I have another point to add about Weierstrass if you guys are still considering it. I would suggest looking into the stronger, Hadamard factorization theorem.
Since we are dealing with an asymptotic half iterate of exp (I'm going to call it \( f \)) I'm going to assume we can say it's order 1?
I.e: Does \( |f(z)| < M e^{\rho|z|} \) for some positive M,rho?
If so then Hadamards factorization theorem guarantees that:
\( f(z) = C e^{Az + B} \prod_{j=1}^\infty (1-\frac{z}{a_j})e^{-\frac{z}{a_j}} \)
for some \( A,B,C \in \mathbb{C} \) where \( a_j \) is all the zeroes counted with multiplicity.
This also implies, quite adequately that at least \( \sum_{j=1}^\infty |a_j|^{-(1+\epsilon)} < \infty \) for \( \epsilon > 0 \).
I think examining the zeroes is really important. I'm going to rifle through hadamard some more and see if theres anything else we can do. I believe, as tommy does, we can probably prove its order 0, so we can probably cut \( A \) out and the exponential factors in the product. This would imply a very plain product.
Sorry if you guys are passed this or if it isn't useful but I realize I could've said it before, I just forgot about Hadamard ^_^. Even though its stronger and requires more work than weierstrass we only ever remember weierstrass. Poor hadamard. ^_^
Since we are dealing with an asymptotic half iterate of exp (I'm going to call it \( f \)) I'm going to assume we can say it's order 1?
I.e: Does \( |f(z)| < M e^{\rho|z|} \) for some positive M,rho?
If so then Hadamards factorization theorem guarantees that:
\( f(z) = C e^{Az + B} \prod_{j=1}^\infty (1-\frac{z}{a_j})e^{-\frac{z}{a_j}} \)
for some \( A,B,C \in \mathbb{C} \) where \( a_j \) is all the zeroes counted with multiplicity.
This also implies, quite adequately that at least \( \sum_{j=1}^\infty |a_j|^{-(1+\epsilon)} < \infty \) for \( \epsilon > 0 \).
I think examining the zeroes is really important. I'm going to rifle through hadamard some more and see if theres anything else we can do. I believe, as tommy does, we can probably prove its order 0, so we can probably cut \( A \) out and the exponential factors in the product. This would imply a very plain product.
Sorry if you guys are passed this or if it isn't useful but I realize I could've said it before, I just forgot about Hadamard ^_^. Even though its stronger and requires more work than weierstrass we only ever remember weierstrass. Poor hadamard. ^_^
