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

41 Views Asked by At

Let $\circ$ and $\Phi$ be functions $\circ:x\rightarrow x, \Phi: x\rightarrow x^2$.

Then let's define operators $\circ, \Phi$ which map onto operators and take an argument $\circ(x):A\times B\rightarrow xA \circ xB$ and $\Phi(x):A\times B\rightarrow xA\Phi(\Phi(xB))$.

We supply each time the argument $2$ to each operator such that: $a\Phi b\rightarrow 2a\Phi4b^2\rightarrow4a\Phi64b^4...$, $a\circ b\rightarrow 2a\circ 2b\rightarrow 4a\circ4b...$

Is there a formal way of defining this?

1

There are 1 best solutions below

0
On BEST ANSWER

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

Similarly we may define $\widetilde\circ$.