Average Error: 40.0 → 0.3
Time: 3.3s
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 r94328 = x;
        double r94329 = exp(r94328);
        double r94330 = 1.0;
        double r94331 = r94329 - r94330;
        double r94332 = r94331 / r94328;
        return r94332;
}
↓
double f(double x) {
        double r94333 = x;
        double r94334 = -9.202491378555593e-05;
        bool r94335 = r94333 <= r94334;
        double r94336 = exp(r94333);
        double r94337 = 3.0;
        double r94338 = pow(r94336, r94337);
        double r94339 = r94338 * r94338;
        double r94340 = 1.0;
        double r94341 = pow(r94340, r94337);
        double r94342 = r94341 * r94341;
        double r94343 = r94339 - r94342;
        double r94344 = r94340 + r94336;
        double r94345 = r94340 * r94344;
        double r94346 = r94333 + r94333;
        double r94347 = exp(r94346);
        double r94348 = r94345 + r94347;
        double r94349 = r94338 + r94341;
        double r94350 = r94348 * r94349;
        double r94351 = r94343 / r94350;
        double r94352 = r94351 / r94333;
        double r94353 = 0.16666666666666666;
        double r94354 = 2.0;
        double r94355 = pow(r94333, r94354);
        double r94356 = r94353 * r94355;
        double r94357 = 0.5;
        double r94358 = r94357 * r94333;
        double r94359 = 1.0;
        double r94360 = r94358 + r94359;
        double r94361 = r94356 + r94360;
        double r94362 = r94335 ? r94352 : r94361;
        return r94362;
}

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))