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Question Number 167183 by alcohol last updated on 09/Mar/22

Commented by ArielVyny last updated on 09/Mar/22

$i l e t q u e s t i o n (f) f o r y o u r a t t e n t i o n$
	Answered by ArielVyny last updated on 09/Mar/22

	$a) s h o w i n g t h a t (U_{n}) a n d (V_{n}) a r e s t r i c t l y$ $p o s i t i v e s e q u e n c e s .$ $r e a s o n n i g b y r e c c u r e n c e o n$ $w e h a v e U_{0} = 1 a n d V_{0} = 2 a l s o a r e s t r i c t l y$ $p o s i t i v e$ $w e s u p p o s e t h a t (U_{n}) a n s (V_{n}) a r e s t r i c t l y$ $p o s i t i v e t h e n w e h a v e U_{n} > 0 a n d V_{n} > 0$ $n o w w e w i l l s b o w i n g t h a t U_{n + 1} a n d V_{n + 1}$ $a r e s t r i c t l y p o s i r i v e$ $V_{n + 1} = \frac{U_{n} + V_{n}}{2} a c c o r d i n g t h a t U_{n} a n d V_{n} a r e$ $s t r i t l y p o s i t i v e a n d 2 > 0 (e v i d e n c e) w e$ $h a v e V_{n + 1} > 0$ $a n d U_{n + 1} = \frac{U_{n} V_{n}}{V_{n + 1}} w e k n o w t h a t V_{n + 1} > 0$ $a n d U_{n} V_{n} i s s t r i c t l y p o s i t i v e b e c a u s e {i t}^{'} s a$ $p r o d u c t o f t w o p o s i t i v e s v a l u e s$
	Answered by ArielVyny last updated on 09/Mar/22

	$(b) s h o s i n g t h a t 0 ⩽ W_{n + 1} ⩽ \frac{1}{2} W_{n}$ $w_{n} = v_{n} - u_{n} \to w_{n + 1} = v_{n + 1} - u_{n + 1}$ $\to w_{n + 1} = \frac{u_{n} + v_{n}}{2} - \frac{2 u_{n} v_{n}}{u_{n} + v_{n}}$ $= \frac{{(u_{n} + v_{n})}^{2}}{2 (u_{n} + v_{n})} - \frac{4 u_{n} v_{n}}{2 (u_{n} + v_{n})}$ $= \frac{u_{n}^{2} + v_{n}^{2} - 2 u_{n} v_{n}}{2 (u_{n} + v_{n})} = \frac{{(u_{n} - v_{n})}^{2}}{2 (u_{n} + v_{n})} ⩾ 0 (q u e s t i o n a)$ $0 ⩽ w_{n + 1}$ $\to w_{n + 1} - \frac{w_{n}}{2} = \frac{u_{n} + v_{n}}{2} - \frac{2 u_{n} v_{n}}{u_{n} + v_{n}} - \frac{v_{n}}{2} + \frac{u_{n}}{2}$ $= u_{n} - \frac{2 u_{n} v_{n}}{u_{n} + v_{n}} = u_{n} (1 - \frac{2 v_{n}}{u_{n} + v_{n}})$ $w_{n + 1} ⩾ 0 \to w_{n} ⩾ 0 a n d v_{n} ⩾ u_{n}$ $\frac{2 v_{n}}{u_{n} + v_{n}} = \frac{v_{n}}{v_{n + 1}} ⩾ \frac{1}{2} \to 2 v_{n} ⩾ v_{n + 1} \to 2 v_{n} - \frac{u_{n}}{2} - \frac{v_{n}}{2} ⩾ 0$ $\to \frac{3}{2} v_{n} - \frac{u_{n}}{2} ⩾ 0 (t r u e) b e c a u s e v_{n} > u_{n}$ $t h e n w e s h o w t h a t 1 - \frac{2 v_{n}}{u_{n} + v_{n}} ⩽ 0 t h e n$ $w_{n + 1} - \frac{w_{n}}{2} ⩽ 0$ $c o n c l u s i o n 0 ⩽ w_{n + 1} ⩽ \frac{w_{n}}{2}$
	Answered by ArielVyny last updated on 09/Mar/22

	$c) s h o w t h a t 0 ⩽ w_{n + 1} ⩽ \frac{1}{2^{n}}$ $b y r e c c u r e n c e o n n$ $0 ⩽ w_{0} ⩽ 1 w i t h w_{0} = 1$ $s u p p o s e t h a n t 0 ⩽ w_{n} ⩽ \frac{1}{2^{n - 1}}$ $k n o w i n g t h a t \frac{1}{2^{n - 1}} ⩾ \frac{1}{2^{n}} a n d w_{n + 1} ⩽ \frac{w_{n}}{2} ⩽ w_{n}$ $0 ⩽ w_{n} ⩽ \frac{1}{2^{n - 1}} \to 0 ⩽ \frac{w_{n}}{2} ⩽ \frac{1}{2^{n}} \to 0 ⩽ w_{n + 1} ⩽ \frac{1}{2^{n}}$ $w d d e d u c e t h a t l i m w_{n + 1} = 0 (b y {g e n d a r m}^{'} s t h e o r e m)$ $a n d l i m w_{n + 1} = l i m w_{n} = 0$
	Answered by ArielVyny last updated on 09/Mar/22

	$d) v_{0} = 2 a n d v_{1} = \frac{1 + 2}{2} = \frac{3}{2}$ $u_{1} = \frac{4}{3} a f t e r c o p u t i n g u_{1} a n d v_{1} w e a d m o t t h a t$ $(v_{n}) d e c r e a s e a n d (u_{n}) i n c r e a s e$ $a n d l i m w_{n} = l i m (v_{n} - u_{n}) = 0$ $a n d s e q u e n s e s (u_{n}) a n (v_{n}) a r e a d j a c e n t$
	Answered by ArielVyny last updated on 09/Mar/22

	$u_{n + 1} = \frac{u_{n} v_{n}}{v_{n + 1}} \to u_{n + 1} v_{n + 1} = u_{n} v_{n} f o r a l l n w e d e d u c e$ $t h a t (u_{n} v_{n}) i s c o n s t a n t$
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