The factorial and the relative sign

For arbitrary smooth input $u$, the scalar response polynomial is

It satisfies

For a free input it reproduces the copied retarded coefficients.

With the free Hessian inverse ${{\Delta }_0}^{\mathrm{R}}=-{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}$, the first response is

No further minus sign is inserted. The second vertex enters once with the same negative response sign. Two successive first responses give a positive iteration.

For the first vertex alone, replacing $\kappa$ by $\lambda$ in $(1+\lambda {{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}{\mathsf{L}}_{1,g}{)}^{-1}u$ shows that the second derivative at zero is twice ${{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}{\mathsf{L}}_{1,g}{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}{\mathsf{L}}_{1,g}u$. This explains the coefficient $\frac{1}{2}{R}_2$. The normalization is dynamical, not a contact chosen after a gauge calculation.