Question-4232

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 2nd part without use of a theorem I posted sometime ago, and generating functions.

$${I}'{m}\:{experiencing}\:{difficulty}\: \\ $$$${to}\:{solve}\:{the}\:\mathrm{2}{nd}\:{part}\:{without}\:{use}\:{of}\:{a}\:{theorem} \\ $$$${I}\:{posted}\:{sometime}\:{ago},\:{and}\: \\ $$$${generating}\:{functions}. \\ $$

Commented by prakash jain last updated on 04/Jan/16

a_n =((1+a_(n−1) ^2 )/a_(n−2) ), a_n =3a_(n−1) −a_(n−2) ,a_0 =1, a_1 =1 a_2 =((1+1^2 )/1)=2, 3a_1 −a_0 =3×1−1=2 assume a_n =3a_(n−1) −a_(n−2) is true for some n. a_(n+1) =((1+a_n ^2 )/a_(n−1) )=((1+(3a_(n−1) −a_(n−2) )^2 )/a_(n−1) ) =((1+a_n (3a_(n−1) −a_(n−2) ))/a_(n−1) )=3a_n +((1−a_n a_(n−2) )/a_(n−1) )=3a_n −a_(n−1) the result is valid for all n by induction. given ((1+a_(n−1) ^2 )/a_(n−2) )=a_n ⇒1−a_n a_(n−2) =−a_(n−1) ^2

$${a}_{{n}} =\frac{\mathrm{1}+{a}_{{n}−\mathrm{1}} ^{\mathrm{2}} }{{a}_{{n}−\mathrm{2}} },\:{a}_{{n}} =\mathrm{3}{a}_{{n}−\mathrm{1}} −{a}_{{n}−\mathrm{2}} ,{a}_{\mathrm{0}} =\mathrm{1},\:{a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{\mathrm{2}} =\frac{\mathrm{1}+\mathrm{1}^{\mathrm{2}} }{\mathrm{1}}=\mathrm{2},\:\mathrm{3}{a}_{\mathrm{1}} −{a}_{\mathrm{0}} =\mathrm{3}×\mathrm{1}−\mathrm{1}=\mathrm{2} \\ $$$${assume}\:{a}_{{n}} =\mathrm{3}{a}_{{n}−\mathrm{1}} −{a}_{{n}−\mathrm{2}} \:{is}\:{true}\:{for}\:{some}\:{n}. \\ $$$${a}_{{n}+\mathrm{1}} =\frac{\mathrm{1}+{a}_{{n}} ^{\mathrm{2}} }{{a}_{{n}−\mathrm{1}} }=\frac{\mathrm{1}+\left(\mathrm{3}{a}_{{n}−\mathrm{1}} −{a}_{{n}−\mathrm{2}} \right)^{\mathrm{2}} }{{a}_{{n}−\mathrm{1}} } \\ $$$$=\frac{\mathrm{1}+{a}_{{n}} \left(\mathrm{3}{a}_{{n}−\mathrm{1}} −{a}_{{n}−\mathrm{2}} \right)}{{a}_{{n}−\mathrm{1}} }=\mathrm{3}{a}_{{n}} +\frac{\mathrm{1}−{a}_{{n}} {a}_{{n}−\mathrm{2}} }{{a}_{{n}−\mathrm{1}} }=\mathrm{3}{a}_{{n}} −{a}_{{n}−\mathrm{1}} \\ $$$${the}\:{result}\:{is}\:{valid}\:{for}\:{all}\:{n}\:{by}\:{induction}. \\ $$$${given}\:\frac{\mathrm{1}+{a}_{{n}−\mathrm{1}} ^{\mathrm{2}} }{{a}_{{n}−\mathrm{2}} }={a}_{{n}} \Rightarrow\mathrm{1}−{a}_{{n}} {a}_{{n}−\mathrm{2}} =−{a}_{{n}−\mathrm{1}} ^{\mathrm{2}} \\ $$

Answered by prakash jain last updated on 04/Jan/16

a_n =3a_(n−1) −a_(n−2) a_n =x^n x^n =3x^(n−1) −x^(n−2) x^2 −3x+1=0 x=((3±(√5))/2) a_n =k_1 x_1 ^n +k_2 x_2 ^n

$${a}_{{n}} =\mathrm{3}{a}_{{n}−\mathrm{1}} −{a}_{{n}−\mathrm{2}} \\ $$$${a}_{{n}} ={x}^{{n}} \\ $$$${x}^{{n}} =\mathrm{3}{x}^{{n}−\mathrm{1}} −{x}^{{n}−\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${a}_{{n}} ={k}_{\mathrm{1}} {x}_{\mathrm{1}} ^{{n}} +{k}_{\mathrm{2}} {x}_{\mathrm{2}} ^{{n}} \\ $$

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 ⇒ a_1 =x whereas a_1 =1 is given.

$${Why}\:\:{a}_{{n}} ={x}^{{n}} \:? \\ $$$${If}\:\:{a}_{{n}} ={x}^{{n}} \:\Rightarrow\:{a}_{\mathrm{1}} ={x} \\ $$$${whereas}\:{a}_{\mathrm{1}} =\mathrm{1}\:{is}\:{given}. \\ $$

Commented by prakash jain last updated on 16/Jan/16

To solve a linear difference equation. You try to find a suitable solution to eliminate n.

$$\mathrm{To}\:\mathrm{solve}\:\mathrm{a}\:\mathrm{linear}\:\mathrm{difference}\:\mathrm{equation}.\:\mathrm{You} \\ $$$$\mathrm{try}\:\mathrm{to}\:\mathrm{find}\:\mathrm{a}\:\mathrm{suitable}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{eliminate} \\ $$$${n}.\: \\ $$

Commented by Rasheed Soomro last updated on 17/Jan/16

ThαnX! I need to study about difference equation.

$$\mathbb{T}\mathrm{h}\alpha\mathrm{n}\mathbb{X}!\:\mathrm{I}\:\mathrm{need}\:\mathrm{to}\:\mathrm{study}\:\mathrm{about}\:\:\mathrm{difference}\:\mathrm{equation}. \\ $$

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