sexp by continuum product ?

[JmsNxn]

That's interesting, especially about the values converging to the gamma function. That pretty much sets me in that the method works. The obvious open question though is how to exactly apply it to tetration.

I see your argument that:
\( \frac{d}{dx} \ln^{\circ 0.5}(x) = \frac{1}{\prod_{k=0}^{-0.5} \ln^{\circ k}(x)} \)

or generally:
\( \frac{d}{dx} \ln^{\circ r}(x) = \frac{1}{\prod_{k=0}^{r-1} \ln^{\circ k}(x)} \)


Hmm, just doing some algebra:
\( \frac{d}{dx}\ln^{\circ r}(\ln^{\circ -r}(x)) = \frac{\frac{d}{dx}(x) \ln^{\circ -r}(x)}{\prod_{k=0}^{r-1} \ln^{\circ k-r}(x)} = 1 \)

which means:
\( \frac{d}{dx} \ln^{\circ -r}(x) = \prod_{k=0}^{r-1} \ln^{\circ k-r}(x) \)

which means:
\( \exp^{\circ r}(x) = \int e^{\sum_{k=0}^{r-1} \ln^{\circ k-r+1}(x)} dx \)

this contradicts
\( \exp^{\circ -r}(x) = \int \frac{dx}{e^{\sum_{k=0}^{r-1} \ln^{\circ k+1}(x)} \)

so i guess we have restrictions on r,

unless, we have the equivalency:
\( e^{\sum_{k=0}^{-r-1} \ln^{\circ k+r+1}(x)} = \frac{1}{e^{\sum_{k=0}^{r-1} \ln^{\circ k+1}(x)} \)

\( \sum_{k=0}^{-r-1} \ln^{\circ k+r+1}(x) = -1 \cdot (\sum_{k=0}^{r-1} \ln^{\circ k+1}(x)) \)

which does not hold for r = -1; and probably not for all r < 0

I wonder for what values it does hold, if any, if it does for r > 0, it already produces some interesting, though probably false identities:

r = 1
\( \sum_{k=0}^{-2} \ln^{\circ k+2}(x) = -1 \cdot (\sum_{k=0}^{0} \ln^{\circ k+1}(x)) = -\ln(x) \)

and
r = 0
\( \sum_{k=0}^{-1} \ln^{\circ k+1}(x) = -1 \cdot (\sum_{k=0}^{-1} \ln^{\circ k+1}(x)) = 0 \) (0 since only zero times negative one is equal itself.)

I'm a little iffy on the restrictions to applying this continuum sum methods. I may just be doing something horribly wrong . Still though, this method looks very interesting.