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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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