Thanks, Daniel. I really felt I was being a tad harsh, thank you for saying that I wasn't; because I honestly felt I was.
What you are doing is absolutely certainly possible, and I agree with your general conclusion. There just seem to be some gaps.
I had an idea not too dissimilar to what you are writing, that I produced mostly for fun as evidence of a sequence of entire functions that satisfy,
\(
F_n(z) : \mathbb{C} \to \mathbb{C}\,\,\text{for}\,\,n\ge 1\\
F_n(F_{n+1}(z)) = F_n(z+1)\\
\)
This was done by instead of insisting \( F_n(0) =1 \) but rather that, \( F_n(L) = L \) for a complex fixed point of \( e^z \), and an insistence that \( |F'_n(L)| > 1 \). Then starting with \( F_1(z) = e^z \), one can construct a sequence of entire functions \( F_n \), of what I called whacky hyper-operators. Where they satisfied the superfunction identity \( F_n(F_{n+1}(z)) = F_{n+1}(z+1) \) but instead of being normalized with \( F_n(0) = 1 \) were normalized to \( F_n(L) = L \) (which proves to be so much simpler!). I only did this to use as evidence that a chain of entire super functions can exist. The way I did it is largely similar to what you are doing, and based on the same principles. So I'm pretty sure your method will work.
I'm excited to see how you're going to make this work. I'm glad you're aware of the anomalies I pointed out--I kind of figured you were. I'm still curious though, is there any kind of functional equation in the matrix equation. I'm consistently visualizing this as,
\(
e \uparrow^n e\uparrow^{n+1} A = e \uparrow^{n+1} T A\\
\)
For invertible square matrices \( A,T \). I'm wondering if at all something like this happens, or if we are really just doing a plug and play with the flow. I didn't see much about the matrix element in your paper, except for the formal identification of \( e \uparrow^n A \) with a formal Taylor series in A.
Anyway, that was still a really interesting paper,
Regards, James.
P.S. Oh and don't worry about a lack of formal education. I had to leave university for health reasons; but kept on learning, and when I came back I'd gawk at most formally educated people. So, don't feel ashamed in anyway. We're both in the same boat.
What you are doing is absolutely certainly possible, and I agree with your general conclusion. There just seem to be some gaps.
I had an idea not too dissimilar to what you are writing, that I produced mostly for fun as evidence of a sequence of entire functions that satisfy,
\(
F_n(z) : \mathbb{C} \to \mathbb{C}\,\,\text{for}\,\,n\ge 1\\
F_n(F_{n+1}(z)) = F_n(z+1)\\
\)
This was done by instead of insisting \( F_n(0) =1 \) but rather that, \( F_n(L) = L \) for a complex fixed point of \( e^z \), and an insistence that \( |F'_n(L)| > 1 \). Then starting with \( F_1(z) = e^z \), one can construct a sequence of entire functions \( F_n \), of what I called whacky hyper-operators. Where they satisfied the superfunction identity \( F_n(F_{n+1}(z)) = F_{n+1}(z+1) \) but instead of being normalized with \( F_n(0) = 1 \) were normalized to \( F_n(L) = L \) (which proves to be so much simpler!). I only did this to use as evidence that a chain of entire super functions can exist. The way I did it is largely similar to what you are doing, and based on the same principles. So I'm pretty sure your method will work.
I'm excited to see how you're going to make this work. I'm glad you're aware of the anomalies I pointed out--I kind of figured you were. I'm still curious though, is there any kind of functional equation in the matrix equation. I'm consistently visualizing this as,
\(
e \uparrow^n e\uparrow^{n+1} A = e \uparrow^{n+1} T A\\
\)
For invertible square matrices \( A,T \). I'm wondering if at all something like this happens, or if we are really just doing a plug and play with the flow. I didn't see much about the matrix element in your paper, except for the formal identification of \( e \uparrow^n A \) with a formal Taylor series in A.
Anyway, that was still a really interesting paper,
Regards, James.
P.S. Oh and don't worry about a lack of formal education. I had to leave university for health reasons; but kept on learning, and when I came back I'd gawk at most formally educated people. So, don't feel ashamed in anyway. We're both in the same boat.