Average Error: 39.7 → 0.3
Time: 2.9s
Precision: binary64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \leq -0.0012102062965107401:\\ \;\;\;\;\frac{\frac{\frac{{\left({\left(e^{x}\right)}^{3}\right)}^{3} - 1}{{\left(e^{x}\right)}^{3} \cdot {\left(e^{x}\right)}^{3} + \left({\left(e^{x}\right)}^{3} + 1\right)}}{{\left(\sqrt[3]{e^{x}}\right)}^{4} \cdot {\left(\sqrt[3]{e^{x}}\right)}^{2} + \left(e^{x} + 1\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(0.5 + x \cdot \left(0.16666666666666666 + x \cdot 0.041666666666666664\right)\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \leq -0.0012102062965107401:\\
\;\;\;\;\frac{\frac{\frac{{\left({\left(e^{x}\right)}^{3}\right)}^{3} - 1}{{\left(e^{x}\right)}^{3} \cdot {\left(e^{x}\right)}^{3} + \left({\left(e^{x}\right)}^{3} + 1\right)}}{{\left(\sqrt[3]{e^{x}}\right)}^{4} \cdot {\left(\sqrt[3]{e^{x}}\right)}^{2} + \left(e^{x} + 1\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 -0.0012102062965107401)
   (/
    (/
     (/
      (- (pow (pow (exp x) 3.0) 3.0) 1.0)
      (+ (* (pow (exp x) 3.0) (pow (exp x) 3.0)) (+ (pow (exp x) 3.0) 1.0)))
     (+ (* (pow (cbrt (exp x)) 4.0) (pow (cbrt (exp x)) 2.0)) (+ (exp x) 1.0)))
    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 <= -0.0012102062965107401) {
		tmp = (((pow(pow(exp(x), 3.0), 3.0) - 1.0) / ((pow(exp(x), 3.0) * pow(exp(x), 3.0)) + (pow(exp(x), 3.0) + 1.0))) / ((pow(cbrt(exp(x)), 4.0) * pow(cbrt(exp(x)), 2.0)) + (exp(x) + 1.0))) / 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

Original39.7
Target40.2
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 < -0.0012102062965107401

    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. Simplified0.0

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

    5. Simplified0.0

      \[\leadsto \frac{\frac{{\left(e^{x}\right)}^{3} - 1}{\color{blue}{{\left(e^{x}\right)}^{2} + \left(e^{x} + 1\right)}}}{x}\]
    6. Using strategy rm

    7. Applied add-cube-cbrt_binary640.0

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

    8. Applied unpow-prod-down_binary640.0

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

    9. Simplified0.0

      \[\leadsto \frac{\frac{{\left(e^{x}\right)}^{3} - 1}{\color{blue}{{\left(\sqrt[3]{e^{x}}\right)}^{4}} \cdot {\left(\sqrt[3]{e^{x}}\right)}^{2} + \left(e^{x} + 1\right)}}{x}\]
    10. Using strategy rm

    11. Applied flip3--_binary640.0

      \[\leadsto \frac{\frac{\color{blue}{\frac{{\left({\left(e^{x}\right)}^{3}\right)}^{3} - {1}^{3}}{{\left(e^{x}\right)}^{3} \cdot {\left(e^{x}\right)}^{3} + \left(1 \cdot 1 + {\left(e^{x}\right)}^{3} \cdot 1\right)}}}{{\left(\sqrt[3]{e^{x}}\right)}^{4} \cdot {\left(\sqrt[3]{e^{x}}\right)}^{2} + \left(e^{x} + 1\right)}}{x}\]

    if -0.0012102062965107401 < x

    1. Initial program 60.0

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

    2. Taylor expanded around 0 0.4

      \[\leadsto \frac{\color{blue}{x + \left(0.5 \cdot {x}^{2} + \left(0.16666666666666666 \cdot {x}^{3} + 0.041666666666666664 \cdot {x}^{4}\right)\right)}}{x}\]

    3. Simplified0.4

      \[\leadsto \frac{\color{blue}{x + x \cdot \left(x \cdot \left(0.5 + x \cdot \left(0.16666666666666666 + x \cdot 0.041666666666666664\right)\right)\right)}}{x}\]

    4. Taylor expanded around 0 0.4

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

    5. Simplified0.4

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq -0.0012102062965107401:\\ \;\;\;\;\frac{\frac{\frac{{\left({\left(e^{x}\right)}^{3}\right)}^{3} - 1}{{\left(e^{x}\right)}^{3} \cdot {\left(e^{x}\right)}^{3} + \left({\left(e^{x}\right)}^{3} + 1\right)}}{{\left(\sqrt[3]{e^{x}}\right)}^{4} \cdot {\left(\sqrt[3]{e^{x}}\right)}^{2} + \left(e^{x} + 1\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(0.5 + x \cdot \left(0.16666666666666666 + x \cdot 0.041666666666666664\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2021176 
(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))