equate like powers of $\Delta K$ term-by-term, mindful of the fact that the left-hand side includes a product over $I$. For example, the first-order term will be $$\tilde{C}_1 = i \sum_{I=0}^{M-1}\Delta K
\gamma_1(K_I)$$ the second will be $$\tilde{C}_2 = - \sum_{I=0}^{M-1}\Delta K [\gamma_2(K_I) - \gamma_1(K_I)^2]$$ with $\gamma_i(K) = \langle \Psi_0(K) | \partial^i_K |
\Psi_0(K)\rangle$. Straightforward algebra and taking the continuous limit ($\Delta K \rightarrow 0$, $M\rightarrow \infty$) gives up to the fourth order term, $$\begin{aligned}
\label{eqn:cmlnts}
C_1 &=& i \frac{L}{2\pi} \int_{-\frac{\pi}{L}}^{\frac{\pi}{L}} d K \gamma_1 \\ \nonumber 
C_2 &=& -\frac{L}{2\pi} \int_{-\frac{\pi}{L}}^{\frac{\pi}{L}} d K [\gamma_2 - \gamma_1^2] \\ \nonumber
C_3 &=& -i \frac{L}{2\pi} \int_{-\frac{\pi}{L}}^{\frac{\pi}{L}} d K [\gamma_3 -3 \gamma_2 \gamma_1+ 2\gamma_1^3] \\ \nonumber 
C_4 &=& \frac{L}{2\pi} \int_{-\frac{\pi}{L}}^{\frac{\pi}{L}} d K [\gamma_4 -3 \gamma_2^2 -4\gamma_3\gamma_1 + 12 \gamma_1^2\gamma_2 -6\gamma_1^4]\end{aligned}$$ The quantities $C_n$ in Eq. (\[eqn:cmlnts\]) are the GICs associated with the Zak phase (the Zak phase itself being equal to $C_1$). The difference between $\tilde{C}_i$ and $C_i$ is the multiplicative factor $L/2\pi$, which is also how the phase is defined by Zak [@Zak89]. This assures that the first moment corresponds to the average position associated with square modulus of the Wannier function (Eq. (10) in Ref.  [@Zak89]). When the underlying probability distribution is well 