Average Error: 39.4 → 0.3
Time: 2.2s
Precision: binary64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.10683631921556827 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{e^{x + x} - 1 \cdot 1}{1 + e^{x}}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{x + x \cdot \left(x \cdot \left(\frac{1}{2} + x \cdot \frac{1}{6}\right)\right)}{x}\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.10683631921556827 \cdot 10^{-4}:\\
\;\;\;\;\frac{\frac{e^{x + x} - 1 \cdot 1}{1 + e^{x}}}{x}\\

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

\end{array}
double code(double x) {
	return ((double) (((double) (((double) exp(x)) - 1.0)) / x));
}
↓
double code(double x) {
	double VAR;
	if ((x <= -0.00011068363192155683)) {
		VAR = ((double) (((double) (((double) (((double) exp(((double) (x + x)))) - ((double) (1.0 * 1.0)))) / ((double) (1.0 + ((double) exp(x)))))) / x));
	} else {
		VAR = ((double) (((double) (x + ((double) (x * ((double) (x * ((double) (0.5 + ((double) (x * 0.16666666666666666)))))))))) / x));
	}
	return VAR;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.4
Target39.8
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 < -1.10683631921556827e-4

    1. Initial program 0.1

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

    3. Applied flip--0.0

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

    4. Simplified0.0

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

    6. Applied pow-exp0.0

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

    7. Simplified0.0

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

    if -1.10683631921556827e-4 < x

    1. Initial program 60.0

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

    2. Taylor expanded around 0 0.5

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

    3. Simplified0.5

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.10683631921556827 \cdot 10^{-4}:\\ \;\;\;\;\frac{\frac{e^{x + x} - 1 \cdot 1}{1 + e^{x}}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{x + x \cdot \left(x \cdot \left(\frac{1}{2} + x \cdot \frac{1}{6}\right)\right)}{x}\\ \end{array}\]

Reproduce

herbie shell --seed 2020181 
(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))