Write $W$ in terms of $U$ and $V$ where $\phi_W=\phi_U+\phi_V$ and $\tanh(i)=\phi_i$ for all $i$.

Question

Write $W$ in terms of $U$ and $V$ where $\phi_W=\phi_U+\phi_V$ and $\tanh(\phi_i)=i$ for all $i$, using the fact that $$\tanh^{-1}(x)=\frac{1}{2}ln\frac{1+x}{1-x}$$

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Where I'm at: $$\tanh(\phi_i)=i$$ $$\tanh^{-1}(i)=\phi_i$$

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$$W$$ $$=\tanh(\phi_W)$$ $$=\tanh(\phi_U+\phi_V)$$ $$=\tanh(\tanh^{-1}(U)+\tanh^{-1}(V))$$ $$=\tanh(\frac{1}{2}ln\frac{1+U}{1-U}+\frac{1}{2}ln\frac{1+V}{1-V})$$ $$=\tanh(\frac{1}{2}ln\frac{(1+U)(1+V)}{(1-U)(1-V)})$$ $$=\tanh(\frac{1}{2}ln\frac{1+U+V+UV}{1-U-V+UV})$$

If only the denominator was equal to $1-U-V-UV$ We could let $x=U+V+UV$ in $\tanh^{-1}(x)=\frac{1}{2}ln\frac{1+x}{1-x}$.

Answer 1

You got as far as \begin{eqnarray*} =\tanh(\frac{1}{2}ln\frac{1+U+V+UV}{1-U-V+UV}) \end{eqnarray*} It can be rewriten as \begin{eqnarray*} =\tanh(\frac{1}{2}ln\frac{1+UV+ (U+V)}{1+UV-(U+V)}). \end{eqnarray*} Now divide top & bottom of the fraction by $\color{blue}{1+UV}$ \begin{eqnarray*} =\tanh\left(\frac{1}{2}ln\frac{1+ \frac{U+V}{1+UV}}{1-\frac{U+V}{1+UV}}\right)=\color{red}{\frac{U+V}{1+UV}}. \end{eqnarray*}