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sum of log of eigenvalues of Carleman matrix
#5
(08/28/2009, 09:10 PM)Gottfried Wrote:
(08/28/2009, 08:58 PM)bo198214 Wrote: Hm, I dont know whether it has a deeper meaning its just an observation Smile

Hmm, this means, the determinant of this matrix is 1 for finite dimension - which extends then also for iterates/powers.
That the determinant is 1 for finite dimension results also from the determinants of its factors, the stirling- ind the bpascal-matrix. Both are triangular and have units on the diagonal, so
det(fS2F) = det(P) = 1 and det(B) = det(fS2F * P~) = det(fS2F)*det(P~) = 1 *1.
It is also interesting in contrast to the version, which has the powerseries developed at the first complex fixpoint for exp(x), x0 = 0.318131505205 + 1.33723570143*I .
With dim=64 I get -at least for the integer iterates -1,1,2 the expected values to 10 digits accuracy - however, the sum of the logarithms of the eigenvalues should be simply, but consequently,
log(x0^0) + log(x0^1) + log(x0^2) + ...
= log(x0)*(0 + 1 + 2 + 3 + ...) =???= log(x0) * zeta(-1) =-0.0265109587671 - 0.111436308453*I

....
well, better not to ride the horse to death... Wink

Gottfried
Gottfried Helms, Kassel
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Messages In This Thread
RE: sum of log of eigenvalues of Carleman matrix - by Gottfried - 08/28/2009, 09:34 PM

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