Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In the context of iteration of functions I look at the eigenvalues of the associated matrixoperator/Carleman-matrix .

If a function $\small f(x)$ has a negative eigenvalue in its associated carleman-matrix, then the definition of a half-iterate must handle the squareroot of that negative number (and fractional iteration in general). If I create the function $\small g(x)$ by taking the absolute value of that eigenvalue, then $\small f(f(x)) = g(g(x))$ , and moreover $\small f(g(x)) = g(f(x)) $ . (I also assume, that the commutativity makes the solution unique). Then I can do fractional iteration on $ \small g(x) $ and I'm interested in the general relation of that two functions.

Is there a name for that functional relation (for instance "g is the dual of f" or something else so I have a keyword for search) and/or some study online available?


[update 2] There was some discussion related to this, where I also was involved answering: "do complex iterates (..) have any meaning?" Here the change-of-sign of the schröder-value, which means an imaginary iteration "height", resembles in some way the change-of-sign of the eigenvalue in the way I looked at in the current question. However, I thought there might be some wider discussion (and possibly a common keyword) to that specific case and the indicated function with "complementary"(?) eigenvalue here.


[update 1] After the remark of Gerald Edgar it might be instructive to show at least one example. Motivated by some other question regarding the inversion of $ \small \zeta$ I constructed a carleman-matrix for the zeta-function. Recentering at one fixpoint ( $ \small \sim -0.29590...$ this matrix could be made triangular (call it ZT) and the basic eigenvalue $\small \lambda_1 \sim -0.51273 $ and the others the sequence $\small \lambda_1^0,\lambda_1, \lambda_1^2,\lambda_1^3, \ldots $ Then using diagonalization I replaced the set of eigenvalues by their absolute values, which can also be understood as taking the matrix-squareroot of the square of the matrix ZT, call it ZTA. The entries in ZTA give the coefficients for the function $\small g(x)$ ; the radius of convergence is about $ \small -1.3 < x <0.7$ In this range we find that $\zeta(\zeta(x))=g(g(x))$ at least with visible accuracy..
Here is a picture picture .

The range nearer to $\small 1-$ must be determined by other means; the twofold iteration of the zeta in this area becomes chaotic and can be seen by the wolfram-alpha call

share|improve this question
    
Why the "nt.number-theory" tag? –  David Loeffler Aug 27 '11 at 9:34
    
@David: I was baffled, that the usual automatic list for the tags didn't open and was a bit helpless. At least "nt.number-theory" was mentioned in the near of the form and so it was the only one which I knew that it was allowed and where I could see the correct spelling. Feel free to adapt the tag. –  Gottfried Helms Aug 27 '11 at 10:07
    
And a generic $n\times n$ complex matrix has $2^n$ square-roots. So I doubt "dual" is used for their relationship. –  Gerald Edgar Aug 27 '11 at 13:12
    
Hmm, at least triangular Bell-matrices have the sequence of eigenvalues $ \small \lambda ^0, \lambda ^1, \lambda ^2 ,\ldots $ so if I use the negative squareroot of a negative $ \lambda ^1 $ for the definition of the function g(x) then this inherits to the following eigenvalues accordingly (and I do not have $ \small 2^n$ or infinitely many choices). I suppose, that is the same with the square carleman-matrices, too. (But this is taken of a view on a very limited set of functions, possibly this view is too restricted in general) –  Gottfried Helms Aug 27 '11 at 14:19
    
Correction: "negative squareroot of a negative $\lambda ^1$ " is a bit messed. I meant, if I take the (positive) absolute value of $ \lambda ^1$ instead of the negative, then I have also to change the signs of all subsequent odd powers, and the Bell-matrix for the function g(x) has only positive eigenvalues –  Gottfried Helms Aug 27 '11 at 16:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.