Question Number 4232 by Yozzii last updated on 03/Jan/16
Commented by Yozzii last updated on 04/Jan/16
$${I}'{m}\:{experiencing}\:{difficulty}\: \\ $$$${to}\:{solve}\:{the}\:\mathrm{2}{nd}\:{part}\:{without}\:{use}\:{of}\:{a}\:{theorem} \\ $$$${I}\:{posted}\:{sometime}\:{ago},\:{and}\: \\ $$$${generating}\:{functions}. \\ $$
Commented by prakash jain last updated on 04/Jan/16
$${a}_{{n}} =\frac{\mathrm{1}+{a}_{{n}−\mathrm{1}} ^{\mathrm{2}} }{{a}_{{n}−\mathrm{2}} },\:{a}_{{n}} =\mathrm{3}{a}_{{n}−\mathrm{1}} −{a}_{{n}−\mathrm{2}} ,{a}_{\mathrm{0}} =\mathrm{1},\:{a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{\mathrm{2}} =\frac{\mathrm{1}+\mathrm{1}^{\mathrm{2}} }{\mathrm{1}}=\mathrm{2},\:\mathrm{3}{a}_{\mathrm{1}} −{a}_{\mathrm{0}} =\mathrm{3}×\mathrm{1}−\mathrm{1}=\mathrm{2} \\ $$$${assume}\:{a}_{{n}} =\mathrm{3}{a}_{{n}−\mathrm{1}} −{a}_{{n}−\mathrm{2}} \:{is}\:{true}\:{for}\:{some}\:{n}. \\ $$$${a}_{{n}+\mathrm{1}} =\frac{\mathrm{1}+{a}_{{n}} ^{\mathrm{2}} }{{a}_{{n}−\mathrm{1}} }=\frac{\mathrm{1}+\left(\mathrm{3}{a}_{{n}−\mathrm{1}} −{a}_{{n}−\mathrm{2}} \right)^{\mathrm{2}} }{{a}_{{n}−\mathrm{1}} } \\ $$$$=\frac{\mathrm{1}+{a}_{{n}} \left(\mathrm{3}{a}_{{n}−\mathrm{1}} −{a}_{{n}−\mathrm{2}} \right)}{{a}_{{n}−\mathrm{1}} }=\mathrm{3}{a}_{{n}} +\frac{\mathrm{1}−{a}_{{n}} {a}_{{n}−\mathrm{2}} }{{a}_{{n}−\mathrm{1}} }=\mathrm{3}{a}_{{n}} −{a}_{{n}−\mathrm{1}} \\ $$$${the}\:{result}\:{is}\:{valid}\:{for}\:{all}\:{n}\:{by}\:{induction}. \\ $$$${given}\:\frac{\mathrm{1}+{a}_{{n}−\mathrm{1}} ^{\mathrm{2}} }{{a}_{{n}−\mathrm{2}} }={a}_{{n}} \Rightarrow\mathrm{1}−{a}_{{n}} {a}_{{n}−\mathrm{2}} =−{a}_{{n}−\mathrm{1}} ^{\mathrm{2}} \\ $$
Answered by prakash jain last updated on 04/Jan/16
$${a}_{{n}} =\mathrm{3}{a}_{{n}−\mathrm{1}} −{a}_{{n}−\mathrm{2}} \\ $$$${a}_{{n}} ={x}^{{n}} \\ $$$${x}^{{n}} =\mathrm{3}{x}^{{n}−\mathrm{1}} −{x}^{{n}−\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${a}_{{n}} ={k}_{\mathrm{1}} {x}_{\mathrm{1}} ^{{n}} +{k}_{\mathrm{2}} {x}_{\mathrm{2}} ^{{n}} \\ $$
Commented by Yozzii last updated on 07/Jan/16
$${Can}\:{you}\:{acquire}\:{a}_{{n}} \:{in}\:{the}\:{form} \\ $$$${indicated}\:{in}\:{the}\:{quesion}? \\ $$
Commented by Rasheed Soomro last updated on 15/Jan/16
$${Why}\:\:{a}_{{n}} ={x}^{{n}} \:? \\ $$$${If}\:\:{a}_{{n}} ={x}^{{n}} \:\Rightarrow\:{a}_{\mathrm{1}} ={x} \\ $$$${whereas}\:{a}_{\mathrm{1}} =\mathrm{1}\:{is}\:{given}. \\ $$
Commented by prakash jain last updated on 16/Jan/16
$$\mathrm{To}\:\mathrm{solve}\:\mathrm{a}\:\mathrm{linear}\:\mathrm{difference}\:\mathrm{equation}.\:\mathrm{You} \\ $$$$\mathrm{try}\:\mathrm{to}\:\mathrm{find}\:\mathrm{a}\:\mathrm{suitable}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{eliminate} \\ $$$${n}.\: \\ $$
Commented by Rasheed Soomro last updated on 17/Jan/16
$$\mathbb{T}\mathrm{h}\alpha\mathrm{n}\mathbb{X}!\:\mathrm{I}\:\mathrm{need}\:\mathrm{to}\:\mathrm{study}\:\mathrm{about}\:\:\mathrm{difference}\:\mathrm{equation}. \\ $$