The reverse recursion is the exact gradient of a finite state-step objective

By the chain rule [F4] applied to $x_{n+1}(\theta )={\mathrm{\Psi }}_n(x_n(\theta ),\theta )$, $S_{n+1}=A_n{S}_n+B_n$; differentiating the initial map gives $S_0=D{x}_0(\theta )$. These are the forward tangent equations for this finite composition.

Apply the reverse accumulation rule [F2] to the finite graph of states and objective summands in [F1]. The terminal contribution seeds $p_N=D_x{\varphi }^T$; at each state node $x_n$, its local running-cost edge contributes $D_x{\mathrm{\ell }}_n^T$, while its state-transition edge contributes $A_n^T{p}_{n+1}$. Thus the accumulated state cotangent is exactly $p_n$ as defined in the Statement, and the direct parameter edges contribute $D_{\theta }{\varphi }^T$ and $D_{\theta }{\mathrm{\ell }}_n^T+B_n^T{p}_{n+1}$. By [F3], reverse mode returns the exact derivative of this finite C1 graph.

Differentiate the objective in [F1] using the forward sensitivities in step 1.1. The chain rule gives ${\mathrm{\nabla }}_{\theta }J=D_{\theta }{\varphi }^T+S_N^T{D}_x{\varphi }^T+{\sum }_{n=0}^{N-1}(D_{\theta }{\mathrm{\ell }}_n^T+S_n^T{D}_x{\mathrm{\ell }}_n^T)\mathrm{.}$ This is the direct tangent expression, with all derivatives evaluated along the trajectory.

The adjoint equation gives $p_n^T{S}_n=D_x{\mathrm{\ell }}_n{S}_n+p_{n+1}^T{A}_n{S}_n$. Since $S_{n+1}=A_n{S}_n+B_n$, rearranging yields $D_x{\mathrm{\ell }}_n{S}_n=p_n^T{S}_n-p_{n+1}^T{S}_{n+1}+p_{n+1}^T{B}_n\mathrm{.}$ Summing over $n$ telescopes; $p_N=D_x{\varphi }^T$ cancels the terminal tangent term in step 2.1 and leaves $S_0^T{p}_0+{\sum }_n{B}_n^T{p}_{n+1}$. Adding the direct parameter derivatives from step 2.1 gives the displayed adjoint formula.