Personal Scratchpad

[jaydfox]

Now for E(z):
\(
\begin{eqnarray}
\exp_b^{\circ t}(z)
& = & \exp_b^{\circ (t{\small +1})}(\log_b(z)) \\
& = & \exp_b\left(\exp_b^{\circ t}(\log_b(z))\right) \\
\mathcal{E}_{[b,t]}(z) & = & \exp_b\left(\mathcal{E}_{[b,t]}(\log_b(z))\right)
\end{eqnarray}
\)

Derivative of E with respect to z:
\(
\begin{eqnarray}
\text{D}_z\left(\mathcal{E}_{[b,t]}(z)\right) & = & \text{D}_z\left[\exp_b\left(\mathcal{E}_{[b,t]}\left(\log_b(z)\right)\right)\right] \\
\mathcal{E}_{[b,t]}^{'}(z) & = & \ln(b)\exp_b\left(\mathcal{E}_{[b,t]}\left(\log_b(z)\right)\right) \text{D}_z\left[\mathcal{E}_{[b,t]}\left(\log_b(z)\right)\right] \\
& = & \ln(b)\exp_b\left(\mathcal{E}_{[b,t]}\left(\log_b(z)\right)\right) \mathcal{E}_{[b,t]}^{'}\left(\log_b(z)\right)\text{D}_z\left[\log_b(z)\right] \\
& = & \mathcal{E}_{[b,t]}(z)\left(\frac{\mathcal{E}_{[b,t]}^{'}\left(\log_b(z)\right)}{\Large z}\right)
\end{eqnarray}
\)


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


Going the other way:
\(
\begin{eqnarray}
\exp_b^{\circ t}(z)
& = & \exp_b^{\circ (t{\small -1})}(\exp_b(z)) \\
& = & \log_b\left(\exp_b^{\circ t}(\exp_b(z))\right) \\
\mathcal{E}_{[b,t]}(z) & = & \log_b\left(\mathcal{E}_{[b,t]}(\exp_b(z))\right)
\end{eqnarray}
\)

And the derivative:
\(
\begin{eqnarray}
\text{D}_z\left(\mathcal{E}_{[b,t]}(z)\right) & = & \text{D}_z\left[\log_b\left(\mathcal{E}_{[b,t]}\left(\exp_b(z)\right)\right)\right] \\
\mathcal{E}_{[b,t]}^{'}(z) & = & \left(\ln(b)\mathcal{E}_{[b,t]}\left(\exp_b(z)\right)\right)^{-1} \text{D}_z\left[\mathcal{E}_{[b,t]}\left(\exp_b(z)\right)\right] \\
& = & \left(\ln(b)\mathcal{E}_{[b,t]}\left(\exp_b(z)\right)\right)^{-1} \mathcal{E}_{[b,t]}^{'}\left(\exp_b(z)\right)\text{D}_z\left[\exp_b(z)\right] \\
& = & \left(\ln(b)\mathcal{E}_{[b,t]}\left(\exp_b(z)\right)\right)^{-1} \mathcal{E}_{[b,t]}^{'}\left(\exp_b(z)\right)\exp_b(z)\ln(b) \\
& = & \frac{\mathcal{E}_{[b,t]}^{'}\left(\exp_b(z)\right) \exp_b(z)}{ \mathcal{E}_{[b,t]}\left(\exp_b(z)\right)} \\
\end{eqnarray}
\)

Finally, transforming:
\(
\mathcal{E}_{[b,t]}\left(\exp_b(z)\right)\ =\ \frac{\exp_b(z) \mathcal{E}_{[b,t]}^{'}\left(\exp_b(z)\right)}{\mathcal{E}_{[b,t]}^{'}(z)}
\)