Disproof of a conjecture
Disproof of a conjecture
复制标题
反驳一个猜想
DOI：
--
复制
发表时间：
1982
期刊：
影响因子：
--
通讯作者：
F. Calogero
中科院分区：
文献类型：
--
作者：
F. Calogero
摘要设n阶对称矩阵A由下式定义：如果n个数xj是n阶Hermite多项式Hn（x）j = 0的n个零点，则矩阵A的特征值为0，1，2，.，n - 1。已经证明，矩阵A具有前n个非负整数作为其特征值的要求意味着n个数字xj在平移之前与n阶Hermite多项式的零重合。这个猜想是伪造的，forn = 4。
SummaryLet the symmetrical matrixA, of ordern, be defined in terms of then numbers xj, by the formula It is known that, if the n numbers xj are the n zeros of the Hermite polynomial of ordern,Hn(x)j = 0, the matrix A has the eigenvalues 0, 1, 2, …,n - 1. It has been conjectured that the requirement that the matrix A have the first n nonnegative integers as its eigenvalues implies that the n numbers xj coincide, up to a translation, with then zeroes of the Hermite polynomial of ordern. This conjecture is falsified, forn = 4.