Thread Rating:
  • 0 Vote(s) - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
help with a distributivity law
#1
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.
Reply
#2
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 ...
Reply
#3
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/tetrationforu...hp?tid=125

regards

tommy1729
Reply
#4
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.
Reply


Possibly Related Threads...
Thread Author Replies Views Last Post
  exponential distributivity bo198214 4 6,247 09/22/2011, 03:27 PM
Last Post: JmsNxn



Users browsing this thread: 1 Guest(s)