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Question Number 210868 by universe last updated on 20/Aug/24
let p ,q be reals such that p>q>0 define the sequence {x_n } where x_1 = p+q and x_n = x_1 −((pq)/x_(n−1) ) for n≥2 for all n then x_n = ??
$$\mathrm{let}\:\mathrm{p}\:,\mathrm{q}\:\mathrm{be}\:\mathrm{reals}\:\mathrm{such}\:\mathrm{that}\:\mathrm{p}>\mathrm{q}>\mathrm{0}\:\mathrm{define} \\ $$$$\mathrm{the}\:\mathrm{sequence}\:\left\{\mathrm{x}_{\mathrm{n}} \right\}\:\mathrm{where}\:\mathrm{x}_{\mathrm{1}} =\:\mathrm{p}+\mathrm{q}\:\mathrm{and} \\ $$$$\mathrm{x}_{\mathrm{n}} \:=\:\mathrm{x}_{\mathrm{1}} −\frac{\mathrm{pq}}{\mathrm{x}_{\mathrm{n}−\mathrm{1}} }\:\mathrm{for}\:\mathrm{n}\geqslant\mathrm{2}\:\mathrm{for}\:\mathrm{all}\:\mathrm{n}\:\mathrm{then}\:\mathrm{x}_{\mathrm{n}} \:=\:?? \\ $$
Commented by Ghisom last updated on 20/Aug/24
after calculating a few x_k I get x_n =((p^(n+1) −q^(n+1) )/(p^n −q^n ))
$$\mathrm{after}\:\mathrm{calculating}\:\mathrm{a}\:\mathrm{few}\:{x}_{{k}} \:\mathrm{I}\:\mathrm{get} \\ $$$${x}_{{n}} =\frac{{p}^{{n}+\mathrm{1}} −{q}^{{n}+\mathrm{1}} }{{p}^{{n}} −{q}^{{n}} } \\ $$
Commented by universe last updated on 20/Aug/24
can u send your solution
$${can}\:{u}\:{send}\:{your}\:{solution} \\ $$
Commented by mr W last updated on 21/Aug/24
i found my own method for solving x_n =a+(b/x_(n−1) ) and got for this question also x_n =((p^(n+1) −q^(n+1) )/(p^n −q^n )).
$${i}\:{found}\:{my}\:{own}\:{method}\:{for}\:{solving} \\ $$$${x}_{{n}} ={a}+\frac{{b}}{{x}_{{n}−\mathrm{1}} }\:{and}\:{got}\:{for}\:{this}\:{question} \\ $$$${also}\:{x}_{{n}} =\frac{{p}^{{n}+\mathrm{1}} −{q}^{{n}+\mathrm{1}} }{{p}^{{n}} −{q}^{{n}} }. \\ $$
Commented by universe last updated on 21/Aug/24
sir please send your solution
$${sir}\:{please}\:{send}\:{your}\:{solution} \\ $$
Commented by mm1342 last updated on 21/Aug/24
by induction for n=1⇒x_1 =((p^2 −q^2 )/(p−q))=p+q ✓ n=k → x_k =((p^(k+1) −q^k )/(p^k −q^k )) n=k+1 → x_(k+1) =p+q−((pq)/x_k ) =p+q−((pq(p^k −q^k ))/(p^(k+1) −q^(k+1) )) =((p^(k+2) −pq^(k+1) +qp^(k+1) −q^(k+2) −p^(k+1) q+pq^(k+1) )/(p^(k+1) −q^(k+1) )) =((p^(k+2) −q^(k+2) )/(p^(k+1) −q^(k+1) )) ✓
$${by}\:{induction} \\ $$$${for}\:{n}=\mathrm{1}\Rightarrow{x}_{\mathrm{1}} =\frac{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }{{p}−{q}}={p}+{q}\:\checkmark \\ $$$${n}={k}\:\rightarrow\:{x}_{{k}} =\frac{{p}^{{k}+\mathrm{1}} −{q}^{{k}} \:\:\:}{{p}^{{k}} −{q}^{{k}} } \\ $$$${n}={k}+\mathrm{1}\:\rightarrow\:{x}_{{k}+\mathrm{1}} ={p}+{q}−\frac{{pq}}{{x}_{{k}} } \\ $$$$={p}+{q}−\frac{{pq}\left({p}^{{k}} −{q}^{{k}} \right)}{{p}^{{k}+\mathrm{1}} −{q}^{{k}+\mathrm{1}} } \\ $$$$=\frac{{p}^{{k}+\mathrm{2}} −{pq}^{{k}+\mathrm{1}} +{qp}^{{k}+\mathrm{1}} −{q}^{{k}+\mathrm{2}} −{p}^{{k}+\mathrm{1}} {q}+{pq}^{{k}+\mathrm{1}} }{{p}^{{k}+\mathrm{1}} −{q}^{{k}+\mathrm{1}} } \\ $$$$=\frac{{p}^{{k}+\mathrm{2}} −{q}^{{k}+\mathrm{2}} }{{p}^{{k}+\mathrm{1}} −{q}^{{k}+\mathrm{1}} }\:\:\:\checkmark \\ $$$$ \\ $$
Answered by mr W last updated on 21/Aug/24
this is my method: x_n =p+q−((pq)/x_(n−1) )=k−(h/x_(n−1) ) let x_n =(a_n /b_n ) (a_n /b_n )=k−((hb_(n−1) )/a_(n−1) )=((ka_(n−1) −hb_(n−1) )/a_(n−1) ) b_n =a_(n−1) a_n =ka_(n−1) −hb_(n−1) =ka_(n−1) −ha_(n−2) ⇒a_n −ka_(n−1) +ha_(n−2) =0 this is a recurrence relation. its corresponding characteristic equation is r^2 −kr+h=0 ⇒r^2 −(p+q)r+pq=0 ⇒r=p, q general solution for a_n is a_n =Ap^n +Bq^n with constants A, B ⇒b_n =a_(n−1) =Ap^(n−1) +Bq^(n−1) general solution for x_n is then x_n =(a_n /b_n )=((Ap^n +Bq^n )/(Ap^(n−1) +Bq^(n−1) ))=((Cp^n +q^n )/(Cp^(n−1) +q^(n−1) )) given: x_1 =((Cp+q)/(C+1))=p+q ⇒p+Cq=0 ⇒C=−(p/q) ⇒x_n =(((−(p/q))p^n +q^n )/((−(p/q))p^(n−1) +q^(n−1) )) or x_n =((p^(n+1) −q^(n+1) )/(p^n −q^n )) ✓
$${this}\:{is}\:{my}\:{method}: \\ $$$${x}_{{n}} ={p}+{q}−\frac{{pq}}{{x}_{{n}−\mathrm{1}} }={k}−\frac{{h}}{{x}_{{n}−\mathrm{1}} } \\ $$$${let}\:{x}_{{n}} =\frac{{a}_{{n}} }{{b}_{{n}} } \\ $$$$\frac{{a}_{{n}} }{{b}_{{n}} }={k}−\frac{{hb}_{{n}−\mathrm{1}} }{{a}_{{n}−\mathrm{1}} }=\frac{{ka}_{{n}−\mathrm{1}} −{hb}_{{n}−\mathrm{1}} }{{a}_{{n}−\mathrm{1}} } \\ $$$${b}_{{n}} ={a}_{{n}−\mathrm{1}} \\ $$$${a}_{{n}} ={ka}_{{n}−\mathrm{1}} −{hb}_{{n}−\mathrm{1}} ={ka}_{{n}−\mathrm{1}} −{ha}_{{n}−\mathrm{2}} \\ $$$$\Rightarrow{a}_{{n}} −{ka}_{{n}−\mathrm{1}} +{ha}_{{n}−\mathrm{2}} =\mathrm{0}\:\:\: \\ $$$${this}\:{is}\:{a}\:{recurrence}\:{relation}. \\ $$$${its}\:{corresponding}\:{characteristic}\: \\ $$$${equation}\:{is} \\ $$$${r}^{\mathrm{2}} −{kr}+{h}=\mathrm{0}\:\Rightarrow{r}^{\mathrm{2}} −\left({p}+{q}\right){r}+{pq}=\mathrm{0} \\ $$$$\Rightarrow{r}={p},\:{q} \\ $$$${general}\:{solution}\:{for}\:{a}_{{n}} \:{is} \\ $$$${a}_{{n}} ={Ap}^{{n}} +{Bq}^{{n}} \:{with}\:{constants}\:{A},\:{B} \\ $$$$\Rightarrow{b}_{{n}} ={a}_{{n}−\mathrm{1}} ={Ap}^{{n}−\mathrm{1}} +{Bq}^{{n}−\mathrm{1}} \\ $$$${general}\:{solution}\:{for}\:{x}_{{n}} \:{is}\:{then} \\ $$$${x}_{{n}} =\frac{{a}_{{n}} }{{b}_{{n}} }=\frac{{Ap}^{{n}} +{Bq}^{{n}} }{{Ap}^{{n}−\mathrm{1}} +{Bq}^{{n}−\mathrm{1}} }=\frac{{Cp}^{{n}} +{q}^{{n}} }{{Cp}^{{n}−\mathrm{1}} +{q}^{{n}−\mathrm{1}} } \\ $$$${given}:\:{x}_{\mathrm{1}} =\frac{{Cp}+{q}}{{C}+\mathrm{1}}={p}+{q} \\ $$$$\Rightarrow{p}+{Cq}=\mathrm{0}\:\Rightarrow{C}=−\frac{{p}}{{q}} \\ $$$$\Rightarrow{x}_{{n}} =\frac{\left(−\frac{{p}}{{q}}\right){p}^{{n}} +{q}^{{n}} }{\left(−\frac{{p}}{{q}}\right){p}^{{n}−\mathrm{1}} +{q}^{{n}−\mathrm{1}} } \\ $$$${or}\:{x}_{{n}} =\frac{{p}^{{n}+\mathrm{1}} −{q}^{{n}+\mathrm{1}} }{{p}^{{n}} −{q}^{{n}} }\:\checkmark \\ $$
Commented by mr W last updated on 21/Aug/24
this is a general method which can solve recurrence relations like x_n =a+(b/x_(n−1) ).
$${this}\:{is}\:{a}\:{general}\:{method}\:{which}\:{can} \\ $$$${solve}\:{recurrence}\:{relations}\:{like} \\ $$$${x}_{{n}} ={a}+\frac{{b}}{{x}_{{n}−\mathrm{1}} }. \\ $$
Commented by universe last updated on 21/Aug/24
thank you so much sir
$${thank}\:{you}\:{so}\:{much}\:{sir} \\ $$

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