Missing term in path integral calculation?

Question

Following Takhtajan (Quantum Mechanics for Mathematicians, AMS 2008, chp. 5), I am trying to calculate the propagator associated with the one-dimensional quantum harmonic oscillator. At one point, it becomes necessary to simplify the expression $$S = \sum_{s=0}^{j-1} [(q_{s+1} -q_s)^2 - \varepsilon^2q_s^2 ]. \tag 1$$ The author provides $$ S = \mathbf q \cdot \mathsf A_{j-1} \mathbf q - 2\tilde{\mathbf q }\cdot\mathbf q + q_j^2+q_0^2, \tag 2$$ where $\mathbf q = (q_1,\dots,q_{j-1})$, $\tilde {\mathbf q} = (q_0,0,\dots,0,q_j) \in \mathbb R^{j-1}$, and $$\mathsf A_{j-1} = \begin{pmatrix} 2-\varepsilon^2 & -1 & \\ -1 & 2-\varepsilon^2 & -1 \\ &-1 & 2-\varepsilon^2 & \\ & & & \cdots\end{pmatrix} \in \operatorname{Mat}_{j-1}(\mathbb R). $$ I managed to retrieve most of $(1)$ by expanding $(2)$, but I think a term $-\varepsilon^2 q_0^2$ is missing in the latter. I'd be certain of this, were it not for the fact that in all the calculations that follow, Takhtajan employs $(2)$ as is, and arrives at the correct formula for the propagator.

Answer 1

This seems to be an inconsequential mistake in the proof, as $\varepsilon = \varepsilon_j = {\omega(t'-t)}/{j}$, and in order to calculate the propagator for the one-dimensional QHO one has to take the limit as $j\to\infty$. In Takhtajan's proof (op. cit., pp. 257-8), one should substitute $(1-\varepsilon_j^2)q_0^2$ in place of $q_0^2$ where necessary; e.g., $$ \sum_{s=0}^{j-1}\left[\left(q_{s+1}-q_{s}\right)^{2}-\varepsilon_j^{2} q_{s}^{2}\right] = \mathbf{q} \cdot \mathsf{A}_{j-1} \mathbf{q}-2 \tilde{\mathbf q} \cdot \mathbf{q} + \color{red}{(1-\varepsilon_j^2)}q_{0}^{2} + q_{j}^{2}, $$ so that $$ K_\omega(q',t';q,t) = \lim_{j\to\infty} e^{-\frac{i\pi}{2}\nu_{j-1}} \sqrt{\frac{mj}{2\pi i\hbar(t'-t)|\det\mathsf A_{j-1}|}} \exp\left[\frac{imj}{2\hbar(t'-t)}\left( \color{red}{(1-\varepsilon_j^2)}q_0^2+q_j^2 - \tilde{\mathbf q} \cdot \mathsf A_{j-1}^{-1}\tilde{\mathbf q} \right)\right]$$ and so on.