New tetration method based on continuum sum and exp-series

[tommy1729]

Hi.

We have

\( f(z) = \sum_{n=0}^{\infty} a_n b^{nz} \)

or, even better

\( f(z) = \sum_{n=-\infty}^{\infty} a_n b^{nz} \).

Then,

\( \sum_{n=0}^{z-1} f(n) = \sum_{n=-\infty}^{\infty} a_n \frac{b^{nz} - 1}{b^n - 1} \)

with the term at \( n = 0 \) interpreted as \( a_0 z \), so,

\( \sum_{n=0}^{z-1} f(n) = \left(\sum_{n=-\infty}^{-1} \frac{a_n}{b^n - 1}\right) + \left(\sum_{n=1}^{\infty} \frac{a_n}{b^n - 1}\right) + a_0 z + \left(\sum_{n=-\infty}^{-1} \frac{a_n}{b^n - 1} b^{nz}\right) + \left(\sum_{n=1}^{\infty} \frac{a_n}{b^n - 1} b^{nz}\right) \).

Thus it seems this continuum sum is recovering all the expected sums and extensions. So the question comes up: what happens if we use it on Tetration, to sum up Ansus' continuum sum formula

\( \log_b\left(\frac{\mathrm{tet}'_b(z)}{\mathrm{tet}'_b(0) \log(b)^z}\right) = \sum_{n=0}^{z-1} \mathrm{tet}_b(n) \)

?

I don't yet have a really fast and efficient numerical program ready to go, but the idea behind the algorithm I'm using and the current code I can post in the Computation forum if you'd like.

but you dont have the coefficients of ansus tet_b , so no taylor series ?

and tet_b is not periodic either ?

yes plz explain how you compute it !

regards

tommy1729