Personal Scratchpad

[jaydfox]

The half-iterate:
\(
\begin{eqnarray}
\exp_b(z) & = & \exp_b^{\circ {\small \frac{1}{2}}}\left(\exp_b^{\circ {\small \frac{1}{2}}}(z)\right) \\
\exp_b(z) & = & \mathcal{E}_{[b,{\small \frac{1}{2}}]}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right)
\end{eqnarray}
\)

Differentiate:
\(
\begin{eqnarray}
\text{D}_z\left[ \exp_b(z) \right] & = & \text{D}_z\left[\mathcal{E}_{[b,{\small \frac{1}{2}}]}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right)\right] \\
\ln(b) \exp_b(z)
& = & \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right) \text{D}_z\left[\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right] \\
& = & \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right) \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}(z)
\end{eqnarray}
\)

Which leads to:
\(
\begin{eqnarray}
\exp_b(z) & = & \frac{1}{\ln(b)} \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right) \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}(z) \\

\mathcal{E}_{[b,{\small \frac{1}{2}}]}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right) & = & \frac{1}{\ln(b)} \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right) \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}(z) \\

\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z) & = & \frac{1}{\ln(b)} \left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}(z)\right) \left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}\left( \mathcal{E}_{[b,{\small -\frac{1}{2}}]}(z) \right)\right)
\end{eqnarray}
\)

(Yes, I'm on a fishing expedition.)