Average Error: 40.0 → 0.3
Time: 3.2s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -9.202491378555592886016173403973539279832 \cdot 10^{-5}:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{3} \cdot {\left(e^{x}\right)}^{3} - {1}^{3} \cdot {1}^{3}}{\left(1 \cdot \left(1 + e^{x}\right) + e^{x + x}\right) \cdot \left({\left(e^{x}\right)}^{3} + {1}^{3}\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -9.202491378555592886016173403973539279832 \cdot 10^{-5}:\\
\;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{3} \cdot {\left(e^{x}\right)}^{3} - {1}^{3} \cdot {1}^{3}}{\left(1 \cdot \left(1 + e^{x}\right) + e^{x + x}\right) \cdot \left({\left(e^{x}\right)}^{3} + {1}^{3}\right)}}{x}\\

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

\end{array}
double f(double x) {
        double r87538 = x;
        double r87539 = exp(r87538);
        double r87540 = 1.0;
        double r87541 = r87539 - r87540;
        double r87542 = r87541 / r87538;
        return r87542;
}
↓
double f(double x) {
        double r87543 = x;
        double r87544 = -9.202491378555593e-05;
        bool r87545 = r87543 <= r87544;
        double r87546 = exp(r87543);
        double r87547 = 3.0;
        double r87548 = pow(r87546, r87547);
        double r87549 = r87548 * r87548;
        double r87550 = 1.0;
        double r87551 = pow(r87550, r87547);
        double r87552 = r87551 * r87551;
        double r87553 = r87549 - r87552;
        double r87554 = r87550 + r87546;
        double r87555 = r87550 * r87554;
        double r87556 = r87543 + r87543;
        double r87557 = exp(r87556);
        double r87558 = r87555 + r87557;
        double r87559 = r87548 + r87551;
        double r87560 = r87558 * r87559;
        double r87561 = r87553 / r87560;
        double r87562 = r87561 / r87543;
        double r87563 = 0.16666666666666666;
        double r87564 = 2.0;
        double r87565 = pow(r87543, r87564);
        double r87566 = r87563 * r87565;
        double r87567 = 0.5;
        double r87568 = r87567 * r87543;
        double r87569 = 1.0;
        double r87570 = r87568 + r87569;
        double r87571 = r87566 + r87570;
        double r87572 = r87545 ? r87562 : r87571;
        return r87572;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original40.0
Target40.4
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 < -9.202491378555593e-05

    1. Initial program 0.0

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

    3. Applied flip3--0.0

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

    4. Simplified0.0

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

    6. Applied flip--0.0

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

    7. Applied associate-/l/0.0

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

    if -9.202491378555593e-05 < 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

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -9.202491378555592886016173403973539279832 \cdot 10^{-5}:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{3} \cdot {\left(e^{x}\right)}^{3} - {1}^{3} \cdot {1}^{3}}{\left(1 \cdot \left(1 + e^{x}\right) + e^{x + x}\right) \cdot \left({\left(e^{x}\right)}^{3} + {1}^{3}\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019353 
(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))