\[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]

↓

\[\begin{array}{l} \mathbf{if}\;x \le -1.3921498547628525 \cdot 10^{-5}:\\ \;\;\;\;\sqrt{\frac{e^{2 \cdot x} - 1}{\mathsf{fma}\left(-1, 1, e^{x + x}\right)} \cdot \left(e^{x} + 1\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\mathsf{fma}\left(0.5, {x}^{2}, \mathsf{fma}\left(1, x, 2\right)\right)}\\ \end{array}\]

\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}

↓

\begin{array}{l}
\mathbf{if}\;x \le -1.3921498547628525 \cdot 10^{-5}:\\
\;\;\;\;\sqrt{\frac{e^{2 \cdot x} - 1}{\mathsf{fma}\left(-1, 1, e^{x + x}\right)} \cdot \left(e^{x} + 1\right)}\\

\mathbf{else}:\\
\;\;\;\;\sqrt{\mathsf{fma}\left(0.5, {x}^{2}, \mathsf{fma}\left(1, x, 2\right)\right)}\\

\end{array}

double f(double x) {
        double r8630 = 2.0;
        double r8631 = x;
        double r8632 = r8630 * r8631;
        double r8633 = exp(r8632);
        double r8634 = 1.0;
        double r8635 = r8633 - r8634;
        double r8636 = exp(r8631);
        double r8637 = r8636 - r8634;
        double r8638 = r8635 / r8637;
        double r8639 = sqrt(r8638);
        return r8639;
}

↓

double f(double x) {
        double r8640 = x;
        double r8641 = -1.3921498547628525e-05;
        bool r8642 = r8640 <= r8641;
        double r8643 = 2.0;
        double r8644 = r8643 * r8640;
        double r8645 = exp(r8644);
        double r8646 = 1.0;
        double r8647 = r8645 - r8646;
        double r8648 = -r8646;
        double r8649 = r8640 + r8640;
        double r8650 = exp(r8649);
        double r8651 = fma(r8648, r8646, r8650);
        double r8652 = r8647 / r8651;
        double r8653 = exp(r8640);
        double r8654 = r8653 + r8646;
        double r8655 = r8652 * r8654;
        double r8656 = sqrt(r8655);
        double r8657 = 0.5;
        double r8658 = 2.0;
        double r8659 = pow(r8640, r8658);
        double r8660 = fma(r8646, r8640, r8643);
        double r8661 = fma(r8657, r8659, r8660);
        double r8662 = sqrt(r8661);
        double r8663 = r8642 ? r8656 : r8662;
        return r8663;
}

Derivation

Split input into 2 regimes
if x < -1.3921498547628525e-05
1. Initial program 0.1
  \[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]
2. Using strategy rm
3. Applied flip--0.0
  \[\leadsto \sqrt{\frac{e^{2 \cdot x} - 1}{\color{blue}{\frac{e^{x} \cdot e^{x} - 1 \cdot 1}{e^{x} + 1}}}}\]
4. Applied associate-/r/0.0
  \[\leadsto \sqrt{\color{blue}{\frac{e^{2 \cdot x} - 1}{e^{x} \cdot e^{x} - 1 \cdot 1} \cdot \left(e^{x} + 1\right)}}\]
5. Simplified0.0
  \[\leadsto \sqrt{\color{blue}{\frac{e^{2 \cdot x} - 1}{\mathsf{fma}\left(-1, 1, e^{x + x}\right)}} \cdot \left(e^{x} + 1\right)}\]
if -1.3921498547628525e-05 < x
1. Initial program 34.0
  \[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]
2. Taylor expanded around 0 6.0
  \[\leadsto \sqrt{\color{blue}{0.5 \cdot {x}^{2} + \left(1 \cdot x + 2\right)}}\]
3. Simplified6.0
  \[\leadsto \sqrt{\color{blue}{\mathsf{fma}\left(0.5, {x}^{2}, \mathsf{fma}\left(1, x, 2\right)\right)}}\]
Recombined 2 regimes into one program.
Final simplification0.8
\[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.3921498547628525 \cdot 10^{-5}:\\ \;\;\;\;\sqrt{\frac{e^{2 \cdot x} - 1}{\mathsf{fma}\left(-1, 1, e^{x + x}\right)} \cdot \left(e^{x} + 1\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\mathsf{fma}\left(0.5, {x}^{2}, \mathsf{fma}\left(1, x, 2\right)\right)}\\ \end{array}\]

could not determine a ground truth for program body (more)

Error

Derivation

`if x < -1.3921498547628525e-05`

`if -1.3921498547628525e-05 < x`

Reproduce