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Question Number 210439 by universe last updated on 09/Aug/24

Commented by universe last updated on 09/Aug/24

$p r o v e t h a t t h e s e q u e n c e i s b o u n d e d a n d$ $c o n v e r g e n t$
	Answered by aleks041103 last updated on 09/Aug/24

	$f (x) = (x + 2) 2^{- x}$ $f^{'} (x) = 2^{- x} - l n (2) (x + 2) 2^{- x} =$ $= (1 - l n (2) (x + 2)) 2^{- x} =$ $= - l n (2) 2^{- x} (x + (2 - \frac{1}{l n (2)}))$ $\Rightarrow x < \frac{1}{l n (2)} - 2 \to f (x) ↗$ $x > \frac{1}{l n (2)} - 2 \to f (x) ↘$ $\Rightarrow f (x) ⩽ f (\frac{1}{l n (2)} - 2) = \frac{1}{l n (2)} 2^{2 - \frac{1}{l n (2)}} =$ $= \frac{4}{e l n (2)} < \frac{4}{\frac{5}{2} \frac{2}{3}} = 2.4$ $\Rightarrow x_{n + 1} = 3 - f (x_{n}) > 3 - 2.4 > 0$ $x_{1} = 1 > 0$ $\Rightarrow x_{n} > 0$ $f (x_{n} > 0) < f (0) = 2 \Rightarrow f (x_{n}) < 2 \Rightarrow x_{n + 1} > 1$ $f (x_{n} > 1) > \underset{x \to \infty}{l i m f} (x) = 0 \Rightarrow f (x_{n} > 1) > 0$ $\Rightarrow x_{n} < 3$ $x_{n} \in [1, 3)$ $b y v i r t u e o f t h e g r a p h o f t h e f u n c t i o n w e c a n$ $a r g u e t h a t x_{n} \to 2$ $t o p r o v e i t w e w i l l u s e t h e f i x e d p o i n t t h e o r e m$ $2 - x_{n + 1} = 2 - (3 - \frac{x_{n} + 2}{2^{x_{n}}}) = \frac{x_{n} + 2}{2^{x_{n}}} - 1 =$ $= \frac{4 - (2 - x_{n})}{2^{2 - (2 - x_{n})}} - 1$ $l e t a_{n} = 2 - x_{n}, t h e n a_{n} \in (- 1, 1]$ $\Rightarrow a_{n + 1} = \frac{4 - a_{n}}{4} 2^{a_{n}} - 1 = 2^{a_{n}} - 1 - \frac{1}{4} a_{n} 2^{a_{n}} =$ $= a_{n} (\frac{2^{a_{n}} - 1}{a_{n}} - \frac{2^{a_{n}}}{4})$ $g (x) = \frac{2^{x} - 1}{x} - \frac{2^{x}}{4}$ $i t i s k n o w n f r o m t h e s e r i e s e x p a n s i o n o f$ $2^{x} t h a t \frac{2^{x} - 1}{x} i s m o n o t o n i c a l l y i n c r e a s i n g .$ $f o r x \in (- 1, 1) w e h a v e$ $g (x) > \frac{2^{- 1} - 1}{- 1} - \frac{2^{x}}{4} = \frac{1}{2} - \frac{2^{x}}{4} > \frac{1}{2} - \frac{2}{4} = 0$ $a l s o$ $g (x) < \frac{2^{1} - 1}{1} - \frac{2^{x}}{4} = 1 - \frac{2^{x}}{4} < 1 - \frac{2^{- 1}}{4} = \frac{7}{8}$ $\Rightarrow∣ a_{n + 1} ∣⩽ \frac{7}{8} ∣ a_{n} ∣\Rightarrow∣ a_{n} ∣⩽∣ a_{1} ∣ {(\frac{7}{8})}^{n - 1} \to 0$ $\Rightarrow a_{n} \to 0$ $\Rightarrow x_{n} \to 2$
	Commented by universe last updated on 09/Aug/24

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