线性离散系统的controllable abundance(能控丰富性、能控充裕性)的递推计算

         本人的文章arXiv1705.08064(On Controllable Abundance Of Saturated-input Linear Discrete Systems) 里定义了线性离散系统的controllable abundance(能控丰富性、能控充裕性)如下：

$v_{c,N}=\mathrm{Vol}(R_{c,N})" original="http://latex.codecogs.com/gif.latex?v_{c,N}=\mathrm{Vol}(R_{c,N})" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;$

其中 $R_{c,N}" original="http://latex.codecogs.com/gif.latex?R_{c,N}" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;$ 为系统的状态能控域。这里体积计算可转化为 由 $n\times n" original="http://latex.codecogs.com/gif.latex?n\times n" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;$ 维矩阵 $A" original="http://latex.codecogs.com/gif.latex?A" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;$ 和 $n\times r" original="http://latex.codecogs.com/gif.latex?n\times r" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;$ 维矩阵 $B" original="http://latex.codecogs.com/gif.latex?B" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;$ 生成的矩阵 $G_{0,N-1}=\{B,AB,\cdots,A^{N-1}B\}(N\geq n+1)" original="http://latex.codecogs.com/gif.latex?G_{0,N-1}=\{B,AB,\cdots,A^{N-1}B\}(N\geq n+1)" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;$ 中的向量 $\{g_{i},i=1,2,\cdots,r\times N\}" original="http://latex.codecogs.com/gif.latex?\{g_{i},i=1,2,\cdots,r\times N\}" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;$ 所生成的平行多面体 $C_{n}(G{}_{0,N-1}))" original="http://latex.codecogs.com/gif.latex?C_{n}(G{}_{0,N-1}))" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;$ 的体积计算,其计算公式为

$V_{n}(C_{n}(G_{0,N-1}))=\sum_{(i_{1},i_{2},\cdots,i_{n})\in\hat{\Omega}_{0,N-1}^{n}}\left|\mathrm{det}([g_{i_{1}},g_{i_{2}},\cdots,g_{i_{n}}])\right|" original="http://latex.codecogs.com/gif.latex?V_{n}(C_{n}(G_{0,N-1}))=\sum_{(i_{1},i_{2},\cdots,i_{n})\in\hat{\Omega}_{0,N-1}^{n}}\left|\mathrm{det}([g_{i_{1}},g_{i_{2}},\cdots,g_{i_{n}}])\right|" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;$

其中 $\hat{\Omega}_{0,N-1}^{n}" original="http://latex.codecogs.com/gif.latex? \hat{\Omega}_{0,N-1}^{n}" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;$ 为由 $\{1,2,\cdots,r\times N\}" original="http://latex.codecogs.com/gif.latex?\{1,2,\cdots,r\times N\}" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;$ 中任意挑 $n" original="http://latex.codecogs.com/gif.latex?n" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;$ 个不同的数并按数的大小组成的排列 $(i_{1},i_{2},\cdots,i_{n})" original="http://latex.codecogs.com/gif.latex?(i_{1},i_{2},\cdots,i_{n})" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;$ 的所有可能组成的集合。对上述的特殊体积可递推计算如下

$V_{n}(C_{n}(G_{0,N-1}))&=\left(1+\left|\mathrm{det}(A)\right|\right)V_{n}(C_{n}(G_{0,N-2}))-\left|\mathrm{det}(A)\right|V_{n}(C_{n}(G_{0,N-3}))$

          $+\sum_{j=1}^{r}\sum_{k=1}^{r}\sum_{(i_{1},\cdots,i_{j})\in\hat{\Omega}_{0,0}^{j}}\sum_{(i_{j+1},\cdots,i_{n-k})\in\hat{\Omega}_{1,N-2}^{n-j-k}}\sum_{(i_{n-k+1},\cdots,i_{n})\in\hat{\Omega}_{N-1,N-1}^{k}}\left|\mathrm{det}([g_{i_{1}},g_{i_{2}},\cdots,g_{i_{n}}])\right|$

当 $r=1" original="http://latex.codecogs.com/gif.latex?r=1" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;display:inline;$ 时,此时 $B=[b]" original="http://latex.codecogs.com/gif.latex?B=[b]" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;display:inline;$ ,上述递推公式可简化表示为

  $V_{n}(C_{n}(G_{0,N-1}))&=\left(1+\left|\mathrm{det}(A)\right|\right)V_{n}(C_{n}(G_{0,N-2}))-\left|\mathrm{det}(A)\right|V_{n}(C_{n}(G_{0,N-3}))$

                                      $\qquad+\sum_{(i_{2},\cdots,i_{n-1})\in\hat{\Omega}_{1,N-2}^{n-2}}\left|\mathrm{det}([b,A^{i_{2}}b,\cdots,A^{i_{n-1}}b,A^{N-1}b])\right|" original="http://latex.codecogs.com/gif.latex?\qquad+\sum_{(i_{2},\cdots,i_{n-1})\in\hat{\Omega}_{1,N-2}^{n-2}}\left|\mathrm{det}([b,A^{i_{2}}b,\cdots,A^{i_{n-1}}b,A^{N-1}b])\right|" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;display:inline;$

上述递推式于2017年5月推导并给出.