[x^2]a^{<-1>}(x) = \frac{1}{2} [x^1] (\frac{x}{a(x)})^2 = \frac{1}{2} [x^1] (\beta+\gamma x)^{-2} =\frac{1}{2} [x^1] \sum_{j \geq 0} {-2 \choose j} (\gamma x)^j \beta^{-2-j}