Right-sign denominator correction in neutral-meson mixing

If a wrong-sign amplitude is $r{e}^{-i\delta }{g}_{+}+g_{-}$, the same final state's right-sign amplitude is $g_{+}+r{e}^{-i\delta }{g}_{-}$. Their ratio is not exactly obtained by replacing the right-sign rate with a pure exponential. For series $N=n_0+n_{1}t+n_2{t}^2$ and $D=1+d_{1}t+d_2{t}^2$, division gives $N/D=n_0+(n_1-n_0{d}_1)t+[n_2-n_1{d}_1+n_0(d_1^2-d_2)]t^2+O(t^3)$.