0-lt-c-lt-1-such-that-the-recursive-sequence-a-n-defined-by-setting-a-1-c-2-a-n-1-1-2-c-a-n-2-for-n-N-monotonic-and-convergent-

Question Number 213548 by universe last updated on 08/Nov/24

0 < c < 1 such that the recursive sequence

{a_{n}} defined by setting

a_{1} = \frac{c}{2}, a_{n + 1} = \frac{1}{2} (c + a_{n}^{2}) for n \in N

monotonic and convergent

Answered by Berbere last updated on 10/Nov/24

L e t f (z) = \frac{1}{2} (c + z^{2}) z \in R_{+}; f^{'} (z) = z ⩾ 0

c l a i m \forall n \in N a_{n} \in [0, 1]

a_{n} = f o f \dots . . f (a_{1}) n t i m e z

f [0, 1] = [f (0), f (1)] = [\frac{c}{2}, \frac{1 + c}{2}] \subseteq [0, 1]; 0 ⩽ c ⩽ 1

\Rightarrow f o \dots . . f ([0, 1]) \subseteq [0, 1] \Rightarrow \forall n \in N a_{n} \in [0, 1]

a_{2} = \frac{c}{2} [1 + \frac{c}{4}] ⩾ \frac{c}{2} \Rightarrow a_{2} ⩾ a_{1} s i n c e f i n c r e s e

\Rightarrow \forall n \in N a_{n + 1} ⩾ a_{n} \Rightarrow a_{n} i n c r e a s e b o u n d e d c v t o f i x p i n t o f f

l = \lim_{n \to \infty} a_{n} i s S o l u t i o n o f f (x) = x \Leftrightarrow x^{2} - 2 x + c = 0

x = 1 + \sqrt{1 - c} > 1; 1 - \sqrt{1 - c}

l = 1 - \sqrt{1 - c}

Commented by universe last updated on 10/Nov/24

nice solution sir

thank you so much