if-x-x-x-x-x-2022-find-x-

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 + \dots}}}

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 + \dots_{(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 + \dots + \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 + \dots}}}}}

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} + \dots + \sqrt{x_{n}}}}} = 2022

a n d \sqrt{x_{\infty} + \sqrt{x_{\infty} + \sqrt{x_{\infty} + \dots}}} = 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 + \dots + \sqrt{x}}}}} = k \in N \land k ⩾ 1

t h e n [x] = (k - 1) k

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