Average Error: 39.4 → 0.4
Time: 14.4s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -0.00017573658557392354:\\ \;\;\;\;\frac{1}{\frac{x}{e^{x} - 1}}\\ \mathbf{else}:\\ \;\;\;\;1 + \left(\frac{1}{6} \cdot x + \frac{1}{2}\right) \cdot x\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -0.00017573658557392354:\\
\;\;\;\;\frac{1}{\frac{x}{e^{x} - 1}}\\

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

\end{array}
double f(double x) {
        double r3613374 = x;
        double r3613375 = exp(r3613374);
        double r3613376 = 1.0;
        double r3613377 = r3613375 - r3613376;
        double r3613378 = r3613377 / r3613374;
        return r3613378;
}
↓
double f(double x) {
        double r3613379 = x;
        double r3613380 = -0.00017573658557392354;
        bool r3613381 = r3613379 <= r3613380;
        double r3613382 = 1.0;
        double r3613383 = exp(r3613379);
        double r3613384 = r3613383 - r3613382;
        double r3613385 = r3613379 / r3613384;
        double r3613386 = r3613382 / r3613385;
        double r3613387 = 0.16666666666666666;
        double r3613388 = r3613387 * r3613379;
        double r3613389 = 0.5;
        double r3613390 = r3613388 + r3613389;
        double r3613391 = r3613390 * r3613379;
        double r3613392 = r3613382 + r3613391;
        double r3613393 = r3613381 ? r3613386 : r3613392;
        return r3613393;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.4
Target38.6
Herbie0.4
\[\begin{array}{l} \mathbf{if}\;x \lt 1 \land x \gt -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -0.00017573658557392354

    1. Initial program 0.0

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

    3. Applied clear-num0.0

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

    if -0.00017573658557392354 < x

    1. Initial program 59.9

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

    2. Taylor expanded around 0 0.6

      \[\leadsto \color{blue}{\frac{1}{2} \cdot x + \left(\frac{1}{6} \cdot {x}^{2} + 1\right)}\]

    3. Simplified0.6

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -0.00017573658557392354:\\ \;\;\;\;\frac{1}{\frac{x}{e^{x} - 1}}\\ \mathbf{else}:\\ \;\;\;\;1 + \left(\frac{1}{6} \cdot x + \frac{1}{2}\right) \cdot x\\ \end{array}\]

Reproduce

herbie shell --seed 2019163 
(FPCore (x)
  :name "Kahan's exp quotient"

  :herbie-target
  (if (and (< x 1) (> x -1)) (/ (- (exp x) 1) (log (exp x))) (/ (- (exp x) 1) x))

  (/ (- (exp x) 1) x))