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 Understanding Abel/Schroeder with matrix-expression Gottfried Ultimate Fellow Posts: 787 Threads: 121 Joined: Aug 2007 05/26/2008, 07:02 PM (This post was last modified: 05/26/2008, 07:06 PM by Gottfried.) Here I'm considering fractional heights h for the function Ut°h(x) of complex base t, which implemets x -> t^x -1 and hope to get to the source of the persisting problem. Note, that with the previously described method everything is fine with integral heights, so the problem should be somehow systematic. Remember the Schröder-function, whose powerseries-coefficients I find in the 2'nd column of the eigenvectormatrix of the operator Ut, such that $\hspace{24} U_t = W * ^dV(u) * WI$ where u=log(t) and WI is the inverse of W. Then the Schröder-function sigma for parameter t is $\hspace{24} \sigma_t(x) = \sum_{k=0}^{\infty} W_{k,1}*x^k$ and the inverse takes the coefficients of WI instead: $\hspace{24} \sigma_t^{-1}(x) = \sum_{k=0}^{\infty} WI_{k,1}*x^k$ To compute Ut°1(x) we simply do $\hspace{24} U_t^{{^o}1}(x) = \sigma_t^{-1}(u*\sigma_t(x))$ or, for integer h $\hspace{24} U_t^{{^o}h}(x) = \sigma_t^{-1}(u^h*\sigma_t(x))$ As I said, this works fine even for complex t, checked with some t where abs(t)<1, if heights are integral. But we'll see, that for fractional h this does not work. With this simple process, for instance t=1/2 + 1/2 I there is also no severe problem with convergence or even divergence - things sum up in a completely easy manner - but anyway, the fractional h in u^h produce obviously complete crap. Currently I suspect, that there must be some conjugacy be involved - but don't have a working idea, where. Gottfried Helms, Kassel bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 05/26/2008, 07:36 PM Gottfried Wrote:With this simple process, for instance t=1/2 + 1/2 I there is also no severe problem with convergence or even divergence - things sum up in a completely easy manner - but anyway, the fractional h in u^h produce obviously complete crap. Currently I suspect, that there must be some conjugacy be involved - but don't have a working idea, where. Apart from the Schroeder function, does the fractional Matrix power return the proper fractional iteration? And what means crap? I am not able to guess your computations and where there occured a problem. Gottfried Ultimate Fellow Posts: 787 Threads: 121 Joined: Aug 2007 05/26/2008, 08:45 PM (This post was last modified: 05/26/2008, 10:07 PM by Gottfried.) bo198214 Wrote:Apart from the Schroeder function, does the fractional Matrix power return the proper fractional iteration?Well, the fractional matrix-power will be computed by that same formula; empirically, the eigensystemsolver of Pari/Gp gives exactly the same results (resp.rescaling and reordering) as the analytically constructed Ut-Eigenmatrices. bo198214 Wrote:And what means crap? I am not able to guess your computations and where there occured a problem.Really crap... But you made me recheck the problem and I found the error getting now correct results at least for this base. I was multiplying dV(sigma)*dV(u^h) , but for fractional powers of complex u this is obviously different from dV(sigma*u^h), so , for instance for the fourth entry I had sigma^3 * (u^0.5^3) where I should have had (sigma*u^0.5)^3 which obviously are different from each other if u is complex and the power of u is fractional. I had to use the second version. [update] To make it more precise: the difference occurs, since for complex exponentiation a) (u^3)^0.5 can be different from b) (u^0.5)^3, so commutativity with multiplication in the exponent is not generally given. What is surprising - considering the principles of diagonalization and computing powers of matrices - that one would expect the correct way to compute the h'th power of a matrix using its eigenvalues would be a) dV(u)^h , but the example indicates that we have to compute b) dV(u^h) where b) gives then (u^h)^0, (u^h)^1, (u^h)^2,... which gives the correct result in my example computation. This is really surprising. [/update] Stupid error - all the half year where I dealt with that problem I didn't catch that bug. Well - I'll recompute my old examples; hope I get consistent results now. New energy! Gottfried Helms, Kassel « Next Oldest | Next Newest »

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