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Question Number 32996 by abdo imad last updated on 09/Apr/18

find the sequence (v_n ) wich verify v_(n+2) =(√(v_n .v_(n+1) )) .

$${find}\:{the}\:{sequence}\:\left({v}_{{n}} \right)\:{wich}\:{verify}\:\:{v}_{{n}+\mathrm{2}} \:=\sqrt{{v}_{{n}} \:.{v}_{{n}+\mathrm{1}} }\:. \\ $$

Commented by prof Abdo imad last updated on 11/Apr/18

v_n must be positif and ln(v_(n+2) )=(1/2)ln(v_n )+(1/2)ln(v_(n+1) ) ley put u_n = ln(v_n ) ⇒u_(n+2) = (1/2) u_n +(1/(2 ))u_(n+1) ⇒ 2u_(n+2) −u_(n+1) −u_n =0 carscteristic equation 2x^2 −x −1 =0 Δ=1−4(2)(−1)= 9 x_1 =((1+3)/4) =1 and x_2 =((1−3)/4) =−(1/2) ⇒u_n = α +β(−(1/2))^n u_0 = α +β and u_1 =α −(β/2) ⇒u_0 −u_1 =(3/2)β ⇒ β =(2/3)(u_0 −u_1 ) α =u_0 −β=u_0 −((2u_0 −2u_1 )/3) = ((u_(0 ) +2u_1 )/3) ⇒ u_n = ((u_0 +2u_1 )/3) + ((2u_0 −2u_1 )/3) (−(1/2))^n v_n =e^u_n .

$${v}_{{n}} \:{must}\:{be}\:{positif}\:{and}\:{ln}\left({v}_{{n}+\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({v}_{{n}} \right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({v}_{{n}+\mathrm{1}} \right) \\ $$$${ley}\:{put}\:{u}_{{n}} =\:{ln}\left({v}_{{n}} \right)\:\Rightarrow{u}_{{n}+\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\:{u}_{{n}} \:+\frac{\mathrm{1}}{\mathrm{2}\:}{u}_{{n}+\mathrm{1}} \:\Rightarrow \\ $$$$\mathrm{2}{u}_{{n}+\mathrm{2}} \:−{u}_{{n}+\mathrm{1}} \:−{u}_{{n}} =\mathrm{0}\:\:\:{carscteristic}\:{equation} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} \:−{x}\:−\mathrm{1}\:=\mathrm{0}\:\:\Delta=\mathrm{1}−\mathrm{4}\left(\mathrm{2}\right)\left(−\mathrm{1}\right)=\:\mathrm{9} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{1}+\mathrm{3}}{\mathrm{4}}\:=\mathrm{1}\:{and}\:{x}_{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{3}}{\mathrm{4}}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{u}_{{n}} =\:\alpha\:+\beta\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$${u}_{\mathrm{0}} =\:\alpha\:+\beta\:\:{and}\:{u}_{\mathrm{1}} \:=\alpha\:−\frac{\beta}{\mathrm{2}}\:\Rightarrow{u}_{\mathrm{0}} \:−{u}_{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{2}}\beta\:\Rightarrow \\ $$$$\beta\:=\frac{\mathrm{2}}{\mathrm{3}}\left({u}_{\mathrm{0}} \:−{u}_{\mathrm{1}} \right) \\ $$$$\alpha\:={u}_{\mathrm{0}} \:−\beta={u}_{\mathrm{0}} \:\:−\frac{\mathrm{2}{u}_{\mathrm{0}} \:−\mathrm{2}{u}_{\mathrm{1}} }{\mathrm{3}}\:=\:\frac{{u}_{\mathrm{0}\:} \:\:+\mathrm{2}{u}_{\mathrm{1}} }{\mathrm{3}}\:\Rightarrow \\ $$$${u}_{{n}} =\:\frac{{u}_{\mathrm{0}} \:\:+\mathrm{2}{u}_{\mathrm{1}} }{\mathrm{3}}\:\:+\:\frac{\mathrm{2}{u}_{\mathrm{0}} \:−\mathrm{2}{u}_{\mathrm{1}} }{\mathrm{3}}\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$${v}_{{n}} ={e}^{{u}_{{n}} } \:\:\:\:. \\ $$

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