Average Error: 39.7 → 0.4
Time: 4.2s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.5819896650017998 \cdot 10^{-4}:\\ \;\;\;\;\frac{\log \left(e^{e^{x} - 1}\right)}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{{\left(\frac{1}{6} \cdot {x}^{2}\right)}^{3} + {\left(\frac{1}{2} \cdot x + 1\right)}^{3}}{\left(\frac{1}{2} \cdot x + 1\right) \cdot \left(\left(\frac{1}{2} \cdot x + 1\right) - \frac{1}{6} \cdot {x}^{2}\right) + \frac{1}{36} \cdot \left({x}^{2} \cdot {x}^{2}\right)}\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.5819896650017998 \cdot 10^{-4}:\\
\;\;\;\;\frac{\log \left(e^{e^{x} - 1}\right)}{x}\\

\mathbf{else}:\\
\;\;\;\;\frac{{\left(\frac{1}{6} \cdot {x}^{2}\right)}^{3} + {\left(\frac{1}{2} \cdot x + 1\right)}^{3}}{\left(\frac{1}{2} \cdot x + 1\right) \cdot \left(\left(\frac{1}{2} \cdot x + 1\right) - \frac{1}{6} \cdot {x}^{2}\right) + \frac{1}{36} \cdot \left({x}^{2} \cdot {x}^{2}\right)}\\

\end{array}
double f(double x) {
        double r109508 = x;
        double r109509 = exp(r109508);
        double r109510 = 1.0;
        double r109511 = r109509 - r109510;
        double r109512 = r109511 / r109508;
        return r109512;
}
↓
double f(double x) {
        double r109513 = x;
        double r109514 = -0.00015819896650017998;
        bool r109515 = r109513 <= r109514;
        double r109516 = exp(r109513);
        double r109517 = 1.0;
        double r109518 = r109516 - r109517;
        double r109519 = exp(r109518);
        double r109520 = log(r109519);
        double r109521 = r109520 / r109513;
        double r109522 = 0.16666666666666666;
        double r109523 = 2.0;
        double r109524 = pow(r109513, r109523);
        double r109525 = r109522 * r109524;
        double r109526 = 3.0;
        double r109527 = pow(r109525, r109526);
        double r109528 = 0.5;
        double r109529 = r109528 * r109513;
        double r109530 = 1.0;
        double r109531 = r109529 + r109530;
        double r109532 = pow(r109531, r109526);
        double r109533 = r109527 + r109532;
        double r109534 = r109531 - r109525;
        double r109535 = r109531 * r109534;
        double r109536 = 0.027777777777777776;
        double r109537 = r109524 * r109524;
        double r109538 = r109536 * r109537;
        double r109539 = r109535 + r109538;
        double r109540 = r109533 / r109539;
        double r109541 = r109515 ? r109521 : r109540;
        return r109541;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.7
Target40.0
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 < -0.00015819896650017998

    1. Initial program 0.0

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

    3. Applied add-log-exp0.0

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

    4. Applied add-log-exp0.0

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

    5. Applied diff-log0.0

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

    6. Simplified0.0

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

    if -0.00015819896650017998 < x

    1. Initial program 59.8

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

    2. Taylor expanded around 0 0.6

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

    4. Applied flip3-+0.6

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

    5. Simplified0.6

      \[\leadsto \frac{{\left(\frac{1}{6} \cdot {x}^{2}\right)}^{3} + {\left(\frac{1}{2} \cdot x + 1\right)}^{3}}{\color{blue}{\left(\frac{1}{2} \cdot x + 1\right) \cdot \left(\left(\frac{1}{2} \cdot x + 1\right) - \frac{1}{6} \cdot {x}^{2}\right) + \frac{1}{36} \cdot \left({x}^{2} \cdot {x}^{2}\right)}}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.5819896650017998 \cdot 10^{-4}:\\ \;\;\;\;\frac{\log \left(e^{e^{x} - 1}\right)}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{{\left(\frac{1}{6} \cdot {x}^{2}\right)}^{3} + {\left(\frac{1}{2} \cdot x + 1\right)}^{3}}{\left(\frac{1}{2} \cdot x + 1\right) \cdot \left(\left(\frac{1}{2} \cdot x + 1\right) - \frac{1}{6} \cdot {x}^{2}\right) + \frac{1}{36} \cdot \left({x}^{2} \cdot {x}^{2}\right)}\\ \end{array}\]

Reproduce

herbie shell --seed 2020034 
(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))