Average Error: 40.0 → 0.4
Time: 3.2s
Precision: binary64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.76806490805435493 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{1 \cdot \left(1 + e^{x}\right) + e^{x + x}}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{{x}^{2} \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right) + x}{x}\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.76806490805435493 \cdot 10^{-4}:\\
\;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{1 \cdot \left(1 + e^{x}\right) + e^{x + x}}}{x}\\

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

\end{array}
double code(double x) {
	return ((double) (((double) (((double) exp(x)) - 1.0)) / x));
}
↓
double code(double x) {
	double VAR;
	if ((x <= -0.0001768064908054355)) {
		VAR = ((double) (((double) (((double) (((double) pow(((double) exp(x)), 3.0)) - ((double) pow(1.0, 3.0)))) / ((double) (((double) (1.0 * ((double) (1.0 + ((double) exp(x)))))) + ((double) exp(((double) (x + x)))))))) / x));
	} else {
		VAR = ((double) (((double) (((double) (((double) pow(x, 2.0)) * ((double) (((double) (x * 0.16666666666666666)) + 0.5)))) + x)) / x));
	}
	return VAR;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original40.0
Target40.4
Herbie0.4
\[\begin{array}{l} \mathbf{if}\;x \lt 1 \land x \gt -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -1.76806490805435493e-4

    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

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

    if -1.76806490805435493e-4 < x

    1. Initial program 59.8

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

    2. Taylor expanded around 0 0.6

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

    3. Simplified0.6

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.76806490805435493 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{1 \cdot \left(1 + e^{x}\right) + e^{x + x}}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{{x}^{2} \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right) + x}{x}\\ \end{array}\]

Reproduce

herbie shell --seed 2020162 
(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))