Iterative Convergence Formulation for Linear Fractional Transformation with Rational Coefficients

Question

All numbers discussed here are rational (we have not yet constructed the real numbers).

Let $S > 0$, $K > 0$ with $K^2 > S$

Set $$F(x) = \frac{S + Kx}{K + x}$$

Let $p > 0$ such that $p^2 < S$

Define $p_{k+1} = F(p_{k})$, with $p_{1} = F(p)$

Let $q > 0$ such that $q^2 > S$

Define $q_{k+1} = F(q_{k})$, with $q_{1} = F(q)$

Give a direct proof (no matrix theory, etc) of the following:

PROPOSITION: $p_{k}$ < $q_{k}$ for all $k>0$ The intersection of the open intervals of rational numbers ($p_{k}$, $q_{k}$) for $k$ greater than $0$ is either empty or contains a single point. The intersection contains a single point iff S has a rational square root, and the single point is then $\sqrt{S}$.

Bonus Problem: Describe what happens when $K > 0$ but $K^2 < S$.

For motivation/grounding see

How to prove that the set $A = \left\{ {p:{p^2} < 2,p \in {\Bbb Q^+}} \right\}$ has no greatest element?

Answer 1

To answer the question, I feel compelled to develop what I call Simple LFT Theory. As far as I know, this is original research.

You can skip this and proceed directly to subtitle "ANSWER" below if you wish.

For fixed $S > 0$, let

$\mathfrak S =$ $\{$ $F_R(x): \frac{S + Rx}{R + x}\; |\; R > 0 $ $\}$

Under functional composition, $\,\mathfrak S$ forms a commutative semigroup:

(1) $F_M \circ F_N = F_\frac{S + MN}{M + N}$

***Note that $F_{\sqrt S}$ is the 'zero' element. This is the constant function sending all $x$ to $\sqrt S$. It 'absorbs' all elements in $\mathfrak S$, since the linear fractional transformations in this semigroup all have $\sqrt S$ as their one and only positive fixed point. So (1) is only true when

$M = \sqrt S$ implies $N = \sqrt S$ and $N = \sqrt S$ implies $M = \sqrt S$.

---

In particular,

$F_K \circ F_K = F_\frac{S + K^2}{2K}$

Proposition 1: If M and N are two positive numbers and the square of neither one of them is equal to $S$ then

$(\frac{S + MN}{M + N})^2 > S$ when ($M^2 > S$ and $N^2 > S$) OR ($M^2 < S$ and $N^2 < S$)

otherwise $(\frac{S + MN}{M + N})^2 < S$

Also, if, say $M^2 > S$, then $\frac{S + MN}{M + N}$ < M.

Proof: Expanding, we have to check

$S^2 + 2MNS + (MN)^2 > M^2S + 2MNS + N^2S$, or

$S^2 - (M^2 + N^2)S + (MN)^2 > 0$

Considering this as a quadratic equation in $S$ will be helpful here. You will get a pleasant surprise when you look at the discriminant, which is positive and equal to $(M^2 - N^2)^2$. The last statement is easy to see. QED

The subset $\mathfrak K$ of $F_K(x)$ with $K^2 > S$ of $\mathfrak S$ is a subsemigroup, closed under functional composition.

Proposition 2: For $S > 0, K > 0$ with $K^2 > S$. Set $F^1 = F_K$. The set (sequence) $\{F^1, F^2, F^3, ..., F^n, ...\}$ of LFTs form a subsemigroup of $\mathfrak K$. The corresponding sequence $\{K, K_2, K_3, ..., K_n, ...\}$ is decreasing.

Proof:Just apply Proposition 1 by setting both $M$ and $N$ to $K$. QED

Proposition 3: The sequence $(K_n)^2$ converges to $S$.

Proof: The decreasing sequence $\{K, K_2, K_3, ..., K_n, ...\}$ is bounded below. Is is obvious that the gaps between two consecutive $K's$ must go to zero, otherwise the $K's$ would 'march' right past this lower bound.

Take an arbitrarily small number of the form

$\frac{\epsilon}{2K}$

We can find an $n$ where $K_n - K_{n+1} < \frac{\epsilon}{2K}$

But since $F^{n+1} = F F^n$ and using (1), we can write

$K_n - \frac{S + KK_n}{K + K_n} < \frac{\epsilon}{2K}$. Since $K + K_n < 2K$, when we multiply both sides by $K + K_n$, you can see that the inequality

${K_n}^2 - S < \epsilon$

holds. QED

We have two corresponding propositions when $K^2 < S$.

Proposition 2': For $S > 0, K > 0$ with $K^2 < S$. Set $F^1 = F_K$. The sequence $(F^1, F^2, F^3, ..., F^n, ...)$ of LFTs naturally spits into odd and even subsequences with corresponding "K" subsequences. Odd: $\{K, K_3, K_5, ..., K_{2n+1}, ...\}$ is increasing. The square of any number in this subsequence is less than $S$. Even: $\{K_2, K_4, K_6, ..., K_{2n}, ...\}$ is decreasing. The square of any number in this subsequence is greater than $S$.

Proof: We leave the proof to the reader.

Proposition 3': The odd subssequence $({K_{2n+1}}^2)$ converges to $S$ and the even subssequence $({K_{2n}}^2)$ converges to $S$.

Proof: Utilizing Proposition 3 one can readily see that the even subsequence converges. Note that the odd subsequence (excluding the first number K) is obtained as follows,

$K_{2n} \to (\frac{S + KK_{2n}}{K + K_{2n}}) = K_{2n+1} $

You now want

$S - (\frac{S + KK_{2n}}{K + K_{2n}})^2 < \epsilon$ as $n \to \infty$.

But by squaring $K_{2n+1}$ and simplifying you can show convergence.

QED

ANSWER:

Using the techniques found in "finding a better $p$", you can show that

$p < p_1$ and $p_1^2 < S$

and

$q > q_1$ and $q_1^2 > S$

So we have

$p < p_1 < p_2 < ... < p_n < ... < ... < q_n < ... < q_3 < q_2 < q_1 < q$

We will now show that the square of the $p's$ converge to $S$.

Take an arbitrarily small number of the form

$\frac{\epsilon}{2K}$

Since there is an upper bound for this increasing sequence, it is obvious that the gaps between two consecutive $p's$ must go to zero, for otherwise the $p's$ would 'march' right past this upper bound. So, for some $n > 1$,

$F(p_n) - p_n < \frac{\epsilon}{2K}$

or

$S - {p_n}^2 < (K + p)\frac{\epsilon}{2K}$

and since $K + p < 2K$, we have

$S - {p_n}^2 < \epsilon$

We leave it to the reader to show that the square of the $q's$ converges to $S$.

We have proven the proposition posed in the question.