Originally Posted by

**arbolis** Thanks for the quick response.

Oh... I didn't realize that I could use $\displaystyle \frac{1}{1+u} = 1 - u + u^2 - u^3 \cdots$.

A stupid question from my part : from your formula, is $\displaystyle P_0(x,y)=1$ and $\displaystyle P_1(x,y)=1-xy$, $\displaystyle P_2(x,y)=1-xy+x^2y^2$, etc...?

If so then I was right by saying that $\displaystyle P_{2n}(0,0)=1$. (I don't think I was right since it would imply that $\displaystyle P_n(0,0)=1$, no matter what $\displaystyle n$ is).