Consider the sequence (x_n) defined by the recurrence relation x_{n+1} = c*x_n^2 - 2, where c > 0. Suppose there exists a non-empty, open interval (a, b) such that for all x_0 satisfying a < x_0 < b, the sequence converges to a limit. The sequence converges to the value:
A. (1 + sqrt(1 + 8c)) / (2c)
B. (1 - sqrt(1 + 8c)) / (2c)
C. 2
D. 1/sqrt(c)
GATE 2007 · Discrete Mathematics · Recurrence Relation · medium
Answer: The sequence converges to L_- = (1 - sqrt(1+8c)) / (2c). The answer is option B.
Find fixed points by solving L = c*L^2 - 2: c*L^2 - L - 2 = 0. By quadratic formula with a=c, b=-1, d=-2: L = (1 +/- sqrt(1 + 8c)) / (2c). Two roots: L_+ = (1 + sqrt(1+8c))/(2c) and L_- = (1 - sqrt(1+8c))/(2c).
Check stability: compute |f'| at each fixed point: f'(x) = 2cx. At L_+: |2c*L_+| = |1 + sqrt(1+8c)| > 1 always. Unstable. At L_-: |2c*L_-| = |1 - sqrt(1+8c)| = sqrt(1+8c) - 1. For c < 3/8, sqrt(1+8c) < 2 so this is < 1. L_- is attracting for 0 < c < 3/8.