sequence {X_n } is defined by X(n+1) = 2X(n)-X^2(n). If 00

Question

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 · Answer

$$X _{n}$$ 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 · Answer

@IrishBoy123 wanna give me some hints with this one? i have tried using the same method..

irishboy123 · Answer

try 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 · Answer

Something else you can try:

Let $f(x)=2x-x^2$. Then $f'(x)=2-2x$. 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.