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

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

$${if}\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}}}=\mathrm{2022} \\ $$$${find}\:\left[{x}\right]=? \\ $$

Answered by TheHoneyCat last updated on 30/Sep/22

if x exists, squaring the equlity implies that: x+2022=(2022)^2 ie x=2022×2021 Now we just need to proove that x exists ie that their exists a value x such that denoting f_x (y):=(√(x+y)) (f_x ^( ○n) (0))_(n∈N) converges to 2022 it turns out that if x>4 then (√x)<x/2 and (√x)>2 so x>2 and y>2 ⇒ (√(x+y))<(x+y)/2 and still (√(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) =((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_∞ its limit x+f_∞ =f_∞ ^( 2) which admits solutions for any values. including f_∞ =2022 _■ Note that I have assumed that (√(x+(√(x+(√(x+(√(x+x)))))))) meant (√(x+(√(x+(√(x+...)))))) if you actually did meant your question to be about a finite amount of square roots, it is equivalent to solving an equation of degree 8. Wich (in general is not solvable) The polynomial is not symetric (so you cannot reduce it to deg=4) the coeficients are either divisible by 2022 (that gives the hope of looking for integer sokutions throught congruences) or ±1 (so I guess congruences are not an option). Your polynomial can be constructed recursively by composition (it′s like the inverse of my “f^( ○n) ”) so I′ve tried solving the smallers n values in the search of a patern to use recursively. But found none. I fear that the precise question you have asked has no explicit answere. However, I might be wrong on that. If someone knows I′m wrong, could you please comment the answere please? Thanks.

$$\mathrm{if}\:{x}\:\mathrm{exists},\:\mathrm{squaring}\:\mathrm{the}\:\mathrm{equlity}\:\mathrm{implies}\:\mathrm{that}: \\ $$$${x}+\mathrm{2022}=\left(\mathrm{2022}\right)^{\mathrm{2}} \\ $$$${ie}\:{x}=\mathrm{2022}×\mathrm{2021} \\ $$$$ \\ $$$$\mathrm{Now}\:\mathrm{we}\:\mathrm{just}\:\mathrm{need}\:\mathrm{to}\:\mathrm{proove}\:\mathrm{that}\:{x}\:\mathrm{exists} \\ $$$${ie}\:\mathrm{that}\:\mathrm{their}\:\mathrm{exists}\:\mathrm{a}\:\mathrm{value}\:{x}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{denoting}\:{f}_{{x}} \left({y}\right):=\sqrt{{x}+{y}} \\ $$$$\left({f}_{{x}} ^{\:\circ{n}} \left(\mathrm{0}\right)\right)_{{n}\in\mathbb{N}} \:\mathrm{converges}\:\mathrm{to}\:\mathrm{2022} \\ $$$$ \\ $$$$\mathrm{it}\:\mathrm{turns}\:\mathrm{out}\:\mathrm{that}\:\mathrm{if}\:{x}>\mathrm{4}\: \\ $$$$\mathrm{then}\:\sqrt{{x}}<{x}/\mathrm{2}\:\mathrm{and}\:\sqrt{{x}}>\mathrm{2} \\ $$$$\mathrm{so}\:{x}>\mathrm{2}\:\mathrm{and}\:{y}>\mathrm{2}\:\Rightarrow\:\sqrt{{x}+{y}}<\left({x}+{y}\right)/\mathrm{2} \\ $$$$\mathrm{and}\:\mathrm{still}\:\sqrt{{x}+{y}}\:>\mathrm{2} \\ $$$$ \\ $$$$\mathrm{so}\:\mathrm{if}\:{x}>\mathrm{4} \\ $$$$\left(\mathrm{so}\:\mathrm{that}\:\mathrm{the}\:\mathrm{second}\:\mathrm{term}\:\mathrm{is}\:\mathrm{superior}\:\mathrm{to}\:\mathrm{2}\right) \\ $$$$\mathrm{the}\:\mathrm{sequence}\:\mathrm{is}\:\mathrm{always}\:\mathrm{over}\:\mathrm{2}\:\mathrm{and}\:\mathrm{always} \\ $$$$\mathrm{under}\:{U}_{{n}+\mathrm{1}} =\frac{{x}+{U}_{{n}} }{\mathrm{2}}\:\left(\mathrm{with}\:{U}_{\mathrm{1}} ={x}\right) \\ $$$$ \\ $$$$\mathrm{basically}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{is}\:\mathrm{bounded} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{trivial}\:\mathrm{to}\:\mathrm{see}\:\mathrm{that}\:\mathrm{it}\:\mathrm{is}\:\mathrm{monotonous},\:\mathrm{so} \\ $$$$\mathrm{it}\:\mathrm{converges}. \\ $$$$\mathrm{Denoting}\:\mathrm{by}\:{f}_{\infty} \:\mathrm{its}\:\mathrm{limit} \\ $$$${x}+{f}_{\infty} ={f}_{\infty} ^{\:\mathrm{2}} \\ $$$$\mathrm{which}\:\mathrm{admits}\:\mathrm{solutions}\:\mathrm{for}\:\mathrm{any}\:\mathrm{values}. \\ $$$$\mathrm{including}\:{f}_{\infty} =\mathrm{2022}\:_{\blacksquare} \\ $$$$ \\ $$$$ \\ $$$${Note}\:{that}\:{I}\:{have}\:{assumed}\:{that} \\ $$$$\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+{x}}}}}\:{meant}\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+...}}} \\ $$$$ \\ $$$${if}\:{you}\:{actually}\:{did}\:{meant}\:{your}\:{question} \\ $$$${to}\:{be}\:{about}\:{a}\:{finite}\:{amount}\:{of}\: \\ $$$${square}\:{roots},\:{it}\:{is}\:{equivalent}\:{to}\:{solving} \\ $$$${an}\:{equation}\:{of}\:{degree}\:\mathrm{8}. \\ $$$${Wich}\:\left({in}\:{general}\:{is}\:{not}\:{solvable}\right) \\ $$$${The}\:{polynomial}\:{is}\:{not}\:{symetric}\:\left({so}\:{you}\right. \\ $$$$\left.{cannot}\:{reduce}\:{it}\:{to}\:{deg}=\mathrm{4}\right) \\ $$$${the}\:{coeficients}\:{are}\:{either}\:{divisible}\:{by}\:\mathrm{2022} \\ $$$$\left({that}\:{gives}\:{the}\:{hope}\:{of}\:{looking}\:{for}\:{integer}\right. \\ $$$$\left.{sokutions}\:{throught}\:{congruences}\right)\:{or}\:\pm\mathrm{1}\: \\ $$$$\left({so}\:{I}\:{guess}\:{congruences}\:{are}\:{not}\:{an}\:{option}\right). \\ $$$${Your}\:{polynomial}\:{can}\:{be}\:{constructed} \\ $$$${recursively}\:{by}\:{composition}\:\left({it}'{s}\:{like}\:{the}\right. \\ $$$$\left.{inverse}\:{of}\:{my}\:``{f}^{\:\circ{n}} ''\right)\:{so}\:{I}'{ve}\:{tried}\:{solving} \\ $$$${the}\:{smallers}\:{n}\:{values}\:{in}\:{the}\:{search}\:{of}\:{a}\: \\ $$$${patern}\:{to}\:{use}\:{recursively}.\:{But}\:{found}\:{none}. \\ $$$$ \\ $$$${I}\:{fear}\:{that}\:{the}\:{precise}\:{question}\:{you}\:{have}\: \\ $$$${asked}\:{has}\:{no}\:{explicit}\:{answere}. \\ $$$${However},\:{I}\:{might}\:{be}\:{wrong}\:{on}\:{that}. \\ $$$${If}\:{someone}\:{knows}\:{I}'{m}\:{wrong},\:{could}\:{you} \\ $$$${please}\:{comment}\:{the}\:{answere}\:{please}? \\ $$$${Thanks}. \\ $$

Commented by mr W last updated on 01/Oct/22

thanks sir! the LHS is an expression with exactly 5 square roots, not infinite ones. btw, the question is not to find x, but to find [x]. as you said, it′s not solvable for x, but maybe solvable for [x], the integer part of x.

$${thanks}\:{sir}! \\ $$$${the}\:{LHS}\:{is}\:{an}\:{expression}\:{with}\:{exactly} \\ $$$$\mathrm{5}\:{square}\:{roots},\:{not}\:{infinite}\:{ones}. \\ $$$${btw},\:{the}\:{question}\:{is}\:{not}\:{to}\:{find}\:{x},\:{but} \\ $$$${to}\:{find}\:\left[{x}\right].\: \\ $$$${as}\:{you}\:{said},\:{it}'{s}\:{not}\:{solvable}\:{for}\:{x},\: \\ $$$${but}\:{maybe}\:{solvable}\:{for}\:\left[{x}\right],\:{the}\:{integer} \\ $$$${part}\:{of}\:{x}. \\ $$

Commented by Tawa11 last updated on 02/Oct/22

Great sirs

$$\mathrm{Great}\:\mathrm{sirs} \\ $$

Answered by Frix last updated on 30/Sep/22

(√(x+(√(x+... _((n times)) ))))=y ⇒ x=y(y−1)+q_n with q_1 =y and q_∞ =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: (√(x+(√x)))=5 ⇒ x≈4(4+1)+.475 (√(x+(√(x+(√x)))))=5 ⇒ x≈4(4+1)+.0477 (√(x+(√x)))=50 ⇒ x≈49(49+1)+.4975 (√(x+(√(x+(√x)))))=50 ⇒ x≈49(49+1)+.004975 we see that with higher y q_2 →.5 and q_3 →0 (√(x+(√x)))=2022 ⇒ q_2 ≈499938 q_3 ≈.000046 ⇒ ⌊x⌋=2021(2021+1)=4086462

$$\sqrt{{x}+\sqrt{{x}+...\:_{\left({n}\:\mathrm{times}\right)} }}={y}\:\Rightarrow\:{x}={y}\left({y}−\mathrm{1}\right)+{q}_{{n}} \\ $$$$\mathrm{with}\:{q}_{\mathrm{1}} ={y}\:\mathrm{and}\:{q}_{\infty} =\mathrm{0} \\ $$$$\mathrm{this}\:\mathrm{means}\:{x}\:\mathrm{starts}\:\mathrm{at}\:{y}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{never}\:\mathrm{gets}\:\mathrm{lower} \\ $$$$\mathrm{than}\:\left({y}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)={y}\left({y}−\mathrm{1}\right) \\ $$$$\mathrm{looking}\:\mathrm{at}\:\mathrm{the}\:\mathrm{first}\:\mathrm{equations}\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve} \\ $$$$\mathrm{for}\:\mathrm{given}\:{y}: \\ $$$$\sqrt{{x}+\sqrt{{x}}}=\mathrm{5}\:\Rightarrow\:{x}\approx\mathrm{4}\left(\mathrm{4}+\mathrm{1}\right)+.\mathrm{475} \\ $$$$\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}=\mathrm{5}\:\Rightarrow\:{x}\approx\mathrm{4}\left(\mathrm{4}+\mathrm{1}\right)+.\mathrm{0477} \\ $$$$ \\ $$$$\sqrt{{x}+\sqrt{{x}}}=\mathrm{50}\:\Rightarrow\:{x}\approx\mathrm{49}\left(\mathrm{49}+\mathrm{1}\right)+.\mathrm{4975} \\ $$$$\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}=\mathrm{50}\:\Rightarrow\:{x}\approx\mathrm{49}\left(\mathrm{49}+\mathrm{1}\right)+.\mathrm{004975} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{see}\:\mathrm{that}\:\mathrm{with}\:\mathrm{higher}\:{y}\:{q}_{\mathrm{2}} \rightarrow.\mathrm{5}\:\mathrm{and}\:{q}_{\mathrm{3}} \rightarrow\mathrm{0} \\ $$$$\sqrt{{x}+\sqrt{{x}}}=\mathrm{2022}\:\Rightarrow\:{q}_{\mathrm{2}} \approx\mathrm{499938}\:{q}_{\mathrm{3}} \approx.\mathrm{000046} \\ $$$$\Rightarrow\:\lfloor{x}\rfloor=\mathrm{2021}\left(\mathrm{2021}+\mathrm{1}\right)=\mathrm{4086462} \\ $$

Commented by mr W last updated on 30/Sep/22

thanks sir for the right answer!

$${thanks}\:{sir}\:{for}\:{the}\:{right}\:{answer}! \\ $$

Answered by mr W last updated on 01/Oct/22

we have (√(x+(√x)))_(2 times roots) ≤(√(x+(√(x+(√(x+...+(√x)))))))_(n times roots) <(√(x+(√(x+(√(x+(√(x+...))))))))_(infinite times roots) with n∈N and n≥2 if (√(x_2 +(√x_2 )))=2022 and (√(x_n +(√(x_n +(√(x_n +...+(√x_n )))))))=2022 and (√(x_∞ +(√(x_∞ +(√(x_∞ +...))))))=2022 then we have x_∞ <x_n ≤x_2 (√(x_2 +(√x_2 )))=2022 x_2 +(√x_2 )=2022^2 x_2 =(2022^2 −x_2 )^2 x_2 =2022^4 −2×2022^2 x_2 +x_2 ^2 x_2 ^2 −(2×2022^2 +1)x_2 +2022^4 =0 x_2 =((2×2022^2 +1−(√((2×2022^2 +1)^2 −4×2022^4 )))/2) =((2×2022^2 +1−(√(4×2022^2 +1)))/2) <((2×2022^2 +1−(√(4×2022^2 )))/2) =((2×2022^2 +1−2×2022)/2) =2022^2 −2022+(1/2) =2021×2022+(1/2) (√(x_∞ +2022))=2022 ⇒x_∞ =2022^2 −2022=2021×2022 from x_∞ <x_n ≤x_2 we get 2021×2022<x_n <2021×2022+(1/2) that means [x_n ]=2021×2022=4086462 generally if (√(x+(√(x+(√(x+...+(√x)))))))_(2 or more, but finite roots) =k ∈ N ∧ k≥1 then [x]=(k−1)k

$${we}\:{have} \\ $$$$\:\underset{\mathrm{2}\:{times}\:{roots}} {\sqrt{{x}+\sqrt{{x}}}}\leqslant\underset{{n}\:{times}\:{roots}} {\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+...+\sqrt{{x}}}}}}<\underset{{infinite}\:{times}\:{roots}} {\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+...}}}}} \\ $$$${with}\:{n}\in{N}\:{and}\:{n}\geqslant\mathrm{2} \\ $$$$ \\ $$$${if}\:\sqrt{{x}_{\mathrm{2}} +\sqrt{{x}_{\mathrm{2}} }}=\mathrm{2022} \\ $$$${and}\:\sqrt{{x}_{{n}} +\sqrt{{x}_{{n}} +\sqrt{{x}_{{n}} +...+\sqrt{{x}_{{n}} }}}}=\mathrm{2022} \\ $$$${and}\:\sqrt{{x}_{\infty} +\sqrt{{x}_{\infty} +\sqrt{{x}_{\infty} +...}}}=\mathrm{2022} \\ $$$${then}\:{we}\:{have} \\ $$$${x}_{\infty} <{x}_{{n}} \leqslant{x}_{\mathrm{2}} \\ $$$$ \\ $$$$\sqrt{{x}_{\mathrm{2}} +\sqrt{{x}_{\mathrm{2}} }}=\mathrm{2022} \\ $$$${x}_{\mathrm{2}} +\sqrt{{x}_{\mathrm{2}} }=\mathrm{2022}^{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} =\left(\mathrm{2022}^{\mathrm{2}} −{x}_{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} =\mathrm{2022}^{\mathrm{4}} −\mathrm{2}×\mathrm{2022}^{\mathrm{2}} {x}_{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} ^{\mathrm{2}} −\left(\mathrm{2}×\mathrm{2022}^{\mathrm{2}} +\mathrm{1}\right){x}_{\mathrm{2}} +\mathrm{2022}^{\mathrm{4}} =\mathrm{0} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{2}×\mathrm{2022}^{\mathrm{2}} +\mathrm{1}−\sqrt{\left(\mathrm{2}×\mathrm{2022}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{2022}^{\mathrm{4}} }}{\mathrm{2}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{2}×\mathrm{2022}^{\mathrm{2}} +\mathrm{1}−\sqrt{\mathrm{4}×\mathrm{2022}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}} \\ $$$$\:\:\:\:<\frac{\mathrm{2}×\mathrm{2022}^{\mathrm{2}} +\mathrm{1}−\sqrt{\mathrm{4}×\mathrm{2022}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{2}×\mathrm{2022}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}×\mathrm{2022}}{\mathrm{2}} \\ $$$$\:\:\:\:\:=\mathrm{2022}^{\mathrm{2}} −\mathrm{2022}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:=\mathrm{2021}×\mathrm{2022}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$\sqrt{{x}_{\infty} +\mathrm{2022}}=\mathrm{2022}\: \\ $$$$\Rightarrow{x}_{\infty} =\mathrm{2022}^{\mathrm{2}} −\mathrm{2022}=\mathrm{2021}×\mathrm{2022} \\ $$$$ \\ $$$${from}\:{x}_{\infty} <{x}_{{n}} \leqslant{x}_{\mathrm{2}} \:{we}\:{get} \\ $$$$\mathrm{2021}×\mathrm{2022}<{x}_{{n}} <\mathrm{2021}×\mathrm{2022}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${that}\:{means}\: \\ $$$$\left[{x}_{{n}} \right]=\mathrm{2021}×\mathrm{2022}=\mathrm{4086462} \\ $$$$ \\ $$$${generally} \\ $$$${if}\:\underset{\mathrm{2}\:{or}\:{more},\:{but}\:{finite}\:{roots}} {\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+...+\sqrt{{x}}}}}}={k}\:\in\:{N}\:\wedge\:{k}\geqslant\mathrm{1} \\ $$$${then}\:\left[{x}\right]=\left({k}−\mathrm{1}\right){k} \\ $$

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