I've been doing a lot of research in areas around linear operators and I've found the following theorem. If it's useful I'll prove how I got it. If not I won't.

We can express the following:

\( \bf{E}f(s) = f(e^s) = \frac{1}{\pi} \int_1^{\infty} f(-u) \int_{-\infty}^{\infty} sin{\pi t}\,\cdot\, e^{-\pi i t}\,\cdot\,\frac{s^t}{t^2}\,\cdot\,\ln(u)^{-t}\,\partial t \partial u \)

\( \bf{E}^{-1}f(s) = f(\ln(s)) = \frac{-1}{\pi} \int_{-\infty}^{\infty} f(u) \int_{-\infty}^{\infty} sin{\pi t}\,\cdot\, e^{(u-\pi i )t}\,\cdot\,\frac{s^t}{t^2}\,\cdot\,\partial t \partial u \)

And so therefore we can write:

\( \exp^{\circ\,y}(s) = \bf{E}^{y-1} e^s \)

If you can't notice \( \bf{E} \) is a linear operator; so:

\( \bf{E}(\alpha f + \beta g) = \alpha \bf{E} f + \beta \bf{E} g \)

This result is quite elaborate to prove and requires knowledge of Hilbert spaces. I just found this expression recently of a more general result that I am more interested in. We must remember this is right hand composition.

Nonetheless; is it easier to iterate a linear operator than how we usually do it? I can apply these methods for pentation and every hyper operator; formally; without considering convergence of the integrals. I'm still in the baby steps.

Questions, comments?

I can do everything I just did for iteration of any base as well. Not sure about convergence though. We can actually turn every super-function into iteration of a linear transformation. However; again; formally; not sure about convergence. I'm working on a paper that proves all of this but feedback helps; maybe someone's seen this.

Edit: What's cool about this is we do not require a fixpoint!

We can express the following:

\( \bf{E}f(s) = f(e^s) = \frac{1}{\pi} \int_1^{\infty} f(-u) \int_{-\infty}^{\infty} sin{\pi t}\,\cdot\, e^{-\pi i t}\,\cdot\,\frac{s^t}{t^2}\,\cdot\,\ln(u)^{-t}\,\partial t \partial u \)

\( \bf{E}^{-1}f(s) = f(\ln(s)) = \frac{-1}{\pi} \int_{-\infty}^{\infty} f(u) \int_{-\infty}^{\infty} sin{\pi t}\,\cdot\, e^{(u-\pi i )t}\,\cdot\,\frac{s^t}{t^2}\,\cdot\,\partial t \partial u \)

And so therefore we can write:

\( \exp^{\circ\,y}(s) = \bf{E}^{y-1} e^s \)

If you can't notice \( \bf{E} \) is a linear operator; so:

\( \bf{E}(\alpha f + \beta g) = \alpha \bf{E} f + \beta \bf{E} g \)

This result is quite elaborate to prove and requires knowledge of Hilbert spaces. I just found this expression recently of a more general result that I am more interested in. We must remember this is right hand composition.

Nonetheless; is it easier to iterate a linear operator than how we usually do it? I can apply these methods for pentation and every hyper operator; formally; without considering convergence of the integrals. I'm still in the baby steps.

Questions, comments?

I can do everything I just did for iteration of any base as well. Not sure about convergence though. We can actually turn every super-function into iteration of a linear transformation. However; again; formally; not sure about convergence. I'm working on a paper that proves all of this but feedback helps; maybe someone's seen this.

Edit: What's cool about this is we do not require a fixpoint!