Question-89852

Question Number 89852 by M±th+et£s last updated on 19/Apr/20

Answered by mr W last updated on 19/Apr/20

F_n =(1/( (√5)))(ϕ^n −ψ^n ) with ϕ=((1+(√5))/2), ψ=((1−(√5))/2)=1−ϕ=−(1/ϕ) x^n F_n =(1/( (√5)))[(xϕ)^n −(xψ)^n ] Σ_(n=1) ^∞ x^n F_n =(1/( (√5)))[Σ_(n=1) ^∞ (xϕ)^n −Σ_(n=1) ^∞ (xψ)^n ] =(1/( (√5)))[((xϕ)/(1−xϕ))−((xψ)/(1−xψ))] =(x/( (√5)))[(ϕ/(1−xϕ))+(1/(ϕ+x))]=1 ⇒(ϕ/((1/x)−ϕ))+(1/((ϕ/x)+1))=(√5) let t=(1/x) ⇒(ϕ/(t−ϕ))+(1/(ϕt+1))=(√5) t^2 +(((√5)(1−ϕ^2 )−1−ϕ^2 )/( (√5)ϕ))t−1=0 t^2 −((((√5)+1)ϕ+2)/( (√5)ϕ))t−1=0 t^2 −2t−1=0 ⇒t=1+(√2) (x>0) ⇒x=(1/t)=(1/(1+(√2)))=(√2)−1

$${F}_{{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\varphi^{{n}} −\psi^{{n}} \right)\:{with} \\ $$$$\varphi=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}},\:\psi=\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}=\mathrm{1}−\varphi=−\frac{\mathrm{1}}{\varphi} \\ $$$${x}^{{n}} {F}_{{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left[\left({x}\varphi\right)^{{n}} −\left({x}\psi\right)^{{n}} \right] \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{{n}} {F}_{{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left[\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({x}\varphi\right)^{{n}} −\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({x}\psi\right)^{{n}} \right] \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left[\frac{{x}\varphi}{\mathrm{1}−{x}\varphi}−\frac{{x}\psi}{\mathrm{1}−{x}\psi}\right] \\ $$$$=\frac{{x}}{\:\sqrt{\mathrm{5}}}\left[\frac{\varphi}{\mathrm{1}−{x}\varphi}+\frac{\mathrm{1}}{\varphi+{x}}\right]=\mathrm{1} \\ $$$$\Rightarrow\frac{\varphi}{\frac{\mathrm{1}}{{x}}−\varphi}+\frac{\mathrm{1}}{\frac{\varphi}{{x}}+\mathrm{1}}=\sqrt{\mathrm{5}} \\ $$$${let}\:{t}=\frac{\mathrm{1}}{{x}} \\ $$$$\Rightarrow\frac{\varphi}{{t}−\varphi}+\frac{\mathrm{1}}{\varphi{t}+\mathrm{1}}=\sqrt{\mathrm{5}} \\ $$$${t}^{\mathrm{2}} +\frac{\sqrt{\mathrm{5}}\left(\mathrm{1}−\varphi^{\mathrm{2}} \right)−\mathrm{1}−\varphi^{\mathrm{2}} }{\:\sqrt{\mathrm{5}}\varphi}{t}−\mathrm{1}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} −\frac{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)\varphi+\mathrm{2}}{\:\sqrt{\mathrm{5}}\varphi}{t}−\mathrm{1}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{1}+\sqrt{\mathrm{2}}\:\:\:\:\:\:\left({x}>\mathrm{0}\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{{t}}=\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{2}}}=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$

Commented by M±th+et£s last updated on 19/Apr/20