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Recently I discovered Tetration, and was wondering about having tetration with fractional "tetronents", take the example $$^{7/2}3\;\Bbb{or}\;3\uparrow\uparrow{\frac72}$$Initially it seems difficult to conceptualize, as we end up with something like the following: $$\underbrace{3^{3^{3^{\cdots}}}}_{7/2\;\Bbb{times}}$$ But equally fractional exponents make little sense either: $\underbrace{3\times3\times\cdots\times3}_{7/2\;\Bbb{times}}$ as you can't really find the product of three and a half threes, yet we know that the answer is $\sqrt[2]{3^7}=46.765...$, so how would one go about getting a value for the first example?

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  • $\begingroup$ Give a reference, because this concepts looks like an OVNI in my mathematical references. $\endgroup$ – Jean Marie May 9 '16 at 18:47
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    $\begingroup$ In this 8 April 2002 sci.math post I considered this question in the section called "hyper-roots" (a term I made up when I posted this). Others had previously considered the idea (I give some references), and several people worked on this after my post appeared -- much in sci.math, where it's hard to find, but I believe several people have incorporated most of it into their web pages and Wikipedia. See also How to evaluate fractional tetrations?. $\endgroup$ – Dave L. Renfro May 9 '16 at 19:01
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In short: I don't see any possibility to introduce associativity between iteration-height (here $h=7/2$) and base (here $b=3$) such that $^7(^{1/2}3)$ and $ (^{7 \cdot {1/2}})3$ (or even $ ^{1/2}(^73)$) relate easily.


But there is an ansatz to assume the notation $$ z_h = \exp^{°h}_b(z_0) $$ as an iteration with fractional iteration-"height" $h$ of the exponentiation with base $b$ , so for integer iteration heights $h=1$ one has $z_1 = \exp_b(z_0) = b^{z_0} $, for $h=2$ one has $z_2 = \exp_b(z_1) = b^{z_1} = { b^{b^{\large z_{\Tiny 0}}}} $ , and so on.

With this, there are some general solutions proposed how to actually compute this for some arithmetically expressed $h=j+k$ such that $ z_h = z_{j+k} = \exp_b^{°k}(z_j) = \exp_b^{°k}(\exp_b^{°j}(z_0))= \exp_b^{°j+k}(z_0)$ . In that solutions the arithmetic in the iteration-height-argument can also include operations like $j=3, k=0.5$ or $j=4,k=-0.5$ , both giving $h=7/2 = 3.5$ .

The key is here, that the base-parameter for the exponentiation $b$ is constant for the whole part of arithmetic in this expressions - no "superroot" is required/involved.


[update] A rough numerical approximation might be the following:
$$ z_0=1 \\ z_{0.5} = \exp_3^{°0.5}(z_0) \approx 1.70683220772 \\ z_{1.5} = 3^{z_{0.5}} \approx 6.52176075446 \\ z_{2.5} = 3^{z_{1.5}} \approx 1293.21482372 \\ z_{3.5} = 3^{z_{2.5}} \approx 1.04780222080 E617$$ This approximation (of $z_{0.5}$) was done by a simple approximate Carlemanmatrix-approach, and likely this $z_{0.5}$ is to a handful of digits approximate to a full-fledged Kneser-type approximation, which would possibly the best estimate in a real-to-real sense.

[update2] A program for the Kneser-type approximation has been given in the tetration-forum (Sheldon Levenstein, also on MSE). It gives for $z_{0.5} \approx 1.70683\;1091 $ and then for $z_{2.5} \approx 1293.2\;2610395 $ and then $z_{3.5} \approx 1.0\;60867555 \; \text E617 $ and which seem usually consistent to more than $20$ digits.


If the older discussions here in MSE(and even in MathOverflow) are not instructive enough for you, I think wikipedia has some references to such proposed solutions; a somehow prominent one is that of Hellmuth Kneser who showed a real-to-real solution for the fixed base $b=\exp(1)$ and real fractional $h$ and real $z_0$. Another (published) proposal is by D. Kousnetzov (extending this to complex $b$, $h$, $z_0$) and some various attempts promoted/discussed in the "tetration-forum" - the entry in wikipedia gives some references at its end.
A visualization and rough comparision of five such attempts for a real base $b=4$ and complex heights $h$ are shown in my essay "5 methods"

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I would start by doing a similar analogy to exponentiation. $3^{\frac{7}{2}} = 3 \times 3 \times 3 \times \sqrt{3} = 46.765... $ so we could easily take it to the next level. I define the second tetrational root as $strt(x) = y \text{ where } y^{y} = x$. Assuming that the answer is 2, you can do trial and improvement on a calculator to determine that strt(3) = 1.825455022925... Now for the answer! $3 \uparrow\uparrow \frac{7}{2} = 3^{3^{3^{strt(3)}}} = 3^{3^{3^{1.825455022925}}} = 5.6803367136 \times 10^{1,672}$. That is a big number.

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According to the Wikipedia section https://en.wikipedia.org/wiki/Tetration#Complex_heights, tetration to the base 2 can be defined for all complex numbers other than those that are real numbers $\leq$ -2 using the real number relation $\leq$, and then can be defined on the for real numbers from by resricting the operation on complex numbers to real numbers. I'm not sure whether its definition for complex numbers is undefined for real numbers $\leq$ -2 but if it is defined for some of them, it probably wouldn't be defined using the real number tetration to the base 2 operation. I haven't learned the exact definitions but I'm pretty sure I'm right about those properties of the way it was defined. Those who insist on being so careful not to say anything wrong probably end up almost never saying anything and contributing so little to research. Although you probably can't compute in terms of an arbitrary real number > -2 its tetration to the base 2 just by applying elementry operations, I'm pretty sure there is a way to compute it.

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