Average Error: 40.0 → 0.2
Time: 2.9s
Precision: binary64
\[\frac{e^{x} - 1}{x}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \leq -0.00015326377826995708:\\ \;\;\;\;\left(e^{x} \cdot e^{x} - 1\right) \cdot \frac{1}{x \cdot \left(e^{x} + 1\right)}\\ \mathbf{else}:\\ \;\;\;\;1 + \log \left(e^{x \cdot \left(x \cdot 0.16666666666666666 + 0.5\right)}\right)\\ \end{array}\]
\frac{e^{x} - 1}{x}
↓
\begin{array}{l}
\mathbf{if}\;x \leq -0.00015326377826995708:\\
\;\;\;\;\left(e^{x} \cdot e^{x} - 1\right) \cdot \frac{1}{x \cdot \left(e^{x} + 1\right)}\\

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

\end{array}
(FPCore (x) :precision binary64 (/ (- (exp x) 1.0) x))
↓
(FPCore (x)
 :precision binary64
 (if (<= x -0.00015326377826995708)
   (* (- (* (exp x) (exp x)) 1.0) (/ 1.0 (* x (+ (exp x) 1.0))))
   (+ 1.0 (log (exp (* x (+ (* x 0.16666666666666666) 0.5)))))))
double code(double x) {
	return (exp(x) - 1.0) / x;
}
↓
double code(double x) {
	double tmp;
	if (x <= -0.00015326377826995708) {
		tmp = ((exp(x) * exp(x)) - 1.0) * (1.0 / (x * (exp(x) + 1.0)));
	} else {
		tmp = 1.0 + log(exp(x * ((x * 0.16666666666666666) + 0.5)));
	}
	return tmp;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original40.0
Target40.4
Herbie0.2
\[\begin{array}{l} \mathbf{if}\;x < 1 \land x > -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -1.53263778269957079e-4

    1. Initial program 0.1

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

    3. Applied flip--_binary64_14170.1

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

    4. Applied associate-/l/_binary64_13890.1

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

    6. Applied div-inv_binary64_14390.1

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

    if -1.53263778269957079e-4 < x

    1. Initial program 60.3

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

    2. Taylor expanded around 0 0.3

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

    3. Simplified0.3

      \[\leadsto \color{blue}{1 + x \cdot \left(0.5 + x \cdot 0.16666666666666666\right)}\]
    4. Using strategy rm

    5. Applied add-log-exp_binary64_14810.3

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

    6. Simplified0.3

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq -0.00015326377826995708:\\ \;\;\;\;\left(e^{x} \cdot e^{x} - 1\right) \cdot \frac{1}{x \cdot \left(e^{x} + 1\right)}\\ \mathbf{else}:\\ \;\;\;\;1 + \log \left(e^{x \cdot \left(x \cdot 0.16666666666666666 + 0.5\right)}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020295 
(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))