- Split input into 2 regimes
if x < -9.922346769656687e-05
Initial program 0.1
\[\frac{e^{x} - 1}{x}\]
- Using strategy
rm Applied flip--0.1
\[\leadsto \frac{\color{blue}{\frac{e^{x} \cdot e^{x} - 1 \cdot 1}{e^{x} + 1}}}{x}\]
Applied associate-/l/0.1
\[\leadsto \color{blue}{\frac{e^{x} \cdot e^{x} - 1 \cdot 1}{x \cdot \left(e^{x} + 1\right)}}\]
Simplified0.1
\[\leadsto \frac{\color{blue}{-1 + e^{x} \cdot e^{x}}}{x \cdot \left(e^{x} + 1\right)}\]
- Using strategy
rm Applied flip3-+0.1
\[\leadsto \frac{\color{blue}{\frac{{-1}^{3} + {\left(e^{x} \cdot e^{x}\right)}^{3}}{-1 \cdot -1 + \left(\left(e^{x} \cdot e^{x}\right) \cdot \left(e^{x} \cdot e^{x}\right) - -1 \cdot \left(e^{x} \cdot e^{x}\right)\right)}}}{x \cdot \left(e^{x} + 1\right)}\]
Applied associate-/l/0.1
\[\leadsto \color{blue}{\frac{{-1}^{3} + {\left(e^{x} \cdot e^{x}\right)}^{3}}{\left(x \cdot \left(e^{x} + 1\right)\right) \cdot \left(-1 \cdot -1 + \left(\left(e^{x} \cdot e^{x}\right) \cdot \left(e^{x} \cdot e^{x}\right) - -1 \cdot \left(e^{x} \cdot e^{x}\right)\right)\right)}}\]
Simplified0.0
\[\leadsto \frac{\color{blue}{-1 + e^{\left(x + x\right) + \left(\left(x + x\right) + \left(x + x\right)\right)}}}{\left(x \cdot \left(e^{x} + 1\right)\right) \cdot \left(-1 \cdot -1 + \left(\left(e^{x} \cdot e^{x}\right) \cdot \left(e^{x} \cdot e^{x}\right) - -1 \cdot \left(e^{x} \cdot e^{x}\right)\right)\right)}\]
if -9.922346769656687e-05 < x
Initial program 60.3
\[\frac{e^{x} - 1}{x}\]
Taylor expanded around 0 0.4
\[\leadsto \color{blue}{\frac{1}{2} \cdot x + \left(\frac{1}{6} \cdot {x}^{2} + 1\right)}\]
Simplified0.4
\[\leadsto \color{blue}{x \cdot \left(\frac{1}{6} \cdot x + \frac{1}{2}\right) + 1}\]
- Recombined 2 regimes into one program.
Final simplification0.3
\[\leadsto \begin{array}{l}
\mathbf{if}\;x \le -9.922346769656687 \cdot 10^{-05}:\\
\;\;\;\;\frac{-1 + e^{\left(\left(x + x\right) + \left(x + x\right)\right) + \left(x + x\right)}}{\left(\left(\left(e^{x} \cdot e^{x}\right) \cdot \left(e^{x} \cdot e^{x}\right) - \left(-e^{x} \cdot e^{x}\right)\right) + 1\right) \cdot \left(\left(1 + e^{x}\right) \cdot x\right)}\\
\mathbf{else}:\\
\;\;\;\;1 + x \cdot \left(\frac{1}{2} + \frac{1}{6} \cdot x\right)\\
\end{array}\]