03/25/2009, 07:06 PM
Ansus Wrote:What is in this formula?
is the fixed point: .
regular slog

03/25/2009, 07:06 PM
Ansus Wrote:What is in this formula? is the fixed point: .
03/25/2009, 11:16 PM
Ansus Wrote:Is this correct: I am only familiar with Maple and Sage, so I can not help you with this. However in Maple the formula works.
03/26/2009, 01:38 PM
Ansus Wrote: Maybe you have to specify the proper branch. (But as I told I can not test because I dont have Mathematica available.)
03/26/2009, 04:55 PM
Ansus Wrote:Anyway with any value of a I cannot get anything close to what expected. What shall I say? It worked for me. For base the fixed point is , thatswhy this base is so preferred, you dont need to compute the fixed point seperately.
03/27/2009, 08:03 AM
Ansus Wrote:Great! Now it works, but only for a limited range of bases. Particularly it works for the base . I used this formula:
04/02/2009, 01:20 AM
bo198214 Wrote:The Abel function has also a singularity at 0.Just realized, this is only if the fixed point is 0. bo198214 Wrote:This should be , which means you can't simplify the matrix like you did. The formula you give is a matrix representation of if those are Bell matrices, or if those are Carleman matrices. Andrew Robbins
04/02/2009, 02:31 PM
andydude Wrote:otherwise at the fixed point. The regular iteration theory always assumes the fixed point at 0. If not one just considers the function where is the fixed point.bo198214 Wrote:The Abel function has also a singularity at 0.Just realized, this is only if the fixed point is 0. Quote:actually thats also wrong. However it is only an intermediate error in my derivation.bo198214 Wrote:This should be , Lets show the correct equations: or, with : if we take the Bell matrices: where is the Bell matrix of . This is the diagonal matrix: I think Gottfried calls this the Vandermonde matrix. The right multiplication of this matrix multiplies each th column with . If we truncate to its first column we get hence: and this can then be transformed to: which I used for my further derivations. (10/07/2007, 10:30 PM)bo198214 Wrote: Now there is the the so called principal Schroeder function of a function with fixed point 0 with slope , given by: Sometimes a thing needs a whole life to be recognized... In the matrixmethod I dealt with the eigendecomposition of the (triangular) dxp_t() Bellmatrix U_t to satisfy the relation While the recursion to compute W and W^1 efficiently is easy and is working well, I did not have a deeper idea about the structure of the columns in W. Now I found, it just agrees with the above formula: which is exactly the above formula; we even can write this, if we refer to the second column of U_t^h as F°h, the second column of W as S, and s = F[1] while F°h[1] = F[1]^h =s^h , then we have Something *very* stupid ... <sigh> But, well, now also this detail is explained for me. <Hmmm I don't know why the forum software merges my two replies (to two previous posts of Henryk) into one So here is the second post> (04/02/2009, 02:31 PM)bo198214 Wrote: This is the diagonal matrix:Not exactly. I call the Vandermonde*matrix* VZ (and ZV=VZ~) the *collection* of consecutive vandermonde V(x)vectors ´ VZ = [V(0), V(1) , V(2), V(3), ...] \\ Vandermondematrix Your M is just c*dV( c) in my notation: the vandermondevector V( c) used as diagonalmatrix (and since the first entry is not c^0 I noted the additional factor c) Gottfried
Gottfried Helms, Kassel
07/31/2009, 08:55 AM

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