- Split input into 2 regimes
if (+ 1 (exp (* -2 x))) < 1.9332077667223637 or 2.0000647776950835 < (+ 1 (exp (* -2 x)))
Initial program 0.0
\[\frac{2}{1 + e^{-2 \cdot x}} - 1\]
- Using strategy
rm Applied flip3-+31.8
\[\leadsto \frac{2}{\color{blue}{\frac{{1}^{3} + {\left(e^{-2 \cdot x}\right)}^{3}}{1 \cdot 1 + \left(e^{-2 \cdot x} \cdot e^{-2 \cdot x} - 1 \cdot e^{-2 \cdot x}\right)}}} - 1\]
Applied associate-/r/31.8
\[\leadsto \color{blue}{\frac{2}{{1}^{3} + {\left(e^{-2 \cdot x}\right)}^{3}} \cdot \left(1 \cdot 1 + \left(e^{-2 \cdot x} \cdot e^{-2 \cdot x} - 1 \cdot e^{-2 \cdot x}\right)\right)} - 1\]
Applied simplify31.8
\[\leadsto \color{blue}{\frac{2}{{\left(e^{x \cdot -2}\right)}^{3} + 1}} \cdot \left(1 \cdot 1 + \left(e^{-2 \cdot x} \cdot e^{-2 \cdot x} - 1 \cdot e^{-2 \cdot x}\right)\right) - 1\]
- Using strategy
rm Applied add-log-exp31.8
\[\leadsto \color{blue}{\log \left(e^{\frac{2}{{\left(e^{x \cdot -2}\right)}^{3} + 1} \cdot \left(1 \cdot 1 + \left(e^{-2 \cdot x} \cdot e^{-2 \cdot x} - 1 \cdot e^{-2 \cdot x}\right)\right) - 1}\right)}\]
Applied simplify0.1
\[\leadsto \log \color{blue}{\left(\frac{{\left(e \cdot {\left(e^{e^{-2 \cdot x}}\right)}^{\left(e^{-2 \cdot x} - 1\right)}\right)}^{\left(\frac{2}{{\left(e^{-2 \cdot x}\right)}^{3} + 1}\right)}}{e}\right)}\]
if 1.9332077667223637 < (+ 1 (exp (* -2 x))) < 2.0000647776950835
Initial program 59.0
\[\frac{2}{1 + e^{-2 \cdot x}} - 1\]
Taylor expanded around 0 0.0
\[\leadsto \color{blue}{\left(\frac{2}{15} \cdot {x}^{5} + x\right) - \frac{1}{3} \cdot {x}^{3}}\]
- Recombined 2 regimes into one program.
Applied simplify0.1
\[\leadsto \color{blue}{\begin{array}{l}
\mathbf{if}\;e^{x \cdot -2} + 1 \le 1.9332077667223637 \lor \neg \left(e^{x \cdot -2} + 1 \le 2.0000647776950835\right):\\
\;\;\;\;\log \left(\frac{{\left(e \cdot {\left(e^{e^{x \cdot -2}}\right)}^{\left(e^{x \cdot -2} - 1\right)}\right)}^{\left(\frac{2}{{\left(e^{x \cdot -2}\right)}^{3} + 1}\right)}}{e}\right)\\
\mathbf{else}:\\
\;\;\;\;\left(\frac{2}{15} \cdot {x}^{5} + x\right) - {x}^{3} \cdot \frac{1}{3}\\
\end{array}}\]