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Question Number 3565 by Yozzii last updated on 15/Dec/15
Define the sequence {a_n } by the recurrence equation a_(n+1) =pa_n +qa_(n−1) (n≥1) where p,q∈C−{0} and a_0 =α , a_1 =β α,β∈C. Find a_n in terms of n.
$${Define}\:{the}\:{sequence}\:\left\{{a}_{{n}} \right\}\:{by}\:{the} \\ $$$${recurrence}\:{equation}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}_{{n}+\mathrm{1}} ={pa}_{{n}} +{qa}_{{n}−\mathrm{1}} \:\:\left({n}\geqslant\mathrm{1}\right) \\ $$$${where}\:{p},{q}\in\mathbb{C}−\left\{\mathrm{0}\right\}\:{and}\: \\ $$$${a}_{\mathrm{0}} =\alpha\:,\:{a}_{\mathrm{1}} =\beta\:\: \\ $$$$\alpha,\beta\in\mathbb{C}. \\ $$$${Find}\:{a}_{{n}} \:{in}\:{terms}\:{of}\:{n}.\: \\ $$$$ \\ $$$$ \\ $$
Answered by prakash jain last updated on 16/Dec/15
a_0 =α a_1 =β a_n =pa_(n−1) +qa_(n−2) x^2 −px−q=0⇒x=((p±(√(p^2 +4q)))/2) assume p^2 +4q>0 a_n =k_1 (((p+(√(p^2 +4q)))/2))^n +k_2 (((p−(√(p^2 +4q)))/2))^n k_1 +k_2 =α⇒k_2 =α−k_1 k_1 (((p+(√(p^2 +4q)))/2))+k_2 (((p−(√(p^2 +4q)))/2))=β k_1 (((p+(√(p^2 +4q)))/2))−k_1 (((p−(√(p^2 +4q)))/2))+α(((p−(√(p^2 +4q)))/2))=β k_1 (√(p^2 +4q)) =β−α(((p−(√(p^2 +4q)))/2)) k_1 =(β/( (√(p^2 +4q))))−((αp)/(2(√(p^2 +4a))))+(α/2) k_2 =(α/2)−(β/( (√(p^2 +4q))))+((αp)/(2(√(p^2 +4a)))) a_n =k_1 (((p+(√(p^2 +4q)))/2))^n +k_2 (((p−(√(p^2 +4q)))/2))^n
$${a}_{\mathrm{0}} =\alpha \\ $$$${a}_{\mathrm{1}} =\beta \\ $$$${a}_{{n}} ={pa}_{{n}−\mathrm{1}} +{qa}_{{n}−\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −{px}−{q}=\mathrm{0}\Rightarrow{x}=\frac{{p}\pm\sqrt{{p}^{\mathrm{2}} +\mathrm{4}{q}}}{\mathrm{2}} \\ $$$${assume}\:{p}^{\mathrm{2}} +\mathrm{4}{q}>\mathrm{0} \\ $$$${a}_{{n}} ={k}_{\mathrm{1}} \left(\frac{{p}+\sqrt{{p}^{\mathrm{2}} +\mathrm{4}{q}}}{\mathrm{2}}\right)^{{n}} +{k}_{\mathrm{2}} \left(\frac{{p}−\sqrt{{p}^{\mathrm{2}} +\mathrm{4}{q}}}{\mathrm{2}}\right)^{{n}} \\ $$$${k}_{\mathrm{1}} +{k}_{\mathrm{2}} =\alpha\Rightarrow{k}_{\mathrm{2}} =\alpha−{k}_{\mathrm{1}} \\ $$$${k}_{\mathrm{1}} \left(\frac{{p}+\sqrt{{p}^{\mathrm{2}} +\mathrm{4}{q}}}{\mathrm{2}}\right)+{k}_{\mathrm{2}} \left(\frac{{p}−\sqrt{{p}^{\mathrm{2}} +\mathrm{4}{q}}}{\mathrm{2}}\right)=\beta \\ $$$${k}_{\mathrm{1}} \left(\frac{{p}+\sqrt{{p}^{\mathrm{2}} +\mathrm{4}{q}}}{\mathrm{2}}\right)−{k}_{\mathrm{1}} \left(\frac{{p}−\sqrt{{p}^{\mathrm{2}} +\mathrm{4}{q}}}{\mathrm{2}}\right)+\alpha\left(\frac{{p}−\sqrt{{p}^{\mathrm{2}} +\mathrm{4}{q}}}{\mathrm{2}}\right)=\beta \\ $$$${k}_{\mathrm{1}} \sqrt{{p}^{\mathrm{2}} +\mathrm{4}{q}}\:=\beta−\alpha\left(\frac{{p}−\sqrt{{p}^{\mathrm{2}} +\mathrm{4}{q}}}{\mathrm{2}}\right) \\ $$$${k}_{\mathrm{1}} =\frac{\beta}{\:\sqrt{{p}^{\mathrm{2}} +\mathrm{4}{q}}}−\frac{\alpha{p}}{\mathrm{2}\sqrt{{p}^{\mathrm{2}} +\mathrm{4}{a}}}+\frac{\alpha}{\mathrm{2}} \\ $$$${k}_{\mathrm{2}} =\frac{\alpha}{\mathrm{2}}−\frac{\beta}{\:\sqrt{{p}^{\mathrm{2}} +\mathrm{4}{q}}}+\frac{\alpha{p}}{\mathrm{2}\sqrt{{p}^{\mathrm{2}} +\mathrm{4}{a}}} \\ $$$${a}_{{n}} ={k}_{\mathrm{1}} \left(\frac{{p}+\sqrt{{p}^{\mathrm{2}} +\mathrm{4}{q}}}{\mathrm{2}}\right)^{{n}} +{k}_{\mathrm{2}} \left(\frac{{p}−\sqrt{{p}^{\mathrm{2}} +\mathrm{4}{q}}}{\mathrm{2}}\right)^{{n}} \\ $$
Commented by Yozzii last updated on 15/Dec/15
A generating function approach can be used?
$${A}\:{generating}\:{function}\:{approach}\:{can} \\ $$$${be}\:{used}? \\ $$
Commented by prakash jain last updated on 16/Dec/15
Thanks for the hint. I didn′t think about it yet.
$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{the}\:\mathrm{hint}.\:\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{think}\:\mathrm{about}\:\mathrm{it} \\ $$$$\mathrm{yet}.\: \\ $$
Commented by Rasheed Soomro last updated on 17/Dec/15
Where has x^2 −px−q=0 come?
$${Where}\:{has}\:{x}^{\mathrm{2}} −{px}−{q}=\mathrm{0}\:{come}? \\ $$
Commented by Rasheed Soomro last updated on 17/Dec/15
Is tbis ′ generating function approach ′ ? I didn′t understand.
$${Is}\:{tbis}\:'\:{generating}\:{function}\:{approach}\:'\:? \\ $$$${I}\:{didn}'{t}\:{understand}. \\ $$
Commented by Yozzii last updated on 17/Dec/15
No. This solution is based on a theorem on difference equations of the form ta_(n+1) +ra_n +ca_(n−1) =0.
$${No}.\:{This}\:{solution}\:{is}\:{based}\:{on}\:{a}\: \\ $$$${theorem}\:{on}\:{difference}\:{equations}\: \\ $$$${of}\:{the}\:{form}\: \\ $$$$\:\:\:\:{ta}_{{n}+\mathrm{1}} +{ra}_{{n}} +{ca}_{{n}−\mathrm{1}} =\mathrm{0}. \\ $$
Commented by Yozzii last updated on 17/Dec/15
Let t,r,c be constants, t≠0, in the recurrence relation ta_(n+1) +ra_n +ca_(n−1) =0 (n≥1) defining the sequence {a_n } with a_0 =f and a_1 =h. Consider the auxiliary equation tλ^2 +rλ+c=0 whose roots are α and β. The general term a_n of the sequence is then given by a_n =Aα^n +Bβ^n (n≥0) if α≠β. If α=β≠0, a_n =(An+B)α^n (n≥0). The values of A and B are uniquely determined by the values of a_1 and a_0 .
$${Let}\:{t},{r},{c}\:{be}\:{constants},\:{t}\neq\mathrm{0},\:{in}\:{the} \\ $$$${recurrence}\:{relation}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:{ta}_{{n}+\mathrm{1}} +{ra}_{{n}} +{ca}_{{n}−\mathrm{1}} =\mathrm{0}\:\:\:\:\left({n}\geqslant\mathrm{1}\right) \\ $$$${defining}\:{the}\:{sequence}\:\left\{{a}_{{n}} \right\}\:{with}\: \\ $$$${a}_{\mathrm{0}} ={f}\:{and}\:{a}_{\mathrm{1}} ={h}. \\ $$$${Consider}\:{the}\:{auxiliary}\:{equation}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{t}\lambda^{\mathrm{2}} +{r}\lambda+{c}=\mathrm{0} \\ $$$${whose}\:{roots}\:{are}\:\alpha\:{and}\:\beta. \\ $$$${The}\:{general}\:{term}\:{a}_{{n}} \:{of}\:{the}\:{sequence} \\ $$$${is}\:{then}\:{given}\:{by}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}_{{n}} ={A}\alpha^{{n}} +{B}\beta^{{n}} \:\:\left({n}\geqslant\mathrm{0}\right) \\ $$$${if}\:\alpha\neq\beta. \\ $$$${If}\:\:\alpha=\beta\neq\mathrm{0},\:{a}_{{n}} =\left({An}+{B}\right)\alpha^{{n}} \:\:\:\left({n}\geqslant\mathrm{0}\right). \\ $$$${The}\:{values}\:{of}\:{A}\:{and}\:{B}\:{are}\:{uniquely} \\ $$$${determined}\:{by}\:{the}\:{values}\:{of}\:{a}_{\mathrm{1}} \:{and}\:{a}_{\mathrm{0}} . \\ $$$$ \\ $$
Commented by Yozzii last updated on 17/Dec/15
Can′t p^2 +4q<0 ?
$${Can}'{t}\:{p}^{\mathrm{2}} +\mathrm{4}{q}<\mathrm{0}\:? \\ $$
Commented by prakash jain last updated on 17/Dec/15
p^2 +4q can be less than 0 real or complex. I just solved for one case >0. As you mentioned if p^2 +4q=0, both roots are equal say r a_n =k_1 r^n +k_2 nr^n for unequal roots previous result is correct.
$${p}^{\mathrm{2}} +\mathrm{4}{q}\:\mathrm{can}\:\mathrm{be}\:\mathrm{less}\:\mathrm{than}\:\mathrm{0}\:{real}\:{or}\:{complex}. \\ $$$${I}\:\mathrm{just}\:\mathrm{solved}\:\mathrm{for}\:\mathrm{one}\:\mathrm{case}\:>\mathrm{0}. \\ $$$$\mathrm{As}\:\mathrm{you}\:\mathrm{mentioned}\:\mathrm{if}\:{p}^{\mathrm{2}} +\mathrm{4}{q}=\mathrm{0},\:\mathrm{both}\:\mathrm{roots} \\ $$$$\mathrm{are}\:\mathrm{equal}\:\mathrm{say}\:{r} \\ $$$${a}_{{n}} ={k}_{\mathrm{1}} {r}^{{n}} +{k}_{\mathrm{2}} {nr}^{{n}} \\ $$$${for}\:{unequal}\:{roots}\:{previous}\:{result}\:{is}\:{correct}. \\ $$

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