Average Error: 39.5 → 0.5
Time: 2.9s
Precision: binary64
\[\frac{e^{x} - 1}{x} \]
↓
\[\begin{array}{l} \mathbf{if}\;\frac{e^{x} - 1}{x} \leq 0:\\ \;\;\;\;\frac{x + \left(x \cdot x\right) \cdot \left(0.5 + x \cdot 0.16666666666666666\right)}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x}}{x} - \frac{1}{x}\\ \end{array} \]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;\frac{e^{x} - 1}{x} \leq 0:\\
\;\;\;\;\frac{x + \left(x \cdot x\right) \cdot \left(0.5 + x \cdot 0.16666666666666666\right)}{x}\\

\mathbf{else}:\\
\;\;\;\;\frac{e^{x}}{x} - \frac{1}{x}\\


\end{array}

(FPCore (x) :precision binary64 (/ (- (exp x) 1.0) x))
↓
(FPCore (x)
 :precision binary64
 (if (<= (/ (- (exp x) 1.0) x) 0.0)
   (/ (+ x (* (* x x) (+ 0.5 (* x 0.16666666666666666)))) x)
   (- (/ (exp x) x) (/ 1.0 x))))
double code(double x) {
	return (exp(x) - 1.0) / x;
}
↓
double code(double x) {
	double tmp;
	if (((exp(x) - 1.0) / x) <= 0.0) {
		tmp = (x + ((x * x) * (0.5 + (x * 0.16666666666666666)))) / x;
	} else {
		tmp = (exp(x) / x) - (1.0 / x);
	}
	return tmp;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.5
Target40.0
Herbie0.5
\[\begin{array}{l} \mathbf{if}\;x < 1 \land x > -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array} \]

Derivation

  1. Split input into 2 regimes

  2. if (/.f64 (-.f64 (exp.f64 x) 1) x) < -0.0

    1. Initial program 62.0

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

    2. Taylor expanded around 0 0

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

    3. Simplified0

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

    if -0.0 < (/.f64 (-.f64 (exp.f64 x) 1) x)

    1. Initial program 2.1

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

    3. Applied div-sub_binary641.4

      \[\leadsto \color{blue}{\frac{e^{x}}{x} - \frac{1}{x}} \]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.5

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{e^{x} - 1}{x} \leq 0:\\ \;\;\;\;\frac{x + \left(x \cdot x\right) \cdot \left(0.5 + x \cdot 0.16666666666666666\right)}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x}}{x} - \frac{1}{x}\\ \end{array} \]

Reproduce

herbie shell --seed 2021196 
(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))