r3725508 = x;
        double r3725509 = -0.00019562550875617412;
        bool r3725510 = r3725508 <= r3725509;
        double r3725511 = 1.0;
        double r3725512 = exp(r3725508);
        double r3725513 = r3725511 + r3725512;
        double r3725514 = r3725513 * r3725508;
        double r3725515 = r3725512 * r3725512;
        double r3725516 = r3725515 - r3725511;
        double r3725517 = r3725514 / r3725516;
        double r3725518 = r3725511 / r3725517;
        double r3725519 = 0.16666666666666666;
        double r3725520 = r3725519 * r3725508;
        double r3725521 = 0.5;
        double r3725522 = r3725520 + r3725521;
        double r3725523 = r3725522 * r3725508;
        double r3725524 = r3725523 + r3725511;
        double r3725525 = r3725510 ? r3725518 : r3725524;
        return r3725525;
}

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 < -0.00019562550875617412

    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. Applied associate-/l/0.1

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

    6. Applied clear-num0.1

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

    if -0.00019562550875617412 < x

    1. Initial program 60.0

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

    2. Taylor expanded around 0 0.5

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

    3. Simplified0.5

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -0.00019562550875617412:\\ \;\;\;\;\frac{1}{\frac{\left(1 + e^{x}\right) \cdot x}{e^{x} \cdot e^{x} - 1}}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{6} \cdot x + \frac{1}{2}\right) \cdot x + 1\\ \end{array}\]

Reproduce

herbie shell --seed 2019162 
(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))