Average Error: 40.6 → 0.1
Time: 15.8s
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 r2076621 = x;
        double r2076622 = exp(r2076621);
        double r2076623 = 1.0;
        double r2076624 = r2076622 - r2076623;
        double r2076625 = r2076624 / r2076621;
        return r2076625;
}
↓
double f(double x) {
        double r2076626 = 1.0;
        double r2076627 = x;
        double r2076628 = r2076626 / r2076627;
        double r2076629 = expm1(r2076627);
        double r2076630 = r2076628 * r2076629;
        return r2076630;
}

Error

Bits error versus x

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Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original40.6
Target39.7
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 40.6

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

  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 2019158 +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))