Average Error: 39.7 → 0.1
Time: 10.5s
Precision: 64
\[\frac{e^{x} - 1}{x}\]
↓
\[\frac{1}{x} \cdot \mathsf{expm1}\left(x\right)\]
\frac{e^{x} - 1}{x}
↓
\frac{1}{x} \cdot \mathsf{expm1}\left(x\right)
double f(double x) {
        double r2299054 = x;
        double r2299055 = exp(r2299054);
        double r2299056 = 1.0;
        double r2299057 = r2299055 - r2299056;
        double r2299058 = r2299057 / r2299054;
        return r2299058;
}
↓
double f(double x) {
        double r2299059 = 1.0;
        double r2299060 = x;
        double r2299061 = r2299059 / r2299060;
        double r2299062 = expm1(r2299060);
        double r2299063 = r2299061 * r2299062;
        return r2299063;
}

Error

Bits error versus x

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Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.7
Target39.0
Herbie0.1
\[\begin{array}{l} \mathbf{if}\;x \lt 1 \land x \gt -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array}\]

Derivation

  1. Initial program 39.7

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

  2. Simplified0.0

    \[\leadsto \color{blue}{\frac{\mathsf{expm1}\left(x\right)}{x}}\]
  3. Using strategy rm

  4. Applied div-inv0.1

    \[\leadsto \color{blue}{\mathsf{expm1}\left(x\right) \cdot \frac{1}{x}}\]

  5. Final simplification0.1

    \[\leadsto \frac{1}{x} \cdot \mathsf{expm1}\left(x\right)\]

Reproduce

herbie shell --seed 2019133 +o rules:numerics
(FPCore (x)
  :name "Kahan's exp quotient"

  :herbie-target
  (if (and (< x 1) (> x -1)) (/ (- (exp x) 1) (log (exp x))) (/ (- (exp x) 1) x))

  (/ (- (exp x) 1) x))