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?

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 $$(x_j, t_k)$$.