\[e^{x} - 1\]

↓

\[\begin{array}{l} \mathbf{if}\;x \le 0.00016358488783184289:\\ \;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{1}{6} \cdot x + \frac{1}{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} \cdot e^{x} - 1}{1 + e^{x}}\\ \end{array}\]

double f(double x) {
        double r6655685 = x;
        double r6655686 = exp(r6655685);
        double r6655687 = 1.0;
        double r6655688 = r6655686 - r6655687;
        return r6655688;
}

↓

double f(double x) {
        double r6655689 = x;
        double r6655690 = 0.00016358488783184289;
        bool r6655691 = r6655689 <= r6655690;
        double r6655692 = r6655689 * r6655689;
        double r6655693 = 0.16666666666666666;
        double r6655694 = r6655693 * r6655689;
        double r6655695 = 0.5;
        double r6655696 = r6655694 + r6655695;
        double r6655697 = r6655692 * r6655696;
        double r6655698 = r6655689 + r6655697;
        double r6655699 = exp(r6655689);
        double r6655700 = r6655699 * r6655699;
        double r6655701 = 1.0;
        double r6655702 = r6655700 - r6655701;
        double r6655703 = r6655701 + r6655699;
        double r6655704 = r6655702 / r6655703;
        double r6655705 = r6655691 ? r6655698 : r6655704;
        return r6655705;
}

e^{x} - 1

↓

\begin{array}{l}
\mathbf{if}\;x \le 0.00016358488783184289:\\
\;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{1}{6} \cdot x + \frac{1}{2}\right)\\

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

\end{array}

Original	58.9
Target	0.4
Herbie	0.1

Derivation

Split input into 2 regimes
if x < 0.00016358488783184289
1. Initial program 59.4
  \[e^{x} - 1\]
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 0.00016358488783184289 < x
1. Initial program 2.0
  \[e^{x} - 1\]
2. Using strategy rm
3. Applied flip--6.0
  \[\leadsto \color{blue}{\frac{e^{x} \cdot e^{x} - 1 \cdot 1}{e^{x} + 1}}\]
Recombined 2 regimes into one program.
Final simplification0.1
\[\leadsto \begin{array}{l} \mathbf{if}\;x \le 0.00016358488783184289:\\ \;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{1}{6} \cdot x + \frac{1}{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} \cdot e^{x} - 1}{1 + e^{x}}\\ \end{array}\]

Reproduce

herbie shell --seed 2019102 
(FPCore (x)
  :name "expm1 (example 3.7)"
  :pre (< -0.00017 x)

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

  (- (exp x) 1))

Error

Target

Derivation

`if x < 0.00016358488783184289`

`if 0.00016358488783184289 < x`

Reproduce