Average Error: 40.0 → 0.3
Time: 8.6s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.6604344095812133123325760042376941783 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{{1}^{3}}{\frac{\mathsf{fma}\left(1, e^{x} + 1, e^{x + x}\right)}{{\left(e^{x}\right)}^{3} - {1}^{3}}}}{x}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.6604344095812133123325760042376941783 \cdot 10^{-4}:\\
\;\;\;\;\frac{\frac{{1}^{3}}{\frac{\mathsf{fma}\left(1, e^{x} + 1, e^{x + x}\right)}{{\left(e^{x}\right)}^{3} - {1}^{3}}}}{x}\\

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

\end{array}
double f(double x) {
        double r116974 = x;
        double r116975 = exp(r116974);
        double r116976 = 1.0;
        double r116977 = r116975 - r116976;
        double r116978 = r116977 / r116974;
        return r116978;
}
↓
double f(double x) {
        double r116979 = x;
        double r116980 = -0.00016604344095812133;
        bool r116981 = r116979 <= r116980;
        double r116982 = 1.0;
        double r116983 = 3.0;
        double r116984 = pow(r116982, r116983);
        double r116985 = 1.0;
        double r116986 = exp(r116979);
        double r116987 = r116986 + r116985;
        double r116988 = r116979 + r116979;
        double r116989 = exp(r116988);
        double r116990 = fma(r116985, r116987, r116989);
        double r116991 = pow(r116986, r116983);
        double r116992 = pow(r116985, r116983);
        double r116993 = r116991 - r116992;
        double r116994 = r116990 / r116993;
        double r116995 = r116984 / r116994;
        double r116996 = r116995 / r116979;
        double r116997 = 0.16666666666666666;
        double r116998 = 2.0;
        double r116999 = pow(r116979, r116998);
        double r117000 = 0.5;
        double r117001 = fma(r117000, r116979, r116982);
        double r117002 = fma(r116997, r116999, r117001);
        double r117003 = r116981 ? r116996 : r117002;
        return r117003;
}

Error

Bits error versus x

Target

Original40.0
Target40.4
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 < -0.00016604344095812133

    1. Initial program 0.0

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

    3. Applied flip3--0.0

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

    4. Simplified0.0

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

    6. Applied *-un-lft-identity0.0

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

    7. Applied unpow-prod-down0.0

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

    8. Applied *-un-lft-identity0.0

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

    9. Applied unpow-prod-down0.0

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

    10. Applied distribute-lft-out--0.0

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

    11. Applied associate-/l*0.0

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

    if -0.00016604344095812133 < x

    1. Initial program 60.3

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

    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. Simplified0.4

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.6604344095812133123325760042376941783 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{{1}^{3}}{\frac{\mathsf{fma}\left(1, e^{x} + 1, e^{x + x}\right)}{{\left(e^{x}\right)}^{3} - {1}^{3}}}}{x}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019353 +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))