Re: functions not polynomials

[Vaughan Pratt <pratt@cs.stanford.edu>]

Paul Taylor's example of an f(x,y) that is polynomial separately in x
and y but not jointly was sum_n (x,n)(y,n) (where (x,n) denotes the
binomial coefficient x!/(n!(x-n)!)).  After mulling over that example
some more it occurred to me that it can be analyzed via the observation
that W^{-1} maps the polynomial P_n(x) = (x(x-1)(x-2)...(x-(n-1)))^2 to
the polynomial xy(x-1)(y-1)(x-2)(y-2)...(x-(n-1))(y-(n-1)).  This
contradicts my earlier claim that the only polynomials in x that W^{-1}
maps to polynomials in x and y are the linear combinations of 1 and x^2.
  These two are easily seen to be the only *monomials* so mapped, but
(the linearity of W^{-1} notwithstanding) it does not follow that the
only *polynomials* so mapped are the linear combinations of these two
monomials.

W^{-1} maps sum_n P_n(x)/(n!)^2 to Paul's example.  The coefficient
1/(n!)^2 of P_n(x) seems to play no role here, and any coefficients
should do as long as infinitely many are nonzero (to make f(x,y) not a
polynomial).  To extend the example (as a function on N^2) directly (via
the constituent polynomials) to a function on the positive reals
however, the coefficients would need to grow somewhat slower than 4^n,
|P_n(x)| being bounded above by at best about 1/4^n for 0 < x < n (the
half-integer points for x in that range give a good approximation of the
bound).  1/(n!)^2 is more than slow enough for this purpose.

A simpler example is g(x) = (2x,x) (again the binomial coefficient),
which W^{-1} maps to f(x,y) = (x+y,x), polynomial separately in x and y
but not jointly.  That is, Pascal's triangle is a sufficient
counterexample for Paul's purposes.  Moreover the Gamma function gives a
nicer (log-convex in fact) extension of f(x,y) to the upper right
quadrant of R^2.

Vaughan Pratt