Average Error: 40.2 → 0.3
Time: 3.0s
Precision: binary64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \leq -0.000170846800793159:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{2} + -1}{e^{x} + 1}}{x}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(0.5 + x \cdot 0.16666666666666666\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \leq -0.000170846800793159:\\
\;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{2} + -1}{e^{x} + 1}}{x}\\

\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 -0.000170846800793159)
   (/ (/ (+ (pow (exp x) 2.0) -1.0) (+ (exp x) 1.0)) x)
   (+ 1.0 (* x (+ 0.5 (* x 0.16666666666666666))))))
double code(double x) {
	return (exp(x) - 1.0) / x;
}
↓
double code(double x) {
	double tmp;
	if (x <= -0.000170846800793159) {
		tmp = ((pow(exp(x), 2.0) + -1.0) / (exp(x) + 1.0)) / x;
	} else {
		tmp = 1.0 + (x * (0.5 + (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

Original40.2
Target40.6
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 < -1.7084680079315899e-4

    1. Initial program 0.1

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

    3. Applied flip--_binary64_21180.1

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

    4. Simplified0.1

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

    if -1.7084680079315899e-4 < x

    1. Initial program 60.2

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

    2. Taylor expanded around 0 0.4

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

    3. 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

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq -0.000170846800793159:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{2} + -1}{e^{x} + 1}}{x}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(0.5 + x \cdot 0.16666666666666666\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020289 
(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))