Average Error: 39.7 → 0.2
Time: 2.4s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.81440388437081299 \cdot 10^{-4}:\\ \;\;\;\;\frac{\sqrt[3]{\log \left(e^{e^{x} - 1}\right)} \cdot \sqrt[3]{\log \left(e^{e^{x} - 1}\right)}}{\frac{x}{\sqrt[3]{e^{x} - 1}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.81440388437081299 \cdot 10^{-4}:\\
\;\;\;\;\frac{\sqrt[3]{\log \left(e^{e^{x} - 1}\right)} \cdot \sqrt[3]{\log \left(e^{e^{x} - 1}\right)}}{\frac{x}{\sqrt[3]{e^{x} - 1}}}\\

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

\end{array}
double code(double x) {
	return ((double) (((double) (((double) exp(x)) - 1.0)) / x));
}
↓
double code(double x) {
	double VAR;
	if ((x <= -0.0001814403884370813)) {
		VAR = ((double) (((double) (((double) cbrt(((double) log(((double) exp(((double) (((double) exp(x)) - 1.0)))))))) * ((double) cbrt(((double) log(((double) exp(((double) (((double) exp(x)) - 1.0)))))))))) / ((double) (x / ((double) cbrt(((double) (((double) exp(x)) - 1.0))))))));
	} else {
		VAR = ((double) (((double) (0.16666666666666666 * ((double) pow(x, 2.0)))) + ((double) (((double) (0.5 * x)) + 1.0))));
	}
	return VAR;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.7
Target40.2
Herbie0.2
\[\begin{array}{l} \mathbf{if}\;x \lt 1 \land x \gt -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -0.0001814403884370813

    1. Initial program 0.1

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

    3. Applied add-log-exp0.1

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

    4. Applied add-log-exp0.1

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

    5. Applied diff-log0.1

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

    6. Simplified0.1

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

    8. Applied add-cube-cbrt0.1

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

    9. Applied associate-/l*0.1

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

    10. Simplified0.1

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

    if -0.0001814403884370813 < x

    1. Initial program 60.2

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

    2. Taylor expanded around 0 0.3

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.81440388437081299 \cdot 10^{-4}:\\ \;\;\;\;\frac{\sqrt[3]{\log \left(e^{e^{x} - 1}\right)} \cdot \sqrt[3]{\log \left(e^{e^{x} - 1}\right)}}{\frac{x}{\sqrt[3]{e^{x} - 1}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020121 
(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))