Average Error: 40.0 → 0.3
Time: 4.3s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -9.202491378555592886016173403973539279832 \cdot 10^{-5}:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{3} \cdot {\left(e^{x}\right)}^{3} - {1}^{3} \cdot {1}^{3}}{\mathsf{fma}\left(1, e^{x} + 1, e^{x + x}\right) \cdot \left({\left(e^{x}\right)}^{3} + {1}^{3}\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -9.202491378555592886016173403973539279832 \cdot 10^{-5}:\\
\;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{3} \cdot {\left(e^{x}\right)}^{3} - {1}^{3} \cdot {1}^{3}}{\mathsf{fma}\left(1, e^{x} + 1, e^{x + x}\right) \cdot \left({\left(e^{x}\right)}^{3} + {1}^{3}\right)}}{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 r83122 = x;
        double r83123 = exp(r83122);
        double r83124 = 1.0;
        double r83125 = r83123 - r83124;
        double r83126 = r83125 / r83122;
        return r83126;
}
↓
double f(double x) {
        double r83127 = x;
        double r83128 = -9.202491378555593e-05;
        bool r83129 = r83127 <= r83128;
        double r83130 = exp(r83127);
        double r83131 = 3.0;
        double r83132 = pow(r83130, r83131);
        double r83133 = r83132 * r83132;
        double r83134 = 1.0;
        double r83135 = pow(r83134, r83131);
        double r83136 = r83135 * r83135;
        double r83137 = r83133 - r83136;
        double r83138 = r83130 + r83134;
        double r83139 = r83127 + r83127;
        double r83140 = exp(r83139);
        double r83141 = fma(r83134, r83138, r83140);
        double r83142 = r83132 + r83135;
        double r83143 = r83141 * r83142;
        double r83144 = r83137 / r83143;
        double r83145 = r83144 / r83127;
        double r83146 = 0.16666666666666666;
        double r83147 = 2.0;
        double r83148 = pow(r83127, r83147);
        double r83149 = 0.5;
        double r83150 = 1.0;
        double r83151 = fma(r83149, r83127, r83150);
        double r83152 = fma(r83146, r83148, r83151);
        double r83153 = r83129 ? r83145 : r83152;
        return r83153;
}

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 < -9.202491378555593e-05

    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 flip--0.0

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

    7. Applied associate-/l/0.0

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

    if -9.202491378555593e-05 < 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

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -9.202491378555592886016173403973539279832 \cdot 10^{-5}:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{3} \cdot {\left(e^{x}\right)}^{3} - {1}^{3} \cdot {1}^{3}}{\mathsf{fma}\left(1, e^{x} + 1, e^{x + x}\right) \cdot \left({\left(e^{x}\right)}^{3} + {1}^{3}\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)\\ \end{array}\]

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))