(f^n)' = nf^(n-1)f'=nf^(n-1)(1+f^2) = nf^(n-1)+nf^(n+1).

You should now see why the CA rule arises.

(Why n,n instead of (n-1),(n+1) as in the rule I gave? Because this is the 'push' version rather than the 'pull' version of the rule that I gave. If you imagine differentiation acting as a linear operation on the vectors with respect to the basis (1,f,f^2,f^3,...) then the 'push' and 'pull' versions are described by matrices that are transposes of each other.)

I was petty excited by this and tried applying it to all sorts of other functions. But actually it was hard to make it work. Maybe the fact that there is a nice formula for the tangent function is just an accident. Anyway, Jan Rutten, whose paper I got this from, has lots of nice papers on coalgebras and infinite streams. I think I've now learned enough about coalgebras to move onto the next section of my category theory book: Adjoints.

## 3 comments:

Good luck! Adjoints broke my brain when I first encountered them.

Eventually I managed to internalise the concept by repeating the mantra "An adjoint is something which looks like a free group construction" over and over until it made sense. :-)

Adjoints look so innocent. Just that tiny little commutative triangle. Over the years I thought that I would just pick them up by osmosis. But I've finally realised that there's no easy way. I just have to take some examples and unpack all of the definitions. And it's incredible how much unpacking you can do from that little triangle...

I din't understand the difference between 'push' and 'pull'.

I only have an introductory knowledge of Linear Algebra. Is push and pull beyond that?

Post a Comment