Average Error: 40.2 → 0.3
Time: 2.3s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.55293744862148258 \cdot 10^{-4}:\\ \;\;\;\;-1 \cdot \frac{1 - e^{x}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.55293744862148258 \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 r95302 = x;
        double r95303 = exp(r95302);
        double r95304 = 1.0;
        double r95305 = r95303 - r95304;
        double r95306 = r95305 / r95302;
        return r95306;
}
↓
double f(double x) {
        double r95307 = x;
        double r95308 = -0.00015529374486214826;
        bool r95309 = r95307 <= r95308;
        double r95310 = -1.0;
        double r95311 = 1.0;
        double r95312 = exp(r95307);
        double r95313 = r95311 - r95312;
        double r95314 = r95313 / r95307;
        double r95315 = r95310 * r95314;
        double r95316 = 0.16666666666666666;
        double r95317 = 2.0;
        double r95318 = pow(r95307, r95317);
        double r95319 = r95316 * r95318;
        double r95320 = 0.5;
        double r95321 = r95320 * r95307;
        double r95322 = 1.0;
        double r95323 = r95321 + r95322;
        double r95324 = r95319 + r95323;
        double r95325 = r95309 ? r95315 : r95324;
        return r95325;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original40.2
Target40.7
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 < -0.00015529374486214826

    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 -0.00015529374486214826 < x

    1. Initial program 60.2

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

    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. Recombined 2 regimes into one program.

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.55293744862148258 \cdot 10^{-4}:\\ \;\;\;\;-1 \cdot \frac{1 - e^{x}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020003 
(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))