the distributive property
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the distributive property
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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
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
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
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 
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 Bennet Hyperoperations are a special case of these (with the natural base and ------------------------ 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 If you quickly plot the graps of Notice that the identity element of -------------------------- 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) MSE MphLee Mother Law \((\sigma+1)0=\sigma (\sigma+1)\) S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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
05/25/2014, 09:35 AM
Interesting thing. for integers ranks
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 Probably JmsNxn knows more about it. MSE MphLee Mother Law \((\sigma+1)0=\sigma (\sigma+1)\) S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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
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 imho we should just see if MSE MphLee Mother Law \((\sigma+1)0=\sigma (\sigma+1)\) S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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 |
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