let-0-lt-u-0-lt-1-and-u-n-1-1-u-n-2-study-the-convergence-of-u-n-

Question Number 37818 by prof Abdo imad last updated on 17/Jun/18

l e t 0 < u_{0} < 1 a n d u_{n + 1} = \sqrt{\frac{1 + u_{n}}{2}}

s t u d y t h e c o n v e r g e n c e o f u_{n}

Commented by math khazana by abdo last updated on 19/Jun/18

l e t f (x) = \sqrt{\frac{1 + x}{2}} w i t h x ⩾ 0

f (x) = x \Rightarrow \frac{1 + x}{2} = x^{2} \Rightarrow 1 + x = 2 x^{2} \Rightarrow

2 x^{2} - x - 1 = 0 Δ = 1 + 8 = 9

x_{1} = \frac{1 + 3}{4} = 1 a n d x_{2} = \frac{1 - 3}{4} = - \frac{1}{2} w e h a v e

u_{n + 1} = f (u_{n}) a n d f i s c o n t i n u e \Rightarrow {l i m}_{n \to + \infty} u_{n} = 1

l e t f i n d a n o t h e r f o r m o f u_{n} w e p u t

u_{0} = c o s α \Rightarrow u_{1} = \sqrt{\frac{1 + c o s (α)}{2}} = c o s (\frac{α}{2})

l e t p r o v e t h a t u_{n} = c o s (\frac{α}{2^{n}})

r e l a t i o n t r u e f o r n = 0

l e t s u p p o s e u_{n} = c o s (\frac{α}{2^{n}}) \Rightarrow

u_{n + 1} = \sqrt{\frac{1 + u_{n}}{2}} = \sqrt{\frac{1 + c o s (\frac{α}{2^{n}})}{2}}

= c o s (\frac{α}{2^{n + 1}}) . i t s n o w c l e a r t h a t

{l i m}_{n \to + \infty} u_{n} = 1 .