Average Error: 39.6 → 0.3
Time: 4.1s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.2034034236075878 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{\frac{{\left(e^{x \cdot 3}\right)}^{3} - {\left({1}^{3}\right)}^{3}}{\left({\left(e^{x}\right)}^{6} + e^{x \cdot 3} \cdot {1}^{3}\right) + {1}^{6}}}{1 \cdot \left(1 + e^{x}\right) + e^{x + x}}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.2034034236075878 \cdot 10^{-4}:\\
\;\;\;\;\frac{\frac{\frac{{\left(e^{x \cdot 3}\right)}^{3} - {\left({1}^{3}\right)}^{3}}{\left({\left(e^{x}\right)}^{6} + e^{x \cdot 3} \cdot {1}^{3}\right) + {1}^{6}}}{1 \cdot \left(1 + e^{x}\right) + e^{x + x}}}{x}\\

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

\end{array}
double f(double x) {
        double r82407 = x;
        double r82408 = exp(r82407);
        double r82409 = 1.0;
        double r82410 = r82408 - r82409;
        double r82411 = r82410 / r82407;
        return r82411;
}
↓
double f(double x) {
        double r82412 = x;
        double r82413 = -0.00012034034236075878;
        bool r82414 = r82412 <= r82413;
        double r82415 = 3.0;
        double r82416 = r82412 * r82415;
        double r82417 = exp(r82416);
        double r82418 = pow(r82417, r82415);
        double r82419 = 1.0;
        double r82420 = pow(r82419, r82415);
        double r82421 = pow(r82420, r82415);
        double r82422 = r82418 - r82421;
        double r82423 = exp(r82412);
        double r82424 = 6.0;
        double r82425 = pow(r82423, r82424);
        double r82426 = r82417 * r82420;
        double r82427 = r82425 + r82426;
        double r82428 = pow(r82419, r82424);
        double r82429 = r82427 + r82428;
        double r82430 = r82422 / r82429;
        double r82431 = r82419 + r82423;
        double r82432 = r82419 * r82431;
        double r82433 = r82412 + r82412;
        double r82434 = exp(r82433);
        double r82435 = r82432 + r82434;
        double r82436 = r82430 / r82435;
        double r82437 = r82436 / r82412;
        double r82438 = 0.16666666666666666;
        double r82439 = 2.0;
        double r82440 = pow(r82412, r82439);
        double r82441 = r82438 * r82440;
        double r82442 = 0.5;
        double r82443 = r82442 * r82412;
        double r82444 = 1.0;
        double r82445 = r82443 + r82444;
        double r82446 = r82441 + r82445;
        double r82447 = r82414 ? r82437 : r82446;
        return r82447;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.6
Target40.1
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 < -0.00012034034236075878

    1. Initial program 0.1

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

    3. Applied flip3--0.1

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

    4. Simplified0.1

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

    6. Applied pow-exp0.0

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

    8. Applied flip3--0.0

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

    9. Simplified0.0

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

    if -0.00012034034236075878 < x

    1. Initial program 60.0

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

    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

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.2034034236075878 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{\frac{{\left(e^{x \cdot 3}\right)}^{3} - {\left({1}^{3}\right)}^{3}}{\left({\left(e^{x}\right)}^{6} + e^{x \cdot 3} \cdot {1}^{3}\right) + {1}^{6}}}{1 \cdot \left(1 + e^{x}\right) + e^{x + x}}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020056 
(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))