Keiji wrote:Though, if we only consider tetration to evaluate from right to left, i.e. x#4 = x^(x^(x^x)), can you suggest what could be done about a fractional, irrational, negative, zero or complex "tetrapower"?

There are already several suggestions for such an extension to the reals available (see my

article in sci.math.research). It might be also very interesting to search for tetration at sci.math.research.

The problem of those extensions really is to find uniqueness criterions. Similarely there are no uniqueness criterions for fractional iteration of iterated exponentiation e<sup>e<sup>e<sup>....x</sup></sup></sup>.

The problem of uniqueness indeed does not occur for fractional trees, they are somehow already designed to be the unique extension.

Keiji wrote:I tried to read your PDF file, but I got lost around page 2. Maybe I'll try reading it again in a few years, when I understand more math.

Yeah, I mean its nearly graduate stuff. But hey, though the topic is difficult, the main idea of the fractional trees is not that difficult to grasp.

We start with some simple expressions (without inversion):

for example

x<sup>x</sup> corresponds to [1]

x<sup>xx</sup>=x<sup>x<sup>2</sup></sup> corresponds to [1,1]

x<sup>x<sup>xx</sup>x</sup> corresponds to [[1,1],1]

In this way we can encode each expression we get from x and application of exponentiation. Especially x corresponds to []=1. We write this correspondence with the operation %, for example [[1],1]%x = x<sup>x<sup>x</sup>x</sup>

We see that we can rearrange the contents between the brackets without changing the meaning, for example [a,b,c]=[b,c,a]. Such a structure is called multiset, or bag in computer science. The difference to a set is that an element can occur multiple times, for example [a,a,a,b]. And indeed a=b (recursively as multisets), if a%x=b%x for all x (which is not exactly easy to show).

Till now there is nothing fractional about it. But we can define already an addition that should correspond to exponentiation:

(a+b)%x = (b%x)<sup>a%x</sup>

and a multiplication which corresponds to function composition:

(a.b)%x = a%(b%x)

On the multisets this looks as follows

a + [b<sub>1</sub>,...,b<sub>k</sub>] = [a,b<sub>1</sub>,...,b<sub>k</sub>]

[a<sub>1</sub>,...,a<sub>m</sub>].[b<sub>1</sub>,...,b<sub>n</sub>] = [a<sub>1</sub>.b,...,a<sub>m</sub>.b,b<sub>1</sub>,...,b<sub>n</sub>]

for example

([1,1]%x) <sup>[1]%x</sup> = (x<sup>xx</sup>)<sup>x<sup>x</sup></sup> = x<sup>xxx<sup>x</sup></sup> = [1,1,[1]] % x

and

[1]%([1,1]%x) = [1]%(x<sup>xx</sup>) = (x<sup>xx</sup>)<sup>x<sup>xx</sup></sup> = [[1,1],1,1]%x = ([1].[1,1])%x

But now we know that all those functions built up by x and exponentiation of terms are continuous and strictly increasing, at least for x>1, moreover they reach every value y>1. This means they have a unique inversion, for each value y>1 there is exactly one value x>1 such that a%x=y. Note however, that the demand x>1 is really necessary, because for example x<sup>x</sup> has a minimum at 1/e and so every y-value in (1/e)<sup>1/e</sup>..1 is assumed twice in the interval (0..1). Its a bit similar to the monomials x<sup>n</sup>: For even numbers n they have two solutions x<sup>n</sup>=y for y>0 and no solution for y<0. While for uneven n they have exactly one solution in (-oo,oo). In the same way standard tetration for even exponents has either no solution, one solution, or two solutions in (0..1) while for uneven exponents there is exactly one solution in the (0..oo).

So we can simply add inversions, and additions and multiplications of inversions, and their inversions, etc. to our multisets.

For example [1,1]<sup>-1</sup>%x would then denote the inverse function of x<sup>xx</sup>. And [1,[1]<sup>-1</sup>]%x = <sup>xf(x)</sup> where f(x) is the inverse function of x<sup>x</sup>. Etc. . Of course I am a bit sloppy here because one have to show that all those new functions are again invertible for x>1, so that the inversion is indeed applicable to all terms.

And the interesting thing is now that that each such term* with inversions can be brought into the form* [a<sub>1</sub>,...,a<sub>m</sub>]<sup>-1</sup>.[b<sub>1</sub>,...,b<sub>n</sub>] where the a<sub>i</sub> and the b<sub>i</sub> are again of this form* (or empty). But this is the form* of a fraction: a<sup>-1</sup>.b, isnt it?! And that's why the name fractional trees. And there is also a cancellation rule which though I will not explicate this time for brevity. There are rules for multiplication, division and addition. You can calculate with them similar to fractional numbers, though non-commutativity of multiplication makes everything quite harder. For calculating/experimenting a bit with those fractional trees, I provided already the link to the

fractional tree calculator.