 Hyperoperators - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Hyperoperators (/showthread.php?tid=454) Hyperoperators - Mr. Pig - 06/10/2010 When I first learned about tetration and hyperoperators, I wondered if each operator was set at a number. For example, + could be written as /1\, * as /2\, ^ as /3\ and so on. would be written as a/3\b and a/0\b would be b+1 (I think, feel free to correct me on anything if I'm wrong). I found the equation and did the math and plugged in 0 to get which simplifies to . According to this, a/-n\b=a/0\b=b+1 (although I could be completely wrong on this) So after that, I wondered if was possible to have an operator that wasn't an integer. For example, 2/e\5 or 3/2i+3\2 and that's where I got stuck. So I would like you guys to help me figure this out. Thanks! RE: Hyperoperators - bo198214 - 06/10/2010 Actually there were some heated debates on this forum about zeration. I dont recollect all the posts but this thread or ths thread may give a good impression. I fully agree with your derivation that all operations are just the increment operation. It follows from the general rule you give to build the hyperoperations ladder. Actually not much result results were collected about operations with fractional ranks or even complex ranks. Your question appeared already in this thread. My idea about it is (I posted it once but couldnt find it anymore) to express the hyperoperations as repetition of an operator and then take fractional powers of that operator. There is a theory about fractional powers of operators, see here. For example if we only care about the superfunctions x -> b [n] x, then as you mentioned we have the equation b [n+1] (x+1) = b [n] (b [n+1] x) that means to get the n+1-th superfunction g(x) = b [n+1] x from f(x) = b [n] x we create some operator S (successor operator) that satisfies S[f](x+1)=f(S[f](x)). Now this operator S may be expressed as an infinite matrix transforming the powerseries coefficients of f, or whatever, and we can take powers of it: ... and then we can also take perhaps fractional or complex powers . RE: Hyperoperators - 73939 - 06/20/2010 What about defining the inverses of operations? So, for example, we could write a-b as A(a,-b,1) (or whatever other notation you wish to use), but if we wanted to evaluate A(a,-b,1) purely in terms of its recursive definitions, we'd never reach a value. Perhaps it could then be written so that a-b = A(a,b,k) for some k? Don't know what that k would be though. An intuitive answer would be something like -1, but A(a,b,-1) is simply b+1, so that doesn't work. The same problem may arise for a/b, log[a](b), etc. Especially log[a](b) - whereas the others can be expressed in terms of their inverse function, log[a](b) is independent of a^b. Sorry if these are old questions - I'm new here, and these things have been troubling me for a while. :/ By the way, how does one write in equation mode? RE: Hyperoperators - andydude - 06/20/2010 first of all, all hyperops are bivariate, which means they have 2 inverses, which means you can't encode both with a nwgative sign. second, yes, -n is always the same as successor when using certain rules... for more rules search this forum for "mother". lastly, there are two systems here for writing in equations... using brackets like ab and texmode which you write Code:$$\exp x$$. Andrew Robbins RE: Hyperoperators - bo198214 - 06/20/2010 (06/20/2010, 02:41 AM)73939 Wrote: By the way, how does one write in equation mode? For the equation mode see this announcement. And, as hint, if you reply to a post with an equation you can see how it was encoded in the preinserted text.