Why tetration breaks.

[Keiji]

I was thinking about this. Anyway, recall how you were taught the definition of exponentiation:

a¹ = a
a² = aa
a³ = aaa
a⁴ = aaaa
etc.

Now, we can say a^b+c = a^ba^c. Why? Here's why:

a²a² = (aa)(aa) = aaaa = a⁴
a³a⁴ = (aaa)(aaaa) = aaaaaaa = a⁷
etc.

Now this may seem blindingly obvious, but we cannot say the same for tetration. Using # to indicate tetration:

a#1 = a
a#2 = a^a
a#3 = a^{a^a}
a#4 = a^{a^{a^a}}
etc.

So:

(a#2)^(a#3) = (a^a)^{(a^{a^a})} = a^{aa^{a^a}} ≠ a^{a^{a^{a^a}}}

Hence (a#2)^(a#3) ≠ a#5, hence (a#b)^(a#c) ≠ a#(b+c) - proof by counterexample.

For this reason, we CANNOT define a#0 = a constant (even if you do assume that a#(2b) = (a#b)^(a#b), you'll get different results for different values of b), and we cannot define what a#b is where b is negative.

----------------------------

For integers b, c: a^b/c = c¬a^b, where a¬b means a'th root of b. Why? Taking integer values of b/c, examples work:

a² = a^4/2 = 2¬a⁴ = 2¬(aaaa) = 2¬((aa)(aa)) = aa
a³ = a^9/3 = 3¬a⁹ = 3¬(aaaaaaaaa) = 3¬((aaa)(aaa)(aaa)) = aaa
a⁵ = a^15/3 = 3¬a¹⁵ = 3¬(aaaaaaaaaaaaaaa) = 3¬((aaaaa)(aaaaa)(aaaaa)) = aaaaa
etc.

Now, let us assume a#(b/c) = c$(a#b), where a$b means a'th tetraroot of b.

We know that 2$(a^{a^aa}) = 2$(a^a)^{(a^a)} = a^a = a#2

Now 2$(a#4) = 2$(a^{a^{a^a}}) ≠ 2$(a^{a^aa}), so 2$(a#4) ≠ a#2, so 2$(a#4) ≠ a#(4/2), so a#(b/c) ≠ c$(a#b) - proof by counterexample again.

For this reason, we cannot define a#b for a rational non-integer b.

I suggest we find some way of defining a#b for rational non-integer or or zero or negative b.

[bo198214]

Yes, very well observed (sorry for my late reply, I missed this thread).

But the interesting thing we can do for non-associative operations (associative means (a*b)*c = a*(b*c) ) is using binary trees instead of natural numbers. The binary trees code the bracketing, for example

(1,(1,1)) *' x = x * (x * x)
((1,1),(1,1)) *' x = (x*x) * (x*x)

Where * is some (possibly non-associative operation) and *' is its successor operation.
If the operation * is associative then the bracketing does not matter and we can use natural numbers instead of binary trees, for example

3 .' x = (1+(1+1)) .' x = x.(x.x) = x³ = (x.x).x = ((1+1)+1) .' x
So to say, the natural numbers are the associative binary trees.

First we notice that we can define all higher (Ackerman/hyper) operations also on the binary trees.
We start with the addition a + b := (a,b)
And for each operation we define its successor operation as above and hence have arbitrary high operations on the binary trees.

Especially we have a multiplication
a.b := a +' b
which we can also imagine as hooking the tree b into each 1 of the tree a.
For example
(1,(1,1)).b=(1,(1,1))+' b = b+(b+b) = (b,(b,b))

Interestingly the multiplication on the binary trees is associative.
So for the next higher operation we merely need natural numbered exponents.
Let |a| denote the number of leaves of a, for example
|(1,(1,1))|=3
so to say the associative version of the tree. Then we can easily verify
|a+b|=|a|+|b|
|a.b|=|a|.|b|
|a.'b|=|b...b|=|b|^|a|

And the good thing is now that for all higher operations * on the binary trees (except +) the following equation is valid:

1*x=x
(a.b)*x = a*(b*x)

And this is a key point to uniquely define inverse operations. On the natural numbers it is given for multiplication and exponentiation:
1.x = x
(a.b).x = a.(b.x)
and
x¹ = x
x^a.b = (x^a)^b

but as we have seen, it is no more valid for the higher operation tetration on the natural numbers.

So the idea is now to extend the binary trees by division in the same way as the natural numbers are extended to the fractional numbers. And further extend it similar to the real numbers and there define unambigously all higher operations. I wrote down the first part here and also figured out the second part, though its not written yet and there seems also no interest in the scientific community for it.

[bo198214]

For the purpose of tetration it even suffice to only regard those extended binary trees as (tetration)exponents while the base remains a real number. Moreover a specialized variant which I call fractional trees. You have then the usual computation laws as for the exponents of powers, except that multiplication is no more commutative.

There is a straight forward definition of tetration t#x with fractional
trees t as (tetration)exponents and real numbers x>1 as base.
In which for example the following laws hold.

(a+b) # x = (b#x)^a#x
a.b # x = a # ( b # x )
(a^-1) # (a # x) = a # ( a^-1 # x ) = x

Those fractional trees are an intricate and intrigued structure and
you can find a computation experimentation page
here. Basicly however they are recursive fractions.

[Keiji]

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.

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"?

[bo198214]

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^{e^{e^....x}}.

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

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^x corresponds to [1]
x^xx=x^x² corresponds to [1,1]
x^{x^xxx} 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^{x^xx}

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)^a%x

and a multiplication which corresponds to function composition:

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

On the multisets this looks as follows
a + [b₁,...,b_k] = [a,b₁,...,b_k]
[a₁,...,a_m].[b₁,...,b_n] = [a₁.b,...,a_m.b,b₁,...,b_n]

for example
([1,1]%x) ^[1]%x = (x^xx)^{x^x} = x^{xxx^x} = [1,1,[1]] % x
and
[1]%([1,1]%x) = [1]%(x^xx) = (x^xx)^{x^xx} = [[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^x has a minimum at 1/e and so every y-value in (1/e)^1/e..1 is assumed twice in the interval (0..1). Its a bit similar to the monomials xⁿ: For even numbers n they have two solutions xⁿ=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]^-1%x would then denote the inverse function of x^xx. And [1,[1]^-1]%x = ^xf(x) where f(x) is the inverse function of x^x. 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* [a₁,...,a_m]^-1.[b₁,...,b_n] where the a_i and the b_i are again of this form* (or empty). But this is the form* of a fraction: a^-1.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.

[The_Tydra]

I believe I have the answer to your questions:

Left to right tetration is actually very easy to define, it doesn't even require a new operation. The reason for this is because of the fact that (x^a)^b=x^(a*b). Taking this into account something like x^x^x^x can be reduced to x^(x*x*x), which can again be reduced to x^(x^3). So, in the tetration problem x#y, you can easily change it to x^(x^(y-1)), which allows calculations to be extended to just about any value.

On the right to left tetration (for which I will use the symbol $ to represent). There's still a lot of mystery as far as tetration by non-interger values goes, to tell the truth I don't even know where to start. Tetration by negative values fairly simple up to a point. You see, by definition, x$y=logx(x$(y+1)). Hence if we started out with 2$1 you could define this as log2(4), which confirms that 2$1 is 2. Then we need only take the log2(2), which equals 1, then log2(1) which equals 0. This gives us values for x$0 and x$-1, however it breaks down at x$-2 as there is no defined value for log2(0).

That's all I've got, if anyone has a way of calculating values for anything like x$.5 or x$i I'd love to see it.

[bo198214]

I believe I have the answer to your questions

So what answer to what question do you have?

That's all I've got, if anyone has a way of calculating values for anything like x$.5 or x$i I'd love to see it.

As I already told somewhere in the forum, there are various different continuations of tetration, mostly to mention Galidakis'. The problem is of finding "the unqiue" one.