Studies-of-convergences-the-numbers-real-sequence-x-n-with-x-1-1-and-x-n-1-x-n-2-2-2x-n-n-1-

Question Number 55641 by gunawan last updated on 01/Mar/19

Studies of convergences

the numbers real sequence {x_{n}},

with x_{1} = 1 and x_{n + 1} = \frac{x_{n}^{2} + 2}{2 x_{n}}, n ⩾ 1

Commented by maxmathsup by imad last updated on 01/Mar/19

w e h a v e x_{n + 1} = f (x_{n}) w i t h f (x) = \frac{x^{2} + 2}{2 x} = \frac{x}{2} + \frac{1}{x} w i t h x > 0

w e {h a v e f}^{^{'}} (x) = \frac{1}{2} - \frac{1}{x^{2}} = \frac{x^{2} - 2}{2 x^{2}} a n d f^{^{'}} (x) = 0 \Leftrightarrow x = \bar{+} \sqrt{2}

x 0 \sqrt{2} + \infty

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

f (x) + \infty d e c c f (\sqrt{2}) i n c r e + \infty

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

f i s c o n t i n u e o n] 0, + \infty [l e t f i n d t h e f i x p o i n t

f (x) = x \Leftrightarrow \frac{x}{2} + \frac{1}{x} = x \Leftrightarrow \frac{1}{x} = \frac{x}{2} \Rightarrow 2 = x^{2} \Rightarrow x = \sqrt{2} b e c a u s e x > 0

(x_{n}) i s c o n v e r g e n t a n d {l i m}_{n \to + \infty} x_{n} = \sqrt{2} .