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 rules for mult, exp, tetr of tetrations Matt D Junior Fellow Posts: 7 Threads: 3 Joined: Oct 2007 10/23/2007, 04:22 AM 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 Member Posts: 174 Threads: 4 Joined: Aug 2007 10/23/2007, 11:07 AM 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 Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/23/2007, 06:24 PM 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 bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 11/02/2007, 06:23 PM 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* « Next Oldest | Next Newest »

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