Truncation error when applying a finite difference scheme to solve $u_x +Au_t = 0$

The wave equation in one space dimension is given as $$u_t + Au_x = 0$$ where $$u := \begin{bmatrix} v(x,\, t) \\ w(x,\, t) \end{bmatrix}, \quad A = \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}.$$ We are also given the following finite difference scheme, where $$V_j^k$$ is the approximation to $$v(x_j,\, t_k)$$: \begin{align} V_j^{k+1} &= V_j^k + \frac{1}{2} p \left( W_{j+1}^k - W_{j-1}^k \right), \\ W_j^{k+1} &= W_j^k + \frac{1}{2} p \left( V_{j+1}^{k+1} - V_{j-1}^{k+1} \right), \end{align} where $$p = \Delta t / \Delta x$$. My task is to find the leading error term in the local truncation error. I am given the answer, but no thurrow explanation. The answer is as follows:

Using the Taylor expansion and the fact that $$v_t=w_x$$, $$w_t = v_x$$, we find \begin{align} \tau_j^k(v) &= \frac{1}{2}t \partial_x^2 v_j^k - \frac{1}{6} \Delta x^2 \partial_x^3 w_j^k + \mathcal{O}(\Delta t^2 + \Delta x^3), \\ \tau_j^k(w) &= -\frac{1}{2}t \partial_x^2 w_j^k - \frac{1}{6} \Delta x^2 \partial_x^3 v_j^k + \mathcal{O}(\Delta t^2 + \Delta x^3). \end{align}

However, I do not understand which Taylor expansions have been applied; obviously $$v$$ and $$w$$ must be approximated, but at which points are we to approximate them? And where do we center the approximations?

Thiank you in advance for your help and your time.

The clue is in the answer: Only terms $$v_j^k=v(x_j,t_k)$$ and $$w_j^k=w(x_j,t_k)$$ appear in the right-hand sides so the expansions must have been made around the space-time point $$(x_j, t_k)$$.