In adiabatic quantum optimization we start with an initial Hamiltonian $H_0$ and then adiabatically evolve from $H_0$ to $H_P$ (problem hamiltonian) for a time $T$ according to \begin{equation}\label{eq:adiabatic hamiltonian} H(t) = A(t) \cdot H_0 + B(t) \cdot H_P, \end{equation} where $A(t)$ evolves from $1$ to $0$ during $T$ and $B(t)$ vice versa. In general, $H_0$ and $H_P$ do not commute.
From here we come to QAOA by this:
The unitary evolution of the Hamiltonian H(t) is approximated by the Trotter-Suzuki decomposition formula \begin{equation} U(0, T) = \mathcal{T} e^{\int_0^T A(t) \cdot H_0 + B(t) \cdot H_Pdt} \approx e^{-iH((2N-1)\delta/2)\delta}\ldots e^{-iH(3\delta/2)\delta}e^{-iH(\delta)\delta} = \\ e^{A_NH_0 + B_NH_P}\ldots e^{A_2H_0 + B_2H_P}e^{A_1H_0 + B_1H_P} \end{equation} with $A_N = -i\delta A(\frac{2N-1}{2}\delta)$, $B_N= -i\delta B(\frac{2N-1}{2}\delta)$ and $N\delta=T$. $ \mathcal{T}$ is the time-ordering operator. The unitary in the QAOA looks like this: \begin{equation} e^{\alpha_NH_0}e^{\beta_NH_P}\ldots e^{\alpha_2H_0}e^{\beta_2H_P}e^{\alpha_1H_0}e^{\beta_1H_P} \end{equation}
So, how do I come from $e^{A_NH_0 + B_NH_P}$ to $e^{\alpha_NH_0}e^{\beta_NH_P}$, what is the relation between $A_N$ and $\alpha_N$?
note: I reformulated my question inspired by the answer of DaftWullie. However, it is basically the same question as before.