If U_n = [U_{n-1}]² + a U_{n-1} + 1, find lim{n->∞} U_1 U_2 U_3 ... U_{n-1} / U_n.?

Question

If U_n = [U_{n-1}]² + a U_{n-1} + 1

(n = 1, 2, 3, 4, ...)

find lim{n->∞} U_1 * U_2 * U_3 ... U_{n-1} / U_n.

Answer 1

Write P_n = U_1 * U_2 * U_3 ... U_{n-1} / U_n

a) Consider the case that a = 3, U_1 = -1.

U_2 = -1^2 - 3 + 1 = -1

So U_n = -1 for all n. Thus

P_n = -1^n

The sequence of numbers oscillates between -1 and 1, so the limit does not exist.

b) consider the case a = -2, U_1 = 0,

U_2 = 0 - 0 + 1 = 1

U_3 = 1 - 2 + 1 = 0

So for N even, P_N = 0, but for N odd, P_N is not well defined.

c) Okay, let's restrict |a| < 2 so that U_{n-1}^2 + a U_{n-1} + 1 cannot possibly have real roots, so we can avoid the case where U_n = 0 and thus P_N being undefined. The minimum value attained by the polynomial x^2 + ax + 1 is attained at x = -a / 2, with value 1-a^2/4

c1) If U_1 = 0, then clearly P_N = 0 for all N > 1, then the sequence converges to 0. Consider the case where U_1 is not zero.

c2) Now, if a < -1, then U_n = U_{n-1} has solutions

x^2 + (a-1)x + 1 => x = (1-a)/2 +/- sqrt( a^2 - 2a -3 )

One of them will be greater than 1 (call it x+), the other less than 1 (call it x-). If U_1 equals the former, than P_n = (x+)^(n-2) -> infinity as n goes to infinity. If U_1 equals the latter, than P_N = (x-)^(n-2) -> 0

Furthermore, these two fixed points are stable, so for a range of starting values (which I will not bother to write down here), we must have P_n >= C (x+ - ε)^(n-2) for all sufficiently large n, for some constant C, and for a particular ε such that x+ - ε > 1. So for these range of starting values, P_n goes to infinity. Similarly, there is a range of starting values for which P_n goes to 0.

If the sequence U_n does not converge to a finite value, then the sequence U_n must escape to positive infinity (U_n is unbounded, so |U_n| > 4 for some n, but if |U_n| > 4, U_{n+1} must be positive and > 2, and we must have U_m > 2 for all m > n, therefore U_m must scape to positive, and not negative, infinity). We'll treat this case in the following section.

c3) If a > -1, x < x^2 + ax + 1, so the sequence U_n is strictly increasing, and must escape to positive infinity. So here we'll treat the case where U_n escapes. [hum... my original method has a flaw... let me think more about this.]

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Even with out the complete solution, I think it is clear that the there isn't a unique limit for the U_n. The limit depends both on a and on U_1.

Answer 2

If

L(u1) = lim{n->∞} U_1 * U_2 * U_3 ... U_{n-1} / U_n

then:

L(u1) = u1*L(u1^2+a*u1+1)

I don't think it's possible to find this function explicitly, because for a>2 it has so much zeroes.

Anyway, this is how it looks like:

http://i524.photobucket.com/albums/cc321...

very interesting!