rules for mult, exp, tetr of tetrations

[Matt D]

This has probably been addressed already, but...

we have the rules for exponentiation like

e^x*e^y = e^(x+y) and
(e^x)^y = e^(x*y) and
(e^x)^(e^y) = e^(xe^y)

so for tetration we might expect

(where ^xe means e tetrated to x) (and * means multiplication)

^xe*^ye = ^(x*y)e
^y(^xe) = ^(x^y)e
^(^ye)(^xe) = ^(x^(^ye))e

translated:

e tetrated to x times y tetrated to e = (x times y) tetrated to e

(e tetrated to x) tetrated to y = e tetrated to (x to the y)

(e tetrated to x) tetrated to (e tetrated to y) = e tetrated to (x to the (e tetrated to y))

does this sound reasonable?

[GFR]

Well, you write:
e#x*e#y , which means (e#x)*(e#y) and you are right because bracketing is not necessary, due to the (hyper)operation's prìorities, and then if you mean (which is not exactly what you wrote):

e#x*e#y = e#(x*y) as a clone of e^x*e^y = e^(x+y) unfortunately, it's wrong
(e#x)#y = e#(x^y) as a clone of (e^x)^y = e^(x*y) it is also wrong
(e#x)#((x^e)#y) as a clone of (e^x)^(e^y) = e^((x*e)^y)) ... idem .... !!

No luck! This is the question. The problem is much more complicated. The hyper-exponents of tetration don't follow the same or similar rules of the lower rank (hyper)-operations.

No problem, we shall overcome!

GFR

[andydude]

Indeed, the only law of tetration is:



which is also the definition of tetration, so it is more a matter of where this definition leads, and it has been found that it leads to a contradiction when your equations are also assumed. So we must conclude that those equations do not hold for tetration.

Andrew Robbins

[bo198214]

e#x*e#y = e#(x*y) as a clone of e^x*e^y = e^(x+y) unfortunately, it's wrong
(e#x)#y = e#(x^y) as a clone of (e^x)^y = e^(x*y) it is also wrong
(e#x)#((x^e)#y) as a clone of (e^x)^(e^y) = e^((x*e)^y)) ... idem .... !!

No luck!

As this board is open to any kind of higher (or super-, or hyper-) operations, why always assume the standard right-bracketed 4th operation. Lets see what we can make of the rules:

1. e#x*e#y = e#(x*y) as a clone of e^x*e^y = e^(x+y)
Here it is well known that any continuous solution of the functional equation
must be of the form or . The proof goes like this:
First we see by substituting that either or .
Then for show by induction that for natural . Then extend this law to integers by and rational numbers by . So that we have for any fraction and by continuity this is even valid for each real number .

Then every positive real number can be expressed as and hence .

So this law nearly defines e#x to be a power (as the law plus continuity forces , proof left to the reader.) and is surely of no use for defining hyper operations.

2. (e#x)#y = e#(x^y) as a clone of (e^x)^y = e^(x*y)
This seems to me being a wrong generalization, as the multiplicative law is also valid one step lower:
. So we rather would demand that for all higher operations (e#x)#y=e#(x*y).

And this even possible, with the balanced tetration: We first define
a#n for n being a power of 2 by
a#(2^0)=a
a#(2^(n+1))=(a#(2^n)) ^ (a#(2^n))

For example a#2=a^a and a#4=(a^a)^(a^a)=a^(a*a^a)

And this can be easily handled by regular frational iterations. Let then a#(2^n)=. The function is analytic, has a fixed point at 1, so it is possible to compute the unique regular iteration at this fixed point and then to define:

a#x=.
So it is possible to (even uniquely) define a tetration based on the above multiplicative law.

3. (e#x)#((x^e)#y) as a clone of (e^x)^(e^y) = e^((x*e)^y))
seems to ugly for me *gg*