JmsNxn,
I am a little suspicious of this method. In particular, I'm not sure the integral
\( I = \int_{0}^{\infty} \vartheta(x) x^{z-1} dx \), \( 0 < z < 1 \)
converges. I have done some numerical tests and it seems that the sup of \( |\vartheta(x)| \) for \( 0 < x < X \) grows to infinity as \( X \rightarrow \infty \) and grows fast enough that the decay of \( x^{z-1} \) 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:
http://math.eretrandre.org/tetrationforu...p?pid=5019
an alternate tetration function which decays to a different, non-principal set of fixed points of the logarithm at \( \pm i \infty \). Such a function (or its reciprocal \( \frac{1}{\mathrm{tet}_{\mathrm{alt}}(-z)} \)) has the asymptotic behavior you want, yet is different than the "usual" tetrational. That is, both \( \mathrm{tet} \) and the alternate \( \mathrm{tet}_{\mathrm{alt}} \) constructed have the property that their mirrored reciprocals \( \frac{1}{F(-z)} \) satisfy \( |\frac{1}{F(-z)}| < Ce^{a|\Im(z)|} \) for \( 0 < a < \pi/2 \) in a vertical strip with \( \Re(z) < 1 \) which does not include points arbitrarily close to \( 1 \) 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 \( F(z) \) should satisfy
\( |\frac{1}{F(-z)}| < Ce^{a|\Im(z)| + \rho|\Re(z)|} \) for some \( 0 < a < \pi/2 \) and \( \rho \ge 0 \) with \( \Re(z) < 1 \)
? I believe that any tetration function \( F(z) \) which is holomorphic for \( \Re(z) \ge 0 \) must take on values arbitrarily close to \( 0 \), 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 \( e^z \). A tetration function (more correctly, a superfunction of the exponential function) satisfies \( F(z+1) = e^{F(z)} \). The exponential map is topologically transitive, which means that if we have an open set \( A \), and another \( B \), we can find an integer n such that \( \exp^n(A) \) contains at least one point of \( B \). In particular, if we let \( A = \{ F(z) : z \in \mathbb{C}\ \mathrm{and}\ 0 < \Re(z) < 1 \} \), which is open by the openness of the strip and the open mapping theorem, we have for any open \( \epsilon \)-disc \( D(\epsilon) \) around \( 0 \), no matter how small \( \epsilon \), there is an n such that \( \exp^n(A) \), and thus \( F \) of the strip \( n < \Re(z) < n + 1 \) contains a point in that disc, hence within \( \epsilon \) of 0, and so the reciprocal \( \frac{1}{F(z)} \) must be unbounded on \( \Re(z) > 0 \), thus \( \frac{1}{F(-z)} \) on \( \Re(z) < 0 \) and \( \Re(z) < 1 \) is also so unbounded.
I am a little suspicious of this method. In particular, I'm not sure the integral
\( I = \int_{0}^{\infty} \vartheta(x) x^{z-1} dx \), \( 0 < z < 1 \)
converges. I have done some numerical tests and it seems that the sup of \( |\vartheta(x)| \) for \( 0 < x < X \) grows to infinity as \( X \rightarrow \infty \) and grows fast enough that the decay of \( x^{z-1} \) 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:
http://math.eretrandre.org/tetrationforu...p?pid=5019
an alternate tetration function which decays to a different, non-principal set of fixed points of the logarithm at \( \pm i \infty \). Such a function (or its reciprocal \( \frac{1}{\mathrm{tet}_{\mathrm{alt}}(-z)} \)) has the asymptotic behavior you want, yet is different than the "usual" tetrational. That is, both \( \mathrm{tet} \) and the alternate \( \mathrm{tet}_{\mathrm{alt}} \) constructed have the property that their mirrored reciprocals \( \frac{1}{F(-z)} \) satisfy \( |\frac{1}{F(-z)}| < Ce^{a|\Im(z)|} \) for \( 0 < a < \pi/2 \) in a vertical strip with \( \Re(z) < 1 \) which does not include points arbitrarily close to \( 1 \) 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 \( F(z) \) should satisfy
\( |\frac{1}{F(-z)}| < Ce^{a|\Im(z)| + \rho|\Re(z)|} \) for some \( 0 < a < \pi/2 \) and \( \rho \ge 0 \) with \( \Re(z) < 1 \)
? I believe that any tetration function \( F(z) \) which is holomorphic for \( \Re(z) \ge 0 \) must take on values arbitrarily close to \( 0 \), 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 \( e^z \). A tetration function (more correctly, a superfunction of the exponential function) satisfies \( F(z+1) = e^{F(z)} \). The exponential map is topologically transitive, which means that if we have an open set \( A \), and another \( B \), we can find an integer n such that \( \exp^n(A) \) contains at least one point of \( B \). In particular, if we let \( A = \{ F(z) : z \in \mathbb{C}\ \mathrm{and}\ 0 < \Re(z) < 1 \} \), which is open by the openness of the strip and the open mapping theorem, we have for any open \( \epsilon \)-disc \( D(\epsilon) \) around \( 0 \), no matter how small \( \epsilon \), there is an n such that \( \exp^n(A) \), and thus \( F \) of the strip \( n < \Re(z) < n + 1 \) contains a point in that disc, hence within \( \epsilon \) of 0, and so the reciprocal \( \frac{1}{F(z)} \) must be unbounded on \( \Re(z) > 0 \), thus \( \frac{1}{F(-z)} \) on \( \Re(z) < 0 \) and \( \Re(z) < 1 \) is also so unbounded.
