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摘要: 张量指数函数已经广泛应用于工程领域.本文得到了一种有效的张量广义逆, 并以此为基础构造了广义逆张量Padé逼近的一种$\varepsilon$-算法.该算法可以编程实施递推的计算, 其特点是, 在计算过程中, 不必计算张量的乘积, 也不必计算张量的逆.给出的计算张量指数函数的数值实验显示, 将本文的方法与目前通常使用的截断法进行比较, 在不降低逼近阶的条件下, $\varepsilon$-算法能很好地降低计算复杂度, 尤其是在张量的维数比较大的时候.
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关键词:
- 张量指数函数 /
- 广义逆张量Padé逼近 /
- 张量$\varepsilon$-算法 /
- 张量指数函数的截断法
Abstract: Tensor exponent functions have been widely used in engineering. In this paper, an effective tensor generalized inverse is obtained, and based on it, an $\varepsilon$-algorithm of generalized inverse tensor Padéapproximation is presented. The algorithm can be programmed to implement recursive calculations, which are characterized by the fact that it is not necessary to calculate the product of the tensors in the calculation nor to calculate the inverse of tensors. The numerical experiment of calculating tensor exponential function shows that compared with the commonly used truncation method, the $\varepsilon$-algorithm can reduce computational complexity well without reducing the approximation order, especially when the dimension of tensor is relatively large.-
Key words:
- Tensor exponential functions /
- generalized inverse tensor Pad$\acute{e}$ approximation /
- tensor $\varepsilon$-algorithm /
- truncation method
1) 本文责任编委 黎铭 -
图 1 不同维数下两种算法运行时间直方图(s)
Fig. 1 The time-consuming comparison histogram of two algorithms in different dimensions (s)
表 1 $[\frac{4}{4}]$型GITPA-算法数值实验
Table 1 The numerical experiment of $[\frac{4}{4}]$ type GITPA-algorithm
$x$ $(1, 2, 1)$ $(2, 2, 1)$ $(1, 2, 2)$ $(2, 2, 2)$ $ RES $ 0.2 $E$值 0.08766299 0.87955329 0.12044671 -0.08766299 $5.69\times10^{-13}$ $A$值 0.08766327 0.87955283 0.12044717 -0.08766327 0.4 $E$值 0.15420167 0.78130960 0.21869040 -0.15420167 $3.74\times10^{-10}$ $A$值 0.15420895 0.78129804 0.21870196 -0.15420895 0.6 $E$值 0.20408121 0.70078192 0.29921808 -0.20408121 $1.40\times10^{-8}$ $A$值 0.20412606 0.70071136 0.29928864 -0.20412606 0.8 $E$值 0.24081224 0.63444735 0.36555265 -0.24081224 $1.63\times10^{-7}$ $A$值 0.24096630 0.63420702 0.36579298 -0.24096630 1 $E$值 0.26715410 0.57953894 0.42046106 -0.26715410 $1.01\times10^{-6}$ $A$值 0.26753925 0.57894247 0.42105753 -0.26753925 
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表 2 算法1和算法2的数值实验比较
Table 2 The numerical experiment comparison of Algorithm 1 and Algorithm 2
$[\frac{j+2k}{2k}]$ 张量$\varepsilon$-算法 $n_{\rm{max}}$ $\sum^{n_{\rm{max}}}_{n=0}\frac{1}{n!}\mathcal{A}^nt^n $ $a_{121}$ $a_{221}$ $a_{122}$ $a_{222}$ $a_{121}$ $a_{221}$ $a_{122}$ $a_{222}$ $[\frac{2}{2}]$ 0.4235 0.3513 0.6487 -0.4235 1 1.0000 -0.3333 1.3333 -1.0000 $[\frac{4}{4}]$ 0.3049 0.4141 0.5859 -0.3049 2 -0.3333 1.0556 -0.0556 0.3333 $[\frac{6}{6}]$ 0.3098 0.4068 0.5932 -0.3098 3 0.7222 -0.0062 1.0062 -0.7222 - - - - 4 0.1049 0.6116 0.3884 -0.1049 - - - - 5 0.3931 0.3234 0.6766 -0.3931 - - - - 6 0.2810 0.4355 0.5645 -0.2810 - - - - 7 0.3184 0.3981 0.6019 -0.3184 - - - - 8 0.3075 0.4090 0.5910 -0.3075 - - - - 9 0.3103 0.4062 0.5938 -0.3103 - - - - 10 0.3097 0.4069 0.5931 -0.3097 - - - - 11 0.3098 0.4067 0.5933 -0.3098 - - - - 12 0.3098 0.4068 0.5932 -0.3098
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表 3 算法1和算法2的计算复杂度分析
Table 3 The analysis of computational complexity of Algorithm 1 and Algorithm 2
计算一次 $\varepsilon$-算法: $[\frac{6}{6}]$型 截断法:取13项 复杂度 计算 计算 $t$-积运算 $l^3n^2$ 5 11 数乘运算 $l^2n$ 21 12 范数运算 $l^2n$ 21 0 总和 $5l^3n^2+42l^2n$ $11l^3n^2+12l^2n$
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表 4 不同维数下两种算法的运行时间表(s)
Table 4 The consuming time of two algorithms in different dimensions (s)
张量维数 张量$\varepsilon$-算法运行时间 截断法运行时间 $ 3~\times ~3\times ~3$ 1.755262 1.103832 $10\times 10\times 10$ 2.272663 2.164860 $20\times 20\times 20$ 3.831094 4.805505 $30\times 30\times 30$ 6.419004 10.545785 $40\times 40\times 40$ 15.814063 30.012744
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