Average Error: 39.9 → 0.3
Time: 13.2s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -0.00010944133059785852:\\ \;\;\;\;\frac{\frac{-1 + e^{\left(\left(x + x\right) + x\right) + \left(\left(x + x\right) + x\right)}}{\left(1 + e^{x + x}\right) + e^{x + x} \cdot e^{x + x}}}{x \cdot \left(1 + e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -0.00010944133059785852:\\
\;\;\;\;\frac{\frac{-1 + e^{\left(\left(x + x\right) + x\right) + \left(\left(x + x\right) + x\right)}}{\left(1 + e^{x + x}\right) + e^{x + x} \cdot e^{x + x}}}{x \cdot \left(1 + e^{x}\right)}\\

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

\end{array}
double f(double x) {
        double r3545371 = x;
        double r3545372 = exp(r3545371);
        double r3545373 = 1.0;
        double r3545374 = r3545372 - r3545373;
        double r3545375 = r3545374 / r3545371;
        return r3545375;
}
↓
double f(double x) {
        double r3545376 = x;
        double r3545377 = -0.00010944133059785852;
        bool r3545378 = r3545376 <= r3545377;
        double r3545379 = -1.0;
        double r3545380 = r3545376 + r3545376;
        double r3545381 = r3545380 + r3545376;
        double r3545382 = r3545381 + r3545381;
        double r3545383 = exp(r3545382);
        double r3545384 = r3545379 + r3545383;
        double r3545385 = 1.0;
        double r3545386 = exp(r3545380);
        double r3545387 = r3545385 + r3545386;
        double r3545388 = r3545386 * r3545386;
        double r3545389 = r3545387 + r3545388;
        double r3545390 = r3545384 / r3545389;
        double r3545391 = exp(r3545376);
        double r3545392 = r3545385 + r3545391;
        double r3545393 = r3545376 * r3545392;
        double r3545394 = r3545390 / r3545393;
        double r3545395 = 0.16666666666666666;
        double r3545396 = r3545376 * r3545395;
        double r3545397 = 0.5;
        double r3545398 = r3545396 + r3545397;
        double r3545399 = r3545376 * r3545398;
        double r3545400 = r3545385 + r3545399;
        double r3545401 = r3545378 ? r3545394 : r3545400;
        return r3545401;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.9
Target39.1
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 < -0.00010944133059785852

    1. Initial program 0.1

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

    3. Applied flip--0.1

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

    4. Applied associate-/l/0.1

      \[\leadsto \color{blue}{\frac{e^{x} \cdot e^{x} - 1 \cdot 1}{x \cdot \left(e^{x} + 1\right)}}\]
    5. Using strategy rm

    6. Applied add-exp-log0.1

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

    7. Simplified0.1

      \[\leadsto \frac{e^{\color{blue}{x + x}} - 1 \cdot 1}{x \cdot \left(e^{x} + 1\right)}\]
    8. Using strategy rm

    9. Applied flip3--0.1

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

    10. Simplified0.0

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

    11. Simplified0.0

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

    if -0.00010944133059785852 < x

    1. Initial program 60.2

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

    2. Taylor expanded around 0 0.4

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

    3. Simplified0.4

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -0.00010944133059785852:\\ \;\;\;\;\frac{\frac{-1 + e^{\left(\left(x + x\right) + x\right) + \left(\left(x + x\right) + x\right)}}{\left(1 + e^{x + x}\right) + e^{x + x} \cdot e^{x + x}}}{x \cdot \left(1 + e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019130 
(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))