03/25/2015, 07:29 AM
That is a fantastic paper!
I have an idea on how to extend this to real bases greater than eta
Define, for integer \( n \), the super root \( \text{srt}_n(z) \) as the inverse of tetration in the base so that we have \( ^n\text{srt}_n(z) = z \)
Then we simply apply your method to factor this in \( n \):
\( \vartheta(z,w) = \sum_{n=0}^{\infty} \text{srt}_{n+1}(z)\frac{w^n}{n!} \)
\( \text{srt}_{n+1}(z) = \frac{\mathrm{d}^n }{\mathrm{d} w^n} |_{w=0} \vartheta(z,w) \)
Then we can simply invert in z.
...unfortunately, there does not seem to be a nice recursion relation between super roots that would force the result to be a tetration.
But it seems to converge numerically to the super root for your tetration.
I have an idea on how to extend this to real bases greater than eta
Define, for integer \( n \), the super root \( \text{srt}_n(z) \) as the inverse of tetration in the base so that we have \( ^n\text{srt}_n(z) = z \)
Then we simply apply your method to factor this in \( n \):
\( \vartheta(z,w) = \sum_{n=0}^{\infty} \text{srt}_{n+1}(z)\frac{w^n}{n!} \)
\( \text{srt}_{n+1}(z) = \frac{\mathrm{d}^n }{\mathrm{d} w^n} |_{w=0} \vartheta(z,w) \)
Then we can simply invert in z.
...unfortunately, there does not seem to be a nice recursion relation between super roots that would force the result to be a tetration.
But it seems to converge numerically to the super root for your tetration.
