Fractional tetration method

[Koha]

\( x = e \uparrow \uparrow \frac{1}{2} = \operatorname{ssrt}(e) \iff x \uparrow \uparrow 2 = x^{x} = e \implies \ln(x^{x}) = x \ln x = e^{\ln x} \ln x = \ln e = 1 \implies W(e^{\ln x}) = \ln x = W(1) = \Omega \implies x = \boxed{e \uparrow \uparrow \frac{1}{2} = e^{\Omega}} \)

Is this method known? This is just an example.



My theorem:


\( \textbf{Notation} \)

\( \text{Superroot of orden \(n\):} \quad x = \sqrt[b]{a}_{\mathrm{s}^{n-1}} \overset{\text{def}}{\iff} x \uparrow^n b = a \)
\( \text{Superlogarithm of order \(n\):} \quad x = \ln(a)_{\mathrm{s}^{n-1}} \overset{\text{def}}{\iff} e \uparrow^n x = a \)


\( \textbf{Theorem} \)

\( Y\big(\ln(y \uparrow^n q)_{\mathrm{s}^{n-2}} \, ; q, n\big) \overset{\text{def}}{=} \ln(y)_{\mathrm{s}^{n-2}} \)

\( \forall a \in \mathbb{C}^*,\ n \in \mathbb{N}_{\ge 2},\ p \in \mathbb{N}_0,\ q \in \mathbb{Z}^*, \)

\( \boxed{a \uparrow^n \frac{p}{q} = \sqrt[q]{a \uparrow^n p}_{\mathrm{s}^{n-1}} = e \uparrow^{n-1} Y\big(\ln(a \uparrow^n p)_{\mathrm{s}^{n-2}} \, ; q, n\big)} \)


\( \textbf{First corollary} \)

\( Y(z \, ; 2, 2) \equiv W(z) \)

I know! The notation is (probably) terrible, but the flavor is in the logic behind it, or so I think. I have been doing some calculations and... yes, it works, although there may be some error in how I wrote it.


PD: I have used \( \operatorname{ssrt}(z) \) as \( \sqrt[2]{z}_{\mathrm{s}} \).

Thanks a lot.

[MphLee]

Your very first assumption seems flawed. I believe it was prooven inconsistent years ago. I don't remember the prof nor the thread.
Anyways you should check if your definition \( ^{\frac{1}{n}}b:={\rm sqrt}_n (b)\) is coherent with the law \( ^{q+1}b =\exp_b(^{q}b) \) for each \(q\) rational number..

I have the feeling that you will find a contraddiction somwhere.

[Pentalogue]

\( x = e \uparrow \uparrow \frac{1}{2} = \operatorname{ssrt}(e) \iff x \uparrow \uparrow 2 = x^{x} = e \implies \ln(x^{x}) = x \ln x = e^{\ln x} \ln x = \ln e = 1 \implies W(e^{\ln x}) = \ln x = W(1) = \Omega \implies x = \boxed{e \uparrow \uparrow \frac{1}{2} = e^{\Omega}} \)

Is this method known? This is just an example.



My theorem:


\( \textbf{Notation} \)

\( \text{Superroot of orden \(n\):} \quad x = \sqrt[b]{a}_{\mathrm{s}^{n-1}} \overset{\text{def}}{\iff} x \uparrow^n b = a \)
\( \text{Superlogarithm of order \(n\):} \quad x = \ln(a)_{\mathrm{s}^{n-1}} \overset{\text{def}}{\iff} e \uparrow^n x = a \)


\( \textbf{Theorem} \)

\( Y\big(\ln(y \uparrow^n q)_{\mathrm{s}^{n-2}} \, ; q, n\big) \overset{\text{def}}{=} \ln(y)_{\mathrm{s}^{n-2}} \)

\( \forall a \in \mathbb{C}^*,\ n \in \mathbb{N}_{\ge 2},\ p \in \mathbb{N}_0,\ q \in \mathbb{Z}^*, \)

\( \boxed{a \uparrow^n \frac{p}{q} = \sqrt[q]{a \uparrow^n p}_{\mathrm{s}^{n-1}} = e \uparrow^{n-1} Y\big(\ln(a \uparrow^n p)_{\mathrm{s}^{n-2}} \, ; q, n\big)} \)


\( \textbf{First corollary} \)

\( Y(z \, ; 2, 2) \equiv W(z) \)

I know! The notation is (probably) terrible, but the flavor is in the logic behind it, or so I think. I have been doing some calculations and... yes, it works, although there may be some error in how I wrote it.


PD: I have used \( \operatorname{ssrt}(z) \) as \( \sqrt[2]{z}_{\mathrm{s}} \).

Thanks a lot.

The point is that the tetration root of the value B from the number A finds such a number C that, when raised to the tetration of B, gives the same number A.
The tetration root of the value A is not equal to the tetration with the index 1/A, since these two functions are not communicative, unlike the usual raising to a power and the root.