Prove that the sequence $a_{n+1} = 2a_{n} - (a_{n})^2$ is bounded. Prove that the sequence $a_{0} = \frac{1}{2}, a_{n+1} = 2a_{n} - (a_{n})^2$ is bounded.
Assume that $0 < a_{n} < 1$ for every $n$ and $a_{0} = \frac{1}{2}$.
Prof. used induction to prove that the sequence is bounded.
Base case $n = 1: 1 > \frac{1}{2}$ holds true. Then for the induction step $(n \rightarrow n+1)$ he wrote:
$a_{n+1} = a_{n}(2 - a_{n})$, so $a_{n+1} > 0$. 
$1 - a_{n+1} = (a_{n})^2 - 2a_{n} + 1 = (a_{n} - 1)^2 > 0$
And therefore the sequence is bounded. So he manipulated the expression so that it can be written as $(a_{n} - 1)^2 > 0$, right? What was the point of doing so? So what is the sequence bounded by? The definition of boundedness is there exists some $M \in \mathbb{R}$ such that $|a_{n}| \leq M$. But what's the $M$ in this case? Is it $0$?
 A: You do not assume that $0<a_n<1$, that is what you need to prove by induction. The only thing that you assume is that $a_{0} = \frac{1}{2}$, and $a_{n+1} = 2a_{n} - (a_{n})^2$ for all $n$. 
So professor proved $a_{n+1}>0$ i.e. $0<a_{n+1}$, and 
$1-a_{n+1}>0$, i.e. $a_{n+1}<1$, these two taken together give you that 
$0<a_{n+1}<1$. That is starting with $0<a_0=\frac12<1$ and taking 
$0<a_n<1$ as the induction hypothesis you prove that $0<a_{n+1}<1$. 
It follows that $0<a_n<1$ for all $n=0,1,2,...$. 
So your sequence is bounded between $0$ and $1$. There are different ways to express that a sequence is bounded, one is to say that all its members belong to some intervals $[p,q]$, e.g. all elements of the above sequence belong to the interval $[0,1]$. Another way (looks different, but amounts to the same) is to say there is $M$ with $|a_n|\le M$ for all $n$, well in our case take $M=1$, it works. Indeed $0<a_n<1$ so then it also holds that $-1<a_n<1$ that is 
$|a_n|<1$ for all $n$. 
A: $(1-a_{n+1})=(1-a_n)^2$
$u_n=(1-a_n) \implies u_{n+1}=u_n^2=u_{n-1}^4=u_{n-i}^{2^{i+1}}=\ldots=u_0^{2^{n+1}}=(1-a_0)^{2^{n+1}}=\left(\dfrac{1}{2}\right)^{2^{n+1}}$ $\therefore a_n=1-\left(\dfrac{1}{2}\right)^{2^{n}}$
