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+--- Thread: rules for mult, exp, tetr of tetrations (/showthread.php?tid=77)
rules for mult, exp, tetr of tetrations - Matt D - 10/23/2007
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?
RE: rules for mult, exp, tetr of tetrations - GFR - 10/23/2007
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
RE: rules for mult, exp, tetr of tetrations - andydude - 10/23/2007
Indeed, the only law of tetration is:
\( {}^{y}x = x^{\left({}^{(y-1)}x\right)} \)
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
RE: rules for mult, exp, tetr of tetrations - bo198214 - 11/02/2007
GFR 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!
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
\( f(x)\cdot f(y)=f(x\cdot y) \) must be of the form \( f(x)=x^c \) or \( f(x)\equiv 0 \). The proof goes like this:
First we see by substituting \( y=1 \) that either \( f(1)=1 \) or \( f(x)\equiv 0 \).
Then for \( f(x)\not\equiv 0 \) show by induction that \( f(x^n)=f(x)^n \) for natural \( n \). Then extend this law to integers by \( 1=f(x^{-n+n})=f(x^{-n})f(x^n) \) and rational numbers by \( f((x^{1/n})^n)=f(x^{1/n})^n \). So that we have \( f(x^q)=f(x)^q \) for any fraction \( q \) and by continuity this is even valid for each real number \( q \).
Then every positive real number can be expressed as \( x=e^{\ln(x)} \) and hence \( f(x)=f(e)^{\ln(x)}=x^{\ln(f(e))}=x^c \).
So this law nearly defines e#x to be a power (as the law \( f(x+y)=f(x)f(y) \) plus continuity forces \( f(x)=f(1)^x \), 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:
\( (e\cdot x)\cdot y=e\cdot(x\cdot y) \). 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 \( f(x)=x^x \) then a#(2^n)=\( f^{\circ n}(a) \). The function \( f \) 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=\( f^{\circ \log_2(x)}(a) \).
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*