Green Eggs and HAM: Tetration for ALL bases, real and complex, now possible?

[mike3]

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where \( b_n \) are properly-chosen "basis functions". I have thought about the Kneser-mapping solution (i.e. regular iteration warped with theta mapping) as a possible set of basis functions... but the problem is this only covers half of the plane (as given, the upper half-plane), and the Cauchy equations require both halves of the plane.
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Do you, perhaps, have any ideas as to how this could be done? The form of solution need not converge on the entire plane, only on and perhaps near the imaginary axis.

You could try mapping the two unit infinite strip to the unit circle using,
\( f(z) = \frac{4}{\pi}\tan^{-1}(z) \)
Here, f(z) maps the unit circle to an infinite strip, from -1 to +1. The left side of the unit circle is mapped to -1+iz, and the right side of the circle is mapped to 1+iz, where iz varies from \( +/-\Im\infty \) I don't know if that would help or not, or whether the singularity at +/-I would be fairly mild, or not, given that the tetration solution also converges to a fixed point at \( +/-\Im\infty \).
- Sheldon

I'm not sure how that'd be useful, since what I need is a representation of the tetrational at the imaginary axis, that is, at \( it \) where \( t \) goes between \( \pm \infty \), not \( -1 + it \) or \( 1 + it \) (these are obtained by exp/log of the function's values at the imaginary axis within the integral equation under the integral sign, while on the other side of the equation (without the integral) is the function on the imaginary axis). So I don't see how this makes for a set of basis functions for the function on the imaginary axis.

Edit: Although, I suppose that could work since you could just drop an "exp" or a "log" on the side of the integral equation that wants values at the imaginary axis. However, now that side of the integral equation can only be valid for input to the function in half the range of what the right side uses, namely the half of the unit circle we use to retrieve the values at the imaginary axis. So I'm not sure this'll work, as the integral equation, as specified, is for the whole function.