Average Error: 39.6 → 0.2
Time: 5.3s
Precision: binary64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \leq -0.001450981745007271:\\ \;\;\;\;\frac{\log \left(\frac{e^{e^{x}}}{e}\right)}{x}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(0.5 + x \cdot \left(0.16666666666666666 + x \cdot 0.041666666666666664\right)\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \leq -0.001450981745007271:\\
\;\;\;\;\frac{\log \left(\frac{e^{e^{x}}}{e}\right)}{x}\\

\mathbf{else}:\\
\;\;\;\;1 + x \cdot \left(0.5 + x \cdot \left(0.16666666666666666 + x \cdot 0.041666666666666664\right)\right)\\

\end{array}
(FPCore (x) :precision binary64 (/ (- (exp x) 1.0) x))
↓
(FPCore (x)
 :precision binary64
 (if (<= x -0.001450981745007271)
   (/ (log (/ (exp (exp x)) E)) x)
   (+
    1.0
    (* x (+ 0.5 (* x (+ 0.16666666666666666 (* x 0.041666666666666664))))))))
double code(double x) {
	return (exp(x) - 1.0) / x;
}
↓
double code(double x) {
	double tmp;
	if (x <= -0.001450981745007271) {
		tmp = log(exp(exp(x)) / ((double) M_E)) / x;
	} else {
		tmp = 1.0 + (x * (0.5 + (x * (0.16666666666666666 + (x * 0.041666666666666664)))));
	}
	return tmp;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.6
Target40.0
Herbie0.2
\[\begin{array}{l} \mathbf{if}\;x < 1 \land x > -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -0.00145098174500727099

    1. Initial program 0.0

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

    3. Applied add-log-exp_binary640.0

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

    4. Applied add-log-exp_binary640.0

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

    5. Applied diff-log_binary640.0

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

    if -0.00145098174500727099 < x

    1. Initial program 60.1

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

    2. Taylor expanded around 0 0.4

      \[\leadsto \color{blue}{0.5 \cdot x + \left(0.16666666666666666 \cdot {x}^{2} + \left(0.041666666666666664 \cdot {x}^{3} + 1\right)\right)}\]

    3. Simplified0.3

      \[\leadsto \color{blue}{1 + x \cdot \left(0.5 + x \cdot \left(0.16666666666666666 + x \cdot 0.041666666666666664\right)\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq -0.001450981745007271:\\ \;\;\;\;\frac{\log \left(\frac{e^{e^{x}}}{e}\right)}{x}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(0.5 + x \cdot \left(0.16666666666666666 + x \cdot 0.041666666666666664\right)\right)\\ \end{array}\]

Alternatives

Reproduce

herbie shell --seed 2021113 
(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))