Question Number 82985 by M±th+et£s last updated on 26/Feb/20
Answered by ~blr237~ last updated on 26/Feb/20
![let n≥2 , b_n ≠0 cause a+b≠0 and 0<a<b we have a_n =(b_n ^2 /b_(n−1) ) and b_(n−1) = 2a_n −a_(n−1) so we got b_(n−1) =((2b_n ^2 )/(b_(n−1) )) −(b_(n−1) ^2 /b_(n−2) ) ⇔ 1=2((b_n /b_(n−1) ))^2 −((b_(n−1) /b_(n−2) )) let state c_n =(b_n /b_(n−1) ) we can prove by induction that ∀ n≥2 0<a_n ≤ b_n we have c_n =(√((1+c_(n−1) )/2)) (1) and c_n ∈[0,1] cause c_n = (b_n /b_(n−1) )=(a_n /b_n ) so there exist ∀ n≥2 θ_n ∈[0,(π/2)] such as c_n =cosθ_n Using (1) we get cosθ_n =cos((θ_(n−1) /2)) cause 0<x<(π/2) → cosx>0 such as ∀ n≥2 θ_n ∈[0,(π/2)] we get θ_n =(θ_(n−1) /2) .After θ_n =θ_2 ((1/2))^(n−2) ∀ n≥2 , for k≥2 we have (b_k /b_(k−1) )=cos( θ_2 ((1/2))^(k−2) ) b_n =b_1 Π_(k=2) ^n (cos((θ_2 /2^(k−2) ))) b_n =b_1 cos(θ_2 )Π_(k=3) ^n (((sin((θ_2 /2^(k−3) )))/(2sin((θ_2 /2^(k−2) ))))) using sin(2x)=2sinxcosx and iteration we got b_n = ((b_1 cos(θ_2 )sin(θ_2 ))/(2^(n−1) sin((θ_2 /2^(n−2) )))) for n≥2 Now lim_(n→∞) b_n = ((b_1 cos(θ_2 )sin(θ_2 ))/(2θ_2 )) cause lim_(n→∞) ((sin((θ_2 /2^(n−2) )))/(1/2^(n−2) ))=θ_2 using c_2 =cosθ_2 and c_2 =(b_2 /(b_1 )) (b_2 /b_1 )=(√((a_2 /b_1 ) )) =(√((1/2)+(a_1 /(2b_1 )) _ )) =(√((1/2)+(1/2)(√(a_1 /b)))) c_2 =(√((1/2)+(1/2)(√((1/2)+(a/(2b))))))](Q83018.png)
$$\:{let}\:{n}\geqslant\mathrm{2}\:,\:\:{b}_{{n}} \neq\mathrm{0}\:\:\:{cause}\:\:{a}+{b}\neq\mathrm{0}\:\:{and}\:\:\mathrm{0}<{a}<{b} \\ $$$$\:{we}\:{have}\:\:\:{a}_{{n}} =\frac{{b}_{{n}} ^{\mathrm{2}} }{{b}_{{n}−\mathrm{1}} }\:\:{and}\:\:{b}_{{n}−\mathrm{1}} =\:\mathrm{2}{a}_{{n}} −{a}_{{n}−\mathrm{1}} \: \\ $$$${so}\:\:{we}\:{got}\:\:\:{b}_{{n}−\mathrm{1}} =\frac{\mathrm{2}{b}_{{n}} ^{\mathrm{2}} }{{b}_{{n}−\mathrm{1}} \:}\:−\frac{{b}_{{n}−\mathrm{1}} ^{\mathrm{2}} }{{b}_{{n}−\mathrm{2}} }\:\:\Leftrightarrow\:\mathrm{1}=\mathrm{2}\left(\frac{{b}_{{n}} }{{b}_{{n}−\mathrm{1}} }\right)^{\mathrm{2}} −\left(\frac{{b}_{{n}−\mathrm{1}} }{{b}_{{n}−\mathrm{2}} }\right) \\ $$$${let}\:{state}\:\:\:{c}_{{n}} =\frac{{b}_{{n}} }{{b}_{{n}−\mathrm{1}} }\:\:\:\:\:{we}\:{can}\:{prove}\:{by}\:{induction}\:{that}\:\forall\:{n}\geqslant\mathrm{2}\:\:\:\mathrm{0}<{a}_{{n}} \leqslant\:{b}_{{n}} \:\: \\ $$$${we}\:\:{have}\:\:\:{c}_{{n}} =\sqrt{\frac{\mathrm{1}+{c}_{{n}−\mathrm{1}} }{\mathrm{2}}}\:\:\:\left(\mathrm{1}\right)\:\:\:\:{and}\:\:{c}_{{n}} \in\left[\mathrm{0},\mathrm{1}\right]\:\:{cause}\:{c}_{{n}} =\:\frac{{b}_{{n}} }{{b}_{{n}−\mathrm{1}} }=\frac{{a}_{{n}} }{{b}_{{n}} } \\ $$$${so}\:{there}\:{exist}\:\:\forall\:\:{n}\geqslant\mathrm{2}\:\:\:\:\theta_{{n}} \in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right]\:\:{such}\:{as}\:\:\:\:{c}_{{n}} ={cos}\theta_{{n}} \:\: \\ $$$${Using}\:\:\left(\mathrm{1}\right)\:\:\:\:{we}\:{get}\:\:\:{cos}\theta_{{n}} ={cos}\left(\frac{\theta_{{n}−\mathrm{1}} }{\mathrm{2}}\right)\:\:{cause}\:\:\:\mathrm{0}<{x}<\frac{\pi}{\mathrm{2}}\:\rightarrow\:{cosx}>\mathrm{0} \\ $$$${such}\:{as}\:\:\forall\:{n}\geqslant\mathrm{2}\:\:\:\theta_{{n}} \in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right]\:\:\:\:{we}\:{get}\:\:\theta_{{n}} =\frac{\theta_{{n}−\mathrm{1}} }{\mathrm{2}}\:.{After}\:\:\:\theta_{{n}} =\theta_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}−\mathrm{2}} \:\:\forall\:\:{n}\geqslant\mathrm{2} \\ $$$$,\:\:\:{for}\:\:{k}\geqslant\mathrm{2}\:\:{we}\:{have}\:\:\:\:\frac{{b}_{{k}} }{{b}_{{k}−\mathrm{1}} }={cos}\left(\:\theta_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{k}−\mathrm{2}} \right) \\ $$$${b}_{{n}} ={b}_{\mathrm{1}} \:\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\left({cos}\left(\frac{\theta_{\mathrm{2}} }{\mathrm{2}^{{k}−\mathrm{2}} }\right)\right) \\ $$$${b}_{{n}} ={b}_{\mathrm{1}} {cos}\left(\theta_{\mathrm{2}} \right)\underset{{k}=\mathrm{3}} {\overset{{n}} {\prod}}\left(\frac{{sin}\left(\frac{\theta_{\mathrm{2}} }{\mathrm{2}^{{k}−\mathrm{3}} }\right)}{\mathrm{2}{sin}\left(\frac{\theta_{\mathrm{2}} }{\mathrm{2}^{{k}−\mathrm{2}} }\right)}\right)\:\: \\ $$$${using}\:\:\:{sin}\left(\mathrm{2}{x}\right)=\mathrm{2}{sinxcosx}\:\:{and}\:\:{iteration}\:\:{we}\:{got} \\ $$$${b}_{{n}} =\:\frac{{b}_{\mathrm{1}} {cos}\left(\theta_{\mathrm{2}} \right){sin}\left(\theta_{\mathrm{2}} \right)}{\mathrm{2}^{{n}−\mathrm{1}} {sin}\left(\frac{\theta_{\mathrm{2}} }{\mathrm{2}^{{n}−\mathrm{2}} }\right)}\:\:\:\:{for}\:\:{n}\geqslant\mathrm{2}\:\: \\ $$$${Now}\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{b}_{{n}} =\:\frac{{b}_{\mathrm{1}} {cos}\left(\theta_{\mathrm{2}} \right){sin}\left(\theta_{\mathrm{2}} \right)}{\mathrm{2}\theta_{\mathrm{2}} }\:\:\:\:{cause}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{sin}\left(\frac{\theta_{\mathrm{2}} }{\mathrm{2}^{{n}−\mathrm{2}} }\right)}{\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{2}} }}=\theta_{\mathrm{2}} \: \\ $$$${using}\:\:\:\:\:{c}_{\mathrm{2}} ={cos}\theta_{\mathrm{2}} \:\:{and}\:\:\:{c}_{\mathrm{2}} =\frac{{b}_{\mathrm{2}} }{{b}_{\mathrm{1}} \:}\: \\ $$$$\frac{{b}_{\mathrm{2}} }{{b}_{\mathrm{1}} }=\sqrt{\frac{{a}_{\mathrm{2}} }{{b}_{\mathrm{1}} }\:}\:=\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{{a}_{\mathrm{1}} }{\mathrm{2}{b}_{\mathrm{1}} }\:_{} }\:=\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{{a}_{\mathrm{1}} }{{b}}}}\:\:\: \\ $$$${c}_{\mathrm{2}} =\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{{a}}{\mathrm{2}{b}}}}\:\:\: \\ $$$$ \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 27/Feb/20
$${thank}\:{you}\:{so}\:{much}\:{sir}\: \\ $$$$ \\ $$