Personal Scratchpad

[jaydfox]

Inverse:
\(
\begin{eqnarray}
z & = & \exp_b^{\circ ({\small -}t)}\left(\exp_b^{\circ t}(z)\right) \\
z & = & \mathcal{E}_{[b,-t]}\left(\mathcal{E}_{[b,t]}(z)\right)
\end{eqnarray}
\)

Differentiate:
\(
\begin{eqnarray}
\text{D}_z\left[ z \right] & = & \text{D}_z\left[\mathcal{E}_{[b,{\small -}t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right] \\
1 & = & \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right) \text{D}_z\left[\mathcal{E}_{[b,t]}(z)\right] \\
& = & \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right) \mathcal{E}_{[b,t]}^{'}(z) \\
\end{eqnarray}
\)

Transforming:
\( \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right)\ =\ \frac{1}{\mathcal{E}_{[b,t]}^{'}(z)} \)


Self-extraction:
\(
\begin{eqnarray}
\exp_b^{\circ t}(z)
& = & \exp_b^{\circ (t{\small -}t)}\left(\exp_b^{\circ t}(z)\right) \\
& = & \log_b^{\circ t}\left(\exp_b^{\circ t}\left(\exp_b^{\circ t}(z)\right) \right) \\
\mathcal{E}_{[b,t]}(z) & = & \mathcal{E}_{[b,-t]}\left(\mathcal{E}_{[b,t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right)
\end{eqnarray}
\)

Differentiate:
\(
\begin{eqnarray}
\text{D}_z\left[\mathcal{E}_{[b,t]}(z)\right] & = & \text{D}_z\left[\mathcal{E}_{[b,{\small -}t]}\left(\mathcal{E}_{[b,t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right)\right] \\
\mathcal{E}_{[b,t]}^{'}(z)
& = & \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right) \text{D}_z\left[\mathcal{E}_{[b,t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right] \\
& = & \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right) \mathcal{E}_{[b,t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right) \text{D}_z\left[\mathcal{E}_{[b,t]}(z)\right] \\
& = & \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right) \mathcal{E}_{[b,t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right) \mathcal{E}_{[b,t]}^{'}(z) \\
& = & \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,2t]}(z)\right) \mathcal{E}_{[b,t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right) \mathcal{E}_{[b,t]}^{'}(z) \\
\end{eqnarray}
\)

Cancelling \( \mathcal{E}_{[b,t]}^{'}(z) \) from both sides:
\( \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,2t]}(z)\right) \mathcal{E}_{[b,t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right)\ =\ 1 \)

In hindsight, that was obvious from the inverse.