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 Distributive "tetration", is this possible JmsNxn Long Time Fellow Posts: 503 Threads: 89 Joined: Dec 2010 08/29/2011, 10:52 PM (This post was last modified: 08/30/2011, 12:38 AM by JmsNxn.) We all know of right hand tetration, evaluated right to left, and left hand tetration, evaluated left to right. which are both defined with different purposes in mind. Right handed tetration is the superfunction of exponentiation, left handed tetration is analytic very simply. But what about an operator that continues the distributivity found across (exponentiation, multiplication) and (multiplication, addition). Or in other words, can such an operator $\otimes_\mu$ exist? if: $(a^b)\,\otimes_\mu\, c\,=\, (a\,\otimes_\mu\,c)^{(b\,\otimes_\mu\,c)}$ And if such is the case, what is it's operator generating function $\mu$ in the sense that: $a\,\otimes_\mu\,b = \mu(\mu^{-1}(a)+\mu^{-1}(b))$ $\mu(x)$ would then be the superfunction of $\mu(1)\,\otimes_\mu\,x$. this would make our distributive tetration commutative, if of course, such a function $\mu$ exists. The term "operator generating function" is novel and is generalized from the fact that exponentiation is the operator generating function of multiplication, insofar as: $a \cdot b = \exp(\exp^{-1}(a) + \exp^{-1}(b))$ and the identity function is the operator generating function of addition. $a + b = \text{id}(\text{id}^{-1}(a) + \text{id}^{-1}(b))$ Interesting things occur with this new operator. I'll start by listing them now: $(a^a)\,\otimes_\mu\,b = (a\,\otimes_\mu\,b)^{(a\,\otimes_\mu\,b)} = (b^b)\,\otimes_\mu\,a$ which occurs more generally as: $^ba \,\otimes_\mu\,c = \,^bc\,\otimes_\mu\,a$ which means if this operator turns out to be consistent, then we have a very handy tool for base conversion. and it also means: $^c [\mu(\mu^{-1}(a) + \mu^{-1}( b ) ) ]\,=\,\mu(\mu^{-1}(^c a) + \mu^{-1} ( b )) =\, \mu(\mu^{-1}(a) + \mu^{-1}(^c b))$ which says more simply: $^c (a\,\otimes_\mu\,b) = (^c a\,\otimes_\mu\, b) = (a\,\otimes_\mu\,^c b)$ which says tetration spreads across $\otimes_\mu$ the same way differentiation spreads across convolution. We also have weird results for $\otimes_\mu$ when 1 is an argument: $(a^1)\,\otimes_\mu\,c = (a\,\otimes_\mu\,c)^{1\,\otimes_\mu\,c}$ and therefore: $1\,\otimes_\mu\,c = 1$ This gives the limit: $\lim_{a\to\infty} \mu(a) = 1$ which is a law given by all operator generating functions. we also find that: $(a^0)\,\otimes_\mu\,c =\,(a\,\otimes_\mu\,c)^{(0\,\otimes_\mu\, c)} = 1$ which means $(0\,\otimes_\mu\, c) = 0$ So therefore the only way to keep this working is to have: $\lim_{a\to -\infty} \mu(a) = 0$ Therefore $\mu$'s inverse is either multivalued, or on the real axis $\mu$ increases from zero to one from across negative infinity to infinity. I'm hoping then that complex values in the domain open up the range a whole lot more. the identity of $\,\otimes_\mu\,$ is $\mu(0)$ If we define the inverse of $\,\otimes_\mu\,$ as $a\,\oslash_\mu\,b = \mu(\mu^{-1}(a) - \mu^{-1}(b))$ and, if $\otimes_\mu$ is the superfunction of $\oplus_\mu$: then we are given: $a\,\oplus_\mu\,b = \,a\,\otimes(b\,\oslash_\mu\, a + 1)$ so that: $a\,\otimes_\mu\, 2 = a\, \oplus_\mu\, a$ $a\,\otimes_\mu\, 3 = a\, \oplus_\mu\, a\,\oplus_\mu$ $(a+b)\,\otimes_\mu\,c = (a\,\otimes_\mu\,c)\,\oplus_\mu\,(b\,\otimes_\mu\,c)$ and $(a\,\oplus_\mu\,b)\,\otimes_\mu\,c = (a\,\otimes_\mu\,c)+(b\,\otimes_\mu\,c)$ which is a general law for all operators with a operator generating function. Anyway, my question to all of you, is if any of this is consistent. I always have trouble poking holes in the ideas I have, and other people seem to be a lot sharper at doing it. Therefore I ask you guys, my fellow tetrationers. regards, James tommy1729 Ultimate Fellow Posts: 1,464 Threads: 351 Joined: Feb 2009 08/31/2011, 09:34 PM im afraid it cannot all be true james. the basic problem is that some equations require your ' operator ' to be commutative while others forbid it. try to solve the first equation first without many fantasy equations. that might save the idea for positive reals with some luck. regards tommy1729 JmsNxn Long Time Fellow Posts: 503 Threads: 89 Joined: Dec 2010 09/03/2011, 07:22 PM Hmm yes, I think you're right. I think it may be pushing it to have it be commutative. Sadly we lose it's base changing properties. But the original equation is intriguing, I'd really like to see if it would even be possible. Off the bat we can rearrange it to make one the identity: $c\, \otimes_\mu\, 1 = c$ and $1\,\otimes_\mu\, c = 1$ we'd also get $0\,\otimes_\mu\, c = 0$ I'd like to continue looking at this. tommy1729 Ultimate Fellow Posts: 1,464 Threads: 351 Joined: Feb 2009 09/05/2011, 11:30 PM http://math.eretrandre.org/tetrationforu...hp?tid=520 did you consider this ? JmsNxn Long Time Fellow Posts: 503 Threads: 89 Joined: Dec 2010 09/05/2011, 11:52 PM I've considered those operators before, I don't think they are related to this operator however. I think the best way to start with this operator, is by observing it's base changing properties. If $a\, \otimes_\mu\, f(a, b) = b$ then $^t a\, \otimes_\mu\, f(a,b) =\, ^t b$ So by observing the original base change formula, we should be on our way to finding a possible solution. I think if we want to disprove its existence we'd best look at this relation: $(^t a)\,\otimes_\mu\, b = \,^t(a\, \otimes_\mu\,b)$ this seems like the most unlikely property for any operator to have. tommy1729 Ultimate Fellow Posts: 1,464 Threads: 351 Joined: Feb 2009 09/05/2011, 11:55 PM (09/05/2011, 11:52 PM)JmsNxn Wrote: I've considered those operators before, I don't think they are related to this operator however. I think the best way to start with this operator, is by observing it's base changing properties. If $a\, \otimes_\mu\, f(a, b) = b$ then $^t a\, \otimes_\mu\, f(a,b) =\, ^t b$ So by observing the original base change formula, we should be on our way to finding a possible solution. I think if we want to disprove its existence we'd best look at this relation: $(^t a)\,\otimes_\mu\, b = \,^t(a\, \otimes_\mu\,b)$ this seems like the most unlikely property for any operator to have. yes , try a = sqrt(2) and t = oo regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,464 Threads: 351 Joined: Feb 2009 09/05/2011, 11:58 PM (09/05/2011, 11:55 PM)tommy1729 Wrote: (09/05/2011, 11:52 PM)JmsNxn Wrote: I think if we want to disprove its existence we'd best look at this relation: $(^t a)\,\otimes_\mu\, b = \,^t(a\, \otimes_\mu\,b)$ this seems like the most unlikely property for any operator to have. yes , try a = sqrt(2) and t = oo regards tommy1729 -- unless you use my distributive formula -- « Next Oldest | Next Newest »

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