Average Error: 39.8 → 0.3
Time: 2.5s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.16324499920971258 \cdot 10^{-4}:\\ \;\;\;\;\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{x \cdot \mathsf{fma}\left(1, e^{x} + 1, e^{x + x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{6}, {x}^{2}, \mathsf{fma}\left(\frac{1}{2}, x, 1\right)\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.16324499920971258 \cdot 10^{-4}:\\
\;\;\;\;\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{x \cdot \mathsf{fma}\left(1, e^{x} + 1, e^{x + x}\right)}\\

\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 r64474 = x;
        double r64475 = exp(r64474);
        double r64476 = 1.0;
        double r64477 = r64475 - r64476;
        double r64478 = r64477 / r64474;
        return r64478;
}
↓
double f(double x) {
        double r64479 = x;
        double r64480 = -0.00011632449992097126;
        bool r64481 = r64479 <= r64480;
        double r64482 = exp(r64479);
        double r64483 = 3.0;
        double r64484 = pow(r64482, r64483);
        double r64485 = 1.0;
        double r64486 = pow(r64485, r64483);
        double r64487 = r64484 - r64486;
        double r64488 = r64482 + r64485;
        double r64489 = r64479 + r64479;
        double r64490 = exp(r64489);
        double r64491 = fma(r64485, r64488, r64490);
        double r64492 = r64479 * r64491;
        double r64493 = r64487 / r64492;
        double r64494 = 0.16666666666666666;
        double r64495 = 2.0;
        double r64496 = pow(r64479, r64495);
        double r64497 = 0.5;
        double r64498 = 1.0;
        double r64499 = fma(r64497, r64479, r64498);
        double r64500 = fma(r64494, r64496, r64499);
        double r64501 = r64481 ? r64493 : r64500;
        return r64501;
}

Error

Bits error versus x

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

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

    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. 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

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

Reproduce

herbie shell --seed 2020018 +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))