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