# Tetration and Fractions

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 $\sqrt{3^7}=46.765...$, so how would one go about getting a value for the first example?

• Give a reference, because this concepts looks like an OVNI in my mathematical references. – Jean Marie May 9 '16 at 18:47
• 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?. – Dave L. Renfro May 9 '16 at 19:01

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"

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.