Average Error: 40.2 → 0.3
Time: 2.9s
Precision: binary64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.51964413078298249 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{1 \cdot \left(1 + e^{x}\right) + e^{x + x}}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.51964413078298249 \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{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 <= -0.00015196441307829825)) {
		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) (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

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 < -1.51964413078298249e-4

    1. Initial program 0.1

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

    3. Applied flip3--0.1

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

    4. Simplified0.1

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

    if -1.51964413078298249e-4 < x

    1. Initial program 60.2

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

    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

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.51964413078298249 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{1 \cdot \left(1 + e^{x}\right) + e^{x + x}}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{6} \cdot {x}^{2} + \left(\frac{1}{2} \cdot x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020161 
(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))