这是一份kcl伦敦大学学院 7CCMMS20作业代写的成功案
$$
|y|+\sum_{i=1}^{r_{1}}\left|y_{i}\right|+2 \sum_{j=1}^{r_{2}}\left|z_{j}\right| \leq t
$$
equivalently by
$$
\sum_{i=1}^{r_{1}}\left|y_{i}\right|+2 \sum_{j=1}^{r_{2}}\left|z_{j}\right| \leq t-|y|
$$
Now if $|y|>t$, then $B_{t}$ is empty. For smaller values of $|y|$, suppose we change $y$ to $y+d y$. This creates a box in $(n+1)$-space with $d y$ as one of the dimensions. The volume of the box is $V\left(r_{1}, r_{2}, t-y\right) d y$. Thus
$$
V\left(r_{1}+1, r_{2}, t\right)=\int_{-t}^{t} V\left(r_{1}, r_{2}, t-|y|\right) d y
$$
which by the induction hypothesis is $2 \int_{0}^{t} 2^{r_{1}}(\pi / 2)^{r_{2}}\left[(t-y)^{n} / n !\right] d y$. Evaluating the integral, we obtain $2^{r+1}(\pi / 2)^{r_{2}} t^{n+1} /(n+1)$ !, as desired.
Finally, $V\left(r_{1}, r_{2}+1, t\right)$ is the volume of the set described by
$$
\sum^{r_{1}}\left|y_{i}\right|+2 \sum^{r_{2}}\left|z_{j}\right|+2|z| \leq t
$$
7CCMMS03 COURSE NOTES ：
Proof. If $x \in B^{*}$, then by (6.1.1) and (2.1.6),
$$
\pm 1=N(x)=\prod_{i=1}^{n} \sigma_{i}(x)=\prod_{i=1}^{r_{1}} \sigma_{i}(x) \prod_{j=r_{1}+1}^{r_{1}+r_{2}} \sigma_{j}(x) \overline{\sigma_{j}(x)}
$$
Take absolute values and apply the logarithmic embedding to conclude that $\lambda(x)=$ $\left(y_{1}, \ldots, y_{r_{1}+r_{2}}\right)$ lies in the hyperplane $W$ whose equation is
$$
\sum_{i=1}^{r_{1}} y_{i}+2 \sum_{j=r_{1}+1}^{r_{1}+r_{2}} y_{j}=0
$$