Average Error: 39.8 → 0.3
Time: 3.0s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.16324499920971258 \cdot 10^{-4}:\\ \;\;\;\;\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{\left(1 \cdot \left(1 + e^{x}\right) + e^{x + x}\right) \cdot x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.16324499920971258 \cdot 10^{-4}:\\
\;\;\;\;\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{\left(1 \cdot \left(1 + e^{x}\right) + e^{x + x}\right) \cdot x}\\

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

\end{array}
double f(double x) {
        double r73724 = x;
        double r73725 = exp(r73724);
        double r73726 = 1.0;
        double r73727 = r73725 - r73726;
        double r73728 = r73727 / r73724;
        return r73728;
}
↓
double f(double x) {
        double r73729 = x;
        double r73730 = -0.00011632449992097126;
        bool r73731 = r73729 <= r73730;
        double r73732 = exp(r73729);
        double r73733 = 3.0;
        double r73734 = pow(r73732, r73733);
        double r73735 = 1.0;
        double r73736 = pow(r73735, r73733);
        double r73737 = r73734 - r73736;
        double r73738 = r73735 + r73732;
        double r73739 = r73735 * r73738;
        double r73740 = r73729 + r73729;
        double r73741 = exp(r73740);
        double r73742 = r73739 + r73741;
        double r73743 = r73742 * r73729;
        double r73744 = r73737 / r73743;
        double r73745 = 0.16666666666666666;
        double r73746 = 2.0;
        double r73747 = pow(r73729, r73746);
        double r73748 = r73745 * r73747;
        double r73749 = 0.5;
        double r73750 = r73749 * r73729;
        double r73751 = 1.0;
        double r73752 = r73750 + r73751;
        double r73753 = r73748 + r73752;
        double r73754 = r73731 ? r73744 : r73753;
        return r73754;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.8
Target40.2
Herbie0.3
\[\begin{array}{l} \mathbf{if}\;x \lt 1 \land x \gt -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -0.00011632449992097126

    1. Initial program 0.1

      \[\frac{e^{x} - 1}{x}\]
    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. Applied associate-/l/0.1

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

    5. Simplified0.1

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

    if -0.00011632449992097126 < x

    1. Initial program 60.2

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

    2. Taylor expanded around 0 0.4

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.16324499920971258 \cdot 10^{-4}:\\ \;\;\;\;\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{\left(1 \cdot \left(1 + e^{x}\right) + e^{x + x}\right) \cdot x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020018 
(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))