Borel的一个经典性定理是,如果两组整函数G_i(Z)(i=1,2,…,n)和H_i(Z)(i=1,2,…n)满足恒等式sum from j=1 to n G_i(Z)e^Hj^(Z)≡0 并且如果G_i(1≤i≤n)的增长性,在某种意义下,较慢于e^Hj^(-H)k(1≤j,k≤n,j≠k)的增长性,则G_i(Z)≡0 (...Borel的一个经典性定理是,如果两组整函数G_i(Z)(i=1,2,…,n)和H_i(Z)(i=1,2,…n)满足恒等式sum from j=1 to n G_i(Z)e^Hj^(Z)≡0 并且如果G_i(1≤i≤n)的增长性,在某种意义下,较慢于e^Hj^(-H)k(1≤j,k≤n,j≠k)的增长性,则G_i(Z)≡0 (i=1,2,…,n),在本文中得出了这个定理的几个推广。展开更多
We find that there is a serious mistake in the proof of Lemma 3 of paper [1]. The author wants to prove, copying the original method in [2], the global Borel theorem. But a Sample example shows that this is impossible.
文摘Borel的一个经典性定理是,如果两组整函数G_i(Z)(i=1,2,…,n)和H_i(Z)(i=1,2,…n)满足恒等式sum from j=1 to n G_i(Z)e^Hj^(Z)≡0 并且如果G_i(1≤i≤n)的增长性,在某种意义下,较慢于e^Hj^(-H)k(1≤j,k≤n,j≠k)的增长性,则G_i(Z)≡0 (i=1,2,…,n),在本文中得出了这个定理的几个推广。
文摘We find that there is a serious mistake in the proof of Lemma 3 of paper [1]. The author wants to prove, copying the original method in [2], the global Borel theorem. But a Sample example shows that this is impossible.
