Average Error: 58.8 → 0.0
Time: 13.2s
Precision: 64
\[-0.00017 \lt x\]
\[e^{x} - 1\]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \le 1.0001162075892607:\\ \;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{1}{2} + \frac{1}{6} \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;-1 + e^{x}\\ \end{array}\]
e^{x} - 1
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \le 1.0001162075892607:\\
\;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{1}{2} + \frac{1}{6} \cdot x\right)\\

\mathbf{else}:\\
\;\;\;\;-1 + e^{x}\\

\end{array}
double f(double x) {
        double r2297754 = x;
        double r2297755 = exp(r2297754);
        double r2297756 = 1.0;
        double r2297757 = r2297755 - r2297756;
        return r2297757;
}
↓
double f(double x) {
        double r2297758 = x;
        double r2297759 = exp(r2297758);
        double r2297760 = 1.0001162075892607;
        bool r2297761 = r2297759 <= r2297760;
        double r2297762 = r2297758 * r2297758;
        double r2297763 = 0.5;
        double r2297764 = 0.16666666666666666;
        double r2297765 = r2297764 * r2297758;
        double r2297766 = r2297763 + r2297765;
        double r2297767 = r2297762 * r2297766;
        double r2297768 = r2297758 + r2297767;
        double r2297769 = -1.0;
        double r2297770 = r2297769 + r2297759;
        double r2297771 = r2297761 ? r2297768 : r2297770;
        return r2297771;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original58.8
Target0.4
Herbie0.0
\[x \cdot \left(\left(1 + \frac{x}{2}\right) + \frac{x \cdot x}{6}\right)\]

Derivation

  1. Split input into 2 regimes

  2. if (exp x) < 1.0001162075892607

    1. Initial program 59.3

      \[e^{x} - 1\]

    2. Taylor expanded around 0 0.0

      \[\leadsto \color{blue}{x + \left(\frac{1}{6} \cdot {x}^{3} + \frac{1}{2} \cdot {x}^{2}\right)}\]

    3. Simplified0.0

      \[\leadsto \color{blue}{x + \left(x \cdot x\right) \cdot \left(x \cdot \frac{1}{6} + \frac{1}{2}\right)}\]

    if 1.0001162075892607 < (exp x)

    1. Initial program 2.7

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

    3. Applied sub-neg2.7

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

    4. Simplified2.7

      \[\leadsto e^{x} + \color{blue}{-1}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.0

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \le 1.0001162075892607:\\ \;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{1}{2} + \frac{1}{6} \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;-1 + e^{x}\\ \end{array}\]

Reproduce

herbie shell --seed 2019168 
(FPCore (x)
  :name "expm1 (example 3.7)"
  :pre (< -0.00017 x)

  :herbie-target
  (* x (+ (+ 1 (/ x 2)) (/ (* x x) 6)))

  (- (exp x) 1))