Menu Close

let-p-q-be-reals-such-that-p-gt-q-gt-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-




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  = ??
letp,qberealssuchthatp>q>0definethesequence{xn}wherex1=p+qandxn=x1pqxn1forn2forallnthenxn=??
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 ))
aftercalculatingafewxkIgetxn=pn+1qn+1pnqn
Commented by universe last updated on 20/Aug/24
can u send your solution
canusendyoursolution
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 )).
ifoundmyownmethodforsolvingxn=a+bxn1andgotforthisquestionalsoxn=pn+1qn+1pnqn.
Commented by universe last updated on 21/Aug/24
sir please send your solution
sirpleasesendyoursolution
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) ))   ✓
byinductionforn=1x1=p2q2pq=p+qn=kxk=pk+1qkpkqkn=k+1xk+1=p+qpqxk=p+qpq(pkqk)pk+1qk+1=pk+2pqk+1+qpk+1qk+2pk+1q+pqk+1pk+1qk+1=pk+2qk+2pk+1qk+1
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 )) ✓
thisismymethod:xn=p+qpqxn1=khxn1letxn=anbnanbn=khbn1an1=kan1hbn1an1bn=an1an=kan1hbn1=kan1han2ankan1+han2=0thisisarecurrencerelation.itscorrespondingcharacteristicequationisr2kr+h=0r2(p+q)r+pq=0r=p,qgeneralsolutionforanisan=Apn+BqnwithconstantsA,Bbn=an1=Apn1+Bqn1generalsolutionforxnisthenxn=anbn=Apn+BqnApn1+Bqn1=Cpn+qnCpn1+qn1given:x1=Cp+qC+1=p+qp+Cq=0C=pqxn=(pq)pn+qn(pq)pn1+qn1orxn=pn+1qn+1pnqn
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) ).
thisisageneralmethodwhichcansolverecurrencerelationslikexn=a+bxn1.
Commented by universe last updated on 21/Aug/24
thank you so much sir
thankyousomuchsir

Leave a Reply

Your email address will not be published. Required fields are marked *