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Could be tetration if this integral converges

I am a little suspicious of this method. In particular, I'm not sure the integral


converges. I have done some numerical tests and it seems that the sup of for grows to infinity as and grows fast enough that the decay of does not suppress it. This is just an experimental result, I'm not sure of the formal proof of divergence yet.

In addition to this, I am suspect of the method used to show uniqueness, in particular, the use of the 1-cyclic warping of the tetration. sheldonison had constructed here:

an alternate tetration function which decays to a different, non-principal set of fixed points of the logarithm at . Such a function (or its reciprocal ) has the asymptotic behavior you want, yet is different than the "usual" tetrational. That is, both and the alternate constructed have the property that their mirrored reciprocals satisfy for in a vertical strip with which does not include points arbitrarily close to since they are bounded in such a strip.

I suspect these are related by a 1-cyclic mapping which is such that it is not entire but instead has branch singularities, and so the warping requires a more careful treatment. In particular, you can get the warping by restricting to the real axis, then applying the 1-cyclic map, then analytically continuing again to the plane. The branch nature precludes a simple substitution on the whole plane.

Also, are you sure you have that right, that a tetration function should satisfy

for some and with

? I believe that any tetration function which is holomorphic for must take on values arbitrarily close to , so that its reciprocal is unbounded, and thus cannot satisfy the exponential bound on an entire half-plane (but it can on a strip, of course).

This can be shown from the chaos of the exponential map . A tetration function (more correctly, a superfunction of the exponential function) satisfies . The exponential map is topologically transitive, which means that if we have an open set , and another , we can find an integer n such that contains at least one point of . In particular, if we let , which is open by the openness of the strip and the open mapping theorem, we have for any open -disc around , no matter how small , there is an n such that , and thus of the strip contains a point in that disc, hence within of 0, and so the reciprocal must be unbounded on , thus on and is also so unbounded.

Messages In This Thread
RE: Could be tetration if this integral converges - by mike3 - 05/03/2014, 01:19 AM

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