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 \( \Psi(n,x) \) to be the n'th super root

\( ^n \Psi(n,x) = x \)

is the function

\( g(u,x) = \sum_{n=0}^\infty \Psi(n+1,x)\frac{u^n}{n!} \)

differintegrable? And if differintegrated, does it equal the super root for arbitrary \( z \)?

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 \( 0 < \sigma < 1 \) we do have

\( \int_0^\infty |g(-u,x)|u^{-\sigma}\,du < \infty \)

Therein defining the function for \( \Re(z) > 0 \)

\( \Psi(z,x) = \frac{1}{\Gamma(1-z)}(\sum_{n=0}^\infty \Psi(n+1,x)\frac{(-1)^n}{n!(n+1-z)} + \int_1^\infty g(-u,x)u^{-z}\,du) \)

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

If I can show that \( \frac{d}{dx}\Psi(z,x) \neq 0 \) then we can invert this function across \( x \), to get \( F \) the function \( \Psi(z,F(z,x)) = x \).

Taking this function observe the following

\( r(z) = \Psi(z,x)^{F(z,\Psi(z,x))} = \Psi(z,x)^x \) is factorizable in \( z \) for x close enough to 1.

Just as well

\( a(z) = F(z+1,\Psi(z,x)) \) is factorizable in \( z \) for x close enough to 1.

Factorizable implies we can use the natural identity theorem. This means if \( a|_{\mathbb{N}} = r|_{\mathbb{N}} \) then \( a = r \)

Well...

\( r(n) = \Psi(n,x)^x \)

and

\( a(n) = F(n+1,\Psi(n,x)) = (^{n+1}\Psi(n,x)) = \Psi(n,x)^{^n\Psi(n,x)} = \Psi(n,x)^x \)

and therefore

\( r(z) = a(z) \)

What this means, unraveling the cryptic writing, is if \( y = \Psi(z,x) \) then

\( y^{F(z,y)} = F(z+1,y) \)

TETRATION!

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

\( \frac{d}{dx}\Psi(z,x) \neq 0 \)

for \( \Re(z) > M \) 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 \( \Psi(n,x) \) to be the n'th super root

\( ^n \Psi(n,x) = x \)

is the function

\( g(u,x) = \sum_{n=0}^\infty \Psi(n+1,x)\frac{u^n}{n!} \)

differintegrable? And if differintegrated, does it equal the super root for arbitrary \( z \)?

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 \( 0 < \sigma < 1 \) we do have

\( \int_0^\infty |g(-u,x)|u^{-\sigma}\,du < \infty \)

Therein defining the function for \( \Re(z) > 0 \)

\( \Psi(z,x) = \frac{1}{\Gamma(1-z)}(\sum_{n=0}^\infty \Psi(n+1,x)\frac{(-1)^n}{n!(n+1-z)} + \int_1^\infty g(-u,x)u^{-z}\,du) \)

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

If I can show that \( \frac{d}{dx}\Psi(z,x) \neq 0 \) then we can invert this function across \( x \), to get \( F \) the function \( \Psi(z,F(z,x)) = x \).

Taking this function observe the following

\( r(z) = \Psi(z,x)^{F(z,\Psi(z,x))} = \Psi(z,x)^x \) is factorizable in \( z \) for x close enough to 1.

Just as well

\( a(z) = F(z+1,\Psi(z,x)) \) is factorizable in \( z \) for x close enough to 1.

Factorizable implies we can use the natural identity theorem. This means if \( a|_{\mathbb{N}} = r|_{\mathbb{N}} \) then \( a = r \)

Well...

\( r(n) = \Psi(n,x)^x \)

and

\( a(n) = F(n+1,\Psi(n,x)) = (^{n+1}\Psi(n,x)) = \Psi(n,x)^{^n\Psi(n,x)} = \Psi(n,x)^x \)

and therefore

\( r(z) = a(z) \)

What this means, unraveling the cryptic writing, is if \( y = \Psi(z,x) \) then

\( y^{F(z,y)} = F(z+1,y) \)

TETRATION!

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

\( \frac{d}{dx}\Psi(z,x) \neq 0 \)

for \( \Re(z) > M \) 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.