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 Personal Scratchpad jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/01/2007, 08:25 PM (This post was last modified: 09/01/2007, 08:28 PM by 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.) ~ Jay Daniel Fox « Next Oldest | Next Newest »