# Tetration Forum

Full Version: Between the ops
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I suppose that between evertyhing is there something. E. g. xy = x*y (multiplication).
I suppose that between every operators must be there a multiplication, let us call it operational multiplication: □
For example:
xy = x*y
123 = 1*100 + 2*10 + 3
¬x = ¬ □ x
¬¬x = ¬ □ ¬ □ x
(x = y) = (x □ = □ y)
etc ...
And as the "normal" multiplication has power and functional multiplication (f o c = f( c )) has functional power (f^oN = f o f o ... o f) as the operational multiplication has operational power and roots: O □ O □ ... □ O = O^□N
For instance:
(¬¬x)^□0.5 = x or ¬x, because id id x = x and ¬¬x = x, right?
What do you think, is it exist or not? Can we substitute operational multiplication with other multiplication, like the functional or not?
Hmm, Xorter, I'm not sure I do actually understand what you're after. But well, in case this is in line...

I've one time considered the relation between $a + b$ and $a * b$ as interpolatable. The key was that $a * b = \log( \exp(a) + \exp(b) )$   and the addition can be written as $\log^0( \exp^0(a) + \exp^0(b) )$ compared with the multiplication $a * b = \log^1( \exp^1(a) + \exp^1(b) )$ . So the idea was to define a continuum of fractional-order operators between "+" and "*" based on fractional iterates of $\log()$ and $\exp()$ .
Well, this has surely a lot of issues, for instance do we want to have that fractional-order operations associative, commutative and so on, remembering, that such properties are reduced when we extrapolate to higher/lower orders by higher/lower iterates of the $\log()$ and $\exp()$ -functions. Also I didn't find this really promising/interesting so I did no more engage much in this ansatz.

Gottfried
(03/30/2017, 07:39 PM)Gottfried Wrote: [ -> ]Hmm, Xorter, I'm not sure I do actually understand what you're after. But well, in case this is in line...

I've one time considered the relation between $a + b$ and $a * b$ as interpolatable. The key was that $a * b = \log( \exp(a) + \exp(b) )$   and the addition can be written as $\log^0( \exp^0(a) + \exp^0(b) )$ compared with the multiplication $a * b = \log^1( \exp^1(a) + \exp^1(b) )$ . So the idea was to define a continuum of fractional-order operators between "+" and "*" based on fractional iterates of $\log()$ and $\exp()$ .
Well, this has surely a lot of issues, for instance do we want to have that fractional-order operations associative, commutative and so on, remembering, that such properties are reduced when we extrapolate to higher/lower orders by higher/lower iterates of the $\log()$ and $\exp()$ -functions. Also I didn't find this really promising/interesting so I did no more engage much in this ansatz.

Gottfried

Gottfried!
I am afraid your topic is not so relevant or close to mine, to this one. But never mind (but please, open a new topic for it). Maybe I was not so clear.
Of course, I am answering for you according to my knowledge.
If I understand your point, then x[.5]y should be equal to log^o.5(exp^o.5(x)+exp^o.5(y)) (between the addition and multiplication), right?
3+3 = 6
3*3 = 9
And look at this function:
https://www.dropbox.com/s/xr0swdwug44abt...5.jpg?dl=0

The fractional (half-)iterate of exp and log functions according to whose graph I created 3[.5]3 is about 4.62197 < 6 which is a paradox if we suppose that x[z]y < x[w]y if z<w, right?
(2exp^o.5(3) = ~11.9136)
I am afraid the formula we would like to use is wrong.
To be honest I thought of the operators between operators, like between two negations...
Upp, so I seem to have missed the point completely. Sorry!

Gottfried
(04/01/2017, 12:10 PM)Gottfried Wrote: [ -> ]Upp, so I seem to have missed the point completely. Sorry!

Gottfried

No problem, my friend.
This is why I am trying to explain it clearlier.
Well, I suppose that an operator must exist between the others (other ops and numbers). For instance: between two tensors: O O x = O^2 x = O □  O x, but what is this? Is it the same as which is between a number and the '='?: So 2  =
This is not so clear like the fractional iterates of hyper ops. But, I think this way is more closeable/easier.