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 the distributive property tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 09/25/2010, 12:05 AM a [r] b = b [r] a = exp^[r]( exp^[-r](a) + exp^[-r](b) ) now the extended distributive property is a [r] ( b [r-1] c ) = ( a [r] b ) [r-1] ( a [r] c ) and once again tetration plays an underestimated important role tommy1729 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 09/28/2010, 12:22 PM (This post was last modified: 09/28/2010, 12:28 PM by tommy1729.) it gets really intresting when we set a = b = x and then take the superfunction. x + x = 2x => 2^r x x * x = x^2 => x^(2^r) x ^ ln(x) => x^(ln(x)^(2^r - 1)) and notice the similarity between the last two !!! ( the first two are similar too but less intresting and better known ) regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 10/13/2010, 11:07 PM perhaps it wasnt clear from the OP but i was playing with the idea of " uniqueness condition " for tetration regarding this " formula " tommy1729 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 05/23/2014, 11:12 PM More precisely : Uniqueness or at least a strong condition : Let B2 > B1 > e^(1/e). Base independance : r > 0 Define for base C : exp_C(z) = c^z a [r]_C b = b [r]_C a = exp_C^[r]( exp_C^[-r](a) + exp_C^[-r](b) ) Conjecture : a [r]_B2 b = b [r]_B2 a = exp_B2^[r]( exp_B2^[-r](a) + exp_B2^[-r](b) ) = a [r]_B1 b = b [r]_B1 a = exp_B1^[r]( exp_B1^[-r](a) + exp_B1^[-r](b) ) Or simply : a [r]_B2 b = a [r]_B1 b. Remark : its known to be true for r a positive integer. (it fails for r = -1 though ) Unfortunately I lost some data concerning this idea. So if you see it mentioned here before , plz let me know. I am also intrested in other " base independance " equations. https://sites.google.com/site/tommy1729/...e-property regards tommy1729 MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 05/24/2014, 09:46 AM As you know these are a generalization of the Bennet Hyperoperations (Commutative Hyperoperations). I usually use this notation for them because i find it very confortable $a \odot_r^k b:={\exp}_k^{\circ r}(log_k^{\circ r}(a)+log_k^{\circ r}(b))$ Bennet Hyperoperations are a special case of these (with the natural base $a \odot_r^e b$) $\odot_0^{K} =+$ $\odot_1^{K} =\cdot$ and $a \odot_{-1}^{+\infty} b$ is the max operation while $a \odot_{-1}^{0} b$ is the min operation(this limit process is related with the litinov-maslov dequantization of the semifield of non-negative real numbers in to the Tropical semifield $\mathbb{T}_{max}$) ------------------------ I apologize in advance if I did not understand your conjecture but i think that is not true at all. In general we have that the operations are different when base changes $a \odot_r^{K} b=a \odot_r^{J} b$ only if $r=0,1$ If you quickly plot the graps of $a \odot_2^{K} x$ for differents bases $K$ you see that the the result changes. Notice that the identity element of $\odot_2^{K}$ is $K$ -------------------------- About tetration: yes! There is an interesting link because an extension of tetration will bring to us the fractional rank operations of this Hyperoperation family (JmsNxn already worked on something similar in his thread on logarithmic semi-operators http://math.eretrandre.org/tetrationforu...+operators) MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 05/25/2014, 12:16 AM Right sorry , I meant for 0 < r < 1. I havent given it any time yet , but what happens if the base goes to 1 from above ? In a limit way ofcourse. regards tommy1729 MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 05/25/2014, 09:35 AM Interesting thing. for integers ranks $r=0$ and $r=1$ the limit $lim_{K\rightarrow 1^+}\odot_{r}^K$ is still addition and multiplication, but between I don't know (but I doubt... I'd except some dequantization phenomenon like for the tropical operations max and min)... If someone has an extension for tetration to the reals he could check if for that extension the operations $\odot_{r \in ]0,1[}^K$ satisfie you conjecture. Probably JmsNxn knows more about it. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 05/26/2014, 11:37 PM Finally , it all comes down to solving : f ( f^[-1](a) + f^[-1](b) ) = a + b. I can explain that if you want, but I think most here understand it by just seeing it. regards tommy1729 MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 05/27/2014, 07:43 AM (This post was last modified: 05/27/2014, 12:13 PM by MphLee.) In fact I don't get it at all, probably is related with iteration theory or idk.. How can help us finding such automorphism of addition? Anyways I guess that probably there are infinite automorphisms...so I do not know where you would end up for example $f=id$ is a trivial solution. imho we should just see if $^{(slog_J(a)+0,5)}J + ^{(slog_J(b)+0,5)}J = { ^{(slog_K(a)+0,5)}K + ^{(slog_K(b)+0,5)}K}$ MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 05/27/2014, 12:23 PM Well we get exp(x) replaced by exp(f(x)) and ln(x) replaced by f^[-1](ln(x)). Hence we arrive at f ( f^[-1](a) + f^[-1](b) ) = a + b ( with f at least C^3 ) now differentiate with respect to a : f ' (f^[-1](a) + f^[-1](b)) / f ' (f^[-1](a)) = 1 substitute X = f(a) , Y = f^[-1](b) : f ' ( a + Y) = f ' (a)^2 Now rewrite f ' = G and differentiate with respect to Y : G ' (Y + a) = 0 Hence G is constant and thus f = lineair. Since f(0) = 0 we conclude f(x) = C x. thus it only holds for r = 0,1 QED regards tommy1729 « Next Oldest | Next Newest »

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