Gottfried Wrote:bo198214 Wrote:Hm, thanks for the link however do we need the matrix-exponential?Well, Andrew asked for such a routine, if I recall right
I now see that the matrix logarithm could be the iterative logarithm, up to multiplicative constant perhaps.
If we define \( (\text{ilog}(f))_n=\log(CM(f))_{1,n} \) then exactly \( \text{ilog}(f^{\circ t})=\log(CM(f)^t)_1=t\text{ilog}(f) \) is satisfied.