Solving tetration for base 0 < b < e^-e
#16
(09/13/2009, 09:57 AM)Gottfried Wrote: The problem with the real fixpoint is, that the triangular Bell-matrices have (alternating signed) units on its diagonal. This prevents the computation of a matrix-logarithm as well of the diagonalization - at least in my implementations.

That should not pose a problem. You need the first row of \( B^t \), where \( B \) is the Bell matrix. You make a Jordan decomposition \( B = S J S^{-1} \) and then \( B^t = S J^t S^{-1} \) where for each Jordanblock \( J_m \) for eigenvalue \( \lambd_m \) with multiplicity \( M_m \) one sets \( J_m^t = \sum_{n=0}^{M_m} \left(t\\n\right) (J_m-\lambda_m I)^n \). The sum is finite because \( J_m-\lambda_m I \) is nilpotent: \( J_m^{M_m}=0 \).

Unfortunately the Jordan decompostion is flawed in Sage so I could not try it myself.

Quote:But well, let's see. It's surely not the highest summit of wisdom... and we also have the Newton-binomial-formula and others...

I dont think that the Newton formula helps. It is real-valued and hence can not return a suitable solution for a decreasing base function.
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RE: Solving tetration for base 0 < b < e^-e - by bo198214 - 09/13/2009, 11:23 AM

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