Average Error: 39.8 → 0.3
Time: 2.0s
Precision: binary64
\[\frac{e^{x} - 1}{x}\]
\[\begin{array}{l} \mathbf{if}\;x \leq -0.00017862444022647447:\\ \;\;\;\;\frac{\sqrt[3]{e^{x} - 1} \cdot \left(\sqrt[3]{e^{x} - 1} \cdot \sqrt[3]{e^{x} - 1}\right)}{x}\\ \mathbf{else}:\\ \;\;\;\;x \cdot 0.5 + \left(1 + 0.16666666666666666 \cdot {x}^{2}\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
\begin{array}{l}
\mathbf{if}\;x \leq -0.00017862444022647447:\\
\;\;\;\;\frac{\sqrt[3]{e^{x} - 1} \cdot \left(\sqrt[3]{e^{x} - 1} \cdot \sqrt[3]{e^{x} - 1}\right)}{x}\\

\mathbf{else}:\\
\;\;\;\;x \cdot 0.5 + \left(1 + 0.16666666666666666 \cdot {x}^{2}\right)\\

\end{array}
(FPCore (x) :precision binary64 (/ (- (exp x) 1.0) x))
(FPCore (x)
 :precision binary64
 (if (<= x -0.00017862444022647447)
   (/
    (*
     (cbrt (- (exp x) 1.0))
     (* (cbrt (- (exp x) 1.0)) (cbrt (- (exp x) 1.0))))
    x)
   (+ (* x 0.5) (+ 1.0 (* 0.16666666666666666 (pow x 2.0))))))
double code(double x) {
	return (exp(x) - 1.0) / x;
}
double code(double x) {
	double tmp;
	if (x <= -0.00017862444022647447) {
		tmp = (cbrt(exp(x) - 1.0) * (cbrt(exp(x) - 1.0) * cbrt(exp(x) - 1.0))) / x;
	} else {
		tmp = (x * 0.5) + (1.0 + (0.16666666666666666 * pow(x, 2.0)));
	}
	return tmp;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.8
Target40.3
Herbie0.3
\[\begin{array}{l} \mathbf{if}\;x < 1 \land x > -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes
  2. if x < -1.7862444022647447e-4

    1. Initial program 0.1

      \[\frac{e^{x} - 1}{x}\]
    2. Using strategy rm
    3. Applied add-cube-cbrt_binary64_21590.1

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

    if -1.7862444022647447e-4 < x

    1. Initial program 60.3

      \[\frac{e^{x} - 1}{x}\]
    2. Taylor expanded around 0 0.4

      \[\leadsto \color{blue}{0.5 \cdot x + \left(0.16666666666666666 \cdot {x}^{2} + 1\right)}\]
  3. Recombined 2 regimes into one program.
  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq -0.00017862444022647447:\\ \;\;\;\;\frac{\sqrt[3]{e^{x} - 1} \cdot \left(\sqrt[3]{e^{x} - 1} \cdot \sqrt[3]{e^{x} - 1}\right)}{x}\\ \mathbf{else}:\\ \;\;\;\;x \cdot 0.5 + \left(1 + 0.16666666666666666 \cdot {x}^{2}\right)\\ \end{array}\]

Alternatives

Reproduce

herbie shell --seed 2021102 
(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))