 Schur specialization by a generating function

Given $F(t)={\sum }_{r\ge 0}{f}_r{t}^r$ with $f_0=1$, put $f_r=0$ for $r<0$. The Cauchy identity for symmetric functions and Jacobi–Trudi identity give ${\prod }_{i}F(t_i)={\sum }_{\lambda }{s}_{\lambda }^F{s}_{\lambda }(t)$, degree by degree in the completion. For $F(t)={\prod }_j(1-x_{j}t{)}^{-1}$ the coefficients are $s_{\lambda }(x)$; for $F(t)={\prod }_j(1+x_{j}t)$ they are $s_{{\lambda }^{\prime }}(x)$.