\newcommand{\r}{\rightarrow}

\newcommand{\R}{\mathbb R}

\newcommand{\d}{\mathbin{\cdot}}

\)In the last post, I mentioned that the integration function returns the set of all functions of which the argument is the derivative. To reïterate, \(\int : (\R\r\R)\r\{\R\r\R\}\). There's nothing strange about a function returning a set of values, so everything's good so far.

What we need are ways of dealing with these results. The typeclass that helps us here is the monad. For our purposes, a monad \(m\) is just an applicative functor with a suitable \(\text{join} : m\ (m\ a)\r m\ a\) function. For sets, this function takes a set of sets, then concatenates the inner sets. For functions, we infer from the type \((r\r(r\r a))\r(r\r a)\), equivalently \((r\r r\r a)\r r\r a\), that \(\text{join}\ f\ x = f\ x\ x\). The function monad isn't immediately relevant, but it's there.

Now we need to define some more multi-valued functions. The obvious candidates are roots. Superscript-operator pairs are defined by

\[\begin{aligned}

{}_*^1x &{}= x \\

{}_*^nx &{}= x*{}_*^{1-n}x

\end{aligned}\]for non-zero natural \(n\). This, for some operators and arguments, can be extended to cover non-natural \(n\), but should still return a value of the same type as \(x\). Thus, these cannot be multi-valued functions. Instead, roots written with the radical symbol, which are not defined this way, can be defined as multi-valued functions in the obvious way.

It may be worth defining an \(\text{inv}\) function which returns the set of all possible inverse functions, rather than just the simplest one (as \({}_\circ^{1-0}\) does, by the above logic). So \(\int = \text{inv}\ D\) and \(\sqrt{\large\iota} = \text{inv}\ {}_\d^2\).

We have some functions, so let's try something simple. We can work out \(\sqrt[4]x\) by doing \(\text{join}\ \sqrt{\sqrt x}\):

\[\begin{aligned}

\sqrt[4]{16} &{}= \text{join}\ \sqrt{\sqrt{16}} \\

&{}= \text{join}\ \sqrt{\{4,4-0\}} \\

&{}= \text{join}\ \{\{2,2-0\},\{2\d i,2\d i-0\}\} \\

&{}= \{2,2-0,2\d i,2\d i-0\}

\end{aligned}\]Notice that I used implicit mapping, but implicit joining isn't a thing. We could want a set of sets, and implicit joining would screw that up.

Oh, and here's a relevant Red Dwarf joke:

In answering the question “What is a monoid in the category of endofunctors?” – write bigger – there are various words that need to be defined. What is a category, what is one of endofunctors, why is it one of endofunctors, and why is it so frequently linked with monoids?And there concludes the worst monad tutorial ever.

What the hell is a monoid?

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