a_n =a_(n−1) +((x−a_(n−1) )/t) lim_(n→∞) a


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Question Number 214216 by MathematicalUser2357 last updated on 01/Dec/24

$a_{n} = a_{n - 1} + \frac{x - a_{n - 1}}{t}$ $\lim_{n \to \infty} a_{n} = ?$
	Answered by mr W last updated on 01/Dec/24

	$a_{n} + (\frac{1}{t} - 1) a_{n - 1} - \frac{x}{t} = 0$ $l e t a_{n} = {A r}^{n} + B$ ${A r}^{n} + B + (\frac{1}{t} - 1) ({A r}^{n - 1} + B) - \frac{x}{t} = 0$ $(r + \frac{1}{t} - 1) {A r}^{n - 1} + \frac{B}{t} - \frac{x}{t} = 0$ $\Rightarrow \frac{B}{t} - \frac{x}{t} = 0 \Rightarrow B = x$ $\Rightarrow r + \frac{1}{t} - 1 = 0 \Rightarrow r = 1 - \frac{1}{t}$ $\Rightarrow a_{n} = A {(1 - \frac{1}{t})}^{n} + x$ $i f - 1 < 1 - \frac{1}{t} < 1, i . e . t > \frac{1}{2},$ $\lim_{n \to \infty} a_{n} = x$ $o t h e r w i s e$ $\lim_{n \to \infty} a_{n} = \infty$
	Commented by mr W last updated on 01/Dec/24
	이것이 당신이 기대했던 것입니까?
	Commented by issac last updated on 03/Dec/24
	한국인임?
	Commented by mr W last updated on 03/Dec/24

	$m a y b e$
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