Average Error: 39.8 → 0.4
Time: 17.9s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -9.443138140720151 \cdot 10^{-05}:\\ \;\;\;\;\frac{\frac{-1 + {\left(e^{3 \cdot x}\right)}^{3}}{e^{x} \cdot \left(\left(e^{3 \cdot x} \cdot e^{3 \cdot x} + \left(1 + e^{3 \cdot x}\right)\right) \cdot \left(1 + e^{x}\right)\right) + \left(e^{3 \cdot x} \cdot e^{3 \cdot x} + \left(1 + e^{3 \cdot x}\right)\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;\log \left(e^{x \cdot \left(\frac{1}{2} + \frac{1}{6} \cdot x\right)}\right) + 1\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -9.443138140720151 \cdot 10^{-05}:\\
\;\;\;\;\frac{\frac{-1 + {\left(e^{3 \cdot x}\right)}^{3}}{e^{x} \cdot \left(\left(e^{3 \cdot x} \cdot e^{3 \cdot x} + \left(1 + e^{3 \cdot x}\right)\right) \cdot \left(1 + e^{x}\right)\right) + \left(e^{3 \cdot x} \cdot e^{3 \cdot x} + \left(1 + e^{3 \cdot x}\right)\right)}}{x}\\

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

\end{array}
double f(double x) {
        double r3897594 = x;
        double r3897595 = exp(r3897594);
        double r3897596 = 1.0;
        double r3897597 = r3897595 - r3897596;
        double r3897598 = r3897597 / r3897594;
        return r3897598;
}
↓
double f(double x) {
        double r3897599 = x;
        double r3897600 = -9.443138140720151e-05;
        bool r3897601 = r3897599 <= r3897600;
        double r3897602 = -1.0;
        double r3897603 = 3.0;
        double r3897604 = r3897603 * r3897599;
        double r3897605 = exp(r3897604);
        double r3897606 = pow(r3897605, r3897603);
        double r3897607 = r3897602 + r3897606;
        double r3897608 = exp(r3897599);
        double r3897609 = r3897605 * r3897605;
        double r3897610 = 1.0;
        double r3897611 = r3897610 + r3897605;
        double r3897612 = r3897609 + r3897611;
        double r3897613 = r3897610 + r3897608;
        double r3897614 = r3897612 * r3897613;
        double r3897615 = r3897608 * r3897614;
        double r3897616 = r3897615 + r3897612;
        double r3897617 = r3897607 / r3897616;
        double r3897618 = r3897617 / r3897599;
        double r3897619 = 0.5;
        double r3897620 = 0.16666666666666666;
        double r3897621 = r3897620 * r3897599;
        double r3897622 = r3897619 + r3897621;
        double r3897623 = r3897599 * r3897622;
        double r3897624 = exp(r3897623);
        double r3897625 = log(r3897624);
        double r3897626 = r3897625 + r3897610;
        double r3897627 = r3897601 ? r3897618 : r3897626;
        return r3897627;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.8
Target38.9
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 < -9.443138140720151e-05

    1. Initial program 0.0

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

    3. Applied flip3--0.1

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

    4. Simplified0.0

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

    5. Simplified0.0

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

    7. Applied flip3-+0.0

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

    8. Applied associate-/l/0.0

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

    9. Simplified0.0

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

    if -9.443138140720151e-05 < 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

      \[\leadsto \color{blue}{1 + x \cdot \left(\frac{1}{6} \cdot x + \frac{1}{2}\right)}\]
    4. Using strategy rm

    5. Applied add-log-exp0.5

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -9.443138140720151 \cdot 10^{-05}:\\ \;\;\;\;\frac{\frac{-1 + {\left(e^{3 \cdot x}\right)}^{3}}{e^{x} \cdot \left(\left(e^{3 \cdot x} \cdot e^{3 \cdot x} + \left(1 + e^{3 \cdot x}\right)\right) \cdot \left(1 + e^{x}\right)\right) + \left(e^{3 \cdot x} \cdot e^{3 \cdot x} + \left(1 + e^{3 \cdot x}\right)\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;\log \left(e^{x \cdot \left(\frac{1}{2} + \frac{1}{6} \cdot x\right)}\right) + 1\\ \end{array}\]

Reproduce

herbie shell --seed 2019151 
(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))