Average Error: 39.8 → 0.3
Time: 2.8s
Precision: binary64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.7550882047531985 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{e^{x} \cdot e^{x} - 1 \cdot 1}{e^{x} + 1}}{x}\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(\frac{1}{2} + x \cdot \frac{1}{6}\right) + 1\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.7550882047531985 \cdot 10^{-4}:\\
\;\;\;\;\frac{\frac{e^{x} \cdot e^{x} - 1 \cdot 1}{e^{x} + 1}}{x}\\

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

\end{array}
double code(double x) {
	return ((double) (((double) (((double) exp(x)) - 1.0)) / x));
}
↓
double code(double x) {
	double VAR;
	if ((x <= -0.00017550882047531985)) {
		VAR = ((double) (((double) (((double) (((double) (((double) exp(x)) * ((double) exp(x)))) - ((double) (1.0 * 1.0)))) / ((double) (((double) exp(x)) + 1.0)))) / x));
	} else {
		VAR = ((double) (((double) (x * ((double) (0.5 + ((double) (x * 0.16666666666666666)))))) + 1.0));
	}
	return VAR;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.8
Target40.3
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 < -1.7550882047531985e-4

    1. Initial program 0.0

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

    3. Applied flip--0.0

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

    if -1.7550882047531985e-4 < 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. Using strategy rm

    4. Applied associate-+r+0.5

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

    5. Simplified0.5

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.7550882047531985 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{e^{x} \cdot e^{x} - 1 \cdot 1}{e^{x} + 1}}{x}\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(\frac{1}{2} + x \cdot \frac{1}{6}\right) + 1\\ \end{array}\]

Reproduce

herbie shell --seed 2020147 
(FPCore (x)
  :name "Kahan's exp quotient"
  :precision binary64

  :herbie-target
  (if (and (< x 1.0) (> x -1.0)) (/ (- (exp x) 1.0) (log (exp x))) (/ (- (exp x) 1.0) x))

  (/ (- (exp x) 1.0) x))