difference equation

題目只是要求解所給予的差分方程和計算最後的limit，所以只需按標準方法(characteristic function+initial condition)就可以了。至於如何得出該方程，不需要花時間去想，因為那條方程本身是錯的(所以題目才會強調CONJECTURE)。

2009-03-09 01:51:30 補充：
Tn=4T(n-1)-2T(n-2)
characteristic equation is
x^2 = 4x - 2
x= 2+sqrt(2) or 2-sqrt(2)
So Tn = A(2+sqrt(2))^n + B(2-sqrt(2))^n.
Put T0=1, T1=2, we get
A=B=1/2. ie, Tn = [(2+sqrt(2))^n + (2-sqrt(2))^n]/2.

For the limit, observe that
(2+sqrt(2))^n /2< Tn < (2+sqrt(2))^n
So ln(2+sqrt(2)) - (ln2)/n < ln(Tn)/n < ln(2+sqrt(2))
By sandwich theorem, when n tends to infinity,
lim(lnTn)/n = ln(2+sqrt(2))

can u solve it for me PLEASE?

2009-03-10 04:52:21 補充：
where is the sketchof the arrangements for the case n=2 ...?
thanks