Average Error: 39.2 → 0.3
Time: 8.6s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.66490381328880487 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{4} - {1}^{4}}{\mathsf{fma}\left(1, 1, e^{2 \cdot x}\right)}}{x \cdot \left(e^{x} + 1\right)}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(x, \mathsf{fma}\left(x, \frac{1}{6}, \frac{1}{2}\right), 1\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.66490381328880487 \cdot 10^{-4}:\\
\;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{4} - {1}^{4}}{\mathsf{fma}\left(1, 1, e^{2 \cdot x}\right)}}{x \cdot \left(e^{x} + 1\right)}\\

\mathbf{else}:\\
\;\;\;\;\mathsf{fma}\left(x, \mathsf{fma}\left(x, \frac{1}{6}, \frac{1}{2}\right), 1\right)\\

\end{array}
double f(double x) {
        double r120075 = x;
        double r120076 = exp(r120075);
        double r120077 = 1.0;
        double r120078 = r120076 - r120077;
        double r120079 = r120078 / r120075;
        return r120079;
}
↓
double f(double x) {
        double r120080 = x;
        double r120081 = -0.0001664903813288805;
        bool r120082 = r120080 <= r120081;
        double r120083 = exp(r120080);
        double r120084 = 4.0;
        double r120085 = pow(r120083, r120084);
        double r120086 = 1.0;
        double r120087 = pow(r120086, r120084);
        double r120088 = r120085 - r120087;
        double r120089 = 2.0;
        double r120090 = r120089 * r120080;
        double r120091 = exp(r120090);
        double r120092 = fma(r120086, r120086, r120091);
        double r120093 = r120088 / r120092;
        double r120094 = r120083 + r120086;
        double r120095 = r120080 * r120094;
        double r120096 = r120093 / r120095;
        double r120097 = 0.16666666666666666;
        double r120098 = 0.5;
        double r120099 = fma(r120080, r120097, r120098);
        double r120100 = 1.0;
        double r120101 = fma(r120080, r120099, r120100);
        double r120102 = r120082 ? r120096 : r120101;
        return r120102;
}

Error

Bits error versus x

Target

Original39.2
Target39.6
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 < -0.0001664903813288805

    1. Initial program 0.0

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

    3. Applied flip--0.0

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

    4. Applied associate-/l/0.0

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

    6. Applied flip--0.0

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

    7. Simplified0.0

      \[\leadsto \frac{\frac{\color{blue}{{\left(e^{x}\right)}^{4} - {1}^{4}}}{e^{x} \cdot e^{x} + 1 \cdot 1}}{x \cdot \left(e^{x} + 1\right)}\]

    8. Simplified0.0

      \[\leadsto \frac{\frac{{\left(e^{x}\right)}^{4} - {1}^{4}}{\color{blue}{\mathsf{fma}\left(1, 1, e^{2 \cdot x}\right)}}}{x \cdot \left(e^{x} + 1\right)}\]

    if -0.0001664903813288805 < x

    1. Initial program 60.0

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

    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. Simplified0.5

      \[\leadsto \color{blue}{\mathsf{fma}\left(\frac{1}{6}, x \cdot x, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)}\]

    4. Taylor expanded around 0 0.5

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

    5. Simplified0.5

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.66490381328880487 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{4} - {1}^{4}}{\mathsf{fma}\left(1, 1, e^{2 \cdot x}\right)}}{x \cdot \left(e^{x} + 1\right)}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(x, \mathsf{fma}\left(x, \frac{1}{6}, \frac{1}{2}\right), 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020043 +o rules:numerics
(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))