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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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