 Formal monodromy correction to an entire-system Stokes product

When Stokes factors act on solution coefficient vectors, a full positive turn gives their ordered product equal to the inverse of formal monodromy for a polynomial system with no finite singularities. The true fundamental matrix is entire, so its actual monodromy is trivial, but its normalized formal branch changes by $M_f$. Indeed, write $Y_{j+1}=Y_j{S}_j^{-1}$; then $Y_{N+1}=Y_1(S_N\cdots S_1{)}^{-1}$, whereas normalized continuation after the positive turn gives $Y_{N+1}=Y_1{M}_f$. Cancelling the invertible fundamental matrix proves the identity. The uncorrected product is identity only when $M_f=I$. For a cubic-exponential two-by-two system with $U=1,V=-1,P=1,R=2,d=3/2$, the formal exponent is $\theta =1/2$, so $M_f=-I$ and the Stokes product is $-I$.