Average Error: 39.6 → 0.4
Time: 3.1s
Precision: binary64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \leq -0.00015264209877748183:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{2}}{e^{x} + 1} - 1 \cdot \frac{1}{e^{x} + 1}}{x}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(0.5 + x \cdot 0.16666666666666666\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \leq -0.00015264209877748183:\\
\;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{2}}{e^{x} + 1} - 1 \cdot \frac{1}{e^{x} + 1}}{x}\\

\mathbf{else}:\\
\;\;\;\;1 + x \cdot \left(0.5 + x \cdot 0.16666666666666666\right)\\

\end{array}
double code(double x) {
	return (((double) (((double) exp(x)) - 1.0)) / x);
}
↓
double code(double x) {
	double VAR;
	if ((x <= -0.00015264209877748183)) {
		VAR = (((double) ((((double) pow(((double) exp(x)), 2.0)) / ((double) (((double) exp(x)) + 1.0))) - ((double) (1.0 * (1.0 / ((double) (((double) exp(x)) + 1.0))))))) / x);
	} else {
		VAR = ((double) (1.0 + ((double) (x * ((double) (0.5 + ((double) (x * 0.16666666666666666))))))));
	}
	return VAR;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.6
Target40.0
Herbie0.4
\[\begin{array}{l} \mathbf{if}\;x < 1 \land x > -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -1.5264209877748183e-4

    1. Initial program 0.0

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

    3. Applied flip--0.0

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

    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 div-sub0.0

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

    7. Simplified0.0

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

    if -1.5264209877748183e-4 < x

    1. Initial program 60.0

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

    2. Taylor expanded around 0 0.5

      \[\leadsto \color{blue}{0.16666666666666666 \cdot {x}^{2} + \left(0.5 \cdot x + 1\right)}\]

    3. Simplified0.5

      \[\leadsto \color{blue}{1 + x \cdot \left(0.5 + x \cdot 0.16666666666666666\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq -0.00015264209877748183:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{2}}{e^{x} + 1} - 1 \cdot \frac{1}{e^{x} + 1}}{x}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(0.5 + x \cdot 0.16666666666666666\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020196 
(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))