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Question Number 227393 by fantastic2 last updated on 20/Jan/26
x^2 +x+1=1+(1/(1+(1/(1+(1/(1+(1/(...)))))))) x=?
$${x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{…}}}} \\ $$$${x}=? \\ $$
Answered by Ghisom_ last updated on 20/Jan/26
x^2 +x+1=y y=1+(1/y) x^2 +x+1=1+(1/(x^2 +x+1)) x^4 +2x^3 +2x^2 +x−1=0 (x+(1/2))^4 +(1/2)(x+(1/2))^2 −((19)/(16))=0 (x+(1/2))^2 =−(1/4)±((√5)/2) x=−(1/2)±((√(−1+2(√5)))/2)∨x=−(1/2)±((√(1+2(√5)))/2)i
$${x}^{\mathrm{2}} +{x}+\mathrm{1}={y} \\ $$$${y}=\mathrm{1}+\frac{\mathrm{1}}{{y}} \\ $$$${x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}} \\ $$$${x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{1}=\mathrm{0} \\ $$$$\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{19}}{\mathrm{16}}=\mathrm{0} \\ $$$$\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{4}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}\vee{x}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}\mathrm{i} \\ $$
Commented by fantastic2 last updated on 22/Jan/26
thanks
$${thanks} \\ $$
Answered by Kassista last updated on 22/Jan/26
RHS = 1+(1/(1+(1/(1+(1/(1+(1/(...)))))))) = y ⇒ 1+(1/y) = y, y^2 −y−1=0 ∴ y = ((1±(√(1−4(1)(−1))))/2) = ((1±(√5))/2) ⇒^(y>0) y=ϕ ⇒x^2 +x+1=y, x^2 +x+(1−ϕ)=0 ∴ x=((−1±(√(1−4(1)(1−ϕ))))/2) x = ((−1±(√(4ϕ−3)))/2)
$$ \\ $$$${RHS}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{…}}}}\:=\:{y}\:\Rightarrow\:\mathrm{1}+\frac{\mathrm{1}}{{y}}\:=\:{y},\:{y}^{\mathrm{2}} −{y}−\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$$$\therefore\:{y}\:=\:\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}\left(\mathrm{1}\right)\left(−\mathrm{1}\right)}}{\mathrm{2}}\:=\:\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\:\overset{{y}>\mathrm{0}} {\Rightarrow}\:{y}=\varphi \\ $$$$ \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{x}+\mathrm{1}={y},\:{x}^{\mathrm{2}} +{x}+\left(\mathrm{1}−\varphi\right)=\mathrm{0}\:\therefore\:{x}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{1}−\varphi\right)}}{\mathrm{2}} \\ $$$${x}\:=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{4}\varphi−\mathrm{3}}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by fantastic2 last updated on 22/Jan/26
great
$${great} \\ $$
Commented by Ghisom_ last updated on 22/Jan/26
can you prove that y=1+(1/(1+(1/(1+...))))>0 btw how to decide the value of z: z=1−(1/(1−(1/(1−...))))
$$\mathrm{can}\:\mathrm{you}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$${y}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+…}}>\mathrm{0} \\ $$$$ \\ $$$$\mathrm{btw}\:\mathrm{how}\:\mathrm{to}\:\mathrm{decide}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{z}: \\ $$$${z}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−…}} \\ $$
Commented by Kassista last updated on 22/Jan/26
first, let (y_n ) be a sequence s.t. y_(n+1) =1+(1/y_n ) and y_1 =1 State that: y_n >0, ∀n∈N By induction, we know y_1 =1 >0✓ and suppose y_k >0 therefore: y_(k+1) =1+(1/y_k ) >^((1/y_k )>0) 1> 0✓ ∴ y_n >0, ∀n∈N y = lim_(n→∞) (y_n )>0 _■ About the value of z: Suppose that z is a finite real number then, z must satisfy z=1−(1/(z )) ⇒z^2 −z+1=0 However, Discriminant=1−4(1)(1)=−3<0 ∴ z∉R →←
$$ \\ $$$${first},\:{let}\:\left({y}_{{n}} \right)\:{be}\:{a}\:{sequence}\:{s}.{t}.\:{y}_{{n}+\mathrm{1}} =\mathrm{1}+\frac{\mathrm{1}}{{y}_{{n}} }\:{and}\:{y}_{\mathrm{1}} =\mathrm{1} \\ $$$${State}\:{that}:\:{y}_{{n}} >\mathrm{0},\:\forall{n}\in\mathbb{N} \\ $$$${By}\:{induction},\:{we}\:{know}\:\:{y}_{\mathrm{1}} =\mathrm{1}\:>\mathrm{0}\checkmark\:{and}\:{suppose}\:{y}_{{k}} >\mathrm{0} \\ $$$${therefore}:\:{y}_{{k}+\mathrm{1}} =\mathrm{1}+\frac{\mathrm{1}}{{y}_{{k}} }\:\overset{\frac{\mathrm{1}}{{y}_{{k}} }>\mathrm{0}} {>}\mathrm{1}>\:\mathrm{0}\checkmark\:\therefore\:{y}_{{n}} >\mathrm{0},\:\forall{n}\in\mathbb{N} \\ $$$${y}\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left({y}_{{n}} \right)>\mathrm{0}\:\:\:_{\blacksquare} \\ $$$$ \\ $$$${About}\:{the}\:{value}\:{of}\:{z}:\:{Suppose}\:{that}\:{z}\:{is}\:{a}\:{finite}\:{real}\:{number} \\ $$$${then},\:{z}\:{must}\:{satisfy}\:{z}=\mathrm{1}−\frac{\mathrm{1}}{{z}\:}\:\Rightarrow{z}^{\mathrm{2}} −{z}+\mathrm{1}=\mathrm{0} \\ $$$${However},\:{Discriminant}=\mathrm{1}−\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{1}\right)=−\mathrm{3}<\mathrm{0}\: \\ $$$$\therefore\:{z}\notin\mathbb{R}\:\:\rightarrow\leftarrow \\ $$
Commented by dionigi last updated on 25/Jan/26
it could be completed that for the 2nd solution y = ((1−(√5))/2) = −(1/ϕ) there is no solution for x^2 + x + 1+ (1/ϕ) = 0 with 1−4(1)(1+(1/ϕ)) < 0
$${it}\:{could}\:{be}\:{completed}\:{that}\: \\ $$$${for}\:{the}\:\mathrm{2}{nd}\:{solution}\:{y}\:=\:\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:=\:−\frac{\mathrm{1}}{\varphi} \\ $$$${there}\:{is}\:{no}\:{solution}\:{for}\:{x}^{\mathrm{2}} +\:{x}\:+\:\mathrm{1}+\:\frac{\mathrm{1}}{\varphi}\:=\:\mathrm{0} \\ $$$${with}\:\mathrm{1}−\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\varphi}\right)\:<\:\mathrm{0} \\ $$
Commented by Ghisom_ last updated on 27/Jan/26
it′s not stated that the solutions must be real
$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{stated}\:\mathrm{that}\:\mathrm{the}\:\mathrm{solutions}\:\mathrm{must}\:\mathrm{be} \\ $$$$\mathrm{real} \\ $$

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