Average Error: 40.0 → 0.4
Time: 2.4s
Precision: binary64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \leq -0.0001638550454587521:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{3} - 1}{1 + e^{x} \cdot \left(e^{x} + 1\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(0.5 + x \cdot 0.16666666666666666\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \leq -0.0001638550454587521:\\
\;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{3} - 1}{1 + e^{x} \cdot \left(e^{x} + 1\right)}}{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 -0.0001638550454587521)
   (/ (/ (- (pow (exp x) 3.0) 1.0) (+ 1.0 (* (exp x) (+ (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 <= -0.0001638550454587521) {
		tmp = ((pow(exp(x), 3.0) - 1.0) / (1.0 + (exp(x) * (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

Original40.0
Target40.5
Herbie0.4
\[\begin{array}{l} \mathbf{if}\;x < 1 \land x > -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -1.63855045458752101e-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. 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

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

    if -1.63855045458752101e-4 < x

    1. Initial program 60.0

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

    2. Taylor expanded around 0 0.5

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

    3. Simplified0.5

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq -0.0001638550454587521:\\ \;\;\;\;\frac{\frac{{\left(e^{x}\right)}^{3} - 1}{1 + e^{x} \cdot \left(e^{x} + 1\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot \left(0.5 + x \cdot 0.16666666666666666\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020268 
(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))