Average Error: 39.6 → 0.3
Time: 2.2s
Precision: binary64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \leq -0.00016629416446113426:\\ \;\;\;\;\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{x \cdot \left({\left(e^{x}\right)}^{2} + 1 \cdot \left(e^{x} + 1\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(0.5 + x \cdot 0.16666666666666666\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \leq -0.00016629416446113426:\\
\;\;\;\;\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{x \cdot \left({\left(e^{x}\right)}^{2} + 1 \cdot \left(e^{x} + 1\right)\right)}\\

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

\end{array}
(FPCore (x) :precision binary64 (/ (- (exp x) 1.0) x))
↓
(FPCore (x)
 :precision binary64
 (if (<= x -0.00016629416446113426)
   (/
    (- (pow (exp x) 3.0) (pow 1.0 3.0))
    (* x (+ (pow (exp x) 2.0) (* 1.0 (+ (exp x) 1.0)))))
   (+ 1.0 (* x (+ 0.5 (* x 0.16666666666666666))))))
double code(double x) {
	return (((double) (((double) exp(x)) - 1.0)) / x);
}
↓
double code(double x) {
	double tmp;
	if ((x <= -0.00016629416446113426)) {
		tmp = (((double) (((double) pow(((double) exp(x)), 3.0)) - ((double) pow(1.0, 3.0)))) / ((double) (x * ((double) (((double) pow(((double) exp(x)), 2.0)) + ((double) (1.0 * ((double) (((double) exp(x)) + 1.0)))))))));
	} else {
		tmp = ((double) (1.0 + ((double) (x * ((double) (0.5 + ((double) (x * 0.16666666666666666))))))));
	}
	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
Herbie0.3
\[\begin{array}{l} \mathbf{if}\;x < 1 \land x > -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -1.6629416446113426e-4

    1. Initial program 0.0

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

    3. Applied flip3--_binary640.0

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

    4. Applied associate-/l/_binary640.0

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

    5. Simplified0.0

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

    if -1.6629416446113426e-4 < x

    1. Initial program 60.2

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

    2. Taylor expanded around 0 0.4

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

    3. Simplified0.4

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq -0.00016629416446113426:\\ \;\;\;\;\frac{{\left(e^{x}\right)}^{3} - {1}^{3}}{x \cdot \left({\left(e^{x}\right)}^{2} + 1 \cdot \left(e^{x} + 1\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(0.5 + x \cdot 0.16666666666666666\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020205 
(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))