07/09/2008, 02:38 PM
Gottfried Wrote:Henryk's formula is (I inserted (x) at f°t)
Hmm ... you mean Woon's formula ...
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matrix function like iteration without power series expansion
matrix function like iteration without power series expansion
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matrix function like iteration without power series expansion
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 Hmm ... you mean Woon's formula ...
07/09/2008, 02:57 PM
I guess I should be more specific. The key observation to show the equivalence is that
which means Woon's formula simplifies to: and letting and assuming and combining (-1) terms, this simplifies to and that is, as you call it, "Henryk's formula" from Woon's formula. Andrew Robbins 
07/09/2008, 06:18 PM
andydude Wrote:Gottfried Wrote:Henryk's formula is (I inserted (x) at f°t) You can also call it Newton's formula: Woon just applied this to linear operators However we cant calculate with iterations as with powers, because the composition is no more right distributive. The interesting thing however is that both expansions together are then again valid: And I think this is new. Especially that the iteration by this formula is the same as regular iteration (which is sure for elliptic iteration, other cases are still to prove). 
I've got better convergence for the binomial (Woon-) method when computing f°0.5(x) by
(but not yet for the Stirling-transformation. This became even worse) Also, the value, computed by diagonalization is verified (and seems to be the best approximation, see below) Here is the list of partial sums of the series for different k+0.5 (with a slight Euler-acceleration) Table of partial sums of series-terms for y_k = f°{k+0.5}(1), f(x)= sqrt(2)^x using 96 terms for all computations Code: k+0.5= 0.5 ... 5.5 6.5 7.5 8.5 9.5We see the improving of convergence for higher k. Here are the results of the second part of computation Table for f°0.5(1) computed by f°{-k}(y_k) Code: using value forComputed by Diagonalization method (with fixpoint-shift to fixpoint 2) Code: diag: 1.24362162766852180429509898361Code: _0.5: 1.243620...
Gottfried Helms, Kassel
07/14/2008, 06:21 PM
bo198214 Wrote:And I think this is new. Especially that the iteration by this formula is the same as regular iteration (which is sure for elliptic iteration, other cases are still to prove). Hmm. Awhile back, I did some test with parabolic iteration, and convinced myself that Woon's formula and Jabotinsky's formula produce exactly the same results for parabolic iteration (for symbolic coefficients). So I could write up something about this if needed. Andrew Robbins
07/14/2008, 09:55 PM
andydude Wrote:Awhile back, I did some test with parabolic iteration, and convinced myself that Woon's formula and Jabotinsky's formula produce exactly the same results for parabolic iteration (for symbolic coefficients). So I could write up something about this if needed. The problematic thing about parabolic iteration is convergence. You usually only have an asymptotic development at the fixed point. I.e. the series is not developable at the fixed point however in every neighborhood and all the coefficients converge (towards the fixed point in a certain sector) to that of the asymptotic development. I also see that I made mistake in my first post about the convergence of that series. It is not true that the series always converges if the function has a finite attracting fixed point. |
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