A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
sequence {X_n } is defined by X(n+1) = 2X(n)X^2(n). If 0<X(0)<1 so is 0<X(n)<1 that also for all integers n>0
anonymous
 one year ago
sequence {X_n } is defined by X(n+1) = 2X(n)X^2(n). If 0<X(0)<1 so is 0<X(n)<1 that also for all integers n>0

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[X _{n}\] is defined by \[X _{n+1}=2X _{n}X ^{2}_{n}\] show that if \[0<X _{0}<1\] so is \[0<X _{n}<1\] for all integers n>0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@IrishBoy123 wanna give me some hints with this one? i have tried using the same method..

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0try something like \(X_n = 2  \frac{X_{n+1}}{X_{n}} \) as \(0<X_n <1\) then \(0< 2  \frac{X_{n+1}}{X_{n}}<1\) do each inequality separately see if it leads somewhere...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Something else you can try: Let \(f(x)=2xx^2\). Then \(f'(x)=22x\). For \(0<x<1\), you have that \(f'(x)>0\), which would suggest that \(\{X_n\}\) is an increasing sequence (at least if \(0<X_n<1\)). This smells like an induction proof waiting to happen.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.