06/06/2008, 06:56 PM
bo198214 Wrote:If we define \( (\text{ilog}(f))_n=\log(CM(f))_{1,n} \) then exactly \( \text{ilog}(f^{\circ t})=\log(CM(f)^t)=t\text{ilog}(f) \) is satisfied.
Do you mean \( \text{ilog}(f^{\circ t})=(\log(CM(f)^t))_1=t\text{ilog}(f) \)? Then yes, that is very nice!

Andrew Robbins

