Average Error: 40.2 → 0.3
Time: 2.7s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.4843984217390087 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{{\left(e^{2}\right)}^{x} - 1 \cdot 1}{e^{x} + 1}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.4843984217390087 \cdot 10^{-4}:\\
\;\;\;\;\frac{\frac{{\left(e^{2}\right)}^{x} - 1 \cdot 1}{e^{x} + 1}}{x}\\

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

\end{array}
double code(double x) {
	return ((exp(x) - 1.0) / x);
}
↓
double code(double x) {
	double VAR;
	if ((x <= -0.00014843984217390087)) {
		VAR = (((pow(exp(2.0), x) - (1.0 * 1.0)) / (exp(x) + 1.0)) / x);
	} else {
		VAR = ((0.16666666666666666 * pow(x, 2.0)) + ((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

Original40.2
Target40.7
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 < -0.00014843984217390087

    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. Using strategy rm

    5. Applied *-un-lft-identity0.1

      \[\leadsto \frac{\frac{e^{x} \cdot e^{\color{blue}{1 \cdot x}} - 1 \cdot 1}{e^{x} + 1}}{x}\]

    6. Applied exp-prod0.1

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

    7. Applied *-un-lft-identity0.1

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

    8. Applied exp-prod0.1

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

    9. Applied pow-prod-down0.0

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

    10. Simplified0.0

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

    if -0.00014843984217390087 < x

    1. Initial program 60.1

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

    2. Taylor expanded around 0 0.5

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.4843984217390087 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{{\left(e^{2}\right)}^{x} - 1 \cdot 1}{e^{x} + 1}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020091 
(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))