这是一份leeds利兹大学MATH5104M01作业代写的成功案例

where $A=\left|\mathrm{a}{1} \times \mathrm{a}{2}\right|$ denotes the area of the unit cell. For analytic purposes, it is convenient to use the values $\mathbf{p}=0, z=\sqrt{A}=d$ (remember that $d$ is the pitch of the direct array). The above formula is characterized by faster convergence via integration with respect to $z$. The integer parameter $q$ gives the number of times the convergence of the lattice sums has been accelerated through integration and is thus called convergence acceleration index. The reciprocal unit cell is defined by the vectors[1]
$$
\mathbf{a}^{1}=2 \pi \frac{\mathbf{a}{2} \times \mathbf{e}{2}}{A}, \quad \mathbf{a}^{2}=2 \pi \frac{\mathbf{e}{2} \times \mathbf{a}{1}}{A},
$$
with the reciprocal lattice vectors
$$
\mathbf{Q}{\mathrm{p}}=p{1} \mathbf{a}^{1}+p_{2} \mathbf{a}^{2}+\mathbf{k}, \quad \theta_{\mathrm{p}}=\arg \left(\mathbf{Q}{\mathrm{p}}\right) . $$ The lattice sums satisfy the identity $$ S{-l}^{Y}\left(k_{\perp}, \mathbf{k}\right)=\overline{S_{l}^{Y}}\left(k_{\perp}, \mathbf{k}\right),
$$

MATH5104M01 COURSE NOTES ：

$$
M_{n}^{\xi \xi}=\mathcal{O}\left(\Gamma^{2}(n) n\left(\frac{1}{2} k_{\perp} r_{c}\right)^{-2 r}\right)
$$
Similarly, one can show that, for the lattice sums,
$$
S_{l}^{Y}\left(k_{\perp}, \mathbf{k}\right)=\mathcal{O}\left(\Gamma(l)\left(\frac{1}{2} k_{\perp} d\right)^{-l}\right), \quad \text { as } l \rightarrow+\infty .
$$
This causes numerical difficulties when
$$
\frac{k_{\perp} d}{2} \leqslant 1,
$$
since the off-diagonal terms increase extremely rapidly with index $l$ :
$$
z_{l}^{\xi}+\sum_{m=-\infty}^{+\infty} D_{l m}^{\xi \xi} z_{m}^{\xi}=0, \quad \forall \xi \in{E, H},
$$
where
$$
z_{l}^{\xi}=b_{l}^{\xi} \sqrt{\left|M_{l}^{\xi \xi}\right|}
$$