## Thursday, March 16, 2006

The answer to that cellular automaton puzzle was that it generates the derivatives of the tan function at zero. It uses the fact that if f=tan x, then f'=1+f^2. We can now repeatedly use this to compute higher derivatives, eg.

(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. David R. MacIver said...

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. :-)

Thursday, 16 March, 2006 sigfpe said...

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...

Thursday, 16 March, 2006 Edwin said...

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?

Saturday, 01 March, 2008