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help with a distributivity law - JmsNxn - 09/18/2011
Well I've been working on trying to generalize a set of operators that are commutative and associative, and everything works fairly fine until I come to the distributivity law across addition and f-addition. The concepts are very simple and the algebra is very simple, but the contradiction just keeps me bewildered. Any help would be greatly appreciated. We start off by defining f-multiplication: right off the bat we see is commutative and associative. We can create it's super-operator, called f-exponentiation: which is distributive across f-multiplication: The trouble comes when we try to develop f-addition, which is defined such that f-multiplication is it's super operator. For this definition we first need to define f-division, which is the inverse of f-multiplication: Therefore f-addition is given by the usual super function/abel function equation: furthermore we also know this extends more generally to: That is, if we give the condition that which I do. and here's where we get our contradiction. if we let we instantly see: and now if we let which is the identity of , we get: This is obviously false so I'm wondering, can we simply not have the distributivity law? And if so, why? What exact step am I doing that is inconsistent? furthermore if we have the distributive law, we also get another distributive law thanks to the commutative and associative nature of so it's like f-multiplication converts f-addition and addition back and forth. Which is consistent when we set and f-multiplication becomes multiplication. but otherwise it becomes oddly inconsistent. again, anyhelp would be greatly appreciated. RE: help with a distributivity law - tommy1729 - 09/21/2011
hmm although you made a typo , your derivation seems correct this time. well to avoid " loss of information " we always need an ordinary addition somewhere - possibly hidden - you have assumed an identity of f-multiplication. i think the identity causes the problem. if you could avoid exp type solutions and identities you might have some luck ... however i think that would require non-complex numbers ... RE: help with a distributivity law - tommy1729 - 09/21/2011
as a sidenote to the above : notice that if a function has no identity , this often results in the equivalent of a value that cannot be attained. if f(z) is entire and has values that cannot be attained ; it is of type exp(entire(z)) + Constant. hence we tend to have solutions of type exp^[a](exp^[b](x) + exp^[b](y)). but more can be said. you might wanna read this old but good thread containing an intresting proof by bo : http://math.eretrandre.org/tetrationforum/showthread.php?tid=125 regards tommy1729 RE: help with a distributivity law - JmsNxn - 09/22/2011
Yes I saw the little mistake I made where I assumed . The reason is because I've only really been observing functions which meet that requirement. I think I'm not doing anything inconsistent but instead we have to create the law, which is not dissimilar to division by zero: or that the distributive law fails when f-multiplied by the identity. I looked at that other thread too, very interesting. I had of hunch bo's proof but it's nice to see it proved. but furthermore, this gives some very strange laws for multiplication: which means for exponentiation: which means f-multiplying a number to the power of another number we convert exponentiation to f-exponentiation: which again is very very inconsistent. I must be doing something incorrect. I think we cannot give the distribution law, but that's not enough for me. I'd really like to know why. I'm absolutely puzzled. |