Operator which takes an argument and maps onto another operator, is it formally possible?

Let ∘\circ and Φ\Phi be functions ∘:x→x,Φ:x→x2\circ:x\rightarrow x, \Phi: x\rightarrow x^2.

Then let’s define operators ∘,Φ\circ, \Phi which map onto operators and take an argument ∘(x):A×B→xA∘xB\circ(x):A\times B\rightarrow xA \circ xB and Φ(x):A×B→xAΦ(Φ(xB))\Phi(x):A\times B\rightarrow xA\Phi(\Phi(xB)).

We supply each time the argument 22 to each operator such that: aΦb→2aΦ4b2→4aΦ64b4…a\Phi b\rightarrow 2a\Phi4b^2\rightarrow4a\Phi64b^4…, a∘b→2a∘2b→4a∘4b…a\circ b\rightarrow 2a\circ 2b\rightarrow 4a\circ4b…

Is there a formal way of defining this?

=================

  

 

The grammar of your question doesn’t quite make sense: you start off by saying that ∘\circ and Φ\Phi are functions on some set, but then in the definitions of the operators (by the same names), you’re assuming that the original functions are acting on a Cartesian product of that set with itself. I think the question almost makes sense if the first line is removed entirely; what did you intend by it? (Also, is 64 correct? Why isn’t it 16?)
– Eric Stucky
2 days ago

  

 

I was hesitating to clarify this when posting the question, so I mean that ∘,Φ\circ, \Phi can be read both as a function or as an operator depending on the context. Since we supply 2 as an argument to the Φ\Phi-operator we get 2aΦ4b2→2∗2aΦ(Φ(2∗4b2))2a\Phi4b^2\rightarrow 2*2a\Phi(\Phi(2*4b^2)) since Φ(2∗4b2)=(8b2)2=64b4\Phi(2*4b^2)=(8b^2)^2=64b^4 we get 2aΦ4b2→4aΦ64b42a\Phi4b^2\rightarrow 4a\Phi 64b^4.
– Bright
2 days ago

  

 

Yes, I did understand that. I don’t understand what 2aΦ4b22a\Phi 4b^2 is supposed to mean: is this just an odd way of writing an ordered pair? Or do you mean that Φ\Phi is supposed to be operating on 2a2a and 4b24b^2. If the latter, why did you write arrows in your chains, instead of equals signs?
– Eric Stucky
2 days ago

  

 

So we could write this as aΦb(2)=2aΦ4b2a\Phi b(2)=2a\Phi 4b^2, or xΦx(1)=xΦx2x\Phi x(1)=x\Phi x^2. Does this answer your question? So the →\rightarrow-chain would be equivalent to f→g→hf\rightarrow g\rightarrow h where f,g,hf,g,h would be functions.
– Bright
2 days ago

  

 

Does it make any sense?
– Bright
2 days ago

=================

1 Answer
1

=================

Here’s my best guess at what you might be looking for: Recall that for sets UU and VV, the notation VUV^U means the set of all functions U→VU\to V. Then define ˜Φ:N→(A×B)A×B\widetilde\Phi:\Bbb N\to (A\times B)^{A\times B}
˜Φ:n↦[(a,b)↦(na,Φ(nb))]\widetilde\Phi: n\mapsto \Big[(a,b)\mapsto (na,\Phi(nb))\Big]
and then your Φ\Phi-chain is then obtained by taking iterated compositions of ˜Φ(2)\widetilde\Phi(2):
(a,b)→˜Φ(2)(a,b)→(˜Φ(2)∘˜Φ(2))(a,b)(a,b)\to\widetilde\Phi(2)(a,b)\to\left(\widetilde\Phi(2)\circ\widetilde\Phi(2)\right)(a,b)

Similarly we may define ˜∘\widetilde\circ.

  

 

Thanks a lot! I will digest it once I get some time.
– Bright
2 days ago