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