2

The first operation is addition, the second multiplication, third exponentiation, fourth tetration, etc. Is there a meaningful way to define a $1.5^{th}$ operation, something between addition and multiplication. How about a $-1^{th}$ operation. Thanks!

mtheorylord
  • 4,274
  • I think one could equivalently ask: can we extend the Ackermann function to negative- or rational-valued inputs? – Patrick Stevens Dec 31 '16 at 20:36
  • Yes. I wonder how that would relate to the extended operations. – mtheorylord Dec 31 '16 at 20:40
  • http://mathoverflow.net/questions/146786/ackermanns-function-over-the-reals – Patrick Stevens Dec 31 '16 at 20:41
  • These are known as the hyperoperations, and are defined as $\begin{align}f_k(x,y)=\begin{cases}xy;&k=1\x;&y=1\f_{k-1}(f_k(x,y-1),y)&k>1\text{ and }y>1\end{cases}\end{align}$

    I think this grows faster than Ackermann if you consider it as a three-argument function $f(k, x, y)$.

     – Challenger5 Dec 31 '16 at 20:44
  • $x,y\mapsto \frac{xy+x+y}{2}$ would be "between" addition and multiplication in at least one sense. Can you think of a property you want your 1.5th operation to have but this one doesn't? – hmakholm left over Monica Dec 31 '16 at 20:45
  • Is it associative? – mtheorylord Dec 31 '16 at 20:46
  • If I plug in k = 1.5 it wouldn't terminate. – mtheorylord Dec 31 '16 at 20:47
  • Take a look at commutative hyper operations. They look easier to define on all reals. But I'm still not sure. – mtheorylord Dec 31 '16 at 20:48
  • Whoops. My bad. Edited the question – mtheorylord Dec 31 '16 at 22:29
  • I suspect that in order to give a good answer to this question, you will need to find some properties that the $q$th operation should satisfy, at least for $q$ rational. Another point is that it's odd to have the index of the operations more general than the inputs, so I think a first step before any of this is to figure out how to feed the $n$th operation (for $n\in\mathbb{N}$) rational values. Already at tetration, there isn't (to the best of my (extremely limited) knowledge) a good way to do this. – Noah Schweber Dec 31 '16 at 22:36
  • There is a way for it to be done but it is complicated for higher powers. What properties do you suggest? – mtheorylord Dec 31 '16 at 22:38
  • I don't know any good properties off the top of my head - but my point is that without some further restrictions, the problem is trivially solvable ("let every non-positive-integrally-indexed operation be the constant $0$ function"). So some things need to be specified. Personally, I am extremely skeptical that this can be done in a satisfactory way. – Noah Schweber Dec 31 '16 at 23:04
  • Incidentally, I've asked here whether exponentiation satisfies any nontrivial identities on its own (that is, excluding identities which bring other operations like $+$ and $\cdot$ into the picture); my suspicion is that it does not. – Noah Schweber Dec 31 '16 at 23:07
  • Well, I'm a dummy - obviously it does, e.g. $(x^y)^z=(x^z)^y$. Still, I'd like to know more about what identities it satisfies. – Noah Schweber Dec 31 '16 at 23:23

0 Answers0