Average Error: 39.8 → 0.3
Time: 3.2s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.3427464677546878 \cdot 10^{-4}:\\ \;\;\;\;-1 \cdot \frac{1 - e^{x}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.3427464677546878 \cdot 10^{-4}:\\
\;\;\;\;-1 \cdot \frac{1 - e^{x}}{x}\\

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

\end{array}
double f(double x) {
        double r90016 = x;
        double r90017 = exp(r90016);
        double r90018 = 1.0;
        double r90019 = r90017 - r90018;
        double r90020 = r90019 / r90016;
        return r90020;
}
↓
double f(double x) {
        double r90021 = x;
        double r90022 = -0.00013427464677546878;
        bool r90023 = r90021 <= r90022;
        double r90024 = -1.0;
        double r90025 = 1.0;
        double r90026 = exp(r90021);
        double r90027 = r90025 - r90026;
        double r90028 = r90027 / r90021;
        double r90029 = r90024 * r90028;
        double r90030 = 0.16666666666666666;
        double r90031 = 2.0;
        double r90032 = pow(r90021, r90031);
        double r90033 = r90030 * r90032;
        double r90034 = 0.5;
        double r90035 = r90034 * r90021;
        double r90036 = 1.0;
        double r90037 = r90035 + r90036;
        double r90038 = r90033 + r90037;
        double r90039 = r90023 ? r90029 : r90038;
        return r90039;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.8
Target40.3
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 < -0.00013427464677546878

    1. Initial program 0.1

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

    2. Taylor expanded around -inf 0.1

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

    if -0.00013427464677546878 < 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. Recombined 2 regimes into one program.

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.3427464677546878 \cdot 10^{-4}:\\ \;\;\;\;-1 \cdot \frac{1 - e^{x}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020081 
(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))