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Question Number 148645 by metamorfose last updated on 29/Jul/21

u_(n+3) =((u_(n+2) +u_(n+1) +u_n )/3) , ∀n∈IN find u_n

$${u}_{{n}+\mathrm{3}} =\frac{{u}_{{n}+\mathrm{2}} +{u}_{{n}+\mathrm{1}} +{u}_{{n}} }{\mathrm{3}}\:,\:\forall{n}\in{IN} \\ $$$${find}\:{u}_{{n}} \: \\ $$

Answered by Olaf_Thorendsen last updated on 29/Jul/21

u_(n+3) = ((u_(n+2) +u_(n+1) +u_3 )/3) 3u_(n+3) −u_(n+2) −u_(n+1) −u_n = 0 We solve 3r^3 −r^2 −r−1 = 0 3(r−1)(r^2 +(2/3)r+(1/3)) = 0 3(r−1)(r−r_1 )(r−r_2 ) = 0 r_(1,2) = −((1±i(√2))/3) u_n = λ.1^n +μr_1 ^n +γr_2 ^n u_n = λ+μr_1 ^n +γr_2 ^n We find λ, μ, γ with u_0 , u_1 , u_2 . In particular for μ = γ = 0, u_n = λ All constant sequences are solutions.

$${u}_{{n}+\mathrm{3}} \:=\:\frac{{u}_{{n}+\mathrm{2}} +{u}_{{n}+\mathrm{1}} +{u}_{\mathrm{3}} }{\mathrm{3}} \\ $$$$\mathrm{3}{u}_{{n}+\mathrm{3}} −{u}_{{n}+\mathrm{2}} −{u}_{{n}+\mathrm{1}} −{u}_{{n}} \:=\:\mathrm{0} \\ $$$$\mathrm{We}\:\mathrm{solve}\:\mathrm{3}{r}^{\mathrm{3}} −{r}^{\mathrm{2}} −{r}−\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{3}\left({r}−\mathrm{1}\right)\left({r}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{3}}{r}+\frac{\mathrm{1}}{\mathrm{3}}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{3}\left({r}−\mathrm{1}\right)\left({r}−{r}_{\mathrm{1}} \right)\left({r}−{r}_{\mathrm{2}} \right)\:=\:\mathrm{0} \\ $$$${r}_{\mathrm{1},\mathrm{2}} \:=\:−\frac{\mathrm{1}\pm{i}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$${u}_{{n}} \:=\:\lambda.\mathrm{1}^{{n}} +\mu{r}_{\mathrm{1}} ^{{n}} +\gamma{r}_{\mathrm{2}} ^{{n}} \\ $$$${u}_{{n}} \:=\:\lambda+\mu{r}_{\mathrm{1}} ^{{n}} +\gamma{r}_{\mathrm{2}} ^{{n}} \\ $$$$\mathrm{We}\:\mathrm{find}\:\lambda,\:\mu,\:\gamma\:\mathrm{with}\:{u}_{\mathrm{0}} ,\:{u}_{\mathrm{1}} ,\:{u}_{\mathrm{2}} . \\ $$$$\mathrm{In}\:\mathrm{particular}\:\mathrm{for}\:\mu\:=\:\gamma\:=\:\mathrm{0},\:{u}_{{n}} \:=\:\lambda \\ $$$$\mathrm{All}\:\mathrm{constant}\:\mathrm{sequences}\:\mathrm{are}\:\mathrm{solutions}. \\ $$$$ \\ $$

Commented by Olaf_Thorendsen last updated on 29/Jul/21

r_1 ^n = (−((1+i(√2))/3))^n r_1 ^n = (((−1)^n )/3^n )3^n e^(inarctan(√2)) r_0 ^n = (−1)^n e^(inarctan(√2)) idem with r_2 ^n So we can write the expression with (−1)^n cos(narctan(√2)) but we cannot simplify a lot.

$${r}_{\mathrm{1}} ^{{n}} \:=\:\left(−\frac{\mathrm{1}+{i}\sqrt{\mathrm{2}}}{\mathrm{3}}\right)^{{n}} \\ $$$${r}_{\mathrm{1}} ^{{n}} \:=\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}^{{n}} }\mathrm{3}^{{n}} {e}^{{in}\mathrm{arctan}\sqrt{\mathrm{2}}} \\ $$$${r}_{\mathrm{0}} ^{{n}} \:=\:\left(−\mathrm{1}\right)^{{n}} {e}^{{in}\mathrm{arctan}\sqrt{\mathrm{2}}} \\ $$$$\mathrm{idem}\:\mathrm{with}\:{r}_{\mathrm{2}} ^{{n}} \\ $$$$\mathrm{So}\:\mathrm{we}\:\mathrm{can}\:\mathrm{write}\:\mathrm{the}\:\mathrm{expression}\:\mathrm{with} \\ $$$$\:\left(−\mathrm{1}\right)^{{n}} \mathrm{cos}\left({n}\mathrm{arctan}\sqrt{\mathrm{2}}\right) \\ $$$$\mathrm{but}\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{simplify}\:\mathrm{a}\:\mathrm{lot}. \\ $$

Commented by metamorfose last updated on 29/Jul/21

thank u sir.

$${thank}\:{u}\:{sir}. \\ $$

Commented by metamorfose last updated on 29/Jul/21

i thought the expression would be with cos

$${i}\:{thought}\:{the}\:{expression}\:{would}\:{be}\:{with}\:{cos} \\ $$

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