Average Error: 40.1 → 0.3
Time: 2.5s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.34542002814569053 \cdot 10^{-4}:\\ \;\;\;\;\left(e^{x} - 1\right) \cdot \frac{1}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{{x}^{2} \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right) + x}{x}\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.34542002814569053 \cdot 10^{-4}:\\
\;\;\;\;\left(e^{x} - 1\right) \cdot \frac{1}{x}\\

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

\end{array}
double f(double x) {
        double r66420 = x;
        double r66421 = exp(r66420);
        double r66422 = 1.0;
        double r66423 = r66421 - r66422;
        double r66424 = r66423 / r66420;
        return r66424;
}
↓
double f(double x) {
        double r66425 = x;
        double r66426 = -0.00013454200281456905;
        bool r66427 = r66425 <= r66426;
        double r66428 = exp(r66425);
        double r66429 = 1.0;
        double r66430 = r66428 - r66429;
        double r66431 = 1.0;
        double r66432 = r66431 / r66425;
        double r66433 = r66430 * r66432;
        double r66434 = 2.0;
        double r66435 = pow(r66425, r66434);
        double r66436 = 0.16666666666666666;
        double r66437 = r66425 * r66436;
        double r66438 = 0.5;
        double r66439 = r66437 + r66438;
        double r66440 = r66435 * r66439;
        double r66441 = r66440 + r66425;
        double r66442 = r66441 / r66425;
        double r66443 = r66427 ? r66433 : r66442;
        return r66443;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original40.1
Target40.6
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 < -0.00013454200281456905

    1. Initial program 0.0

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

    3. Applied div-inv0.0

      \[\leadsto \color{blue}{\left(e^{x} - 1\right) \cdot \frac{1}{x}}\]

    if -0.00013454200281456905 < x

    1. Initial program 60.3

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

    2. Taylor expanded around 0 0.4

      \[\leadsto \frac{\color{blue}{\frac{1}{2} \cdot {x}^{2} + \left(\frac{1}{6} \cdot {x}^{3} + x\right)}}{x}\]

    3. Simplified0.4

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.34542002814569053 \cdot 10^{-4}:\\ \;\;\;\;\left(e^{x} - 1\right) \cdot \frac{1}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{{x}^{2} \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right) + x}{x}\\ \end{array}\]

Reproduce

herbie shell --seed 2020059 
(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))