Zeration = inconsistant ? - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Zeration = inconsistant ? (/showthread.php?tid=926) Pages: 1 2 3 RE: Zeration = inconsistant ? - GFR - 10/03/2014 (10/02/2014, 10:58 PM)tommy1729 Wrote: It feels a bit strange ... A so called new concept " zeration " being almost equivalent to max[a,b]. Max[a,b] does not seem so intresting as a function. Max[12,100] = Max[13,100] = Max[91,100] Nothing special. ............... Also there is not much algebra or geometry about Max. Max+ algebra being a big exception. ............... Or ... Finding addional must-have properties. ................ Im sure there are many nice algorithms that use max. ( even without max+ algebra ) Maybe one of those could help us out. --- regards tommy1729 Well, ... why not! As a matter of fact 'Tropical Algebra' and 'Tropical Geometry' seem to use Max-Plus algebra, where the max operation such as: c = a[max]b is one of the basic operations. Together with: d = a[+]b they are used to build an 'Idempotent Semi-Ring'. Not a 'Field', because with [max] we cannot create a 'group' (-oo is the unity element, but there is no 'inverse element'), but a 'monoid'. Nevertheless, the result of this formalism is extremely important, because a lot of successful applications from this new discipline are expected in fields such as the Petri-nets, machine scheduling, discrete event processes (DEP), industrial manufacturing systems, telecommunication networks, parallel processing, coding/decoding systems, traffic control and, last but not least, cellular automata (NKS). Amazing. We need to think about that. Best regards. RE: Zeration = inconsistant ? - GFR - 10/03/2014 (10/02/2014, 11:11 PM)tommy1729 Wrote: But what bothers me most is that zeration , unlike addition and multiplication does not have an inverse !!! a - b , a / b .............. Take the equations : a + b = c a = c - b a * b = c a = c/b max(a,b) + 1 = c a = max^-1(c - 1,b) ?? fail. tommy1729 As a matter of fact, Max-Plus algebra, applied to the 'Reals', is not a 'Field' but an 'Idempotent Semi-Ring', where a[max]b has some 'peculiarities': a[max](-oo) = (-oo)[max]a = a ; -oo is the 'unity element' a[max]a = a ; [max] is idempotent and there is NO max-inverse number. For zeration, this is only partially true. In fact a o (-oo) = (-oo) o a = a ; but: a o a = a + 2. We should discuss here the fact that Rubtsov thinks (and I agree) that zeration HAS an inverse operation, that he calls 'Deltation', producing a new set of numbers, called the 'Delta Numbers', which can be put in correlation with the log of negative numbers (multi-valued Complex numbers, but also ... transfinite extension of the 'Reals'). This would need a separate detailed discussion. A similar position is taken by Cesco Reale, with his definition of 'incrementation' and 'decrementation', giving an extension of the Reals that he calls 'Stigma-Reals'. Also to be seen. Regards! RE: Zeration = inconsistant ? - GFR - 10/04/2014 (10/02/2014, 09:27 PM)tommy1729 Wrote: Hi. Thanks for your swift reply. ....... ... I tried to reverse everything : define zeration in terms of higher operators , instead of lower , and define distributivity in the REVERSE way. It was therefore done intentionally ! NOT to try to ridicule or dismiss zeration but rather as an attempt to " fix it ". I understand it looks like a silly mistake but it was not. Have a bit more confidence in me ...... regards tommy1729 No problem. I trust you, Tommy! As a matter of fact, I think that the higher rank operation should always distribute over the lower rank, if distributivity holds. I mean: a + (b o c) = (a + b) o (a + c), like: a * (b + c) = ab + ac. You tried the reverse way and you didn't succeed. I think that it was normal! Keep it going! Best regards. GFR RE: Zeration = inconsistant ? - GFR - 10/04/2014 (10/01/2014, 08:40 AM)tommy1729 Wrote: ........ To understand tetration you need to understand +,*,^ at least. To understand ^ you need to understand *. To understand * you need to understand +. Defining and understanding a hyperoperator seems to require the lower hyperoperators. But with zeration that is A PROBLEM. ..... ..... properties like distributive require use of a lower hyperoperator : a * (b+c) = a*b + a*c. .... Zeration seems to lack properties and therefore consistancy. regards tommy1729 As a matter of fact, if you apply these principles, you would indeed discover a problem with ... 'Addition', because: "To understand + you (would) need to understand o!". In fact, we should have (sorry for 'o' and, please, forget 'some' useless brackets): a * a = a ^ 2; a * (a * a) = a ^ 3 a + a = a * 2; a + (a + a) = a * 3 a o a = a + 2; a o (a o a) = a + 3. And, moreover, as we have seen: a + (b o c) = (a + b) o (a + c), i.e. the 'descending' distributivity seems to hold! And ... to gain consistency! And, don't forget: 2o2 = 2+2 = 2*2 = 2^2 = 2#2 = .... 2[s]2 = 4. The 'Magic Four' relation! Thank you for your interest. Best regards. GFR RE: Zeration = inconsistant ? - MphLee - 10/04/2014 (10/03/2014, 09:32 PM)tommy1729 Wrote: Thanks MphLee. How was zeration for complex numbers defined ??SEE: http://math.eretrandre.org/tetrationforum/showthread.php?tid=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... RE: Zeration = inconsistant ? - tommy1729 - 10/04/2014 (10/03/2014, 11:29 PM)GFR Wrote: (10/02/2014, 10:58 PM)tommy1729 Wrote: It feels a bit strange ... A so called new concept " zeration " being almost equivalent to max[a,b]. Max[a,b] does not seem so intresting as a function. Max[12,100] = Max[13,100] = Max[91,100] Nothing special. ............... Also there is not much algebra or geometry about Max. Max+ algebra being a big exception. ............... Or ... Finding addional must-have properties. ................ Im sure there are many nice algorithms that use max. ( even without max+ algebra ) Maybe one of those could help us out. --- regards tommy1729 Well, ... why not! As a matter of fact 'Tropical Algebra' and 'Tropical Geometry' seem to use Max-Plus algebra, where the max operation such as: c = a[max]b is one of the basic operations. Together with: d = a[+]b they are used to build an 'Idempotent Semi-Ring'. Not a 'Field', because with [max] we cannot create a 'group' (-oo is the unity element, but there is no 'inverse element'), but a 'monoid'. Nevertheless, the result of this formalism is extremely important, because a lot of successful applications from this new discipline are expected in fields such as the Petri-nets, machine scheduling, discrete event processes (DEP), industrial manufacturing systems, telecommunication networks, parallel processing, coding/decoding systems, traffic control and, last but not least, cellular automata (NKS). Amazing. We need to think about that. Best regards. The max+ algebra is used to find the " shortest path ". Kinda like a discrete version of the calculus of variations , based on isomorphisms with matrices. " shortest paths " are naturally useful in some discrete problems , optimization , graph theory , economics and planning. However saying it is very useful for cellular automatons seems like advertising to me. Afterall many problems in cellular automaton theory , If we can even speak of theory already , do not resemble " shortest paths " but are rather of type halting problems complexity measures randomness measures construction problems problems equivalent to number theory etc. If you can show me how a max+ algebra is used for anything that is quite different from a " shortest path " type of problem that would be appreciated and surprising. regards tommy1729 RE: Zeration = inconsistant ? - tommy1729 - 10/04/2014 (10/03/2014, 11:39 PM)GFR Wrote: (10/02/2014, 11:11 PM)tommy1729 Wrote: But what bothers me most is that zeration , unlike addition and multiplication does not have an inverse !!! a - b , a / b .............. Take the equations : a + b = c a = c - b a * b = c a = c/b max(a,b) + 1 = c a = max^-1(c - 1,b) ?? fail. tommy1729 As a matter of fact, Max-Plus algebra, applied to the 'Reals', is not a 'Field' but an 'Idempotent Semi-Ring', where a[max]b has some 'peculiarities': a[max](-oo) = (-oo)[max]a = a ; -oo is the 'unity element' a[max]a = a ; [max] is idempotent and there is NO max-inverse number. For zeration, this is only partially true. In fact a o (-oo) = (-oo) o a = a ; but: a o a = a + 2. We should discuss here the fact that Rubtsov thinks (and I agree) that zeration HAS an inverse operation, that he calls 'Deltation', producing a new set of numbers, called the 'Delta Numbers', which can be put in correlation with the log of negative numbers (multi-valued Complex numbers, but also ... transfinite extension of the 'Reals'). This would need a separate detailed discussion. A similar position is taken by Cesco Reale, with his definition of 'incrementation' and 'decrementation', giving an extension of the Reals that he calls 'Stigma-Reals'. Also to be seen. Regards! Im skeptical about deltation. regards tommy1729 RE: Zeration = inconsistant ? - tommy1729 - 10/04/2014 (10/04/2014, 11:24 AM)MphLee Wrote: (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. I already made the choice : a [0] b = max(a,b) + 1 + kroneckerdelta(a,b) regards tommy1729 RE: Zeration = inconsistant ? - MphLee - 10/04/2014 (10/04/2014, 11:24 AM)MphLee Wrote: 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... I looked at it better and is easy to write. The solution of the equation $x \odot_e^{a}x=b$ is $x=b\oslash_e^{1+a} {\exp}^{\circ a}(2)$ where $\oslash_e^{1+a}$ is the inverse operation of $\odot_e^{a+1}$ so we should have $x \odot_e^{a}x=x\odot_e^{a+1}{\exp}^{\circ a}(2)$ Anyways i'm not 100% sure. I have to chek it with calm. (10/04/2014, 12:16 PM)tommy1729 Wrote: I already made the choice : a [0] b = max(a,b) + 1 + kroneckerdelta(a,b) regards tommy1729 Ok i get it... Well first of all $(-1)$-ation should have a non-empty intersection with RR-Zeration RR-Zeration: $a [0]_{RR} b = max(a,b) + 1 + \delta_{ab}$ $[0]_{RR} \cap [-1]\neq \emptyset$ Their intersection has to contain a segment of the trivial zeration (the successor of the second argument) because in that segment $(-1)$-ation is the subfunction of RR-Zeration. Anyways I think that if we chose RR Zeration $(-1)$-ation its gonna be a multivalued oepration (see Hyperstructures theory and multimaps and this Brief introduction by Viro, comment by Mphlee) Why? Because its translations are not invertible functions. Anyways I guess that using the Litinov-Maslov's Limit Process we could find a formula that show to us the real shape/behaviour of the set $[-1]$. RE: Zeration = inconsistant ? - tommy1729 - 10/04/2014 Example : Solve exp^[2]( 2 ln^[2](x) ) = 7 This is equivalent to x^ln(x) = 7. Take some real a,b >= exp(1). a_0 = a^{ 1 / ln(b) } b_0 = 7^{ ln(b) / ln(a) } replacement rules : --- a ' = a * 7^{ ln(b) } b ' = a * b --- a_1 = a ' ^{ 1/ln(b ') } b_1 = 7^{ ln(b ') / ln(a ') } repeat forever lim n-> oo a_n/b_n = x. this gives x = exp( sqrt( ln(7) ) ) as it should. Numerically we get x = 4.0348084730118923250275859453110072467762717139110... Notice 1/x is also a solution. If we take 0 < a,b < 1/exp(1) ( =exp(-1) ) we probably achieve that. This numerical algorithm can probably be improved with adding some + operators at the right places ... Still investigating. Notice I did not use a derivative anywhere like most numerical methods do. Also the method can probably be extended nicely to all interpretations of hyperoperators. Extending towards complex numbers however seems difficult. (unless we split up by variables for real and imag part ) Can zeration inprove this algoritm ? regards tommy1729 " So he can probably do math better than most cows , so what ? " Ullrich mocking me on sci.math. " Truth is that what does not go away when you stop believing in it " tommy1729