Math Question and Answers


Question and Answers Forum

All Questions Topic List
Matrices and Determinants Questions
Previous in All Question Next in All Question
Previous in Matrices and Determinants Next in Matrices and Determinants
Question Number 208359 by alcohol last updated on 13/Jun/24

Commented by alcohol last updated on 13/Jun/24

$p l e a s e h e l p$
	Answered by Berbere last updated on 17/Jun/24

	$f (x) = x^{n} + x - 1 \Rightarrow f^{'} (x) = {n x}^{n - 1} + 1 ⩾ 1; f i n c r e a s e$ $f (0) = - 1, f (1) = 1 \exists c_{n} ∣ f (x_{n}) = 0$ $x_{n}^{n} + x - 1 = 0$ $f_{n + 1} (x_{n}) = x_{n}^{n + 1} + x_{n} - 1 ⩽ x_{n}^{n} + x_{n} - 1 = 0$ $x_{n + 1} > x_{n} \Rightarrow x_{n} c v x_{n} = l$ $i f l < 1 \Rightarrow \lim_{n \to \infty} l^{n} = 0 \Rightarrow l - 1 = 0 \Rightarrow l = 1 a b s u r d$ $\Rightarrow l = 1 s i n c e 0 ⩽ x_{n} ⩽ 1$ $1 - x_{n} = x_{n}^{n} = u_{n} \Rightarrow u_{n} \to 0 \Rightarrow$ $1 - x_{n} = x_{n}^{n} \Rightarrow n l n (x_{n}) - l n (1 - x_{n}) = 0$ $x_{n} = x_{n} = 1 - u_{n}$ $n l n (1 - u_{n}) - l n (u_{n}) = 0 \Rightarrow u_{n} s o l u t i o n o f f (x) = n l n (1 - x) - l n (x)$ $u_{n} \in] 0, 1 [$ $f^{'} (x) = \frac{- n}{1 - x} - \frac{1}{x} < 0$ $f (\frac{l n (n)}{2 n}) = n l n (1 - \frac{l n (n)}{2 n}) - l n (\frac{l n (n)}{2 n})$ $\sim - \frac{l n (n)}{2} - l n (l n (n)) + l n (2 n)$ $= \frac{l n (n)}{2} - l n (l n (n)) > 0 f o r n e n o u g h b i g$ $f (2 \frac{l n (n)}{n}) = n l n (1 - \frac{2 l n (n)}{n}) - l n (\frac{2 l n (n)}{n})$ $\sim - 2 l n (n) - l n (2 l n (n) + l n (n) \sim - l n (n) < 0$ $f o r n e n o u g h b i g$ $f (\frac{l n (n)}{2 n}) >; f (\frac{2 l n (n)}{n}) > 0 f (u_{n}) = 0 f d e c r e a s e$ $\Rightarrow \frac{l n (n)}{2 n} < u_{n} < \frac{2 l n (n)}{n}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum