三对角矩阵特征值的收敛半径
Convergence Radii for Eigenvalues of Tri--diagonal Matrices
摘要：
考虑一族形式为$L+ zB,$的无限三对角矩阵，其中矩阵$L$是对角的，元素为$L_{kk}= k^2,$，矩阵$B$是非对角的，元素为非零$B_{k,{k+1}}=B_{{k+1},k}= k^α, 0 \leq α< 2.$。$L+ zB$的谱是离散的。对于小$|z|$的$n$ -th特征值$E_n (z), E_n (0) = n^2,$是一个定义良好的解析函数。设$R_n$是其泰勒级数关于$z= 0.$的收敛半径，证明了$$ R_n \leq C(α) n^{2-α} \quad \text{if} 0 \leq α<11/6.$$
英文摘要：
Consider a family of infinite tri--diagonal matrices of the form $L+ zB,$ where the matrix $L$ is diagonal with entries $L_{kk}= k^2,$ and the matrix $B$ is off--diagonal, with nonzero entries $B_{k,{k+1}}=B_{{k+1},k}= k^α, 0 \leq α< 2.$ The spectrum of $L+ zB$ is discrete. For small $|z|$ the $n$-th eigenvalue $E_n (z), E_n (0) = n^2,$ is a well--defined analytic function. Let $R_n$ be the convergence radius of its Taylor's series about $z= 0.$ It is proved that $$ R_n \leq C(α) n^{2-α} \quad \text{if} 0 \leq α<11/6.$$
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