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 Zeration = inconsistant ? MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 10/04/2014, 11:24 AM (10/03/2014, 09:32 PM)tommy1729 Wrote: Thanks MphLee. How was zeration for complex numbers defined ??SEE: http://math.eretrandre.org/tetrationforu...122&page=3 (02/24/2008, 11:02 AM)bo198214 Wrote: I just wanted to show that from the by you given conditions, i.e. aoa=a+2 and ao(aoa)=a+3, etc, from which follows ao(a+n)=a+n+1 for n>1, even if we enhance these conditions, by letting n be real, the by you given zeration is not a consequence but rather an example that satisfies these conditions. However that the condition does not contain a complex but only real n, does not mean that zeration itself can not be defined on the complex numbers (though you did not define it on the complex numbers). This would be *my* definition for *complex* $a,b$: $a\circ b = \begin{cases}b+1 & : \Re(b)>\Re(a)+1\\a+2&: \Re(b)\le \Re(a)+1\end{cases}$ Let us verify the conditions: $a\circ a = a+2$ because $\Re(a)\le \Re(a)+1$, let $x>1$ then $a\circ(a+x)=a+x+1$ because $\Re(a+x)>\Re(a)+1$ and for natural $x$ the other conditions follow $a\circ (a\circ a)=a\circ(a+2)=a+3$, etc. (10/03/2014, 09:32 PM)tommy1729 Wrote: Question : a [-1] b = ?? a [0] ( b [-1] c ) = (a [0] b) [-1] (a [0] c) downation ? regards tommy1729I don't think that this question really makes sense. If we want to define it we should start from some concept of Zeration, thus, once we obtain $(-1)$-ation it will be only relative to the Zeration concept chosen. It makes even less sense if we think that we don't even know if Zeartion should be distributive: at this point we can safely say that is matter of choice IMHO. (10/03/2014, 09:41 PM)tommy1729 Wrote: As for the exp^[a]( ln^[a](x) + ln^[a](y) ) case , I think I have something intresting ; numerical methods independant of a. Need more research. Looks promising. Ok ok , I say a bit more : solving for x when given a and b : exp^[a]( 2 ln^[a] (x) ) = b seems to have a method. regards tommy1729That would be pretty awesome. This means tha you could solve the "square-root" problem for all the Commutative Hyperoperations. $a \odot_e^{a}b:=\exp^{\circ a}( ln^{\circ a}(x) + ln^{\circ a}(y) )$ the solution of the equation $x \odot_e^{a}x=b$ should be, as you say $\exp^{\circ a}( 2 ln^{\circ a} (x) ) = b$ $x={\exp}^{\circ a}(\frac{ln^{\circ a}(b)}{2})$ I guess is possible to do more here... I'll try to find a forumula only involving homomorphic operators defined via exponential... MathStackExchange account:MphLee « Next Oldest | Next Newest »

 Messages In This Thread Zeration = inconsistant ? - by tommy1729 - 10/01/2014, 08:40 AM RE: Zeration = inconsistant ? - by MphLee - 10/01/2014, 11:27 AM RE: Zeration = inconsistant ? - by GFR - 10/02/2014, 02:44 PM RE: Zeration = inconsistant ? - by tommy1729 - 10/02/2014, 09:27 PM RE: Zeration = inconsistant ? - by MphLee - 10/02/2014, 10:02 PM RE: Zeration = inconsistant ? - by GFR - 10/04/2014, 09:58 AM RE: Zeration = inconsistant ? - by tommy1729 - 10/02/2014, 10:58 PM RE: Zeration = inconsistant ? - by GFR - 10/03/2014, 11:29 PM RE: Zeration = inconsistant ? - by tommy1729 - 10/04/2014, 12:11 PM RE: Zeration = inconsistant ? - by tommy1729 - 10/02/2014, 11:11 PM RE: Zeration = inconsistant ? - by GFR - 10/03/2014, 11:39 PM RE: Zeration = inconsistant ? - by tommy1729 - 10/04/2014, 12:12 PM RE: Zeration = inconsistant ? - by MphLee - 10/03/2014, 09:20 AM RE: Zeration = inconsistant ? - by tommy1729 - 10/03/2014, 09:32 PM RE: Zeration = inconsistant ? - by tommy1729 - 10/03/2014, 09:41 PM RE: Zeration = inconsistant ? - by GFR - 10/04/2014, 10:19 AM RE: Zeration = inconsistant ? - by MphLee - 10/04/2014, 11:24 AM RE: Zeration = inconsistant ? - by tommy1729 - 10/04/2014, 12:16 PM RE: Zeration = inconsistant ? - by MphLee - 10/04/2014, 12:58 PM RE: Zeration = inconsistant ? - by tommy1729 - 10/04/2014, 10:20 PM RE: Zeration = inconsistant ? - by MphLee - 10/05/2014, 03:36 PM

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