study-the-convergence-of-u-n-1-2-1-u-n-2-u-n-1-with-u-0-0-

Question Number 56828 by maxmathsup by imad last updated on 24/Mar/19

s t u d y t h e c o n v e r g e n c e o f u_{n + 1} = 2 \sqrt{1 + u_{n}^{2}} - u_{n} - 1 w i t h u_{0} = 0

Commented by maxmathsup by imad last updated on 16/Apr/19

w e h a v e u_{n + 1} = f (u_{n}) w i t h f (x) = 2 \sqrt{1 + x^{2}} - x - 1

l e t p r o v e t h a t u_{n} > 0 w e h a v e u_{1} = 2 - 1 = 1 > 0 l e t s u p p o s e u_{n} > 0

2 \sqrt{1 + u_{n}^{2}} - u_{n} - 1 = 2 \sqrt{1 + u_{n}^{2}} - (u_{n} + 1) a n d 4 (1 + u_{n}^{2}) - {(u_{n} + 1)}^{2}

= 4 u_{n}^{2} + 4 - u_{n}^{2} - 2 u_{n} - 1 = 2 u_{n}^{2} - 2 u_{n} + 3 \to Δ = 4 - 4 (2) . 3 < 0 (a = 2 > 0) \Rightarrow

2 u_{n}^{2} - u_{n} + 3 f o r t h a t l e t s t u d y t h e v a r i a t i o n o f f o n [0, + \infty [

f^{^{'}} (x) = 2 \frac{2 x}{2 \sqrt{1 + x^{2}}} - 1 = \frac{2 x}{\sqrt{1 + x^{2}}} - 1 = \frac{2 x - \sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}} = \frac{4 x^{2} - 1 - x^{2}}{\sqrt{1 + x^{2}} (2 x + \sqrt{1 + x^{2}}})

= \frac{3 x^{2} - 1}{\sqrt{1 + x^{2} (2 x + \sqrt{1 + x^{2})}}} s o f^{^{'}} (x) = 0 \Leftrightarrow x = \frac{1}{\sqrt{3}}

f (\frac{1}{\sqrt{3}}) = 2 \sqrt{1 + \frac{1}{3}} - \frac{1}{\sqrt{3}} - 1 = \frac{4}{\sqrt{3}} - \frac{1}{\sqrt{3}} - 1 = \sqrt{3} - 1

x 0 \frac{1}{\sqrt{3}} + \infty

f^{^{'}} (x) - +

f (x) 1 d e c r \sqrt{3} - 1 i n c r + \infty

l e t d e t e r m i n e t h e f i x e d p o i n t

f (x) = x \Rightarrow 2 \sqrt{1 + x^{2}} - x - 1 = x \Rightarrow 2 \sqrt{1 + x^{2}} = 2 x + 1 \Rightarrow 4 (1 + x^{2}) = 4 x^{2} + 4 x + 1 \Rightarrow

4 = 4 x + 1 \Rightarrow 4 x = 3 \Rightarrow x = \frac{3}{4} f i s c o n t i n u e o n [0, + \infty [s o (u_{n}) c o n v e r g e s a n d

{l i m}_{n \to + \infty} u_{n} = \frac{3}{4} .

Answered by kaivan.ahmadi last updated on 25/Mar/19

l = 2 \sqrt{1 + l^{2}} - l - 1 \Rightarrow 2 l + 1 = 2 \sqrt{1 + l^{2}} \Rightarrow

4 l^{2} + 4 l + 1 = 4 + 4 l^{2} \Rightarrow 4 l = 3 \Rightarrow l = \frac{3}{4}