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8+4*sqrt(11)

2010-02-21 18:29:58 補充：
Set a(1)=4√7, a(n+1)=4√[7+a(n)], n=1,2,3,...
(1) a(n)<8+4√11
(M.I.) a(n+1)<= 4√(7+8+4√11) =8+4√11
(2) a(n)> 8 (trivial)
(3) a(n+1)> a(n)
a(n+1)^2 - a(n)^2 = 16[7+a(n)] - a(n)^2
= - [a(n)-8]^2+ 176> -[ 4√11]^2 + 176=0
and a(n)>0, so, a(n+1)>a(n)

From (1),(3), we know that the sequence { a(n) } is increasing
and bounded above, hence, lim(n->inf.) a(n) exist.
Let lim(n->inf.) a(n)=x, then x=4√(7+x).
x^2= 16(7+x), x= 8 +/- 4√11 (8-4√11 rejects)
Ans: 8+4√11

Let x be 4√[7+4[√7+4[√7+4[√7+4]]]]...
x = 4√[7+4[√7+4[√7+4[√7+4]]]]... ---------(1)
x^2 = 7+4[√7+4[√7+4[√7+4]]]]... -----------(2)
(2) - (1), x^2 - x = 7
x^2 - x - 7 = 0
x = [1 +/- √29]/2
= (1+√29)/2 or (1-√29)/2

好問題,有無高手可以指教?
假設題目加多句if exists

是不是要先證limit存在?

設 x = 4√(7+4√(7...........

所以 x = 4√(7+x)

x^2 - 16x - 112 = 0

x = 8 ± √704 /2

=21.267 or = -5.267