New tetration method based on continuum sum and exp-series

[bo198214]

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

Really really interesting.

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

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

However, when viewed in the complex plane, we see that exp-series are just Fourier series

\( f(z) = \sum_{n=-\infty}^{\infty} a_n e^{i \frac{2\pi}{P} n z} \).

which represent a periodic function with period \( P \).

Thus it would seem that only periodic functions can be continuum-summed this way. Tetration is not periodic, so how could this help?

This is not completely true. The regular tetration in the base range \( (1,e^{1/e}) \) has the form
\( \sigma(z)=\eta(se^{\kappa z}) + \lambda \)
where \( \eta \) is a holomorphic function with \( \eta(0)=0 \) and \( \eta'(0)=1 \) (this is the inverse of the Schröder function), \( \lambda \) is the fixed point and \( \kappa=\ln(f'(\lambda)) \) (and s is some arbitrary constant which you would choose to ascertain that \( \sigma(0)=1 \).

So it is \( 2\pi i/\kappa=2\pi i/\ln(\ln \lambda) \) periodic.

If I consider Kneser's approach for \( b>e^{1/e} \) we can obtain a related representation.
Kneser's approach also starts with the regular iteration at a complex fixed point. Lets call the corresponding superfunction \( \sigma_0 \) which is not real on the real axis. If we call the real superfunction \( \sigma \) then we get that \( \theta(z)=\sigma^{-1}_0(\sigma(z))-z \) is a 1-periodic function, i.e.

\( \sigma(z)=\sigma_0(z+\theta(z))=\eta(s e^{\kappa(z+\theta(z))})+\lambda \)

This is not periodic, nonetheless we can expand it similar to a periodic function.
\( \sigma(z)=\lambda+\sum_{n=1}^\infty \eta_n e^{n\kappa z} e^{n\kappa \theta(z)} \) (for simplicity consdier \( s^n \) contained in \( \eta_n \)).
The last factor is a 1-periodic function:
\( e^{n\kappa\theta(z)}=\rho(z)^n=\sum_{k=-\infty}^{\infty} \rho^n_k e^{2\pi i k z} \)
If we insert this into our previous equation:
\( \sigma(z)-\lambda=\sum_{n=1}^{\infty}\sum_{k=-\infty}^{\infty} \eta_n \rho^n_k e^{(\kappa n + 2\pi i k) z}=\sum_{n=1}^{\infty}\sum_{k=-\infty}^{\infty} \sigma_{n,k} e^{(\kappa n + 2\pi i k) z} \)

So we have a double exponential series instead of a single series, but nevertheless you again can apply your exponential summation. Though I in the moment have not the time to carry it out myself (so either you do it or I do it later).