0<c<1 such that the recursive sequence {a_n } defined by setting a_(1 ) = (c/2) , a_(n+1) = (1/2)(c+a


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Question Number 213548 by universe last updated on 08/Nov/24
$0<c<1 such that the recursive sequence {a_n } defined by setting a_(1 ) = (c/2) , a_(n+1) = (1/2)(c+a_n ^2 ) for n∈ N monotonic and convergent$
$0 < c < 1 such that the recursive sequence$ ${a_{n}} defined by setting$ $a_{1} = \frac{c}{2}, a_{n + 1} = \frac{1}{2} (c + a_{n}^{2}) for n \in N$ $monotonic and convergent$
	Answered by Berbere last updated on 10/Nov/24

	$L e t f (z) = \frac{1}{2} (c + z^{2}) z \in R_{+}; f^{'} (z) = z ⩾ 0$ $c l a i m \forall n \in N a_{n} \in [0, 1]$ $a_{n} = f o f . . . . . f (a_{1}) n t i m e z$ $f [0, 1] = [f (0), f (1)] = [\frac{c}{2}, \frac{1 + c}{2}] \subseteq [0, 1]; 0 ⩽ c ⩽ 1$ $\Rightarrow f o . . . . . f ([0, 1]) \subseteq [0, 1] \Rightarrow \forall n \in N a_{n} \in [0, 1]$ $a_{2} = \frac{c}{2} [1 + \frac{c}{4}] ⩾ \frac{c}{2} \Rightarrow a_{2} ⩾ a_{1} s i n c e f i n c r e s e$ $\Rightarrow \forall n \in N a_{n + 1} ⩾ a_{n} \Rightarrow a_{n} i n c r e a s e b o u n d e d c v t o f i x p i n t o f f$ $l = \lim_{n \to \infty} a_{n} i s S o l u t i o n o f f (x) = x \Leftrightarrow x^{2} - 2 x + c = 0$ $x = 1 + \sqrt{1 - c} > 1; 1 - \sqrt{1 - c}$ $l = 1 - \sqrt{1 - c}$
	Commented by universe last updated on 10/Nov/24

	$nice solution sir$ $thank you so much$
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