# Sequences with induction and proving. Polynomial and rational functions

$1.$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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This was asked very recently, and repeatedly. I wrote out a solution, and there were other at least part solutions (not everyone asked the whole question). Will try to find link. No more than $2$ days ago. Upvote for calling it homework! FOUND IT. There are a few others, I think this was the first one. –  André Nicolas Nov 22 '12 at 16:04
Although this is the same as the linked question, the answer for (e) is confusing, the answer for (f) is missing and in the linked answer there is no 2. in the question. –  Matt Nov 23 '12 at 3:15
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 $4$. –  André Nicolas Nov 23 '12 at 3:24

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?