Real-analytic tetration uniqueness criterion?

[sheldonison]

Also is there a value \( C \) such that for \( x > C \) we have \( D^n tet(x) >= 0 \) for all positive integer \( n \) ?
Maybe Im confused ...

I once read about the same conditions as in the OP but for another function then \( tet(x) \). So there must be some theory I think.

I changed my mind, I think it might be true.

regards

tommy1729

For tetration at the real axis, the nearest singularity is at x=-2, and there are no other singularities to the right of that anywhere in the complex plane. So, for all points on the real axis, the derivatives must eventually be governed by log(x+2), for large enough derivatives! And the even derivatives of log(x+2) are all negative. So I think that is a proof that for any point on the real axis, no matter how large, eventually the (2n)th derivative must be negative for large enough values of n. However, all of the odd derivatives are always positive.

Now the other side of the conjecture is that for tet(z+theta(z)/k), eventually for large enough odd derivatives misbehave, and the odd derivatives are no longer all positive. I believe a possible avenue to proving this is to imagine a circle from (-1) to z0, where we are looking at the odd derivatives at a point (z0/2-1/2). Without the theta perturbation, the tetration function on the boundary of that circle has its maximum positive value at z0. Especially if we are dealing with a simple case like theta(z)=sin(2piz), then with the z+theta(z), permutation, we can prove that eventually as the circle gets bigger, the maximum absolute value occurs somewhere else on the circle, instead of at z0, since theta(z) grows exponentially and that exponential growth eventually overcomes the "k" in the denominator. Somewhere near or past that point, as the derivatives are now more or less randomly controlled by the phase of the maximum, and that means some of the odd derivatives must also misbehave. edit Making this is probably much more complicated (I don't know how yet). We may need to know how to approximate some of the derivatives at z0/2-1/2, which I'm also working on....

Another problem I'm working is how to actually approximate the very large derivatives for Tetration, that can't be easy calculated by Cauchy integral, so we can tell when the derivatives go from being dominated by super-exponential growth, to the derivatives being dominated by the singularity at x=-2, so that the even derivatives become negative. typos new prediction the 4904th derivative of sexp(z) at z=0.7 is the first negative derivative at z=0.7, with a Taylor series coefficient for a_4904 = -4.371521861406 E-2122. I need to post the algorithm later. Unfortunately, this is outside of the range that I can verify numericaly against a Cauchy series, so I tried again with z=0.5. This time the prediction was that the the 406th derivative would be the first negative derivative; with a Taylor series coefficient of a_406~=-9.3E-166. This is within the range that I could verify with a "\p 522" kneser Taylor series at 0.5, and the results mach with a_406 the first negative coefficient=-9.535 E-166. The algorithm is based on the algorithm in post#16, I used for the Taylor series of the entire half sexp approximation thread; it will take me a little while to write up the equations for this algorithm. But basically, we divide the sexp(z) into two parts and estimate the logarithmic singularity at -2 plus a second part, sexp(z)=exp(sexp(z-1)), where we look at the first and second derivative and values at z-1 to estimate the behavior at z, and use this to approximate the higher derivatives. Equations will be posted later, when I have time; plus I'm still working on simplifications to the algorithm, so I need more time.