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$1.$We define a sequence of rational number {$a_n$} by putting $a_1 =3,\;\text{ and}\;\; a_{n+1} = 4 - \frac{2}{a_n} \text{ for all}\; n \in \mathbb{R}.\;\text{ Put}\;\; \alpha = 2 + \sqrt{2}.$

$(a)$ Calculate $a_1,\ a_2,\ a_3,\ a_4,\ a_5$, and $a_6.$ Determine the decimal expansion of $a_6$ and $\alpha$ on your calculator.

$(b)$ Prove, by induction on $n$ that $3 \leq a_n \leq 4$ for all $n \in \mathbb{N}.$

$(c)$ Show that $\displaystyle 3 \leq \alpha \leq 4$ and $\alpha = 4 - \frac{2}{\alpha}$.

$(d)$ Show that $\displaystyle a_{n+1} - \alpha = \frac{2(a_n-\alpha)}{\alpha a_n}$ for all $n \in \mathbb{N}.$

$(e)$ Prove, by induction on $n$, that $\displaystyle |a_n - \alpha| \leq \frac{|a_1 - \alpha|}{4^{n-1}}$ for all $n \in \mathbb{N}$.

$(f)$ Deduce that $a_n \rightarrow \alpha$ as $n \rightarrow \infty$.


$2.$Find a $rational$ function $f: \mathbb{R} \longrightarrow \mathbb{R}$ with range $f(\mathbb{R}) = [-1,\ 1].$ (Thus $\displaystyle f(x) = \frac{P(x)}{Q(x)}$ for all $x \in \mathbb{R}$ for suitable polynomials $P$ and $Q$ where $Q$ has no real root).


$1(a-d)$ are completed but $1(e)$, $1(f)$ and $2$ are still confusing me.


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    In one of the repeats of this question, someone asked me in a comment about what he/she called (e), but which was your (f). I answered in a comment, giving some detail. You can probably find it, the question was after the one I liked to. I think I never wrote out a solution to the estimate, what you call (e), because I wanted to leave the OP something to do. It is straightforward, each time our difference shrinks by a factor of at least $4$.2012-11-23

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Here's one way to do (2). You want $Q$ to have no real root. The simplest non-constant polynomial with no real root is $x^2+1$, so let's try to use that. To make life as simple as possible, what if we make $P$ a constant? In fact, what if we make it the constant $1$ --- can't get too much simpler than that.

So, we're looking at the function $f(x)={1\over x^2+1}$ Does it work? Not quite --- it's range is $(0,1]$. We could do a little jiggling, look at $2f(x)-1$, which has range $(-1,1]$. Hmm, missed it by one point. Back to the drawing board.

We can still look for something that goes to zero as $x\to\pm\infty$, but it has to have maximum value $1$ and minimum value $-1$. Constant $P$ won't do, so try the next simplest thing, linear $P$. Can you find real numbers $a,b$ such that $f(x)={ax+b\over x^2+1}$ works?