This is a MathJax test: \[\Huge e^{i\pi}+1=0\]

\begin{align*} 1\phantom{^2}:\qquad\quad\mathbf{d}_{ij}^{(0)}\cdot\mathbf{d}_{ij}^{(0)}&=\phantom{-}l_{ij}^2\\ t\phantom{^2}:\qquad\quad\mathbf{d}_{ij}^{(0)}\cdot\mathbf{d}_{ij}^{(1)}&=\phantom{-}0\\ t^2:\qquad\quad\mathbf{d}_{ij}^{(0)}\cdot\mathbf{d}_{ij}^{(2)}&=- \frac{1}{2} \mathbf{d}_{ij}^{(1)}\cdot\mathbf{d}_{ij}^{(1)}\\ t^3:\qquad\quad\mathbf{d}_{ij}^{(0)}\cdot\mathbf{d}_{ij}^{(3)}&=-\phantom{\frac{1}{2}}\mathbf{d}_{ij}^{(1)}\cdot\mathbf{d}_{ij}^{(2)}\\ t^4:\qquad\quad\mathbf{d}_{ij}^{(0)}\cdot\mathbf{d}_{ij}^{(4)}&=-\phantom{\frac{1}{2}}\mathbf{d}_{ij}^{(1)}\cdot\mathbf{d}_{ij}^{(3)}- \frac{1}{2}\mathbf{d}_{ij}^{(2)}\cdot\mathbf{d}_{ij}^{(2)}\\ &\hspace{0.5em}\vdots\\ t^m:\qquad\quad\!\!\mathbf{d}_{ij}^{(0)}\cdot\mathbf{d}_{ij}^{(m)}&= -\sum_{k=1}^{\lfloor\frac{m}{2}\rfloor}\frac{\mathbf{d}_{ij}^{(k)}\cdot\mathbf{d}_{ij}^{(m-k)}}{1+\delta_{k,m-k}} =Q_{ij}^{(m)}(\mathbf{d}_{ij}^{(1)},\dots,\mathbf{d}_{ij}^{(m-1)})\\ \end{align*}