Well about a year ago I posted my paper on the bounded analytic hyper-operators. The result was straight forward and involved expressing functions using the Mellin transform (or the exponential differintegral as I expressed it). One question that arose is the following,

If we define to be the n'th super root

is the function

differintegrable? And if differintegrated, does it equal the super root for arbitrary ?

Well I recently thought more about this question, and using an invaluable Lemma I proved I've found that in fact, for x>1 and we do have

Therein defining the function for

Now here's the really beautiful part, and sadly where I am stuck.

If I can show that then we can invert this function across , to get the function .

Taking this function observe the following

is factorizable in for x close enough to 1.

Just as well

is factorizable in for x close enough to 1.

Factorizable implies we can use the natural identity theorem. This means if then

Well...

and

and therefore

What this means, unraveling the cryptic writing, is if then

TETRATION!

Therefore the requirement boils down into showing the rather painful fact that

for for some M > 0. This will successfully construct a tetration function. I'm at a loss frankly on how to show this. The rest fell into place rather easily, nothing too exhaustive was required.

It is obviously true for most x>1, but not for all. There may be exceptional values. Therefore we do have a tetration solution using the differintegral given some conditions... Numerical evidence seems to support the result too but I've only tried a little bit.

If we define to be the n'th super root

is the function

differintegrable? And if differintegrated, does it equal the super root for arbitrary ?

Well I recently thought more about this question, and using an invaluable Lemma I proved I've found that in fact, for x>1 and we do have

Therein defining the function for

Now here's the really beautiful part, and sadly where I am stuck.

If I can show that then we can invert this function across , to get the function .

Taking this function observe the following

is factorizable in for x close enough to 1.

Just as well

is factorizable in for x close enough to 1.

Factorizable implies we can use the natural identity theorem. This means if then

Well...

and

and therefore

What this means, unraveling the cryptic writing, is if then

TETRATION!

Therefore the requirement boils down into showing the rather painful fact that

for for some M > 0. This will successfully construct a tetration function. I'm at a loss frankly on how to show this. The rest fell into place rather easily, nothing too exhaustive was required.

It is obviously true for most x>1, but not for all. There may be exceptional values. Therefore we do have a tetration solution using the differintegral given some conditions... Numerical evidence seems to support the result too but I've only tried a little bit.