Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 106794 by bobhans last updated on 07/Aug/20

repost old question unanswer Given → { ((((4x^2 )/(1+4x^2 )) = y)),((((4y^2 )/(1+4y^2 )) = z)),((((4z^2 )/(1+4z^2 )) = x)) :}

$$\mathrm{repost}\:\mathrm{old}\:\mathrm{question}\:\mathrm{unanswer} \\ $$$$\mathcal{G}\mathrm{iven}\:\rightarrow\begin{cases}{\frac{\mathrm{4x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4x}^{\mathrm{2}} }\:=\:\mathrm{y}}\\{\frac{\mathrm{4y}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4y}^{\mathrm{2}} }\:=\:\mathrm{z}}\\{\frac{\mathrm{4z}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4z}^{\mathrm{2}} }\:=\:\mathrm{x}}\end{cases} \\ $$

Answered by thearith last updated on 07/Aug/20

$$ \\ $$

Answered by john santu last updated on 07/Aug/20

^(@JS@) { ((1+(1/(4x^2 )) = (1/y))),((1+(1/(4y^2 )) = (1/z))),((1+(1/(4z^2 )) = (1/x))) :} ⇒(1/(4x^2 ))+(1/(4y^2 ))+(1/(4z^2 ))+3 = (1/x)+(1/y)+(1/z) ⇒((1/(4x^2 ))−(1/x)+1)+((1/(4y^2 ))−(1/y)+1)+((1/(4z^2 ))−(1/z)+1)=0 (((4x^2 −4x+1)/(4x^2 )))+(((4y^2 −4y+1)/(4y^2 )))+(((4z^2 −4z+1)/(4z^2 )))=0 (((2x−1)/(2x)))^2 +(((2y−1)/(2y)))^2 +(((2z−1)/(2z)))^2 =0 { ((x = (1/2))),((y=(1/2))),((z=(1/2))) :}.

$$\:\:\:\:^{@\mathrm{JS}@} \\ $$$$\begin{cases}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4x}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{y}}}\\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4y}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{z}}}\\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4z}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{x}}}\end{cases} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4y}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4z}^{\mathrm{2}} }+\mathrm{3}\:=\:\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{y}}+\frac{\mathrm{1}}{\mathrm{z}} \\ $$$$\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{4x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{x}}+\mathrm{1}\right)+\left(\frac{\mathrm{1}}{\mathrm{4y}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{y}}+\mathrm{1}\right)+\left(\frac{\mathrm{1}}{\mathrm{4z}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{z}}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(\frac{\mathrm{4x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{1}}{\mathrm{4x}^{\mathrm{2}} }\right)+\left(\frac{\mathrm{4y}^{\mathrm{2}} −\mathrm{4y}+\mathrm{1}}{\mathrm{4y}^{\mathrm{2}} }\right)+\left(\frac{\mathrm{4z}^{\mathrm{2}} −\mathrm{4z}+\mathrm{1}}{\mathrm{4z}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\left(\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{2x}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2y}−\mathrm{1}}{\mathrm{2y}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2z}−\mathrm{1}}{\mathrm{2z}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\begin{cases}{\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{y}=\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{z}=\frac{\mathrm{1}}{\mathrm{2}}}\end{cases}. \\ $$

Commented by bemath last updated on 07/Aug/20

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Commented by bobhans last updated on 07/Aug/20

and the other solution { ((x=0)),((y=0)),((z=0)) :}

$$\mathrm{and}\:\mathrm{the}\:\mathrm{other}\:\mathrm{solution}\:\begin{cases}{\mathrm{x}=\mathrm{0}}\\{\mathrm{y}=\mathrm{0}}\\{\mathrm{z}=\mathrm{0}}\end{cases} \\ $$

Answered by 1549442205PVT last updated on 07/Aug/20

WLOG suppose that x≥y≥z≥0.Then from the hypothesis we have ((4z^2 )/(1+4z^2 ))≥((4x^2 )/(1+4x^2 )) ≥((4y^2 )/(1+4y^2 )) ⇒((4z^2 )/(1+4z^2 ))≥((4x^2 )/(1+4x^2 )) ⇔4z^2 (1+4x^2 )≥4x^2 (1+4z^2 )⇔z^2 ≥x^2 ⇒x=y=z.Replace into (1) we get 4x^2 =x(1+4x^2 )⇔4x^3 −4x^2 +x=0 ⇔x(4x^2 −4x+1)=0⇔x(2x−1)^2 =0 ⇔x∈{0;(1/2)} Hence,the given system has two solutions as {(x,y,z)∈{(0,0,0);((1/2);(1/2),(1/2))}

$$\mathrm{WLOG}\:\mathrm{suppose}\:\mathrm{that}\:\mathrm{x}\geqslant\mathrm{y}\geqslant\mathrm{z}\geqslant\mathrm{0}.\mathrm{Then} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{hypothesis}\:\mathrm{we}\:\mathrm{have} \\ $$$$\frac{\mathrm{4z}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4z}^{\mathrm{2}} }\geqslant\frac{\mathrm{4x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4x}^{\mathrm{2}} }\:\geqslant\frac{\mathrm{4y}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4y}^{\mathrm{2}} }\:\:\:\:\:\Rightarrow\frac{\mathrm{4z}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4z}^{\mathrm{2}} }\geqslant\frac{\mathrm{4x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4x}^{\mathrm{2}} }\: \\ $$$$\Leftrightarrow\mathrm{4z}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{4x}^{\mathrm{2}} \right)\geqslant\mathrm{4x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{4z}^{\mathrm{2}} \right)\Leftrightarrow\mathrm{z}^{\mathrm{2}} \geqslant\mathrm{x}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{y}=\mathrm{z}.\mathrm{Replace}\:\mathrm{into}\:\left(\mathrm{1}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{4x}^{\mathrm{2}} =\mathrm{x}\left(\mathrm{1}+\mathrm{4x}^{\mathrm{2}} \right)\Leftrightarrow\mathrm{4x}^{\mathrm{3}} −\mathrm{4x}^{\mathrm{2}} +\mathrm{x}=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{x}\left(\mathrm{4x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{1}\right)=\mathrm{0}\Leftrightarrow\mathrm{x}\left(\mathrm{2x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Leftrightarrow\boldsymbol{\mathrm{x}}\in\left\{\mathrm{0};\frac{\mathrm{1}}{\mathrm{2}}\right\}\:\mathrm{Hence},\mathrm{the}\:\mathrm{given}\:\mathrm{system} \\ $$$$\mathrm{has}\:\mathrm{two}\:\mathrm{solutions}\:\mathrm{as} \\ $$$$\left\{\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}},\boldsymbol{\mathrm{z}}\right)\in\left\{\left(\mathrm{0},\mathrm{0},\mathrm{0}\right);\left(\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)\right\}\right. \\ $$

Commented by bemath last updated on 07/Aug/20

sir what is WLOG

$$\mathrm{sir}\:\mathrm{what}\:\mathrm{is}\:\mathrm{WLOG} \\ $$

Commented by 1549442205PVT last updated on 07/Aug/20

it is the brief writting of phrase “without loss the generality”is used in the math. proof

$$\mathrm{it}\:\mathrm{is}\:\mathrm{the}\:\mathrm{brief}\:\mathrm{writting}\:\mathrm{of}\:\mathrm{phrase} \\ $$$$``\mathrm{without}\:\mathrm{loss}\:\mathrm{the}\:\mathrm{generality}''\mathrm{is}\:\mathrm{used} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{math}.\:\mathrm{proof}\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com