Average Error: 39.8 → 0.3
Time: 3.1min
Precision: binary64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.8926949204835697 \cdot 10^{-4}:\\ \;\;\;\;\log \left(e^{e^{x} - 1}\right) \cdot \frac{1}{x}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.8926949204835697 \cdot 10^{-4}:\\
\;\;\;\;\log \left(e^{e^{x} - 1}\right) \cdot \frac{1}{x}\\

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

\end{array}
double code(double x) {
	return (((double) (((double) exp(x)) - 1.0)) / x);
}
↓
double code(double x) {
	double VAR;
	if ((x <= -0.00018926949204835697)) {
		VAR = ((double) (((double) log(((double) exp(((double) (((double) exp(x)) - 1.0)))))) * (1.0 / x)));
	} else {
		VAR = ((double) (1.0 + ((double) (x * ((double) (((double) (x * 0.16666666666666666)) + 0.5))))));
	}
	return VAR;
}

Error

Bits error versus x

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Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.8
Target40.2
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 < -1.8926949204835697e-4

    1. Initial program 0.0

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

    3. Applied add-log-exp0.0

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

    4. Applied add-log-exp0.1

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

    5. Applied diff-log0.1

      \[\leadsto \frac{\color{blue}{\log \left(\frac{e^{e^{x}}}{e^{1}}\right)}}{x}\]

    6. Simplified0.0

      \[\leadsto \frac{\log \color{blue}{\left(e^{e^{x} - 1}\right)}}{x}\]
    7. Using strategy rm

    8. Applied div-inv0.0

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

    if -1.8926949204835697e-4 < x

    1. Initial program 60.0

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

    2. Taylor expanded around 0 0.5

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

    3. Simplified0.5

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.8926949204835697 \cdot 10^{-4}:\\ \;\;\;\;\log \left(e^{e^{x} - 1}\right) \cdot \frac{1}{x}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right)\\ \end{array}\]

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))