Question-99025

Question Number 99025 by joki last updated on 18/Jun/20

Commented by bobhans last updated on 18/Jun/20

set 2207−(1/(2207−(1/(2207−(1/(2207−...)))))) = q 2207−(1/q) = q ⇒q^2 −2207q+1=0 q= ((2207 ± (√(4870845)))/2) = ((2207 ± 2207)/2) case(1) q = 2207 ⇒((2207−(1/(2207−(1/(2207−...))))))^(1/(8 )) = ((2207))^(1/(8 )) ≈ 2.618034056

$$\mathrm{set}\:\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{2207}−…}}}\:=\:{q} \\ $$$$\mathrm{2207}−\frac{\mathrm{1}}{{q}}\:=\:{q}\:\Rightarrow{q}^{\mathrm{2}} −\mathrm{2207}{q}+\mathrm{1}=\mathrm{0} \\ $$$${q}=\:\frac{\mathrm{2207}\:\pm\:\sqrt{\mathrm{4870845}}}{\mathrm{2}}\:=\:\frac{\mathrm{2207}\:\pm\:\mathrm{2207}}{\mathrm{2}} \\ $$$$\mathrm{case}\left(\mathrm{1}\right)\:{q}\:=\:\mathrm{2207}\:\Rightarrow\sqrt[{\mathrm{8}\:\:}]{\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{2207}−…}}}=\:\sqrt[{\mathrm{8}\:\:\:}]{\mathrm{2207}}\:\approx\:\mathrm{2}.\mathrm{618034056} \\ $$

Commented by joki last updated on 18/Jun/20

$${thanks}\:{sir} \\ $$

Commented by MJS last updated on 18/Jun/20

(√(4870845))≠2207 ⇒ answer is only an approximation

$$\sqrt{\mathrm{4870845}}\neq\mathrm{2207}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{only}\:\mathrm{an}\:\mathrm{approximation} \\ $$

Commented by bobhans last updated on 18/Jun/20

$$\mathrm{yes}\:\mathrm{sir}.\:\mathrm{my}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{approximation} \\ $$

Answered by bemath last updated on 18/Jun/20

let ((2207−(1/(2207−(1/(2207))−...))))^(1/(8 )) = w w^8 = 2207−(1/w^8 ) ⇒ w^(16) −2207w^8 + 1=0

$$\mathrm{let}\:\sqrt[{\mathrm{8}\:\:}]{\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{2207}}−…}}\:=\:\mathrm{w} \\ $$$$\mathrm{w}^{\mathrm{8}} \:=\:\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{w}^{\mathrm{8}} }\:\Rightarrow\:\mathrm{w}^{\mathrm{16}} \:−\mathrm{2207w}^{\mathrm{8}} \:+\:\mathrm{1}=\mathrm{0} \\ $$

Commented by mr W last updated on 18/Jun/20

wrong sir! should be w^8 = 2207−(1/w^8 )

$${wrong}\:{sir}! \\ $$$${should}\:{be} \\ $$$$\mathrm{w}^{\mathrm{8}} \:=\:\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{w}^{\mathrm{8}} } \\ $$

Commented by joki last updated on 18/Jun/20

$${have}\:{done}?{what}\:{is}\:{the}\:{value}\:{w}? \\ $$

Commented by Rasheed.Sindhi last updated on 18/Jun/20

Sir bemath I′ve tried to solve your q#98806 Please say something whether it′s right or wrong.

$${Sir}\:{bemath} \\ $$$$\:\:\:{I}'{ve}\:{tried}\:{to}\:{solve}\:{your}\:{q}#\mathrm{98806} \\ $$$${Please}\:{say}\:{something}\:{whether} \\ $$$${it}'{s}\:{right}\:{or}\:{wrong}. \\ $$

Commented by bemath last updated on 18/Jun/20

sorry sir. i forgot. yes sir your answer is correct

$$\mathrm{sorry}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{forgot}.\:\mathrm{yes}\:\mathrm{sir}\:\mathrm{your}\:\mathrm{answer}\: \\ $$$$\mathrm{is}\:\mathrm{correct} \\ $$

Commented by Rasheed.Sindhi last updated on 18/Jun/20

$${Thanx}\:{sir}! \\ $$

Answered by MJS last updated on 18/Jun/20

x=2207−(1/x) x^2 −2207x+1=0 x=((2207±987(√5))/2); x≈2207 ⇒ x=((2207+987(√5))/2) (x)^(1/8) =(√(√(√x))) (√((2207+987(√5))/2))=((47+21(√5))/2) (√((47+21(√5))/2))=((7+3(√5))/2) (√((7+3(√5))/2))=((3+(√5))/2)

$${x}=\mathrm{2207}−\frac{\mathrm{1}}{{x}} \\ $$$${x}^{\mathrm{2}} −\mathrm{2207}{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{2207}\pm\mathrm{987}\sqrt{\mathrm{5}}}{\mathrm{2}};\:{x}\approx\mathrm{2207}\:\Rightarrow\:{x}=\frac{\mathrm{2207}+\mathrm{987}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\sqrt[{\mathrm{8}}]{{x}}=\sqrt{\sqrt{\sqrt{{x}}}} \\ $$$$\sqrt{\frac{\mathrm{2207}+\mathrm{987}\sqrt{\mathrm{5}}}{\mathrm{2}}}=\frac{\mathrm{47}+\mathrm{21}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\sqrt{\frac{\mathrm{47}+\mathrm{21}\sqrt{\mathrm{5}}}{\mathrm{2}}}=\frac{\mathrm{7}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\sqrt{\frac{\mathrm{7}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}}=\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$ \\ $$

Commented by mr W last updated on 18/Jun/20

$${x}=\frac{\mathrm{2207}−\mathrm{987}\sqrt{\mathrm{5}}}{\mathrm{2}}\:{is}\:{also}\:{solution}? \\ $$

Commented by MJS last updated on 18/Jun/20

solution of the polynome, yes. but ((2207−987(√5))/2)≠2207−(1/(2207−(1/(2207...))))

$$\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{polynome},\:\mathrm{yes}.\:\mathrm{but} \\ $$$$\frac{\mathrm{2207}−\mathrm{987}\sqrt{\mathrm{5}}}{\mathrm{2}}\neq\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{2207}…}} \\ $$

Commented by bobhans last updated on 18/Jun/20

$$\mathrm{great}\:\mathrm{sir}… \\ $$

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