What factor does to unknown functions' arguments



PS Thanks to Barton for looking at this with me.

On Thu, Aug 25, 2011 at 14:55, Stavros Macrakis <macrakis at alum.mit.edu>wrote:

> factor( f(x-1/x) ) => f( (x^2-1)/x)
>
> That is, factor calls sratsimp (with ratfac=T) on the arguments of unknown
> functions, even though it doesn't actually factor them.
>
> This seems peculiar.  Either it should map over all the arguments of f,
> giving f((x-1)*(x+1)/x), or it should leave them completely untouched,
> returning f(x-1/x).
>
> Is there any disagreement about this?
>
>              -s
>