Average Error: 39.7 → 0.3
Time: 14.3s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.188228550013960803117402109663203191303 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{e^{x \cdot 6} - {1}^{6}}{\left(e^{x} + 1\right) \cdot \left(e^{x \cdot 4} + \left(1 \cdot 1\right) \cdot \left(e^{x + x} + 1 \cdot 1\right)\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{{x}^{2} \cdot \left(\frac{1}{6} \cdot x + \frac{1}{2}\right) + x}{x}\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.188228550013960803117402109663203191303 \cdot 10^{-4}:\\
\;\;\;\;\frac{\frac{e^{x \cdot 6} - {1}^{6}}{\left(e^{x} + 1\right) \cdot \left(e^{x \cdot 4} + \left(1 \cdot 1\right) \cdot \left(e^{x + x} + 1 \cdot 1\right)\right)}}{x}\\

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

\end{array}
double f(double x) {
        double r83144 = x;
        double r83145 = exp(r83144);
        double r83146 = 1.0;
        double r83147 = r83145 - r83146;
        double r83148 = r83147 / r83144;
        return r83148;
}
↓
double f(double x) {
        double r83149 = x;
        double r83150 = -0.00011882285500139608;
        bool r83151 = r83149 <= r83150;
        double r83152 = 6.0;
        double r83153 = r83149 * r83152;
        double r83154 = exp(r83153);
        double r83155 = 1.0;
        double r83156 = pow(r83155, r83152);
        double r83157 = r83154 - r83156;
        double r83158 = exp(r83149);
        double r83159 = r83158 + r83155;
        double r83160 = 4.0;
        double r83161 = r83149 * r83160;
        double r83162 = exp(r83161);
        double r83163 = r83155 * r83155;
        double r83164 = r83149 + r83149;
        double r83165 = exp(r83164);
        double r83166 = r83165 + r83163;
        double r83167 = r83163 * r83166;
        double r83168 = r83162 + r83167;
        double r83169 = r83159 * r83168;
        double r83170 = r83157 / r83169;
        double r83171 = r83170 / r83149;
        double r83172 = 2.0;
        double r83173 = pow(r83149, r83172);
        double r83174 = 0.16666666666666666;
        double r83175 = r83174 * r83149;
        double r83176 = 0.5;
        double r83177 = r83175 + r83176;
        double r83178 = r83173 * r83177;
        double r83179 = r83178 + r83149;
        double r83180 = r83179 / r83149;
        double r83181 = r83151 ? r83171 : r83180;
        return r83181;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.7
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 < -0.00011882285500139608

    1. Initial program 0.0

      \[\frac{e^{x} - 1}{x}\]
    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. Simplified0.0

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

    6. Applied div-inv0.0

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

    8. Applied flip3--0.0

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

    9. Applied frac-times0.0

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

    10. Simplified0.0

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

    11. Simplified0.0

      \[\leadsto \frac{\frac{e^{x \cdot 6} - {1}^{6}}{\color{blue}{\left(e^{x} + 1\right) \cdot \left(e^{x \cdot 4} + \left(1 \cdot 1\right) \cdot \left(e^{x + x} + 1 \cdot 1\right)\right)}}}{x}\]

    if -0.00011882285500139608 < x

    1. Initial program 60.1

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

    2. Taylor expanded around 0 0.5

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

    3. Simplified0.5

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.188228550013960803117402109663203191303 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{e^{x \cdot 6} - {1}^{6}}{\left(e^{x} + 1\right) \cdot \left(e^{x \cdot 4} + \left(1 \cdot 1\right) \cdot \left(e^{x + x} + 1 \cdot 1\right)\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{{x}^{2} \cdot \left(\frac{1}{6} \cdot x + \frac{1}{2}\right) + x}{x}\\ \end{array}\]

Reproduce

herbie shell --seed 2019322 
(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))