Natural Tetration

[pat]

I was just playing around with a different approach to tetration. What I've said below can apply to any ordered field (a field with a well-defined "less-than-or-equal-to" operation), I think. But, for those not too savvy with fields and groups and such, I'm going to focus on the real numbers, which is what we're after anyhow. I will parenthetically throw in stuff that you can ignore if you're not into group theory.

Let's start with (an ordered field, for instance) the real numbers. We can look at the relationship between addition and multiplication. We can then try to define a relationship on the positive numbers which has the same relationship to multiplication that multiplication had to addition.

Because phpbb makes it annoying to use so many of the useful HTML character entities, I'm going to just use + for addition, * for multiplication and # for my tetration. So, we're looking for an operation # so that # is to * as * is to + except that we're only going to use the positive numbers for the first half of the analogy.

Let's start by enumerating the laws of (fields like) the real numbers. First, the addition laws:

• a + b = b + a
• a + ( b + c ) = ( a + b ) + c
• There exists a 0 so that a + 0 = a for all a
• For every a, there exists a unique number -a so that a + -a = 0

Then, the multiplication laws (for non-zero numbers):

• a * b = b * a
• a * ( b * c ) = ( a * b ) * c
• There exists a 1 so that a * 1 = a for all a
• For every a, there exists a unique number a^-1 so that a * a^-1 = 1

And a distributive law:

• a * ( b + c ) = ( a * b ) + ( a * c )

From the above, we can prove that a * 0 = 0 for all a. So, except for the inverses, we can do all of the above with 0, too.

Now, by analogy, the tetration laws must be (for positive numbers):

• a # b = b # a
• a # ( b # c ) = ( a # b ) # c
• There exists an E so that a # E = a for all a
• For every a not equal to 1, there exists a unique number a so that a # a = E

And a distributive law:

• a # ( b * c ) = ( a # b ) * ( a # c )

By analogy with the a * 0 = 0, we have that a # 1 = 1 for all positive a.

From that, we get some fun things like: ( a # b )^-1 = ( a # b^-1 ). The multiplicative inverse of the tetration of a and b is the tetration of a with b's multiplicative inverse. And, because tetration is symmetric, we can do it again on the a side to find that ( a # b ) = ( a^-1 # b^-1 ). The tetration of a and b is equal to the tetration of the inverses of a and b. This is analagous to the fact that ( a * b ) = ( (-a) * (-b) ).

I then started playing around with positive integer powers trying to evaluate ( a^m # bⁿ ) in terms of ( a # b ). It turns out that, because of the distributive law, we must have that ( a^m # bⁿ ) = ( a # b )^m*n. I can prove that by induction in a separate reply if people want to see it. But, for now, suffice it to say that it is analogous to the fact that (a*m) * (b*n) = (a*b) * (m*n) when we replace multiplication by addition and tetration by multiplication. And, we can even extend the powers to negative numbers because of the previous paragraph.

So, given that, I started playing around. The first operation that I hit on which works (and, I'm pretty sure it's unique up to the choice (and, yes, I do mean "choice") of E) was:

• ( a # b ) = E^{log_Ea * log_Eb} = a^log_Eb

We can pick pretty much any positive number except 1 to be E. (Actually, there may be some lurking reason why E must be greater than one, too... I haven't checked on that yet.) (In a finite field, we have to pick E to be a generator of the multiplicative group, otherwise we won't be able to calculate the logarithms.) Since we're concerned with the real numbers, let's just let E be e, the base of the natural logarithms (hence my title "Natural Tetration"). So, we can write:

• ( a # b ) = e^{ln(a) * ln(b)} = a^ln(b)

It is clearly symmetric, so we know that (a # b) = (b # a). And, we can show it's associative:

• a # ( b # c ) = a # ( e^{ln(b) * ln(c)} )
  = e^{ln(a) * ln( e[sup]ln(b) * ln(c)} )[/sup]
  = e^{ln(a) * ln(b) * ln(c)}
  = ( e^{ln(a) * ln(b)} ) # c = ( a # b ) # c

For the tetrational inverses, we can see that:

• a = e^{1 / ln(a)} for a not equal to 1

That is, the tetrational inverse of a is the ln(a)-th root of e. ...say that five times fast...

The distributive law is also easy:

• a # ( b * c ) = a^{ln( b * c )}
  = a^{ln(b) + ln(c)}
  = a^ln(b) * a^ln(c)
  = ( a # b ) * ( a # c )

So, there you have it. Tetration that works, naturally. It gets around the problems raised in the "Why tetration breaks" thread.

As mentioned above, it works for any choice of E. For example, if we prefer base-10 logarithms, we can instead let E = 10. If we do that, then:

• ( a # 10 ) = a for all a
• ( 100 # 100 ) = 10,000
• ( 1000 # 100 ) = ( 100 # 1000 ) = 1,000,000 ... 1000 squared is 100 cubed
• ( 1000 # 1000 ) = 1,000,000,000
• ( 10,000 # 1000 ) = ( 1000 # 10,000 ) = 1,000,000,000,000
• ( 10,000 # 10,000 ) = 10,000,000,000,000,000
• ( 2 # 2 ) = 1.232023688689006...
• ( 20 # 20 ) = 49.2809475475602...

[bo198214]

There are some flaws in your consideration.

1. it is not tetration - the 4th operation - but it is a "triation" - the 3rd operation (i.e. the one that follows the multiplication).

That does of course not really matter because you can define arbitrary high operations by this approach, for example

x ## y = exp(exp( ln(ln(x))*ln(ln(y)) ))
x ### y = exp(exp(exp( ln(ln(ln(x)))*ln(ln(ln(y))) )))

or you could even define negative operations

x -# y = log(exp(x)+exp(y))
x -## y = log(log( exp(exp(x)) + exp(exp(y)) ))

One can show that (R,+,*) is isomorphic to to (R₊,*,#) via exp.
Where we could replace *,# by any two successive operations of that hierarchy and R₊ by the corresponding exp(exp(....(exp(R))...)).

2. Unfortunately it is not what was intended. So to say the distributive law is wrong. Look for example at the normal 3rd operation, the exponentiation **. And compare your and the normal distributive law:

x ** (y+z) = (x ** y) * (x ** z)
x # (y*z ) = (x # y) * (x # z)

distribution is over + and not * and changes then into the previous operation. In the same way one would try to define tetration ***:

x *** (y+z) = (x***y) ** (x***z)

But this is of course not possible because + is associative and commutative and ** neither. So the compromise is to use a certain bracketing for example:

x *** (1+z) = (x***1) ** (x***z) = x ** ( x***z )

which is the normal (inductive) definition of tetration for natural tetr.-exponents.

(The other sided bracketing:
x *** (y+1) = (x***y) ** (x***1) = (x***y) ** x = x ** ( x ** y)
reduces to just two exponentiations and is thatswhy not so interesting.)

3. To retain the desired distribution law x *** (y+z) = (x***y) ** (x***z) we can change to a number domain where + is neither associative nor commutative. This then leads to my arborescent numbers, where though the multiplication is again associative.

[Keiji]

Lots of words

That's not tetration, in fact it's no operation generalized under the hyper function. You're just making some completely arbitrary operation.

Exponentiation is defined as multiplying a number x times by itself.
Tetration is defined as taking a number x times to its own power.

The operation you've come up with is to do with neither of those...

[pat]

Okay, so I shouldn't have used the word "tetration" unless I understood it better. You're right, my ( a # n ) does not mean a raised to the a raised to the a ... so there are n a's.

For me, the goal was to find an operation that shared the same relationship to multiplication as multiplication had to addition.

I'll have to go back and look at what operations "the hyper function" generalizes.

[bo198214]

As a side note, the operation you describe or better the sequence of operations (k) defined inductively by

x (0) y = x + y
x (k+1) y = exp( log(x) (k) log(y) )

especially x(2)y is your x # y

was already considered by A. A. Benett 1915 in a 3 page article "Note on an operation of the third grade".
His question there is how we could define x(n)y also for real/complex values n. And he naturally ends up thinking of how to define continuously iterated exp. Though nothing substantial is contained in that article, its interesting that it was already considered at that time.

As a second side note, my master's theses advisor showed in his school time (there was a program called "youth is researching" (in translation)), that (R,+,*) is isomorphic to (R₊,*,#).

[Keiji]

I'll have to go back and look at what operations "the hyper function" generalizes.

hyper₁(a,b) = a + b
hyper_n(a,2) = hyper_n-1(a, a)
hyper_n(a,b) = hyper_n-1(a, hyper_n-1(a, b-1))

[bo198214]

Originally what you call hyper function, is called orignal Ackermann function (which though starts at index 0 not at 1).

A_n(x,y)

There are surely many other ways (due to bracketing) how to define hyperfunctions. My favorite (which I will call here #₂):

x #₂ 0 = x
x #₂ n = (x #₂ (n-1) ) ^ (x #₂ (n-1))

We see that in the expression of x #₂ n are 2ⁿ x's contained. If we had a continuation of #₂ to real x, we could define an alternate tetration x # n that only contains n x's, by:

x # n = x #₂ (log₂ n)

Interestingly for #₂ we have some compatibility in the exponent, i.e.:

x #₂ (n+m) = (x #₂ n) #₂ m

that means we can trace it back to iteration of functions, i.e. let f(x) = x^x
then x #₂ n = fⁿ(x).

Now there is a unique analytic iteration of f(x)=x^x and in the end we can define a continuous tetration as:

x # y = f^{log₂ y}(x)

The difference to our normal tetration $ is the bracketing, for example we have:

x $ 4 = x^(x^(x^x))
x # 4 = (x^x)^(x^x)

[PWrong]

For me, the goal was to find an operation that shared the same relationship to multiplication as multiplication had to addition.

That's exponentiation, it's already sorted out. What we want is an operation that continues the sequence addition, multiplication, exponentiation...

[bo198214]

For me, the goal was to find an operation that shared the same relationship to multiplication as multiplication had to addition.

That's exponentiation, it's already sorted out.

No, in his sense it is not exponentiation. His third operation # satisfies
x # (a*b) = (x#a)*(x#b) and not
x ^ (a+b) = (x^a)*(x^b)

I mean it is well known that the functional equation
f(a*b)=f(a)*f(b) has the only solutions f_p(x) = x^p and the identical to 0 function. Excluding the trivial case by demanding x # 1 = 1, i.e. f(1)=1, we get

x # y = y^F(x) for some function F.
If we now additionally demand the other sided distribution law
(x * y) # a = (x#a) * (y#a)
i.e.
a^F(x*y) = a^F(x) a^F(y)
we have
F(x*y)=F(x)+F(y)

which is well known to have the only solutions F_c(x) = log_c(x)

hence the by pat designed operations of the third grade are the only possible operations that satisfy his equations.

x #_c y = y^log_c(x) = c^{log_c(x) * log_c(y)}