Yes I noticed that it wasn't exactly lower hyper operators, but it's closer to it than anything else.
It depends on a lot of factors but the summation formula I gave may or may not work.
I think the recursion is more aptly written by the general law:
\( a [x] \alpha + \beta = (a[x] \alpha)[x-1](a[x] \beta) \)
So it's almost like it's necessarily satisfying the exponential law. Which is really cool now that I think about it.
Let's see. I'm thinking there's trouble evaluating \( a[n]k \) for arbitrary natural n and arbitrary natural k and complex a. So let's be a bit more modest with our findings. So let us fix a to N for some natural N
\( N [m] k \) is definitely calculatable for m, k natural as well.
Now let us suppose that \( 1/N [m] s \) is holo in s for m greater than or equal to exponentiation
Define:
\( 2/N [m] s = \frac{1}{\Gamma(-s)} \sum_{k=0}^\infty \frac{(-1)^k}{n! N[m]k (k-s)} + \sum_{k=0}\frac{a_k}{\Gamma(k-s+1)} \)
Where
\( a_k = \sum_{j=0}^\infty \frac{(-1)^j}{j!N[m](k+j)} \)
Now do the same trick (hope fully you get what I mean so that I don't have to reexplain) so that we get:
\( 2 / w[m]s = \frac{1}{\Gamma(-w)} \sum_{k=0}^\infty \frac{(-1)^k}{k!(k[m]s)(k-w)} + \sum_{k=0}^\infty \frac{b_k}{\Gamma(k-w+1)} \)
And do the same trick one more time to get, \( w + s \neq 0 \), and \( w \cdot s \neq 0 \):
\( 2 / w [ z] s = \frac{1}{\Gamma(-z)} \sum_{k=0}^\infty \frac{(-1)^k}{k!(w[k]s)(k-z)} + \sum_{k=0}^\infty \frac{c_k}{\Gamma(k-z+1)} \)
This may or may not work depending on if the functions you defined are holo morphic. Mine are.
They should obey the recursion by some fractional calculus theorems I've given, it's a little clunky. but Not very hard to show. They may not equal your operators though, you'll have to do the calculations :p
I don't really want to spend the time to rigourously justify this. It's very time consuming and it requires a lot of finesse to pull off. But I can see the cauchy integrals coming out as they should after about 3 pages of equations for one identity. lol
It depends on a lot of factors but the summation formula I gave may or may not work.
I think the recursion is more aptly written by the general law:
\( a [x] \alpha + \beta = (a[x] \alpha)[x-1](a[x] \beta) \)
So it's almost like it's necessarily satisfying the exponential law. Which is really cool now that I think about it.
Let's see. I'm thinking there's trouble evaluating \( a[n]k \) for arbitrary natural n and arbitrary natural k and complex a. So let's be a bit more modest with our findings. So let us fix a to N for some natural N
\( N [m] k \) is definitely calculatable for m, k natural as well.
Now let us suppose that \( 1/N [m] s \) is holo in s for m greater than or equal to exponentiation
Define:
\( 2/N [m] s = \frac{1}{\Gamma(-s)} \sum_{k=0}^\infty \frac{(-1)^k}{n! N[m]k (k-s)} + \sum_{k=0}\frac{a_k}{\Gamma(k-s+1)} \)
Where
\( a_k = \sum_{j=0}^\infty \frac{(-1)^j}{j!N[m](k+j)} \)
Now do the same trick (hope fully you get what I mean so that I don't have to reexplain) so that we get:
\( 2 / w[m]s = \frac{1}{\Gamma(-w)} \sum_{k=0}^\infty \frac{(-1)^k}{k!(k[m]s)(k-w)} + \sum_{k=0}^\infty \frac{b_k}{\Gamma(k-w+1)} \)
And do the same trick one more time to get, \( w + s \neq 0 \), and \( w \cdot s \neq 0 \):
\( 2 / w [ z] s = \frac{1}{\Gamma(-z)} \sum_{k=0}^\infty \frac{(-1)^k}{k!(w[k]s)(k-z)} + \sum_{k=0}^\infty \frac{c_k}{\Gamma(k-z+1)} \)
This may or may not work depending on if the functions you defined are holo morphic. Mine are.
They should obey the recursion by some fractional calculus theorems I've given, it's a little clunky. but Not very hard to show. They may not equal your operators though, you'll have to do the calculations :p
I don't really want to spend the time to rigourously justify this. It's very time consuming and it requires a lot of finesse to pull off. But I can see the cauchy integrals coming out as they should after about 3 pages of equations for one identity. lol