(10/24/2009, 08:13 PM)bo198214 Wrote:(10/24/2009, 08:01 PM)mike3 Wrote: Actually it does seem to converge. The problem is that it seems to converge to the same value for every z in such cases. I.e., converging to a constant function.
Interesting!
Hmm. Sounds like it's time for a graph... I'll see if I can prepare a 2D one looking at the values of various branches on the real axis (can't do 3D with anything I've got).
(10/24/2009, 08:13 PM)bo198214 Wrote:Quote: There are uncountably many such limit values, yet as constant functions they are "analytically incompatible" (is that a real term?) with the function (you can't analytically continue a constant function to tetration!)
Well, each constant fixed point of b^x is a tetration! I.e. it satisfies c(z+1)=b^c(z).
Does it converge to fixed points?
But it's a constant function, so it cannot be interpreted as analytic continuation of the specific function \( \mathrm{tet}_b(z) \) to another branch (think about the problem in "reverse": how would you analytically continue from this constant function to a non-constant one? You can't). And not all branches of \( \mathrm{tet}_b(z) \) satisfy \( \mathrm{tet}_b(z+1) = b^{\mathrm{tet}_b(z)} \). Consider the branch \( \mathrm{tet}_b_{[1]}(z) = \mathrm{tet}_b(z) + \omega \). We get \( b^{\mathrm{tet}_b_{[1]}(z)} = b^{\mathrm{tet}_b(z) + \omega} = b^{\mathrm{tet}_b(z)} * b^{\omega} = b^{\mathrm{tet}_b(z)} * 1 = b^{\mathrm{tet}_b(z)} = \mathrm{tet}_b(z+1) \ne \mathrm{tet}_b(z+1) + \omega = \mathrm{tet}_b_{[1]}(z+1) \). Note that it takes us back to the principal branch. The equation \( \mathrm{tet}_b(z+1) = b^{\mathrm{tet}_b(z)} \) seems to only hold for all \( z \) when using the principal branch (though there may be some freedom in the choice of cut of course), if we are interpreting the symbol \( \mathrm{tet}_b(z) \) as a specific single-valued branch. As you can see above, though, if we interpret it in a multivalued sense, that values on some branches of \( \mathrm{tet}_b(z+1) \) equal \( b^{\mathrm{tet}_b(z)} \) for values on some branches of \( \mathrm{tet}_b(z) \), then it is true, as can be seen from the example I just showed. The situation is similar to that with \( \log(z) \) and \( \exp(z) \): \( \log(\exp(z)) = z \) for some branch of log for any given \( z \) but which branch that is will depend on what \( z \) is. "Multivalued functions" are funny things, you know?