Average Error: 39.8 → 0.4
Time: 2.2s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.9141971874233585 \cdot 10^{-4}:\\ \;\;\;\;\frac{\log \left(e^{e^{x} - 1}\right)}{x}\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(\frac{1}{2} + x \cdot \frac{1}{6}\right) + 1\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.9141971874233585 \cdot 10^{-4}:\\
\;\;\;\;\frac{\log \left(e^{e^{x} - 1}\right)}{x}\\

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

\end{array}
double f(double x) {
        double r60579 = x;
        double r60580 = exp(r60579);
        double r60581 = 1.0;
        double r60582 = r60580 - r60581;
        double r60583 = r60582 / r60579;
        return r60583;
}
↓
double f(double x) {
        double r60584 = x;
        double r60585 = -0.00019141971874233585;
        bool r60586 = r60584 <= r60585;
        double r60587 = exp(r60584);
        double r60588 = 1.0;
        double r60589 = r60587 - r60588;
        double r60590 = exp(r60589);
        double r60591 = log(r60590);
        double r60592 = r60591 / r60584;
        double r60593 = 0.5;
        double r60594 = 0.16666666666666666;
        double r60595 = r60584 * r60594;
        double r60596 = r60593 + r60595;
        double r60597 = r60584 * r60596;
        double r60598 = 1.0;
        double r60599 = r60597 + r60598;
        double r60600 = r60586 ? r60592 : r60599;
        return r60600;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.8
Target40.2
Herbie0.4
\[\begin{array}{l} \mathbf{if}\;x \lt 1 \land x \gt -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -0.00019141971874233585

    1. Initial program 0.0

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

    3. Applied add-log-exp0.0

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

    4. Applied add-log-exp0.1

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

    5. Applied diff-log0.1

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

    6. Simplified0.1

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

    if -0.00019141971874233585 < x

    1. Initial program 60.1

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

    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. Using strategy rm

    4. Applied associate-+r+0.5

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

    5. Simplified0.5

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.9141971874233585 \cdot 10^{-4}:\\ \;\;\;\;\frac{\log \left(e^{e^{x} - 1}\right)}{x}\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(\frac{1}{2} + x \cdot \frac{1}{6}\right) + 1\\ \end{array}\]

Reproduce

herbie shell --seed 2020060 
(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))