设a>0,{Xn}满足X0>0,Xn+1=1/2(Xn+a/Xn) ,n+1是下标,n=0,1,2...,证明:{Xn}收敛,求(n趋向无穷) lim Xn
问题描述:
设a>0,{Xn}满足X0>0,Xn+1=1/2(Xn+a/Xn) ,n+1是下标,n=0,1,2...,证明:{Xn}收敛,求(n趋向无穷) lim Xn
答
证明:∵x(0)>0且x(n+1)=[x(n)+a/x(n)]/2∴x(n)>0∴由均值不等式知[x(n)+a/x(n)]/2≥√a即x(n+1)≥√a∴数列{x(n)}有下界.(1)又x(n+1)/x(n)=[x(n)+a/x(n)]/[2x(n)]=[1+a/x²(n)]/2x(n+1)≥√a∴x(n)≥√a∴a/x...