if (√(x+(√(x+(√(x+(√(x+(√x)))))))))=2022 find [x]=?


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Question Number 177030 by mr W last updated on 01/Oct/22

$i f \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x}}}}} = 2022$ $f i n d [x] = ?$
	Answered by TheHoneyCat last updated on 30/Sep/22

	$if x exists, squaring the equlity implies that :$ $x + 2022 = {(2022)}^{2}$ $i e x = 2022 \times 2021$ $Now we just need to proove that x exists$ $i e that their exists a value x such that$ $denoting f_{x} (y) := \sqrt{x + y}$ ${(f_{x}^{\circ n} (0))}_{n \in N} converges to 2022$ $it turns out that if x > 4$ $then \sqrt{x} < x / 2 and \sqrt{x} > 2$ $so x > 2 and y > 2 \Rightarrow \sqrt{x + y} < (x + y) / 2$ $and still \sqrt{x + y} > 2$ $so if x > 4$ $(so that the second term is superior to 2)$ $the sequence is always over 2 and always$ $under U_{n + 1} = \frac{x + U_{n}}{2} (with U_{1} = x)$ $basically the sequence is bounded$ $it is trivial to see that it is monotonous, so$ $it converges .$ $Denoting by f_{\infty} its limit$ $x + f_{\infty} = f_{\infty}^{2}$ $which admits solutions for any values .$ $including f_{\infty} = 2022_{◼}$ $N o t e t h a t I h a v e a s s u m e d t h a t$ $\sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + x}}}} m e a n t \sqrt{x + \sqrt{x + \sqrt{x + . . .}}}$ $i f y o u a c t u a l l y d i d m e a n t y o u r q u e s t i o n$ $t o b e a b o u t a f i n i t e a m o u n t o f$ $s q u a r e r o o t s, i t i s e q u i v a l e n t t o s o l v i n g$ $a n e q u a t i o n o f d e g r e e 8 .$ $W i c h (i n g e n e r a l i s n o t s o l v a b l e)$ $T h e p o l y n o m i a l i s n o t s y m e t r i c (s o y o u$ $c a n n o t r e d u c e i t t o d e g = 4)$ $t h e c o e f i c i e n t s a r e e i t h e r d i v i s i b l e b y 2022$ $(t h a t g i v e s t h e h o p e o f l o o k i n g f o r i n t e g e r$ $s o k u t i o n s t h r o u g h t c o n g r u e n c e s) o r \pm 1$ $(s o I g u e s s c o n g r u e n c e s a r e n o t a n o p t i o n) .$ $Y o u r p o l y n o m i a l c a n b e c o n s t r u c t e d$ $r e c u r s i v e l y b y c o m p o s i t i o n ({i t}^{'} s l i k e t h e$ $Prime causes double exponent: use braces to clarify$ $t h e s m a l l e r s n v a l u e s i n t h e s e a r c h o f a$ $p a t e r n t o u s e r e c u r s i v e l y . B u t f o u n d n o n e .$ $I f e a r t h a t t h e p r e c i s e q u e s t i o n y o u h a v e$ $a s k e d h a s n o e x p l i c i t a n s w e r e .$ $H o w e v e r, I m i g h t b e w r o n g o n t h a t .$ $I f s o m e o n e k n o w s I^{'} m w r o n g, c o u l d y o u$ $p l e a s e c o m m e n t t h e a n s w e r e p l e a s e ?$ $T h a n k s .$
	Commented by mr W last updated on 01/Oct/22

	$t h a n k s s i r!$ $t h e L H S i s a n e x p r e s s i o n w i t h e x a c t l y$ $5 s q u a r e r o o t s, n o t i n f i n i t e o n e s .$ $b t w, t h e q u e s t i o n i s n o t t o f i n d x, b u t$ $t o f i n d [x] .$ $a s y o u s a i d, {i t}^{'} s n o t s o l v a b l e f o r x,$ $b u t m a y b e s o l v a b l e f o r [x], t h e i n t e g e r$ $p a r t o f x .$
	Commented by Tawa11 last updated on 02/Oct/22

	$Great sirs$
	Answered by Frix last updated on 30/Sep/22

	$\sqrt{x + \sqrt{x + . . ._{(n times)}}} = y \Rightarrow x = y (y - 1) + q_{n}$ $with q_{1} = y and q_{\infty} = 0$ $this means x starts at y^{2} and never gets lower$ $than {(y - 1)}^{2} + (y - 1) = y (y - 1)$ $looking at the first equations we can solve$ $for given y :$ $\sqrt{x + \sqrt{x}} = 5 \Rightarrow x \approx 4 (4 + 1) + . 475$ $\sqrt{x + \sqrt{x + \sqrt{x}}} = 5 \Rightarrow x \approx 4 (4 + 1) + . 0477$ $\sqrt{x + \sqrt{x}} = 50 \Rightarrow x \approx 49 (49 + 1) + . 4975$ $\sqrt{x + \sqrt{x + \sqrt{x}}} = 50 \Rightarrow x \approx 49 (49 + 1) + . 004975$ $we see that with higher y q_{2} \to . 5 and q_{3} \to 0$ $\sqrt{x + \sqrt{x}} = 2022 \Rightarrow q_{2} \approx 499938 q_{3} \approx . 000046$ $\Rightarrow ⌊ x ⌋ = 2021 (2021 + 1) = 4086462$
	Commented by mr W last updated on 30/Sep/22

	$t h a n k s s i r f o r t h e r i g h t a n s w e r!$
	Answered by mr W last updated on 01/Oct/22

	$w e h a v e$ $\underset{2 t i m e s r o o t s}{\sqrt{x + \sqrt{x}}} ⩽ \underset{n t i m e s r o o t s}{\sqrt{x + \sqrt{x + \sqrt{x + . . . + \sqrt{x}}}}} < \underset{i n f i n i t e t i m e s r o o t s}{\sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + . . .}}}}}$ $w i t h n \in N a n d n ⩾ 2$ $i f \sqrt{x_{2} + \sqrt{x_{2}}} = 2022$ $a n d \sqrt{x_{n} + \sqrt{x_{n} + \sqrt{x_{n} + . . . + \sqrt{x_{n}}}}} = 2022$ $a n d \sqrt{x_{\infty} + \sqrt{x_{\infty} + \sqrt{x_{\infty} + . . .}}} = 2022$ $t h e n w e h a v e$ $x_{\infty} < x_{n} ⩽ x_{2}$ $\sqrt{x_{2} + \sqrt{x_{2}}} = 2022$ $x_{2} + \sqrt{x_{2}} = 2022^{2}$ $x_{2} = {(2022^{2} - x_{2})}^{2}$ $x_{2} = 2022^{4} - 2 \times 2022^{2} x_{2} + x_{2}^{2}$ $x_{2}^{2} - (2 \times 2022^{2} + 1) x_{2} + 2022^{4} = 0$ $x_{2} = \frac{2 \times 2022^{2} + 1 - \sqrt{{(2 \times 2022^{2} + 1)}^{2} - 4 \times 2022^{4}}}{2}$ $= \frac{2 \times 2022^{2} + 1 - \sqrt{4 \times 2022^{2} + 1}}{2}$ $< \frac{2 \times 2022^{2} + 1 - \sqrt{4 \times 2022^{2}}}{2}$ $= \frac{2 \times 2022^{2} + 1 - 2 \times 2022}{2}$ $= 2022^{2} - 2022 + \frac{1}{2}$ $= 2021 \times 2022 + \frac{1}{2}$ $\sqrt{x_{\infty} + 2022} = 2022$ $\Rightarrow x_{\infty} = 2022^{2} - 2022 = 2021 \times 2022$ $f r o m x_{\infty} < x_{n} ⩽ x_{2} w e g e t$ $2021 \times 2022 < x_{n} < 2021 \times 2022 + \frac{1}{2}$ $t h a t m e a n s$ $[x_{n}] = 2021 \times 2022 = 4086462$ $g e n e r a l l y$ $i f \underset{2 o r m o r e, b u t f i n i t e r o o t s}{\sqrt{x + \sqrt{x + \sqrt{x + . . . + \sqrt{x}}}}} = k \in N \land k ⩾ 1$ $t h e n [x] = (k - 1) k$
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