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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

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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
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1 Answer 1

There are plenty of pairs of functions with the property $f \circ f = g \circ g$. For example, any two involutions (i.e. $f = f^{-1}$, $g = g^{-1}$) will satisfy it. So will any two functions whose kernel contains their image (i.e. $f \circ f = g \circ g = 0$). Of course, these are but two equivalence classes out of many.

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Dear @Ilmari, possibly I made myself not clear enough. After the carleman-matrix is diagonalized, following my model there is exactly one option to change; in particular to invert the sign of the basic eigenvalue (and to adapt the signs of all the following eigenvalues which are just the consecutive powers of the "basic" one. After that the new function is uniquely determined - just recompose the new diagonal with the eigenvectors. Or did I miss your point? Because of this tight relation of the two functions I thought there might be some specific examination of this somewhere available. –  Gottfried Helms Aug 29 '11 at 14:32
    
One more property, which might make the wished function g(x) possibly unique, is commutativity: we have additionally $f\circ g = g \circ f$ –  Gottfried Helms Aug 29 '11 at 21:51
    
OK, fair enough. I just tried to answer the literal question as asked; I'm not familiar enough with fractional iteration or Carleman matrices to say anything much about them. –  Ilmari Karonen Aug 30 '11 at 1:07
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