Average Error: 39.4 → 0.4
Time: 14.9s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -0.00017573658557392354:\\ \;\;\;\;\frac{1}{\frac{x}{e^{x} - 1}}\\ \mathbf{else}:\\ \;\;\;\;1 + \left(\frac{1}{6} \cdot x + \frac{1}{2}\right) \cdot x\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -0.00017573658557392354:\\
\;\;\;\;\frac{1}{\frac{x}{e^{x} - 1}}\\

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

\end{array}
double f(double x) {
        double r4029615 = x;
        double r4029616 = exp(r4029615);
        double r4029617 = 1.0;
        double r4029618 = r4029616 - r4029617;
        double r4029619 = r4029618 / r4029615;
        return r4029619;
}
↓
double f(double x) {
        double r4029620 = x;
        double r4029621 = -0.00017573658557392354;
        bool r4029622 = r4029620 <= r4029621;
        double r4029623 = 1.0;
        double r4029624 = exp(r4029620);
        double r4029625 = r4029624 - r4029623;
        double r4029626 = r4029620 / r4029625;
        double r4029627 = r4029623 / r4029626;
        double r4029628 = 0.16666666666666666;
        double r4029629 = r4029628 * r4029620;
        double r4029630 = 0.5;
        double r4029631 = r4029629 + r4029630;
        double r4029632 = r4029631 * r4029620;
        double r4029633 = r4029623 + r4029632;
        double r4029634 = r4029622 ? r4029627 : r4029633;
        return r4029634;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.4
Target38.6
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 < -0.00017573658557392354

    1. Initial program 0.0

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

    3. Applied clear-num0.0

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

    if -0.00017573658557392354 < x

    1. Initial program 59.9

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

    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

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -0.00017573658557392354:\\ \;\;\;\;\frac{1}{\frac{x}{e^{x} - 1}}\\ \mathbf{else}:\\ \;\;\;\;1 + \left(\frac{1}{6} \cdot x + \frac{1}{2}\right) \cdot x\\ \end{array}\]

Reproduce

herbie shell --seed 2019163 
(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))