一題數學：數列之極限值

a(1)=2, a(n+1)= [a(n)+ 2/a(n)]/2, then
by induction, a(n) are rational number ,√2<=a(n) and a(n+1)<a(n) for n>0
(since, a(n+1)=[a(n)+2/a(n)]/2 >= √[a(n)*2/a(n)]=√2
and 2[a(n+1)-a(n)]= 2/a(n) - a(n)= [2-a(n)^2]/a(n) < 0 )
so that the sequence {a(n)} is decreasing and bounded below by √2
then lim(n->inf) a(n) exists.
Let x=lim(n->inf) a(n), then
lim(n->inf) a(n+1)=lim(n->inf) [a(n)+2/a(n)]/2
x=[x+2/x]/2, then x^2=2, x=√2

2010-06-23 22:36:51 補充：
Solve x^2=2 by the Newton's method and x1=2.

2010-06-26 09:03:15 補充：
Newton's method是指用切線方法求方程根的近似值
公式如下:解f(x)=0, 先猜(預測)a(1)
再用遞迴a(n+1)=a(n)- f(a(n))/f'(a(n))
本題: a(1)=2, a(n+1)=a(n)-[a(n)^2-2]/[2a(n)]=[a(n)+2/a(n)]/2